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 |
|---|---|---|---|---|
496,488 | <p>I have few questions to the proof on Universality of linearly ordered sets. Could anyone advise please? Thank you.</p>
<p>Lemma: Suppose $(A,<_{1})$ is a linearly ordered set and $(B,<_{2})$ is a dense linearly ordered set without end points. Assume $F\subseteq A$ and $E\subseteq B$ are finite and $h:F \to E$ is an isomorphism from $(F,<_{1})$ to $(E,<_{2})$. If $a\in A-F$, then $\exists b\in B-E$ such that $h\cup \left\{(a,b)\right\}$ is an isomorphism from $F \cup \left\{a\right\}$ to $E \cup \left\{b\right\}$.</p>
<p>Theorem: Every countable linearly ordered set is embeddable into the linearly ordered set $(\mathbb{Q},<)$</p>
<p>Proof: Let $(A,<_{1})$ be countably linearly ordered set. Let $f:\mathbb{N}\to A$ be a bijection. Fix bijection $g:\mathbb{N} \to \mathbb{Q}$. Define an embedding $h:A\to \mathbb{Q}$. Let $h(f(0))=g(0).$ </p>
<p>The Induction hypothesis:$h:\left\{f(i): i<n\right\} \to \left\{h(f(i)): i<n\right\}$ is an isomorphism. Let $a=f(n)$, $F=\left\{f(i): i<n\right\},$ $E= \left\{h(f(i)): i<n\right\} ,D=h:\left\{m\in \mathbb{N}: g(m) \not\in E \wedge h\cup\left\{(a,g(m)) \right\}\right\}$, where $ h\cup\left\{(a,g(m)) \right\}$ is an isomorphism. By Lemma, $D\neq\varnothing$. Let $m*=min(D)$ and define $h(f(n))=g(m*)$. done</p>
<p>My questions:
What is the role of $h(f(0))=g(0) ?$ How does it follow that $h: A =F \cup \left\{f(n)\right\} \cup (A-F \cup \left\{f(n)\right\}) \to \mathbb{Q}=E\cup\left\{g(m*)\right\} \cup (\mathbb{Q}-(E \cup \left\{g(m*)\right\}))$ is an isomorphism? </p>
| user43208 | 43,208 | <p>There is no particular role of $h(f(0)) = g(0)$ except to start the inductive argument, which traditionally begins with a base case $n=1$. (Although one <em>could</em> start with the case $n=0$, where the base case starts with the function on the empty set. Possibly the author and/or instructor thought that starting off with a vacuous case would be distracting or confusing, and so opted to go with $n=1$.) </p>
<p>What this proof does is construct a injective function that preserves the order (a bijective function to the image $I$ that preserves the order); given that the domain and the image are totally ordered, it may be shown that the inverse function from image to domain must also be order-preserving. To show it is injective and order-preserving, take any two elements $a, b \in A$ with $a < b$ and verify that $f(a) < f(b)$. But since $a, b$ belong to some finite subset $\{f(i): i < n\}$ (by taking $n$ large enough), this should be clear from the inductive argument. </p>
|
4,604,730 | <p>Given a function <span class="math-container">$f$</span> from vectors to scalars, and a vector <span class="math-container">$\vec v$</span>, the directional derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$\vec v$</span> is defined as <span class="math-container">$\nabla_{\vec v} f = \lim_{h\rightarrow 0} \frac{f(\vec x + h \vec v) - f(\vec x)}{h}$</span>. It can also be computed as <span class="math-container">$\nabla_{\vec v} f = \vec v \cdot \nabla f$</span>. I find it very unintuitive that these two things are equivalent. I've looked at some proofs of this, but they seem to be using concepts that I don't yet understand.</p>
<p>For example, in two dimensions, if <span class="math-container">$\vec v = \pmatrix{1\\1}$</span>, then the directional derivative is <span class="math-container">$\nabla_{\vec v} f = f_x + f_y$</span>. Expanding the definitions, this gives
<span class="math-container">$$\lim_{h \rightarrow 0} \frac{f(x+h, y+h) - f(x,y)}{h} = \lim_{h \rightarrow 0} \frac{f(x+h, y) - f(x,y)}{h} + \lim_{h \rightarrow 0} \frac{f(x, y+h) - f(x,y)}{h}$$</span></p>
<p>This seems to imply that, for small values of <span class="math-container">$h \neq 0$</span>, we can make the approximation
<span class="math-container">$$\frac{f(x+h, y+h) - f(x,y)}{h} \approx \frac{f(x+h, y) - f(x,y)}{h} + \frac{f(x, y+h) - f(x,y)}{h}$$</span>
which simplifies to <span class="math-container">$$f(x+h,y+h) \approx f(x+h,y) + f(x,y+h) - f(x,y)$$</span></p>
<p>Why can we make this approximation? I can see that this approximation is exact when <span class="math-container">$f$</span> is a linear function. But for general <span class="math-container">$f$</span>, why can we get any information for <span class="math-container">$f(\_,\_)$</span> from values of <span class="math-container">$f(x,\_)$</span> and <span class="math-container">$f(\_,y)$</span>? Knowing the values of <span class="math-container">$f$</span> for some inputs give us no information for the values of <span class="math-container">$f$</span> for other inputs, right?</p>
| Toby Bartels | 63,003 | <p>You're right that the approximation
<span class="math-container">$$ f ( x + h , y + h ) \approx f ( x + h , y ) + f ( x , y + h ) - f ( x , y ) $$</span>
is not always valid. On the other hand, if <span class="math-container">$ f $</span> is linear (even in the affine sense, that is of the form <span class="math-container">$ f ( x , y ) = a x + b y + c $</span> but not necessarily <span class="math-container">$ f ( x , y ) = a x + b y $</span>), then the approximation is not only valid but exact. Indeed, in that case we even have the more general version
<span class="math-container">$$ f ( x + h , y + k ) \approx f ( x + h , y ) + f ( x , y + k ) - f ( x , y ) $$</span>
(where some of the <span class="math-container">$ h $</span>s are now <span class="math-container">$ k $</span>s, so I'm no longer necessarily using your particular vector <span class="math-container">$ \vec v $</span>).</p>
<p>In general, this approximation is valid whenever <span class="math-container">$ f $</span> is <em>approximately</em> linear, or more precisely whenever <span class="math-container">$ f $</span> has a good linear approximation near any point. In other words, the approximation is valid when <span class="math-container">$ f $</span> is <em>differentiable</em>. Since not every function is differentiable, this is not a trivial or automatic thing! We have a lot of theorems showing that various functions are differentiable, but in general, you do have to check this. And so this equivalence between the limit and the dot product is only true for differentiable functions.</p>
|
467,574 | <p>Using permutation or otherwise, prove that $\displaystyle \frac{(n^2)!}{(n!)^n}$ is an integer,where $n$ is a positive integer.</p>
<p>I have no idea how to prove this..!!I am not able to even start this Can u give some hints or the solution.!cheers.!!</p>
| André Nicolas | 6,312 | <p>One can do a little better. We have $n^2$ people, to be divided into $n$ teams, each of which has $n$ people. Let $w$ be the number of ways to do the job.</p>
<p>Do the division into teams this way. Line up the $n^2$ people, put the first $n$ into a team, then the next $n$, and the next, and so on.</p>
<p>Permuting the people in each group of $n$ does not change the division into teams. This permuting can be done in $(n!)^n$ ways. Also, permuting the groups does not change the division into teams. It follows that
$$w=\frac{(n^2)!}{(n!)^{n+1}}.$$
So we get the stronger result that $(n!)^{n+1}$ divides $(n^2)!$.</p>
<p><strong>Remark:</strong> In the same way, one can show that $n!(m!)^n$ divides $(mn)!$. </p>
|
1,290,111 | <p>How one can prove the following statement:</p>
<p>$k(n-1)<n^2-2n$ for all odd $n$ and $k<n$</p>
<p><em>Tried so far</em>: induction on $n$, graphing, and rewriting $n^2−2n$ as $(n−1)^2−1$.</p>
| MCT | 92,774 | <p>Following the hint that $n^2 - 2n = (n-1)^2 - 1$</p>
<p>$k(n-1) < (n-1)^2 - 1$</p>
<p>Since $k < n$, we can take $k = n-1$ to derive a counter example.</p>
<p>$(n-1)^2 < (n-1)^2 - 1$</p>
<p>$1 < 0$.</p>
<p>So, taking $k=n-1$ always gives a counter example.</p>
|
1,776,726 | <p>I'm trying to determine whether or not </p>
<blockquote>
<p>$$\sum_{k=1}^\infty \frac{2+\cos k}{\sqrt{k+1}}$$ </p>
</blockquote>
<p>converges or not. </p>
<p>I have tried using the ratio test but this isn't getting me very far. Is this a sensible way to go about it or should I be doing something else?</p>
| Stefan4024 | 67,746 | <p>Minimizing $1 - \frac{1}{(t - \frac 4t +1)^2 + 2}$ is equivalent to minimizing $(t - \frac 4t +1)^2 + 2$. But obviously the minimal value for this is $2$, as the square of a number is always bigger than $0$. To find the value which minimizes it just solve $t- \frac4t + 1 = 0$</p>
|
908,309 | <p>I'm finding the principal value of $$ i^{2i} $$</p>
<p>And I know it's solved like this:</p>
<p>$$ (e^{ i\pi /2})^{2i} $$ </p>
<p>$$ e^{ i^{2} \pi} $$</p>
<p>$$ e^{- \pi} $$</p>
<p>I understand the process but I don't understand for example where does the $ i $ in $ 2i $ go?</p>
<p>Is this some kind of a property of Euler's number? if so please explain to me. </p>
| marty cohen | 13,079 | <p>Also remember that
$i = e^{\pi i/2+2\pi i k}
=e^{\pi i(\frac12+2k)}
$
for any integer $k$,
since $e^{2 \pi i k} = 1$.
$k=0$ gives the usual principal value.</p>
<p>Therefore,
$i^{2i}
=(e^{\pi i(\frac12+2k)})^{2i}
=e^{2\pi i^2(\frac12+2k)}
=e^{-2\pi (\frac12+2k)}
=e^{-\pi (1+4k)}
$.
So,
thanks to the joy of
infinite valued complex logarithms,
your expression has an infinite number of
distinct values.</p>
<p>Note that,
for large negative $k$,
the value is quite large.</p>
|
1,947,082 | <p>Let us have sum of sequence (I'm not sure how this properly called in English): $$X(n) = \frac{1}{2} + \frac{3}{4}+\frac{5}{8}+...+\frac{2n-1}{2^n}$$</p>
<p>We need</p>
<p>$$\lim_{n \to\infty }X(n)$$</p>
<p>I have a solution, but was unable to find right answer or solution on the internet.</p>
<p>My idea:</p>
<p>This can be represented as $$ \frac{1}{2} + \frac{1}{4} + \frac{2}{4} + \frac{3}{8}+\frac{2}{8}+\frac{5}{16} + \frac{2}{16} ... + ...$$</p>
<p>Which is basically </p>
<p>$\frac{1}{2} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8}$ - 1/2 + geometric progression + our initial sum divided by 2.</p>
<p>And then I thought: hey, so I can figure out one part of this sum, and second is twice smaller, and then it forms a cycle! (I suppose).</p>
<p>So it would be $B_1 = 1/2 + b_1/(1-1/2) = 3/2$
$$lim_{n \to\infty }X(n) = B_1/(1-1/2) = 3$$</p>
<p>Is this correct?</p>
| Gordon | 169,372 | <p>Let
\begin{align*}
f(x) &= \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2^n}\\
&=\frac{x}{2}\sum_{n=0}^{\infty}\left(\frac{x^2}{2}\right)^n\\
&=\frac{x}{2-x^2}.
\end{align*}
Then
\begin{align*}
f'(x) = \frac{2+x^2}{(2-x^2)^2}.
\end{align*}
Therefore,
\begin{align*}
\lim_{n\rightarrow \infty}X(n) = f'(1)=3.
\end{align*}</p>
|
3,562,202 | <p>Show that <span class="math-container">$$\int_0^{2\pi}\arctan\Bigl( \frac {\sin x} {\cos x -2 }\Bigl) \, dx=0.$$</span> I can write this as <span class="math-container">$$\mathrm {Im}\int_0^{2\pi}\log( e^{ix}-2)\,dx.$$</span> Making the substitution <span class="math-container">$z =e^{ix} $</span> leads to <span class="math-container">$\operatorname{Im}\int_C\log( z-2)\frac {dz}{iz}$</span>. This should be equal to <span class="math-container">$2\pi \log (-2)$</span>. However the imaginary part of <span class="math-container">$\log (-2)$</span> is not <span class="math-container">$0,$</span> so I must be wrong somewhere. Thanks for any clarify</p>
| Eugene | 726,796 | <p><span class="math-container">$$
\begin{aligned}
\int_0^{2\pi}\arctan\left(\frac{\sin x}{\cos x - 2}\right)dx &=
\int_0^{\pi}\arctan\left(\frac{\sin x}{\cos x - 2}\right)dx +
\int_{\pi}^{2\pi}\arctan\left(\frac{\sin x}{\cos x - 2}\right)dx = \\
|\text{in a second integral}: x = 2\pi-y| &\Leftrightarrow \\
|y \text{ changes from }\pi \text{ to } 0; dx = -dy| &= \\
&=\int_0^{\pi}\arctan\left(\frac{\sin x}{\cos x - 2}\right)dx +
\int_{\pi}^{0}\arctan\left(\frac{\sin(2\pi-y)}{\cos(2\pi-y) - 2}\right)(-dy) = \\
|\sin(2\pi-y) = -\sin y| \\
|\cos(2\pi-y) = \cos y| \\
&=\int_0^{\pi}\arctan\left(\frac{\sin x}{\cos x - 2}\right)dx +
\int_{0}^{\pi}\arctan\left(\frac{-\sin y}{\cos y - 2}\right)dy = \\
|\arctan(-x) = -\arctan x| \\
|\text{in a second integral, let} y = x| \\
&=\int_0^{\pi}\arctan\left(\frac{\sin x}{\cos x - 2}\right)dx -
\int_{0}^{\pi}\arctan\left(\frac{\sin x}{\cos x - 2}\right)dx = 0.
\end{aligned}
$$</span></p>
|
2,461,820 | <p>I am a bit confused with the following question, I get that P(T|D) = 0.95 and P(D) = 0.0001 however because i'm unable to work out P(T|~D) i'm struggling to apply the theorem, am i missing something? Also i'm unsure about what to do with the information relating to testing negative when you don't have the disease correctly 95% of the time</p>
<p>You take a test T to tell whether you have a disease D. The test comes back positive. You know that test is
95% accurate (the probability of testing positive when you do have the disease is 0.95, and the probability of
testing negative when you don’t have the disease is also 0.95). You also know that the disease is rare, only 1
person is 10,000 gets the disease.
What is the probability that you have the disease?
How would this change if the disease was more common, say affecting 1 person in 100?</p>
| Siong Thye Goh | 306,553 | <p>$\ln e = 1$.</p>
<p>You just have to evaluate $\ln 1$.</p>
<p>The input of a natural log has to be positive, the output need not be positive.</p>
|
2,461,820 | <p>I am a bit confused with the following question, I get that P(T|D) = 0.95 and P(D) = 0.0001 however because i'm unable to work out P(T|~D) i'm struggling to apply the theorem, am i missing something? Also i'm unsure about what to do with the information relating to testing negative when you don't have the disease correctly 95% of the time</p>
<p>You take a test T to tell whether you have a disease D. The test comes back positive. You know that test is
95% accurate (the probability of testing positive when you do have the disease is 0.95, and the probability of
testing negative when you don’t have the disease is also 0.95). You also know that the disease is rare, only 1
person is 10,000 gets the disease.
What is the probability that you have the disease?
How would this change if the disease was more common, say affecting 1 person in 100?</p>
| XRBtoTheMOON | 478,087 | <p>The natural log of a number is whatever you have to raise $e$ to, in order to get that number. So for example $\ln(10)$ is saying, "what do I need to raise $e$ to, in order to get 10?. So you could plug that in your calculator, and see $\ln(10) \approx 2.3025$ and then you could check it by seeing what $e^{2.3025}$ is.</p>
<p>In your case, $\ln(e)$ is saying, "what do I need to raise $e$ to, in order to get $e$? well obviously that's just 1 since $e^1 = e$. So then we are left with $\ln(1)$. "What do I need to raise $e$ to in order to get 1?". Well raising anything to the 0 power gives you 1. So the answer of what you need to raise $e$ to is, zero.</p>
<p>Thus $\ln(\ln(e)) = 0$</p>
|
2,461,820 | <p>I am a bit confused with the following question, I get that P(T|D) = 0.95 and P(D) = 0.0001 however because i'm unable to work out P(T|~D) i'm struggling to apply the theorem, am i missing something? Also i'm unsure about what to do with the information relating to testing negative when you don't have the disease correctly 95% of the time</p>
<p>You take a test T to tell whether you have a disease D. The test comes back positive. You know that test is
95% accurate (the probability of testing positive when you do have the disease is 0.95, and the probability of
testing negative when you don’t have the disease is also 0.95). You also know that the disease is rare, only 1
person is 10,000 gets the disease.
What is the probability that you have the disease?
How would this change if the disease was more common, say affecting 1 person in 100?</p>
| Guillemus Callelus | 361,108 | <p>$$\ln(\ln e)=\ln 1= 0$$
The logarithm on base $a$ of $a$ is always $1$, since $a^{1} = a$. In addition, it has to be $a> 0$ and $a$ other than $1$, so that there are no problems.</p>
|
3,337,382 | <p>I want to find how many roots of the equation <span class="math-container">$z^4+z^3+1=0$</span> lies in the first quadrant.</p>
<p>Using Rouche's Theorem how to find ?</p>
| Lutz Lehmann | 115,115 | <p>Look at the family of polynomials <span class="math-container">$z^4+tz^3+1$</span>. For <span class="math-container">$t=0$</span> we know the solutions <span class="math-container">$z=\sqrt{\frac12}(\pm1\pm i)$</span> which has one root per quadrant. We additionally know that the set of roots is continuous in the coefficients of the polynomial.</p>
<p>Now if changing <span class="math-container">$t$</span> from <span class="math-container">$0$</span> to <span class="math-container">$1$</span> were to change the number of roots in the first quadrant, one of the other roots would have to pass the positive <span class="math-container">$x$</span> or <span class="math-container">$y$</span> axis. However, on the positive <span class="math-container">$x$</span> axis the real part <span class="math-container">$1+x^3+x^4$</span> and on the <span class="math-container">$y$</span> axis the real part <span class="math-container">$1+y^4$</span> are never zero. Thus <span class="math-container">$$|z^4+tz^3+1|\ge |z^4+1|-t|z|^3>0$$</span> on the boundary of the first quadrant for <span class="math-container">$t\in [0,1]$</span>. There is no change in the number of roots over this homotopy.</p>
<p><a href="https://i.stack.imgur.com/0TUOx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0TUOx.png" alt="plot of root paths"></a></p>
<p><sub><em>roots of <span class="math-container">$z^4+tz^3+1$</span> for red: <span class="math-container">$t=0$</span> over blue to green: <span class="math-container">$t=1$</span></em></sub></p>
|
2,793,077 | <blockquote>
<p>Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$.
Also prove that $R$ is commutative. Give an example (with explanation)
of a Boolean ring.</p>
</blockquote>
<p>From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.</p>
<p>Under addition a ring is a commutative group. </p>
<p>$a + b = b + a$ (commutative)</p>
<p>$(a + b) + c = a + (b + c)$ (associative)</p>
<p>$a + (-a) = 0$ (inverse exists for every element)</p>
<p>$a + 0 = a$ (identity exists)</p>
<p>Where $a,b,c \in R$</p>
<p>But I'm not really sure how to proceed with the proof from here. Any idea?</p>
| Travis Willse | 155,629 | <p><strong>Hint</strong> Applying the definition of <em>Boolean</em> gives $$a + a = (a + a)^2 = a^2 + a^2 + a^2 + a^2 .$$ What does applying it again give?</p>
|
683,149 | <p>So I am just learning about bijections and I am having difficulty figuring out if these three problems are bijections and how to prove them. </p>
<ol>
<li>$f(x)=x/2$</li>
<li>$f(x)=2x^2$</li>
<li>$f(x)=\lfloor x\rfloor$</li>
</ol>
<p>Sorry I forgot to add the entire question. It is "Determine whether each of these functions is a bijection from $\Bbb R$ to $\Bbb R$."</p>
| Raven | 130,125 | <p>One way to determine whether f is bijective or not by drawing a figure (if possible). If a line parallel to x cuts f at more than one point, then f is not bijective.</p>
|
406,437 | <p>Calculus the extreme value of the $f(x,y)=x^{2}+y^{2}+xy+\dfrac{1}{x}+\dfrac{1}{y}$</p>
<p>pleasee help me.</p>
| Ink | 34,881 | <p>Given an abelian group $M$, let $\operatorname{End}(M)$ denote the set of all homomorphisms $M \to M$ (i.e endomorphisms). This set becomes a ring under pointwise addition and composition. </p>
<p>To see that $\operatorname{End}(M)$ may not be a commutative ring, choose another noncommutative ring $R$ (you already know one). Left multiplication by elements of $R$ are endomorphisms of the underlying abelian group. Since there are elements $a, b \in R$ such that $ab \ne ba$, $\operatorname{End}(R, +)$ is noncommutative.</p>
|
406,437 | <p>Calculus the extreme value of the $f(x,y)=x^{2}+y^{2}+xy+\dfrac{1}{x}+\dfrac{1}{y}$</p>
<p>pleasee help me.</p>
| Key Ideas | 78,535 | <p>Fairly concrete examples of noncommutative rings arise in calculus when studying differential equations using operator algebra. For example, consider the ring of linear operators generated by the derivative $\, D = d/dx\,$ and the operation of multiplication by $\,x,\,$ i.e. $\, f\mapsto xf,\,$ where $\,(L+M)f = Lf + Mf\, $ and $\,(LM)(f) = L(Mf)),\,$ i.e. multiplication is <em>composition</em> of operators. This bivariate ring of differential polynomials $\,\Bbb R\langle x,D\rangle$ is noncommutative since
$$ Dx = xD + 1\ \ \ {\rm i.e.}\ \ (Dx)(f) = D(xf) = x Df + f = (xD + 1) f$$</p>
<p>There is also a discrete analog for difference equations (recurrences), consisting of polynomials in the shift operator $\,Sf(n) = f(n\!+\!1),\,$ and multiplication by $\,n,\,$ i.e. $\, f\mapsto nf.\,$ Then</p>
<p>$$ S n = (n\!+\!1) S\ \ \ {\rm i.e.}\ \ \ (Sn)(f(n)) = S(nf(n)) = (n\!+\!1)f(n\!+\!1) = ((n\!+\!1)S)(f(n))$$</p>
<p>Both of these operator algebras prove useful because we can sometimes employ noncomutative analogs of familiar polynomial arithmetic, e.g. factoring operator polynomials in order to solve differential and difference equations, e.g. solving the Fibonacci recurrence by factoring it as $\,(S-\phi)(S-\bar \phi) f_n = 0,\,$ so $\,f_n = c\phi^n + d\bar \phi^n,\,$ or the <a href="https://math.stackexchange.com/a/17723/78535">noncommutative example here</a>
$$ (n\!-\!1)\ S^2\! - (3n\!-\!2)\ S + 2\,n\, =\, ((n\!-\!1)\, S - n)\ (S - 2)$$
Special cases go by various names, e.g the method of characteristic polynomials, or the classic Heaviside operator calculus, etc.</p>
|
2,825,237 | <p>\begin{cases}
4x \equiv 14 \pmod m \\
3x \equiv 2 \pmod 5
\end{cases}</p>
<p>I want to prove that for $m \in 4\mathbb{Z}$ there are no solutions(1). Moreover, I want to determine all m for which I have solutions(2). </p>
<p>First of all, the second equation is equivalent to $ x \equiv 4$ (mod 5).</p>
<p>If $m$ and $5$ are coprime, the Chinese remainder theorem states that I have solution, so I pick $m$ and $5$ not coprime. In this case, $m = 5t$ for some $t \in \mathbb{Z}$.
