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 |
|---|---|---|---|---|
153,186 | <p>How can I prove that the double negation elimination is not provable in constructive logic?<br>
To clarify, double negation elimination is the following statement: </p>
<p>$$\neg\neg q \rightarrow q$$</p>
| Joseph Vidal-Rosset | 62,171 | <p>A very simple and efficient tree proof method (i.e. a tableau method without signed formulae) for intuitionistic logic has been published by Bell, DeVidi and Solomon in "Logical Options: An Introduction to Classical and Alternative Logics"
<a href="http://books.google.fr/books/about/Logical_Options.html?id=zUVYx-bTLgMC&redir_esc=y" rel="nofollow noreferrer">http://books.google.fr/books/about/Logical_Options.html?id=zUVYx-bTLgMC&redir_esc=y</a></p>
<p>Here is the tree counter-model for the formula $\neg \neg p \to p$, à la Bell $et{}~ al.$:</p>
<p>$$\underline{?(\neg \neg p \to p)}^{\surd}$$
$$ ? p $$
$$ \neg \neg p$$
$$ \underline{? \neg p }^{\surd}$$
$$ p $$
The tree show a Kripke counter-model where, in a locality expressed by a space between two horizontal lines, $p$ is not known to be true (i.e. is false in the locality), while $\neg \neg p$ is proved, i.e. is known to be true. The symbol $?$ says that the formula is not known to be true, and it sticks the formula in the locality. The symbol $\surd$ says that the formula is deactivated. Every formula without $\surd$ or without $?$ can pass through any horizontal line (truth is persistent). For more details, see Bell et al.'s book.</p>
<p>This is very simple proof method to echo to the very nice explanation given by Zhen Lin. (I wish to thank Zhen Lin warmly for his post.)</p>
<p>Reading this page again, I must add that, contrarily to what Ben Millwood claimed, double negation elimination is <em>not</em> equivalent to the law of excluded middle, because if it is true that $(\neg A \lor A)$ implies $(\neg \neg A \to A)$ in intuitionistic logic, the converse is <em>not</em> intuitionistically provable. The fact that $(\neg \neg A \to A) \to (\neg A \lor A)$ is <em>not</em> an intuitionistic theorem contradicts the claim that double negation elimination is equivalent to the law of excluded middle.</p>
|
130,235 | <p>I need to find </p>
<p>$$f(n) = \int^\infty_0 t^{n-1} e^{-t} dt$$</p>
<p>So I think I find the indefinate integral first? But what do I do with $n$, since I am integrating with respect to $t$?</p>
| Pedro | 23,350 | <p>This integral can be easily computed by means of integration by parts. Let's call it $$\Gamma(n) =\int\limits_0^\infty t^{n-1}e^{-t}dt$$</p>
<p>We want to show that $\Gamma(n)$ follows some law. </p>
<p>Let's integrate by parts, with $u=t^{n-1}$. Then </p>
<p>$$\int\limits_0^\infty t^{n-1}e^{-t}dt=\left.-e^{-t}t^{n-1}\right|_0^\infty+(n-1)\int\limits_0^\infty t^{n-2}e^{-t}dt$$</p>
<p>but since $\left.-e^{-t}t^{n-1}\right|_0^\infty$ vanishes, we have that</p>
<p>$$\int\limits_0^\infty t^{n-1}e^{-t}dt=(n-1)\int\limits_0^\infty t^{n-2}e^{-t}dt$$ or that</p>
<p>$$\Gamma(n)=(n-1)\Gamma(n-1)$$</p>
<p>We can go on and use the relation to get</p>
<p>$$\Gamma(n)=(n-1)!$$</p>
<p>Another approach is copper's:</p>
<p>Define $$I(x) = \int\limits_0^\infty e^{-tx} dt $$</p>
<p>By means of Leibniz' rule, we have that</p>
<p>$$I^{(n)}(x) = \int\limits_0^\infty (-t)^ne^{-tx} dt $$</p>
<p>So we're interested in</p>
<p>$$(-1)^nI^{(n)}(1) = \int\limits_0^\infty t^n e^{-t} dt $$</p>
<p>But since</p>
<p>$$I(x) = \int\limits_0^\infty e^{-tx} dt=\frac{1}{x}$$ and we can simply show that</p>
<p>$$I^{(n)}(x)=\frac{(-1)^n n!}{x^{n+1}}$$ it follows that</p>
<p>$$(-1)^nI^{(n)}(1)=\frac{ n!}{1^{n+1}}=n!$$ as desired.</p>
|
1,615,201 | <p>I just solved a long problem for my physics w/calculus homework that required a simplification using a quadratic formula. The "textbook" (flipItPhysics) came up with a different simplification than mine but it turns out they are equivalent. I can't, for the life of me, figure out how to simplify mine into theirs.</p>
<h2>Mine:</h2>
<p>$$
x = \frac{d\left(q\pm\sqrt{Qq}\right)}{q - Q}
$$</p>
<h2>Theirs:</h2>
<p>$$
x = d \cdot \left(\frac{q}{q- Q}\right) \left(1 \pm\sqrt{\frac{Q}{q}}\right)
$$</p>
<p>Can someone show me how to get from mine to theirs? I'm specifically confused about how to make the inside of the square root a division instead of a multiplication.</p>
| mweiss | 124,095 | <p>Notice that both expressions have a factor of $d$ (you have it in the numerator, and the textbook has it as the first factor), and both expressions have a factor of $q-Q$ in the denominator. So let's ignore those and compare what's left. We need to show that
$$q \pm \sqrt{Qq}$$
and
$$q \cdot \left(1 \pm \sqrt{\frac{Q}{q}} \right)$$
are equal.</p>
<p>It should be obvious that if you distribute the $q$ in the second expression through the parentheses, you end up with the $q$ in the first term of the first expression. Really, then, the only things we have to compare are $\sqrt{Qq}$ and $q\sqrt{\frac{Q}{q}}$. They both contain a factor of $\sqrt{Q}$, so let's ignore that. The only real mystery is, why is $\sqrt{q}$ equal to $q\sqrt{\frac{1}{q}}$?</p>
<p>This turns out to be one of those properties that's really simple to understand once you see it a few times, but for some reason it is rarely taught in high school. It comes down to the fact that $$q\sqrt{\frac{1}{q}}=\frac{q}{\sqrt{q}}=\frac{\sqrt{q}\sqrt{q}}{\sqrt{q}}=\sqrt{q}$$</p>
<p>I think the real question here is why the textbook writes the answer in what is (in my opinion at least) less simplified than the form of the answer you came up with.</p>
|
2,599,978 | <p>This is a homework problem I have.</p>
<blockquote>
<p>Let a random variable <span class="math-container">$x$</span> follow a Gaussian distribution with mean <span class="math-container">$\mu = 10$</span> and variance <span class="math-container">$1$</span>.</p>
<p>Which of the following samples of size 4 has the largest probability
of being generated from this distribution?</p>
<p>i. 9, 9.1, 10, 11</p>
<p>ii. 1, 2, 0, 11</p>
<p>iii. 9.9, 10.1, 9.8, 10.3</p>
<p>iv. 11, 9, 12, 8</p>
<p>v. 10, 10, 10, 1</p>
</blockquote>
<p>My friend argues that iii. is the right answer because <span class="math-container">$\Pi_i \, p(x_i;\mu=10,\sigma^2 =1)$</span> would be higher for it. But I don't think that is true. If it were, a sample size of a million with all values equal to 10 would become most likely to be generated.</p>
<p>In my opinion, we need to take sample mean and variance, and find their likelihood from the distribution of sample mean and variance.</p>
<p>Please help clear my conceptual confusion.</p>
| Jack D'Aurizio | 44,121 | <p>$$f(a,b,c)=\int_{-\pi/2}^{\pi/2}\left[\sin(x)-ax^2-bx-c\right]^2\cos(x)\,dx $$
is a regular function. By equating to zero $\frac{\partial f}{\partial b}$ you get $b=\frac{\pi}{2(\pi^2-8)}$, then by defining $g(a,c)=f\left(a,\frac{\pi}{2(\pi^2-8)},c\right)$ and equating to zero $\frac{\partial g}{\partial a}$ and $\frac{\partial g}{\partial c}$ you get $a=c=0$, so the optimal approximation is given by $p(x)=\frac{\pi x}{2(\pi^2-8)}$. In such a case</p>
<p>$$ \int_{-\pi/2}^{\pi/2}\left[\sin(x)-p(x)\right]^2\cos(x)\,dx = \frac{13}{24}-\frac{1}{\pi^2-8}=0.00679415598\ldots$$</p>
<p>The solution is greatly simplified by noticing in advance that $p(x)$ has to be an odd function, by symmetry.</p>
|
2,599,978 | <p>This is a homework problem I have.</p>
<blockquote>
<p>Let a random variable <span class="math-container">$x$</span> follow a Gaussian distribution with mean <span class="math-container">$\mu = 10$</span> and variance <span class="math-container">$1$</span>.</p>
<p>Which of the following samples of size 4 has the largest probability
of being generated from this distribution?</p>
<p>i. 9, 9.1, 10, 11</p>
<p>ii. 1, 2, 0, 11</p>
<p>iii. 9.9, 10.1, 9.8, 10.3</p>
<p>iv. 11, 9, 12, 8</p>
<p>v. 10, 10, 10, 1</p>
</blockquote>
<p>My friend argues that iii. is the right answer because <span class="math-container">$\Pi_i \, p(x_i;\mu=10,\sigma^2 =1)$</span> would be higher for it. But I don't think that is true. If it were, a sample size of a million with all values equal to 10 would become most likely to be generated.</p>
<p>In my opinion, we need to take sample mean and variance, and find their likelihood from the distribution of sample mean and variance.</p>
<p>Please help clear my conceptual confusion.</p>
| Disintegrating By Parts | 112,478 | <p>The expression $\langle f,g\rangle = \int_{-\pi/2}^{\pi/2}f\overline{g}\cos x dx$ does define a complex inner product on $C[-\pi,\pi]$. The problem can then be restated as finding scalars $\alpha,\beta,\gamma$ such that $\|\sin x-\alpha -\beta x-\gamma x^2\|$ is minimized, which is achieved iff $(\sin x -\alpha-\beta x-\gamma x^2)\perp 1,x,x^2$. That gives a system of $3$ equations in the $3$ unknowns determined by
$$
\int_{-\pi/2}^{\pi/2}(\sin x-\alpha-\beta x-\gamma x^2)\cos x dx =0 \\
\int_{-\pi/2}^{\pi/2}(\sin x-\alpha-\beta x-\gamma x^2)x\cos xdx = 0
\\
\int_{-\pi/2}^{\pi/2}(\sin x-\alpha-\beta x-\gamma x^2)x^2\cos xdx = 0.
$$</p>
|
2,982,205 | <p>I need help with a question that asks this:</p>
<p>Let <span class="math-container">$f(x)= -1/x$</span> and <span class="math-container">$g(x)=e^x$</span> find the domain , range ,monotonicity intervals. Draw the graph of <span class="math-container">$g(f(x))$</span>. </p>
<p>Everything else except the graph is pretty easy but trying to find the graph has left me pretty confused.</p>
<p>Any hints would be greatly appreciated.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>I have got <span class="math-container">$$\overline{h}h^2+\overline{h}z^2+2h\overline{h}z+h^2\overline{z}+2hz\overline{z}$$</span></p>
|
100,433 | <p>I am trying to export some largish integer arrays to HDF5 and know that every entry in them would fit into an unsigned 8 bit integer array. As the default "DataFormat" for export that Mathematica is using is a 32bit integer array the resulting files are unnecessarily large. Does anyone know the correct syntax to export such an integer array as an 8 bit integer dataset?</p>
<p>I found the following suggestions in this <a href="https://mathematica.stackexchange.com/a/92064/169">answer</a> and a comment to this <a href="https://mathematica.stackexchange.com/a/47861/169">answer</a>:</p>
<pre><code>Export["int8.h5",
RandomInteger[{2^8 - 1}, {100, 100}],
"DataFormat" -> {"UnsignedInteger8"}
]
Export["int8.h5",
RandomInteger[{2^8 - 1}, {100, 100}],
{"UnsignedInteger8", "DataFormat"}
]
</code></pre>
<p>but they both seem to not work at all or not create the expected data representation in the files, which can be checked by either HDFView (which I consider authorative) or Mathematica itself with:</p>
<pre><code>Import["int8.h5","DataFormat"]
</code></pre>
| ScotMartin | 42,448 | <p>You can see:</p>
<p><a href="https://github.com/scotmartin1234/HDF5Mathematica" rel="nofollow">https://github.com/scotmartin1234/HDF5Mathematica</a></p>
<p>The package has the functionality you need for writing at low level. The high level functions are designed more for reading. The package could be expanded as open source project to develop the writing part. It's straight forward to assemble the low level functions into a more Mathematica-friendly format (i.e., the higher level functions), but it would take a volunteer to do so. My efforts and work have been on developing the reading side of the equation into a Mathematica friendly format.</p>
<p>This is version 2.00 (August 2016) of the package that was originally provided as version 1.00 in July 2011</p>
|
3,398,164 | <p>I am working through the math behind homographies, but my math skills are a bit rusty.</p>
<p>A homography can be calculated with 8 corresponding points (4-4) because the homography matrix has 8 degrees of freedom. This is because, <a href="https://docs.opencv.org/master/d9/dab/tutorial_homography.html#lecture_16" rel="nofollow noreferrer">eventhough the 3x3 matrix has 9 variables, one can "normalized to one"</a>
the cited explanation says:</p>
<blockquote>
<p>Note that we can multiply all <span class="math-container">$h_{ij}$</span> by nonzero k without changing the
equations</p>
</blockquote>
<p>and the following explanation is given in "multiple view geometry in computer vision":</p>
<blockquote>
<p>Note that the matrix H occurring in this equation may be changed by
multiplication by an arbitrary non-zero scale factor without altering
the projective transformation. Consequently we say that H is a
homogeneous matrix, since as in the homogeneous representation of a
point, only the ratio of the matrix elements is significant. There are
eight independent ratios amongst the nine elements of H, and it
follows that a projective transformation has eight degrees of freedom.</p>
</blockquote>
<p>I have also read this post: <a href="https://math.stackexchange.com/questions/508668/degree-of-freedom-of-homography-matrix">degree of freedom of Homography matrix</a>, but im afraid i still don't understand.</p>
<p>Can someone explain in detail how this normalization works? Or why the scaling multiplication normalizes the matrix?</p>
<p>thanks a ton!</p>
| dan_fulea | 550,003 | <p>The following discussion may be an explanation, but if it is not, please try to ask where there is something hard to catch.</p>
<p>My strategy is always to simplify things, but keep the problem related (and slightly simpler, but essentially the same). Here, i propose to understand homographies from a (projective) line to an other (projective) line, since planes are too complicated. </p>
<p>For short, a point <span class="math-container">$P$</span> on the <em>affine</em> line <span class="math-container">$L$</span> is given by its coordinate, "<span class="math-container">$x$</span>", say. But we consider also the infinite point, so we have to consider more "complicated point( representation)s". Instead of <span class="math-container">$x$</span> we write <span class="math-container">$[x:1]$</span>, in words, <span class="math-container">$x$</span> divided by one. And the point at infinity is <span class="math-container">$[1:0]$</span>. We can easily "see" these points if we also know the second dimension as follows. Instead of <span class="math-container">$[x:1]$</span> we draw in the plane the point <span class="math-container">$(x,1)$</span>. This is all. In this plane there are also many other points, but from the point of view of the "camera placed in <span class="math-container">$(0,0)$</span>", we can not distinguish two points on the same line, more exactly, on the same <em>ray</em>. For instance, the points <span class="math-container">$(2,1)$</span>, and <span class="math-container">$(4,2)$</span>, and <span class="math-container">$(6,3)$</span>, and ... map in the projective line to the same point <span class="math-container">$[2:1]=[4:2]=[6:3]=\dots$</span> and we will always want this last view. </p>
<p>Why do we like this representation, and also the representation <span class="math-container">$[1:0]$</span> for the "point at infinity"? Because we can take also an other way to view things. Imagine a radar, a sonar, a camera placed in the middle <span class="math-container">$O$</span> of a 2D beach, and the sea starts in some <span class="math-container">$20$</span> meters, say, in front of us, in a point <span class="math-container">$A=[0:1]=0$</span>. The camera should not be able to recognize the depth. Then a full turn of the camera is something that can be easily imagined, and in this full turn, we cover the line of the breakers, of the last bastion of sand, from <span class="math-container">$A$</span> to the right. At some point, after some seconds, we pass through the point of view "opposite" to <span class="math-container">$A$</span>, it has no coordinate <span class="math-container">$x$</span>. And in the next ms we are coming from the left to <span class="math-container">$A$</span>.</p>
<hr>
<p>Now imagine there are some <span class="math-container">$7$</span> surfers playing in the breakers. At coordinates <span class="math-container">$1,3,7,8,9,21,100$</span>. An other camera sees the first three surfers in <span class="math-container">$-3,5,1$</span>. Where should we place the other surfers?</p>
<hr>
<p>This is a similar question to the one in the OP. We need a matrix transformation,
<span class="math-container">$$
\begin{aligned}
\begin{bmatrix}
x\\ 1
\end{bmatrix}
&\to
\begin{bmatrix}
x'\\ 1
\end{bmatrix}
:=
\begin{bmatrix}
a&b \\ c&d
\end{bmatrix}
\begin{bmatrix}
x \\ 1
\end{bmatrix}
\ ,\qquad\text{ so }
\\
x&\to x'=
\frac{ax+b}{cx+d}\ .
\end{aligned}
$$</span>
Notice the above simpler form for the transformation.</p>
<p>It is clear that the simpler form for the homographic transformation
<span class="math-container">$x\to x'=
\frac{ax+b}{cx+d}
$</span> is homogenous in <span class="math-container">$(a,b,c,d)$</span>. Multiplying them in the same time by something <span class="math-container">$\ne 0$</span> would lead to a transformation, which is the same one.</p>
<p>Now we are searching for a specific homography. There are too many (redundant) variables in
<span class="math-container">$$
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
$$</span>
We are free to make one choice, one norming.
Let us say, i would like to norm <span class="math-container">$d=1$</span>. This means, we replace the above by
<span class="math-container">$$
\frac 1d
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
=
\begin{bmatrix}
a/d&b/d\\c/d&d/d
\end{bmatrix}
=
\begin{bmatrix}
a/d&b/d\\c/d&1
\end{bmatrix}\ .
$$</span>
(The one right lower corner entry is now one.)</p>
<hr>
<p>Now we have two cameras. One point gives a boring situation, but if we take some seven points, say, and try to get the right camera transform, things are slightly more complicated. Let us take the (random) points mentioned above. We try to norm <span class="math-container">$d=1$</span>, thus obtaining a system in the other three variables <span class="math-container">$a,b,c$</span>. Conditions:</p>
<ul>
<li><span class="math-container">$1\to(a\cdot1+b)/(c\cdot 1+1)=-3$</span>,</li>
<li><span class="math-container">$3\to(a\cdot3+b)/(c\cdot 3+1)=5$</span>,</li>
<li><span class="math-container">$7\to(a\cdot7+b)/(c\cdot 7+1)=1$</span>.</li>
</ul>
<p>We solve this system getting exactly (with sage, to have a short end):</p>
<pre><code>sage: var('a,b,c');
sage: def T(x): return (a*x+b)/(c*x+1)
sage: solve( [T(1) == -3, T(3) == 5, T(7) == 1], [a,b,c], solution_dict=True )
[{c: -5/11, b: -17/11, a: -1/11}]
</code></pre>
<p>So the solution is:
<span class="math-container">$$
T=
\begin{bmatrix}
a&b\\c&1
\end{bmatrix}
=
\begin{bmatrix}
-1/11&-17/11\\-5/11 &1
\end{bmatrix}
\ .
$$</span>
Humanly we would multiply with <span class="math-container">$-11$</span>, obtaining an other matrix giving the same homographic transformation:
<span class="math-container">$$
U
=
\begin{bmatrix}
1&17\\5 &-11
\end{bmatrix}
\ .
$$</span>
Passing from <span class="math-container">$U$</span> to <span class="math-container">$T$</span> is this step of norming.
(By chance, we have now a normed entry in the <span class="math-container">$a$</span>--place.)</p>
<p>We can (if we can) also try to norm the entry denoted above all the time by <span class="math-container">$b$</span>, the matrix <span class="math-container">$V$</span> obtained implements the same homographic transformation:
<span class="math-container">$$
V
=
\frac 1{17}U
=
\begin{bmatrix}
1/17&1\\5/17 &-11/17
\end{bmatrix}
\ .
$$</span>
I think i should stop here, a pointed question would be now simpler to answer. </p>
|
3,646,694 | <p>Let <span class="math-container">$M$</span> be an elliptic element of <span class="math-container">$SL_2(\mathbb R)$</span>. Then it is conjugate to a rotation <span class="math-container">$R(\theta)$</span>. Note that we can calculate <span class="math-container">$\theta$</span> in terms of the trace of <span class="math-container">$M$</span>; it means that we actually know <span class="math-container">$R(\theta)$</span> and we can write:</p>
<p><span class="math-container">$$M=TR(\theta) T^{-1}$$</span></p>
<p>If <span class="math-container">$S^1$</span> is the unit circle in <span class="math-container">$\mathbb R^2$</span>, it follows that <span class="math-container">$T(S^1)$</span> is the conic section <span class="math-container">$\mathcal C$</span> which is preserved by <span class="math-container">$M$</span>.</p>
<blockquote>
<p>Is there any explicit way to find the equation <span class="math-container">$\mathcal C$</span> in general?</p>
</blockquote>
<p>My procedure is quite uneffective, because one has to find <span class="math-container">$T$</span> first (so non-linear system) and then write down <span class="math-container">$T(S^1)$</span>, which is in general not obvious.</p>
| Quanto | 686,284 | <p>Note </p>
<p><span class="math-container">$$\alpha (\alpha^2+2\alpha -4)\beta =-1$$</span></p>
<p>Thus,</p>
<p><span class="math-container">$$\beta =- \frac1{\alpha (\alpha^2+2\alpha -4)}
=- \frac1{\alpha^3+2\alpha^2-4\alpha}
=- \frac1{5\alpha -1 -4\alpha} =\frac1{1-\alpha}
$$</span></p>
|
569,814 | <p>I was watching a video on YouTube on Quantum Mechanics Concepts and saw that if you wanted to convert a probability amplitude to a probability, you square it. In the video he said that this was equivalent to multiply by it's complex conjugate. So is this correct? Is squaring the same as multiplying by the complex conjugate, or is this just a thing you do in Quantum Mechanics?
Also, I'm not sure if this should be asked on the physics.se instead of math.se, if it should then I'm sorry.</p>
| Clive Newstead | 19,542 | <p>A probability is a <em>real</em> number (between $0$ and $1$). The magnitude of a complex number $z$ is $|z|$, which is real. Now $z \bar z = |z|^2$, so multiplying by the complex conjugate gives the square of its <em>magnitude</em>, not the square of the complex number itself.</p>
<p>For example, $i^2 = -1$ but $|i|^2 = 1$.</p>
|
354,512 | <p>Find the minimal polynomial of the matrix M:
\begin{pmatrix}
0 & 0 & 0 & \dots & 0 & a_{1}\\
1 & 0 & 0 & \dots & 0 & a_{2}\\
0 & 1 & 0 & \dots & 0 & a_{3} \\
\dots & \dots \\
0 & 0 & 0 & \dots & 1 & a_{n}
\end{pmatrix}</p>
<p>Let's take vector $e_{1}$:
\begin{pmatrix}
1 \\
0 \\
\vdots\\
0
\end{pmatrix}</p>
<p>$M e_{1} = e_{2}$, $M e_{2} = e_{3}$, $M e_{3}= e_{4}\dots$.
$M^{n} e_{1}=(a_{1}\dots a_{n})$.
