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 |
|---|---|---|---|---|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| user02138 | 2,720 | <p><span class="math-container">\begin{eqnarray}
1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2}
\end{eqnarray}</span>
More generally, let <span class="math-container">$D_{k} = \{ d\}$</span> be the set of unitary divisors of a positive integer <span class="math-container">$k$</span>, and let <span class="math-container">$\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$</span> denote the number-of-unitary-divisors (arithmetic) function. Then
<span class="math-container">\begin{eqnarray}
\sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2}
\end{eqnarray}</span></p>
<p>Note that <span class="math-container">$\mathsf{d}^{*}(k) = 2^{\omega(k)}$</span>, where <span class="math-container">$\omega(k)$</span> is the number distinct prime divisors of <span class="math-container">$k$</span>.</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Hans Lundmark | 1,242 | <p>$$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Wok | 2,380 | <p><img src="https://i.stack.imgur.com/zBPEc.png" alt="Parallelogram"></p>
<p>$$\left|z+z'\right|^{2}+\left|z-z'\right|^{2}=2\times\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)$$</p>
<blockquote>
<p>The sum of the squares of the sides
equals the sum of the squares of the
diagonals.</p>
</blockquote>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Fred Daniel Kline | 28,555 | <p>By excluding the first two primes, Euler's Prime Product becomes a square:</p>
<p>$$\prod _{n=3}^{\infty } \frac{1}{1-\frac{1}{(p_n)^{2}}}=\frac{\pi ^2}{9}$$</p>
<p>By using multiples of the product of the first two primes, we get the square root:</p>
<p>$$\prod _{n=1}^{\infty } \frac{1}{1-\frac{1}{(n p_1 p_2)^{2}}}=\frac{\pi }{3}$$</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Rick Decker | 36,993 | <p>$$
\frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots
$$
[Nick Pippenger, <em>Amer. Math. Monthly</em>, <strong>87</strong> (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Elias Costa | 19,266 | <p>We have by <a href="http://en.wikipedia.org/wiki/Determinant#Block_matrices">block partition rule for determinant</a>
$$
\det
\left[
\begin{array}{cc}
U & R \\
L & D
\end{array}
\right]
=
\det U\cdot \det ( D-LU^{-1}R)
$$
But if $U,R,L$ and $D$ commute we have that
$$
\det
\left[
\begin{array}{cc}
U & R \\
L & D
\end{array}
\right]
=
\det (UD-LR)
$$</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| user45926 | 45,926 | <p>$(x-a)(x-b)(x-c)\ldots(x-z) = 0$</p>
|
4,388,974 | <p>Let <span class="math-container">$f$</span> be a analytic function in the closed unit circle with its center
at the point <span class="math-container">$\alpha\in\mathbb{R}$</span>, then:
<span class="math-container">\begin{equation*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx=\frac{2\pi}{1-p^2}f(\alpha+p) ,
\end{equation*}</span>
for <span class="math-container">$|p|<1$</span>.</p>
<p>My attempt: By
<span class="math-container">\begin{align*}
\therefore\quad \sum_{n=1}^{\infty}p^{n}\sin(nx)=\frac{p\sin (x)}{1-2p\cos (x)+p^2},\qquad|p|<1
\end{align*}</span>
adjusting <span class="math-container">$p\to -p$</span> and highlighting <span class="math-container">$\displaystyle \frac1{1+2p\cos(x)+p^2}$</span>:
<span class="math-container">\begin{align*}
\frac1{1+2p\cos(x)+p^2}=-\frac{1}{\sin(x)}\sum_{n=1}^\infty(-p)^{n}\sin(nx).
\end{align*}</span>
Thus:
<span class="math-container">\begin{align*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx=-\sum_{n=1}^\infty(-p)^{n}\int_0^\pi\left\{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)\right\}\frac{\sin(nx)}{\sin(x)}\mathrm dx\tag{1}
\end{align*}</span>
by the \textit{Dirichlet Kernel}: <span class="math-container">$\displaystyle \sum_{k=0}^{N-1}e^{2ikx}=e^{(N-1)x}\frac{\sin(Nx)}{\sin(x)}$</span> setting <span class="math-container">$N\to n$</span>, <span class="math-container">$n\in\mathbb{N}$</span> and then taking <span class="math-container">$(1)$</span>, follows that:
<span class="math-container">\begin{align*}
&=-\sum_{n=1}^\infty(-p)^{n}\sum_{k=0}^{n-1}\int_0^\pi\left\{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)\right\}e^{-(n-1)x}e^{2ikx}\mathrm dx,\quad\left(e^{ix}\to z\right)\\
&=...
\end{align*}</span><br />
At this point I'm out of ideas. I would like some light on my last step, or another approach that is similar to this one.</p>
| Random Variable | 16,033 | <p>If we convert the integral into a contour integral, we get</p>
<p><span class="math-container">$$ \begin{align} \int_{0}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx &= \frac{1}{2}\int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx \\ &= \int_{-\pi}^{\pi}\frac{f(\alpha+e^{ix})}{1+2p \cos x +p^{2}} \, \mathrm dx \\&= \int_{-\pi}^{\pi} \frac{f(\alpha +e^{ix})}{(e^{ix}+p)(e^{-ix}+p)} \, \mathrm dx \\ &= \int_{|z|=1} \frac{f(\alpha + z)}{(z+p)(\frac{1}{z}+ p)} \frac{dz}{iz} \\ &= \int_{|z|=1} \frac{f(\alpha +z)}{(z+p)(1+pz)} \, \frac{\mathrm dz}{i}. \end{align}$$</span></p>
<p>Since <span class="math-container">$|p| <1$</span>, the only singularity inside the unit circle is a simple pole at <span class="math-container">$z=-p$</span>.</p>
<p>Therefore, <span class="math-container">$$ \begin{align} \int_{0}^{\pi}\frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos x +p^{2}} \, \mathrm dx &= 2 \pi i \operatorname{Res} \left[ \frac{f(\alpha +z)}{i(z+p)(1+pz)}, -p \right] \\ &= \frac{2\pi f(\alpha \color{red}{-}p)}{1-p^{2}}. \end{align}$$</span></p>
<hr />
<p>If <span class="math-container">$|p| <1$</span> and not equal to zero, we also have <span class="math-container">$$ \begin{align} \int_{0}^{\pi} \frac{f(\alpha+e^{ix})+ f(\alpha+e^{-ix})}{1+2p \cos (x) +p^{2}} \, \cos (x) \, \mathrm dx &= \frac{1}{2}\int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})+f(\alpha+e^{-ix})}{1+2p \cos (x) +p^{2}} \, \cos (x) \, \mathrm dx \\ &=\int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})}{1+2p \cos (x) +p^{2}} \, \cos (x) \, \mathrm dx \\ &= \int_{-\pi}^{\pi} \frac{f(\alpha+e^{ix})}{(e^{ix}+p)(e^{-ix}+p)} \, \frac{\left(e^{ix}+e^{-ix} \right)}{2} \, \mathrm dx \\ &= \int_{|z|=1} \frac{f(\alpha+z)}{(z+p)(\frac{1}{z}+p)} \, \frac{1}{2} \left(z+\frac{1}{z} \right) \, \frac{\mathrm dz}{iz} \\ &= \int_{|z|=1} \frac{f(\alpha+z)}{(z+p)(1+pz)} \, \frac{1+z^{2}}{2iz} \, \mathrm dz \\ &= 2 \pi i \left(\frac{f(\alpha - p)}{1-p^{2}} \frac{1+p^{2}}{-2ip}+ \frac{f(\alpha)}{2ip} \right) \\ &= \frac{\pi}{p} \left(f(\alpha)- \frac{1+p^{2}}{1-p^{2}} \, f(\alpha -p) \right). \end{align}$$</span></p>
<p>At <span class="math-container">$p=0$</span>, we get <span class="math-container">$$ \int_{0}^{\pi} \left(f(\alpha+e^{ix})+f(\alpha+e^{-ix}) \right) \cos(x) \, \mathrm dx = \lim_{p \to 0} \frac{\pi}{p} \left(f(\alpha)- \frac{1+p^{2}}{1-p^{2}} \, f(\alpha -p) \right) = \pi f^{\prime}(\alpha). $$</span></p>
<hr />
<p>Finally, if <span class="math-container">$|p| <1$</span> and not equal to zero, <span class="math-container">$$ \begin{align} \int_{0}^{\pi} \frac{f(\alpha+e^{ix})\color{red}{-} f(\alpha +e^{-ix})}{1+2p \cos(x) + p^{2}} \, \sin(x) \, \mathrm dx &= \int_{|z|=1} \frac{f(\alpha+z)}{(z+p)(1+pz)} \, \frac{1-z^{2}}{2z} \, \mathrm dz \\ &= 2 \pi i \left(\frac{f(\alpha-p)}{1-p^{2}} \frac{1-p^{2}}{-2p} + \frac{f(\alpha)}{2p}\right) \\ &=\frac{\pi i}{p} \left(f(\alpha) - f(\alpha-p) \right). \end{align}$$</span></p>
<p>And at <span class="math-container">$p=0$</span>, we get <span class="math-container">$$ \int_{0}^{\pi}\left(f(\alpha+e^{ix})\color{red}{-} f(\alpha +e^{-ix})\right)\, \sin(x) \mathrm dx = \pi i f^{\prime}(\alpha)$$</span></p>
|
495,069 | <p>Find an equation of the plane.
The plane that passes through the line of intersection of the planes
$x − z = 1$ and $y + 4z = 1$
and is perpendicular to the plane
$x + y − 2z = 2$.</p>
<p>I keep getting the answer of $7x-y+5z=6$ and I am told that it is wrong. I do not understand what I am doing wrong.
I have the 1st normal vector to be $\langle 1,0,-1\rangle$ and the second normal vector to be $\langle0,1,4\rangle$
and when I did a cross product on them, I got $\langle1,-4,1\rangle$ to be the direction of the line.
I then got a 2nd vector parallel to the desired plane as $\langle1,1,-2\rangle$ since its perpendicular to $x+y-2z=2$
I got a normal plane by the cross product of the 2 normal vectors and the result was $\langle 7,-1,5\rangle$
Then I plugged $\langle7,-1,5\rangle$ into the scalar equation of the plane for $\langle a,b,c\rangle$ and used the point $(1,1,0) $
and for my final answer I got $7x-y+5z=6$
I need help figuring out where I went wrong.</p>
| Sharabh | 95,079 | <p>This is actually a 2 line problem which can be solved as follows:
Equation of the plane passing through the planes x−z-1=0 and y+4z-1=0 can be written as
x−z-1 + k(y+4z-1) = 0 for some constant k (which is required to be found out).
i.e., x + ky + (4k-1)z - k - 1 = 0 .... (1)
Now since this plane is perpendicular to the plane x+y−2z=2 so, 1.1 + k.1 + (4k-1).(-2) = 0
which gives k = 3/7. Putting in (1) we get the required equation of the plane as
7x+3y+5z=10. </p>
|
314,239 | <blockquote>
<p>If <span class="math-container">$P_1 , P_2 $</span> are two sylow <span class="math-container">$p$</span>-subgroups of the group <span class="math-container">$G$</span> prove that:</p>
<p><span class="math-container">$ P_1 \bigcap $</span> <span class="math-container">$P_2$</span> = <span class="math-container">$ { 1 } $</span></p>
</blockquote>
<p>I tried to prove it by induction as follows: proved it when <span class="math-container">$P_1 , P_2$</span> have the order p for some prime p then supposed it is true when the sylow p-subgroup has the order <span class="math-container">$p^n$</span> and supposed that there is some element in the intersection ,
made <span class="math-container">$H$</span> = the subgroup generated by this element " say , x "</p>
<p>I proved that H is normal subgroup of <span class="math-container">$P_1$</span> , <span class="math-container">$P_2$</span> , and made the factor group, <span class="math-container">$P_1$</span> mod <span class="math-container">$H $</span> = <span class="math-container">$Q_1$</span> and <span class="math-container">$P_2$</span> mod <span class="math-container">$H$</span> = <span class="math-container">$Q_2$</span></p>
<p>So by the induction, if <span class="math-container">$h$</span> <span class="math-container">$\in$</span> the intersection of <span class="math-container">$Q_1 , Q_2$</span> then <span class="math-container">$Q_1$</span>= <span class="math-container">$Q_2$</span></p>
<p>But, I couldn't determine the element which is in this intersection--I don't know if this element <span class="math-container">$h$</span> must be exist or not -</p>
<p>I don't know what is the next step now; I need some hints to prove this statement:</p>
<p>I found that the text - dummit and foote - use the fact that the intersection of two sylow p-subroup is the identity element, but it didn't prove this fact so I look for a proof.</p>
| Jyrki Lahtonen | 11,619 | <p>This is not true in general. For example the group $S_6$ has the following two Sylow 2-subgroups. Let $P_1$ be generated by $(12),(13)(24)$ and $(56)$ (the first two permutations generate a Sylow 2-subgroup of $S_4$). Let $P_2$ be generated by $(12),(35)(46)$ and $(56)$ (here we have a Sylow 2-subgroup of $S_4$ in a diguise that $S_4$ is now acting on the set $\{3,4,5,6\}$). Obviously $P_1$ and $P_2$ intersect non-trivially as they share two generators. Yet they are not the same subgroup as the orbits of $P_1$ are $\{1,2,3,4\}$ and $\{5,6\}$ whereas the orbits of $P_2$ are $\{1,2\}$ and $\{3,4,5,6\}$. </p>
|
1,310,530 | <p>From: $2015$ Singapore Mathematical Olympiad Secondary 2 (Grade 8) Question 21 Round 1 on 3rd June.</p>
<blockquote>
<p>Find the value of $\sqrt {(98 \times 100+2)(100\times102+2)+(100\times2)^2}$ (No use of calculators)</p>
</blockquote>
<p>My attempt:Special cases formula -> $x^2=(x+1)(x-1)+1$</p>
<p>Therefore,</p>
<p>\begin{align}
\sqrt {(98 \times 100+2)(100\times102+2)+(100\times2)^2}&=\sqrt {(99^2-1)(101^2-1)+(100\times 2)^2}\\
\end{align}</p>
<p>Now take $y$=100</p>
<p>\begin{align}
\sqrt {(99^2-1)(101^2-1)+(100\times 2)^2}&=\sqrt {[(y-1)^2-1][(y+1)^2-1]+(2y)^2}\\
&= \sqrt {[y^2-2(y)(1)+1-1][y^2+2(y)(1)+1-1]+4y^2}\\
&= \sqrt {(y^2-2y)(y^2+2y)+4y^2}\\
&= \sqrt {y^4-4y^2+4y^2}\\
&= \sqrt {y^4}\\
&= y^2
\end{align}</p>
<p>$100^2=10000$</p>
<p>therefore $\sqrt {(98 \times 100+2)(100\times102+2)+(100\times2)^2}=10000$</p>
<p>But using the calculator,the answer is 10002.Where did I go wrong,and is there a simpler way to do this other than using long multiplication?</p>
<p>EDITED ANSWER(Error found by @Mathlove):</p>
<p>\begin{align}
\sqrt {(98 \times 100+2)(100\times102+2)+(100\times2)^2}&=\sqrt {(99^2+1)(101^2+1)+(100\times 2)^2}\\
\end{align}</p>
<p>Now take $y$=100</p>
<p>\begin{align}
\sqrt {(99^2+1)(101^2+1)+(100\times 2)^2}&=\sqrt {[(y-1)^2+1][(y+1)^2+1]+(2y)^2}\\
&= \sqrt {[y^2-2(y)(1)+1+1][y^2+2(y)(1)+1+1]+4y^2}\\
&= \sqrt {(y^2-2y+2)(y^2+2y+2)+4y^2}\\
&= \sqrt {(y^4-4y^2)+2(y^2+2y+2)-4y-2y^2+4y^2+4y^2}\\
&= \sqrt {y^4-4y^2+2y^2+4y+4-4y-2y^2+4y^2+4y^2}\\
&= \sqrt {y^4+4y^2+4}\\
&= \sqrt {(y^2+2)^2}
&= y^2+2
\end{align}</p>
<p>$100^2+2=10002$</p>
<p>therefore $\sqrt {(98 \times 100+2)(100\times102+2)+(100\times2)^2}=10002$</p>
| Mark Bennet | 2,906 | <p>I would do something like you have done, but just put $x=100$ straight away since that is the central number. You can always do the $x^2-1$ simplification afterwards, if it helps, but I would get rid of the big numbers as quickly as possible.</p>
<p>Then $$\left(x(x+2)+2\right)\left(x(x-2)+2\right)+4x^2=x^2(x^2-4)+2x(x+2)+2x(x-2)+4+4x^2=$$$$=x^4-4x^2+2x^2+4x+2x^2-4x+4+4x^2=x^4+4x^2+4$$</p>
<p>So I would work on the part underneath the square root, but I wouldn't write the square root every time.</p>
<p>I find this kind of approach less prone to errors, and more easy to check under test conditions.</p>
|
39,828 | <p>Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.</p>
<p>So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:</p>
<ul>
<li>Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.</li>
<li>The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.</li>
<li>Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.</li>
</ul>
<p>The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.</p>
<p>Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.</p>
<blockquote>
<p>How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?</p>
</blockquote>
<p>Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?</p>
<p>Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.</p>
<p>I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is</p>
<blockquote>
<p>How much would you subscribe to the statement that
EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?</p>
</blockquote>
<p>Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.</p>
<hr>
<p>Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.</p>
<hr>
<p>Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".</p>
| Denis Serre | 8,799 | <p>Jordan algebras were introduced first by P. Jordan and J. von Neumann in order to give a mathematical context for observables in quantum mechanics (say, a structure that generalizes the space of Hermitian matrices). At the end, the classification was disappointing, and Jordan algebras do not play a role any more in QM, but the topic survived in Mathematics until now.</p>
|
39,828 | <p>Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.</p>
<p>So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:</p>
<ul>
<li>Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.</li>
<li>The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.</li>
<li>Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.</li>
</ul>
<p>The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.</p>
<p>Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.</p>
<blockquote>
<p>How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?</p>
</blockquote>
<p>Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?</p>
<p>Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.</p>
<p>I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is</p>
<blockquote>
<p>How much would you subscribe to the statement that
EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?</p>
</blockquote>
<p>Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.</p>
<hr>
<p>Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.</p>
<hr>
<p>Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".</p>
| JSE | 431 | <p>"How much would you subscribe to the statement that studying questions one finds interesting is something established mathematicians do, while younger ones are better off studying questions that the rest of the community finds interesting?"</p>
<p>Not at all. I don't think anyone, young or old, will find success by working on questions other than those they find interesting. Mathematics is just too difficult for that.</p>
<p>Ideally, everyone should work on problems that are interesting to both themselves and the community. Senior mathematicians have the luxury of working on problems whose interest to the community has not been established.</p>
|
39,828 | <p>Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.</p>
<p>So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:</p>
<ul>
<li>Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.</li>
<li>The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.</li>
<li>Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.</li>
</ul>
<p>The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.</p>
<p>Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.</p>
<blockquote>
<p>How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?</p>
</blockquote>
<p>Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?</p>
<p>Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.</p>
<p>I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is</p>
<blockquote>
<p>How much would you subscribe to the statement that
EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?</p>
</blockquote>
<p>Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.</p>
<hr>
<p>Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.</p>
<hr>
<p>Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".</p>
| Scott Carter | 36,108 | <p>I am thinking of specific examples. In much the same way, David Corfield mentioned groupoids. </p>
<p>I am personally not a big fan of the general theory of loops. In part, my own disinterest is because I have not found an application. On the other hand, I have seen enough to believe that Moufang loops are interesting even if I personally don't know a lot about them. Still I like the idea of algebraists thinking about the structures of loops because <em>they</em> find them interesting. </p>
<p>Closer to my own interest is the idea of quandles. These were introduced essentially in the 1940s, then again in the late 1970s and early 1980s, rediscovered, and have only found some greater applicability because quandle cohomology gives interesting topological invariants. The idea, apparently was natural: it was discovered, forgotten, rediscovered, forgotten, and found to be applicable. Nevertheless, some of you might find it to be be a fringe notion. Even knot theorist might believe that there is not much in the quandle concept because the information in the quandle is present in the fundamental group and a peripheral subgroup. </p>
<p>I think Tim's articulation of Ronnie's list should include that the algebraic concept yields a more concise language in which ideas can be expressed. </p>
|
1,729,893 | <p>Let $\nu_1, \nu_2, \nu_3 \in \mathbb{R}$ not all be zero.</p>
<p>I wish to show
$$\nu_1^2 + \nu_1\nu_2 + \nu_2^2+\nu_2\nu_3 + \nu_3^2 > 0\text{.}$$</p>
<p><a href="http://www.wolframalpha.com/input/?i=x%5E2%2Bx*y%2By%5E2%2By*z%2Bz%5E2+%3E+0" rel="nofollow">Wolfram</a> seems to suggest splitting this into cases, but I'm wondering if there's a shorter way to approach this. This expression seems very similar to the binomial expansion, minus the fact that we have three terms that are squared and two cross-terms occurring (rather than one cross-term multiplied by $2$).</p>
<p>As for my work, if any one of these are $0$, (I think) this is a trivial exercise. If any two of these are $0$, this is a trivial exercise (you're left with a squared non-zero term). But if all three are non-zero? Then I'm at a loss on how to pursue this, because there isn't a clean way to deal with three variables (or is there?). I've tried seeing if Wolfram could perhaps factor the above. It can't, but maybe, I thought, we could try working with
$$(\nu_1 + \nu_2 + \nu_3)^2 = (\nu_1^2 + \nu_1\nu_2 + \nu_2^2+\nu_2\nu_3 + \nu_3^2) +2\nu_1\nu_3+\nu_1\nu_2+\nu_2\nu_3$$
but there is no guarantee that this is $> 0$ either (take $\nu_3 = -(\nu_1 + \nu_2)$, for example).</p>
| Ted Shifrin | 71,348 | <p><strong>HINT</strong>: Complete the square. So start with $(v_1+\frac12v_2+\frac12v_3)^2$.</p>
|
239,900 | <p>Hatcher states the following theorem on page 114 of his Algebraic Topology:</p>
<blockquote>
<p>If $X$ is a space and $A$ is a nonempty closed subspace that is a deformation retract of some neighborhood in $X$, then there is an exact sequence
$$...\longrightarrow\widetilde{H}_n(A)\overset{i_*}\longrightarrow \widetilde{H}_n(X)\overset{j_*}\longrightarrow\widetilde{H}_n(X/A)\overset{\partial}\longrightarrow \widetilde{H}_{n-1}(A)\overset{i_*}\longrightarrow... $$</p>
<p>where $i: A\hookrightarrow X$ is the inclusion and $j:X \rightarrow X/A$ is the quotient map. </p>
</blockquote>
<p>Perhaps I am having a brain malfunction at the moment, but what are some interesting nonempty spaces which do not satisfy these criterion? By interesting, I mean something that appears "in nature." </p>
| apurv | 49,824 | <p>One example which occurs a lot "in nature" is when $A$ is a large open subset of $X$. For example, $X$ is a manifold and $A = X - \text{point}$ or when $X$ is the total space of a vector bundle and $A$ is the complement of the <strike>base space</strike> zero section. However, in these cases, it is often possible to "shrink" A so that the condition remains true. </p>
|
2,125,297 | <p><a href="https://i.stack.imgur.com/cBCJe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cBCJe.png" alt="enter image description here"></a></p>
<p>I would like to know how exactly this works. I watched a khan academy video: "Multiplying matrices" but in this case he would've done B*A and had 2 columns, why does this one have 3 columns and 3 rows??</p>
| kotomord | 382,886 | <p>Square of the distance from (1,1,1,1) to W is minimum of </p>
<p>$(1-x)^2 + (1+x)^2 +(1-y)^2 + (1-y)^2 = 2+ 2x^2 + (1-y)^2$ over $\mathbb{R}^2$</p>
<p>Simple to see it is 2 </p>
|
698,474 | <p>I am trying to find the phi(18). Using an online calculator, it says it is 6 but im getting four.
