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H: Why not generalize the concept of prime numbers?
A prime number is a natural number N > 1, that is divisible only by 1 and N (N/1).
The number 1 is given special treatment in this definition, and this can be generalized by extending the set of the specially treated numbers, for example to also include number 2, lets call these secondary numbers:
A secondary number is a natural number N > 2, that is divisible only by 1, N/1, 2, N/2.
This gives us a completely new set of 'prime numbers' that have many qualities of prime number set,
theres infinitely many of them and it seems the fundamental theorem of arithmetics is also true for this set.
Would this set be somehow less special than that of prime numbers?
Did anyone already do this?
AI: Here is one natural generalization along these lines:
A natural number $n$ is $k$-prime when it has exactly $k+1$ positive divisors.
Then $1$-prime is the same as ordinary prime. And $0$-prime is the unit $1$. Also, a $2$-prime is the square of a prime but a $3$-prime is the cube of a prime or the product of two primes.
The problem with this generalization is that $k$-primes do not occur naturally in number theory and so are not very useful.
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H: Would I be correct to assume that the minimum amount of vertices required to have an object with 3 dimensions is 4?
The question seems so simple to me but often the simplest ones are the most complex. My assumption is that the minimum amount of points required to create an object with 3 dimensions is 4?
You would get a tetrahedron as far as I am aware. What I mean exactly is the minimum amount of points / vertices to create an object which has an exterior and interior, or another way of putting it would be that it has a volume. I am not well versed enough in geometry to use the proper terminology but I hope this is clear. Is this correct, that the minimum amount of points for such an object is 4?
AI: If you mean a three-dimensional solid by 'object', then yes, that is correct. Three (non-collinear) points are needed to define a plane in 3D space, so the minimum number of points to do so cannot be 3 or less. Since a tetrahedron has 4 vertices, the minimum number of points for a $3$D object to exist is therefore 4.
The fact that "three points define a plane" comes from the fact that any plane can be written as $ax + by + cz = d$. The rank of the augmented matrix with $n$ rows is at most $n$, so at least $3$ equations are needed for the matrix to have rank $3$ (which is a full rank in $3$D space). For a less technical explanation, see this post.
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H: Let $f: [0, +\infty] \rightarrow \mathbb{R}$ such as f' is decreasing and positive. Prove that $\sum_ {n=1}^\infty f'(n) $ converges iff f is bounded
It seems a lot like this question, but I wasn't able to prove with the same tips
AI: As $f'$ is decreasing, we have
$$f(n)-f(n-1)=\int_{n-1}^nf'(x)\,\mathrm dx \ge \int_{n-1}^nf'(n)\,\mathrm dx=f'(n).$$
Similarly,
$$f(n+1)-f(n)=\int_{n}^{n+1}f'(x)\,\mathrm dx \le \int_{n}^{n+1}f'(n)\,\mathrm dx=f'(n).$$
By adding these inequalities for $n=1,2,\ldots, N$, we get
$$ f(N+1)-f(1)\le \sum_{n=1}^Nf'(n)\le f(N)-f(0).$$
Conclude.
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H: Property of Lebesgue measure in $\mathbb{R}^2$
Let $I=[0,1]\times [0,1]$ and $E\subset \mathbb{R}^2,$ be a set of zero Lebesgue measure. Is it true that $$\overline{I\setminus E}=I?$$
I guess that the counterexample will be some form space filling curve.
AI: Yes, it is true.
Proving that $\overline{I\setminus E}\subseteq I$ is trivial.
For proving $I\subseteq\overline{I\setminus E}$ let $(x,y)\in I$ and assume that $(x,y)\notin\overline{I\setminus E}$.
Then some open set $U$ must exist with $(x,y)\in U$ and $U\cap(I\setminus E)=\varnothing$ or equivalently $U\cap I\subseteq E$.
But $U\cap I$ has positive Lebesgue measure.
So this contradicts that $E$ is a set with Lebesgue measure zero and we conclude that our assumption must be wrong.
That means that $(x,y)\in I$ implies that $(x,y)\in\overline{I\setminus E}$ and we are ready.
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H: Is a column-wise and row-wise Gaussian matrix, jointly Gaussian?
Let $A\in\mathbb{R}^{n\times m}$ be a random matrix that each of its rows or columns is a Gaussian vector with iid components. More formally we have $A_{i,\cdot}\sim\mathcal{N}(0,I_m),A_{\cdot,j}\sim\mathcal{N}(0,I_n)$ for all $i\in[n],j\in[m]$, and furthermore, $\mathrm{law}(A)$ has support almost everywhere. Is $\mathrm{law}(A)$ a multivariate Gaussian?
Example ($m=n=2$): let $A=\left[\begin{matrix}a& b\\ c & d\end{matrix}\right]$. Assuming that $(a,b),(c,d),(a,c),(b,d)\sim\mathrm{N}(0,I_2)$, and $\mathrm{support}(\mathop{law}(A))$ is non-zero almost everywhere. Is $\mathop{law}(a,b,c,d)$ a multivariate Gaussian?
Possible generalization: if each row and column follow a multivariate p-Stable distribution, namely Cauchy for $p=1$, would the matrix follow a joint p-Stable distribution?
AI: No. Answer to a previous version of the problem:
Let $X,Y\sim N(0,1)$ iid, let $S=\pm1$ be independent of $X,Y$ and let $$A=\pmatrix{X&Y\\Y&S X}.$$ Note that $SX\sim N(0,1)$, and $(X,Y)\sim N(0,I_2)$,
$(Y,SX)\sim N(0,I_2)$ but $(X,SX)\nsim N(0,I_2)$.
With the current, changed version of the problem: There are rvs $X,Z$ which are marginally $N(0,1)$ but not jointly Gaussian, whose joint density function is positive everywhere.
(For example, let $f(x,z)=3\phi(x,z)/2$ in the 1st and 3d quadrants and $f(x,z)=\phi(x,z)/2$ in the 2nd and 4th quadrants, where $\phi$ is the density of $N(0,I_2)$.
Now construct $$A=\pmatrix{X&Y_1\\Y_2&Z},$$ where $Y_1,Y_2$ are independent $N(0,1)$ rv.s
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H: How to show that the distance of the points of tangency along a tangent line on two tangent circles with radius $a$ and $b$ is equal to $2\sqrt{ab}$?
How to show that the distance of the points of tangency along a tangent line on two tangent circles with radius $a$ and $b$ is equal to $2\sqrt{ab}$?
Please see the image below. Line $DE$ is tangent to Circles $B$ and $C$ at point $D$ and $E$, respectively. Line $BC$ passes through point $A$, which is the tangent point of the two given circles. I am trying to prove visually that $DE=2\sqrt{(BA)(AC)}$.
Here is my attempt:
I construct a segment from point $B$ perpendicular to radius $CE$ at point $F$. Since quadrilateral $BFED$ is a parallelogram (a rectangle) $BF=DE$.
Applying the Pythagorean Theorem,
$BF=DE=\sqrt{BC^2-CF^2}$
After this, I got stuck. Any comments or suggestions will be much appreciated.
AI: Draw a paralel to $DE$ through $B$ which cuts $CE$ at $F$. If $DE = x$ then we have, by Pythagora theorem $$x^2+FD^2 = CB^2$$
i.e. $$x^2+(a-b)^2=(a+b)^2\implies x^2=4ab$$
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H: arrow category and functor category
Let A be an abelian category and D the category having two objects and only one nonidentity morphism between them.
The functor category A$^D$ is also abelian and it is called an arrow category with objects morphisms in A and morphisms commutative squares.
I cannot see the equivalence between the functor category and the arrow category. I understand arrow category but how it is equivalent to the functor category? Any help would be appreciated!
AI: Let $0$ and $1$ denote the objects of $D$ and write $a:0\rightarrow 1$ for the only non-identity arrow of $D$. To every functor $F:D\to A$ associate the morphism $F(a):F(0)\to F(1)$ in $A$. Conversely, to every morphism $f:X\to Y$ in $A$ associate the functor $\hat f:D\to A$ given by $\hat f(0)=X$, $\hat f(1)=Y$ and $\hat f(a)=f$. Can you continue from here?
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H: Property of Lebesgue measure in $\mathbb{R}^2$, part 2
Let $I=A\times B,$ where $A,B\subset \mathbb{R}$ are closed sets of positive Lebesgue measure, and $E\subset \mathbb{R}^2,$ be a set of zero Lebesgue measure. Is it true that $$\overline{I\setminus E}=I?$$
When $I=[0,1]\times [0,1]$ the answer is true and one can find a solution in the following link.
Property of Lebesgue measure in $\mathbb{R}^2$
AI: No. Set $A=B=[0,1]\cup\{2\}$. Let $E$ be the singleton $\{(2,2)\}$. Then
$\overline{I\setminus E}\neq I$.
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H: Compute $\iint (x+y)\,dx\, dy$ with circle constraint $x^{2}+y^{2}=x+y$
I have a double integral:
$$\iint (x+y)\,dx\, dy$$
with circle constraint:
$$x^{2}+y^{2}=x+y$$
I tried to calculate it with transition to polar coordinates:
$$x^{2}+y^{2}=x+y$$
$$\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$$
In polar coordinates:
$$r^{2}(\cos(\varphi))^{2} + r^{2}(\sin(\varphi))^{2} = r\cos(\varphi) + r\sin(\varphi)$$
$$r = \cos(\varphi) + \sin(\varphi)$$
Graph looks like this:
But i don't understand how to find polar radius change interval here. If i separate circle into two, for first half circle for example it will go from $\textbf{some point}$ to $\frac{\pi}{2}$. I don't understand how to find that $\textbf{some point}$, cause it starts from point ($\frac{1}{2}-\frac{1}{\sqrt{2}} = -0.2071$).
AI: We are asked to evaluate the integral, $I$, of the function $f(x,y)=x+y$ over the disk defined by the boundary circle $x^2+y^2= x+y$. We can express $I$ in Cartesian coordinates as
$$I=\int_{1/2-1/\sqrt2}^{1/2+1/\sqrt2} \int_{1/2-\sqrt{1/2-(y-1/2)^2}}^{1/2+\sqrt{1/2-(y-1/2)^2}} (x+y)\,dx\,dy$$
If we make a brute force transformation to polar coordinates, $(r,\phi)$, then the locus of points on the boundary of the disk are given by $r=\cos(\phi)+\sin(\phi)$ with $\phi\in [-\pi/4,3\pi/4]$ serving as a parameter. Then, we have
$$\begin{align}
I&=\int_{-\pi/4}^{3\pi/4} \int_0^{\cos(\phi)+\sin(\phi)}(r\cos(\phi)+r\sin(\phi))\,r\,dr\,d\phi\\\\
\end{align}$$
Can you finish now?
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H: Intersection point of two curves with errors - covariance matrix
I have measured the parameters for a hyperbola and an ellipse, let us call them
$$
\begin{cases} a^2x^2 + b^2y^2 = 1 \\ c^2x^2 + d^2y^2 = 1 \end{cases}
$$
and I have errors associated with each parameter a, b, c and d ($\pm\delta a$, $\pm\delta b$, $\pm\delta c$, $\pm\delta d$). I want to find the intersection point(s) (due to symmetry there is in practice only one solution), between these two curves, that is, I want to solve the equation:
$$
AX = Y
$$
where I have defined $A=\begin{pmatrix}a^2 & b^2 \\ c^2 & d^2\end{pmatrix}$, $X=\begin{pmatrix}x^2 \\ y^2\end{pmatrix}$, $Y=\begin{pmatrix}1 \\ 1\end{pmatrix}$.
Now, solving the equation is done by simply taking the inverse of A. What I am struggling with however is how to incorporate the errors, such that I derive the covariance matrix for the estimated intersection point $(\hat{x},\hat{y})$.
AI: Let $a,b,c,d$ be the "correct" parameters and $a + \delta a$, ..., $ d + \delta d$ the measured parameters including errors. The resulting matrix is $A + \delta A$ where to first order
$$ \delta A \approx \pmatrix{2 a\; \delta a & 2 b \; \delta b\cr 2 c\; \delta c & 2 d\; \delta d}$$
and you compute $X + \delta X = (A + \delta A)^{-1} Y$.
Again to first order, assuming $A$ is invertible,
$$(A + \delta A)^{-1} \approx A^{-1} + A^{-1} (\delta A) A^{-1}$$
so $$\delta X \approx A^{-1} (\delta A) A^{-1} Y $$
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H: Show $\mathbb{P}[X-m>\alpha]\leq \frac{\sigma^2}{\sigma^2+\alpha^2}$
I found this problem in an old statistics book:
Suppose $X$ is a square integrable random variable with mean $m$ and variance $\sigma^2$. For any $\alpha>0$, show
$$
\mathbb{P}[X-m>\alpha]\leq\frac{\sigma^2}{\sigma^2 +\alpha^2}
$$
At first I thought that the inequality results from direct application of Markov-Chebyshev's inequality, but when I actually tried it I realized it was not so. Does anybody know about this inequality and how to obtain it?
AI: That is is known as Cantelli's inequality. It can be obtained from Chebyshev's but with a twist.
For any $x>0$, $\alpha+x>0$ and so,
$$
\begin{align}
\mathbf{P}[X-m>\alpha]&=\mathbf{P}[X-m+x>x+\alpha]\\
&\leq \frac{\mathbf{E}[(X-m+x)^2]}{(\alpha+x)^2}=\frac{\sigma^2+x^2}{(\alpha +x)^2}=:g(x)
\end{align}
$$
The game now is to find the best $x$. You can use differential Calculus to check that $x=\frac{\sigma^2}{\alpha}$ does the trick.
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H: Reference books on the Baum Connes conjecture
Do there exist readable reference books about Baum-Connes Conjecture for beginners ?
AI: Alain Valette, who is an active user on mathoverflow, has a book called Introduction to the Baum-Connes connecture. I remember reading parts of it a few years ago and finding it quite accessible and nicely written. However, I do not do research in this field, so I cannot say how up to date it is (if that is a concern for you).
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H: Prove if infinite product of $f(x)$ is $0$ then so is infinite product of $f(x\varphi)$
Prove or disprove that if $$\prod\limits_{x=2}^{\infty} f(x)=0$$ and $f(x)\neq0$ for any $x\geq0$ then $$\prod\limits_{x=2}^{\infty} f(x\varphi)=0$$ for any constant $\varphi\geq2$
This seems true but I'm not quite sure how to prove it since the constant is inside a function f.
AI: This is not true. Consider any continuous function defined over $\mathbb{N}$ by
$$f(n)=\cases{\frac12&$n$ odd\\1&$n$ even}$$
An explicit example is given by
$$f(x)=\frac{3+\cos{(\pi x)}}4$$
Then we have
$$\prod_{x=2}^\infty f(x)=0$$
But
$$\prod_{x=2}^\infty f(2x)=1$$
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H: Determining analytically the number of times a line intersects a general 3D surface
Consider a general surface and a line in $\mathbb{R}^3$. Given equations for both the surface and line, is there a way to analytically determine the number of times the line intersects the surface?
I am only interested in the number of intersections, not the precise locations of intersection. I know that there are methods to calculate this numerically by first finding where the intersections are, but I want to know if there is a simple method to just calculate the number of intersections.
The surface might be given implicitly as $F(x, y, z)=0$ or parametrically as $\vec{r}(u, v)=x(u, v) \hat{\imath}+y(u, v)\hat{\jmath}+z(u, v)\hat{k}$, and may or may not be closed. The line might be given as $\vec{r}(t)=\vec{r}_0+t \vec{d}$ where the line passes through the point $\vec{r}_0$ and $\vec{d}$ is the direction vector. The line extends infinitely in both directions, but I can also work with the case in which the line has finite endpoints.
EDIT: As pointed out in comments below, I am specifically working with the surface given by an implicit equation of the form
$$
F(x, y, z)=x^k+y^k+z^k-1
$$
where the terms involving $x$, $y$, and $z$ can have constant coefficients. Specifically, I am interested in the region $x$, $y \geq 0$. $k$ can be a general rational exponent greater than $2$. One solution suggested below involving the Sturm sequence works simply enough for integer $k$, where one can write a polynomial involving just the parametric variable by substituting in the equation of the line.
This surface is part of a shape called a superellipsoid, but just taken for $x$, $y \geq 0$ to simplify $F(x, y, z)$. In my application, I can make some geometric and symmetry arguments so that I can only consider this portion.
AI: Substituting a parametrization of the line into an implicit equation for the surface, the problem becomes one of counting the number of zeros of a function. If the function is rational, you can do this using Sturm's theorem. For more general classes of functions, the problem can be unsolvable: see Richardson's theorem. For example, Laczkovich showed there is no algorithm that, given a function $A(x)$ in the ring generated by integers, $x$, $\sin(x^n)$ and $\sin(\sin(x^n))$, decides whether there is a real solution of $A(x)=0$.
EDIT: If your function involves fractional exponents, it should still be possible to convert the equation to one involving a polynomial. For example, consider the equation
$$ \sqrt{x^2+1} - (x^4+1)^{1/4} - x + 2 = 0 $$
Write this as $f(\alpha,\beta,x) = \alpha - \beta - x + 2 = 0$
where $\alpha^2 - (x^2+1) = 0$ and $\beta^4 - (x^4+1) = 0$.
First take the resultant of $f(\alpha,\beta,x)$ and $\alpha^2 - (x^2+1)$ with respect to $\alpha$, then the resultant of that and
$\beta^4 - (x^4+1)$ with respect to $\beta$, and we get the
polynomial $-15 x^8 + 64 x^7 - 112 x^6 + 112 x^5 + 160 x^4 - 704 x^3 + 752 x^2 - 320 x$. Sturm tells us this has four real roots.
However, we must then look at each root (approximately) to check whether the corresponding $\alpha$ and $\beta$ are the correct (i.e. positive) square root of $x^2+1$ and fourth root of $x^4+1$:
it turns out that only one of the four roots works.
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H: Intersecting diameter and chord
A diameter $AB$ and a chord $CD$ of a circle $k$ intersect at $M.$ $CE$ and $DF$ are perpendiculars from $C$ and $D$ to $AB$. $(A,E,M,F,B$ lie on AB in that order$)$. What is the length of $CD$ if $AE=1,FB=49$ and $MC:MD=2:7$?
How do I approach the given problem? I would be very grateful if you could give me some hints and tips to follow. I see that the triangles $CEM$ and $DFM$ are similar and $\dfrac{MC}{MD}=\dfrac{CE}{DF}=\dfrac{EM}{FM}=\dfrac{2}{7}.$
AI: Let $CM=2x$ and $EM=2y$.
Thus, $$MF=7y,$$ $$MD=7x$$ and since $\measuredangle ACB=90^{\circ}$, we obtain $$CE^2=AE\cdot EB.$$
Also, $$AM\cdot MB=CM\cdot MD$$ and we obtain the following system:
$$(2x)^2-(2y)^2=1\cdot(2y+7y+49)$$ and $$(1+2y)(7y+49)=2x\cdot7x.$$
The last equality it's $$(1+2y)(y+7)=2x^2$$ and we can substitute $2x^2$ in the first equation.
Thus $$2(1+2y)(y+7)-4y^2=9y+49.$$
Can you end it now?
I got $CD=39$.
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H: Proof $\exists\alpha$ s.t. $P(X>\alpha)>0$ if $P(X>0)>0$
For probability triple $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mu)$ prove that for a random variable $X$, if $\mu(X>0)>0$, there must be $\alpha>0$ s.t. $\mu(X>\alpha)>0$.
So if $X$ is a random variable with that property, it means that $\exists $ event $ A \in \mathcal{F}$ and interval $B=(0, \infty)$ s.t.
$$
A:\{\omega \in \mathbb{R}: X^{-1}(B)= A \}, \mu(A)=m>0
$$
Since $A$ is an interval, we can set its upper and lower bounds as $\beta_1, \beta_2$. Since $A \in \mathcal {F}$, we can certainly find number $\alpha^{-1}$ such that there exist two disjoint intervals $A_1 \cup A_2=A$ with the same measure:
$$
A=A_1 \cup A_2, A_1 = [\beta_1, \alpha^{-1}], A_2 = [\alpha^{-1}, \beta_2], \mu(A_1)=\mu(A_2)=\frac{m}{2}
$$
Obvisouly $\alpha^{-1} \in A$, and, since $A$ is a preimage of $B, \alpha^{-1} = X^{-1}(\{\alpha\})$, and $\alpha \in B$. Therefore
$$
A_2 = \{\omega:X^{-1}(\alpha, \infty)\}
$$
and $\mu(X>\alpha) = \mu(A_2)=\frac{m}{2}>0$.
