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 |
|---|---|---|---|---|
3,190,781 | <blockquote>
<p>Suppose the polynomial <span class="math-container">$P(x)$</span> with integer coefficients satisfies the following conditions:<br>
(A) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 − 4x + 3$</span>, the remainder is <span class="math-container">$65x − 68$</span>.<br>
(B) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 + 6x − 7$</span>, the remainder is <span class="math-container">$−5x + a$</span>.<br>
Then we know that <span class="math-container">$a =$</span>?</p>
</blockquote>
<p>I am struggling with this first question from the <a href="https://www.maa.org/sites/default/files/pdf/programs/JUEEDocument.pdf" rel="noreferrer">1990 Japanese University Entrance Examination</a>. Comments from the linked paper mention that this is a basic application of the "remainder theorem". I'm only familiar with the <a href="https://en.wikipedia.org/wiki/Polynomial_remainder_theorem" rel="noreferrer">polynomial remainder theorem</a> but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?</p>
<p>So for some <span class="math-container">$g(x)$</span> and <span class="math-container">$h(x)$</span> we have:
<span class="math-container">$$P(x) = g(x)(x^2-4x+3) + (65x-68),\\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$</span>
which looks to have more unknowns than equations. How should I proceed from here?</p>
| Viki 183 | 664,768 | <p>Since <span class="math-container">$P(x)=g(x)(x^2-4x+3)+(65x-68)$</span>, for <span class="math-container">$x=1$</span> you get <span class="math-container">$P(1)=-3$</span>. This imply that
<span class="math-container">$P(1)=h(1)\cdot 0+(-5+a)$</span>, i.e. <span class="math-container">$a=2$</span>.</p>
|
2,588,225 | <p>To start off, I'm very new to the complex plane and complex numbers.</p>
<p>I'm trying to determine if the series $\sum_{n=1}^\infty\frac{1}{n^2 + i}$ is convergent or divergent.</p>
<p>However, I'm not exactly sure how to express the value of the function. Could I just treat the denominator of the expression as the complex number's modulus and ignore the rotation in the plane? If I do that, I could determine that the series is convergent via a direct comparison test.</p>
<p>Geometrically speaking, it doesn't seem like a rotation would affect the convergence of a series. However, I don't know if I can consider the modulus of the complex number to be equivalent to its "value", and this is the cause of my confusion.</p>
| mechanodroid | 144,766 | <p>When in doubt, you can always decompose it to the real and imaginary part:</p>
<p>$$\frac{1}{n^2 + i} = \frac{n^2-i}{n^4+ 1} = \frac{n^2}{n^4+1}-i\cdot\frac{1}{n^2}$$</p>
<p>Both the real and imaginary parts converge:</p>
<p>$$\sum_{n=1}^\infty\frac{n^2}{n^4+1} \le \sum_{n=1}^\infty\frac{n^2}{n^4} = \sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}< +\infty$$</p>
<p>$$\sum_{n=1}^\infty\frac{-i}{n^2} = -i\sum_{n=1}^\infty\frac1{n^2} = -\frac{i\pi^2}{6}$$</p>
<p>So in particular, $$\sum_{n=1}^\infty \frac{1}{n^2+i} = \sum_{n=1}^\infty\frac{n^2}{n^4+1} + \sum_{n=1}^\infty\frac{-i}{n^2}$$ also converges as a sum of two convergent series.</p>
|
4,155,453 | <p>I am trying to evaluate <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$</span> Now I am aware that clearly <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{-1}(1) = \frac{\pi}{4},$$</span> but I do not know what to do if each sum is already sent to infinity. Im taking a limit of limits. I suppose I could rewrite my limit as <span class="math-container">$$\lim_{n\to \infty} \lim_{m\to \infty} \sum_{k=1}^m \frac{n}{n^2+k^2}?$$</span> But I am unaware if this is helpful at all. Any hints would be appreciated. Obviously, Wolfram calculates this as <span class="math-container">$\frac{\pi}{2}$</span> but I am unaware of the steps and logic to get there.</p>
| Claude Leibovici | 82,404 | <p><em>Another solution.</em></p>
<p><span class="math-container">$$\frac{n}{n^2+k^2}=\frac{n}{(k+in)(k-in)}$$</span> Using partial fraction decomposition and summing
<span class="math-container">$$S_m=\sum_{k=1}^m\frac{n}{n^2+k^2}=\frac{i}{2} (\psi(m+i n+1)-\psi (m-i n+1)+\psi (1-i n)-\psi (1+i n))$$</span>
<span class="math-container">$$\psi (1-i n)-\psi (1+i n)=i \left(\frac{1}{n}-\pi \coth (\pi n)\right)$$</span>
<span class="math-container">$$\psi(m+i n+1)-\psi (m-i n+1)=H_{m+i n}-H_{m-i n}$$</span>
Using the asymptotic of generalized harmonic numbers
<span class="math-container">$$H_{m+i n}-H_{m-i n}=\frac{2 i n}{m}-\frac{i n}{m^2}+O\left(\frac{1}{m^3}\right)$$</span> Combining all the above
<span class="math-container">$$S_m=\frac{\pi n \coth (\pi n)-1}{2 n}-\frac{n}{m}+\frac{n}{2 m^2}+O\left(\frac{1}{m^3}\right)$$</span>
<span class="math-container">$$\lim_{m\to \infty} S_m=\frac{\pi n \coth (\pi n)-1}{2 n}=\frac{\pi}{2} \coth (\pi n)-\frac{1}{2 n}$$</span>
<span class="math-container">$$\lim_{n\to \infty} \lim_{m\to \infty} \sum_{k=1}^m \frac{n}{n^2+k^2}=\lim_{n\to \infty} \frac{\pi n \coth (\pi n)-1}{2 n}=\frac \pi 2$$</span></p>
|
2,335,334 | <p>I'm trying to solve the following ODE:
$$3xy''+(3x+1)y'+y=0$$
So I started by assuming a solution(and its derivates) with the form:</p>
<p>$$
\tag{1}y = \sum_{n=0}^{\infty } a_{n} x^{n + r}$$</p>
<p>$$y' = \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}$$</p>
<p>$$y'' = \sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2}$$
Replacing in original equation:
$$3x\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2} + 3x\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}+\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0$$</p>
<p>$$ 3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0 $$</p>
<p>To transform all $x^{n+r}$ into $x^{n+r-1}$ I've rewritten the equation as:</p>
<p>$$
3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} +\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0 $$</p>
<p>In order to make all sums start from the $n=1$:</p>
<p>$$ \tag{2}
3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=1}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0
$$</p>
<p>If I understood correctly, the indicial equation should then be:
$$ 3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} = 0 $$</p>
<p>and since $x^{r-1} \neq 0 $ and $a_{0} \neq 0 $:</p>
<p>$$ 3r(r-1) + r = 0 $$
$$ r_{1} = 0 $$
$$ r_{2} = \frac{2}{3} $$</p>
<p>From $(2)$:</p>
<p>$$ [3(n+r)(n+r-1)+(n+r)]a_{n} + [3(n+r-1)+1]a_{n-1} = 0
$$
which seems to give for both solutions of $r$:
$$ a_{n} = \frac{-a_{n-1}}{n} $$
It seems to me that this is in terms of $a_{0}$:</p>
<p>$$ a_{n} = \frac{(-1)^{n}}{n!}a_{0} $$</p>
<p>So rewriting $(1)$ now would give me the solution and $y$ would be an infinite sum since all $a_{n} \neq 0$. And I know this can't be true because by deduction one can see that $y = e^{-x}$ is a solution to the ODE, which makes me think I made a mistake in some of the steps shown above, I'd be glad if anyone could show me where I made the mistake and give me some help in finding the correct general solution for this ODE.</p>
| N. F. Taussig | 173,070 | <p>For each of the four questions, there are two possible outcomes. Thus, in total, there are $2^4 = 16$ ways to answer the four questions.<br>
\begin{array}{c c c c}
T & T & T & T\\
T & F & T & T\\
T & T & F & T\\
T & F & F & T\\
T & T & T & F\\
T & F & T & F\\
T & T & F & F\\
T & F & F & F\\
F & T & T & T\\
F & F & T & T\\
F & T & F & T\\
F & F & F & T\\
F & T & T & F\\
F & F & T & F\\
F & T & F & F\\
F & F & F & F
\end{array}</p>
<p>Only one of these $16$ sequences is correct. Assuming each of these $16$ sequences is equally likely to occur (as would result from random guessing), the probability that all four questions are answered correctly is $1/16$.</p>
<p>If we assume that a person is equally likely to guess true or false on each question, then he or she has probability $1/2$ of answering each question correctly. Under the assumption of <a href="https://en.wikipedia.org/wiki/Independence_(probability_theory)" rel="nofollow noreferrer">independence</a>, the probability that all four questions are answered correctly is
$$\left(\frac{1}{2}\right)^4 = \frac{1}{16}$$
We add probabilities of mutually exclusive events (events that cannot occur at the same time). We multiply probabilities of independent events.</p>
|
2,979,755 | <p>I wish to show that <span class="math-container">$2^{2^{2^x}}<100^{100^x}$</span> for <span class="math-container">$x$</span> sufficiently large. I have taken logs (base 10) of both sides to get <span class="math-container">$2^{(2^x-1)}\log_{10} 2$</span> and <span class="math-container">$100^x$</span>. It is not immediately clear how I can prove that the second term here is larger than the first.</p>
<p>Any help appreciated. </p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>Since <span class="math-container">$\log$</span> function is strictly increasing we have</p>
<p><span class="math-container">$$2^{2^{2^x}}<100^{100^x}\iff \log_{10}\left(2^{2^{2^x}}\right)<\log_{10}\left(100^{100^x}\right) \iff 2^{2^x}\log_{10}2<2\cdot 100^x$$</span></p>
<p>then again</p>
<p><span class="math-container">$$2^{2^x}\log_{10}2<2\cdot 100^x \iff \log_{10}\left(2^{2^x}\log_{10}2\right)<\log_{10}\left(2\cdot 100^x\right)$$</span></p>
<p><span class="math-container">$$2^x\log_{10}2 + \log_{10}(\log_{10}2)<\log_{10}2+2x$$</span></p>
|
2,979,755 | <p>I wish to show that <span class="math-container">$2^{2^{2^x}}<100^{100^x}$</span> for <span class="math-container">$x$</span> sufficiently large. I have taken logs (base 10) of both sides to get <span class="math-container">$2^{(2^x-1)}\log_{10} 2$</span> and <span class="math-container">$100^x$</span>. It is not immediately clear how I can prove that the second term here is larger than the first.</p>
<p>Any help appreciated. </p>
| fleablood | 280,126 | <p><span class="math-container">$100^{100^x}= 2^{[\log_2{100}]*100^x} $</span>. Let <span class="math-container">$k = \log_2{100}$</span>. (<span class="math-container">$6< k < 7$</span>)</p>
<p><span class="math-container">$100^{100^x} = 2^{k*2^{kx}}= 2^{2^{\log_2k}*2^{kx}}$</span>. Let <span class="math-container">$m = \log_2 k$</span>. (<span class="math-container">$2< m < 3$</span>)</p>
<p><span class="math-container">$100^{100^x} = 2^{2^m*2^{kx}}=2^{2^{m + kx}}$</span></p>
<p>Now <span class="math-container">$2^{2^{m+kx}}< 2^{2^{2^x}} \iff m + kx < 2^x$</span> for sufficiently large <span class="math-container">$x$</span>.</p>
<p>Which should be enough to be convincing.</p>
<p>But if not:</p>
<p>If <span class="math-container">$x > m$</span> then <span class="math-container">$m + kx < x +kx = (k+1)x< 8x=2^3*x$</span>.</p>
<p>And if we can show <span class="math-container">$x < 2^{x-3}$</span> for sufficiently large <span class="math-container">$x$</span> we are done.</p>
<p>which... of course it is. </p>
<p><span class="math-container">$(x)' = 1$</span> and <span class="math-container">$(2^{x-3})'= \ln 2* 2^{x-3}> 1$</span> for all <span class="math-container">$x > ...$</span> well <span class="math-container">$x > 3 + \log_2 (\frac 1{\ln 2})= 3 + \frac {\ln \frac 1{\ln 2}}{\ln 2}\approx 3.52$</span>. then <span class="math-container">$2^{x-3}$</span> is increasing faster than <span class="math-container">$x$</span>. So at <span class="math-container">$x = 6>3 + \frac {\ln \frac 1{\ln 2}}{\ln 2}$</span> we have <span class="math-container">$x = 6 < 8 = 2^{x-3}$</span> and <span class="math-container">$2^{x-3}$</span> is increasing faster than <span class="math-container">$x$</span> so for <span class="math-container">$x \ge 6$</span> we will have our result. </p>
|
3,579,457 | <p>Last Christmas, a kid in my neighborhood dressed in a pirate suit and knocked my door for "trick or treat". He explained to me that Dec 25, which is 25 written in base 10, is equal to 31 in base 8, which is Oct 31. So ingeniously, Halloween coincides with Christmas. </p>
<p>I really wonder which other two dates would possibly coincides with each other.</p>
<p>Can you help me find them? And explain what kind the two dates are. Your birthday is also appreciated.</p>
| emonHR | 565,609 | <p><strong>which other two dates would possibly coincide with each other</strong> </p>
<blockquote>
<p>There are too many to list out. </p>
</blockquote>
<p>Maybe you could find them using converting decimal number to octal number. <a href="https://www.tutorialspoint.com/how-to-convert-decimal-to-octal" rel="nofollow noreferrer">Here</a> is the rules you can follow to find. My birthday is <code>01 jan</code> which is same in binary, octal and hexadecimal system . </p>
|
1,860,243 | <p>Let $|G|$ be an abelian group and let $H = \{g \in G : |g| \text{ divides } 12 \}.$ Prove that $H$ is a subgroup of $G$.</p>
<p>I know that I have to show that $a,b \in H \Rightarrow ab^{-1} \in H$ or $(ab \in H \land a^{-1} \in H).$ But I can't figure out how $|a|$ and $|b|$ dividing $12$ relates to $|ab|$ or $|ab^{-1}|$ dividing $12$.</p>
| lhf | 589 | <p>The crucial observation is that $|g|$ divides $12$ if and only if $g^{12} = e$.
With this, it is simple to verify the requirements for subgroup:</p>
<ul>
<li><p>$e \in H$ because $e^{12}=e$.</p></li>
<li><p>$a \in H \implies a^{-1} \in H$ because $(a^{-1})^{12}=(a^{12})^{-1}=e^{-1}=e $ (or simply that $|a^{-1}|=|a|$).</p></li>
<li><p>$a,b \in H \implies ab \in H$ because $(ab)^{12}=a^{12} b^{12}=ee=e$.</p></li>
</ul>
<p>The last argument relies crucially on $G$ being abelian.</p>
|
44,310 | <p>I wanted to draw (lattice-aligned) unit cubes inside a sphere with a given radius centered on the origin, so I wrote this program:</p>
<pre><code>Cube[x_, y_, z_] := Cuboid[{x - 1, y - 1, z - 1}, {x, y, z}];
Coords[r_] := {#1, #2, Floor[Sqrt[r^2 - #1^2 - #2^2]]} & @@ # & /@
With[{t = Sqrt[r^2 - 1]},
Select[Flatten[Table[{x, y}, {x, 1, t}, {y, 1, x}], 1],
Norm[#] <= t &]];
Cubes[r_] :=
Cube @@ # & /@
Union[Flatten[Permute[#, SymmetricGroup[3]] & /@ Coords[r], 1]];
Draw[r_] :=
Graphics3D[
Union[Cubes[r], {{Green, Opacity[0.1], Sphere[{0, 0, 0}, r]}}],
PlotRange -> {{0, r}, {0, r}, {0, r}},
ViewPoint -> {r, 3 r/4, 3 r/5}];
Draw /@ Sqrt /@ {50, 100, 250, 500, 1000, 2500, 6054}
</code></pre>
<p>(Note that it shows only the cubes on the outside -- drawing the inner ones takes too long and usually Mathematica runs out of memory.) It gets the job done, but it's pretty ugly. I'm not a regular Mathematica user, and tips for improving my code?</p>
<p>That is, how can I write idiomatically-better Mathematica code?</p>
<hr>
<p>Improved code:</p>
<pre><code>Coords[r_] := {#1, #2, Floor[Sqrt[r^2 - #1^2 - #2^2]]} & @@@
With[{t = Sqrt[r^2 - 1]},
Select[Join @@ Table[{x, y}, {x, t}, {y, 1, x}], Norm[#] <= t &]];
Cubes[r_] := (Cuboid /@ (Union @@ (Permute[#, SymmetricGroup[3]] & /@
Coords[r]) - 1));
Draw[r_] :=
Graphics3D[
Union[Cubes[r], {{Green, Opacity[0.1], Sphere[{0, 0, 0}, r]}}],
PlotRange -> {{0, r}, {0, r}, {0, r}},
ViewPoint -> {r, 3 r/4, 3 r/5}];
Draw /@ Sqrt /@ {50, 100, 250}
</code></pre>
| Kuba | 5,478 | <p><strong>New, not the best but can do</strong></p>
<pre><code>r = 50;
da = Pi/(4. r);
sa = da;
ea = Pi/2. - da;
Composition[
Graphics3D[{#,
GeometricTransformation[#, ReflectionTransform /@ {{0, 0, 1}, {0, 1, 1}}]
}, Axes -> True, PlotRange -> All, ImageSize -> 600] &,
Cuboid /@ # &,
{# - Sign[#], #} & /@ # &,
Select[#, FreeQ[#, 0] &] &,
DeleteDuplicates,
Floor,
Flatten[#, 1] &
][
Table[r {Cos[d] Sin[a], Cos[d] Cos[a], Sin[d]}, {d, sa, ea, da}, {a,
sa, ea, da}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/atlx6.png" alt="enter image description here"></p>
<p><strong>Old and slow</strong> This is not optimal or faster code but it is cleaner:</p>
<pre><code>r = 6;
s = 0;
e = r;
d = .5;
Composition[
Graphics3D[{#}, Axes -> True, PlotRange -> {{0, r}, {0, r}, {0, r}}] &,
Flatten,
Map[If[1 == Times @@ UnitStep[r - Norm /@ #],
{Hue@RandomReal[], Cuboid@#[[{1, -1}]]},
## &[]
] &,
#, {3}] &,
Map[Flatten[#, 2] &, #, {3}] &,
Partition[#, {2, 2, 2}, 1] &
][Table[{i, j, k}, {i, s, e, d}, {j, s, e, d}, {k, s, e, d}]]
</code></pre>
<p>Important part of this code is function <code>Partition[#, {2, 2, 2}, 1] &</code> which does something like <code>{a,b,c} --> {{a,b},{b,c}}</code> but in 3D. It is clean but it is a big choke point. We have to create equally spaced array for this. Including points we know are to far away from origin.
You can add <code>Opacity@.7, Sphere[{0, 0, 0}, r]</code> to <code>Graphics3D</code>, look out, opacity slows down rotation.
<img src="https://i.stack.imgur.com/EceVp.png" alt="enter image description here"></p>
|
1,016,585 | <p>I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$</p>
<p>How do I do this?</p>
<p>It looks like it is close to being an $\operatorname{arcsin}$?</p>
<hr>
<p>I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$</p>
<p>But apparently not. </p>
| Community | -1 | <p>Let us set $c=1$ for a moment.</p>
<p>From hyperbolic trigonometry, we know that $\cosh^2t-1=\sinh^2t$ (equivalent of $1-cos^2=\sin^2t$) and the substitution $x=\cosh t$ seems natural. Then</p>
<p>$$\int\frac{dx}{\sqrt{x^2-1}}=\int\frac{\sinh t\ dt}{\sinh t}=t\text{ (!)}$$</p>
<p>Similarly, set $c=-1$ and use $x=\sinh t$:</p>
<p>$$\int\frac{dx}{\sqrt{x^2+1}}=\int\frac{\cosh t\ dt}{\cosh t}=t\text{ (!)}$$</p>
<p>Scaling $x$ by $1/\sqrt c$ gives you the final solution, one of $\text{arcosh}(x/\sqrt c)$ or $\text{arsinh}(x/\sqrt{-c})$.</p>
|
1,016,585 | <p>I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$</p>
<p>How do I do this?</p>
<p>It looks like it is close to being an $\operatorname{arcsin}$?</p>
<hr>
<p>I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$</p>
<p>But apparently not. </p>
| Vishwa Iyer | 71,281 | <p>To answer this question. You need to use a technique called Trigonometric Substitution (of course there are other ways, but this way is the easiest in my opinion).</p>
<p>First draw a right triangle. Call the height $\sqrt{x^2-c}$, the base $\sqrt{c}$, and the hypotenuse $x$. Call the angle between the base and the hypotenuse $\theta$. We know that $$\sec(\theta) = \frac{x}{\sqrt{c}}$$
$$\sqrt{c}\sec(\theta) = x$$
$$dx = \sqrt{c}\sec(\theta)\tan(\theta)d\theta$$
So we substitute this into the integral, and the integral simplifies to $$\begin{align}
\int \frac{dx}{\sqrt{x^2-c}} &= \int \frac{\sqrt{c}\sec(\theta)\tan(\theta)} {\sqrt{(\sqrt{c}\sec(\theta))^2-c}}d\theta \\
&= \int \frac{\sqrt{c}\sec(\theta)\tan(\theta)}{\sqrt{c}\tan(\theta)}d\theta \\
&= \int \sec(\theta)d\theta \\
&= \ln|\sec(\theta)+\tan(\theta)| + C \\
&= \ln\left|\frac{x}{\sqrt{c}} + \frac{\sqrt{x^2-c}}{\sqrt{c}} \right| + C
\end{align}$$</p>
|
3,249,926 | <p>I’ve tried applying Vandermonde’s identity, but got stuck. Any help would be appreciated!</p>
| Julian Mejia | 452,658 | <p>What happens is that a square is always <span class="math-container">$\geq 0$</span>. What I would do is split in 2 cases (when <span class="math-container">$x$</span> is positive or negative)</p>
<p>If <span class="math-container">$0\leq x\leq 3$</span>, then <span class="math-container">$0\leq x^2\leq 9$</span>.</p>
<p>If <span class="math-container">$-2\leq x\leq 0$</span>, then <span class="math-container">$0\leq x^2\leq 4$</span> (think about it).</p>
<p>So, by taking the "union of the sets" <span class="math-container">$0\leq x^2\leq 9$</span> and <span class="math-container">$0\leq x^2\leq 4$</span>, you get all the possible values of <span class="math-container">$x^2$</span>. You get <span class="math-container">$0\leq x^2\leq 9$</span>.</p>
|
3,949,872 | <p>I am to expand <span class="math-container">$\log_3\frac{7x^2+21x}{7x(x-1)(x-2)}$</span> using the quotient rule of logs. I arrived at:</p>
<p><span class="math-container">$$\log_3(7)+\log_3(4)+\log_3(x)-\log_3(x-1)-\log_3(x-2)$$</span></p>
<p>Whereas my textbook solution is:</p>
<p><span class="math-container">$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$</span></p>
<p>It looks like I got it right for the denominator part but not the numerator. I don't see how to arrive at <span class="math-container">$\log_3(x+3)$</span> from <span class="math-container">$7x^2+21x$</span>?</p>
<p>My train of thought was to divide <span class="math-container">$7x^2$</span> in the numerator by the <span class="math-container">$7x$</span> in the denominator leaving me with <span class="math-container">$7x+21x = 28x$</span> in the numerator and just <span class="math-container">$(x-1)(x-2)$</span> in the denominator.</p>
<p>With <span class="math-container">$28x$</span> in the numerator I expanded out to be <span class="math-container">$$\log_3(7)+\log_3(4)+\log_3(x).$$</span></p>
<p>Where did I go wrong, and how can I arrive at <span class="math-container">$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$</span></p>
| rishav | 842,762 | <p>"My train of thought was to divide 7<span class="math-container">$x^2$</span> in the numerator by the 7 in the denominator leaving me with 7+21=28 in the numerator and just (−1)(−2) in the denominator."</p>
<p>Once re-read your statement!!
when you divide 7<span class="math-container">$x^2$</span> with 7x you end up with x not 7x;</p>
<p>you will be left with (x+3) not (7x+21x) when you divide (7<span class="math-container">$x^2$</span> + 21x)/7x .</p>
|
403,977 | <blockquote>
<p>Let $u_n>u_{n+1}>0$ for all $n \in \Bbb N$, and suppose that $u_2+u_4+u_8+u_{16}+\dots$ diverges. Prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$diverges.</p>
</blockquote>
<p>Please provide me a hint or a full solution.</p>
| marty cohen | 13,079 | <p>$\sum_{n=2^m}^{2^{m+1}-1}\frac{u_n}{n}
>\sum_{n=2^m}^{2^{m+1}-1}\frac{u_{2^{m+1}}}{2^{m+1}}
= 2^m \frac{u_{2^{m+1}}}{2^{m+1}}
=u_{2^{m+1}}/2
$.</p>
<p>Therefore
$\sum_{n=1}^{2^m-1} \frac{u_n}{n}
> \frac1{2}\sum_{k=1}^m u_{2^k}
$
and, if the latter sum diverges, so does the former.</p>
<p>Note that an almost identical argument will show that
if $(a_m)_{m=1}^{\infty}$
is an increasing sequence of positive integers
such that there is a $c < 1$ such that
$\frac{a_{m}}{a_{m+1}} \le c$ for all $m$,
and $u_n \ge u_{n+1}$ for all $n$,
then $\sum_{m=1}^{\infty} u_{a_m}$ diverges
implies that
$\sum_{n=1}^{\infty} \frac{u_n}{n}$ diverges.</p>
<p>Here is the proof by copy and paste:</p>
<p>$\sum_{n=a_m}^{a_{m+1}-1}\frac{u_n}{n}
>\sum_{n=a_m}^{a_{m+1}-1}\frac{u_{a_{m+1}}}{a_{m+1}}
= (a_{m+1}-a_m) \frac{u_{a_{m+1}}}{a_{m+1}}
=(1-\frac{a_{m}}{a_{m+1}})u_{a_{m+1}}
=(1-c)u_{a_{m+1}}
$.