<br/>
The method I am using is by breaking 18 down into primes and then multiplying the phi(primes)</p>
<p>$$=\varphi (18)$$
$$=\varphi (3) \cdot \varphi(3) \cdot \varphi(2)$$
$$= 2 \cdot 2 \cdot 1$$
$$= 4$$</p>
| NasuSama | 67,036 | <p>Remember that you need to determine the prime factorization of $18$. That is, $18 = 3^2 \cdot 2$. Since $18 = 3^2 \cdot 2$, we have</p>
<p>$$\begin{aligned}
\varphi(18) = \varphi(3^2) \cdot \varphi(2) &= (3^2 - 3)(2 - 1) = 6
\end{aligned}$$</p>
<p>So in general, if $k = p_1^{n_1}p_2^{n_2}\cdots p_s^{n_s}$, then we have</p>
<p>$$\varphi(k) = \varphi(p_1^{n_1})\varphi(p_2^{n_2})\cdots \varphi(p_s^{n_s}) = (p_1^{n_1} - p_1^{n_1 - 1})(p_2^{n_2} - p_2^{n_2 - 1})\cdots (p_s^{n_s} - p_s^{n_s - 1})$$</p>
|
1,243,750 | <p>$A \in \mathbb{R}^{n\times n}$, with $A^2 = 1$ and $A\ne\pm1$</p>
<p>Show that the only eigenvalues of $A$ are $1$ and $-1$.</p>
| Thomas Russell | 32,374 | <p>We have that:</p>
<p>$$\mathbf{A}^{2} = \mathbf{I} \implies \mathbf{A} = \mathbf{A}^{-1}$$</p>
<p>But we know that if the eigenvalues of $\mathbf{A}$ are $\lambda_{1},\dots,\lambda_{n}$, then the eigenvalues of $\mathbf{A}^{-1}$ are $\lambda_{1}^{-1},\dots,\lambda_{n}^{-1}$, but if:</p>
<p>$$\lambda_{i}=\lambda_{i}^{-1} \qquad \forall i \in \{1,\dots,n\}$$</p>
<p>Then:</p>
<p>$$\lambda_{i}=\pm 1 \qquad \forall i \in \{1,\dots,n\}$$</p>
|
587,484 | <p>Show that $\left(\displaystyle\frac{\partial (u,v)}{\partial (x,y)}\right)\left(\displaystyle\frac{\partial (x,y)}{\partial (r,s)}\right)=\displaystyle\frac{\partial (u,v)}{\partial (r,s)}$. Thus, prove that $\displaystyle\frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{\displaystyle\frac{\partial (x,y)}{\partial (u,v)}}$ </p>
<hr>
<p>I have no idea on how to get started with this problem. I tried attacking it down with Jacobians, but to no avail. I'm not even sure how did they define the systems in this case.</p>
| user110373 | 110,373 | <p>Lemma: Let $A, B, C$ be $m \times m $ matrices such that $A \times B= C$, then $detC= detA \cdot detB$. </p>
<p>Proof: By lemma, it's sufficient to show that $$ \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial s}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial s} \end{pmatrix}= \begin{pmatrix}\frac{\partial u}{\partial r} & \frac{\partial u}{\partial s}\\ \frac{\partial v}{\partial r}& \frac{\partial v}{\partial s} \end{pmatrix}$$</p>
<p>However, it is directly given by the chain rule, since according to the chain rule, regarding (u,v) as the composition of (x,y) and (r,s), we have, for example
$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$$ And other three can be checked as well.
Therefore we have $$ \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial s}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial s} \end{pmatrix}= \begin{pmatrix}\frac{\partial u}{\partial r} & \frac{\partial u}{\partial s}\\ \frac{\partial v}{\partial r}& \frac{\partial v}{\partial s} \end{pmatrix}$$
Thus $(\displaystyle\frac{\partial (u,v)}{\partial (x,y)})(\displaystyle\frac{\partial (x,y)}{\partial (r,s)})=\displaystyle\frac{\partial (u,v)}{\partial (r,s)}$</p>
|
105,058 | <p>Consider the (cumbersome) statement: "Every integer greater than 1 can be written as a unique product of integers belonging to a certain subset, $S$ of integers.</p>
<p>When $S$ is the set of primes, this is the Fundamental Theorem of Arithmetic. My question is this: Are there any other types of numbers, for which this is true. </p>
<p>EDIT: As the answers show, this obviously cannot be done. What if we relax the integer condition, i.e. can there be <em>any</em> other canonical representation of positive integers using complex numbers?</p>
| J.. | 4,743 | <p>No. Such a set $S$ must include the primes (because they have no other factors). If $s \in S$ is not prime, then it can be written as a product of primes, i.e., as a product of other elements of $S$. This contradicts uniqueness. Therefore the only such set $S$ is the set of the primes.</p>
|
105,058 | <p>Consider the (cumbersome) statement: "Every integer greater than 1 can be written as a unique product of integers belonging to a certain subset, $S$ of integers.</p>
<p>When $S$ is the set of primes, this is the Fundamental Theorem of Arithmetic. My question is this: Are there any other types of numbers, for which this is true. </p>
<p>EDIT: As the answers show, this obviously cannot be done. What if we relax the integer condition, i.e. can there be <em>any</em> other canonical representation of positive integers using complex numbers?</p>
| Steven Stadnicki | 785 | <p>When you relax the integrality condition you start to pick up other solutions - for instance, the set of square roots of primes, $S=\{\sqrt{p}:p\in P\}$, is such that every number $n$ has a unique factorization $n=s_1^{e_1}s_2^{e_2}\ldots$ in terms of the elements of $S$; just consider the breakdown $n=p_1^{d_1}p_2^{d_2}\ldots$ and then set $e_1=2d_1$, etc. On the other hand, this gives up <em>existence</em> : it's no longer the case that every sequence $e_1, e_2, \ldots$ of whole numbers corresponds to a unique integer; instead we have the additional requirement that all the $e_i$ be even.</p>
<p>More generally, any breakdown of primes into products of (non-integral) real factors will give you a solution: if $p_1 = t_{11}\cdot t_{12} \cdot \ldots\cdot t_{1m}$, $p_2 = t_{21}\cdot t_{22}\cdot\ldots\cdot t_{2n}$, and so on, then obviously any factorization of a number $n$ into products of $p_i$ extends to a (unique) factorization of $n$ into products of the $t_{ij}$. What's more, any unique factorization of all numbers into real 'factors' $x_i$ has to come from this kind of breakdown, because obviously the prime numbers all need unique factorizations, and the factorization of a number $n$ into primes then 'extends through' the factorization of those primes into a factorization of $n$ in terms of the $x_i$, which by definition must be <em>the</em> factorization of $n$. So you don't get anything particularly interesting by removing the non-integrality condition, just 'refinements' of the factorizations you already had.</p>
|
2,378,717 | <p>I wondered many times how the editors of mathematical journals know that the content of a paper or article that was submited by an author is original, this is, that the main content of such calculations and reasonings is unpublished.</p>
<blockquote>
<p><strong>Question.</strong> How do editors of mathematical journals know that a submited paper is original, that is unpublished? <strong>Many thanks.</strong></p>
</blockquote>
<p>I would like to know it as curiosity, because I imagine that professors working in mathematical journals are experts in a field of mathematics. Then a day they receive a writing, a paper or a remark, with perhaps several results, and they have to evaluate if this was not published or known previously.
Additionally I imagine that these professors or editors have tools to search information for this purpose, and maybe they have contact with other colleagues who have historical memory about what results, and of what kind, have been published. Am I wrong in these believes? How do they know and they note the originality of such mathematical content?</p>
| Tobias Kildetoft | 2,538 | <p>When submitting a paper for publication, it is not only the editor who looks at it. The editor will (usually) take a first look to determine if the paper looks to be suitable for the journal. If they see that the main results are already well-known this will likely make them reject the paper (especially if the authors have not made any remark on this themselves). </p>
<p>If this is not the case and the paper generally looks fine (this also includes the results actually seeming to be interesting), then the paper will be sent to one or two experts in the field for review. These will (hopefully) take a much closer look at the paper and determine if they think it is suitable for the journal. Once again, this will include trying to determine whether the results are actually new, which is why it is important for the editor to choose reviewers who really are experts, as otherwise they will not be familiar enough with the field to determine this.</p>
<p>Finally, the reviewers will make a recommendation to the editor based on their reading, and using the editor will make a final decision on whether to accept the paper, possibly contingent on certain improvements.</p>
|
2,665,539 | <p>How to solve the following system of equations</p>
<blockquote>
<p>$$
\begin{cases}
a+c=12\\
b+ac+d=86\\
bc+ad=300\\
bd=625\\
\end{cases}
$$</p>
</blockquote>
| Dave | 334,366 | <p>This system of equations corresponds to a factorization of $$f(x)=x^4+12x^3+86x^2+300x+625$$ into two quadratics as
$$f(x)=(x^2+ax+b)(x^2+cx+d)$$
The roots of $f(x)$ are $-3-4i$ and $-3+4i$, each having multiplicity $2$. So what should the quadratic factors of $f(x)$ look like?</p>
|
331,169 | <p>Let $\Omega\subset \mathbb{R}^n$ be bounded and let $X:=H^1(\Omega)$. Let $a:\Omega\times \mathbb{R} \to \mathbb{R}, (x,z)\mapsto a(x,z)$ be a bounded function such that $a(x,.)$ is continuous on $\mathbb{R}$ for every $x$ and $a(.,z)$ is measurable on $\Omega$ for every $z$.</p>
<p>For $v,\phi\in H^1$ one finds in pde the integral
$$\int_\Omega a(x,v(x))\nabla v(x) \cdot \nabla\phi(x) dx$$
All measures should be the Lebesgue measure
My question is: Why is $a(x,v(x))$ a measurable function, why is $a(x,z)$ measurable on $\Omega\times \mathbb{R}$ or why is the integral defined?</p>
<p>Usually one can not conclude from measurability of each compontent to measurability on the product space. Somehow i feel I need more measure theory than i know yet :)</p>
| Julien | 38,053 | <p>Take an element $(n,m)\in\mathbb{Z}^2=\mathbb{Z}\times\mathbb{Z}$. If it is a generator, then $(1,0)$ and $(0,1)$ are in $\mathbb{Z}\cdot(n,m)$ the group generated by $(n,m)$. Write down what this actually means. Can you reach a contradiction?</p>
|
1,843,724 | <p>I know that the Maclaurin expansion of $e^x$ is $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
But i'm not sure how to find the Maclaurin series here
I tried this</p>
<p>$$
f'_{(0)}=-2xe^{-x^2}=0
$$
And that follows to every derivative that follows, so how can I get a power series out of it?</p>
| Daniel | 150,142 | <p>As Dark mentions in the comments,
$$e^y = 1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+\cdots$$
works for <strong>any</strong> $y\in \Bbb R$. In particular, you can substitute $y=-x^2$ and that should give you the result!</p>
<p>$$e^{-x^2} = 1+\left(-x^2\right)+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!}+\cdots$$</p>
<p>There's no need of using a derivative here though, since you're just simply making a substitution in an identity that holds for <strong>any $y$</strong> (here, it is important to notice that the radius of convergence is infinite, we'd need to be more careful if this were not the case). </p>
<p>That's the power of using algebraic expressions. When you say that certain property holds for <em>any $y$</em> it means that it holds for <strong>any</strong> expression you put instead of $y$, because that expression represents <strong>any</strong> number at the very end.</p>
|
3,290,073 | <p>I'm trying to help my daughter learn maths. She is struggling with factors, which is to work out what numbers go into a larger number (division).</p>
<p>I've already learned that by summing numbers, if they make 3, it can divide by 3. I also know the rules for 2, 5, 6, 9 and 10.</p>
<p>I'm trying to see if there is a rule for 4. I'm thinking not.</p>
<p><a href="https://www.quora.com/Why-does-the-divisibility-rule-for-the-number-4-work" rel="nofollow noreferrer">https://www.quora.com/Why-does-the-divisibility-rule-for-the-number-4-work</a> shows the following</p>
<blockquote>
<p>The divisibility rule for 4 is in any large number, if the digits in tens and units places is divisible by then the whole number is divisible by 4.</p>
</blockquote>
<p>This doesn't make sense. 56 divides by 4. However, the 2 numbers add to 11, and so can't be divided by 4.</p>
<p>It may very well get a "no" answer, but is there any pattern/method I can use for determining if a number can be divided by 4 if it is less than 100 (and greater than 4)</p>
| J. W. Tanner | 615,567 | <p>How to make sense of that rule for divisibility by <span class="math-container">$4$</span>:
it's not saying to <em>add</em> the last two digits; it's merely saying to <em>look</em> at the last two digits. Because <span class="math-container">$4$</span> divides <span class="math-container">$100$</span>, a number is divisible by <span class="math-container">$4$</span> if and only if its last two digits (ten's place and one's place) are divisible by <span class="math-container">$4$</span>. <a href="https://math.stackexchange.com/users/8508/robert-israel">Robert Israel</a>'s answer gives a method for determining whether a two-digit number is divisible by <span class="math-container">$4$</span>, and the rule is saying that's essentially all you need.</p>
<p>For example, if you want to know whether <span class="math-container">$2389080349$</span> is divisbile by <span class="math-container">$4$</span>, you merely have to determine whether <span class="math-container">$49$</span> is divisible by <span class="math-container">$4$</span>. (It's not.)</p>
|
2,086,134 | <p>Solve for $x$ if $4^x + x^x = 20$ where $x \in \mathbb{Z}$ </p>
<p>My Attempt: </p>
<p>By inspection,<br>
$x=2$</p>
<p>Now,
$4^2 + 2^2 = 20$</p>
<p>$16+4=20$</p>
<p>$20=20$</p>
<p>Which is True.</p>
<p>But, I Could not do it by any process. Can anyone help me? </p>
<p>Thanks in Advance.</p>
| projectilemotion | 323,432 | <p>There does not exist an explicit solution for $x$ which can be expressed in a finite number of elementary functions.</p>
<p>However, you can solve this using an <a href="https://en.wikipedia.org/wiki/Iterative_method" rel="nofollow noreferrer">iterative method</a>. In order to apply it, we will generalise our possible solutions to $x \in \mathbb{R}$.</p>
<p>We will use the <a href="https://en.wikipedia.org/wiki/Newton's_method" rel="nofollow noreferrer">Newton-Raphson method</a>. The methodology is as follows:</p>
<p>$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \tag{1}$$</p>
<p>We can set a suitable 'guess' in order to obtain a solution to $x$.</p>
<p>For your function, $f(x)=4^x+x^x-20$, we evaluate the derivative to be:
$f'(x)=x^x(\ln{x}+1)+\ln{(4)} \cdot 4^x$</p>
<p>We can substitute our functions into equation $(1)$.</p>
<p>$$x_{n+1}=x_n-\frac{4^{x_n}+{x_n}^{x_n}-20}{{x_n}^{x_n}(\ln{x_n}+1)+\ln{(4)} \cdot 4^{x_n}}$$</p>
<p>Let's use an initial value of $x_0=11$. On $x_{29}$, the iteration is accurate to over 10 decimal places (extremely close to $2$). As $n \to \infty$, $x_n \to 2$.</p>
<p>You can apply this iteration into a spreadsheet, or you can use more sophisticated software such as MATLAB.</p>
<p>As others have noted, there exists only one solution to your equation. If there were more solutions, you can set different values for $x_0$ and see if they converge to different solutions.</p>
|
225,481 | <p>I would like to write a program that gives me the Klein Gordon Equations given a metric. I will explain.</p>
<p>My code is in the following:</p>
<p><strong>I) Standard Quantities</strong></p>
<p>I have no doubt here the first part is <a href="http://web.physics.ucsb.edu/%7Egravitybook/mathematica.html" rel="nofollow noreferrer">given by Hartle's</a>.</p>
<pre><code>Clear[coord, metric, inversemetric, affine, riemann, ricci,
scalar, einstein, t, x, y, z]
n = 4;
coord = {t, r, θ, ϕ};
metric = {{-(1 - ((2*m)/(r))), 0, 0, 0},
{0, (1)/(1 - ((2*m)/(r))), 0, 0},
{0, 0, r^2, 0},
{0, 0, 0, r^2*(Sin[θ]*Sin[θ])}};
inversemetric = Simplify[Inverse[metric]];
Det[metric]
</code></pre>
<p><strong>II) MY TRY</strong></p>
<p>I wrote the components by hand:</p>
<pre><code>KG00 = FullSimplify[((1)/(Sqrt[-Det[metric]]))*
D[(Sqrt[-Det[metric]])*(inversemetric[[1, 1]])*
D[Ξ[t, r, θ, ϕ], t], t]];
KG11 = FullSimplify[((1)/(Sqrt[-Det[metric]]))*
D[(Sqrt[-Det[metric]])*(inversemetric[[2, 2]])*
D[Ξ[t, r, θ, ϕ], r], r]];
KG22 = FullSimplify[((1)/(Sqrt[-Det[metric]]))*
D[(Sqrt[-Det[metric]])*(inversemetric[[3, 3]])*
D[Ξ[t, r, θ, ϕ], θ], θ]];
KG33 = FullSimplify[((1)/(Sqrt[-Det[metric]]))*
D[(Sqrt[-Det[metric]])*(inversemetric[[4, 4]])*
D[Ξ[t, r, θ, ϕ], ϕ], ϕ]];
KG00 + KG11 + KG22 + KG33
</code></pre>
<p><strong>III) What I would Like</strong></p>
<p>I would like to use summation convention on the code of section <strong>II)</strong>, since the Klein-Gordon equations are given by:</p>
<p><span class="math-container">$$ \frac{1}{\sqrt{-g}}\sum_{\mu=1}^{4}\sum_{\nu=1}^{4}\partial_{\mu}\Bigg(\sqrt{-g}g^{\mu\nu}\partial_{\nu} \Psi(r,\theta,\phi,t) \Bigg) \tag{1}$$</span></p>
<p><strong>IV) Hartle's code on summation convention</strong></p>
<p>Actually, Hartle's <span class="math-container">$[1]$</span> gives an way to work with tensor indexes, for instance the Christoffel Symbols are given by:</p>
<p><span class="math-container">$$ \Gamma^{s}_{jk}=\sum_{s=1}^{4}\frac{1}{2}g^{is}\Bigg(g_{sj,k} + g_{sk,j} - g_{jk,s} \Bigg) \tag{2}$$</span></p>
<p>and the code using summation is:</p>
<pre><code>affine :=
affine = Simplify[
Table[(1/2)*
Sum[
inversemetric[[i, s]]*(D[metric[[s, j]], coord[[k]]] +
D[metric[[s, k]], coord[[j]]] -
D[metric[[j, k]], coord[[s]]]),
{s, 1, n}
],
{i, 1, n}, {j, 1, n}, {k, 1, n}
] ];
listaffine :=
Table[
If[UnsameQ[affine[[i, j, k]], 0],
{ToString[Γ[i - 1, j - 1, k - 1]],
affine[[i, j, k]]}
],
{i, 1, n}, {j, 1, n}, {k, 1, j}
];
TableForm[
Partition[DeleteCases[Flatten[listaffine], Null], 2],
TableSpacing -> {2, 2}
]
</code></pre>
| xzczd | 1,871 | <p>You can try my <a href="https://mathematica.stackexchange.com/a/216806/1871"><code>allowtensor</code></a>. Since it's a 4D problem, you need to set the value of global variable <code>$tensordimension</code> to <code>4</code>:</p>
<pre><code>dmetric = Det[metric];
ref = KG00 + KG11 + KG22 + KG33;
$tensordimension = 4;
mykg = allowtensor[
1/Sqrt[-dmetric] D[Sqrt[-dmetric] inversemetric[[i, j]] D[Ξ[t, r, θ, ϕ],
coord[[j]]], coord[[i]]], {i, j}];
mykg == ref // Simplify
(* True *)
</code></pre>
<p>Definitions of <code>metric</code> etc. are the same as yours.</p>
|
3,096,933 | <p>In studying the causes of power failures, the following data have been gathered:
10% are due to a transformer damage, 75% are due to line damage, 5% involve both
problems. Based on these percentages, find the probability that a given power failure
involves:</p>
<p>a) line damage given that there is transformer damage</p>
<p>b) transformer damage given that there is line damage</p>
<p>c) transformer damage but not line damage</p>
<p>d) transformer damage given that there is no line damage</p>
<p>e) transformer damage or line damage.</p>
<p>Now, I am familiar with conditional probabilities ( to some degree at least ) and the first thing that came to my mind for point a) was Bayes </p>
<p>so </p>
<p><span class="math-container">$ T $</span> - it involves transformer damage</p>
<p><span class="math-container">$ L $</span> - it involves line damage</p>
<p><span class="math-container">$ B $</span> - it involves both</p>
<p><span class="math-container">$$ P(L/T)=\frac{P(L)*P(T/L)}{P(T)} $$</span></p>
<p>but the problem is that I am stuck at <span class="math-container">$P(T/L)$</span> I have no idea from were to start and maybe this approach is not even the correct one so I would appreciate some help, maybe a hint on how to proceed...</p>