I think this is correct, but the hint for the problem is to use the continuity of probabilities, which I didn't.
AI: Consider the expanding sequence of sets $A_n = \{ X > \frac{1}{k}\}$. Notice $ \bigcup_{k=1}^\infty \{ X > \frac{1}{k}\} = \{X > 0\}$. Since $A_n$ is an expanding sequence of sets, then by the Monotone Convergence Theorem for sets $$\lim_{n\to\infty} \mu(A_n) = \mu\Big(\bigcup_{k=1}^\infty A_k\Big) = \mu(\{X > 0\}) > 0$$
We then get that $$0 < \mu\Big(\bigcup_{k=1}^\infty A_k\Big)\leq \sum_{k=1}^\infty \mu(A_k)$$
Since the series $\sum \mu(A_k) > 0$ and each term is nonnegative, then at least one term must be positive, for otherwise $\sum \mu(A_k) = 0$.
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H: Show that $\sin\theta \tan\theta <2(1-\cos 3\theta )$
Show that $$\sin\theta \tan\theta <2(1-\cos 3\theta)$$ for $0<\theta<\pi/2$
MY ATTEMPT :
Let $$ t = \cos \theta $$
$$t \times E= (1-t^2) -2t(1-4t^3+3t) $$
$$t \times E= 8t^4-7t^2-2t+1 $$
After this, I'm confused about how to proceed !
Any help ? thank you
AI: As peter's comment explains, your inequality is false because $\sin(\theta)\tan(\theta)$ grows arbitrarily large as $\theta$ approaches $\frac{\pi}{2}$ from the left, while the RHS of the inequality does not.
You can show this by recalling that $\cos(x)$ is bounded between $-1$ and $1$ for all $x \in \mathbb{R}$, in particular, this means you get
\begin{align}
&1 \ge \cos(3\theta) \ge -1 \ \implies \ -1 \le -\cos(3\theta) \le 1 \ \\
\implies \ &0 \le 1-\cos(3\theta) \le 2 \ \implies \ 0 \le 2\left(1-\cos(3\theta)\right) \le 4
\end{align}
so the RHS of your inequality is always bounded between $0$ and $4$.
As for why the LHS blows up, peter again explains that the $\cos(\theta)$ in the denominator is the culprit. You can explicitly see this by taking the limit as $\theta \to \frac{\pi}{2}$ from the left. Doing this you get
$$
\lim_{\theta \to \frac{\pi}{2}^-}\sin(\theta)\tan(\theta) = \lim_{\theta \to \frac{\pi}{2}^-}\sin(\theta)\frac{\sin(\theta)}{\cos(\theta)} =\left[\lim_{\theta \to \frac{\pi}{2}^-} \sin^2(\theta) \right]\left[\lim_{\theta \to \frac{\pi}{2}^-} \frac{1}{\cos(\theta)} \right] = (1)\left[\lim_{\theta \to \frac{\pi}{2}^-} \frac{1}{\cos(\theta)} \right]
$$
since $\sin^2\left(\frac{\pi}{2}\right) = 1$. For the latter limit, recall that since $0 < \theta < \frac{\pi}{2}$, the function $\cos(\theta)$ is positive on this interval, and since $\cos\left(\frac{\pi}{2}\right) =0$, in the latter limit your dividing $1$ by a really small positive number, which results in a really big positive number. Using this, you conclude that
$$
\lim_{\theta \to \frac{\pi}{2}^-}\sin(\theta)\tan(\theta)= \lim_{\theta \to \frac{\pi}{2}^-} \frac{1}{\cos(\theta)} = + \infty
$$
You can also visually see this if you graph the $2$ functions:
where it is clear that $\sin(\theta)\tan(\theta)$ overtakes $2\left(1-\cos(3\theta)\right)$ in the interval we're analyzing.
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H: Equality of an Inequality
I encountered the following problem in my textbook-
Let $a, b, c$ be three arbitrary real numbers. Denote $$ x = \sqrt{b^2-bc+c^2}, y = \sqrt{c^2-ca+a^2}, z = \sqrt{a^2-ab+b^2} $$ Prove that $$ xy+yz+zx \ge a^2+b^2+c^2 $$
Textbook's Solution:
Rewrite x,y in the following forms $$ x = \sqrt{{3c^2\over 4}+\left(b-{c\over 2}\right)^2}, y = \sqrt{{3c^2\over 4}+\left(a-{c\over 2}\right)^2} $$
According to Cauchy-Schwarz inequality, we conclude $$ xy \ge {3c^2\over 4}+{1\over 4}\left( 2b-c\right)\left(2a-c \right) $$
which implies
$$ \sum_{cyc}xy \ge {3\over 4}\sum_{cyc}c^2 +{1\over 4}\sum_{cyc}\left( 2b-c\right)\left(2a-c \right) = \sum_{cyc}a^2. $$
My Approach:
From the given values of $x$ and $y$-
$$ xy = \sqrt{\left( c^2-bc+b^2\right) \left(c^2-ca+a^2 \right)} \ge c^2+c\sqrt{ab}+ab $$
So, $$xy+yz+zx \ge \sum_{cyc}a^2+\sum_{cyc}a\sqrt{bc}+\sum_{cyc}bc$$
And it rests to prove that-
$$ \sum_{cyc}a^2+\sum_{cyc}a\sqrt{bc}+\sum_{cyc}bc \ge \sum_{cyc}a^2 $$
$$ \sum_{cyc}a\sqrt{bc}+\sum_{cyc}bc \ge 0 $$
Now my question is that how to prove this? Or are we done?
And also if both approachs are correct, what is the equality case?
The equality case is not given in the textbook so I ask.
Thanks!
AI: Your first step is wrong.
Try $a=b=c$.
Id est, you got a right inequality, but after a wrong step and the proof of your last inequality is not relevant already.
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H: The meaning of definition written in the form ... is...
I am studying mathematics logic and i am now criticising all the things...
I met some definitions which are in the form of ... is ....
For example, A construction sequence for an expression $\alpha$ is a finite sequence $\alpha_1,...\alpha_n=\alpha$ such that each $\alpha_i$ is member of core set or it is a result of applying some $f\in P$ to $\alpha_i,...\alpha_j$.
My question is, is it true that the set of construction sequence for $\alpha$ is equal to the set of finite sequence $\alpha_1,...\alpha_n=\alpha$ such that each $\alpha_i$ is member of core set or it is a result of applying some $f\in P$ to $\alpha_i,...\alpha_j$?
Moreover, are there any formal way to construct definition in mathematics?
AI: is it true that the set of construction sequence for is equal to the set of ...
Yes, trivially -- they are equal by definition, because the same set is being talked about. There is the same written to the left and right, it just looks different.
A definition is just syntactic sugar for the meta language: It means that instead of writing "a finite sequence $\alpha_1,...\alpha_n=\alpha$ such that each $\alpha_i$ is member of core set or it is a result of applying some $f\in P$ to $\alpha_i,...\alpha_j$", we can write "a construction sequence for $\alpha$". There is nothing deeply set-theoretic going on behind it: A definition is just inventing a new word for some mathematical concept to use as a convenient abbreviation to refer to said concept in texts. The structure underlying the concept already existed before inventing a special word for it, and writing up a definition doesn't create a new kind of entity nor does it perform any sort of set-theoretic operation on objects: It's just giving that thing a short name (the definiendum) so we don't have to spell out its exact description (the definiens) every time we want to mention it in a text; the short name and the long description refer to the same one thing.
If you want to define it formally, you can say something like a definition being an expression in the meta language, "A := B", whose semantics is that the extension of A is identical to the extension of B (and the extension of B again is defined by what the extension of the word "sequence" etc. is, until we come down to words for concepts that can no longer be decomposed into already defined concepts any further, such as the concept of a set). This doesn't entirely capture the idea of a definition; we rather need some kind of operational semantics with meaning assignment rather than merely a statement about equality, but you get the idea.
|
H: Rootspaces are $\mathop{ad}$ nilpotent
$ \DeclareMathOperator{\ad}{ad}$
Let $L$ be a semisimple Lie algebra with root space $L=H \oplus \bigoplus_{\alpha \in \Phi}L_\alpha$. Let $x\in L_\alpha$ with $\alpha\neq 0$. I want to show
Then $\ad x $ is nilpotent.
I know that if $\alpha, \beta\in H^*$ then $[L_\alpha,L_\beta]\subset L_{\alpha+\beta}$. I think I should make if we could show that there are only finitely many non-zero $L_\alpha$, we could use this fact to push $x$ into a trivial root space. Then it should follow that $(\ad x)$ is nilpotent? I am slightly confused as to what it means for $(\ad x)$ to be nilpotent, is it that $(\ad x)^n$ is zero, or is it that $\ad^n x$ is zero?
Note this is a Proposition in Humphreys book but I do not see how it follows directly.
AI: I presume we are in the finite-dimensional case, when there are only finitely
many non-zero $L_\alpha$.
$\text{ad}\, x$ is a map from $L$ to $L$ and then its $n$-th power
$(\text{ad}\, x)^n$ is also a map from $L$ to $L$. If $x\in L_\alpha$ and $y\in L_\beta$
then $(\text{ad}\, x)(y)=[x,y]\in L_{\alpha+\beta}$ and
then $(\text{ad}\, x)^n(y)\in L_{n\alpha+\beta}$. If $\alpha\ne 0$ then there is $n$
large enough so that $L_{n\alpha+\beta}=0$ for all $\beta$ with $L_\beta\ne0$
since there are only finitely many root spaces. Then
$(\text{ad}\, x)^n(y)=0$ for all $y\in L_\beta$ and so
$(\text{ad}\, x)^n(y)=0$ for all $y\in L$.
|
H: If $F, K$ are fields, $F$ algebraically closed, and $F \subseteq K$ then $K = F$?
I want to show that an algebraically closed field $F$ cannot be contained in a larger field $K$. So if $F \subseteq K$ then $F = K$ for all fields $K$.
Here's my attempt at a proof:
For contradiction, Say $F \subsetneq K$. Hence there is an element $k \in K$, $k \not \in F$.
if $k$ is algebraic over $F$, then we adjoin $k$ into $F$. So consider $F(k) \equiv F[X] / (p)$ for some polynomial $p \in F[X]$. Since $F(k)$ should be a field, we need $(p)$ to be maximal [ring quotient maximal ideal is field]. Since we have that $F$ is algebraically closed, irreducible polynomials [which generate maximal ideals] are going to be linear. Hence $(p)$ is maximal and F$(k)$ is a field iff $p = (x - f_\star)$ for some $f_\star \in F$. But $F[X]/ (x- f_\star) \simeq F$, since quotienting by $(x - f_\star)$ keeps only degree 0 polynomials, which are the constant elements. So we have that $F(k) = F$. Hence we cannot have $k \not \in F$, giving us the desired contradiction.
If $k$ is transcendental over $F$, I feel that some argument ought to hold, but I don't know how to proceed.
Is there a counter-example, where $F$ is algebraically closed, while still possessing an extension $K = F(k)$ for some $k$ that is transcendental over $F$? The only algebraically closed field I have experience with, $\mathbb C$, does not allow such a thing to happen as far as I am aware.
AI: The algebraic numbers are an algebraically closed field that is properly contained in $\mathbb{C}$ and many other fields. For example, if $F$ is the field of algebraic numbers, $K=\mathbb{C}$, and $k=\pi$, then we are precisely in the second case of your proof sketch, and we of course cannot show that $k$ is in $F$.
Given any (algebraically closed field) $F$ one can always construct $F(t)$ where $t$ is transcendental with respect to $F$ to get a bigger field. Then you can take the algebraic closure of $F(t)$ to get a bigger algebraically closed field.
By adjoining a lot of transcendentals you can get algebraically closed fields of arbitrarily large cardinality.
I will add some details that have by now been talked about in other comments. Let $F$ be a field and let $X$ be a set of variables. Let $F[X]$ be the ring of polynomials whose variables come from $X$ and coefficients from $F$. Now let $F(X)$ be the field of fractions of $F[X]$ (see: https://en.wikipedia.org/wiki/Field_of_fractions). You can also view $F(X)$ as the field of rational functions in variables from $X$ with coefficients in $F$. Now $F(X)$ is a new field properly containing $F$. Moreover, $|F(X)|=\max\{|F|,|X|,|\mathbb{N}|\}$.
|
H: For what values of $k$ is the following matrix diagonalizable?
The matrix is:
$$\begin{bmatrix}k&1\\0&k^2\end{bmatrix}$$
I am aware that a square matrix is diagonalizable if there exists a matrix $P$ such that $D = P^{-1 }* A * P$ is a diagonal matrix.
AI: You can see that the set of the eigenvalues of the matrix is $\{k,k^2\}$.
If $k=k^2$ (that is, if $k=1$ or $k=0$) then our matrix coincides with his Jordan form. Since a matrix is diagonalizable if and only if his Jordan form is diagonal, you can conclude that our matrix is not diagonalizable.
If $k \neq k^2$, on the other hand, the matrix has distinct eigenvalues and as such is diagonalizable.
Edit: This problem is gonna be very difficult if you approach it using only the definition. You need to know a bit about eigenvalues and eigenspaces to solve it quite easily. It's a bit of theory that isn't very hard but can't be all put in the space of an answer. You can find it in any book on basic linear algebra, though (disregard the observation on the Jordan form, you don't need it to solve this).
Sorry, I'm not able to give you a simpler answer that's not very long. Hope someone else can help you in a better way. In the meantime, notice that there are real values of $k$ such that the matrix (which I'll call $A$) is diagonalizable.
Take $k=2$ and:
$$P =\begin{pmatrix}
1 & 1\\
0 & 2
\end{pmatrix}
$$
I'll leave it to you to check that:
$$\begin{pmatrix}
2 & 0 \\
0 & 4
\end{pmatrix} = P^{-1}AP $$
|
H: Difference in eigenvalue equations
I came across a strange equation in the solution to a problem. It looks like this:$$My - Mx = \lambda x$$
In this problem $M$ is an $n\times n$ (full rank) matrix, and $x$ and $y$ are vectors. ($y$ is given in the problem and I am looking for $x$). Is there a solution for $x$ here?
AI: Suppose that $M+\lambda$ is invertible. Then
$$
My = (M+\lambda)x
$$
which means that
$$
(M+\lambda)^{-1} M y = x
$$
which would then directly give you $x$ if you know $y,\lambda, M$.
|
H: Preservation of convergence in measure by absolutely continuous measures
In a paper on Risk theory that I am reading, it is stated that, unlike convergence in $L_p$, $1\leq p<\infty$, convergence in measure is preserved within a collection of probability measures that are absolutely continuous. That is,
Suppose $\mu$ and $\nu$ are probability measures on a measurable space $(\Omega,\mathcal{F})$ and $\nu\ll \mu$. If the sequence $X_n$ of random variables converges to $X$ in $\mu$-measure, then $X_n$ converges to $X$ in $\nu$-measure.
This seems to be an easy enough problem, but I don't have a clear idea of how to start. I would appreciate any hints.
AI: Hint:
For finite measures, use this wiki fact: A sequence $X_n$ converges to $X$ in measure if and only if for any subsequence $X_{n_k}$ there is a sub-subsequence $X_{n_{k_h}}$ that converges to $X$ almost everywhere.
Then compare (for the relevant sub-subsequence):
$$ \mu(\{\omega \in \Omega: X_{n_{k_h}}(\omega) \mathrm{\; does\; not \; converge \; to \;} X(\omega) \}) $$
and
$$ \nu(\{\omega \in \Omega: X_{n_{k_h}}(\omega) \mathrm{\; does\; not \; converge \; to \;} X(\omega) \}) $$
|
H: Example which proves that a closed subset of an incomplete metric space need not be complete.
We know that a closed subset of a complete metric space is complete.
But I want to find a closed subset $A$ of an incomplete metric space $(X,d)$ such that $A$ is not complete.
AI: Let $x$ be an irrational number and use the interval $(-\infty,x] \cap \mathbf Q$, which is the same as $(-\infty,x) \cap \mathbf Q$ since $x$ is irrational.
|
H: 'Classical' Infinitesimals and Tangent Spaces
I do not know much differential geometry, and was led to this question from complex dynamics. It seems that it is often possible to reason 'infinitesimally' about maps between tangent spaces. For example, quasiconformal maps are typically motivated and thought of as sending infinitesimal circles to infinitesimal ellipses, but then defined via pullbacks of almost-complex structures. Why is this possible? What is the correspondence between 'classical' infinitesimals and maps on tangent spaces?
Put differently, why should "infinitesimal = lives in tangent space"?
AI: Without going into much detail, manifolds have the locally euclidean property which means that in a neighbourhood of a point there is an open set which looks like (is homeomorphic to) an open set in $\mathbb{R}^n$. The idea of infintesimals living in the (co)tangent space is basically that if I zoom into the open neighbourhood the set starts looking like $\mathbb{R}^n$ which is a vector space, so I can sort of make sense of infintesimals living in a vector space. This motivates us to define the tangent space more formally using curves and so on.
|
H: construct matrix group in GAP
I am having trouble construct the following group in GAP. It is a solvable primitive linear group acting on V where |V|=5^8. We know the Fitting subgroup is of order 2^6*4 (central product of extra special group E of order 2^7 with a cyclic group of order 4). On top of E/Z(E) we have a group A of order 6^4 acts on E/Z(E). Here A itself has a normal extra special group D of order 27 and A/D acts on D/Z(D) and A/D \cong GL(2,3). In some sense, G would be a maximal solvable primitive group on V=5^8.
If it is possible, I need similar construction in |V|=7^8.
AI: I cannot easily tell you how I did this calculation, but in case it is helpful anyway, here is the group that you are looking for.
F1 := Identity(GF(5));;
G := Subgroup (GL(8,5), [
F1*[
[ 1, 1, 4, 4, 2, 3, 2, 3 ],
[ 3, 2, 2, 3, 4, 4, 4, 4 ],
[ 3, 3, 3, 3, 1, 4, 4, 1 ],
[ 1, 4, 1, 4, 3, 3, 2, 2 ],
[ 3, 2, 2, 3, 1, 1, 1, 1 ],
[ 1, 1, 4, 4, 3, 2, 3, 2 ],
[ 1, 4, 1, 4, 2, 2, 3, 3 ],
[ 3, 3, 3, 3, 4, 1, 1, 4 ]
],
F1*[
[ 2, 0, 4, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 2, 0, 4, 0 ],
[ 0, 0, 0, 0, 3, 0, 4, 0 ],
[ 2, 0, 1, 0, 0, 0, 0, 0 ],
[ 0, 3, 0, 1, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 2, 0, 4 ],
[ 0, 0, 0, 0, 0, 2, 0, 1 ],
[ 0, 2, 0, 1, 0, 0, 0, 0 ]
]
]);;
gap> Size(G);
331776
gap> StructureDescription(G);
"((((((C2 x ((C4 x C2) : C2)) : C2) : C2) : ((C3 x C3) : C3)) : Q8) : C3) : C4"
|
H: Rolling Dice Game, Probability of Ending on an Even Roll
The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game stops.
What is the probability that the game ends on an even turn when $A$ rolls first?
Now the book gives the answer as $\frac{4}{7}$, however, when try to calculate I end up with $\frac{2}{11}$.
Below is my work:
To calculate this probability, we decompose the event into two disjoint events, (a) the event where $A$ wins on an even roll, and (b) the event where $B$ wins on an even roll.
(a) Now, the probability $A$ wins can be calculated as follows
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{1}{3}\\
= \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{4}{81}\biggr)^k = \frac{2}{27}\cdot \frac{1}{1- \frac{4}{81}} = \frac{6}{77}.
\end{align*}
(b) Similarly we calculate the probability $B$ wins on an even roll as
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot \frac{2}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot\frac{2}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{4}{9}\\
= \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{4}{81}\biggr)^k = \frac{8}{81}\cdot \frac{1}{1- \frac{4}{81}} = \frac{8}{77}.
\end{align*}
Therefore, it follows that the probability of the game ending on an even number of rolls is
\begin{equation*}
\frac{6}{77} + \frac{8}{77} = \frac{2}{11}.