I can write:
$$ 4x = 14 + (5t)c$$
$$ 4x = 4 + 5(2 + tc)$$
$$ 4x = 4$$
$$ x \equiv 1 mod 5$$
However, I also have that $ x \equiv 4 \pmod 5$, so there are no solution. In the proof I did not use the fact that $m$ is a multiple of $4$, so I think the answer for (2) is that we have solution only for $(m,a) = 1$. Is that right?</p>
| Jonathan Dunay | 538,622 | <p>The solutions to the second equation are the integers of the form $4+5k$. So, for a given $m$, having a solution to both equations is equivalent to having a $k$ and a $k_1$ such that $4(5k+4)=14+mk_1$ which is true iff $20k-mk_1=-2$. This has a solution iff $gcd(20,m)|2$.</p>
<p>The Chinese Remainder theorem does not work here since you need the left hand side to be $x$ without a coefficient for all of the congruence relations to apply the theorem.</p>
|
2,825,237 | <p>\begin{cases}
4x \equiv 14 \pmod m \\
3x \equiv 2 \pmod 5
\end{cases}</p>
<p>I want to prove that for $m \in 4\mathbb{Z}$ there are no solutions(1). Moreover, I want to determine all m for which I have solutions(2). </p>
<p>First of all, the second equation is equivalent to $ x \equiv 4$ (mod 5).</p>
<p>If $m$ and $5$ are coprime, the Chinese remainder theorem states that I have solution, so I pick $m$ and $5$ not coprime. In this case, $m = 5t$ for some $t \in \mathbb{Z}$.
I can write:
$$ 4x = 14 + (5t)c$$
$$ 4x = 4 + 5(2 + tc)$$
$$ 4x = 4$$
$$ x \equiv 1 mod 5$$
However, I also have that $ x \equiv 4 \pmod 5$, so there are no solution. In the proof I did not use the fact that $m$ is a multiple of $4$, so I think the answer for (2) is that we have solution only for $(m,a) = 1$. Is that right?</p>
| Steven Alexis Gregory | 75,410 | <p>If $m=4n$, then</p>
<p>\begin{align}
4x \equiv 14 \pmod m
&\implies 4x \equiv 14 \pmod{4n} \\
&\implies 4n \mid 14-4x \\
&\implies 2 \mid 7-2x \\
&\implies 2 \mid 7
\end{align}</p>
<p>which is false.</p>
<hr>
<p>$3x \equiv 2 \pmod 5 \implies x \equiv 4 \pmod 5 \implies x = 4 + 5u$ for some integer, $u$.</p>
<p>\begin{align}
4x \equiv 14 \pmod m
&\implies 4(4 + 5u) \equiv 14 \pmod m \\
&\implies 16 + 20u \equiv 14 \pmod m \\
&\implies 20u \equiv -2 \pmod m \\
&\implies m \mid 2(10u+1)
\end{align}</p>
<p>If $u=0$, we get $x=4$ and $m \mid 2$</p>
<p>If $u=1$ we get $x=9$ and $m \mid 22$</p>
<p>$\dots$</p>
<p>So $x=4+5u$ and $m \mid 2(10u+1)$</p>
<h2>Check</h2>
<p>$3x = 12 + 15u \equiv 2 \pmod 5$</p>
<p>$4x - 14 = 16 + 20u - 14 = 20u + 2 = 2(10u+1)$</p>
<p>So $m \mid 2(10u+1) \implies m \mid 4x-14 \implies 4x \equiv 14 \pmod m$.</p>
|
619,564 | <p>I have to prove if this function is differentiable.</p>
<p>$$f(x,y)= \begin{cases} \frac{\cos x-\cos y}{x-y} \iff x \neq y \\-\sin x \iff x=y \end{cases}$$</p>
<p>if $x \neq y$ it is continuous, but i want to see if it is continuous in x=y too.</p>
<p>i can rewrite f as
$$ f(x,y)= \begin{cases} \frac{g(x)-g(y)}{x-y} \iff x \neq y \\
g'(x)=g'(y) \iff x=y \end{cases}$$</p>
<p>and see that $lim_{xy \to xx} g(x,y)=g'(x)$. THus, it is continuous.
Also, the partial derivatives exist:
$$f_x(x,y)=\begin{cases} \frac{-\sin x(x-y)-\cos x+\cos y}{(x-y)^2} \\ -\cos(x) \end{cases}$$
$$f_y(x,y)=\begin{cases} \frac{\sin y(x-y)+\cos x-\cos y}{(x-y)^2} \\ 0 \end{cases}$$
If I proved that they are continuous, too, for the theorem of the total differential, the function would be differentiable. Still, I'm not sure this is the right way of reasoning.</p>
| John Hughes | 114,036 | <p>This seems like a perfectly OK way to do your work. </p>
<p>ALternatively, you might look at the function </p>
<p>$$
h(u, y) = f(u+y, y).
$$</p>
<p>Writing that out, you get
$$
h(u, y) = f(u+y,y)= \begin{cases} \frac{\cos(u+y)-\cos(y)}{u} \iff u+y \neq y \\
-\sin(y) \iff u = 0 \end{cases}
$$</p>
<p>Which you can rewrite as
$$
h(u, y) = f(u+y,y)= \begin{cases} \frac{\cos(u+y)-\cos(y)}{u} \iff u \neq 0 \\
-\sin(y) \iff u = 0 \end{cases}
$$</p>
<p>In these rotated coordinates, it might be a little easier to prove things. </p>
|
619,564 | <p>I have to prove if this function is differentiable.</p>
<p>$$f(x,y)= \begin{cases} \frac{\cos x-\cos y}{x-y} \iff x \neq y \\-\sin x \iff x=y \end{cases}$$</p>
<p>if $x \neq y$ it is continuous, but i want to see if it is continuous in x=y too.</p>
<p>i can rewrite f as
$$ f(x,y)= \begin{cases} \frac{g(x)-g(y)}{x-y} \iff x \neq y \\
g'(x)=g'(y) \iff x=y \end{cases}$$</p>
<p>and see that $lim_{xy \to xx} g(x,y)=g'(x)$. THus, it is continuous.
Also, the partial derivatives exist:
$$f_x(x,y)=\begin{cases} \frac{-\sin x(x-y)-\cos x+\cos y}{(x-y)^2} \\ -\cos(x) \end{cases}$$
$$f_y(x,y)=\begin{cases} \frac{\sin y(x-y)+\cos x-\cos y}{(x-y)^2} \\ 0 \end{cases}$$
If I proved that they are continuous, too, for the theorem of the total differential, the function would be differentiable. Still, I'm not sure this is the right way of reasoning.</p>
| Lutz Lehmann | 115,115 | <p>$\cos x-\cos y=-2\sin\frac{x+y}2\,\sin\frac{x-y}2$ and $\operatorname {sinc}u=\frac{\sin u}{u}$ is a known analytical function. So $f(x,y)=-\sin\frac{x+y}2\operatorname {sinc}\frac{x-y}2$.</p>
|
4,105,854 | <blockquote>
<p>Solve for integers <span class="math-container">$x, y$</span> and <span class="math-container">$z$</span>:</p>
<p><span class="math-container">$x^2 + y^2 = z^3.$</span></p>
</blockquote>
<p>I tried manipulating by adding and subtracting <span class="math-container">$2xy$</span> , but it didn't give me any other information, except the fact that <span class="math-container">$z^3 - 2xy$</span> and <span class="math-container">$z^3+2xy$</span> are perfect squares.</p>
<p>This doesn't give us much information to work on. I don't know if my steps are correct, I do not know how to approach this problem.</p>
<p>Any help would be appreciated.</p>
| 2'5 9'2 | 11,123 | <p>Factoring over Gaussian integers <span class="math-container">$$(x+iy)(x-iy)=z^3$$</span> so it is sufficient (but not necessary) for <span class="math-container">$x+iy$</span> to be a cube. That is,
<span class="math-container">$$\begin{align}x+iy&=(a+bi)^3\\
&=(a^3-3ab^2)+\left(3a^2b-b^3\right)i
\end{align}$$</span></p>
<p>So choose any <span class="math-container">$a,b$</span> and set <span class="math-container">$x=a^3-3ab^2$</span>, <span class="math-container">$y=3a^2b-b^3$</span>. For example <span class="math-container">$a=5$</span>, <span class="math-container">$b=7$</span> gives <span class="math-container">$x=-610$</span>, <span class="math-container">$y=182$</span>. And indeed, <span class="math-container">$(-610)^2+182^2=74^3$</span>.</p>
<hr />
<p>This recovers some solution families mentioned in the comments. For example with <span class="math-container">$a=-k, b=k$</span>, it recovers <span class="math-container">$x=2k^3,y=2k^3,z=2k^2$</span>.</p>
<p>However it does not produce all solutions. For example in the family <span class="math-container">$x=1+k^2,y=k+k^3$</span>, there is the the solution <span class="math-container">$x=5,y=10$</span>. But the family in this answer will not yield that as a solution.</p>
|
4,105,854 | <blockquote>
<p>Solve for integers <span class="math-container">$x, y$</span> and <span class="math-container">$z$</span>:</p>
<p><span class="math-container">$x^2 + y^2 = z^3.$</span></p>
</blockquote>
<p>I tried manipulating by adding and subtracting <span class="math-container">$2xy$</span> , but it didn't give me any other information, except the fact that <span class="math-container">$z^3 - 2xy$</span> and <span class="math-container">$z^3+2xy$</span> are perfect squares.</p>
<p>This doesn't give us much information to work on. I don't know if my steps are correct, I do not know how to approach this problem.</p>
<p>Any help would be appreciated.</p>
| Servaes | 30,382 | <p>Let <span class="math-container">$C$</span>, <span class="math-container">$D$</span>, <span class="math-container">$S$</span> and <span class="math-container">$T$</span> be integers, and define
<span class="math-container">\begin{eqnarray*}
x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\
y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\
z&=&ab^2Z=(C^2+D^2)(S^2+T^2).
\end{eqnarray*}</span>
Then a routine verification shows that <span class="math-container">$x^2+y^2=z^3$</span>. I will show that every solution is of this form. Moreover, if we require <span class="math-container">$S$</span> and <span class="math-container">$T$</span> to be coprime and nonnegative, every solution will have <em>precisely one</em> such representation, making this a proper parametrization.</p>
<p>Let <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> be integers such that
<span class="math-container">$$x^2+y^2=z^3.$$</span>
First note that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are not both odd, as otherwise we get a contradiction by reducing mod <span class="math-container">$8$</span>.</p>
<p>Let <span class="math-container">$d:=\gcd(x,y)$</span> and let <span class="math-container">$a$</span> and <span class="math-container">$b$</span> be integers such that <span class="math-container">$d=ab^3$</span> and <span class="math-container">$a$</span> is cube-free. Then <span class="math-container">$d^2=a^2b^6$</span> divides <span class="math-container">$z^3$</span> and hence <span class="math-container">$a$</span> divides <span class="math-container">$z$</span>. Writing <span class="math-container">$x=au$</span>, <span class="math-container">$y=av$</span> and <span class="math-container">$z=aw$</span> we see that
<span class="math-container">$$a^3w^3=z^3=x^2+y^2=(ab^3u)^2+(ab^3v)^2=a^2b^6(u^2+v^2),$$</span>
from which it follows that <span class="math-container">$b^2$</span> divides <span class="math-container">$w$</span> because <span class="math-container">$a$</span> is cube-free. So writing <span class="math-container">$x=ab^3X$</span>, <span class="math-container">$y=ab^3Y$</span>, <span class="math-container">$z=ab^2Z$</span> shows that
<span class="math-container">$$X^2+Y^2=aZ^3,$$</span>
where <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are coprime. Factoring in <span class="math-container">$\Bbb{Z}[i]$</span> then shows that
<span class="math-container">$$aZ^3=(X+Yi)(X-Yi),$$</span>
where <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are coprime and not both odd, so the two factors are coprime. Then
<span class="math-container">$$X+Yi=(A+Bi)(U+Vi)^3,$$</span>
for some integers <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$U$</span> and <span class="math-container">$V$</span> such that <span class="math-container">$\gcd(A,B)=\gcd(U,V)=1$</span> and <span class="math-container">$A^2+B^2=a$</span> and <span class="math-container">$U^2+V^2=Z$</span>. Then
<span class="math-container">\begin{eqnarray*}
X&=&AU^3-3BU^2V-3AUV^2+BV^3,\\
Y&=&BU^3+3AU^2V-3BUV^2-AV^3,
\end{eqnarray*}</span>
and hence for <span class="math-container">$C=bA$</span> and <span class="math-container">$D=bB$</span> we find that
<span class="math-container">\begin{eqnarray*}
x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\
y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\
z&=&ab^2Z=(C^2+D^2)(S^2+T^2).
\end{eqnarray*}</span></p>
<hr />
<p>In particular, parametrizations given in the other answers and comments correspond to <span class="math-container">$(C,D,S,T)=$</span>
<span class="math-container">$$(1,0,a,b),\qquad(k,k,1,0),\qquad(1,k,1,0),\qquad(a,b,1,0).$$</span></p>
|
217,244 | <p>I would like to know some of the most important definitions and theorems of definite and semidefinite matrices and their importance in linear algebra.
Thanks for your help</p>
| Aamir M. Khan | 623,862 | <p>Positive definite matrices have applications in various domains like physics, chemistry etc. In CS, optimization problems are often treated as quadratic equations of the form <span class="math-container">$Ax=b$</span> where <span class="math-container">$x$</span> can be any higher degree polynomial. To solve such equations, we need to calculate <span class="math-container">$A^{-1}$</span> which is then used to find out <span class="math-container">$x=A^{-1}.b$</span>.</p>
<p>Computing <span class="math-container">$A^{-1}$</span> is time consuming for complex higher rank matrices. Instead we use Cholesky’s decomposition: <span class="math-container">$A = L.L^T$</span>, if the matrix <span class="math-container">$A$</span> is symmetric, positive definite then we can decompose it into lower triangular matrix <span class="math-container">$L$</span>. Finding the inverse of this lower triangular matrix <span class="math-container">$L$</span> and its transpose <span class="math-container">$L^T$</span> is computationally efficient process. Hence we get huge performance gain on <span class="math-container">$x = (L^T)^{-1}.(L^{-1}).b$</span></p>
|
1,781,227 | <p>For a simple x and y plane (2 dimensional), to find the distance between two points we would use the formula </p>
<p>$$
a^2 +b^2 = c^2
$$</p>
<p>For a slightly more complicated plane; x,y and z (3 dimensional), to find the distance between two points we would use the formula</p>
<p>$$
d^2 = a^2 + b^2 + c^2
$$</p>
<p>My question is this; is it possible to use pythagoras's theorem to find the distance/magnitude/modulus between two points in 4 dimensions? And what format would it be in?
Using graphs of triangles and cuboids, I have proved for the 2 dimensional and 3 dimensional pythagorean theorem usage, but since I do not understand the 4th dimension entirely (for convenience and understanding, I assume that it is time), I cannot picture how to start nor work this.</p>
<p>Note: I am a highschool student, and new. If I have not provided enough information or something is unclear please comment and I will try to change it. Thank you in advance</p>
| Ethan Bolker | 72,858 | <p>Good question.</p>
<p>In the theory of relativity the fourth dimension is time, and the distance formula is weird, as @Arthur comments.</p>
<p>But it's quite possible and (for mathematicians) very natural to study spaces with four (or even $n$) geometric dimensions. You just think of a point as a list of its $n$ coordinates. In the plane points are $(x,y)$. In space they are $(x,y,z)$. In four dimensions they are $(x,y,z,w)$, and the distance $d$ from that point to the origin $(0,0,0,0)$ is, as you might guess, given by the Pythagorean relationship
$$
d^2 = x^2 + y^2 + z^2 + w^2 .
$$</p>
<p>As a high school student you might be able to see how to prove that by analogy with how you proved your formula for three dimensions. Later on in your study of mathematics you'll understand how think of it as the <em>definition</em> of distance.</p>
<p>The wikipedia page <a href="https://en.wikipedia.org/wiki/Four-dimensional_space" rel="nofollow">https://en.wikipedia.org/wiki/Four-dimensional_space</a> is a good place to begin reading about geometry in the four dimensions.</p>
|
2,681,451 | <p>For the following question I'm getting stuck on a proof. Below I've just written out all the things/steps I've tried (even if they might be wrong). Could someone steer me in the right direction?</p>
<blockquote>
<p>Suppose $f$ is a real-valued function $f:\mathbb{R}\to\mathbb{R}$,
which is continuous at $0$, with the property
$\forall_{x,y\in\mathbb{R}}:f(x+y)=f(x)+f(y)$. Show that
$\exists_{\lambda\in\mathbb{R}}\forall_{x\in\mathbb{R}}:f(x)=\lambda
x$.</p>
</blockquote>
<p>So far I've found that $f(0)=0$, as $f(x+0)=f(x)+f(0)=f(x)$.