| user0102 | 322,814 | <p>According to your notation, we have <span class="math-container">$\textbf{P}(T) = 0.1$</span>, <span class="math-container">$\textbf{P}(L) = 0.75$</span> and <span class="math-container">$\textbf{P}(T\cap L) = 0.05$</span>. Therefore:</p>
<p>(a) The sought probability is given by <span class="math-container">$\textbf{P}(L|T)$</span>, which can be rewritten as
<span class="math-container">\begin{align*}
\textbf{P}(L|T) = \frac{\textbf{P}(L\cap T)}{\textbf{P}(T)} = \frac{0.05}{0.1} = 0.5
\end{align*}</span></p>
<p>(b) Analogously, we have
<span class="math-container">\begin{align*}
\textbf{P}(T|L) = \frac{\textbf{P}(T\cap L)}{\textbf{P}(L)} = \frac{0.05}{0.75} = \frac{1}{15}
\end{align*}</span></p>
<p>(c) Here, the event in which we are interested in is described by <span class="math-container">$\textbf{P}(T\cap L^{c})$</span>:
<span class="math-container">\begin{align*}
\textbf{P}(T\cap L^{c}) = \textbf{P}(T) - \textbf{P}(T\cap L) = 0.1 - 0.05 = 0.05
\end{align*}</span></p>
<p>(d) In this case, the target event is described by <span class="math-container">$\textbf{P}(T|L^{c})$</span>:
<span class="math-container">\begin{align*}
\textbf{P}(T|L^{c}) = \frac{\textbf{P}(T\cap L^{c})}{\textbf{P}(L^{c})} = \frac{\textbf{P}(T\cap L^{c})}{1 - \textbf{P}(L)} = \frac{0.05}{1 - 0.75} = \frac{0.05}{0.25} = 0.2
\end{align*}</span></p>
<p>(e) Finally, the last event is given by
<span class="math-container">\begin{align*}
\textbf{P}(T\cup L) = \textbf{P}(T) + \textbf{P}(L) - \textbf{P}(T\cap L) = 0.1 + 0.75 - 0.05 = 0.8
\end{align*}</span></p>
<p>Hope this helps.</p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| Omran Kouba | 140,450 | <p>This might be a possible solution. For a positive integer $n$, let $\nu_2(n)$ be the largest $k$ such that $2^k|n$, and similarly, let $\nu_3(n)$ be the largest $k$ such that $3^k|n$. Finally let
$$h(n)=\frac{n}{3^{\nu_3(n)}2^{1+4\lfloor \nu_2(n)/4\rfloor}}$$
If we consider
$$
a\times ~ b {\buildrel \rm def\over =}~b h(ab)
$$
then
$(k-1)\times k$ coincides with the proposed results for $k=4,5,6,7$ and yields $224$ for $k=8$.</p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| Asinomás | 33,907 | <p>Here is something I did which lead me to an incorrect result, but it is still pretty close.</p>
<p>Since all the values we are given are of the form $a\times (a+1)$, I decided to make the function $f(a)=a\times (a+1)$. Assuming $f$ is a polynomial of grade $4$ or less we obtain $f$ is equal to $\frac{101 x^3}{6}-233 x^2+\frac{6301 x}{6}-1500$ using interpolation.</p>
<p>This function gives us $f(7)=208$, which comes close, but is still not correct.</p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| vuur | 85,875 | <p>One solution is to define the operation $\times$ between two integers as</p>
<p>$m \times n = n \cdot
\left\{
\begin{array}{ll}
\frac{1}{3}\sum_{k=1}^m k &\mbox{if } 3 \mid \sum_{k=1}^m k \\
\sum_{k=1}^m k &\mbox{otherwise.}
\end{array}
\right.$</p>
<p>The point is, that what remains of the RHS after dividing by $n$ can be recognized as the sum of the first $m$ integers, divided by $3$ should that be possible.</p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| user2477732 | 164,475 | <p>numbers sequence</p>
<p>3 . 4 = 4 X 2 = 8</p>
<p>4 . 5 = 5 X 10 = 50</p>
<p>5 . 6 = 6 X 5 = 30</p>
<p>6 . 7 = 7 X 7 = 49</p>
<p>7 . 8 = 8 X 28 = 224</p>
<p>8 . 9 = 9 X 4 = 36</p>
<p>9 . 10 = 10 X 5 = 50</p>
<p>10 . 11 = 11 X 55 = 605</p>
<p>11 . 12 = 12 X 11 = 132</p>
<p>12 . 13 = 13 X 13 = 169</p>
<p>13 . 14 = 14 X 91 = 1274</p>
<p>14 . 15 = 15 X 7 = 105</p>
<p>15 . 16 = 16 X 8 = 128</p>
<p>16 . 17 = 17 X 136 = 2312</p>
<p>17 . 18 = 18 X 17 = 306</p>
<p>18 . 19 = 19 X 19 = 361</p>
<p>19 . 20 = 20 X 190 = 3800</p>
<p>20 . 21 = 21 X 10 = 210</p>
<hr>
<p>a.(a+1) = (a+1)x((a+1)/2) - even number/2</p>
<p>b.( b+1) = (b+1)x(((a+1)/2 )xc)) - middle of the sandwich </p>
<p>c.(c+1) = (c+1)xc </p>
<hr>
<p>d.(d+1) = (d+1)x(d+1) - odd number - copy</p>
<p>e.(e+1) = (e+1)x((d+1)x(f/2))</p>
<p>f.(f+1) = (f+1)x(f/2)</p>
<hr>
<p>a+1=b, b+1=c,…</p>
<hr>
<hr>
<p>The problem is more about finding the patterns and relations between numbers and given equations.
<br></p>
<p>3×4=8<br>
4×5=50<br>
5×6=30<br>
6×7=49<br>
7×8=?<br></p>
<p>There are some assumptions that have to be made:
<br>
1) look at the given equations as a sequences of numbers (sequences is plural - not just one sequence)<br>
2) results on the right side can always by divided by the second number on the left side 8:4=2, 50:5=10, 30:6=5, 49:7=7 => the result of the last equations is therefore multiple of 8 => 7x8=8x?=???<br>
(Note: Why did they use "x" when multiplication is clearly not what is done with those numbers? Why not better use symbol ∘
for unknown operations? My guess is - it's also a hint.... multiplication is necessary in the answer. ....so don't try to come up with solutions that are more complex than that ;-) But that's just my guess)
<br>
3) we can write down what we assume so far:
<br>
3 ∘4 = 4 X 2 = 8<br>
4 ∘ 5 = 5 X 10 = 50<br>
5 ∘ 6 = 6 X 5 = 30<br>
6 ∘ 7 = 7 X 7 = 49<br>
7 ∘8 = 8 X ?=???<br>
5) we can also say that after 7∘8=8x???
some other equations should follow and the patter we know so far is is:<br><br>
8 ∘9 = 9 X ?=???<br>
9 ∘10 = 10 X ?=???<br>
<br>
5) now look at the numbers sequence (fourth number in each equation): 2, 10, 5, 7, ... there are of course many things we can do (2+8=10, 10-5=5, 5+2=7,etc.)... but we also have a possible relation to 3.4, 4.5, 5.6,6.7
<br>
6) the easy patter would be "sandwich"- second number = first*third
<br>
7) how to define first and third number? - check the relation with 3.4 and 5.6 and first number also has a relation to 7.7
<br>
..the rest I already explained in the comment section below ;-)</p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| I. J. Kennedy | 130 | <p>Surely even this interviewer meant for <span class="math-container">$\times$</span> to be commutative, so I propose
<span class="math-container">$$a \times b = \left( \binom{a+1}{3},\binom{b+1}{3} \right) \cdot \max(a,b)$$</span>
where the inner brackets are binomial coefficients, and the outer brackets indicate the gcd.</p>
|
1,401,760 | <p>First I tried to use integration:
$$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$
$$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$
But I could not express it as a <em>riemann integral</em>. Now I am thinking about sandwich theorem.</p>
<p>$$\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}$$
Since $\frac{a}{t+1}>\frac{a}{t+2}>\frac{a}{t+1}>\cdots>\frac{a}{n}$
$$\frac{a^n}{n!}<\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$
since $\frac{a}{t+1}<1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$
Hence, $$\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0$$
And by using sandwich theorem, $y=0$. Is this correct?</p>
| Zach466920 | 219,489 | <p>What is,</p>
<p>$$\lim_{n \to {\infty}} {{a^n} \over {n!}}$$</p>
<p>Well, for large values of $n$, $n!$ can be evaluated with <a href="https://en.wikipedia.org/wiki/Stirling%27s_approximation" rel="nofollow">Stirling's Approximation</a>.</p>
<p>$$\lim_{n \to {\infty}} {{a^n} \over {n!}}=\lim_{n \to {\infty}} {{a^n} \over {\sqrt{2 \pi n} \cdot (n/e)^n}}$$</p>
<p>$${{a^n} \over {\sqrt{2 \pi n} \cdot (n/e)^n}} = {1 \over {\sqrt{2 \pi n}}} \cdot a^n \cdot (e/n)^n = {1 \over {\sqrt{2 \pi n}}} \cdot \left({{a \cdot e } \over {n}} \right)^n$$</p>
<p>The value of the limit is clearly $0$ for any finite $a$,
$$\lim_{n \to {\infty}} {{a^n} \over {\sqrt{2 \pi n} \cdot (n/e)^n}} = 0$$
This is because the square root term goes to zero, and because the term inside the parentheses must be less than 1 since $a$ is finite.</p>
<p>Thus,</p>
<p>$$\lim_{n \to {\infty}} {{a^n} \over {n!}}=0$$</p>
|
1,401,760 | <p>First I tried to use integration:
$$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$
$$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$
But I could not express it as a <em>riemann integral</em>. Now I am thinking about sandwich theorem.</p>
<p>$$\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}$$
Since $\frac{a}{t+1}>\frac{a}{t+2}>\frac{a}{t+1}>\cdots>\frac{a}{n}$
$$\frac{a^n}{n!}<\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$
since $\frac{a}{t+1}<1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$
Hence, $$\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0$$
And by using sandwich theorem, $y=0$. Is this correct?</p>
| idm | 167,226 | <p>Let $x_n=\frac{a^n}{n!}$.</p>
<p>$$\left|\frac{x_{n+1}}{x_n}\right|=\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}=\frac{a^{n+1}n!}{a^n (n+1)!}= \frac{a}{n+1}\underset{n\to \infty }{\longrightarrow }0$$</p>
<p>and thus, by $x_n\to 0$ by <a href="https://en.wikipedia.org/wiki/Ratio_test" rel="nofollow">Ratio test</a>.</p>
|
3,635,266 | <p>As stated in title, I am curious how to construct a sequnce <span class="math-container">$\{a_n\}_{n=1}^{\infty}$</span> such that <span class="math-container">$\sum_{n=1}^{\infty} a_n^2 < +\infty$</span> but <span class="math-container">$\sum_{n=1}^{\infty}\frac{a_n}{\sqrt{n}} = \infty$</span>.</p>
<p>This problem arises when I try to show that <span class="math-container">$f(x) = \sum_{i=1}^{\infty} \frac{x_n}{\sqrt{n}}$</span> defined on <span class="math-container">$l_2$</span> space is not bounded.</p>
| QC_QAOA | 364,346 | <p>Consider</p>
<p><span class="math-container">$$a_n=\frac{1}{\sqrt{n}\ln(n+1)}$$</span></p>
<p>Then</p>
<p><span class="math-container">$$\sum_{n=1}^\infty a_n^2=\sum_{n=1}^\infty\frac{1}{n\ln(n+1)^2}<\infty$$</span></p>
<p>but</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{a_n}{\sqrt{n}}=\sum_{n=1}^\infty\frac{1}{n\ln(n+1)}=\infty$$</span></p>
|
63,641 | <p>Here is the problem: Two mathematicians meet at a bar. They like each other and tend to collaborate. But it is not so clear what problems could be of common interest to both of them. Of course, the traditional way is they keep describing they work or their field in general so that hopefully they catch something at the end. But is there any reference, graph, table or whatever that they can use to help them? This, of course, makes sense only when such a reference keeps updated based on the continuous production in mathematics.</p>
| GH from MO | 11,919 | <p>The Mathieu group $M_{11}$ does not have this property. A quote from Example 2.16 in <a href="http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3895v1.pdf">this paper</a>: "Hence there is no automorphism of $M_{11}$ that maps $x$ to $x^{−1}$."</p>
<p>Background how I found this quote as I am no group theorist: I used Google on "groups with no outer automorphism" which led me to this <a href="http://en.wikipedia.org/wiki/Outer_automorphism_group">Wikipedia article</a>, and from there I jumped to this other <a href="http://en.wikipedia.org/wiki/Mathieu_group">Wikipedia article</a>. So I learned that $M_{11}$ has no outer automorphism. Then I used Google again on "elements conjugate to their inverse in the mathieu group" which led me to the above mentioned paper. </p>
<p><strong>EDIT:</strong> Following Geoff Robinson's comment let me show that any element $x\in M_{11}$ of order 11 has this property, using only basic group theory and the above <a href="http://en.wikipedia.org/wiki/Mathieu_group">Wikipedia article</a>. The article tells us that $M_{11}$ has 7920 elements of which 1440 have order 11. So $M_{11}$ has 1440/10=144 Sylow 11-subgroups, each cyclic of order 11. These subgroups are conjugates to each other by one of the Sylow theorems, so each of them has a normalizer subgroup of order 7920/144=55. In particular, if $x$ and $x^{-1}$ were conjugate to each other, then they were so by an element of odd order. This, however, is impossible as any element of odd order acts trivially on a 2-element set.</p>
|
642,497 | <p>How do we prove by vector method that "if the diagonals of a trapezium have equal length then the non-parallel sides of the trapezium have equal length." ? </p>
<p>(taking $ABCD$ to be the trapezium with
$AD$ || $BC$ and $O$ the intersection point of diagonals , if it can be shown by vector method that
$OB=OC$ , then $AB=CD$ follows ; I can show $OB=OC$ but not by vector method and thus the problem , though any other line of approach is acceptable.) </p>
| egreg | 62,967 | <p>Hint: Set $y=x+2$, so that $x=y-2$; do your math and …</p>
<p>Hint 2: Taylor expansion.</p>
|
398,002 | <p>There is a <a href="https://en.wikipedia.org/wiki/Lie_group%E2%80%93Lie_algebra_correspondence" rel="noreferrer">classical correspondence</a> between Lie algebras (over <span class="math-container">$\mathbb{R}$</span> or <span class="math-container">$\mathbb{C}$</span>) and Lie groups in the finite dimensional case: to every Lie group <span class="math-container">$G$</span> there is an associated Lie algebra <span class="math-container">$\mathfrak{g}$</span>, and conversely. Moreover, this correspondence is one-to-one if one requires <span class="math-container">$G$</span> to be simply connected. One also has maps <span class="math-container">$\exp$</span> and <span class="math-container">$\log$</span> that map between a Lie group and its associated Lie algebra, and the Baker–Campbell–Hausdorff (BCH) formula gives a Lie series <span class="math-container">$z$</span> for given <span class="math-container">$x$</span> and <span class="math-container">$y$</span> in <span class="math-container">$\mathfrak{g}$</span> such that <span class="math-container">$\exp x\exp y=\exp z$</span> (at least for sufficiently small <span class="math-container">$x$</span> and <span class="math-container">$y$</span>).</p>
<p>Every Lie group gives rise to a Lie algebra, but in the infinite-dimensional case there are ‘non-enlargeable’ Lie algebras which don't correspond to a Lie group.</p>
<p>However, I came across a <a href="https://mathoverflow.net/questions/41644/does-a-finite-dimensional-lie-algebra-always-exponentiate-into-a-universal-cover#comment98452_41650">remark</a> in <a href="https://mathoverflow.net/questions/41644/does-a-finite-dimensional-lie-algebra-always-exponentiate-into-a-universal-cover">Does a finite-dimensional Lie algebra always exponentiate into a universal covering group</a> saying that every Lie algebra over <span class="math-container">$\mathbb{R}$</span> can be exponentiated to give an abstract group, but that it is a non-trivial theorem. Rather than poking this old comment to another question (where I may have taken something out of context), I thought this deserves to be explicitly asked.</p>
<blockquote>
<p>Let <span class="math-container">$\mathfrak{g}$</span> be a Lie algebra over a field <span class="math-container">$k$</span>. Is there an abstract group <span class="math-container">$G$</span> with which <span class="math-container">$\mathfrak{g}$</span> is naturally associated? What if we restrict to <span class="math-container">$k$</span> having characteristic zero — or even <span class="math-container">$k=\mathbb{R}$</span> or <span class="math-container">$k=\mathbb{C}$</span>?</p>
</blockquote>
<p>By ‘naturally associated’ I am thinking of a bijective correspondence reminiscent of the classical situation, where the algebra and the group are linked by the BCH formula, and Lie algebra homomorphisms <span class="math-container">$\phi$</span> lift to group homomorphisms <span class="math-container">$\Phi=\exp \phi\log$</span>. However I am not expecting a differentiable structure on <span class="math-container">$G$</span>. (That said, if <span class="math-container">$G$</span> comes with some extra structure for free, I would be interested to hear about it.)</p>
<p>I have looked at other MO questions such as <a href="https://mathoverflow.net/questions/4765/lie-groups-and-lie-algebras">Lie Groups and Lie Algebras</a> but it's not clear to me that infinite dimensional Lie algebras are considered there.</p>
<p>A further observation is that the Lie series encountered in the classical case converge in the usual sense on sufficiently small neighbourhoods. However, in some cases of interest to me, such as free Lie algebras, the Lie algebra can be embedded in an (associative) algebra of formal power series with non-commuting indeterminates, in which case there are no restrictions on when the Lie series converge. So it seems to me that a subcase of the question above corresponding to the situation where all Lie series converge may well have been studied. There is a 1948 Doklady paper <em>On normed Lie algebras over the field of rational numbers</em> by Mal'cev which sounds relevant but has proven resistant to my attempts to track it down.</p>
<p>I would be grateful for any pointers to books or articles that discuss this.</p>
| Robert Bryant | 13,972 | <p>Here is an informative example that illustrates the difficulties: Consider the Lie algebra <span class="math-container">${\frak{g}} = \mathrm{Vect}(\mathbb{S})$</span> of smooth vector fields on the circle <span class="math-container">$\mathbb{S}$</span>. The flow of any vector field <span class="math-container">$X\in{\frak{g}}$</span> is a <span class="math-container">$1$</span>-parameter subgroup of <span class="math-container">$\mathrm{Diff}_+(\mathbb{S})$</span>, the (connected) Lie group of orientation preserving diffeomorphisms of <span class="math-container">$\mathbb{S}$</span>, so this defines an exponential map <span class="math-container">$\exp:{\frak{g}}\to\mathrm{Diff}_+(\mathbb{S})$</span>. Moreover, every <span class="math-container">$1$</span>-parameter subgroup of <span class="math-container">$\mathrm{Diff}_+(\mathbb{S})$</span> is the flow of some smooth vector field on <span class="math-container">$\mathbb{S}$</span>.</p>
<p>Unfortunately, there is also a sequence <span class="math-container">$\phi_k$</span> of diffeomorphisms of <span class="math-container">$\mathbb{S}$</span> that converges to the identity in the <span class="math-container">$C^\infty$</span> topology such that none of the <span class="math-container">$\phi_k$</span> lie in <span class="math-container">$\exp\bigl({\frak{g}}\bigr)$</span>. In particular, the exponential map does not cover a neighborhood of the identity in <span class="math-container">$\mathrm{Diff}_+(\mathbb{S})$</span>. See, for example, R. Hamilton's Bulletin article <a href="https://doi.org/10.1090/S0273-0979-1982-15004-2" rel="noreferrer"><em>The inverse function theorem of Nash and Moser</em></a>.</p>
<p>Meanwhile there is no proper Lie subgroup (in Lie's original sense) of <span class="math-container">$\mathrm{Diff}_+(\mathbb{S})$</span> that contains <span class="math-container">$\exp\bigl({\frak{g}}\bigr)$</span>.</p>
<p>There has been a lot of work on so-called 'infinite dimensional Lie groups'. It might be good to start with J. Milnor's classic paper, <em>Remarks on infinite-dimensional Lie groups</em>, in “Relativité, Groupes et Topologie II,” B. DeWitt and R. Stora (Eds), North-Holland, Amsterdam, 1983, 1007–1057, but you will also want to read some of the more modern literature as well, such as <em>Fundamental Problems in the Theory of Infinite-Dimensional Lie Groups</em>, by H. Glöckner (<a href="https://arxiv.org/abs/math/0602078" rel="noreferrer">https://arxiv.org/abs/math/0602078</a>) and works on this subject by P. Michor and collaborators.</p>
|
2,307,268 | <p>I tried to compute the power of 2^2^2^2 on google calculator and my casio calculator but both are giving different results. same is true for 3^3^3.
