qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,553,233 | <p>I am having difficulty finishing this proof. At first, the proof is easy enough. Here's what I have thus far:<br>
Because <span class="math-container">$5 \nmid n$</span>, we know <span class="math-container">$\exists q \in \mathbb{Z}$</span> such that <span class="math-container">$$n = 5q + r$$</span> where <span class="math-container">$0 < r < 4$</span>. Note <span class="math-container">$r \neq 0$</span> because if <span class="math-container">$r = 0$</span>, then <span class="math-container">$5 \mid n$</span>. Also note that <span class="math-container">$n^2 = 25q^2 + 10qr + r^2$</span>. Then we have four cases: when <span class="math-container">$r=1$</span>, <span class="math-container">$r=2$</span>, <span class="math-container">$r = 3$</span>, and <span class="math-container">$r = 4$</span>. This is where I run into difficulty. In each of these cases, we can prove that either <span class="math-container">$n^2 = 5k + 1$</span> or <span class="math-container">$n^2 = 5k - 1$</span> for some integer <span class="math-container">$k$</span>, but I cannot see how to prove both for each case. Any ideas? </p>
<p>As a side note on how I went to prove each case, I simply plugged <span class="math-container">$r$</span> into the formula <span class="math-container">$n^2 = 25q^2 + 10qr + r^2$</span>. This results in <span class="math-container">$n^2 = 25q^2 + 10q + 1$</span>.
Continuing, we get <span class="math-container">$n^2 = 5(5q^2 + 2q) + 1$</span>, and because <span class="math-container">$5q^2 + 2q$</span> is still an integer, this is of the form <span class="math-container">$n^2 = 5k + 1$</span> for some integer <span class="math-container">$k$</span>. But I cannot find how to make the 1 a negative to prove both cases.</p>
| J. W. Tanner | 615,567 | <p>As you showed, <span class="math-container">$n^2=25q^2+10qr+r^2$</span>, where <span class="math-container">$r=1, 2, 3,$</span> or <span class="math-container">$4$</span>.</p>
<p>When <span class="math-container">$r=1$</span>, <span class="math-container">$n^2=25q^2+10qr+1=5(5q^2+2qr)+1=5k+1.$</span></p>
<p>When <span class="math-container">$r=2$</span>, <span class="math-container">$n^2=25q^2+10qr+4=5(5q^2+2qr+1)-1=5k-1.$</span></p>
<p>When <span class="math-container">$r=3$</span>, <span class="math-container">$n^2=25q^2+10qr+9=5(5q^2+2qr+2)-1=5k-1.$</span></p>
<p>When <span class="math-container">$r=4$</span>, <span class="math-container">$n^2=25q^2+10qr+16=5(5q^3+2qr+3)+1=5k+1.$</span></p>
<p>(Note: I did not mean to imply that <span class="math-container">$k$</span> in one equation is the same as <span class="math-container">$k$</span> in another.)</p>
|
1,334,680 | <p>How to apply principle of inclusion-exclusion to this problem?</p>
<blockquote>
<p>Eight people enter an elevator at the first floor. The elevator
discharges passengers on each successive floor until it empties on the
fifth floor. How many different ways can this happen</p>
</blockquote>
<p>The people are <strong>distinguishable</strong>.</p>
| Vadeem | 249,634 | <p>Set theory is one of the branch of mathematical logic it clearly means not every mathematical logic can be explained by set theory.Mathematical logic is often divided into the fields of set theory, model theory, recursion theory, and proof theory. These areas share basic results on logic, particularly first-order logic, and definability.</p>
<p>One can easily say that at the bottom of mathematics lies elementary arithmetic which is the elementary part of number theory and mathematics in general.Although arithmetic and number theory is axiomized by peano axioms there is a problem with peano axioms The Peano axioms do not define what natural numbers are, they instead describe certain properties of natural numbers that can be used to reason about natural numbers.</p>
<p>You can refer to: <a href="https://en.wikipedia.org/wiki/Arithmetic" rel="nofollow">https://en.wikipedia.org/wiki/Arithmetic</a></p>
|
1,334,680 | <p>How to apply principle of inclusion-exclusion to this problem?</p>
<blockquote>
<p>Eight people enter an elevator at the first floor. The elevator
discharges passengers on each successive floor until it empties on the
fifth floor. How many different ways can this happen</p>
</blockquote>
<p>The people are <strong>distinguishable</strong>.</p>
| Stig Hemmer | 224,001 | <p>There are several ways to build up mathematics. One book may do it one way, another book will do it another.</p>
<p>All these methods suffer from the fact that you must start with normal human language for your first definitions. The author try to make their definitions as rigid as possible.</p>
<p>It is very important both for the writer and the reader to separate the informal language used to explain things and the formal language being defined!</p>
<p>The most common way is to first define a formal logic. This logic is defined <em>without</em> referring to set theory. This logic talks about "propositions" without saying anything about what these propositions are. There will be a few axioms and usually the rule of Modus Ponens.</p>
<p>The next level is set theory. This gives you some propositions to do logic about. There are more axioms describing sets and their members.</p>
<p>So, set theory is based on formal logic... in most text books.</p>
<p>Some books will present these two together since pure logic isn't very interesting without something to reason about.</p>
|
4,189,614 | <p>I am trying to answer the following question...</p>
<blockquote>
<p>Consider a wall made of brick <span class="math-container">$10$</span> centimeters thick, which separates a room in a house from the outside. The room is kept at <span class="math-container">$20$</span> degrees. Initially the outside temperature is <span class="math-container">$10$</span> degrees and the temperature in the wall has reached steady state. Then there is a sudden cold snap and the outside temperature drops to <span class="math-container">$-10$</span> degrees. Find the temperature in the wall as a function of position and time.</p>
</blockquote>
<p>I am okay executing the separation of variables technique, but I can't really reason through how to model this scenario. The solution manual states that the Initial/Boundary Value Problem is...<span class="math-container">$$u_t = ku_{xx},\ u(0,\ t) = 20,\ u(10,\ t) = -10,\ u(x,\ 0) = 20 - x$$</span></p>
<p>This question comes in a section before higher dimensional heat equations are introduced, but to me, it seems that this should be modeled as a three-dimensional heat equation, because thin walls are two-dimensional, and the bricks are prescribed thickness. How can I intuitively reason through this word problem to model it correctly?</p>
| user89699 | 85,392 | <p>Strictly speaking, you are correct. But if you stay away from the corners, then you just have 1d heat conduction across the brick only - the transverse gradients can be neglected. Then, the equations are just as you decribe with an error function solution. HTH</p>
|
13,705 | <p>Let $m$ be a positive integer. Define $N_m:=\{x\in \mathbb{Z}: x>m\}$. I was wondering when does $N_m$ have a "basis" of two elements. I shall clarify what I mean by a basis of two elements: We shall say the positive integers $a,b$ generate $N_m$ and denote $N_m=<a,b>$ if every element $x\in N_m$ can be written as $x=\alpha a+\beta b$ where $\alpha,\beta$ are nonnegative integers (not both zero) <STRIKE>and no nonnegative linear combination (with at least one coefficient nonzero) of $a,b$ gives an element not in $N_m$.</STRIKE></p>
<p>More specifically, I want to understand the set $\{(a,b,m):N_m=<a,b>\}$.</p>
<p>An example would be the triple $(2,3,1)$. Every positive integer greater than 1 can be written as $2\alpha +3\beta$ for nonnegative integers $\alpha,\beta$, for if it is even we can write it as $2\alpha$ for some positive integer $\alpha$ and if it is odd and greater than $3$ then we can write it as $3+2\alpha$ for some positive integer $\alpha$ and of course $3=1\cdot 3$. FInally the smallest integer a positive linear combination of $2,3$ can generate is $2$.</p>
<p>PS: Not quite sure what to tag this. Feel free to retag.</p>
<p>EDIT: After Jason's answer, it seems the first version of this question is more interesting, where the last condition in paragraph 1 is relaxed.</p>
| Robin Chapman | 226 | <p>If $m$ and $n$ are coprime positive integers, then every integer with $>mn-m-n$ is expressible in the form $am+bn$ where $a$ and $b$ are nonnegative integers, but $mn-m-n$ isn't. This is due to <a href="http://en.wikipedia.org/wiki/Coin_problem" rel="nofollow">Sylvester</a>.</p>
<p><strong>Added</strong> The key to the proof is the lemma that if $0\le k<(m-1)(n-1)$ then $k$ is representable iff $mn-n-m-k$ isn't. ISTR this involves a cunning use of the Chinese remainder theorem.</p>
|
69,208 | <p>Consider $f:\{1,\dots,n\} \to \{1,\dots,m\}$ with $m > n$.
Let $\operatorname{Im}(f) = \{f(x)|x \in \{1,\dots,n\}\}$.</p>
<p>a.)
What is the probability that a random function will be a bijection when viewed as
$$f':\{1,\dots,n\} \to \operatorname{Im}(f)?$$</p>
<p>b.)
How many different function f are there for which $$\sum_{i=1}^n f(i)=k?$$
Not sure where to start and have no idea what $\operatorname{Im}(f)$ means. Please help me get started on this.</p>
| marty cohen | 13,079 | <p>If $\sin x$ is approximately $x$, and $\cos x$ is approximately $1−.5x^2$,
then $(\sin x)^2 + (\cos x)^2$ is approximately
$$x^2 + (1-.5 x^2)^2 = x^2 + 1 - x^2 + .25 x^4 = 1 + .25 x^4.$$</p>
<p>Note that the $x^2$ term cancels out, so the result is $1$ with a term of order $x^4$.</p>
<p>As you use more terms of the power series for sin and cos, the result will be closer to 1.</p>
|
1,530,406 | <p>How to multiply Roman numerals? I need an algorithm of multiplication of numbers written in Roman numbers. Help me please. </p>
| John Martin | 301,716 | <p>Take the first number you want to multiply and break it down into parts, any way you choose.
E.G. 43 = XLIII = XL + III, or X + X + X + X + III, etc.
Do the same with the second number in the multiplication.
E.G. 15 = XV = X + V, or V + V + V, etc.</p>
<p>Take any pair of those combinations and set them within a table, with the higher row showing the first number's parts, and the lower row showing the second number's parts.
Multiply starting from the lower right to the upper row, then the lower left to the upper row. Place the result to each equation underneath, vertically.<br />
Total them.</p>
<p><a href="https://i.stack.imgur.com/ztZxn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ztZxn.png" alt="enter image description here" /></a></p>
|
2,327,675 | <p>Using the GPU Gems Article <a href="https://developer.nvidia.com/gpugems/GPUGems/gpugems_ch01.html" rel="nofollow noreferrer" title="Effective Water Simulation From Physical Models">Effective Water Simulation From Physical Models</a> I have implemented Gerstner Waves into UE4, I have built the function both on GPU for the tessellated mesh displacement and in code for the purpose of sampling height of the waves, but the issue I have run into is that the points by nature of the formula move away from their original points, so when using the gerstner function to get the height of a point, it is not necessarily at the point you are sampling so provides an incorrect height for that coordinate.</p>
<p>So somehow I need to solve for the point in the gerstner function that is actually lines up with the world position. I've taught myself enough to implement the waves, but i'm stuck on this solution, though I suspect there is some sort of matrix or way to take the inputs, apply the cos values and use it as a lookup for the actual point to sample. Any ideas and help is greatly appreciated.</p>
<p>For Reference the Gerstner Wave Formula</p>
<p><a href="https://i.stack.imgur.com/am1bR.jpg" rel="nofollow noreferrer">Formula Image</a></p>
<p><code>Q = Slope<br>
L = WaveLength<br>
A = Amplitude<br>
D = Vector2 Direction<br>
x = x world coordinate<br>
y = y world coordinate<br>
t = time<br>
Where Qi = Q/wi * A * NumWaves<br>
Where wi = 2pi/L<br>
Where phase = Speed * 2/L</code></p>
| Daniel Wells | 740,611 | <p>Note I'm ignoring a lot of the variables for brevity
Gerstner Waves vaguely simplify to y=sin(x) x = cos(x)</p>
<p>'Sample a vertical offset and send it somewhere that way horizontally' but we want the vertical offset of the point that landed here. We can guess where that point comes from by going backwards by the horizontal offset of our current position.</p>
<p>y = sin(x-cos(x))</p>
<p>This isn't perfectly accurate, but we can keep doing this to make it more accurate.
But if we 'guess' again based on the position we've arrived at we can continue to get closer.</p>
<p>y = sin(x-cos(x-cos(x-cos(x)) or y = sin(x-cos(x-cos(x-cos(x-cos(x-cos(x)))) and so on</p>
<p>By the fifth time we've gotten a pretty accurate guess and I think five steps is about the right amount of times. <strong>This solution doesn't work if you step back an even number of times.</strong></p>
<p>Obviously, this isn't very performant because it's about 5 times more complex than sampling a single point. But it's accurate enough for buoyancy.</p>
<p>Proof of concept: <a href="https://www.desmos.com/calculator/28taueduzs" rel="nofollow noreferrer">https://www.desmos.com/calculator/28taueduzs</a>
It didn't occur to me to step back multiple times until after I watched . <a href="https://youtu.be/kGEqaX4Y4bQ?t=746" rel="nofollow noreferrer" title="Ocean waves simulation with Fast Fourier transform">this video at timestamp</a>
It's also a good idea to swap from Gerstner Waves to FFT as sampling a texture multiple times over is probably a lot cheaper than calculating the offsets multiple times over. Though FFTs are more complicated concepts to wrap your head around.</p>
|
3,436,430 | <p>Evaluate <span class="math-container">$$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{2-\sqrt{4-x^2-y^2}}^{2+\sqrt{4-x^2-y^2}}(x^2+y^2+z^2)^{3/2} \; dz \; dy \; dx$$</span> by converting to spherical coordinates.<br>
We know that <span class="math-container">$(x^2+y^2+z^2)^{3/2} = (\rho^2)^{3/2} = \rho^3$</span>. The range of <span class="math-container">$y$</span> tells us that we have disk of radius <span class="math-container">$2$</span>. How can we use this find the limits of each integral and ultimately find the solution? Thanks you for the help. </p>
| Doug M | 317,162 | <p>The boundary of our sphere is <span class="math-container">$x^2 + y^2 + z^2 = 4z$</span></p>
<p>Plugging</p>
<p><span class="math-container">$x = \rho\cos\theta \sin\phi\\
y = \rho\sin\theta\sin\phi\\
z = \rho \cos\phi$</span></p>
<p>We get:</p>
<p><span class="math-container">$\rho^2 = 4\rho\cos\phi\\
\rho = 4\cos\phi$</span></p>
<p>And the sphere is above the <span class="math-container">$xy$</span> plane, or <span class="math-container">$\phi \le \frac {\pi}{2}$</span></p>
<p><span class="math-container">$\int_0^{2\pi}\int_0^{\frac {\pi}{2}}\int_0^{4\cos\phi} (\rho^3)(\rho\sin\phi) \ d\rho\ d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} (\frac 15) (4^5) (\cos^5 \phi)(\sin\phi) d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} -(\frac 1{30}) (4^5) \cos^6 \phi\ d\theta\\
(2\pi)(\frac {1}{30})4^5$</span></p>
|