\end{equation*}
Am I missing something?
AI: Thanks to the comment of @JMoravitz I realized my mistake. I was interpreting turns as the rolls $A$ AND $B$, as in $\{A_1,B_1\}, \{A_2,B_2\}, \dots$. In reality the question is merely asking what the probability of $B$ winning if $A$ rolls first.
The work is as follows:
We calculate the probability of $B$ winning. Denote the probability of $B$ winning on their $i$th roll as $S_i$. Now, the probabilities of $B$ winning on her first roll, second roll, third roll, etc., are as follows:
\begin{equation*}
P(S_1) = \biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \quad P(S_2) = \biggr(\frac{2}{3}\biggr)\biggr(\frac{1}{3}\biggr)\biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \quad P(S_3) = \biggr(\biggr(\frac{2}{3}\biggr)\biggr(\frac{1}{3}\biggr)\biggr)^2\biggr(\frac{2}{3}\biggr)\biggr(\frac{2}{3}\biggr), \dots
\end{equation*}
It then follows that in general that $\displaystyle P(S_i) = \biggr(\frac{2}{9}\biggr)^{i-1} \biggr(\frac{4}{9}\biggr).$ Thus, it follows that the probability of $B$ winning is calculated as
\begin{equation*}
P(S) = P\biggr(\bigcup_{i=1}^\infty S_i\biggr) = \sum_{i=1}^\infty P(S_i) = \sum_{i=1}^\infty \biggr(\frac{2}{9}\biggr)^{i-1} \biggr(\frac{4}{9}\biggr) = \frac{4}{9} \sum_{i=1}^\infty \biggr(\frac{2}{9}\biggr)^{i-1} = \frac{4}{9} \cdot \frac{9}{7} = \frac{4}{7}.
\end{equation*}
|
H: Example of absolutely continuous function $f$ with $\sqrt{f}$ not absolutely continuous
I'm looking for an example of a function $f$ that is absolutely continuous, but $\sqrt{f}$ is not absolutely continuous.
I've been playing around with the Cantor-Lebesgue function, but I feel like there should be something simpler.
AI: I believe
$f(x) = x^2 (\cos \frac1x)^4$
is an example on the interval $(0,1)$. While a proof is certainly needed, the key observation is that sum of the infinitely many local maxima of $f$ converges (indeed $f'$ is uniformly bounded), but the sum of the infinitely many local maxima of $\sqrt f$ does not converge.
|
H: What is the problem with this method while integrating $(e^x-(2x+3)^4)^3$?
What is the problem with this method while integrating $(e^x-(2x+3)^4)^3$?
I already know what is its integration. I collected the answers from Quora (black ones) and WolframAlpha website (the red one).
Alright, then I tried to solve the integration in this method. But, the answer appeared different. The mathematical approach seemed very legitimate to me. Where did I go wrong, would you kindly point that out?
here's my answer
now here is the difference between my itegrals(1) graph and the graph that gave wolphrapalpha(2).
I used same method to solve this two functions i found. and it worked.
BUT WHY IT DID NOT WORK FOR THAT ONE, WHERE DID I WENT WRONG?
AI: Your methodology is highly nonstandard and suspicious, but I will try to formalize it.
You appear to deal with integrals of the form
$$
\int f(x)^n dx
$$
I think your general strategy is
Set $y= f(x)$
Integrate $y^n$ to get $y^{n+1}/(n+1)$.
Divide $y^{n+1}/(n+1)$ by $dy$ and then write everything in terms of $x$.
So in the end you are saying that the following is an antiderivative for $f(x)^n$
$$
\frac{f(x)^{n+1}}{(n+1) f'(x)}\tag{1}
$$
So let's take the derivative of the last function and see what we get (we should get $f(x)^n$ if your strategy is correct). Applying the quotient rule and chain rule, we get the derivative of the function in $(1)$ as:
$$
\frac{f(x)^n (f'(x))^2-\frac{1}{n+1}f(x)^{n+1} f''(x)}{(f'(x))^2}\tag{2}
$$
In general, there is no reason to expect that the function in $(2)$ is going to equal $f(x)^n$.
On the other hand, if we are in the special case that $f'(x)$ is a nonzero constant, then the expression in $(2)$ is equal to $f(x)^n$ since in this case $f''(x)=0$. This is precisely what happens in your two easier examples where the strategy works. But this is not the case in the more complicated example.
|
H: Find all functions $f$ that satisfy the following
Let $\Omega $ be an open, bounded, and connected subset of $\mathbb{C}$
Find all functions $f:\bar{\Omega }\rightarrow \mathbb{C}$ that satisfy the following conditions simultaneously :
$f$ is continuous
$f$ is holomorphic on $\Omega $
$f(z)=e^z$ for all $z\in \partial\Omega$
My work: $e^z$ is analytic and $f(z)=e^z$ on $\partial\Omega$ which is closed , then it contains all of its accumulation points.
Then for all $z\in \partial\Omega $ , $f(z)=e^z$ on a neighborhood of $z$ and $Germ(f-e^z,z)=0$
Then by principle of analytic continuation $f(z)=e^z$ on $\bar{\Omega }$
Correct ?
AI: Your argument is not valid because $f$ is not holomorphic in a neighborhood of $z\in \partial\Omega$.
But you can apply the maximum modulus principle to the difference $g(z) = f(z) - e^z$ and conclude that $g$ must be identically zero in $\Omega$.
|
H: Assume $T_n,T$ are bounded bijective linear operators $T_n \to T$ pointwise. Show $T_n^{-1}\to T^{-1}$ pointwise $\iff$ $\|T_n^{-1}\|\leq C$
Assume $T_n,T$ are bounded bijective linear operators $X \to Y$ and $T_n \to T$ pointwise. Show $T_n^{-1}\to T^{-1}$ pointwise $\iff$ $\|T_n^{-1}\|\leq C$
Note: $X,Y$ are banach spaces.
My proof:
Forward direction is uniform boundedness principle. Backwards:
Let us assume that $T_n^{-1}(y)\not \to T^{-1}(y)$ so there is a subsequence and $y$ s.t $\|T_{n_k}^{-1}(y)- T^{-1}(y)\|\geq \epsilon$. Now we know that $T_{n_k}(x)\to T(x)$ and since $T^{-1}_{n}$ are uniformly boundaed we get that $\|T^{-1}_{n_k}(T_{n_k}(x)-T(x))\|<C\|T_{n_k}(x)-T(x)\|\to 0$ Now just let $x=T^{-1}(y)$ and we arrive at contradiction. Is this correct? It seems roundabout in my opinon and there is probably more direct way to do it.
AI: You can do it directly without the contradiction. Let $y\in Y$ be arbitrary. Since $\{T_n\}, T$ are assumed to be bijective functions, there exist $T^{-1}(y) = x\in X$. We also have a sequence of vectors $T^{-1}_n(y) = x_n$. You noticed that $T^{-1}_n(T_n(x)) = x$ and $T^{-1}_n(T(x)) = T^{-1}_n(y) = x_n$. Hence
$$\begin{align*}|| T_n^{-1}(y) - T^{-1}(y) || = ||x_n - x || &= ||T^{-1}_n(T(x)) - T^{-1}_n(T_n(x)) ||\\
&= ||T_n^{-1} (T(x) - T_n(x))||\\
&\leq C ||T(x) -T_n(x)||\to 0 \end{align*}$$
Since $T_n\to T$ pointwise.
|
H: Two sequences $f_n$ and $g_n$ such that $\int_{[0,1]}f_n g_n$ does not go to $0$ as $n\rightarrow\infty$, with these conditions on $f_n$ and $g_n$
Question: Suppose $f_n, g_n:[0,1]\rightarrow\mathbb{R}$ are measurable functions such that $f_n\rightarrow 0$ a.e. on $[0,1]$ and $\sup_n\int_{[0,1]}|g_n|dx<\infty$.
Give an example of two sequences $f_n$ and $g_n$ such that $\int_{[0,1]}f_n g_n$ does not go to $0$ as $n\rightarrow\infty$.
Prove that for any such sequences $f_n$ and $g_n$, and every $\epsilon>0$, there exists a measurable set $E\subset[0,1]$ such that $m(E)>1-\epsilon$ and $\int_Ef_n g_ndx\rightarrow 0$.
My thoughts: I was thinking of doing something like $f_n=n\chi_{(0,\frac{1}{n}]}$, which I believe would converge pointwise to $1$ a.e...I am just having a hard time trying to think of a $g_n$ that would work such that the integral of their product over $[0,1]$ wouldn't go to $0$....
For the second question, I immediately was thinking Egorov, but I haven't quite been able to figure out how to use it here.
Any suggestions, ideas, etc. are appreciated! Thank you.
AI: Your functions $f_n$ work fine if you take $g_n=1$ for all $n$, since
$$
\int_0^1f_n=1
$$
for all $n$.
Given any such pair $\{f_n\}$, $\{g_n\}$, and $\varepsilon>0$, let $k=\sup_n\int_0^1|g_n|$. By Egorov's Theorem there exists $E\subset[0,1]$ with $m(E)>1-\varepsilon$ and $f_n\to0$ uniformly on $E$. . So there exists $n_0$ such that, for all $n>n_0$, we have $|f_n|\leq\varepsilon/k$. Then
$$
\int_E|f_ng_n|\leq\frac\varepsilon k\,\int_E|g_n|\leq\varepsilon.
$$
|
H: Line Integral gives no work done?
For the following question,
$$
\mathbf{F}=\langle-y, x\rangle
$$
For this field:
Compute the line integral along the path that goes from (0,0) to (1,1) by first going along the $x$ -axis to (1,0) and then going up one unit to (1,1) .
I got an answer of $0$, by doing:
But the answer key concludes that the answer is $1$:
To compute $\int_{C} \mathbf{F} \cdot d \mathbf{r}$ we break the curve into two pieces, then add the line integrals along each piece.
First, fix $y=0$ (so $d y=0$ ) and let $x$ range from 0 to 1 .
$$
\int_{x=0}^{x=1} \mathbf{F} \cdot d \mathbf{r}=\int_{x=0}^{x=1}-y d x+x d y=\int_{0}^{1} 0 d x=0
$$
Next, fix $x=1$ (so $d x=0$ ) and let $y$ range from 0 to 1:
$$
\int_{y=0}^{y=1} \mathbf{F} \cdot d \mathbf{r}=\int_{y=0}^{y=1}-y d x+1 d y=1
$$
We conclude that $\int_{C} \mathbf{F} \cdot d \mathbf{r}=1$
I understand the solution from the answer key, but I don't get why my solution doesn't work. Please assist.
AI: Your error is in the second integral.
Along the path from $(0, 0)$ to $(1, 0)$ we can take as parameterization $x= t$ (from $0$ to $1$), $y= 0$ for all $t$. So $F(x,y)= \langle-y, x\rangle= \langle 0,t \rangle$ and the vector differential is $\langle dt, 0 \rangle$ so the integral is $$\int_0^1
\langle 0, t \rangle \cdot \langle dt, 0 \rangle = \int_0^t 0= 0$$ That is what you correctly have.
Along the path from $(1, 0)$ to $(1, 1)$ we can take as parameterization $x= 1$ for all $t$, $y= t$ (from $0$ to $1$). So $F(x,y)= \langle -y, x \rangle= \langle -t, 1 \rangle$ (not $\langle-t, 0 \rangle $ because $x= 1$) and the vector differential is $\langle 0, dy \rangle$ so the integral is $$\int_0^1 \langle -t, 1 \rangle \cdot \langle 0, dy \rangle = \int_0^1 dy= 1$$
So the complete integral is 1.
Again, your error is that on the second line, from $(1, 0)$ to $(1, 1)$ as $x$ is always $1$, not $0$.
|
H: Question involving ratios and Greatest Common Divisors
Consider 6 variables $a,b,c,x,y,z\in\mathbb Z$. We have two ratios, $a:b:c=1:2:3$ and $x:y:z=1:2:3$. We also have that $\gcd(a,x)=2$. What is $\gcd(a+b+c,x+y+z)$? I know that $\gcd(a,x)=2$, which means $a$ and $x$ are even, but I'm not too sure how to use this information to solve the problem. I also tried to use the ratios in some way, but I couldn't get far. I noted that $y=2x$, $z=3x$. $b=2a$, and $c=3a$, so $\gcd(a+b+c,x+y+z)=\gcd(a+2a+3a,x+2x+3x)=\gcd(6a,6x)$, but I'm not sure how to continue from here.
Could someone point me in the right direction to solve this problem? Thanks.
AI: Let's roll out the tanks.
$b = 2a; c=3a; y=2x;z=3x$.
So $a+b + c =6a$ and $x+y+z=6x$.
So you are asked to find the $\gcd(6a,6x)$ give that $\gcd(a,x)=2$.
Can you take it from there?
Use $\gcd(jm,jn) = j\gcd(m,n)$.
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H: Is the barrier problem for a linear program a convex problem?
By applying the barrier method to the linear programming problem min $c^T x, Ax ≥
b$, we can formulate:
$$\underset{x}{\text{minimize}} \hspace{0.5cm} c^T x - \sum_{i=1}^{m} \log(e_i - d_i^T x)$$
But is the sequence of optimization problems solved in the method convex? Why or
why not?
AI: Yes it can because of the following reasons:
$c^Tx$ is affine (and hence is convex).
$-\log x$ is convex for $x>0$
If $f(x)$ is convex for $x\in D\subseteq\Bbb R$, then $f(a+b^Ty)$ is convex where $y\in\{ z:a+b^Tz\in D\}$.
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H: Area between parabola and a line that don't intersect? 0 or infinity
Came across a problem on social media,
Find the area of the region bounded by a parabola, $y = x^2 + 6$ and
line a line $y = 2x + 1$.
I tried to draw it on paper and they didn't seem to intersect. So I drew them online (attached screenshot). My answer was 0, but someone said that we assume they meet at infinity and answer would be infinity. Parallel lines don't diverge like these do, so I think we can assume that they would never interest at infinity.
AI: $$x^2 + 6 = 2x + 1$$
$$x^2 - 2x + 5$$
$$\frac{2 \pm \sqrt{4 - 4(5)}}{2}$$
As you can see by analyzing the discriminant, this quadratic has no real roots, so there are no points at which the two curves intersect. You could say that the area between the curves tends to infinity. As was stated in the comments, whoever posted this most likely intended to include more information/restrictions.
Also, these two curves will not "meet at infinity." Both diverge as $x$ gets arbitrarily large
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H: Why is identity map on a separable Hilbert space not compact? False proof.
Why is identity map on a separable Hilbert space not compact? False proof.
Let $e_n$ be the orthonormal basis. Then the projection map onto $H$ is defined by $\sum(x,e_k)e_k$. What is stopping us from taking a finite part of this series. This gives us a compact operators, that converge to the identity map. Clearly something is wrong here not all Hilbert spaces are finite dimensional.
AI: Beacause that sequence of operators does not converge to the identity: if $n\in\Bbb N$,$$\left\lVert e_{n+1}-\sum_{j=1}^n\langle e_{n+1},e_k\rangle e_k\right\rVert=\lVert e_{n+1}\rVert=1$$and therefore$$\left\lVert\operatorname{Id}-\sum_{k=1}^n\langle\cdot,e_k\rangle e_k\right\rVert\geqslant1.$$
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H: Finding a general way to construct least degree polynomial having rational coefficient having irrational roots
Let p(x) be the least degree polynomial equation having $\sqrt[3]{7}$ + $\sqrt[3]{49}$ as one of it's roots, Then product of all roots of p(x) is ?
Following from the 'irrational roots occurring in pair for quadratics'
eg: $ (x^2 -2) = (x + \sqrt{2} ) ( x- \sqrt{2})$
I hypothesize that cubic roots would occur in triplets ( I look for proof of this and maybe generalization)
taking the generic cubic,
$$ p = ax^3 + bx^2 + cx + d $$
Evaluating this cubic at the root given would be a pain, but I assume we could get a system of two equations from equating rational and irrational part separately to 0.
Find the polynomial equation of the lowest degree with rational coefficients whose one root is.......?
I saw this question, but one thing bothered me, if we plug one of the factors of coefficent as 'x' we get a completely rational polynomial but not otherwise. My basic question was this trick swapping coefficent with 'x' only possible because roots were equivalent? Or could we do this if roots were not equal?
The reason I am very suspicious with this method, suppose
$$ p = (x-a)^2 = x^2 + a^2 -2ax$$
and, since [x=a] (? this won't work at non root points)
$$ p=x^2 +a^2 -2a^2 = (x-a)(x+a)$$?
AI: Here is an explanation using basics of number fields.
The number $x = \sqrt[3]7 + \sqrt[3]{49}$ lives in a cubic number field, namely $\Bbb Q[\sqrt[3]{7}]$. This field has three embeddings into $\Bbb C$: you can send $\sqrt[3]7$ to $\sqrt[3]7$, $\sqrt[3]7 \omega$ or $\sqrt[3]7 \omega^2$, where $\omega = \frac{-1+\sqrt 3 i}2$ is a primitive third root of unity.
This means that the three conjugates of $x$ are:$$\sqrt[3]7 + \sqrt[3]{49}, \sqrt[3]7\omega + \sqrt[3]{49}\omega^2, \sqrt[3]7 \omega^2 + \sqrt[3]{49} \omega.$$
They are the roots of the minimal polynomial that you are looking for.
It is easy to see that their product equals $56$.
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H: Are $\mathbb{C}-\mathbb{R}$ imaginary numbers?
Background
I am teaching senior high school students about the structure of numbers.
Start from defining $\mathbb{Q}$ and $\mathbb{R}$ as the rational and real numbers respectively, we can define $\mathbb{R}-\mathbb{Q}$ as the irrational numbers.
I am trying to use the same logic to define imaginary numbers by making use of the relationship between $\mathbb{R}$ and $\mathbb{C}$. Another definition for imaginary numbers is
numbers that become negative under squaring operation.
Let $\mathbb{C}$ and $\mathbb{R}$ be the complex and real number sets respectively. Are $\mathbb{C}-\mathbb{R}$ imaginary numbers?
AI: Imaginary numbers are real multiples of $\mathrm{i}$. So the complex number $1+\mathrm{i} \in \Bbb{C} \smallsetminus \Bbb{R}$ is neither real nor imaginary.
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H: For a function $f: X \to Y$, if $Y-V$ is finite, when is $X - f^{-1}(V)$ finite?
I apologize if this is a silly question but I just do not know enough set theory (i.e., sizes) to understand if it's even silly.
My question is
Let $f: X \to Y$ be a function of sets. Suppose $V \subset Y$, and that $Y - V$ is finite (countable). Is $X - f^{-1}(V)$ finite (countable)? What conditions do we need to place on $f$ to guarantee it will be finite (countable)?
So I kind of have two questions. But I will take either.
Why I care:
I'm asking this because I want to know the following. Recall that for a set $X$, we can endow $X$ with the finite complement topology where a set $U \subset X$ is open if $X - U$ is finite. Denote this topology on a set $X$ as $FC_X$.
Now suppose $f: X \to Y$ is a function. As $X$ and $Y$ are sets, we may ask:
Does the function $f: X \to Y$ extend to a continuous function $f: (X, FC_X) \to (Y, FC_Y)$?
For such a function to be continuous, we need that $f^{-1}(U)$ is open if $U$ is open. In this case, that means that if $Y - U$ is finite then $X - f^{-1}(U)$ must be finite. But I do not know when that last statement holds, and hence it's my question above. My guess is that if $f$ is injective, then it will be true, but I don't really know if that's true. I lack the set theory knowledge to really attack such a problem.
Finally, if the above answer is true, it tells me that I've got a functor $F: \textbf{Set} \to \textbf{Top}$ where $X$ is sent to the finite complement topology. I'm also interested in the countable case, since that would give me another different functor. But that's besides the question.
AI: As pointed out by @Cronus in the comments, it is sufficient that $f$ has the property that the preimage of every point is a finite set. I believe it is also a necessary condition.
Let's just consider surjective functions, just for convenience (it really doesn't change anything since any function $f:X \to Y$ can be 'made surjective' by replacing its codomain with its image.)
To see that the condition is sufficient, note that if some $f: X \to Y$ has the property, then for finite $Y \setminus V$,
$$ X \setminus f^{-1}(V) = f^{-1}(Y \setminus V) = \bigcup_{y \in Y \setminus V} f^{-1}(y).$$
In other words, $X \setminus f^{-1}(V)$ is the union of a finite number of finite sets.