Therefore
$\sum_{n=1}^{a_m-1} \frac{u_n}{n}
> (1-c)\sum_{k=1}^m u_{a_k}
$
and, if the latter sum diverges, so does the former.</p>
<p>The fact that $1-\frac{a_{m}}{a_{m+1}}$
occurs leads to the requirement on $a_m$.</p>
<p>We can also go the other way
and see what is needed.</p>
<p>$\sum_{n=a_m}^{a_{m+1}-1}\frac{u_n}{n}
<\sum_{n=a_m}^{a_{m+1}-1}\frac{u_{a_{m}}}{a_{m}}
= (a_{m+1}-a_m) \frac{u_{a_{m}}}{a_m}
=(\frac{a_{m+1}}{a_m}-1)u_{a_{m}}
$.</p>
<p>From this we see that if
$\frac{a_{m+1}}{a_m}-1 < d$
for some real $d$ and all $m$,
then
$u_{a_{m}} > \frac1{d}\sum_{n=a_m}^{a_{m+1}-1}\frac{u_n}{n}$,
so that
$\sum \frac{u_n}{n}$ diverges
implies that
$\sum u_{a_m}$ diverges.</p>
<p>Combining these two,
we get this:</p>
<p>If $(a_m)_{m=1}^{\infty}$
is an increasing sequence of positive integers
such that there exists positive reals $c$ and $d$ such that
$c \le \frac{a_{m+1}}{a_{m}}-1 \le d$ for all $m$,
and $u_n \ge u_{n+1}$ for all $n$,
then $\sum_{m=1}^{\infty} u_{a_m}$ converges
if and only if
$\sum_{n=1}^{\infty} \frac{u_n}{n}$ converges.</p>
|
2,228,913 | <p>For some geometric problem that I am trying to solve, I am interested
in characterizing the cross product between a Bezier curve and its second order derivative.</p>
<p>So, let's $\mathbf{c}(t)$ be a Bezier curve, and $\ddot{\mathbf{c}}(t)$ it's derivative.</p>
<p>Is $\mathbf{w}(t) = \mathbf{c}(t) \times \ddot{\mathbf{c}}(t)$ a Bezier curve?</p>
<p>If it is the case, is it also the case in general ?</p>
<p>Thanks a lot for your time and answers. Sincerely,</p>
<p>Steve</p>
| hmakholm left over Monica | 14,366 | <p><em>Every</em> parametric curve where each coordinate is a polynomial in the parameter can be expressed as a Bézier curve.</p>
<p>Thus also the cross product of Bézier curves. The <em>order</em> of the resulting curve will in general (except in degenerate cases) be the sum of the orders of the two curves you multiply -- so you can't, for example, multiply two cubic curves and expect to get a cubic curve back.</p>
|
1,531,525 | <p>I would like to compute the following sum:</p>
<p>$$\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}$$</p>
<p>I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $\cos \theta = {e^{i\theta} + e^{-i\theta}\over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated. </p>
| QiLin Xue | 624,739 | <p>I know I'm late but there's a slightly different solution I want to present that doesn't involve any exponentiation.</p>
<p>We can write the summation as the real part of</p>
<p><span class="math-container">$$\sum_{n=0}^{\infty} \frac{\cos n\theta + i\sin m\theta}{2^n}$$</span></p>
<p>Using De Moivre's theorem:</p>
<p><span class="math-container">$$\sum_{n=0}^{\infty} \left(\frac{\cos \theta + i\sin \theta}{2}\right)^n$$</span></p>
<p>We can then calculate the real portion of the infinite sum.</p>
<p><span class="math-container">\begin{align*}
\operatorname{Re}\left(\frac{1}{1-r}\right)
&= \operatorname{Re}\left(\frac{1}{\frac{2-(\cos \theta + i\sin \theta)}{2}}\right) \\
&= \operatorname{Re}\left(\frac{2}{(2-\cos\theta) - i\sin\theta}\right) \\
&= \operatorname{Re}\left(\frac{2((2-\cos\theta) + i\sin\theta)}{4-4\cos\theta+1}\right) \\
&= \frac{4-2\cos\theta}{4-4\cos\theta+1}
\end{align*}</span></p>
<p>Hope this helps!</p>
|
159,777 | <p>In his <a href="http://www.ams.org/mathscinet-getitem?mr=44124">short paper</a> (1951) and <a href="http://www.ams.org/mathscinet-getitem?mr=74819">longer monograph</a> (1955), Postnikov introduced what I believe are now called <a href="http://en.wikipedia.org/wiki/Postnikov_system"><em>Postnikov systems</em> or <em>towers</em></a>. It is my understanding that Postnikov systems have since then been widely adopted, as a way of totally encoding the homotopy type of a topological space. Expressed algebraically, a Postnikov system of a topological space $X$ consists of its homotopy groups $\pi_i = \pi_i(X)$, where each of the higher groups $\pi_i$ ($i\ge 2$) has the structure of a $\pi_1$-module, together with a group cohomology class $[k_1] \in H^{3}(\pi_1,\pi_2)$ and higher "cohomology classes" (that are somewhat more difficult to describe algebraically) $[k_i]$ with coefficients in $\pi_{i+1}$ for each $i\ge 2$.</p>
<p>The main application of these systems, in Postnikov's original work, was to reconstruct the cohomology of the space $X$ from its homotopy invariants in a purely algebraic way. The reconstruction allowed also for cohomology with coefficients in a <a href="http://en.wikipedia.org/wiki/Local_system">local system</a>, with the local system algebraically presented as some $\pi_1$-module. In that work, this result was presented as a generalization of the earlier work (1945) of <a href="http://www.ams.org/mathscinet-getitem?mr=13312">Eilenberg and MacLane</a> (and of course others) on the algebraic reconstruction of the cohomology of spaces that are aspherical except in one degree.</p>
<p>Now my question. <strong>Is there a modern reference for the reconstruction of the cohomology (desirably with local coefficients) of a topological space from its Postnikov sequence?</strong> While Postnikov sequences themselves are treated in many places, I've not been able to find the cohomology reconstruction theorem anywhere except in Postnikov's original <a href="http://mi.mathnet.ru/tm1182">longer monograph</a> (in Russian).</p>
| Ronnie Brown | 19,949 | <p>One question which has puzzled me is, just to take the <span class="math-container">$2$</span>-type, how does one specify an element of <span class="math-container">$H^3(\pi_1,\pi_2)$</span>?</p>
<p>In 1972, Philip Higgins and I discussed with Saunders Mac Lane the possibility of a <span class="math-container">$2$</span>-dimensional van Kampen theorem to calculate the homotopy <span class="math-container">$2$</span>-type of a union, and he explained why he had decided this was impossible. Calculating <span class="math-container">$\pi_1$</span> is OK, by the van Kampen theorem. Then you have to calculate <span class="math-container">$\pi_2$</span>, as a <span class="math-container">$\pi_1$</span>-module; then the <span class="math-container">$k$</span>-invariant! Absurd!</p>
<p>Now the <span class="math-container">$2$</span>-type is also described by a crossed module, by a theorem of Mac Lane and Whitehead. Philip and I proved in 1975, in a <a href="http://groupoids.org.uk/pdffiles/brown-higgins-connection.pdf" rel="nofollow noreferrer">paper</a> published in 1978, that under reasonable circumstances, the crossed module of a union is given by a pushout of crossed modules. That seems some kind of answer!</p>
<p>Of course, there is still a problem of calculating <span class="math-container">$\pi_2$</span> from this pushout, and even more of calculating the <span class="math-container">$k$</span>-invariant. But at least the topology has been translated into algebra; the theorem generalises strongly a theorem of Whitehead on free crossed modules, and has enabled specific, even computer, calculations of <span class="math-container">$2$</span>-types. For more detail, see this 2011 EMS <a href="http://groupoids.org.uk/nonab-a-t.html" rel="nofollow noreferrer">Tract vol 15</a>.</p>
<p>In effect, crossed modules are a useful <strong>algebraic model</strong> of homotopy <span class="math-container">$2$</span>-types, and it is certainly useful to consider what one wants from an algebraic model.</p>
<p>Thus Loday worried that his cat<span class="math-container">$^n$</span>-group model (called by him <span class="math-container">$n$</span>-cat groups in his 1982 paper) of homotopy <span class="math-container">$(n+1)$</span>-types was "purely formal", so was very happy with our van Kampen theorem for <span class="math-container">$n$</span>-cubes of spaces which enabled some new computations, for example of homotopy <span class="math-container">$3$</span>-types of a suspension of a <span class="math-container">$K(G,1)$</span>. More applications and related algebra were developed by Ellis and Steiner, in a 1987 JPAA paper.</p>
<p>It also seems true that strict <span class="math-container">$n$</span>-fold groupoids model weak homotopy <span class="math-container">$n$</span>-types. How does one use this? and how is it related to Postnikov systems?</p>
<p>To add a bit to a possible answer to the question on cohomology, Graham Ellis has a paper on the cohomology of crossed modules.</p>
<p>Let the debate continue!</p>
<p>May 7, 2019 Looking at my answer, I first updated two links and then realised there is more to be said on modelling homotopy types, part of which is in <a href="https://arxiv.org/abs/1610.07421" rel="nofollow noreferrer">this paper</a>, published in 2018.</p>
<p>The general problem of relating cohomology to models of homotopy types seems very difficult.</p>
<p>May 23, 2019 I should also have referred you to Loday's paper ‘Spaces with finitely many nontrivial homotopy groups’. J. Pure Appl. Algebra
24 (2) (1982) 179–202. and Ellis, G. J. and Steiner, R. ‘Higher-dimensional crossed modules and the homotopy groups of
(n + 1)-ads’. J. Pure Appl. Algebra 46 (2-3) (1987) 117–136.</p>
|
495,863 | <p>I am trying to show that this sequence $\{a_n\}$ = (2n+1)/(3n+5) does not converge to $42$ if it is not bounded above. I have already showed that it converges to $2/3$. For this I want to use a proof by contradiction,i.e, I assume that the sequence does converge to $42$, which will lead to a contradiction to the initial assumption that is the sequence will be bounded above. Any ideas or theorems that can help me solve this question? </p>
| André Nicolas | 6,312 | <p>We will show that there is an $\epsilon\gt 0$ such that
$$\left|\frac{2n+1}{3n+5}-42\right|\gt \epsilon$$
for infinitely many $n$. </p>
<p>Pick $\epsilon=1$. It is obvious that for positive $n$, we have $\frac{2n+1}{3n+5}\lt 1$. It follows that for all positive $n$, we have $\left|\frac{2n+1}{3n+5}-42\right|\gt 41\gt \epsilon$. </p>
|
3,926,580 | <p>I'm trying to prove with squeeze theorem that the limit of the following series equals 1:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the left side of the inequality I did:</p>
<p><span class="math-container">$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} < \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the right side, at first I did the following:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} < \frac{n\sqrt[n]{n}}{n}$$</span></p>
<p>But then I realized it wasn't true and that the direction of this inequality is the opposite.</p>
<p>Do you have any idea which series with limit 1 is bigger from the original series?</p>
<p>Thanks!</p>
| Misha Lavrov | 383,078 | <p>You'll want to know some things about how big <span class="math-container">$\sqrt[n]{n}$</span> is. The key facts to prove are:</p>
<ul>
<li>For <span class="math-container">$n$</span> a positive real number, it's increasing when <span class="math-container">$n < e$</span> and decreasing when <span class="math-container">$n>e$</span>. For integers, <span class="math-container">$3^{1/3} \approx 1.44$</span> is the largest value, with <span class="math-container">$2^{1/2} \approx 1.41$</span> taking second place.</li>
<li>As <span class="math-container">$n \to \infty$</span>, <span class="math-container">$\sqrt[n]{n} \to 1$</span>. A more precise estimate of <span class="math-container">$\sqrt[n]{n}$</span> as <span class="math-container">$n \to \infty$</span> is <span class="math-container">$1 + \frac{\log n}{n}$</span>, but we won't need it.</li>
</ul>
<p>So we are averaging a few large terms, and many many terms close to <span class="math-container">$1$</span>. One good way to deal with a situation like that with the squeeze theorem is to separate into two parts:
<span class="math-container">$$
\frac1n \sum_{k=1}^n \sqrt[k]{k} = \frac1n \sum_{k=1}^{\sqrt n}\sqrt[k]k + \frac1n \sum_{k=\sqrt{n}+1}^{n}\sqrt[k]k.
$$</span>
What can we say about these two parts?</p>
<ul>
<li>In the first sum, we have <span class="math-container">$\sqrt n$</span> terms, each of which is at most <span class="math-container">$3^{1/3}$</span>. So the sum is at most <span class="math-container">$3^{1/3} \sqrt n$</span>, and we're dividing by <span class="math-container">$n$</span>. This sum goes to <span class="math-container">$0$</span>.</li>
<li>In the second sum, we have nearly <span class="math-container">$n$</span> terms, each of which is less than <span class="math-container">$\sqrt[k]{k}$</span> for <span class="math-container">$k = \sqrt n$</span>. So they add up to less than <span class="math-container">$n \sqrt[k]{k}$</span>. When we divide by <span class="math-container">$n$</span>, we get <span class="math-container">$\sqrt[k]{k}$</span> where <span class="math-container">$k=\sqrt n$</span>, and this approaches <span class="math-container">$1$</span> as <span class="math-container">$n \to \infty$</span>.</li>
</ul>
<p>(The specific cutoff of <span class="math-container">$\sqrt n$</span> is very flexible: any function <span class="math-container">$1 \ll f(n) \ll n$</span> would do.)</p>
|
3,926,580 | <p>I'm trying to prove with squeeze theorem that the limit of the following series equals 1:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the left side of the inequality I did:</p>
<p><span class="math-container">$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} < \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the right side, at first I did the following:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} < \frac{n\sqrt[n]{n}}{n}$$</span></p>
<p>But then I realized it wasn't true and that the direction of this inequality is the opposite.</p>
<p>Do you have any idea which series with limit 1 is bigger from the original series?</p>
<p>Thanks!</p>
| CHAMSI | 758,100 | <p>First of all we have, for any <span class="math-container">$ n\in\mathbb{N}^{*} $</span>, the following : <span class="math-container">$$ \sqrt[n]{n}=1+\frac{\ln{n}}{n}\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x} $$</span></p>
<p>Since : <span class="math-container">\begin{aligned}0\leq\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x}&\leq n^{\frac{1}{n}}\\ &\leq 2\end{aligned}</span></p>
<p>We have : <span class="math-container">\begin{aligned} 1\leq \sqrt[n]{n}\leq 1+\frac{2\ln{n}}{n}&=1+\frac{4\ln{\sqrt{n}}}{n}\\ &\leq 1+\frac{4\sqrt{n}}{n}= 1+\frac{8}{2\sqrt{n}}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq1+\frac{8}{\sqrt{n}+\sqrt{n-1}} \end{aligned}</span></p>
<p>That remains true for any <span class="math-container">$ n\in\mathbb{N}^{*} $</span>, which means given <span class="math-container">$ n\in\mathbb{N}^{*} $</span>, we have : <span class="math-container">\begin{aligned} 1\leq\frac{1}{n}\sum_{k=1}^{n}{\sqrt[k]{k}}&\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\frac{1}{\sqrt{k}+\sqrt{k-1}}} \\ &\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\left(\sqrt{k}-\sqrt{k-1}\right)}\\ &\leq 1+\frac{8}{\sqrt{n}} \end{aligned}</span></p>
<p>Thus, using the squeezing theorem, the limit would be <span class="math-container">$ 1 \cdot$</span></p>
|
1,221,206 | <p>In this computer, numbers are stored in <span class="math-container">$12$</span>-bits. We will also assume
that for a floating point (real) number, <span class="math-container">$6$</span> bits of these bits are reserved for
the mantissa (or significand) with <span class="math-container">$2^{k-1}-1$</span> as the exponent bias (where
<span class="math-container">$k$</span> is the number of bits for the characteristic).</p>
<p><span class="math-container">$011100100110010111110011$</span></p>
<p>What pair of floating point numbers could be represented by these
<span class="math-container">$24$</span>-bits?</p>
<p>I have gone this far:</p>
<p>As described above that each number is of <span class="math-container">$12$</span> bit so we get each number </p>
<p><span class="math-container">$011100100110$</span></p>
<p>First one is <span class="math-container">$0$</span> bit so it is positive and </p>
<p>Mantissa will be <span class="math-container">$100110$</span></p>
<p>Exponent will be <span class="math-container">$11100b=28$</span></p>
<p>my unbiased exponent will be <span class="math-container">$2^{28-15}=2^{13}$</span></p>
<p>How to find the floating point number from here?</p>
| Olivier Oloa | 118,798 | <p>One may observe that, by expanding and noticing the cancellations,
$$
(1-x)(1+x+x^2+...+x^n)=1-x^{n+1},
$$
giving $$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad x \neq 1, \tag1
$$ and then put $x=\dfrac14$ in $(1)$.</p>
|
15,875 | <p>It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is <em>continuously</em> differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or <a href="http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations" rel="noreferrer">Wikipedia</a>). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below. </p>
<p>Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then<br>
\begin{equation*}
u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|)
\end{equation*}
and
\begin{equation*}
v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|).
\end{equation*}
Multiplying the second equation by $i$ and adding the two together gives
\begin{align*}
(u+iv)(z+h)-(u+iv)(z) &= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\
&= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|).
\end{align*}
Now dividing by $h$ gives us the desired result. </p>
<p>Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function? </p>
| NebulousReveal | 2,548 | <p>See <a href="http://www.jstor.org/stable/2321164" rel="noreferrer"><em>When is a Function that Satisfies the Cauchy-Riemann Equations Analytic?</em></a> J. D. Gray and S. A. Morris
The American Mathematical Monthly
Vol. 85, No. 4 (Apr., 1978), pp. 246-256. </p>
|
1,482,564 | <p>In school I'm learning about concavity and finding points of inflection using the second-derivative test. A typical question will look like</p>
<blockquote>
<p><strong>Determine the open intervals on which the graph is concave upward or concave downward.</strong></p>
</blockquote>
<p>Here is how I would solve a problem like that.</p>
<ol>
<li><p>Find the derivative, and then second derivative of f.</p>
</li>
<li><p>Find the critical numbers of f'(x) by setting f''(x) = 0 and f''(x) is undefined, then simplify. Let's say I get the critical numbers <strong>a</strong> and <strong>b</strong> (a < b).</p>
</li>
<li><p>Take f'' of a number in the open interval <span class="math-container">$(-\infty, a)$</span> If it is negative, f is concave downward on <span class="math-container">$(-\infty, a)$</span> if it is positive, it is concave upward. Repeat this process for a number in <span class="math-container">$(a, b)$</span> and in <span class="math-container">$(b, \infty)$</span></p>
</li>
</ol>
<p>In general, I will get a result like</p>
<blockquote>
<p>f is concave upward on <span class="math-container">$(-\infty, a) \cup (b, \infty)$</span></p>
<p>and concave downward on <span class="math-container">$(a, b)$</span></p>
</blockquote>
<p>But now I'm wondering, is it possible for the concavity to stay the same even on an interval containing a critical number of f prime? For example, if f'(x) is increasing when x < a, f'(x) = 0 @ x = a, and f'(x) is still increasing when a < x < b? Does that mean that f is concave upward on <span class="math-container">$(-\infty, b)$</span> Can this even happen?</p>
| Paul Sinclair | 258,282 | <p>Concavity generally stays the same the same on intervals containing critical numbers. The exception is when the critical number is an inflection point instead of a maximum or minimum. Think about it: consider a typical maximum. The curve reaches a high point and then heads back down. Does it switch from concave down to concave up here? No. It is concave down on both sides. A typical minimum is the just the opposite. It will be concave up on both sides. I don't know where you got the idea that $f''$ is undefined at critical points. There is no particular reason for it to be.</p>
<p>For a function $f(x)$ whose 2nd derivative is continuous:</p>
<ul>
<li>A <em>critical point</em> is a point $x$ where $f'(x) = 0$. If $f''(x) > 0$, then the point is a minimum. If $f''(x) < 0$, then the point is a maximum, and if $f''(0) = 0$, then it could be any of the three (maximum, minimum, or inflection point).</li>
<li>An <em>inflection point</em> is a point $x$ where $f''$ switches from positive to negative, or vice versa. Since $f''$ is continuous, this necessarily means that $f''(x) = 0$, but the inverse is not necessarily true: $f''(x)$ can be $0$ without it being an inflection point ($f(x) = x^4$ at $0$, for example).</li>
</ul>
<p>As a specific example following your pattern, consider $f(x) = x^3 - 3x$. Then $f'(x) = 3x^2 - 3$ has roots at $-1$ and $1$. By your pattern, you could look at $f''(x) = 6x$ at say $-2, 1/2, 2$ and claim that $f$ is concave downward on $(-\infty, -1)$ and concave up on $(-1,1)$ and on $(1, \infty)$. But if you chose $-1/2$ instead of $1/2$, you would have it concave down on $(-1, 1)$.</p>
<p>The truth is that $f''(x) = 6x = 0$ at $x = 0$. Further $f''(x) < 0$ for $x \le 0$ and $f''(x) > 0$ for $x > 0$. So $f$ is concave down on $(-\infty, 0)$ and concave up on $(0, \infty)$.</p>
|
4,551,543 | <p>You have ten coins, each with a bias for head drawn from a uniform distribution [0,1]. For example, a coin with a 0.7 bias means that the probability of flipping a head is 0.7. You are allowed 100 flips and you are free to choose any coins to flip. Each time you flip a head, you get $1. How much money would you pay to play this game and what is your strategy?</p>
<p>Here's what I have in mind so far: The most basic strategy is to each time pick a coin at random and flip it. The expected payoff is $50. To do better than that, I need to know which coins have biases bigger than 0.5. It seems that I could use a Bayesian approach to update my belief about each coin, but doing so would require multiple rounds, and I'm not sure how to actually quantify this.</p>
| quasi | 400,434 | <p>Let <span class="math-container">$C$</span> be the number of coins, and let <span class="math-container">$N$</span> be the number of flips (e.g., <span class="math-container">$C=10,\,N=100$</span>).</p>
<p>
Here's a proposed strategy which I think is pretty good, although I'm fairly certain it's not optimal . . .