3,660,101 | <p>I want to determine if the series <span class="math-container">$ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{\left(-1\right)^{n}+n} $</span> converge/diverge. the sequence in the denominator is not monotinic, so I cant use Dirichlet's or Abel's tests. My intuition is that this series converge, becuase its looks close to <span class="math-container">$ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n} $</span> but im not sure how to prove. Any ideas will help, thanks.</p>
| Barry Cipra | 86,747 | <p>It sometimes helps to write out the first few terms, in order to see what you're dealing with, and possibly spot a useful pattern. In this case we have</p>
<p><span class="math-container">$$\begin{align}
\sum_{n=2}^\infty{(-1)^n\over(-1)^n+n}
&={1\over3}-{1\over2}+{1\over5}-{1\over4}+{1\over7}-{1\over6}+\cdots\\
&=-\left({1\over2}-{1\over3}+{1\over4}-{1\over5}+{1\over6}-{1\over7}+\cdots \right)
\end{align}$$</span></p>
<p>Now depending on your standard of rigor, that may already by enough to prove conditional convergence. If you need to be more finicky, then a careful examination of the expansion tells us</p>
<p><span class="math-container">$$\sum_{n=2}^N{(-1)^n\over(-1)^n+n}=-\sum_{n=2}^N{(-1)^n\over n}+
\begin{cases}
0&\text{if $N$ is odd}\\
\displaystyle{1\over N}-{1\over N+1}&\text{if $N$ is even}
\end{cases}$$</span></p>
<p>Since <span class="math-container">$\sum(-1)^n/n$</span> is conditionally convergent by the familiar tests, and since <span class="math-container">${1\over N}-{1\over N+1}\to0$</span> as <span class="math-container">$N\to\infty$</span>, the given series converges (conditionally) as well.</p>
|
93,458 | <blockquote>
<p>Let <span class="math-container">$n$</span> be a nonnegative integer. Show that <span class="math-container">$\lfloor (2+\sqrt{3})^n \rfloor $</span> is odd and that <span class="math-container">$2^{n+1}$</span> divides <span class="math-container">$\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $</span>.</p>
</blockquote>
<p>My attempt:</p>
<p><span class="math-container">$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$</span></p>
<p><span class="math-container">$$ 0\leq (2-\sqrt{3})^n \leq1$$</span></p>
<p><span class="math-container">$$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $$</span></p>
<p><span class="math-container">$$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $$</span></p>
<p><span class="math-container">$$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $$</span></p>
| Henry | 6,460 | <p>You can use recurrences such as $$f(n)=4f(n-1)-f(n-2)+2$$ or $$f(n)=5f(n-1)-5f(n-2)+f(n-3)$$ starting at $f(0)=1, f(1)=3$.</p>
<p>Then show the various results by induction. </p>
|
1,018,270 | <p>While I was studying about finite differences I came across an article that says "computers can't deal with limit of $\Delta x \to 0$ " in <a href="http://1drv.ms/1sB5P1B" rel="nofollow">finite differences</a>.But if computers can't deal with these equations does anybody know how they compute $ \frac {d}{dx}$ of $x^2$ and other such equations.Or whether these equations are pre-written.</p>
| Claude Leibovici | 82,404 | <p>This is not an answer but an illustration of automatic differentiation of a formula (I used Tapenade software online).</p>
<p>The source code I submitted is</p>
<pre><code> SUBROUTINE DUMMY(X,Y)
Y = X ** 2
END
</code></pre>
<p>which was interpreted as</p>
<pre><code> SUBROUTINE DUMMY(x, y)
IMPLICIT NONE
REAL x , y
y = x**2
END
</code></pre>
<p>and what I received is </p>
<pre><code> SUBROUTINE DUMMY_D(x, xd, y, yd)
IMPLICIT NONE
REAL x , xd , y , yd
yd = 2*x*xd
y = x**2
END
</code></pre>
<p>which now computes both the function $y$ and its derivative $\frac{dy}{dx}$ ($xd=x$).</p>
|
168,118 | <p>I'm attempting to differentiate an equation in the form</p>
<pre><code>D[sqrt((2*(((a*b*c+Pi*d*e^2+Pi*f*g^2+h*i*j+Pi*k*l^2)/(a*b*c+Pi*d*e^2+Pi*f*g^2))-1)*m)/(n^2 - o^2)/p), a]
</code></pre>
<p>in order to do an error propagation analysis. So I need to differentiate it against a, against b, against c and so on.</p>
<p>All my values will be positive (they reflect physical dimensions of my design) and the value of n will always be greater than o.</p>
<p>Is there a way to define all my variables as real and positive before I differentiate? And to define n greater than o?</p>
<p>The best I've found is $Assumptions = _Symbol [Element] Reals but this only gets me part of the way.</p>
| kcr | 49,048 | <p>This is making the assumption that a is Real and positive, and then it calculates the derivative. But the result is zero.</p>
<pre><code>Assuming[ a \[Element] Reals && a > 0,
D[sqrt ((2*(((abc + Pide^2 + Pifg^2 + hij + Pikl^2)/(abc + Pide^2 +
Pifg^2)) - 1)*m)/(n^2 - o^2)/p), a]]
</code></pre>
<p>Hope it helps. </p>
|
2,106,662 | <p>I'm trying to show that the barycentric coordinate of excenter of triangle ABC, where BC=a, AC=b, and AB=c, and excenter opposite vertex A is Ia, is Ia=(-a:b:c). I've gotten to the point where after a lot of ratio bashing I have that it's (ab/(b+c)):CP:BP, where P is the incenter, but I have no idea where to go from there. What would be the proof?</p>
| marty cohen | 13,079 | <p>Here is my solution
to get the generating functions.</p>
<p>I have shown every step
so any errors
can be readily found.</p>
<p>The method should be valid
even if there are errors.</p>
<p>Let
$A(x)
=\sum_{n=0}^{\infty} a_n x^n
$
and
$B(x)
=\sum_{n=0}^{\infty} b_n x^n
$.</p>
<p>Then
$A(x)-a_0
=\sum_{n=1}^{\infty} a_n x^n
=x\sum_{n=0}^{\infty} a_{n+1} x^n
$
so
$(A(x)-a_0)/x
=\sum_{n=0}^{\infty} a_{n+1} x^n
$.</p>
<p>Similarly,
$(B(x)-a_0)/x
=\sum_{n=0}^{\infty} b_{n+1} x^n
$.</p>
<p>Therefore,
from the recurrances,</p>
<p>$(A(x)-a_0)/x
=2A(x)-B(x)
$
and
$(B(x)-b_0)/x
=A(x)+4B(x)
$.</p>
<p>From the first,
$A(x)-a_0
=2xA(x)-xB(x)
$
or
$B(x)
=(A(x)(1-2x)-a_0)/(2x)
$.</p>
<p>From the second,
$A(x)
=(B(x)(1-4x)-b_0)/x
$.</p>
<p>Therefore</p>
<p>$\begin{array}\\
A(x)
&=(B(x)(1-4x)-b_0)/x\\
&=(((A(x)(1-2x)-a_0)/(2x))(1-4x)-b_0)/x\\
\text{or}\\
xA(x)
&=((A(x)(1-2x)-a_0)/(2x))(1-4x)-b_0\\
&=(A(x)(1-2x)(1-4x)-a_0(1-4x))/(2x))-b_0\\
\text{or}\\
(xA(x)+b_0)(2x)
&=A(x)(1-2x)(1-4x)-a_0(1-4x)\\
\text{or}\\
2x^2A(x)+2xb_0
&=A(x)(1-2x)(1-4x)-a_0(1-4x)\\
\text{or}\\
(2x^2-(1-2x)(1-4x))A(x)
&=-a_0(1-4x)-2xb_0\\
\text{or}\\
(2x^2-(1-6x+8x^2))A(x)
&=-a_0+x(4a_0-2b_0)\\
\text{or}\\
(-6x^2-1+6x)A(x)
&=-a_0+x(4a_0-2b_0)\\
\text{or}\\
(6x^2-6x+1)A(x)
&=a_0+x(-4a_0+2b_0)\\
\end{array}
$</p>
<p>Putting in the initial values,
$(6x^2-6x+1)A(x)
=a_0+x(-4a_0+2b_0)
=2-6x
$
so that
$A(x)
=\dfrac{2-6x}{6x^2-6x+1}
$.</p>
<p>$B(x)$
can then be gotten,
but I'll stop here.</p>
|
853,308 | <p>I want to calculate the expected number of steps (transitions) needed for absorbtion. So the transition probability matrix $P$ has exactly one (lets say it is the first one) column with a $1$ and the rest of that column $0$ as entries.</p>
<p>$P = \begin{bmatrix} 1 & * & \cdots & * \\ 0 & \vdots & \ddots & \vdots \\ \vdots & & & \\ 0 & & & \end{bmatrix} \qquad s_0 = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \\0 \\ \vdots \\ \\ \end{bmatrix}$</p>
<p>How can I now find the expected (<i>mean</i>) number of steps needed for the absorbtion for a given initial state $s_0$?</p>
<p>EDIT: An explicit example here:</p>
<p>$P = \begin{bmatrix} 1 & 0.1 & 0.8 \\ 0 & 0.7 & 0.2 \\ 0 & 0.2 & 0 \end{bmatrix}
\qquad s_0 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \implies s_1 = \begin{bmatrix} 0.8 \\ 0.2 \\ 0 \end{bmatrix} \implies s_2 = \begin{bmatrix} 0.82 \\ 0.14 \\ 0.04 \end{bmatrix} \ldots $</p>
| Jakub Konieczny | 10,674 | <p>We do it like that because we have a very good, rigorous grasp on what a set is, so we can speak of a set of pairs representing a function with no ambiguity at all. This is precisely the reason why we often construct <em>everything</em> out of sets - the integers, rationals, reals, and so on. This is mundane, and not very enlightening, but if you ever have a doubt if, say, $1/2$ and $2/4$ are the same fraction, or if $0.999...$ and $1.000$ are the same reals, you have a rigorous way of answering this question.</p>
<p>On the other hand, we all <em>know</em> what that a function $f\colon X \to Y$ is "some sort of rule" for assigning a value $f(x)$ to any $x$. But try to give a rigorous definition to "some sort of rule" in such a way that it won't be possible to misunderstand it... </p>
|
2,108,558 | <p>We are started Linear programming problem question. Questions provided in examples are easy. And in exercise starting questions are easy to solve. As we have given conditions to form equations and solve them.</p>
<p>But this question little difficult.</p>
<p>Question - </p>
<p><strong>An aeroplane can carry a maximum of 200 passengers. A profit of Rs. 1000 is made on each executive class ticket and a profit of Rs. 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?</strong></p>
<p>I tried it as let x passenger travel in executive class then 200-x travel in economy class.</p>
<p>But after then I am confused what to do. Can anyone please provide solution for this with explanation.</p>
| Arnaud D. | 245,577 | <p><strong>Hint :</strong> Let $P_A(t)$ be the characteristic polynomial of $A$. $P_A(t)$ can be written as a product of $n$ linear factors $(t-\lambda_i)$; since $A$ is not diagonalizable one of the factors has to appear twice.</p>
|
2,390,538 | <p>The problem says:</p>
<p>If every closed ball in a metric space $X$ is compact, show that $X$ is
separable.</p>
<p>I'm trying to use an equivalence in metric spaces that tells us: let X be the matric space, the following are equivalent</p>
<p>X is 2nd countable</p>
<p>X is Lindeloff</p>
<p>X is separable</p>
<p>I also thought about taking balls of a "big" radius and that they are disjoint, but I do not see how to make the set of balls is countable.</p>
| Henno Brandsma | 4,280 | <p>$X = \cup_{n \in \mathbb{N}} D(p, n)$, for any $p \in X$ (where $D(x,r)$ denotes the closed ball around $p$ of radius $r$). </p>
<p>So $X$ is $\sigma$-compact hence Lindelöf hence separable. </p>
|
3,528,000 | <p>Given <span class="math-container">$f:\mathbb{R}\rightarrow\mathbb{R}^n$</span> differential and for every <span class="math-container">$x\in\mathbb{R}$</span> <span class="math-container">$\|f\|=1$</span>. </p>
<blockquote>
<p>Prove <span class="math-container">$\langle\,f(x),f'(x)\,\rangle=0$</span> for every <span class="math-container">$x\in\mathbb{R}$</span>. </p>
</blockquote>
<p>I've got the idea that the function is on the unit circle and the tangent space is vertical but I've got stuck after trying to assume that the inner product isn't <span class="math-container">$0$</span>.</p>
| José Carlos Santos | 446,262 | <p>Note that<span class="math-container">\begin{align}\lVert f\rVert=1&\iff\lVert f\rVert^2=1\\&\iff\langle f,f\rangle=1.\end{align}</span>So, if you differentiate <span class="math-container">$\langle f,f\rangle$</span>, you get <span class="math-container">$0$</span>. But<span class="math-container">\begin{align}\langle f,f\rangle'&=\langle f,f'\rangle+\langle f',f\rangle\\&=2\langle f,f'\rangle.\end{align}</span></p>
|
3,528,000 | <p>Given <span class="math-container">$f:\mathbb{R}\rightarrow\mathbb{R}^n$</span> differential and for every <span class="math-container">$x\in\mathbb{R}$</span> <span class="math-container">$\|f\|=1$</span>. </p>
<blockquote>
<p>Prove <span class="math-container">$\langle\,f(x),f'(x)\,\rangle=0$</span> for every <span class="math-container">$x\in\mathbb{R}$</span>. </p>
</blockquote>
<p>I've got the idea that the function is on the unit circle and the tangent space is vertical but I've got stuck after trying to assume that the inner product isn't <span class="math-container">$0$</span>.</p>
| BCLC | 140,308 | <p>Let <span class="math-container">$f(x)=[f_1(x), ..., f_n(x)]$</span>, for all <span class="math-container">$x \in \mathbb R$</span> for some unique (scalar) functions <span class="math-container">$f_i: \mathbb R \to \mathbb R$</span>, <span class="math-container">$i=1,...,n$</span></p>
<p>We're given that <span class="math-container">$f_1^2(x) + ... + f_n^2(x) = 1$</span> for all <span class="math-container">$x \in \mathbb R$</span>. We want to show that <span class="math-container">$f_1(x)f_1'(x) + ... + f_n(x)f_n'(x) = 0$</span> for all <span class="math-container">$x \in \mathbb R$</span></p>
<p>Differentiate both sides of <span class="math-container">$f_1^2 + ... + f_n^2 = 1$</span> to get</p>
<p><span class="math-container">$$2f_1f_1' + ... + 2f_nf_n' = 0$$</span></p>
<p>Equivalently, </p>
<p><span class="math-container">$$f_1f_1' + ... + f_nf_n' = 0$$</span></p>
|
1,904,553 | <blockquote>
<p>$$\displaystyle\lim_{x\to0}\left(\frac{1}{x^5}\int_0^xe^{-t^2}\,dt-\frac{1}{x^4}+\frac{1}{3x^2}\right)$$</p>
</blockquote>
<p>I have this limit to be calculated. Since the first term takes the form $\frac 00$, I apply the L'Hospital rule. But after that all the terms are taking the form $\frac 10$. So, according to me the limit is $ ∞$. But in my book it is given 1/10. How should I solve it? </p>
| N. S. | 9,176 | <p>By bringing the fractions to the same denominator, start by writing the limit as
$$\displaystyle\lim_{x\to0}\frac{3\int_0^xe^{-t^2}\,dt -3x+x^3 }{3x^5}$$</p>
<p>Now, since this is of the form $0/0$ by L'H and FTC you get
$$\displaystyle\lim_{x\to0}\frac{3e^{-x^2}-3+x^2 }{15x^4}$$</p>
<p>From here it is easy.</p>
|
2,466,855 | <p>We were given this question for homework that the professor couldn't explain how to solve (even in class he had trouble working it out). I'm only aware that we should be using the law of large numbers but I'm not sure how to apply it as the book for the course provides no examples. The answer in class was 10 and the book gave us an 8. Any help would be appreciated.</p>
| Dr Potato | 622,884 | <p>Another approach: Consider this sequence as the direct product of the reciprocal of the factorial function with itself.</p>
<p>Given that the exponential function is the <a href="https://mathworld.wolfram.com/GeneratingFunction.html" rel="nofollow noreferrer">generating function</a> of the reciprocal of the factorial
<span class="math-container">$$e^z=\sum_{n\in\mathbb{N}} \frac{z^n}{n!},$$</span>
take
<span class="math-container">$$f(z)=\sum_{n\in\mathbb{N}}\frac{z^n}{(n!)^2}$$</span>
as the generating function of the termwise product of the former sequence with itself and use the <a href="https://math.stackexchange.com/q/3526636/622884">Hadamard product of two generating functions</a> formula:
<span class="math-container">$$f(z)= e^z\cdot e^z = \sum_{n\in\mathbb{N}} \frac{1}{n!}\frac{1}{n!}z^n
= \frac{1}{2\pi i}\oint_{x=0} e^{x\sqrt{z}}e^{\frac{\sqrt{z}}{x}} \frac{dx}{x},$$</span>
so
<span class="math-container">$$\boxed{\sum_{n\in\mathbb{N}}\frac{z^n}{(n!)^2} = \frac{1}{2\pi i}\oint_{x=0} e^{\sqrt{z}(x+\frac{1}{x})} \frac{dx}{x}}$$</span>
which indeed matches with the <span class="math-container">$0$</span>-th <a href="https://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html" rel="nofollow noreferrer">Modified Bessel Function of the First Kind</a> asserted in the previous answer.</p>
<p>Now the task is to find a closed form of this function or to prove that this modified Bessel function of the first kind has no closed expression in terms of the standard elementary functions.</p>
|
4,600,131 | <blockquote>
<p>If <span class="math-container">$$f(x)=\binom{n}{1}(x-1)^2-\binom{n}{2}(x-2)^2+\cdots+(-1)^{n-1}\binom{n}{n}(x-n)^2$$</span>
Find the value of <span class="math-container">$$\int_0^1f(x)dx$$</span></p>
</blockquote>
<p>I rewrote this into a compact form.