I believe this condition is also necessary, and here is a proof that seems to be correct. Assume that $f$ doesn't have the property, and thus there exists some $y$ with $f^{-1}(y)$ not finite. Taking $V = Y \setminus \{ y \}$, one then concludes that
$$ X \setminus f^{-1}(V) = f^{-1}(y),$$
which is not finite. Importantly, we have that $f^{-1}(V)$ does not intersect $f^{-1}(y)$ since no $x \in X$ can be mapped both into $V$ and into $\{ y \}$. Thus by contradiction we have that $f$ must have the finite-to-one property.
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H: Root space decomposition of $C_n=\mathfrak{sp}(2n,F)$
I want to find the root space decomposition of the symplectic lie algebra $\mathfrak{sp}(2n,F)=C_n$.
I use the notation from Humphreys.
The root space decomposition of a semisimple lie algebra $L$ is $L=H\oplus \bigoplus_{\alpha \in \Phi} L_\alpha$. Where $H$ is a maximal total subalgebra (this is more typically called cartan subalgebra). The $L_\alpha$ are the root spaces, and $\Phi$ is the root system.
First we must determine a suitable $H$. For this it seems we can pick the diagonal matrices in $C_n$. I think that this is toral since all its elements are diagonal and hence semisimple? To see that it is maximal, suppose not i.e. $H\subset H'$ where $H'$ is the maximal toral subalgebra. There there must be some $a\in H'$ that commutes with every $ha=ah$ for every $h\in H$. But I think by picking some $h$'s cleverly this implies that $a$ must also be diagonal.
My main confusion is about trying to find the roots, and then the root spaces. The roots are the $\alpha$ such that $L_\alpha$ is non zero. How are we supposed to find which $L_\alpha$ are non zero before finding the $\alpha$'s?
If we try to work directly from the definition we have $L_\alpha=\{x\in L \,|\, [h,x]=\alpha(h)x \quad \forall h \in H \}$, we are left with quite a complicated eigen value equation to solve. I think if we had an intuition for for the spaces should look like, we could use the fact that root spaces are one dimensional.
I have done this calculation for $\mathfrak{sl}(n,F)$ but that feel too prototypical to help get a feel for doing these. I would like to complete this calculation for $\mathfrak{sp}(2n,F)$ and then try again myself to do the other classical lie algebras.
AI: You're right, $H$, the subalgebra consisting of the diagonal matrices in ${sp}(2n, \mathbb C)$ do form a Cartan subalgebra. Clearly, $H$ is an abelian subalgebra consisting of diagonalisable elements.
But how can we see there is no bigger abelian subalgebra consisting of diagonalisable elements? I suggest you don't worry about showing that this straight away. Instead, I suggest you go ahead and show that the $L = sp(2n, \mathbb C)$ has a decomposition
$$ L = H \oplus \bigoplus_{\alpha} L_\alpha,$$
where each $\alpha \in H^\star$ is non-zero, and where $[h , x] = \alpha(h) x$ for $h \in H $ and $x \in L_\alpha$. Once you know that the $\alpha$'s are non-zero, then you know that no element outside of $H$ commutes with all the elements in $H$, which means that you can't make $H$ bigger and have it remain abelian.
So how do we find generators for these $L_\alpha$'s? I agree that it's a hard eigenvalue equation to solve. But actually, it's not that hard to guess a set of generators for the $L_\alpha$'s. In fact, the most natural vector-space basis for $L$ that you can think of gives you a set of generators for these $L_\alpha$'s!
I'll do the case of $sp(4, \mathbb C)$. A convenient basis for $H$ is
$$ H_1 := \begin{bmatrix} 1 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &0 &-1 & 0 \\ 0 &0 &0 &0\end{bmatrix}, \ \ H_2 := \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 1 & 0 & 0 \\ 0 &0 & 0 & 0 \\ 0 &0 &0 &-1\end{bmatrix}$$
And then, the $L_\alpha$'s are generated by
$$ X_{1, 2} = \begin{bmatrix} 0 & 1 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &0 &0 & 0 \\ 0 &0 &-1 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 1, \ \alpha(H_2) = -1)$$
$$ X_{2, 1} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 1 & 0 & 0 & 0 \\ 0 &0 &0 & -1 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = -1, \ \alpha(H_2) = 1)$$
$$ Y_{1, 2} = \begin{bmatrix} 0 & 0 & 0 &1 \\ 0 & 0 & 1 & 0 \\ 0 &0 &0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 1, \ \alpha(H_2) = 1)$$
$$ Z_{1, 2} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &1 &0 & 0 \\ 1 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = -1, \ \alpha(H_2) = -1)$$
$$ U_{1} = \begin{bmatrix} 0 & 0 &1 & 0\\ 0 & 0 & 0 & 0 \\ 0 &0 &0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 2, \ \alpha(H_2) = 0)$$
$$ U_{2} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 1 \\ 0 &0 &0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 0, \ \alpha(H_2) = 2)$$
$$ V_{1} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 &0 &0 &0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = -2, \ \alpha(H_2) = 0)$$
$$ V_{2} = \begin{bmatrix} 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 \\ 0 &0 &0 & 0 \\ 0 &1 &0 & 0\end{bmatrix} \ \ \ \ ({\rm with \ } \alpha(H_1) = 0, \ \alpha(H_2) = -2)$$
It shouldn't be too hard to generalise this to higher $n$!
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H: Conditions on inequalities $a>b$ and $b
If $a>b$ and $b<c$, and $a$ and $c$ are positive, under what conditions is $a<c$? I am just curious to know. I know that the following are true $\frac ab>1$, $\frac bc<1$, and $\frac ac<1$. Any ideas on what next?
AI: This may be what you are looking for.
$$\begin{cases} a>b \\ b<c \\ a=\min \left\{a,c\right\} \\ a≠c \end{cases} \Longrightarrow a<c$$
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H: Finding the local extrema of $f(x, y) = \sin(x) + \sin(y) + \sin(x+y)$ on the domain $(0, 2 \pi) \times (0, 2 \pi)$
I am trying to find the relative extrema of
$$f(x, y) = \sin(x) + \sin(y) + \sin(x+y), \text{ where } (x, y) \in (0, 2 \pi) \times (0, 2 \pi)$$
Setting the partial derivatives equal to zero gives
$$\frac{\partial f}{\partial x}(x, y) = \cos(x) + \cos(x+ y) = 0$$
$$\frac{\partial f}{\partial y}(x, y) = \cos(y) + \cos(x+ y) = 0$$
Subtracting the equations gives $\cos(x) = \cos (y)$, and since $0 < x, y < 2 \pi,$ we can see from the unit circle that this equation holds $\iff y = 2\pi - x \iff x+y = 2\pi$. Now using this information in the two equations above, we get
$$\cos(x) + \cos(2 \pi) = 0 \implies x = \pi$$
$$\cos(y) + \cos(2 \pi) = 0 \implies y = \pi$$
However, I graphed $f$, and this seems incorrect. In the pictures, we can see that there appears to be a local maximum around $(1, 1)$ and around $(5.5, 5.5)$. Could someone please tell me my mistake?
AI: We have that
$$\cos(x) = \cos (y) \implies x=y \:\lor\: x=2\pi-y$$
it seems you only have considered the second condition.
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H: Find all integer values of $a$ such that $a^2 - 4a$ is a perfect square
I am trying to determine a systematic way to find all the integer vales of $a$ such that $a^2 - 4a$ is a perfect square. If it helps, I already know that the two solutions to this equation are 0 and 4.
Furthermore, I wonder how I can prove that these are all the solutions.
AI: If $m\ge 0$ is such that $m^2 = a^2 - 4a$ we can notice that $m^2 = a^2 - 4a< a^2 -4a + 4 =(a-2)^2 = m^2 + 4$.
So $m < a-2$. Le $(a-2)- m = k$ so that $m+k = a-2$.
so $(m+k)^2 = (a-2)^2$
$m^2 + 2mk + k^2 = a^2 -4a + 4 = m^2 + 4$
$2mk + k^2 = 4$.
but $m\ge 0$ and $k>0$. There's just not that many choices! $2mk \ge 0$ so $k^2 = 4-2mk \le 4$ so $k\le 2$ so $k = 1, 2$ but $k=1$ means $2m + 1=4$ so $m=\frac 32$ is not an integer. So $k =2$ and $m =0$.
That's it $m = 0$ and $m+k = a-2$ so $0+2 = a-2$ so $a=4$.
==== or ....======
If $m^2 = a^2 -4a < a^2$ then $m^2 +4 = a^2 -4a + 4 = (a-2)^2$ is also a perfect square.
Is there ever a case of two perfect squares being exact $4$ apart?
There's something that comes to mind but... I'll ignore it because clever tricks are tricks and if you don't see them right away you should be able to work it out anyway....
But if $m^2 + 4=k^2=(a-2)^2$ then ... two ideas:
$k^2 - m^2 = (k-m)(k+m) = 4$ so either $k-m =k+m=2$ and $m=0$ and $k=2$ and $k=a-2$ so $a = 4$.
or $k-m =1$ and $k+m=4$ and $m=\frac 32$ and $k=\frac 52$ but that's not a perfect integer. Although we do have $(\frac 32)^2 = (\frac 92)^2 -4\cdot \frac {9}2$....
Le $k=m+i$ and so $m^2 + 4 = k^2 =m^2 +2mi + i^2$ so $2mi+i^2 = 4$ and the only options are $i=2$ and $m=0$ so $k =2$ and $a = k+2 =4$.
.....
Clever trick that I mentioned but didn't want to use...
$n^2 = 1 + 3 + 5 + ...... +(2n-1)$ so if $k^2 - m^2 = 4$ we have a sequence of consecutive odd numbers that add to $4$. And can only be $1+3 = 4$ so $m^2 =0$ and $k^2= 1+3 = 4 = 2^2$ so $m = 0; k=2; a=k+2=4$.
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H: Show that $\|uv^T-wz^T\|_F^2\le \|u-w\|_2^2+\|v-z\|_2^2$
Show that $\|uv^T-wz^T\|_F^2\le \|u-w\|_2^2+\|v-z\|_2^2$, assuming $u,v,w,z$ are all unit vectors.
AI: Another approach: note that for orthogonal matrices $U,V,$ we have
$$
\|uv^T - wz^T\|_F^2 = \|U(uv^T - wz^T)V\|_F^2 = \|(Uu)(Vv)^T - (Uw)(Vz)^T\|_F^2.
$$
So without loss of generality, we can assume that $u = v = (1,0,\dots,0)^T$, so $uv^T$ is the matrix with a $1$ as its $1,1$ entry and zeros elsewhere. The left-hand side is then given by
$$
\|uv^T - wz^T\|_F^2 = \|wz^T\|_F^2 + [(1 - w_1z_1)^2 - (w_1z_1)^2] \\
= (w^Tz)(z^Tw) + [1 - 2w_1 z_1] = 2 - 2w_1z_1.
$$
The right hand size is given by
$$
\|u - w\|^2 + \|v-z\|^2 = \|w\|^2 + [(1 - w_1)^2 - w_1^2] + \|z\|^2 + [(1 - z_1)^2 - z_1^2]
\\ = 2 - 2w_1 + 2 - 2z_1 = 4 - 2(w_1 + z_1),
$$
and from there the reasoning is similar.
Another approach for expanding the exact expression: note that
$$
M = uv^T - wz^T = \pmatrix{u & w} \pmatrix{v & -z}^T,
$$
so that
$$
\operatorname{tr}(MM^T) =
\operatorname{tr}[\pmatrix{u & w} \pmatrix{v & -z}^T\pmatrix{v & -z}\pmatrix{u & w}^T]
\\ =
\operatorname{tr}[\pmatrix{v & -z}^T\pmatrix{v & -z}\pmatrix{u & w}^T\pmatrix{u & w}]
\\ =
\operatorname{tr}\left[\pmatrix{1 & -v^Tz\\ -v^Tz & 1}\pmatrix{1 & u^Tw\\u^Tw & 1}\right]
$$
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H: Are all complex functions onto?
I am not sure whether this question even makes sense. But I was just wondering whether all inverse operations of functions defined in complex numbers will stay inside complex numbers. (i.e. we don't have to extend the complex number system):
$x^2$ is a well-defined function of real numbers alone and yet there is no real number such that $x^2 = -1$, i.e. there is no inverse, for $-1$ (and so we need complex numbers).
Is there a theorem that says that this kind of thing cannot happen with complex numbers? Maybe all continuous complex functions are onto? Or, maybe all taylor series with complex coefficients are onto?
AI: Since I feel like your question is trying to address finding some kind of generalisation of the fundamental theorem of algebra (which can be rephrased as saying any nonconstant complex polynomial is surjective), I think one of the best things you can get is Picard's little theorem.
First, let me mention that continuous complex functions are not necessarily surjective, even if they aren't constant: for example, the absolute value function $|\cdot|:\Bbb C\to\Bbb R_{\geq0}\subset\Bbb C$ is certainly continuous, but is also certainly not surjective.
Therefore, just having continuity is not enough, so if we want to remedy this by making the functions in question look "more like polynomials", then we ought to make them smoother; that is, (complex) differentiable.
It turns out being complex differentiable is quite a bit to ask: unlike in the real case, a function that is complex differentiable will automatically be analytic; that is, it will have a Taylor series expansion at the point where it is differentiable. Therefore, differentiable complex functions can be thought of as "infinite degree polynomials", and we can go back to the question of: does the fundamental theorem of algebra somehow generalise to this setting?
Cutting to the chase, one place we might end up is Picard's little theorem, which says that if our complex function is differentiable everywhere and also not a constant, then its image will be just about surjective; that is, its image will be $\Bbb C$ except possibly a single point.
Therefore, given a complex function $f:\Bbb C\to\Bbb C$ that is differentiable, you will be able to solve $f(z)=a$ for all $a\in\Bbb C$ with at most one exception.
For the record, an example of an entire function whose image is missing a point would be the exponential map $\exp:\Bbb C\to\Bbb C$, whose image is $\Bbb C\setminus\{0\}$.
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H: Suppose a pair of random variable is independent from another pair, does it mean that each random variable is independent from the other?
Let $(X_1, Y_1)$, and $(X_2, Y_2)$ be two pairs of random variables, and they are assumed to be independent.
Does it mean that:
$X_1$ is independent from $X_2$?
$X_1$ is independent from $Y_2$?
$Y_1$ is independent from $X_1$?
$Y_1$ is independent from $Y_2$?
Motivation:
I had this question because in a lot of applications in statistics, you would like to find a parameter such that the joint distribution
$$P_{Y_1, Y_2}(y_1, y_2| x_1, x_2; \theta)$$ is maximized. And there is often the assumption that $(X_1, Y_1)$, and $(X_2, Y_2)$ are independent and identically distributed.
Using this assumption ,
$$P_{Y_1, Y_2}(y_1, y_2| x_1, x_2; \theta) = P_{Y_1}(y_1| x_1; \theta)P_{Y_1}(y_2| x_2; \theta) = \prod_{i = 1}^2 P_{Y_i}(y_i| x_i; \theta)$$
However, for this calculation to work, you must be able to show,
$$P_{Y_1, Y_2}(y_1, y_2| x_1, x_2; \theta) = \dfrac{\Pr(Y_1 = y_1, Y_2 = y_2, X_1 = x_1, X_2 = x_2; \theta)}{\Pr(X_1 = x_1, X_2 = x_2; \theta)}$$
which means you must split the denominator $\Pr(X_1 = x_1, X_2 = x_2; \theta) = \Pr(X_1 = x_1; \theta)\Pr(X_2 = x_2; \theta)$. However, the independence of $X_1, X_2$ is not explicitly stated. So it is implicit in the assumption? Or is an assumption missing?
AI: YES, YES, NO and YES. If two sigma algebras are independent then any sub-sigma algebras of these are also independent. This proves 1) 2) and 4).
For 3) take two independent (non-constant) random variables $U$ and $V$ and take $X_1=Y_1=U, X_2=Y_2=V$.
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H: Evaluate $\int_0^1\frac{\mathrm{e}^{12x}-\mathrm{e}^{-12x}}{\mathrm{e}^{12x}+\mathrm{e}^{-12x}}\,\mathrm{d}x$
My work so far
$${\displaystyle\int_0^1}\dfrac{\mathrm{e}^{12x}-\mathrm{e}^{-12x}}{\mathrm{e}^{12x}+\mathrm{e}^{-12x}}\,\mathrm{d}x$$
using substitution with $u=\mathrm{e}^{12x}+\mathrm{e}^{-12x}$
$$={\displaystyle\int}\dfrac{1}{12u}\,\mathrm{d}u$$
$$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{12}}}{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$$
$${\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$$
which is the standard integral:
$$=\ln\left(u\right)$$
$$=\dfrac{\ln\left(u\right)}{12}$$
and using substitution with $u=\mathrm{e}^{12x}+\mathrm{e}^{-12x}$
$$=\dfrac{\ln\left(\mathrm{e}^{12x}+\mathrm{e}^{-12x}\right)}{12}+C$$
$$=\dfrac{\ln\left(\mathrm{e}^{-24}\left(\mathrm{e}^{24}+1\right)\right)}{12}-\dfrac{\ln\left(2\right)}{12}+1$$
which is rewritten as
$$\dfrac{\ln\left(\mathrm{e}^{-24}\left(\mathrm{e}^{24}+1\right)\right)-\ln\left(2\right)+12}{12}$$
Is my work correct so far? Also, I thought I was done at the last step, but I noticed this integral can be approximated to $0.9422377349564838$. How would I go about doing this part?
AI: If you want to approximate this numerically, we'll have to resort to series
$$\frac{\log \cosh(12)}{12} = \frac{\log\left(\frac{e^{12}+e^{-12}}{2e^{12}}\right)+12}{12} \approx \frac{\log(1-\frac{1}{2})}{12}+1 \approx 1 - \frac{1}{24} - \frac{1}{96}= \frac{91}{96} \sim 0.948$$
You won't need to consider the $e^{-12}$ until you want around $8$ figures of precision.
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H: Prove this tabulated integral $\int_0^\infty x^ne^{-\alpha x} \, dx=\frac{n!}{\alpha^{n+1}}$
I ran into this problem where I needed to use the following integral equality in my physics textbook.
$$\int_0^\infty x^ne^{-\alpha x} \, dx=\frac{n!}{\alpha^{n+1}}$$
where $n$ is a positive integer and $\alpha$ is a positive constant.
I was just wondering how one arrives at this equality.
AI: \begin{align}
& \int_0^\infty x^ne^{-\alpha x} \, dx \\[8pt]
= {} & \frac 1 {\alpha^{n+1}} \int_0^\infty (\alpha x)^n e^{-\alpha x} \, (\alpha\, dx) \\[8pt]
= {} & \frac 1 {\alpha^{n+1}} \int_0^\infty y^n \big( e^{-y} \, dy\big) \\[8pt]
= {} & \frac 1 {\alpha^{n+1}} \int u\, dv \\[8pt]
& \text{where } u = y^n \text{ and } dv = e^{-y}\, dy.
\end{align}
From here you integrate by parts: $\displaystyle \int u\, dv = uv - \int v\,du.$
In the $uv$ term, the value when $y=0$ will be $0$ except when $n=0.$
The value when $y\to\infty$ can be found via L'Hopital's rule to be $0.$
You should end up with $\displaystyle \frac 1 {\alpha^{n+1}} \cdot n \int_0^\infty y^{n-1} e^{-y} \, dy.$
So doing the same thing again will give you $\displaystyle \frac 1 {\alpha^{n+1}} \cdot n(n-1) \int_0^\infty y^{n-2} e^{-y} \, dy.$
And again, and you get: $\displaystyle \frac 1 {\alpha^{n+1}} \cdot n(n-1)(n-2) \int_0^\infty y^{n-3} e^{-y} \, dy.$
And so on, so you just need to recognize the pattern. Or to put it another way, use mathematical induction on $n.$
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H: About holomorph of a finite group being the normalizer of regular image
Here is part of Exercise 5.5.19 in Dummit & Foote's Abstract Algebra:
Let $H$ be a group of order $n$, let $K=\operatorname{Aut}(H)$ and $G=\operatorname{Hol}(H)=H\rtimes K$ (where $\varphi$ is the identity homomorphism). Let $G$ act by left multiplication on the left cosets of $K$ in $G$ and let $\pi$ be the associated permutation representation $\pi:G\to S_n$.