<p>
When there are <span class="math-container">$n$</span> flips remaining, choose a coin for which
<span class="math-container">$$
\frac{h+1+\min(1,n-1)}{h+t+2+\min(1,n-1)}
$$</span>
is greatest, (where <span class="math-container">$h,t$</span> are the prior number of heads, tails, respectively for that coin), and if there's more than one such coin, choose one for which <span class="math-container">$h$</span> is greatest.
<p>
Let <span class="math-container">$e(C,N)$</span> be the value of the game, assuming the above strategy.
<p>
For relatively small values of <span class="math-container">$C,N$</span>, the exact value of <span class="math-container">$e(C,N)$</span> can be computed by recursion.
<p>
For example, a Maple implementation yields
<span class="math-container">$$
e(4,12)
=
\frac{1042111771}{129729600}
\approx
8.032952934
$$</span>
For larger values of <span class="math-container">$C,N$</span>, the value of <span class="math-container">$e(C,N)$</span> can approximated via simulation.
<p>
For example, simulation yields <span class="math-container">$82 < e(10,100) < 83$</span>.
<p>
For reference, here's the Maple code which I used to compute <span class="math-container">$e(4,12)$</span>.
<p><a href="https://i.stack.imgur.com/YoZzt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YoZzt.png" alt="enter image description here" /></a></p>
|
2,491,943 | <p>In general, for two sets $A$ and $B$, $|A^B|$=the number of functions from $B$ into $A$. </p>
<p>Can someone please show that $|\{1,2\}^{\{3,4,5\}}|=8$ by showing that there are eight functions from $\{3,4,5\}$ to $\{1,2\}$?</p>
| Olivier Oloa | 118,798 | <p><strong>Hint</strong>. By using polar coordinates one gets, for $(x,y)\ne (0,0)$,
$$
g(x,y)=\frac{x^3+xy^2+2x^2+2y^2}{x^2+y^2}=\frac{r^3\cos^3 \theta+r^3\cos\theta \sin^2\theta +2r^2}{r^2},\quad r\ne0,
$$ that is
$$
g(x,y)=r\cos^3 \theta+r\cos\theta \sin^2\theta +2,\quad r\ne0,
$$ then this tends to $2$ as $r \to 0^+$.</p>
<p>Can you take it from here?</p>
|
2,491,943 | <p>In general, for two sets $A$ and $B$, $|A^B|$=the number of functions from $B$ into $A$. </p>
<p>Can someone please show that $|\{1,2\}^{\{3,4,5\}}|=8$ by showing that there are eight functions from $\{3,4,5\}$ to $\{1,2\}$?</p>
| Ahmed S. Attaalla | 229,023 | <p>For $(x,y) \neq (0,0)$, with some simplifications we may get:</p>
<p>$$g(x,y)=x+2$$</p>
<p>In order for the function to be continuous at the origin, $g(0,0)$ must exist and we must have that $\lim_{x \to (0,0)} g(x,y)=g(0,0)$.</p>
<p>When is that the case? </p>
<blockquote class="spoiler">
<p> When $c=g(0,0)=\lim_{(x,y) \to (0,0)} g(x,y)=2$</p>
</blockquote>
|
800,802 | <p>Can someone give me a reference in which I can find the following result </p>
<p>Let $C$ be a curve, then $$g(C)=\frac{(n-1)(n-2)}{2}-s.$$</p>
<p>where $g=$ genus of $C,$ $n=$ degree of curve, $s=$ number of singular points.</p>
<p>Any help would be appreciated.</p>
| Hoot | 127,490 | <p>I know this is done in a series of exercises (this may be good news or not) at the end of Appendix A in Arbarello, Cornalba, Griffiths, and Harris, <em>Geometry of Algebraic Curves, Volume I</em>. They call it the "Gorenstein relation".</p>
|
978,202 | <p>When I use $\sin x \sim x$ , the answer is $1$ , is the answer correct?</p>
| John | 105,625 | <p>No. $|sinx^2| \leq 1$, then apply squeeze theorem, you get 0 </p>
<p>And you can only use $x$ to approximate $sinx$ when $x$ is close to 0. This is because $\lim_{x\to 0} \frac{sinx}{x}=1$. It's not true when $x$ is large.</p>
|
978,202 | <p>When I use $\sin x \sim x$ , the answer is $1$ , is the answer correct?</p>
| Hakim | 85,969 | <p><strong>Hint:</strong> $$-\dfrac{1}{x^2+x}\leqslant\dfrac{\sin x^2}{x^2+x}\leqslant\dfrac{1}{x^2+x},\tag{$x\gt0$}$$ then squeeze everything.</p>
|
4,426,699 | <p>I want to solve this functional</p>
<p><span class="math-container">\begin{equation}
L(y)=\int_{-1}^7\sqrt{1+y'^2} \ dx
\end{equation}</span></p>
<p>with IC: <span class="math-container">$y(0)=1,\ y(1)=2$</span></p>
<p>I start using the Euler Lagrange equation</p>
<p><span class="math-container">\begin{equation}
\frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0
\end{equation}</span></p>
<p>But I encounter immediately a problem, namely that</p>
<p><span class="math-container">$\frac{d}{dx}\frac{\partial F}{\partial y'}=0$</span>, since <span class="math-container">$\frac{\partial F}{\partial y'}=\frac{y'}{\sqrt{1+y'^2}}$</span> which has no x-variable. Also <span class="math-container">$\frac{\partial F}{\partial y}=0$</span>, hence it does not make sense.</p>
<p>Alterantively, I set this equal to a constant instead, and get:</p>
<p><span class="math-container">\begin{equation}
\begin{array}
f\frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}}=C \\
\frac{y'}{\sqrt{1+y'^2}}=C_1x + C_3\\
\frac{y'}{\sqrt{1+y'^2}}=C_2x+C_4
\end{array}
\end{equation}</span></p>
<p>How do I solve this?</p>
<p>Thanks</p>
| Dr. Sundar | 1,040,807 | <p>Method I:</p>
<p>Obviously, the functional
<span class="math-container">$$
L(y)=\int_{a}^b \ \sqrt{1+y'^2} \ dx
$$</span>
represents the arc-length of a curve, and the variational problem reduces to finding a curve of shortest arc-length connecting two points on the plane. This is one of the classical problems in the <em>Calculus of Variations</em>.</p>
<p>Geometrically, the well-known answer is a straight line, <span class="math-container">$$y = m x + c.$$</span></p>
<p>The constants <span class="math-container">$m$</span> and <span class="math-container">$c$</span> can be found from the given boundary conditions. (we need two conditions basically)</p>
<p>You have given the conditions in the problem as
<span class="math-container">$$
y(0) = 1, \ \ y(1) = 2
$$</span></p>
<p>Substituting, we get
<span class="math-container">$$
1 = m (0) + c \ \ \mbox{or} \ \ c = 1
$$</span>
<span class="math-container">$$
2 = m(1) + c \ \ \mbox{or} \ \ m = 2 - c = 2 - 1 = 1
$$</span></p>
<p>Hence, the integral of the given variational problem is
<span class="math-container">$$
y = x + 1
$$</span>
(Clearly, it satisfies the conditions <span class="math-container">$y(0) = 1, y(1) = 2$</span>.</p>
<p>Method II:</p>
<p>Using the Euler-Lagrange equation
<span class="math-container">$$
{\partial f \over \partial y} -
{d \over dx}\left({\partial f \over \partial y'}
\right) = 0 \tag{1}
$$</span></p>
<p>Since <span class="math-container">$f$</span> is purely a function of <span class="math-container">$y'$</span>, we have
<span class="math-container">$$
{\partial f \over \partial y} = 0
$$</span></p>
<p>Thus, (1) simplifies to
<span class="math-container">$$
{d \over dx}\left({\partial f \over \partial y'}
\right) = 0
$$</span>
or
<span class="math-container">$$
{\partial f \over \partial y'} = \mbox{constant}
$$</span></p>
<p>Thus, we get
<span class="math-container">$$
{1 \over 2 \sqrt{1 + y'^2}} (2 y') = k_1
$$</span>
or
<span class="math-container">$$
{y' \over \sqrt{1 + y'^2}} = k_1
$$</span></p>
<p>Simplifying, we get
<span class="math-container">$$
1 + y'^2 = k_1 y'^2
$$</span>
where <span class="math-container">$k_1 = {1 \over c_1^2} = $</span>constant</p>
<p>Simplifying again, we get
<span class="math-container">$$
y' = \mbox{constant} = m (\mbox{say})
$$</span></p>
<p>Integrating, we get
<span class="math-container">$$
y = m x + c
$$</span>
which is a straight line.</p>
<p>The constants <span class="math-container">$m$</span> and <span class="math-container">$c$</span> can be found as in Method I.</p>
<p>Finally, we arrive at the optimal solution as
<span class="math-container">$$
y = x + 1
$$</span></p>
|
2,193,306 | <p>I have to evaluate the integral of the $\sec x$ function and I do it as follows $$\int\sec xdx=\int\frac{dx}{\cos x}=\int\frac{\cos^2 x+\sin^2 x}{\cos x}dx=\sin x+\int\frac{\sin^2x}{\cos x}dx$$ Now we make a change of variables $\cos x=t$ so our integral becomes $$\sin x-\int\frac{\sqrt{1-t^2}}{t}dt$$ Now we make another change of variable $\sqrt{1-t^2}=z$ so our integral becomes $$\sin x-\int\frac{z^2-1+1}{z^2-1}dz=\sin x-\int\left(1+\frac{1}{z^2-1}\right)dz=$$ $$=\sin x-\sqrt{1-\cos^2x}+\frac12\ln\left|\frac{1+\sqrt{1-\cos^2x}{}}{1-\sqrt{1-\cos^2x}}\right|+C=\frac12\ln|\tan^2x+\sec^2x|+C$$ The <a href="http://www.integral-calculator.com/" rel="nofollow noreferrer">calculator</a> however evaluates this integral as $$\ln|\tan x+\sec x|+C$$
but I can't figure out where I made a mistake in my calculations.</p>
| Community | -1 | <p><strong>The fast way</strong>:</p>
<p>$$\int\frac{dx}{\cos x}=\int\frac{\cos x\,dx}{\cos^2x}=\int\frac{\sin'x\,dx}{1-\sin^2x}=\text{artanh}(\sin x).$$</p>
|
3,995,954 | <p>Today during a Calc I exam we(me and my classmates) have been ask to prove <span class="math-container">$$-a\cdot e\cdot \ln(x)\le x^{-a}$$</span>, <span class="math-container">$\forall x>0$</span> and <span class="math-container">$\forall a\in \mathbb{R}$</span>. But, noneone in the room has known how to do it. Anyone knows how to?</p>
<p>Thanks in advance</p>
| Barry Cipra | 86,747 | <p>The inequality is obviously true if <span class="math-container">$a=0$</span>. If <span class="math-container">$a\not=0$</span>, let <span class="math-container">$x=e^{-u/a}$</span>. The inequality to prove becomes <span class="math-container">$eu\le e^u$</span> for all <span class="math-container">$u\in\mathbb{R}$</span>. Now <span class="math-container">$f(u)=e^u-eu$</span> has a unique critical point at <span class="math-container">$u=1$</span>, that being where <span class="math-container">$f'(u)=e^u-e=0$</span>. Since <span class="math-container">$f''(u)=e^u\gt0$</span> for all <span class="math-container">$u$</span>, the critical point at <span class="math-container">$u=1$</span> is a global minimum, and thus</p>
<p><span class="math-container">$$e^u-eu=f(u)\ge f(1)=e^1-e\cdot1=0$$</span></p>
<p>for all <span class="math-container">$u$</span>, which proves the desired inequality.</p>
|
3,984,652 | <p>Let <span class="math-container">$(X_n,d_n) $</span> a family of metric spaces and <span class="math-container">$(X=\prod_{n \in \mathbb N} X_n,\mathcal d ) $</span> a metric space with <span class="math-container">$\mathcal T_p$</span> the product topology is equal to the topology defined by the distance <span class="math-container">$d$</span>, such that for all <span class="math-container">$x=(x_n)_n, y=(y_n)_n\in X$</span> we have <span class="math-container">$d(x,y) = \sum_{n=0}^{\infty} \frac{1}{2^n}d_n(x_n,y_n) $</span> and <span class="math-container">$d_n(x,y) \leq M$</span> for all <span class="math-container">$x, y\in X_n$</span>.</p>
<p>Show that <span class="math-container">$X$</span> is compact iff <span class="math-container">$X_n$</span> compact for all <span class="math-container">$n\in \mathbb N$</span>.</p>
<hr />
<p>For the first implication, I have used the the fact that <span class="math-container">$X$</span> is compact and the projection maps are continuous. For the other one, I get stuck with it. any help is appreciated.</p>
| Siong Thye Goh | 306,553 | <p>I think you have an indexing problem, if you start indexing from <span class="math-container">$0$</span> and in total there are <span class="math-container">$n$</span> digits, the last index should be <span class="math-container">$n-1$</span>.</p>
<p><span class="math-container">$$N = \sum_{i=0}^{n-1} a_i \cdot 10^i, \text{ where } 0 \le a_i \le 9$$</span></p>
<p>For example, the number <span class="math-container">$123$</span> has <span class="math-container">$a_0=3, a_1=2, a_2=1$</span>.</p>
<p>In contrast, if you start your index from <span class="math-container">$1$</span>, then the last index should be <span class="math-container">$n$</span>.</p>
<p><span class="math-container">$$N = \sum_{i=1}^{n} b_i \cdot 10^{i-1}, \text{ where } 0 \le b_i \le 9$$</span></p>
<p>For example, the number <span class="math-container">$123$</span> has <span class="math-container">$b_1=3, b_2=2, b_3=1$</span>.</p>
|
1,373,759 | <p>Given the natural number $n$,who is in the form $p^2 \cdot q^2$,with $p$,$q$ prime numbers.Also $φ(n)$ is given.Describe a fast algorithm(polynomial time) that calculates the $p$ and $q$.Apply your algorithm to calculate the $p$,$q$ when $n=34969$ and $φ(n)=29920$</p>
<p>I found this problem on a mathematical competition on cryptography.I tried to find a solution alone and in the internet and i didn't conclude anywhere, can we find a solution?</p>
| callculus42 | 144,421 | <p>I post the derivative w.r.t T. You can first seperate the variable T from the others:</p>
<p><span class="math-container">$n=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \sqrt{T}$</span></p>
<p>Then you write <span class="math-container">$\sqrt T$</span> as <span class="math-container">$T^{1/2}$</span></p>
<p><span class="math-container">$n=\underbrace{\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}}_{\large{\begin{matrix} \texttt{constant in case of} \\ \texttt{ differentiation w.r.t. T} \end{matrix}}}\cdot T^{1/2}$</span></p>
<p>Therefore you need the derivative of <span class="math-container">$T^{1/2}$</span></p>
<p><span class="math-container">$\frac{dn}{dT}=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \frac{1}{2}\cdot T^{1/2-1}=\frac{1}{DL}\cdot \sqrt{\frac{g}{\pi \sigma}}\cdot \frac{1}{2}\cdot T^{-1/2}$</span></p>
<p>And <span class="math-container">$T^{-1/2}=\frac{1}{\sqrt{T}}=\sqrt{\frac{1}{T}}$</span></p>
<p>Now you can put all the terms under the radical together.</p>
<p><span class="math-container">$\frac{dn}{dT}=\frac{1}{DL}\cdot \sqrt{\frac{g}{T\pi \sigma}}\cdot \frac{1}{2}=\frac{dn}{dT}=\frac{1}{2DL}\cdot \sqrt{\frac{g}{T\pi \sigma}}$</span></p>
<p>You can try the other derivative for yourself.</p>
|
4,143,759 | <p><strong>What I understand so far :</strong></p>
<p>Intuitively this question gives me a yuuuuuge hint, it tells me at t=0, t = -4. So that let's me know that I have a phase shift of -4 immediately.</p>
<p>I understand what a period is, I understand what a phase shift is, I understand what amplitude and vertical displacement are and how they work.</p>
<p>I'm just not sure how to build a function even with that info at my disposal.</p>
<p><em><strong>My ultimate question here isn't to find an answer, but to be given a hint or a pointing in the right direction. Is there anything that you, my acquaintances at stackexchange have to offer?</strong></em></p>
| MPW | 113,214 | <p><strong>Hint:</strong> Think of <span class="math-container">$V$</span> and <span class="math-container">$r$</span> as being functions of <span class="math-container">$t$</span> (time), so that
<span class="math-container">$$V(t) = \frac43\pi \cdot r(t)^3$$</span>
and therefore by the chain rule
<span class="math-container">$$V'(t) = 4\pi\cdot r(t)^2 \cdot r'(t)$$</span>
You know <span class="math-container">$r(t_0)$</span> and <span class="math-container">$V'(t_0)$</span> at the instant <span class="math-container">$t_0$</span> in question, so this will determine <span class="math-container">$r'(t_0)$</span> at that instant as well.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Georges Elencwajg | 450 | <p>[In front of a blackboard, in an office at Real College] </p>
<p>Skeptic: And why should I care about holomorphic functions?</p>
<p>Holomorphic enthusiast:$\;$ Can you compute $\quad$ $\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}$ ? Here $a$ is one of your cherished real numbers, but not an integer.</p>
<p>Skeptic: Well, hm...</p>
<p>Holomorphic enthusiast, nonchalantly: Oh, you just get</p>
<p>$$\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}=\pi^2 cosec^2 \pi a $$</p>
<p>It's easy using residues.</p>
<p>Skeptic: Well, maybe I should have a look at these "residues".</p>
<p>Holomorphic enthusiast (generously): Let me lend you this introduction to Complex Analysis by Remmert, this one by Lang and this oldie by Titchmarsh. As Hadamard said: "Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe".You can look for a translation at Mathoverflow. They have a nice list of mathematical quotations, following a question there.</p>
<p>Skeptic: Mathoverflow ??</p>
<p>Holomorphic enthusiast (looking a bit depressed) : I think we should have a nice long walk together now.
[Exeunt] </p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Georges Elencwajg | 450 | <p>Let us call "division algebra over $\mathbb R$" a finite-dimensional vector space $A$ equipped with a bilinear map $A \times A \to A: (a,b) \mapsto a \bullet b$ , such that $a\bullet b=0$ implies $a=0$ or $b=0$. ( Associativity is not required).</p>
<p>Examples : the reals, the complexes, the real quaternions and the octonions of Graves-Cayley.</p>
<p>Any such division algebra must necessarily have dimension 1,2,4 or 8 (as in the examples).
This was proved indepently in 1958 by Kervaire and Milnor using Bott's periodicity theorem, a fantastic result in algebraic topolgy which had just been proved.</p>
<p>To the best of my knowledge there is still no purely algebraic proof of this theorem on possible dimensions of real division algebras, although the statement is completely algebraic and elementary.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Mariano Suárez-Álvarez | 1,409 | <p>One problem that one can solve with Fourier analysis very easily is the isoperimetrical inequality and the corresponding characterization of the circle. Of course, this can also be done in many. many other ways, but using Fourier series it becomes particularly simple.</p>
|
3,808,111 | <p>Let <span class="math-container">$(X_n)_{n \geq 1}$</span> be a sequence of pairwise independent random variables such that :</p>
<p><span class="math-container">$$\sum_{n=1}^{\infty} n^{-1} P\left\{\max _{1 \leq m \leq n}\left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|>\varepsilon n\right\}<\infty$$</span></p>
<p>show that <span class="math-container">$n^{-1} \sum_{k=1}^{n}\left(X_{k}-E X_{k}\right) \rightarrow 0$</span> almost surely.</p>
<p>I'm fairly certain Borel Cantelli Lemma for pairwise independent random variables is to be used here but I dont know how to get rid of the <span class="math-container">$n^{-1}$</span> inside the series.</p>
| user | 505,767 | <p>Yes it can be either a maximum/minimum or it could be an inflection point.</p>
<p>Let consider for example</p>
<p><span class="math-container">$$f(x)=x^3 \implies f'(x)=3x^2\quad f''(x)=6x$$</span></p>
<p>and in this case <span class="math-container">$x=0$</span> is an inflection point.</p>
<p>For the case</p>
<p><span class="math-container">$$f(x)=x^4 \implies f'(x)=4x^3\quad f''(x)=12x^2$$</span></p>
<p>to check the minimum by derivatives we need to consider the fourth derivative which is indeed positive.</p>
<p>More in general the nature of the point is given by the sign of the first even derivative not equal to zero at that point.</p>
|
106,466 | <ol>
<li>I know that rational numbers are order-isomorphic to real algebraic numbers. Does it imply that irrational numbers are order-isomorphic to real transcendental numbers?</li>
<li>I know that the order type of rationals $\eta$ is a homogeneous order type (meaning that for any two its elements there is an automorphism that sends one to another). Are the order types of irrationals and real transcendental numbers homogeneous as well?</li>
</ol>
| Michael Hardy | 11,667 | <p>Suppose $f$ is a strictly increasing bijection function from the linearly ordered set of all rational numbers to the linearly ordered set of all real algebraic numbers. Let $g:\mathbb R \to \mathbb R$ be defined as follows:</p>
<p>$$
g(x) = \sup \{ f(u) : u \in \mathbb Q \text{ and }u < x \}.