Why does $x^{n}-(a_{1}+a_{2}x+\dots+a_{n}x^{n-1})=0$ be the minimal polynomial of this matrix? How does it connect with the dimension of image? </p>
| user1551 | 1,551 | <p>Consider a monic polynomial $q(x) = b_nx^n + b_{n-1}x^{n-1} + b_{n-2}x^{n-2} + ... + b_1x + b_0$ of $M$ of degree $m\le n$ (if $m<n$, we have $b_m=1$ and $b_{m+1}=b_{m+2}=\cdots=b_n=0$) and let $\{e_1,e_2,\ldots,e_n\}$ be the standard basis of $\mathbb{R}^n$. If $q(M)=0$, we have, in particular,
$$q(M)e_1=b_n\sum_{i=1}^na_ie_i + b_{n-1}e_n + b_{n-2}e_{n-1} + ... + b_1e_2 + b_0e_1 = 0.\tag{1}$$
Note that $b_n\neq0$, or else $(1)$ would imply that all $b_i$s are zero, which is impossible. Therefore $b_n=1$ ($q$ is monic by definition). By comparing coefficients on both sides of $(1)$, we see that $q$ must be in the form of
$$q(x)=x^n-(a_{1}+a_{2}x+\dots+a_{n}x^{n-1}).\tag{2}$$
Now it is easy to verify that $q(M)e_i=0$ for all $i$. Hence the minimal polynomial of $M$ is given by $(2)$.</p>
|
468,510 | <p>Given $n^2$, how many fourth powers $(x^4)$ are between 0 and $n^2$? </p>
<p>$n,x\in \mathbb{Z}$</p>
<p>Does this just reduce down to how many squares are below $n$?</p>
| Brian M. Scott | 12,042 | <p>Yes, it does: assuming that $x$ is a non-negative integer, $0\le x^4\le n^2$ if and only if $0\le x^2\le n$. You can express the number of squares exactly using the floor (greatest integer) function; I’ve left the answer spoiler-protected below.</p>
<blockquote class="spoiler">
<p> $\left\lfloor\sqrt{n}\right\rfloor+1$, where the $+1$ accounts for $0^2$.</p>
</blockquote>
|
1,650,638 | <p>I have a sequence of reals $S = s_1,s_2,\dots,s_n$ such that $s_i-s_{i-1}$ is a Gaussian distribution. From histogram of sequence $S$ (10000 elements) it appears that it is uniform distribution. Is it true ? If yes, can we prove it ? If no, what can we say about $s_i$ ?</p>
| Laurent Duval | 257,503 | <p>My interpretation is that diagonalization, when possible, simplifies the interpretation (and subsequent computations) of a matrix. In the eigenbasis, operations can be performed "coordinate-wise", or basis vector by basis vector, almost independently. </p>
<p>Suppose you have a vector $x^a = (x_1,x_2,\dots,x_n)$, and you already have computed $y^a = Ax^a$. If you modify only one component of $x^a$, for instance $x^b = (x_1,7\times x_2,\dots,x_n)$, you might be forced to recompute $y^b = Ax^b$ in a non-eigenbasis. In an eigenbasis, you only need the recompute the second coordinate: $y^b_2 = 7\times y^a_2$, the others remaining unchanged. </p>
<p>This is one of the reasons why one sometimes first diagonalizes operators (or nearly diagonalizes), to enable faster and more accurate subsequent computations. An iportant example is related to the Fourier transform. A linear and time invariant system can be represented by a convolution operator (see <a href="https://math.stackexchange.com/questions/918345/fourier-transform-as-diagonalization-of-convolution">Fourier transform as diagonalization of convolution</a>). A convolution is not fully easy to interpret. Yet, complex exponentials (cisoids, <a href="https://en.wikipedia.org/wiki/Euler%27s_formula" rel="nofollow noreferrer">cis($x$): "cosine plus i sine"</a>) are eigenfunctions for such systems.</p>
<p>So, if you choose an appropriate basis of cisoids, the convolution operator is diagonalized. Consequently, after the discovery of fast implementations for the Fourier transform (such as the <a href="https://en.wikipedia.org/wiki/Fast_Fourier_transform" rel="nofollow noreferrer">FFT</a>), due to the simplicity of operations in the Fourier basis (mere products instead of convolutions), many signal analysis and image filtering operations have been implemented in the Fourier domain. Think about mp3, JPEG, <a href="https://en.wikipedia.org/wiki/Shazam_(service)" rel="nofollow noreferrer">Shazam</a>, which are by-products of the FFT.</p>
<p>In short, diagonalization is at the heard of the digital revolution.</p>
|
1,373,927 | <p>We know that a tree with n edges have n+1 nodes.So if $|B_{n+1}|$ is the number of all possible ordered trees with n+1 nodes then its true that $C_{n+1} = |B_{n+1}|$ where $C$ is the Catalan number.Let it be $|L_k|$ the number of all possible ordered trees with n+1 nodes and k leaves and Its true that
$|B_{n+1}| = \displaystyle\sum_{i=1}^{n}|L_i|$.</p>
<hr>
<p>If my thinking is correct how can I continue or if you have a better idea please let me know.</p>
| Marko Riedel | 44,883 | <p>Here is a contribution using basic complex variables.
<P></p>
<p>We will compute the number of trees on <span class="math-container">$n$</span> nodes and having <span class="math-container">$q$</span>
leaves.</p>
<p>The combinatorial class equation for ordered rooted trees with leaves marked is
<span class="math-container">$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T} = \mathcal{Z}\times \mathcal{U}
+ \mathcal{Z} \times \textsc{SEQ}_{\ge 1}(\mathcal{T})
\quad\text{or}\quad
\mathcal{T} = \mathcal{Z}\times\mathcal{U}
+ \mathcal{Z} \times \sum_{p\ge 1} \mathcal{T}^p.$$</span></p>
<p>This yields the functional equation for the generating function <span class="math-container">$T(z)$</span>
<span class="math-container">$$T(z) = zu + z\frac{T(z)}{1-T(z)}$$</span>
or <span class="math-container">$$z = \frac{T(z)}{u+T(z)/(1-T(z))}
= \frac{T(z)(1-T(z))}{T(z)+u(1-T(z))}.$$</span></p>
<p>Note that leaves in addition to being marked as such also carry the
node marker so that the total number of nodes includes the leaves. If
this is not desired subtract the number of leaves from the number of
nodes to get the count of genuine internal nodes.</p>
<p><P>
Starting the computation we seek</p>
<p><span class="math-container">$$T_n(u) = \frac{1}{2\pi i}
\int_{|z|=\gamma} \frac{1}{z^{n+1}} T(z) \; dz.$$</span>
and will compute this by Lagrange inversion.</p>
<p>Put <span class="math-container">$w=T(z)$</span> so that
<span class="math-container">$$dz =
\left(\frac{1-2w}{w+u(1-w)}
- \frac{w(1-w)}{(w+u(1-w))^2} (1-u) \right) dw.$$</span></p>
<p>This yields the two integrals
<span class="math-container">$$A = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(w+u(1-w))^{n+1}}{(w(1-w))^{n+1}} w
\frac{1-2w}{w+u(1-w)} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(w+u(1-w))^{n}}{w^n (1-w)^{n+1}} (-w+(1-w))\;dw
\\ = - \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(w+u(1-w))^{n}}{w^{n-1} (1-w)^{n+1}} \; dw
\\ + \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(w+u(1-w))^{n}}{w^n (1-w)^{n}} \; dw.$$</span></p>
<p>and
<span class="math-container">$$B = -\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(w+u(1-w))^{n+1}}{(w(1-w))^{n+1}} w
\frac{w(1-w)}{(w+u(1-w))^2} (1-u) \; dw
\\ = - (1-u)\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(w+u(1-w))^{n-1}}{w^{n-1} (1-w)^{n}} \; dw.$$</span></p>
<p>We extract the coeffcient in <span class="math-container">$u$</span> first. The integral <span class="math-container">$A$</span> gives two
pieces
<span class="math-container">$$-{n\choose q}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{w^{n-q}(1-w)^q}{w^{n-1} (1-w)^{n+1}} \; dw
\\ = -{n\choose q}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{q-1} (1-w)^{n-q+1}} \; dw
= -{n\choose q} {n-2\choose n-q}$$</span></p>
<p>and
<span class="math-container">$${n\choose q}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{w^{n-q}(1-w)^q}{w^{n} (1-w)^{n}} \; dw
\\ = {n\choose q}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{q} (1-w)^{n-q}} \; dw
= {n\choose q} {n-2\choose n-q-1}.$$</span></p>
<p>The integral in <span class="math-container">$B$</span> also gives two pieces
<span class="math-container">$$-{n-1\choose q} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{w^{n-1-q} (1-w)^{q}}{w^{n-1} (1-w)^{n}} \; dw
\\ = -{n-1\choose q} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{q} (1-w)^{n-q}} \; dw
= -{n-1\choose q} {n-2\choose n-q-1}.$$</span></p>
<p>and
<span class="math-container">$${n-1\choose q-1} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{w^{n-q} (1-w)^{q-1}}{w^{n-1} (1-w)^{n}} \; dw
\\ = {n-1\choose q-1} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{q-1} (1-w)^{n-q+1}} \; dw
= {n-1\choose q-1} {n-2\choose n-q}.$$</span></p>
<p>This yields the following answer before simplification:
<span class="math-container">$${n\choose q} {n-2\choose n-q-1}-{n\choose q} {n-2\choose n-q}
+ {n-1\choose q-1} {n-2\choose n-q}
- {n-1\choose q} {n-2\choose n-q-1}.$$</span></p>
<p>This simplifies to
<span class="math-container">$${n\choose q} {n-2\choose n-q}
\left(\frac{n-q}{q-1} - 1
+ \frac{q}{n}
- \frac{n-q}{n} \frac{n-q}{q-1}\right)
\\ = {n\choose q} {n-2\choose n-q}
\frac{n-q}{n} \frac{1}{q-1}
= \frac{1}{q-1} {n-1\choose q} {n-2\choose n-q}
\\ = \frac{1}{n-1} {n-1\choose q} {n-1\choose n-q}.$$</span></p>
<p>The generating function <span class="math-container">$T_n(u)$</span> can be verified using Maple's
<strong>combstruct</strong> package. This is the code.</p>
<pre>
with(combstruct);
gf_cs :=
proc(n)
option remember;
local trees, leaves;
trees := { T=Union(Prod(Z, U),
Prod(Z, Sequence(T, 1<= card))),
Z=Atom, U=Epsilon };
leaves :=
proc(struct)
if type(struct, function) then
return add(leaves(op(q, struct)), q=1..nops(struct));
fi;
if struct = Z then return 0 fi;
return 1;
end;
add(u^leaves(t), t in allstructs([T, trees], size=n));
end;
CF := (n,q) -> 1/(n-1)*binomial(n-1,q)*binomial(n-1,n-q);
gf_verif := n -> add(CF(n,q)*u^q, q=1..n-1);
</pre>
<p>This will produce e.g. for <span class="math-container">$T_9(u)$</span> the generating function
<span class="math-container">$${u}^{8}+28\,{u}^{7}+196\,{u}^{6}+490\,{u}^{5}+490\,{u}^{4}
+196\,{u}^{3}+28\,{u}^{2}+u,$$</span></p>
<p>which matches the binomial coefficient formula.</p>
<p><strong>Addendum.</strong> We show that the counts of the number of trees
classified according the number of leaves does indeed add up to the
Catalan numbers in order to verify the above computation.</p>
<p>We have the sum
<span class="math-container">$$\frac{1}{n-1}\sum_{q=1}^{n-1} {n-1\choose q} {n-1\choose n-q}.$$</span></p>
<p>We can extend this to include <span class="math-container">$q=0$</span> because the second binomial
coefficient is zero in that case:
<span class="math-container">$$\frac{1}{n-1}\sum_{q=0}^{n-1} {n-1\choose q} {n-1\choose n-q}.$$</span></p>
<p>Put <span class="math-container">$${n-1\choose n-q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n-q+1}} \; dz.$$</span></p>
<p>This yields for the sum
<span class="math-container">$$\frac{1}{n-1}\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n+1}}
\sum_{q=0}^{n-1} {n-1\choose q} z^q \; dz
\\ = \frac{1}{n-1}\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n+1}}
(1+z)^{n-1} \; dz
\\ = \frac{1}{n-1}\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n-2}}{z^{n+1}} \; dz
\\ = \frac{1}{n-1}
{2n-2\choose n}.$$</span></p>
<p>This is
<span class="math-container">$$\frac{(2n-2)!}{n!\times (n-1)!}
= \frac{1}{n} {2n-2\choose n-1}.$$</span></p>
<p>We now recognize the <a href="https://oeis.org/A000108" rel="nofollow noreferrer">Catalan number</a>
formula shifted by one, obtaining
<span class="math-container">$$ 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, \ldots $$</span>
from <span class="math-container">$n=2$</span> on.</p>
<p><strong>Addendum.</strong> The above admits radical simplification, wich can be found at this <a href="https://math.stackexchange.com/questions/3108340/">MSE link</a>.</p>
|
2,006,437 | <p>Is it possible to get an example of two matrices $A,B\in M_4(\mathbb{R})$ both having $rank<2$ but $\det(A-\lambda B)\ne 0$ i.e it is not identically a zero polynomial. where $\lambda$ is indeterminate, I mean a variable. I want to say $(A-\lambda B)$ is of full rank matrix, assuming entries of $A-\lambda B$ is a polynomial matrix with linear polynomial.</p>
| Daniel McLaury | 3,296 | <p>No. If $\operatorname{rank} A < 2$ then either $A = 0$ or $\operatorname{rank} A = 1$, and similarly for $B$. Obviously if $A = 0$ then
$$\det(A - \lambda B) = - \lambda \det B = 0$$
since $B$ is not full-rank, and similarly if $B = 0$.</p>
<p>So suppose $\operatorname{rank} A = \operatorname{rank} B = 1$. Then the image of $A$ is some one-dimensional space $\operatorname{span} \{ v\}$ and the image of $B$ is some one-dimensional space $\operatorname{span} \{ w\}$. It follows that for any $u \in \mathbb{R}^4$
$$(A - \lambda B) u = A u - \lambda B u = c v + d w \in \operatorname{span}\{v,w\}$$
for some $c, d$. In particular, $\operatorname{rank} (A - \lambda B)$ is at most two, and thus $\det (A - \lambda B)$ is always zero.</p>
<hr>
<p>If you meant to say $\operatorname{rank} A \leq 2$ and $\operatorname{rank} B \leq 2$ instead, then the answer is yes; just take
$$A = \begin{pmatrix}
1 \\
& 1 \\
& & 0 \\
& & & 0 \end{pmatrix}, \qquad B = \begin{pmatrix}
0 \\
& 0 \\
& & 1 \\
& & & 1 \end{pmatrix}$$
for instance. In fact, any random pair of rank-two matrices will have this property with 100% probability.</p>
|
3,143,532 | <blockquote>
<p>Find <span class="math-container">$\det A$</span> and <span class="math-container">$\text{Tr} A$</span> for the matrix <span class="math-container">$A\in M_n(\mathbb{Q})$</span> such that <span class="math-container">$\sqrt[n]{p}$</span> is an eigenvalue of <span class="math-container">$A$</span>, where <span class="math-container">$p$</span> is a prime number or a positive integer such that the square root is irrational number.</p>
</blockquote>
<p>My attempt is described below. From hypothesis we know that
<span class="math-container">$$ \det(A-\sqrt[n]{p}I)=0,
$$</span> hence <span class="math-container">$ \det(A+\sqrt[n]{p}I)=0$</span> because the characteristic polynomial of matrix <span class="math-container">$A$</span> has rational coefficients. Multiplying these two relation we get <span class="math-container">$ \det(A^2-\sqrt[n]{p^2}I)=0$</span> and so on. </p>
<p>From here, I don't find something. How to proceed? Thanks.</p>
| Zacky | 515,527 | <p>We start off by some <span class="math-container">$x\rightarrow \frac{1}{x}$</span> substitutions while derivating under the integral sign: <span class="math-container">$$I(\theta)=\int_0^\infty \log \left(1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx\overset{x\rightarrow \frac{1}{x}}=\int_0^\infty \frac{\ln(1- 2\cos(2\theta) x^2 +x^4)}{x^2}dx$$</span>
<span class="math-container">$$I'(\theta)=4\int_0^\infty \frac{\sin(2\theta)}{x^4-2\cos(2\theta)x^2+1}dx\overset{x\rightarrow \frac{1}{x}}=4\int_0^\infty \frac{\sin(2\theta)x^2}{x^4-2\cos(2\theta)x^2+1}dx$$</span>
Now summing up the two integrals from above gives us:
<span class="math-container">$$\Rightarrow 2I'(\theta)=4\int_0^\infty \frac{\sin(2\theta)(1+x^2)}{x^4-2\cos(2\theta)x^2+1}dx=4\int_0^\infty \frac{\sin(2\theta)\left(\frac{1}{x^2}+1\right)}{x^2+\frac{1}{x^2}-2\cos(2\theta)}dx$$</span>
<span class="math-container">$$\Rightarrow I'(\theta)=2\int_0^\infty \frac{\sin(2\theta)\left(x-\frac{1}{x}\right)'}{\left(x-\frac{1}{x}\right)^2 +2(1-\cos(2\theta))}dx\overset{\large x- \frac{1}{x}=t}=2\int_{-\infty}^\infty \frac{\sin(2\theta)}{t^2 +4\sin^2 (\theta)}dt$$</span>
<span class="math-container">$$=2 \frac{\sin(2\theta)}{2\sin(\theta)}\arctan\left(\frac{t}{2\sin(\theta)}\right)\bigg|_{-\infty}^\infty=2\cos(\theta) \cdot \pi$$</span>
<span class="math-container">$$\Rightarrow I(\theta) = 2\pi \int \cos(\theta) d\theta =2\pi \sin \theta +C$$</span>
But <span class="math-container">$I(0)=0$</span> (see J.G. answer), thus:<span class="math-container">$$I(0)=0+C\Rightarrow C=0 \Rightarrow \boxed{I(\theta)=2\pi\sin(\theta)}$$</span></p>
|
3,143,532 | <blockquote>
<p>Find <span class="math-container">$\det A$</span> and <span class="math-container">$\text{Tr} A$</span> for the matrix <span class="math-container">$A\in M_n(\mathbb{Q})$</span> such that <span class="math-container">$\sqrt[n]{p}$</span> is an eigenvalue of <span class="math-container">$A$</span>, where <span class="math-container">$p$</span> is a prime number or a positive integer such that the square root is irrational number.</p>
</blockquote>
<p>My attempt is described below. From hypothesis we know that
<span class="math-container">$$ \det(A-\sqrt[n]{p}I)=0,
$$</span> hence <span class="math-container">$ \det(A+\sqrt[n]{p}I)=0$</span> because the characteristic polynomial of matrix <span class="math-container">$A$</span> has rational coefficients. Multiplying these two relation we get <span class="math-container">$ \det(A^2-\sqrt[n]{p^2}I)=0$</span> and so on. </p>
<p>From here, I don't find something. How to proceed? Thanks.</p>
| Random Variable | 16,033 | <p>Assume that <span class="math-container">$\theta \in (0, \pi).$</span></p>
<p>Then</p>
<p><span class="math-container">$$\begin{align}\int_{0}^{\infty} \log \left(1 -\frac{2 \cos (2\theta)}{x^{2}} +\frac{1}{x^{4}} \right) \, \mathrm dx &= \int_{0}^{\infty} \frac{ \log \left(1- 2 \cos(2 \theta) u^{2} +u^{4}\right)}{u^{2}} \, \mathrm du \\ &= 2\, \Re \int_{0}^{\infty} \frac{ \log \left( 1-u^{2}e^{2 i \theta} \right)}{u^{2}} \, \mathrm du \\ &= -4 \, \Re \int_{0}^{\infty} \frac{e^{2 i \theta}}{1-u^{2}e^{2i \theta}} \, \mathrm du \tag{1}\\ &= -4 \, \Re \left( e^{i \theta} \operatorname{artanh} \left(ue^{i \theta} \right) \Big|_{0}^{\infty} \right) \\ &=-4 \, \Re \left(e^{i \theta} \, \frac{i \pi}{2} \right) \tag{2} \\ &= 2 \pi \sin(\theta). \end{align}$$</span></p>
<hr />
<p><span class="math-container">$(1)$</span> Integrate by parts.</p>
<p><span class="math-container">$(2)$</span> If <span class="math-container">$ \theta \in (0, \pi )$</span>, then <span class="math-container">$$\lim_{u \to +\infty} \operatorname{artanh}(ue^{i \theta})=\lim_{u \to +\infty}\frac{1}{2} \, \log \left( \frac{1 +ue^{i \theta}}{1-ue^{i \theta}} \right) = \lim_{u \to +\infty}\frac{1}{2} \, \log \left( \frac{\frac{1}{u} +e^{i \theta}}{\frac{1}{u}-e^{i \theta}} \right) = \frac{i\pi}{2}$$</span> because the expression inside the <span class="math-container">$\log$</span> is approaching <span class="math-container">$-1$</span> from above the branch cut.</p>
|
4,367,330 | <p>Let <span class="math-container">$X_1, X_2,...$</span> be a sequence of independent uniform random variables on <span class="math-container">$(0,1)$</span>. Define: <span class="math-container">$$N := \text{min} \{n\geq 2: X_n < X_{n-1}\}.$$</span> Calculate <span class="math-container">$E(N)$</span>.</p>
<p>I think this problem asked about the expectations of the first dropping entry of the series, I also did a simulation and I think the answer is <span class="math-container">$e$</span>? But I'm not sure how to compute it. I tried using the definition of expectations and I compute that <span class="math-container">$P(N=2)$</span> is <span class="math-container">$1/2$</span>, but I stuck with computing <span class="math-container">$P(N=3)$</span>. Can anyone tell me how to do this?</p>
| drhab | 75,923 | <p>Observe that: <span class="math-container">$$N>n\iff X_1\leq X_2\leq\cdots\leq X_n\text{ for }n=2,3,\dots$$</span>so that by symmetry: <span class="math-container">$$\Pr(N>n)=\frac1{n!}\text{ for }n=2,3,\dots$$</span>
Now apply:<span class="math-container">$$\mathbb EN=\sum_{n=0}^{\infty}\Pr(N>n)$$</span></p>
|
412,051 | <p>Suppose
$X_i \sim N(0,1)$ (independent, identical normal distributions) </p>
<p>Then by Law of large number,
$$
\sqrt{1-\delta} \frac{1}{n}\sum_i^\infty e^{\frac{\delta}{2}X_i^2} \rightarrow \sqrt{1-\delta} \int e^{\frac{\delta}{2}x^2}\frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}x^2}dx =1
$$</p>
<p>However, according to simulations, this approximation doesn't seem to work when $\delta$ close to one. Is that ture or just need to run larger samples? Thanks !</p>
<p>Update (6/6): As sos440 mentioned, there's a typo and now fixed.</p>
| Avitus | 80,800 | <p>Your general solution is not correct.
The differential equation you are dealing with is the Laplace equation.</p>
<p>For the general theory of Laplace equation, please have a look at the introductory exposition in</p>
<p><a href="http://en.wikipedia.org/wiki/Laplace" rel="nofollow">http://en.wikipedia.org/wiki/Laplace</a>'s_equation</p>
<p>Once the setting becomes familiar, you should get used to the method called "separation of variables" and the Fourier series expansions of periodic functions.</p>
<p>Some good references (to start with):</p>
<p><a href="http://en.wikipedia.org/wiki/Separation_of_variables" rel="nofollow">http://en.wikipedia.org/wiki/Separation_of_variables</a> (paragraph: PDEs)</p>
<p><a href="http://en.wikipedia.org/wiki/Fourier_series" rel="nofollow">http://en.wikipedia.org/wiki/Fourier_series</a></p>
<p><em>Edit</em> I would add some explicit fomulae. Let us assume a separable solution
$u(x,y)=X(x)Y(y)$: the Laplace equation leads to $X''(x)=k X(x)$ and $Y''(y)=-k Y(y)$
for some real constant $k$. Depending whether $k<0, k=0$ or $k>0$ the solutions to the ordinary diff. eqs for $X$ and $Y$ are different. If we had some extra boundary conditions (periodic for example) we could select a choice for $k$ at this stage. In our case, however, we should consider all the combinations of $u(x,y)=X(x)Y(y)$ for $k<0, k=0$ or $k>0$. Any of these choices must satisfy the boundary condition $\frac{\partial u(x,0)}{\partial y}=h(x)$, for all $x\in\mathbb R$. Depending on the form of $h(x)$ we could, in general, determine the "right" $k$. As you can see I am pretty vague, as Iwe do not know anything about $h(x)$ and we cannot choose $k$ using other boundary conditions.</p>
<p>Let us move to a more explicit example. We select the case $k:=\omega^2>0$, obtaining through superposition the separable solution $u(x,y)=\sum_{i}(A_ie^{\omega_i x}+B_ie^{-\omega_i x})(C_ie^{i\omega_i y}+D_ie^{-i\omega_i y})$, for real coefficients $A_i,B_i,C_i,D_i$. We have to impose the boundary condition obtaining</p>
<p>$\sum_i(A_ie^{\omega_i x}+B_ie^{-\omega_i x})(C_i-D_i)i\omega_i=h(x)$,</p>
<p>for all $x\in \mathbb R$. At this stage a Fourier analysis of $h(x)$ is needed. Note that the Fourier analysis is done using</p>
<p>$\sum_i(\tilde{A}_ie^{\omega_i x}+\tilde{B}_ie^{-\omega_i x})=h(x)$,</p>
<p>where $\tilde{A}_i=i\omega_iA_i(C_i-D_i)$ and $\tilde{B}_i=i\omega_iB_i(C_i-D_i)$.</p>
|
222,312 | <p>How to prove that if you have $x^*$ such that $x^*=\text{psuedoinverse}(A) b$, and $Ay=b$, then
$$\Vert x^* \Vert_2 \leq \Vert y \Vert_2$$</p>
| chaohuang | 27,973 | <p>Let $A: \mathbb{U} \mapsto \mathbb{V}$, then in terms of SVD, we can write $A$ as $$A=\sum_{n=1}^R\sigma_nv_nu_n^{\dagger},$$ where $\sigma_n$ is a nonzero singular value; $u_n$ and $v_n$ are the right and left singular vector, respectively; $R$ is the rank of $A$.</p>
<p>Since {$u_n$} form an orthonormal basis of $\mathbb{U}$, we can expand $y$ as $$y=\sum_{m=1}^M\alpha_m u_m,$$ where $M\ge R$ is the dimension of $\mathbb{U}$.</p>
<p>So $$b=Af=\sum_{n=1}^R\sigma_nv_nu_n^{\dagger}\sum_{m=1}^M\alpha_mu_m=\sum_{n=1}^R\alpha_n\sigma_nv_n$$</p>
<p>Also, the pseudoinverse is $$A^+=\sum_{m=1}^R\frac{1}{\sigma_m}u_mv_m^{\dagger}$$</p>
<p>Then $$x^*=A^+b=\sum_{m=1}^R\frac{1}{\sigma_m}u_mv_m^{\dagger}\sum_{n=1}^R\alpha_n\sigma_nv_n=\sum_{m=1}^R\alpha_mu_m$$.</p>
<p>Finally, we can see that $$\|y||_2=(\sum_{m=1}^M\alpha_m^2)^{1/2},$$ $$\|x^*||_2=(\sum_{m=1}^R\alpha_m^2)^{1/2}.$$
Therefore, $$\|x^*||_2 \le \|y||_2,$$
where equality holds when $M=R$ or $\alpha_m=0$ for $m=R+1, \cdots, M$.</p>
<p>In other words, if we define $$\|y_{null}\|_2=(\sum_{m=R+1}^M\alpha_m^2)^{1/2},$$ we have $$\|y\|_2^2=\|x^*||_2^2+\|y_{null}\|_2^2.$$</p>
|
577,490 | <p>I wonder if I have approached this in the right way. I'm not sure if I have interpreted the question correctly, or made correct use of the successor function. Thank you in advance</p>
<blockquote>
<p><strong>Question:</strong> Prove that $¬∃n∅ = n^+$</p>
</blockquote>
<p>My approach:</p>
<p>Prove that $¬∃n∅ = n^+$</p>
<p>(i) In other words, prove (by counterexample) that $∅ ≠ n^+$.</p>
<p>(ii) Let $n^+ = 1 \Rightarrow n^+ = ∅^+ \Rightarrow n^+ = ∅^+ = \{∅\}$.</p>
<p>(iii) But $∅ ≠ \{∅\} \Rightarrow ¬∃n∅ = n^+$.</p>
| MarnixKlooster ReinstateMonica | 11,994 | <p>Here is how I would approach this, resulting in a proof which is in essence the same as the one from the first answer.</p>
<p>In a slightly different notation, you're asked to prove $\;\lnot \langle \exists n :: n^+ = \emptyset \rangle \;$.</p>
<p>Let's see what we can do with $\;n^+ = \emptyset\;$ :
\begin{align}
& n^+ = \emptyset \\
\equiv & \;\;\;\;\;\text{"basic property of $\;\emptyset\;$"} \\
& \langle \forall x :: \lnot(x \in n^+) \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;^+\;$"} \\
& \langle \forall x :: \lnot(x \in n \cup \{n\}) \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\cup\;$; definition of $\;\{\cdot\}\;$"} \\
& \langle \forall x :: \lnot(x \in n \;\lor\; x = n) \rangle \\
\equiv & \;\;\;\;\;\text{"DeMorgan"} \\
& \langle \forall x :: x \not\in n \;\land\; x \ne n \rangle \\
\Rightarrow & \;\;\;\;\;\text{"choose $\;x := n\;$; $\;n \ne n\;$ is false"} \\
& \text{false} \\
\end{align}
This proves that there is no $\;n\;$ such that $\;n^+ = \emptyset\;$, or in other words, it proves the original statement.</p>
|
2,433,765 | <p><a href="https://i.stack.imgur.com/AolxA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AolxA.png" alt="enter image description here"></a></p>
<p>In the following diagram of a triangle, $\overline{AB} = \overline{BC} = \overline{CD}$ and $\overline{AD} = \overline{BD}$. Find the measure of angle $D$.</p>
<p>I know this should be easy but I am stuck. I started by saying angle $\widehat{ACB} = \theta$ and that the supplement angle $\widehat{BCD} = 180^\circ-\theta$. I know that angle $\widehat{CAB}=\theta$ as well and that angle $\widehat{ABC} = 180^\circ-2\theta$. In addition, angles $\widehat{CDB}$ and $\widehat{CBD}$ are equal. I am not sure how to solve for angle $\widehat{CDB}$ ... is it possible to find an exact numerical measure? I hate overlooking something obvious. Thank you for your help.</p>
| Edward Porcella | 403,946 | <p>$\triangle ABD$ is isosceles with each base angle double the angle at the vertex. The angle at D is therefore $36^o$. Euclid constructs this triangle in <em>Elements</em>,IV,10, as necessary for construction of the regular pentagon. </p>
|
778,140 | <p>I want to draw a box plot, which requires that I know the median, the lower and upper quartiles, and the minimum and maximum values of my data.</p>
<p>I understand that the quartiles are simply the value on certainly "percentile" of the cumulative frequency of the data.</p>
<p>So lower quartile = the value of the observation on the 25th percentile of the data.