<span class="math-container">$$\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}$$</span>
Now,
<span class="math-container">$$\int_0^1\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}dx$$</span>
<span class="math-container">$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}-\sum_{k=1}^n\binom{n}{k}\frac{(-k)^3}{3}(-1)^{k-1}$$</span>
<span class="math-container">$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}+\sum_{k=1}^n\binom{n}{k}\frac{k^3}{3}(-1)^{k-1}$$</span>
After this, I took <span class="math-container">$\dfrac13$</span> common and did some simplifications but nothing useful came out.</p>
<p>Any help is greatly appreciated.</p>
| Dr. Wolfgang Hintze | 198,592 | <p>We consider the more general function</p>
<p><span class="math-container">$$f(x,m,n) =\sum _{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^m\tag{1}$$</span></p>
<p>with <span class="math-container">$m=0,1,2,...$</span> and <span class="math-container">$n\ge 1$</span>. The case of interest here has <span class="math-container">$m=2$</span>.</p>
<p>It turns out (*) that <span class="math-container">$f(x,m,n\ge m+1) = x^m$</span> so that the integral becomes in this case</p>
<p><span class="math-container">$$i(m,n\ge m+1) = \int_{0}^{1} f(x,m,n\ge m+1)\;dx = \frac {1}{m+1}\tag{2}$$</span></p>
<p>For any <span class="math-container">$n\ge 1$</span>, including <span class="math-container">$n \le m$</span>, we do the <span class="math-container">$x$</span>-integral and obtain for the integral in question this expression</p>
<p><span class="math-container">$$i(m,n) = \frac{1}{m+1}i_1 (m,n)\tag{3a}$$</span></p>
<p>where the numerator is</p>
<p><span class="math-container">$$i_1 (m,n) = \sum _{k=1}^n (-1)^{k-1} \binom{n}{k}\left(k (-k)^m+(1-k)^{m+1}\right)\tag {3b}$$</span></p>
<p>For <span class="math-container">$m=2$</span> we get</p>
<p><span class="math-container">$$i(2,1) = \frac{1}{3}\\
i(2,2) = -\frac{5}{3}\\
i(2,n\ge 3) = \frac{1}{3} $$</span></p>
<p>In general the start of the matrix of the numerator <span class="math-container">$i_1(m,n)$</span> reads</p>
<p><span class="math-container">$$i_1(m,n)=\left(
\begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & 1 \\
-1 & 1 & 1 & 1 & 1 & 1 \\
1 & -5 & 1 & 1 & 1 & 1 \\
-1 & 13 & -23 & 1 & 1 & 1 \\
1 & -29 & 121 & -119 & 1 & 1 \\
-1 & 61 & -479 & 1081 & -719 & 1 \\
\end{array}
\right)\tag{4}$$</span></p>
<p>The second column of this matrix in listed in OEIS as <a href="http://oeis.org/A036563" rel="nofollow noreferrer">http://oeis.org/A036563</a>, the third and forthcoming columns are not contained in OEIS.</p>
<p>Notice that appearantly</p>
<p><span class="math-container">$$i_1(m, n\ge m+1) = 1\tag{*}$$</span></p>
<p>In other words, if there are sufficiently many summands in <span class="math-container">$i_1$</span> (<span class="math-container">$n \ge m+1$</span>) all terms add up to unity.</p>
<p>Unfortunately, I haven't yet found a prrof for (*).</p>
<p>A possibly useful intermediate result is the generating function of <span class="math-container">$h_1$</span> with respect to <span class="math-container">$m$</span>:</p>
<p>Replacing the kernel</p>
<p><span class="math-container">$$\sum _{m=0}^{\infty } \left(k (-k)^m+(1-k)^{m+1}\right) t^m=\frac{1}{(k t+1) (k t-t+1)}$$</span></p>
<p>in <span class="math-container">$i_1$</span> we obtain</p>
<p><span class="math-container">$$g_1(t,n) := \sum _{m=0}^{\infty } t^m i_1(m,n) \\
= \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n}{k}}{(k t+1) (k t-t+1)}\\=
\frac{\Gamma (n+2) \Gamma \left(\frac{t+1}{t}\right)}{(t-1) \Gamma \left(\frac{n t+t+1}{t}\right)}-\frac{1}{t-1}\\
=\frac{1}{t-1}\left(\frac{\Gamma (n+2) \Gamma \left(\frac{1}{t}+1\right)}{ \Gamma \left(\frac{1}{t}+n+1\right)}-1 \right)
\tag{5}$$</span></p>
<p>(proof of (*) still to be done)</p>
|
2,643,900 | <p>for the problem </p>
<p>$$(1-2x)y'=y$$</p>
<p>the BC'S are $y(0)=-1$ and $y(1)=1$ and $0\leq x\leq 1$.</p>
<p>I solved this and got $\ln y =\ln\left(\dfrac 2 {1-2x}\right)+c$.</p>
<p>How do we determine the constant such that $y$ is real and finite everywhere from $0$ to $1$ (both limits included)?</p>
| Dylan | 135,643 | <p>Let's go through the solution again. The equation separates to
$$ \frac{y'}{y} = \frac{1}{1-2x} $$</p>
<p>The equation in undefined on $y=0$ and $x=\frac12$. This means there are 4 possible regions in $\Bbb R^2$ where the solution can exist.</p>
<p>$$ \begin{cases}
\ln y = -\frac12 \ln (1-2x) + c_1, && x < \frac12, \ y > 0 \\
\ln (-y) = -\frac12 \ln (1-2x) + c_2, && x < \frac12, \ y < 0 \\
\ln y = -\frac12\ln(2x-1) + c_3, && x > \frac12, \ y > 0 \\
\ln (-y) = -\frac12\ln(2x-1) + c_4, && x > \frac12, \ y < 0
\end{cases} $$</p>
<p>If one is not looking for a specific solution that satisfies an initial condition, then you may write the general solution as
$$ \ln |y| = -\frac12\ln|1-2x| + C $$</p>
<p>or
$$ |y| = A |1-2x|^{-1/2} $$</p>
<p>That's the short answer for why we need absolute values when integrating the log function. The absolute values go away the moment an initial condition is assigned.</p>
<p>Back to our function. Your two initial conditions allow solutions in the second and third regions, respectively. The condition $y(0)=-1$ gives $c_2 = 0$. The condition $y(1)=1$ gives $c_3=0$. Putting it altogether and taking the exponential, we obtain the piecewise final solution</p>
<p>$$ y(x) = \begin{cases} -\dfrac{1}{\sqrt{1-2x}}, && x < \frac12 \\
\dfrac{1}{\sqrt{2x-1}}, && x > \frac12 \end{cases} $$</p>
<hr>
<p>As for your last question, there are no such solution (except $y \equiv 0$) that is finite on $(0,1)$, again due to the singular point.</p>
|
1,699,627 | <p>Why is the following equation true?$$\int_0^\infty e^{-nx}x^{s-1}dx = \frac {\Gamma(s)}{n^s}$$
I know what the Gamma function is, but why does dividing by $n^s$ turn the $e^{-x}$ in the integrand into $e^{-nx}$? I tried writing out both sides in their integral forms but $n^{-s}$ and $e^{-x}$ don't mix into $e^{-nx}$. I tried using the function's property $\Gamma (s+1)=s\Gamma (s)$ but I still don't know how to turn it into the above equation. What properties do I need?</p>
| DeepSea | 101,504 | <p>The number of ways to select $5$ cards that have $0,1,2,3$ kings are: $ \binom{48}{5}, \binom{4}{1}\binom{48}{4}, \binom{4}{2}\binom{48}{3}, \binom{4}{3}\binom{48}{2}$ respectively. Thus the probability of at most $3$ kings is: $Pr(x \leq 3) =\dfrac{\binom{48}{5}+\binom{4}{1}\binom{48}{4}+\binom{4}{2}\binom{48}{3}+\binom{4}{3}\binom{48}{2}}{\binom{52}{5}}$</p>
|
1,320,112 | <p>Using the following identity $$\int_{0}^{\infty}\frac{f\left ( t \right )}{t}dt= \int_{0}^{\infty}\mathcal{L}\left \{ f\left ( t \right ) \right \}\left ( u \right )du$$
I rewrote $$\int_{0}^{\infty}\frac{\sin^{2}\left ( t \right )}{t^{2}}dt$$
as $$\int_{0}^{\infty}\frac{\sin^{2}\left ( t \right )}{t\cdot t}dt$$
And thus the initial integral should be easily evaluated as $$\int_{0}^{\infty}\mathcal{L}\left \{ \frac{\sin^{2}t}{t} \right \}\left ( u \right )du$$
According to my calculations, this is equal to $$\int_{0}^{\infty}\left ( \frac{1}{4}\ln \left ( u^{2}+4 \right )-\frac{u^{2}}{4} \right )du$$
Which evaluates to $\frac{\pi}{16}$. Being that this acutally a well-known integral and that its value is actually $\frac{\pi}{2}$ I think that I made a crucial mistake somewhere. Any ideas?</p>
| Gappy Hilmore | 219,783 | <p>You seemed to have confused logarithmic identities. The integrand should be</p>
<p>$$\frac{1}{4} \log \left(\frac{4}{s^2}+1\right)$$</p>
<p>which indeed evaluates as $\pi/2$. I am guessing you wrongly factored out an $s^2$</p>
|
118,545 | <p>I gather that the question whether the Bruck-Chowla-Ryser condition was sufficient used to top the list, but now that that's settled - what is considered the most interesting open question?</p>
| Yuichiro Fujiwara | 27,829 | <p><strong>Edit</strong>: Since this just became available on arXiv, <a href="http://arxiv.org/pdf/1401.3665.pdf" rel="nofollow noreferrer"><strong><em>Peter Keevash solved the existence conjecture of Steiner $t$-designs</em></strong></a>, which means that what I wrote below a year ago as one of the most important open problems in design theory is now settled.</p>
<p>Here's a brief summary of the history of progress on Gil Kalai's blog "Combinatorics and more":</p>
<p><a href="http://gilkalai.wordpress.com/2014/01/16/amazing-peter-keevash-constructed-general-steiner-systems-and-designs/" rel="nofollow noreferrer">http://gilkalai.wordpress.com/2014/01/16/amazing-peter-keevash-constructed-general-steiner-systems-and-designs/</a></p>
<p>Congratulations, Peter Keevash!</p>
<hr>
<p>I think this kind of question is inherently subjective, but I'll try my best to be as objective as possible by limiting myself to the classical kind of stuff that is expected to be covered in a well-rounded graduate level textbook if published in 2013. I also deliberately avoid repeating what is already said in <a href="http://press.princeton.edu/titles/8350.html" rel="nofollow noreferrer">The Princeton Companion to Mathematics</a>. Also omitted are things that are clearly mentioned by Colbourn's "Opening the Door" and Anderson, Colbourn, Dinitz, Griggs' "Design Theory: Antiquity to 1950" in the <a href="http://www.emba.uvm.edu/~jdinitz/hcd.html" rel="nofollow noreferrer">Handbook of Combinatorial Designs</a>.</p>
<p>One more caveat is that I personally don't see some branches of mathematics like coding theory and finite geometry any different than design theory. So my view must be biased toward interactions with those adjacent fileds, and I wouldn't be surprised if hardcore design theorists disagree with me at every turn in my post.</p>
<p>To clarify what I mean by the above caveat, for instance, take the Bruck-Chowla-Ryser theorem you brought up. If you see it from a finite geometric viewpoint, you can rule out the existence of finite projective planes of order $6$ and $14$. But as you mentioned, Lam proved the nonexistence of such planes of order $10$. So, to my eye, it's definitely one of the most interesting problems whether finite projective planes exist only for prime and prime power orders. Similarly, because many problems in coding theory are of design theory to me, a large chunk of <a href="https://mathoverflow.net/questions/92192/hot-topics-in-error-correcting-coding-related-to-interesting-math/118041#118041">my answer to "What's hot topics in error-correcting codes?"</a> asked by Alex on MO counts as one of the more interesting problems in design theory. But I don't think every design theorist agrees with me here.</p>
<p>Anyway, one of the most important open problems about objects with the word "design" in their names is, in my opinion, the existence of Steiner $t$-designs. For total newbies, a <em>Steiner</em> $t$-design <em>block size</em> $k$ is an ordered pair $(V, \mathcal{B})$ of finite sets $V$ and $\mathcal{B}$, where $\mathcal{B}$ is a set of $k$-subsets of $V$ such that any $t$-subset of $V$ is contained in exactly one element of $\mathcal{B}$. The cardinality $\vert V \vert$ is the <em>order</em> of the design. A Steiner $t$-design of order $v$ and block size $k$ is written as $S(t,k,v)$. (Infinite Steiner designs have also been considered in design theory, but they haven't caught on, at least not yet.)</p>
<p>One of the main goals is to understand the spectrum of those parameters that admit the existence of an $S(t,k,v)$. The problem is said to originate Kirkman's study in the 19th century, and elementary combinatorics works relatively well for small $t$. The problem becomes extremely difficult for large $t$.</p>
<p>The most successful methods for finding $S(t,k,v)$s with larger $t$ is investigating interesting finite groups. For instance, as I said in the linked answer to Alex's question about coding theory, the known $S(5,8,24)$ has the the Mathieu group $M_{24}$ as its automorphism group. It is very interesting and often powerful to see interactions between $S(t,k,v)$s and, say, sporadic simple groups. But this line of work soon hits the wall because, for example, you would like to use a $t$-transitive group to find an $S(t,k,v)$ in its orbit and, alas, there are no $6$-transitive groups (other than the symmetric groups and the alternating groups). In fact, there are no known nontrivial $S(t,k,v)$ for $t \geq 6$ when apparently there isn't any plausible argument against their existence.</p>
<p>Wilson's existence theory (aka <em>Wilson's Fundamental Construction</em> published as a series of papers in J. Combin. Theory Ser. A) takes a different approach and proves, as its corollary, that the necessary conditions for the existence of an $S(2,k,v)$ (which can be obtained by a simple counting argument) are always sufficient for sufficiently large $v$. So, for $t=2$, the problem is solved in an asymptotic sense. This approach is very design theoretic (or the definition of "design theoretic" had changed a little by this), and there have been effort to generalize it to the case $t \geq 3$. Unfortunately, we've seen rather limited success, and while the case when $t=3$ has made some progress, only finitely many $S(t,k,v)$s are known for $t \geq 4$.</p>
<p>Steiner $t$-designs have numerous applications to other fields both inside and outside mathematics. This one-century-and-a-half old study gave a new, solid foundation to design theory by tidying up the mess of ad hoc techniques. So, I vote for "developing a general theory of existence of $S(t,k,v)$s" as one of the most interesting problems in design theory. (If you're familiar with design theory and are wondering why I put Wilson's existence theory only in the context of $S(t,k,v)s$ and not of $t$-designs of larger index $\lambda$, that's because such designs exist if $\lambda$ is sufficiently large, and also because the larger $\lambda$ case is mentioned in the intro section of the Handbook, which disqualifies the problem.)</p>
<p>But there are many other important and intriguing problems out there in design theory. For instance, the existence of difference sets is classical and has always been in the center of design theory because of its intrinsic importance (perceived by many) and connections to other fields. The recent revival of additive combinatorics/number theory may be a big push to the study of difference sets as well. It is closely related to the study of symmetric designs, which is already mentioned by Chris. Those classical problems like MOLS and others are often related and inter-winded. So, any of them is equally interesting, at least to my eye.</p>
<p>Also, as is the case with any other branch of math, design theory has also been driven by interactions with other fields. For instance, <a href="http://en.wikipedia.org/wiki/Spherical_design" rel="nofollow noreferrer">spherical $t$-designs</a> are gaining more and more attention not only because of its combinatorial appeal but also because of its applications to numerical analysis, quantum information, and so on. So, the study of these types of designs may have a different flavor than traditional design theory like the one stemming from the Bruck-Chowla-Ryser theorem, but I think it's definitely among the most important subfields of combinatorial design theory.</p>
|
118,545 | <p>I gather that the question whether the Bruck-Chowla-Ryser condition was sufficient used to top the list, but now that that's settled - what is considered the most interesting open question?</p>
| Shahrooz | 19,885 | <p>These four below problems are interesting and still open conjecture in design theory and its related topics:</p>
<p>$1.$ There exist Ucycles for $k$-subsets of $[n]$, provided $k$ divides $Cr(n-1,k-1)$ and $n$ is sufficiently large.</p>
<p>$2.$ For $n≥6$ even, it is not possible to have a length $\frac{n^2}{2}$ cyclic covering word for the $(n−2)$-subsets of an $n$-set.</p>
<p>$3.$ For each $k≥3$, there exists a constant $c_k$ such that for all $v≥c_k$ and $λ≥1$, the $1$ block-intersection graph of any $BIBD(v,k,λ)$ is Hamiltonian.</p>
<p>$4.$ For each $k≥3$, there exists a constant $c_k$ such that for all $v≥c_k$ and $λ≥1$, the ${1,2}$-block-intersection graph of any $BIBD(v,k,λ)$ is Hamiltonian.</p>
<p>These four conjectures are respectively from:</p>
<p>$1.$ Chung, F., Diaconis, P., Graham, R.: Universal cycles for combinatorial structures. Discrete Math.110, 43–59 (1992)</p>
<p>$2.$ Stevens, B., Buskell, P., Ecimovic, P., Ivanescu, C., Malik, A., Savu, A., Vassilev, T., Verrall, H., Yang, B., Zhao, Z.: Solution of an outstanding conjecture: the non-existence of universal cycles withk=n−2. Discrete Math.258, 193–204 (2002)</p>
<p>$3.$ Jesso, Andrew T.(3-NF); Pike, David A.(3-NF); Shalaby, Nabil(3-NF) Hamilton cycles in
restricted block-intersection graphs. (English summary) Des. Codes Cryptogr.61(3), 345–353</p>
<p>$Also,$ I believe that Hadamard conjecture is a nice diamond in the conjectures land. </p>
|
320,355 | <p>Show that $$\nabla\cdot (\nabla f\times \nabla h)=0,$$
where $f = f(x,y,z)$ and $h = h(x,y,z)$.</p>
<p>I have tried but I just keep getting a mess that I cannot simplify. I also need to show that </p>
<p>$$\nabla \cdot (\nabla f \times r) = 0$$</p>
<p>using the first result.</p>
<p>Thanks in advance for any help</p>
| Shuhao Cao | 7,200 | <p>Suppose everything is smooth: consider an arbitrary smooth simply connected domain $D\subset \mathbb{R}^3$, divergence theorem reads:</p>
<p>$$
\iiint_D \nabla \cdot (\nabla f\times \nabla h) \,dV = \iint_{\partial D} \nabla f\times \nabla h\cdot d\mathbf{S} = \iint_{\partial D} \nabla f\times \nabla h\cdot\mathbf{n} \,dS
$$
Since everything is smooth, we can use triple product to switch positions:
$$
\iint_{\partial D} \nabla f\times \nabla h\cdot\mathbf{n} \,dS = \iint_{\partial D} \nabla h\times \mathbf{n}\cdot \nabla f\, dS =
\iiint_D\Big(\nabla h\cdot \nabla\times(\nabla f) - \nabla f\cdot \nabla \times(\nabla h)\Big)dV = 0
$$
by the curl of an irrotational vector field is always zero. For $D$ is arbitrary, then
$$
\nabla \cdot (\nabla f\times \nabla h) = 0.
$$ </p>
<p>Looks like cheated though, but the intuition behind is that:</p>
<blockquote>
<p>The cross product of two irrotational vector fields: $\mathbf{v} = \nabla h\times\nabla f$, must be perpendicular to both on any surface they span: $\mathbf{v}\perp \nabla h$ and $\mathbf{v}\perp \nabla f$, hence it is a solenoidal (rotational or harmonic) vector fields, and have zero divergence.</p>
</blockquote>
|
2,658,627 | <p>I am trying to solve the following:</p>
<blockquote>
<p>Suppose the bank paid 12 % per year, but compounded that interest
monthly. That is, suppose 1 % interest was added to your account every
month. Then how much would you have after 30 years and after 60 years
if you started with $100?</p>
</blockquote>
<p>What I did is uisng this formula:</p>
<p><span class="math-container">$$ y[n]=(1.001^{12})*y[0] $$</span></p>
<p>where <span class="math-container">$ y[0] = 100\$ $</span></p>
<p>The answer for 30 years was <span class="math-container">$~143\$ $</span></p>
<p>but it is wrong as I've been said.</p>
<blockquote>
<p>When I compound your interest monthly this means that I give you a
1/12th of the total years rate every single month. In this specific
example we have a 12% interest rate so that would mean that every
single month your account is going to grow by 1%. Now, we are doing
this 12 times a year for 30 years</p>
</blockquote>
<p>Still complicating for me.</p>
| Andrew Li | 344,419 | <p>Remember that the compound interest formula is $C = P\left(1+\dfrac{r}{n}\right)^{nt}$ where $r$ is the rate, $n$ is the amount of times during the year it's compounded, $t$ is the number of years, and $P$ is the principle, or initial starting value.</p>
<p>Your rate $r$ is $0.12$, or $12%$. When you divide that by the $12$ months you're compounding interest, you should get $0.01$, not $0.001$.</p>
<p>Also notice the exponent, which should be $nt$. Since you're compounding $12$ times a year, and for $30$ years, you exponent should be $360$:</p>
<p>$$C = 100\left(1+\dfrac{0.12}{12}\right)^{12\cdot30} = 100(1.01)^{360}\approx3594.96$$ </p>
|
609,770 | <p>We have an empty container and $n$ cups of water and $m$ empty cups. Suppose we want to find out how many ways we can add the cups of water to the bucket and remove them with the empty cups. You can use each cup once but the cups are unique. </p>
<p>The question: In how many ways can you perform this operation.</p>
<p>Example: Let's take $n = 3$ and $m = 2$.</p>
<p>For the first step we can only add water to the bucket so we have 3 choices.
For the second step we can both add another cup or remove a cup of water.
So for the first 2 steps we have $3\times(5-1) = 12$ possibilities.
For the third step it gets more difficult because this step depends on the previous step.