$$</p>
<p>If I'm not mistaken, one can show by definition-chasing that the restriction of $g$ to the irrationals is a strictly increasing bijection from that set to the real transcendental numbers.</p>
<p>To see that it's surjective, suppose $y$ is a real transcendental number and let</p>
<p>$$x=\sup\{ f^{-1}(w) : w \in \mathbb Q\text{ and }w < y\}.$$</p>
<p>Since $f$ is a strictly increasing bijection, the set whose sup is taken is a non-empty initial segment of $\mathbb Q$. Since some members of $\mathbb Q$ are $>y$, the complement of the set whose sup is taken is not empty. Hence the set has an upper bound in $\mathbb R$; hence the sup exists. The number $x$ cannot be rational since then $f(x)$ would be a rational number $<y$ and other rationals would be between that number and $y$, and their inverse-images under $f$ would be greater than $x$; but that contradicts the definition of $x$.</p>
|
106,466 | <ol>
<li>I know that rational numbers are order-isomorphic to real algebraic numbers. Does it imply that irrational numbers are order-isomorphic to real transcendental numbers?</li>
<li>I know that the order type of rationals $\eta$ is a homogeneous order type (meaning that for any two its elements there is an automorphism that sends one to another). Are the order types of irrationals and real transcendental numbers homogeneous as well?</li>
</ol>
| Brian M. Scott | 12,042 | <p>Here’s an answer to part of (2).</p>
<p>The order type of $\mathbb{P}$, the irrationals, is homogeneous, because $\langle\mathbb{P},\le\rangle$ is order-isomorphic to $\langle\mathbb{Z}^\omega,\preceq\rangle$, where $\preceq$ is the lexicographic order, which is order-homogeneous.</p>
<p>To construct the order-isomorphism, enumerate the rationals as $\mathbb{Q}=\{q_n:n\in\omega\}$. Recursively construct open intervals $I_\sigma$ for $\sigma\in\mathbb{Z}^{<\omega}$ to satisfy the following conditions.</p>
<ol>
<li>$I_{\langle\rangle}=\mathbb{R}$. </li>
<li>If $\langle\rangle\ne\sigma\in\mathbb{Z}^{<\omega}$, $I_\sigma$ is a non-empty open interval with rational endpoints. </li>
<li>For each $n\in\omega$, $q_n$ is an endpoint of some $I_\sigma$ with $|\sigma|\le n+1$. </li>
<li>For each $\sigma\in\mathbb{Z}^{<\omega}$ and $n\in\mathbb{Z}$, $I_{\sigma^\frown n}\subseteq I_\sigma$. </li>
<li>For each $\sigma\in\mathbb{Z}^{<\omega}$ and $n\in\mathbb{Z}$, the right endpoint of $I_{\sigma^\frown n}$ is the left endpoint of $I_{\sigma^\frown (n+1)}$. </li>
<li>For each $\sigma\in\mathbb{Z}^{<\omega}$, $\{I_{\sigma^\frown n}:n\in\mathbb{Z}\}$ covers all of $I_\sigma$ except the endpoints of the intervals $I_{\sigma^\frown n}$. </li>
<li>If $\langle\rangle\ne\sigma\in\mathbb{Z}^{<\omega}$, the length of $I_\sigma$ is less than $2^{-|\sigma|}$.</li>
</ol>
<p>Clauses (4)-(6) ensure that for each $\sigma\in\mathbb{Z}^{<\omega}$ and $n\in\mathbb{Z}$, $\operatorname{cl}I_{\sigma^\frown n}\subseteq I_\sigma$, so for each $\sigma\in\mathbb{Z}^\omega$, $$\bigcap\limits_{n\in\omega}I_{\sigma\upharpoonright n}\ne\varnothing\;.\tag{1}$$ Clause (7) ensures that the intersection in $(1)$ is at most a singleton, so for each $\sigma\in\mathbb{Z}^\omega$ there is a unique $h(\sigma)\in\mathbb{R}$ such that $$\bigcap\limits_{n\in\omega}I_{\sigma\upharpoonright n}=\{h(\sigma)\}\;.$$ Finally, clause (3) ensures that $h(\sigma)\in\mathbb{P}$, so $h:\mathbb{Z}^\omega\to\mathbb{P}$.</p>
<p>To see that $h$ is an injection, suppose that $\sigma,\tau\in\mathbb{Z}^\omega$, and $\sigma\ne\tau$. Let $n\in\omega$ be minimal such that $\sigma\upharpoonright n\ne\tau\upharpoonright n$; then by construction $I_{\sigma\upharpoonright n}\cap I_{\tau\upharpoonright n}=\varnothing$, so $h(\sigma)\ne h(\tau)$. To see that $h$ is surjective, simply observe that for any $x\in\mathbb{P}$ and any $n\in\omega$ there is a unique $\sigma\in\mathbb{Z}^n$ such that $x\in I_\sigma$. Thus, $h$ is a bijection, and it only remains to check that $h$ is order-preserving. </p>
<p>Suppose that $\sigma,\tau\in\mathbb{Z}^\omega$ with $\sigma\prec\tau$. Let $n\in\omega$ be minimal such that $\sigma(n)\ne\tau(n)$, and let $\varphi=\sigma\upharpoonright n=\tau\upharpoonright n$. Then $h(\sigma),h(\tau)\in I_\varphi$, $h(\sigma)\in I_{\varphi^\frown \sigma(n)}$, $h(\tau)\in I_{\varphi^\frown \tau(n)}$, and $\sigma(n)<\tau(n)$, so $h(\sigma)<h(\tau)$ by clause (5).</p>
<p>Finally, to see that $\langle\mathbb{Z}^\omega,\preceq\rangle$ is order-homogeneous, let $\sigma,\tau\in\mathbb{Z}^\omega$, and define $\delta\in\mathbb{Z}^\omega$ by $\delta(n)=\tau(n)-\sigma(n)$. Then the shift $$s:\mathbb{Z}^\omega\to\mathbb{Z}^\omega:\varphi\mapsto\langle\varphi(n)+\delta(n):n\in\omega\rangle$$ is an order-automorphism of $\langle\mathbb{Z}^\omega,\preceq\rangle$ taking $\sigma$ to $\tau$.</p>
|
1,117,924 | <p>How can I show with linear approximation that $y \approx x$ for small x? I know the rule $$f(x) \approx f(a) + f^{\prime}(a) (x-a),$$ but I don't know how to put it to use in this case.</p>
| Null_Space | 798,900 | <p>You are looking at the first degree <strong>Maclaurin series</strong> of <span class="math-container">$\ln{(1+x)}$</span></p>
|
569,960 | <p>Suppose $\mathcal{A}$ and $\mathcal{B}$ are categories with products, and $T$ a functor between them. If $X$ and $Y$ are objects in $\mathcal{A}$, what does it mean when we say there is a <em>natural</em> morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$?</p>
<p>In $\mathcal{A}$, we have the product $X\times Y$, with corresponding morphisms $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$. Under $T$, we get a diagram of objects in $\mathcal{B}$ of morphisms $T(\pi_1):T(X\times Y)\to T(X)$ and $T(\pi_2):T(X\times Y)\to T(Y)$. </p>
<p>Since products exist in $\mathcal{B}$, we have a product $(T(X)\times T(Y),p_1,p_2)$ such that there is a unique morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$ such that $p_1f=T(\pi_1)$ and $p_2f=T(\pi_2)$. </p>
<p>My guess is that this $f$ is the so called natural morphism, but I don't know how to verify that because I don't know what it means. I've only heard of natural transformations/isomorphisms between functors, but not natural morphisms between objects. Can anyone clarify?</p>
| Sig TM | 49,336 | <p>Consider the categories $\mathcal B^2$ and $\mathcal C^2$ of pairs of objects in $\mathcal B$ and $\mathcal C$, respectively. The Cartesian product on $\mathcal B$ is a functor $\times_\mathcal B$ from $\mathcal B^2$ to $\mathcal B$ and similarly for $\mathcal C$. There is also the functor $T^2:\mathcal B^2\to\mathcal C^2$ which applies $T$ to each object in a pair. The "natural morphism" you describe is a natural transformation from the functor $T\circ\times_\mathcal B$ to $\times_\mathcal C\circ T^2$.</p>
|
569,960 | <p>Suppose $\mathcal{A}$ and $\mathcal{B}$ are categories with products, and $T$ a functor between them. If $X$ and $Y$ are objects in $\mathcal{A}$, what does it mean when we say there is a <em>natural</em> morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$?</p>
<p>In $\mathcal{A}$, we have the product $X\times Y$, with corresponding morphisms $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$. Under $T$, we get a diagram of objects in $\mathcal{B}$ of morphisms $T(\pi_1):T(X\times Y)\to T(X)$ and $T(\pi_2):T(X\times Y)\to T(Y)$. </p>
<p>Since products exist in $\mathcal{B}$, we have a product $(T(X)\times T(Y),p_1,p_2)$ such that there is a unique morphism $f\colon T(X\times Y)\to T(X)\times T(Y)$ such that $p_1f=T(\pi_1)$ and $p_2f=T(\pi_2)$. </p>
<p>My guess is that this $f$ is the so called natural morphism, but I don't know how to verify that because I don't know what it means. I've only heard of natural transformations/isomorphisms between functors, but not natural morphisms between objects. Can anyone clarify?</p>
| Abolfazl Tarizadeh | 269,262 | <p>In (modern) mathematics those morphisms which arise from the definitions or the universal properties are called the $\mathbf{canonical}$ morphisms. The functorial morphisms are called the $\mathbf{natural}$ morphisms; for instance if $\phi:F\rightarrow G$ is a natural transformation between the two functors $F,G:\mathscr{C}\rightarrow\mathscr{D}$ then for each object $C$ of $\mathscr{C}$ the morphism $\phi_{C}:F(C)\rightarrow G(C)$ is called a natural morphism. In Grothendieck's style of mathematics, all of the morphisms (which appear in practice) are either canonical or natural.</p>
|
1,258,414 | <p>I want to prove that $$c = \{ (x_n)_{n\geq 0} \mid x_n \in \Bbb C \text{ and the sequence converges} \}$$ with the norm $$ \left\|(x_n)_{n\geq 0}\right\|_{\infty} =\sup_{n\geq 0} |x_n|$$ is a Banach space. I have done some work, but I am having trouble concluding.</p>
<p>So far: let $(\xi_n)_{n\geq 0} = \left( (x_k^{(n)})_{k\geq 0} \right)_{n\geq 0}$ be a $\|\cdot\|_{\infty}$-Cauchy sequence. Let $\epsilon > 0$, there exists $n_0 \in \Bbb N$ such that: $$\begin{align} \| \xi_n - \xi_m \|_{\infty} &< \epsilon, \quad \forall\, n,m > n_0 \\ \sup_{k \geq 0} |x_k^{(n)}-x_k^{(m)}| &< \epsilon, \quad \forall\,n,m > n_0 \\ |x_k^{(n)} - x_k^{(m)}| &< \epsilon, \quad \forall\, k\geq 0, \quad \forall\,n,m > n_0 \end{align}$$ So fixed $k$, $(x_k^{(n)})_{n \geq 0}$ is a $|\cdot |$-Cauchy sequence, and since $\Bbb C$ is Banach, there exists a limit $\lim_{n\to \infty} x_k^{(n)} =: x_k$. Then define $\xi = (x_k)_{k \geq 0}$. </p>
<p>Now I understand I have two things to do:</p>
<ul>
<li><p>prove that $\xi_n \stackrel{\|\cdot \|_{\infty}}{\longrightarrow} \xi $: we proceed as before. Let $\epsilon > 0$. Now, there is $n_0 \in \Bbb N$ such that: $$\begin{align} \| \xi_n - \xi_m \|_{\infty} &< \epsilon, \quad \forall\, n,m > n_0 \\ \sup_{k \geq 0} |x_k^{(n)}-x_k^{(m)}| &< \epsilon, \quad \forall\,n,m > n_0 \\ |x_k^{(n)} - x_k^{(m)}| &< \epsilon, \quad \forall\, k\geq 0, \quad \forall\,n,m > n_0 \\ \lim_{m \to \infty} |x_k^{(n)} - x_k^{(m)}| &\leq \epsilon, \quad \forall\,k\geq 0, \quad \forall\,n >n_0 \\ |x_k^{(n)} - x_k| &\leq \epsilon, \quad \forall\,k\geq 0 \quad \forall\, n>n_0 \\ \sup_{k\geq 0} |x_k^{(n)} - x_k| &\leq \epsilon,\quad \forall\, n > n_0 \\ \| \xi_n - \xi\|_{\infty} &\leq\epsilon, \quad \forall\,n>n_0, \end{align}$$ so ok.</p></li>
<li><p>prove that $\xi \in c$. I'm not sure of how to do this. I think I must use that every $(x_k^{(n)})_{k \geq 0}$ converge, because I don't seem to have used this yet. </p></li>
</ul>
<blockquote>
<p>How can I prove that $\xi \in c $?</p>
</blockquote>
<p>Thanks.</p>
| Tim.ev | 291,929 | <p>Here is an alternative proof, using as a starting point that $\ell^{\infty}$ is a Banach space. As $c$ is a subspace of $\ell^{\infty}$, it suffices to show that $c$ is closed in $\ell^{\infty}$.</p>
<p>To see this let $\mathbf{x}=(x_1,x_2,\dotsc)\in\ell^{\infty}$, and suppose that $\{\mathbf{x}^n\}_{n=1}^{\infty}$ is a sequence in $c$ converging to $\ell^{\infty}$ in the $\|\,.\|_{\infty}$ norm. For $n\in\mathbb{N}$ write $\mathbf{x}^n=(x_1^n,x_2^n,\dotsc)$ so that $x_i^n$ is the $i$-th term of the sequence $\mathbf{x}^n$ in $c$.</p>
<p>Let $\varepsilon>0$. As $\mathbf{x}^n\to\mathbf{x}$ as $n\to\infty$, there exists $N\in\mathbb{N}$ such that $\|\mathbf{x}^N-\mathbf{x}\|_{\infty}<\varepsilon/3$. As $\mathbf{x}^N$ is in $c$, it is Cauchy, so there exists $K\in\mathbb{N}$ such that $|x_i^N-x_j^N|<\varepsilon/3$ for all $i,j\geq K$. But then for such $i,j$ we have</p>
<p>\begin{align*}
|x_i-x_j|&\leq |x_i-x_i^N|+|x_i^N-x_j^N|+|x_j^N-x_j|\\
&\leq \|\mathbf{x}-\mathbf{x}^N\|_{\infty}+|x_i^N-x_j^N|+\|\mathbf{x}^N-\mathbf{x}\|_{\infty}\\
&<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon,
\end{align*}</p>
<p>which proves that $\mathbf{x}$ is cauchy and hence it is in $c$, so $c$ is a Banach space.</p>
<p>On a side note, using that $c$ is a Banach space gives a very quick proof that $c_0=\{(x_n)_{n\geq 0}\in\ell^{\infty}:\lim_{n\to\infty}x_n=0\}$ is a Banach space: one simply observes that we have a bounded linear functional $T:c\to\mathbb{C}$ given by $(x_n)\mapsto \lim_{n\to\infty}x_n$ (check). Then $c_0=\mathrm{ker}(T)$ is a closed linear subspace of $c$, and hence $c_0$ is a Banach space as well.</p>
|
4,415,254 | <p>When is <span class="math-container">$n!>x^n$</span>? assuming that x is a fixed positive. I know we can take the log of both sides and use the following formula:</p>
<p><span class="math-container">$$n\log(x) = \log(x^n) < \log(n!) = \sum_{i = 1}^n\log(i)$$</span>.
But this still gives me difficulty trying to find a specific N that makes the right side larger. Is there another formula I should be using?</p>
| Claude Leibovici | 82,404 | <p>There is an almost exact solution for the equation <span class="math-container">$$n!=x^n\tag 1$$</span> Have a look at @robjohn's answer to <a href="https://math.stackexchange.com/questions/1333449/could-this-approximation-be-made-simpler-solve-n-an-10k">this question</a> of mine.</p>
<p>Adapted to <span class="math-container">$(1)$</span> (that is to say <span class="math-container">$a=x$</span> and <span class="math-container">$k=0$</span> in my post), it gives (as a real)
<span class="math-container">$$n\sim x\, e^{1+W(t)}-\frac 12 \quad \text{where}\qquad t=-\frac{\log (2 \pi x)}{2 e x}\tag 2$$</span> <span class="math-container">$W(t)$</span> being <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert function</a>.</p>
<p>Suppose <span class="math-container">$x=1234$</span>. The above will give <span class="math-container">$n=3349.378835$</span> while the exact solution is <span class="math-container">$3349.378848$</span>. As usual, you will use <span class="math-container">$\lceil n \rceil$</span>.</p>
<p>If <span class="math-container">$x$</span> is "large", <span class="math-container">$t$</span> is small and you could use the approximation
<span class="math-container">$$W(t) \sim t \, \frac{1+\frac{19 }{10}t+\frac{17 }{60} t^2} {1+\frac{29 }{10}t+\frac{101 }{60}t^2 }+O\left(t^6\right)$$</span></p>
<p>Using it for the worked case, it would give ... the same.</p>
<p><strong>Edit</strong></p>
<p>If you want a shortcut approximation for <span class="math-container">$n \leq 1000$</span>, you could use the empiriccal
<span class="math-container">$$n \approx a\,x^b-c$$</span> obtained by a quick and dirty regression <span class="math-container">$(R^2>0.999999)$</span>.
<span class="math-container">$$\begin{array}{clclclclc}
\text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
a & 2.68239 & 0.00045 & \{2.68151,2.68327\} \\
b & 1.00195 & 0.00003 & \{1.00190,1.00201\} \\
c & 2.53581 & 0.01065 & \{2.51491,2.55671\} \\
\end{array}$$</span></p>
<p>It gives a maximum error of <span class="math-container">$\sim 0.1$</span> in the range.</p>
|
1,923,226 | <p>I have been trying to prove this with some continuity theorems but haven't put together a good proof yet. </p>
| celtschk | 34,930 | <p>Let's take the definition that a function is continuous if all pre-images of open sets are open. Now $\mathbb Z$ has the discrete topology, and therefore <em>all</em> its subsets are open, including the pre-images of open sets under $f$.</p>
|
1,515,817 | <p>I conjecture that in a consecutive sequence of $n$ natural numbers all greater than $n$, there exists at least one number which is not divisible by any prime number less than or equal to $n/2$.</p>
<p>Can any one prove or disprove this?</p>
| Konstantinos Gaitanas | 99,437 | <p>The conjecture is false.<br>
We can cover $O(n^2)$ natural numbers using the primes not exceeding $n$.<br>
Check the <a href="http://oeis.org/wiki/Jacobsthal_function." rel="nofollow">Jacobsthal function</a></p>
|
2,229,532 | <p>I have a circle which has a triangle inscribed in it.</p>
<p>The circle radius R = 4</p>
<p>The triangle ABC vertices divide circle into 3 arcs in 1:2:3 ratio</p>
<p>Find the perimeter and area of triangle.</p>
<p>Can you guys help me with this one? </p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>If the three arcs are in $1:2:3$ ratio ( and the sum is $2 \pi$), than the first arc goes form $0$ to $\pi/3$, the second from $\pi/3$ to $\pi$ and the third from $\pi$ to $2\pi$.</p>
<p>Can you do from this? ( see the figure)</p>
<p><a href="https://i.stack.imgur.com/lbERW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lbERW.jpg" alt="enter image description here"></a></p>
|
1,401,661 | <blockquote>
<p>Dice are cubes with pips (small dots) on their sides, representing
numbers 1 through 6. Two dice are considered the same if they can be
rotated and placed in such a way that they present matching numbers on
the top, bottom, left, right, front, and back sides.</p>
<p>Below is an example of two dice that can be rotated to show that they
are the same if the 2-pip and 4-pip sides are opposite and the 3-pip
and 5-pip sides are also opposite.
<a href="https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0" rel="nofollow">https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0</a></p>
<p>How many different dice exist? That is, how many ways can you make
distinct dice that cannot be rotated to show they are the same? Note:
This problem does not involve rolling the dice or the probability of
roll outcomes.</p>
</blockquote>
<p>I'm having trouble understanding exactly what is being asked in this question. I understand that I have to find how many different ways the dice can be placed to show that they are the same, but saying they cannot be rotated confuses me.</p>
<p>Could somebody make an attempt at rewording this? Or walking me through how to solve this?</p>
| mjqxxxx | 5,546 | <p>With no loss of generality, place the <span class="math-container">$1$</span> on top. Any of the <span class="math-container">$5$</span> remaining numbers could be on the bottom (opposite the <span class="math-container">$1$</span>). For each of these cases, there are <span class="math-container">$4$</span> numbers to arrange along the sides. Rotate the die (again, w.n.l.g.) so the smallest of these <span class="math-container">$4$</span> numbers is in the front. There are then <span class="math-container">$3!=6$</span> ways that the remaining <span class="math-container">$3$</span> could be arranged, for a grand total of <span class="math-container">$5\times 3! = 30$</span> distinguishable <span class="math-container">$6$</span>-sided dice.</p>
<p>A more general way to get the same result is to divide the number of ways to label the <span class="math-container">$6$</span> faces (<span class="math-container">$6!=720$</span>) by the number of ways a die can be rotated (<span class="math-container">$6\times 4=24$</span>), yielding <span class="math-container">$720 / 24 = 30$</span>.</p>
|
553,103 | <p>consider the series of function $\sum f_n$ with $f_n(x)=\frac{x}{x^2+n^2}$.
It is easy to see that there is pointwise convergence on $\mathbb{R}$ (to a function that we'll call $f$) but not normal convergence. I want to know whether there is uniform convergence towards $f$ or not. </p>
<p>I tried to disprove uniform convergence by looking at $|\sum_{k=0}^n f_k(n)-f(n)|$ : the sum is unbounded and equivalent to $\ln n$. The problem is that I don't know how to estimate $f(n)$. </p>
<p>I'm sure I'm missing something not too difficult here, so help would be appreciated.</p>
| imranfat | 64,546 | <p>You can also replace the quadratic trig terms with the double angle formula involving $cos2x$ Your denominator then will consists of a constant and a $cos2x$ term. Then sub away $2x=v$ and then use the Weierstrass substitution converting the integral into a rational function. This is hilarious of course, but what the heck?</p>
|
17,398 | <p>I am teaching my daughter, who is currently about <span class="math-container">$46$</span> months old, additions. She is very curious and asks a lot of <em>good</em> questions. For example, when I told her that <span class="math-container">$2+6=8$</span> and <span class="math-container">$4+4=8$</span>, she asked me the following question:</p>
<blockquote>
<p>Why are they the same?</p>
</blockquote>
<p>Surely I know the logical answer this question using <a href="https://en.wikipedia.org/wiki/Peano_axioms" rel="noreferrer">Peano Axioms</a> and the definition of natural numbers. But she has not learnt the Peano Axiom at this age.</p>
<p>So how to answer my three-year-old daughter's question that </p>
<blockquote>
<p>Why <span class="math-container">$2+6$</span> is the same as <span class="math-container">$4+4$</span>?</p>
</blockquote>
| Sue VanHattum | 60 | <p>Blocks work well for thinking about addition. Have her count out 8 blocks, and then ask her about all the addition problems that have 8 blocks as the answer.</p>
<p>A lovely children's book which looks at all the sum pairs for 7 is <em>Quack and Count</em>, by Keith Baker. (You can <a href="https://www.abebooks.com/servlet/SearchResults?isbn=0152928588&cm_sp=mbc-_-ISBN-_-all" rel="noreferrer">buy it used here</a>.) It has luscious pictures, a driving rhythm, and a lovely storyline. (“Slipping, sliding, having fun, 7 ducklings, 6 plus 1.”) </p>
|
510,672 | <p>My prof taught us that during Gaussian Elimination, we can perform three elementary operations to transform the matrix:</p>
<p>1) Multiple both sides of a row by a non-zero constant
2) Add or subtract rows
3) Interchanging rows</p>
<p>In addition to those, why isn't removing zero rows an elementary operation? It doesn't affect the system in any way. Define zero rows to be a row with no leading variables.</p>
<p>For example isn't $\begin{bmatrix}a & b & k\\c & d & m\end{bmatrix} \rightarrow
\begin{bmatrix}a & b & k\\c & d & m\\0 & 0 & 0\end{bmatrix}$</p>
| egreg | 62,967 | <p>Doing an elementary row operation on the matrix $A$ can be seen as multiplying $A$ by a suitable invertible matrix, call it $E_1$. So the whole process leads to writing
$$
U=E_{k}\dots E_{2}E_{1}A
$$
where $E_j$ $(j=1,2,\dots,k)$ correspond to the elementary row operation we performed and $U$ is a row reduced matrix (for instance, the row reduced echelon form). Since each $E_j$ is invertible, we can write
$$
A=FU
$$
where $F=E_1^{-1}E_2^{-1}\dots E_{k}^{-1}$ and, if one performs all row swaps at the start, the matrix $F$ can be written even without any computation.</p>
<p>Of course no operation of this kind changes the shape of the matrices we successively get and the final matrix $U$ will have zero or more “zero rows” at the bottom.</p>
<p>At this point we <em>can</em> remove those “zero rows” from $U$; if there are $l$ of them, we can also remove the $l$ rightmost columns of $F$. Calling $F_0$ and $U_0$ the matrices so obtained we still have
$$
A=F_0U_0
$$
and both $F_0$ and $U_0$ have their rank equal to the rank of $A$. This is called a <em>full rank decomposition</em>, because those matrices have the maximum rank they possibly have, based on their shape; say that $A$ is $m\times n$ and its rank is $r$. Then $F_0$ will be $m\times r$ $(r\le m)$ and $U_0$ will be $r\times n$ ($r\le n$).</p>
<p>In particular $F_0^HF_0$ ($H$ denotes the hermitian transpose, the simple transpose when the matrices are over the reals) is $r\times r$ invertible and
$$
(F_0^HF_0)^{-1}F_0^H
$$
not only is a left inverse of $F_0$ but is its Moore-Penrose pseudoinverse $F_0^+$. Similarly, $U_0U_0^H$ is invertible and $U_0^H(U_0U_0^H)^{-1}=U_0^+$. Moreover the Moore-Penrose pseudoinverse of $A$ is
$$
A^+=U_0^+F_0^+
$$
Indeed, any time we find a full rank decomposition $A=BC$, we can write $A^+=C^+B^+$ (and the pseudoinverse of $B$ and $C$ are computed as above).</p>
<p>The Moore-Penrose pseudoinverse is useful in computing the least squares solutions of $Ax=b$ and this is an efficient algorithm for finding it.</p>
|
4,014,144 | <p>Let <span class="math-container">$f(x) = \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$</span>.
where <span class="math-container">$x, c_1, c_2, \dots, c_k \in \mathbb R^n$</span>.