Now my question (for AQA GCSE prep) is - what if taking 25% of my data ends up in a decimal number, let's say, $3.5$. And my data consists of classes in a grouped frequency table. And two of my classes are:</p>
<p>$ class 1$ || $ 2 <= h < 3.5$</p>
<p>$ class 2$ || $ 3.5 <= h < 5$</p>
<p>So when I take 25% of 3.5 falls in between two classes. Which value should I choose as the lower quartile? Should it be $class 1$, or $class 2$? Should my rounding of 3.5 be the same as regular rounding is done, i.e. just rounding up to 4 (hence selecting $class 2$)? Or should I round choose $class 1$ for some reason?</p>
| Thomas Rasberry | 265,575 | <p>Intro textbooks tend to use one of two different methods for boxplot/five number summary construction, in my experience. </p>
<p>The first method is to apply your percentile to the total number of observations; for example, the first quartile of 14 data points ordered least to greatest is the data value located in position $14*(0.25)= 3.5;$ that is, your fourth quartile is the value in the "3.5th place." </p>
<p>Well, there's no such thing as the "three point fifth place," so the first method has you round up to the nearest <em>larger</em> integer; this means you round this "place value calculation" up when you have a nonzero decimal in your computation (even if normal rounding rules would have you round down). In this case, 3.5 rounds up to 4, so the first quartile is whatever data value is in 4th place in the list of 14 data values ordered least to greatest. This method, if defined this way in your text, is good to use for finding <em>any</em> percentile: for instance, the 80th percentile is $(.8)(14) = 11.2,$ or the 12th place, and the median or 50th percentile is $(.5)(14) = 7,$ which lacks a nonzero fractional part, indicating that the median is to be taken as the average of the 7th place and its "next door neighbor," the data value in 8th place. </p>
<p>The second method is slightly different; after finding median, you find the first quartile by taking the median of the values to the left of your median placement in your ordered list and you find the third quartile by taking the median of the data place values beyond the place value of your median. In this case, the boxplot for your 14-point data set would not change, but if it had, say, 13 elements instead, the median would be in place value $(0.5)(13) = 6.5$ or the 7th place, the first quartile is the median, then, of the first six values in the ordered list, meaning it is the average of the 3rd and 4th values in the ordered list (in the previous paragraph, the quartile would have been the value in place number $13*.25 = 3.25 \to 4$th place). </p>
<p>Double-check your text to see which applies to you, but it is likely one of these two ways unless I am expressing something in error above. </p>
|
17,103 | <p>I'm reading some very old papers (by Birch et al) on quadratic forms and I don't get the following point: </p>
<blockquote>
<p>If <span class="math-container">$f$</span> is a quadratic form in <span class="math-container">$X_1,X_2,\cdots,X_n$</span> over a
finite field, then one can change variables such that <span class="math-container">$f$</span> can be written as <span class="math-container">$\sum_{i = 1}^s Y_{2i - 1}Y_{2i} + g$</span>, where <span class="math-container">$g$</span> is a quadratic form which involves
variables other than
<span class="math-container">$Y_1,Y_2,\cdots,Y_{2s}$</span> and has order at
most 2 (i.e. can be written using at
most two linear forms).</p>
</blockquote>
<p>So either this is a well-known result - but I don't find a reference - or either this is easy to see, but in that case I'm just missing the point. By the way, is this really true in characteristic 2? </p>
<p>(And in fact I'm not sure what role is played by the fact that the field is finite...)</p>
| Pete L. Clark | 1,149 | <p>Let me try a similar explanation with different words. (Note that my explanation does not cover characteristic <span class="math-container">$2$</span>.)</p>
<p>A quadratic form is <strong>nondegenerate</strong> if any of its associated symmetric matrices has nonzero determinant. (Alternately, if the associated bilinear form <span class="math-container">$B(x,y) = q(x+y) - q(x) - q(y)$</span> is nondegenerate in the usual sense: <span class="math-container">$B(x,y) = 0 \ \forall y \in K \implies x = 0$</span>.)</p>
<p>Let <span class="math-container">$K$</span> be a field of characteristic different from <span class="math-container">$2$</span>. The <strong>hyperbolic plane</strong> is the special quadratic form</p>
<p>H(x,y) = xy.</p>
<p>(As with any quadratic form over <span class="math-container">$K$</span>, it can be diagonalized: <span class="math-container">$\frac{1}{2} x^2 - \frac{1}{2} y^2$</span>.)</p>
<p>A nondegenerate quadratic form <span class="math-container">$q(x_1,\ldots,x_n)$</span> is <strong>isotropic</strong> if there exist <span class="math-container">$a_1,\ldots,a_n \in K$</span>, not all <span class="math-container">$0$</span>, such that <span class="math-container">$q(a_1,\ldots,a_n) = 0$</span> and otherwise <strong>anisotropic</strong>.</p>
<p>Witt Decomposition Theorem: Any quadratic form <span class="math-container">$q$</span> can be written as an orthogonal direct sum of an identically zero quadratic form, an anistropic quadratic form, and some number of hyperbolic planes. In particular, any isotropic quadratic form <span class="math-container">$q(x_1,...,x_n)$</span> can be written, after a linear change of variables, as <span class="math-container">$x_1 x_2 + q(x_3,...,x_n)$</span>.</p>
<p>For your purposes, you might as well assume your quadratic form is nondegenerate -- otherwise, it simply involves more variables than actually appear!</p>
<p>Now over a finite field, the Chevalley-Warning theorem implies that any nondegenerate quadratic form in at least three variables is isotropic, so that by Witt Decomposition, you can split off a hyperbolic plane. If you still have at least three variables, you can do this again. Repeated application gives your result.</p>
<p>References:</p>
<p>For Chevalley-Warning:</p>
<p><a href="http://alpha.math.uga.edu/%7Epete/4400ChevalleyWarning.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/4400ChevalleyWarning.pdf</a></p>
<p>For Witt Decomposition:</p>
<p><a href="http://alpha.math.uga.edu/%7Epete/quadraticforms.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/quadraticforms.pdf</a></p>
|
2,552,720 | <p>How would you prove algebraically that the assertion $\exists (x,y) \in (0,1)^2. x +y > 1 + xy$ <a href="http://www.wolframalpha.com/input/?i=x%20%2B%20y%20%3E%20%201%20%2B%20xy,%200%20%3C%20x%20%3C%201,%200%20%3C%20y%20%3C%201" rel="nofollow noreferrer">is false</a>?</p>
| nonuser | 463,553 | <p>$$L-R =x+y-1-xy = x(1-y)-(1-y)= (1-y)(x-1)<0$$</p>
|
2,552,720 | <p>How would you prove algebraically that the assertion $\exists (x,y) \in (0,1)^2. x +y > 1 + xy$ <a href="http://www.wolframalpha.com/input/?i=x%20%2B%20y%20%3E%20%201%20%2B%20xy,%200%20%3C%20x%20%3C%201,%200%20%3C%20y%20%3C%201" rel="nofollow noreferrer">is false</a>?</p>
| PunkZebra | 120,252 | <p>I'll just work assuming that $x$ and $y$ can't equal $1$, otherwise the inequality isn't true for $x=0$ and $y=1$.</p>
<p>Squaring them both gets rid of the $xy$ term
$$x+y>1+xy$$
$$x^2+2xy+y^2>1+2xy+x^2y^2$$
$$x^2+y^2>1+x^2y^2$$
Now $x^2+y^2$ is a positive term less or equal to $1$ since it's inside the unit circle. Now since $x^2y^2$ is a positive term the inequality is not true since the left side is less than $1$ and the right is greater than $1$</p>
|
2,895,129 | <blockquote>
<p>Let $f(x) \leq g(x)$ for all $x$ in an open set $D$, and let $a$ be an interior point of $D$ such that $f(a) = g(a)$ and both $f, g$ are differentiable at $a$.
Does it follows that $f'(a)=g'(a)$?</p>
</blockquote>
<p>Is that right, and if so, why? If not then please give examples.</p>
| Jack M | 30,481 | <p>Fundamentally, the intuition here is that if two curves touch each other ($f(a)=g(a)$) and have different derivatives at that point, then one has to actually <em>cross</em> the other. If we are to have $f\leq g$ on the entire interval, then the only way those two curves can touch is if they just "kiss", tangent to each other, and then separate again without actually crossing.</p>
<p>For example, if $f'(a)>g'(a)$, then $f$ is steeper than $g$ at $a$. Therefore, "obviously", at the point $a$, $f$ is cutting up across the curve of $g$ from below, and we will end up with $f(a+h)>g(a)$ for sufficiently small $h$. Now, the question is, how do we turn this visual intuition into a mathematical proof. There are many ways. One is to just look at the definition of the derivative directly. For sufficiently small $h$, we will have:</p>
<p>$$\frac{f(a+h)-f(a)}{h}>\frac{g(a+h)-g(a)}{h}$$</p>
<p>This is true because it's true in the limit $h\to 0$, which means it has to be true for sufficiently small $h$. In detail: if we call the left hand side $A(h)$ and the right hand side $B(h)$, we have $\lim_{h\to 0} A(h)>\lim_{h\to 0} B(h)$, and you can prove from the definition of a limit that this implies $A(h)>B(h)$ for sufficiently small $h$.</p>
<p>Anyway, going back to the inequality, we can cancel out $f(a)$ and $g(a)$ since they're equal to one another and we can multiply both sides by $h$ to get, for sufficiently small $h$:</p>
<p>$$f(a+h)>g(a+h)$$</p>
<p>Which proves exactly what we said: because $f$ is steeper than $g$ at the point $a$, $f$ will be higher than $g$ immediately after $a$.</p>
<p>The case $f'(a)<g'(a)$ can be handled in a similar way.</p>
|
3,621,836 | <p>Let's say I have a large determinant with scalar elements:</p>
<p><span class="math-container">$$\begin{vmatrix} x\cdot a & x\cdot b & x\cdot c & x\cdot d \\ x\cdot e & x\cdot f & x\cdot g & x\cdot h \\ x\cdot i & x\cdot g & x\cdot k & x\cdot l \\ x\cdot m & x\cdot n & x\cdot o & x\cdot p\end{vmatrix}$$</span></p>
<p>Is it valid to factor out a term that's common to every element of the determinant? Is the following true:</p>
<p><span class="math-container">$$\begin{vmatrix} x\cdot a & x\cdot b & x\cdot c & x\cdot d \\ x\cdot e & x\cdot f & x\cdot g & x\cdot h \\ x\cdot i & x\cdot g & x\cdot k & x\cdot l \\ x\cdot m & x\cdot n & x\cdot o & x\cdot p\end{vmatrix} = x \cdot \begin{vmatrix} a & b & c & d \\ e & f & g & h \\ i & g & k & l \\ m & n & o & p\end{vmatrix}$$</span></p>
| lhf | 589 | <p>Since <span class="math-container">$\omega^3=1$</span> and <span class="math-container">$\omega^2+\omega+1=0$</span>, we have
<span class="math-container">$$
\begin{align}
(2-\omega)(2-\omega^2)(2-\omega^{19})(2-\omega^{23})
&=
(2-\omega)(2-\omega^2)(2-\omega^{1})(2-\omega^{2})
\\&=
((2-\omega)(2-\omega^2))^2
\\&=
(-2 \omega^2 - 2 \omega + 5)^2
\\&=
(-2(\omega^2 + \omega) + 5)^2
\\&=
(-7)^2
=49
\end{align}
$$</span></p>
|
77,290 | <p>How would you determine all integers $m$ such that the following is true? </p>
<p>$$\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} .$$</p>
<p>Note that $\lfloor \cdot \rfloor$ means the greatest integer function. Also, $x$ must be a positive real number.</p>
| Robert Israel | 8,508 | <p>If $k + j/10 \le x < k + (j+1)/10$ where $k$ and $j$ are nonnegative integers and $j \le 9$, $\lfloor 2x \rfloor = \begin{cases} 2k & 0 \le j \le 4 \\ 2k+1 & 5 \le j \le 9 \end{cases}$ while $\lfloor 5x \rfloor = \begin{cases} 5k + j/2 & j \ \text{even} \\ 5k + (j-1)/2 & j \ \text{odd} \end{cases}$.</p>
<p>Going over the various cases $j = 0$ to $9$, I find two cases where $f(x) = \frac{1}{1/\lfloor 2x \rfloor + 1/\lfloor 5x \rfloor} $ is an integer:</p>
<p>1) if $j = 0$ or $1$ and $k > 0$ is divisible by 7, $f(x) = 10 (k/7)$.</p>
<p>2) if $k=2$ and $j=4$, $f(x) = 3$.</p>
<p>Case (2) comes about as follows: if $j = 4$, $f(x) = \frac{1}{1/(2k) + 1/(5k+2)} = \frac{(2k)(5k+2)}{7k+2}$. Note that $\gcd(k,7k+2) = \gcd(k,2) = 1$ or $2 $ while $\gcd(5k+2,7k+2) = \gcd(5k+2,2k) = \gcd(k+2,2k) =1$, $2$ or $4$. The largest possible value of the denominator that could divide the numerator is thus $2 \times 2 \times 4 = 16$, which occurs for $k=2$.</p>
|
2,082,062 | <p>In Avner Friendman's Mondern Analysis book, he makes a statement that has stumped me when proving that for $\{f_n\}$ a sequence of measurable functions, $\sup_n f_n$ And $\inf f_n$ Are measurable.</p>
<blockquote>
<p>The assertion for $\inf_n f_n $ follows from $\inf (f_n )=- \sup (-f_n)$ .</p>
</blockquote>
<p>I really struggle with the concept of inf and sup of a sequence of functions, so I do not see why this statement is true. Why does $\inf (f_n )=- \sup (-f_n)$? Thank you!</p>
| copper.hat | 27,978 | <p>It is straightforward to check that if $L_1, L_2$ are linear operators then
so is $L_1 \circ L_2$.</p>
<p>It is straightforward to check that $n(x) = -x$ is a linear functional.</p>
<p>Hence given some linear $L$, the map $n \circ L$ is linear.</p>
|
254,926 | <p>I was working on a problem involving perturbation methods and it asked me to sketch the graph of $\ln(x) = \epsilon x$ and explain why it must have 2 solutions. Clearly there is a solution near $x=1$ which depends on the value of $\epsilon$, but I fail to see why there must be a solution near $x \rightarrow \infty$. It was my understanding that $\ln x$ has no horizontal asymptote and continues to grow indefinitely, where for really small values of $\epsilon, \epsilon x$ should grow incredibly slowly. How can I 'see' that there are two solutions?</p>
<p>Thanks!</p>
| Spenser | 39,285 | <p>Right after the first solution, you have that $\ln x>\epsilon x \Rightarrow \ln x-\epsilon x>0$. However,
$$\lim_{x\to\infty}(\ln x-\epsilon x)=\lim_{x\to\infty}\ln\frac{x}{e^{\epsilon x}}\to-\infty
$$
since $\frac{x}{e^{\epsilon x}}\to 0$. (Assuming $\epsilon>0$)</p>
<p>Hence, $\ln x-\epsilon x$, must cross the $x$-axis a second time which gives your second solution.</p>
|
4,517,597 | <p>Suppose <span class="math-container">$\{f_n\}_{n \in \mathbb{N}}$</span> is a family of bounded, differentiable, monotone increasing functions on <span class="math-container">$[0,1]$</span>, which converge uniformly to a limit <span class="math-container">$f$</span>. Also, suppose we know that <span class="math-container">$f_n'$</span> is <span class="math-container">$\alpha$</span>-Lipschitz continuous for some constant <span class="math-container">$\alpha$</span> (not depending on <span class="math-container">$n$</span>). I want to analyze the differentiability of <span class="math-container">$f$</span> and the convergence of <span class="math-container">$f_n'$</span>, if this is even possible.</p>
<p>Of course, the monotone increasing property of <span class="math-container">$f_n$</span> implies that <span class="math-container">$f$</span> is also monotone increasing, so it is differentiable Lebesgue almost surely on <span class="math-container">$[0,1]$</span>. Is it possible to say that <span class="math-container">$f_n'$</span> converges to <span class="math-container">$f'$</span> pointwise, wherever <span class="math-container">$f'$</span> exists (even if only on a subsequence)?</p>
| Jacob Manaker | 330,413 | <p>Yes. I will prove this via functional analysis; I do not know if there is a simpler method.</p>
<ol>
<li>Note that <span class="math-container">$f_n$</span> are uniformly bounded, since they converge uniformly.</li>
<li>Suppose <span class="math-container">$m\leq f_n\leq M$</span> for any <span class="math-container">$n$</span>. By the intermediate value theorem, there is some point <span class="math-container">$a_n\in[0,1]$</span> and <span class="math-container">$b_n\in[0,M-m]$</span> such that <span class="math-container">$f_n'(a_n)=b_n$</span>.</li>
<li>Each <span class="math-container">$f_n'$</span> is uniformly bounded in <span class="math-container">$L^{\infty}$</span>, since <span class="math-container">$\{f_n'\}_n$</span> are equi-Lipschitz on a compact domain, and attain the similar values of the <span class="math-container">$b_n$</span>. The domain <span class="math-container">$[0,1]$</span> has finite measure; thus <span class="math-container">$f_n'$</span> are uniformly bounded in any <span class="math-container">$L^p$</span> for <span class="math-container">$p\in[1,\infty]$</span>.</li>
<li>For any <span class="math-container">$p\in(1,\infty)$</span>, the space <span class="math-container">$L^p$</span> is reflexive. Pick your favorite such <span class="math-container">$p$</span> (mine is <span class="math-container">$p=2$</span>); then there is a subsequence (which I will also call <span class="math-container">$f_n'$</span>) that converges weakly, by separable Banach-Alaoglu. Call the limit <span class="math-container">$g$</span>, but note that <span class="math-container">$g$</span> is only well-defined up to a.e. equivalence.</li>
<li>Since <span class="math-container">$[0,1]$</span> has finite measure, if <span class="math-container">$\phi\in L^\infty$</span>, then <span class="math-container">$\phi\in L^2$</span>. Thus <span class="math-container">$$\int{\phi f_n'}\to\int{\phi g}$$</span> by weak convergence in <span class="math-container">$L^2$</span>. So actually <span class="math-container">$f_n'\rightharpoonup g$</span> in <span class="math-container">$L^1$</span>.</li>
<li>Fix <span class="math-container">$\epsilon$</span> and let <span class="math-container">$U(\epsilon)_n=\{x:f_n'(x)-g(x)\geq\epsilon\}$</span> and <span class="math-container">$V(\epsilon)_n=\{x:g(x)-f_n'(x)\geq\epsilon\}$</span>. Then <span class="math-container">$$0\leq\epsilon|U(\epsilon)_n|\leq\int{1_{U(\epsilon)_n}(f_n'-g)}\to0$$</span> Likewise <span class="math-container">$|V(\epsilon)_n|\to0$</span>, and so <span class="math-container">$f_n'\to g$</span> in measure as well as weakly.</li>
<li>Convergence in measure implies convergence a.e. up to a subsequence. Again, I will call that subsequence <span class="math-container">$f_n'$</span>.</li>
<li>Let <span class="math-container">$U$</span> be a dense set on which <span class="math-container">$f_n'\to g$</span> pointwise. A pointwise limit of uniformly Lipschitz functions is also Lipschitz (exercise); thus any representative of the a.e.-equivalence class <span class="math-container">$g$</span> is uniformly continuous on <span class="math-container">$U$</span>. A uniformly continuous function on a dense set has a unique continuous extension to that set's closure; thus <span class="math-container">$g$</span> has representatives that are continuous on <span class="math-container">$[0,1]$</span>. Pick one such, and also call it <span class="math-container">$g$</span>.</li>
<li>Pick <span class="math-container">$x\in[0,1]$</span>. Then there exists <span class="math-container">$\{x_t\}_{t=0}^\infty\in U^\omega$</span> converging to <span class="math-container">$x$</span>. For any <span class="math-container">$t$</span>, <span class="math-container">\begin{align*}
\limsup_{n\to\infty}{|g(x)-f_n'(x)|}&\leq\limsup_{n\to\infty}{|g(x)-g(x_t)|+|g(x_t)-f_n'(x_t)|+|f_n'(x_t)-f_n'(x)|} \\
&\leq|g(x)-g(x_t)|+0+\alpha|x_t-x| \\
&\to0
\end{align*}</span> as <span class="math-container">$t\to\infty$</span>. Thus <span class="math-container">$f_n'\to g$</span> pointwise everywhere.</li>
<li>For any set <span class="math-container">$[0,x]$</span>, we have <span class="math-container">$$\int_0^x{g(x)\,dx}=\int{1_{[0,x]}g}=\lim_{n\to\infty}{\int{1_{[0,x]}f_n'}}=\lim_{n\to\infty}{(f_n(x)-f_n(0))}=f(x)-f(0)$$</span></li>
<li>Since <span class="math-container">$g$</span> is continuous (everywhere), <span class="math-container">$f'=g$</span> everywhere by Lebesgue's differentiation theorem.</li>
</ol>
|
1,136,109 | <p>Suppose that $a,b$ are reals such that the roots of $ax^3-x^2+bx-1=0$ are all positive real numbers. Prove that: </p>
<p>$(i)~~0\le 3ab\le 1$<br>
$(ii)~~b\ge \sqrt3$. </p>
<p>My attempt: </p>
<p>I could solve the first part by Vieta's theorem. But, I am stuck on the second part. Please help. Thank you.</p>
| Macavity | 58,320 | <p>Let $x, y, z > 0$ be the three roots. Then, $x+y+z = xyz = \dfrac1a$ and $xy+yz+zx = \dfrac{b}a$</p>
<p>$(i),\quad $ Clearly, $a, b > 0$. Also $(x+y+z)^2 \ge 3(xy+yz+zx) \implies \dfrac1{a^2} \ge 3\dfrac{b}a \implies 1 \ge 3ab$. </p>
<p>For $(ii),\quad (xy+yz+zx)^2 \ge 3xyz(x+y+z) \implies \dfrac{b^2}{a^2} \ge 3\dfrac{1}{a^2} \implies b \ge \sqrt3$.</p>
<hr>
<p>P.S. In case the second inequality used is not familiar, you can show that it is equivalent to the following rearrangement:
$$(xy)^2+(yz)^2+(zx)^2 \ge (xy)(yz)+(yz)(zx)+(zx)(xy)$$</p>
|
478,912 | <p>Ok so I get the basics of this, I just can't put it all together. $U$ is contained in the cartesian product of $\mathbb R^n \times \mathbb R$. What is $\mathbb R^n \times \mathbb R$ though? I know what a cartesian product is, but this seems weird to me. Someone please help explain this part, thanks!</p>
| azimut | 61,691 | <p>$\mathbb R^n\times \mathbb R$ is the set of all pairs
whose first component comes from the set $\mathbb R^n$ and the second component comes from the set $\mathbb R$.