There are two scenarios after the second step. 1: The bucket either contains 2 cups of water or 2: The bucket contains no water at all.</p>
<p>1) We can both add or subtract a cup of water
2) We have to add a cup of water</p>
<p>So after step 3 we have $3\times(2\cdot3 + 2\cdot2) = 30$ combinations.</p>
<p>etc.</p>
<p>I hope I stated this question clearly enough since this is my first post. This is not a homework assignment just personal curiosity. </p>
| parsiad | 64,601 | <p>Take $\epsilon=1$. Let $\delta>0$. We can always choose $n$ sufficiently large and $x=\frac{1}{n}$ and $y=\frac{1}{n+1}$ so that
$$
x-y=\frac{1}{n\left(n+1\right)}<\delta.
$$
However,
$$
\left|f\left(\frac{1}{n}\right)-f\left(\frac{1}{n+1}\right)\right|=\left|n-\left(n+1\right)\right|=1.
$$</p>
|
613,961 | <p>I got the following problem:</p>
<p>Let $V$ be a real vector space and let $q: V \to \mathbb R$ be a real quadratic form,<br/> Prove that if the set $L = \{v \in V | q(v) \ge 0\}$ forms a subspace of $V$
then q is definite (meaning $q$ is positive definite, positive semidefinite, negative definite or negative semidefinite)</p>
<p>I don't know where to begin</p>
| Eric Auld | 76,333 | <p><strong>Hint:</strong> You want to show that if $L=\{x \mid q(x)\geq 0\}$ is a subspace, then $L=0$ or $L=V$</p>
|
817,680 | <p><strong>Question:</strong></p>
<blockquote>
<p>Assume that $a_{n}>0,n\in N^{+}$, and that $$\sum_{n=1}^{\infty}a_{n}$$ is convergent. Show that
$$\sum_{n=1}^{\infty}\dfrac{a_{n}}{(n+1)a_{n+1}}$$ is divergent?</p>
</blockquote>
<p>My idea: since
$\sum_{n=1}^{\infty}a_{n}$ converges, then there exists $M>0$ such
$$a_{1}+a_{2}+\cdots+a_{n}<M$$
then I can't continue. Thank you.</p>
| math110 | 58,742 | <p>Longtime ago,I post my methods</p>
<p>proof:Assmue that: $\displaystyle\sum_{n=1}^{\infty}\dfrac{a_{n}}{(n+1)a_{n+1}}$,then for any $p\in N^{+}$,have
$$\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{(k+1)a_{k+1}}<\dfrac{1}{4}\Longrightarrow\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{(n+p)a_{k+1}}<\dfrac{1}{4}$$
then
$$\dfrac{1}{p}\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{a_{k+1}}<\dfrac{1}{4}\cdot\dfrac{n+p}{p}<\dfrac{1}{2}(p>n\mbox{时})$$
other hand,By AM-GM
$$\dfrac{1}{p}\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{a_{k+1}}=\dfrac{1}{p}\left(\dfrac{a_{n}}{a_{n+1}}+\dfrac{a_{n+1}}{a_{n+2}}+\cdots+\dfrac{a_{n+p-1}}{a_{n+p}}\right)$$
so $$\sqrt[n]{\dfrac{a_{n}}{a_{n+p}}}<\dfrac{1}{2}\Longrightarrow a_{n+p}>2^na_{n}$$
contradiction</p>
|
1,391,214 | <p>I've been thinking about the differences in numbers so for example: </p>
<p>$\begin{array}{ccccccc} &&&0&&&\\ &&1&&1&&\\&1&&2&&3&\\0&&1&&3&&6
\end{array}$</p>
<p>or with absolute differences:</p>
<p>$\begin{array}{ccccccccccc}
&&&\vdots&&\vdots&&\vdots\\
&&1&&0&&2&&2\\
&1&&2&&2&&4&&2\\
2&&3&&5&&7&&11&&13
\end{array}$</p>
<p>so I found this and I wanted to know if this has an actual mathematical formula: </p>
<p>$\begin{array}{cccccccc}
&&&&\vdots\\
&&&3&\cdots&\vdots\\
&&2&&5&\cdots&\vdots\\
&1&&3&&8&\cdots&\vdots\\
0&&1&&4&&12&\cdots
\end{array}$</p>
<p>and the sequence continues 0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688, 245760,... If you couldn't tell what I was doing to generate them the first diagonal going up from the 0 increased by one. I wasn't able to find the formula, but I have found some other diagonals have properties like, the second (1,3,5...) is the odd numbers, all of those diagonals are linear.</p>
| rela589n | 892,830 | <p>It seems to be the number of edges in N-dimensional cube:</p>
<pre><code>4 = 1 * 2 + 2 (square)
12 = 4 * 2 + 4 (ordinary cube)
32 = 12 * 2 + 8 (4-D cube)
80 = 32 * 2 + 16
192 = 80 * 2 + 32
448 = 192 * 2 + 64
1024 = 448 * 2 + 128
2304 = 1024 * 2 + 256
5120 = 2304 * 2 + 512
11264 = 5120 * 2 + 1024
24576 = 11264 * 2 + 2048
53248 = 24576 * 2 + 4096
114688 = 53248 * 2 + 8192
245760 = 114688 * 2 + 16384
524288 = 245760 * 2 + 32768
</code></pre>
|
2,915,735 | <p>So I just got done showing explicitly that an isomorphism exists between these two rings if the $\gcd(m, n) = 1$, and I did not have much trouble with that. For some reason I'm having a lot harder of a time showing that the result <em>is not</em> true if $m$ and $n$ are not relatively prime. Can somebody help me out here? Thanks.</p>
| fleablood | 280,126 | <p>If $m$ and $n$ are not relatively prime then they have a least common multiple $L$ which is <em>less</em> than $mn$. (Namely $\frac {mn}{\gcd(n,m)}$)</p>
<p>And if you take any element of $(a,b)\in \mathbb Z_m\times \mathbb Z_n$ and add it to itself $L$ times you get the identity[1]. So there is no element with order $mn$. So it can not be isomorphic to $\mathbb Z_{mn}$ as $1\in \mathbb Z_{mn}$ has order $mn$.</p>
<p>.....</p>
<p>[1] If $a \in \mathbb Z_m$ then $(km)*a = 0$ for all integers $k$ and if $b \in \mathbb Z_n$ then $(kn)*b = 0$ for all integers $k$. So if $k = L = m*\frac n{\gcd(n,m)} = n*\frac m{\gcd(m,n)}$ then $L*(a,b) = (0,0)$.</p>
|
9,416 | <p>Say I pass 512 samples into my FFT</p>
<p>My microphone spits out data at 10KHz, so this represents 1/20s.</p>
<p>(So the lowest frequency FFT would pick up would be 40Hz).</p>
<p>The FFT will return an array of 512 frequency bins
- bin 0: [0 - 40Hz)
- bin 1: [40 - 80Hz)
etc</p>
<p>So if my original sound contained energy at say 115Hz, how can I accurately retrieve this frequency?</p>
<p>That is going to lie in bin #2, but very close to bin #3. so I would expect both bins to contain something nonzero.</p>
<p>Question: how about the bins either side of this? Would they be guaranteed to be zero if there are no other frequencies close in the original signal?</p>
<p>Main question: is there some algorithm for deciphering the original frequency given the relative bin strengths?</p>
| endolith | 2,206 | <p>Generally you need to apply a <a href="http://en.wikipedia.org/wiki/Window_function" rel="noreferrer">windowing function</a> to your signal before performing the FFT, to get clean spikes for the frequency components (without "skirts"). (The only exception is if your frequency components <a href="https://gist.github.com/236567" rel="noreferrer">fit <em>exactly</em> into the length of the FFT</a>, so that they each fill exactly one bin.) </p>
<p>If you use a Gaussian windowing function, and then fit a parabola to the highest three points in your FFT, you can get a theoretically exact result for the frequency:</p>
<p><a href="https://ccrma.stanford.edu/~jos/sasp/Quadratic_Interpolation_Spectral_Peaks.html" rel="noreferrer">Quadratic Interpolation of Spectral Peaks</a></p>
|
9,416 | <p>Say I pass 512 samples into my FFT</p>
<p>My microphone spits out data at 10KHz, so this represents 1/20s.</p>
<p>(So the lowest frequency FFT would pick up would be 40Hz).</p>
<p>The FFT will return an array of 512 frequency bins
- bin 0: [0 - 40Hz)
- bin 1: [40 - 80Hz)
etc</p>
<p>So if my original sound contained energy at say 115Hz, how can I accurately retrieve this frequency?</p>
<p>That is going to lie in bin #2, but very close to bin #3. so I would expect both bins to contain something nonzero.</p>
<p>Question: how about the bins either side of this? Would they be guaranteed to be zero if there are no other frequencies close in the original signal?</p>
<p>Main question: is there some algorithm for deciphering the original frequency given the relative bin strengths?</p>
| Mykie | 832 | <p>You can try a zero-padded FFT ( add zeros to the end of your signal and then take FFT).</p>
|
4,536,320 | <p>Let <span class="math-container">$R$</span> be a ring and <span class="math-container">$A \subseteq R$</span> be finite, say <span class="math-container">$A = \{a\}$</span>. The set <span class="math-container">$$RaR = \{ras\:\: : r,s\in R\}$$</span> Why is this not necessarily closed under addition?</p>
<p>Take <span class="math-container">$r_1as_1$</span> and <span class="math-container">$r_2as_2$</span>. Is it because the ring must be commutative in order to guarantee that we can add these two elements and still stay within <span class="math-container">$RaR$</span>?</p>
<p>Even though I see why commutativity will make this possible, I'm still a bit confused because the very definition of <span class="math-container">$RaR$</span> includes all possible combinations of elements of the form <span class="math-container">$ras$</span>.</p>
| cluelessmathematician | 798,206 | <p>Brute forcing this (i.e. go through all <span class="math-container">$21^2$</span> combinations of <span class="math-container">$(m,a)\in\{0,...,20\}^2$</span> and tally up the answer) is likely feasible, but tedious and would not work on an exam, especially if the professor chooses anything much higher than 20.</p>
<p>An alternative approach would be as follows: for each <em>fixed</em> <span class="math-container">$a\in\{0,...,20\}$</span>, how many <span class="math-container">$m$</span> in <span class="math-container">$\{0,...,20\}$</span> satisfy <span class="math-container">$m+a=2k$</span>? Call that number <span class="math-container">$n_a$</span>. After you determine <span class="math-container">$\{n_a\}_{a=0}^{20}$</span>, calculate <span class="math-container">$\sum_{a=0}^{20} n_a$</span>. Note that you could go through each <span class="math-container">$a\in\{0,...,20\}$</span> one by one, but this may not be necessary. (Hint, what if <span class="math-container">$a$</span> is even? odd?)</p>
<p>Other (more elegant approaches) are surely possible.</p>
|
1,088,973 | <p>I am posed with the following problem:</p>
<blockquote>
<p>Suppose that $V$ is finite-dimensional and that $S,T\in \mathcal{L}(V)$. Prove that $ST=I$ if and only if $TS=I$, where $I$ is the identity map/operator.</p>
</blockquote>
<p>I have attempted it in the following way.</p>
<p>Consider the proof in one direction, starting at the equality $ST=I$. If we multiply by $S$ on both sides (which means take the composition of $S$ on both sides), we get</p>
<p>$$(ST)S=IS=S$$</p>
<p>However, the composition of operators is associative, so </p>
<p>$$(ST)S=S(TS)$$</p>
<p>Which means we have, from what we just got in the last two equalities,</p>
<p>$$S(TS)=S$$</p>
<p>And since this should work for all $v\in V$, this means that $TS=I$. Thus, $ST=I \implies TS=I$.</p>
<p>For the proof in the other direction, just replace every $S$ with a $T$ and vice versa. $\blacksquare$.</p>
<hr>
<p>Is this proof valid? Is there a shorter or easier proof? Please be very holistic.</p>
| Ilmari Karonen | 9,602 | <p>Hint: Rewrite your integral as:</p>
<p>$$\int_0^1 \int_y^1 \int_0^y f(x,y,z) \;dz\,dx\,dy
= \int_0^1 \int_0^1 \int_0^1 f(x,y,z) \,\mathbf 1_{x \ge y \ge z} \;dz\,dx\,dy,$$</p>
<p>where the <a href="//en.wikipedia.org/wiki/Indicator_function" rel="nofollow">indicator function</a> $\mathbf 1_{x \ge y \ge z}$ equals $1$ if ${x \ge y \ge z}$, and $0$ otherwise.</p>
<p>You can now easily rearrange the order of integration. Then adjust the limits for the inner integrals so that you only integrate over the volume where the $\mathbf 1_{x \ge y \ge z} = 1$ (i.e. where $x \ge y \ge z$), and drop the indicator function from the integrand again (since it's now, once more, identically equal to $1$ over the entire volume of integration).</p>
<p>For example, for the $dy\,dz\,dx$ order, by noting that $x \ge y \ge z \iff x \ge y,\ x \ge z,\ y \ge z$, we get:</p>
<p>$$\int_0^1 \int_0^1 \int_0^1 f(x,y,z) \,\mathbf 1_{x \ge y \ge z} \;dy\,dz\,dx
= \int_0^1 \int_0^x \int_z^x f(x,y,z) \;dy\,dz\,dx.$$</p>
<p>You should be able to deal with the last case ($dy\,dx\,dz$) in the same manner.</p>
<hr>
<p>More generally, the important thing to realize is that your original integral, where the limits of the inner integrals depend on the value of the outer integration variables, is really just a <a href="//en.wikipedia.org/wiki/Volume_integral" rel="nofollow">volume integral</a> over a subset $A = \{(x,y,z) \in \mathbb R^3 \mid 1 \le z \le y \le x \le 1\} \subset \mathbb R^3$ of the whole space. Thus, we can rewrite it as:</p>
<p>$$\iiint\limits_A f(x,y,z) \;dx\,dy\,dz
= \iiint\limits_{\mathbb R^3} f(x,y,z) \,\mathbf 1_A \;dx\,dy\,dz,$$</p>
<p>where the order of integration no longer matters.</p>
|
1,088,973 | <p>I am posed with the following problem:</p>
<blockquote>
<p>Suppose that $V$ is finite-dimensional and that $S,T\in \mathcal{L}(V)$. Prove that $ST=I$ if and only if $TS=I$, where $I$ is the identity map/operator.</p>
</blockquote>
<p>I have attempted it in the following way.</p>
<p>Consider the proof in one direction, starting at the equality $ST=I$. If we multiply by $S$ on both sides (which means take the composition of $S$ on both sides), we get</p>
<p>$$(ST)S=IS=S$$</p>
<p>However, the composition of operators is associative, so </p>
<p>$$(ST)S=S(TS)$$</p>
<p>Which means we have, from what we just got in the last two equalities,</p>
<p>$$S(TS)=S$$</p>
<p>And since this should work for all $v\in V$, this means that $TS=I$. Thus, $ST=I \implies TS=I$.</p>
<p>For the proof in the other direction, just replace every $S$ with a $T$ and vice versa. $\blacksquare$.</p>
<hr>
<p>Is this proof valid? Is there a shorter or easier proof? Please be very holistic.</p>
| Mark McClure | 21,361 | <p>The bounds of integration determine equations that bound a solid $S$ in three-dimensional space. After you integrate with respect to the first variable, you should orthogonally project $S$ along the axis specified by that first variable onto the plane spanned by the other two variables. That projection then determines a two-dimensional region which is the domain over which you integrate next.</p>
<p>In your original integral, you have
$$\int_{1}^{0}\int_{y}^{1}\int_{0}^{y} f(x,y,z)\,dz\,dx\,dy = -\int_{0}^{1}\int_{y}^{1}\int_{0}^{y} f(x,y,z)\,dz\,dx\,dy.$$
Note that I have switched the outer bounds of integration and changed the sign of the integral to alleviate the minor annoyance that $1$ is not less than $0$. Now, the bounds of integration imply three inequalities that specify the domain of integration.
$$
\begin{array}{l}
0<z<y \\
y<x<1 \\
0<y<1.
\end{array}
$$
Now, the solid, together with the projections of interest, looks like so:</p>
<p><img src="https://i.stack.imgur.com/O925f.png" alt="enter image description here"></p>
<p>Now, I guess the reason that the $yz$ and $xy$ projections are relatively easy is that those projections correspond to cross-sections of the solid, i.e. $z=0$ for the $xy$ projection and $x=1$ for the $yz$ projection. This allows us to just set the variable that we integrate with respect to first to that constant. It's even easier here, since neither $x$ nor $z$ appear explicitly in the bounds. We can't do that here but, when we think of it as a projection as we have here, we can see that it's just the triangle
$$\left\{(x,z): 0<x<1 \text{ and } 0<z<x \right\}.$$</p>
|
1,532,202 | <p>I want to find out $$\mathcal{L^{-1}}\{\frac{e^{-\sqrt{s+2}}}{s}\}$$
How do you find the inverse Laplace? </p>
<p>thanks</p>
| Jan Eerland | 226,665 | <p>Mathematica 10.0 gives this output:</p>
<p>$$\mathcal{L}_{s}^{-1}\left[\frac{e^{-\sqrt{s+2}}}{s}\right]_{(t)}=\frac{1}{2}e^{-\sqrt{2}}\left(\text{erfc}\left(\frac{1-2t\sqrt{2}}{2\sqrt{t}}\right)+e^{e\sqrt{2}}\space\text{erfc}\left(\frac{1+2t\sqrt{2}}{2\sqrt{t}}\right)\right)$$</p>
|
1,564,962 | <p>I need to calculate the following:</p>
<p>$$x=8 \pmod{9}$$
$$x=9 \pmod{10}$$
$$x=0 \pmod{11}$$</p>
<p>I am using the chinese remainder theorem as follows:</p>
<p>Step 1:</p>
<p>$$m=9\cdot10\cdot11 = 990$$</p>
<p>Step 2:</p>
<p>$$M_1 = \frac{m}{9} = 110$$</p>
<p>$$M_2 = \frac{m}{10} = 99$$</p>
<p>$$M_3 = \frac{m}{11} = 90$$</p>
<p>Step 3:</p>
<p>$$x=8\cdot110\cdot2 + 9\cdot99\cdot9 + 0\cdot90\cdot2 = 9779 = 869\mod 990$$</p>
<p>I have used online calculators to check this result and I know it is wrong (it should be 539 I think) but cannot find out what am I doing wrong. Can you help me?</p>
<p>Thanks</p>
| Bernard | 202,857 | <p>Start solving the first two congruences: a <em>Bézout's relation</em> between $9$ and $10$ is $\;9\cdot 9-8\cdot 10=1$, hence
$$\begin{cases}m\equiv\color{red}{8}\mod9 \\ m\equiv \color{red}{9}\mod 10\end{cases}\iff m\equiv\color{red}{9}\cdot 9\cdot 9-\color{red}{8}\cdot8\cdot 10=\color{red}{89}\mod 90 $$
Now solve for
$\begin{cases}x\equiv\color{red}{89}\mod90, \\ m\equiv \color{red}{0}\mod 11.\end{cases} $</p>
<p>We might proceed the same, from a Bézout's relation, but it is simpler, due to the value $0\mod 11$, to consider it means $\;89+90k$ is divisible by $11$. Now
$$89+90k=88+1+88k+2k\equiv 2k+1\mod 11, $$
so it means $2k+1$ is an odd multiple of $11$, e. g. $k=5$. Thus we obtain tha basic solution:
$$x=89+5\cdot 90=\color{red}{539}.$$</p>
|
2,665,723 | <p>Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.</p>
<p>solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$</p>
<p>$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$</p>
<p>$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k}}+\sqrt{2z+k}=\frac{12+2z+k}{\sqrt{2z+k}}$</p>
<p>$\displaystyle 2\sqrt{6z^2+6z+3kz+3k}=\sqrt{12+2z+k}$</p>
<p>$\displaystyle 4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$</p>
<p>Help me how i solve it after that point </p>
| Robert Z | 299,698 | <p>We have that
$$n(n-1)=\frac{n^3-(n-1)^3}{3}-\frac{1}{3}.$$
Therefore
$$\begin{align}
(1\cdot 2)+(3\cdot 4)+(5\cdot 6)+\dots +(99\cdot 100)&=\sum_{n=1}^{50}(2n-1)(2n)=4\sum_{n=1}^{50}n(n-1)+2\sum_{n=1}^{50}n\\
&=4\sum_{n=1}^{50}\frac{n^3-(n-1)^3}{3}-\frac{4\cdot 50}{3}+50\cdot 51\\
&=\frac{4}{3}\left(50^3-0^3\right)-\frac{200}{3}+50\cdot 51=169150
\end{align}$$
where we noted that the last sum on the right is telescopic.</p>
|
2,665,723 | <p>Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.</p>
<p>solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$</p>
<p>$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$</p>
<p>$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k}}+\sqrt{2z+k}=\frac{12+2z+k}{\sqrt{2z+k}}$</p>
<p>$\displaystyle 2\sqrt{6z^2+6z+3kz+3k}=\sqrt{12+2z+k}$</p>
<p>$\displaystyle 4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$</p>
<p>Help me how i solve it after that point </p>
| Sri-Amirthan Theivendran | 302,692 | <p>Observe that your sum may be written as
$$
\sum_{n=0}^{99}n(n+1)=2\sum_{n=0}^{99}\binom{n+1}{2}=2\sum_{i=2}^{100}\binom{i}{2}.