What fast iterative methods are available for finding the (approximate) min of <span class="math-container">$f$</span> with the constraint <span class="math-container">$\lVert x \rVert_2 = 1$</span>?</p>
<p>Notes:</p>
<ol>
<li><span class="math-container">$f$</span> is convex and and non negative, that is, <span class="math-container">$c_i$</span> positively span <span class="math-container">$\mathbb R^n$</span></li>
<li>For my use case <span class="math-container">$k \gg n$</span> and (rougly) <span class="math-container">$3 \le n \le 50$</span> and <span class="math-container">$100 \le k \le 10000$</span>.
I tried <a href="https://en.wikipedia.org/wiki/Subgradient_method" rel="nofollow noreferrer">projected subgradients</a> (projected to the sphere) but it can be slow to converge and improving the initial guess doesn't seem to accelerate the method. From the literature, this seems to be equivalent to Riemannian manifold subgradient methods which use the exponential map. I haven't tried bundle methods yet but I am investigating them now. They seem like they might be too slow.</li>
</ol>
<p>I've also tried approximating <span class="math-container">$f$</span> with a smooth maximum <a href="https://en.wikipedia.org/wiki/LogSumExp" rel="nofollow noreferrer">LogSumExp</a> and doing gradient descent with projection. I also tried unconstrained gradient descent with a <a href="https://en.wikipedia.org/wiki/Penalty_method" rel="nofollow noreferrer">quadratic penalty</a>. The approximation is too inaccurate and the evaluation of exp is too slow for my use case. I'm not interested in <a href="https://en.wikipedia.org/wiki/Sequential_quadratic_programming" rel="nofollow noreferrer">SQP</a> because I suspect whatever method used to smooth (e.g. LogSumExp, hyperbolic) will be too costly/inaccurate to evaluate.</p>
<p>Edit:
I believe this problem is equivalent to a linear programming problem with quadratic <strong>equality</strong> constraints. I haven't been able to to find much literature on this type of problem. Perhaps I can use a <a href="https://en.wikipedia.org/wiki/Semidefinite_programming" rel="nofollow noreferrer">SDP</a> method but I'm not sure.</p>
| V.S.e.H. | 443,030 | <p>EDIT: I have included all of the MATLAB code here: <a href="https://gitlab.com/TheRTGuy/point-cloud-support-function-minimization" rel="nofollow noreferrer">https://gitlab.com/TheRTGuy/point-cloud-support-function-minimization</a>. There I also have other evaluations.</p>
<p>Given the assumption that <span class="math-container">$f$</span> is always nonnegative, then this implies that the convex hull of <span class="math-container">$C=\{c_1,\ldots,c_k\}$</span> contains the origin. Were this not the case, then <span class="math-container">$f$</span> is always negative, and the origin is separable from the point cloud. In this case, minimizing <span class="math-container">$f$</span> just amounts to finding the supporting hyperplane that maximizes the distance from the origin, a problem that is solved easily using convex optimization (e.g. solving the dual of the primal problem).</p>
<p>The problem, as some pointed out in the comments, boils down to enumerating the facets of the convex hull of <span class="math-container">$C$</span>, and finding the one closest to the origin. This is a very hard problem, see <a href="https://www.cs.mcgill.ca/%7Efukuda/soft/polyfaq/node22.html#polytope:Vredundancy" rel="nofollow noreferrer">https://www.cs.mcgill.ca/~fukuda/soft/polyfaq/node22.html#polytope:Vredundancy</a>, since the number of facets is exp. in <span class="math-container">$n$</span> and <span class="math-container">$k$</span>. So, forget about an "efficient" exact solution.</p>
<p>However, in my research I needed a way to compute a tight overapproximation of the convex hull of a point cloud. Here is a simplified adaptation, of the approach I used, to OP's problem:</p>
<p>The approach:</p>
<ol>
<li>Generate vertices of a regular simplex, <span class="math-container">$\{h_1,\ldots,h_{n+1}\}$</span>, and compute <span class="math-container">$$b_0 = \begin{pmatrix}\max_{c\in C} h_1^\top c \\ \vdots \\ \max_{c\in C}h_{n+1}^\top c\end{pmatrix}.$$</span></li>
<li>Form the new simplex <span class="math-container">$S_0 = \{x\in\mathbb{R}^n~\vert~ H_0x \leq b_0\} = \operatorname{Conv}(\{s^1_0,\ldots,s^{n+1}_0\})$</span>, where <span class="math-container">$$H_0 = \begin{pmatrix}h_1^\top\\\vdots \\h_{n+1}^\top\end{pmatrix},$$</span>
and <span class="math-container">$s^i_0$</span> is the <span class="math-container">$i$</span>-th vertex of the simplex. This simplex tightly bounds the set <span class="math-container">$C$</span>.</li>
<li>Optionally, perturb the simplex by a random rotation.</li>
<li>Set <span class="math-container">$i=1$</span>, and <span class="math-container">$f_0 = \min_j [b_0]_j$</span>, where <span class="math-container">$[\cdot]_j$</span> is the <span class="math-container">$j$</span>-th component of a vector.</li>
<li>For each vertex <span class="math-container">$s^j_{i-1}$</span> solve the LP
<span class="math-container">$$
x_j=\arg\min_x ~ g(x)= x^\top s^j_{i-1} \quad \\s.t. \\\quad (c_l - s^j_{i-1})^\top x \leq -1, ~~~\forall l\in\{1,\ldots,k\}, \\
\quad (s^l_{i-1}- s^j_{i-1})^\top x \leq -1, ~~~\forall l\in \{1,\ldots,n+1\}\setminus\{j\},\\
\quad x^\top s^j_{i-1} \geq -1.
$$</span>
The LP basically finds a supporting hyperplane of <span class="math-container">$C$</span> that separates it from <span class="math-container">$s^j_{i-1}$</span>. This plane is often a facet of <span class="math-container">$\operatorname{Conv}(C)$</span>.</li>
<li>Set
<span class="math-container">$$
H_i = \begin{pmatrix}\frac{x_1^\top}{\lVert x_1\rVert}\\\vdots\\\frac{x_{n+1}^\top}{\lVert x_1\rVert}\end{pmatrix},\qquad b_i = \begin{pmatrix}\frac{x_1^\top s^1_{i-1}-1}{\lVert x\rVert} \\ \vdots \\ \frac{x_{n+1}^\top s^{n+1}_{i-1}-1}{\lVert x\rVert} \end{pmatrix}.
$$</span>
This forms a new simplex <span class="math-container">$S_i = \{x\in\mathbb{R}^n~\vert~ H_ix \leq b_i\} = \operatorname{Conv}(\{s^1_i,\ldots,s^{n+1}_i\})$</span>.</li>
<li>Set <span class="math-container">$f_i = \min\{\min_j[b_i]_j , f_{i-1}\}$</span>, <span class="math-container">$i:=i+1$</span>, and repeat steps 5-7 for <span class="math-container">$N$</span> iterations, or until <span class="math-container">$f_i = f_{i-1}$</span> for some consecutive iterations.</li>
<li>Set set the optimal value to <span class="math-container">$f^* = f_i$</span>.</li>
</ol>
<p>Evaluation:
The implementation I have is for MATLAB.</p>
<ol>
<li>I evaluated the approach for <span class="math-container">$n=2,\ldots,20$</span>, with <span class="math-container">$k=1000$</span>. For each <span class="math-container">$n$</span> I ran the algorithm 3 times on a uniformly generated set <span class="math-container">$C$</span>, to which I also apply a random rotation, and time each run. The number of iterations is fixed to <span class="math-container">$N=5$</span>. In case you are curious, this is the set of MATLAB commands executed to generate the point cloud:</li>
</ol>
<pre><code>%Randomly rotated uniform
C = 5*rand(n,N)-2.5;
%Indexing pattern for off-diagonal entries of matrix
idx = nchoosek(1:n,2);
dU = zeros(n);
dth = pi*(2*rand(n*(n-1)/2,1) - 1); %Uniform
dU(sub2ind([n,n],idx(:,1),idx(:,2))) = dth;
dU(sub2ind([n,n],idx(:,2),idx(:,1))) = -dth;
C = expm(dU) * C;
</code></pre>
<ol start="2">
<li>To evaluate how good the solution is, for each run I generate 1000000 points, <span class="math-container">$\{x_k\}$</span>, normally distributed on the unit sphere in <span class="math-container">$\mathbb{R}^n$</span>, and evaluate <span class="math-container">$f$</span> at each point. Then, I compute <span class="math-container">$d = \min_k\{f(x_k)\} - f^*$</span> and plot it. Positive means that the solution is good, negative means the solution is bad. Of course, if positive, this does not mean the solution is the best. Furthermore, the higher the dimension, the lower the quality of this evaluation. Ideally, we want to compare it with another algorithm, which I don't have.</li>
</ol>
<p>Results:</p>
<p>The first plot shows the execution times vs. <span class="math-container">$n$</span>. The second plot shows the difference <span class="math-container">$d$</span> vs <span class="math-container">$n$</span>. The final plot is just an arbitrary run for <span class="math-container">$n=20$</span>, which shows <span class="math-container">$f(x_k)$</span> and <span class="math-container">$f^*$</span>.</p>
<p><a href="https://i.stack.imgur.com/5Fb8a.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Fb8a.jpg" alt="Timing" /></a></p>
<p><a href="https://i.stack.imgur.com/fPlfA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fPlfA.jpg" alt="Solution quality" /></a></p>
<p><a href="https://i.stack.imgur.com/1Died.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Died.jpg" alt="Example for <span class="math-container">$n=20$</span>" /></a></p>
<p>Conclusion:</p>
<p>On first glance, the approach seems to perform very well. Only a small number of iteratios is necessary to get a good suboptimal solution. Also, solving LPs is very efficient, even for large <span class="math-container">$n$</span> and <span class="math-container">$k$</span>. However, there is no rigorous tight bound for optimality yet. More rigorous analysis is needed. Furthermore, evaluating the approach with normally distributed points on the unit sphere is obviously not the best way, but at least it gives a sense of optimality. This is clearly seen from the second plot where the gap is increasing. If the OP is interested in further evaluation, then I would be willing to share some code.</p>
<p>The intuition is simple: we try to "squeeze" the points by simplices as much as possible, by using the vertices of previous simplices as anchors. The facets of the simplices often coincide with the faces of <span class="math-container">$\operatorname{Conv}(\{c_i\})$</span>.</p>
|
3,423,138 | <p>A) Always positive </p>
<p>B) Always negative </p>
<p>C)Always non-negative </p>
<p>D) Always zero </p>
<p>Obviously, this isn’t a solving question, so I can’t really show any working here (I would have otherwise). I didn’t really understand the question, so help would be greatly appreciated </p>
<p>Thanks!</p>
| clathratus | 583,016 | <p>In case you're interested in a general formula, <a href="https://en.wikipedia.org/wiki/Polylogarithm" rel="nofollow noreferrer">Wikipedia</a> gives
<span class="math-container">$$\begin{align}
\mathrm{Li}_{-n}(x)&=\left(z\frac{\partial}{\partial z}\right)^n\frac{z}{1-z}\\
&=(-1)^{n+1}\sum_{k=0}^{n}\frac{\left\{{{n+1}\atop{k+1}}\right\}k!}{(1-z)^{k+1}}\\
&=\frac{1}{(1-z)^{n+1}}\sum_{k=0}^{n-1}\left<{{n}\atop{k}}\right>z^{n-k},
\end{align}$$</span>
where <span class="math-container">$$\left\{{{n}\atop{k}}\right\}=\frac1{k!}\sum_{i=0}^{k}(-1)^i{k\choose i}(k-i)^n$$</span>
are the Stirling numbers of the second kind, and
<span class="math-container">$$\left<{{n}\atop{k}}\right>=\sum_{i=0}^{k}(-1)^i{{n+1}\choose i}(k+1-i)^n$$</span>
are the Eularian numbers.</p>
|
1,794 | <p>I have always loved the beauty of mathematics and physics. However I'm severely dyslexic and find it hard to keep numbers in my head, any more than 4 numbers at a time and they melt together and lose their meaning. (You know that feeling when you wake up after a lucid dream but as you think about it, it just evaporates? Well that's what it's like.) However, I'm pretty good at understanding concepts like Cantor Sets, mathematical logic, etc.</p>
<p>I have degrees in the philosophy of science and have studied epistemology, so I'm comfortable with abstract concepts BUT, I want to do some mathematics, I want to try algebra and trigonometry, to play with equations but I just don't know how to start.</p>
<p>Are there any books designed for the likes of me? Can someone who finds it hard to add two two-digit numbers together in his head, and even takes a second to decipher a 4-digit number on paper be taught to do advanced math? Is it like teaching a 1-legged man to run or a pig to fly?</p>
<p>I'm interested in hearing the opinions of people who love math, who can do math and maybe understand what a lot of my friends don't - namely why I at least want to try. Thanks.</p>
<hr>
<p>Thanks for your comments,</p>
<p>Let me rephrase my original question if it helps.</p>
<ol>
<li><p>Being dyslexic. I have a very poor working memory (like a computer with very little ram if you like) the processing power required to recognise the shape of numbers written down as characters (i.e. 1, 5, 9, 3, 2, 8, etc) and then understand what they represent requires all my processing power. That means if I want to calculate two numbers in my head, 7+5 for example it takes a second but I can do it. However, if I want to add 14+63 my brain will spin and crash. Ive not been able to move past what most 7 year olds can do in math class.</p></li>
<li><p>That said, in common with many dyslexics I’m very good at understanding complex and abstract theories and ideas (spatial reasoning). I have degrees (BA and MA form Durham) in subjects that can broadly be described as philosophy. I have been fascinated with subjects like mathematics and physics since I was a child and I have always read popular books on the subject. I can understand, for example, ‘set theory’ .. at a conceptual level, mathematic proof, .. at a conceptual level. </p></li>
</ol>
<p>What you have written is encouraging, but math - even at a basic level is a foreign land to me. I would love to do some sort of higher math (no, I don’t really know what that means). but I would love to try to understand equations. But can I learn to do algebra or trigonometry and beyond when I can’t recite the 6x table or add 34 and 58 in my head? If I can, and I think you are saying that I might just be able to, where do I start? </p>
| Benjamin Dickman | 262 | <p>I'm not exactly sure what's being asked here, since you say you want to to learn how "to do advanced math," but also that you "want to try algebra and trigonometry, to play with equations." I can't tell exactly what you mean by <em>do advanced math</em>, but you say:</p>
<blockquote>
<p>I'm pretty good at understanding concepts like Cantor Sets, mathematical logic, etc.</p>
</blockquote>
<p><strong>So:</strong> It seems to me like you have <em>already</em> "done [some] advanced math." Furthermore, you write:</p>
<blockquote>
<p>I have degrees in the philosophy of science and have studied epistemology, so I'm comfortable with abstract concepts</p>
</blockquote>
<p><strong>So:</strong> Sure, if those degrees are legitimate, and your study of [some area of] epistemology was more than superficial, then I expect you could handle abstract mathematics as well.</p>
<p><strong>That said:</strong> I get the sense that you are concerned about a combination of dyslexia and dyscalculia interfering with your learning. To this extent, there are some comments by Poincare and Hadamard about how neither of them was a particular capable "calculator" (in the sense of doing mental computations). The <a href="http://en.wikipedia.org/wiki/Jacques_Hadamard">latter</a> goes on to remark in his book "The Psychology of Invention in the Mathematical Field" (1945) that <em>very few</em> of the mathematicians who he knows think in words or even algebraic symbols. </p>
<p>More precisely, Hadamard writes that nearly all of the mathematicians who he knows </p>
<blockquote>
<p>"avoid not only the use of mental words but also, just as I do, the mental use of algebraic or any other precise signs; also as in my case, they use vague images" (p. 84). </p>
</blockquote>
<p>Finally, Hadamard discusses physical intuition and mental pictures of a kinetic sort as being relevant to mathematicians' ways of thinking. </p>
<p><strong>Thus:</strong> If you can think abstractly but are tangled up by words and symbols, then you could still be in good mathematical company.</p>
<hr>
<p><strong>For completeness:</strong> Hadamard does write about a few notable exceptions.</p>
<p>He contrasts the use of image-based thinking with the following three mathematicians:</p>
<blockquote>
<p>George D. Birkhoff, "who is accustomed to visualize algebraic symbols and to work with them mentally," Norbert Wiener, who "happens to think either with or without words," and Jesse Douglas, whose "research thought is in connection with words, but only with their rhythm, a kind of Morse language where only the numbers of syllables of some words appear" (p. 84). </p>
</blockquote>
<p>Lastly, he mentions the rare exception of George Polya, whose thoughts rely in some large part on words and letters. Hadamard quotes Polya describing his own thinking: </p>
<blockquote>
<p>"I believe… that the decisive idea which brings the solution of a problem is rather often connected with a well-turned word or sentence… The right word, the subtly appropriate word, helps us to recall the mathematical idea…" (p. 84). </p>
</blockquote>
<p>Hadamard himself continues: </p>
<blockquote>
<p>"Moreover, [Polya] finds that a proper notation – that is, a properly chosen letter to denote a mathematical quantity – can give him similar help; and some kind of puns, whether of good or poor quality, may be useful for that purpose" (p. 85). </p>
</blockquote>
<p>(In case it is unclear: I highly recommend this book!)</p>
|
1,794 | <p>I have always loved the beauty of mathematics and physics. However I'm severely dyslexic and find it hard to keep numbers in my head, any more than 4 numbers at a time and they melt together and lose their meaning. (You know that feeling when you wake up after a lucid dream but as you think about it, it just evaporates? Well that's what it's like.) However, I'm pretty good at understanding concepts like Cantor Sets, mathematical logic, etc.</p>
<p>I have degrees in the philosophy of science and have studied epistemology, so I'm comfortable with abstract concepts BUT, I want to do some mathematics, I want to try algebra and trigonometry, to play with equations but I just don't know how to start.</p>
<p>Are there any books designed for the likes of me? Can someone who finds it hard to add two two-digit numbers together in his head, and even takes a second to decipher a 4-digit number on paper be taught to do advanced math? Is it like teaching a 1-legged man to run or a pig to fly?</p>
<p>I'm interested in hearing the opinions of people who love math, who can do math and maybe understand what a lot of my friends don't - namely why I at least want to try. Thanks.</p>
<hr>
<p>Thanks for your comments,</p>
<p>Let me rephrase my original question if it helps.</p>
<ol>
<li><p>Being dyslexic. I have a very poor working memory (like a computer with very little ram if you like) the processing power required to recognise the shape of numbers written down as characters (i.e. 1, 5, 9, 3, 2, 8, etc) and then understand what they represent requires all my processing power. That means if I want to calculate two numbers in my head, 7+5 for example it takes a second but I can do it. However, if I want to add 14+63 my brain will spin and crash. Ive not been able to move past what most 7 year olds can do in math class.</p></li>
<li><p>That said, in common with many dyslexics I’m very good at understanding complex and abstract theories and ideas (spatial reasoning). I have degrees (BA and MA form Durham) in subjects that can broadly be described as philosophy. I have been fascinated with subjects like mathematics and physics since I was a child and I have always read popular books on the subject. I can understand, for example, ‘set theory’ .. at a conceptual level, mathematic proof, .. at a conceptual level. </p></li>
</ol>
<p>What you have written is encouraging, but math - even at a basic level is a foreign land to me. I would love to do some sort of higher math (no, I don’t really know what that means). but I would love to try to understand equations. But can I learn to do algebra or trigonometry and beyond when I can’t recite the 6x table or add 34 and 58 in my head? If I can, and I think you are saying that I might just be able to, where do I start? </p>
| Jessica B | 4,746 | <p>I think the short answer to your question is that you are (like many people) confusing the arithmetic that you called 'mathematics' at school with the mathematics that mathematicians do. I very rarely add up numbers above 10, or write down ones with several digits. I think I possibly do own a calculator... somewhere.</p>
<p>I would suggest starting with some popular-maths books, and see how you get on. Since you've already done some philosopy, maybe try some logic. (I was going to suggest a book I liked, but I've completely forgotten what it was.)</p>
|
3,459,178 | <p>Is there a general way to factor <span class="math-container">$ax^2 + bx$</span> into <span class="math-container">$($</span>some expression<span class="math-container">$)^2$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are constants?</p>
<p>The reason I am asking is to understand how the integral <span class="math-container">$\int_{-\infty}^{\infty} e^{ax^2+bx}\ dx$</span> is turned into the form <span class="math-container">$\int_{-\infty}^{\infty} e^{-u^2}\ du$</span> which is an important step in solving it.</p>
| Saketh Malyala | 250,220 | <p><span class="math-container">$\displaystyle ax^2+bx=a\left(x^2+\frac{b}{a}x\right)=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\right)-\frac{b^2}{4a}=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}$</span>.</p>
<p>Then, <span class="math-container">$$\displaystyle \int_{-\infty}^{\infty}e^{ax^2+bx}\,dx$$</span> <span class="math-container">$$\displaystyle=\int_{-\infty}^{\infty}e^{a(x+\frac{b}{2a})^2-\frac{b^2}{4a}}\,dx$$</span>
<span class="math-container">$$\displaystyle=e^{-\frac{b^2}{4a}}\int_{-\infty}^{\infty}e^{a(x+\frac{b}{2a})^2}\,dx$$</span></p>
<p>We now make the assumption <span class="math-container">$a$</span> is negative.