So
$$U = (\mathbf v, x)$$
with $\mathbf v\in\mathbb R^n$ and $x\in \mathbb R$.</p>
<p>Since $\mathbb R^n$ is the set of all $n$-tuples over $\mathbb R$, the $n$-tuple $\mathbf v$ has the form $\mathbf v = (v_1,v_2,\ldots,v_n)$. So $U$ can also be written down as
$$
U = ((v_1,v_2,\ldots,v_n),x)
$$
with $v_1,v_2,\ldots,x\in\mathbb R$.
This is occasionally abbreviated as
$$
U = (v_1,v_2,\ldots,v_n;x)
$$
or even
$$
U = (v_1,v_2,\ldots,v_n,x).
$$</p>
|
28,348 | <p>Thomson et al. provide a proof that <span class="math-container">$\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$</span> in <a href="http://classicalrealanalysis.info/documents/TBB-AllChapters-Landscape.pdf#page=95" rel="noreferrer">this book (page 73)</a>. It has to do with using an inequality that relies on the binomial theorem:
<a href="https://i.stack.imgur.com/n3dxw.png" rel="noreferrer"><img src="https://i.stack.imgur.com/n3dxw.png" alt="enter image description here" /></a></p>
<p>I have an alternative proof that I know (from elsewhere) as follows.</p>
<hr />
<p><strong>Proof</strong>.</p>
<p><span class="math-container">\begin{align}
\lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0
\end{align}</span></p>
<p>Then using this, I can instead prove:
<span class="math-container">\begin{align}
\lim_{n\rightarrow \infty} \sqrt[n]{n} &= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline
& = \exp{0} \newline
& = 1
\end{align}</span></p>
<hr />
<p>On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is:
<span class="math-container">\begin{align}
\lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}}
\end{align}</span></p>
<p>I know such an identity would hold for bounded <span class="math-container">$n$</span> but I'm not sure I can use this identity when <span class="math-container">$n\rightarrow \infty$</span>.</p>
<p><strong>Question:</strong></p>
<p>If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence <span class="math-container">$x_n$</span>, can I always assume:
<span class="math-container">\begin{align}
\lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n)
\end{align}</span>
Or are there sequences that invalidate that identity?</p>
<hr />
<p>(Edited to expand the last question)
given any sequence <span class="math-container">$x_n$</span>, can I always assume:
<span class="math-container">\begin{align}
\lim_{n\rightarrow \infty} x_n &= \exp(\log \lim_{n\rightarrow \infty} x_n) \newline
&= \exp(\lim_{n\rightarrow \infty} \log x_n) \newline
&= \lim_{n\rightarrow \infty} \exp( \log x_n)
\end{align}</span>
Or are there sequences that invalidate any of the above identities?</p>
<p>(Edited to repurpose this question).
Please also feel free to add different proofs of <span class="math-container">$\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$</span>.</p>
| robjohn | 13,854 | <p>Let $n$ be an integer $n>2$ and real $x>0$, the binomial theorem says
$$
(1+x)^n>1+nx+\frac{n(n-1)}{2}x^2
$$
Let $N(x)=\max(2,1+\frac{2}{x^2})$. For $n>N(x)$, we get that $\frac{n(n-1)}{2}x^2>n$. Thus, for any $x>0$, we get that for $n>N(x)$
$$
1<\sqrt[n]{n}<1+x
$$
Thus, we have
$$
1\le\liminf_{n\to\infty}\sqrt[n]{n}\le\limsup_{n\to\infty}\sqrt[n]{n}\le 1+x
$$
Since this is true for any $x>0$, we must have
$$
\lim_{n\to\infty}\sqrt[n]{n}=1
$$</p>
|
2,694,122 | <blockquote>
<p>Given that a function $f$ is defined as $$f(x)=1+2x+3x^2+4x^3+...$$. We have to prove that $f$ is continuous on $[0,\frac{1}{8}]$ and evaluate $$\int_{0}^{\frac{1}{8}}f(x)dx$$ </p>
</blockquote>
<p>I am not sure in a particular area. I have proved $f$ is continuous on $[0,\frac{1}{8}]$. But the evaluation of integral gives </p>
<p>$$\int_{0}^{\frac{1}{8}}f(x)dx=\sum_{n=1}^{\infty}\frac{1}{8^{n}}=\frac{8}{7}(1-({\frac{1}{8})^n})= \frac{8}{7}(1-\frac{1}{8^n})$$. </p>
<p>I think I am wrong somewhere in calculations. Please guide me where I am wrong. Any help or suggestion will be precious.</p>
| operatorerror | 210,391 | <p>Another approach: Evaluate the series first, then integrate
$$
f(x)=\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n+
\sum_{n=0}^\infty x^n\\
=x\sum_{n=0}^\infty nx^{n-1}+\frac{1}{1-x}\\
=x\frac{1}{(1-x)^2}+\frac{1}{1-x}
$$
Now both integrating and checking continuity (it's the sum of two rational functions, neither singularity is contained in the interval $[0,1/8]$) isn't too bad. Indeed,
$$
\int f(x)\mathrm dx=\int \frac{x}{(1-x)^2}\mathrm dx-\ln|1-x|\\
\stackrel{u=1-x}=-\ln|1-x|-\int\frac{1}{u^2}\mathrm du+\int\frac1u\mathrm du\\
=\frac{1}{1-x}+C
$$
Now plugging in we find
$$
\int_0^{1/8}f(x)\mathrm dx=\frac{1}{1-1/8}-1=1/7
$$</p>
|
3,477,532 | <p>In the cyclic quadrilateral <span class="math-container">$ABCD$</span>, <span class="math-container">$AB:BC:CD:DA=1:9:9:8$</span>, <span class="math-container">$AC$</span> intersects <span class="math-container">$BD$</span> at <span class="math-container">$P$</span>, what is <span class="math-container">$S_{\triangle PAB}:S_{\triangle PBC}:S_{\triangle PCD}:S_{\triangle PDA}$</span>?</p>
<p><a href="https://i.stack.imgur.com/RBUH0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RBUH0.png" alt="Diagram"></a></p>
<p>I have no idea how to start this question; how do I get the areas of the triangles when only given side lengths? Please help me out.</p>
| Quanto | 686,284 | <p><a href="https://i.stack.imgur.com/9UAPd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9UAPd.png" alt="enter image description here"></a></p>
<p>Note,</p>
<p><span class="math-container">$$\frac{PB}{PD}=\frac{S_{\triangle ABC}}{S_{\triangle ACD}}
= \frac{\frac12\cdot AB\cdot BC\cdot \sin B}{\frac12\cdot AD\cdot DC\cdot \sin D}
= \frac{ AB\cdot BC}{ AD\cdot DC}=\frac{1}{8}$$</span></p>
<p>Similarly,</p>
<p><span class="math-container">$$\frac{PA}{PC}=\frac{ AB\cdot AD}{ BC\cdot DC}=\frac{8}{81}$$</span></p>
<p>Recognize</p>
<p><span class="math-container">$$\frac{S_{\triangle PAB}}{S_{\triangle PAD}}
=\frac{S_{\triangle PBC}}{S_{\triangle PCD}} =\frac{PB}{PD} = \frac18,\>\>\>\>\>
\frac{S_{\triangle PAB}}{S_{\triangle PBC}}=\frac{PA}{PC} = \frac8{81} $$</span></p>
<p>to obtain,</p>
<p><span class="math-container">$$S_{\triangle PAB}:S_{\triangle PBC}:S_{\triangle PCD}:S_{\triangle PDA}
=8:81:648:64$$</span></p>
|
2,268,225 | <p>I want to investigate the convergence behavior of $$\int_{0}^{\infty} \cos(x^r)\, dx \hspace{5mm} \textrm{and} \hspace{5mm} \int_{\pi}^{\infty} \left(\frac{\cos x}{\log x}\right)\arctan\lfloor x\rfloor dx.$$ My theoretical tools are Abel's test and Dirichlet's test:
Say I have an integral of the form $$\int_{a}^{b}f\cdot g \hspace{1.5mm} dx$$
with improperness (vertical or horizontal asymptote) at $b$.</p>
<p>Abel's test guarantees convergence if $g$ is monotone and bounded on $(a,b)$, and $\int_{a}^{b}f $ converges. Dirichlet's test guarantees convergence if $g$ is monotone on $(a,b)$ and $\displaystyle\lim_{x\to b} g(x) = 0$, and $\displaystyle\lim_{\beta \to b}$ $\int_{a}^{\beta}f $ is bounded.</p>
<p>For the first integral $\displaystyle\int_{0}^{\infty} \cos(x^r)\, dx $ I'm guessing a substitution $t = x^r $ will give me an expression of the form $f\cdot g$ with $\cos(t)$ as my $f$. For the second integral $\displaystyle\int_{\pi}^{\infty} \dfrac{\cos x}{\log x}\arctan\lfloor x\rfloor\, dx$, I'm (even more) clueless. Help please?</p>
| Hypergeometricx | 168,053 | <p>$$\begin{align}
S&=\qquad \frac 14+\frac 28+\frac 3{16}+\frac 4{32}+\frac 5{64}+\cdots\tag{1}\\
2S&=\frac 12+\frac 24+\frac 38+\frac 4{16}+\frac 5{32}\cdots\tag{2}\\
(2)-(1):\qquad\\
S&=\frac 12+\frac 14+\frac 18+\frac 1{16}+\frac 1{32}\cdots\\
&=\frac {\frac 12}{1-\frac 12}\\
&=\color{red}1
\end{align}$$</p>
|
2,240,616 | <p>Let $f:[0,1] \to \mathbb{R}$ be a function such that for every $a \in [0,1)$ and $b \in (0,1]$ the one-sided limits $$f(a^+)=\lim _{x\to a^+}f(x) \in \mathbb{R}$$ $$f(b^-)=\lim _{x \to b^-} f(x) \in \mathbb {R}$$ exist. </p>
<p>A) Show that $f$ is bounded. </p>
<p>B) Does $f$ necessarily achieve its maximum at some $x \in [0,1]$?</p>
<p>C) Suppose further that $f$ is continuous at $0$ and $1$, and that $f(0) f(1)<0$. Prove that there exists some point $p \in (0,1)$ such that $f(p^-)f(p^+) \leq 0$. </p>
<p>Intuitively, I can see why part A is true, but I am not sure how to prove this formally. For part B, I think the answer is no, but I haven't yet come up with a counterexample. My initial thoughts on part C are to somehow apply the intermediate value theorem, but I am not sure if this is the correct approach or not.</p>
| Community | -1 | <p>Such an $f$ is called a regulated function. The space of step functions is dense in the space of regulated functions for $\|\cdot\|_{\infty}$, so $\exists \phi$ a step function $[0,1] \to \Bbb R$ such that $\|\phi - f\|_{\infty} \le 1$. Thus $\|f\|_{\infty} \le 1 + \|\phi\|_{\infty} < \infty$.</p>
<p>For $B$, consider the function:</p>
<p>$$f(x) = \begin{cases} x, 0\le x < 1/2 \\ 0, 1/2 \le x \le 1 \end{cases}$$</p>
<p>For $C$, WLOG say $f(0) < 0$ and $f(1) > 0$. As $f$ is continuous at $0$, $\exists 0<\delta < 1$ such that $f(x) < 0$ for all $0\le x \le \delta$. Similarly, there is $\epsilon \in (0,1)$ such that $f(x) > 0$ for all $x \in [1-\epsilon, 1]$. </p>
<p>Let: $$S =\{s \in [0,1]: \text{for all but finitely many } t\in [0,s], f(t) < 0\}$$</p>
<p>Let $p = \sup S$. We have $0<p<1$ since $\delta \in S$ and $S \subset [0,1-\epsilon]$. Clearly $f(p-) < 0$. Now suppose that $f(p+)<0$. We have $f(x) < \frac{f(p+)}{2} < 0$ for all $x \in (p, p+\eta)$ for some small $\eta > 0$. It follows, thus, that $p+\eta/2 \in S$, so $p+\eta/2 \le p$, a contradiction. Therefore $f(p+) \ge 0$, and we get $f(p-)f(p+) \le 0$.</p>
|
4,517,063 | <p>Prove that there are exactly <span class="math-container">$8100$</span> different ways of distributing <span class="math-container">$4$</span> indistinguishable black marbles and <span class="math-container">$6$</span> distinguishable coloured marbles ( none of them black) into <span class="math-container">$5$</span> distinguishable boxes in such a way that each box contains exactly <span class="math-container">$2$</span> marbles.</p>
<hr />
<p>I have done problems involving indistinguishable balls and indistinguishable/distinguishable boxes, distinguishable balls and indistinguishable/distinguishable boxes.</p>
<p>I am confused about how to handle the situation when both indistinguishable and distinguishable balls are given at the same time.</p>
<hr />
<p>Any hints will be helpful.</p>
| Cathedral | 933,575 | <p>We can break this problem down into simply arranging the distinguishable marbles in 5 distinguishable boxes(such that no box gets more than 2 marbles), and then using the indistinguishable ones a "filler" of sorts.</p>
<p>We can do this by considering that the distinguishable marbles can be arranged in 3 ways,</p>
<ol>
<li><span class="math-container">$4$</span> boxes with one marble, <span class="math-container">$1$</span> with two marbles: This can be done in <span class="math-container">$\binom51\frac{6!}{2!}=1800$</span> ways.</li>
<li><span class="math-container">$2$</span> boxes with one marble, <span class="math-container">$2$</span> with two marbles and 1 with none: <span class="math-container">$\binom5{2,2,1}\frac{6!}{2!2!}=5400$</span> ways.</li>
<li><span class="math-container">$3$</span> boxes with two marbles, <span class="math-container">$2$</span> with none: <span class="math-container">$\binom{5}{3}\frac{6!}{2!2!2!}=900$</span> ways.</li>
</ol>
<p>Thus, we have a total of <span class="math-container">$1800+5400+900=\bbox[5px,border:2px solid #C0A000]{8100\text{ ways.}}$</span></p>
|
211,297 | <p>Let $ \mathbb{F} $ be an uncountable field. Suppose that $ f: \mathbb{F}^{2} \rightarrow \mathbb{F} $ satisfies the following two properties:</p>
<ol>
<li>For each $ x \in \mathbb{F} $, the function $ f(x,\cdot): y \mapsto f(x,y) $ is a polynomial function on $ \mathbb{F} $.</li>
<li>For each $ y \in \mathbb{F} $, the function $ f(\cdot,y): x \mapsto f(x,y) $ is a polynomial function on $ \mathbb{F} $.</li>
</ol>
<p>Is it necessarily true that $ f $ is a bivariate polynomial function on $ \mathbb{F}^{2} $? What if $ \mathbb{F} $ is merely countably infinite?</p>
| Ewan Delanoy | 15,381 | <p>As shown is Gerry Myerson’s answer, the answer is NO when $\mathbb F$ is countably infinite.</p>
<p>The answer is YES when $\mathbb F$ is uncountable, however.</p>
<p>Sketch of proof : since there are only countably many degrees, the polynomials will share a common degree on an uncountable set. This bound on the degree allows one to use interpolation, and to retrieve the whole of $f$.</p>
<p>More detailed proof : Denote by $d(x)$ the degree of the univariate polynomial $f(x,.)$ for $x\in {\mathbb F}$ (recall that the degree of the zero polynomial is $-\infty$), and put $U_d=\lbrace x \in {\mathbb F} | d(x)=d\rbrace$ for $d\in \lbrace -\infty \rbrace \cup {\mathbb N}$. Then the $U_d$ form a countable partition of $\mathbb F$, so at least one of the $U_d$, say $U_{n}$, is uncountable. </p>
<p>We may assume that $n>0$, as the cases $n=-\infty$ and $n=0$ are similar and simpler. Let $y_0,y_1, \ldots y_{n}$ be $n+1$ distinct values in $\mathbb F$, this is possible because $\mathbb F$ is uncountable. (if the characteristic of $\mathbb F$ is zero, we can simply take $y_i=i$). Using Lagrange interpolation, let us put</p>
<p>$$L_k(y)=\frac{\prod_{j \neq k}{(x-x_j)}}{\prod_{j \neq k}{(x_k-x_j)}}$$ </p>
<p>for $0 \leq k \leq n$. Then one has, for any polynomial $P$ of degree $\leq n$ and any $y\in{\mathbb F}$,</p>
<p>$$
P(y)=P(y_0)L_0(y)+P(y_1)L_1(y)+ \ldots +P(y_n)L_n(y)
$$</p>
<p>In particular, one has for any $(x,y)\in U_n \times {\mathbb F}$,</p>
<p>$$
(1) \ f(x,y)=f(x,y_0)L_0(y)+f(x,y_1)L_1(y)+f(x,y_2)L_2(y)+ \ldots +f(x,y_n)L_n(y)
$$
The right-hand side is a fixed bivariate polynomial, let us denote it by $Q(x,y)$. Let $y\in {\mathbb F}$. Then the two univariate polynomials $f(.,y)$ and $Q(.,y)$ coincide on the uncountable set $U_n$, so they must coincide everywhere.
Finally $f=Q$ everywhere and we are done. </p>
|
2,417,029 | <p>I have been told that a line segment is a set of points.
How can even infinitely many point, each of length zero, can make a line of positive length?</p>
<p>Edit:
As an undergraduate I assumed it was due to having uncountably many points.
But the Cantor set has uncountably many elements and it has measure $0$.</p>
<p>So having uncountably many points on a line is not sufficient for the measure to be positive.</p>
<p>My question was: what else is needed? It appears from the answers I've seen that the additional thing needed is the topology and/or the sigma algebra within which the points are set.</p>
<p>My thanks to those who have helped me figure out where to look for full answers to my question.</p>
| Chappers | 221,811 | <p>This may seem rather a strange thing to say, but I don't think it's helpful to think of lines as made up of points: the "lininess" of a line is an inherent property that points don't have, so it has some extra qualities that points don't, such as length.</p>
<p>The real numbers are basically the answer to the question "How can I augment the set of rational numbers so that I don't have to worry about whether limits that ought to exist really do exist?", from which one can then do calculus. One can wheel out $\sqrt{2}$, $\pi$ and so on if one so desires as an obvious example of a point where one needs this.</p>
<p>Perhaps a more helpful introduction of the real numbers is to say "I want to know how far I am along this line." You then say "Am I halfway?" "Am I a quarter of the way?" "Am I 3/8ths of the way?", and so on. This gives you a way of producing binary expansions using closed intervals, and you can then introduce the idea of asking infinitely many of these questions (which will obviously be necessary, since $1/3$ has an infinite binary expansion), and the object in which the infinite intersection of the decreasing family of closed intervals with rational endpoints constructed by answering the sequence of questions contains precisely one point is called the real numbers. Hence one ends up with the real numbers as describing locations on the line, while not actually being the line itself.</p>
<p>In fact, the construction of the real numbers also gives you some "lininess" as baggage from the construction: you produce a <em>topology</em>, which tells you about locations being close to one another. This gives the real numbers more "substance" than just being ordered and containing the rationals. One can define topologies on the rationals, but the real numbers' completeness in their topological construction is the key. Completeness forces there to be "too many" real numbers to be covered by arbitrarily small sets. (Obviously countable is too small since the rationals don't work, but the Cantor set shows that one can produce uncountable sets with zero "length".)</p>
<p>One large hole in this so far is what "length" actually is. To do things this way, one is forced to introduce a definition of the length of a rational interval $[p,q]$, which must of course be $q-p$. Since one is not concerned at that point about the interval actually being "full" of points, one can simply introduce this as an axiom of the theory: all of us at some point have owned a ruler and know how they work with integers and small fractions, and it's not too much of a stretch to stipulate that one can have a ruler with as small a rational subdivision as required, without having to resort to infinite subdivision. (Which is another point worth emphasising: without infinite processes, there is no need for the real numbers <em>in toto</em>: one can simply introduce "enough" rationals for the precision one requires, and work modulo this "smallest length".)</p>
<p>This way, one starts with "length" and ends up with "real numbers", rather than trying to go the other way, which is theoretically difficult and mentally taxing and counterintuitive (besides all the Cantorian stuff).</p>
|
2,417,029 | <p>I have been told that a line segment is a set of points.
How can even infinitely many point, each of length zero, can make a line of positive length?</p>
<p>Edit:
As an undergraduate I assumed it was due to having uncountably many points.
But the Cantor set has uncountably many elements and it has measure $0$.</p>
<p>So having uncountably many points on a line is not sufficient for the measure to be positive.</p>
<p>My question was: what else is needed? It appears from the answers I've seen that the additional thing needed is the topology and/or the sigma algebra within which the points are set.</p>
<p>My thanks to those who have helped me figure out where to look for full answers to my question.</p>
| Readin | 278,069 | <p>If you have a full period to use, and your students are bright, you could explain the why integers are countably infinite while irrationals are uncountably infinite. Once you'v done that, you can point out how adding up the lengths of a countably infinite number of points does indeed get you zero (0+0+0..=0) but that a line segment doesn't have a countably infinite number of points - it has an uncountably infinite number of points.</p>
<p>That doesn't explain how the points create a line, but it does explain why the main objection - adding zero length to zero length gets you zero - doesn't apply.</p>
|
1,072,302 | <p>let $f$ and $g$ be two functions from $[0,1]$ to $[0,1]$ with $f$ strictly increasing. Which of the follwing is true?</p>
<blockquote>
<blockquote>
<p>(a). If $g$ is continuous, then $f\circ g$ is continuous.</p>
<p>(b). If $f$ is continuous, then $f\circ g$ is continuous.</p>
<p>(c). If $f$ and $f\circ g$ are continuous, then $g$ is continuous.</p>
<p>(d). If $g$ and $f\circ g$ are continuous, then $f$ is continuous.</p>
</blockquote>
</blockquote>
<p><strong>I guessed this</strong></p>
<p>$f$ is strictly increasing $\implies$ $f$ is continuous on $[0,1]$ So, If $g$ is continuous then $f\circ g$ is continuous. Is my approach is correct?
If i am right, why the others are wrong? can you give a counter examples for that?</p>
| ajotatxe | 132,456 | <p>If $f$ is strictly increasing, we can guarantee that $f$ has, at most, countably infinitely many points of discontinuity. But strict monotonicity does not imply continuity at all. Consider $f(x)=x/2$ for $x<1/2$ and $f(x)=(1+x)/2$ for $x\ge1/2$.</p>
|
1,072,302 | <p>let $f$ and $g$ be two functions from $[0,1]$ to $[0,1]$ with $f$ strictly increasing. Which of the follwing is true?</p>
<blockquote>
<blockquote>
<p>(a). If $g$ is continuous, then $f\circ g$ is continuous.</p>
<p>(b). If $f$ is continuous, then $f\circ g$ is continuous.</p>
<p>(c). If $f$ and $f\circ g$ are continuous, then $g$ is continuous.</p>
<p>(d). If $g$ and $f\circ g$ are continuous, then $f$ is continuous.</p>
</blockquote>
</blockquote>
<p><strong>I guessed this</strong></p>
<p>$f$ is strictly increasing $\implies$ $f$ is continuous on $[0,1]$ So, If $g$ is continuous then $f\circ g$ is continuous. Is my approach is correct?