As $f$ is continuous at $0$:
$$\forall_{\epsilon >0}\exists_{\delta >0}:|x-0|<\delta\implies|f(x)-f(0)|<\epsilon \ \ \iff \ \ \ |x|<\delta\implies|f(x)|<\epsilon$$</p>
<p>I suspect that $f|x|=|f(x)|$, which I have yet to prove or find. I'm also afraid the next step I tried is illegal: $|x|<\delta\implies f|x|<f|\delta|$, cause this way, you can compare it to $\epsilon$ by $|f(x)|=f|x|$. This didn't really take me anywhere but I thought I'd just write out everything I tried so far.</p>
<p>I also though that $f$ is continuous everywhere: $f(x-y)=f(x)-f(y)$ gives:
$$|x|<\delta\implies|f(x)|<\epsilon \ \ \iff \ \ \ |a-b|<\delta\implies |f(a-b)|=|f(a)-f(b)|<\epsilon$$
as we substitute $x=a-b$ for arbitrary $a,b\in\mathbb{R}$.</p>
| fleablood | 280,126 | <p>Outline:</p>
<p>Let $\lambda = f(1)$</p>
<p>1) Since $f(x + y) = f(x) + f(y)$ then prove for any $r \in \mathbb Q$ then $f(r) = r*f(1) = r*\lambda$. (This follows inductively.)</p>
<p>2) Since $f$ is continuous prove that $f(x) = x*f(1) = x*\lambda$ by considering a sequence of rations $q_i \to x$. As $f$ is continuous $f(x) = \lim f(q_i) = \lim q_i*\lambda = \lambda * \lim q_i = \lambda x$.</p>
<p>.... to put it simply...</p>
<p>$\mathbb Q$ is the smallest ordered field. $\mathbb Q$ is generated inductively from $1$. As $f(x+y)= f(x) + f(y)$ we can generate a field from $f(1)$ where $f(q) = q*f(1)$. </p>
<p>However we can't generate any "larger" field from thes alone as for $x \not \in \mathbb Q$ we can not get any linear combination of rationals to result in $x$.</p>
<p>But the way we construct the reals from the rationals is to let $\mathbb R$ have the least upper bound property so $\mathbb R = \mathbb Q \cup \{$ all possible limits of cauchy sequences of rational numbers$\}$.</p>
<p>By requiring $f$ to be continuous we must have $\lim f(q_i) = f(\lim q_i) = f(x)$ for any cauchy sequence $q_i \to x$. So the forces us to extend our field for $\langle q*f(1)\rangle; q \in \mathbb Q$ to $\langle x*f(1) \rangle; x \in \mathbb R$.</p>
<p>Okay, maybe that wasn't "simply" but that was the ... "gist".</p>
<p>========</p>
<p>Pf:</p>
<p>1) $f(0) = f(0+0) = f(0) + f(0)$ so $f(0) - f(0) = f(0) + f(0) = f(0)$ so $0 = f(0)$.</p>
<p>$0 = f(x + (-x)) = f(x) + f(-x)$ so $f(-x) = -f(x)$ for all $x\in \mathbb R$.</p>
<p>For $n \in \mathbb N$ then $f(nx) = f(x+x+x+...+x) = f(x) + f(x) + f(x)+...+f(x) = n*f(x)$ and $f(n) = f(n*1) = n*f(1) = \lambda n$.</p>
<p>For $a,b \in \mathbb N$ then $f(\frac ab) = f(a*\frac 1b) = a*f(\frac 1b)$ and $b*\frac{1b} = f(\frac 1b) + .... + f(\frac 1b) = f(\frac 1b + \frac 1b + ....+\frac 1b) = f(b*\frac 1b) = f(1) = \lambda$ so $f(\frac 1b) = \frac {\lambda}b$ and $f(\frac ab) = a*f(\frac 1b) = \frac ab*\lambda$.</p>
<p>So for $q \in \mathbb Q$ then $f(q) = q*\lambda$.</p>
<p>2)Let $x$ be irrational. Then there exists a sequence of $q_i$ so the $q_i \to x$ as $i \to \infty$ (that's the <em>definition</em> of the reals). And $f$ is continuous so $f(x) = \lim_{i\to \infty} f(q_i) = \lim_{i\to\infty}q_i*\lambda = \lambda \lim_{i\to \infty} q_i = \lambda*x$.</p>
|
3,339,647 | <p>I had a class in algebraic topology, our main book is Allen Hatcher, our professor defined a term called "Exponential Law" as the following:</p>
<p><span class="math-container">$Hom (X \times Y, Z) \cong Hom (X, Hom (Y, Z))$</span> </p>
<p><span class="math-container">$\alpha : X \times Y \rightarrow Z $</span></p>
<p><span class="math-container">$\tilde{\alpha} : X \rightarrow Hom (Y, Z)$</span></p>
<p><span class="math-container">$\tilde{\alpha} (x)(y) = \alpha (x, y) $</span></p>
<p>(I may have errors in copying after my professor, forgive me if I have).</p>
<p><strong>My questions are:</strong></p>
<p>1-Where can I find this title in Allen Hatcher or any other book (Actually I asked my professor and he/she said that I may find it in Munkres under the title of "Mapping spaces" and I assumed that he/she means Munkres of general topology and also I did not find this exponential law ), could anyone help me in this please?</p>
<p>2-Why it is called exponential law?</p>
| Wlod AA | 490,755 | <p><strong><em>Why is it called</em> exponential law?</strong></p>
<p>Let <span class="math-container">$|X|$</span> be the cardinality of set <span class="math-container">$X$</span> if <span class="math-container">$X$</span> is considered to be a set.</p>
<p>We have, in the <strong>category of sets</strong>,</p>
<p><span class="math-container">$$ | Hom(X\ Y)|\ =\ |Y|^{|X|} $$</span></p>
<p>Also</p>
<p><span class="math-container">$$ |X\times Y|\ =\ |X|\cdot|Y| $$</span></p>
<p>Hence</p>
<p><span class="math-container">$$ |Hom(X\!\times\! Y\,\ Z)|\ =\ |Z|^{|X\times Y|}\ =\ |Z|^{|X|\cdot|Y|}
\ =\ (|Z|^{|Y|})^{|X|}\ =\ |Hom(X\,\ H(Y\ Z))| $$</span></p>
<p>This is why the bijection <span class="math-container">$\ Hom(X\!\times\! Y\,\ Z)\rightarrow Hom(X\,\ Hom(Y\ Z))\ $</span>
is called the exponential law for the category of sets; and that's why this bijection
is called the exponential law for the arbitrary category for which it is true, whenever it is true.</p>
|
2,787,227 | <p>I'm trying to compute two integrals involving the Dirac delta, namely
\begin{align}
I_1&=\int^{1}_0\!\!\!\! dx_1\!\cdots\!\int^{1}_0\!\!\!\! dx_{8}\,\delta(x_1-x_2+x_3-x_4+x_5-x_6+x_7-x_8)\,,\\
I_2&=\int^{1}_0\!\!\!\! dx_1\!\cdots\!\int^{1}_0\!\!\!\! dx_{8}\,\delta(x_1-x_2+x_4-x_5)\delta(x_3-x_4+x_6-x_7)\,\delta(x_5-x_6+x_8-x_1)\,,
\end{align}
but I don't seem to have the right approach. I try to do case differentiations to find the individual contributions, but I havne't made much progress this way.</p>
<p>Is there a systematic method to evaluate such integrals? I also tried to evaluate them in Mathematica, but I didn't succeed with getting the exact fraction - however, I could approximate the integrals numerically and found $I_1\approx .50\pm.02$ and $I_2\approx .38\pm.02$.</p>
<p>I'd be happy about any suggestions!</p>
| greg | 357,854 | <p>Define the matrix
$$\eqalign{
A &= \sqrt{I+\alpha xx^T} \cr
A^2 &= I+\alpha xx^T \cr
}$$
and define separate variables for the length and direction of the $x$ vector
$$\eqalign{
\lambda &= \|x\| \cr
n &= \lambda^{-1}x \cr
}$$</p>
<p>We can create an ortho-projector $P$ with $n$ in its nullspace
$$\eqalign{
P &= I - nn^T \cr
P^2 &= P^T = P,\,\,\,\,\,P n &= 0 \cr
}$$
Notice that
$$\eqalign{
(P+\beta nn^T)^2
&= P + \beta^2 nn^T \cr
&= I + (\beta^2-1) nn^T \cr
}$$
By choosing $\,\,\beta=\sqrt{1+\alpha\lambda^2}\,\,\,$ we can equate this to $A^2$, which then yields<br>
$$\eqalign{
A &= P+\beta nn^T \cr
}$$
Now take the derivative wrt $\alpha$
$$\eqalign{
\frac{dA}{d\alpha}
&= \frac{d\beta}{d\alpha}\, nn^T \cr
&= \frac{\lambda^2}{2\beta}\,nn^T = \frac{1}{2\beta}xx^T \cr
}$$
To find the gradient wrt $x$, start by finding differentials of the relevant quantities
$$\eqalign{
d\lambda &= \lambda^{-1}x^Tdx = n^Tdx \cr
d\beta &= \frac{\alpha\lambda\,d\lambda}{\beta}
= \frac{\alpha\lambda}{\beta}\,n^Tdx \cr
dn &= \lambda^{-1}P\,dx \cr
dP &= -(dn\,n^T+n\,dn^T)\cr
}$$
So the differential of $A$ is easy enough to find
$$\eqalign{
dA &= nn^T d\beta + \beta\,d(nn^T) + dP \cr
&= nn^T d\beta + (\beta-1)(dn\,n^T+n\,dn^T) \cr
&= nn^T\,\frac{\alpha\lambda}{\beta}\,(n^T\,dx)
+ \frac{\beta-\!1}{\lambda}(P\,dx\,n^T+n\,dx^TP) \cr
}$$
but casting this into the form of gradient (i.e. a 3rd order tensor) using standard matrix notation is impossible. You must introduce special (isotropic) higher-order tensors, or resort to vectorization, or to index notation in order to proceed. Index notation is the best route.</p>
<p>If you contract the differential with matrices from the standard basis $E_{ij}=e_ie_j^T$ you can find the gradient on a component-wise basis</p>
<p>$$\eqalign{
E_{ij}:dA = dA_{ij}
&= n_in_j\,\frac{\alpha\lambda}{\beta}\,(n^T\,dx)
+ \frac{\beta-\!1}{\lambda}(n_jp_i^T\,dx + n_ip_j^Tdx) \cr
\frac{\partial A_{ij}}{\partial x}
&= n_in_j\,\frac{\alpha\lambda}{\beta}\,n
+ \frac{\beta-\!1}{\lambda}(n_jp_i + n_ip_j) \cr
}$$
where $p_k=Pe_k$ is the $k^{th}$ column of $P$ and $n_k$ is the $k^{th}$ element of $n$.</p>
<p>However, if you just want to calculate how $A$ will change in response to a change in $x$, using the differential is sufficient.</p>
|
850,046 | <p>When we calculate sine/cos/tan etc. of a number what exactly are we doing in terms of elementary mathematical concept, please try to explain in an intuitive and theoretical manner and as much as possible explain in the most basic mathematical way.</p>
| 2'5 9'2 | 11,123 | <p>If you have a unit-radius circle centered at the origin, place yourself at $(1,0)$. Now to calculate $\sin(x)$ for the given number $x$, move counter-clockwise around the circle until you have traveled a distance $x$. Wherever you land, the $y$-coordinate is $\sin(x)$. And the $x$-coordinate is $\cos(x)$. The slope of the line connecting the origin to wherever you are is $\tan(x)$.</p>
|
850,046 | <p>When we calculate sine/cos/tan etc. of a number what exactly are we doing in terms of elementary mathematical concept, please try to explain in an intuitive and theoretical manner and as much as possible explain in the most basic mathematical way.</p>
| Tony Piccolo | 71,180 | <p>It is not the whole story, anyhow you could have a "aha!" moment.</p>
<p>I refer to G. A. Jennings $\,$ <em>Modern Geometry with Applications</em> $\,$ (1994), pp. 25-26, paraphrasing the text.</p>
<p>To solve practical problems, initially they drew and measured scale models of triangles; then someone realized that it could be useful to have a table of triangles and their dimensions, putting in the table <strong>the dimensions of the right triangles with the hypotenuse of length one</strong>. <em>Sines</em> and <em>cosines</em> are the lengths of the legs of those right triangles. Then the dimensions of any right triangle can be obtained by similarity multiplying the table entries by a suitable factor.<br>
It is enough to consider only right triangles because a generic triangle can be seen as the sum or the difference of two right triangles. </p>
<p>Summing up, trigonometric tables are a substitute for scale drawings.</p>
|
3,162,056 | <p>In Kleene's "Mathematical Logic" and "Introduction to Metamathematics" for a classical predicate calculus the following two rules of inference are chosen.</p>
<p>If <span class="math-container">$A(x) \Rightarrow C$</span> then <span class="math-container">$(\exists xA(x)) \Rightarrow (C)$</span> and
if <span class="math-container">$C \Rightarrow A(x)$</span> then <span class="math-container">$(C) \Rightarrow (\forall xA(x))$</span> where <span class="math-container">$C$</span> does not contain variable <span class="math-container">$x$</span> free. </p>
<p>I tried motivating these choices but unfortunately I could not. Because it is a classical predicate calculus I tried considering truth table semantics to somehow see why these results should be valid, but what I found (not sure if correct) is that the following results are semantically valid as well.</p>
<p>If <span class="math-container">$A(x) \Rightarrow C$</span> then <span class="math-container">$(\forall x(A(x)) \Rightarrow (C)$</span> and also if <span class="math-container">$C \Rightarrow A(x)$</span> then <span class="math-container">$(C) \Rightarrow (\exists x A(x))$</span> where <span class="math-container">$C$</span> again does not contain variable <span class="math-container">$x$</span> free. </p>
<p>If this is indeed true then I am confused as one sees that <span class="math-container">$\exists$</span> and <span class="math-container">$\forall$</span> act in inference rules in exactly the same way while intuitively I would think that these two logical symbols should act differently.</p>
<p>I would appreciate your advice or thoughts about this.</p>
| Alex Kruckman | 7,062 | <p>The additional rules you give are indeed valid (on non-empty domains). The reason for this is that the implication <span class="math-container">$\forall x\, A(x) \Rightarrow \exists x\, A(x)$</span> is valid (on non-empty domains) - if I recall correctly, this is sometimes called "existential import". </p>
<p>So if <span class="math-container">$A(x) \Rightarrow C$</span>, then we have both <span class="math-container">$\forall x\, A(x)\Rightarrow \exists x\, A(x)$</span> and <span class="math-container">$\exists x\, A(x)\Rightarrow C$</span>, so <span class="math-container">$\forall x\, A(x)\Rightarrow C$</span>. </p>
<p>And dually, if <span class="math-container">$C\Rightarrow A(x)$</span>, then we have both <span class="math-container">$C\Rightarrow \forall x\, A(x)$</span> and <span class="math-container">$\forall x\, A(x)\Rightarrow \exists x\, A(x)$</span>, so <span class="math-container">$C\Rightarrow \exists x\, A(x)$</span>. </p>
<p>But we don't get from this that <span class="math-container">$\exists$</span> and <span class="math-container">$\forall$</span> "act in exactly the same way", because the rules you've given in your question don't totally capture the meaning of <span class="math-container">$\exists$</span> and <span class="math-container">$\forall$</span>. You need more rules. </p>
<p>I'm not sure what rules Kleene uses, but there are two common approaches: </p>
<ol>
<li><p>One approach is to also give the converses of your rules: If <span class="math-container">$\exists x\, A(x)\Rightarrow C$</span>, then <span class="math-container">$A(x)\Rightarrow C$</span>, and if <span class="math-container">$C\Rightarrow \forall x\, A(x)$</span>, then <span class="math-container">$C\Rightarrow A(x)$</span> (where <span class="math-container">$x$</span> is not free in <span class="math-container">$C$</span>). </p></li>
<li><p>Another approach is to include the axioms <span class="math-container">$A(t)\Rightarrow \exists x\, A(x)$</span> and <span class="math-container">$\forall x\, A(x)\Rightarrow A(t)$</span>, where <span class="math-container">$t$</span> is a term substitutable for <span class="math-container">$x$</span>. </p></li>
</ol>
<p>In either of these approaches, you'll notice that if you try to replace <span class="math-container">$\exists$</span> by <span class="math-container">$\forall$</span> or vice versa, the rules are obviously not sound. </p>
<hr>
<p>One more comment: The classical approach to first-order logic assumes that every structure has non-empty domain. But this assumption is not universal. If our semantics for first-order logic allows empty domains, then existential import fails: If there are no <span class="math-container">$x$</span>s, then <span class="math-container">$\forall x\, A(x)$</span> is vacuously true, while <span class="math-container">$\exists x\, A(x)$</span> is false. </p>
<p>And similarly, if we allow empty domains, then your versions of the rules where we swap <span class="math-container">$\exists$</span> and <span class="math-container">$\forall$</span> are no longer valid. Frabala's answer gives a very nice intuitive example of this. </p>
|
2,445,855 | <p>1)How is this letter or symbol pronounced mathematically?</p>
<p>$$\overline k$$</p>
<p>2) $'$ is this sign just a symbol of derivative? For example: </p>
<p>$$k'$$ Do we only understand this as a derivative?</p>
| gen-ℤ ready to perish | 347,062 | <ol>
<li>“Kay bar” (or maybe “bar kay” if $k$ is replaced with something long)</li>
<li>“Kay prime” could hypothetically be used to notate anything, depending on the author and field. However, $k’$ would generally be accepted as any of the following:
<ul>
<li>the derivative of a function $k$</li>
<li>the complement of an event $k$ (in which case it would probably be pronounced “kay complement”)</li>
<li>a symbol that has been distinguished from an otherwise defined but related $k$ (e.g., spacetime dilation, integration dummy variables)</li>
</ul></li>
</ol>
|
4,535,612 | <blockquote>
<p>Is a group of order <span class="math-container">$2^kp$</span> not simple, where <span class="math-container">$p$</span> is a prime and <span class="math-container">$k$</span> is an positive integer?</p>
</blockquote>
<p>I did this for the groups of order <span class="math-container">$2^k 3$</span>. Here the intersection of two distinct Sylow <span class="math-container">$2$</span>-subgroups (if number of sylow <span class="math-container">$2$</span>-subgroup is more than <span class="math-container">$1$</span>) is a nontrivial proper normal subgroup of the group.
So group is not simple.</p>
<p>But if <span class="math-container">$p$</span> is any prime greater than <span class="math-container">$3$</span> then is this always not simple?</p>
<p>Please give some hint to solve this?</p>
<p>For <span class="math-container">$\;p=3\;$</span> I solved like this: <a href="https://math.stackexchange.com/a/2525990/934187">https://math.stackexchange.com/a/2525990/934187</a></p>
| orangeskid | 168,051 | <p>Here is a standard way to get an equation for the image of a plane parametrized curve:</p>
<p><span class="math-container">$$x = \phi(t) \\
y = \psi(t)$$</span></p>
<p>From the first equality obtain a polynomial equation for <span class="math-container">$t$</span> with coefficients dependent on <span class="math-container">$x$</span>, and similarly from the second:</p>
<p><span class="math-container">$$P(t, x) = 0\\
Q(t, y) = 0$$</span></p>
<p>Now, impose the condition on the polynomials <span class="math-container">$P(t)$</span>, <span class="math-container">$Q(t)$</span> in <span class="math-container">$t$</span> to have a common root <span class="math-container">$t$</span>. This involves the <a href="https://en.wikipedia.org/wiki/Resultant" rel="nofollow noreferrer">resultant</a>:</p>
<p><span class="math-container">$$\operatorname{Res}(P, Q) = 0$$</span></p>
<p>The above will be an equation in <span class="math-container">$x$</span>, <span class="math-container">$y$</span>.</p>
<p>Let's take a simple example</p>
<p><span class="math-container">$$x = \frac{1- t^2}{1+t^2} \\
y = \frac{2 t}{1+t^2}$$</span></p>
<p>Then</p>
<p><span class="math-container">$$P(t, x) = (1+t^2) x - (1-t^2)\\
Q(t,x) = (1+t^2) y - 2 t$$</span>
With WA we <a href="https://www.wolframalpha.com/input?i=Resultant%5B%20%281%2Bt%5E2%29%20x%20-%20%281-t%5E2%29%2C%20%281%2Bt%5E2%29%20y%20-%202%20t%2C%20t%5D" rel="nofollow noreferrer">get</a></p>
<p><span class="math-container">$$\operatorname{Res}(P(t), Q(t)) = 4(x^2 + y^2-1)$$</span></p>
|
626,095 | <blockquote>
<p>$R = \mathbb{F}_3[x]/\langle X^3+X^2+1\rangle$ and $\alpha=[X]$ in $R$. How do you prove that the group $R^*$ is not cyclic?</p>
</blockquote>
<p>We have shown that $\alpha$ is a unit in $R$ with order $8$ and that $\alpha^4$ and $-\alpha^4$ are two different elements in $R^*$ both with order 2.</p>
| Josué Tonelli-Cueto | 15,330 | <p>Hint: How many elements of order two have a cyclic group? If $R^*$ has two or more, is it compatible with it being cyclic?</p>
|
1,435,603 | <blockquote>
<p>Let $S$ be a subset of $\mathbb{R}$ with strictly positive Lebesgue measure. Prove that almost every (with respect to Lebesgue measure) real number can be written as the sum of an element of $S$ and an element of $\mathbb{Q}$. </p>
</blockquote>
<p>So, I remember proving a long time ago that $S-S$ contains a neighborhood of 0, and I thought that it would be possible to use this to prove the problem, but I was unable to do so.<br>
I know that $S + \mathbb{Q}$ is dense in the reals and has infinite measure, but I can't figure out why it must be $\mathbb{R}$ minus a measure zero set.<br>
Thanks in advance :).</p>
| Eugene Zhang | 215,082 | <p>This can be formulated and proved as the following theorem.</p>
<p><strong>Theorem:</strong> Let <span class="math-container">$S$</span> be a Lebesgue measurable set on <span class="math-container">$[a,b]$</span> with <span class="math-container">$m(S)>0$</span>. Then<br />
<span class="math-container">$$
\bigcup_{q \in \mathbb{Q}}(q+S)=\Bbb{R}-N
$$</span>
where <span class="math-container">$m(N)=0$</span>.</p>
<p>Proof: For any small <span class="math-container">$\epsilon>0$</span>, there is an interval <span class="math-container">$I$</span> such that
<span class="math-container">$$
m(S\cap I)>(1-\epsilon)m(I)\tag1
$$</span>
For if not, there is a <span class="math-container">$\epsilon>0$</span> that for any interval <span class="math-container">$I$</span> that
<span class="math-container">$$
m(S\cap I)\leqslant(1-\epsilon)m(I)
$$</span>
Then for <span class="math-container">$x\in S$</span>
<span class="math-container">$$
1_S(x)=\lim_{\delta\to0}\frac{m(S\cap(x-\delta,x+\delta))}{m((x-\delta,x+\delta))}\leqslant 1-\epsilon<1
$$</span>
This means <span class="math-container">$m(S)=0$</span> by Lebesgue density theorem, contradicting <span class="math-container">$m(S)>0$</span>.</p>
<p>Let <span class="math-container">$B=\bigcup_{q \in \mathbb{Q}}(q+S)$</span>. Clearly <span class="math-container">$m(B)>m(S)>0$</span>. Let <span class="math-container">$B^c=\Bbb{R}-B$</span>. If <span class="math-container">$m(B^c)=0$</span>, then done. So assume <span class="math-container">$m(B^c)>0$</span>. By <span class="math-container">$(1)$</span>, there are intervals <span class="math-container">$I,J$</span> and <span class="math-container">$m(I)=m(J)$</span> such that
<span class="math-container">$$
m(B\cap I)>r_1m(I)\quad\text{and }\quad m(B^c\cap J)>r_2m(J)
$$</span>
where <span class="math-container">$r_1>2/3,\: r_2>2/3$</span>. Then there exists a <span class="math-container">$p\in \mathbb{Q}$</span> that <span class="math-container">$I+p=J$</span>. So
<span class="math-container">$$
m(B+p\cap J)=m(B+p\cap I+p)=m(B\cap I)>r_1m(I)=r_1m(J)
$$</span>
But since <span class="math-container">$B+p\subset B$</span>, there is
<span class="math-container">$$
m(B\cap J)\geqslant m(B+p\cap J)>r_1m(J)
$$</span>
And since <span class="math-container">$(B\cap J)\cap (B^c\cap J)=\varnothing$</span>
<span class="math-container">$$
m(J)=m((B\cap J)\cup (B^c\cap J))=m(B\cap J)+m(B^c\cap J)>(r_1+r_2)m(J)>m(J)
$$</span>
which is a contradiction. So <span class="math-container">$m(\Bbb{R}-B)=m(B^c)=0$</span>.</p>
|
151,076 | <p>If A and B are partially-ordered-sets, such that there are injective order-preserving maps from A to B and from B to A, is there necessarily an order-preserving bijection between A and B ?</p>
| Brian M. Scott | 12,042 | <p>No. Let $P$ be the disjoint union of chains of every odd length, and let $Q$ be the disjoint union of chains of every positive even length. Clearly each can be embedded in the other, but the two are not isomorphic.</p>
<p>More formally, let $$P=\{\langle 2n+1,k\rangle:n\in\Bbb N\text{ and }1\le k\le 2n+1\}$$ and $$Q=\{\langle 2n,k\rangle:n\in\Bbb Z^+\text{ and }1\le k\le 2n\}\;,$$</p>
<p>each ordered by the relation $\langle m,k\rangle\preceq\langle n,\ell\rangle$ iff $m=n$ and $k\le\ell$. The map $$f:P\to Q:\langle 2n+1,k\rangle\mapsto\langle 2n+2,k\rangle$$ embeds $P$ into $Q$, and the map $$g:Q\to P:\langle 2n,k\rangle\mapsto\langle 2n+1,k\rangle$$ embeds $Q$ into $P$, but $\langle P,\preceq\rangle$ and $\langle Q,\preceq\rangle$ are clearly not isomorphic, as they contain maximal chains of different lengths.</p>
<p>This shows that there is no such result even for well-founded partial orders.</p>
|
4,467,763 | <p>I have the equation</p>
<p><span class="math-container">$A\vec{x} = \vec{b} \tag{1}.$</span></p>
<p>where <span class="math-container">$A$</span> is an <span class="math-container">$m\times n$</span> matrix of rank <span class="math-container">$m$</span>, so that <span class="math-container">$m<n$</span> and the system is underdetermined. As I understand it, I can get the minimum L2 norm solution of the system by premultiplying <span class="math-container">$\vec{b}$</span> with the Moore-Penrose right inverse:</p>
<p><span class="math-container">$\vec{x} = A^T(AA^T)^{-1}\vec{b}\tag{2}.$</span></p>
<p>What I don't understand is how I get from (1) to (2), i.e. what are the algebraic steps? I'm specifically interested in knowing how to do it with just basic matrix manipulations, without calculus.</p>
| Charlie Vanaret | 741,916 | <p>A more geometrical interpretation is to project <span class="math-container">$b$</span> onto the column space of <span class="math-container">$A$</span>. This projection is a linear combination of the columns of <span class="math-container">$A$</span>, given by <span class="math-container">$Ax$</span>. Then the "error" <span class="math-container">$b - Ax$</span> is orthogonal to the column space of <span class="math-container">$A$</span>, so in particular to the columns of <span class="math-container">$A$</span>. In other words, the dot product between each column of <span class="math-container">$A$</span> and <span class="math-container">$b - Ax$</span> is <span class="math-container">$0$</span>, that is <span class="math-container">$A^T (b - Ax) = 0$</span>. Therefore <span class="math-container">$A^T Ax = A^T b$</span>.</p>
<p>I can recommend the online MIT class given by Gilbert Strang, it's absolutely brilliant to develop intuition.</p>
|
3,741,222 | <p>I'm reading the Thompson's book about lattices and sphere packing and got stuck by a sentence of a kind of <span class="math-container">$Z_8$</span> he introduced to reach 2 pages later the full <span class="math-container">$E_8$</span> lattice.