<span class="math-container">$$\displaystyle =e^{-\frac{b^2}{4a}}\int_{-\infty}^{\infty}e^{ax^2}\,dx = e^{-\frac{b^2}{4a}}\int_{-\infty}^{\infty}e^{(-a)(-x^2)}\,dx\frac{e^{-\frac{b^2}{4a}}}{\sqrt{-a}}\int_{-\infty}^{\infty}\sqrt{-a}e^{(\sqrt{ -a} \,x)^2}\,dx$$</span></p>
<p><span class="math-container">$$\displaystyle =\frac{e^{-\frac{b^2}{4a}}}{\sqrt{-a}}\int_{-\infty}^{\infty}e^{-x^2}\,dx$$</span></p>
|
2,501,450 | <p>I was trying for a while to prove non-existence of theu following limit:</p>
<p>$$\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}$$</p>
<p>Unfortunately, with no results.</p>
<p>My hope was to show, that:</p>
<p>$$\lim_{n\to\infty}\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}\not=0$$</p>
<p>But showing that was harder than I thought.</p>
<p>Can anyone show me how to solve this problem?</p>
| Community | -1 | <p>$$\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}=\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n}\frac1{\sqrt[n]{5}-1}$$</p>
<p>is of the form $\dfrac10$.</p>
|
235,425 | <p>Ok, here is what I have for the proof of this conjecture. Let me know if I'm on the right path? all input appreciated.</p>
<p>There exist integers $j$, $k$, and $m$, such that,
$b = aj $ and $ c = ajk.$ Then $c = ajk $ (substituting $aj$ for $b$)
let $m = jk$, then $c = ma, => a|c.$ </p>
| MarnixKlooster ReinstateMonica | 11,994 | <p><em>(In this answer all variables are integers, i.e., elements of $\mathbb{Z}$.)</em></p>
<p>By the definition of divisibility we are given that $\;n * a = b\;$ and $\;m * b = c\;$ for some $\;n\;$ and $\;m\;$. Now we are asked to find a $\;k\;$ which makes $\;k*a = c\;$:</p>
<p>\begin{align}
& k*a = c \\
\equiv & \;\;\;\;\;\text{"use the only fact we know about $\;c\;$"} \\
& k*a = m*b \\
\equiv & \;\;\;\;\;\text{"use the other fact"} \\
& k*a = m*n*a \\
\Leftarrow & \;\;\;\;\;\text{"weaken using Leibniz' rule -- to achieve our goal"} \\
& k = m*n \\
\end{align}</p>
<p>Therefore we have found such a $\;k\;$, and hence proved $\;a|c\;$.</p>
|
223,521 | <p>Suppose that $a^2-b^2 =x$ where $a,b,x$ are natural numbers.</p>
<p>Suppose $x$ is fixed. If there is one $(a,b)$ found, can there be another $(a,b)$?</p>
<p>Also, would there be a way to know how many such $(a,b)$ exists?</p>
| Harish Kayarohanam | 30,423 | <p>Any number N can be represented as product of two no's .</p>
<p>$$ N= a*b =({a+b\over 2})^2 - ({a-b\over 2})^2 $$</p>
<p>$$ To\ be\ difference\ of\ perfect\ squares\ the\ condition\ is\ ({a+b\over 2}) and \ ({a-b\over 2}) \ should\ be\ integers\ .$$ </p>
<p>This happens only iff </p>
<p>1) a and b are odd</p>
<p>or</p>
<p>2) a and b are even</p>
<p>eg. $$ 100 = 2^2 * 5^2 $$
So number of factors in 100= (2+1)(2+1) =9</p>
<p>They are 1,2,4,5,10,20,25,50,100 </p>
<p>To find pairs of numbers which gives product 100 ,(trick is take corresponding numbers comming from opposite ends from the list above )</p>
<p>so 100 = 1*100 , 2*50 , 4*25 , 5*20 ( dont take the middle number in case the number is itself as square like 100 = 10^2)</p>
<p>here the pair which satisfies the condition is</p>
<p>2*50 ( even * even) , so a=50 , b=2
now calculate ,$$ ({a+b\over 2})\ and\ ({a-b\over 2}) $$</p>
<p>So </p>
<p>$$ 100 = 26^2 - 24^2 $$ is the only possibilty </p>
|
591,014 | <p>I need help with this proof:</p>
<p>$f: X\rightarrow Y$</p>
<p>$C,D\subseteq Y$</p>
<p>$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$</p>
<p>Thanks.</p>
| amWhy | 9,003 | <p>To show set <em>equality</em>,$$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$$ we typically proceed by showing that both the following inclusions hold:</p>
<p>$$f^{-1}(C \cap D) \quad \subseteq \quad f^{-1}(C) \cap f^{-1}(D)\tag{1}$$
$$f^{-1}(C) \cap f^{-1}(D) \quad \subseteq \quad f^{-1}(C \cap D)\tag{2}$$</p>
<p>For each of $(1), (2),$ we can use "element chasing": In general, to show $A \subseteq B$, it suffices to show that $a \in A \implies a \in B$.</p>
<p>For $(2)$, to show $$f^{-1}(C) \cap f^{-1}(D) \subseteq f^{-1}(C \cap D)\tag{2}$$</p>
<p>We start by assuming $x \in f^{-1}(C) \cap f^{-1}(D)$. Then, by the definition of set intersection, we have that $x \in f^{-1}(C)$ AND $x \in f^{-1}(D)$. This means $x \in C$ AND $x \in D$, which means, by definition, that $x \in C\cap D$. Now, $x\in C\cap D$ implies that $x \in f^{-1}(C\cap D).$ Hence, we've shown that $$f^{-1}(C) \cap f^{-1}(D) \subseteq f^{-1}(C \cap D)\tag{2}$$</p>
<p>Use the very same strategy for approaching the first inclusion $(1)$. When you show that inclusion $(1)$ also holds, then you can assert what you set out to prove: $$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$$</p>
|
2,327,181 | <p>I'm a bit confused about the definition of finite sets/intervals. I know that a set S is called finite when it has a finite number of elements, or formally, when there exists a bijection $f:S\to\{1,...,n\}$ for some natural number n. </p>
<p>However, the interval $(1,2)$ is called finite. I don't understand why; $(1,2)$ is not even countable, and there definitely does not exist a bijection $f:(1,2)\to\{1,...,n\}$. </p>
<p>Why do we call $(a,b)$ with $a,b\in\mathbb{R}$, finite? Did we just agree to do so, or is it incorrect to use the interval $(a,b)$ as a set like I did in the definition of finiteness above?</p>
<p>Thanks!</p>
| Woett | 65,418 | <p>With 'finite' the length is meant. So a finite interval is an interval with finite length, where the length (or measure) of an interval $(a,b)$ is generally defined as $b-a$.</p>
|
2,327,181 | <p>I'm a bit confused about the definition of finite sets/intervals. I know that a set S is called finite when it has a finite number of elements, or formally, when there exists a bijection $f:S\to\{1,...,n\}$ for some natural number n. </p>
<p>However, the interval $(1,2)$ is called finite. I don't understand why; $(1,2)$ is not even countable, and there definitely does not exist a bijection $f:(1,2)\to\{1,...,n\}$. </p>
<p>Why do we call $(a,b)$ with $a,b\in\mathbb{R}$, finite? Did we just agree to do so, or is it incorrect to use the interval $(a,b)$ as a set like I did in the definition of finiteness above?</p>
<p>Thanks!</p>
| JanneK | 342,700 | <p>The limits of (1,2) are very clear. Hence not infinite.
The paradox of 'Achilles and the tortoise'can be applied here to clear up the difference between a limit and infinity.</p>
|
2,661,718 | <p>My basic understanding is that each time
I have 30% chance of winning the prize
so between 3 and 4 tries I should win it</p>
<p>cause .3 +.3 +.3 = 90% as I need to win only once with 3 try
.3+.3+.3+.3 = 120% with 4 try</p>
<p>I do remember a formula saying 1-(0.7)^3 = 65.7% chance but I dont remember what that number mean exactly</p>
<p>thanks for answering this noob question my math are way behind me now</p>
| sku | 341,324 | <p>Let $y = x+1$</p>
<p>In general, $y^y = e^{y\ln y} \implies e^y <= y^y <= e^{y\ln y} <= e^{y^2}$ </p>
<p>Hence, $y^{y\ln y} <= e^{y(\ln y)^2} <= e^{y^2 \ln y}$ </p>
<p>we need to prove that $e^{y(\ln y)^2} <= e^{y^2}$ which can be trivially proven. </p>
|
2,752,421 | <p>a) Consider the equation $x_1 + x_2 + x_3 + x_4 = 35$. How many different solutions does this equation have
if all the variables must be positive integers? Enter the exact numeric answer.</p>
<p>b) Suppose that a license plate consists of three letters followed by three digits. How many different license plates start with the letter A if letters and digits cannot be repeated? Enter the exact numeric answer.</p>
<p><strong>My work</strong></p>
<p>a) $C(38,35)$</p>
<p>b)$1*P(26,2)*P(10,3)$</p>
<p>A_ _ _ _ _ . The last three number should not be repeated, so $P(10,3)$ and then the letters can be chosen randomly so $C(26,2)$ </p>
| Tal-Botvinnik | 331,471 | <p>Hint for a)</p>
<p>Write
$$35=1+1+\dots+1$$
Now, in how many ways can you place three commas?
For instance $1,1+1,1+1+1,1+\dots+1$.</p>
<p>For b), you are not considering that the digits come after the letters.</p>
|
2,061,655 | <p>Could we compute the limits
$$\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3} \\ \lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$$
without using the l'Hospital rule and the Taylor expansion?</p>
| Olivier Oloa | 118,798 | <p><strong>Hint</strong>. If one recalls that, for any function $f$ differentiable near $0$,
$$
\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=f'(0) \tag1
$$ then one may write
$$
\begin{align}
l:=\lim_{x \to 0}\frac{e^x-\sin (x)-1}{x^2}=\lim_{x \to 0}\:\frac1x\left(\frac{e^x-1}{x} -\frac{\sin x}x\right)
\end{align}
$$ then one may apply $(1)$ to
$$
f(x)=\frac{e^x-1}{x} -\frac{\sin x}x,\quad f(0)=0,
$$$$
f'(x)=-\frac{e^x-\sin (x)-1}{x^2}+\frac{e^x}{x} -\frac{\cos x}x,
$$ getting
$$
l=-l+\lim_{x \to 0}\frac{e^x-1}{x}-\lim_{x \to 0}\frac{\cos x-1}{x}
$$ and
$$
l=-l+1-0
$$ that is</p>
<blockquote>
<p>$$
l=\lim_{x \to 0}\frac{e^x-\sin (x)-1}{x^2}=\frac12.
$$</p>
</blockquote>
<p>Can you take it from here applying it to the first limit?</p>
|
1,083,141 | <p>According to Zeidler, 1995, in his book "Applied Functional Analysis: Application to Mathematical Physics".</p>
<p>Dirichlet problem is a problem to minimize $$F(u)=\frac{1}{2}\int_G(u')^2\ dx-\int_G fu\ dx,\qquad u=g \textrm{ in } bd(G).$$
Then, He says</p>
<blockquote>
<p>If $G$ is open, bounded, nonempty subset of $\mathbb{R}$ and given
continuous functions $f:G\rightarrow\mathbb{R}$ and
$g:bd(G)\rightarrow\mathbb{R}$. Then, for $u\in C^2(\overline{G})$:</p>
<p>$1.$ If $u$ is solution to Dirichlet problem then $u$ is also a
solution to generalized boundary value problem $$\int_G u'v'\ dx=\int_G fv\ dx,\qquad\forall v\in C_0^\infty(G)$$ $2.$ $u$ is a
solution to boundary value problem $$-u''=f,\qquad \textrm{in } G$$
$$u=g,\qquad \textrm{in }\partial G$$ if and only if $u$ is a solution
to generalized boundary value problem $$\int_G u'v'\ dx=\int_G fv\ dx,\qquad\forall v\in C_0^\infty(G)$$</p>
</blockquote>
<p>After that, he gives explanation that there is a Dirichlet problem which doesn't have solution (so Dirichlet Principle can't be justified). What I know about this, later on, is that the solution space needs to be complete so the Dirichlet Principle can be justified.</p>
<p>Therefore, the Dirichlet problem needs to be generalized by introducing Sobolev Space $W_2^1(G)$. The generalized Dirichlet Problem is to minimize the function $$\frac{1}{2}\int_G (\partial u)^2\ dx-\int_G fu\ dx, \qquad u-g\in \overset{\circ}{W^1_2}(G),$$ where $\partial u$ denotes weak derivatives of $u.$</p>
<p>The Dirichlet Principle is stated as below:</p>
<blockquote>
<p>If $G$ is open, bounded, nonempty subset of $\mathbb{R}$. Given two
functions $f\in L_2(G)$ and $g\in W_2^1(G)$. Then, these are true:</p>
<p>$1.$ The Generalized Dirichlet Problem has a unique solution $u\in W_2^1(G).$</p>
<p>$2.$ Solution to Generalized Dirichlet Problem is also a solution to
generalized boundary value problem $$\int_G \partial u \partial v \ dx=\int_G fv \ dx,\qquad \forall v\in \overset{\circ}{W^1_2}(G), \textrm{ and }u-g\in \overset{\circ}{W^1_2}(G)$$</p>
</blockquote>
<p>My questions are:</p>
<ol>
<li><p>Why the solution space $C^2(G)$ is generalized to $W_2^1(G)$? $C^2(G)$ is SECOND order continuously differentiable functions space while $W_2^1(G)$ is FIRST order weak derivable functions space.</p></li>
<li><p>Why the given function $g$ is generalized from only in $C(bd(G))$ to $W_2^1(G)$? The domain is extended from $bd(G)$ to $G$.</p></li>
<li><p>When proving the Dirichlet Principle, Zeidler, uses a theorem called Theorem in Quadratic Variational Problem, which only solved the minimizing problem and not the boundary condition, Why?</p></li>
</ol>
<p><strong>Note: Please help me. This is my undergraduate thesis topic and my undergraduate thesis defense seminar will be conducted on next Tuesday. Nobody familiar with this topic in my campus. So please, I'm desperately need an answer. Anything. Any explanation.</strong></p>
| Siminore | 29,672 | <p>Let me try to explain. If you have a solution $u \in C^2(G)$ of the equation $-u''=f$ in $G$, then for any $v \in C_c^\infty(G)$ an integration by parts shows that $$\int_G u' v' = \int_G f v.$$ Notice that there is no reference to the value of $u$ on $\partial G$, since $v$ vanishes on $\partial G$ and the boundary terms in the integration by parts disappear no matter how $u$ in defined on the boundary of $G$.</p>
<p>Now you <em>define</em> weak solution any element $u \in X$, $X$ being a suitable function space, such that
$$\int_G u' v' = \int_G f v$$
for any $v \in C_c^\infty(G)$. Now, how do you choose $X$? You have much freedom, but $u'$ should be meaningful (in some sense), and the boundary condition $u=g$ on $\partial G$ should be included.
It turns out that $X=g+ W_0^{1,2}(G)$, often written $g+H_0^1(G)$, is a good choice. As Zeidler says, everything works.
Your questions remain valid, in particular the second one: why should we extend $g$ to the whole $\bar{G}$, if we need it only on the boundary of $G$? You are right, but an answer would rely on the theory of traces in Sobolev spaces, which is a difficult topic.</p>
<p>Moreover, it can be proved that any weak solution is a classical $C^2$ solution as soon as $f$ is continuous up to the boundary of $G$. It is rather easy in 1D (i.e. in $\mathbb{R}), while it is really hard in higher dimension. I suggest that you read carefully the chapter devoted to Sobolev spaces in 1D in the book by Haim Brezis, <em>Functional analysis</em>. You'll find a more detailed explanation of the road from classical to weak solutions and viceversa.</p>
|
1,647,399 | <p>Suppose that $(X,d)$ is a compact metric space and $(E_n)$ is any sequence of nonempty closed subsets of $X$ with $E_{n+1}\subset E_n$ for all $n\in\mathbb{N}$. Show that $\cap_{n=1}^\infty E_n$ is nonempty.</p>
<p><strong>What I know:</strong> Since $X$ is compact, we need to use the fact that every sequence in $X$ has subsequence which converges in $X$. $E_n$ is nonempty and closed, that means it contains all its limit points. And $E_n$ is also compact right? </p>
<p><strong>My doubts are:</strong> How can we construct a sequence <strong>that has a subsequence that has a limit in $X$</strong>? I don't think we should use prove by contradiction here. My guess is to show the limit that we have from the fact that $X$ is compact is in the intersection. But I don't even have the sequence yet. </p>
<p>Can anyone please give some clue on the line of reasoning of the proof? Many thanks!</p>
| Sam | 189,840 | <p>For each $n$ choose some $x_n \in E_n$ to get a sequence $\{ x_n \}$. By passing to a subsequence if necessary we may assume that $x_n \to x$ for some $x \in X$. Now for any $n$ all but the first $n$ terms (at most) of this sequence lie in $E_n$. If we ignore these first few terms we still have a sequence in $E_n$ converging to $x$. Since $E_n$ is closed, $x \in E_n$. This holds for any $n$, so $x \in \cap_{n=1}^{\infty} E_n$.</p>
|
1,327,755 | <p>Prove that the antipodal mapping $A: S^n \to S^n$ given by $A(p)=-p$ is an isometry. </p>
<p>I know that in order to prove that a map $f$ is an isometry of a smooth manifold $M$ it must hold true that
$$\langle v,u\rangle_p = \langle d_f v, d_fu \rangle_{f(p)}$$</p>
<p>Now I found online <a href="http://image.sciencenet.cn/olddata/kexue.com.cn/upload/blog/file/2009/9/2009941352126256.pdf" rel="noreferrer">this document</a> that proves this exercise (Exercise 1.1.). I cannot follow it though. It basically says that in order to prove this we first need to claim that $T_{p}S^n = T_{A(p)}S^n$ and for this it is enough to prove that $T_{p}S^n \subset T_{A(p)}S^n$. <strong>Why is this true?</strong> In order to show this they claim that $T_{A(p)}S^n \subset \subset T_{A \circ A(p)}S^n = T_pS^n$. <strong>What does the $\subset \subset$ mean</strong> and <strong>why the last relation holds true, i.e. why $T_{A(p)S^n}=T_pS^n$ since these tangent spaces are defined in different points?</strong></p>
<p>Finally, <strong>why they later claim that $dA_pv = -v$?</strong> I do not understand why this also holds true.</p>
| Amitai Yuval | 166,201 | <p>As user 7530 suggests, the antipodal map on the sphere is the restriction to the sphere of the $-id$ map on $\mathbb{R}^{n+1}$. Note that $-id$ is linear, and hence, its derivative at a point is also equal to $-id$. More explicitly, for a point $p\in\mathbb{R}^{n+1}$, identify the tangent space $T_p\mathbb{R}^{n+1}$ with $\mathbb{R}^{n+1}$. Then for any $v\in T_p\mathbb{R}^{n+1}$ we have$$d(-id)_p(v)=-v\in T_{-p}\mathbb{R}^{n+1}.$$Consequently, for any $u,v\in T_p\mathbb{R}^{n+1}$ we have$$\langle d(-id)_p(u),d(-id)_p(v)\rangle=\langle-u,-v\rangle=\langle u,v\rangle,$$and so $-id$ is an isometry.</p>
<p><strong>Edit:</strong> We show $T_pS^n=T_{-p}S^n$. The argument is essentially the one used in Lee's answer, but spelled out differently. Let $\alpha:(-\epsilon,\epsilon)\to S^n$ satisfy $\alpha(0)=p$. We have$$\langle\alpha(t),\alpha(t)\rangle=const,$$and differentiating at $t=0$, using the Leibniz rule, we obtain$$\langle\dot{\alpha}(0),p\rangle=0.$$Hence,$$T_pS^n\subset p^\bot.$$It now follows from dimension consideration that in fact$$T_pS^n=p^\bot.$$Since $p^\bot=(-p)^\bot$, we are done.</p>
<p><strong>Remarak:</strong> As commented bellow, the fact that $T_pS^n=T_{-p}S^n$ may be interesting in general, but is not crucial for this question. Every $M\in O_n(\mathbb{R})$ restricts to an isometry of $S^n$, and the antipodal map is just a particular case.</p>
|
1,098,028 | <p>Let $T: \mathbb{R}_{\leq 3}[X] \rightarrow \mathbb{R}_{\leq 2}[X]$ be a linear map defined as $T(f(x)) = f'(x)$, and let $\beta$ and $\gamma$ be the standard ordered bases for resp. $\mathbb{R}_{\leq 3}[X]$ and $\mathbb{R}_{\leq 2}[X]$. Then the matrixrepresentation of the linear map is given as \begin{align*} A = [T]_{\beta}^{\gamma} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}. \end{align*} Let $\phi_{\beta}: \mathbb{R}_{\leq 3}[X] \rightarrow \mathbb{R}^4$ and $\phi_{\gamma}: \mathbb{R}_{\leq 2}[X] \rightarrow \mathbb{R}^3$ be linear maps that represent de coordinates of the respective polynomials as columnvectors. We search a relationship between $L_A, \phi_{\beta}, \phi_{\gamma}$ and $T$, where $L_A$ is the left-multiplication transformation. Choose the polynomial $g(x)$, defined as $g(x) = 2 + x - 3x^2 + 5x^3 $. Then $\phi_{\beta} = \begin{pmatrix} 2 \\ 1 \\ -3 \\ 5 \end{pmatrix}$. Hence we have \begin{align*} L_A \circ \phi_{\beta} (g(x)) = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -6 \\ 15 \end{pmatrix}. \end{align*} Because $T(g(x)) = g'(x) = 1 -6x + 15x^2$, we have also \begin{align*} \phi_{\gamma}\circ (T(g(x))) = \phi_{\gamma} (g'(x)) = \begin{pmatrix} 1 \\ -6 \\ 15 \end{pmatrix}. \end{align*} So we can conclude that \begin{align*} L_A \circ \phi_{\beta} = \phi_{\gamma} \circ T. \end{align*} How should I understand this now? And how can I prove it in a rigorous manner? Some explanation would be helpful.</p>
| Emily | 31,475 | <p>First condition (given): $f(-2) = 0$.</p>
<p>Second condition (from evenness, may not be useful!): $f(2) = 0$.</p>
<p>Third condition: $f'(x_0) = 6$.</p>
<p>Fourth condition: $f(x_0) = 6-6x_0$.</p>
<p>Fifth condition: At some $x_1$, $f'(x_1) = 0$ and $g(x_1) = f(x_1)$.</p>
<p>Sixth condition: $g'(0.5) = 0, g(0.5) = 5.25$</p>
<p>Seventh condition (since $g$ has a <em>maximum</em> it must open downwards): $b_1 < 0$.</p>
<p>Can you use these to find what you need?</p>
|
3,467,203 | <p><span class="math-container">$$\frac{(-1)^{k+1}k}{3^k}$$</span></p>
<p>Correct me if I am wrong but with <span class="math-container">$(-1)^{k+1}$</span> this makes it an alternating series test therefore i think I need to show that <span class="math-container">$\frac{k}{3^k}$</span> is decreasing and non-negative. But the problem with that is that there is no restriction on bounds. So maybe Direcheltes test would be more appropriate and let <span class="math-container">$a_k=(-1)^{k+1}$</span> which is bounded I believe and let <span class="math-container">$b_k = \frac{k}{3^k}$</span> and show that it is approaching 0 by taking the limit as <span class="math-container">$n \to \infty$</span></p>
| Peter Szilas | 408,605 | <p>1)<span class="math-container">$a_k= \dfrac{k}{3^k} >0$</span> is decreasing.</p>
<p><span class="math-container">$\dfrac{a_{k+1}}{a_k}=\dfrac{(k+1)3^k}{k3^{k+1}}=$</span></p>
<p><span class="math-container">$(1/3)\dfrac{k+1}{k} =(1/3)(1+1/k) <1$</span> for <span class="math-container">$k=1,2,3....$</span></p>
<p>2)<span class="math-container">$\lim_{k \rightarrow \infty}a_k=0$</span>.</p>
<p>3) Leibniz alternating series test.</p>
|
3,415,211 | <p>I was solving my Rubik's cube when I though that if i toss my cube what is the probability of getting red colour twice in a row. So I calculate that it is 1/6 for first time, 1/6 possibility for second time getting a red and 1/6 * 1/6 = 1/36 is the answer.</p>
<p>But then I thought if the colour is unspecified i.e the colour that come when i toss my cube one time is taken to be the input colour to come twice in a row. </p>
<p>If i toss my cube then a random colour, it can be red, white etc. come then for the second time it has to be same, would it change the probability. </p>
<p>In short, will the probability of getting a random colour twice on tossing a cube be same a getting as specified colour?</p>
<p>(English is not my native language so it might not make sense)</p>
| user | 505,767 | <p>Note that more in general for any <span class="math-container">$p\le 1$</span> the integral</p>
<p><span class="math-container">$$\int_0^{1} \frac{1}{x (\ln x)^p} dx$$</span></p>
<p>diverges, indeed by <span class="math-container">$y=\ln x \implies dy = \frac1x dx$</span> we have</p>
<p><span class="math-container">$$\int_0^{1} \frac{1}{x (\ln x)^p} dx=\int_{-\infty}^{0} \frac{1}{y^p} dy$$</span></p>
|
1,764,162 | <p>I have $$f(x) = x \left| x - 1 \right|$$</p>
<p>Here my given value for $x$ is 1</p>
<p>And I need to test if the function is "continuous" when $x$ is $1$ and also when reaching
$$ f(1)$$</p>
<p>$$ \lim\limits_{x \to 1} f(x)$$</p>
<p>So I plug in 1 to the equation to find out the value of $f(1)$</p>
<p>$$ f(1) = 1|1-1| = 0 $$
Then I check the $lim$ to see if I get $0$ as well.</p>
<p>$$ \lim\limits_{x \to 1} f(x) = x|x-1| = 1|1-1| = 0$$</p>
<p>What I am not understanding is why I need to even check for $lim$ even from the first place because the process looks identical. But I am taught to do it anyway.</p>
<p>Could an equation not have the same value for $f(AnyValue)$ and $\lim\limits_{x \to AnyValue} f(x)$ ?</p>
| A.Riesen | 329,330 | <p>The specific value matters for the function might be continuous at some values and not at others. For example $|x|/x=f(x)$ is continuous everywhere except $x=0$ for if you approach zero from the left you will get -1 but if you approach it from the right you will get 1. Also you can not just plug in x=0 for f(x) since the expression would have no meaning thus you have to check the limit.</p>
|
3,592,368 | <p>I managed to solve this question using trigonometry. But I wondered if there'd be anyway of doing it using only synthetic geometry. Here it is.</p>
<blockquote>
<p>Let <span class="math-container">$ABC$</span> be a right isosceles triangle of hypotenuse <span class="math-container">$AB$</span>. Let also <span class="math-container">$\Gamma$</span> be the semicircle whose diameter is the line segment <span class="math-container">$AC$</span> such that <span class="math-container">$\Gamma\cap\overline{AB} = \{A\}$</span>. Consider <span class="math-container">$P\in\Gamma$</span> with <span class="math-container">$PC = k$</span>, with <span class="math-container">$k \leq AC$</span>. Find the area of triangle <span class="math-container">$PBC$</span>.</p>