If i am right, why the others are wrong? can you give a counter examples for that?</p>
| Ben Grossmann | 81,360 | <p>Only statement (c) is true.</p>
<p><strong>Hint:</strong> if $f$ is continuous and strictly increasing, it has a continuous (left-)inverse $f^{-1}$. Consider $f^{-1} \circ f \circ g$.</p>
<hr>
<p>Counterexamples, in no particular order:</p>
<ul>
<li>$f(x) = x$, $g(x)$ is discontinuous</li>
<li>$f$ is increasing but discontinuous, $g$ is a constant function</li>
<li>$f$ is increasing but discontinuous, $g(x) = x$</li>
</ul>
<hr>
<p>An example of a strictly increasing discontinuous function:
$$
f(x) =
\begin{cases}
x/3 & x \in [0,1/2)\\
(x+1)/3 & x \in [1/2,1]
\end{cases}
$$</p>
|
545,771 | <blockquote>
<p>let sequence $\{a_{n}\}$,such $a_{1}=0,a_{2}=2,a_{3}=5$,and for $n\in N^{+}$,such
$$\begin{cases}
a_{2n}=2n+2a_{n}\\
a_{2n+1}=2n+1+a_{n}+a_{n+1}
\end{cases}$$
How can I find the closed form of $\{a_{n}\}$?</p>
</blockquote>
<p>My try:
we have</p>
<blockquote>
<p>$$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
so if $n=2k$,then
$$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
$$a_{2n}-a_{2n-1}=a_{n}-a_{n-1}+1$$
$\cdots\cdots\cdots$
$$a_{3}-a_{2}=a_{2}-a_{1}+1$$
so add all this equation,we have
$$a_{2n+1}-a_{2}=a_{n+1}-a_{1}+(2n-1)$$
then
$$a_{2n+1}=a_{n+1}+2n+1$$
<strong>My idea is true? and How solve this problem,and this This problem background is china comption today</strong></p>
</blockquote>
| gvo | 78,233 | <p>Take care, if I had one more line, you'll see that we cannot sum like you did : </p>
<p>$$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
$$a_{2n}-a_{2n-1}=a_{n}-a_{n-1}+1$$
$$a_{2n-1}-a_{2n-2}=a_{n}-a_{n-1}+1$$ and not
$$a_{2n-1}-a_{2n-2}= a_{n-1}-a_{n-2}+1$$
$\cdots\cdots\cdots$
$$a_{3}-a_{2}=a_{2}-a_{1}+1$$</p>
<p>And the equation you end up with is $a_{2n+1}=2n+1+a_{n}+a_{n+1}$, which unfortunately is something you already knew.</p>
|
545,771 | <blockquote>
<p>let sequence $\{a_{n}\}$,such $a_{1}=0,a_{2}=2,a_{3}=5$,and for $n\in N^{+}$,such
$$\begin{cases}
a_{2n}=2n+2a_{n}\\
a_{2n+1}=2n+1+a_{n}+a_{n+1}
\end{cases}$$
How can I find the closed form of $\{a_{n}\}$?</p>
</blockquote>
<p>My try:
we have</p>
<blockquote>
<p>$$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
so if $n=2k$,then
$$a_{2n+1}-a_{2n}=a_{n+1}-a_{n}+1$$
$$a_{2n}-a_{2n-1}=a_{n}-a_{n-1}+1$$
$\cdots\cdots\cdots$
$$a_{3}-a_{2}=a_{2}-a_{1}+1$$
so add all this equation,we have
$$a_{2n+1}-a_{2}=a_{n+1}-a_{1}+(2n-1)$$
then
$$a_{2n+1}=a_{n+1}+2n+1$$
<strong>My idea is true? and How solve this problem,and this This problem background is china comption today</strong></p>
</blockquote>
| Greg Martin | 16,078 | <p>By calculating the first few dozen values by hand, one notices the pattern that
$$
a_{n+1} = a_n + (k+2) \quad\text{for all } 2^k \le n < 2^{k+1}.
$$
This is quickly verified by induction on $k$ from the original recursion; for example,
$$
a_{2n+1}-a_{2n} = (2n+1+a_n+a_{n+1}) - (2n+2a_n) = 1 + (a_{n+1}-a_n).
$$
From here, it is not hard to prove by induction that $a_{2^k} = k2^k$ for all $k\ge0$; combining these "landmark" values with the new pattern above will provide a closed form for every $a_n$.</p>
|
2,938,491 | <p>I have been pondering this question for a little while, and unfortunately the Google has not given me an answer.</p>
<p>I understand that for example you had a table or graph that crosses or touches the x axis at say x= -2, 0, 3 you could form an equation as</p>
<p>f(x) = ax(x+2)(x-3)</p>
<p>Then solve for a and you have your function.</p>
<p>I have considered transforming a graph to force zeros, and in the couple of attempts I made, it was successful, but I am unsure if this would be the mathmatically proper way to do so.</p>
<p>So my question is if you have a graph or table of coordinates similar to my example above, but the points never cross zerI, what would be the proper Mathmatics procedure to find the equation.</p>
<p>UPDATE</p>
<p>y = x^2. Vertex = 0,0 and zero = 0 </p>
<p>In comparison to:</p>
<p>y = (x-1)^2+1 vertex = 1,1 and zero = null</p>
<p>Its the same form but in a different position. In this situation the functions were provided, but for clarification of what I am looking for, I thought this would help. </p>
<p>Thank you.</p>
| Doug M | 317,162 | <p><span class="math-container">$\sum_\limits{i=0}^n 2(i + 1)^2 = 2\sum_\limits{i=1}^{n+1} i^2$</span></p>
<p>Which gets to the meat of the question, what is <span class="math-container">$\sum_\limits{i=1}^n i^2$</span>?</p>
<p>There are a few ways to do this. I think that this one is intuitive.
<a href="https://i.stack.imgur.com/Hpobh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hpobh.png" alt="enter image description here"></a></p>
<p>In the first triangle, the sum of <span class="math-container">$i^{th}$</span> row equals <span class="math-container">$i^2$</span></p>
<p>The next two triangles are identical to the first but rotated 120 degrees in each direction.</p>
<p>Adding corresponding entries we get a triangle with <span class="math-container">$2n+1$</span> in every entry. What is the <span class="math-container">$n^{th}$</span> triangular number?</p>
<p><span class="math-container">$3\sum_\limits{i=1}^n i^2 = (2n+1)\frac {n(n+1)}{2}\\
\sum_\limits{i=1}^n i^2 = \frac {n(n+1)(2n+1)}{6}$</span></p>
<p>To find: <span class="math-container">$\sum_\limits{i=1}^{n+1} i^2 $</span>, sub <span class="math-container">$n+1$</span> in for <span class="math-container">$n$</span> in the formula above.</p>
<p><span class="math-container">$\sum_\limits{i=0}^n 2(i + 1)^2 = \frac {(n+1)(n+2)(2n+3)}{3}$</span></p>
<p>Another approach is to assume that <span class="math-container">$S_n$</span> can be expressed as a degree <span class="math-container">$3$</span> polynomial. This should seem plausible</p>
<p><span class="math-container">$S(n) = a_0 + a_1 n + a_2 n^2 + a_3n^3\\
S(n+1) = S(n) + 2(n+2)^2\\
S(n+1) - S_n = 2(n+2)^2\\
S(n+1) = a_0 + a_1 (n+1) + a_2 (n+1)^2 + a_3(n+1)^3\\
a_0 + a_1 n+a_1 + a_2 n^2 + 2a_2n+a_21 + a_3n^3 + 3a_3n^2 + 3a_3n + 1\\
S(n+1) - S(n) = (a_1 + a_2 + a_3) + (2a_2 + 3a_3) n + 3a_3 n^2 = 2n^2 + 4n + 2$</span></p>
<p>giving a system of equations:</p>
<p><span class="math-container">$a_1 + a_2 + a_3 = 2\\
2a_2 + 3a_3 = 4\\
3a_3 = 1\\
a_0 = S(0)$</span></p>
|
2,843,337 | <blockquote>
<p>If $0 \lt x \lt \dfrac{\pi}{2}$, prove that
$$x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$$</p>
</blockquote>
<p>This question must be done without calculus. First, I tried splitting it into the intervals $(0,\pi/4)$ and $(\pi/4, \pi/2)$, hoping that, $\sin x$ was bound tightly enough on the interval that it'd be less than 3 even if $\cos x = 1$ (which doesn't work -- letting $\sin x = \dfrac{1}{\sqrt{2}}$ and $\cos x = 1$ produces a result greater than 3).</p>
<p>The other thing I noticed was that inside the square root sign, we have $\sqrt{9-x^3} = \sqrt{(3-x^{3/2})(3+x^{3/2})}$, and an $x^{3/2}$ appears in the first term, but I'm not sure how useful the similarity there is.</p>
<p>Advice on how to proceed?</p>
| mengdie1982 | 560,634 | <p>Let $$\alpha=\sqrt{x^3},~~~~\beta=\sqrt{9-x^3};~~~~u=\sin x,~~~~ v=\cos x; ~~~~~~~~\.$$</p>
<p>You may see that $\alpha,\beta,u,v>0$ when $x \in (0,\pi/2)$, and$$\alpha^2+\beta^2=9;~~~~~u^2+v^2=1.$$</p>
<p>Thus, by Cauchy's inequality,$$x^{3/2}\sin x + \sqrt{9-x^3}\cos x=\alpha u+\beta v \leq \sqrt{(\alpha^2+\beta^2)(u^2+v^2)}=3.$$</p>
<p>Till here, you have to show that the equality could hold. Notice that Cauchy's inequality require that the equality holds if and only if $$\frac{\alpha}{u}=\frac{\beta}{v}.$$</p>
<p>How to show this would hold?</p>
|
394,489 | <p>$$\int^L_{-L} x \sin(\frac{\pi nx}{L})$$</p>
<p>I've seen something like this in Fourier theory, but I'm still not sure how to approach this integral. Wolfram Alpha gives me the answer, but no method. Integrate by parts? Substitution?</p>
| gt6989b | 16,192 | <p>Integrate by parts letting $u = x, dv = \sin(ax) dx$, and so $du = dx, v = -\cos (ax)/a$ and you have</p>
<p>$$\int x \sin(ax) dx = \int u \cdot dv = uv - \int v \cdot du
= \frac{-x\cos(ax)}{a} + \frac{1}{a} \int \cos(ax) dx,$$</p>
<p>which should be much easier.</p>
|
148,612 | <p>You are given a rectangular paper sheet. The diagonal vertices of the sheet are brought together and folded so that a line (mark) is formed on the sheet. If this mark length is same as the length of the sheet, what is the ratio of length to breadth of the sheet?</p>
<p>This is my first question on this site, so if this is not a good question please help.</p>
| Community | -1 | <p>$\hskip 2.2in$ <img src="https://i.stack.imgur.com/vLPeY.png" alt="enter image description here"></p>
<p>The above figure was done using grapher on mac osx.</p>
<p>Let $l$ be the length (i.e. the sides $AD$ and $BC$) and $b$ be the breadth (i.e. the sides $AB$ and $CD$). Once you get the diagonal vertices together, i.e. when $D$ coincides with $B$, the length $EB$ is the same as the length $ED = l-a$.</p>
<p>Hence, for the right triangle, we have that $$a^2 + b^2 = (l-a)^2\\ b^2 = l^2 - 2al\\ a = \frac{l^2 - b^2}{2l}$$
The length of $BF$ is $l-a$ and is given by $$l-a = l - \frac{l^2 - b^2}{2l} = \frac{2l^2 - l^2 + b^2}{2l} = \frac{l^2 + b^2}{2l}$$
The distance between the two points is $$EF^2 = r^2 = b^2 + (l-2a)^2 = b^2 + \left(l - \frac{l^2 - b^2}{l} \right)^2 = b^2 + \frac{b^4}{l^2}$$
This is so since the vertical distance between $E$ and $F$ is $l-2a$.</p>
<p>You are given that $r = l$ and hence $$l^2 = b^2 + \frac{b^4}{l^2}$$If we let $$\frac{l}{b} = x,$$ then we get that $$x^2 = 1 + \frac1{x^2}\\ x^4 = x^2 + 1$$ which gives us that $$x = \sqrt{\frac{1}{2} (1+\sqrt{5})} = \sqrt{\phi}$$ where $\phi$ is the golden ratio.</p>
|
162,364 | <p>Assuming we have a list below, it has real number elements and complex numbers, how can I quickly find if the list has any Real number that its value is less than 1.0? </p>
<pre><code> lis = {3 Sqrt[354], Sqrt[2962], Sqrt[2746], 3 Sqrt[282], Sqrt[2338],
Sqrt[2146], 3 Sqrt[218], Sqrt[1786], Sqrt[1618], 27 Sqrt[2], Sqrt[
1306], Sqrt[1162], 3 Sqrt[114], Sqrt[898], Sqrt[778], 3 Sqrt[74],
Sqrt[562], Sqrt[466], 3 Sqrt[42], Sqrt[298], Sqrt[226], 9 Sqrt[2],
Sqrt[106], Sqrt[58], 3 Sqrt[2], I Sqrt[14], I Sqrt[38], 3 I Sqrt[6],
I Sqrt[62], I Sqrt[62], 3 I Sqrt[6], I Sqrt[38], I Sqrt[14], Sqrt[
1.2], Sqrt[58], Sqrt[1.06], 9 Sqrt[2], Sqrt[226]}
</code></pre>
<p>I am also considering if we can find methods to filter any real or complex numbers with specific values in any types list (e.g. there are some string elements mixed with real and complex number elements in a given list). But this could be another question and isn't necessary for my example. Please leave some advice if you interested! Thanks in advance!</p>
| kglr | 125 | <pre><code>Pick[lis, Internal`RealValuedNumericQ@# && # < 1 & /@ lis]
</code></pre>
<blockquote>
<p>{}</p>
</blockquote>
<pre><code>Pick[lis, Internal`RealValuedNumericQ @ # && # < 5 & /@ lis]
</code></pre>
<blockquote>
<p>{3 Sqrt[2], 1.09545, 1.02956}</p>
</blockquote>
<p><code>Pick</code> combined with @aardvark2012's selector approach and exploiting the fact that <code>Negative</code>, <code>Positive</code>, <code>NonPositive</code> ... are <code>Listable</code>:</p>
<pre><code>Pick[lis, NonPositive[lis - 5]]
</code></pre>
<blockquote>
<p>{3 Sqrt[2], 1.09545, 1.02956}</p>
</blockquote>
<p>This seems to be the fastest of methods posted so far.</p>
|
407,289 | <p>Is it true that in the category of connected smooth manifolds equipped with a compatible field structure (all six operations are smooth) there are only two objects (up to isomorphism) - <span class="math-container">$\mathbb{R}$</span> and <span class="math-container">$\mathbb{C}$</span>?</p>
| Arthur Parker | 436,511 | <p>Finite-dimensional manifolds are locally compact, and the only non-discrete locally compact topological fields are the reals and the complex numbers. So the answer is yes.</p>
|
1,143,161 | <p>My question is inspired by the structure of Royden's <em>Real Analysis</em>, which introduces measure theory and Lebesgue integration for $\mathbb{R}$ in its Part I and then reconstructs large portions of those mathematical apparatuses in greater generality by constructing (versions of) them for general metric spaces.</p>
<p>I'll suggest a first, elementary example: In $\mathbb{R}^n$, every convergent sequence is Cauchy, and vice versa. In a general metric space, every convergent sequence is Cauchy, but a Cauchy sequence need not converge. To provide an example, a Cauchy sequence in $\mathbb{Q}$ (using the standard metric) can converge to an irrational number, which is of course not in $\mathbb{Q}$.</p>
<p>What other results in analysis hold for $\mathbb{R}^n$ but not for general metric spaces?</p>
| PVAL-inactive | 83,337 | <p>For $\Bbb R^n$ with the Euclidean metric the open $\epsilon$-ball around a point $p$, $$B=\{x\in \Bbb R^n\ |\ d(x,p)<\epsilon\}$$
has closure
$$\bar B= \{x\in \Bbb R^n\ |\ d(x,p)\leq \epsilon\}$$ </p>
<p>The analogous statement when $\Bbb R^n$ is replaced by a general metric space is false even for relatively nice subsets of $\Bbb R^2$ (say polygonal simple closed curves or even finite subsets).</p>
|
1,143,161 | <p>My question is inspired by the structure of Royden's <em>Real Analysis</em>, which introduces measure theory and Lebesgue integration for $\mathbb{R}$ in its Part I and then reconstructs large portions of those mathematical apparatuses in greater generality by constructing (versions of) them for general metric spaces.</p>
<p>I'll suggest a first, elementary example: In $\mathbb{R}^n$, every convergent sequence is Cauchy, and vice versa. In a general metric space, every convergent sequence is Cauchy, but a Cauchy sequence need not converge. To provide an example, a Cauchy sequence in $\mathbb{Q}$ (using the standard metric) can converge to an irrational number, which is of course not in $\mathbb{Q}$.</p>
<p>What other results in analysis hold for $\mathbb{R}^n$ but not for general metric spaces?</p>
| spin | 12,623 | <p>$\mathbb{R}^n$ is a complete inner product space where closed balls are compact.</p>
<p>Topologically it is path-connected, separable, locally compact, non-compact and non-discrete.</p>
<p>All of these properties can fail to hold for metric spaces.</p>
|
68,147 | <p>While I was investigating some specific types of prime numbers I have faced with the following infinite sequence :</p>
<p>$1,2,8,9,15,20,26,38,45,65,112,244,303,393,560,....$</p>
<p>I tried to find recursive formula using Maple and it's listtorec command, so up to $393$ I got the next output:</p>
<p>$ f(n+3) = ((-10604990407411886564453040+8614360900967683126093782*n$ $-1437788330056801496567841*n^2-20019334790519891406942*n^3$ $+10676199651161684501481*n^4)*f(n+1)$ $+(-1637719982644311036922320-2457276199701830407970234*n$ $-480059310080505210547097*n^2+383671472063948372228234*n^3$ $-33849767081583104776903*n^4)*f(n+2))$ $/(-936042047504931985146406*n -3812415630664251269364960$ $+337414858035611215686569*n^2+50641450188283496191324*n^3$ $-8211420729473965803551*n^4) $</p>
<p>but when I added $560$ to list Maple sent me message FAIL.</p>
<p>So, my question is : how can I find pattern for this sequence if it exists ?</p>
| JavaMan | 6,491 | <p>Try the <a href="http://oeis.org/search?q=,2,8,9,15,20,26,38,45,65,112,244,303,393,560&language=english&go=Search" rel="nofollow">Online Encyclopedia of Integer Sequences</a>.</p>
|
4,543,203 | <p>I am going through Achim Klenke's Probability Theory textbook. In Section 7.4, he discusses what it means for two measures to be singular to one another and gives the following example.</p>
<p>Let <span class="math-container">$\Omega = \{0, 1\}^\mathbb{N}$</span> and let <span class="math-container">$(\mathrm{Ber}_p)^{\otimes \mathbb{N}}$</span> and <span class="math-container">$(\mathrm{Ber}_q)^{\otimes \mathbb{N}}$</span> be the infinite product measures with parameters <span class="math-container">$p$</span> and <span class="math-container">$q$</span>, respectively. For <span class="math-container">$n \in \mathbb{N}$</span>, let <span class="math-container">$X_n$</span> be the <span class="math-container">$n$</span>-th coordinate map. Then under <span class="math-container">$(\mathrm{Ber}_r)^{\otimes \mathbb{N}}$</span>, <span class="math-container">$(X_n)_{n \in \mathbb{N}}$</span> is independent and Bernoulli distributed with parameter <span class="math-container">$r$</span>.</p>
<p>Here is the step where I get confused:</p>
<p>Klenke states that one can apply the strong law of large numbers such that for any <span class="math-container">$r \in \{p, q\}$</span>, there exists a measurable set <span class="math-container">$A_r \subset \Omega$</span> with <span class="math-container">$(\mathrm{Ber}_r)^{\otimes \mathbb{N}}(\Omega \backslash A_r) = 0$</span> and <span class="math-container">$\lim_{n \to \infty} n^{-1} \sum_1^n X_i(\omega) = r$</span> for all <span class="math-container">$\omega \in A_r$</span> and therefore in particular <span class="math-container">$A_p \cap A_q = \emptyset$</span> if <span class="math-container">$p \not = q$</span>, and thus <span class="math-container">$(\mathrm{Ber}_p)^{\otimes \mathbb{N}}$</span> and <span class="math-container">$(\mathrm{Ber}_q)^{\otimes \mathbb{N}}$</span> are singular in that case.</p>
<p>Now I am completely confused by the last paragraph. First off, what guarantees the existence of such a measurable set <span class="math-container">$A_r$</span>? Then, how does it particular follow that <span class="math-container">$A_p \cap A_q = \emptyset$</span> if <span class="math-container">$p \not = q$</span>? And finally, how does the last imply singularity of the two measures? A nice and clear explanation would be greatly appreciated, thanks!</p>
| Mason | 752,243 | <p>Consider the sequence space under the measure <span class="math-container">$P_r = \text{Ber}(r)^{\otimes \mathbb{N}}$</span>. By the strong law of large numbers, <span class="math-container">$$P_r(\limsup_{n \to \infty}\frac{X_1 + \dots + X_n}{n} = r) = 1.$$</span>
So <span class="math-container">$A_r = \{\limsup_{n \to \infty}\frac{X_1 + \dots + X_n}{n} = r\}$</span> is the set Klenke is talking about. The above shows that <span class="math-container">$P_r$</span> assigns mass <span class="math-container">$1$</span> to <span class="math-container">$A_r$</span>. Clearly if <span class="math-container">$p \neq q$</span> then <span class="math-container">$A_p \cap A_q = \emptyset$</span> because <span class="math-container">$\limsup_{n \to \infty}\frac{X_1 + \dots + X_n}{n}$</span> cannot simultaneously equal <span class="math-container">$p$</span> and <span class="math-container">$q$</span>.</p>
|
183,237 | <blockquote>
<p>Does</p>
<p><span class="math-container">$$\int_{-\infty}^\infty \text{e}^{\ a\ (x+b)^2}\ \text dx=\int_{-\infty}^\infty \text{e}^{\ a\ x^2}\ \text dx\ \ \ \ \ ?$$</span></p>
<p>hold, even if the imaginary part of <span class="math-container">$b$</span> is nonzero?</p>
</blockquote>
<p>What I really want to understand is what the phrase "<a href="http://en.wikipedia.org/wiki/Common_integrals_in_quantum_field_theory#Integrals_with_a_complex_argument_of_the_exponent" rel="nofollow noreferrer">By analogy with the previous integrals</a>" means in that link. There, the expression <span class="math-container">$\frac{J}{a}$</span> is complex but they seem to imply the integral can be solved like above anyway.</p>
<p>The reusult tells us that the integral is really independend of <span class="math-container">$J$</span>, which is assumed to be real here. I wonder if we can also generalize this integral to include complex <span class="math-container">$J$</span>. In case that the shift above is possible, this should work out.</p>
<p>But even if the idea is here to perform that substitution, how to get rid of the complex <span class="math-container">$a$</span> to obtain the result. If everything is purely real or imaginary, then <a href="https://math.stackexchange.com/questions/163946/are-complex-substitutions-legal-in-integration/166359#166359">this</a> solves the rest of the problem.</p>
| Fabian | 7,266 | <p>Let us write $b= r+it$. The real part of $b$ does not matter as you have already proven yourself. So wlog $r=0$.</p>
<p>For shifting along the imaginary axis, we have to employ the residue theorem. We have
$$
\begin{align}
\int_{-\infty}^\infty f(x+i t) \,dx&- \int_{-\infty}^\infty f(x)\, dx\\
&=\int_{-\infty-it}^{\infty-it} f(x) \,dx- \int_{-\infty}^\infty f(x)\, dx \\
&= 2\pi i \sum \text{Res}(f)+ \int_{\infty-it}^{\infty} f(x) \,dx
- \int_{-\infty-it}^{-\infty} f(x) \,dx
\end{align},$$
where $\sum \text{Res}(f)$ is the sum over the residues of $f$ in the area $z\in \mathbb{C}$ with $-t<\text{Im}\, z<0$.</p>
<p>So the two integrals are the same if there are no residues and if the two integral at $\pm \infty$ vanish (both of which is the case for your example as long as $\text{Re}\,a <0$).</p>
|
481,017 | <blockquote>
<p>Find all $f(x)$ satisfying $f(f(x)) = x^2 - 2$.</p>
</blockquote>
<p>Presumably $f(x)$ is supposed to be a function from $\mathbb R$ to $\mathbb R$ with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further. </p>
<p><strong>Possibly Helpful Links:</strong> Information on similar problems can be found <a href="https://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx">here</a> and <a href="https://mathoverflow.net/questions/17614/solving-ffx-gx">here</a>.</p>
<p><strong>Source:</strong> <a href="https://math.stackexchange.com/questions/481000/some-old-russian-problems">This question.</a> It is about to be closed for containing too many problems in one question. I'm posting each problem separately. </p>
| Christian Blatter | 1,303 | <p>We are looking for maps $f$ satisfying $$f\circ f=h\ ,\tag{1}$$ where $h$ is given by
$$h:\quad x\mapsto h(x):=x^2-2\ .$$
The map $h$ has the two fixed points $p=-1$, $q=2$ with $h'(p)=-2$, $h'(q)=4$.</p>
<p>Writing $x=2+t$ with a new coordinate $t$ the map $h$ assumes the form $$h:\ t\mapsto 4t+t^2\ .\tag{2}$$
By Koenig's theorem (see John Milnor: <em>Dynamics in one complex variable,</em> Theorem 8.2) one can replace $t$ in the neighborhood of $t=0$ by a new local variable $\tau=\phi(t)$ such that $h$ now appears as
$$h:\quad \tau\mapsto 4\tau\ .$$
Coming back to $f$ there might be solutions of $(1)$ with $f(2)=c\ne2$, $f(c)=2$. In any case we now can look for solutions $f$ for which $2$ is a fixed point as well. In terms of the variable $\tau$ these would satisfy
$$f\bigl(f(\tau)\bigr)\equiv4\tau\ ,\tag{3}$$
and it should not be difficult to show that the only analytical solutions to $(3)$ are $f(\tau)=\pm 2\tau$. When we want $f$ in terms of $t$ we have to write
$$f(t):=\pm2t+\sum_{k=2}^\infty a_k t^k$$ and to determine the coefficients $a_k$ from $(2)$, i.e. using
$$f\bigl(f(t)\bigr)\equiv 4t + t^2\ .$$
A similar analysis can be done with the fixed point $p=-1$ of $h$. There we would write $x=-1+t$ and obtain two $f$'s of the form
$$f(t)=\pm \sqrt{2} i\> t +{\rm higher\ terms}\ .$$</p>
|
481,017 | <blockquote>
<p>Find all $f(x)$ satisfying $f(f(x)) = x^2 - 2$.</p>
</blockquote>
<p>Presumably $f(x)$ is supposed to be a function from $\mathbb R$ to $\mathbb R$ with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further. </p>
<p><strong>Possibly Helpful Links:</strong> Information on similar problems can be found <a href="https://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx">here</a> and <a href="https://mathoverflow.net/questions/17614/solving-ffx-gx">here</a>.</p>
<p><strong>Source:</strong> <a href="https://math.stackexchange.com/questions/481000/some-old-russian-problems">This question.</a> It is about to be closed for containing too many problems in one question. I'm posting each problem separately. </p>
| user3791713 | 519,190 | <p>The solution that I have seen, I think due <a href="https://en.wikipedia.org/wiki/John_Horton_Conway" rel="nofollow noreferrer">to John Horton Conway</a>, is
$$f(x) = 2 \cos\Big(\sqrt2 \arccos\frac x2\Big)$$
This was alleged to work for $-2\leqslant x\leqslant 2$.</p>
<p>$$\begin{align*}
f(f(x))&= 2 \cos\Big(\sqrt2 \arccos\frac12\Big(2 \cos\big(\sqrt2 \arccos\frac x2\big)\Big)\Big)\\
&= 2 \cos\Big(\sqrt2 \arccos\cos\big(\sqrt2 \arccos\frac x2\big)\Big)\\
&= 2 \cos\Big(\sqrt2 \sqrt2 \arccos\frac x2\Big)\\
&= 2 \cos\Big(2 \arccos\frac x2\Big)\\
&= 2\Big(2 \cos^2\arccos\frac x2- 1\Big) \qquad{[\cos 2\theta=2\cos^2\theta-1]}\\
&= 2\Big(2\Big(\frac x2\Big)^2- 1\Big)\\
&= 2(x^2/2 - 1)\\
&=x^2- 2
\end{align*}$$
It looks right, but, as I discovered many years later when trying to draw the graphs of $f(x)$ and $f(f(x))$, it isn't. The problem is that $\arccos\theta$ isn't single-valued, so the simplification of
$\arccos\cos(\sqrt2 \arccos(x/2))$ to $\sqrt2 \arccos(x/2)$ isn't valid. You have to choose a branch of $\arccos\theta$, and it isn't possible to consistently choose a branch which makes it work over the domain specified. In fact, if we impose the reasonable restrictions:-</p>
<ul>
<li>$f(x)$ has to be continuous</li>
<li>for some domain $D$ of $x$, if $x\in D$, $f(x)\in D$</li>
</ul>
<p>then it impossible to pick a domain of $x$ for which the solution given above works. I believe that with these restrictions there is no solution at all.</p>
<p>If you don't require the domain of $x$ to be continuous,
$$f(x) = 2 \cosh\Big(\sqrt2 \operatorname{arccosh}\frac{|x|}2\Big)$$</p>
<p>works for $|x|\geqslant 2$. Of course $\operatorname{arccosh}\theta$ isn't single-valued either, but it doesn't matter which branch you choose because $\cosh \theta = \cosh -\theta$. For $|x| < 2$, $\operatorname{arccosh} x/2$ is undefined, so $f(x)$ isn't defined over this domain either.</p>
|
3,671,941 | <p>In <a href="https://sites.math.northwestern.edu/~mpopa/571/chapter3.pdf" rel="nofollow noreferrer">this chapter</a>, Example 2.1 notes that we can write <span class="math-container">$\mathbb{Z}_p$</span> as <span class="math-container">$$p\mathbb{Z}_p\cup(1+p\mathbb{Z}_p)\cup\cdots\cup (p-1+p\mathbb{Z}_p).$$</span> I understand that <span class="math-container">$\mathbb{Z}_p=\{x\in\mathbb{Q}_p:|x|_p\leq 1\}$</span> and that <span class="math-container">$p\mathbb{Z}_p=\{x\in\mathbb{Q}_p : |x|_p<1\}=\{x\in\mathbb{Q}_p : |x|_p\leq p^{-1}\}$</span>, but I'm having trouble seeing why <span class="math-container">$\mathbb{Z}_p\subset \{a+p\mathbb{Z}_p\}$</span> for <span class="math-container">$a$</span> ranging between <span class="math-container">$0$</span> and <span class="math-container">$p-1$</span>. I understand that each of these is the closed ball of radius <span class="math-container">$p^{-1}$</span> centered at <span class="math-container">$a$</span>, but I don't see how that leads to these balls being a cover for <span class="math-container">$\mathbb{Z}_p$</span>. It must be the case that any point in <span class="math-container">$\mathbb{Z}_p$</span> can be written as <span class="math-container">$a+x$</span> for <span class="math-container">$a$</span> between <span class="math-container">$0$</span> and <span class="math-container">$p-1$</span> and <span class="math-container">$|x|_p\leq p^{-1}$</span>, but I don't see why that's true. While showing this mathematically is helpful, I'm trying to understand it from a conceptual point of view.</p>
| Lubin | 17,760 | <p>I think that in this case, a very good way to think is that <span class="math-container">$p\Bbb Z_p$</span> is a <em>subgroup</em> of <span class="math-container">$\Bbb Z_p$</span>, indeed of index <span class="math-container">$p$</span>, since <span class="math-container">$\Bbb Z_p/p\Bbb Z_p\cong\Bbb F_p$</span>, the field with <span class="math-container">$p$</span> elements. But the cosets of a subgroup are pairwise disjoint, so all you need to do is find a representative of the <span class="math-container">$p$</span> cosets. The choice is up to you, but there are natural candidates, the ones you know.</p>
<p>(I always recommend the way of looking at <span class="math-container">$p$</span>-adic numbers in the way that @Rob has in his answer, so I certainly do not deprecate what he had to say.)</p>
|
28,166 | <p>With the initial conditions: $a>b>0$;</p>
<p>I need to find $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$$</p>
<p>I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theorem.</p>
| Isaac | 72 | <p>Suppose the limit exists and is $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L.$$ Then,
$$\begin{align}
\log L
&=\log\lim_{n\to\infty}\sqrt[n]{a^n-b^n}
\\
&=\lim_{n\to\infty}\log\sqrt[n]{a^n-b^n}
\\
&=\lim_{n\to\infty}\frac{1}{n}\log(a^n-b^n)
\\
&=\lim_{n\to\infty}\frac{\log(a^n-b^n)}{n}.