You can find this lattice defined at pages 73-74 and it's basically. To resume it, it's a lattice packing with 16 closest point to the origin having shape: <span class="math-container">$$(\pm2, 0^7)$$</span>
The packing radius is then <span class="math-container">$1/2$</span> of their distance from the origin, i.e. <span class="math-container">$\rho=1$</span>. So far, so good.</p>
<p>My problem is when he tries to compute the center density of the lattice. Notice this center density can be interpreted as the "real" sphere center density since <span class="math-container">$\rho=1$</span>, as claimed in SPLAG when describing the formula for <span class="math-container">$\delta$</span>.</p>
<p>Instead of using any formula, Thompson uses a clever idea to estimate it, which sound like this:</p>
<blockquote>
<p>The center density is fairly easy to calculate. If all coordinate entries were written in binary form, then the lattice, by definition, would contain only those coordinates whose ones digits were either all 0's or 1's. In this case the only two out of every <span class="math-container">$2^8$</span> points with integer coordinates are acceptable. Thus, the center density = <span class="math-container">$1/2^7$</span></p>
</blockquote>
<p>I've got 2 problems with this result.</p>
<p>The first is I can choose for this lattice a generating matrix made by only 2 in all diagonal entries, i.e. twice the identity matrix.
The determinant of this would then <span class="math-container">$2^8$</span>. Using SPLAG formula for center density, and keeping <span class="math-container">$\rho=1$</span>, I would get <span class="math-container">$\delta=1/2^8$</span>, which is smaller by a factor of 2 compared to the one claimed by Thompson.</p>
<p>To confirm this latest sentence: as far as I can see, the lattice defined above can be seen as <span class="math-container">$Z_8$</span> lattice, which density is (always from SPLAG), <span class="math-container">$\delta=1/2^8$</span></p>
<p>However Thompson is using this <span class="math-container">$1/2^7$</span> to derive the full <span class="math-container">$E_8$</span> lattice, so I'm not claiming it's wrong a priori. But I'd like to understand where my reasoning is wrong and how to express coordinates in that binary format (I'm a programmer, so used to binary digits) to emulates Thompson's idea.</p>
<p>Thanks in advance</p>
| riccardoventrella | 264,042 | <p>I think to have found the explanation of the binary trick. First of all, let's consider this lattice has all coordinates even or all odd, by definition.
The idea was suggested to me in the way in which Leech enriched the extended Golay code <span class="math-container">$C_{24}$</span>.
Let's write a generic point stacking the binary digits in vertical (let's consider positive numbers up to 15 for the sake of simplicity and without loss of generality).</p>
<p><span class="math-container">$$(5, 9, 1, 3, 1,....) = \begin{bmatrix}0 & 1 & 0 & 0 & 0 &...\\1 & 0 & 0 & 0 & 0 &...\\0 & 0 & 0 & 1 & 0 &...\\1 & 1 & 1 & 1 & 1 &...\end{bmatrix}$$</span></p>
<p>, where the top row is the 8s digit while the bottom row is the 1s digit.</p>
<p>It's easy to see, for the case of 8 coordinates like for this lattice, the bottom row will be made by these couple of octuplets ONLY: <span class="math-container">$(0, 0, 0, 0, 0, 0, 0, 0)$</span> and <span class="math-container">$(1, 1, 1, 1, 1, 1, 1, 1)$</span>.
This is because we have claimed above the coordinates for this lattice are all odd or all even.
If we consider 8 generic integer coordinates, it's easy to see the bottom row could span from all 0's to all 1's, passing through all the <span class="math-container">$2^8=256$</span> possibilities.
As claimed by the book author, then, the density will be the ratio of the couple combinations over the overall possible ones, i.e. <span class="math-container">$\delta=\frac{2}{2^8}=\frac{1}{2^7}$</span>.
It's easy to extend this proof encompassing integral numbers > 15 (so having any kind of row numbers) and negative (considering a kind of 2-complement representation <span class="math-container">$-6=2(mod 8)$</span>, like Leech did).
Finally, we can use the same trick to estimate quickly the center density of the <span class="math-container">$D_4$</span> checkerboard lattice (i.e. the densest packing in 4 dimensions), which has the same property of having all coordinates even or all odd.
Here we will have 4 coordinates to be represented in binary form, which forms an overall <span class="math-container">$2^4$</span> possible combinations for their 1s digit. But only the <span class="math-container">$(0, 0, 0, 0)$</span> and <span class="math-container">$(1, 1, 1, 1)$</span> out of them will host the <span class="math-container">$D_4$</span> lattice points, which nails the center density to <span class="math-container">$\delta=\frac{2}{2^4}=\frac{1}{8}$</span>.
This matches exactly the center density enumerated in SPLAG 1.2 Table pag.15.</p>
|
1,513,056 | <p>Why is a limit of an <strong>integer</strong> function $f(x)$ also integer? For example, a function that's defined on interval $[a, \infty)$ and the limit is $L$</p>
| Lutz Lehmann | 115,115 | <p>Do you mean $f:[a,\infty)\to\Bbb Z$ as integer function? In $\Bbb R$, the set $\Bbb Z$ consists only of isolated points.</p>
|
3,987,357 | <p>Let
<span class="math-container">$$
\psi(x)=
\left\{
\begin{array}{cll}
x \sin\Big(\dfrac{1}{x}\Big) & \text{if} & x\in (0,1],\, \\
0 & \text{if} & x=0,
\end{array}
\right.
$$</span>
and let <span class="math-container">$f:[-1,1]\rightarrow \mathbb{R}$</span> be Riemann integrable.</p>
<p>How can I show that <span class="math-container">$f\circ \psi$</span> is Riemann integrable?</p>
<p>I have several theorems in my book that could work if <span class="math-container">$\psi$</span> were a <span class="math-container">$C^1$</span> diffeomorphism or a homeomorphism with inverse satisfying Lipschitz condition. But <span class="math-container">$\psi$</span> is clearly neither of those. Any help with a zero set argument or reverting to definitions?</p>
| zhw. | 228,045 | <p>i) If <span class="math-container">$\psi:[\alpha,\beta]\to [a,b]$</span> is <span class="math-container">$C^1$</span> and <span class="math-container">$\psi'$</span> has only finitely many zeros, then <span class="math-container">$f\circ \psi$</span> is RI on <span class="math-container">$[\alpha,\beta]$</span> for every <span class="math-container">$f$</span> RI on <span class="math-container">$[a,b].$</span> I'll take this one as known.</p>
<p>ii) Let's generalize: Now suppose <span class="math-container">$\psi:[\alpha,\beta]\to [a,b]$</span> is <span class="math-container">$C^1$</span> and <span class="math-container">$\psi'$</span> has finitely many zeros in every closed interval contained in <span class="math-container">$(\alpha,\beta).$</span> Then <span class="math-container">$f\circ \psi$</span> is RI on <span class="math-container">$[\alpha,\beta]$</span> for every <span class="math-container">$f$</span> RI on <span class="math-container">$[a,b].$</span> To prove this, let <span class="math-container">$f$</span> be RI 0n <span class="math-container">$[a,b].$</span> Let <span class="math-container">$\epsilon>0$</span> be small. Then from i) we have <span class="math-container">$f\circ \psi$</span> RI on <span class="math-container">$[\alpha+\epsilon,\beta-\epsilon].$</span> Letting <span class="math-container">$\epsilon\to 0^+,$</span> we get <span class="math-container">$f\circ \psi$</span> is RI on <span class="math-container">$[\alpha,\beta]$</span> for every <span class="math-container">$f$</span> RI on <span class="math-container">$[a,b].$</span></p>
<p>Corollary: Suppose <span class="math-container">$\psi:[\alpha,\beta]\to [a,b]$</span> is real analytic on <span class="math-container">$(a,b).$</span> Then <span class="math-container">$f\circ \psi$</span> is RI on <span class="math-container">$[a,b]$</span> for all <span class="math-container">$f$</span> RI on <span class="math-container">$[a,b].$</span> Proof: If <span class="math-container">$\psi$</span> is constant there is nothing to prove. If <span class="math-container">$\psi$</span> is nonconstant, then <span class="math-container">$f'$</span> is real analytic and not <span class="math-container">$\equiv 0$</span> on <span class="math-container">$(a,b).$</span> By the identity principle, <span class="math-container">$\psi$</span> has finitely many zeros in every closed interval contained in <span class="math-container">$(\alpha,\beta).$</span> From ii) we then have the desired result.</p>
<p>The corollary gives a quick non computational proof of the problem at hand, since <span class="math-container">$x\sin(1/x)$</span> is real analytic on <span class="math-container">$(0,1).$</span></p>
|
791,020 | <p>From my textbook. </p>
<p>$$\sum\limits_{k=0}^\infty (-\frac{1}{5})^k$$</p>
<p>My work:</p>
<p>So a constant greater than or equal to $1$ raised to ∞ is ∞.</p>
<p>A number $n$ for $0<n<1$ is $0$. So when taking the limit of this series you get 0 but when formatting the problem a different way $(-1)^k/(5^k)$ it seems like an alternating series. Can someone help me figure this out?</p>
| Pavelshu | 82,360 | <p>If what you meant to ask about was a binomial expansion, then the approach is as follows: </p>
<p>First, note that the binomial theorem states that
\begin{align*}
(a+b)^n = \sum_{k=0}^{\infty}\binom{n}{k}a^kb^{n-k}
\end{align*}
Thus, by substituting for the above values, we have
\begin{align*}
(x-2)^\frac{1}{2} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k(-2)^{\frac{1}{2}-k}
\end{align*}</p>
|
1,844,894 | <p>To explain my question, here is an example.</p>
<p>Below is an AP:</p>
<p>2, 6, 10, 14....n</p>
<p>Calculating the nth term in this sequence is easy because we have a formula. The common difference (d = 4) in AP is constant and that's why the formula is applicable, I think.</p>
<p>But what about this sequence:</p>
<p>5, 12, 21, 32....n</p>
<p>Here, the difference between two consecutive elements is not constant, but it too has a pattern which all of you may have guessed. Taking the differences between its consecutive elements and formimg a sequence results in an AP. For the above example, the AP looks like this:</p>
<p>5, 7, 9, 11.....n</p>
<p>So given a sequence with "uniformly varying common difference" , is there any formula to calculate the nth term of this sequence?</p>
| parsiad | 64,601 | <p>I will change your notation a bit and use $x_{-1}\equiv n$ instead.</p>
<p><strong>Hint</strong>: for $m>0$,
$$
\sum_{x_{0}=0}^{x_{-1}-1}\cdots\sum_{x_{m}=0}^{x_{m-1}-1}1=\sum_{x_{0}=0}^{x_{-1}-1}\left[\sum_{x_{1}=0}^{x_{0}-1}\cdots\sum_{x_{m}=0}^{x_{m-1}-1}1\right]=\cdots
$$
You should use an induction hypothesis on the inner sum at this point.</p>
|
983,200 | <p>Find the derivative of the following: $$f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$$ Would I use the chain rule and product rule?
So far I have:</p>
<p>$$\begin{align}g(x)=x^3-4x+6
\\g'(x)=2x^2-4\end{align}$$</p>
<p>would $h(x)$ be $\ln(x^4-6x^2+9)$?
If so, how would I find $h'(x)$?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>given $f(x)=(x^3-4*x+6)\ln(x^4-6x^2+9)$ we find by the product and chaine rule
$f'(x)=(3x^2-4)\ln(x^4-6x^2+9)+(x^3-4x+6)\frac{4x^3-12x}{x^4-6x^2+9}$
simplifying this in a few minutes
simplifying this i got
$f'(x)=\frac{4 x^4-16 x^2+3 x^4 \ln \left(x^4-6 x^2+9\right)-13 x^2 \ln \left(x^4-6
x^2+9\right)+12 \ln \left(x^4-6 x^2+9\right)+24 x}{x^2-3}$</p>
|
1,523,427 | <p>Is it possible to cover all of $\mathbb{R}^2$ using balls $\{ B(x_n,n^{-1/2})\}_{n=1}^\infty$ of decreasing radius $n^{-1/2}$? I know that if we chose e.g. radius $n^{-1}$ it could never work because $\sum \pi (n^{-1})^2 < \infty$. But in this case the balls cover an infinite amount of area, so it seems that it may be possible to construct.</p>
<p>Edit: It can definitely be done with balls of radius $n^{-1/2+\epsilon}$ for any $\epsilon > 0$.</p>
| detnvvp | 85,818 | <p>Yes, this can be done.</p>
<p>Let $n\in\mathbb N$, consider $m\in\{1,\dots n\}$, and set $$a(n,m)=n^2+(m-1)n.$$ Note then that $$\sum_{k=a(n,m)+1}^{a(n,m+1)}\frac{1}{\sqrt{k}}\geq\sum_{k=a(n,m)+1}^{a(n,m+1)}\frac{1}{\sqrt{n^2+mn}}=\frac{n}{\sqrt{n^2+mn}}\geq\frac{n}{\sqrt{n^2+n^2}}=\frac{1}{\sqrt{2}}.$$ Therefore, if we consider adjacent squares, this shows that we can cover the strip $\left[0,\frac{1}{\sqrt{2}}\right]\times\left[0,\frac{1}{n\sqrt{2}}\right]$ by squares of the form $$S\left(x_{a(n,m)+1},\frac{1}{\sqrt{a(n,m)+1}}\right),S\left(x_{a(n,m)+2},\frac{1}{\sqrt{a(n,m)+2}}\right),\dots,$$$$S\left(x_{a(n,m+1)},\frac{1}{\sqrt{a(n,m+1)}}\right),$$ for any fixed $m\in\{1,\dots n\}$. If we now repeat this process for the strips $$\left[0,\frac{1}{\sqrt{2}}\right]\times\left[\frac{1}{n\sqrt{2}},\frac{2}{n\sqrt{2}}\right],\,\,\left[0,\frac{1}{\sqrt{2}}\right]\times\left[\frac{2}{n\sqrt{2}},\frac{3}{n\sqrt{2}}\right],\dots\left[0,\frac{1}{\sqrt{2}}\right]\times\left[\frac{n-1}{n\sqrt{2}},\frac{n}{n\sqrt{2}}\right],$$ this shows that we can always cover the square $\left[0,\frac{1}{\sqrt{2}}\right]^2$ by squares of the form $S(x_i,i^{-1/2})$, for $i\in\{n^2,n^2+1,\dots 2n^2\}$.</p>
<p>Repeating this process for all $n\in\mathbb N$ and moving the squares on $\mathbb R^2$, we actually get a covering of all of $\mathbb R^2$ in that way. We then get a covering by balls by circumscribing a ball outside each one of the squares.</p>
|
354,986 | <p>I have to show that $h$ is measurable as well as $\int h d(\mu \times \nu) < \infty$ .</p>
<p>I tried showing by contradiction that $\int h$ had to be finite but I'm stuck with showing how it is measurable.</p>
| icurays1 | 49,070 | <p>Hint: show that $\int f(x)g(y) d(\mu\times \nu)=\int f(x)(\int g(y)d\nu)d\mu$.</p>
|
1,985,653 | <p>I want to show that {$\sqrt 2$} $\cup$ {$\sqrt 2$ + $1/n$ : $n$ $∈$ $\mathbb{N}$} is closed. I'm having trouble. I've been trying to show that the complement is open, but the presence of {$\sqrt 2$} is confusing me.</p>
| 5xum | 112,884 | <p>Let's call your set $A$.</p>
<hr>
<p>The easiest way to show that $A$ is closed is by seeing that the complement of $A$ is the set</p>
<p>$$(-\infty, \sqrt{2})\cup (\sqrt 2+1,\infty)\cup \bigcup_{n=1}^\infty \left(\sqrt 2+ \frac{1}{n+1}, \sqrt 2 + \frac1n\right)$$</p>
<p>which is pretty clearly open. Of course, you still have to show that what I wrote is in fact the complement of $A$, but that part is also not difficult.</p>
<hr>
<p>Another simple way to show $A$ is closed is by seeing that the only limit point of $A$ is $\sqrt{2}$, and since $\sqrt2\in A$, $A$ contains all limit points and is closed.</p>
<p>You still have to prove that $\sqrt 2$ is the only limit point, and while this isn't entirely trivial, it is not difficult.</p>
<hr>
<p>However, if you want to go the long way round, you can take any $x$ from the complement. Then you have three options:</p>
<ol>
<li>$x<\sqrt{2}$. In this case, you can easily find some $\epsilon$ such that $(x-\epsilon, x+\epsilon)$ does not intersect with your set.</li>
<li>$x>\sqrt 2 + 1$. Similar case as above.</li>
<li>$\sqrt 2 < x < \sqrt 2 + 1$. In this case, there exists some $n\in\mathbb N$ such that $\sqrt 2 + \frac 1{n+1} < x < \sqrt 2 + \frac 1n$ (<strong>WHY?</strong>). Then there exists some $\epsilon$ for which $\sqrt{2}+\frac1{n+1} < x-\epsilon < x+\epsilon < \sqrt 2 + \frac 1n$ (you can calculate it directly from $x$!) and you are done.</li>
</ol>
|
1,511,733 | <p>B = matrix given below. I is identity matrix.</p>
<pre><code> [1 2 3 4]
[3 2 4 3]
[1 3 2 4]
[5 4 3 7]
</code></pre>
<p>So What will be the relation between the matrices A and C if AB = I and BC = I?</p>
<p>I think that A = C because both AB and BC have B in common and both of their product is an identity matrix but I'm not sure. </p>
| egreg | 62,967 | <p>Suppose $AB=I$ and $BC=I$; then
$$
A=AI=A(BC)=(AB)C=IC=C
$$</p>
|
4,274,085 | <p>A linear equation in one variable <span class="math-container">$x$</span>, <span class="math-container">$$ax+b=k$$</span> has only one non-negative integer solution. For example, <span class="math-container">$2x+3=85$</span> has a solution 41.</p>
<p>How to find the number of non-negative integer solutions of a linear equation in two variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, for example, <span class="math-container">$5x+3y=100$</span> and in general, <span class="math-container">$$ax+by=k$$</span></p>
<p>By brute-force search, I found 7 non-negative integer solutions of <span class="math-container">$5x+3y=100$</span>. They are <span class="math-container">$(2, 30), (5, 25), (8, 20), (11, 15), (14, 10), (17, 5), \text{ and }(20, 0)$</span>.</p>
<p><em>P.S.</em> I came to know that a linear equation in two variables is also know as linear Diophantine equation. I also came across <a href="https://en.wikipedia.org/wiki/Ku%E1%B9%AD%E1%B9%ADaka" rel="nofollow noreferrer">Kuṭṭaka</a>, Aryabhata's algorithm for solving linear Diophantine equations in two variables.</p>
| Chris Langfield | 978,915 | <p>Given <span class="math-container">$a, b, k$</span>, your linear equation defines a line on the plane: <span class="math-container">$y = k - \frac{a}{b} x$</span></p>
<p>The solutions will be limited to <span class="math-container">$x$</span> less than the <span class="math-container">$x$</span>-intercept, and <span class="math-container">$y$</span> less than the <span class="math-container">$y$</span> intercept, i.e. <span class="math-container">$0 < x < \frac{bk}{a}$</span> and <span class="math-container">$0 < y < k$</span>. So your answer will be bounded by <span class="math-container">$\mathrm{min} ( \lfloor \frac{bk}{a}\rfloor, \lfloor k \rfloor$</span>) if both intercepts are positive, and zero otherwise. This reduces the search space for brute force. I'm not sure if there is an analytic solution.</p>
|
335,483 | <p>Let $N$ be a set of non-negative integers. Of course we know that $a+b=0$ implies that $a=b=0$ for $a, b \in N$.</p>
<p>How do (or can) we prove this fact if we don't know the subtraction or order?</p>
<p>In other words, we can only use the addition and multiplication.</p>
<p>Please give me advise.</p>
<p>EDIT</p>
<p>The addition law mean that for $a, b \in N$, there is an element $a+b$ in $N$ and this operation is associative.
The multiplication law means that for $a, b \in N$, there is an element $ab$ in $N$ and this operation is associative.
Also the distribution laws hold.</p>
<hr>
<p>EDIT2</p>
<p>Let me rephrase the question since I don't want arguments on orders.</p>
<p>Let $N$ be a set with operation $+$ and $\times$.</p>
<p>$N$ is a monoid with the operation $+$ and $\times$ respectively. There is an unit element $0\in N$.</p>
<p>The distribution laws hold as in the case of the set of integers.</p>
<p>Can we prove the fact above with this assumption?</p>
| marty cohen | 13,079 | <p>What is the addition law?</p>
<p>If it the one from Peano arithmetic,
it is
$x+0=x$ and $x+S(y) = S(x+y)$,
where $S(x)$ is the successor of $x$.</p>
<p>If $x+y=0$,
suppose $y$ was not zero.
Then there is a $z$ such that
$S(z) = y$.</p>
<p>Then $0 = x+y = x+S(z) = S(x+z)$
which is a contradiction, since
$0$ is not the successor of anything.</p>
<p>Therefore $y=0$.