(a) Prove the elements of $H$ are coset representatives for the left cosets of $K$ in $G$ and with this choice of coset representatives $\pi$ restricted to $H$ is the regular representation of $H$.
(b) Prove $\pi(G)$ is the normalizer in $S_n$ of $\pi(H)$.
Deduce that under the regular representation of any finite group $H$ of order $n$, the normalizer in $S_n$ of the image of $H$ is isomorphic to $\operatorname{Hol}(H)$. [Show $|G|=|N_{S_n}(\pi(H))|$.]
I could easily show (a), but even before attempting to prove part (b), I was puzzled: Does it imply that $\pi$ is injective? The restriction of $\pi$ to $H$ is injective by part (a), but I don't think $\pi$ is injective in general.
For example, $D_8\simeq\operatorname{Aut}(D_8)$, so if $H=D_8$, then $K$ is a normal subgroup of $G$ (being of index 2), so $ker\pi=K\neq1$.
Am I wrong somewhere in my deduction?
AI: In the case of $H=D_8$, it's not true $K$ is index $2$; actually $K$ is index $|H|$ in $G=H\rtimes K$. And $K$ is never normal in $G$, unless it's trivial of course.
The best way to think about $G$ is as an "affine group" of $H$. Indeed, if $H=\mathbb{Z}_p^n$ then $G$ is literally the affine group of $H$ as a vector space. In general, we can think of $G$ as the subset of $S_H$ comprised of "affine functions" of the form $x\mapsto \alpha(x)b$ where $\alpha\in\mathrm{Aut}(H)$ and $b\in H$.
Conjugating $K$ by $H$ yields functions of the form $\alpha(xb^{-1})b=\alpha(x)\alpha(b)^{-1}b$; to be an automorphism of $H$ (element of $K$) it must preserve $e\in H$ as a function, which requires $\alpha(b)=b$ (which, conversely, is sufficient), and this is only true for all $b$ if $\alpha$ is the identity automorphism. On the other hand, conjugating $H$ by $K$ yields functions $\alpha(\alpha^{-1}(x)b)=x\alpha(b)$, which are still elements of $H$, so $H$ is normal in $G$.
Note $H$ is a transversal for $G/K$, and indeed $G$ acts on $H$ representing $G/K$ matches $G$ acting on $H$ by affine functions in the way I described above. You want to show $N_{S_H}(H)=G$ here.
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H: Prove that the functional in $C_c^0(\Omega)$ is a Radon measure
Let $\Omega \subset \mathbb{R}^n$ be an arbitrary open set and $(x_n)_{n \in\mathbb{N}} \subset \Omega$ a sequence. Let $(a_n)_{n \in\mathbb{N}} \subset \mathbb{C}$ be a sequence such that
$$\sum_{j=1}^{\infty} |a_j| < \infty.$$
I want to prove that the functional $T:C_c^0(\Omega) \longrightarrow \mathbb{C}$
given by
$$T(\varphi)=\sum_{j=1}^{\infty} a_j\varphi(x_j),\; \forall \; \varphi \in C_c^0(\Omega)$$
is a Radon meausure in $\Omega$, that is, I want to prove that $T$ is continuous when $C_c^0(\Omega)$ is equipped with the topology inductive limit of the spaces $C_c^0(K)$, here $K \subset \Omega$ is an arbitrary compact subset of $\Omega$.
I thought of the following: it is enough I prove that $ T $ is continuous in $C_c^0(K)$, for all $K \subset \Omega$ compact, where the space $C_c^0(K)$ is (in particular) a Banach space with the norm
$$|f|_K=\sup_{x \in K} |f(x)|, \; \forall \; f \in C_c^0(K).$$
That's what I thought true? This is the way?
AI: Your thought is in the right direction.
You can also consider $b_n=\max(a_n,0)$ and $c_n=-\min(a_n,0)$.
Then $T_+\phi=\sum_n b_n\phi(x_n)$ and $T_- \phi=\sum_n c_n\phi(x_n)$. Both $T_+$ and $T_-$ are nonenagtive bounded operators on $C_{00}(\Omega)$. The Reisz-Markov representation implies that $T_+$ and $T_-$ can be represented as positive finite measures on $(\Omega,\mathscr{B}(\Omega))$. $T=T_+-T_-$ and $|T|=T_+ +T_-$ (this last part requires a small justification, but it is not complicated), thus $T$ is represented by a Radon measure ($|T|$ is the variation measure of $T$).
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H: How can I evaluate ${\lim_{h\to 0}\frac{\cos(\pi + h) + 1}{h}}$?
I'm supposed to evaluate the following limit using the cosine of a sum and one of the "special limits" which are ${\lim_{x\to 0}\frac{\sin(x)}{x}=1}$ and ${\lim_{x\to 0}\frac{1-\cos(x)}{x}=0}$.
The limit is : ${\lim_{h\to 0}\frac{\cos(\pi + h) + 1}{h}}$
I currently have : ${\frac{\cos(\pi)\cos(h) - \sin(\pi)\sin(h) + 1}{h}}$
I'm not sure where to go from here and have been stuck for awhile, help would be very much appreciated.
AI: You have written it correctly. Now because ${\sin(\pi)=0}$ and ${\cos(\pi)=-1}$ it simplifies further to
$${\frac{\cos(\pi)\cos(h) - \sin(\pi)\sin(h) + 1}{h}=\frac{-1\times \cos(h) - 0\times \sin(h)+1}{h}=\frac{-\cos(h) + 1}{h}}$$
can you take it from here?
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H: Is eigenvalue multiplied by constant also an eigenvalue?
Let $A$ be an $n × n$ matrix.
If $\lambda$ is an eigenvalue of $A$ and $c$ is a nonzero scalar, then $c\lambda$ is another eigenvalue of $A$.
I found this on "Linear Algebra and its applications (Jim Defranza)", summary of Chapter 5.
It is acceptable, that eigenvectors multiplied by constant is ok, cause $A(cv) = cAv = c\lambda v = \lambda(cv)$.
But I don't understand $c\lambda$ is also an eigenvalue of $A$.
Thank you.
AI: Great question, and good catch!
It's a typo/brain-o. The author surely meant to write "If $v$ is an eigenvector of $A$ and ..."
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H: If $f ∈ C^∞(M)$ has vanishing first-order Taylor polynomial at $p$, is it a finite sum of $gh$ for $g, h ∈ C^∞(M)$ that vanish at $p$?
This is 11-4(a) in Lee's "Introduction to Smooth Manifolds":
Let $M$ be a smooth manifold with or without boundary and $p$ be a point of $M$. Let $\mathcal{I}_p$ denote the subspace of $C^\infty(M)$ consisting of smooth functions that vanish at $p$, and let $\mathcal{I}_p^2$ be the subspace of $\mathcal{I}_p$ spanned by functions of the form $fg$ for some $f, g \in \mathcal{I}_p$.
(a) Show that $f \in \mathcal{I}_p^2$ if and only if in any smooth local coordinates, its first-order Taylor polynomial at $p$ is zero.
The $\Rightarrow$ direction is easy enough, but the $\Leftarrow$ direction has a complication in that Taylor's theorem only tells you that $f$ is a finite sum of products of pairs of functions in $C^\infty(U)$ for some open $U$ around $p$. How can I extend those functions so that $f$ is a finite sum of products of pairs of functions in $C^\infty(M)$?
Concretely,
$$
f(x) = \sum_{i,j} c_{ij}(x) (x^i - p^i) (x^j - p^j)
$$
for some smooth functions $c_{ij} \colon U \to \mathbb{R}$. I tried using a smooth bump function to have $c_{ij}(x) (x^i - p^i)$ and $x^j - p^j$ go to $0$ outside of $U$ except for $x^0 - p^0$ which goes to $1$, and then try to extend $c_{00}(x)(x^0 - p^0)$ to be $f$ outside of $U$, but ran into difficulties getting things to equal when the bump function is between $0$ and $1$. It feel like this is the wrong track.
On the other hand, the usual formulation of this problem has $\mathcal{I}_p$ be an ideal of $\mathcal{O}_p$, the ring of germs of functions at $p$, in which case there's no need to extend the functions globally. I'm wondering if the $\Leftarrow$ direction is even true as stated.
Am I missing something?
(This question was asked before, but the answers only work in a single chart.)
AI: You only need a single chart.
By standard construction, there is a smooth function $\varphi\colon M\to[0,1]$ which is supported on a coordinate neighbourhood of $p$, $\varphi(x)=1$ for all $x$ near $p$, namely, doing the construction on a coordinate neighbourhood of $p$ and extend by 0 outside that neighbourhood. You can write
$$
f=\varphi f+(1-\varphi)f.
$$
The $(1-\varphi)f$ is already in $\mathcal{I}_p^2$ since $1-\varphi$ and $f$ both vanish at $p$ by supposition. The function $\varphi f$ is identical to $f$ near $p$, so you reduce to the case of $\mathbb{R}^n$.
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H: Evaluate the following series $ \sum_{n=0}^{\infty} \big(e^{(4n+1)\pi\sqrt{3}}+2+e^{-(4n+1)\pi\sqrt{3}}\big)^{-1} $
I found this convergent series while solving a calculus problem $$ \sum_{n=0}^{\infty} \big(e^{(4n+1)\pi\sqrt{3}}+2+e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}=A $$How can I evaluate A?
My Attempt
$$\sum_{n=0}^{\infty} \big(e^{(4n+1)\pi\sqrt{3}}+2+e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}=0.00429610616+1.5281084 \times 10^{-12}+5.3886194 \times 10^{-22}+...$$
The summation appears to converge at around 0.004308 using excel but how can I evaluate this sum analytically?
AI: Consider the partial sum
$$S_p=\frac{1}{2}\sum_{n=0}^{+\infty}\frac{1}{1+\text{cosh}[(4n+1)\pi\sqrt{3}]}$$ As mentioned in comments, a CAS gives
$$S_p=\frac{\psi _{e^{4 \sqrt{3} \pi }}^{(1)}\left(p-\frac{i}{4
\sqrt{3}}+\frac{5}{4}\right)-\psi _{e^{4 \sqrt{3} \pi }}^{(1)}\left(\frac{1}{12}
\left(3-i \sqrt{3}\right)\right)}{48 \pi ^2}$$ where appear the q-polygamma function.
If $p\to \infty$
$$\frac{1}{4 \sqrt{3} \pi }-\frac{\psi _{e^{4 \sqrt{3} \pi }}^{(1)}\left(-\frac{i \pi
-\sqrt{3} \pi }{4 \sqrt{3} \pi }\right)}{48 \pi ^2}\approx 0.00429610616779$$
As you noticed, the convergence is very fast
$$\left(
\begin{array}{cc}
p & S_p \\
0 &\color{red}{ 0.00429610616}62624620267200340161868494728474773150101 \\
1 &\color{red}{ 0.0042961061677905704599958358}661023829062918667327659 \\
2 &\color{red}{ 0.0042961061677905704605346978063372}090930922497948456 \\
3 &\color{red}{ 0.0042961061677905704605346978065272297646}032916399424 \\
4 &\color{red}{ 0.0042961061677905704605346978065272297646702992589}733 \\
5 &\color{red}{ 0.0042961061677905704605346978065272297646702992589970}
\end{array}
\right)$$ which is normal since
$$a_n=\frac{1}{1+\text{cosh}[(4n+1)\pi\sqrt{3}]} \sim e^{-\sqrt{3} \pi (4 n+1)}$$
$$\frac{a_{n+1}}{a_n} \sim e^{-4 \sqrt{3} \pi }\approx 3.53 \times 10^{-10}$$
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H: Transformation of function $y = x^2 $
I want to transform function $y = x^2$ to $y = 4 x^2 $. Now this transformed function can be thought of as $ y/4 = x^2$ or $ y = (2x)^2$. If it is $ y/4 = x^2$, then this is a vertical stretch by a factor of 4 and if it is $ y = (2x)^2 $, then this is a horizontal compression by a factor of 2. Now, point $(1,1)$ is on the original curve. If we think of vertical stretch, then $(1,1)$ is transformed into $(1,4)$ and if we think of horizontal compression, then $(1,1)$ is transformed into $(\frac{1}{2}, 1) $. So, where does point $(1,1)$ transform to ? It can not go into two different points depending upon how we think of the transformation of the function.
AI: Yes, it can. The transformations are different. Both take $(1,1)$ to a point on the curve. You can find some intermediate transformations, like $y/2=(\sqrt 2 x)^2$. This will take $(1,1)$ to $(\sqrt 2, 2)$, which is again on the new curve. There are many continuous transformations between the two curves.
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H: Specific Probability Question
So I had come up with this problem during my spare time and was wondering if my answers to the questions were correct? The problem is:
Let's say an individual does not go out that much from their house and lives in the basement, and so they decide to check the outside weather every time when that individual's friend/roommate goes out of the house. If the friend/roommate is wet when they go downstairs to the basement, then the weather is rainy. Moreover, if the friend/roommate is dry, then the weather is sunny. However, here is the catch:
The roommate/friend decides to flip a fair two-sided coin every time when he reaches back. If the coin reveals tails and the weather is rainy, then the roommate/friend decides to dry himself before going down to the basement, hence being dry. If the coin reveals tails and the weather is sunny, then the roommate/friend decides to take a shower before going down to the basement, hence being wet. Hence, the individual may predict the weather incorrectly if the coin lands on tails.
Now, what would the probability be such that the individual in the basement predicts that it is a rainy day and it is indeed rainy outside (finding P(Predicts rainy AND is actually rainy))? Also, what would the probability be such that the individual in the basement predicts that it is a rainy day given that it is actually rainy (P(Predicts rainy | actually rainy))? Using these probabilities, can we deduce if the individual's prediction of rain is independent of the weather actually being rainy that day?
My approach to this question was that we have 4 possible outcomes, which are:
Individual predicts rainy, roommate lands on tails, weather is actually dry
Individual predicts rainy, roommate lands on heads, weather is actually rainy
Individual predicts sunny, roommate lands on tails, weather is actually rainy
Individual predicts sunny, roommate lands on heads, weather is actually dry
Using this, P(Predicts rainy AND is actually rainy) = 1 outcome from 4, hence 0.25. And P(Predicts rainy | actually rainy) = 1 outcome from 2, hence 0.5. But what confuses me is finding the independence, since I would need the probability of the weather being rainy/sunny to decide the independence, but I have not included any probability within the question. So are my 3 answers to this problem correct? If not, what was wrong with my answer exactly? Thank you so much :)
AI: You are doing a great job doing a sanity check on the question and information and thinking about your answer.
Your answers are correct if you assume that there is an equal probability between it being sunny and rainy. In this case, the 4 outcomes you suggested are all equally likely.
However, if that is not the case, you will get different answers. For example, say there is a 10% chance of it raining. Then the probability of it being rainy will be greatly reduced, and instead of a uniform distribution over your 4 options, you would instead get something like this:
45%
5%
5%
45%
Hopefully you can see where I got these numbers. In this case, the answer to the second question is still 50%, but the first is instead 5%.
I hope that was helpful. Let me know if you have any questions.
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H: Evaluating $\iint dx\,dy$ over the region bounded by $y^2=x$ and $x^2+y^2=2x$ in the first quadrant
Identify the region bounded by the curves $y^2=x$ and $x^2+y^2=2x$, that lies in the first quadrant and evaluate $\iint dx\,dy$ over this region.
In my book the solution is like:
$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}} \, dx \, dy\\
&=\int_{x=0}^1 \big[y\big]_{\sqrt x}^{\sqrt{2x-x^2}}\,dx\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&{\begin{aligned}\\
=\int_0^1\sqrt{1-x^2}\,dx-\int_0^1\sqrt{x}&\,dx\text{(applying} \int_0^af(x)\,dx=&\int_0^af(a-x)\,dx \text{ in the first part)}\\
\end{aligned}\\}\\
&=\left[\frac{\sqrt{1-x^2}}{2}+\sin^{-1}x\right]_0^1-\left[\frac{x^{\frac{3}{2}}}{\frac32}\right]_0^1\\
&=\frac{\pi}{2}-\frac12-\frac23(1-0)\\
&=\frac{\pi}{2}-\frac76\\
\end{align}\\
$$
And I did it like:
$$\begin{align}\\
\iint dx\,dy &=\int_{x=0}^1\int_{y=\sqrt x}^{\sqrt{2x-x^2}}dx\,dy\\
&=\int_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx\\
&=\int_0^1\sqrt{1-(x-1)^2}\,dx-\int_0^1\sqrt{x}\,dx\\
&{\begin{aligned}\\
=&\left[\frac{x-1}{2}\sqrt{1-(x-1)^2}+\frac12\sin^{-1}(x-1)\right]_0^1&-\left[\frac23x^{\frac32}\right]_0^1\\
\end{aligned}\\}\\
&=-\frac{\pi}{4}-\frac23\\
\end{align}\\
$$
Which one is correct?
AI: Clearly, your area cannot be negative, so your result is immediately incorrect.
The system $$x = y^2 \\ x^2 + y^2 = 2x$$ is readily solved by substitution. We have $$\begin{align}
0 &= x^2 + y^2 - 2x \\
&= x^2 + x - 2x \\
&= x^2 - x \\
&= x(x-1).
\end{align}$$ Hence $x \in \{0, 1\}$ and the full solution set is $$(x,y) \in \{(0,0), (1, -1), (1, 1)\}.$$ In the first quadrant, the area of interest may be expressed as
$$\begin{align}
\int_{x = 0}^1 \int_{y = \sqrt{x}}^\sqrt{2x-x^2} \, dy \, dx
&= \int_{x=0}^1 \sqrt{2x - x^2} - \sqrt{x} \, dx \\
&= \int_{x=0}^1 \sqrt{1 - (1-x)^2} - \int_{x=0}^1 \sqrt{x} \, dx \\
&= \int_{u=0}^1 \sqrt{1-u^2} \, du - \left[\frac{2}{3}x^{3/2}\right]_{x=0}^1 \\
&= \int_{\theta = 0}^{\pi/2} \sqrt{1 - \sin^2 \theta} \cos \theta \, d\theta - \frac{2}{3} \\
&= \int_{\theta = 0}^{\pi/2} \cos^2 \theta \, d \theta - \frac{2}{3} \\
&= \int_{\theta = 0}^{\pi/2} \frac{1 + \cos 2\theta}{2} \, d\theta - \frac{2}{3} \\
&= \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\theta = 0}^{\pi/2} - \frac{2}{3} \\
&= \left(\frac{\pi}{4} + 0 - 0 + 0\right) - \frac{2}{3} \\
&= \frac{\pi}{4} - \frac{2}{3}.
\end{align}$$ This step-by-step calculation should resolve all doubt. This is because for a fixed $x \in [0,1]$, we note $$y = \sqrt{x} \le \sqrt{2x-x^2}.$$ Alternatively, we may change the order of integration, but this requires us to solve the equation for the circle in terms of $x$. We can do this by completing the square: $x^2 - 2x + y^2 = 0$ implies $$1-y^2 = x^2 - 2x + 1 = (x-1)^2,$$ hence $$x = 1 \pm \sqrt{1-y^2},$$ and we choose the negative root because we require $x < 1$. Therefore, the area can be expressed as $$\int_{y=0}^1 \int_{x=1 - \sqrt{1-y^2}}^{y^2} \, dx \, dy.$$ Both integrals evaluate to $$\frac{\pi}{4} - \frac{2}{3}.$$
As has already been noted, the figure is misleading because the point $(1,1)$ lies directly above the center of the circle at $(1,0)$.
We can also check our solution by noting that the desired area is equal to the area under a parabola $y = x^2$ on $x \in [0,1]$, minus the area of a unit square from which a quarter of a unit circle has been cut out; i.e., this is simply $$\int_{x=0}^1 x^2 \, dx - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{2}{3}.$$
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H: Property of a positive Lebesgue measure set in $\mathbb{R}^n$
Let $E\subset\mathbb{R}^n$ be a positive Lebesgue measure set. Can we always find a $x\in E$ such that for any $r>0$, $B(x,r)\cap E$ is positive Lebesgue measure in $\mathbb{R}^n$?