$$
Moreover, using Pascal's identity and telescoping, we obtain
$$
2\sum_{i=2}^{100}\left[\binom{i+1}{3}-\binom{i}{3}\right]
=2\binom{101}{3}=\frac{2(101)(100)(99)}{6}=333300.
$$</p>
|
2,665,723 | <p>Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.</p>
<p>solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$</p>
<p>$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$</p>
<p>$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k}}+\sqrt{2z+k}=\frac{12+2z+k}{\sqrt{2z+k}}$</p>
<p>$\displaystyle 2\sqrt{6z^2+6z+3kz+3k}=\sqrt{12+2z+k}$</p>
<p>$\displaystyle 4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$</p>
<p>Help me how i solve it after that point </p>
| robjohn | 13,854 | <p>$$
\begin{align}
\sum_{k=1}^n(2k-1)2k
&=\sum_{k=1}^n\left[8\binom{k}{2}+2\binom{k}{1}\right]\\
&=8\binom{n+1}{3}+2\binom{n+1}{2}\\
&=\frac43(n+1)n(n-1)+(n+1)n\\[3pt]
&=\frac{n(n+1)(4n-1)}3
\end{align}
$$
Plug in $n=50$ to get
$$
\sum_{k=1}^{50}(2k-1)2k=169150
$$</p>
|
713,732 | <p>I want to know how one would go about solving an <em>unfactorable cubic</em>. I know how to factor cubics to solve them, but I do not know what to do if I cannot factor it. For example, if I have to solve for $x$ in the cubic equation:
$$2x^3+6x^2-x+4=0$$
how would I do it?</p>
<p>Edit: I have heard people telling me to convert it into a <em>depressed cubic</em> (where the $x^2$ term disappears), but I have no idea how to do that.</p>
<p>Edit 2: I am aware that there is a <em>cubic formula</em>, that for any cubic equation $ax^3+bx^2+cx+d$, it's roots are:
$$x = \sqrt[3]{-\dfrac{b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a} + \sqrt{\left(-\dfrac{b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} + \sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a} - \sqrt{\left(-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}$$
This formula is way too complicated so I do not even bother memorizing it or using it.</p>
| Batman | 127,428 | <p>Normally, you want to work with codes in standard form - that is, you want to write the generator matrix in the form $[I | P]$. Then, a parity check matrix corresponding to this code is $[-P^T | I]$ (i.e. it is a generator of the dual code - the code consisting of all vectors orthogonal to the original code). </p>
|
1,043 | <p>Hi all,</p>
<p>The short-time fourier transform decomposes a signal window into a sin/cosine series.</p>
<p>How would one approximate a signal in the same way, but using a set of arbitrary basis functions instead of sin/cos? These arbitrary basis functions are likely in my case to be very small discrete chunks of a 1-dimensional non-periodic waveform. </p>
<p>Is this something wavelets are used for? </p>
<p>Please excuse my tag, 'signal-analysis' does not exist and I can not create it. </p>
| Michael Hoffman | 429 | <p>Wavelets are generally used for nonperiodic signals. They're often used in earthquake detection and things like that. There are many books on the subject, a quick look for "Wavelets" in amazon.com should reveal many.</p>
<p>The Haar Wavelet and Daubchies Wavelet might be good choices. Haar may be better if you don't need a smooth decomposition, the wavelet decomposition their is much easier.</p>
<p>Google scholar (scholar.google.com) may be a good place to look, just look up "Wavelet decomposition" and your particular topic.</p>
<p>You can make your own wavelets per your particular needs, but it's not particularly easy, they need to fulfill certain conditions</p>
<p>This may be useful</p>
<p><a href="https://doi.org/10.1007/BF01257191" rel="nofollow noreferrer">https://doi.org/10.1007/BF01257191</a></p>
<p>Generalized multi-resolution analyses and a construction procedure for all wavelet sets in R^n</p>
<p>If you can get this down to 1-d you may have your answer.</p>
|
2,963,324 | <p>I want to prove <span class="math-container">$a \equiv b\;(\text{mod} \;n)$</span> is an equivalence relation then would it be ok to write,</p>
<p>Reflexive as, for all <span class="math-container">$a$</span>, <span class="math-container">$a \equiv a\;(\text{mod} \;n)$</span></p>
<p>Symmetric as, <span class="math-container">$a \equiv b\;(\text{mod} \;n)$</span> which implies <span class="math-container">$b \equiv a\;(\text{mod} \;n)$</span></p>
<p>Transitive as if <span class="math-container">$a \equiv b\;(\text{mod} \;n)$</span> and <span class="math-container">$b \equiv c\;(\text{mod} \;n)$</span> this implies <span class="math-container">$a \equiv c\;(\text{mod} \;n)$</span></p>
<p>I dont know if this has actually proved equivalence. Also the set on which this relatiton acts on was not specified so for reflexivity is it ok to say for all <span class="math-container">$a$</span>. Thanks.</p>
| Mark | 470,733 | <p>It is not ok, because what you did is just wrote what you have to prove, not really proved it. First of all, the relation is on <span class="math-container">$\mathbb{Z}$</span>. Next, what is the relation? <span class="math-container">$a\equiv b$</span>(mod <span class="math-container">$n$</span>) by definition means that <span class="math-container">$n|(a-b)$</span>. So that is the relation: <span class="math-container">$a,b\in\mathbb{Z}$</span> are related if <span class="math-container">$n$</span> divides <span class="math-container">$a-b$</span>. Now you have to prove it is an equivalence relation. For example, it is reflexive because for each <span class="math-container">$a\in\mathbb{Z}$</span> we have <span class="math-container">$a-a=0$</span> and obviously <span class="math-container">$n|0$</span> because <span class="math-container">$0\times n=0$</span>. Now try to show the other properties. </p>
|
1,573,113 | <p>Find the derivative of $\tan$($\sqrt{1 -x}$)</p>
<p>So I know I have to apply the product rule so wouldn't it be </p>
<p>$\sec^x(\sqrt{1-x})+ \sqrt{1-x}$
$\frac{tan(x)}{2}$
but the final answer says $\frac{-sec^2(\sqrt{1-x})}{2\sqrt{1-x}}$.</p>
| Claude Leibovici | 82,404 | <p>It is not the product rule but the chain rule which must be used.</p>
<p>Consider $A=\tan(u)$ with $u=\sqrt{1-x}$. So $$\frac{dA}{dx}=\frac{dA}{du}\times \frac{du}{dx}$$ with $$\frac{dA}{du}=\sec ^2(u)=\sec ^2(\sqrt{1-x})$$ $$\frac{du}{dx}=-\frac{1}{2 \sqrt{1-x}}$$ and finally $$\frac{dA}{dx}=-\frac{\sec ^2\left(\sqrt{1-x}\right)}{2 \sqrt{1-x}}$$</p>
|
1,573,113 | <p>Find the derivative of $\tan$($\sqrt{1 -x}$)</p>
<p>So I know I have to apply the product rule so wouldn't it be </p>
<p>$\sec^x(\sqrt{1-x})+ \sqrt{1-x}$
$\frac{tan(x)}{2}$
but the final answer says $\frac{-sec^2(\sqrt{1-x})}{2\sqrt{1-x}}$.</p>
| John Joy | 140,156 | <p>Same as the other answer, with a bit of difference in style.</p>
<p>$$\begin{array}{lll}
\frac{d}{dx}\tan\sqrt{1-x} &=& \frac{d\tan\sqrt{1-x}}{d\sqrt{1-x}}\cdot\frac{d\sqrt{1-x}}{d (1-x)}\cdot\frac{d(1-x)}{dx}\\
&=&\frac{\sec^2\sqrt{1-x}}{1}\cdot\frac{1}{2\sqrt{1-x}}\cdot(-1)\\
&=&\frac{-\sec^2\sqrt{1-x}}{2\sqrt{1-x}}\end{array}$$</p>
<p>Note that the substitutions $v=1-x$ and $u=\sqrt{v}$ were't made explicitily, but they were understood.</p>
|
2,498,123 | <blockquote>
<p>Given a $2 \times 2$ matrix $B$ that satisfies $B^2=3B-2I$, find the eigenvalues of $B$.</p>
</blockquote>
<p>My attempt: </p>
<p>Let $v$ be an eigenvector for B, and $\lambda$ it's corresponding eigenvalue. Also, let $T$ be the linear transformation (not that this is exactly necessary for the question, but just added it in for my understanding.) Therefore, </p>
<p>$$T(v) = Bv = \lambda v$$</p>
<p>Now I'm unsure how to incorporate this information into the quadratic equation given above since by matrix / vector arithmetic isn't extremely solid. Thanks!</p>
| Jonas Meyer | 1,424 | <p>Suppose $\lambda $ is an eigenvalue for $B$, with eigenvector $v$. Note that $B^2v=BBv=B(\lambda v)=\lambda^2v$. Apply each side of your equation $B^2=3B-2I$ to the vector $v$ to get $\lambda^2 v= (3\lambda -2)v$, or $(\lambda^2-3\lambda +2)v=0$. If a scalar times a nonzero vector is the zero vector, then the scalar is $0$, so $\lambda^2-3\lambda +2=(\lambda -1)(\lambda -2) =0$. This means that the set of eigenvalues of $B$ is a subset of $\{1,2\}.$ It is impossible to determine which subset from the information given, as diagonal matrix examples show.</p>
|
3,118 | <p>Can anyone help me out here? Can't seem to find the right rules of divisibility to show this:</p>
<p>If $a \mid m$ and $(a + 1) \mid m$, then $a(a + 1) \mid m$.</p>
| David E Speyer | 448 | <p>An alternate route: We show that, if $ax+by=1$ and $m$ is divisible by $a$ and $b$ then $m$ is divisible by $ab$. (Then apply this with $b=a+1$, $x=-1$ and $y=1$.) </p>
<p><strong>Proof:</strong> Let $m=ak=bl$. Then $ab(xl+ky)=(ax+by)m=m$. QED</p>
<p>The point here is that the hypothesis $\exists_{x,y}: ax+by=1$ is often easier to use than $GCD(a,b)=1$. The equivalence between these two is basically equivalent to unique factorization, and you can often dodge unique factorization by figuring out which of these two you really need.</p>
|
3,385,921 | <p>What does this mean? </p>
<blockquote>
<p>A matrix is diagonizable if and only if its eigenvectors are invertible.</p>
</blockquote>
| Mike | 544,150 | <p>An <span class="math-container">$n \times n$</span> matrix <span class="math-container">$A$</span> is diagonizable i.e., there exists a matrix <span class="math-container">$P$</span> such that <span class="math-container">$A = P^{-1}DP$</span> where <span class="math-container">$D$</span> is a diagonal matrix; iff there are <span class="math-container">$n$</span> linearly independent eigenvectors.</p>
<p>A matrix can be diagonizable but not invertible i.e., have eigenvectors with eigenvalues 0, and a matrix can be invertible but no diagonizable.</p>
|
2,043,429 | <p>In my textbook, it states that the general formula for the partial sum </p>
<p>$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$</p>
<p>My question is, if I have the following sum instead:</p>
<p>$$\sum_{i=1}^n \frac{1}{i^2}$$ </p>
<p>Can I just flip the general formula to get this:</p>
<p>$$\sum_{i=1}^n \frac{1}{i^2} = \frac{6}{n(n+1)(2n+1)} $$</p>
<p>Or does it not work like that? Thank you!</p>
| Vidyanshu Mishra | 363,566 | <p>The way you are saying it, I remembered the mistake I often used to do when I was in fifth standerd. I used to convert $\frac{1}{\frac{1}{a}+\frac{1}{b}}=a+b$. That is what you are doing here. Just take deep breath and look at what you have written make sense or not.</p>
|
805,341 | <p>Need a bit of help with this question. </p>
<p>We're given two invertible square $n\times n$ matrices $A$ and $B$ with entries in the reals.</p>
<p>We have to show that $AB$ is also invertible and then express $(AB)^{-1}$ in terms of $A$ and $B$. </p>
<p>I've managed to get the first part out. </p>
<p>Since $A$ is invertible, $Det(A)$ $\neq$ $0$. Similarily for $B, Det(B) \neq 0$</p>
<p>Also from the properties of Determinants: $Det(A)Det(B) = Det(AB)$
Hence $Det(AB) \neq 0$ and so $AB$ is invertible. </p>
<p>It's the second part that I need help. Thanks. </p>
| Cousin | 54,755 | <p>Consider $B^{-1}A^{-1}$. Multiply with this matrix on both sides of $AB$. You'll get the identity each time. Note that since $A$ and $B$ are invertible $A^{-1}$ and $B^{-1}$ are well defined.</p>
|
805,341 | <p>Need a bit of help with this question. </p>
<p>We're given two invertible square $n\times n$ matrices $A$ and $B$ with entries in the reals.</p>
<p>We have to show that $AB$ is also invertible and then express $(AB)^{-1}$ in terms of $A$ and $B$. </p>
<p>I've managed to get the first part out. </p>
<p>Since $A$ is invertible, $Det(A)$ $\neq$ $0$. Similarily for $B, Det(B) \neq 0$</p>
<p>Also from the properties of Determinants: $Det(A)Det(B) = Det(AB)$
Hence $Det(AB) \neq 0$ and so $AB$ is invertible. </p>
<p>It's the second part that I need help. Thanks. </p>
| tpb261 | 125,795 | <p>Here's another way:
let C be the inverse of AB. then <code>ABC = I</code> and <code>CAB = I</code>. Now try to get C by multiplying on the left and/or multiplying or the right so that you are finally left with expression of the form <code>C = f(A, B)</code>. This is the mathematical form of <a href="https://math.stackexchange.com/users/58401/andreas-caranti">Andreas Caranti</a>'s answer.</p>
|
1,101,104 | <p>Let $f: A \to B$. How can I show that $f$ is surjective if and only if (for every $C$ and every pair of functions $g, h: B \to C$) when there is the following implication?</p>
<p>$$
g \circ f = h\circ f \to g=h
$$</p>
| Richard | 19,749 | <p>When you multiply the series for $sin(x)$ by $i$, and add it to the series for $cos(x)$, you get the series for $exp(ix)$. In other words, $cos(x)$, is the real part of $exp(ix)$. </p>
<p>Therefore, $cos(x+y)$ is the real part of $exp(i(x+y))$. By laws of exponents, this is $$real( exp(ix)exp(iy))$$</p>
<p>By the first observation, this is $$real( (cox(x)+i sin(x)) (cox(y)+i sin(y)))$$</p>
<p>Multiply it out and select the real part: $$cos(x)cos(y) - sin(x)sin(y)$$</p>
|
1,101,104 | <p>Let $f: A \to B$. How can I show that $f$ is surjective if and only if (for every $C$ and every pair of functions $g, h: B \to C$) when there is the following implication?</p>
<p>$$
g \circ f = h\circ f \to g=h
$$</p>
| Gabriel | 209,805 | <p>Very bad answer. One cannot use complex analysis to prove the sum of angles formula, one must in fact use the sum of angles formula to prove all of the useful properties of the "complex exponential" (which is based on the definitions of sine and cosine!). Otherwise you are just playing with definitions and giving circular reasoning. The proof should be self-contained, involving multiplying the partial sums of the series and re-arranging the terms in a clever way.</p>
|
41,888 | <p>So I've got this line that contains the solution of a partial, non-extraordinary differential equation (because Mathematica doesn't handle extraordinary partial differential equations):</p>
<pre><code>phi6m = NDSolveValue[{D[u[t, x], t, t] -
D[u[t, x], x, x] == -6 u[t, x]^5 + 10.5 u[t, x]^3 - 4.5 u[t, x], u[0, x] == Tanh[x], Derivative[1, 0][u][0, x] == 0, u[t, -7] == -1,
u[t, 7] == 1}, u, {t, 0, 6}, {x, -7, 7}]
</code></pre>
<p>And now I wish to compute the following as a function of time, as well as its derivative with respect to time:</p>
<p>$ E(t)=\int_{-7}^{7} dx (\frac{\partial u}{\partial x})^2+(u(t,x)^2-1)^2(u(t,x)^2-0.625)$</p>
<p>where u is the solution given by the aforementionned line of code. But now I wish to take the function, as well as its derivative, and plot it into a graph... what I tried allowed me to take the function at a fixed time value, here t=0.1:</p>
<pre><code>phi62 = phi6m[0.1, x];
energiephi6 = D[phi62, x]^2 + (phi62^2 - 1)^2 (phi62^2 - 0.625);
energietotale = NIntegrate[energiephi6, {x, -7, 7}]
</code></pre>
<p>But not to make something plottable as a function of time, much less a derivative of the aforementionned function with respect to time. Is there anything else I can do to resolve the issue?</p>
| Szabolcs | 12 | <p>There is <code>ValueFunction</code>, documented <a href="http://reference.wolfram.com/mathematica/Experimental/ref/ValueFunction.html">here</a>.</p>
<p>It allows detecting value changes for given symbols.</p>
<p>For example,</p>
<pre><code>In[1]:= Experimental`ValueFunction[x] := Print["x changed"]
In[2]:= x = 6
During evaluation of In[10]:= x changed
Out[2]= 6
</code></pre>
|
1,439,920 | <blockquote>
<p>So, the question is:<br>
Calculate the probability that 10 dice give more than 2 6s.</p>
</blockquote>
<p>I've calculated that the probability for throwing 3 6s is 1/216.</p>
<p>And by that logic: 1/216 + 1/216 + .. + 1/216 = 10/216.</p>
<p>But I've been told that this isn't the proper way set it up.</p>
<p>Anyone having a good way to calculate this?</p>
| Zach466920 | 219,489 | <p>$$x^5 − 1102 \cdot x^4 − 2015 \cdot x = 0$$</p>
<p>This factors into,</p>
<p>$$\left( x^4 − 1102x^3 − 2015 \right) \cdot x = 0$$</p>
<p>Therefore $x=0$ is a root. Keep simplifying,</p>
<p>$$x^4 − 1102x^3 − 2015=0$$</p>
<p>Set this equal to $f(x)$,</p>
<p>$$f(x)=x^4 − 1102x^3 − 2015$$</p>
<p>$f(-1)=-912$ and $f(-2)=6817$, thus by the <a href="https://en.wikipedia.org/wiki/Intermediate_value_theorem" rel="nofollow">Intermediate Value Theorem</a>, there is a zero in-between $-2$ and $-1$.</p>
<p>The same thing holds for $f(1100)$ and $f(1110)$</p>
<p>Now, in the spirit of fairness, let's quickly come up with the method that will allow us to "guess" where the roots of $f(x)$ lie. We'll use <a href="https://en.wikipedia.org/wiki/Newton%27s_method" rel="nofollow">Newton's Method</a>.</p>
<p>We get,</p>
<p>$$x_{n+1}=x_{n}-{{x^4 − 1102x^3 − 2015} \over {4 \cdot x^3-2204 \cdot x^2}}$$</p>
<p>Now, the property we'll use is the fact that the convergence for the method is oscillatory. That means if you guess too low, then the next guess will be to high $^1$. Applying this principle results in a guess of $1000$ for the root resulting in a new guess of $1170$ for the next root. Once again, the Intermediate Value Theorem applies.</p>
<p>$^1$ (There are subtleties about convergence and when this works and doesn't work, but generally speaking, this is the case if you pick a reasonable guess)</p>
|
1,439,920 | <blockquote>
<p>So, the question is:<br>
Calculate the probability that 10 dice give more than 2 6s.</p>
</blockquote>
<p>I've calculated that the probability for throwing 3 6s is 1/216.</p>