Substituting this in $x+y=0$
and using $x+0=x$, we get $x=0$.</p>
<p>Is this what you wanted?</p>
|
3,156,962 | <blockquote>
<p>Find general term of <span class="math-container">$1+\frac{2!}{3}+\frac{3!}{11}+\frac{4!}{43}+\frac{5!}{171}+....$</span></p>
</blockquote>
<p>However it has been ask to check convergence but how can i do that before knowing the general term. I can't see any pattern,comment quickly!</p>
| John Omielan | 602,049 | <p>As you've already shown, factoring out the cube of the GCD of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> gives a new equation</p>
<p><span class="math-container">$$e = \gcd\left(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2 \right) \tag{1}\label{eq1}$$</span></p>
<p>where</p>
<p><span class="math-container">$$\gcd\left(\alpha,\beta \right) = 1 \tag{2}\label{eq2}$$</span></p>
<p><strong>Update</strong>: Here is a simpler solution than what I originally wrote. First, note that no factor of <span class="math-container">$e$</span> may divide <span class="math-container">$\alpha$</span> or <span class="math-container">$\beta$</span>. If any do, let's say <span class="math-container">$\alpha$</span>, then it must divide <span class="math-container">$\beta^3 - 3\beta\alpha^2$</span> and, thus, must divide <span class="math-container">$\beta^3$</span>, which is not possible due to \eqref{eq2}. Thus, from the first term of \eqref{eq1}, since <span class="math-container">$\alpha^3-3\alpha \beta^2 = \alpha\left(\alpha^2 - 3\beta^2\right)$</span>, this means that <span class="math-container">$e \mid \alpha^2 - 3\beta^2$</span>. Similarly, for the second term, <span class="math-container">$\beta^3-3\beta \alpha^2 = \beta\left(\beta^2 - 3\alpha^2\right)$</span> gives that <span class="math-container">$e \mid \beta^2 - 3\alpha^2$</span>. Also, <span class="math-container">$e$</span> must divide any linear combination of these values, including <span class="math-container">$e | \alpha^2 - 3\beta^2 + 3\left(\beta^2 - 3\alpha^2\right) = -8\alpha^2$</span>. Thus, <span class="math-container">$e$</span> can only be a power of <span class="math-container">$2$</span>. To finish this off, go to the second last paragraph. Otherwise, for the rest of the original, longer solution, continue reading.</p>
<p><span class="math-container">$ $</span></p>
<p>Next, note that if <span class="math-container">$f = \gcd(g,h)$</span>, then <span class="math-container">$f$</span> divides <span class="math-container">$g$</span> and <span class="math-container">$h$</span> and, thus, will also divide any linear combination of <span class="math-container">$g$</span> and <span class="math-container">$h$</span>, including their sum & difference. From \eqref{eq1}, first check the sum of the <span class="math-container">$2$</span> inside values:</p>
<p><span class="math-container">\begin{align}
\alpha^3-3\alpha \beta^2 + \beta^3-3\beta \alpha^2 & = \alpha^3 + \beta^3 -3\left(\alpha \beta\right)\beta - 3\left(\alpha \beta\right)\alpha \\
& = \left(\alpha + \beta\right)\left(\alpha^2 - \alpha\beta + \beta^2\right) - 3\alpha\beta\left(\alpha + \beta\right) \\
& = \left(\alpha + \beta\right)\left(\alpha^2 - 4\alpha\beta + \beta^2\right) \tag{3}\label{eq3}
\end{align}</span></p>
<p>Suppose there's a factor <span class="math-container">$m \gt 1$</span> which divides <span class="math-container">$e$</span> and <span class="math-container">$\alpha + \beta$</span>. Then, <span class="math-container">$\alpha \equiv -\beta \pmod m$</span>, so <span class="math-container">$\alpha^3-3\alpha \beta^2 \equiv 2\beta^3 \pmod m$</span>. From \eqref{eq2}, this means that <span class="math-container">$m = 2$</span>, and that any other factors of <span class="math-container">$e$</span> must divide <span class="math-container">$\alpha^2 - 4\alpha\beta + \beta^2$</span>.</p>
<p>From \eqref{eq1}, next check the difference of the <span class="math-container">$2$</span> inside values:</p>
<p><span class="math-container">\begin{align}
\alpha^3-3\alpha \beta^2 - \beta^3 + 3\beta \alpha^2 & = \alpha^3 - \beta^3 -3\left(\alpha \beta\right)\beta + 3\left(\alpha \beta\right)\alpha \\
& = \left(\alpha - \beta\right)\left(\alpha^2 + \alpha\beta + \beta^2\right) + 3\alpha\beta\left(\alpha - \beta\right) \\
& = \left(\alpha - \beta\right)\left(\alpha^2 + 4\alpha\beta + \beta^2\right) \tag{4}\label{eq4}
\end{align}</span></p>
<p>Suppose there's a factor <span class="math-container">$n \gt 1$</span> which divides <span class="math-container">$e$</span> and <span class="math-container">$\alpha - \beta$</span>. Then, <span class="math-container">$\alpha \equiv \beta \pmod n$</span>, so <span class="math-container">$\alpha^3-3\alpha \beta^2 \equiv -2\beta^3 \pmod m$</span>. From \eqref{eq2}, this means that <span class="math-container">$n = 2$</span>, and that any other factors of <span class="math-container">$e$</span> must divide <span class="math-container">$\alpha^2 + 4\alpha\beta + \beta^2$</span>.</p>
<p>This shows that any factor, other than <span class="math-container">$2$</span>, which divides <span class="math-container">$e$</span> must divide both <span class="math-container">$\alpha^2 - 4\alpha\beta + \beta^2$</span> and <span class="math-container">$\alpha^2 + 4\alpha\beta + \beta^2$</span>. Thus, it must also divide their difference, which is <span class="math-container">$8\alpha\beta$</span>. This can only be true for <span class="math-container">$2$</span>, <span class="math-container">$4$</span> or <span class="math-container">$8$</span>.</p>
<p>At this shows overall, only powers of <span class="math-container">$2$</span> may possibly divide <span class="math-container">$e$</span>. Since <span class="math-container">$e$</span> is relatively prime to <span class="math-container">$\alpha$</span> & <span class="math-container">$\beta$</span>, this means they must both be odd. From <span class="math-container">$\alpha^3 - 3\alpha\beta^2 = \alpha\left(\alpha^2 - 3\beta^2\right)$</span>, note that <span class="math-container">$\alpha^2 \equiv \beta^2 \equiv 1 \pmod 4$</span>, so <span class="math-container">$\alpha^2 - 3\beta^2 \equiv -2 \pmod 4$</span>. In other words, there will only be <span class="math-container">$1$</span> factor of <span class="math-container">$2$</span>.</p>
<p>In summary, with your original equation of <span class="math-container">$d = \gcd\left(a,b\right)$</span>, we get that <span class="math-container">$\gcd\left(a^3-3ab^2, b^3-3ba^2\right)$</span> is <span class="math-container">$2d^3$</span> if both <span class="math-container">$\frac{a}{d}$</span> and <span class="math-container">$\frac{b}{d}$</span> are odd, else it's <span class="math-container">$d^3$</span>.</p>
|
3,155,463 | <blockquote>
<p><span class="math-container">$$
\lim _{n \rightarrow \infty}\left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^{n}\right] \text { equals }\_\_\_\_
$$</span></p>
</blockquote>
<p>I tried to expand the term in power using binomial theorem but still could not obtain the limit. </p>
| MarcM | 655,795 | <p>The problem is that you cannot use Binomial Theorem because you cannot control the behaviour of <span class="math-container">$\binom{n}{k}\frac{1}{n^k}$</span> when <span class="math-container">$n$</span> and <span class="math-container">$k$</span> increase towards infinity. One must expand <span class="math-container">$(1+\frac{1}{n})^n$</span> using Taylor Series, it's more straight to the point.</p>
|
98,196 | <p>Ok, so I am running a classic linear regression where betahat = (X'X)^-1X'y</p>
<p>Due to performance issues, I would like to estimate betahat with an additional data point (x1,x2,x3,x4,...,y) without recalculating based on the whole history. </p>
<p>Could I do some sort of multiplication based on Xmu or std deviation of the variables? And then what is the probability that betahat is predicting the true beta from this point? This would be used so that I can judge when I need to do a full recalculation.</p>
<p>Or am I thinking about this entirely wrong?</p>
<p>Thank you for any help or suggestions,
Jeremy</p>
| Brian Borchers | 9,022 | <p>There's a very large literature on updating solutions to least squares problems as new data are added. The naive formula $\hat{\beta}=(X^{T}X)^{-1}X^{T}y$ can be problematic in practice because of numerical issues with ill-conditioning. The QR factorization is typically a much better choice in practice. There are lots of papers on updating the QR factorization as data are added. See for example:</p>
<p><a href="http://eprints.ma.man.ac.uk/1192/01/covered/MIMS_ep2008_111.pdf" rel="nofollow">http://eprints.ma.man.ac.uk/1192/01/covered/MIMS_ep2008_111.pdf</a></p>
|
98,196 | <p>Ok, so I am running a classic linear regression where betahat = (X'X)^-1X'y</p>
<p>Due to performance issues, I would like to estimate betahat with an additional data point (x1,x2,x3,x4,...,y) without recalculating based on the whole history. </p>
<p>Could I do some sort of multiplication based on Xmu or std deviation of the variables? And then what is the probability that betahat is predicting the true beta from this point? This would be used so that I can judge when I need to do a full recalculation.</p>
<p>Or am I thinking about this entirely wrong?</p>
<p>Thank you for any help or suggestions,
Jeremy</p>
| user44099 | 44,099 | <p>You might look into the Kalman filter <a href="http://www.lce.hut.fi/~ssarkka/course_k2012/handout2.pdf" rel="nofollow">http://www.lce.hut.fi/~ssarkka/course_k2012/handout2.pdf</a>
to do the update.</p>
|
877,687 | <p>So my textbook's explanation of the derivative of e is very sketchy. They used lots of approximations and plugging things into the calculator. Basically I want to know how you can work out as h approaches 0</p>
<p>$$
\lim_{h\to0}\frac {10^{x+h}-10^x }h
$$</p>
| mvw | 86,776 | <p>We try to model the family of lines and then try to infer the envelope.</p>
<p>The guiding lines (left and right arms of the V shape) are
$$
g(t) = u_g \, (1-t) + v_g \, t \quad h(t) = u_h \, (1-t) + v_h \, t
$$
for $t \in [0, 1]$, where $u$ is the start point and $v$ the end point of that line.</p>
<p>A line $f_r$ of the family starts on $g$ and ends on $h$:
$$
f_r(t) = g(r) \, (1-t) + h(r) \, t
$$
again with the parameter $t \in [0, 1]$.</p>
<p>The envelope is approximated by points where two of the family lines intersect:
$$
\begin{align}
f_r(t) &= f_s(t') \iff \\
g(r) \, (1-t) + h(r) \, t &= g(s) \, (1-t') + h(s) \,t'
\end{align}
$$</p>
<p>This gives the matrix equation
$$
\left[
\begin{matrix}
h(r) - g(r) & -(h(s) - g(s))
\end{matrix}
\right]
\left[
\begin{matrix}
t \\
t'
\end{matrix}
\right]
=
g(s) - g(r) \quad (*)
$$</p>
<p><strong>Example:</strong>
$$
u_g =
\left[
\begin{matrix}
-1 \\
1
\end{matrix}
\right]
\quad
v_g =
\left[
\begin{matrix}
0 \\
-1
\end{matrix}
\right]
\quad
u_h =
\left[
\begin{matrix}
0 \\
-1
\end{matrix}
\right]
\quad
v_h =
\left[
\begin{matrix}
1 \\
1
\end{matrix}
\right]
$$</p>
<p>This gives
$$
g(r) =
\left[
\begin{matrix}
r - 1 \\
1 - 2r
\end{matrix}
\right]
\quad
h(r) =
\left[
\begin{matrix}
r \\
2r - 1
\end{matrix}
\right]
\quad
g(s) =
\left[
\begin{matrix}
s - 1 \\
1 - 2s
\end{matrix}
\right]
\quad
h(s) =
\left[
\begin{matrix}
s \\
2s - 1
\end{matrix}
\right]
$$
and thus for two different family lines ($r \ne s$):
$$
\left[
\begin{matrix}
1 & -1 \\
4r-2 & 2 - 4s
\end{matrix}
\right]
\left[
\begin{matrix}
t \\
t'
\end{matrix}
\right]
=
\left[
\begin{matrix}
s - r \\
2 (r - s)
\end{matrix}
\right]
$$
The first row gives $t' = t + r - s$ and the second row
$$
2(r-s) = (4r-2)t + (2-4s)(t+r-s) = 4(r-s) t + (2-4s)(r-s) \iff \\
4s(r-s) = 4(r-s) t \iff \\
t = s
$$</p>
<p>and thus $t' = r$. The intersection point results to
$$
p(r,s) =
f_r(s) =
f_s(r)
$$
however we want $r \to s$ and thus:
$$
p(r) =
f_r(r) =
\left[
\begin{matrix}
(r-1)(1-r) + r^2 \\
(1-2r)(1-r) + (2r-1) r
\end{matrix}
\right]
=
\left[
\begin{matrix}
2r - 1 \\
4r^2 -4r + 1
\end{matrix}
\right]
$$
Substituting $x = 2 r - 1 \iff r = (x + 1) / 2$ and remembering $r \in [0, 1]$ we get
$$
p(x) =
\left[
\begin{matrix}
x \\
4\left(\frac{x+1}{2}\right)^2 -4 \frac{x+1}{2} + 1
\end{matrix}
\right]
=
\left[
\begin{matrix}
x \\
x^2
\end{matrix}
\right]
$$
which is the graph of a parabola.</p>
<p><img src="https://i.stack.imgur.com/POrFq.png" alt="Graph of the parabola as parametric plot"></p>
<p><strong>Note:</strong></p>
<p>I picked the example to correspond to a nice symmetric V shape.
Equation (*) holds for general situations, e.g. a V with a longer and a shorter arm. In those cases, other envelopes, if at all, arise!</p>
<p>E.g. it is possible to arrange the guiding lines as a | | shape, where the result should look like a butterfly or a (boring) stripes pattern, depending on the relative orientations of the guiding lines. </p>
|
1,610,616 | <blockquote>
<p>$6$ letters are to be posted in three letter boxes.The number of ways
of posting the letters when no letter box remains empty is?</p>
</blockquote>
<p>I solved the sum like dividing into possibilities $(4,1,1),(3,2,1)$ and $(2,2,2)$ and calculated the three cases separately getting $90,360$ and $90$ respectively.</p>
<p>What I wanted to know is there any shortcut method to solve this problem faster?Can stars and bars be used?How?</p>
<p>Or can the method of coefficients be used like say finding coefficient of $t^6$ in $(t+t^2+t^3+t^4)^3$.Will that method work here?</p>
| Henry | 6,460 | <p><a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow">Stirling numbers of the second kind</a>: your particular example of six letters and 3 letterboxes has a solution of $3! \,S_2(6,3) = 6 \times 90 =540$ possible surjections, the same as the result you found</p>
<p>Using exponential generating functions, you could find the coefficient of $x^6$ in the expansion of $(e^x-1)^3$ and multiply that by $6!$ to get $\frac34 \times 720= 540$ </p>
|
1,518,697 | <blockquote>
<p>For this homework exercise, we are asked to show that the ideal $I=(3,1+\sqrt{-5})$ is a flat $\mathbb{Z}[\sqrt{-5}]$-module. The hint is to show that $I$ becomes principal (and thus free as a module) when we invert $2$ <strong>or</strong> $3$, so that $I$ is locally flat.</p>
</blockquote>
<p>I'm having trouble understanding what happens to $I$ when we invert $2$ <strong>or</strong> $3$. I would say that if we invert $3$, then $3$ becomes a unit. How does this make $I$ principal? If we invert $2$, I don't see how this gets us anywhere.</p>
<p>Furthermore, I don't understand why it helps to prove that $I$ is locally flat. There is a lemma in our course notes saying that if $M$ is a finitely generated $R$-module in a local ring $R$, then $M$ is flat iff $M$ is free. So if $I$ is locally flat, why is it flat as well?</p>
<p>I hope my confusion is coming across. <strong>Any</strong> help relating to these questions is appreciated, and if anything is unclear, please tell me.</p>
| B. Pasternak | 283,676 | <p><em>So, a couple days later, a little wiser. Here it goes.</em></p>
<p>Indeed, as Gerry suggested, inverting $(2)$ makes $I$ equal to $(\tfrac{1}{2}(1+\sqrt{-5}))$ and inverting $(3)$ makes $I$ equal to the whole ring since $I$ contains a unit, so in both cases $I$ is principal and thus free as a module.</p>
<p><strong>Theorem: Let $R$ be a ring and $M$ a finitely generated $R$-module. Then $M$ is flat if and only if for every prime ideal $\mathfrak{p}\subset R$, the localization $M_{\mathfrak{p}}$ is free as an $R_{\mathfrak{p}}$-module.</strong></p>
<p>Note that either $(2)$ <strong>or</strong> $(3)$ is not contained in every prime ideal of $\mathbb{Z}[\sqrt{-5}]$, so for every prime ideal $\mathfrak{p}$ we have inverted either $2$ or $3$ in $R_{\mathfrak{p}}$. This implies that $M_{\mathfrak{p}}$ is free as an $R_{\mathfrak{p}}$-module for every prime ideal of $R$, so by the theorem, $M$ is flat.</p>
|
3,256,881 | <p>How to prove that?
I try to use the comparasion test, but i don't know with that function compare.</p>
| azif00 | 680,927 | <p><span class="math-container">$$-1\le \sin\left(x+\frac{1}{x}\right)\le1$$</span>
And, therefore
<span class="math-container">$$-1\le \int_0^1\sin\left(x+\frac{1}{x}\right)\mathrm{d}x\le1$$</span></p>
|
3,501,052 | <p>I want to find the number of real roots of the polynomial <span class="math-container">$x^3+7x^2+6x+5$</span>.
Using Descartes rule, this polynomial has either only one real root or 3 real roots (all are negetive). How will we conclude one answer without doing some long process?</p>
| Community | -1 | <h2>Use the Discriminant formula for a 3rd ordered polynomial.Click <a href="https://en.wikipedia.org/wiki/Discriminant" rel="nofollow noreferrer" title="this text appears when you mouse over">here</a>!</h2>
<p><strong>N.B</strong> <em>The discriminant is zero if and only if at least two roots are equal. If the coefficients are real numbers, and the discriminant is not zero, the discriminant is positive if the roots are three distinct real numbers, and negative if there is one real root and two complex conjugate roots</em></p>
<p><strong>Coming to the ans of your question as discriminant is coming negative , hence there is one real root and two complex conjugate roots</strong></p>
|
2,869,031 | <p>Let $A$ a domain, i.e. $ab\in A\implies a=0$ or $b=0$. It's written that all domains are commutative. Is it by definition, or can we prove that domains are commutative? I mean, do we only consider domaind for commutative rings, or is a ring that is a domain then commutative?</p>
| quid | 85,306 | <p>It is "by definition" though, as the other answer indicates, some might disagree with the terminology. </p>
<p>It is perfectly possible to have a non-commutative ring that has no zero-divisors (except for $0$). Examples would be, for example, non-commutative <a href="https://en.wikipedia.org/wiki/Division_ring" rel="noreferrer">division rings</a>, also called skew fields. (These are rings where every non-zero element has a multiplicative inverse.)</p>
<p>As an aside, for division rings there is an interesting result that all finite division rings in fact are commutative. I mention this as an example where commutativity does follow from at first glance unrelated properties. </p>
<p>Maybe even more to the point, and as mentioned on the page lined in the other answer, as in a finite ring (with identity) without non-trivial zero-divisors every non-zero element has an inverse for finite rings indeed commutativity follows from the property you mentioned. </p>
|
941,632 | <p>Is the Set $$S=\{e^{2x},e^{3x}\}$$ linearly independent?? And answer says Linearly independent over any interval $(a,b)$,only when $0$ doesnot belong to $(a,b)$</p>
<p>How do I proceed??</p>
<p>Thanks for the help!!</p>
| tattwamasi amrutam | 90,328 | <p>Consider $$c_1e^{2x}+c_2e^{3x}=0$$. Then Differentiating once you will get $$2c_1e^{2x}+3c_2e^{3x}=0$$. Again differentiate it to get $$4c_1e^{2x}+9c_2e^{3x}=0$$. Plug in a value $x_0$ such that $x_0$ belongs to $(a,b)$. Then the equation reduces to </p>
<p>$$2c_1e^{2x_0}+3c_2e^{3x_0}=0$$ and $$4c_1e^{2x_0}+9c_2e^{3x_0}=0$$. This is a linear system of equations with determinant = $6e^{5x_0}$ which is never zero. Hence $c_1=c_2=0$.</p>
|
941,632 | <p>Is the Set $$S=\{e^{2x},e^{3x}\}$$ linearly independent?? And answer says Linearly independent over any interval $(a,b)$,only when $0$ doesnot belong to $(a,b)$</p>
<p>How do I proceed??</p>
<p>Thanks for the help!!</p>
| Belgi | 21,335 | <p>Hint: Consider the Wronskian.