Please explain me the difference between two expressions.</p>
<p><a href="https://i.stack.imgur.com/2CMDt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2CMDt.jpg" alt="enter image description here"></a></p>
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| John | 7,163 | <p>The difference lies in the applied order of operations.</p>
<p>The Casio calculator does operations strictly left to right unless you break it with parentheses:</p>
<blockquote>
<p>2^2^2^2 = 4^2^2 = 16^2 = 256</p>
</blockquote>
<p>But Google evaluates the entire expression (correctly) using right-to-left order or operations for the stacked powers:</p>
<blockquote>
<p>2^2^2^2 = 2^2^4 = 2^16 = 65536</p>
</blockquote>
|
1,201,352 | <p>I have a very simple question which is bugging me.</p>
<p>We are 3 roommates and our total electricity bill is $61 this month,
I was home for the whole month,
Friend X was home for 15 days,
Friend Y was home for 20 days</p>
<p>now the easy question, how much would each person would pay?</p>
<p>My calculation is </p>
<pre><code>61/3=20,33
Guy X : 20,33 / 2 = 10.16
Guy Y : (20,33 * 2) / 3 = 13.55
Me : 61 - (10.16+13.55) = 37.29 which doesn't make sense at all!!!
</code></pre>
<p>Help me!!!</p>
| Community | -1 | <p>n = $61/(30.5 + 15 + 20)</p>
<p>30.5 X n = $28.40</p>
<p>15 X n = $13.97</p>
<p>20 X n = $18.63</p>
<p>The individual amounts total $61.</p>
|
1,920,781 | <p>Let's say I have a basic equation (here a linear combination):
$$ax_1 + bx_2=y$$</p>
<p>Is there any way to get the value of the sum $x_1+x_2$ if $a$, $b$ and $y$ is known?</p>
| MPW | 113,214 | <p>In the absence of any additional information, no.</p>
<p>For example, there are lots of pairs of numbers that differ by $1$, but their sum could be anything. Consider the pairs (0,1), (10,11), (1000, 1001).</p>
|
3,041,841 | <p>Let <span class="math-container">$f(z)$</span> be holomorphic function on an open unit disk such that <span class="math-container">$\lim_{z \to 1} f(z)$</span> doesn't exists. Let <span class="math-container">$f(z) = \Sigma_{i=1}^{\infty} a_iz^i$</span> be its Taylor series around <span class="math-container">$0$</span>. Then the radius of convergence of <span class="math-container">$f(z)$</span> is?</p>
<p>Option </p>
<p>1) <span class="math-container">$R = 0$</span></p>
<p>2) <span class="math-container">$0 < R < 1$</span></p>
<p>3) <span class="math-container">$R = 1$</span></p>
<p>4) <span class="math-container">$R > 1$</span>.</p>
<p>My attempt:</p>
<p>I considered the function <span class="math-container">$f(z) = \frac{1}{1 - z} = \Sigma_{i = 0}^{\infty}z^i$</span>. This satisfies the above hypothesis, and its radius of convergence is <span class="math-container">$R < 1$</span>, hence option 2) is correct. But the answer key says option 4) is correct? So can anyone explain me the reason?</p>
<p>Reference:
CSIR NET DEC 2017 Qno 34, Paper A
<a href="http://csirhrdg.res.in/mathA_Dec2017.pdf" rel="nofollow noreferrer">http://csirhrdg.res.in/mathA_Dec2017.pdf</a>
<a href="http://csirhrdg.res.in/Mathkey_Dec2017.pdf" rel="nofollow noreferrer">http://csirhrdg.res.in/Mathkey_Dec2017.pdf</a></p>
| nmasanta | 623,924 | <p>Given that <span class="math-container">$f(z)$</span> be holomorphic function on an open unit disk, so
let <span class="math-container">$$f(z)=\frac{1}{1-z}$$</span>
then <span class="math-container">$\quad\lim_{z \to 1} f(z)\quad$</span> does not exist.</p>
<p>Taylor series of <span class="math-container">$f(z)$</span> about <span class="math-container">$z=0$</span> is <span class="math-container">$$f(z)=(1-z)^{-1}=1+z+z^2+z^3+. . . $$</span>
<span class="math-container">$$=\sum_{n=0}^{\infty}a_nz^n\qquad \text{where}\quad a_n=1\quad \forall n=0, 1, 2, . . . $$</span>
Radius of convergence <span class="math-container">$$R=\lim_{n \to \infty}\frac{a_n}{a_{n+1}}=\lim_{n \to \infty}\frac{1}{1}=1$$</span></p>
<p><strong>Only "option <span class="math-container">$3$</span>" is correct</strong>.</p>
|
3,609,791 | <p><span class="math-container">$f(x) = \frac{x}{(1+x^2)}$</span></p>
<p>I need to find the range of function.</p>
<p>My method-</p>
<p>Let <span class="math-container">$f(x)=y$</span> </p>
<p>Then <span class="math-container">$x²y-x+y=0$</span></p>
<p><span class="math-container">$$x=1\pm\frac{\sqrt{(1-4y²)}}{2y}$$</span></p>
<p>For <span class="math-container">$x$</span> to be real , </p>
<p><span class="math-container">$1-4y^2\ge0$</span> <strong>and</strong> <span class="math-container">$y\neq 0$</span></p>
<p><span class="math-container">$(2y+1)(2y-1)\ge0$</span> <strong>and</strong> <span class="math-container">$y\neq0$</span></p>
<p>Hence <span class="math-container">$y \in [-1/2,1/2] -\{0\}$</span></p>
<p>So far so good.</p>
<p>But if I put <span class="math-container">$0$</span> in the function,</p>
<p>Then <span class="math-container">$f(0)= 0/1+0 =0$</span></p>
<p>While my solution says that <span class="math-container">$y$</span> cannot be zero. </p>
<p>Where am I going wrong?</p>
| Luca Goldoni Ph.D. | 264,269 | <p>When you have a second degree equation <span class="math-container">$ax^2+bx+c=0$</span> in order to use the formula
<span class="math-container">$$
x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}
{{2a}}
$$</span>
it must be <span class="math-container">$a \neq 0$</span>. As <span class="math-container">$a=0$</span> your equation turn to be <span class="math-container">$bx+c=0$</span> which get of course
<span class="math-container">$x=-c/b$</span> (as long as <span class="math-container">$b\neq 0$</span>). In your case, you have that, if <span class="math-container">$y=0$</span> your equation
is <span class="math-container">$0x^2+x+0=0$</span> so <span class="math-container">$x=0$</span>.</p>
|
2,072,061 | <p>My Working: </p>
<p>\begin{align}
a_0=&-0+2=2\\
a_1=&-1+2=1\\
a_2=&-2+2=0.
\end{align}
and $$a_2=a_{2-1}+2a_{2-2}+2(2)-9=-4.$$</p>
<p>I don't know what's wrong with my solution. I used this method on various questions but now I'm just ending up with the wrong answer.</p>
| Arnaldo | 391,612 | <p>We have:</p>
<p>$$a_n=-n+2$$
$$a_{n-1}=-(n-1)+2=-n+3$$
$$a_{n-2}=-(n-2)+2=-n+4$$</p>
<p>And then </p>
<p>$$a_n-a_{n-1}-2a_{n-2}=-n+2-(-n+3)-2(-n+4)= 2n-9$$</p>
|
812,263 | <p>To further explain the title:</p>
<p>Is there a probabilistic reason as to why a 6-sided die has the opposing sides suming to 7?</p>
<p>My argument begun when a friend decided to use <a href="http://ecx.images-amazon.com/images/I/51OvRGphnOL.jpg" rel="nofollow">this die</a> instead of <a href="http://nerdywithchildren.com/wp-content/uploads/2013/05/d20.jpg" rel="nofollow">this die</a>.</p>
<p>I understand that having 20 sides, each is as likely to come up, but does the different pattern affect the subsequent rolls?</p>
<p>Thanks in advance!</p>
| Tara | 150,046 | <p>In the 20-sided dice linked, there is actually a good reason not to use the die with numbers of equal size clustered closely together.</p>
<p>Just as in a roulette wheel, it is harder to cheat if a slight difference in movement totally changes the number, and in addition, it is much more suspenseful to watch if you can only tell at the very end which size/color the number has.</p>
<p>However, for an ordinary die, it is impossible to avoid neighbouring sides having neighbouring numbers, so it makes sense to go for a symmetric arrangement instead.</p>
|
3,804,195 | <p>I am currently reworking the following exercise that was worked during my previous class's lecture on normed vector spaces:</p>
<blockquote>
<p>Show that <span class="math-container">$||x||_{\infty} \leq ||x||_{2} \leq ||x||_{1}$</span> for any <span class="math-container">$x \in \mathbb{R}^n$</span>.</p>
</blockquote>
<p>After class, I asked if this is the same as proving that <span class="math-container">$||a||_p\le ||a||_1$</span> for any <span class="math-container">$p \leq 1$</span> or more generally, <span class="math-container">$||a||_q\le ||a||_p$</span> whenever <span class="math-container">$p \leq q$</span> . My professor then asked me to finish the following computation and report back my findings:</p>
<p><span class="math-container">$(\sum_{i=0}^n |a_i|^p)^{1/p}\le (\sum_{i=0}^{n-1} |a_i|^p)^{1/p} +(|a_n|^p)^{1/p}\\\le (\sum_{i=0}^{n-2} |a_i|^p)^{1/p} +(|a_{n-1}|^p)^{1/p} + (|a_n|^p)^{1/p} \\ ...\\...\\$</span></p>
<p>However, as I have made progress I have a few questions.</p>
<p><span class="math-container">$\bullet$</span> Is this using Minkowski's inequality repeatedly?</p>
<p><span class="math-container">$\bullet$</span> I don't fully understand the step <span class="math-container">$(\sum_{i=0}^n |a_i|^p)^{1/p}\le (\sum_{i=0}^{n-1} |a_i|^p)^{1/p} +(|a_n|^p)^{1/p}$</span>. Why are we adding the <span class="math-container">$p$</span> norm of the <span class="math-container">$n^{th}$</span> term to (<span class="math-container">$\sum_{i=0}^{n-1} |a_i|^p)^{1/p}$</span>?</p>
| RobPratt | 683,666 | <p>Yes, you can use linear programming to find such <span class="math-container">$p_k$</span>. For each <span class="math-container">$k$</span>, let <span class="math-container">$p_k$</span> be the desired weight. The objective is arbitrary, so you can minimize the constant <span class="math-container">$0$</span> function. The (linear) constraints are:
<span class="math-container">\begin{align}
\sum_k x^k_i p_k &= x_i &&\text{for $i \in \{1,\dots,n\}$} \\
\sum_k p_k &= 1 \\
0 \le p_k &\le 1 &&\text{for all $k$}\\
\end{align}</span></p>
|
2,131,224 | <blockquote>
<p>Say I have vectors $x, y$, then is $\text{proj }_x y $ a scalar multiple of $x$?</p>
</blockquote>
<p>I have a book saying that it is, but I have no clue why this true. Is this really true?</p>
| Martin Argerami | 22,857 | <p>Of course. What else could the projection be? It has to be an element of the subspace spanned by $y$, which consists precisely of the scalar multiples of $x$. </p>
|
457,427 | <p>Find the derivative of the functions:<br>
$$\int_{x^2}^{\sin(x)}\sqrt{1+t^4}dt$$<br></p>
<p>In class we had the following solution:<br>
By the fundamental theorem of calculus we know that <br>
$$\left(\int_a^xf(t)dt\right)'=f(x)$$ So<br>
$$\int_{x^2}^0\sqrt{1+t^4}dt+\int_0^{\sin(x)}\sqrt{1+t^4}dt=$$<br>
$$\int_0^{\sin(x)}\sqrt{1+t^4}dt-\int_0^{x^2}\sqrt{1+t^4}dt=$$<br>
Letting $g(x)=\sqrt{1+t^4} $<br>
$$g(\sin(x))(\sin(x))'-g(x^2)(x^2)'=$$<br>
$$\sqrt{1+\sin(x)^4}\cdot \cos(x)-\sqrt{1+x^8} \cdot 2x$$<br></p>
<p>However, if we have that $\left(\int_a^xf(t)dt \right)'=f(x)$ wouldn't the answer just be <br>
$$\sqrt{1+\sin(x)^4}-\sqrt{1+x^8}?$$</p>
| Mikasa | 8,581 | <p>Generally, the way @Ron walked, is known as <a href="http://en.wikipedia.org/wiki/Leibniz_integral_rule" rel="nofollow">Leibniz integral rule</a> which follows from the chain rule:
$$ {d\over dx} \left( \int_{f(x)}^{g(x)} \phi(t) \,dt \right )= \phi\left(g(x)\right) {g'(x)} - \phi(f(x)) {f'(x)} $$</p>
|
33,424 | <p>Suppose $f(z) = P(z)e^{Q(z)}$ where $P,Q$ are real polynomials. What is the number of non-real zeros of $f^{(k)}$ as $k$ increases?</p>
<p>We know that $f''$ has $\geq m$ zeros where $m$ depends on $Q(z)$. </p>
| Pietro Majer | 6,101 | <p>Just some hints. The functions $f^{\,(k)}$ have the same zeros of the polynomials $P_k:=f^{\,(k)}\exp(-Q)$, that satisfy $P_0:=P$ and $P_{k+1}=P_k'+P_kQ'$. In particular $P_k$ has degree $\deg(P)+k\left(\deg(Q)-1\right)$, and this is also the total number of zeros of $f^{\,(k)}$. They may be all real: for instance if $Q:=-x^2$ and $P:=1$ one finds the Hermite polynomials, that are orthogonal, hence have all zeros real and simple. </p>
|
2,038,617 | <p>The question asks to show that $$\sum_{k=0}^{3n}(-3)^k \binom{6n}{2k}=2^{6n}$$ by considering the binomial expansion </p>
<p>I thought about the use of $$(1+z)^n=\sum_{k=0}^{n}\binom{n}{k}z^k$$
with suitable complex number $z$, as the formula shows the $(-3)^k$ term might suggest the use of complex number, which take the imaginary part of the expansion</p>
<p>However, I cannot find such $z$ that makes the sum to $2^{6n}$</p>
<p>Any hints are appreciated!</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 0}^{3n}\pars{-3}^{k}{6n \choose 2k} & =
\sum_{k = 0}^{\infty}{6n \choose 2k}\pars{\root{3}\ic}^{2k} =
\sum_{k = 0}^{\infty}{6n \choose k}\pars{\root{3}\ic}^{k}
\,\,{1 + \pars{-1}^{k} \over 2} \\[5mm] & =
{1 \over 2}\sum_{k = 0}^{\infty}{6n \choose k}\pars{\root{3}\ic}^{k} +
{1 \over 2}\sum_{k = 0}^{\infty}{6n \choose k}\pars{-\root{3}\ic}^{k}
\\[5mm] & =
\Re\pars{\sum_{k = 0}^{\infty}{6n \choose k}\bracks{\root{3}\ic}^{k}}
=
\Re\pars{\bracks{1 + \root{3}\ic}^{6n}}= \pars{1 + 3}^{3n}
\\[5mm] & =
\bbx{\ds{2^{6n}}}
\end{align}</p>
|
2,737,144 | <p>I tried to prove it by contradiction. </p>
<p>Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.</p>
<p>$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$</p>
<p>${3-(x(x-2)\gt0}$</p>
<p>${3-x^2-2x\gt0}$</p>
<p>${-x^2-2x+3\gt0}$</p>
<p>${-1(x^2+2x-3)\gt0}$</p>
<p>$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$</p>
<p>${(x-1)(x+3)\lt0}$</p>
<p>At this point I really do not know what to do after this point or if I really even went about it the right way. Thank you for the help.</p>
| TheSimpliFire | 471,884 | <p>Note that if $x>3\implies x-2>1$ then $\dfrac1x<\dfrac13$ and $\dfrac1{x-2}<1$ so $$\frac3{x(x-2)}<\frac3{3(3-2)}=1$$</p>
|
2,737,144 | <p>I tried to prove it by contradiction. </p>
<p>Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.</p>
<p>$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$</p>
<p>${3-(x(x-2)\gt0}$</p>
<p>${3-x^2-2x\gt0}$</p>
<p>${-x^2-2x+3\gt0}$</p>
<p>${-1(x^2+2x-3)\gt0}$</p>
<p>$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$</p>
<p>${(x-1)(x+3)\lt0}$</p>
<p>At this point I really do not know what to do after this point or if I really even went about it the right way. Thank you for the help.</p>
| Bernard | 202,857 | <p>Why do you make things more complex than they are? </p>
<p>The function $x(x-2)$ is <em>increasing</em> and <em>positive</em> on $(2,+\infty)$, hence
$\;\dfrac2{x(x-2)}\;$ is <em>decreasing</em> and <em>positive</em> on this interval.