<p>And by that logic: 1/216 + 1/216 + .. + 1/216 = 10/216.</p>
<p>But I've been told that this isn't the proper way set it up.</p>
<p>Anyone having a good way to calculate this?</p>
| Christian Blatter | 1,303 | <p>It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\lim_{x\to\pm\infty} q(x)=+\infty$ guarantee two more real zeros.</p>
|
785,314 | <p>let $$I_n = \int_{\pi/2}^{x} \frac{\cos^{2n+1}t}{\sin(t)} \ dt, n \geq 0$$</p>
<p>show $$2(n+1)I_{n+1} = 2(n+1)I_n +\cos^{2n+2}x$$</p>
<p>I showed the result by considering $I_{n+1} - I_n$ but I'm wondering how could I do it using integration by parts?</p>
<p>Similarly for $J_n = \int_0^x \frac{\sinh^{2n+1}t}{\cosh(t)} \ dt$</p>
| Pranav Arora | 117,767 | <p>$$I_n=\int_{\pi/2}^x \frac{\cos^{2n+1}t}{\sin t}\,dt=\int_{\pi/2}^x \frac{\cos^{2n-1}t (1-\sin^2 t)}{\sin t}\,dt= I_{n-1}-\int_{\pi/2}^x \cos^{2n-1}t\sin t\,dt $$</p>
<p>Use the substitution $\cos t=y \Rightarrow -\sin t\,dt=dy$ to get:
$$\int_{\pi/2}^x \cos^{2n-1}t\sin t\,dt =-\int_0^{\cos x} t^{2n-1}\,dt=-\frac{\cos^{2n}x}{2n}$$
Hence,
$$I_n=I_{n-1}+\frac{\cos^{2n}x}{2n}$$
Replace $n$ with $n+1$ to get the final result.</p>
<p>You should be able to proceed in a similar manner for $J_n$.</p>
|
16,754 | <p>Let $c$ be an integer, not necessarily positive and not a square. Let $R=\mathbb{Z}[\sqrt{c}]$
denote the set of numbers of the form $$a+b\sqrt{c}, a,b \in \mathbb{Z}.$$
Then $R$ is a subring of $\mathbb{C}$ under the usual addition and multiplication.</p>
<p>My question is: if $R$ is a UFD (unique factorization domain), does it follow that
it is also a PID (principal ideal domain)?</p>
| Akhil Mathew | 536 | <p>Yes. If it is a UFD, then it is integrally closed, hence it is a Dedekind domain because it is of dimension one (being contained in the integral closure of $\mathbb{Z}$ in some finite extension of $\mathbb{Q}$). A Dedekind domain is a UFD iff it is a PID: indeed, this is equivalent to every non-zero prime being principal. (A noetherian domain is a UFD iff every height one prime is principal. So if a Dedekind domain is a UFD, then all its primes are principal, so by factorization of ideals, every ideal is principal.)</p>
<p>A simple example of a UFD that is not a PID is the polynomial ring $\mathbb{C}[x,y]$.</p>
|
748,489 | <p>I have a measure ($x$) which the domain is $[0, +\infty)$ and measure some sort of variability. I want to create a new measure ($y$) that represents regularity and is inverse related to $x$.</p>
<p>It is easy, just make $y = -x$. However I want this new measure to be positive, to make it more interpretable. Linearly, it is impossible, because I would have to map 0 to $+\infty$. However I can create a non-linear measure that follows:</p>
<p>$$
f(x) = y
$$
$$
f(0) = 1
$$
$$
x \to +\infty => y \to 0
$$</p>
<p>The logarithmic transformation almost do this, but it does not inverse the relation. What function would do this mapping?</p>
<p>Would be interesting if I could set a "precision", like if $x > 100$, the step in $f(x)$ can be smaller than 0.01.</p>
| Henry | 6,460 | <p>Try something like $y=f(x)= \dfrac{1}{1+x}$.</p>
<p>This has the properties:</p>
<ul>
<li>$f(x)$ decreases as positive $x$ increases </li>
<li>$f(0)=1$</li>
<li>$f(1)=\frac12$</li>
<li>$f(x)\lt 0.01$ for $x \gt 99$ and so also for $x \ge 100$</li>
<li>$f(x) \to 0$ as $x \to +\infty$</li>
</ul>
<p>Its inverse is $x=g(y)=\dfrac{1}{y}-1$. </p>
|
3,362,916 | <p>I'm trying to graph <span class="math-container">$|x+y|+|x-y|=4$</span>. I rewrote the expression as follows to get a function that resembles the direction of unit vectors at <span class="math-container">$\pi/4$</span> to the horizontal axis (take it to be <span class="math-container">$x$</span>)<span class="math-container">$$\biggl|\dfrac{x+y}{\sqrt{2}}\biggr|+\biggl|\dfrac{x-y}{\sqrt{2}}\biggr|=2\sqrt{2}$$</span></p>
<p>However, I'm not able to proceed further. Any hints are appreciated. Notice this is an exam problem, so time-efficient methods are key. Please provide any hints accordingly.</p>
| Z Ahmed | 671,540 | <p>It is a square with sides as <span class="math-container">$x=\pm 2$</span> and <span class="math-container">$y=\pm 2$</span>,and diagonals as <span class="math-container">$y=\pm x$</span>. for the explanation see my answer in MSE below.</p>
<p><a href="https://math.stackexchange.com/questions/3329862/how-to-draw-graph-of-xy-1-2x-y-1-1/3329924#3329924">How to draw graph of $|x+y-1| + |2x + y + 1|=1$?</a></p>
|
1,386,307 | <p>If you consider that you have a coin, head or tails, and let's say tails equals winning the lottery. If I participate in one such event, I may not get tails. It's roughly 50%. But if a hundred people are standing with a coin and I or them get to flip it, my chances of having gotten a tail after these ten attempts, is higher, is it not? Way higher than 50% though I'm not sure how to calculate it.</p>
<p>So why is it different for lotteries? Or is it? I was once told that in a certain lottery, I had a one in 12 million chance of winning. And like the coin toss, each lottery is different with different odds, but would the accumulated odds be way higher if I participate, be it in this same lottery over a thousand times, or this lottery and thousand other lotteries around country, thereby increasing my chances of getting a win, a tail? </p>
<p>I appreciate a response, especially at level of high school or first year university (did not do math past first year university). Thank you. </p>
| karmalu | 258,239 | <p>From a probability point there is no difference between lottery and coin toss, but there is when you compute the number. Tossing ten coin you have $\frac{1}{2^{10}}=\frac{1}{1024}$ probability of not winning which is $\frac{1023}{1024}$ probability of winning. If you partecipate in ten lotteries, say that in each of them you have a 1 on a million probability of winning(i think in real lotteries this is lower). Then you have $(\frac{999999}{1000000})^{10}$ probability of losing which is still almost zero probability of winning</p>
|
2,109,347 | <p>My statistics note states that the variance of the empirical distribution is
$v= \sum_{i=1}^{n}(x_i-\bar x )^2\frac {1} {n}$ which the author then re-writes as
$v= \sum_{i=1}^{n}x_i^2 (\frac {1} {n}) - \bar x^2$. How is this achieved?</p>
| zhoraster | 262,269 | <p>A useful idea is to consider sample mean, variance, moments, quantiles etc as the mean, variance etc with respect to the <em>empirical measure</em>. Namely, the sample mean $\overline x$ is the expectation of empirical measure, which assigns probabilities $1/n$ to the sampled values $\{x_1,x_2,\dots,x_n\}$:
$$
\overline x = \hat{\mathbb{E}}_n X,
$$
where $\hat{\mathbb{P}}_n(X = x_k) = 1/n$, $k=1,\dots,n$. Similarly, the sample variance
$$
\hat\sigma^2 = \frac1n \sum_{k=1}^n (x_k-\overline x)^2 = \hat{\mathbb{E}}_n (X -\hat{\mathbb{E}}_n X)^2 = \hat{\operatorname{var}}_n(X)
$$
is the variance of $X$ with respect to $\hat{\mathbb{P}}_n$. Thus, the formula
$$
\hat\sigma^2 = \overline{x^2} - (\overline x)^2 = \hat{\mathbb{E}}_n X^2 - (\hat{\mathbb{E}}_n X)^2
$$
is a particular case of the general formula for variance.</p>
|
2,779,379 | <blockquote>
<p>Let $f:\mathbb R\to \mathbb R$ be a continuous function and $\Phi(x)=\int_0^x (x-t)f(t)\,dt$. Justify that $\Phi(x)$ is twice differentiable and calculate $\Phi''(x)$.</p>
</blockquote>
<p>I'm having a hard time finding the first derivative of $\Phi(x)$. Here's what I tried so far:</p>
<p>Since $f$ is a continuous function and $x-t$ is a polynomial function, thus continuous, $f(t)(x-t)$ is the product of two continuous functions and is also continuous. Since $x$ and $0$ are differentiable functions, by the Fundamental Theorem of Calculus </p>
<p>$\Phi'(x)= (x-x)f(x)x' - (x-0)f(0)0'=0$</p>
<p>I checked the solution and this is wrong, the solution goes like this:
$\Phi'(x) = (x\int_0^xf(t)dt - \int_0^xtf(t)dt)' = \int_0^xf(t)dt + xf(x) - xf(x) = \int_0^xf(t)dt$</p>
<p>So I tried to do it their way, expanding $(x-t)f(t) to xf(t) - tf(t)$ and I got this:</p>
<p>$\Phi'(x) = (\int_0^x (x-t)f(t)dt)' = (\int_0^x xf(t) - tf(t)dt)' = (x\int_0^xf(t)dt - \int_0^xtf(t)dt)' = xf(x)x' - xf(0)0' - (xf(x)x' - 0f(0)0') = xf(x) - xf(x) = 0$</p>
<p>$0$ again.</p>
<p>Another thing I didn't understand is why they put the $x$ outside the integral, I thought we were only supposed to do that with constants. As in, why is $\int_0^x xf(t)dt = x\int_0^x f(t)\,dt$</p>
<p>I understand the rest of the exercise, I just can't get this derivative right with the Fundamental Theorem of Calculus. The version I'm using says</p>
<p>Let $f$ be a continuous function and $a(x)$ and $b(x)$ be differentiable functions. If $$F(x) = \int_{a(x)}^{b(x)} f(t) \,dt$$ then $F'(x) = f(b(x))b'(x) - f(a(x))a'(x)$</p>
<p>Is this correct? Because if so I don't understand how the derivative of this exercise works.</p>
| chilliefiber | 499,334 | <p>So, this was a silly mistake, but I was really having trouble with it. I'm still not sure I understand exactly what is going on, I'm a bit confused because there are 2 variables (x and t), but we have a multivariable function in $\Phi(x)=\int_0^x (x-t)f(t)\,dt$, so I don't think I can apply the Fundamental Theorem of Calculus directly to it as I did at first, because maybe it only holds true for single variable functions?</p>
<p>In the second approach, the error was in the derivative of $x\int_0^xf(t)dt$. Applying the product rule to both functions with respect to x you get $x'\int_0^xf(t)dt + x(\int_0^xf(t)dt)' = \int_0^xf(t)dt + xf(x)$ and then the result is correct.</p>
|
3,295,318 | <p><span class="math-container">$$\int _{ c } \frac { \cos(iz) }{ { z }^{ 2 }({ z }^{ 2 }+2i) } dz$$</span></p>
<p>I used residue caculus for solving the problem but i am not pretty sure if the approach is right.</p>
<p>The attempt has been annexed in the pictures.</p>
<p><a href="https://i.stack.imgur.com/ZSv27.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZSv27.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/v533k.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v533k.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/ldJpM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ldJpM.jpg" alt="enter image description here" /></a></p>
| herb steinberg | 501,262 | <p>correction:</p>
<p>Let <span class="math-container">$m$</span> be the number. Find the smallest prime factor <span class="math-container">$n$</span> of <span class="math-container">$m$</span> and let <span class="math-container">$k=\frac{m}{n}$</span> The sequence will be <span class="math-container">$k-n+1,k-n+3,.....,k-2,k,k+2,...,k+n-3,k+n-1$</span>.</p>
<p>Note: The list is of length <span class="math-container">$n$</span>, which is the smallest possible.</p>
<p>Note: The method will work with any factor <span class="math-container">$\le \sqrt{m}$</span>. My solution may be the easiest, but not necessarily.</p>
|
2,919,841 | <blockquote>
<p><span class="math-container">$$\Large\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k=\bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg)$$</span></p>
</blockquote>
<hr />
<p><strong>My attempt:</strong></p>
<p><span class="math-container">$\large x\in\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k\iff (\exists k\in\bigcup\limits_{i\in I}J_i)(x\in A_k)\iff [(\exists i\in I) (k\in J_i)](x\in A_k)$</span></p>
<p><span class="math-container">$\large x\in \bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)(x\in \bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$</span></p>
<p>I think the equality holds if and only if we show that <span class="math-container">$$[(\exists i\in I) (k\in J_i)](x\in A_k) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$$</span></p>
<hr />
<blockquote>
<p>My questions:</p>
<ol>
<li><p>Are my above transformations fine?</p>
</li>
<li><p>How do I proceed to prove the last statement?</p>
</li>
</ol>
<p>Many thanks for your help!</p>
</blockquote>
| marty cohen | 13,079 | <p>If you assume
the binomial theorem,
you can argue directly
like this:</p>
<p>If
$n >m > a+1$
and
$c = 1+h$
then</p>
<p>$\begin{array}\\
c^n
&=(1+h)^n\\
&=\sum_{k=0}^n \binom{n}{k}h^k\\
&>\binom{n}{m}h^{m}\\
&=\dfrac{\prod_{j=0}^{m-1}(n-j)}{m!}h^{m}\\
&=\dfrac{n^m\prod_{j=0}^{m-1}(1-j/n)}{m!}h^{m}\\
\text{so}\\
\dfrac{n^a}{c^n}
&\lt \dfrac{n^a}{\dfrac{n^m\prod_{j=0}^{m-1}(1-j/n)}{m!}h^{m}}\\
&\lt \dfrac{m!}{n^{m-a}h^m\prod_{j=0}^{m-1}(1-j/n)}\\
\end{array}
$</p>
<p>For fixed $m$ and $h$,
$\dfrac{m!}{h^m}
$
is fixed and
$\prod_{j=0}^{m-1}(1-j/n)
$
is an increasing function of $n$,
so
$\prod_{j=0}^{m-1}(1-j/n)
\gt \prod_{j=0}^{m-1}(1-j/(m+1))
$
so
$\dfrac{n^a}{c^n}
\lt \dfrac{g(m, h)}{n^{m-a}}
\lt \dfrac{g(m, h)}{n}
$
where
$g(m, h)$
is a function of $m$ and $h$,
so
$\dfrac{n^a}{c^n}
\to 0$.</p>
|
2,919,841 | <blockquote>
<p><span class="math-container">$$\Large\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k=\bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg)$$</span></p>
</blockquote>
<hr />
<p><strong>My attempt:</strong></p>
<p><span class="math-container">$\large x\in\bigcup\limits_{k\in\bigcup\limits_{i\in I}J_i}A_k\iff (\exists k\in\bigcup\limits_{i\in I}J_i)(x\in A_k)\iff [(\exists i\in I) (k\in J_i)](x\in A_k)$</span></p>
<p><span class="math-container">$\large x\in \bigcup\limits_{i\in I}\bigg(\bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)(x\in \bigcup\limits_{k\in J_i}A_k\bigg) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$</span></p>
<p>I think the equality holds if and only if we show that <span class="math-container">$$[(\exists i\in I) (k\in J_i)](x\in A_k) \iff (\exists i\in I)[(\exists k\in J_i)(x\in A_k)]$$</span></p>
<hr />
<blockquote>
<p>My questions:</p>
<ol>
<li><p>Are my above transformations fine?</p>
</li>
<li><p>How do I proceed to prove the last statement?</p>
</li>
</ol>
<p>Many thanks for your help!</p>
</blockquote>
| user | 505,767 | <p>That nice, as an alternative without binomial theorem we can show that</p>
<p>$$\frac{n^{\alpha}}{c^n}=e^{\alpha \log n-n\log c} \to 0$$</p>
<p>indeed since $\frac{\log x}x\to 0 \implies \frac{\log n}n\to 0$ we have</p>
<p>$$\alpha \log n-n\log c = n\left(\alpha\frac{\log n}n-\log c\right) \to-\infty$$</p>
<p>and to prove $\frac{\log x}x\to 0$ assuming $x=e^y \to \infty$ since eventually $e^y>y^2$ we have</p>
<p>$$\frac{\log x}x=\frac{y}{e^y}<\frac1y\to 0$$</p>
|
2,231,487 | <p>In [Mathematical Logic] by Chiswell and Hodges, within the context of natural deduction and the language of propositions LP (basically like <a href="http://www.cs.cornell.edu/courses/cs3110/2011sp/lectures/lec13-logic/logic.htm" rel="nofollow noreferrer">here</a>) it is asked to show, by counter-example that a certain 'sequent rule' is 'unacceptable'.</p>
<p>I suppose the proof should follow an example a few pages earlier that shows that the sequent rule $$(p_0 \to p_1 \vdash p_1)$$ is unacceptable due to the following counter-example: let both $p_0$ and $p_1$ mean $(2=3)$. The book argues that indeed if $(2=3)$ then $(2=3)$, so the left side is true, but the right side: $(2=3)$ is false. The conclusion is that we found a counter-examples and the sequent rule is unacceptable.</p>
<p>It is now asked to prove that this sequent rule is unacceptable: </p>
<p>If $$(\Gamma \vdash (\phi \lor \psi))$$ is correct (i.e has a derivation), then at least one of $$(\Gamma \vdash (\phi))$$ and $$(\Gamma \vdash (\psi))$$ is also correct. </p>
<p>The book hints that one should first try to give a counter-example 'for both sequents $(\vdash p_0)$ and $(\vdash \lnot p_0)$'. What could such a $p_0$ be? Does the book mean using as $p_0$ something like 'red is square'? Or some liar kind of sentence 'this sentence is false'? Or some known undecidable statement (which I doubt due to the level of this book)? Or am I totally off and misunderstand something?</p>
| Noah Schweber | 28,111 | <p>There is in fact a small taste of undecidability lurking here, but it's in a very weak form: we're looking at sentences which are undecidable <em>from the empty theory</em>.</p>
<p>The point is that no matter what $p$ is, we can prove - from no axioms at all! - the sentence "$p\vee\neg p$". That is, the sequent $$\vdash p\vee\neg p$$ is derivable (at least, in any decent sequent calculus).</p>
<p>But think about what "$\vdash p$" means - it means that $p$ is a tautology. Similarly, $\vdash \neg p$ means that $\neg p$ is a tautology. So it's enough to find a single sentence which is neither a tautology nor a contradiction.</p>
<p>In natural language, a good example is something like "Blue is my favorite color." Maybe it is! Maybe it isn't! Certainly we can't prove what my favorite color is just using the axioms of logic - we would need some <em>assumptions</em> about me - but we <em>can</em> prove "Either blue is my favorite color or it isn't."</p>
<p>In propositional logic, we could just take $p$ to be some propositional atom. Then $p\mapsto\top$ makes $\neg p$ false, while $p\mapsto\bot$ makes $p$ false; so neither $p$ nor $\neg p$ is true under every evaluation. That is, neither $\vdash p$ nor $\vdash \neg p$ is acceptable.</p>
<p>In predicate (=first-order) logic, we could take $p$ to be something like "$U(c)$", where $U$ is a predicate symbol and $c$ is a constant symbol. This sentence is true in some structures, and false in others. So again neither $\vdash p$ nor $\vdash \neg p$ is acceptable.</p>