For more details see <a href="http://en.wikipedia.org/wiki/Wronskian" rel="nofollow">http://en.wikipedia.org/wiki/Wronskian</a></p>
|
4,518,908 | <p>For sufficiently large integer <span class="math-container">$m$</span>, in order to prove</p>
<p><span class="math-container">$\frac{(m+1)}{m}<\log(m)$</span></p>
<p>is it sufficient to point out that</p>
<p><span class="math-container">$ \displaystyle\lim_{m \to \infty} \frac{(m+1)}{m}=1 $</span></p>
<p>while</p>
<p><span class="math-container">$ \displaystyle\lim_{m \to \infty} \log(m)=\infty $</span>?</p>
| ryang | 21,813 | <blockquote>
<p>Suppose that <span class="math-container">$$\forall x {\in} \mathbb{R} \begin{bmatrix} P(x) \implies \begin{bmatrix} \forall y {\in}\mathbb{R} \;P(x + y) \end{bmatrix} \end{bmatrix}.$$</span> Is this sufficient to conclude that <span class="math-container">$$\begin{bmatrix}\forall x {\in}\mathbb{R}\; P(x)\end{bmatrix} \text{ or } \begin{bmatrix}\forall x {\in}\mathbb{R}\;\lnot P(x)\end{bmatrix}?$$</span></p>
</blockquote>
<p>Yes, this argument is valid. In words:</p>
<p>For each number that satisfies P, P continues to be true however we modify that number.</p>
<p>Therefore, if some number satisfies P, then every number satisfies P; on the other hand, it may be that no number satisfies P.</p>
<p>Hence, P must either be universally true or be universally false.</p>
|
273,308 | <p>Find the standard form of the conic section $x^2-3x+4xy+y^2+21y-15=0$.</p>
<p>I understand the approach in trying to solve these problems. But the $4xy$ is confusing me. I am not sure of where to start on this one.</p>
<p><strong>UPDATE:</strong></p>
<p>i have found a way of doing this with reference to <a href="https://math.stackexchange.com/questions/195690/canonical-form-of-conic-section?rq=1">Canonical form of conic section</a></p>
<p><strong>Solution:</strong></p>
<p>$x^2-3x+4xy+y^2+21y-15=(x+2y)^2-3y^2-3x+21y$</p>
<p>Let: </p>
<p>$X = x+2y$</p>
<p>$Y=y$</p>
<p>$x = X - 2Y$</p>
<p>Now we substitute:</p>
<p>$X^2-3Y^2-3X+27Y-15=0$</p>
<p>so it is a hyperbola?</p>
<p>Please check.</p>
| lab bhattacharjee | 33,337 | <p>So, $$m=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot5}}2=-2\pm i$$</p>
<p>Then as the roots are unequal, $$y=Ae^{(-2+i)x}+Be^{(-2-i)x}$$ where $A,B$ are arbitrary constants.</p>
<p>$$y=e^{-2x}(Ae^{ix}+Be^{-ix})$$
$$=e^{-2x}\{(A+B)\cos x+i(A-B)\sin x\}$$ using <a href="http://mathworld.wolfram.com/EulerFormula.html" rel="nofollow">Euler's identity</a>.</p>
<p>or, $$y=e^{-2x}(c_1\cos x+c_2\sin x)$$ where $c_1=A+B,c_2=i(A-B)$ are arbitrary constants.</p>
|
273,308 | <p>Find the standard form of the conic section $x^2-3x+4xy+y^2+21y-15=0$.</p>
<p>I understand the approach in trying to solve these problems. But the $4xy$ is confusing me. I am not sure of where to start on this one.</p>
<p><strong>UPDATE:</strong></p>
<p>i have found a way of doing this with reference to <a href="https://math.stackexchange.com/questions/195690/canonical-form-of-conic-section?rq=1">Canonical form of conic section</a></p>
<p><strong>Solution:</strong></p>
<p>$x^2-3x+4xy+y^2+21y-15=(x+2y)^2-3y^2-3x+21y$</p>
<p>Let: </p>
<p>$X = x+2y$</p>
<p>$Y=y$</p>
<p>$x = X - 2Y$</p>
<p>Now we substitute:</p>
<p>$X^2-3Y^2-3X+27Y-15=0$</p>
<p>so it is a hyperbola?</p>
<p>Please check.</p>
| Community | -1 | <p>If you want all solution real: </p>
<p>$m^2+4m+5=0$ $\implies$ $m=-2\pm i$.</p>
<p>Then $e^{(-2+i)t}=e^{-2t}(\cos{t}+i\sin{t}) $ and $e^{(-2-i)t}=e^{-2t}(\cos{t}-i\sin{t})$ are two solutions.</p>
<p>Clearly linear combinations these solutions still are solutions. Then</p>
<p>$\phi(t)=\dfrac{e^{(-2+i)t}+e^{(-2-i)t}}{2}=\dfrac{e^{-2t}(\cos{t}+i\sin{t})+e^{-2t}(\cos{t}-i\sin{t})}{2}=e^{-2t}\cos{t}$</p>
<p>and </p>
<p>$\xi(t)==\dfrac{e^{(-2+i)t}-e^{(-2-i)t}}{2i}=\dfrac{e^{-2t}(\cos{t}+i\sin{t})-e^{-2t}(\cos{t}-i\sin{t})}{2i}=e^{-2t}\sin{t}$</p>
<p>We have $\phi$ and $\xi$ are two solutions linearly independent. Therefore</p>
<p>all real solution are $C_1e^{-2t}\cos{t}+C_2e^{-2t}\sin{t}$ for $C_1,C_2\in \mathbb R$ </p>
|
3,777,042 | <p>What is the particular solution to <span class="math-container">$ \frac{dy}{dx} =\cos(x^2)$</span> with the initial condition <span class="math-container">$y\left( \sqrt{\frac{\pi}{2}}\right)=3$</span>?</p>
<p>A.) <span class="math-container">$y = 3 + \int_{0}^{x}\cos(t^2)dt$</span></p>
<p>B.) <span class="math-container">$y = 3 + \int_{\sqrt \frac{\pi}{2}}^{x}\cos(t^2)dt$</span></p>
<p>C.) <span class="math-container">$y= \frac{\sin(x^2)}{2x} + 3 - \frac{1}{2\sqrt \frac{\pi}{2}}$</span></p>
<p>D.) <span class="math-container">$y= \sin(x^2) + 2 $</span></p>
<p>So, I thought that the answer was D, as that looks like the derivative of that will be the equation that's given in the question. That's not correct though, and the integral of <span class="math-container">$\cos x^2$</span> looks like it should be sin<span class="math-container">$x^2$</span>, but as said, D is not correct. Very confused on what to do.</p>
| Community | -1 | <p>I'll give a outline about it problem.
<span class="math-container">$$\frac{dy}{dx}=\cos(x^2) \implies y=\int \cos(x^2)dx +K$$</span>
But we know that <span class="math-container">$$\int \cos(x^2)dx=\sqrt{\frac{\pi}{2}}C\left( \sqrt{\frac{2}{\pi}}x\right)+K$$</span> where <span class="math-container">$C(x)$</span> is the Fresnel <span class="math-container">$C$</span> integral.
Also we know that <span class="math-container">$y\left(\sqrt{\frac{\pi}{2}}\right)=3$</span>, so <span class="math-container">$$K=3-
\sqrt{\frac{\pi}{2}}C\left( \sqrt{\frac{2}{\pi}}\sqrt{\frac{\pi}{2}}\right)=3-\sqrt{\frac{\pi}{2}}\int_{0}^{1}\cos \left( \frac{\pi x^{2}}{2}\right) dx$$</span>
Finally <span class="math-container">$$\boxed{y=\sqrt{\frac{\pi}{2}}C\left( \sqrt{\frac{2}{\pi}}x\right)+3-\sqrt{\frac{\pi}{2}}\int_{0}^{1}\cos \left( \frac{\pi x^{2}}{2}\right) dx}$$</span>I'll finish the details and the final answer later.</p>
|
3,777,042 | <p>What is the particular solution to <span class="math-container">$ \frac{dy}{dx} =\cos(x^2)$</span> with the initial condition <span class="math-container">$y\left( \sqrt{\frac{\pi}{2}}\right)=3$</span>?</p>
<p>A.) <span class="math-container">$y = 3 + \int_{0}^{x}\cos(t^2)dt$</span></p>
<p>B.) <span class="math-container">$y = 3 + \int_{\sqrt \frac{\pi}{2}}^{x}\cos(t^2)dt$</span></p>
<p>C.) <span class="math-container">$y= \frac{\sin(x^2)}{2x} + 3 - \frac{1}{2\sqrt \frac{\pi}{2}}$</span></p>
<p>D.) <span class="math-container">$y= \sin(x^2) + 2 $</span></p>
<p>So, I thought that the answer was D, as that looks like the derivative of that will be the equation that's given in the question. That's not correct though, and the integral of <span class="math-container">$\cos x^2$</span> looks like it should be sin<span class="math-container">$x^2$</span>, but as said, D is not correct. Very confused on what to do.</p>
| Digitallis | 741,526 | <p>For each proposed solution we need to check if <span class="math-container">$\frac{dy}{dx} = \cos(x^2)$</span> and <span class="math-container">$y\left(\sqrt \frac \pi 2 \right) = 3$</span>. The fundamental theorem of calculus will come in handy !</p>
<p>Let <span class="math-container">$x_0 = \sqrt \frac \pi 2$</span></p>
<p>You stated in the comments that you hesitated between <span class="math-container">$A$</span> and <span class="math-container">$B$</span>.</p>
<p>For <span class="math-container">$A$</span> we have</p>
<p><span class="math-container">$$\frac{dy}{dx} = 0 + \cos(x^2), \quad y(x_0) = 3 + \int_0^{\sqrt \frac \pi 2} \;\cos(t^2) dt.$$</span></p>
<p>Since <span class="math-container">$\cos(t^2) > 0$</span> on <span class="math-container">$\left(0,\sqrt \frac \pi 2 \right)$</span> we have</p>
<p><span class="math-container">$$ y(x_0) = 3 + \int_0^{\sqrt \frac \pi 2} \;\cos(t^2) \; dt > 3$$</span></p>
<p>This means that <span class="math-container">$y(x_0) \neq 3$</span> and that <span class="math-container">$A$</span> is not the correct choice.</p>
<p>For <span class="math-container">$B$</span> we have</p>
<p><span class="math-container">$$\frac{dy}{dx} = 0 + \cos(x^2), \quad y(x_0) = 3 + \int_{\sqrt \frac \pi 2}^{\sqrt \frac \pi 2} \;\cos(t^2) dt = 3.$$</span></p>
<p>So <span class="math-container">$B$</span> is what you're looking for. There are no other correct choices since <span class="math-container">$C$</span> and <span class="math-container">$D$</span> don't verify <span class="math-container">$\frac{dy}{dx} = \cos(x^2).$</span></p>
<p>Finally you say that the integral of <span class="math-container">$\cos(x^2)$</span> should be <span class="math-container">$\sin(x^2)$</span> this is not the case. This is a big mistake. It looks like you're doing something like this in your head :</p>
<p>"Let <span class="math-container">$u = x^2$</span> then the integral of <span class="math-container">$\cos(x^2)$</span> is the integral of <span class="math-container">$\cos(u)$</span> which is <span class="math-container">$\sin(u)$</span>. Replace <span class="math-container">$u = x^2$</span> and the integral of <span class="math-container">$\cos(x^2)$</span> is <span class="math-container">$\sin(x^2)$</span> ! "</p>
<p>This is false !!! Why ? By the the same reasoning you could say that the integral of <span class="math-container">$\cos(g(x))$</span> is <span class="math-container">$\sin(g(x))$</span> but by the chain rule</p>
<p><span class="math-container">$$ \frac{d}{dx}\sin(g(x)) = \cos(g(x)) g'(x) \neq \cos(g(x))$$</span></p>
|
2,069,001 | <p>A team of seven netballers is to be chosen from a squad of twelve players A,B,D, E, F, G, H, I, J, K, L. In how many ways can they be chosen:<br>
a) with no restriction
This is fairly easy. 12C7 = 792</p>
<p>b) if the captain C is to be included
11C6 = 462</p>
<p>c) If J and K are both to be excluded
10C7 = 120</p>
<p>d) If A is included but H is not
10C6 = 210</p>
<p>e) if one of F and L is to be included and the other excluded.</p>
<p>This one I'm having trouble with.
I'm not 100% about the others either.</p>
| Robert Z | 299,698 | <p>As you noted in d) if F is included but L is not then the number of teams is $\binom{10}{6}$.</p>
<p>Similarly if L is included but F is not then the number is $\binom{10}{6}$.</p>
<p>These two sets of teams are disjoint therefore the total number is $$\binom{10}{6}+\binom{10}{6}=420.$$.</p>
|
1,204,864 | <blockquote>
<p>$$\text{Find }\,\dfrac{d}{dx}\Big(\cos^2(5x+1)\Big).$$</p>
</blockquote>
<p>I have tried using the rules outlined in my standard derivatives notes but I've failed to find the point of application.</p>
| Jordan Glen | 225,803 | <p>$$y = \cos^2(5x+1)=\Big(\cos(5x+1)\Big)^2$$</p>
<p>$$\frac{dy}{dx} = 2(\cos(5x+1))\cdot (-\sin(5x+1)) \cdot \frac{d}{dx}(5x+1)$$</p>
<p>$$=-10\cos(5x+1) \sin (5x+1)$$
We applied the power-rule first, then used the chain-rule, twice.</p>
|
3,224,102 | <p>For a given curve: <span class="math-container">$$C: \frac {ax^2+bx+c}{dx+e} $$</span> where <span class="math-container">$a,b,c,d,e$</span> are integers. Let <strong><span class="math-container">$f(x)=ax^2+bx+c$</span></strong> .</p>
<hr>
<p>Oblique asymptote can be found by long division of numerator by denominator.Here oblique asymptote is <strong>y=<span class="math-container">$\frac {a}{d}x$</span>+ <span class="math-container">$\frac {b}{d}$</span></strong>. </p>
<hr>
<h2>Now If I were to multiply <strong><span class="math-container">$y$</span></strong> with <strong><span class="math-container">$dx+e$</span></strong>. I will get <strong><span class="math-container">$ax^2+bx+q$</span></strong> , where <strong>q<span class="math-container">$\not=$</span>c</strong> which is close to my f(x) but this doesn't get me to my original <strong>f(x)</strong> where a constant value is different.So what needs to be done to retain exact f(x) part of a curve?Please help!!</h2>
| Claude Leibovici | 82,404 | <p>For large values of <span class="math-container">$x$</span>, write <span class="math-container">$$y=\frac {ax^2+bx+c}{dx+e}=m+n x+\epsilon$$</span> and cross multiply
<span class="math-container">$${ax^2+bx+c}=(m+n x+\epsilon)(d x+e)=d n x^2+x (d m+e n +d \epsilon) +e( m+ \epsilon)$$</span> </p>
<p>Compare the coefficients :
<span class="math-container">$$a=d n\implies n=\frac a d$$</span>
<span class="math-container">$$b=d m+e n+d\epsilon\implies m=\frac{bd-a e}{d^2}-\epsilon $$</span> Since <span class="math-container">$\epsilon$</span> is very small, then the equation of the oblique asymptote is
<span class="math-container">$$y=\frac{bd-a e}{d^2}+\frac a d x$$</span></p>
|
2,485,276 | <p>Is the statement true? if it is, how to prove it?</p>
<p>If $\binom{p}{k} \mod p=0 $ for $k=1,2,..,p−1$ then $p$ is prime.</p>
| Ricardo Largaespada | 305,474 | <p>Hint.$${p \choose k} = \frac{p \cdot (p-1) \cdot (p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$$</p>
|
29,155 | <p>Do we have a pullback operation on singular simplicial chains,that is if f:X-->Y is a continuous map between topological space X and Y,and C is a singular simplicial chain on Y,then do we have a singular simplicial chain on X which is the pullback of C along f?</p>
| Daniel Litt | 6,950 | <p>No, there is a pullback on singular cochains, given by composition.</p>
|
2,210,531 | <blockquote>
<p>Prove that the limit is zero:
$$ F(x, y)= \frac{x^2y^3}{3x^2+ 2y^3}$$</p>
</blockquote>
<p>Definition1. Let $ U ⊂ R^n$ be an open set and letf : $U→R^m$ beafunction with domain U. Let x0 be a vector in U or on the boundary of U. Let b∈Rm. We say that the limit of f as x approaches x0 is b, written
lim f(x) = b, x→x0
provided that for every ε > 0 there is a δ > 0 so that ||f(x)−b|| < ε whenever 0 < ||x − x0|| < δ and x ∈ U.</p>
<p>First:</p>
<p>$ 0\le\sqrt {x^2+y^2} <\delta $ and $ |F(x,y) -0| < \epsilon$</p>
<p>$|\frac{x^2y^3}{3x^2+ 2y^3}| = \frac{x^2|y^3|}{|3x^2+ 2y^3|} = ...$ how can i solve this? Please, help me.</p>
| Mark Viola | 218,419 | <p><strong>HINT:</strong></p>
<p>The limit, $\lim_{(x,y)\to(0,0)}\left(\frac{x^2y^3}{3x^2+2y^3} \right)$,fails to exist.</p>
<p>Examine the limit along the curve described parametrically by $x=t$ and $2y^3=-3t^2+t^a$, $a\ge4$. Then, note that </p>
<p>$$\left|\frac{x^2y^3}{3x^2+2y^3} \right|=\left|\frac{t^{a-2}+3}{2t^{a-4}} \right|$$</p>
<p>What can one conclude as $t\to 0$?</p>
|
178,342 | <p>This is an exercise from Kunen's book.</p>
<p>Write a formula expressing $z=\langle \langle x,y\rangle, \langle u,v\rangle \rangle$ using just $\in$ and $=$.</p>
<p>What I've tried: because $\langle x,y\rangle= \{\{x\},\{x,y\}\}$ and $\langle u,v\rangle= \{\{u\},\{u,v\}\}$, and hence $z=\{\{\langle x,y\rangle\},\{\langle x,y\rangle,\langle u,v\rangle\}\}= \{\{\{\{x\},\{x,y\}\}\} ,\{\{\{x\},\{x,y\}\},\{\{u\},\{u,v\}\}\}\}$. </p>
<blockquote>
<p>So the formula expressing is this: $z_1\in z$, then $z_1= \{\{x\},\{x,y\}\}$ or $z_1= \{\{u\},\{u,v\}\}$</p>
</blockquote>
<p>Am I right? Thanks ahead:)</p>
| Asaf Karagila | 622 | <p>Write a formula:</p>
<p>$$\varphi(u,v,z)=
\forall x\bigg(x\in z\leftrightarrow
\underbrace{\forall y(y\in x\leftrightarrow y=u\lor y=v)}_{\Large x=\{u,v\}}\lor\underbrace{\forall y(y\in x\leftrightarrow y=u)}_{\Large x=\{u\}}\bigg)$$</p>
<p>Note that $\varphi(u,v,z)$ holds if and only if $z=\langle u,v\rangle$.</p>
<p>Now define the following formula:</p>
<p>$$\psi(x,y,u,v,z)=\exists a\exists b(\varphi(x,y,a)\land\varphi(u,v,b)\land\varphi(a,b,z))$$</p>
<p>Namely, $z$ is the ordered pair $\langle a,b\rangle$, and $a,b$ are both the ordered pairs $\langle x,y\rangle,\langle u,v\rangle$ respectively.</p>
<p>Note that not only that you can "expand" $\psi$ by replacing the instances of $\varphi$ by its explicit formulation; you can replace equality by $\in$, and use the axiom of extensionality to prove it is the same thing.</p>
|
483,442 | <p>I am trying to learn about velocity vectors but this word problem is confusing me.</p>
<p>A boat is going 20 mph north east, the velocity u of the boat is the durection of the boats motion, and length is 20, the boat's speed. If the positive y axis represents north and x is east the boats direction makes an angle of 45 degrees. You can compute the components of u by using trig</p>
<p>$$u_1 = 20 \cos 45$$
$$u_2 = 20 \sin 45$$</p>
<p>Why? How did this happen? Why sin and why cos? What does this represent? Why two points? What are these two points? It says that these are $R^2$ which I am not sure what that means and my book does not explain. I think t he R means all real numbers and the squared is referencing 2d maybe so x and y but the book doesn't say so I am not so sure. My book mentions none of these things.</p>
| Caleb Stanford | 68,107 | <p>Hints:</p>
<ul>
<li>In how many ways can you pull 6 socks out, one at a time, so that there are <strong>no</strong> matching pairs among them? (How many choices do you have for the first sock? The second? The third?)</li>
<li>In how many ways total can you pull 6 socks out, one at a time?</li>
<li>What does this mean about the probability that there <strong>is</strong> a matching pair?</li>
</ul>
|
1,279,328 | <p>Can someone please explain to me if a 2/3D Poisson's equation is separable or non separable?
Thank you</p>
| ApproximatelyTrue | 240,066 | <p>For a homogenous linear PDE to be separable, you must be able to write the differential operator as the sum of two or more differential operators involving non-empty, non-overlapping subsets of the variables in the original PDE. Then, if you solve the eigenvalue problem for these differential operators (which should be easier since the number of variables in each has reduced - in many cases, we simply get ODEs), and assuming appropriate conditions (e.g. the differential operators are Hermitian, the eigenfunctions satisfy appropriate completeness relations, etc.), you can write the solution to the original PDE as a linear combination of products of the separated eigenfunctions with appropriate coefficients. Then, the original PDE, together with boundary conditions, constrain the allowed eigenvalues, any constants of integration in the eigenfunctions and the aforementioned coefficients of the linear combination, hopefully giving enough information to specify a unique solution.</p>
<p>For instance, the Laplace equation is separable since we can decompose the Laplacian into the sum of the second derivatives with respect to $x,y,z$ in Cartesian coordinates, or a radial and an angular part in spherical coordinates. In the first case the eigenfunctions are complex exponentials (or, equivalently sines and cosines) in $x,y,z$ separately, whereas in the second case we get eigenfunctions are complex exponentials (or, equivalently sines and cosines) in $r$ and the famous spherical harmonics $Y_{l,m}$ in $\theta,\phi$. The choice of decomposition here is suggested by the boundary conditions - if we know our solution on the walls of a rectangular box, we should probably choose Cartesian coordinates; if we know them on the surface of a sphere, we should probably go with spherical coordinates.</p>
<p>I'm a little unsure what is meant by separable in the case of an inhomogenous linear PDE such as the Poisson equation, but I suppose it just means that the underlying homogenous linear PDE, in this case the Laplace equation is separable: since, in this case, we could expand the inhomogenous term in terms of the eigenfunctions from the homogenous case (in the case of the Laplace equation and Cartesian coordinates, this is just a 3D Fourier series expansion), and then we still get the desired simplification of the problem as we did in the homogenous case.</p>
<p>Thus, I would probably categorise the 3D Poisson equation as separable. I would go ahead and put it into the software like this. You can always go back and check that the differential equation and boundary conditions are indeed satisfied (approximately satisfied, if this is a numerical solver).</p>
|
2,206,938 | <p>Context: <a href="http://www.hairer.org/notes/Regularity.pdf" rel="nofollow noreferrer">http://www.hairer.org/notes/Regularity.pdf</a>, section 4.1 (pages 15-16)</p>
<blockquote>
<p>Define
$$(\Pi_x\Xi^0)(y)=1 \qquad (\Pi_x\Xi)(y)=0 \qquad (\Pi_x\Xi^2)(y)=c$$
and
$$(\Pi^{(n)}_x\Xi^0)(y)=1 \qquad (\Pi^{(n)}_x\Xi)(y)=\sqrt{2c}\sin(nx) \qquad (\Pi^{(n)}_x\Xi^2)(y)=2c\sin^2(nx).$$
As a model, $\Pi^{(n)}$ converges to $\Pi$.</p>
</blockquote>
<p>I don't see how this convergence is supposed to take place. Isn't the limit of $\sin(nx)$, as $n\to \infty$, undefined? </p>
| Mohsen Shahriari | 229,831 | <p>As it's noted by @5xum, the concepts of Lipschitz continuity and Hölder continuity are related to global behaviour of the function, but I suppose you have an intuition about local properties of <span class="math-container">$ f $</span>.