($B(x,r)$ denotes the ball of radius $r$ with centre at $x$ in $\mathbb{R}^n$)
AI: For ae. $x \in E$ we have $\lim_n {1 \over m B(x,{1 \over n}) } \int_{B(x,{1 \over n})} 1_E(t)dt = 1_E(x)$.
Pick any $x$ for which we have equality, then ${m (E \cap B(x,{1 \over n})) \over m B(x,{1 \over n}) } \to 1$.
In particular, for some $N$, if $n \ge N$ then $m (E \cap B(x,{1 \over n})) \ge {1 \over 2}{m B(x,{1 \over n}) } >0$.
Since $r \mapsto mB(x,r)$ is increasing we see that $m (E \cap B(x,r)) >0$ for all $r>0$.
Alternative:
We can prove this directly. Since the Lebesgue measure is inner regular, we can assume that $E$ is compact.
Now suppose that for all $x \in E$ there is some $r>0$ such that $m(E \cap B(x,r)) =0 $.
The $B(x,r)$ form an open cover of $E$ and hence there is a finite subcover
$B(x_k,r_k)$. Then
$mE \le \sum_k m (E \cap B(x,r)) = 0$.
Hence there is some $x$ such that for all $r >0$ we have $m(E \cap B(x,r)) >0 $.
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H: In factoring consecutive numbers, how soon do we expect to see the smallest prime not yet seen?
For $A\subseteq\{1,2,3,\ldots\}$ one can take $\Pr(A)$ to mean $\displaystyle \lim_{n\,\to\,\infty} \frac{|A\cap\{1,\ldots,n\}|} n,$ if one doesn't insist on probability being countably additive.
Suppose we have a long sequence of positive integers $m, m+1, m+2,\ldots, m+k$ and all of them are${}\ll\sqrt N.$ Among prime numbers${}<\sqrt N,$ suppose there are many that are not among the factors of any numbers in this long sequence and many that are. ("Many" must mean at least one, but typically many more.)
Can anything of interest be said about the probability that among the prime factors of $m+k+1,$ the smallest prime number not yet occurring in the factorizations of numbers in $[m,m+k]$ will appear?
AI: All primes up to $k+1$ divide one of $m,m+1,\dots,m+k$, so the primes that might divide $m+k+1$ are the primes greater than $k+1$. For each such prime $p$, the probability that it divides none of $m,\dots,m+k$ is presumably $1-(k+1)/p$, while the probability that it divides $m+k+1$ is presumably $1/p$. (I say "presumably" because I don't know how $m$ and $k$ are chosen; if $k$ is fixed then these probabilities are accurate.) In particular, the probability that $p$ is the smallest prime not dividing $m(m+1)\cdots(m+k)$ and that it does divide $m+k+1$ is
$$
\bigg( \prod_{k+1<q<p} \frac{k+1}q \bigg) \frac1p,
$$
where the product is over primes $q$ in the given range (each factor is the probability that $q$ does divide $m(m+1)\cdots(m+k)$).
Therefore the probability that the smallest prime not dividing $m(m+1)\cdots(m+k)$ does divide $m+k+1$ should be
$$
\sum_{p>k+1} \bigg( \prod_{k+1<q<p} \frac{k+1}q \bigg) \frac1p.
$$
Numerical calculations of this expression up to $k=8000$ suggest that this probability decays like, maybe, $1/(\sqrt k(\log k)^2)$? Very hard to guess exactly; I'm sure the asymptotics could be worked out with enough pain.
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H: Inherited Riemannian metric on a submanifold
I am a beginner in differential geometry and I am reading chapter 1 of Differential Geometry of Loring Tu. For a smooth manifold $M$, a Riemannian metric on $M$ is an assignment that assigns $p\in M$ to an inner product on $T_pM$, such that for any smooth vector fields $X,Y$ on $M$, the map $p\mapsto \langle X_p,Y_p \rangle$ is a smooth function on $M$.
Let $(M,\langle,\rangle_M)$ be a Riemannian manifold and $N$ be a submanifold. Then for each $p\in N$, $T_pN$ is a subspace of $T_pM$, so we can naturally define a Riemannian metric on $N$ by letting $\langle v,w\rangle_N=\langle v,w\rangle_M$ for $v,w\in T_pN, p\in N$. But how can we show that this Riemannian metric on $N$ satisfy the smoothness condition? I.e., for any smooth vector fields $X,Y$ on $N$, how can we show the map $p\mapsto \langle X_p,Y_p \rangle$ is a smooth function on $N$?
AI: Smoothness is a local property, so consider an open neighborhood $U\subseteq M$ of $p \in M$, and extend $X$ and $Y$ to vector fields $\widetilde{X}$ and $\widetilde{Y}$ on $U$. So the function $U \cap N \ni q \mapsto \langle X_q,Y_q\rangle_N$ is the restriction of the smooth map $U \ni q \mapsto \langle \widetilde{X}_q,\widetilde{Y}_q\rangle_M$, which by assumption is smooth. Restrictions of smooth maps are smooth, and so you are done.
If you're a more coordinate-oriented person, you can take coordinates $(x^j, y^\mu)$ adapted to $N$, i.e., such that $N$ is described by $y^\mu = 0$, so that the tangent spaces to $N$ are spanned by the first coordinate fields $\partial_j$. Then the metric matrix has the block form $$\begin{pmatrix} (g_{jk}) & (g_{j\lambda}) \\ (g_{\lambda k}) & (g_{\mu\nu})\end{pmatrix}$$where all the entries are smooth. In particular the first block $(g_{jk})$ is smooth, so you're done.
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H: Proof of equation
Suppose, $E_1, E_2, E_3$ and $E_4$ are subsets of a universal set $E$ such that $E_1\subseteq E_2$ and $E_4\subseteq E_3$. Then, $(E_1\cap E_3) \cup (E_2\cap E_4)=E_2\cap E_3$.
It is observed that any counter examples we take sayisfy this equation, but unable to establish a formal proof of this.
Edit: the above equation is incorrect but $(E_1\cap E_3)\cup (E_2\cap E_4)\subseteq E_2 \cap E_3$ holds. So, the question should have been under what conditions, the above equality will hold?
AI: $(E_1\cap E_3)\cup (E_2\cap E_4)\subseteq E_1 \cup E_4.$
So for $(E_1\cap E_3)\cup (E_2\cap E_4)= E_2 \cap E_3$ to hold, $E_2 \cap E_3 \subseteq E_1 \cup E_4 $ must hold.
In addition. $E_1 \subseteq E_2$ and $E_4 \subseteq E_3$,
$E_1 \cap E_4 \subseteq E_1\cap E_3$ and $E_1 \cap E_4 \subseteq E_2\cap E_4$
So $E_1 \cap E_4 \subseteq (E_1\cap E_3)\cup (E_2\cap E_4)$
Therefore if $E_2 \cap E_3 = E_1 \cap E_4 $ holds, then $(E_1\cap E_3)\cup (E_2\cap E_4)= E_2 \cap E_3$ also holds.
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H: True or false questions about divisibility and non-divisibility in the integers. I'm confused.
For $a$, $b$, c $\in \mathbb{Z}$.
True or false that:
If $a$ doesn't divide $b$ and $b$ doesn't divide $c$ then $a$ doesn't divide $c$
If $a$<$b$ then $a$ divide $b$
If $a$ doesn't divide $b$ and $b$ doesn't divide $c$ then $a$ doesn't divide $b+c$
If $a$ divide $b$ and $b$ divide $a$ then $a$=$b$
I'm very confused about divisibility and non-divisibility
AI: False. E.g. $a = 3, b = 4, c = 6$. Meanwhile if the "doesn't" changes to "does" then the statement becomes True.
False. E.g. $a = 3, b = 4$. Meanwhile if $a$ divides $b$ then $a \leq b$ ($a = b$ is okay).
False. E.g. $a = 3, b = 4, c = 5$.
False if either $a$ or $b$ can be negative, but if both must be positive or negative at the same time then True. This depends on the definition of "divisor" you need.
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H: The number of ways to represent a natural number as the sum of three different natural numbers
Prove that the number of ways to represent a natural number $n$ as the sum of three different natural numbers is equal to $$\left[\frac{n^2-6n+12}{12}\right].$$
It was in our meeting a year ago, but I forgot, how I proved it.
Let the needed number be $a_n$, where $n\geq6$ and let $b_n$ be number of ways to represent a natural number $n$ as the sum of two different natural numbers.
Thus, $a_n=b_{n-3}+b_{n-6}+...$ because we can go from $(a,b)$, where $a<b$, to $(1,a+1,b+1)$, $(2,a+2,b+2)$...
Thank you for your help!
AI: If $n=r+s+t$ is a representation of a positive integer as a sum of integers which $r>s>t>0$, then
$n-6=(r-3)+(s-2)+(t-1)$ is a representation of $n-6$ as a sum of integers $(r-3)
\ge(s-2)\ge(t-1)\ge0$, that is a partition of $n-6$ into at most three parts.
Therefore $a_n=c_{n-6}$ where $c_n$ is the number of partitions of $n$ into at most three
parts.
By conjugation of partitions, $c_n$ is the number of partitions of $n$ into parts
of size at most $3$. So the generating function of the $c_n$ is
$$C(x)=\sum_{n=0}^\infty c_nx^n=\frac1{(1-x)(1-x^2)(1-x^3)}.$$
Now use the normal manoeuvres with rational functions to find the $n$-th term:
write in partial fractions
$$C(x)=\frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{(1-x)^3}+
\frac{D}{1+x}+\frac{E+Fx}{1+x+x^2}$$
and go from there.
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H: Probability of picking buttons from a bag
A bag contains $30$ buttons that are colored either blue, red or yellow. There are the same number of each color ($10$ each). A total $4$ buttons are drawn from the bag. Compute the followings:
Find $n(\Omega)$.
The probability that at least $3$ of them are red?
The probability that there is at least one of each color?
This seems like a basic problem but my professor and I cannot agree on an answer.
I think the probabilities are $2/21$ and $100/203$ for parts $2$ and $3$ respectively. I used combinations to calculate the probabilities.
My professor said $n(A)/n(\Omega)$ is $3/15$ for both so that is the answer for both $2$ and $3$.
AI: A bag contains $30$ buttons that are colored either blue, red or yellow. There is the same number of each color ($10$ each). A total of $4$ buttons are drawn from the bag. Compute the following:
$1$. Find $n(\Omega)$.
$2$. The probability that at least $3$ of them are red.
$$\frac{{10\choose3}{20\choose1}+{10\choose4}}{{30\choose4}}=\frac{2610}{27405}=\frac{2\cdot3^2\cdot5\cdot29}{3^3\cdot5\cdot7\cdot29}=\frac 2{21}$$
$3$. The probability that there is at least one of each color.
$$\frac 12\times\frac{{10\choose1}{10\choose1}{10\choose1}{27\choose1}}{{30\choose4}}=\frac 12\times\frac{10\cdot10\cdot10\cdot27}{27405}=\frac 12\times\frac{2^3\cdot3^3\cdot5^3}{3^3\cdot5\cdot7\cdot29}=\frac {100}{203}$$
where the factor of $\frac12$ was to cancel double-counting, in that every combination like $(\underbrace{R_i}_{{10\choose1}}\underbrace{B_i}_{{10\choose1}}\underbrace{Y_i}_{{10\choose1}}\underbrace{R_j}_{{27\choose1}})$ is also counted in $(R_jB_iY_iR_i)$.
You are correct.
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H: To show a function with a singularity is holomorphic by proving it is bounded.
If $f\in H(A)$, where $A=\text{Ann($z_0;r_1,r_2$)}$.
Let $z\in \text{Ann($z_0;s_1,s_2$)}$, where $r_1<s_1<s_2<r_2$, define $g$ by
$$g(\zeta)=\begin{cases}
\frac{f(\zeta)-f(z)}{\zeta-z} & \text{if }\zeta\neq z \\
\ f'(z) & \text{if }\zeta=z
\end{cases} $$
,where $\zeta \in \text{Ann($z_0;s_1,s_2$)}$.
I want to show that $g$ is bounded, so $z$ is a removable singularity of $\frac{f(\zeta)-f(z)}{\zeta-z}$ and $g$ is holomorphic on $\text{Ann($z_0;s_1,s_2$)}$.
Since $f'(z)=l_0$ exists, $$ \forall \varepsilon>0 \;\;\exists \delta>0 \;\;\text{such taht } \forall \zeta\in D(z,\delta), \;\;| \frac{f(\zeta)-f(z)}{\zeta-z}|< \varepsilon+l_0$$
Can I use the definition of limit to show the boundedness of $g$? Or there is another way to show that.
Since $\text{Ann($z_0;s_1,s_2$)}$ is not compact, I think I can't say that $g$ is continuous and has maximum value.
Thanks for helping!!
AI: You have almost answered it yourself. $|g(\zeta)| \leq (\epsilon+l_0) $ for $|\zeta -z| <\delta$. Boundedness in some disk around $z$ is enough to prove that the singularity is removable.
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H: What is the deep meaning of this quotes acording to Sir David Hilbert logics?
One of the famous mathematician David Hilbert quotes:
“Wir müssen wissen. Wir werden wissen."
(We must know. We will know.)
What is the deep meaning of this quotes acording to Sir David Hilbert logics ?
AI: That was in Hilbert's radio speech. He wanted to axiomatize the whole mathematics, and believed that every axiomatic system is decidable. So, in particular, every well formulated math problem has a solution. It was before G$\ddot{o}$del.
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H: Let $h:[0,1] \times [0,1] \rightarrow \mathbb{R}$ be the function $h(x,y)=f(x)g(y)$. Show h is integrable.
Let $f,g:[0,1] \rightarrow R$ be bounded, nonnegative, and nondecreasing $f(x_1) \leq f(x_2)$ for all $x_1 \leq x_2$ functions. Let $h:[0,1] \times [0,1] \rightarrow \mathbb{R}$ be the function $h(x,y)=f(x)g(y)$. Show h is integrable.
Theorem: Let Q be a rectangle, and let $f: Q \rightarrow \mathbb{R}$ be a bounded function. Then $\underline{\int_Q} f \leq \overline{\int_Q}f$; equality holds if and only if given $\epsilon>0$, $\exists$ a corresponding partition P of Q for which $U(f,P)-L(f,P)<\epsilon$.
Lemma: Let $Q$ be a rectangle; ;et $f: Q \rightarrow \mathbb{R}$ be a bounded function. If P and P' are any two partitions of Q, then $L(f,P) \leq U(f,P')$.
Corollary: If $f,g: Q \rightarrow \mathbb{R}$ are bounded functions on a rectangle Q such that $\{x \in Q: f(x) \neq g(x) \}$is a finite set then f is integrable if and only if g is integrable. In this case $\int_Q f=\int_Q g$.
I don't have clue so far for this question, so I am trying to list some potentially useful theorem/lemma/corollary and wonder if someone can help out. Appreciate it.
AI: We have $|f(x)| \leqslant M_f$ and $|g(y)| \leqslant M_g$ for all $x,y \in [0,1]$. Assume $M_f, M_g > 0$. (Otherwise we have the trivial case $h = 0$).
Since $f$ and $g$ are each bounded and monotone and, hence, Riemann integrable on $[0,1]$, there exist partitions $P' = (x_0,x_1,\ldots, x_n)$ and $P''= (y_0,y_1,\ldots, y_m)$ such that
$$U(P',f) - L(P',f) < \frac{\epsilon}{2M_g}, \quad U(P'',g) - L(P'',g) < \frac{\epsilon}{2M_f}$$
Leaving some steps for you to complete ...
(1) Forming the partition $P = (P',P,'')$ of $[0,1]^2$ show that
$$U(P,h) - L(P,h)= [U(P',f)- L(P',f)] U(P'',g) + [U(P'',g)- L(P'',g)] L(P',f)$$
(2) Show that $|U(P'',g)| \leqslant M_g$ and $|L(P',f)| \leqslant M_f$, and
$$U(P,h) - L(P,h) < \epsilon$$
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H: Deriving the Cauchy integral formula from the residue theorem
I'm currently going through complex analysis, and I'm trying to grasp the concept of the whole residue theorem and so on.
I followed the derivation of the residue theorem from the Cauchy integral theorem, and I think I kind of understand what is going on there.
I thought about whether it's possible to derive the Cauchy integral formula from the residue theorem since I read somewhere that the integral formula is just a special case of the residue theorem.
I tried looking it up somewhere but didn't find anything.
AI: Sure it's a special case, since the residue of $g(z):=\dfrac{f(z)}{z-a}$ at $z=a$ is $f(a)$.
This is fairly trivial, but to elaborate a little, recall that $f$ is holomorphic. So we can write $f(z)=f(a)+f'(a)(z-a)+f''(a)/2(z-a)^2+\dots$, and the residue is easy to get.
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H: Proof Verification: $(x^n)_{n=1}^{\infty}$ diverges when $x>1$
Proof
Suppose for the sake of contradiction that $(x^n)_{n=1}^{\infty}$ converges to some limit $L$. Consider the identity $(1/x)^n (x^n) = 1$. Since this holds for all $n \in \mathbb{N}$, $\lim_{n \rightarrow \infty}(1/x)^n \lim_{n \rightarrow \infty} x^n = 1$, and $\lim_{n \rightarrow \infty} (1/x)^n = 1/L$. But $\lim_{n \rightarrow \infty} (1/x)^n = 0$ if $x > 1$, a contradiction. As such, $\lim_{n \rightarrow \infty}(x^n)_{n=1}^{\infty}$ diverges when $x>1$.
AI: A more reasonable argument would be to use the inequaiity $x^{n}=(1+(x-1))^{n} >1+n(x-1) \to \infty$.
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H: If $p,q$ are prime, $pq\pm 2$, $pq\pm 4$, $pq\pm 6$ cannot be all primes
Let $p,q$ be distinct primes. Prove that the six integers $pq-2$, $pq+2$, $pq-4$, $pq+4$,$pq-6$, $pq+6$ cannot be all primes.
This is Exercise 5.60 in Chartrand's Mathematical Proofs. The claim is intriguing as for $p=3$ and $q=5$, the four integers $pq-2$, $pq+2$, $pq-4$, $pq+4$ are all primes. I haven't been able to explain why it stops working when $pq-6$ and $pq+6$ are added in the hypothesis.
I think I should look for a proof by contradiction. I don't know where to use the assumption that $p$ and $q$ are distinct primes.
I wonder if there is something peculiar about the prime gaps in this problem. I don't have any other clue.
AI: Let's assume that $pq+2$, $pq+4$ and $pq+6$ are all primes. I'm not considering the other 3 because the same logic will work for them too. One of these three numbers $pq+2$, $pq+3$ and $pq+4$ must be a multiple of 3. But since $pq+2$ and $pq+4$ already are primes therefore $pq+3$ is a multiple of 3. So, $pq+3=3k$ for some natural number $k$.
which in turn implies that $pq+6$ which is just $(pq+3)+3 = 3(k+1)$ is also a multiple of $3$. Which contradicts our assumption that all three of these numbers are primes.
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H: Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$
I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\frac{a-ax^3/2-3bx^2}{(1+x^3)^{3/2}}$$
Finally by setting $a=2,b=0$, we get $$I(x)=\frac{2x}{(1+x^3)^{1/2}}+C$$
The question is: How to do it otherswise?
AI: $$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{2x^{-3}-1}{(x^{-2}+x)^{3/2}} dx$$
Now substitute $t=x^{-2}+x$.
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H: Show the function $f(x)=\begin{cases}|x|^x, &x\neq0 \\ 1, &x=0\end{cases}$ is not differentiable at zero.
By the definition of differentiability, we need to show the following limit doesn't exist:
$$\lim_{x \to 0}{\frac{|x|^x-1}{x}}$$
I've shown $\displaystyle \lim_{x \to 0}{|x|^x}=1$, but couldn't proceed since L'Hospital's rule can't be used (cannot differentiate |x| since it is not differentible at $x=0$).
Is there any smarter way to find this limit?
AI: If one knows that
$$
e^u=1+u+O(u^2), \qquad u \to0,
$$ then one may write
$$
|x|^x=e^{x\ln|x|}=1+x\ln|x|+O(x^2\ln^2|x|),\qquad x\to 0,
$$ giving
$$
\frac{|x|^x-1}{x}=\ln|x|+O(x\ln^2|x|),\qquad x\to 0,
$$
the conclusion being direct.
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H: $4$-element subset of $\{1..6\}$ that includes $1$ or $4$, and $2$ or $5$, and $3$ or $6$?