\end{align}$$
If $a>b>1$ (<strong>note</strong>: $1$, not $0$—I am not sure offhand how to handle the case where $b$ and possibly $a$ are less than 1), $a^n-b^n\to\infty$, so $\log(a^n-b^n)\to\infty$, so the limit is of the indeterminate form $\frac{\infty}{\infty}$ and L'Hôpital's rule applies.
$$\begin{align}
\lim_{n\to\infty}\frac{\log(a^n-b^n)}{n}
&=\lim_{n\to\infty}\frac{\frac{d}{dn}\log(a^n-b^n)}{\frac{d}{dn}n}
\\
&=\lim_{n\to\infty}\frac{\frac{1}{a^n-b^n}\cdot\frac{d}{dn}(a^n-b^n)}{1}
\\
&=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot(a^n\log a-b^n\log b)
\\
&=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot(a^n\log a-b^n\log a+b^n\log a-b^n\log b)
\\
&=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot((a^n-b^n)\log a+b^n(\log a-\log b))
\\
&=\lim_{n\to\infty}\left(\log a+\frac{b^n}{a^n-b^n}(\log a-\log b)\right)
\\
&=\log a+(\log a-\log b)\lim_{n\to\infty}\frac{1}{(\frac{a}{b})^n-1}
\end{align}$$
and since $a>b>1$, $\frac{a}{b}>1$, so $(\frac{a}{b})^n\to\infty$ and $\lim_{n\to\infty}\frac{1}{(\frac{a}{b})^n-1}=0$, so $$\log L=\log a$$ and $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L=a.$$</p>
|
28,166 | <p>With the initial conditions: $a>b>0$;</p>
<p>I need to find $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$$</p>
<p>I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theorem.</p>
| AD - Stop Putin - | 1,154 | <p>Here is a short solution based on standard inequalities.</p>
<p>Our first inequality is obvious since $b^n>0$
$$(1)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\leq a^n.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$ </p>
<p>Next we note that
$$a^n-b^n = a\cdot a^{n-1}-b\cdot a^{n-1}+ b\cdot a^{n-1}- b\cdot b^{n-1}
=(a-b)a^{n-1} + b(a^{n-1}-b^{n-1})$$
which together with $a^{n-1}- b^{n-1}\ge0$ leads to
$$(2)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\geq (a-b)a^{n-1}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $$ </p>
<p>Combining (1) with (2) and taking the $n$:th-root we get
$$
a \ge (a^n-b^n)^{1/n}\ge(a-b)^{1/n}\cdot a^{(n-1)/n}= a\cdot(a-b)^{1/n}\cdot a^{-1/n}
$$
where the right hand side tends to $a$ as $n\to\infty$, and hence
we reach $$\lim_{n\to\infty} (a^n-b^n)^{1/n}=a.$$</p>
|
3,489,376 | <p>Let <span class="math-container">$C$</span> be a category. Then by definition, for very ordered triple <span class="math-container">$A,B,C$</span> of objects, there is a law of composition of morphisms, i.e., a map <span class="math-container">$$Hom_C(A,B)\times Hom_C(B,C)\longrightarrow Hom_C(A,C)$$</span> where <span class="math-container">$(f,g)\mapsto gf$</span>.</p>
<p>I was wondering if it is possible in the definition that <span class="math-container">$Hom_C(A,C)=\emptyset$</span> when the other two are not empty.</p>
| SCappella | 199,074 | <p>No. If <span class="math-container">$Hom_C(A,B)$</span> and <span class="math-container">$Hom_C(B,C)$</span> are inhabited, then <span class="math-container">$Hom_C(A,B) \times Hom_C(B,C)$</span> is inhabited. This forces <span class="math-container">$Hom_C(A,C)$</span> to be inhabited as well. If <span class="math-container">$f \in Hom_C(A, B)$</span> and <span class="math-container">$g \in Hom_C(B, C)$</span> then <span class="math-container">$g \circ f \in Hom_C(A, C)$</span>.</p>
|
2,145,412 | <p>If you have a vector line m with equation $r = i + j + 2k + s( 3i + j - k)$
how do you find the normal?</p>
<p>If you presume that normal is $n = ( n i + o j + p k )$ then</p>
<p>$$3n + o -p = 0 $$</p>
<p>but this leaves you with too many options $\left \lbrace (0,0,0), (1,0,3), (0,1,1), ....\right \rbrace$ is it possible to find <em>the</em> normal?</p>
<p>(This is asked in relation to when I am trying to find the equation of a plane and have a line m and a point A, if I can find the normal to line $m$ I can use $n \cdot a=n \cdot r.$)</p>
| Adren | 405,819 | <p>Let $\mathbb{P}$ the set of all prime numbers.</p>
<p>It can be shown that if $2^p-1\in\mathbb{P}$ (which implies that $p\in\mathbb{P}$), then $2^{p-1}(2^p-1)$ is perfect. This was proved by Euclid (and those numbers are sometime called Euclid's numbers).</p>
<p>Conversely, any <em>even</em> perfect number is an Euclid's number : this was proved about 20 centuries later by Euler.</p>
<p>Until today, no odd perfect number is known ...</p>
<p>So detecting even perfect numbers is closely related to the question of detecting Mersenne primes (that is : primes of the form $2^p-1$).</p>
<p>A fast algorithm exists for this : it's called the Luca's test (see <a href="https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test" rel="nofollow noreferrer">here</a>)</p>
<p>It should also be worth looking at <a href="http://www.mersenne.org/" rel="nofollow noreferrer">this site</a></p>
|
2,525,903 | <blockquote>
<p>Given $x^2+x-a<3$, solve for $a$ so that there is at least one solution $<0$</p>
</blockquote>
<p>I have the idea how to solve this problem but my solution seems weird and unfinished.</p>
<p>Firstly, I move $3$ to the left hand side.</p>
<p>The graphic of the solution is open upwards and has an axis of symmetry at $-1/2$. This means that in order for the inequality to have any solution, its discriminant must be $>0$. So $4a+8>0$, which means $a>-2$. But that seems to be it. Due to the axis and the graph, there must be at least one negative solution. Am I missing something?</p>
| Nosrati | 108,128 | <p>\begin{align}
\int\sec^3x\,dx
&= \int\dfrac{1}{\cos^3x}\,dx \\
&= \int\dfrac{\cos x}{(1-\sin^2x)^2}\,dx \hspace{0.5cm};\hspace{0.5cm} \sin x=u\\
&= \int\dfrac{1}{(1-u^2)^2}\,dx \\
&= \dfrac{u}{2(1-u^2)}+\dfrac{1}{4}\ln\dfrac{1+u}{1-u}+C\\
&= \dfrac{u}{2(1-u^2)}+\dfrac{1}{4}\ln\dfrac{1+u}{1-u}+C\\
&= \dfrac{\sin x}{2\cos^2x}+\dfrac{1}{4}\ln\dfrac{1+\sin x}{1-\sin x}+C
\end{align}</p>
|
2,525,903 | <blockquote>
<p>Given $x^2+x-a<3$, solve for $a$ so that there is at least one solution $<0$</p>
</blockquote>
<p>I have the idea how to solve this problem but my solution seems weird and unfinished.</p>
<p>Firstly, I move $3$ to the left hand side.</p>
<p>The graphic of the solution is open upwards and has an axis of symmetry at $-1/2$. This means that in order for the inequality to have any solution, its discriminant must be $>0$. So $4a+8>0$, which means $a>-2$. But that seems to be it. Due to the axis and the graph, there must be at least one negative solution. Am I missing something?</p>
| Michael Hardy | 11,667 | <p>Here's one way:
\begin{align}
\int \sec^3 x \, dx = {} & \int (\sec x) \Big( \sec^2 x \, dx\Big) = \overbrace{\int u\, dv = uv - \int v \, du}^{\text{integration by parts}} \\[10pt]
= {} & \sec x \tan x - \int (\tan x) \big( \sec x\tan x\, dx\big) \\[10pt]
= {} & \sec x \tan x - \int(\sec x)(\tan^2 x) \, dx \\[10pt]
= {} & \sec x \tan x - \int(\sec x) (\sec^2 x - 1) \,dx \\[10pt]
= {} & \sec x\tan x + \int \sec x\,dx - \int\sec^3 x\,dx. \\[10pt]
& \text{Then adding $\int\sec^3 x\,dx$ to both sides, we get} \\[10pt]
2\int\sec^3 x\, dx & = \sec x \tan x + \int \sec x\, dx. \\[10pt]
& \text{Hence} \\[10pt]
\int \sec^3 x\, dx & = \frac 1 2 \sec x \tan x + \frac 1 2 \int \sec x\, dx = \cdots
\end{align}</p>
|
2,840,091 | <blockquote>
<p>Consider the linear map$:T:\mathbb{R}^3 → \mathbb{R}$ with
$$T\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right)=x-2y-3z$$
Find the basis of its kernel.</p>
</blockquote>
<p><strong>My try</strong></p>
<p>Since the plane is the nullspace of the matrix
$$A=\begin{bmatrix} 1 & -2 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$</p>
<p>But I am stuck here. Can anyone explain this furthur</p>
| Surb | 154,545 | <p>Set $x=1$ and solve $0=x-2y-3z$ for $y$ with $z=0$ and vice versa. Once you have found $y',z'$ satisfying $0=1-2y'$ and $0=1-3z'$, then you know that $f(1,y',0)=f(1,0,z')=0$ and so you have two vectors in the kernel of $f$. Verify that they are linearly independent. Finally, as $\dim(\Bbb R^3)=\dim(Im(f))+\dim(\ker(f))$, we have $\dim(\ker(f))=2$ and so the vectors form a basis.</p>
|
1,773,060 | <p>Is every (non-trivial) quotient of a Boolean algebra isomorphic to a subalgebra of that Boolean algebra? And conversely is every subalgebra isomorphic to a quotient algebra?</p>
| Doug McLellan | 250,373 | <p>This question gets right to the inner workings of Boolean algebras so I'm surprised nobody has answered it. I don't know the answer in general and would love to see a proof of a positive answer, or a counterexample!</p>
<p>For what it's worth I think I can show the answer is "yes" in both directions for <em>finite</em> Boolean algebras, and that something stronger than isomorphism holds. If $A$ is a Boolean algebra, and $Q$ is a quotient algebra of $A$ by some ideal, and $S$ is a subalgebra of $A$, let us say, for lack of a better word, that $Q$ "subsumes" $S$ when each $Q$-element (which is of course an equivalence class of $A$-elements) has exactly one element from $S$. If $Q$ subsumes $S$, the map taking each $Q$-member to its respective $S$-member is a (boolean) isomorphism onto $S$.</p>
<p>Claim: when $A$ is finite, every quotient algebra $Q$ subsumes some subalgebra $S$, and conversely every $S$ is subsumed by some $Q$.</p>
<p>First note that any finite boolean algebra $A$ is isomorphic to the power set of its atoms: the function mapping each $p \in A$ to the set of atoms $\leq p$ is a bijection, and the boolean operations on $A$ agree respectively with union, intersection, and complement on the power set of $A$'s atoms. Thus the finite boolean algebras can be characterized up to isomorphism as $\mathcal{P}(\{1, ..., n\})$ where $n$ can be any natural number $> 0$. (I am ignoring the one-element "degenerate" algebra.)</p>
<p>Second, note that a subalgebra of $\mathcal{P}(\{1,...,n\})$ is uniquely determined by its atoms, which together necessarily form a partition of $\{1,...,n\}$; and any such partition can serve as atoms for a subalgebra. A quotient of $\mathcal{P}(\{1,...,n\})$ by a nontrivial ideal is uniquely determined by that ideal's greatest element, which can be any nonempty subset of $\{1,...,n\}$.</p>
<p>Now let $Q$ be the quotient of $\mathcal{P}(\{1,...,n\})$ by some given ideal, which by the preceding paragraph must have a greatest element which can by any nonempty subset $X \subseteq \{1,...,n\}$. Let $Z$ be any partition of $\{1,...,n\}$ such that each $e \in Z$ has exactly one member that is not in $X$. Then I think it should be easy to show that $Z$ is the set of atoms of a subalgebra $S$ that is subsumed by $Q$. Note that since $Z$ need not be unique in general for a given $Q$, $S$ need not be unique in general.</p>
<p>Conversely, when $S$ is any subalgebra of $\mathcal{P}(\{1,...,n\})$, and $Z$ is its set of atoms (which again can be any partition of $\{1,...,n\}$), let $X$ be any subset of $\{1,...,n\}$ such that for all $e \in Z$, $|X \cap e| = |e| - 1$. Then when $Q$ is the quotient by the ideal whose greatest element is $X$, $Q$ subsumes $S$.</p>
|
2,497,799 | <h2>Question</h2>
<p>While Solving a recursive equation , i am stuck at this summation and unable to move forward.Summation is </p>
<blockquote>
<p>$$\sum_{j=0}^{n-2}2^j (n-j)$$</p>
</blockquote>
<h2>My Approach</h2>
<blockquote>
<p>$$\sum_{j=0}^{n-2}2^j (n-j) = \sum_{j=0}^{n-2}2^j \times n-\sum_{j=0}^{n-2}
2^{j} \times j$$</p>
</blockquote>
<p>$$=n \times (2^{n-1}-1)-\sum_{j=0}^{n-2}
2^{j} \times j$$</p>
<p>I am unable to move forward , please help me out!</p>
| Claude Leibovici | 82,404 | <p><strong>Hint</strong></p>
<p>Consider $$\sum_{i=0}^p i x^i=x\sum_{i=0}^p i x^{i-1}=x\left(\sum_{i=0}^p x^i \right)'$$</p>
|
1,413,955 | <p>I get confused when I put the following three notes together:</p>
<ol>
<li>Power set of any set is a $\sigma$-algebra.</li>
<li>If $X$ is a set and $\Sigma$ is a $\sigma$-algebra over $X$, then the pair $(X, \Sigma)$ is a measurable space.</li>
<li><a href="https://en.wikipedia.org/wiki/Vitali_set" rel="nofollow">Vitali</a> set is known as a counterexample that there is no measure on all the subsets of $\mathbb{R}$.</li>
</ol>
<p>By (1) and (2), one may think that $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ is a measurable space, which intuitively concludes that there must be a measure on all the subsets of $\mathbb{R}$. However, (3) says the opposite. Can anyone help me understand what is going on?</p>
| supinf | 168,859 | <p>the reason for your confusion could be, that you forgot some conditions for 3.</p>
<p>Vitali sets are a counterexample that there is no measure on all the subsets of $\mathbb R$ with the following conditions:</p>
<ol>
<li>the measure of an open interval (a,b) should be b-a</li>
<li>the measure should be invariant w.r.t to translation</li>
<li>the measure is countable additive</li>
</ol>
|
1,842,448 | <p>$F(x)= 2x^2 -3x$. </p>
<p>find the range of $x$ to check whether the function the is strictly increasing and strictly decreasing.</p>
| haqnatural | 247,767 | <p><strong>Hint</strong> :
Write it as follows $$ 2{ x }^{ 2 }-3x=2{ \left( x-\frac { 3 }{ 4 } \right) }^{ 2 }-\frac { 9 }{ 8 } $$</p>
|
2,127,110 | <p>I am working on this differetiation problem:</p>
<p>$ \frac{d}{dx}x(1-\frac{2}{x})$</p>
<p>and I am currently stuck at this point:</p>
<p>$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x$</p>
<p>Symbolab tells me this simplifies to $1$ but I do not understand how. I am under the impression that;</p>
<p>$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x \equiv 1- 2x^{-x^2}-2^{-x}$</p>
| Kanwaljit Singh | 401,635 | <p>$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x $</p>
<p>= $1\cdot 1- 1 \cdot \frac{2}{x} + \frac{2}{x}$</p>
<p>= $1- \frac{2}{x} + \frac{2}{x}$</p>
<p>=$1$</p>
|
588,228 | <p>A,B,C are distinct digits of a three digit number such that </p>
<pre><code> A B C
B A C
+ C A B
____________
A B B C
</code></pre>
<p>Find the value of A+B+C.</p>
<p>a) 16 b) 17 c) 18 d) 19</p>
<p>I tried it out by using the digits 16 17 18 19 by breaking them in three numbers but due to so large number of ways of breaking I cannot help my cause.</p>
| Community | -1 | <p><em>EDIT</em> I just realized this question was asked over a year ago, my bad. </p>
<p>I assume that the question was a typo, at least I hope it was. </p>
<pre><code> A B C
B C A
C A B
___________
A B B C
</code></pre>
<p>instead of </p>
<pre><code> A B C
B A C
C A B
___________
A B B C
</code></pre>
<p>In this case, we have the easy problem:</p>
<p>$$111(A+B+C)=1000A+110B+C$$</p>
<p>We want $1000A+110B+C$ to be divisible by $111$, and that is just a matter of trial and error. The result is $A+B+C=18$.</p>
|
549,505 | <p>I am trying to show the open mapping theorem. As I was trying to prove it, I made the following conjecture:</p>
<blockquote>
<p>Let $X$ be a complete metric space. Let $V_1,V_2,...$ be a sequence of
open sets in $X$ such that: $$V_1\supseteq \overline{V_2}\supseteq
V_2\supseteq \overline{V_3}\supseteq V_3\supseteq
\overline{V_4}\supseteq V_4\supseteq .... $$Then
$\cap_{n=1}^{\infty}V_n\not=\emptyset$</p>
</blockquote>
<p><strong>Question 1:</strong> Is the above conjecture true ?</p>
<p>I would be able to prove the open mapping theorem if the conjecture is true, but I am curious about this question as well:</p>
<p><strong>Question 2:</strong> Does the conjecture hold even if we don't require $X$ to be complete ?</p>
<p>Thank you</p>
<hr>
<p><strong>Edit:</strong> Neal answered both questions. There is still one more question that Ii formulated after seeing the counterexamples of Neal.</p>
<p><strong>Question 3:</strong> Does the conjecture hold if $X$ is a complete bounded metric space ?</p>
| Neal | 20,569 | <p>No, the conjecture need not be true. Consider $V_i = (i,\infty)\subset\mathbb{R}$. Then:
$$(1,\infty)\supseteq [2,\infty)\supseteq (3,\infty)\supseteq \cdots$$
but (by the Archimedean principle) their intersection is empty.</p>
<p>If $X$ is bounded but not complete, the conjecture still does not hold: Consider the punctured interval $X = [-1,0)\cup(0,1]$ and a $$V_i = \bigg(-\frac{1}{i},\frac{1}{i}\bigg)\cap X.$$</p>
<hr>
<p>If $X$ is just bounded and complete, I think it won't work. We'll take a disjoint union of countably many little balls and then let $V_i$ be the union of all but finitely many of them.</p>
<p>Consider a separable infinite-dimensional Hilbert space with orthonormal basis $e_i$. Define auxiliary sets
$$E_{i,\epsilon} = B(e_i,\epsilon).$$
Choose $\epsilon$ small enough so that the $E_{i,\epsilon}$ are pairwise disjoint and put
$$V_i = \bigcup_{j=i}^\infty E_{j,\epsilon/i}$$
Now for each $i$, we've $\bar{V_i}\subseteq V_{i-1}$, and yet the intersection of all $V_i$ is empty.</p>
<p>Since $V_1$ is contained in the ball of radius $1 + \epsilon$, it is a subset of the bounded, closed metric space $\overline{B(0,1+\epsilon)}$. </p>
<p>This example may be easier to visualize if you consider the simpler space defined as a countably infinite collection of intervals $[0,1]$ wedged at $0$, i.e.,
$$\bigsqcup_{i\in\mathbb{N}} [0,1]_i/(\forall i,j\ \ 0_i\sim 0_j).$$</p>
<p>Let $E_{(j,\epsilon)} = (1-\epsilon, 1]_j$, the open $\epsilon$-neighborhood of the endpoint of the $j^{th}$ interval, and $V_i$ the union of all but the first $i$ of the $E_{j,\epsilon/i}$s. As $i$ increases, the size of each $E_j$ shrinks, so $\overline{V_{i+1}}\subseteq V_i$ for all $i$. But any $x\in V_1$ is only contained in finitely many $\overline{V_i}$, so the intersection is empty.</p>
<hr>
<p>However, if $X$ is compact, you can get your result by arguing that
$$V_1\cap \bar{V_2}\cap V_3\cap\cdots = \bar{V_1}\cap\bar{V_2}\cap\bar{V_3}\cap\cdots.$$</p>
|
549,505 | <p>I am trying to show the open mapping theorem. As I was trying to prove it, I made the following conjecture:</p>
<blockquote>
<p>Let $X$ be a complete metric space. Let $V_1,V_2,...$ be a sequence of
open sets in $X$ such that: $$V_1\supseteq \overline{V_2}\supseteq
V_2\supseteq \overline{V_3}\supseteq V_3\supseteq
\overline{V_4}\supseteq V_4\supseteq .... $$Then
$\cap_{n=1}^{\infty}V_n\not=\emptyset$</p>
</blockquote>
<p><strong>Question 1:</strong> Is the above conjecture true ?</p>
<p>I would be able to prove the open mapping theorem if the conjecture is true, but I am curious about this question as well:</p>
<p><strong>Question 2:</strong> Does the conjecture hold even if we don't require $X$ to be complete ?</p>
<p>Thank you</p>
<hr>
<p><strong>Edit:</strong> Neal answered both questions. There is still one more question that Ii formulated after seeing the counterexamples of Neal.</p>
<p><strong>Question 3:</strong> Does the conjecture hold if $X$ is a complete bounded metric space ?</p>
| Stefan Smith | 55,689 | <p>Question 1 : false . take $X = \mathbb{R}$, $V_n = (-\infty, -n)$.</p>
|
1,515,645 | <p>The question is to find the values of a real number $\lambda$ for which the following equation is satisfied for all real values of $\alpha$ which are not integral multiples of $\pi/2$
$${\sin\lambda\alpha\over \sin\alpha}-{\cos\lambda\alpha\over \cos\alpha}=\lambda-1$$</p>
<p>All I could do was to guess some values that just came to mind by observation, like $-1,1,3$</p>
<p>What should be a more mathematical way to find all possible values of $\lambda$? </p>
<p>SOURCE: KVPY 2015 SB stream </p>
| james_d | 248,598 | <p>$$
1111111111111111 = \sum_0^{15} 10^ i
$$</p>
<p>Since 17 is prime, the 16 values $10^0$, $10^1$, ..., $10^{15}$ are distinct and non-zero mod 17. (Otherwise we have $10^m = 17k$ for some integers $k$, $0\leq m < 17$ in which 17 is a factor of the RHS but not of the LHS.)</p>
<p>Thus </p>
<p>$$
1111111111111111 = \sum_0^{15} 10^i
$$</p>
<p>$$
= 1 + 2 + 3 + ... + 16 \mod 17
$$</p>
<p>$$
= (1 + 16) + (2 + 15) + ... + (8 + 9) \mod 17
$$</p>
<p>$$
= 8 \times 17 \mod 17
$$</p>
<p>$$
= 0 \mod 17
$$</p>
|
259,431 | <p>In the book of Richard Hammack, I come accross with the following question:</p>
<blockquote>
<p>There are two different equivalence relations on the set $A = \{a,b\}$.