So I give a hint to make a counterexample for a local version of your question.</p>
<p>It's possible that <span class="math-container">$ f $</span> is not Hölder continuous when restricted to any neighborhood of some point. To make a counterexample, you can construct a function that grows faster than any <span class="math-container">$ x ^ \alpha $</span> near <span class="math-container">$ 0 $</span>. Such function can be defined using <span class="math-container">$ C ^ \infty $</span> functions that are not analytic.</p>
<p>If you think this hint in not helpful enough, feel free to ask me to elaborate more.</p>
|
3,916,130 | <p>Assume the weight of a person follows a normal distribution N(71,7). What is the probability of 4 people weighing more than 300kg?</p>
<p>I tried solving this by multiplying the values by 4, so it'd be N(284,28). I converted that into <span class="math-container">$x=284+28z$</span> which lead to <span class="math-container">$z=\frac{x-284}{24}$</span>.
To solve for <span class="math-container">$P(x>300)$</span>, I converted it around to <span class="math-container">$ 1-P(x<300)$</span> <span class="math-container">$$ 1-P\left(\frac{300-284}{24}\right) $$</span>
<span class="math-container">$$ 1-P\left(z<\frac{2}{3}\right) $$</span>
Then i used the normal distribution table and got the following result: <span class="math-container">$1-0.2546 = 0.745$</span>, however I have a suspicion I started from the wrong track and this result isn't correct.</p>
<p>This was an exam question at a college statistics class</p>
| J.G. | 56,861 | <p>Your mistake is that variances add, not standard deviations. If <span class="math-container">$7$</span> is the one-person variance, work with <span class="math-container">$300=284+\sqrt{28}z$</span>; if <span class="math-container">$7$</span> (<span class="math-container">$49$</span>) is the one-person srandard deviation (variance), work with <span class="math-container">$300=284+\sqrt{4\times49}z=284+14z$</span>.</p>
|
167,848 | <p>I am trying to solve numerically an equation and generate some results. I use the following code </p>
<pre><code>u[c_] := (c^(1 - σ) - 1)/(1 - σ)
f[s_] := g s (1 - s/sbar1)
h[s_] := (2 hbar)/(1 + Exp[η (s/sbar - 1)])
co[a_] := ϕ (a^2)/2
ψ[k_] := wbar (ω + (1 - ω) Exp[-γ k])
</code></pre>
<p>The equation I try to solve is the following</p>
<pre><code>adap[k_, s_] := (ρ + δ) u'[f[s] - priceadap δ k] co'[δ k] + ψ'[k] h[s]
</code></pre>
<p>I have the following constant parameter set </p>
<pre><code>paramFinal2 = {σ -> 1.7, ρ -> 0.025, g -> 0.05, sbar -> 10, η -> 11, hbar -> 0.5, priceadap -> 0.0006, γ -> 0.6, χ -> 1000, ϕ -> 0.05, ω -> 0.35, β -> 0.8, δ -> 0.065, sbar1 -> 10, wbar -> 1000};
</code></pre>
<p>So, for different values of $s$, I try to generate the corresponding values of $k$.</p>
<p>For this, I use the following code </p>
<pre><code>tmax1 = 10;
solK[i_] := Solve[adap[k, i] == 0 /. paramFinal2, k];
Table[solK[i], {i, 1, tmax1}];
</code></pre>
<p>Unfortunately, this does not give any result. Mathematica is always on mode "Running...".</p>
<p>P.S I am using Mathematica 9.0</p>
| george2079 | 2,079 | <p>one approach here is to use <code>ContourPlot</code> </p>
<pre><code>p = ContourPlot[(adap[k, s] /. paramFinal2) == 0, {s, 0, 10},
{k, 0, 30}, PlotPoints -> 100]
</code></pre>
<p><a href="https://i.stack.imgur.com/TlfZA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TlfZA.png" alt="enter image description here"></a></p>
<p>interpolate the curve and extract specific values:</p>
<pre><code>g = Interpolation[First@Cases[Normal@p, Line[x_] :> x, Infinity]];
Table[{s, g[s]}, {s, 9}]
</code></pre>
<blockquote>
<p>{{1, 10.7315}, {2, 12.1675}, {3, 12.8475}, {4, 13.1713}, {5,
13.2854}, {6, 13.1591}, {7, 12.7972}, {8, 12.0124}, {9, 10.3233}}</p>
</blockquote>
<p>note <code>ContourPlot</code> only finds points to adequate precison for plotting purposes. If you need better precision use as initial guesses for <code>FindRoot</code>:</p>
<pre><code>Table[{s, k /. FindRoot[adap[k, s] /. paramFinal2, {k, g[s]}]},
{s,9}]
</code></pre>
<blockquote>
<p>{{1, 10.7363}, {2, 12.1674}, {3, 12.8497}, {4, 13.185}, {5,
13.2841}, {6, 13.169}, {7, 12.797}, {8, 12.0139}, {9, 10.3238}}</p>
</blockquote>
|
45,211 | <p>Instances of SAT induce a bipartite graph between clauses vertices and variable vertices, and for planar 3SAT, the resulting bipartite graph is planar. </p>
<p>It would be very convenient if there was a planar layout that had all the variable vertices in one line and all the clause vertices in a straight line. This can't be done because such a graph would be outerplanar, and $K_{2,3}$ isn't. </p>
<p>But maybe a weaker layout is possible. </p>
<blockquote>
<p>Is it possible to lay out any planar
bipartite graph $G = (A \cup B, E)$
such that</p>
<ul>
<li>All vertices of $B$ are on a straight line</li>
<li>A can be partitioned into $A_1 \cup A_2$ such that all vertices of $A_1$
are on a parallel straight line to the
left of $B$, and all vertices of $A_2$
are on a parellel straight line to the
right of $B$.</li>
</ul>
</blockquote>
<p>This seems to relate to <a href="http://mashfiquirabbi.110mb.com/files/thesis.pdf">track drawings of planar graphs</a>. </p>
| Louigi Addario-Berry | 3,401 | <p><b>Edit:</b> When I posted this I was assuming you also wanted a straight-line drawing, which I now realize you did not say. The below relates only to straight-line drawings. </p>
<hr>
<p>This is not possible. The $3$-cube is already a counterexample. Viewing the cube as the Hamming cube, up to symmetries there is only one way to place the middle two layers in the manner you suggest -- one must take $\{100,010,001\}\subset B$, $\{110,011\} \subset A_1$ and $\{101\}\subset A_2$. But then it is impossible to put $111$ in either $A_1$ or $A_2$ without creating crossing edges. </p>
<p>More generally, a counting argument should quite straightforwardly show that for large $n$, the proportion of planar graphs that satisfy your criteria is asymptotically small. (Using the fact that the number of labeled planar graphs on $n$ vertices is asymptotically $n! \cdot (27.22687\ldots)^n$ times lower order terms, which is a result of Gimenez and Noy.)</p>
|
248,182 | <p>Some textbooks I've seen declare inequalities such as $-2>x>2$ to have no solution, or to be ill-defined, which I disagree with. I'm curious to know if anyone else thinks the same.</p>
<p>Inequalities can always be written two ways. For example, $x>2$ is the same as $2<x$. So far as I understand, the same applies to compound inequalities; for example, everyone would regard $-3<x<3$ to be well-defined, and it can be written "backwards" as $3>x>-3$.</p>
<p>When someone interprets $-3<x<3$, upon reflection, it is understood that there is an implicit intersection behind the scenes, as it can be read out-loud as "$-3<x$ and $x<3$." And when they interpret $3>x>-3$, it is the "backwards" version of $-3<x<3$. Both are two different, compact ways of expressing {$ x<3 $} $\cap$ {$ x>-3 $}.</p>
<p>So when I look at an inequality such as $-2>x>2$, I take it to mean there is an implicit union behind the scenes. In other words, $-2>x>2$ and $2<x<-2$ both refer to the same thing, namely {$ x<-2 $} $\cup$ {$ x>2 $}. Were I to read $-2>x>2$ out-loud, I would read it as "$-2>x$ or $x>2$."</p>
<p>Am I crazy, or is there something wrong with this interpretation?</p>
<p>It seems to offer some advantages. For example, it makes the solution of certain absolute value inequalities very easy and natural.</p>
| Michael Hardy | 11,667 | <p>I wouldn't say the system of inequalities $2<x<-2$ is "ill defined", but certainly it has no solutions: there is no number that is both bigger than $2$ and less than $-2$.</p>
|
3,346,676 | <blockquote>
<p><strong>Question.</strong> Find a divergent sequence <span class="math-container">$\{X_n\}$</span> in <span class="math-container">$\mathbb{R}$</span> such that for any <span class="math-container">$m\in\mathbb{N}$</span>,
<span class="math-container">$$\lim_{n\to\infty}|X_{n+m}-X_n|=0$$</span></p>
</blockquote>
<p>I don't really know, if someone could walk me through this it'd be really appreciated.
Edit: I'm dumb af ignore what I said before I deleted it. lmao</p>
| Peter Foreman | 631,494 | <p>Let <span class="math-container">$X_n=\sqrt{n}$</span> then we have
<span class="math-container">$$\begin{align}
\lim_{n\to\infty}(\sqrt{n+m}-\sqrt{n})
&=\lim_{n\to\infty}\left(\sqrt{n}\left(\sqrt{1+\frac{m}n}-1\right)\right)\\
&=\lim_{n\to\infty}\left(\sqrt{n}\left(1+\frac{m}{2n}+o\left(\frac1n\right)-1\right)\right)\\
&=\lim_{n\to\infty}\left(\sqrt{n}\left(\frac{m}{2n}+o\left(\frac1n\right)\right)\right)\\
&=\lim_{n\to\infty}\left(\frac{m}{2\sqrt{n}}+o\left(\frac1{\sqrt{n}}\right)\right)\\
&=0\\
\end{align}$$</span>
and clearly <span class="math-container">$X_n$</span> is divergent. Any strictly increasing concave function should work for <span class="math-container">$X_n$</span>.</p>
|
1,530,057 | <p>At my multivariable calculus class we gave this definition for the limit of a function:</p>
<blockquote>
<p><em>Definition:</em></p>
<p>Let <span class="math-container">$ \mathbb{R}^n \supset A $</span> be a open set , let <span class="math-container">$f:A \to\mathbb{R}^m $</span> be a function, let <span class="math-container">${\bf x_0}$</span> be a point of <span class="math-container">$A$</span> and <span class="math-container">${\bf P}$</span> a point of <span class="math-container">$\mathbb{R}^m$</span>.</p>
<p>To say that <span class="math-container">$f$</span> has limit <span class="math-container">$\bf{P}$</span> at <span class="math-container">$ {\bf x_0} \in A$</span>,</p>
<p> is difined to mean</p>
<p><span class="math-container">$\forall \, \varepsilon>0$</span>, <span class="math-container">$\exists \, \delta(\varepsilon)=\delta >0 : ( \forall \, {\bf x} \in A: \left\lVert {\bf x} - {\bf x_0} \right\rVert_{\mathbb{R}^n}< \delta \Rightarrow \left\lVert f({\bf x}) - {\bf P} \right\rVert_{\mathbb{R}^m}< \varepsilon )$</span></p>
</blockquote>
<p>So I have a question. Why the set <span class="math-container">$A$</span> has to be open? It seems that the problem is the way that the <span class="math-container">$ {\bf x}$</span> will approach the <span class="math-container">${\bf x_0}$</span>.</p>
<p>What happens when the domain of <span class="math-container">$f$</span> is a closed set or neither open nor closed?</p>
<p>Is there another more 'general' definition?</p>
| BCLC | 140,308 | <p>$Y|X=x$ is a uniformly distributed random variable on the interval $(0,x)$</p>
<p>Hence, $$E[Y|X=x] = \int_{0}^{x} y\frac{1}{x} dy$$</p>
<p>$$= y^2\frac{1}{2x} |_{0}^{x}$$</p>
<p>$$= x^2\frac{1}{2x}$$</p>
<p>$$= \frac{x}{2}$$</p>
|
767,304 | <p>Prove that there are no real numbers $x$ such that</p>
<p>$$\sum_{n\,=\,0}^\infty \frac {(-1)^{n + 1}} {n^x} = 0$$</p>
<p>Can I have a hint please?</p>
| Daniel Fischer | 83,702 | <p>For $x \leqslant 0$, the terms of the series don't converge to $0$, hence the series diverges then. Therefore, we need only consider $x > 0$.</p>
<p>For $x > 0$, the sequence $\left(\frac{1}{n^x}\right)_{n\in\mathbb{Z}^+}$ is strictly decreasing and converges to $0$, thus by Leibniz' criterion</p>
<p>$$\eta(x) := \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^x}$$</p>
<p>converges.</p>
<p>Now, if $(a_n)_{n\in\mathbb{Z}^+}$ is any monotonically non-increasing sequence converging to $0$, we can consider the partial sums of an odd and an even number of terms separately,</p>
<p>$$s_{2p} = \sum_{n=1}^{2p} (-1)^{n+1} a_n;\qquad s_{2p+1} = \sum_{n=1}^{2p+1} (-1)^{n+1} a_n.$$</p>
<p>We find</p>
<p>$$\begin{align}
s_{2p+3} - s_{2p+1} &= (-1)^{2p+3}a_{2p+2} + (-1)^{2p+4}a_{2p+3} = a_{2p+3} - a_{2p+2} \leqslant 0,\\
s_{2p+2} - s_{2p} &= (-1)^{2p+3} a_{2p+2} + (-1)^{2p+2} a_{2p+1} = a_{2p+1} - a_{2p+2} \geqslant 0,\\
s_{2p+1} - s_{2p} &= (-1)^{2p+2}a_{2p+1} = a_{2p+1} \geqslant 0.
\end{align}$$</p>
<p>So</p>
<ul>
<li>the sequence of partial sums of an odd number of terms is monotonically non-increasing,</li>
<li>the sequence of partial sums of an even number of terms is monotonically non-decreasing, and</li>
<li>the partial sum of an odd number of terms is never smaller than the partial sum of an even number of terms.</li>
</ul>
<p>If - like for the specific sequence under consideration - the sequence $(a_n)$ is strictly monotonically decreasing, all inequalities above are strict.</p>
<p>It is straightforward to deduce from that that $\eta(x) > 0$ for all $x > 0$.</p>
|
3,006,046 | <p>How to find the Newton polygon of the polynomial product <span class="math-container">$ \ \prod_{i=1}^{p^2} (1-iX)$</span> ?</p>
<p><strong>Answer:</strong></p>
<p>Let <span class="math-container">$ \ f(X)=\prod_{i=1}^{p^2} (1-iX)=(1-X)(1-2X) \cdots (1-pX) \cdots (1-p^2X).$</span></p>
<p>If I multiply , then we will get a polynomial of degree <span class="math-container">$p^2$</span>.</p>
<p>But it is complicated to express it as a polynomial form.</p>
<p>So it is complicated to calculate the vertices <span class="math-container">$ (0, ord_p(a_0)), \ (1, ord_p(a_1)), \ (2, ord_p(a_2)), \ \cdots \cdots$</span> </p>
<p>of the above product.</p>
<p>Help me doing this</p>
| Lubin | 17,760 | <p>It’s really quite simple. There are <span class="math-container">$p^2-p$</span> roots <span class="math-container">$\rho$</span> with <span class="math-container">$v(\rho)=0$</span>, <span class="math-container">$p-1$</span> roots with <span class="math-container">$v(\rho)=-1$</span>, and one root with <span class="math-container">$v(\rho)=-2$</span>. Consequently, there is one segment of the polygon with slope <span class="math-container">$0$</span> and width <span class="math-container">$p^2-p$</span>, one segment with slope <span class="math-container">$1$</span> and width <span class="math-container">$p-1$</span>, and one segment with slope <span class="math-container">$2$</span> and width <span class="math-container">$1$</span>.</p>
<p>Thus, the vertices are <span class="math-container">$(0,0)$</span>, <span class="math-container">$(p^2-p,0)$</span>, <span class="math-container">$(p^2-1,p-1)$</span>, and <span class="math-container">$(p^2,p+1)$</span>.</p>
|
4,069,185 | <p>I got stuck here:</p>
<p>The probability that it will rain today is that it did not rain in the previous two days is <span class="math-container">$0.3$</span>, but if it rained in one of the last two days then the probability of rain today is <span class="math-container">$0.6$</span>.</p>
<p><span class="math-container">$W(n)$</span> is a random variable that receives the value <span class="math-container">$1$</span> if in the <span class="math-container">$n\ge1$</span> day it rained, and the value <span class="math-container">$0$</span>, if it's not.</p>
<p>I need to explain why the series <span class="math-container">$(W(n))_{n\ge 0}$</span> is not a Markov chain, but when they define <span class="math-container">$X_n=(W_n,W_{n-1})$</span>, so <span class="math-container">$(W(n))_{n\ge 0}$</span> it is a Markov chain!</p>
| Kavi Rama Murthy | 142,385 | <p>If <span class="math-container">$0$</span> is not an eigenvalue then <span class="math-container">$T$</span> is invertible and so is any power of <span class="math-container">$T$</span>. Hence, <span class="math-container">$ker (T^{n-2})=ker(T^{n-1})=\{0\}$</span>.</p>
|
4,044,708 | <p>[3,4] is closed in R <-- R-[3,4] is open</p>
<p>[5,6] is closed in R <-- R-[5,6] is open</p>
<p>Show that [3,4] x [5,6] is closed in R x R by writing it as the complement of the intersection of two open sets in R x R.</p>
<p>(R - [3,4]) x (R - [5,6]) not equal R x R - [3,4] x [5,6]</p>
| Peter Morfe | 711,689 | <p>Martin gives a very reasonable answer extending the question's second argument; here's how to extend the first one.</p>
<p>If <span class="math-container">$u: \mathbb{R}^{d} \to \mathbb{R}$</span> is continuous and convex, then mollification of <span class="math-container">$u$</span> gives a family of smooth functions <span class="math-container">$(u^{\epsilon})_{\epsilon > 0}$</span> such that <span class="math-container">$u^{\epsilon} \to u$</span> locally uniformly in <span class="math-container">$\mathbb{R}^{d}$</span>. (Recall that mollification means that we define <span class="math-container">$u^{\epsilon} = \rho^{\epsilon} * u$</span>, where <span class="math-container">$\rho^{\epsilon}(x) = \epsilon^{-d} \rho(\epsilon^{-1} x)$</span> and <span class="math-container">$\rho \in C^{\infty}_{c}(\mathbb{R}^{d})$</span> is a non-negative, even function with <span class="math-container">$\int_{\mathbb{R}^{d}} \rho(x) \, dx = 1$</span>.)</p>
<p>It is not hard to show that, for each <span class="math-container">$\epsilon > 0$</span>, <span class="math-container">$u^{\epsilon}$</span> is convex. Indeed, if <span class="math-container">$x, y \in \mathbb{R}^{d}$</span> and <span class="math-container">$\lambda \in [0,1]$</span>, then
<span class="math-container">\begin{align*}
u^{\epsilon}((1- \lambda)x + \lambda y) &= \int_{\mathbb{R}^{d}} u((1 - \lambda)x + \lambda y - \xi) \rho^{\epsilon}(\xi) \, d \xi \\
&\leq \int_{\mathbb{R}^{d}} ((1 - \lambda) u(x - \xi) + \lambda u(y - \xi)) \rho^{\epsilon}(\xi) \, d \xi \\
&= (1 - \lambda) u^{\epsilon}(x) + \lambda u^{\epsilon}(y).
\end{align*}</span>
Thus, <span class="math-container">$u^{\epsilon}$</span> is subharmonic.</p>
<p>Now if <span class="math-container">$x \in \mathbb{R}^{d}$</span> and <span class="math-container">$r > 0$</span>, then
<span class="math-container">\begin{equation*}
u(x) = \lim_{\epsilon \to 0^{+}} u^{\epsilon}(x) \leq \lim_{\epsilon \to 0^{+}} \frac{1}{|\partial B_{r}(x)|} \int_{\partial B_{r}(x)} u^{\epsilon}(y) \, dy = \frac{1}{|\partial B_{r}(x)|} \int_{\partial B_{r}(x)} u(y) \, dy.
\end{equation*}</span>
Therefore, <span class="math-container">$u$</span> is also subharmonic.</p>
|
2,131,679 | <p>Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and $D \subset \mathbb{R}$ be a dense subset of $\mathbb{R}$. Furthermore, $\forall y_1,y_2 \in D \ f(y_1)=f(y_2)$. Should $f$ be a constant function?</p>
<p>My attempt:
Since $f$ is continuous
$$\forall x_0 \ \forall \varepsilon >0 \ \exists \delta>0 \ \forall x \in \mathbb{R} \ \left(|x-x_0|<\delta \Longrightarrow |f(x)-f(x_0)|<\varepsilon \right)$$
Let $f$ be non-constant function.
Since $D$ is dense $\exists x_1 \in (x_0-\delta, x_0+\delta) \ : \ x_1 \in D$.
Let's take $x_2 \in (x_0-\delta, x_0+\delta)$ such that $f(x_2) \ne f(x_1)$.
Let $\varepsilon = \frac{|f(x_2)-f(x_1)|}{2}>0$.
Therefore, we have
$$|f(x_1)-f(x_0)|<\frac{|f(x_2)-f(x_1)|}{2} \ \ \ |f(x_2)-f(x_0)|<\frac{|f(x_2)-f(x_1)|}{2}$$
Adding the expressions above, we obtain
$$|f(x_2)-f(x_1)|\le |f(x_1)-f(x_0)|+|f(x_2)-f(x_0)|<|f(x_2)-f(x_1)|$$
what is the contradiction.