I am stuck on the following question which was asked in our combinatorics exam:
Let us consider the set $S=\{1,2,3,4,5,6\}$.
We want to find a set $A\subset S$ such that $A$ must have the following property:
either $1$ or $4$ must belong to $A$.
either $2$ or $5$ must belong to $A$.
either $3$ or $6$ must belong to $A$.
Also $A$ must have $4$ elements in it.
What are the possible ways to form the set $A$?
My try:
For the 1st element of $A$ we have $2$ choices namely $1,4$, for second we have $2$ choices namely $2,5$ and for $3rd$ element we have $2$ choices namely $3,6$.
Thus there are in total $2\times 2\times 2=8$ choices for the first 3 elements of $A$.
Now for the 4th element we have $3$ choices. So there will be $\binom{3}{1}=3$ choices.
So the total ways of forming $A$ is $8\times 3=24$ .
There was also a 2nd question which asked what if $A$ had $5$ or $6$ elements?
If $A$ had $5$ elements I answered that there will be $8\times \binom{3}{2}=24$ choices.
Now if $A$ had $6$ elements I answered that there will be $8\times \binom{3}{3}=8$ choices.
My both answers had gone wrong.Can someone please help me figure it out.
I dont understand why my answer is wrong and how can I get the correct answer.
Is it possible for someone to please help me out?
AI: It is never a good idea to start by fulfilling the condition and then letting the rest of the elements do whatever they want. You overcounting part of the results in this way. For example, the set $\{1,2,3,4\}$ is counted twice -- once when $1$ was selected in step 1 and once when it was $4$.
The correct answer will be to determine which couple is entirely in $A$. You have 3 ways to choose this couple, then 2 ways to choose a representative from each of the other couples. In total: $3\times 2 \times 2=12$.
Similarly, if $A$ had to have 5 elements, you need to choose which couple is not represented, then choose one element of this couple (and the 2 others come in completely). Total: $3\times 2=6$.
If $A$ to have 6 elements, there is one option.
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H: Compress a three digit number into a single number
If I have a three digit number like 293 Is there a method to compress it into one digit. Can any three digit number be rewritten into a single digit with some formula and then back to its original form?
AI: No, assuming by three digit number you mean $100\leq n\leq 999$, then there are $999-100+1=900$ three digit numbers. But there are only $10$ single digit numbers (the integers $0$ through $9$). Since sets with different cardinalities can not have a bijection between them, there is no invertible function which maps three digit numbers to single digit numbers.
Of course, if you do not need the map to be invertible, then there are any number of functions you can use. For example
$$f(x)=\left\lfloor\frac{x}{100}\right\rfloor$$
simply maps every three digit number to its first digit (i.e. $f(293)=2$).
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H: How to represent the given information correctly to solve for a particular solution to a differential equation?
Lupita's lawn is left unattended so that an infestation of weeds begins to take over. The rate of growth of the weeds is proportional to the area of lawn not yet invaded by weeds.
a) If $W \space m^2$ is the area of lawn taken over by weeds after t weeks, and the lawn has a total area of $A \space m^2$, write the differential equation that models the situation.
b) The area taken over by weeds grows from one quarter to one half the total area of the lawn in $T$ weeks. If $t = 0$ when $W = \frac{1}{4}A$, solve the differential euqation in part a).
First I write down the differential equation, which is: $\frac{dW}{dt} = k(A - W)$. I proceed and find the general solution for this differential equation which is: $W = A - Be^{-kt}$ for that I let $e^c = B$
.
I continue this with the information I know to interpret; that is when $t = 0$, $W = \frac{1}{4}A$.
This gives:
$\frac{1}{4}A = A - Be^{-k(0)}$
$\frac{3}{4}A = B$. Substituting this back to the equation:
$W = A - \frac{3}{4}Ae^{-kt}$
Now this is where I am stuck. I am unsure how to represent "The area taken over by weeds grows from one quarter to one half the total area of the lawn in $T$ weeks" mathematically. I have the proportional constant $k$ left to solve.
Do I have to do something like:$ \int _{\frac{1}{4}A}^{\frac{1}{2}A}\:\frac{1}{A-W}dW\:=\:\int _0^T\:k\:dt$ How can I solve for $k$, hence the equation?
FYI the answer is:
$W\:=\:A\:-\frac{3}{4}Ae^{ln\left(\frac{2}{3}\right)t}$
Thanks.
AI: These are first $T$ weeks (since the initial condition is $W_0 = \frac{1}{4} A$), you have:
$$ \frac{1}{4} A = A(1-\frac{3}{4} e^{k*0}) $$
and
$$ \frac{1}{2} A = A (1-\frac{3}{4} e^{k*T}) $$
Dividing the second one with the first one:
$$\begin{align}
2 &= \frac{(1-\frac{3}{4} e^{kT})}{(1-\frac{3}{4} e^{k*0})} \\
1 &= 2 - 2 \frac{3}{4} e^{kT} \\
\frac{1}{2} &= \frac{3}{4} e^{kT} \\
\ln(\frac{2}{3}) &= kT \\
k &= \frac{\ln(\frac{2}{3})}{T}
\end{align} $$
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H: In how many ways can a group of six people be divided into: 2 equal groups? 2 unequal groups, if there must be at least one person in each group?
In how many ways can a group of six people be divided into:
a) two equal groups
I have $^6C_3 \times \space ^3C_3 = 20$
So, to choose the first group I have $6$ possibilities of which I am choosing $3$. For the second group, I have $3$ remaining people of which $3$ must be chosen -> hence $^6C_3 \times ^3C_3 = 20$.
But the answer is $\frac{^6C_3}{2}$ but I don't understand why you divide by $2$.
b) two unequal groups, if there must be at least one person in each
group?
Applying the same logic as before, I got:
$$(^6C_2 \times ^4C_4) + (^6C_1 \times ^6C_5) = 51$$
But the answer is $^6C_1 + \space ^6C_2 = 21$
Could anyone explain how to solve these/the intuition behind it? Thanks in advance!
AI: If you're dividing 6 people into two groups $x$ and $y$, then $^6C_3$ would give you the total number of pairs of $(x,y)$ as well as $(y,x)$. To avoid the repetition of the pair $(y,x)$, you divide by two.
The same logic applies here: when you split into groups, the remaining people whom you didn't select to form a group form the second group. To avoid repetition, the number of groups is $^6C_1 + \space ^6C_2 = 21$.
If you're still confused, the best way to grasp this concept is to take 4 people $a,b,c,d$ and see in how many ways you can split them into groups of 2 manually (by writing down all cases).
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H: Is this series $\sum_{n=0}^\infty(\frac{n}{2n^2-1})^2$ convergent?
Is the series $\sum_{n=0}^\infty(\frac{n}{2n^2-1})^2$ convergent?
My attempt:
The series is convergent iff $\sum_{n=0}^\infty 2^ka_{2^k}$ convergent.
Then we need to show that $\sum_{n=0}^\infty 2^k (\frac{2^k}{2\times2^k-1})^2$ converges.
I want to find a series that bigger than the above series, but don't know what to do.
Can someone help me with that?
Or does root test works in this case?
AI: Yes, of course because for $n\rightarrow+\infty$ $$\frac{\frac{n^2}{(2n^2-1)^2}}{\frac{1}{n^2}}\rightarrow\frac{1}{4}$$ and $\sum\limits_{n=1}^{+\infty}\frac{1}{n^2}$ converges (to $\frac{\pi^2}{6}$).
Also, we have $$\sum_{n=2}\frac{n^2}{(2n^2-1)^2}<\sum_{n=2}^{+\infty}\frac{1}{n(n+1)}=\sum_{n=2}^{+\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}$$
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H: If $φ(p) > φ(k)$ for $k
So I was playing around with the Euler totient function on desmos, and found that whenever the function "spikes", we can add $1$ to it and I always found a prime number. With very powerful computers or software why can't we use this for finding prime numbers?
It's my first time on this site and the question maybe stupid but can someone can please explain? Thanks in advance!
AI: Just for fun, let's rephrase this into a theorem:
Theorem: if $\phi(n)>\phi(k)$ for all $k<n$ then $\phi(n)+1$ is prime.
Lemma: if $p$ is prime then $\phi(p)>\phi(k)$ for all $k<p$
Proof:
Let $C(m,n)=1$ if $\gcd(m,n)=1$ and $C(m,n)=0$ if $\gcd(m,n)\neq1$
Therefore $$\phi(x)=\sum_{n=1}^{x-1}C(x,n)$$
Since $p$ being prime implies $\gcd(p,k)=1$
$$\implies\phi(p)=\sum_{n=1}^{x-1}\gcd(p,n)=\sum_{n=1}^{x-1}1=p-1$$
Since that is the maximal possible sum, then $\phi(p)>\phi(k)$ for all $k<p$
Therefore $\phi(n)>\phi(k)$ for all $k<n$ implies $n$ is prime.
$n$ being prime implies $\phi(n)=n-1$, therefore $\phi(n)+1$ is prime.
QED
As for using this to find more primes. It's no more efficient than a prime sieve. Specifically, as user21820 pointed out
$\gcd(m,n)$ with $m≤n$ takes $O(\log n)$ multiplications and divisions on operands of bit-length $O(\log n)$, and each operation on b-bit integers takes $O(b^2)$ time using schoolbook multiplication, or $O(b\log b)$ time even with state-of-the-art algorithms. So $\gcd(m,n)$ would take $O((\log n)^2⋅\log(\log n))$ time using best known algorithms. It suffices for you to just say that the summation takes $Ω(n)$ time, which is silly because prime factorization would take $O(\sqrt{n}(\log n)^2)$ time even with schoolbook algorithms.
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H: Trivial Fundamental Group and Orientation
Maybe it is an easy question but I cannot figure out.
If the fundamental group of (you may assume compact) an $n$-dimensional manifold $M$ is trivial, i.e., $$\pi_1(M)=0,$$ then can we conclude that $M$ is orientable?
AI: If $M$ is non-orientable, the orientable double cover of $M$ gives you a connected cover space of $M$ where each point has a two-point fiber. Thus the fundamental group of $M$ cannot be trivial.
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H: Proving that $g\circ f$ is injective if $f$ and $g$ are injective. [Verification]
Let $f:X\rightarrow Y, g:Y\rightarrow Z$. Show that if $f$ and $g$ are injective, then $g \circ f$ is also injective.
My attempt:
$x \neq x’ \implies f(x) \neq f(x’)$
$f(x) \neq f(x’) \implies g(f(x)) \neq g(f(x’))$
$(A\implies B) \wedge (B\implies C) \implies (A\implies C)$
Therefore: $x\neq x’ \implies g(f(x))\neq g(f(x’))$
Hence $g \circ f$ is injective.
AI: Perfectly correct. You can also proceed as follows:
Let $x,x' \in X$. Then
$$(g \circ f)(x) = (g\circ f)(x')\implies g(f(x)) = g(f'(x)) \implies f(x) = f'(x) \implies x = x'$$
where the second implication uses that $g$ is injective and the third (and last implication) that $f$ is injective.
A good follow up exercise: show that the same holds if you replace 'injective' by 'surjective'.
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H: Let $f(x)=\lim_{n\to \infty} (1-\sin x + e^{\frac 1n} \sin x)^n)$. If ....
Let $f(x)=\lim_{n\to \infty} (1-\sin x + e^{\frac 1n} \sin x)^n)$. If $\lim_{u\to 0} (1+u \log (1+k^2))^{\frac 1u}=2k\log^2 (f(x))$, for $k>0$ and $x\in (0,\pi)$, find $x+k$
Log is to the base $e$
$$f(x)=e^{\sin x}$$
And $$\lim_{u\to 0} (1+u\log (1+k^2))^u=e^{\log (1+k^2)}= 1+k^2$$
So $$1+k^2=2k(\sin^2x)$$
How do I find $x$ and $k$ from here?
AI: Now, $$ 0=k^2-2k\sin^2x+1=(k-1)^2+2k\cos^2x,$$ which gives $k=1$ and $x=\frac{\pi}{2}.$
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H: Prove that if $f$ is bijective, then $f^{-1}$ is bijective. [Verification]
Let $f: X \to Y$ be bijective, and let $f^{-1}: Y \to X$ be it's inverse. Conclude that $f^{-1}$ is also invertible.
Suppose that $f^{-1}(f(x)) = f^{-1}(f(x')) \nRightarrow x=x'$ (not injective), then $x=x' \nRightarrow x=x'$ which is a contradiction. Hence it is injective.
For any $x$ there exists an $f(x)$. Suppose that there exists an $x$ such that $\nexists x' \in X: f^{-1}(f(x'))=x.$ But that means that for some $x$, $\nexists x'\in X: x'=x$. But that $x'$ is simply $x$. This means that for every $x$, there is a corresponding $x$ value that satisfies surjectivity.
AI: Injectivety:
Your solution is not correct, you have to suppose
$f^{-1}(x)=f^{-1}(x')$ and not $f^{-1}(f(x))=f^{-1}(f(x'))$. Here is correct procedure:
Suppose we have $x$ and $x'$ such that $f^{-1}(x)=f^{-1}(x')$. Then we have: $$f(f^{-1}(x))=f(f^{-1}(x'))\implies x=x'$$
and thus a conclusion.
Surjectivity:
There is no need to suppose not existence of $x'$ and involving $f$ in first place with $f^{-1}(f(x')) =x$. Remember you have $b$ and you have to find $a$ such that $f^{-1}(a)=b$. Here is faster solution:
Take any $b$, and let $a=f(b)$. Then $$f^{-1}(a) = f^{-1}(f(b)) =b$$ and thus a conclusion.
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H: Intuitive explanation of what happens when we remove functions from integrals by exploiting bounds
Suppose I have an integral like,
$$ P = \int_{0}^{\pi} x \sin x \cos^4 x dx$$
by the property,
$$ \int_{0}^{a} f(a-x) dx = \int_{0}^{a} f(x) dx$$
And, we do
$$ P = \int_{0}^{\pi} (\pi -x) \sin x \cos^4 x dx $$
Now if we add the first integral and this new one, we get
$$ P= \frac{\pi}{2} \int_{0}^{\pi} \sin x \cos^4 x dx$$
Now, why did the integrand dependent becoming independent of it at the end? why does the $\pi$ factor account for the removal of 'x'? I have seen many integrals where this trick was used to remove some functio in integrand to make it simpler but I never understood why it works.
The kind of answer I am looking for: A geometrical / graphical approach (if possible)
AI: You might like to think of $\int x dx$ first. We get
$$
P = \int_0^1x dx \quad (1)\\
= \int_0^1(1 - x)dx \\
= \int_0^1 1dx - \int_0^1x dx \quad [(2) - (3)] \\
= 1 - P \\
\therefore P = 1/2
$$
We can draw this pictorially as:
The core idea is the duality of $g(x) = (1- x)$ when $f(x) = 1-x$:
We can consider $\int (1-x)dx$ as performing the original integral "right to left"
But we can also interpret it as doing the original integral "top to bottom"
When doing as "top to bottom", we wind up recovering the original term $\int f(x) dx$ as being subtracted from the full area of $\int 1 dx$. This lets us get a "copy" of the integral $P$.
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H: Means of neighboring squares
mn squares of equal size are arranged to form a rectangle of dimension m by n where m and n are natural numbers. Two squares will be called 'neighbours' if they have exactly one common side. A natural number is written in each square such that the number in written any square is the arithmetic mean of the numbers written in its neighbouring squares. Show that this is possible only if all the numbers used are equal.
We can assume this
B
C A D
E
Therefore $A= (B+C+D+E)/4$
$4A = B+C+D+E$
What does mean ? Please solve the question.
AI: Take the minimal number $A$ in the rectangle.
Thus, by the given any neighbour numbers are equal to $A$.
Can you end it now?
About your new question. We can not solve this equation.
For example, $$4\cdot5=1+2+3+14=2+4+5+9=...$$
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H: What is the minimal Hamming distance of concatenation of some word and a part of it encoded using Hamming codes?
A word $M$ is of $n^2$ bits, $n>3$ is arranged in a $n\times n$ matrix. $A$ is the main diagonal of the matrix (that is elements $a_{i,j}$, $i=j$) and is encoded with Hamming code, resulting in $B$. The final result transmitted is word $M$ concatenated with $B$ (let us call it $R$).
What is the minimal Hamming distance of $R$?
The minimal Hamming distance of $B$ is $3$ as holds with any Hamming code. $M$ is not encoded at all so its Hamming distance is $1$. Although $B$ is part of $R$ but $M$ is also part of $R$ and we have no way of knowing if a potential error comes from $B$ or $M$ so the minimal Hamming distance of $R$ is $1$ as well.
Am I in the right direction?
AI: That is correct. The off-diagonal bits are not protected at all in this scheme, and toggling one of them does not affect any other bits. Therefore the minimum distance is $1$.
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H: Random Walk Around A Circle
I am having difficulties in solving the following problem
I guessed that the chain is irreducible when $\gcd(n,s)=1$. But I'm unable to proceed. Can someone help me? Any hint will be appreciated.
AI: You are right thinking the chain is irreducible iff $n$ and $s$ are coprime:
Suppose $n,s$ coprime, and let $i,j$ be two nodes, we have that exists $r$ such that $sr\equiv j-i \mod n$ and so $$\Bbb P [X_r=j| X_0=i]=\Bbb P [X_r=j, X_0=i]\frac 1n \ge \frac 1n \Bbb P [X_0=i,X_1=i+s,...,X_{r-1} =i+(r-1)s,X_r=j]=\frac 1n \cdot p^r >0$$ and this for all $i,j$ so your chain is irriducible.
Now, observe that if $X_0=i$ and $X_r=j$ for some $i,j$ then necessarily $j-i=as+b(n-s)\equiv (a-b)s \mod n$ and so if the chain is irreducible $\exists a,b$ such that (setting $j=i+1$) $1\equiv (a-b)s\mod n$ and this is possible only if $\gcd (n,s)=1$.
Let’s call $p_i(r)= \Bbb P[X_r=i | X_0=i]$. We know that $i$ has period $\gcd \{r\in\Bbb N|\; p_i(r)>0\}$. Let’s observe also that $p_i(2)>0$ since you can go from $i$ to $i+s$ and then come back to $i$. So $i$ has period $1$ iff $\exists r$ odd such that $p_i(r)>0$.
Let’s write $n=2^k\cdot m$ with $m$ odd. If $2^k$ divides $s$ you have that $n$ divides $ms$ and so $p_i(m)>0$ since you can follow the path $i, i+s, ..., i+ms=i$ and so the chain in aperiodic in i (and so all the chain is aperiodic since we never used really that the node was i). At the same time if exists $r$ odd such that $p_i(r)>0$ then $\exists a,b\in\Bbb N$ such that $\begin{cases} a+b=r\\ i\equiv i+as+b(n-s) \mod n\end{cases}$ i.e. $n$ divides $s(a-b)$ but $a+b$ is odd and then it is also $a-b$, so it must be $2^k| s$.
We conclude that the chain is aperiodic $\iff \frac n{\gcd(n,s)}$ is odd.
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H: Find the value of $a$ for which the two lines are on the same plane
Find the value(s) of $a$ for which the two lines $$L_1:=\left\{\left(-1,0,-1\right)+t\left(a,2,0\right):t \in \mathbb R\right\}$$
$$L_2:=\left\{\left(1,2a,1\right)+t\left(2,3,2\right):t \in \mathbb R\right\}$$
place on the same plane in $\mathbb R^3$.
The equation of a plane is :
$$A\left(x-x_{0}\right)+B\left(y-y_{0}\right)+C\left(z-z_{0}\right)=0$$
Where $n=\left(A,B,C\right)$ is normal to the plane and $P=(x_0,y_0,z_0)$ is on the plane.
Since the two lines are going to be on the plane,then we conclude that the normal to the lines are also normal to the plane.