Describe them.</p>
</blockquote>
<p>OK, I found that the solution is,</p>
<p>$$R_1 = \{(a,a),(b,b),(a,b),(b,a)$$
and
$$R_2 = \{(a,a),(b,b)\}$$</p>
<p>Then I thought two more equivalence classes $R_3 = \{(a,a)\}$, $R_4 = \{(b,b)\}$. But when I looked the answer, I saw that, $R_1$ and $R_2$ are true but others are false. Why is that?</p>
| amWhy | 9,003 | <p>A relation is an equivalence relation if and only if it is reflexive, symmetric, and transitive. Your first two relations are indeed equivalence relations.</p>
<blockquote>
<p>A relation $R$ is reflexive on a set $A$ if and only if for all $x \in A, (x, x) \in R$.</p>
</blockquote>
<p>In $R_3$, we do not have that for $b \in A$, $(b, b) \in R_3$.</p>
<p>And in $R_4$, we do not have that for $a \in A$, $(a, a) \in R_4$.</p>
<p>So neither $R_3$ nor $R_4$ are reflexive, hence neither can be an equivalence relation.</p>
|
1,679,383 | <p>Find numbers $\alpha,\beta, \gamma \in \mathbb{C}$ so that the integral $$\int_{-1}^{1}|x^3-\alpha-\beta x-\gamma x^2|^2dx$$ is minimal.
Here I want to apply Hilbert space and projection lemma by considering the integral as the distance square between a point and a closed linear subspace. So how can I start my work? </p>
| Friedrich Philipp | 303,983 | <p>As the Hilbert space, choose the span of the functions $x^0,x^1,x^2,x^3$ (on $(-1,1)$), equipped with the scalar product
$$
\langle f,g\rangle = \int_{-1}^1 f(x)\overline{g(x)}\,dx.
$$
The subspace $M$ that you want to project onto should be the span of $x^0,x^1,x^2$. In this setting you are looking for that function $f_0$ in $M$ which has the smallest distance to $x^3$.</p>
|
2,711,785 | <p>Having some issues with this proof. Assume we've already proven addition, etc.</p>
<p>Definition of multiplication:</p>
<p>$a \times S(b) = a \times b + a$ (the "definition of multiplication")</p>
<p>$a \times 0 = 0$ (the "zero property of multiplication")</p>
<p>First, some supporting proofs:</p>
<p><strong>Claim</strong>: $0 \times a = 0$</p>
<p><strong>Base Case</strong>: We induct on $a$. Let $a=0$. See that $0 \times 0 = 0$ by definition of zero property of multiplication. </p>
<p><strong>Inductive Step</strong>: Suppose $0 \times a = 0$. We must show that $0 \times S(a) = 0$. By definition of multiplication we have $0 \times S(a) = 0 \times a + 0 = 0 \times a$ which is just $0$ by the inductive hypothesis.</p>
<hr>
<p><strong>Claim</strong>: $a \times b = b \times a$</p>
<p><strong>Base Case</strong>: We induct on $a$. Let $a=0$. See that $0 \times b = 0 = b \times 0$ by zero property of multiplication.</p>
<p><strong>Inductive Step</strong>: Suppose $a \times b = b \times a$. We must show that $S(a) \times b = b \times S(a)$. By definition of multiplication we have $S(a) \times b = a \times b + b$ which is $b \times a + b$ by inductive hypothesis. Then by definition of multiplication $b \times a + b = b \times S(a)$ and we are done.</p>
<hr>
<p>Now technically I used a proof there I haven't derived yet. But that's where I am having trouble.</p>
<hr>
<p><strong>Claim</strong>: $S(a) \times b = a \times b + b$</p>
<p><strong>Base Case</strong>: We induct on $b$. Let $b=0$. See that $S(a) \times 0 = 0 = a \times 0 + 0$ by zero property of multiplication and additive identity.</p>
<p><strong>Inductive Step</strong>: Suppose $S(a) \times b = a \times b + b$. We must show that $S(a) \times S(b) = a \times S(b) + S(b)$. By inductive hypothesis we have $S(a) \times S(b) = a \times S(b) + S(b)$ and we are done.</p>
<p>Is this correct? I feel like I am making a mistake somewhere. I only needed to use the inductive hypothesis to get what I needed?</p>
| Mauro ALLEGRANZA | 108,274 | <p><em>Long comment</em></p>
<p>Last <strong>Claim</strong> : the induction is on $b$ with $S(a)$ fixed. </p>
<p><em>Basis</em> : $b=0$. It's Ok. </p>
<p><em>Induction hypothesis</em> : $S(a)b=ab+b$ and we want to show that $S(a)S(b)=aS(b)+S(b)$. </p>
<p>Thus: $S(a)S(b)=S(a)b+S(a)$ (by axiom for mult) $=(ab+b)+S(a)$ (by ind.hy.) $=ab+(b+S(a))=ab+S(b+a)=ab+S(a+b)=ab+(a+S(b))=(ab+a)+S(b)=aS(b)+S(b).$</p>
|
1,113,934 | <p>How to prove that $\mathrm{Hom}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$ ?</p>
<p>(We treat it as field homomorphisms. )</p>
<p>I know that $\mathrm{Aut}(\mathbb{R})=\{\mathrm{id}\}$ and $\mathrm{Mon}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$.</p>
| Olórin | 187,521 | <p>Let $\varphi : \mathbf{R}\to \mathbf{R}$ be a morphism of fields. The kernel of $\varphi$ is an ideal of the field $\mathbf{R}$, and a (commutative) field $k$ has only two ideal, $(0)$ and $k$, and the kernel of $\varphi$ can't obviously be equal to whole $\mathbf{R}$, as then we would have $\varphi(1)=0$ and at the same time (as $\varphi$ is a morphism of fields) $\varphi (1) = 0$. This shows that the kernel of $\varphi$ is $(0)$, that is, that $\varphi$ is injective.</p>
|
1,583,540 | <p>Is it possible to find an example of an one-sided inverse of a function? other than matrix?</p>
<p>I am trying to find such an example but having no luck. Anybody got an idea about it?</p>
| BrianO | 277,043 | <p>Examples are abundant. Some basic facts may help you find many:</p>
<ol>
<li><p><em>If $\:f\colon X\to Y$ has a left inverse $h\colon Y\to X$ ($h\circ f = id_X$), then $f$ is a injection and $h$ is a surjection.</em></p></li>
<li><p><em>If $\:f\colon X\to Y$ has a right inverse $g\colon Y\to X$ ($f\circ g = id_Y$), then $f$ is a surjection and $g$ is an injection.</em></p>
<p>(<em>Proof of 2.</em>) $f$ is a surjection: if $y\in Y$ but $y\notin range(f)$, then $y\ne f(g(y))$, so $f\circ g \ne id_Y$. </p>
<p>$g$ is an injection: if $y_1, y_2\in Y$ and $g(y_1) = g(y_2)$; then $y_1 = f(g(y_1)) = f(g(y_2)) = y_2$.</p>
<p>(<em>Proof of 1.</em>) Given the hypothesis, $h$ has a right inverse $f$, so by 2., $h$ is a surjection and $f$ is an injection.</p>
<p><em>Note</em>: a right inverse for $f$ is a <em>choice function</em> for the indexed family of nonempty sets $\{f^{-1}(y)\mid y\in Y\}$. An equivalent form of the Axiom of Choice is: <em>Every surjection has a right inverse</em>. By contrast, it's easy to prove in ZF that every injection on a nonempty set has a left inverse.</p></li>
</ol>
<p>Now you can characterize a bijection as a function that has both one-sided inverses:</p>
<ol start="3">
<li><p><em>If $f$ has a right inverse $g$ and a left inverse $h$ then $g = h = f^{-1}$ and $f$ is a bijection.</em></p>
<p>Consider: $h = h\circ id_Y = h\circ (f\circ g) = (h\circ f) \circ g = id_X \circ g = g$.</p></li>
</ol>
<hr>
<p>Matrix multiplication is actually a relevant analogy.
Matrices represent functions, linear maps, and composition of linear maps corresponds to multiplication of their associated matrices. </p>
<p>An $n\times m$ matrix $A$ represents a linear map, a function $T_A\colon\Bbb R^m \to \Bbb R^n$ that computes $\vec{x}\mapsto A\cdot \vec{x}$, multiplication of an m-vector by $A$, yielding an $n$-vector. Thus $T_{AB} = T_A\circ T_B$, and $T_{I_m} = id_{\Bbb R^m}$. </p>
<p>When $m>n$, there are no injective linear maps $\Bbb R^m \to \Bbb R^n$, just as there are no injections from $X\to Y$ if (and only if) $card(Y)< card(X)$. (By definition, $card(X)\le card(Y)$ iff there is an injection $X\to Y$.) So if $Y$ is smaller than $X$, then given $X\stackrel{f}{\to} Y\stackrel{g}{\to} X$, $f$ isn't an injection and $g\circ f \ne id_X$. </p>
<p>For matrices $A$ and $B$ of sizes $n\times m$ and $m\times n$ respectively, both products $AB$ and $BA$ are defined. Both $\operatorname{rank} A$ and $\operatorname{rank} B$ are $\le \min(n,m)$, so if $n\ne m$, then one of the matrices has to lose information — it has a nontrivial kernel, multiplying by it on the left is not 1-1 — and one of the products will not be the identity. </p>
<ol>
<li><p><em>If $A$ is $n\times m$ and $A$ has a left inverse, then $m\le n$ and $T_A$ is an injection.</em> Suppose $BA=I_m$ and $n<m$. Then the rank of $A$ is at most $n$, and then the rank of $BA$ is also at most $n$. But the rank of $I_m$ is $m >n$. (In fact, $\operatorname{rank} A = m$.)</p></li>
<li><p><em>If $A$ is $n\times m$ and $A$ has a right inverse, then $n\le m$ and $T_A$ is a surjection.</em> If $AB = I_m$, then $A$ is a left inverse of $B$, so this follows from the previous result.</p></li>
</ol>
|
2,278,120 | <p>In one of my algorithm courses, there is this: </p>
<blockquote>
<p>A subset $S$ of vertices in a directed graph $G$ is
strongly connected if for each pair of distinct
vertices ($v_i$, $v_j$) in $S$, $v_i$ is connected to $v_j$ and
$v_j$ is connected to $v_i$.</p>
</blockquote>
<p>And then the following example graph is given for this proposition: </p>
<p><a href="https://i.stack.imgur.com/m6PLe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m6PLe.jpg" alt="enter image description here"></a></p>
<p>Maybe i do not understand what that phrase means. What i think it means is this: A node, say $E$ can be a member of an ordered pair $(v_i, v_j)$ only once, ie. if $(E, A)$, then $\lnot (E, [someOtherNode]) $. But we clearly see here, that we have $(E,A)$ and $(E,D)$. </p>
<p>How should i correctly interpret this phrase. What does it mean exactly? Thanks.</p>
| vadim123 | 73,324 | <p>Also, "connected" does not mean by a directed edge, but by a directed <a href="https://en.wikipedia.org/wiki/Path_(graph_theory)" rel="noreferrer">path</a>.</p>
|
195,432 | <p>Mathematica 12 has a new function <code>FunctionCompileExportLibrary</code>, which can export a pure function as a <code>.dll</code> file like this:</p>
<pre><code>FunctionCompileExportLibrary["function.dll", Function[
Typed[arg, "MachineInteger"], arg + 1]]
</code></pre>
<blockquote>
<p><code>"...\\function.dll"</code></p>
</blockquote>
<p>But how to call it in an external program (e.g., visual studio)? I have no <code>.h</code> file, I have no <code>.lib</code> file. I even don't know the function name that I want to call... Can anyone tell me how to use it?</p>
<p>The documentation of <a href="https://reference.wolfram.com/language/ref/FunctionCompileExportLibrary.html" rel="noreferrer"><code>FunctionCompileExportLibrary</code></a> states that</p>
<blockquote>
<p>The library generated by FunctionCompileExportLibrary is suitable for linking into external programs. It can also be loaded into the Wolfram System using LibraryFunctionLoad.</p>
</blockquote>
| yode | 21,532 | <p>As the ilian's answer, I make a note for Visual Studio user here to supplement:</p>
<p><strong>source.cpp file</strong></p>
<pre class="lang-cpp prettyprint-override"><code>#include<iostream>
using namespace std;
extern "C" int Main(int);
int main() {
cout << Main(3) << endl;
return 0;
}
</code></pre>
<p>configure Visual Studio</p>
<ol>
<li>In <strong>Configuration Properties/Debugging/Environment</strong> add</li>
</ol>
<blockquote>
<p>PATH=C:\Program Files\Wolfram
Research\Mathematica\12.0en\SystemFiles\Kernel\Binaries\Windows-x86-64;%appdata%\Mathematica\SystemFiles\LibraryResources\Windows-x86-64;C:\Program
Files\Wolfram
Research\Mathematica\12.0en\SystemFiles\Libraries\Windows-x86-64</p>
</blockquote>
<p>The first directory has some <a href="https://i.stack.imgur.com/oaKqN.png" rel="nofollow noreferrer">necessary .dll file</a>. The second and the three directory have <code>function.dll</code> and <code>WolframRTL.dll</code> respectively. Of course, your can don't set the <code>PATH</code> evironment if you copy all those <code>.dll</code> file into the direcotry with <code>source.cpp</code> or the <code>Configuration Properties/Output Directory</code>.</p>
<ol start="2">
<li>In <strong>Configuration Properties/VC++ Directories/Library Directories</strong> add</li>
</ol>
<blockquote>
<p>%appdata%\Mathematica\SystemFiles\LibraryResources\Windows-x86-64</p>
</blockquote>
<p>This directory include <code>function.lib</code></p>
<ol start="3">
<li>In <strong>Configuration Properties/Linker/Input/Additional Dependencies</strong> add</li>
</ol>
<blockquote>
<p>function.lib</p>
</blockquote>
<p>Then you can run the c++ project now. I think this is a small step for Wolframer, but a giant leap for those men who hope to use Mathematica in his work.</p>
|
2,656,095 | <p>I want to solve the following PDE using canonical form:</p>
<p>$$xu_{xx} + 2x^2u_{xy} = u_x - 1$$</p>
<p>I have found the characteristic curves $\xi = y$ and $\eta = y-x^2$. By computing $u_{xx}$ and $u_{xy}$, I got the following (hopefully correct) relation :
$$4x^3u_{\xi\eta} = -1$$
How do I get the general solution from this?</p>
<p>The correct solution is $u(x,y) = x+f(x^2 - y) + g(y)$ </p>
| ab123 | 454,871 | <p>I have now solved it by putting $x^3 = (\xi - \eta)^{3/2}$ and then integrating the canonical equation will give the given solution.</p>
<p><strong>Edit(Complete solution):</strong>
Comparing the given PDE with the general form $A(x, y)u_{xx} + B(x, y)u_{xy} + C(x, y)u_{yy} = Φ(x, y,u,u_x,u_y)$ we find
$$A = x, B= 2x^2, C = 0$$</p>
<p>Now, the characteristic curves can be found by the equation $\frac{dy}{dx} = \frac{B \pm\sqrt{B^2-4AC}}{2A} = 0, 2x$</p>
<p>Integrating, we find the 2 curves as $y = $constant and $y-x^2 = $constant, so we choose
$$\xi = y,\qquad \eta = y-x^2$$
and apply the relations found by chain rule for differentiating $u$ w.r.t. $\xi$ and $\eta$:</p>
<p>$u_x = u_ξξ_x + u_ηη_x, u_y = u_ξξ_y + u_ηη_y,$</p>
<p>$u_{xx} = u_{ξξ}ξ^2_x + 2u_{ξη}ξ_xη_x + u_{ηη}η^2_x+ u_ξξ_{xx} + u_ηη_{xx}$</p>
<p>$u_{xy} = u_{ξξ}ξ_xξ_y + u_{ξη} (ξ_xη_y + ξ_yη_x) + u_{ηη}η_xη_y + u_ξξ_{xy} + u_ηη_{xy},$</p>
<p>$u_{yy} = u_{ξξ}ξ^2_y + 2u_{ξη}ξ_yη_y + u_{ηη}η^2_y + u_ξξ_{yy} + u_ηη_{yy}$</p>
<p>and substituting in original equation will give $4x^3u_{\xi\eta} = -1$</p>
<p>We want to eliminate $x$ so that equation consists only of $\xi$ and $\eta$, so put $$x^3 = (\xi - \eta)^{3/2}$$ then integrate equation once w.r.t. variable $\xi$ , then with $\eta$ to get the given solution.</p>
|
2,656,095 | <p>I want to solve the following PDE using canonical form:</p>
<p>$$xu_{xx} + 2x^2u_{xy} = u_x - 1$$</p>
<p>I have found the characteristic curves $\xi = y$ and $\eta = y-x^2$. By computing $u_{xx}$ and $u_{xy}$, I got the following (hopefully correct) relation :
$$4x^3u_{\xi\eta} = -1$$
How do I get the general solution from this?</p>
<p>The correct solution is $u(x,y) = x+f(x^2 - y) + g(y)$ </p>
| Aleksas Domarkas | 562,074 | <p>$$xu_{xx} + 2x^2u_{xy} -u_x =- 1$$</p>
<p>Equation is linear. Then
$$u=u_h+u_p$$</p>
<p>where $u_p$ is is particular solution(we can take $\; u_p=x\;$),</p>
<p>$u_h$ is solution homogeneus equation $xu_{xx} + 2x^2u_{xy} -u_x =0$.</p>
<p>After change $\quad \xi = x^2-y,\quad \eta = y\quad$ we get
$$u_{\xi\eta}=0,$$
$$u_h=f(\xi)+g(\eta)=f(x^2-y)+g(y),$$
$$u=u_h+u_p=f(x^2-y)+g(y)+x$$</p>
|
2,126,299 | <p>My goal is to write $\mathbb{Q}(\pi^3+\pi^2, \pi^8+\pi^5)$ as $\mathbb{Q}(f(\pi))$ where $f \in \mathbb Q(X)$.</p>
<p>I have $(\pi^3+\pi^2)^2 = \pi^6+2\pi^5+\pi^4$ and then $2\pi^8 - \pi^6 - \pi^4$ is in my field. But I don't think it generates it. </p>
<p>Thank you for your help!</p>
| Ewan Delanoy | 15,381 | <p>Let $a=\pi^3+\pi^2$ and $b=\pi^8+\pi^5$.
Then</p>
<p>$$
\pi=\frac{3b(1-a)+(a^4+3a^3)}{b(a-3)+(3a^3+3a^2)}
$$</p>
<p>so you can take $f(\pi)=\pi$.