Are my mussings correct?</p>
| Michael Rozenberg | 190,319 | <p>Since $BD=CE=AF=k$, we get $OP=OQ=PQ=\frac{3}{7}k$ and since $AP=PQ$, </p>
<p>we get$AP=PQ=PO$ and we are done!</p>
|
726,574 | <p>Ant stands at the end of a rubber string which has 1km of length. Ant starts going to the other end at speed 1cm/s. Every second the string becomes 1km longer. </p>
<p>For readers from countries where people use imperial system: 1km = 1000m = 100 000cm</p>
<p><strong>Will the ant ever reach the end of the string? But how to explain it.</strong> </p>
<p>I know that yes.</p>
<p>Let :
<code>a</code> - distance covered by ant
<code>d</code> - length of string
<code>c</code> - constant by which the string is extended</p>
<p>The distance covered by ant in second <code>i</code> is <code>a[i] = (a[i-1] + 1)* (d + c)/d</code></p>
<p>I even did computer simulation in microscale where the string is 10cm long and extends by 10cm every second and the ant reaches the end:</p>
<pre><code>public class Mrowka {
public final static double DISTANCE_IN_CM = 10;
public static void main(String[] args) {
double ant = 0;//ants distance
double d = DISTANCE_IN_CM;//length of string
double dLeft = d - ant;//distance left
int i = 0;
while(dLeft > 0){
ant++;
ant = ant * (d + DISTANCE_IN_CM)/d;
d = d + DISTANCE_IN_CM;
dLeft = d - ant;
System.out.println(i + ". Ant distance " + ant +"\t Length of string " + d + " distance left " + dLeft);
i++;
}
System.out.println("end");
}
}
</code></pre>
<p><strong>Output:</strong></p>
<pre><code>0. Ant distance 2.0 Length of string 20.0 distance left 18.0
1. Ant distance 4.5 Length of string 30.0 distance left 25.5
2. Ant distance 7.333333333333333 Length of string 40.0 distance left 32.666666666666664
.....
12364. Ant distance 123658.53192119849 Length of string 123660.0 distance left 1.4680788015102735
12365. Ant distance 123669.5318833464 Length of string 123670.0 distance left 0.46811665360291954
12366. Ant distance 123680.53192635468 Length of string 123680.0 distance left -0.5319263546844013
end
</code></pre>
<p>EDIT:</p>
<p>I think that I need to calculate this <code>a[n] = (a[n-1] + 1)*(1 + 1/(1+n))</code> when <code>n->+oo</code></p>
| Community | -1 | <p>Here's an another method using point slope form of straight line.</p>
<ul>
<li><span class="math-container">$m = \dfrac12$</span></li>
<li><span class="math-container">$(x_1,y_1) = (2,-2)$</span></li>
</ul>
<p>Equation of straight line is given by,</p>
<p><span class="math-container">$(y-y_1) = m(x-x_1)$</span></p>
<p><span class="math-container">$(y-(-2)) = \dfrac12(x-2)$</span></p>
<p><span class="math-container">$ 2(y+2) =x-2 $</span></p>
<p><span class="math-container">$2y + 4 = x-2$</span></p>
<p><span class="math-container">$2y - x + 6 = 0$</span></p>
|
3,088,620 | <p>Let <span class="math-container">$M$</span> be a second countable smooth manifold. When I learned about differential geometry, a side note was made about how if <span class="math-container">$E$</span> is a vector bundle, <span class="math-container">$\Gamma(E)$</span> is a <span class="math-container">$C^\infty(M)$</span>-Module that is not free, but projective. I now realized that I have no idea how to prove that!</p>
<p>My first attempt was to look for torsion elements, but <span class="math-container">$F_R(X)$</span> (the free <span class="math-container">$\operatorname{\underline{R-Mod}}$</span> over the set <span class="math-container">$X$</span>) having no torsion elements is only satisfied if <span class="math-container">$R$</span> is a domain – which is clearly not the case for <span class="math-container">$C^\infty(M)$</span> (take bump functions with different support).</p>
<p>So:</p>
<blockquote>
<ol>
<li>How do you show that <span class="math-container">$\Gamma(E)$</span> is not free as a <span class="math-container">$C^\infty(M)$</span>-Module?</li>
<li>What are good tactics to show that a (non-finitely generated) module is not free if the ring <span class="math-container">$R$</span> is not even a domain?</li>
</ol>
</blockquote>
| anomaly | 156,999 | <p>Let <span class="math-container">$X$</span> be a closed, smooth manifold, and let <span class="math-container">$E\to X$</span> be a vector bundle. Put <span class="math-container">$R = C^\infty(X)$</span>. By induction on <span class="math-container">$\dim E$</span> (and splitting off a trivial subbundle), it's sufficient to prove the case where <span class="math-container">$E$</span> is a line bundle. Then <span class="math-container">$E = Rs$</span> for some section <span class="math-container">$s\in \Gamma(E)$</span>. By local trivialization, <span class="math-container">$\Gamma(E)$</span> contains for each <span class="math-container">$p\in X$</span> a section that is nonzero at <span class="math-container">$p$</span>. Thus <span class="math-container">$s \not = 0$</span> everywhere. But a nowhere-vanishing section is exactly a trivialization of <span class="math-container">$E$</span>.</p>
<p>In the other direction, choose a finite open cover <span class="math-container">$\{U_i\}$</span> of <span class="math-container">$X$</span> with each <span class="math-container">$E\vert U_i$</span> trivial. These restrictions then patch together to give an embedding of <span class="math-container">$E$</span> in the trivial bundle <span class="math-container">$\theta^N \to X$</span> for some large <span class="math-container">$N$</span>. Furthermore, the decomposition <span class="math-container">$\theta^n = E\oplus E^\perp$</span> gives <span class="math-container">$\Gamma(E)$</span> as a direct summand of <span class="math-container">$\Gamma(\theta^N) = R^N$</span>; that is, <span class="math-container">$\Gamma(E)$</span> is a projective <span class="math-container">$R$</span>-module. There are several invariants that can prove that bundles are nontrivial; these are good sources of projective modules, albeit ones over a very complicated ring. </p>
<p>Swan's theorem is an extension of this idea: For <span class="math-container">$X$</span> as above, the map
<span class="math-container">\begin{align*}
\Gamma:\{\text{finite-dimensional vector bundles over $X$}\} \to \{\text{f.g. projective $C^\infty(X)$-modules}\}
\end{align*}</span>
is an equivalence. (There are also variants that lower the restrictions on the left and right sides above.)</p>
|
3,384,280 | <p>I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:</p>
<p><span class="math-container">$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$</span></p>
<p>Here there are my differents methods:</p>
<ol>
<li>assuming <span class="math-container">$f(n)=\sqrt{4n^2+1}, \,$$\ g(n)=2n$</span>, <span class="math-container">$h(n)=\sqrt{n^2-1}$</span>, <span class="math-container">$\ \psi(n)= n$</span> <span class="math-container">$$f(n)-g(n)=\frac{\dfrac{1}{g(n)}-\dfrac{1}{f(n)}}{\dfrac{1}{f(n)\cdot g(n)}}, \quad h(n)-\psi(n)=\frac{\dfrac{1}{\psi(n)}-\dfrac{1}{h(n)}}{\dfrac{1}{h(n)\cdot \psi(n)}}$$</span>
I always have an undetermined form.</li>
<li>I've done some rationalizations:
<span class="math-container">$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{4n^2+1}+2n}{\sqrt{4n^2+1}+2n}$$</span> where to the numerator I find <span class="math-container">$1$</span> and to the denominator an undetermined form. Similar situation considering
<span class="math-container">$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{n^2-1}+n}{\sqrt{n^2-1}+n}$$</span></li>
<li><span class="math-container">$$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{n\left(\sqrt{4+\dfrac{1}{n^2}}-2\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-1\right)}\rightsquigarrow \left(\frac{0}{0}\right)$$</span>
At the moment I am not able to think about other possible simple solutions.</li>
</ol>
| Barry Cipra | 86,747 | <p><strong>Hint</strong>:</p>
<p><span class="math-container">$$\begin{align}
{\sqrt{4n^2+1}-2n\over\sqrt{n^2-1}-n}
&={\sqrt{4n^2+1}-2n\over\sqrt{n^2-1}-n}\cdot{\sqrt{4n^2+1}+2n\over\sqrt{4n^2+1}+2n}\cdot{\sqrt{n^2-1}+n\over\sqrt{n^2-1}+n}\\
&={(4n^2+1)-4n^2\over(n^2-1)-n^2}{\sqrt{n^2-1}+n\over\sqrt{4n^2+1}+2n}
\end{align}$$</span></p>
|
3,384,280 | <p>I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:</p>
<p><span class="math-container">$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$</span></p>
<p>Here there are my differents methods:</p>
<ol>
<li>assuming <span class="math-container">$f(n)=\sqrt{4n^2+1}, \,$$\ g(n)=2n$</span>, <span class="math-container">$h(n)=\sqrt{n^2-1}$</span>, <span class="math-container">$\ \psi(n)= n$</span> <span class="math-container">$$f(n)-g(n)=\frac{\dfrac{1}{g(n)}-\dfrac{1}{f(n)}}{\dfrac{1}{f(n)\cdot g(n)}}, \quad h(n)-\psi(n)=\frac{\dfrac{1}{\psi(n)}-\dfrac{1}{h(n)}}{\dfrac{1}{h(n)\cdot \psi(n)}}$$</span>
I always have an undetermined form.</li>
<li>I've done some rationalizations:
<span class="math-container">$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{4n^2+1}+2n}{\sqrt{4n^2+1}+2n}$$</span> where to the numerator I find <span class="math-container">$1$</span> and to the denominator an undetermined form. Similar situation considering
<span class="math-container">$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{n^2-1}+n}{\sqrt{n^2-1}+n}$$</span></li>
<li><span class="math-container">$$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{n\left(\sqrt{4+\dfrac{1}{n^2}}-2\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-1\right)}\rightsquigarrow \left(\frac{0}{0}\right)$$</span>
At the moment I am not able to think about other possible simple solutions.</li>
</ol>
| xpaul | 66,420 | <p>Note
<span class="math-container">\begin{eqnarray}
\lim_{n\to\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}&=&\lim_{n\to\infty}\frac{(\sqrt{4n^2+1}-2n)(\sqrt{4n^2+1}+2n)}{(\sqrt{n^2-1}-n)(\sqrt{n^2-1}+n)}\cdot\frac{\sqrt{n^2-1}+n}{\sqrt{4n^2+1}+2n}\\
&=&-\lim_{n\to\infty}\frac{\sqrt{n^2-1}+n}{\sqrt{4n^2+1}+2n}\\
&=&-\lim_{n\to\infty}\frac{\sqrt{1-\frac1{n^2}}+1}{\sqrt{4+\frac1{n^2}}+2}\\
&=&-\frac12.
\end{eqnarray}</span></p>
|
2,764,141 | <p>This is what I have tried so far: </p>
<p>Since $g(z)$ is bounded, then $\lim\limits_{z\rightarrow 0} zg(z)=0$ and hence $z=0$ is a removable singularity of $g(z)$. We can define $g(0) = \lim\limits_{z\rightarrow 0} f(z)f(\frac{1}{z})$ and make $g$ entire.</p>
<p>Then $g(z)$ is a bounded entire function and hence $g$ is a constant function. In other words, $f(z)f(\frac{1}{z}) = c$ for some $c\in\mathbb{C}$ and for $z\neq 0$</p>
<p>I don't know how to continue from this step.
I tried to prove that $f(\frac{1}{z})$ has either a pole or a removable singularity at $z=0$ to show first that $f(z)$ is a polynomial or a constant but I failed.</p>
| Chappers | 221,811 | <p>Suppose that $c \neq 0$, or the equation $f(z)f(1/z)=c$ has only zero solutions. $f$ is analytic, so it has a zero of order $n$ at $z=0$, where $n$ can be zero. Since $c \neq 0$, $f \neq 0$ since $f(1)^2=c$, for example.</p>
<p>Then $h(z) = f(z)/z^n$ can be extended to an entire function by adding the value $h(0) = \lim_{z \to 0} f(z)/z^n$. Moreover, it is easy to see that $h(z)$ satisfies the same functional equation as $f$. So
$$ h(z)h(1/z) = c, $$
and $h(z) \to a \neq 0$ as $z \to 0$. But then $h(1/z) = c/h(z) \to c/a$ as $z \to \infty$. But this means that $h$ is an entire function with a finite limit at $\infty$, so it is bounded, and hence constant. So $h(z)=a$, and $f(z)=az^n$.</p>
|
4,241 | <p>I was preparing for an area exam in analysis and came across a problem in the book <em>Real Analysis</em> by Haaser & Sullivan. From p.34 Q 2.4.3, If the field <em>F</em> is isomorphic to the subset <em>S'</em> of <em>F'</em>, show that <em>S'</em> is a subfield of <em>F'</em>. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.</p>
<p>I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.</p>
<p>What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so "restrictive", in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?</p>
| User3568 | 1,767 | <p>Yes, you are right in saying that field axioms are restrictive. Some other examples of the restriction [for finite fields] are:</p>
<p>1) You can't have field of arbitrary order. Only of the order $p^n$ are possible. </p>
<p>2) Non-zero elements of a field form a multiplicative group. When the field is finite, this group is cyclic [it's not straightforward to prove this starting with field axioms but you should try it]</p>
<p>3) Any finite division ring is a field [Wedderburn's theorem]. This is a very surprising result as a division ring need not by commutative. But finiteness imposes it. </p>
<p>For the question that you mentioned [from that book], you should note that when some set with some algebraic structure [like group, ring, field] is isomorphic [as that algebraic structure] to some other set, then that under that isomorphism, we are giving an algebraic structure to that other set. Hence in your case S' is a field. </p>
|
3,520,269 | <p><img src="https://i.stack.imgur.com/mdM8B.png" alt="enter image description here"></p>
<p>I also know that given the length of 2 sides in a kite and the angle of one of the other angles (which aren't included angles), you can find the area by multiplying these two sides and the sine of the angle. I am unable to find a relationship between this situation and the situation posted in the title question.</p>
<p>Also I can't visualize why the situation in the title question is true. Thanks!</p>
| Quanto | 686,284 | <p>The quadrilateral consists of two triangles of the same base <span class="math-container">$d_1$</span>. Therefore the area is </p>
<p><span class="math-container">$$A=\frac12h_1b_1+ \frac12h_2b_1=\frac12(h_1+h_2)d_1$$</span></p>
<p>Let the two segments of <span class="math-container">$d_2=x+y$</span> and recognize that <span class="math-container">$h_1=x \sin\alpha $</span> and <span class="math-container">$h_2=y \sin\alpha $</span>. Then,</p>
<p><span class="math-container">$$A = \frac12 (x\sin\alpha+y\sin\alpha)d_1=\frac12(x+y)\sin\alpha d_1=\frac12d_1d_2\sin\alpha$$</span></p>
|
3,858,962 | <p>given a rectangle <span class="math-container">$ABCD$</span> how to construct a triangle such that <span class="math-container">$\triangle X, \triangle Y$</span> and <span class="math-container">$\triangle Z$</span> have equal areas.i dont know where to start. .i tried some algebra with the area of the triangles and used pythagoras theoram to find the sides of the triangle. i just need a hint.
here is the picture<a href="https://i.stack.imgur.com/B7iSk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B7iSk.png" alt="enter image description here" /></a></p>
| Shreyas Reddy | 834,702 | <p><img src="https://i.stack.imgur.com/lLSyI.png" alt="enter image description here" /></p>
<p>well this is my first answer so i cant upload a picture, so i am using the picture in the above answer.
Consider, AB=CD=a...
BD=AC=b...
CF=c..
CE=d.'' so, AE=b-d, FD=a-c..
as the areas of 3 triangles(right angled) equal,
a(b-d)=(a-c)b...
b-d/b = a-c/a....
d/b=c/a...
that gives you = a/b=c/d...
so to construct take a right angled first take a point on one side and take another point on its adjacent side which accepts this propotionality, draw a triangle and then connect both the points to the point which is diametrically opposite to the right angle of the triangle and you are done. pls upvote if it is useful</p>
|
36,756 | <p>In a class, 18 students like to play chess, 23 like to play soccer, 21 like biking, and 17 like jogging. The number of those who like to play both chess and soccer is 9. We also know
that 7 students like chess and biking, 6 students like chess and jogging, 12 like soccer and
biking, 9 like soccer and jogging, and finally 12 students like biking and jogging. There
are 4 students who like chess, soccer, and biking, 3 who like chess, soccer, and jogging, 5
who like chess, biking, and jogging, and 7 who like soccer, biking, and jogging. Finally,
there are 3 students who like all four activities. In addition, we know that every student
likes at least one of these activities. How many students are there in the class? </p>
<p>I know this is a case of using the inclusion-exclusion principle, but I'm a little overwhelmed, given that there are 4 sets. Can someone please explain this to me? Thanks!</p>
| Arturo Magidin | 742 | <p>Take $S$ to be the set of kids who play soccer; $C$ of kids who play chess; $B$ for biking; and $J$ for jogging.</p>
<p>If you just add up $|S|+|C|+|J|+|B|$, then you are overcounting: any kid who likes more than one sport is getting counted as many times as there are sports he likes. So you need to compensate for that.</p>
<p>You can compensate for those who like exactly two sports by subtracting the six pairwise intersections: $|S\cap C|$, $|S\cap J|$, $|S\cap B|$, $|C\cap J|$, $|C\cap B|$, and $|J\cap B|$. Now, you've counted everyone who likes just one sport once, everyone who likes exactly two sports once (you counted them twice to begin with, and have subtracted them once now).</p>
<p>But you've overcompensated for kids who like three or more sports: for kids who like three sports, you counted them three times when you counted $|S|+|C|+|J|+|B|$, and then you subtracted them <em>three times</em> when you subtracted the pairwise intersections (since they are in all three pairwise intersections). For kids who like all four sports, you counted them 4 times to begin with but now you subtracted them <em>six</em> times... we'll deal with those later...</p>
<p>To compensate for kids who like exactly three sports, which you have now not counted at all, we just need to add the four 3-fold intersections; $|S\cap C\cap J|$, $|S\cap C\cap B|$, $S\cap J\cap B|$ and $|C\cap J\cap B|$. We counted them 3 times first, then subtracted them three times, now add them once. Great.</p>
<p>But now what about the kids who like all four sports? You counted them four times at first; then you subtracted them six times when you dealt with pairs; and now you've added them <em>four</em> times when you counted 3-fold intersections; that means that you've counted them 8 times and subtracted them six times, so they are still overcounted. You need to count how many kids like all four sports and subtract them to get the right total. So you need to subtract $|S\cap C\cap J\cap B|$. </p>
|
1,041,212 | <p>I want to show that the function $f(X) = -log \ det(X)$ is convex on the space $S$ of positive definite matrices. </p>
<p>What I have done:</p>
<p>It seems like this problem could be tackled by considering the restriction of $f(X)$ to a line through a given point $X \in S$ so that $g(\alpha) = f(X + \alpha V)$ for some $V \in S$. So now I am trying to show that g is convex.</p>
| Michael Grant | 52,878 | <p>Since $X$ is positive definite, it admits a symmetric, invertible square root $X^{1/2}$. Then
$$\begin{aligned}
f(X)&=-\log\det(X+tV)\\
&=-\log\det X^{1/2}(I+tX^{-1/2}VX^{-1/2})X^{1/2}\\
&=-\log(\det X^{1/2})^2\det(I+tX^{-1/2}VX^{-1/2}) \\
&=-\log\det X-\log\det(I+t\tilde{V})
\end{aligned}$$
where $\tilde{V}\triangleq X^{-1/2}VX^{-1/2}$. Let $Q\Lambda Q^T$ be a Schur decomposition of $\tilde{V}$, with $\Lambda=\mathop{\textrm{diag}}(\lambda_1,\lambda_2,\dots,\lambda_n)$.
$$\begin{aligned}
f(X)&=-\log\det X-\log\det(I+tQ\Lambda Q^T) \\
&=-\log\det X-\log\det Q(I+t\Lambda)Q^T \\
&=-\log\det X-(\log\det Q)^2-\log\det (I+t\Lambda) \\
&=-\log\det X-0-\log\prod_{i=1}^n(1+t\lambda_i) \\
&=-\log\det X-\sum\log(1+t\lambda_i)
\end{aligned}$$
It is not difficult to show that $g_\lambda(t)=-\log(1+t\lambda)$ is a convex function of $t$ for any fixed $\lambda$. (For instance, $g_\lambda''(t)>0$ will be strictly positive for $t\neq 0$, and identically zero if $t=0$). </p>
<p>Thus we have expressed $f(X)$ as the sum of a constant and $n$ convex functions of $t$.</p>
|
1,569,411 | <p>Express $log_3(a^2 + \sqrt{b})$ in terms of m and k where
$m = log_{3}a$</p>
<p>$k = log_{3}b$</p>
<p>Given this information I made
$a = 3^m$</p>
<p>$b = 3^k$</p>
<p>Therefore
= $log_{3} ((3^m)^2 + (3^k))^{\frac{1}{2}}$</p>
<p>= $log_{3} (3^{2m} + 3^{\frac{k}{2}})$</p>
<p>I don't know if I'm done or there is still more things I can simplify. Can anyone help please, thanks</p>
| Community | -1 | <p>You are correct. If $P$ is a Markov transition matrix, then $P^n$ converges
as $n\to\infty$ if and only if $P$'s only eigenvalue of modulus one is $\lambda=1.$ </p>
|
125,705 | <p>Find an open cover $\mathcal{G}$ of the set $E = \{ 1/n : n \in \mathbb{Z^{+}}\}$ such that any proper subset of $\mathcal{G}$ is not an open cover of $E$.</p>
| David Mitra | 18,986 | <p>Hint: for each positive integer $n$, find an open interval $O_n$ containing $1\over n$ that contains no other point of $E$. (I'm assuming you're working in $\Bbb R$ with the usual topology.) </p>
|
125,705 | <p>Find an open cover $\mathcal{G}$ of the set $E = \{ 1/n : n \in \mathbb{Z^{+}}\}$ such that any proper subset of $\mathcal{G}$ is not an open cover of $E$.</p>
| Brett Frankel | 22,405 | <p>You can cover each point of $E$ with an interval so small that it contains no other points in $E$. Since this looks like it might be homework, I'll leave you to work out the cover explicitly.</p>
|
125,705 | <p>Find an open cover $\mathcal{G}$ of the set $E = \{ 1/n : n \in \mathbb{Z^{+}}\}$ such that any proper subset of $\mathcal{G}$ is not an open cover of $E$.</p>
| Brian M. Scott | 12,042 | <p>The easiest way is to make sure that every member of $\mathcal{G}$ is needed in order to cover $E$. That means that for each $G\in\mathcal{G}$ you want a point $e_G\in E$ that is covered by $G$ and by no other member of $\mathcal{G}$. One easy way to do this is to arrange matters so that each member of $\mathcal{G}$ contains exactly one point of $E$, a different point for each $G\in\mathcal{G}$.</p>
<p>What if $E$ were the set of positive integers? Can you find open sets $G_n$ for $n\in\Bbb Z^+$ such that $G_n\cap\Bbb Z^+=\{n\}$? That shouldn’t be too hard. And once you’ve seen the idea, it shouldn’t be too hard to adapt it to your actual problem.</p>
|
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