Furthermore $f(3)=1$…</p>
|
1,679,772 | <p>A binary relation R on $N×N$ is defined as follows$: (a,b)R(c,d)$ if $a≤c$ or $b≤d$. Consider the following
propositions:</p>
<p>$P: R$ is reflexive</p>
<p>$Q: R$ is transitive</p>
<p>Which one of the following statements is TRUE?</p>
<ol>
<li>Both $P$ and $Q$ are true. </li>
<li>$P$ is true and $Q$ is false. </li>
<li>$P$ is false and $Q$ is true. </li>
<li>Both $P$ and $Q$ are false.</li>
</ol>
<hr>
<hr>
<p>My attempt :</p>
<p>Reflexive$: (a, a) R(a, a)$
Since $a \leq a$, or $a \leq a$</p>
<p>Transitive$: (a, b) R (c, d)$ or $(c, d) R(m, n)$ then $(a, b) R(m, n)$
Suppose $(a, b) R(c, d)$</p>
<p>$\implies a \leq c$ or $b \leq d$
and $(c, d) R(m, n)$</p>
<p>$\implies c \leq m, d \leq n$</p>
<p>Since $a \leq c$, or $c \leq m$ so $a \leq m$</p>
<p>$b \leq d$ or $d \leq n$, so $b \leq n$</p>
<p>$\implies (a, b) R(m, n)$</p>
<blockquote>
<p>Can you explain in formal way, please?</p>
</blockquote>
| Hagen von Eitzen | 39,174 | <p>(I'll use <em>sequence compactness</em> here; for metric spaces this is equivalent to compactness)</p>
<p>Pick $x_n\in K_n$. As $x_n\in K_1$, there is a converging subsequence, so we may assume wlog. that the original sequence is already convergent, say $x_n\to x_\infty$.
For any $i$, we have $x_n\in K_i$ for almost all $n$, hence $x_\infty\in K_i$ for all $i$.</p>
<p>A counterexample for closed sets is $A_n=[n,\infty)$ in $\Bbb R$.</p>
|
870,367 | <p>Suppose $A$ is a $2$-by-$2$ matrix, and $\mathcal{l}$ is an invariant line under $A$, so $(x,mx+c)$ is mapped to $(X,mX+c)$ for some variable $X$ linear in $x$. Then is there a point on the line $\mathcal{l}$ which is a fixed point of $A$, i.e. there is some $x' \in \mathcal{l}$ such that $Ax'=x'$?</p>
<p>The reason I ask is that apparently if $y=mx+c$ is an invariant line under $A$, so is $y=mx$ - the above is equilivant to this by linearity, I believe. Certainly, the case where the invariant line goes through the origin is simple - but unfortunately this isn't the only case, e.g. the matrix
$$ \left(\begin{matrix} 0 & 1 \\ 5 & -4 \end{matrix}\right)$$
has an invariant line $y=-5x+2$ as
$$ \left(\begin{matrix} 0 & 1 \\ 5 & -4 \end{matrix}\right)\left( \begin{matrix} x\\ -5x + 2 \end{matrix} \right) = \left( \begin{matrix} -5x+2\\ -5(-5x+2) + 2 \end{matrix} \right)$$
However, I can't see this example being very illuminating, as it's quite contrived to force $(0,1)$ to be mapped to itself, and I can't imagine the $y$-intercept always being a fixed point in these circumstances. Does anyone have any ideas?</p>
| Juanito | 153,015 | <p>As Andre commented, "at least 11" means, in this case, 11 or 12. So, find the the numbers for 11 and 12, and then add them. Answer${}=\dfrac{^{18}C_{12}+{^{18}C_{11}}\cdot10}{^{28}C_{12}}$</p>
|
14,765 | <p>I like to make the "dominoes" analogy when I teach my students induction.</p>
<p>I recently came across the following video:</p>
<p><a href="https://www.youtube.com/watch?v=-BTWiZ7CYoI" rel="noreferrer">https://www.youtube.com/watch?v=-BTWiZ7CYoI</a></p>
<p>In this video, a sequence of concrete block wall caps are set up like dominoes on the top of a wall. The first wall cap is knocked down, setting off the domino effect. The blocks are spaced so that they are resting on each other when they fall, but just barely. So rather than resting flat each block is supported slightly by its successor. When the last block falls, however, it falls flat (having no subsequent block to rest on). This causes the block behind it to slip off, and lay flat, which causes the brick behind it to slip off and lie flat, until all the blocks are lying flat perfectly end to end.</p>
<p>Is there any instance of a similar phenomena occurring in mathematics? I am thinking of a situation in which you want to prove both <span class="math-container">$P(n)$</span> and <span class="math-container">$Q(n)$</span> for <span class="math-container">$n = 1, 2, 3, \dots, 100$</span> (say). If you are able to prove: </p>
<ol>
<li><span class="math-container">$P(1)$</span></li>
<li><span class="math-container">$\forall k \in \{1,2,3, \dots, 99\} P(k) \implies P(k+1)$</span></li>
<li><span class="math-container">$P(100) \implies Q(100)$</span></li>
<li><span class="math-container">$\forall k \in \{ 100, 99, 98, \dots, 3,2\}, Q(k) \implies Q(k-1)$</span></li>
</ol>
<p>Then it will follow that both <span class="math-container">$P(n)$</span> and <span class="math-container">$Q(n)$</span> are true for <span class="math-container">$n = 1, 2, 3, \dots, 100$</span>.</p>
<p>If an example is found, it could be a great example for teaching because it would force students to think through the logic of why induction works rather than blindly following a certain form of "an induction proof".</p>
| mweiss | 29 | <p>There are possibly two different issues here.</p>
<p>Issue 1 is that some students are under the mistaken impression that the symbol <span class="math-container">$\sqrt{5}$</span> actually designates <em>two different numbers</em>, one positive and one negative. I have even heard high school teachers (in the United States) tell students this; it seems to be a very pervasive belief and you can find many examples of it on MSE. If this is the case, then the problem is not that they don't know that there are two solutions, it's that they don't know the correct way to denote it. Notation, of course, is just a convention, so this would be more of a failure to adapt to the common culture of mathematical writing than it would be a failure of comprehension.</p>
<p>One way to identify if this is, in fact, what's happening would be to give an example in which the square root is a simple whole number. If you give the equation <span class="math-container">$x^2=9$</span>, would they say <span class="math-container">$x=3$</span> or <span class="math-container">$x=\pm 3$</span>? If the issue is purely notation -- that is, if the problem is that they think the <span class="math-container">$\pm$</span> symbol is already "built in" to the radical sign -- then they will include it explicitly when the square root can be evaluated exactly, and drop it when the square root cannot be completely evaluated.</p>
<p>The second issue is this:</p>
<blockquote>
<p>I've told them several time during lectures and exercise sessions. </p>
</blockquote>
<p>Maybe the problem is that you're <em>telling</em> them during exercise sessions. I don't know what the format of your exercise sessions is -- in particular I don't know whether <em>you're</em> presenting the solutions, or having <em>students</em> present solutions -- but if the students are the ones showing their work, and such a mistake occurs, you can stop and <em>ask</em> "Are you sure that's right?" In my experience this is even more effective when the answer is extremely simple. I have had Calculus students go from <span class="math-container">$x^2=25$</span> to <span class="math-container">$x=5$</span>; asking "Are you sure that's right?" typically catches them off-guard, precisely because <em>of course</em> they're sure that <span class="math-container">$5^2=25$</span>. If that doesn't give them enough of a hint, prod them "Is <span class="math-container">$5$</span> the <em>only</em> number whose square is <span class="math-container">$25$</span>?" However you get them there, the goal should be for <em>them</em> to tell <em>you</em> that there's a second, negative solution. If they figure it out themselves once, they will be much more likely to remember it than if they are told it fifty times.</p>
|
126,057 | <p>Hi,</p>
<p>Can one define a Fubini-Study metric/Kaehler metric on the projective space of an infinite dimensional Hilbert space, i.e. using the formula $\partial \bar{\partial} \log |Z|^2$?
This should be very well-known to the experts. Anyhow I don't have much experience with infinite dimension and worried that something may go wrong. I appreciate any comments or references.</p>
| alvarezpaiva | 21,123 | <p>Maybe it's easier to see that the definition extends if you <strong>don't</strong> use a formula:</p>
<p>The unit sphere inherits a Riemannian metric from the Hilbert space in the standard manner and since it is invariant under the circular symmetry $(e^{i\theta},x) \mapsto e^{i\theta} x$, it will
project down to a Riemannian metric on the complex projective space. </p>
|
307,822 | <p>Excuse my incorrect use of terminology, I hope my question is clear:</p>
<p>I am coding a Python module which tests whether a given number is a member of the Fibonacci series. No problem with that. Additionally, should a number not be a member of the series, I would like to test whether it is significantly close to its nearest Fibonacci neighbor. Here an increasing deviation margin of what is considered "close" is needed, along the lines of:</p>
<pre><code>deviation margin (x) increases as given number (n) increases
</code></pre>
<p>For my purposes 9 is significantly close to Fibonacci number 8 but 10 is not. 1600 is significantly close to Fibonacci number 1597 but 1610 is not, etc. So, the test for "significant closeness" is applying a deviation margin of 1 at lower numbers and an increasing deviation margin as the series increments up to infinity.</p>
<p>I figured a logical candidate for inclusion in the test would be Standard Deviation. So I have calculated a coarse margin as follows:</p>
<pre><code>margin = (StdDev / ( n + closestfibneighbor)) * StdDev
</code></pre>
<p>This does not give me good control over the margin rate of increase and I am sure there is a more appropriate function to express the growth in margin caused by increments in x. Please feel free to elaborate on the mathematics of this - I am seeking a general solution and not a Python-specific function.</p>
| Jyrki Lahtonen | 11,619 | <p>The Fibonacci numbers, the larger ones in particular, are very close to being given by the formula
$$
F_n\approx\frac1{\sqrt5} \left(\frac{1+\sqrt5}2\right)^n
$$
($F_n$ is always the closest integer to the r.h.s.).
In light of this I would also consider calculating, given input $x$, how far the ratio
$$
r(x):=\frac{\log(x\cdot\sqrt5)}{\log\big((1+\sqrt5)/2\big)}
$$
is from being an integer. In other words, I might use the number
$$
s(x)=\left|r(x)-round(r(x))\right|
$$
for measuring how far $x$ is from being a Fibonacci number.</p>
<p>The point is that a test based on Binet's formula should be faster than anything recursive. Of course, you need very precise floating point operations to be able to use that as a test, whether an integer $x$ actually is a Fibonacci number: compute $r(x)$, and
find whether
$$
x=round\left(\frac1{\sqrt5} \left(\frac{1+\sqrt5}2\right)^{round(r(x))}\right).
$$</p>
|
3,643,917 | <p>Find the limit of
<span class="math-container">$$\lim_{x\to 0+} \frac{1}{x^2} \int_{0}^{x} t^{1+t} dt$$</span>. </p>
<p>My idea to solve it is to use L'Hospital's rule but I am not sure why I can use it and how should i do it. Many thanks to them who are willing to help. </p>
| Jephph | 777,586 | <p>Let <span class="math-container">$L=\lim_{x \rightarrow 0^+}\frac{1}{x^2}\int_{0}^{x}t^{1+t}dt = \lim_{x \rightarrow 0^+}\frac{\int_{0}^{x}t^{1+t}dt}{x^2}$</span>.</p>
<p>Note that both the numerator and the denominator tend to <span class="math-container">$0$</span> as <span class="math-container">$x$</span> tends to <span class="math-container">$0$</span>. It follows by l'Hopital's Rule that</p>
<p><span class="math-container">$L=\lim_{x \rightarrow 0^+}\frac{\frac{d}{dx}\int_{0}^{x}t^{1+t}dt}{\frac{d}{dx}x^2} = \lim_{x \rightarrow 0^+} \frac{x^{1+x}}{2x}$</span>.</p>
<p>Here we have made use of the fundamental theorem of calculus in the numerator. Carrying on, </p>
<p><span class="math-container">$L=\frac{1}{2} \lim_{x \rightarrow 0^+} x^{1+x}x^{-1}=\frac{1}{2} \lim_{x \rightarrow 0^+} x^{x} = \frac{1}{2}$</span>.</p>
|
197,667 | <p>I have a table with 10 columns, obtained as import from file. I would like to use 4 columns as <code>{x, y, z, u}</code> params for <code>BubbleChart3d</code>, and an additional column to color the points. I'm unable to make this work. The code I have is like this:</p>
<p><code>t</code> is table from a file import. I am using columns 1, 2, 9, 10 for xyz-coordinates and size of the bubble, I an trying to use column 7 for color of the bubble</p>
<pre><code>BubbleChart3D[Map[Function[r, {r[[1]], r[[2]], r[[9]], r[[10]]}]][t],
ColorFunction -> Map[Function [c, RGBColor[1, 0, c[[7]]]]][t]]
</code></pre>
<p>This doesn't work — I get a bubble plot with same color. The 7th column does have differing values. What am I doing wrong?</p>
<p>Example of table t:</p>
<pre><code>{{1, -1, 1, 0, 15, 0, 82899, 177, 1, 0}, {1, -1, 1, 0, 15, 0, 10231991, 177, 1, 0},
{1, 0, 1, 0, 15, 0, 4633, 177, 1, 0}, {2, -1, 2, 0, 0, 0, 10231991, 204, 1, 1},
{2, 1, 2, 0, 0, 0, 0, 204, 0, 1}, {4, 3, 4, 4, 354, 1, 2, 13, 0, 2},
{4, -1, 4, 4, 354, 1, 10231991, 13, 1, 2}, {4, 0, 4, 4, 354, 1, 4633, 13, 1, 2},
{4, 4, 4,4, 354, 1, 0, 13, 0, 2}, {5, 5, 5, 0, 0, 0, 0, 64, 0, 2},
{5, -1, 5, 0, 0, 0, 82899, 64, 1, 2}, {5, -1, 5, 0, 0, 0, 10231991, 64, 1, 2},
{5, 0, 5, 0, 0, 0, 4633, 64, 1, 2}, {5, 6, 5, 0, 0, 0, 0, 64, 0,2},
{6, 7, 6, 2, 0, 0, 0, 519, 0, 3}, {6, 8, 6, 2, 0, 0, 0, 519, 0,3},
{6, -1, 6, 2, 0, 0, 10231991, 519, 1, 3}, {6, 9, 6, 2, 0, 0, 0, 519, 0, 3},
{6, -1, 6, 2, 0, 0, 82899, 519, 1, 3}, {7, -1, 7, 0, 0, 0, 10231991, 0, 1, 0}}
</code></pre>
| Anton.Sakovich | 59,438 | <p>Mathematica does not evaluate limits unless explicitly asked to do so. What it does is simply evaluate the numerator, then evaluate the denominator, and then evaluate the resulting fraction. The only subtlety is that "evaluate the resulting fraction" means a simple syntactic evaluation, without bothering of possible zeros and thus returning an expression which is not fully equivalent to the input expression. Mathematica also doesn't try to factor the numerator/denominator to check for possible cancellations.</p>
<p>If the resulting fraction turns out to be <code>1/0</code>, then it will evaluate to <code>ComplexInfinity</code>. If the resulting fraction is <code>0/0</code>, then it will evaluate to <code>Indeterminate</code>.</p>
<p>Your first example is evaluated as follows:</p>
<ol>
<li><p>Evaluate the numerator: <code>(x + 1) (x + 2)</code> => <code>(x + 1) (x + 2)</code>;</p></li>
<li><p>Evaluate the denominator: <code>(x + 2) (x + 3)</code> => <code>(x + 2) (x + 3)</code>;</p></li>
<li><p>Evaluate the fraction (Mathematica performs simple syntax cancellation of terms): <code>(x + 1) (x + 2) / ((x + 2) (x + 3))</code> => <code>(x + 1) / (x + 3)</code>;</p></li>
<li><p>Substitute <code>-2</code> for <code>x</code>: <code>(x + 1) / (x + 3)</code> => <code>-1 / -1</code> => <code>1</code>.</p></li>
</ol>
<p>For the example form the comments the evaluation chain is as follows:</p>
<ol>
<li><p>Evaluate the numerator: <code>(1 + 2 x + x^2)</code> => <code>(1 + 2 x + x^2)</code>;</p></li>
<li><p>Evaluate the denominator: <code>(x + 1)</code> => <code>(x + 1)</code>;</p></li>
<li><p>Evaluate the fraction: <code>(1 + 2 x + x^2) / (x + 1)</code> => <code>(1 + 2 x + x^2) / (x + 1)</code>;</p></li>
<li><p>Substitute <code>-1</code> for <code>x</code> : <code>(1 + 2 x + x^2) / (x + 1)</code> => <code>0 / 0</code> => <code>Indeterminate</code>.</p></li>
</ol>
|
1,574,290 | <p>How do I prove this? </p>
<p>For the Fibonacci numbers defined by $f_1=1$, $f_2=1$, and $f_n = f_{n-1} + f_{n-2}$ for $n ≥ 3$, prove that $f^2_{n+1} - f_{n+1}f_n - f^2_n = (-1)^n$ for all $n≥ 1$.</p>
| Nathan Marianovsky | 257,054 | <ul>
<li>Base Case:</li>
</ul>
<p>For $n = 1$:</p>
<p>$$f_2^2 - f_2f_1 - f_1^2 = (-1)^1$$</p>
<p>$$(1)^2 - (1)(1) - (1)^2 = -1$$</p>
<p>$$-1 = -1$$</p>
<hr>
<ul>
<li>Induction Step:</li>
</ul>
<p>Assume that the given statement is true. We now try to prove that it holds true for $n+1$:</p>
<p>$$f_{n+2}^2 - f_{n+2}f_{n+1} - f_{n+1}^2 = (-1)^{n+1}$$</p>
<p>Typically you choose one side and try to get to the other side. I will choose the left side:</p>
<p>$$f_{n+2}^2 - f_{n+2}f_{n+1} - f_{n+1}^2 = (f_{n+1} + f_{n})^2 - (f_{n+1} + f_{n})(f_{n+1}) - f_{n+1}^2$$</p>
<p>$$= f_{n+1}^2 + 2f_{n+1}f_{n} + f_{n}^2 - f_{n+1}^2 - f_{n+1}f_{n} - f_{n+1}^2$$</p>
<p>$$= f_{n}^2 + f_{n+1}f_{n} - f_{n+1}^2$$</p>
<p>$$= (-1)(f_{n+1}^2 - f_{n+1}f_{n} - f_{n}^2)$$</p>
<p>$$= (-1)(-1)^n$$</p>
<p>$$= (-1)^{n+1}$$</p>
|
2,423,086 | <p>Show, by listing the elements or how we can list the elements, that </p>
<p>$\mathbb{N}^3$ = $\mathbb{N×N×N}$ is a countable set.</p>
<p><strong>Attempt at a solution:</strong></p>
<p>I was thinking about making use of Cantor's diagonal argument. If this were $\mathbb{N×N}$ I could just do:</p>
<p>(1,1), (1,2), (1,3),...</p>
<p>(2,1), (2,2), (2,3)...</p>
<p>(3,1), (3,2), (3, 3)...</p>
<p>and snake my way through it.</p>
<p>However, I am having trouble finding a way to list $\mathbb{N×N×N}$ such that I can implement the diagonal argument.</p>
| Kenny Lau | 328,173 | <p>Let $S = \Bbb N$ and $T = \Bbb N \times \Bbb N \times \Bbb N$.</p>
<p>Then, $f:S \to T$ defined by $f(n) = (n,0,0)$ is an injection, and so is $g:T \to S$ defined by $g(n,m,k) = 2^n 3^m 5^k$.</p>
<p>Hence, there exists a bijection between $S$ and $T$ per <a href="https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem" rel="nofollow noreferrer">Schröder–Bernstein theorem</a>.</p>
|
2,423,086 | <p>Show, by listing the elements or how we can list the elements, that </p>
<p>$\mathbb{N}^3$ = $\mathbb{N×N×N}$ is a countable set.</p>
<p><strong>Attempt at a solution:</strong></p>
<p>I was thinking about making use of Cantor's diagonal argument. If this were $\mathbb{N×N}$ I could just do:</p>
<p>(1,1), (1,2), (1,3),...</p>
<p>(2,1), (2,2), (2,3)...</p>
<p>(3,1), (3,2), (3, 3)...</p>
<p>and snake my way through it.</p>
<p>However, I am having trouble finding a way to list $\mathbb{N×N×N}$ such that I can implement the diagonal argument.</p>
| David K | 139,123 | <p>If you have already found an explicit bijection $f:\mathbb N\to\mathbb N\times\mathbb N$
defined by
$f(n)\mapsto(g(n), h(n)),$
you can extend this to a bijection
$k:\mathbb N\to\mathbb N\times\mathbb N\times\mathbb N$
by setting
$$
k(n)=(g(n), g(h(n)), h(h(n))).
$$
You can recursively apply this technique as many times as desired to get a bijection to the cross product of any number of copies of $\mathbb N.$</p>
|
3,685,969 | <p>I have been investigating the brachistochrone problem with friction and in my derivations, I would like help solving the Euler-Lagrange equation below</p>
<p><span class="math-container">$\frac{d}{dx}\frac{\partial F}{\partial y'}=\frac{\partial F}{\partial y}$</span>
where <span class="math-container">$F=\sqrt{\frac{1+y'^2}{2g(y-\mu x)}}$</span></p>
<p>I can get up to</p>
<p><span class="math-container">$\frac{d}{dx}\frac{y'}{\sqrt{2g(y-\mu x)(1+y'^2)}}=-\sqrt{\frac{(1+y'^2)}{2g}}\frac{1}{2(y-\mu x)^\frac32}$</span> </p>
<p>But I am unsure how the equation above reduces into
<span class="math-container">$(1+y'^2)(1+\mu y')+2(y-\mu x)y''=0$</span>, as seen in equation (29) of <a href="https://mathworld.wolfram.com/BrachistochroneProblem.html" rel="nofollow noreferrer">this Wolfram page</a>.</p>
<p>I am quite new to calculus and would appreciate a step by step solution.