</blockquote>
<p>Here is my interpretation of the picture:
<a href="https://i.stack.imgur.com/Vz0ck.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vz0ck.png" alt="enter image description here" /></a></p>
<p>I managed to get the solution via trigonometry as below.</p>
<p><a href="https://i.stack.imgur.com/dQ8hU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dQ8hU.png" alt="enter image description here" /></a></p>
<p>Then, the area <span class="math-container">$S$</span> requested is:</p>
<p><span class="math-container">$$\begin{align} S &= \displaystyle\frac{PC\cdot BC\cdot \sin(90^\circ + \beta)}{2}\\
&= \displaystyle\frac{k\cdot d\cdot \cos\beta}{2}\\
&= \displaystyle\frac{k\cdot d\cdot \frac{k}{d}}{2}\\
&= \displaystyle\frac{k^2}{2}.\\
\end{align}$$</span></p>
| Quanto | 686,284 | <p><a href="https://i.stack.imgur.com/fa7mB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fa7mB.png" alt="enter image description here"></a></p>
<p>Draw the parallel line BX || AP and extend PC to meet it at X. Note that <span class="math-container">$\angle XCB = 90 - \beta = \gamma$</span>. Along with CA = CB, the right triangles APC and CBX are congruent, which leads to <span class="math-container">$CP = BX = k$</span>. </p>
<p>Thus, the area of CPB is</p>
<p><span class="math-container">$$Area_{CPB}=\frac12 CP\cdot BX = \frac12k^2$$</span></p>
|
1,369,428 | <p>I am trying to solve the following equation;
$$\int_{-1}^{1}e^{i(x+a\cos x)} \, \mathrm{d}(\cos x)$$ or $$\int_{0}^{\pi}e^{i(x+a\cos x)} \sin x \, \mathrm{d}x$$</p>
<p>I tried this in <em>Wolfram Alpha</em>, but it says that answer cannot be obtained.</p>
| Claude Leibovici | 82,404 | <p>In the same spirit as Josh Broadhurst's answer, you should arrive to $$\int_{0}^{\pi}e^{i(x+a\cos(x))} \sin(x)\,dx=\frac{\pi a J_1(a)-2 a \cos (a)+ 2 \sin (a)}{a^2}\,i$$ I do not find any way to compute the antiderivative itself.</p>
|
2,073,794 | <p>Let's consider the following variant of Collatz (3n+1) : </p>
<p>if $n$ is odd then $n \to 3n+1$</p>
<p>if $n$ is even then you can choose : $n \to n/2$ or $n \to 3n+1$</p>
<p>With this definition, is it possible to construct a cycle other than the trivial one, i.e., $1\to 4 \to 2 \to 1$?</p>
<p>Best regards</p>
| sTertooy | 336,630 | <p>Yes! With the standard Collatz conjecture, every number must eventually end up at the cycle $4 \to 2 \to 1 \to 4 \cdots$ . This has been verified for all numbers up to $2^{60}$.</p>
<p>With your altered definition, you can start at $2$, apply $3n+1$ instead of $n/2$, and then continue like the standard Collatz again. </p>
<p>$$2 \xrightarrow{3n+1} 7 \to 22 \to 11 \to \cdots ,$$
you'll eventually end at $2$ again, since this is one of the verified cases.</p>
|
2,073,794 | <p>Let's consider the following variant of Collatz (3n+1) : </p>
<p>if $n$ is odd then $n \to 3n+1$</p>
<p>if $n$ is even then you can choose : $n \to n/2$ or $n \to 3n+1$</p>
<p>With this definition, is it possible to construct a cycle other than the trivial one, i.e., $1\to 4 \to 2 \to 1$?</p>
<p>Best regards</p>
| goodvibration | 402,209 | <p>$4\to13\to40\to20\to10\to5\to16\to8\to4$</p>
|
1,549,181 | <p><strong>Calculate Using Polar Coordinates</strong></p>
<p>This is a drawing I made to illustrate the problem. <a href="http://tube.geogebra.org/m/ZzvL0a38" rel="nofollow">http://tube.geogebra.org/m/ZzvL0a38</a></p>
<p>$$\int_{\frac 12}^{1} \int_{0}^{\sqrt{1-x^2}} 1 \quad dydx $$</p>
<p>What I am confused about in this problem is how does one redefine the upper and lower bound coordinates? From what I can see this is a type 1 domain. I know that I am supposed to use the unit circle, but I am unsure on how to proceed.</p>
<p>This is what I have so far</p>
<p>$$\iint r dr dϴ $$</p>
| E.H.E | 187,799 | <p>$$\int_{\frac 12}^{1} \int_{0}^{\sqrt{1-x^2}} 1 \quad dydx=\int_{0}^{\frac{\pi}{3}} \int_{1-0.5\sec\theta}^{1} \quad rdrd\theta$$
<a href="https://i.stack.imgur.com/A07HN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A07HN.png" alt="enter image description here"></a>
some details:</p>
<hr>
<p>$$x=0.5$$
$$r\cos\theta=0.5$$
$$r=0.5\sec\theta$$</p>
|
106,853 | <p>I have taken a lot of math in university, but chose to omit differential equations. Unfortunately, now I have to read computer science proofs that use them, mostly ODEs, and this is always a struggle. What textbook(s) should I read to take me from the basics to practical use of the theory?</p>
| Gerry Myerson | 8,269 | <p>Boyce and DiPrima, Elementary Differential Equations, Eighth Edition. <a href="http://bcs.wiley.com/he-bcs/Books?action=index&itemId=047143339X&itemTypeId=BKS&bcsId=2026" rel="nofollow">Website</a>. </p>
|
106,853 | <p>I have taken a lot of math in university, but chose to omit differential equations. Unfortunately, now I have to read computer science proofs that use them, mostly ODEs, and this is always a struggle. What textbook(s) should I read to take me from the basics to practical use of the theory?</p>
| Community | -1 | <p>The books which I referred are:</p>
<ul>
<li><p>Ordinary differential equations by E. Coddington.</p></li>
<li><p>Differential Equations With Applications and Historical Notes by George F. Simmons. This book is a classic and at the end of each chapter has historical notes of Mathematicians which is fun to read.</p></li>
<li><p>An online book published by the AMS is available <a href="http://www.mat.univie.ac.at/~gerald/ftp/book-ode/ode.pdf" rel="nofollow">here</a>. I basically learn't the proofs of Gronwall's equality from here and used it as a reference book.</p></li>
</ul>
|
828,668 | <p>Let $G$ be a finite group. If there exists an $a\in G$ not equal to the identity such that for all $x\in G$,$\phi(x) = axa^{-1}=x^{p+1} $ is an automorphism of $G$ then $G$ is a $p$-group.</p>
<p>This is what I have so for.</p>
<p>The order of $a$ is $p$ since $\phi(a) = a= a^pa\rightarrow a^p=e$ therefore the $order(\phi)|p$</p>
<p>If $order(\phi) = 1$ then for all $ x\in G$ $\phi(x) = x=x^{p+1}\rightarrow x^p=e$ . Thus every element has order $p$ therefore $G$ is a $p-group$</p>
<p>For $order(\phi) = p$ I get stuck:</p>
<p>$\phi^p(x) = x = x^{(p+1)^p}$ using the expansion formula and simplifying i reach that the order of each element in $G$ divides $\displaystyle\sum_{k=1}^n \binom{p}{k}p^k=(p+1)^p-1$. But other primes divide this so I can't easily conclude $G$ is a $p$-group. </p>
<p>I proceeded by contradiction:</p>
<p>Suppose for contradiction that $G$ is not a $p$-group. Let $|G| = kp^n$ where $k$ is not a multiple of $p$ ($p\nmid k$). If $k>1$ then take a $q$ in $k$'s prime factorization. So we have $q|k$ and by Cauchy's Theorem $\exists y \in G$ with $y^q = e$ i.e $order(y) = q$.</p>
<p>If $q<p$, applying $\phi$ to $y$ I get that $\langle y\rangle$ has more than $q$ elements since $\phi(y)=y^{p+1}\in \langle y \rangle$ and there are $p$ distint elements achieved by $\phi$</p>
<p>If $p>q$ then....... I cannot reach a contradiction :'(</p>
<p>Question #2: Show each element of $G$ has order $p$</p>
<p>When $order(\phi) = 1$ i get what I want but I'm also stuck when $order(\phi) =p$</p>
<p>I believe $a\in Z(G)$, is there a way I can show this?</p>
<p>Thank you....This is a pretty hard problem. :'(</p>
| Dan Shved | 47,560 | <p>Here is a plan.</p>
<p>Let $N = \{x \in G \mid x^p = 1\}$.</p>
<ol>
<li><p>Note that $x \in N$ if and only if $x$ commutes with $a$. It follows that $N$ is a subgroup of $G$. It is easy to see that $N$ is characteristic and therefore normal.</p></li>
<li><p>Take an arbitrary $y \in G$. We have $a y a^{-1} = y^{p+1}$ in $G$. Project this equality onto $G/N$, it gives us $yN = y^{p+1}N$ ($a$ is gone because $a \in N$). Therefore, $y^p \in N$, therefore $y^{p^2} = 1$. This shows that $G$ is a $p$-group, even that the order of each element divides $p^2$.</p></li>
</ol>
|
3,828,982 | <blockquote>
<p><strong>Statement</strong></p>
<p>If <span class="math-container">$U$</span> is open and <span class="math-container">$\text{int}(S)\neq\emptyset$</span> then <span class="math-container">$\text{int}(U\cap S)\neq\emptyset$</span> too when <span class="math-container">$U\cap S\neq\emptyset$</span> and when <span class="math-container">$S$</span> is path connected.</p>
</blockquote>
<p>We know that <span class="math-container">$\text{int}(U\cap S)=\text{int}(U)\cap\text{int}(S)=U\cap\text{int}(S)$</span> so that if <span class="math-container">$\text{int}(U\cap S)=\emptyset$</span> and <span class="math-container">$U\cap S\neq\emptyset$</span> then <span class="math-container">$(U\cap S)\subseteq\text{Bd}(S)$</span> but unfortunately I don't see in this any contradiction. So I don't be able to prove the statement. Could be that it is generally false? Anyway it seems to me that in the case where <span class="math-container">$U$</span> and <span class="math-container">$S$</span> are <strong>path connencted</strong> subset of <span class="math-container">$\Bbb R^n$</span> then the statement holds so I think that although it is gerenally false it could be true if we consider some particular case. If it is more difficul understand when the statement is generally true I ask to prove it when <span class="math-container">$U$</span> and <span class="math-container">$S$</span> are subset of <span class="math-container">$\Bbb R^n$</span>.
Finally I point out that the statement is generally false when <span class="math-container">$U$</span> or <span class="math-container">$S$</span> are not path connected: e.g. if you take <span class="math-container">$U=(0,1)$</span> and <span class="math-container">$S=\{\frac 1 2\}\cup(2,3)$</span> or <span class="math-container">$U=\big[(0,1)\cup(2,3)\big]$</span> and <span class="math-container">$S=[1,2]$</span> then the statement is clearly false!</p>
<p>So could someone help me, please?</p>
| riemann_lebesgue | 539,416 | <p>Ok here's a counter example. Below, set A is the interior of the blue ball. Set B is the red line and the interior of the red ball. Clearly, the interior of B is <span class="math-container">$R^2$</span> is the interior of the red ball. And the interior of the blue ball clearly doesn't intersect with that. Making this rigorous wouldn't be tricky. Just set two balls a certain distance apart, and pick a line to go from one into the other.</p>
<p><a href="https://i.stack.imgur.com/s2XXq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s2XXq.png" alt="enter image description here" /></a></p>
<p>Now, your examples looked specifically at <span class="math-container">$R$</span>. But if a set is path connected in R, then visually you can see that there will either be points in the interior either directly to the right of it, or directly to the left of it. That's becauuse, if we take another point y, and we take the continuous path mapping from x to y, by the intermediate value theorem all points inbetween them will be mapped to by our curve. That's why you seem to have trouble finding counter examples in R</p>
|
165,888 | <p>I want to return the final value from the Table.For example something like Table[n-1] value.</p>
| Henrik Schumacher | 38,178 | <p>Not faster, but maybe easier to implement</p>
<pre><code>makeKMat2[k1_, k2_] := Module[{A, m, n},
m = Length[k1];
n = Length[k2];
A = Map[SparseArray[Band[{1, 1}] -> #, {n, n}, 0.] &, k1, {2}];
With[{B = SparseArray@k2}, Do[A[[i, i]] += B, {i, 1, m}]];
ArrayFlatten[A]
]
</code></pre>
<p>Your implementation needs about <code>0.0048</code>s; mine needs <code>0.0051</code>s on my machine.</p>
<p>(<em>b3m2a1 comment</em>)<br>
Note that this has a simple extension to the N-dimensional case:</p>
<pre><code>makeKMat2ND[ks : {__List}] := Fold[makeKMat2, ks]
</code></pre>
|
3,025,375 | <p>What is<span class="math-container">$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$</span>So it is<span class="math-container">$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$</span>
I do not know what to do next, because my resuts is <span class="math-container">$∞$</span> but the answer from book is <span class="math-container">$\dfrac{1}{4\sqrt{2}}$</span>.</p>
| lab bhattacharjee | 33,337 | <p>Let <span class="math-container">$1/n=h$</span></p>
<p><span class="math-container">$$\lim_{h\to0^+}\dfrac{\sqrt{1+\sqrt{1+h^4}}-\sqrt2}{h^4}$$</span></p>
<p><span class="math-container">$$=\lim_{h\to0^+}\dfrac{1+\sqrt{1+h^4}-2}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$</span></p>
<p><span class="math-container">$$=\lim_{h\to0^+}\dfrac{1+h^4-1}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+h^4}+1}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$</span></p>
<p><span class="math-container">$$=\dfrac1{(\sqrt1+1)(\sqrt{1+\sqrt1}+\sqrt2)}$$</span></p>
|
292,041 | <p>The Helly theorem in the Euclidean plane asserts that if $S_1, \dots, S_n$ are $n \ge 3$ convex subsets such that $S_i \cap S_j \cap S_k \ne \emptyset$ for all distinct triples $i,j,k$, then the total intersection $\bigcap_{i = 1}^n S_i$ is also nonempty. </p>
<p>I'm wondering if the same theorem is true in the hyperbolic plane (for concreteness, let's assume the Poincaré disk model). My understanding is that if the analogue of Radon's theorem is true in this setting, then Helly follows axiomatically.</p>
<p>Radon's theorem in the Euclidean plane asserts that given any four points $x_1, \dots, x_4$, there is a partition into two nonempty subsets such that the convex hulls intersect. The proof I know uses the affine structure on the Euclidean plane and so doesn't seem to port directly into hyperbolic space. On the other hand, I can verify that the Radon property holds for all the collections of four points I've looked at... </p>
| Igor Rivin | 11,142 | <p>I don't understand your reference to the model (since the geometry of the hyperbolic plane does not depend on any model), but, in fact, the Beltrami-Klein model demonstrates that any qualitative statement about convex sets in the Euclidean plane holds in the Hyperbolic plane and <em>vice versa</em>, since the model maps convex sets to convex sets.</p>
<p><strong>EDIT</strong> This has (almost) absolutely nothing to do with the above, but the topological version of Helly's theorem goes back to <em>at least</em> Debrunner (and it is a Monthly paper, so is human-readable), no need to allude to Farb's paper.</p>
<p><a href="https://i.stack.imgur.com/wSpBf.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wSpBf.png" alt="enter image description here"></a></p>
|
4,365,005 | <p>The set of all homomorphisms between two spaces <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is denoted as <span class="math-container">$\text{Hom}(X,Y)$</span>, the set of endomorphisms of a space <span class="math-container">$X$</span> is denoted as <span class="math-container">$\text{End}(X)$</span>,... these are standard notations, but what about the set of all maps between two spaces <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> (without any extra structure) Are <span class="math-container">$\text{Map}(X,Y)$</span> or <span class="math-container">$\text{Fun}(X,Y)$</span> standard notations?</p>
<p>This might be an stupid question, but I want to make sure I am using the right notation, or at least the most usual notation for this spaces. Thanks for your help.</p>
| user829347 | 719,291 | <p>Common ways to denote the set of all functions from a set <span class="math-container">$X$</span> to a set <span class="math-container">$Y$</span> that I have encountered (some of which have already been mentioned) are <span class="math-container">$\mathcal{F}(X;Y)$</span>, <span class="math-container">$\text{Map}(X;Y)$</span> and <span class="math-container">$Y^X$</span>. Then you can define <span class="math-container">$\mathcal{F}(X):=\mathcal{F}(X;X)$</span> etc. I don't think there is a 'standard' way of presenting such functions that has anywhere near a consensus. I would personally use whichever notation feels most natural, depending on the context.</p>
<p>The <span class="math-container">$Y^X$</span> notation comes from the fact that you are choosing one value of <span class="math-container">$Y$</span> for each value of <span class="math-container">$X$</span>, so you can kind of think of it is an '<span class="math-container">$X$</span>-fold Cartesian product' in this sense i.e. <span class="math-container">$X$</span> copies of the set <span class="math-container">$Y$</span>.</p>
|
1,633,846 | <p>I'm currently a high school Pre-Calculus student and my textbook presents the following theorem without proof:</p>
<blockquote>
<p>Let <span class="math-container">$f(x)$</span> be a polynomial with real coefficients and a positive leading coefficient.</p>
<p>Let <span class="math-container">$a \geq 0$</span>. Then, if when <span class="math-container">$f(x)$</span> is divided by <span class="math-container">$x-a$</span>, all of the coefficients of the quotient and the remainder are non-negative, <span class="math-container">$a$</span> is an upper bound on the real zeroes of <span class="math-container">$f(x)$</span>.</p>
<p>Now, let <span class="math-container">$a \leq 0$</span>. Then, if when <span class="math-container">$f(x)$</span> is divided by <span class="math-container">$x-a$</span>, all of the coefficients of the quotient and the remainder alternate between non-positive and non-negative, <span class="math-container">$a$</span> is a lower bound on the real zeroes of <span class="math-container">$f(x)$</span>.</p>
</blockquote>
<p>I am trying to find a proof for this theorem and I think I found a proof for it <a href="http://mathweb.scranton.edu/monks/courses/ProblemSolving/POLYTHEOREMS.pdf" rel="nofollow noreferrer">pages 6 and 7 of this PDF file</a>. However, the proof does not seem correct:</p>
<blockquote>
<p>[...] since all of <span class="math-container">$q(x)$</span>'s remaining coefficients are positive [...]</p>
</blockquote>
<p>This is a quote from the proof of the first part of this theorem. Here, <span class="math-container">$q(x)$</span> is the resulting polynomial from dividing <span class="math-container">$f(x)$</span> by <span class="math-container">$x-b$</span> for some root <span class="math-container">$b$</span> for <span class="math-container">$f(x)$</span>. However, there is a clear counter example to this: If <span class="math-container">$a=5$</span> and <span class="math-container">$f(x)=x^2-5x+6$</span>, then <span class="math-container">$f(x)$</span> and <span class="math-container">$a$</span> meet the hypothesis since <span class="math-container">$\frac{f(x)}{x-a}=x+0+\frac{6}{x-a}$</span>, so there are all non-negative coefficients in the quotient and the remainder. Then, <span class="math-container">$b=3$</span> is a root of <span class="math-container">$f(x)$</span>, but, <span class="math-container">$q(x)=\frac{f(x)}{x-b}=x-2$</span> does not have all positive coefficients, contradicting the above.</p>
<p>The proof of the second part of the theorem is also wrong:</p>
<blockquote>
<p>Because <span class="math-container">$a < 0$</span> and the leading term in <span class="math-container">$q(x)$</span> has a positive coefficient, the constant term in <span class="math-container">$q(x)$</span> has the same sign as <span class="math-container">$q(a)$</span>.</p>
</blockquote>
<p>However, if we let <span class="math-container">$a=-4$</span> and <span class="math-container">$f(x)=x^2+3x+2$</span>, then <span class="math-container">$f(x)$</span> and <span class="math-container">$a$</span> meet the hypothesis since <span class="math-container">$\frac{f(x)}{x-a}=x-1+\frac{6}{x-a}$</span>, which alternatives between non-negative and non-positive coefficients in the quotient and remainder. Then, <span class="math-container">$b=2$</span> is a root of <span class="math-container">$f(x)$</span>, but <span class="math-container">$q(x)=\frac{f(x)}{x-b}=x+1$</span> and thus <span class="math-container">$q(a)=-3$</span> while the constant term of <span class="math-container">$q(x)$</span> is <span class="math-container">$1$</span>, which also clearly contradicts the above.</p>
<p>Thus, while I have found proofs for these theorems, I do not think they are valid. Could someone show me valid proofs for these theorems in my Pre-Calc textbook? Thank you!</p>
| Noble Mushtak | 307,483 | <p>Proof of Upper Bound Theorem:</p>
<p>Let <span class="math-container">$f(x)=q(x)*(x-a)+r$</span>. It is given that all of the coefficients of <span class="math-container">$q(x)$</span> and <span class="math-container">$r$</span> are non-negative. Factoring out a <span class="math-container">$x-a$</span> from the right side of the equation yields <span class="math-container">$f(x)=(x-a)(q(x)+\frac{r}{x-a})$</span> for all <span class="math-container">$x \neq a$</span>. Now, let <span class="math-container">$c > a$</span>. Note that <span class="math-container">$c$</span> is positive since <span class="math-container">$a$</span> is non-negative. We will prove that <span class="math-container">$f(c) \neq 0$</span>, which will prove the theorem by showing that <span class="math-container">$f(c)=0$</span> implies <span class="math-container">$c \leq a$</span> and thus <span class="math-container">$a$</span> is an upper bound on all zeroes of <span class="math-container">$f(c)$</span>.</p>
<p>We have already shown that <span class="math-container">$f(c)=(c-a)(q(c)+\frac{r}{c-a})$</span>. Since <span class="math-container">$c \neq a$</span>, <span class="math-container">$c-a \neq 0$</span> and thus <span class="math-container">$q(c)+\frac{r}{c-a} \neq 0$</span> implies <span class="math-container">$f(c) \neq 0$</span>. Therefore, we will prove that <span class="math-container">$q(c)+\frac{r}{c-a} \neq 0$</span>. <span class="math-container">$q(x)=\sum_{i=0}^{n-1}q_i*x^i$</span> where <span class="math-container">$q_i$</span> are the non-negative coefficients of <span class="math-container">$q(x)$</span> and <span class="math-container">$n$</span> is the degree of <span class="math-container">$f(x)$</span>. Since <span class="math-container">$c$</span> is positive, <span class="math-container">$c^i$</span> is also positive for any <span class="math-container">$i \in \mathbb{N}$</span> and thus, since <span class="math-container">$q_i$</span> is non-negative, <span class="math-container">$q_i*c^i$</span> is non-negative. As the sum of non-negative numbers are non-negative and <span class="math-container">$q(c)$</span> is the sum of all <span class="math-container">$q_i*c^i$</span> for <span class="math-container">$i \in \mathbb{N}$</span>, <span class="math-container">$q(c)$</span> is non-negative. Also, since <span class="math-container">$c > a$</span>, <span class="math-container">$c-a$</span> is positive and thus, since <span class="math-container">$r$</span> is non-negative, <span class="math-container">$\frac{r}{c-a}$</span> is non-negative.</p>