|
1,750,104 | <p>I've had this question in my exam, which most of my batch mates couldn't solve it.The question by the way is the Laplace Transform inverse of </p>
<p>$$\frac{\ln s}{(s+1)^2}$$</p>
<p>A Hint was also given, which includes the Laplace Transform of ln t.</p>
| Salihcyilmaz | 227,487 | <p>$f(s,a) = L(t^a) = \frac{\Gamma(a+1)}{s^{a+1}} $ </p>
<p>Differentiating with respect to a, we get</p>
<p>$L(t^a\cdot lnt) = \frac{\Gamma'(a+1) - \Gamma(a+1)\cdot lns}{s^{a+1}} $<br>
set a = 1.</p>
|
3,978,378 | <p>The question asks Find an efficient proof for all the cases at once by first demonstrating</p>
<p><span class="math-container">$$
(a+b)^2 \leq (|a|+|b|)^2
$$</span></p>
<p>My attempt at the proof:</p>
<p>for <span class="math-container">$a,b\in\mathbb{R}$</span></p>
<p><span class="math-container">$$
\begin{align*}
|a+b|^2 =&(a+b)^2 \text{ (Since $\forall x\in\mathbb{R}$, $x^2=|x|^2$)}\\
=& a^2+2ab+b^2\\
\leq&|a|^2+2|a||b|+|b|^2 \text{( Since $\forall x\in\mathbb{R}, x\leq |x|$)}\\
=&(|a| +|b|)^2
\end{align*}
$$</span></p>
<p>But this is where I get stuck, I can arrive at the inequality but I do not know how to continue you from here. The question states that this should be efficient proof for all the cases, but jumping from that step to <span class="math-container">$|a+b|\leq |a|+|b|$</span> seems like a big jump with some steps missing. Any push in the right direction would be appreciated!</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Note that for positive numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, <span class="math-container">$$a^2 \le b^2 \implies a\le b $$</span>
Thus <span class="math-container">$$|a+b|^2 \le (|a|+|b|)^2 \implies |a+b|\le|a|+|b|$$</span></p>
|
496,011 | <p>Suppose that $ p_n(t) $ is the probability of finding n particle at a time t. And the dynamics of the particle is described by this equation : </p>
<p>$$ \frac{d}{dt} p_n(t) = \lambda \Delta p_n(t) $$</p>
<p>Defining one - dimensional lattice translation operator $ E_m = e^{mk} $ with $ km - mk = 1 $ and $\Delta = E_1 + E_{-1} - 2 $ is a lattice laplacian.</p>
<p>So, here are my questions:</p>
<ol>
<li>What is the one-dimensional lattice translation operator ? I think this one is a term from statistical physics, can you give me a simple explanation or a reference ?</li>
<li>What is the lattice laplacian ? Is it similar to discrete laplacian ? Could you give me a reference for this one ?</li>
<li>Is it possible to solve this equation analytically ?</li>
</ol>
<p>Oh, all of this equation is about random walk with a boundary (random walk of a particle)</p>
<p>Thanks</p>
| J.H. | 94,278 | <p>Your substitution should be completed by letting
$$\frac12\int (u-1) \sqrt{u}du$$ </p>
<p>By using integration by parts, this can be solved.</p>
|
540,227 | <p>I was try to understand the following theorem:-</p>
<p><strong>Let $X,Y$ be two path connected spaces which are of the same homotopy type.Then their fundamental groups are isomorphic.</strong></p>
<p><strong>Proof:</strong> The fundamental groups of both the spaces $X$ and $Y$ are independent on the base points since they are path connected. Since $X$ and $Y$ are of the same homotopy type, there exist continuous maps $f:X\to Y $ and $g:Y\to X$ such that $g\circ f\sim I_X$ by a homotopy, say, $F$ and $f\circ g \sim I_Y$ by some homotopy, say $G$. Let $x_0\in X$ be a base point. Let<br>
$$f_\#:\pi_1(X,x_0)\to \pi_1(Y,f(x_0))$$ and $$g_\#:\pi_1(Y,f(x_0))\to \pi_1(X,g(f(x_0)))$$ be the induced homomorphisms. Let $\sigma$ be the path joining $x_0$ to $gf(x_0)$ defined by the homotopy $F$.</p>
<p>After that the author says that $\sigma_\#$ is a isomorphism. obviously $\sigma_\#$ is a homomorphism but I could not understand how it becomes a isomorphism.</p>
<p>Can someone explain me please. thanks for your kind help and time.</p>
| Pece | 73,610 | <p>In my humble opinion, your proof is too much <em>ad hoc</em>. Just show that $f,f' : (X,x) \to (Y,y)$ induced the same homomorphism of groups $\pi_1(X,x) \to \pi_1(Y,y)$ when they are homotopic : this is sufficient, the rest will come from abstract non sense.</p>
|
3,254,191 | <p>Let <span class="math-container">$V$</span> be a vector space. Determine all linear transformations <span class="math-container">$T:V\rightarrow V$</span> such that <span class="math-container">$T=T^2$</span>.</p>
<p>Suppose <span class="math-container">$x\in V$</span>. Then we can write <span class="math-container">$x=T(x)+(x-T(x))$</span>. Then <span class="math-container">$T(x)\in$</span> Range <span class="math-container">$T$</span>, and <span class="math-container">$x-T(x)\in$</span> kernel of <span class="math-container">$T$</span>. From this how I to proceed further? </p>
| Richard Jensen | 658,583 | <p>Given such a <span class="math-container">$T$</span>, notice that <span class="math-container">$T(V)$</span> is a subspace of <span class="math-container">$V$</span>, and that <span class="math-container">$T$</span> acts on <span class="math-container">$T(V)$</span> as the identity. Therefore, such linear transformations are a subset of projections from <span class="math-container">$V$</span> to subsets of <span class="math-container">$V$</span>. Is it all of them?</p>
|
3,254,191 | <p>Let <span class="math-container">$V$</span> be a vector space. Determine all linear transformations <span class="math-container">$T:V\rightarrow V$</span> such that <span class="math-container">$T=T^2$</span>.</p>
<p>Suppose <span class="math-container">$x\in V$</span>. Then we can write <span class="math-container">$x=T(x)+(x-T(x))$</span>. Then <span class="math-container">$T(x)\in$</span> Range <span class="math-container">$T$</span>, and <span class="math-container">$x-T(x)\in$</span> kernel of <span class="math-container">$T$</span>. From this how I to proceed further? </p>
| Hudson Lima | 680,269 | <p>There are lots of such linear maps. Given a decomposition of the space <span class="math-container">$V=R\oplus K$</span>, we have a unique <span class="math-container">$T:V\to V$</span> such that <span class="math-container">$T(V)=R$</span> and <span class="math-container">$\ker(T)=K$</span> (which is the projection determined by the decomposition). </p>
<p>So what you really want to classify are all decompositions of <span class="math-container">$V$</span> as a (ordered) direct sum of subspaces. Given <span class="math-container">$0\leq k\leq \dim V$</span>, the Grassmannian <span class="math-container">$Gr_k(V)$</span> is formed by all subspaces of <span class="math-container">$V$</span> with dimension <span class="math-container">$k$</span>. </p>
<p>On the other hand, there is a nice way of finding all complements of a fixed subspace <span class="math-container">$R\subseteq V$</span>, using a fixed complement <span class="math-container">$R^\perp$</span> (coming from a inner product, for example): every complement of <span class="math-container">$R\subseteq V$</span> is the graphic of a linear map <span class="math-container">$A:R^\perp\to R$</span>.</p>
<p>In conclusion, there is a bijection between maps <span class="math-container">$T\in End(V)$</span> with <span class="math-container">$T^2=T$</span> and the set
<span class="math-container">$$\bigcup_{k=1}^{\dim V}\left(\bigcup_{R\in Gr_k(V)} Hom(R^\perp;R)\right).$$</span></p>
<p>Considering that each <span class="math-container">$Gr_k(V)$</span> is a manifold and each <span class="math-container">$\bigcup_{R\in Gr_k(V)} Hom(R^\perp;R)$</span> is a bundle over <span class="math-container">$Gr_k(V)$</span>, I'd say there are lots of solutions.</p>
|
3,765,225 | <p>I have a matrix:
<span class="math-container">$$\left(\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right)$$</span>
Which I want to change to:
<span class="math-container">$$\left(\begin{array}{lll}
a & 0 & 0 \\
0 & c & 0 \\
0 & 0 & b
\end{array}\right)$$</span>
How can I do that with a unitary transformation?</p>
| BartBog | 493,336 | <p>To the best of our knowledge, the problem MIN-FORMULA is not in NP.
The <strong>complement</strong> of this problem is in NP^SAT, that is a non-deterministic Turing machine with access to a SAT oracle can solve this in polynomial time.</p>
<p>Given input phi, the machine in question would guess a smaller formula psi and then use the SAT oracle to check whether phi and psi are inequivalent (whether there is an assignment on which they disagree). If the outcome of this test is "NO", i.e., they are equivalent, the machine answers "YES" (meaning, the original formula is NOT minimal).</p>
<p>See for instance Sipser's book Example 9.19.</p>
|
29,861 | <p>The meta question <em><a href="https://math.meta.stackexchange.com/q/29857/290189">Not actually a question, just a rant!</a></em> has inspired me to ask for <em>what</em> an answerer can do in case of self-deletion by the question asker while the answerer is typing the answer.</p>
<p>Since per se site is supposed to be a collection of Q&A, can one re-post the deleted question provided that it has good context?</p>
| GNUSupporter 8964民主女神 地下教會 | 290,189 | <p>Due to <a href="https://creativecommons.org/licenses/by-sa/3.0/" rel="nofollow noreferrer">CC-BY-SA 3.0</a> license on SE network, once a post is published to the SE network, then it's released to the community. Therefore, <em>everyone has the right to use the contents of every post provided that the post owner is properly attributed</em> (through a URL).</p>
<p>The self-deletion action <em>doesn't</em> nullify the act of posting, so in principle, the previous paragraph applies to self-deleted posts as well. In practice, 10k users can view <em>all</em> deleted posts by their URL, so it won't be difficult to click "edit" to retrieve the Markdown source code of the self-deleted post and repost.</p>
<p>For other users, they may <em>put the URL of the self-deleted question</em>, which is viewable by 10k users, in order to properly attribute the original answerer. They may omit the original question asker's user name and the original question title since it's <em>covered</em> by a message like "this question has been deleted"---the URL will enable 10k users to find out the original question. Nevertheless, the re-posted question should be <em>complete</em> in the sense that it's <em>answerable on its own</em>.</p>
<hr>
<p>Edit: Creative common licenses are <em>irrevocable</em>. This consolidates my arguments that the act of granting the community the right to reuse the question is <em>irreversible</em>.</p>
<blockquote>
<ol start="3">
<li><p>License Grant. Subject to the terms and conditions of this License, Licensor hereby grants You a worldwide, royalty-free, non-exclusive, <strong>perpetual</strong> (for the duration of the applicable copyright) license to exercise the rights in the Work as stated below:</p>
<ul>
<li>to Reproduce the Work, to incorporate the Work into one or more Collections, and to Reproduce the Work as incorporated in the Collections;</li>
<li>to create and Reproduce Adaptations provided that any such Adaptation, including any translation in any medium, takes reasonable steps to clearly label, demarcate or otherwise identify that changes were made to the original Work. For example, a translation could be marked "The original work was translated from English to Spanish," or a modification could indicate "The original work has been modified.";</li>
<li>to Distribute and Publicly Perform the Work including as incorporated in Collections; and,
-to Distribute and Publicly Perform Adaptations.</li>
</ul></li>
</ol>
<p>--- <a href="https://creativecommons.org/licenses/by-sa/3.0/legalcode" rel="nofollow noreferrer">Creative Commons 3.0 legal code</a></p>
</blockquote>
|
1,343,995 | <p>We know that $i^i$ is real. But how to explain it geometrically maybe in terms of rotation. like we can explain geometrically multiplication of two complex numbers and so on. Can someone show me a little bit about geometric interpretation of $i^i$ and tell me if what I think below is true?</p>
<p>Note :(Some informations):
It is equal to $e^{i \cdot \log(i)}$, but $\log$ is only well-defined up to
adding integer multiples of $2 \pi i$. Thus in the correct setting $i^i$ is
each of the numbers $e^{-\frac{(4n+1)\pi}{2}}$, $n$ an integer.
Each of those are equally valid, thus finding a geometric interpretation
would by a bit silly.</p>
<p>Thank you for any help</p>
| user21820 | 21,820 | <p>Your note is precisely the point. There is a geometric meaning for addition and multiplication and real powers, but imaginary exponents don't have any because they must be defined in terms of $\exp$ and $\ln$ or something equivalent. And there is no geometric interpretation of $\exp$.</p>
|
2,992,416 | <blockquote>
<p>A pendulum of length <span class="math-container">$1$</span> m and mass <span class="math-container">$100$</span> g attached to the end. Another 100 g mass move horizontally with speed 2 m/s. When collision happens this ball sticks with the pendulum and move together. Find the initial linear speed of the block.</p>
</blockquote>
<p>conservation of energy :-</p>
<pre><code>E(initially) = .5*m*v*v = .5*.1*2*2 = .2 joule
E(initially) = E(final) = .5*(.2)*v*v = .2 joule
v=sqrt(2)
</code></pre>
<p>conservation of momentum :-</p>
<pre><code>m1 v1 + m2 v1 = m1 v2 + m2 v2
.1*0 + .1*2 = .1 (2v)`
v=1 m/s
</code></pre>
<p>main question: why they are different <code>1 m/s</code> and <code>sqrt(2)</code></p>
| Arthur | 15,500 | <p>There is a bijection from <span class="math-container">$\Bbb Z^2$</span> to <span class="math-container">$\Bbb N$</span>, inducing a bijection from <span class="math-container">$\{-1,1\}^{\Bbb Z^2}$</span> to <span class="math-container">$\{-1,1\}^{\Bbb N}$</span>. And <span class="math-container">$\{-1,1\}^{\Bbb N}$</span> is famously uncountable by Cantor's diagonal argument:</p>
<p>Assume for contradiction that it is countable. Then it's possible to list <strong>all</strong> functions in <span class="math-container">$\{-1,1\}^{\Bbb N}$</span> as <span class="math-container">$f_1,f_2,f_3,\ldots$</span>. Now consider the function <span class="math-container">$g\in \{ -1,1 \} ^{\Bbb N}$</span> given by
<span class="math-container">$$
g(n)=-f_n(n)
$$</span>
This function cannot be equal to any of the functions <span class="math-container">$f_1,f_2,\ldots$</span>, so it's not in the list, contradicting that our list contained all functions of <span class="math-container">$\{-1,1\}^{\Bbb N}$</span>.</p>
|
2,992,416 | <blockquote>
<p>A pendulum of length <span class="math-container">$1$</span> m and mass <span class="math-container">$100$</span> g attached to the end. Another 100 g mass move horizontally with speed 2 m/s. When collision happens this ball sticks with the pendulum and move together. Find the initial linear speed of the block.</p>
</blockquote>
<p>conservation of energy :-</p>
<pre><code>E(initially) = .5*m*v*v = .5*.1*2*2 = .2 joule
E(initially) = E(final) = .5*(.2)*v*v = .2 joule
v=sqrt(2)
</code></pre>
<p>conservation of momentum :-</p>
<pre><code>m1 v1 + m2 v1 = m1 v2 + m2 v2
.1*0 + .1*2 = .1 (2v)`
v=1 m/s
</code></pre>
<p>main question: why they are different <code>1 m/s</code> and <code>sqrt(2)</code></p>
| badjohn | 332,763 | <p>Consider the simpler and, naively, smaller set <span class="math-container">$\{0, 1\} ^ \mathbb{N}$</span>. There is a clear near bijection to the interval <span class="math-container">$[0, 1]$</span> by writing the reals in binary. So, this might make it more clear that this set is uncountable and yours also. </p>
<p>I say <em>near</em> bijection because you need to consider that the binary representation is not always unique. E.g. <span class="math-container">$0.011111... = 0.1000000...$</span>. You can ignore this if you just want an intuition of the size of the set. To fix the problem, look at <a href="https://en.m.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem" rel="nofollow noreferrer">Schröder–Bernstein theorem</a>.</p>
|
4,168,223 | <p>Let <span class="math-container">$R$</span> be the row reduced echelon form of a <span class="math-container">$4 \times 4$</span> real matrix <span class="math-container">$A$</span> and
let the
third column of <span class="math-container">$R$</span> be <span class="math-container">$\left[\begin{array}{l}0 \\ 1 \\ 0 \\ 0\end{array}\right]$</span>. Then which is true?</p>
<p>P): If <span class="math-container">$\left[\begin{array}{l}\alpha \\ \beta \\ \gamma \\ 0\end{array}\right]$</span> is a solution of <span class="math-container">$A \mathrm{x}=0$</span>, then <span class="math-container">$\boldsymbol{\gamma}=0$</span>.</p>
<p>Q): For all <span class="math-container">$\mathrm{b} \in \mathbb{R}^{4}, \operatorname{rank}[A \mid \mathrm{b}]=\operatorname{rank}[R \mid \mathbf{b}]$</span>.</p>
<p>For <span class="math-container">$P$</span> <span class="math-container">\begin{aligned}
&{\left[\begin{array}{llll}
a & b & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & \phi \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
\alpha \\
\beta \\
\gamma \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
&a \alpha+b \beta+0+0=0 \\
&0+0+\gamma+0=0 \\
&\Rightarrow \gamma=0
\end{aligned}</span></p>
<p>For <span class="math-container">$Q$</span></p>
<p>Rank of a matrix and rank of Row echelon matrix is same . So Q is also correct.