Thanks in advance!</p>
| Claude Leibovici | 82,404 | <p>I suppose that the problem with Mathematica came from the fact that you computed first the antiderivative
<span class="math-container">$$I(x)=\int \frac{\log \left(1-x^2\right)}{(1-\beta x)^2}\,dx$$</span> which gives
<span class="math-container">$$ I(x)=\frac1\beta\left(\frac{\log \left(1-x^2\right)}{1-\beta x}-\frac{2 \log (1-\beta x)}{\beta
^2-1}+\frac{\log (1-x)}{\beta -1}-\frac{\log (x+1)}{\beta +1}\right)$$</span> and that you tried to compute <span class="math-container">$I(1)$</span> and <span class="math-container">$I(-1)$</span> which are undefined.</p>
<p>Asking directly for the definite integral would have given the result.</p>
<p>Starting from the antiderivative, you could have computed <span class="math-container">$I(1-\epsilon)-I(-1+\epsilon)$</span> and developing the result as a Taylor series built around <span class="math-container">$\epsilon=0$</span> would give
<span class="math-container">$$\frac{4}{\beta(1 -\beta ^2) }\left(\beta \log (2)- \tanh ^{-1}(\beta )\right)-\frac{2 \left(\beta ^2+1\right)}{\left(\beta ^2-1\right)^2}\epsilon \log \left(\frac{2 \epsilon }{e}\right)+O\left(\epsilon ^2\right)$$</span> and then the limit
<span class="math-container">$$\int_{-1}^{+1} \frac{\log \left(1-x^2\right)}{(1-\beta x)^2}\,dx=\frac{4}{\beta(1 -\beta ^2) }\left(\beta \log (2)- \tanh ^{-1}(\beta )\right)$$</span></p>
|
1,809,871 | <p>I am trying to solve Poisson equation using FFT. The issue appears at wavenumber $k = 0$ when I want to get inverse Laplacian which means division by zero.</p>
<p>We have </p>
<p>${\nabla ^2}\phi = f$ </p>
<p>Taking FFT from both side we get:</p>
<p>$-k^2\hat\phi = \hat f $</p>
<p>or </p>
<p>$\hat\phi = \frac{\hat f}{-k^2} $</p>
<p>Assuming that we want to solve this equation in periodic domain and using DFT using FFTW package, at $k=0$ we have a division by zero. Anybody knows how to deal with this singularity?</p>
| mathreadler | 213,607 | <p>This is conceptually similar the <strong>integrating constants</strong> that show up when you are solving a differential equation by other methods. Usually these integrating constants are decided by your <strong>boundary conditions</strong>.</p>
<p>Instead of doing any division you can simply rewrite it as a least-norm problem: $$\|-k^2\hat \phi +\hat f\|_2^2+\text{any more terms you may want}$$</p>
<p>If your $\hat \phi,\hat f$ are stored in vectors the $k^2$ in the expression above will be a diagonal weight matrix which multiplies the $\hat \phi$ vector.</p>
<hr>
<p>Oops sorry I forgot where the FFTW comes into place, you can use it to calculate the matrix-vector product, $F$ below is the Fourier transform matrix.</p>
<p>$$\|-k^2 \hat \phi + F f\|_2^2$$</p>
<p>So $Ff$ means multiply $f$ vector by $F$ and it will be the output of calling your FFTW library with $f$ vector as the input.</p>
<p>And for other terms you have it might be $F^{-1}$ you want to multiply with but that is simply the corresponding IFFT routine in your library.</p>
|
1,489,078 | <p>Let $P(S)$ denotes the power set of set $S$. Which of the following is always true$?$</p>
<ol>
<li>$P(P(S)) = P(S)$</li>
<li>$P(S) ∩ P(P(S)) = \{ Ø \}$</li>
<li>$P(S) ∩ S = P(S)$</li>
<li>$S ∉ P(S)$</li>
</ol>
<hr>
<p>I try to explain $:$
If $S$ is the set $\{x, y, z \},$ then the subsets of S are:</p>
<p>${}$ (also denoted $\phi,$ the empty set)
$\{x\},
\{y\},
\{z\},
\{x, y \},
\{x, z \},
\{y, z \},
\{x, y, z \}$
and hence the power set of S is i.e.,</p>
<p>$P(S) =$ $\{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\}$.</p>
<p>Similarly , </p>
<p>$P(P(S))=\{\{\{\}\}, \{\{x\}\}, \{\{y\}\}, \{\{z\}\}, \{\{x, y\}\}, \{\{x, z\}\}, \{\{y, z\}\}, \{\{x, y, z\}\}\}.....\}$</p>
<p>therefore , </p>
<p>$P(S) ∩ P(P(S)) = \{ Ø \}$</p>
<p>Note that $\{ Ø \}$ is always element of powerset of a set , and also $\{ Ø \}$ is the subset of a set , in other words all subset of a set is a powerset .</p>
<hr>
<blockquote>
<p>My question is $:$ both $\{\}$ and $\{\{\}\}$ same element ?</p>
</blockquote>
| wythagoras | 236,048 | <p>No, <span class="math-container">$\{\}$</span> and <span class="math-container">$\{\{\}\}$</span> are not the same. Think of sets as bags. <span class="math-container">$\{\}$</span> is an empty bag. <span class="math-container">$\{\{\}\}$</span> is a bag with an empty bag within it. Therefore the outermost bag is not empty: It has a bag inside it.</p>
<hr />
<p>For the original question, you are wrong, lets consider all four possibilities.</p>
<ol>
<li><p>No. By Cantor's theorem, we always have <span class="math-container">$|P(P(S))| > |P(S)|$</span>. Hence they cannot be equal. Alternatively, let <span class="math-container">$S=\{\}$</span> and write out.</p>
</li>
<li><p>Consider <span class="math-container">$S=\{\{\}\}$</span>. Then <span class="math-container">$P( S)=\{\{\},\{\{\}\}\}$</span> and <span class="math-container">$P(P(S))=\{\{\},\{\{\}\},\{\{\{\}\}\},\{\{\},\{\{\}\}\} \}$</span>, hence <span class="math-container">$P(S) \cap P(P(S)) = \{\{\},\{\{\}\}\}$</span>.</p>
</li>
<li><p>No. Note that <span class="math-container">$A \cap B \subseteq B$</span> for all sets <span class="math-container">$A,B$</span>. Hence <span class="math-container">$S \cap P(S) \subseteq S$</span>, hence <span class="math-container">$$|S \cap P(S)| \leq |S| < |P(S)|$$</span></p>
<p>hence they are not equal.</p>
</li>
<li><p>By definition <span class="math-container">$S \subseteq S$</span> and thus <span class="math-container">$S \in P(S)$</span>.</p>
</li>
</ol>
<p>Therefore, none of the statements is true.</p>
|
1,123,247 | <p>let $a_n$ be a sequence of real numbers such that the series $\sum |a_n|^2$ is convergent.Find range of $p$ such that the series $\sum |a_n|^p$ is convergent.</p>
<p>My try:</p>
<p>To show the series it is necessary to show that sequence of partial sums of $\sum |a_n|^p$ is bounded.</p>
<p>Consider $S_n=|a_1|^p+|a_2|^p+...+|a_n|^p$. How to proceed next? Any inequality which can be used?</p>
| guest | 125,454 | <p>If $p\geq2$ then $\sum|a_n|^p$ converges since $$\lim\limits_{n\to\infty}\frac{|a_n|^p}{|a_n|^2}=\lim\limits_{n\to\infty}|a_n|^{p-2}=0$$ where p>2. On the other hand if $0<p<2$ then there exists a sequcence such that $\sum|a_n|^2$ is convergent but $\sum|a_n|^p$ is divergent. Formally, for all $p\in(0,2)$ there exists $0<\varepsilon<\frac{2}{p}-1$. then if we choose the sequence $a_n=\frac{1}{n^{\frac{1+\varepsilon}{2}}}$ then $\sum|a_n|^2$ converges but $\sum|a_n|^p$ does not converge.</p>
|
1,405,809 | <p>So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L'Hopital's rule.<br>
I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.</p>
| GeorgSaliba | 142,772 | <p>$$x=\log_a(t)$$</p>
<p>$$\lim_{t\rightarrow 1}\frac{t-1}{\log_at}=\lim_{t\rightarrow 1}\frac{(t-1)\log(a)}{\log(t)}=\log(a)$$
EDIT:</p>
<p>$$L=\lim_{x\rightarrow 1}\frac{\log(x)}{x-1}=\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}=\lim_{x\rightarrow 0}\int_0^x \frac 1x\frac{dt}{1+t}$$
Put: $t=xs$ then:
$$L=\lim_{x\rightarrow 0}\int_0^1 \frac{ds}{1+xs}=\int_0^1ds=1$$</p>
<p>Also:
$$L=\lim_{x\rightarrow 0}\frac{\log(1+x)-\log(1)}{(x+1)-1}=(\log(x))'_{x=1}=1$$</p>
|
1,405,809 | <p>So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L'Hopital's rule.<br>
I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.</p>
| Paramanand Singh | 72,031 | <p>In order to evaluate the limit $$\lim_{x \to 0}\frac{a^{x} - 1}{x}\tag{1}$$ for $a > 0$, it is necessary to have a proper definition of $a^{x}$. This has been pointed out in one of the comments to the question, but OP has perhaps not understood the necessity of such a comment. In case you want to evaluate the limit $$\lim_{x \to 1}\frac{\sqrt{x} - 1}{x - 1}$$ it is necessary to have a definition of the symbol $\sqrt{x}$ and it is easy to define it by saying that <em>$\sqrt{x}$ is a non-negative real number which gives $x$ when squared</em>.</p>
<p>The unfortunate part here is that there is no definition of $a^{x}$ which is as simple as the definition of $\sqrt{x}$ and most introductory calculus texts try their best not to define this symbol in proper manner. The <em>real sin</em> is committed when they are able to convince the students that there is no necessity of such a definition. It is one thing to avoid a complicated definition but totally another matter to convince that such a thing does not exist.</p>
<p>There are multiple approaches to define $a^{x}$ and almost all of them are hard enough for a beginner in calculus (see <a href="http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-1.html" rel="nofollow">these</a> <a href="http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-2_10.html" rel="nofollow">posts</a> <a href="http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-3.html" rel="nofollow">here</a> in case you are interested).</p>
<hr>
<p>I prefer the following approach which is suitable for a beginner in calculus:</p>
<p><strong>Theorem</strong>: <em>For a given real number $a > 0$ there exists a unique function $F(x)$ defined and continuous for all real $x$ such that $F(0) = 1, F(1) = a$ and $F(x + y) = F(x)F(y)$ for all $x, y$. Moreover if $x$ is a rational number (say equal to $r \in \mathbb{Q}$) then $F(x) = F(r) = a^{r}$</em>.</p>
<p>This function is called the general exponential function and denoted by symbol $a^{x}$. And then we come to the limit in question:</p>
<p><strong>Theorem</strong>: <em>For any real number $a > 0$ the limit $$\lim_{x \to 0}\frac{a^{x} - 1}{x}$$ exists and hence defines a function of $a$ (say $L(a)$) and this function $L(a)$ satisfies the following properties: $$L(1) = 0, L(xy) = L(x) + L(y)$$ Further $L(x)$ is continuous for all $x > 0$ and $$\lim_{x \to 0}\frac{L(1 + x)}{x} = 1\tag{2}$$</em> The function $L$ is traditionally denoted by $\log x$ and is called the <em>natural logarithm</em> (or just the logarithm) of $x$.</p>
<hr>
<p>Essentially the above approach gives two standard limits $$\lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a,\,\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ and defers their proof (to advanced courses). But it is more honest than supplying <em>vague ideas</em> and <em>intuition</em> about exponential and logarithmic functions which I consider more of a <em><a href="https://en.wikipedia.org/wiki/Handwaving" rel="nofollow">hand waving</a></em>.</p>
|
1,405,809 | <p>So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L'Hopital's rule.<br>
I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.</p>
| aman | 264,003 | <p>Convert it into standard limit as write $a^x$ as $e^{x\ln a}$ and
$\lim_{x \to 0} \dfrac{e^{x-1}}{x}=1$</p>
|
2,140,965 | <p>Arithmetic progression
:$a_n=\{a,a+d,a+2d,...,a+nd,...\}$</p>
<p>Geometric progression
:$b_n=\{b,bq,bq^2,...,bq^n,...\}$</p>
<p>if : $a_r,a_s,a_t,$ be a Geometric progression($b_n$)</p>
<p>then : What is q?</p>
<p>my way :
let :$r>s>t$</p>
$$ a_{r} =a +(r-1)d $$
$$ a_{s} =a +(s-1)d $$
$$ a_{t} =a +(t-1)d $$
<p>$$ a_{s}^{2} =a_{r} a_{t} \Rightarrow a^{2} +2ad(s-1)+ d^{2}(s-1)^{2} =a^{2}+ad(r+t-2)+d^{2}(r-1)(t-1)$$</p>
$$ 2a(s-1)+ d(s-1)^{2} =a(r+t-2)+d(r-1)(t-1)=$$
$$ 2as-ar-at=2sd-dt-dr-d s^{2} +drt \Rightarrow $$
$$a= \frac{d(2s-t-r-s^{2} +rt)}{2s-r-t} $$
$$ \frac{a_{r}}{a_{t}} = \frac{a +(r-1)d}{a +(t-1)d} $$
$$ \frac{d(2s-t-r-s^{2} +rt)}{2s-r-t}+ (r-1)d= $$
$$d \frac{2s-t-r-s^{2} +rt+2rs- r^{2}-rt-2s+r+t }{2s-r-t} = $$
$$ d \frac{-s^{2} -d \frac{ (r-s)^{2}}{2s-r-t} +2rs- r^{2}}{2s-r-t}=-d \frac{ (r-s)^{2}}{2s-r-t} $$
<p>$$ -d \frac{ (s-t)^{2}}{2s-r-t} $$</p>
$$ q^{2}=\frac{a_{r}}{a_{t}}= \frac{(r-s)^{2}}{(s-t)^{2}} $$
<p>There is simpler method?</p>
| ajotatxe | 132,456 | <p>Let $\mathbf w'$ be a vector orthogonal to $\mathbf w$, for example $\mathbf w'=(2,-1)$.</p>
<p>Find coefficients $\alpha$ and $\beta$ such that $\mathbf v=\alpha\mathbf w+\beta\mathbf w'$. The vector $\alpha\mathbf w$ is the projection.</p>
<p>The second question looks the same, but note that $\mathbf v$ and $\mathbf w$ are linealry dependent.</p>
|
411,549 | <p>Here is an modular equation</p>
<p>$$5x \equiv 6 \bmod 4$$</p>
<p>And I can solve it, $x = 2$.</p>
<p>But what if each side of the above equation times <strong>8</strong>, which looks like this</p>
<p>$$40x \equiv 48 \bmod 4$$</p>
<p>Apparently now, $x = 0$. Why is that? Am I not solving the modular equation in a right way, or should I divide both side with their greatest-common-divisor before solving it?</p>
<p>P.S.</p>
<p>To clarify, I was solving a system of modular equations, using <strong>Gaussian Elimination</strong>, and after applying the elimination on the coefficient matrix, the last row of the echelon-form matrix is :</p>
<p>$$0, \dots, 40 | 48$$</p>
<p>but I think each row in the echelon-form should have been divided by its greatest common divisor, that turns it into :</p>
<p>$$0, \dots, 5 | 6$$</p>
<p>But apparently they result into different solution, one is $x = 0,1,2,3....$, the other $x = 2$. And why? Am I applying <strong>Gaussian-Elimination</strong> wrong?</p>
| Alex Wertheim | 73,817 | <p>Modular equivalence classes are multiplicative. Hence, since $40 \equiv 0 (\text{mod } 4)$ and $48 \equiv 0 (\text{mod } 4)$, all you've written there is $0\cdot x \equiv 0 (\text{mod } 4)$, which is true for any $x \in \mathbb{Z}/4\mathbb{Z}$ (the set of equivalence classes mod $4$). $x = 0$ is not the only "solution", but that's because as written the equation is effectively tautological - there is nothing to "solve" for. </p>
|
1,818,557 | <p>I will be teaching some "topology" to high school students. I was wondering how to explain to such a school student that on a sphere the shortest path between 2 points is given by a great circle?</p>
<p>Also, how to explain that if they lived on a sphere they would have no notion of "above" or "below"? I cannot find a nice way to convince them since they see the sphere embedded on 3d? </p>
| Mikhail Katz | 72,694 | <p>I have found it helpful to replace the sphere by an apple and introduce an "internal" observer by placing an ant on the apple. The ant will crawl from point $A$ to point $B$ on the sphere by following the shortest path (the queen can't wait) which is always an arc of great circle.</p>
<p>An additional point that students find illuminating is the phenomenon that a plane on a direct flight from New York to Paris will veer rather far North instead of following the same latitude throughout the flight. This is of course also because the latitude is not a minimizing path (except for the equator).</p>
|
4,358,080 | <p>I'm making a pump and I need to make a circular (from the front perspective) hole in a side of a pipe. I can't use a drill and I have to print out a shape that I will stick onto it and cut and file away. Will this circle projected off center onto a cylinder be an ellipse, or is it not an exact ellipse and I have to use a different shape?</p>
<p>To further clarify, it will look something like this picture if you imagine the orange pipe is a boring bit and is not reaching beyond the centerline of the green pipe.