The parametric equation of the lines is:$$L_1:x=at-1,y=2t,z=-1$$$$L_2:x=2t+1,y=3t+2a,z=2t+1$$
The points $P(a-1,2,-1)$ and $Q(-1,0,-1)$ are on $L_1$ and the point $U(1,2a,1)$ is on $L_2$,we can construct the normal to the plane by computing $n=\vec{ UP} \times \vec{UQ}=(a-2,2-2a,-2) \times (-2,-2a,-2)=(-4,2a,-2a^2+4)$
Since the two points $P,Q$ are on the plane we see that:
$$-4\left(x+1\right)+2a\left(y\right)+\left(-2a^{2}+4\right)\left(z+1\right)=0$$$$-4\left(x-1\right)+2a\left(y-2a\right)+\left(-2a^{2}+4\right)\left(z-1\right)=0$$
$$\iff$$
$$2\left(x+1\right)-a\left(y\right)+\left(a^{2}-2\right)\left(z+1\right)=0$$
$$-2\left(x-1\right)+a\left(y-2a\right)+\left(-a^{2}+2\right)\left(z-1\right)=0$$
But after solving the equations I see that the result is trivially true and is independent of $a$.
So how to get the value of $a$?
AI: Lines $L_1 = \{P+t\vec{u} : t \in \Bbb{R}\}$ and $L_2 = \{Q+t\vec{v} : t \in \Bbb{R}\}$ are coplanar if and only if $$\det(\vec u, \vec v, \overrightarrow{PQ})=0.$$
We have $\vec{PQ} = (1,2a,1)-(-1,0,-1) = (2,2a,2)$ so this determinant is
$$\det(\vec u, \vec v, \overrightarrow{PQ}) = \begin{vmatrix} a & 2 & 0 \\ 2 & 3 & 2 \\ 2 & 2a & 2 \end{vmatrix} = 2a(3-2a)$$
which is zero if and only if $a \in \left\{0,\frac32\right\}$.
When $a = 0$ then both lines are in the plane $x-z=0$, and when $a = \frac32$ then they are in the plane $8x-6y+z+9=0$.
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H: Generating subgroup of $\langle \mathbb{Q} \setminus \{0\},\cdot\rangle$
Does such a subgroup even exist?
My guess is That it is $\mathbb{Z} \setminus \{0\}$
Any help?
AI: A proper subgroup cannot generate the whole group. The set of nonzero integers is a generating set for the group $(\mathbb{Q}\setminus\{0\},{\cdot})$, because every nonzero rational can be written as $ab^{-1}$ with $a,b\in\mathbb{Z}\setminus\{0\}$.
However, there is a more “economical” generating set, namely the set of (positive) primes together with $-1$, because every nonzero rational number is, possibly up to the sign, a product of (positive) primes or reciprocal thereof.
This is a minimal set of generators, because removing even one element yields a set that doesn't generate the group.
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H: Number of wrong answers
In a certain test $a_i$ students gave wrong answers to at least i questions, where $i= 1,2,3......k$ No student gave more than k wrong answers. The total number of wrong answers is
I wasn't able to start solving this. I tried subtracting 2-1 for exactly 1 answer but I didn't understand what was happening. The solution provided uses the same logic from the below sum of Therefore, total number of wrong answers contributed by $2^{n−i} −2^{n−i−1}$ students who answered i questions wrong is $(2^{n−i}−2^{n−i−1})i$. I have no clue how this was derived.
There is a similar question on the site already but it is the other way around and doesn't resolve my doubt.
To find number of questions when number of wrong answers is given
AI: The number who gave at least $k$ wrong answers is $a_k$. Since no student gave more than $k$ wrong answers, the number who gave exactly $k$ wrong answers is $a_k$.
The number who gave at least $k-1$ wrong answers is $a_{k-1}$. Since $a_k$ students gave more than $k-1$ wrong answers, the number who gave exactly $k-1$ wrong answers is $a_{k-1}-a_k$.
The number who gave at least $k-2$ wrong answers is $a_{k-2}$. Since $a_{k-1}$ students gave more than $k-2$ wrong answers, the number who gave exactly $k-2$ wrong answers is $a_{k-2}-a_{k-1}$.
$\cdots$
The number who gave at least $1$ wrong answer is $a_{1}$. Since $a_2$ students gave more than $1$ wrong answer, the number who gave exactly $1$ wrong answers is $a_1-a_2$.
So the total number of wrong answers is $$1(a_1-a_2)+2(a_2-a_3) +\cdots+(k-1)(a_{k-1}-a_k)+k(a_k)$$ $$=a_1 +a_2(2-1)+a_3(3-2)+\cdots+a_k(k-(k-1))$$ $$=a_1+a_2+a_3+...+ a_k$$ $$=\sum\limits_1^k a_n$$
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H: Show that a certain projection is rank one.
Consider the following fragment from Murphy's "$C^*$-algebras and operator theory", namely a part of the proof of Theorem 2.4.8.
Can someone explain why $q_e$ is a rank-one projection? (see marked text). Thanks in advance.
AI: Let $V$ be a closed subspace of a Hilbert space $H$ and let $q$ be the orthogonal projection onto $V$. Then $qK(H)q= K(V)$ where $K(V)$ is embedded into $B(H)$ by extending maps to be $0$ on the orthogonal complement of $V$, this preserves compactness of the map.
By definition $q u q = u$ for all $u\in K(V)$ once you have embedded $u$ into $B(H)$, so it follows that $qK(H)q \supseteq K(V)$. This partial result already implies what is needed: If $q_e$ is not rank $1$ (and not $0$) then $\mathrm{dim}_{\Bbb C}F(\mathrm{im}(q_e))≥2$, which is in contradiction to:
$$\Bbb Cq_e=q_eK(H')q_e \supseteq K(\mathrm{im}(q_e)) \supseteq F(\mathrm{im}(q_e)).$$
For completeness the remaining direction $qK(H)q \subseteq K(V)$: If $quq\in qK(H)q$ then $quq$ is zero on the orthogonal complement of $V$ and also valued in $V$, so it is the extension by zero of a linear operator defined on $V$. The image of the closed unit ball under $quq$ is pre-compact in $H$ and also contained in the closed subspace $V$, so it is also pre-compact in $V$ and $quq$ is actually a compact operator when viewed as a map $V\to V$.
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H: Exercise about a limit with greatest integer function
Let:
$$f(x) := \begin{cases} x+3 &\text{, if } x\in (-2,0) \\ 4 &\text{, if } x=0 \\ 2x+5 &\text{, if } 0<x<1\end{cases}\;.$$
Then find $\lim_{x\to 0^-} f([x-\tan x])$, where $[\cdot]$ is greatest integer.
Since $x$ is approaching zero from the negative side, $x>\tan x$
Therefore $[x-\tan x] =0$
$$\lim_{x\to 0^-}f( [x-\tan x])=[x+\tan x] +3 =3$$
But the right answer is 4.
Basically, I am having a problem in choosing the right function for the given limit, because $f(x)=4$ for $x=0$, but $x$ actually isn’t 0, then how can it be the answer?
AI: For $-\pi/2 < x < 0$ you have $\tan x < x < 0$, hence $x - \tan x > 0$. On the other hand, in a suitable left neighborhood $-\sigma < x <0$ of $0$ (with $\sigma > 0$), you have also $2x < \tan x$, therefore $x - \tan x < -2x$; moreover you can also choose $\sigma$ so small that $-2x < 1$. Thus $-\sigma < x < 0 \Rightarrow 0 < x - \tan x < 1$ and this entails $[x - \tan x] = 0$ everywhere in the small left neighborhood $]-\sigma , 0[$ of $0$.
Finally, you find:
$$\lim_{x \to 0^-} f([x - \tan x]) = \lim_{x\to 0^-} f(0) = f(0)\; ,$$
therefore you have to check the text of your exercise: if $f(0) = 0$ as you wrote at the beginning, the answer is $0$; if $f(0) = 4$ as you wrote at the end, the answer is $4$. ;-)
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H: Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2 (using proof by contradiction).
The question is "Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2."
I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a well written solution. I would like to know if I solved it right and if I did, I would like some advice on how I can improve writing proofs.
My attempt:
Assume to the contrary that solution of this equation is greater then or equal to 2.
Let $x=\frac 2p$.
Then we have
$ (\frac 2p)^5-2(\frac 2p)^3-3=0 $
$ \frac {2^5}{p^5} - \frac {2^4}{p^3} - 3 = 0$
We now consider two case: when $p=1$ and $p<1$.
when $p=1$:
$ 2^5 - 2^4 - 3 = 32 - 16 - 3 = 13$.
Since $x = 2$ is not a solution this is a contradiction.
When $p<1$:
If $p<1$, then we know $1/p>1.$ This implies $\frac {1}{p^{n+1}} > \frac {1}{p^n}.$
Since $\frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$, it is clear that the inequality
$ \frac 1{p^5}(2^5)- \frac 1{p^3} 2^4-3 > \frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$ holds
Since for all $x > 2$ is not a solution this is a contradiction.
Thus, solution $x$ is less then 2.
AI: Let $x$ be a root and $x>2$.
Thus, $$x^5-2x^3-3=x^5-2x^4+2x^4-4x^3+2x^3-3>0,$$ which is a contradiction.
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H: how to find convolution between $f(x)=e^{-2x^2}$ and $g(x)=e^{-2x^2}$?
I wanted to find the convolution between the two function. By definition I get,
$$(f*g)(x)=\int_{\mathbb R}e^{-2(x-y)^2}e^{-2x^2}dy=\int_{-\infty}^{\infty}e^{-4x^2+4xy-2y^2}dy $$
But I am not able to calculate the given integral.
AI: Note that $$(f * g)(x) = e^{-2x^2} \int_{\mathbb R} e^{-\frac{-(y-x)^2}{2 \cdot \frac{1}{4}}}dy = e^{-2x^2} \frac{\sqrt{2\pi}}{2} \cdot \int_{\mathbb R} \frac{2}{\sqrt{2\pi}} e^{-\frac{(y-x)^2}{2 \cdot \frac{1}{4}}}dy = e^{-2x^2}\frac{\sqrt{2\pi}}{2}$$
Where at the last equality we used the fact that measure with density $\frac{1}{\sqrt{2\pi}\sigma} \exp(-\frac{(y-\mu)^2}{2\sigma^2})$ for $\mu \in \mathbb R,\sigma >0$ is a probability measure
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H: Convergence in distribution to normal random random variable
Let $X_n$ be a sequence of random variables on $\mathbb R$ such that $X_n$ converges in distribution to a Gaussian variable $N(0,\sigma^2)$.
Is it true that
$$\mathbb E[e^{-\frac{X_n}{\sqrt n}}]\to 1?$$
Can you help me a bit?
AI: Let $Z\sim\mathcal{N}(0,\sigma^2)$ and $U\sim\mathcal{U}[0,1]$ be independent. Define $X_n$ by
$$ X_n = Z - (\sqrt{n} \log n) \mathbf{1}_{\{U \leq 1/n\}}. $$
Since $X_n \to Z$ $\mathbb{P}$-almost surely, we have $X_n \to Z$ in distribution. However,
$$ \mathbb{E}[e^{-X_n/\sqrt{n}}]
= \mathbb{E}[e^{-Z/\sqrt{n}}]\mathbb{E}[n^{\mathbf{1}_{\{U \leq 1/n\}}}]
= e^{\sigma^2/2n} \left( 2 - \frac{1}{n} \right) \xrightarrow{n\to\infty} 2. $$
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H: Probability density function of s = $u^2 + v^2$ with uniformly distributed u and v
I have a question on the section "Polar form" in the Wikipedia article of the Box-Muller-Scheme. There it is said that if u and v are two independent uniformly distributed random variables in the interval [-1,1] each, then if you sample from them and discard every point (u,v) with $s=u^2 + v^2 \ge 1$ or $s=0$, then the random variable s will be uniformly distributed.
I tried to verify this, but it didn't work. I started by defining the probability densities of u and v as $f_u(u) = \frac{1}{2}$ and $f_v(v) = \frac{1}{2}$. Then I set $\tilde{u} = u^2$ and $\tilde{v} = v^2$ and calculated their probabilities by the following steps (where $F_x$ is the cumulative density function of the variable x):
$F_{\tilde{u}}(U) = P(\tilde{u} \le U) = P(u^2 \le U) = P(-\sqrt{U} \le u \le \sqrt{U}) = P(u \le \sqrt{U}) - P(u \le - \sqrt{U}) = F_u(\sqrt U) - F_u(- \sqrt U) = 0.5 (\sqrt{U} + 1) - 0.5 (-\sqrt{U} +1) = \sqrt{U}$
and then we get the pdf by taking the derivative: $f_{\tilde u}(\tilde u) = \frac{1}{2\sqrt{\tilde u}}$ and analogous for $\tilde v$.
Then the pdf of a sum of two independent random variables, namely $s = \tilde u + \tilde v$ should be the convolution of the pdfs of $\tilde u$ and $\tilde v$:
$$f_s(s) = \int_0^1 f_{\tilde u}(\tilde u) \cdot f_{\tilde v}(s - \tilde u) d\tilde{u} = \frac{1}{4} \int_0^1 \frac{1}{\sqrt{\tilde u (s-\tilde u)}}d\tilde u.$$
I calculated this with WolframAlpha, but it didn't look like a uniform distribution at all. Where did I make a mistake here?
AI: Your mistake is with the limits of the integral in $f_s$ --- $\tilde{u}$ does not take all values $0$ to $1$, but only $0$ to $s$. Then
$$
\int_0^s\frac{1}{\sqrt{\tilde{u} (s-\tilde{u})}}\,\mathrm{d}\tilde{u}=\int_0^1\frac{1}{\sqrt{\frac{\tilde{u}}{s}(1-\frac{\tilde{u}}{s})}}\,\mathrm{d}\left(\frac{\tilde{u}}{s}\right)
$$
therefore does not depend on $s\in(0,1]$ (The integral works out to $\pi$ but it doesn't really matter).
Note that you need to re-normalize $f_s$ because you have thrown out the set $s>1$ (or in other words, conditioned on $s\leq 1$).
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H: Visualizing a lemma in metric spaces and possibly relaxing requirements
I know that the following statement can be proved with relative ease:
Let $ K \subset \mathbb{R}^n $ be a compact (closed and bounded) set in real Euclidean space. Assume also that $ K \subset U $ where $ U \subset \mathbb{R}^n $ is an open set. Then the distance from $ K $ to the complement of $ U $ is positive, meaning $ d = \text{inf} \{ \lVert x-y \rVert \ : x\in K ; y \notin U\} > 0 $.
This is easy to prove using Bolzano-Weierstrass, but I am trying to visualize it: why visually speaking are closedness and boundedness of $ K $ both necessary. I can see why boundedness is necessary as I can come up with an example in $ \mathbb{R}^2 $ where the boundary of an unbounded $ U $ approaches the boundary of the subset $ K $ arbitrarily closely. But visually speaking (using a visual example): what if $ K $ is bounded and open and $ U $ containing it were bounded and open. How can the distance in this case not be strictly positive (how can it be zero)? Can someone please provide a visual example? It seems that all the "nice"/"intuitive" open bounded sets $ K \subset U $ satisfy the conclusion of the statement. Maybe the lemma requirements can be a bit relaxed but I cannot rigorously prove it without $ K $ being compact. Any explanation and visual examples/counterexamples of $ K,U $ open and bounded satisfying the conclusion of the lemma are appreciated.
******** The case of $K=U$ gives zero distance but is trivial. I was hoping for a proper subset $ K \subset U $ where the distance $d$ is zero.
AI: A simple example is $K=(0,1)$ and $U=(0,1)$. If $K$ is not closed, we could e.g. take it equal to $U$ and then there is no room at all between $K$ and $U^\complement$.. Or $K=[0,1)$ and $U=(-2,1)$ as another example: at the "open end" at $x=1$ we also have no distance to the complement of $U$.
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H: How to estimate the number of loop given that the inner loop index always greater than outer loop for Big O complexity estimation?
I have run into a program where there are two nested dependent loops in this manner:
for k=1:K-1
for ell=k+1:K
do something
end
end
How do I calculate the number of total loops this program will run mathematically to estimate the Big O complexity ?
AI: Since the inner loop is in $O(K)$ and it is executed up to $K-1$ times, the big-O cost is quadratic, i.e. $O(K^2)$.
The precise max number of iterations (in the absence of early termination in "do something") is $\displaystyle \sum_{i=1}^{K-1}\ i$, which equals $\displaystyle \frac{K(K-1)}{2}$, confirming the quadratic complexity.
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H: Solution of differential equation ${x^2}\frac{{dy}}{{dx}} + {y^2}{e^{\frac{{x\left( {y - x} \right)}}{y}}} = 2y\left( {x - y} \right)$
Solutions of differential equation
$$
{x^2}\frac{{dy}}{{dx}} + {y^2}{e^{\frac{{x\left( {y - x} \right)}}{y}}} = 2y\left( {x - y} \right)
$$
are given by
(A) $x(x + y) = y \ln (Ce^x– 1)$
(B) $x(x – y) = x \ln (Ce^x– 1)$
(C) $x(x + y) = x\ln (Ce^x + 1) $
(D) $x(x – y) = y \ln (Ce^x –1)$,
where $C$ is constant of integration.
My approach is as follow
\begin{align}
& \Rightarrow \frac{{{x^2}}}{{{y^2}}}\frac{{dy}}{{dx}} + {e^{\frac{{x\left( {y - x} \right)}}{y}}} = \frac{{2y\left( {x - y} \right)}}{{{y^2}}} \\
& t = \frac{1}{y} \\
& dt = - \frac{1}{{{y^2}}}dy \\
& \Rightarrow - \frac{{{x^2}dt}}{{dx}} + {e^{x\left( {1 - tx} \right)}} = 2\left( {tx - 1} \right) \\
& {e^{x\left( {1 - tx} \right)}} = g,\because \frac{{dg}}{{dx}} = g\left( {1 - 2tx - {x^2}\frac{{dt}}{{dx}}} \right) \Rightarrow \frac{{dg}}{{gdx}} = \left( {1 - 2tx - {x^2}\frac{{dt}}{{dx}}} \right) \Rightarrow \\
& \qquad- {x^2}\frac{{dt}}{{dx}} = 2tx - 2 + \frac{{dg}}{{gdx}} + 1 \\
& \Rightarrow 2tx - 2 + \frac{{dg}}{{gdx}} + 1 + g = 2\left( {tx - 1} \right) \Rightarrow \frac{{dg}}{{gdx}} + 1 + g = 0 \Rightarrow \frac{{dg}}{{g\left( {g + 1} \right)}} = - dx \\
& \Rightarrow \frac{{dg}}{g} - \frac{{dg}}{{\left( {g + 1} \right)}} = - x + C \\
& \Rightarrow \ln{e^{x\left( {1 - tx} \right)}} - \ln\left( {{e^{x\left( {1 - tx} \right)}} + 1} \right) = - x + C \Rightarrow \ln\frac{{{e^{x\left( {1 - tx} \right)}}}}{{{e^{x\left( {1 - tx} \right)}} + 1}} = - x + C
\end{align}
I am not able to proceed from here
AI: $${x^2}\frac{{dy}}{{dx}} + {y^2}{e^{\frac{{x\left( {y - x} \right)}}{y}}} = 2y\left( {x - y} \right)$$
Substitute $y=tx$
$$t'x+t+ t^2{e^{x(1-\frac 1t )}} = 2t\left( {1 - t} \right)$$
$$t'x+ t^2{e^{x(1-\frac 1t )}} = t -2 t^2$$
$$-\left(\dfrac 1t \right)'x+ {e^{x(1-\frac 1t )}} = \dfrac 1t -2 $$
$$-\left(\dfrac xt \right)'+ {e^{x(1-\frac 1t )}} = -2 $$
$$\left(\dfrac xt \right)'= {e^{x(1-\frac 1t )}} +2 $$
Substitute $u=\dfrac x t$ .
$$(e^u)'-2e^u=e^x $$
This is easy to solve.
$$(e^ue^{-2x})'=e^{-x}$$
Integrate
$$e^u=-e^{x}+Ce^{2x}$$
$$\dfrac xt=\ln |-e^{x}+Ce^{2x}|$$
$$\dfrac {x^2}y=x+\ln |-1+Ce^{x}|$$
$$x(x-y) =y\ln |-1+Ce^{x}|$$
Solution is option D
You can continue from the last line. Put the denominator on the right side and then factorize
$$z=Ke^{-x}(z+1)$$
$$ \implies z(1-ke^{-x})=ke^{-x}$$
$$e^{x(1-tx)}= \dfrac {ke^{-x}}{1-ke^{-x}}$$
$$e^{x(1-tx)}= \dfrac {ke^{-x}}{1-ke^{-x}}$$
Take log function on both sides and continue. Normally you should end with the right solution.
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