So both the statements are true. Is my approach correct?
Thanks in advance.</p>
| THIRUMAL 5688 | 469,196 | <p>For <span class="math-container">$P$</span> <span class="math-container">\begin{aligned}
&{\left[\begin{array}{llll}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
0 \\
-2 \\
2 \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
\\
\\
&\Rightarrow \gamma=2
\end{aligned}</span></p>
<p>The statement <span class="math-container">$P$</span> is also false.</p>
|
132,591 | <p>Let $f(x)$ be a positive function on $[0,\infty)$ such that $f(x) \leq 100 x^2$. I want to bound $f(x) - f(x-1)$ from above. Of course, we have $$f(x) - f(x-1) \leq f(x) \leq 100 x^2.$$ This is not good for me though. I need a bound which is linear (or at worst linear-times-root) in $x$.</p>
<p>Is there an inequality of the form $f(x) - f(x-1) \leq f^\prime (x)=200 x$?</p>
| Yongyi Chen | 12,810 | <p>There is no such bound. Let $c$ be a real number, and let
$$f(x)=\begin{cases} 0&\text{if $x<c$}\\100x^2&\text{if $x\geq c$}.\end{cases}$$
Then for $x\in [c,c+1)$ the inequality $f(x)-f(x-1)\leq 100x^2$ is the best bound possible. So we cannot make a better bound for a general function satisfying $f(x)\leq 100x^2$ for all $x$.</p>
|
569,300 | <p>Let $G$ be a finite group and assume it has more than one Sylow $p$-subgroup.</p>
<p>It is known that order of intersection of two Sylow p-subgroups may change depending on the pairs of Sylow p-subgroups.</p>
<blockquote>
<p>I wonder whether there is a condition which guarantees that intersection of any two Sylow $p$-subgroups has the same order.</p>
</blockquote>
<p>Thanks for your help.</p>
| Nicky Hekster | 9,605 | <p>I can prove the following, which provides a criterion to start with.<p>
<strong>Theorem</strong> Let $G$ be a finite group. Then the following are equivalent.<p>
$(a)$ For all $S,T \in Syl_p(G)$ with $S \neq T$, $S \cap T=1$.<br>
$(b)$ For each non-trivial $p$-subgroup $P$ of $G$, $N_G(P)$ has a <em>unique</em> Sylow $p$-subgroup.<br>
$(c)$ For each non-trivial $p$-subgroup $P$ of $G$, $P$ is contained in a <em>unique</em> Sylow $p$-subgroup of $G$.</p>
|
1,504,433 | <p>I was wondering if you could help me with this question, in discrete math.</p>
<p>Prove that gcd(m, n) | lcm(n, m) for any non-zero integers m, n</p>
<p>any help is appreciated!</p>
| Ian Miller | 278,461 | <p><strong>General Case</strong></p>
<p>For the general problem (which sounds more what you need) consider several inputs $i_1$, $i_2$, $i_3$, etc. If you use a proportion $p_1$, $p_2$, $p_3$, etc of each then the output, $o$, will be:</p>
<p>$$o=i_1\times p_1+i_2\times p_2+i_3\times p_3+\cdots$$</p>
<p>Also $p_1+p_2+p_3+\cdots=1$ as the parts must add up to a whole.</p>
<p><strong>Two Inputs Case</strong></p>
<p>For only two inputs you will get $p_2=1-p_1$ which leads to:</p>
<p>$o=i_1\times p_1+i_2\times(1-p_1)$</p>
<p>If you know 3 out of 4 of $o$, $i_1$, $i_2$, $p_1$ then you can determine the fourth by rearranging the mathematics. $p_2$ can be found from $p_1$.</p>
|
222,480 | <p>How many $10$-digit numbers have two digits $1$, two digits $2$, three digits $3$, three digits $4$ so that between the two digits $1$ it has at least <strong>other two digits</strong> and between two digits $2$ it has at least <strong>other two digits</strong> (not necessarily distinct)? Thanks!</p>
| Lutz Lehmann | 115,115 | <p>The order reduction method seeks a second basis solution in the form <span class="math-container">$y=y_1u$</span>, where <span class="math-container">$y_1(x)=\frac{x}{(1-x)^2}$</span> is the already found basis solution.
<span class="math-container">$$
x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0
\implies \frac{u''}{u'}=\frac{3y_1-2(1-x)y_1'}{(1-x)y_1}
$$</span>
Insert <span class="math-container">$y_1(x)=\dfrac1{(1-x)^2}-\dfrac1{1-x}$</span>, <span class="math-container">$y_1'=\dfrac2{(1-x)^3}-\dfrac1{(1-x)^2}$</span>, <span class="math-container">$y_1''=\dfrac{6}{(1-x)^4}-\dfrac{2}{(1-x)^3}$</span>
into that formula to find
<span class="math-container">\begin{align}
\frac{u''}{u'}&=\frac{-\frac1{(1-x)^2}-\frac1{1-x}}{\frac{x}{1-x}}=-\frac{2-x}{x(1-x)}=-\frac{2}x+\frac1{1-x}
\\
\implies u'&=\frac1{x^2(1-x)}=\frac{1+x}{x^2}+\frac1{1-x}
\\
\implies u&=-\frac1x+\ln|x|-\ln|1-x|
\end{align}</span>
so that the second basis solution is
<span class="math-container">$$
y_2=\frac{x\ln|x|-x\ln|1-x|-1}{(1-x)^2}
$$</span></p>
|
33,389 | <p>Consider Schrödinger's <em>time-independent</em> equation
$$
-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi.
$$
In typical examples, the potential $V(x)$ has discontinuities, called <em>potential jumps</em>.</p>
<p>Outside these discontinuities of the potential, the wave function is required to be twice differentiable in order to solve Schrödinger's equation.</p>
<p>In order to control what happens at the discontinuities of $V$ the following assumption seems to be standard (see, for instance, Keith Hannabus' <em>An Introduction to Quantum Theory</em>):</p>
<blockquote>
<p><strong>Assumption</strong>: The wave function and its derivative are continuous at a potential jump.</p>
</blockquote>
<p><strong>Questions</strong>:</p>
<p>1) Why is it necessary for a (physically meaningful) solution to fulfill this condition?</p>
<p>2) Why is it, on the other hand, okay to abandon twofold differentiability?</p>
<p>Edit: One thing that just became clear to me is that the above assumption garanties for a well-defined probability/particle current.</p>
| Piero D'Ancona | 7,294 | <p>Since you talk about 'jump' discontinuities, I guess you are interested in a one dimensional Schroedinger equation, i.e., $x\in\mathbb{R}$. In this situation a nice theory can be developed under the sole assumption that $V\in L^1(\mathbb{R})$ (and real valued of course). By a nice theory I mean that the operator $-d^2/dx^2+V(x)$ is selfadjoint, with continuous spectrum the positive real axis, and (possibly) a sequence of negative eigenvalues accumulating at 0. Better behaviour can be produced by requiring that $(1+|x|)^a V(x)$ be integrable (e.g. for $a=1$ the negative eigenvalues are at most finite in number). If you are interested in this point of view, a nice starting point might be the classical paper by Deift and Trubowitz on Communications Pure Appl. Math. 1979. Notice that the solutions are at least $H^1_{loc}$ (hence continuous) and even something more.</p>
<p>A theory for the case $V$ = Dirac delta (or combination of a finite number of deltas) was developed by Albeverio et al.; the definition of the Schroedinger operator must be tweaked a little to make sense of it. This is probably beyond your interests.</p>
<p>Summing up, no differentiability at all is required on the potential to solve the equation in a meaningful way. However, I suspect that this point of view is too mathematical and you are actually more interested in the physical relevance of the assumptions.</p>
|
2,239,203 | <p>Why does $\lim\limits_{x\to\infty}(x!)^{1/x}\neq 1?$</p>
<p>As far as I know, anything to the power of $0$ is $1$.</p>
<p>We have a factorial raised to $1/\infty=0$, but the limit is not $1$? I don't even know what the limit is. But it seems like infinity? Why is this?</p>
<p><a href="https://i.stack.imgur.com/hYnAE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYnAE.png" alt="enter image description here"></a></p>
| Trevor Gunn | 437,127 | <p>That doesn't follow: $1/x$ is not the same as $0$ even for very large values of $x$. Keep in mind that "the limit as $x$ goes to infinity" is not the same thing as "plug in infinity wherever you see $x$". It is about how the function is behaving for larger and larger $x$. No matter how large $x$ gets, $1/x$ is still positive and as such we can find a (possibly huge) real number $\alpha$ such that $\alpha^{1/x}$ is as big as we want. For example, you will agree that</p>
<p>$$\lim_{x \to \infty} (a^x)^{1/x} = a.$$</p>
<p>Combining this with the fact that $x! > a^x$ for any $a$ and sufficiently large $x$, it shouldn't be too surprising that</p>
<p>$$\lim_{x \to \infty} (x!)^{1/x} = \infty.$$</p>
|
2,239,203 | <p>Why does $\lim\limits_{x\to\infty}(x!)^{1/x}\neq 1?$</p>
<p>As far as I know, anything to the power of $0$ is $1$.</p>
<p>We have a factorial raised to $1/\infty=0$, but the limit is not $1$? I don't even know what the limit is. But it seems like infinity? Why is this?</p>
<p><a href="https://i.stack.imgur.com/hYnAE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYnAE.png" alt="enter image description here"></a></p>
| DanielWainfleet | 254,665 | <p>Method (1). Prove the following: If $f(x)\to \infty$ as $x\to \infty$ then $G(n)=\frac {1}{n}\sum_{j=1}^nf(j)\to \infty$ as $n\to \infty.$</p>
<p>With $f(x)=\ln x$ we have $G(n)=\ln (n!^{(1/n)}).$</p>
<p>Method (2). $\ln x$ is monotonic increasing. So for $n\geq 2$ we have $\ln n>\int_{n-1}^n \ln x\;dx .$ So $$\ln n!=\sum_{j=2}^n \ln j>\sum_{j=2}^n\int_{j-1}^j\ln x\;dx=\int_1^n \ln x \;dx=n(\ln n)-n+1.$$ Therefore for n\geq 2 we have $$\ln (n!^{(1/n)})=\frac {1}{n}\ln n!>\frac {1}{n}(n (\ln n)-n+1)=(\ln n)-1+\frac {1}{n}.$$</p>
|
2,239,203 | <p>Why does $\lim\limits_{x\to\infty}(x!)^{1/x}\neq 1?$</p>
<p>As far as I know, anything to the power of $0$ is $1$.</p>
<p>We have a factorial raised to $1/\infty=0$, but the limit is not $1$? I don't even know what the limit is. But it seems like infinity? Why is this?</p>
<p><a href="https://i.stack.imgur.com/hYnAE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYnAE.png" alt="enter image description here"></a></p>
| Claude Leibovici | 82,404 | <p>As Salahamam_ Fatima commented, think about Stirling approximation.</p>
<p>$$y=(x!)^{1/x}\implies \log(y)=\frac 1x \log(x!)$$ Now, using Stirling approximation $$\log(x!)=x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
\left({x}\right)\right)+O\left(\frac{1}{x}\right)$$ $$\log(y)=(\log (x)-1)+\frac{\log (2 \pi )+\log \left({x}\right)}{2 x}+O\left(\frac{1}{x^2}\right)$$ and now, $$y=e^{\log(y)}=\frac{x}{e}+\frac{\log (2 \pi )+\log \left({x}\right)}{2 e}+O\left(\frac{1}{x}\right)$$</p>
|
1,600,051 | <blockquote>
<p>If $x_1,x_2,\ldots,x_n$ are real numbers larger than $1$, prove that $$\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} \geq \dfrac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}$$</p>
</blockquote>
<p><strong>Attempt</strong></p>
<p>AM-GM doesn't work here since we will get an upper bound. I don't see Cauchy-Schwarz working either. Thus, I think a substitution might work, but I am unsure of which one to use.</p>
| Community | -1 | <p>Consider the function $f(x)=\frac{1}{1+e^x}$ which is convex for $x>0$ .</p>
<p>Now use Jensen's inequality : </p>
<p>$$f( \ln x_1)+f( \ln x_2)+\ldots+f( \ln x_n) \geq n f \left (\frac{\ln x_1+\ln x_2+\ldots+\ln x_n}{n} \right)$$ </p>
<p>This is exactly your inequality : </p>
<p>$$\frac{1}{1+x_1}+\frac{1}{1+x_2}+\ldots+\frac{1}{1+x_n} \geq \frac{n}{1+\sqrt[n]{x_1x_2\ldots x_n}}$$</p>
|
1,600,051 | <blockquote>
<p>If $x_1,x_2,\ldots,x_n$ are real numbers larger than $1$, prove that $$\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} \geq \dfrac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}$$</p>
</blockquote>
<p><strong>Attempt</strong></p>
<p>AM-GM doesn't work here since we will get an upper bound. I don't see Cauchy-Schwarz working either. Thus, I think a substitution might work, but I am unsure of which one to use.</p>
| openspace | 243,510 | <p>Use Cauchy :
$$\frac{1}{1+x_{1}} + ... >= \frac{n}{((1+x_{1})...)^{\frac{1}{n}}}$$</p>
<p>Consider the:
$$(1+x_{1})...)^{\frac{1}{n}} <= (x_{1}...)^{\frac{1}{n}}+1$$</p>
<p>The last one you could prove by yourself(use induction).</p>
|
1,717,149 | <p>Is it true or false that if $V$ is a vector space and $T:V \to W$ is a linear transformation such that $T^2 = 0$, then $Im(T) \subseteq Ker(T)$ ?<br>
I don't understand it that much. It doesn't seem related... I can have a vector $v$ from $V$ that its power by 2 equals zero but $T(v) \neq 0_{v}$ </p>
| lhf | 589 | <p>Try <em>The Fascination of Groups</em> by Budden.</p>
|
1,717,149 | <p>Is it true or false that if $V$ is a vector space and $T:V \to W$ is a linear transformation such that $T^2 = 0$, then $Im(T) \subseteq Ker(T)$ ?<br>
I don't understand it that much. It doesn't seem related... I can have a vector $v$ from $V$ that its power by 2 equals zero but $T(v) \neq 0_{v}$ </p>
| p Groups | 301,282 | <p><em>Groups and Symmetry: A Guide to <strong>Discovering Mathematics</em></strong>, by David W. Farmer.</p>
<p>The highlighted title may convince that it assumes not too much mathematics for the learner. It is very little book, not of the type <em>Definition-Theorem-Proof</em>.</p>
<p>At least (in on-line preview) I don't find a single concrete mathematical statement, but always beautiful pictures and only that.</p>
<p><a href="https://i.stack.imgur.com/rqrBr.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/rqrBr.jpg" alt="enter image description here"></a></p>
|
1,095,918 | <p><img src="https://i.stack.imgur.com/EST8r.jpg" alt="my problem is in prop 27, cannot prove it. Can use definition before. Notice that p-closure is the closure of G in the open point topolgy"></p>
<p>For extra notations: C(E,F) is the set of all continuous functions from E to F (topological spaces).
Can anybody help me prove the proposition?</p>
| Robert Israel | 8,508 | <p>What is $p$-closure? If a neighbourhood $V$ of $x$ works for $G$, does it also work for $\overline{G}$?</p>
|
468 | <p>Textbook writers are blessed with only solving problems with neat answers. Numerical coefficients are small integers, many terms cancel, polynomials split into simple factors, angles have trigonometric functions with known values. Pure bliss.</p>
<p>The "real life" is different (as any of us knows).</p>
<p>Giving such questions for homework runs the risk that the student knows she is wrong when some crooked formula/value shows up. On the other hand, checking messy results (unless perhaps directly numerical values) is harder.</p>
<p>What do you think about posing questions which don't have neat derivations/results? I presume the answer could depend on the subject matter, the level of the students, and perhaps on exactly what the question should teach.</p>
| Confutus | 40 | <p>It may be useful to present at least one application that is solvable but does not have a simple answer. As an example, quadratic equations with rational coefficients that cannot be easily solved by factoring and require use of the quadratic formula arise in chemistry.</p>
|
4,551,407 | <p>Here's the question:<br />
If we have m loaves of bread and want to divide them between n people equally what is the minimum number of cuts we should make?<br />
example:<br />
3 loaves of bread and 15 people the answer is 12 cuts.<br />
6 loaves of bread and 10 people the answer is 8 cuts.</p>
<p>for example 1, I found that I should cut each piece of bread 4 times so that we can have 15 pieces in total, but I can't find an algorithm for it. Any help would be appreciated.</p>
| user577215664 | 475,762 | <p><span class="math-container">$$x^2y''+4xy'+2y=f(x)$$</span>
Rewrite the DE as:
<span class="math-container">$$(x^2y)''=f(x)$$</span></p>
|
1,851,209 | <p>Let $L:X\to Y$ an linear operator. I saw that $L$ is bounded if $$\|Lu\|_Y\leq C\|u\|_X$$
for a suitable $C>0$. This definition looks really weird to me since such application is in fact not necessary bounded as $f:\mathbb R\to \mathbb R$ defined by $f(x)=x$. So, is there an error in <a href="https://en.wikipedia.org/wiki/Bounded_operator" rel="nofollow">wikipedia definition</a> ? And if the are true (what I supposed, since it's wikipedia), what is the interest to defined boundness as Lipschitz condition ? (it make no sense in fact...) </p>
| Thomas | 128,832 | <p>The term "bounded" has a special meaning (the one you wrote down) when it comes to linear operators on topological vector spaces (like normed spaces). It's a common definition in Functional analysis. It is not equivalent to the usual notion of a bounded map. (And it's not really helping to say this that it's an unfortunate term, since it is widely used in the pertaining literature).</p>
<p>See also <a href="https://en.wikipedia.org/wiki/Bounded_operator" rel="nofollow">here</a></p>
|
1,366,372 | <p>In this <a href="https://math.stackexchange.com/questions/1365989/testing-pab-using-2-dice">question</a>
: </p>
<p>$$ P_r(a\cap b)=P_r(a,b)=P_r(a)P_r(b)$$</p>
<p>However in this <a href="https://stats.stackexchange.com/questions/156852/what-do-did-you-do-to-remember-bayes-rule/156866#156866">question</a>: </p>
<p>$$p(a,b) = p(a|b)p(b) = p(b|a)p(a)$$</p>
<p>Is this a contradiction as $P_r(a)P_r(b) \ne p(b|a)p(a)$ ? </p>
| Conrad Turner | 201,962 | <p>No because the two dice are independent the outcome on one has no effect on the other so: $p(a,b)=p(a)p(b)=p(b|a)p(a)$ hence $p(b|a)=p(b)$ (note the first equality is the definition of independence in this context: RVs $A$ and $B$ are independent iff $\mbox{prob}(A=a,B=b)=\mbox{prob}(A=a)\times \mbox{prob}(B=b)$ )</p>
|
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