<img src="https://i.stack.imgur.com/eVA8G.jpg" alt="enter image description here" /></p>
| Jean Marie | 305,862 | <p>A) Let us consider first the case of intersecting axes (even, as Jean-Claude Arbaut has pointed to, they aren't intersecting in the question). In fact, the equation of the unwrapped intersection curve is a general oval curve (see equation (*)) and even more complicated curves with equation (**) which are not ellipses.</p>
<p><a href="https://i.stack.imgur.com/WPZrj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WPZrj.jpg" alt="enter image description here" /></a></p>
<p>It is possible to play on the <span class="math-container">$r$</span> slider (and also the <span class="math-container">$s$</span> slider, see below) on the Geogebra figure:</p>
<p><a href="https://www.geogebra.org/calculator/eugrnnec" rel="nofollow noreferrer">https://www.geogebra.org/calculator/eugrnnec</a></p>
<p>(Look in particular how close to a circle is the curve when <span class="math-container">$r$</span> is small).</p>
<p>Its equation is (assuming the radius of the initial cylinder is <span class="math-container">$1$</span> and the boring cylinder has radius <span class="math-container">$r$</span> with the <span class="math-container">$x$</span> axis taken as its horizontal axis):</p>
<p><span class="math-container">$$z=f_r(t)=r \sin(\cos^{-1}(\frac1r \sin(t)))\tag{*}$$</span></p>
<p>in a <span class="math-container">$(t,z)$</span> coordinate system .</p>
<p>Explanation:</p>
<p><span class="math-container">$$\begin{cases}x&=&\cos(t)& \ Eq. 1a\\y&=&\sin(t)& \ Eq. 1b\end{cases}\tag{1}$$</span> and <span class="math-container">$$\begin{cases}y&=&r \cos(u)& \ Eq. 2a\\z&=&r\sin(u)& \ Eq. 2b\end{cases}\tag{2}$$</span></p>
<p>Indeed, knowing that unrolling the big cylinder is like taking angle <span class="math-container">$t$</span> as the new abscissa, it suffices to be able to express height <span class="math-container">$z$</span> as a function of <span class="math-container">$t$</span> as given by (*). This will be done in two steps, starting from the equality of Eq. 1b and Eq. 2b :</p>
<p><span class="math-container">$$r\cos u= \sin t \implies u= \cos^{-1}(\frac1r \sin t),$$</span></p>
<p>and then plugging this expression into Eq. 2b.</p>
<p>B) Now, the non interesting case is a little more complicated because (2) has to be replaced by</p>
<p><span class="math-container">$$\begin{cases}y&=&r \cos(u)+s& \ Eq. 2a\\z&=&r\sin(u)& \ Eq. 2b\end{cases}\tag{2}$$</span></p>
<p>where <span class="math-container">$s$</span> is the shift amount, finally giving:</p>
<p><span class="math-container">$$z=f_{r,s}(t)=r \sin(\cos^{-1}(\frac1r (\sin(t)-s)))\tag{**}$$</span></p>
<p>Remark: Playing with the <span class="math-container">$s$</span> slider, you will see the interesting case of lemniscates (looking like an <span class="math-container">$\infty$</span> sign).</p>
|
702,506 | <p>We have an exam in $3$ hours and I need help how to solve such trigonometric equations for intervals.</p>
<p>How to solve</p>
<p>$$\sin x - \cos x = -1$$</p>
<p>for the interval $(0, 2\pi)$.</p>
| Shuchang | 91,982 | <p><strong>HINT:</strong> Square both sides and you obtain $\sin 2x=0$ and then...</p>
|
702,506 | <p>We have an exam in $3$ hours and I need help how to solve such trigonometric equations for intervals.</p>
<p>How to solve</p>
<p>$$\sin x - \cos x = -1$$</p>
<p>for the interval $(0, 2\pi)$.</p>
| MPW | 113,214 | <p>Use the fact that $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ to rewrite the equation as (after simplifying)
$$e^{2ix} -(1-i)e^{ix}-(1+i)=0$$
Viewing this as a quadratic equation in $e^{ix}$, use the quadratic formula to get
$$e^{ix}= \frac{(1-i)\pm(1+i)}{2}$$
$$e^{ix} = 1 \text{ or } e^{ix} = -i$$
Take logs to get all solutions
$$x= 2\pi k \text{ or } x=\frac{3\pi}{2}+ 2\pi k$$
for $k\in\mathbb Z$. Restricting to $(0,2\pi)$ as required in the problem statement, we get only the solution
$$\boxed{x=\dfrac{3\pi}{2}}.$$</p>
|
716,767 | <p>I feel like an idiot for asking this but i can't get my formula to work with negative numbers</p>
<p>assume you want to know the percentage of an increase/decrease between numbers</p>
<pre><code>2.39 1.79 =100-(1.79/2.39*100)=> which is 25.1% decrease
</code></pre>
<p>but how would i change this formula when there are some negative numbers?</p>
<pre><code>6.11 -3.73 =100-(-3.73/6.11*100) which is 161% but should be -161%
</code></pre>
<p>the negative sign is lost.. what I am missing here?</p>
<p>also</p>
<pre><code>-2.1 0.6 =100-(-3.73/6.11*100) which is 128.6% ??? is it?
</code></pre>
| amWhy | 9,003 | <p>Perhaps this "formula" will be easier to understand (this formula is equivalent to your formula - each can be derived from the other):</p>
<p>$$\dfrac{\text{original value} \;- \;\text{final value}}{\text{original value}} \times 100\% = \text{percent change}$$</p>
<p>That change will be </p>
<ul>
<li><p>an <em>increase</em> if the original value is <em>less than</em> the final value,</p></li>
<li><p>a <em>decrease</em> if the original value is <em>greater than</em> the final value.</p></li>
</ul>
<hr>
<p>Original value $6.11$, final value $-3.73$:</p>
<p>$$\dfrac{6.11 -(-3.73)}{6.11}\times 100\% \approx 161\% \;\;\text{DECREASE}$$</p>
<hr>
<p>Original value $-2.1$, final value $0.6$:</p>
<p>$$\dfrac{-2.1 - 0.6}{-2.1}\times 100\% \approx 128.6\% \;\;\text{INCREASE}$$</p>
|
2,317,406 | <p>How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of $N$ is a perfect square?
Any help would be much appreciated.</p>
| Atul Mishra | 396,163 | <p>See the two numbers whose sum you want</p>
<p>That is $10a+b$ and $10b+a$ </p>
<p>Their sum is $11(a+b)$</p>
<p>Since $11$ is a prime ,, so $(a+b)$ must equal to $11$ for being a perfect square</p>
<p>So the numbers whose sum of digits is $11$ will satisfy your condition</p>
<p>Numbers are $29,92, 38,83,47,74,56,65$</p>
|
4,107,656 | <p>Let <span class="math-container">$G$</span> be a finite abelian group with a neutral element <span class="math-container">$e$</span>. Prove that for any element <span class="math-container">$g$</span> of <span class="math-container">$G$</span>: <span class="math-container">$g^{|G|}=e$</span>. <span class="math-container">$|G|$</span> shall be the number of elements within <span class="math-container">$G$</span>.</p>
<p>Use and show that the multiplication with <span class="math-container">$g$</span> defines a bijection on <span class="math-container">$G$</span>. Therefore: <span class="math-container">$\prod_{h\in G}h=\prod_{h\in G}gh$</span> applies!</p>
<p>I do not have any Idea on how to solve it, as I do not understand the idea behind all this. I found out that it has to do with the Lagrange Theorem but we did not learn that in class. Every single hint is welcome!</p>
<p>Thank you very much!</p>
| Vercingetorix | 848,746 | <p>So first we show that multiplication with <span class="math-container">$g$</span> is a bijection, that is, the map <span class="math-container">$h \mapsto gh$</span> is a bijection. We leverage basic properties of groups here:</p>
<p>First suppose there is <span class="math-container">$h_1, h_2$</span> such that <span class="math-container">$gh_1 = gh_2$</span>. But then multiply both sides on the left with <span class="math-container">$g^{-1}$</span> to get <span class="math-container">$h_1 = h_2$</span>. This gives us injectivity.</p>
<p>Next pick any <span class="math-container">$h \in G$</span>. Notice then that by applying the function on <span class="math-container">$g^{-1}h$</span> we get that <span class="math-container">$gg^{-1}h = h$</span> as desired. This gives us surjectivity.</p>
<p>So multiplication with <span class="math-container">$g$</span> is in fact a bijection.</p>
<p>So why does this mean <span class="math-container">$\prod h = \prod gh$</span>? Well simply because the factors in each product line up: Suppose <span class="math-container">$\prod h = h_1 \times h_2 \times \cdots \times h_{|G|}$</span>. Then for each <span class="math-container">$h_i$</span>, since <span class="math-container">$h \mapsto gh$</span> is a bijection, there is one and exactly one <span class="math-container">$h_j$</span> such that <span class="math-container">$h_i = gh_j$</span>. Call this <span class="math-container">$h_j$</span> as <span class="math-container">$h_i'$</span> instead. So by doing this <span class="math-container">$|G|$</span> many times, we can rewrite <span class="math-container">$\prod h = h_1 \times \cdots \times h_{|G|} = gh_1' \times \cdots \times gh_{|G|}' = \prod gh$</span></p>
<p>Thus, <span class="math-container">$\prod h = \prod gh = g^{|G|}\prod h$</span> which implies <span class="math-container">$g^{|G|} = e$</span>.</p>
|
4,473,264 | <p>I have part of a circle described by three two dimensional vectors.</p>
<ul>
<li>start point <code>s1</code></li>
<li>center point <code>c1</code></li>
<li>end point <code>e</code></li>
</ul>
<p>I move the start point <code>s1</code> by <code>m1</code>, which is a <strong>known</strong> two dimensional vector. My question is: Can I calculate the new center point <code>c2</code> from the data I have? And if so, how?</p>
<p>Problem</p>
<p><a href="https://i.stack.imgur.com/3wWOr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3wWOr.jpg" alt="enter image description here" /></a></p>
<p>I'm creating a svg-manuipulation-app (drawing-app) in javascript where I want to edit one point of an arc, but keep the shape of the arc intact by appropriately moving the center of the arc.</p>
<p>It only looks like I want to keep the <code>x</code> value the same. Small coincidence I didn't realised. The question should cover any vector <code>m1</code>, no matter where the new center <code>c2</code> would end up.</p>
| Daniel R. Collins | 266,243 | <p>To answer the question in the title (completely separate from the question of elementary-function antiderivatives):</p>
<blockquote>
<p>Does every function have an integral?</p>
</blockquote>
<p>No, there are functions that do not have any defined integral at all. But the exact category of such functions depends on which definition of integral you're using: Riemann, Lebesgue, or less commonly, <a href="https://en.wikipedia.org/wiki/Integral#Other_integrals" rel="nofollow noreferrer">something else</a>.</p>
<p>For example, the characteristic function of the rational numbers (function with value 1 at a rational number, 0 elsewhere) is not Riemann integrable, but it is Lebesgue integrable. On the other hand, the <a href="https://en.wikipedia.org/wiki/Sinc_function" rel="nofollow noreferrer">sinc function</a> taken over the entire real number line is not Lebesgue integrable.</p>
<p>If you're interested in pursuing that more, consider reading the Wikipedia article on the <a href="https://en.wikipedia.org/wiki/Lebesgue_integration" rel="nofollow noreferrer">Lebesgue integral</a>, particularly sections on limitations of the various integral definitions.</p>
|
2,783,129 | <p>How many different value of x from 0° to 180° for the equation $(2\sin x-1)(\cos x+1) = 0$?</p>
<p>The solution shows that one of these is true:</p>
<p>$\sin x = \frac12$ and thus $x = 30^\circ$ or $120^\circ$ </p>
<p>$\cos x = -1$ and thus $x = 180^\circ$</p>
<p><strong>Question:</strong> Inserting the $\arcsin$ of $1/2$ will yield to $30°$, how do I get $120^\circ$? and what is that $120^\circ$, why is there $2$ value but when you substitute $\frac12$ as $x$, you'll only get $1$ value which is the $30^\circ$?</p>
<p>Also, when I do it inversely: $\sin(30^\circ)$ will result to 1/2 which is true as $\arcsin$ of $1/2$ is $30^\circ$. But when you do $\sin(120^\circ)$, it will be $\frac{\sqrt{3}}{2}$, and when you calculate the $\arcsin$ of $\frac{\sqrt{3}}{2}$, it will result to $60^\circ$ and not $120^\circ$. Why?</p>
| Meow Mix | 193,986 | <p>The function $\arcsin$ returns the angle <em>within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$</em>. In fact, there are infinitely many angles whose sine is $1/2$. Since a function can't spit out more than one number however, we define the $\arcsin$ function to return the inverse of the sine on that interval.</p>
<p>Anyways, the true answer here is actually $30$ degrees and $150$ degrees (with also $180$ from the cosine expression). How would you find this? Just look at our unit circle.</p>
<p><a href="https://www.mathsisfun.com/geometry/images/circle-unit-radians.gif" rel="nofollow noreferrer"><img src="https://www.mathsisfun.com/geometry/images/circle-unit-radians.gif" alt="enter image description here"></a></p>
<p>The part we're restricted to is the top half of the circle going counterclockwise (0 to 180). What values of $x$ spit out a y-coordinate (sine) of $1/2$? Well, $\pi/6$ and $5\pi/6$ (or 30 and 150). There's your answer!</p>
|
1,870,305 | <p>I am trying to calculate the Jacobian of a function that has quaternions and 3D points in it. I refer to quaternions as <span class="math-container">$q$</span> and 3D points as <span class="math-container">$p$</span>
<span class="math-container">$$h_1(q)=A C(q)p $$</span>
<span class="math-container">$$h_2(q)=q_1\otimes q \otimes q_2 $$</span></p>
<p>where <span class="math-container">$A\in R^{3x3}$</span> and <span class="math-container">$C(q)$</span> is <em>Direction cosine matrix</em>.</p>
<p>I am using the Hamilton form for the quaternions.</p>
<p>I would like to calculate the following Jacobians:
<span class="math-container">$$H_1 = \frac{\partial h_1(q)}{\partial q} $$</span>
<span class="math-container">$$H_2 = \frac{\partial h_2(q)}{\partial q} $$</span></p>
<p>Following <a href="http://www.iri.upc.edu/people/jsola/JoanSola/objectes/notes/kinematics.pdf" rel="nofollow noreferrer">Joan Solà's reference</a> eq. 18 what I have is </p>
<p><span class="math-container">$$H_1 = A^TC(q)^T[p]_x $$</span>
<span class="math-container">$$H_2 = [q_1]_L[q_2]_R $$</span></p>
<p>Where <span class="math-container">$[q]_R$</span> and <span class="math-container">$[q]_L$</span> are the right and left handed conversion of quaternion to matrix form as defined in <a href="http://www.iri.upc.edu/people/jsola/JoanSola/objectes/notes/kinematics.pdf" rel="nofollow noreferrer">Joan Solà's reference</a> eq. 18.</p>
<p>All rotations are body centric.</p>
<p>Is this correct?
Is there a better way to do this?
Can the expression be easily simplified?</p>
| cactus314 | 4,997 | <p>Eq (163) in your <a href="http://www.iri.upc.edu/people/jsola/JoanSola/objectes/notes/kinematics.pdf" rel="nofollow">notes</a> has the really amazing equation:</p>
<p>$$ \frac{\partial(\mathbf{q} \otimes \mathbf{a} \otimes \mathbf{q\ast})}{\partial \mathbf{q}} =
\frac{\partial(\mathbf{R} \mathbf{a})}{\partial \mathbf{q}} =
2 \big[
w \mathbf{a} +
\mathbf{v} \times \mathbf{a} +
\mathbf{v}^\top \mathbf{a} \mathbf{I} +
\mathbf{v}\mathbf{a}^\top - \mathbf{a}\mathbf{v}^\top - w[\mathbf{a}]_\times \big] $$</p>
<p>These engineering symbols for quaternions are quite fascinating. I have not digested all the symbols. In math we simply note that $x \mapsto q x q^{-1}$ (which might be called $q\ast$) is a Rotation.</p>
|
1,870,305 | <p>I am trying to calculate the Jacobian of a function that has quaternions and 3D points in it. I refer to quaternions as <span class="math-container">$q$</span> and 3D points as <span class="math-container">$p$</span>
<span class="math-container">$$h_1(q)=A C(q)p $$</span>
<span class="math-container">$$h_2(q)=q_1\otimes q \otimes q_2 $$</span></p>
<p>where <span class="math-container">$A\in R^{3x3}$</span> and <span class="math-container">$C(q)$</span> is <em>Direction cosine matrix</em>.</p>
<p>I am using the Hamilton form for the quaternions.</p>
<p>I would like to calculate the following Jacobians:
<span class="math-container">$$H_1 = \frac{\partial h_1(q)}{\partial q} $$</span>
<span class="math-container">$$H_2 = \frac{\partial h_2(q)}{\partial q} $$</span></p>
<p>Following <a href="http://www.iri.upc.edu/people/jsola/JoanSola/objectes/notes/kinematics.pdf" rel="nofollow noreferrer">Joan Solà's reference</a> eq. 18 what I have is </p>
<p><span class="math-container">$$H_1 = A^TC(q)^T[p]_x $$</span>
<span class="math-container">$$H_2 = [q_1]_L[q_2]_R $$</span></p>
<p>Where <span class="math-container">$[q]_R$</span> and <span class="math-container">$[q]_L$</span> are the right and left handed conversion of quaternion to matrix form as defined in <a href="http://www.iri.upc.edu/people/jsola/JoanSola/objectes/notes/kinematics.pdf" rel="nofollow noreferrer">Joan Solà's reference</a> eq. 18.</p>
<p>All rotations are body centric.</p>
<p>Is this correct?
Is there a better way to do this?
Can the expression be easily simplified?</p>
| Joan Solà | 229,373 | <p>There are different ways to answer your question, but you probably want one of these two: </p>
<ol>
<li><p>You want the derivative with respect to the 4 components of the quaternion q=w+ix+iy+iz, that is, with respect to a 4 vector <span class="math-container">$v_q=(w,x,y,z)^\top\in R^4$</span>.
We have the derivative of the rotation wrt this vector q as:
<span class="math-container">$$
\frac{\partial q\otimes p \otimes q^*}{\partial v_q} = 2[wp+v×p , v^\top p I + vp^\top−pv^\top−w[p]_\times] \in R^{3\times4}
$$</span>
where:</p>
<ul>
<li><span class="math-container">$I$</span> is the 3x3 identity matrix.</li>
<li><span class="math-container">$[p]_\times$</span> is the skew symmetric matrix fromed from <span class="math-container">$p$</span>.</li>
<li><span class="math-container">$\times$</span> is the cross product</li>
<li><span class="math-container">$\otimes$</span> is the quaternion product. This sign can be omitted since it's a product and write simply <span class="math-container">$qpq^*$</span>.</li>
</ul>
<p>So assuming A is a constant matrix, and knowing that <span class="math-container">$C(q)p = qpq^*$</span>, your derivative is
<span class="math-container">$$
\frac{\partial (AC(q)p)}{\partial v_q} = 2A[wp+v×p , v^\top p I + vp^\top−pv^\top−w[p]_\times] \in R^{3\times4}
$$</span>
This is the derivative that e.g. Ceres is going to compute should you use automatic differentiation of your function using <code>#include Jet.h</code>.</p></li>
<li><p>You want the derivative with respect to the rotation itself seen as a 3-vector of the Lie algebra of the rotation group. The Lie-theory defines two Jacobians, left and right, for this, depending on whether you perturb the rotation on the right, <span class="math-container">$\tilde R=R\exp([\theta]_\times)$</span>, or on the left, <span class="math-container">$\tilde R=\exp([\theta]_\times)R$</span>. </p>
<p>The right Jacobian of the rotation is:
<span class="math-container">$$
\frac{\partial C(q)p}{\partial C} = -C(q)[p]_\times \in R^{3\times 3}
$$</span>
and so your full Jacobian is
<span class="math-container">$$
\frac{\partial (AC(q)p)}{\partial C} = -AC(q)[p]_\times \in R^{3\times 3}
$$</span></p></li>
<li><p>The left Jacobian of the rotation is
<span class="math-container">$$
\frac{\partial C(q)p}{\partial C} = -[C(q)p]_\times \in R^{3\times 3}
$$</span>
and so your full Jacobian is
<span class="math-container">$$
\frac{\partial (AC(q)p)}{\partial C} = -A[C(q)p]_\times \in R^{3\times 3}
$$</span></p></li>
</ol>
|
1,217,771 | <p>Let $x,y \in R$.
If $0 \leq y < x$ for all $x > 0$, then $y=0$.</p>
<p>Proof by contradiction: </p>
<p>Assume the opposite that is; "If $0 \leq y < x$ for all $x > 0$, then $y\neq0$".
Subtract $x$ from each part of the inequality to get,
$0-x \leq y-x < 0$
Then multiply through by -1 to get,
$x \geq y+ x > 0$
Since $x>0$, this implies a contradiction of the original statement, therefore we conclude that if $0 < y < x$ for all $x > 0$, then $y=0$.