<p>Therefore, since both <span class="math-container">$q(c)$</span> and <span class="math-container">$\frac{r}{c-a}$</span> are non-negative, the only way their sum could be <span class="math-container">$0$</span> is if both of them are <span class="math-container">$0$</span>. However, if both <span class="math-container">$q(c)$</span> is <span class="math-container">$0$</span>, then all of the coefficients of <span class="math-container">$q(x)$</span> are <span class="math-container">$0$</span>, meaning <span class="math-container">$q(x)=0$</span>. Also, if <span class="math-container">$\frac{r}{c-a}=0$</span>, since <span class="math-container">$c-a$</span> is positive, <span class="math-container">$r=0$</span>. If both <span class="math-container">$q(x)$</span> and <span class="math-container">$r$</span> are <span class="math-container">$0$</span>, since <span class="math-container">$f(x)=q(x)*(x-a)+r$</span>, <span class="math-container">$f(x)=0$</span>. However, <span class="math-container">$f(x) \neq 0$</span> since it has a leading positive coefficient. Therefore, <span class="math-container">$q(c)+\frac{r}{c-a} \neq 0$</span> because otherwise, <span class="math-container">$f(x)$</span> would not have a leading positive coefficient. Thus, we have proven the theorem.</p>
<p>Proof of Lower Bound Theorem:</p>
<p>Let <span class="math-container">$f(x)=q(x)*(x-a)+r$</span> and <span class="math-container">$q(x)=\sum_{i=0}^{n-1}q_i*x^i$</span> where <span class="math-container">$q_i$</span> are the coefficients of <span class="math-container">$q(x)$</span> and <span class="math-container">$n$</span> is the degree of <span class="math-container">$f(x)$</span>. Since <span class="math-container">$f(x)$</span> has a positive leading coefficient, <span class="math-container">$q(x)$</span> must have a non-negative leading coefficient and thus <span class="math-container">$q_{n-1} \geq 0$</span>. Also, let <span class="math-container">$c < a$</span>. Note that <span class="math-container">$c$</span> is negative since <span class="math-container">$a$</span> is non-positive. We can use the argument from the previous theorem to show that all we need to prove is that <span class="math-container">$q(c)+\frac{r}{c-a} \neq 0$</span> in order to prove the whole theorem. We will now split this proof up into two cases: <span class="math-container">$n$</span> is odd and <span class="math-container">$n$</span> is even.</p>
<p>Case 1: <span class="math-container">$n$</span> is Odd</p>
<p>If <span class="math-container">$n$</span> is odd, then <span class="math-container">$n-1$</span> is even and since all of the coefficients alternate between non-negative and non-positive, all of the coefficients of even powers of <span class="math-container">$x$</span> must be the same sign as <span class="math-container">$q_{n-1}$</span>, or non-negative, and all of the coefficients of odd powers of <span class="math-container">$x$</span> must be the opposite sign as <span class="math-container">$q_{n-1}$</span>, or non-positive. Let <span class="math-container">$i \in \mathbb{N}$</span>. If <span class="math-container">$i$</span> is odd, then <span class="math-container">$q_i$</span> is non-positive, as we have just shown, and since <span class="math-container">$c$</span> is negative, <span class="math-container">$c^i$</span> is negative, so <span class="math-container">$q_i*c^i$</span> is non-negative. Otherwise, <span class="math-container">$i$</span> is even, so <span class="math-container">$q_i$</span> is non-negative and <span class="math-container">$c^i$</span> is positive, making <span class="math-container">$q_i*c^i$</span> still non-negative. Thus, since the sum of non-negative numbers is non-negative and <span class="math-container">$q(c)=\sum_{i=0}^{n-1}q_i*c^i$</span>, <span class="math-container">$q(c)$</span> is non-negative.</p>
<p>Also, since <span class="math-container">$q_0$</span> is non-negative (as <span class="math-container">$0$</span> is even) and <span class="math-container">$r$</span> alternates in sign with <span class="math-container">$q_0$</span>, <span class="math-container">$r$</span> is non-positive. Since <span class="math-container">$c < a$</span>, <span class="math-container">$c-a$</span> is negative. Therefore, <span class="math-container">$\frac{r}{c-a}$</span> is non-negative.</p>
<p>Thus, we have shown that <span class="math-container">$q(c)$</span> and <span class="math-container">$\frac{r}{c-a}$</span> are both non-negative, meaning <span class="math-container">$q(c)+\frac{r}{c-a}=0$</span> only if <span class="math-container">$q(c)=0$</span> and <span class="math-container">$\frac{r}{c-a}=0$</span>. However, we have already shown in the argument of the above theorem that this contradicts <span class="math-container">$f(x)$</span>'s positive leading coefficient. Therefore, <span class="math-container">$q(c)+\frac{r}{c-a} \neq 0$</span>, proving the theorem for this case.</p>
<p>Case 2: <span class="math-container">$n$</span> is Even</p>
<p>The proof of this case is very similar to the proof of Case 1. Simply switch "odd" with "even" and "positive" with "negative" from the proof of the previous case (except where it says "If <span class="math-container">$i$</span> is odd", "since <span class="math-container">$c$</span> is negative, <span class="math-container">$c^i$</span> is negative", "Otherwise, <span class="math-container">$i$</span> is even", "<span class="math-container">$c^i$</span> is positive", and "because <span class="math-container">$0$</span> is even") to prove this new case.</p>
<p>Thus, since we have proven both cases of this theorem, we have proven the whole theorem.</p>
|
1,234,320 | <p>It is given that $4 x^4 + 9 y^4 = 64$.</p>
<p>Then what will be the maximum value of $x^2 + y^2$?</p>
<p>I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .</p>
| Community | -1 | <p>If $64= (3y^2)^2 + 4x^4$, then $3y^2 =\sqrt{ 64 - 4x^4}$. Plugging that to $x^2+y^2$ we get:
$$x^2+\frac{\sqrt{64 - 4x^4}}{3}.$$</p>
<p>Now you have to maximalize a function of one variable. Remember that if $64 = 4x^4+9y^4$, then (setting $y = 0$) we obtain $16 \ge x^4$, therefore $|x| \le 2$.</p>
|
760,216 | <p>The sewing pattern on a basketball is composed of two great circles and a single curve that intersects each great circle twice. Does this curve have a name? Are there any parametric descriptions of the curve?</p>
<p>The same question can apply to the sewing pattern on a tennis ball.</p>
<p>Answers about why this sewing pattern came to be used are also welcome.</p>
| hmakholm left over Monica | 14,366 | <p>Actually I thought the seams on a basketball were two great circles and <em>two ovals</em> that each intersect the same great circle twice. And so apparently do many people who make basketball texture maps one can find on the web. (<a href="http://www.xkcd.com/1365/" rel="nofollow noreferrer">Randall Munroe too</a>).</p>
<p>It is difficult to settle the question without having a physical ball to inspect; a usual photograph doesn't show both poles of the ball which is one needs. However, here's a photo showing that at least some basketballs are constructed like you describe:</p>
<p><img src="https://i.stack.imgur.com/klwMG.jpg" alt="picture of basketball"></p>
<p>I don't think the shape of the curved seam has a name -- judging from the photos I've found its precise shape can vary from ball to ball. The most realistic-looking images look like it's made up of four about 150° arcs of great circles, joined by curved segments near the poles, but how tight the joining curves are seems to vary.</p>
<p>From a construction point of view, the one-curve design does make more sense, because then the seams divide the surface of the ball into eight <em>congruent</em> pieces, each of which can be cut out of flat material without too much distortion (at compared to dividing into eights by three great circles), and also without needing eight-way meets at the poles. With two great circles and two ovals, the pieces would have two different shapes.</p>
<p>Incidentally, I don't think they're actually seams with modern manufacturing methods -- just a decoration that's put on the ball for reasons of tradition, and possibly to make it easier for players to perceive its spin.</p>
|
4,183,734 | <p>I just started trigonometry, and I came across this problem. I know that this problem would be very simple with a calculator, but without one, I'm lost. How would you determine if <span class="math-container">$\sin 4$</span> is bigger than <span class="math-container">$\sin 3$</span> in negativity or positivity? How would you even determine if they are positive or negative in the first place, and how would you know which one is bigger?</p>
| Ritam_Dasgupta | 925,091 | <p>Here's a possible method:
<span class="math-container">$$\sin 3+\sin 4=2\sin\left(\frac {3+4}{2}\right) \cos\left(\frac {3-4}{2}\right)=2\sin\left(\frac 72\right)\cos\left(\frac 12\right)$$</span>
Now, we know that <span class="math-container">$0<\frac 12<\frac {\pi}{2}$</span>, hence the cosine is positive. Also, <span class="math-container">$\frac {3\pi}{2}>\frac 72>\pi$</span>, so the sine part is negative. Hence the product is negative, and consequently, the sum is negative too.</p>
<p><strong>Notes:</strong></p>
<p>●Here the formula I've used is:
<span class="math-container">$$\sin x+\sin y=2\sin\left(\frac {x+y}{2}\right) \cos \left(\frac {x-y}{2}\right)$$</span>
●Also note how I replaced <span class="math-container">$\cos \left(-\frac 12\right)$</span> with <span class="math-container">$\cos\left( \frac 12\right)$</span>, this is due to the fact that <span class="math-container">$\cos$</span> is an even function, and hence, <span class="math-container">$\cos (-x)=\cos x$</span>.</p>
<p>●Observe how this method gives you the general process for such questions, where it is possible that neither angle is very close to multiples of <span class="math-container">$\pi$</span>.</p>
|
2,738,023 | <p>We know we can check if two vectors are 'orthogonal' by doing an inner product.</p>
<p>$a*b=0$</p>
<p>tells us that these two vectors are orthogonal</p>
<p>here comes the question:</p>
<p>if there a way to compute if they are 'parallel'? i.e., they are pointing at the same direction. </p>
| Anastassis Kapetanakis | 342,024 | <p>You can check the followings:</p>
<p>1) Find their slope if you have their coordinates. The slope for a vector $\vec v$ is $λ=\frac{y_v}{x_v}$. If the slope of $\vec a$ and $\vec b$ are equal, then they are parallel.</p>
<p>2) Find the if $\vec a= k\vec b$ where $k \in \mathbb{R}$. If there is a value that satisfies the above equation, then they are parallel.</p>
|
4,240,892 | <p><strong>My attempt</strong>:
<span class="math-container">$f(x+y)=f(x)f(y)$</span></p>
<p>Differenting wrt <span class="math-container">$y$</span>,</p>
<p><span class="math-container">$$f'(x+y)=f(x)f'(y)$$</span></p>
<p>Putting <span class="math-container">$y=0$</span>,</p>
<p><span class="math-container">$$f'(x)=f(x)f'(0)=kf(x)$$</span></p>
<p><span class="math-container">$$\frac{dy}{dx}=ky$$</span></p>
<p>We know, <span class="math-container">$y \neq 0$</span> because <span class="math-container">$y$</span> is different everywhere so dividing by y and integrating both sides</p>
<p><span class="math-container">$$\int \frac{dy}{y}= \int dx$$</span></p>
<p><span class="math-container">$$\ln \lvert y\rvert = kx + c$$</span></p>
<p>(We can get <span class="math-container">$c=0$</span> by putting <span class="math-container">$x=y=0$</span> in the first equation)</p>
<p><span class="math-container">$$y = \pm {e}^{kx}$$</span></p>
<p><strong>Doubt</strong>: Shouldn't we get <span class="math-container">$y = \pm {a}^{kx}$</span> as solution where <span class="math-container">$a$</span> <span class="math-container">$\in$</span> <span class="math-container">$R$</span> except <span class="math-container">$0$</span> ?</p>
| Infinity_hunter | 826,797 | <p>Actually, it does not matter as long as we are dealing with exponential functions.</p>
<p>You can still write the solution in terms of <span class="math-container">$a^{jx}$</span> where <span class="math-container">$a$</span> is a non-negative real number.</p>
<p>Take <span class="math-container">$j = \frac{k}{\log a}$</span> then you see that <span class="math-container">$e^k = a^j$</span>, so they are interconvertible.</p>
|
673,229 | <blockquote>
<p>The medians to the two legs of a right angled triangle are $10$ and $4\sqrt{10}$. Find the hypotenuse of the right angled triangle.</p>
</blockquote>
<p>I'm confused. Can somebody please explain to me how to do this step by step? Not just the answer I want to know what you did to get the answer and why you did it please. Thank you!</p>
| Mick | 42,351 | <p>No one area in mathematics can be considered as fully developed. This is because some day, some one will come up with some brilliant ideas that will</p>
<p>extend our research to a new dimension; or even disprove what we have accepted as must-be-true.</p>
|
673,229 | <blockquote>
<p>The medians to the two legs of a right angled triangle are $10$ and $4\sqrt{10}$. Find the hypotenuse of the right angled triangle.</p>
</blockquote>
<p>I'm confused. Can somebody please explain to me how to do this step by step? Not just the answer I want to know what you did to get the answer and why you did it please. Thank you!</p>
| Moishe Kohan | 84,907 | <p>Here are some "fully developed areas" (and not having PhDs awarded in), which once upon a time were active research areas:</p>
<ol>
<li><p>Squaring the circle and trisecting the angle using compass and ruler. </p></li>
<li><p>Proving the 5th postulate from the rest of Euclidean axioms. </p></li>
<li><p>Proving Fermat's last theorem using elementary algebra/number theory. </p></li>
<li><p>More recent: constructing homotopy 3-spheres which are not homeomorphic to the standard 3-sphere. </p></li>
</ol>
<p>(Of course, areas described in items 1, 2 and 3 are very still active among math cranks.) </p>
<p>I can go on with examples, but I think, you got my point by now. </p>
|
3,245,838 | <p>We have <span class="math-container">$n$</span> voltages <span class="math-container">$V_1, V_2,\dots , V_n$</span> that are received in a condensator or sum, such that <span class="math-container">$V=\sum V_i$</span> is the sum of received voltages in that point. Every voltage <span class="math-container">$V_i$</span> is a random variable uniformly distributed in the interval <span class="math-container">$[0, 10].$</span></p>
<ol>
<li>Calculate expected values and standard deviation of the voltages <span class="math-container">$V_i.$</span></li>
<li>Calculate probability that the total voltage entrance overpases <span class="math-container">$105$</span> volts, for <span class="math-container">$n = 20, 50, 100.$</span></li>
</ol>
<p>I dont need much help with point <span class="math-container">$2,$</span> I just need to use the central limit theorem but I need an expected value and standard deviation of point <span class="math-container">$1.$</span> I thought in using theorem of big numbers but I am missing something cause I need to get constants for expected and standard deviation, please help.</p>
| José Carlos Santos | 446,262 | <p>Yes, it is fine, except that you forgot to multiply by <span class="math-container">$2\pi i$</span> at the last line. The answer is <span class="math-container">$\pi\bigl(\cos(4+i)-\cos(4-i)\bigr)$</span>.</p>
|
1,620,186 | <p>I have to find an example of a surface of revolution excluding a sphere and a cone. </p>
<p>Is $\sigma(x,y)=(\cos x, 5, x^2+y^2)$ such an example? </p>
<p>$$$$ </p>
<p>I also have to find an example of a surface the image of which is not the graph of a smooth function $z=f(x,y)$. </p>
<p>Is $\sigma(x,y)=(3\sqrt{3}, 10\sqrt{y}, 0)$ such an example? </p>
| Rafael | 303,887 | <p><strong>1.</strong> There are $p,q\in\mathbb{Z}$, with $gcd(p,q)=1$ such that $\frac{x+a}{x+b}=\frac{p}{q}$. Then $$(x+a)q=(x+b)p\Leftrightarrow x(q-p)=bp-aq\Leftrightarrow x=\frac{bp-aq}{q-p}\in\mathbb{Q}\Leftrightarrow x=0.$$ This is absurd. Therefore $p=q$ and $x+a=x+b$, so $a=b$.</p>
<p><strong>2.</strong> Let $r\in\mathbb{Q}$ such that $\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r$ and suppose that $\frac{x}{y}$ is a irreducible fraction. Then $$\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r\Leftrightarrow x²+x+\sqrt{2}=(y²+y+\sqrt{2})r\Leftrightarrow x²+x-y²r-yr=\sqrt{2}(r-1).$$ If $r\neq 1$ we have $\sqrt{2}=\frac{x²+x-y²r-yr}{r-1}\in\mathbb{Q}$, but this is absurd. It follows that $r=1$ and $$x²+x+\sqrt{2}=y²+y+\sqrt{2}\Rightarrow y(y+1)=x(x+1).$$ If $x=-1$, we have $y(y+1)=0$ and so $y=0=x+1$ or $y=x=-1$. If $x\neq -1$ we have $$x=y(\frac{y+1}{x+1}).$$ If $y=-1$, it is analogous. We can suppose that $y\neq -1$. If $y=0$, $x=0$. Suppose $-1\neq y\neq 0$. We have $$\frac{x}{y}=\frac{y+1}{x+1}.$$ Since $\frac{x}{y}$ is irreducible, follows the last equality that $x=y$.</p>
|
621,438 | <p>Is there any Characterization for all measurable sets in $\mathbb{R}$?
Can I say that a set is measurable if an only if it has the property of Baire? (differs from an open set by a first category set). If the answer is no, Is there an example of a measurable set that does not have the property of Baire?</p>
<p>Thank you!
Shir </p>
| Listing | 3,123 | <p>To answer your second question, it is <a href="https://math.stackexchange.com/questions/62560/does-every-lebesgue-measurable-set-have-the-baire-property">not true</a> that a set is measurable iff it has the Baire property.</p>
<p>There is a way to see how measurable subsets $U\subset \mathbb{R}$ look like though, it is called <a href="https://math.stackexchange.com/questions/66803/characterization-of-measurable-sets-e-with-e-e-infty">Littlewood's first principle</a> and states:</p>
<p>If the outer measure $\mu^*(U)<\infty$ then $U$ is measurable if and only if for all $\epsilon > 0$ there is a finite union of intervals $J$ such that $\mu^*(U \triangle J)< \epsilon$. </p>
|
3,799,888 | <p>So I was attempting to solve a topology exercise when the following question came to me.</p>
<p>The objective was trying to fine a bijection <span class="math-container">$f$</span> between the following disks in <span class="math-container">$\Bbb R^2$</span>:</p>
<p><span class="math-container">$$D_1 :=\{(x,y)\in \Bbb R^2: x^2 + y^2 \leq1\}$$</span>
<span class="math-container">$$D_2 :=\{(x,y)\in \Bbb R^2: x^2 + y^2 \leq 4\}$$</span></p>
<p>So <span class="math-container">$D_1$</span> is a disk of radius <span class="math-container">$1$</span> centered at the origin, and <span class="math-container">$D_2$</span> is a disk of radius <span class="math-container">$2$</span> centered at the origin.</p>
<p>The first thing that came to my mind was to use polar coordinates, so let's redefine both disks as:</p>
<p><span class="math-container">$$D_1 :=\{(r,\varphi): r\in [0,1] \wedge \varphi \in [0,2\pi]\}$$</span>
<span class="math-container">$$D_2 :=\{(r,\varphi): r\in [0,2] \wedge \varphi \in [0,2\pi]\}$$</span></p>
<p>Now we can just simply scale disk 1 into disk 2:</p>
<p><span class="math-container">$$f:D_1\to D_2$$</span>
<span class="math-container">$$f(r,\varphi)=(2r,\varphi)$$</span></p>
<p>My question about the injectivity of this function, more concretely in the center of the disks.</p>
<p>Let <span class="math-container">$\varphi_1,\varphi_2 \in [0,2\pi]$</span>, with <span class="math-container">$\varphi_1 \neq \varphi_2$</span>. Then how do we treat points like <span class="math-container">$(0,\varphi_1)$</span> and <span class="math-container">$(0,\varphi_2)$</span>. In the disk they represent the same point: the center of the disk. But, when learning about double integrals with polar coordinates, my teacher taught us that when we use polar coordinates to describe a disk we are just defining an rectangle in the <span class="math-container">$rO\varphi$</span> plane, instead of in the <span class="math-container">$xOy$</span> plane:</p>
<p><a href="https://i.stack.imgur.com/NVi2y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NVi2y.jpg" alt="enter image description here" /></a></p>
<p>And all the points of the form <span class="math-container">$(0,\varphi)$</span> are in that line in <span class="math-container">$\varphi-$</span>axis and are indeed different points.</p>
<p>So how do we treat the points with this form? Are they considered all the same point and thus <span class="math-container">$(0,\varphi_1) = (0,\varphi_2)$</span>? Or are they considered different points as seen in the <span class="math-container">$rO\varphi$</span> plane?</p>
| José Carlos Santos | 446,262 | <p>Your function is simply <span class="math-container">$(x,y)\mapsto2(x,y)$</span>, and therefore it is injective (indeed, bijective).</p>
<p>Concerning your final questions, all the pairs <span class="math-container">$(0,\varphi_1)$</span>, with <span class="math-container">$\varphi\in[0,2\pi]$</span>, describe the same point (the origin), and therefore there is no problem there.</p>
|
3,799,888 | <p>So I was attempting to solve a topology exercise when the following question came to me.</p>
<p>The objective was trying to fine a bijection <span class="math-container">$f$</span> between the following disks in <span class="math-container">$\Bbb R^2$</span>:</p>
<p><span class="math-container">$$D_1 :=\{(x,y)\in \Bbb R^2: x^2 + y^2 \leq1\}$$</span>
<span class="math-container">$$D_2 :=\{(x,y)\in \Bbb R^2: x^2 + y^2 \leq 4\}$$</span></p>
<p>So <span class="math-container">$D_1$</span> is a disk of radius <span class="math-container">$1$</span> centered at the origin, and <span class="math-container">$D_2$</span> is a disk of radius <span class="math-container">$2$</span> centered at the origin.</p>
<p>The first thing that came to my mind was to use polar coordinates, so let's redefine both disks as:</p>
<p><span class="math-container">$$D_1 :=\{(r,\varphi): r\in [0,1] \wedge \varphi \in [0,2\pi]\}$$</span>
<span class="math-container">$$D_2 :=\{(r,\varphi): r\in [0,2] \wedge \varphi \in [0,2\pi]\}$$</span></p>
<p>Now we can just simply scale disk 1 into disk 2:</p>
<p><span class="math-container">$$f:D_1\to D_2$$</span>
<span class="math-container">$$f(r,\varphi)=(2r,\varphi)$$</span></p>
<p>My question about the injectivity of this function, more concretely in the center of the disks.</p>
<p>Let <span class="math-container">$\varphi_1,\varphi_2 \in [0,2\pi]$</span>, with <span class="math-container">$\varphi_1 \neq \varphi_2$</span>. Then how do we treat points like <span class="math-container">$(0,\varphi_1)$</span> and <span class="math-container">$(0,\varphi_2)$</span>. In the disk they represent the same point: the center of the disk. But, when learning about double integrals with polar coordinates, my teacher taught us that when we use polar coordinates to describe a disk we are just defining an rectangle in the <span class="math-container">$rO\varphi$</span> plane, instead of in the <span class="math-container">$xOy$</span> plane:</p>
<p><a href="https://i.stack.imgur.com/NVi2y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NVi2y.jpg" alt="enter image description here" /></a></p>
<p>And all the points of the form <span class="math-container">$(0,\varphi)$</span> are in that line in <span class="math-container">$\varphi-$</span>axis and are indeed different points.</p>
<p>So how do we treat the points with this form? Are they considered all the same point and thus <span class="math-container">$(0,\varphi_1) = (0,\varphi_2)$</span>? Or are they considered different points as seen in the <span class="math-container">$rO\varphi$</span> plane?</p>
| PeteBabe | 815,764 | <p>The correct way to define polar co-ordidinates is using the domain <span class="math-container">$r > 0$</span> , <span class="math-container">$\varphi$</span> in <span class="math-container">$(0, 2\pi)$</span> so that the transformation between cartesian and polar co-ordinates is a diffeomorphism (has a differentiable inverse).</p>
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