How did I obtain this formula ? I tried to find the minimal
polynomial of $\pi$ over ${\mathbb Q}(a,b)$, by applying the Euclidean
algorithm to $X^3+X^2-a$ and $X^8+X^5-b$.</p>
<p>The last nonzero term produced by this algorithm is $3b(1-a)+(a^4+3a^3)-X(b(a-3)+(3a^3+3a^2))$.</p>
|
2,126,299 | <p>My goal is to write $\mathbb{Q}(\pi^3+\pi^2, \pi^8+\pi^5)$ as $\mathbb{Q}(f(\pi))$ where $f \in \mathbb Q(X)$.</p>
<p>I have $(\pi^3+\pi^2)^2 = \pi^6+2\pi^5+\pi^4$ and then $2\pi^8 - \pi^6 - \pi^4$ is in my field. But I don't think it generates it. </p>
<p>Thank you for your help!</p>
| 1Emax | 324,326 | <p>Let $\alpha = \pi^3+\pi^2$ and $\beta = \pi^8+\pi^5$. It can be shown that if $f = \frac{p}{q}$ where $\gcd(p,q)=1$, we have $$[\mathbb{Q}(\pi):\mathbb{Q}(f(\pi))] = \max\{\deg(p), \deg(q)\}.$$ Therefore $[\mathbb{Q}(\pi):\mathbb{Q}(\pi^3+\pi^2)]=3$, and if $\pi \notin \mathbb{Q(\alpha, \beta)}$, we must have $\mathbb{Q(\pi)}:\mathbb{Q(\alpha, \beta)}=3$, since
$$[\mathbb{Q}(\pi):\mathbb{Q}(\alpha)] = [\mathbb{Q}(\pi):\mathbb{Q}(\alpha, \beta)][\mathbb{Q}(\alpha, \beta):\mathbb{Q}(\alpha)].$$</p>
<p>This means that $\beta \in \mathbb{Q}(\alpha)$, or $\pi^8+\pi^5 = \frac{g}{h}$ where $g,h \in \mathbb{Q}[\alpha]$ and are relatively prime. So $\pi^5|g$ which is impossible, because if $\pi^k |g$, $k$ must be even. So $\pi \in \mathbb{Q}(\alpha, \beta)$.</p>
|
187,511 | <p>There are two questions:</p>
<ol>
<li><p>How to prove that in general </p>
<p>$[\hat{A}(\mathbb HP^m)]_{4m} = 0$</p>
<p>It is possible to verify it for low values of $m$.</p></li>
<li><p>How to prove that in general</p>
<p>$\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$</p>
<p>where $\hat{M}(\mathbb HP^m)$ is the Mayer class defined by</p>
<p>$\hat{M}(V) = \prod _{i=1}^{s}\cosh \left( \frac{y_{{i}}}{2} \right)$</p>
<p>with</p>
<p>$p(V) =\prod _{i=1}^{s}(1+{y_{i}}^2)$.</p>
<p>It is possible to verify it for low values of $m$.</p></li>
</ol>
| Juan Ospina | 61,860 | <p>I think that using the method of Professor Ebert is possible to prove the question 2 according with the following procedure.</p>
<p>For the expression ${\frac {\hat{A}}{\hat{M}}}$ the corresponding $F(x)$ is</p>
<p>$F(x) = \frac{x/2}{\sinh(x/2) \cosh(x/2)} = \frac{x}{\sinh(x)} $</p>
<p>Then we have</p>
<p>$\frac{1}{2 \pi i}\int \frac{F(x)^2}{F(2x)x} \frac{1}{1- \frac{F(x)t}{x}} dx = \frac{1}{2 \pi i}\int{\frac {\cosh \left( x \right) }{\sinh \left( x \right) -t}}dx$</p>
<p>With the substitution $u=\sinh \left( x \right)$, the integral is reduced to</p>
<p>$\frac{1}{2 \pi i}\int \frac{1}{u-t} du$</p>
<p>This last integral is independent of $t$, then $G(t) \equiv 1$ and then we have that</p>
<p>$[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }]_{4m} = 0$</p>
<p>Do you agree?</p>
|
2,111,579 | <p>A question:</p>
<p>If $\displaystyle{m}^{2}={1}-{m},{\quad\text{and}\quad}{n}^{2}={1}-{n},{\quad\text{and}\quad}{n}\ne{m};$</p>
<p>Proof that $\displaystyle{m}^{7}+{n}^{7}+{30}={1}$</p>
<p>Without finding the roots of equation $\displaystyle{x}^{2}+{x}-{1}={0}$.</p>
<p>Is there such a shortcut solution?</p>
| G Cab | 317,234 | <p>Just to show a different approach.<br>
Consider the matrix $\mathbf E$, having $1$ only on the first subdiagonal
$$
\mathbf{E} = \left\| {\,e_{\,n,\,m} = \left\{ {\begin{array}{*{20}c}
1 & {n = m + 1} \\
0 & {n \ne m + 1} \\
\end{array} } \right.\;} \right\| = \left\| {\,\left( \begin{gathered}
0 \\
n - m - 1 \\
\end{gathered} \right)\;} \right\|
$$
Multiply it by $\mathbf A$ , and it is easily seen that $$\mathbf E \, \mathbf A=\mathbf A-\mathbf I \quad \Rightarrow \quad \left( \mathbf I -\mathbf E \right)\,\mathbf A=\mathbf I$$</p>
<p>In another way, consider that the powers of $\mathbf E$ are readily found and have a simple formulation
$$
\begin{gathered}
\mathbf{E}^{\,\mathbf{2}} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered}
0 \\
n - k - 1 \\
\end{gathered} \right)\left( \begin{gathered}
0 \\
k - m - 1 \\
\end{gathered} \right)} \;} \right\| = \left\| {\,\left( \begin{gathered}
0 \\
n - m - 2 \\
\end{gathered} \right)\;} \right\| = \left\| {\,\left\{ {\begin{array}{*{20}c}
1 & {n = m + 2} \\
0 & {n \ne m + 2} \\
\end{array} } \right.\;} \right\| \hfill \\
\quad \vdots \hfill \\
\mathbf{E}^{\,\mathbf{q}} = \mathbf{E}^{\,\mathbf{q} - \mathbf{1}} \,\mathbf{E} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered}
0 \\
n - k - \left( {q - 1} \right) \\
\end{gathered} \right)\left( \begin{gathered}
0 \\
k - m - 1 \\
\end{gathered} \right)} \;} \right\| = \hfill \\
= \left\| {\,\left( \begin{gathered}
0 \\
n - m - q \\
\end{gathered} \right)\;} \right\|\quad \left| {\;0 \leqslant \text{integer}\;q} \right. \hfill \\
\end{gathered}
$$
Therefore
$$
\mathbf{A} = \sum\limits_{0\, \leqslant \,j} {\mathbf{E}^{\,\mathbf{j}} } = \frac{\mathbf{I}}
{{\mathbf{I} - \mathbf{E}}}
$$
To connect to the answer of <em>Jean Marie</em>, note that
$$
\left( {\mathbf{I} - \mathbf{E}} \right)\left\| {\,\begin{array}{*{20}c}
{x_{\,0} } \\
{x_{\,1} } \\
\vdots \\
{x_{\,n} } \\
\end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
{x_{\,0} ( - 0)} \\
{x_{\,1} - x_{\,0} } \\
\vdots \\
{x_{\,n} - x_{\,n - 1} } \\
\end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
{\nabla x_{\,0} \;\left| {x_{\, - 1} = 0} \right.} \\
{\nabla x_{\,1} } \\
\vdots \\
{\nabla x_{\,n} } \\
\end{array} \;} \right\|
$$</p>
|
3,411,982 | <p>I know the answer using
Total-none of it is <span class="math-container">$6 = 900−648=252$</span></p>
<p>My doubt is that if we you it like</p>
<p>when unit digit is <span class="math-container">$6 = 9\cdot10\cdot1 = 90$</span></p>
<p>when tens place is <span class="math-container">$6 = 9\cdot1\cdot10 = 90$</span></p>
<p>when hundredth place is <span class="math-container">$6 = 1\cdot10\cdot10 = 100$</span></p>
<p>So total <span class="math-container">$= 90 + 90 + 100 = 280$</span></p>
<p>why my answer is not the same </p>
| Shashi kant | 1,027,485 | <p>your answer is different in second method because in first case when you have counted many no. 2 or 3 times like 666 is counted all three times so that ways counted multiple times must be subtracted and your answer will match</p>
|
2,278,436 | <p>Does there exist integers $a, n > 1$ such that $1 + \frac{1}{1 + a} + \frac{1}{1 + 2a} + ... + \frac{1}{1 + na}$ is an integer? I have no clue how to begin. I've tried to simplify this somehow, but with no effect.</p>
| vrugtehagel | 304,329 | <p>This is not a complete answer (yet). I'm posting it because it may be useful to people who are thinking about this problem.</p>
<hr>
<p>Note that</p>
<p>$$\gcd(1+a,1+ka)=\gcd(1+a,(k-1)a)=\gcd(1+a,k-1)$$</p>
<p>Let $p\mid a+1$ (where $p$ is prime). For $\frac1{1+a}+\frac1{1+2a}+\cdots+\frac1{1+na}$ to be an integer (we leave the $1+$ at the start away since that's an integer already), we need</p>
<p>$$(1+a)(1+2a)\cdots(1+na)\ \ \mid \ \ \sum_{k=1}^n\prod_{\substack{1\le i\le n\\ i\neq k}} (1+ai)$$</p>
<p>Since $p$ divides the left-hand side, it must divide the right-hand side; so,</p>
<p>$$p\ \ \mid \ \ \sum_{k=1}^n\prod_{\substack{1\le i\le n\\ i\neq k}} (1+ai)$$</p>
<p>But each of the terms on the left has a $1+a$ in it, except when $k=1$; so, $p\mid \prod_{i=2}^n(1+ai)$, which means $p\mid 1+ai$ for some $2\le i\le n$. But we've seen that $\gcd(1+a,1+ia)=\gcd(1+a,k-1)$, so that $p\mid i-1$ for some $2\le i\le n$. Now we can do the same trick with $1+2a$ instead of $1+2a$; let $q\mid 1+2a$, similarly derive that $q\mid 1+ia$ for some $i\in\{1,3,4,\cdots,n\}$, which means $q\mid i-2$. This we can do for every $1+ka$; we obtain:</p>
<blockquote>
<p>Let $p$ be a prime such that $p\mid 1+ak$. Then there exists a number $i\in\{1,2,\cdots,n\}\backslash\{k\}$ such that $p\mid i-k$.</p>
</blockquote>
<p><hr>
This tells us that the numbers $1+a,1+2a,\cdots,1+na$ all need to have prime factors smaller than $n$.</p>
<p>Let now $p<n$ be a prime such that $p\not\mid a$. Then we know for $i=1,2,\cdots,p$, that all $1+ia$ are distinct $\mod p$ (since $1+ai\equiv 1+aj\mod p$ implies $i\equiv j\mod p$). This in turn means there's a $1\le i\le p$ such that $1+ai\equiv 0\mod p$, so that $p$ is a factor of $1+ai$.
<hr>
We've now determined exactly what the prime factors of $(1+a)(1+2a)\cdots(1+na)$ are; they are all primes less than $n$ that don't divide $a$.</p>
|
2,278,436 | <p>Does there exist integers $a, n > 1$ such that $1 + \frac{1}{1 + a} + \frac{1}{1 + 2a} + ... + \frac{1}{1 + na}$ is an integer? I have no clue how to begin. I've tried to simplify this somehow, but with no effect.</p>
| sku | 341,324 | <p>Let $X = (1+a)(1+2a)\cdots(1+na)$</p>
<p>Let $Y = \sum_{k = 1}^{n}\frac{X}{1+ka}$</p>
<p>Let $Z = \frac{Y}{X}$</p>
<p>Let $p^r$ be the highest power of a prime $p$ that divides $X$</p>
<p>Let $p^s$ be the highest power of $p$ that divides $1+ka$</p>
<p>So the power of $p$ that divides $Y$ is at least $p^r/p^s = p^{r-s}$</p>
<p>So $p^r$ which divides $X$ does not divide $Y \implies Z$ is not an integer. </p>
|
4,155,453 | <p>I am trying to evaluate <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$</span> Now I am aware that clearly <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{-1}(1) = \frac{\pi}{4},$$</span> but I do not know what to do if each sum is already sent to infinity. Im taking a limit of limits. I suppose I could rewrite my limit as <span class="math-container">$$\lim_{n\to \infty} \lim_{m\to \infty} \sum_{k=1}^m \frac{n}{n^2+k^2}?$$</span> But I am unaware if this is helpful at all. Any hints would be appreciated. Obviously, Wolfram calculates this as <span class="math-container">$\frac{\pi}{2}$</span> but I am unaware of the steps and logic to get there.</p>
| Paresseux Nguyen | 758,600 | <p>Let <span class="math-container">$$f(x):= \frac{1}{1+x^2}$$</span>
then
<span class="math-container">$$\sum_{k=1}^{\infty} \frac{n}{n^2+x^2} = \int_{0}^{+\infty} f\left( \frac{ \lceil nx \rceil }{n} \right) dx$$</span>
From this you can do whatever you like to conclude, maybe using DCT while knowing that <span class="math-container">$$f\left( \frac{ \lceil nx \rceil }{n} \right) \le f(x) \quad \forall n \quad \text{and } \int_{0}^{\infty} f(x)dx <+\infty$$</span>
for a direct conclusion.</p>
<p><strong>Edit</strong>: It's probable that you haven't learnt DCT yet, so here is an alternative</p>
<p>If <span class="math-container">$0 \le x \le y$</span>, we see that:
<span class="math-container">$$ 0 \le f(x)-f(y) = \frac{(y-x)(y+x)}{(1+y^2)(1+x^2)} \le \frac{y-x}{1+x^2}=(y-x)f(x)$$</span>
Thus for all <span class="math-container">$x>0$</span>
<span class="math-container">$$0 \le f(x)- f\left( \frac{ \lceil nx \rceil }{n} \right) \le \left( \frac{ \lceil nx \rceil }{n} -x\right)f(x) \le \frac{1}{n}f(x) $$</span>
So
<span class="math-container">$$\int_{0}^{\infty} f(x)dx \ge \int_{0}^{\infty} f\left( \frac{ \lceil nx \rceil }{n} \right)dx \ge (1-\frac{1}{n})\int_{0}^{\infty} f(x)dx $$</span>
Hence forth the conclusion</p>
|
4,155,453 | <p>I am trying to evaluate <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$</span> Now I am aware that clearly <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{-1}(1) = \frac{\pi}{4},$$</span> but I do not know what to do if each sum is already sent to infinity. Im taking a limit of limits. I suppose I could rewrite my limit as <span class="math-container">$$\lim_{n\to \infty} \lim_{m\to \infty} \sum_{k=1}^m \frac{n}{n^2+k^2}?$$</span> But I am unaware if this is helpful at all. Any hints would be appreciated. Obviously, Wolfram calculates this as <span class="math-container">$\frac{\pi}{2}$</span> but I am unaware of the steps and logic to get there.</p>
| zhw. | 228,045 | <p>Let <span class="math-container">$f(x)= \dfrac{1}{1+x^2}.$</span> Note that our sum equals</p>
<p><span class="math-container">$$\sum_{k=1}^{\infty}f\left(\frac{k}{n}\right)\frac{1}{n}.$$</span></p>
<p>Now <span class="math-container">$f$</span> decreases from <span class="math-container">$1$</span> to <span class="math-container">$0$</span> on <span class="math-container">$[0,\infty).$</span> Thus, for <span class="math-container">$k=1,2,\dots,$</span></p>
<p><span class="math-container">$$\int_{k/n}^{(k+1)/n}f\le f\left(\frac{k}{n}\right)\frac{1}{n} \le \int_{(k-1)/n}^{k/n}f.$$</span></p>
<p>Summing on <span class="math-container">$k$</span> then gives</p>
<p><span class="math-container">$$\int_{1/n}^\infty f(x)\,dx \le \sum_{k=1}^{\infty}f\left(\frac{k}{n}\right)\frac{1}{n}\le \int_0^\infty f(x)\,dx.$$</span></p>
<p>This is true for any <span class="math-container">$n.$</span> As <span class="math-container">$n\to \infty,$</span> the integral on the left approaches <span class="math-container">$\int_0^\infty f(x)\,dx.$</span> By the squeeze theorem, the limit of our sum equals the value of this integral, which is <span class="math-container">$\pi/2.$</span></p>
|
3,133,152 | <blockquote>
<p><span class="math-container">$g(x)=|\sin{x}-1|+|3-\cos{x}-\sin{x}|+2\sin{x}$</span></p>
<p>Answer: Above equality is simplified to <span class="math-container">$$1-\sin{x}+3-\cos{x}-\sin{x}+2\sin{x}=4-\cos{x}$$</span></p>
</blockquote>
<p><span class="math-container">$$-1 \le\sin{x}\le1$$</span></p>
<p>So , I know that <span class="math-container">$f(x)=|\sin{x}-1|$</span> will be equal to <span class="math-container">$1-\sin{x}$</span> when <span class="math-container">$-1 \le\sin{x}\lt1$</span>.</p>
<p>But what if <span class="math-container">$\sin{x}$</span> is exactly <span class="math-container">$1$</span>, then wouldn't the expression be <span class="math-container">$\sin{x}-1$</span>?</p>
<p>I always considered <span class="math-container">$|x|=x$</span>, whenever <span class="math-container">$x$</span> is <em>equal to</em> or greater than <span class="math-container">$0$</span>.</p>
| Robert Lewis | 67,071 | <p><span class="math-container">$\vert \sin x - 1 \vert = \vert 1 - \sin x \vert, \; \forall x \in \Bbb R; \tag 1$</span></p>
<p>since</p>
<p><span class="math-container">$1 - \sin x \ge 0, \; \forall x \in \Bbb R, \tag 2$</span></p>
<p><span class="math-container">$\vert \sin x - 1 \vert = \vert 1 - \sin x \vert = 1 - \sin x, \; \forall x \in \Bbb R. \tag 3$</span></p>
|
868,663 | <blockquote>
<p>Prove or disprove that $$7^{8}+8^{9}+9^{7}+1$$ is a prime number, without using a computer.</p>
</blockquote>
<p>I tried to transform $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, unsuccessfully, no useful conclusion.</p>
| Community | -1 | <p>This is in answer to the edits (I'm going to start with your second question):</p>
<p><strong>#2: How we know that we can always represent a number by adding factors of base-10 numbers</strong></p>
<p>There is no proof needed. Decimal notation is simply a shorthand for expressing numbers in terms of base-10 numbers. $342$ is DEFINED to mean $(3×10^2)+(4×10^1)+(2×10^0)$, where we take the numbers $0 - 10$ and the operations $\times, +$, and ^ (exponentiation) as "primitives" (things we don't NEED to define because we all hopefully have the same idea in mind as to what they mean). Because of this definition it doesn't matter how big the number is, you can ALWAYS reduce it to this form (okay, not ALWAYS, but if we assume that a number HAS a decimal representation, then it can be reduced to this form). For instance $ 12345678909876543210.54321 := 1 \times 10^{19} + 2 \times 10^{18} + 3 \times 10^{17} + 4 \times 10^{16} + 5 \times 10^{15} + 6 \times 10^{14} + 7 \times 10^{13} + 8 \times 10^{12} + 9 \times 10^{11} + 0 \times 10^{10} + 9 \times 10^9 + 8 \times 10^8 + 7 \times 10^7 + 6 \times 10^6 + 5 \times 10^5 + 4 \times 10^4 + 3 \times 10^3 + 2 \times 10^2 + 1 \times 10^1 + 0 \times 10^0 + 5 \times 10^{-1} + 4 \times 10^{-2} + 3 \times 10^{-3} + 2 \times 10^{-4} + 1 \times 10^{-5}$, where $:=$ means "is defined as".</p>
<p><strong>#1: How multiplication works</strong></p>
<p>Say you want to multiply 345 by 678. The "usual method" is:
<br>$\ \ \ \ \ \ 345$<br>$\ \ \ \ \ \ 678$ <br> $\ \ \ \ \overline{2760}$ <br>$\ \ 2415$<br> $2070$<br>$\overline{233910}$</p>
<p>Here's how it actually works:</p>
<p>1) First we write out our numbers explicitly in base-10 notation:</p>
<p>$$345 \times 678 = [(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)] \times [(6 \times 10^2) + (7 \times 10^1) + (8 \times 10^0)]$$</p>
<p>2) Then we <a href="http://en.wikipedia.org/wiki/Distributivity" rel="nofollow">distribute</a>:</p>
<p>$$= \left(\left[(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)\right] \times (6 \times 10^2)\right) + \left(\left[(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)\right] \times (7 \times 10^1)\right) + \left(\left[(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)\right] \times (8 \times 10^0)\right)$$
$$ = \left(\left[(3 \times 10^2) \times (6 \times 10^2)\right] + \left[(4 \times 10^1) \times (6 \times 10^2)\right] + \left[(5 \times 10^0) \times (6 \times 10^2)\right]\right) + \left(\left[(3 \times 10^2) \times (7 \times 10^1)\right] + \left[(4 \times 10^1) \times (7 \times 10^1)\right] + \left[(5 \times 10^0) \times (7 \times 10^1)\right]\right) + \left(\left[(3 \times 10^2) \times (8 \times 10^0)\right] + \left[(4 \times 10^1) \times (8 \times 10^0)\right] + \left[(5 \times 10^0) \times (8 \times 10^0)\right]\right)$$</p>
<p>3) Now to save space, I'm only going to work on those last 3 terms which will produce the first line of the solution of the "usual method". I will now use the <a href="http://en.wikipedia.org/wiki/Commutativity" rel="nofollow">commutativity</a> of addition to reorder the terms:</p>
<p>$$= \left[(5 \times 10^0) \times (8 \times 10^0)\right] + \left[(4 \times 10^1) \times (8 \times 10^0)\right] + \left[(3 \times 10^2) \times (8 \times 10^0)\right] + \cdots$$</p>
<p>4) Now I'll use the commutativity and <a href="http://en.wikipedia.org/wiki/Associativity" rel="nofollow">associativity</a> of multiplication to reorder the factors in each term:</p>
<p>$$= \left[(5 \times 8) \times (10^0 \times 10^0)\right] + \left[(4 \times 8) \times (10^1 \times 10^0)\right] + \left[(3 \times 8) \times (10^2 \times 10^0)\right] + \cdots$$</p>
<p>5) Use our exponent rules:</p>
<p>$$= \left[(5 \times 8) \times 10^0\right] + \left[(4 \times 8) \times 10^1\right] + \left[(3 \times 8) \times 10^2\right] + \cdots$$</p>
<p>6) Here is where our multiplication table comes in:</p>
<p>$$= \left[(40) \times 10^0\right] + \left[(32) \times 10^1\right] + \left[(24) \times 10^2\right] + \cdots$$</p>
<p>7) Recognize that $40$ is shorthand for $(4 \times 10^1) + (0 \times 10^0)$ and so on to get:</p>
<p>$$= \left[(4 \times 10^1 + 0 \times 10^0) \times 10^0\right] + \left[(3 \times 10^1 + 2 \times 10^0) \times 10^1\right] + \left[(2 \times 10^1 + 4 \times 10^0) \times 10^2\right] + \cdots$$</p>
<p>8) From distributivity and our exponent rules:</p>
<p>$$= \left[4 \times 10^1 + 0 \times 10^0\right] + \left[3 \times 10^2 + 2 \times 10^1\right] + \left[2 \times 10^3 + 4 \times 10^2\right] + \cdots$$</p>
<p>9) From commutativity:</p>
<p>$$= \left[0 \times 10^0 + 4 \times 10^1\right] + \left[2 \times 10^1 + 3 \times 10^2\right] + \left[4 \times 10^2 + 2 \times 10^3\right] + \cdots$$</p>
<p>10) From associativity:</p>
<p>$$= (0 \times 10^0) + (4 \times 10^1 + 2 \times 10^1) + (3 \times 10^2 + 4 \times 10^2) + (2 \times 10^3) + \cdots$$</p>
<p>11) From distributivity:</p>
<p>$$= (0) \times 10^0 + (4 +2) \times 10^1 + (3 + 4) \times 10^2 + (2) \times 10^3 + \cdots$$</p>
<p>12) And we finally end up with the first term in the solution of the "usual method" as promised:</p>
<p>$$= (0) \times 10^0 + (6) \times 10^1 + (7) \times 10^2 + (2) \times 10^3 + \cdots = 2760 + \cdots$$</p>
<p>At this point, if I had written out all the other terms instead of just writing "$\cdots$" we could break each term up into base-10 parts and add.</p>
<p>As you can see, the "usual method" has a LOT going on under the surface that (thankfully) you don't normally need to think about.</p>
|
1,860,243 | <p>Let $|G|$ be an abelian group and let $H = \{g \in G : |g| \text{ divides } 12 \}.$ Prove that $H$ is a subgroup of $G$.</p>
<p>I know that I have to show that $a,b \in H \Rightarrow ab^{-1} \in H$ or $(ab \in H \land a^{-1} \in H).$ But I can't figure out how $|a|$ and $|b|$ dividing $12$ relates to $|ab|$ or $|ab^{-1}|$ dividing $12$.</p>
| Jendrik Stelzner | 300,783 | <p>Notice that $|g|$ divides $12$ if and only if $12g = 0$.
Because the map $f \colon G \to G$, $g \mapsto 12g$ is a group homomorphism it follows that $H = \ker f$ is a subgroup.</p>
|
4,244,565 | <p>In resources like Wikipedia the following equation holds for <span class="math-container">$k>0$</span>.
<span class="math-container">\begin{equation}
\sum_{n=0}^{k} [x^n]=\frac{1-x^{k+1}}{1-x}\hspace{1cm}x\neq1
\end{equation}</span>
If you put the right hand side in a limit it holds for <span class="math-container">$k\geq0$</span>. But why do we rarely show the bounds of the k value and why not put the k in an absolute value on the right hand side (Other than the aversion to absolute values)?</p>
| user247327 | 247,327 | <p>(You mean that |X|= |Y|= m or cardinality of X= cardinality of Y= m. X and Y themselves are sets so not equal to a number.)</p>
<p>I don't know what you mean by "<span class="math-container">$f(x_1)= y_1, y_2,..., y_m$</span>". <span class="math-container">$f(x_1)$</span> is a single element of X (I suppose- you don't actually say what these x's or y's are) not all members of Y.</p>
<p>You then talk about <span class="math-container">$f(X_1)$</span>, <span class="math-container">$f(X_2)$</span>, ... without any previous mention of <span class="math-container">$X_1$</span>, etc. Do you mean <span class="math-container">$x_1$</span>, etc.?</p>
<p>Finally, and most egregious, you write "f(X_1)= m", "f(X_2)= m- 1", "f(X_2)= m- (m-1)", etc. which make no sense at all. You are told that f is a function from set X to set Y. Nothing at all is said about Y being a set of numbers or any set on which "1" or subtraction are defined!</p>
<p>I would do this: Since X is finite we can call its cardinality "n" and order the members of X calling them <span class="math-container">$x_1$</span>, <span class="math-container">$x_2$</span>, ..., <span class="math-container">$x_n$</span>. Since Y has the same cardinality as X we can call its members <span class="math-container">$y_1$</span>, <span class="math-container">$y_2$</span>, ..., <span class="math-container">$y_n$</span>. Now there exist n! possible permutationa of Y. And each bijective function can be written as mapping <span class="math-container">$\{x_1, x_2, ..., x_n\}$</span> to the "y"s in one of those permutations.</p>
|
30,970 | <p>I try to filter sublists of a list which match a pattern. </p>
<pre><code>test = {{"String1", "a"}, {"String2", "b"}, {"String3",
"a"}, {"String4", "a"}};
</code></pre>
<p>The result should be:</p>
<pre><code>result = {{"String1", "a"}, {"String3", "a"}, {"String4", "a"}}
</code></pre>
<p>That means the first entry should be any String and the second should be "a".</p>
<p>I tried:</p>
<pre><code>Select[test, (# == {_, "a"}) &]
</code></pre>
<p>Which evaluates to {}. </p>
| M6299 | 7,115 | <p>Another solution:</p>
<pre><code>Select[test, Function[{x}, StringQ[x[[1]]] && x[[2]] == "a"]]
</code></pre>
<blockquote>
<pre><code>{{"String1","a"},{"String3","a"},{"String4","a"}}
</code></pre>
</blockquote>
|
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