Is my reasoning correct or is there something I can improve upon?</p>
| user21820 | 21,820 | <p><strong>Useful inequality chain</strong></p>
<p>Given $a \ge 0$ and $b > -\frac{1}{2}a$:</p>
<p>  $a+b-2\frac{b^2}{a} \le a+b-\frac{b^2}{a+b} = \frac{a^2+2ab}{a+b} \le \sqrt{a^2+2ab} \le a+b-\frac{1}{2}\frac{b^2}{a+b} \le a+b$</p>
<p><strong>Solution</strong></p>
<p>Given $x \ge 1$:</p>
<p>  $\sqrt{x}+\frac{1}{2}-\frac{1}{4\sqrt{x}+2} \le \sqrt{x+\sqrt{x}} \le \sqrt{x}+\frac{1}{2}$.</p>
<p>  $\sqrt{x}-\frac{1}{2\sqrt{x}}-\frac{1}{4x\sqrt{x}} \le \sqrt{x-1} \le \sqrt{x}-\frac{1}{2\sqrt{x}}$.</p>
<p>Therefore $\sqrt{x+\sqrt{x}}-\sqrt{x-1} \to \frac{1}{2}$ as $x \to \infty$.</p>
<p><strong>Notes</strong></p>
<p>This kind of inequalities are useful when we do not want to use asymptotic expansion but otherwise as shown by <em>abel</em> expansion is the most widely applicable.</p>
|
138,866 | <p>I have data in a csv file. The first row has labels, and the first column, too.</p>
<pre><code>Datos = Import["C:\\Users\\jodom\\Desktop\\Data.csv"]
</code></pre>
<p>Tha data in the csv file is that:</p>
<pre><code>{{"No", "Vol", "Vel"}, {1, 500, 45}, {2, 700, 67}, {3, 350, 87}, {4,
123, 23}, {5, 587, 45}, {6, 435, 89}, {7, 896, 65}, {8, 125,
45}, {9, 476, 27}, {10, 987, 80}}
</code></pre>
<p>I put those csv data into a dataset:</p>
<pre><code>B = Dataset[Datos]
</code></pre>
<p>You can check it out as an image here,on how it has seen on wolfram after the import:
<a href="https://drive.google.com/file/d/0B56r_V66BiodQUhUMWNHcHZFOWc/view?usp=sharing" rel="noreferrer">https://drive.google.com/file/d/0B56r_V66BiodQUhUMWNHcHZFOWc/view?usp=sharing</a></p>
<p>Now I want to convert the first row that has the labels, into a head or label of the dataset, and the first column into a label column, so I can get data from this dataset, like </p>
<pre><code>Dataset[labelrow, labelcolumn]
</code></pre>
| jodomofo | 46,969 | <p>I have already do something.... </p>
<pre><code>data1 = {{"No", "Vol", "Vel"}, {1, 500, 45}, {2, 700, 67}, {3, 350, 87}, {4, 123, 23}, {5, 587, 45}, {6, 435, 89}, {7, 896, 65}, {8, 125, 45}, {9, 476, 27}, {10, 987, 80}}
data2 = Drop[data1, 1]
data3 = Dataset[data2]
data4 = data3[All, <|"No" -> 1, "Vol" -> 2, "Vel" -> 3|>]
</code></pre>
<p>But if you see, i have to delete the first row, to create the labels....there is a way to do that without deleting the first row? another thing....can i do the samething with the rows></p>
|
150,084 | <p>I need to say whether or not $f_n(x)=n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ is uniformly convergent on $(0,\infty)$.</p>
<p>I've found that the function is locally convergent to $f(x)=\frac{1}{2\sqrt{x}}$ and was trying to find $\sup{|f_n(x)-f(x)|}$.</p>
<p>I got the derivative $f_n'(x)= \frac{2nx\left(x-\sqrt x\sqrt{x+\frac{1}{n}}\right)+\sqrt{x}\sqrt{x+\frac{1}{n}}}{...}$ and could not find $x$ so that $f_n'(x)=0$</p>
<p>Any ideas?</p>
| Zarrax | 3,035 | <p>By the definition of uniform convergence, if they did uniformly converge to a limit function $f$ there would be some $N$ such that if $n \geq N$ then $|f_n(x) - f(x)| < 1$ for all $x$. In particular this would hold for $n = N$ itself. Since $f_N(x)$ is a bounded function, this means so is $f(x)$. So we can let $M$ be such that $|f(x)| < M$ for all $x$. Thus by the above, for all $n > N$ and all $x \in (0,\infty)$ we have
$$|f_n(x)| \leq |f_n(x) - f(x)| + |f(x)|$$
$$ < M + 1$$
But $f_n({1 \over n}) = (\sqrt{2} - 1)\sqrt{n}$. For $n$ large enough this will be greater than $M + 1$, a contradiction. So the functions don't converge uniformly.</p>
|
150,084 | <p>I need to say whether or not $f_n(x)=n\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$ is uniformly convergent on $(0,\infty)$.</p>
<p>I've found that the function is locally convergent to $f(x)=\frac{1}{2\sqrt{x}}$ and was trying to find $\sup{|f_n(x)-f(x)|}$.</p>
<p>I got the derivative $f_n'(x)= \frac{2nx\left(x-\sqrt x\sqrt{x+\frac{1}{n}}\right)+\sqrt{x}\sqrt{x+\frac{1}{n}}}{...}$ and could not find $x$ so that $f_n'(x)=0$</p>
<p>Any ideas?</p>
| mrf | 19,440 | <p>As already pointed out by several others, the convergence is not uniform on $(0,\infty)$ but it may be worth noting that it is uniform on every subinterval $[c,\infty)$ for $c > 0$.</p>
<p>This follows from the following computation:</p>
<p>$$
\begin{split}
|f_n(x) - f (x)| &= \left|\frac{1}{\sqrt{x+\dfrac1n}+\sqrt{x}} - \frac1{2\sqrt{x}}\right| \\
&= \left|\frac{\sqrt{x}-\sqrt{x+\dfrac1n}}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)} \right|\\
&= \left|\frac{\dfrac1n}{2\sqrt{x}\left( \sqrt{x+\dfrac1n}+\sqrt{x} \right)^2} \right|
\le \frac1n \cdot \frac{1}{8c\sqrt{c}}.
\end{split}
$$</p>
|
701,176 | <p>A function $f$ is differentiable over its domain and has the following properties:</p>
<ol>
<li><p>$\displaystyle f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$</p></li>
<li><p>$\lim_{h \to 0} f(h) = 0$</p></li>
<li><p>$\lim_{h \to 0} f(h)/h = 1$</p></li>
</ol>
<p>i) Show that $f(0)=0$</p>
<p>ii) show that $f'(x)=1+[f(x)]^2$ by using the def of derivatives Show how the above properties are involved.</p>
<p>iii) find $f(x)$ by finding the antiderivative. Use the boundary condition from part (i).</p>
<hr>
<p>So basically I think I found out how to do part 1 because if $x+y=0$ then the top part of the fraction will always have to be zero.</p>
<p>part 2 and 3 are giving me trouble. The definition is the limit $(f(x+h)-f(x))/h$</p>
<p>So I can set $x+y=h$ and make the numerator equal to $f(h)$?</p>
<p>Thanks for all who help</p>
| Lord Soth | 70,323 | <p>Perhaps not the shortest or the greatest argument, but it may be helpful to the OP. Let $f(a) = a^3-6a^2-2$. We take the derivative to obtain $f'(a) = 3a^2-12a$. The roots of the derivative are $0$ and $4$. Looking at the sign of the derivative, we observe that $f$ is increasing on $(-\infty,0)$, decreasing on $[0,4)$, and again increasing on $[4,\infty)$. Since $f(0) = -2$, and $f$ is increasing on $(-\infty,0)$, we know there cannot be any roots that are less or equal to $0$. Now, we also know $f$ is decreasing on $[0,4]$ so there can be roots in this interval either. Now we know that the only root is (if it exists) greater than or equal to $4$. Again the fact that $f$ is decreasing on $[0,4]$ shows that $f(4)<0$ (without calculation). Since $f$ grows to infinity as its argument grows to infinity, there is thus one real root greater than $4$. We now observe $f(6) = -2$ and $f(7)>0$. Hence the one real root is in the interval $(6,7)$ which contains no integers.</p>
|
235,319 | <p>The problem states: Suppose $f'(b) = M$ and $M <0$. Find $\delta>0$ so that if $x\in (b-\delta, b)$, then $f(x) > f(b).$</p>
<p>This intuitively makes sense, but I am not exactly sure how to find $\delta$. I greatly appreciate any help I can receive. </p>
| Gerry Myerson | 8,269 | <p>The answer is the coefficient of $x^{20}$ in $$(x+x^2+\cdots+x^7)(1+x+\cdots+x^7)^5$$ Do you see why? Then the first bracket is $x(1-x^7)/(1-x)$, and then second is the 5th power of $(1-x^8)/(1-x)$. OK? So now you need to be able to expand $x(1-x^7)(1-x^8)^5(1-x)^{-6}$, and pick out the coefficient of $x^{20}$. Can you use the Binomial Theorem to do that?</p>
|
383,735 | <p>Given this (very) tricky determinant, how can we calculate it easily?</p>
<p>$$\begin{pmatrix} \alpha + \beta & \alpha \beta & 0 & ... & ... & 0 \\ 1 & \alpha + \beta & \alpha \beta & 0 & ... & 0 \\ 0 & 1 & \alpha + \beta & \alpha \beta & ... & ... \\ ... & ... & ... & ... & ... & 0 \\ ... & ... & .... & ... & ... & \alpha \beta \\ 0 & 0 & 0 & ... & 1 & \alpha + \beta \\ \end{pmatrix} \in M_{n\times n}$$</p>
<p>EDIT:</p>
<blockquote>
<p>I have to prove it is equal to $\frac{{\alpha}^{n+1} - {\beta}^{n+1}}{\alpha - \beta}$</p>
</blockquote>
<p>Any help is appreciated, I just could not find a trick to ease it up!</p>
| vadim123 | 73,324 | <p>Let $D_n$ represent the determinant of this $n\times n$ matrix. Expanding on the first column, we see that $D_n=(\alpha+\beta)D_{n-1}-\alpha\beta D_{n-2}$, where the second is found by expanding on the first row of the resulting minor. The recurrence begins with $D_1=\alpha+\beta, D_2=(\alpha+\beta)^2-\alpha\beta=\alpha^2+\alpha\beta+\beta^2$.</p>
<p>To prove that this recurrence equals $\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$ one can use induction.</p>
|
670,570 | <p>Four gallons of yellow paint plus two gallons of red paint make orange paint. I assume this makes six gallons. So the ratio is 4:2, or 2:1.</p>
<p>Question: how many gallons of yellow paint, and how many gallons or red paint, to make two gallons of orange paint?</p>
<pre><code>2y + r = 2
2y + y = 2
3y = 2
y = 2/3
</code></pre>
<p>or</p>
<pre><code>4y + 2r = 6
(4y + 2r)/3 = 2
</code></pre>
<p>so I get 4/3 and 2/3.</p>
<p>However, in this section of <a href="http://www.pearsonschool.com/index.cfm?locator=PS2cJa&PMDBSUBCATEGORYID=25741&PMDBSITEID=2781&PMDBSUBSOLUTIONID=&PMDBSOLUTIONID=6724&PMDBSUBJECTAREAID=&PMDBCATEGORYID=806&PMDbProgramID=32310&elementType=attribute&elementID=1" rel="nofollow">the text book</a>, I'm not sure that it's "allowed" to do any of that. Is it possible to solve this just with cross-multiplying a ratio?</p>
<p>Their examples setup a ratio with an unknown <code>n</code>, cross multiply and solve for <code>n</code>. I don't see how to solve this word problem with that technique.</p>
| André Nicolas | 6,312 | <p>It is not hard to see that $x^5-1$ divides $x^{55}-1$. </p>
<p>Thus $(x-1)(x^4+x^3+x^2+x+1)$ divides $(x^{11}-1)(x^{44}+x^{33}+x^{22}+x^{11}+1)$ and therefore </p>
<p>$x^4+x^3+x^2+x+1$ divides $(x^{10}+x^9+\cdots +x+1)( x^{44}+x^{33}+x^{22}+x^{11}+1)$.</p>
<p>But $x^4+x^3+x^2+x+1$ and $x^{10}+x^9+\cdots +x+1$ are relatively prime (use the Euclidean Algorithm). </p>
<p>The result follows. </p>
|
670,570 | <p>Four gallons of yellow paint plus two gallons of red paint make orange paint. I assume this makes six gallons. So the ratio is 4:2, or 2:1.</p>
<p>Question: how many gallons of yellow paint, and how many gallons or red paint, to make two gallons of orange paint?</p>
<pre><code>2y + r = 2
2y + y = 2
3y = 2
y = 2/3
</code></pre>
<p>or</p>
<pre><code>4y + 2r = 6
(4y + 2r)/3 = 2
</code></pre>
<p>so I get 4/3 and 2/3.</p>
<p>However, in this section of <a href="http://www.pearsonschool.com/index.cfm?locator=PS2cJa&PMDBSUBCATEGORYID=25741&PMDBSITEID=2781&PMDBSUBSOLUTIONID=&PMDBSOLUTIONID=6724&PMDBSUBJECTAREAID=&PMDBCATEGORYID=806&PMDbProgramID=32310&elementType=attribute&elementID=1" rel="nofollow">the text book</a>, I'm not sure that it's "allowed" to do any of that. Is it possible to solve this just with cross-multiplying a ratio?</p>
<p>Their examples setup a ratio with an unknown <code>n</code>, cross multiply and solve for <code>n</code>. I don't see how to solve this word problem with that technique.</p>
| Bill Dubuque | 242 | <p>${\rm mod} \dfrac{(x^5\!-\!1)}{(x\!-\!1)}:\,\ \color{#c00}{x^5\equiv 1}\,\Rightarrow\,x^{11n}\equiv x^n (\color{#c00}{x^5})^{2n}\equiv x^n\Rightarrow\,x^{44}\!+\!x^{33}\!+\cdots+1\equiv x^4\!+\!x^3\!+\cdots+1\equiv 0$</p>
|
3,310,812 | <p>Solve the equation :
<span class="math-container">$$p^k=kl+1, $$</span> with <span class="math-container">$p$</span> a prime number and <span class="math-container">$k,l\ge 1$</span> two integers.</p>
<p>I know that <span class="math-container">$(p,k,l)=(3,1,2)$</span> is a solution, but can we find all solutions to the equation ?</p>
| Henry | 6,460 | <p>Intuitively I feel that the heights and widths should each be in the ratio <span class="math-container">$\sqrt{a}:\sqrt{b}:\sqrt{c}$</span></p>
<p>and so the widths should be <span class="math-container">$\sqrt{a}\frac{W}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$</span>, <span class="math-container">$\sqrt{b}\frac{W}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$</span>, <span class="math-container">$\sqrt{c}\frac{W}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$</span> </p>
<p>so the heights should be <span class="math-container">$\sqrt{a}\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{W}$</span> , <span class="math-container">$\sqrt{b}\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{W}$</span> , <span class="math-container">$\sqrt{c}\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{W}$</span> </p>
<p>leading to the minimum overall height being <span class="math-container">$\dfrac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{W}$</span> </p>
<p>and empirical experimentation suggests this is correct</p>
<hr>
<p>As for a proof, let's initially forget <span class="math-container">$c$</span> and try to minimise the height <span class="math-container">$H=h_a+h_b$</span> of <span class="math-container">$2\times 2$</span> boxes knowing <span class="math-container">$w_a+w_b=W$</span> and <span class="math-container">$h_aw_a=a$</span> and <span class="math-container">$h_bw_b=b$</span> </p>
<p>We can say that <span class="math-container">$H = \frac{a}{W-w_b}+\frac{b}{w_b}$</span> </p>
<p>so <span class="math-container">$\frac{dH}{dw_b} = \frac{a}{(W-w_b)^2}-\frac{b}{w_b^2} = \frac{a}{w_a^2} -\frac{b}{w_b^2}$</span> and it is not difficult to see that this is zero when <span class="math-container">$w_a^2: w_b^2$</span> is the same as <span class="math-container">$a:b$</span> which is <span class="math-container">$h_aw_a : h_bw_b$</span>, i.e. when <span class="math-container">$w_a: w_b$</span> is the same as <span class="math-container">$ h_a : h_b$</span> and so the same as <span class="math-container">$\sqrt{a}:\sqrt{b}$</span>, and that this minimises the height of this <span class="math-container">$2\times 2$</span> selection</p>
<p>Thus we know that this bit of the height is minimised when the respective widths and heights are each in the ratio <span class="math-container">$\sqrt{a}:\sqrt{b}$</span> and any other possibility is not optimal. The same is true with any other similar pairs, so the widths and heights of second and third columns and rows need to be in the ratio <span class="math-container">$\sqrt{b}:\sqrt{c}$</span> and the widths and heights of first and third columns and rows need to be in the ratio <span class="math-container">$\sqrt{a}:\sqrt{c}$</span>. The only way to do this is in the way I initially explained. It is easily extended to other similar <span class="math-container">$n \times n$</span> questions</p>
|
738,934 | <p>Write an equation for the line tangent to the graph of x=y^2+4 at the point (5,1).</p>
<p>I have found the derivative y'=1/2y but I do not know what to do next.</p>
| Community | -1 | <p><strong>Hint:</strong> In general, the tangent to the curve $y=f(x)$ at $(x_1,y_1)$ is
$$y=y_1+\left.\dfrac{dy}{dx}\right|_{x=x_1}(x-x_1)=y_1+f^\prime(x_1)(x-x_1)$$</p>
|
738,934 | <p>Write an equation for the line tangent to the graph of x=y^2+4 at the point (5,1).</p>
<p>I have found the derivative y'=1/2y but I do not know what to do next.</p>
| Berci | 41,488 | <p>The curve is $x=y^2+4$, it indeed contains $(5,1)$. Its differential is
$$dx=2y\,dy$$
(Consequently, $\displaystyle\frac{dy}{dx}\ =\ \frac1{2y}\ $ indeed.)</p>
<p>Now, at $y=1$, it gives $dy/dx=1/2$, this is the slope of the curve at that point (note that value of $y$ uniquely determines value of $x$).</p>
<p>Finally, the equation of the tangent will be of the form
$$y=\frac12\,x+c$$
and it goes through $(5,1)$, so $x=5$ and $y=1$ should satisfy it, this gives you the constant part $c$:
$$1=\frac52\,+c\,.$$</p>
|
1,830,867 | <p>I'm trying to self-study some algebraic topology, reading Hatcher. His questions seem much less straightforwardly worded than Munkres - with Munkres it was always clear that you weren't expected to know much coming in, whereas Hatcher seems to assume a lot of knowledge. Often simple seeming statements of his just seem vague and unclear to me, even though I also think they ought to be basic.</p>
<p>Right now I'm stuck deciphering exercise 1.2:</p>
<blockquote>
<p>Show that the change-of-basepoint homomorphism $\beta_h$ depends only on the homotopy class of $h$.</p>
</blockquote>
<p>At a guess, this means that given any two homotopic paths $h_1$ and $h_2$, for all loops $f$, there exists a homotopy between $\beta_{h_1}(f)$ and $\beta_{h_2}(f)$ ? I'm uncertain, as though I'm guessing at what Hatcher means rather than actually understanding him. Is this the correct interpretation, or am I off? And in general, anything I could do to more easily understand Hatcher? Thanks!</p>
| Ronnie Brown | 28,586 | <p>The natural environment for "change of base point" is that of groupoid and the <strong>fundamental groupoid</strong> $\pi_1 X$ of a space $X$. In a groupoid $G$, we have "object groups" $G(x)$for each object $x$ of $G$, and if $a: x \to y$ in $G$, then $a$ determines an isomorphism, "conjugation by $a$", $a_\#: G(x) \to G(y)$. This is dealt with in Chapter 6 of <a href="http://groupoids.org.uk/topgpds.html" rel="nofollow">Topology and Groupoids</a>, and in Philip Higgins' downloadable 1971 book <a href="http://www.tac.mta.ca/tac/reprints/articles/7/tr7abs.html" rel="nofollow">Categories and Groupoids</a>.</p>
|
555,051 | <p>I am having trouble understanding the following homework question,</p>
<p>Let $H=\{0,\pm 3, \pm6, \pm9,\ldots\}$ Find all the left cosets of $H$ in $\Bbb Z$.</p>
<p>I know the answer is $H$, $1+H$, and $2+H$ but I am having difficulty understanding why.</p>
<p>Thank you!</p>
| Pedro | 23,350 | <p>Observe that $H=3\Bbb Z$. The fact that $H,1+H,2+H$ are the only cosets is a restatement of the fact any integer $n$ is of the form $3k+b$ with $b=0,1,2$.</p>
|
1,953,628 | <p>0 choices for the 1st person.
17 choices for the 2nd person (must exclude 1st and his/her two neighbours) </p>
<p>For 2 of these choices of 2nd person, there is one shared neighbour, so 15 remaining choices. (e.g. if they are numbered 1 to 20 in a circle, 1st person is #1, 2nd is #3, then people 20,1,2,3,4 are excluded).
For the other 15 choices of 2nd person, there are no shared neighbors, so 14 remaining choices. </p>
<p>So if order matters, total is $20 \cdot (2 \cdot 15 + 15 \cdot 14) $
but since order does not matter, divide by $3! = 6$ to account for the permutations in order of the 3 people.
So total = $20 \cdot \frac{2 \cdot 15 + 15 \cdot 14}{6}$</p>
<p>just redid it; does this make any sense?</p>
| Community | -1 | <p>Why not again Stars and Bars theorem 1 with the extension of 2 gaps on the bounds. Theorem 1 because we don't want consecutive bars. Then $n=17, k_{max}=3$. </p>
<p>We don't take two on the bounds because we wrap around a circular table !</p>
<ul>
<li>|xxxxxx|xxxxxx|x : $\binom{17-1}{2}$ , two inside, one at the beginning</li>
<li>x|xxxxx|xxxxxxx| : $\binom{17-1}{2}$ , two inside, one at the end</li>
<li>x|xxxxx|xxxxxx|x : $\binom{17-1}{3}$ , three inside, none on the bounds</li>
</ul>
<p>$$\binom{17-1}{2}+\binom{17-1}{2}+\binom{17-1}{3} = 800$$</p>
<p>Helped by @N. F. Taussig, I corrected my first answer which was false</p>
|
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