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H: Functions satisfying differential equation of the Weierstrass elliptic function $\wp$
The Weierstrass elliptic function $\wp$ satisfies the following differential equation: $${\wp'}^2 = 4 \wp^3 - g_2 \wp - g_3$$
Which other functions do?
Are the solutions always elliptic functions?
AI: Given theorem on existence and uniquenes of solutions to ordinary differential equations (Picard's theorem) there is only one local solution if the Lipschitz condition is satisfied, in fact it is when we take $x_1$,$x_2$ in the same parallelogram (determined with $(g_2,g_3)$) away from the lattice points. Since $\wp$ is doubly periodic it determines the solution everywhere (the lattice points are the second order poles).
Rewriting the equation one can reformulate the definition of the Weierstrass elliptic function, i.e. $\wp(x;g_2,g_3)$ yields the value $z$ for which
$$x=\int_{\infty}^{z} \frac{d t}{\sqrt{4t^3-g_2 t-g_3}} $$
Reassuming there is only one solution if an initial condition is prescribed appropriately and it is just $\wp(x\pm c;g_2,g_3)$, where $c$ can be determined from the initial condition. If we prescribe e.g. this condition $f(\infty)=0$, then there are no solutions unless $\;g_2=0=g_3$.
In a degenerate case $g_2=g_3=0$ the only solution is $f(x)=\frac{1}{\left(x \pm c\right)^2}$ and this may be considered as non-elliptic function since it is not periodic, elliptic functions are doubly periodic meromorphic functions in the complex plane. |
H: How to solve $x^2 + \ln(x) = 0$
I was just investigating $y = f(x) = e^{-x^2}$ and then went ahead to plot $x=f(y), y=-f(x), and x=-f(y)$, and what I got was interest rounded square shape, and I think we can calculate this area using integration.
However to get the bounds I must solve the equation $x^2 + \ln(x) = 0$. Is there a way to solve this equation without using the graph?
Shape
AI: Not really. Just as you can't solve $2^x=3$ using algebra (you need to create a new kind of function; logarithms), you can't solve tihs one without creating a new kind of special function. The one we have for this is called the Lambert W function, the inverse function of $xe^x$. Then we can solve
$$ x^2+\ln x=0 $$
$$ x^2+\frac{1}{2}\ln x^2=0 $$
$$ 2x^2+\ln x^2=0 $$
$$ 2x^2+\ln(2x^2)=\ln(2) $$
$$ 2x^2e^{2x^2}=2 $$
$$ 2x^2=W(2) $$
$$ x=\sqrt{\frac{W(2)}{2}}. $$ |
H: $L^p$-norm diverges for a sequence of functions
Let $\{f_n \}_n$ real valued functions such that:
\begin{cases}
f_n \geq 0 \\
\int_{\mathbb{R}} f_n =1\\
\forall \delta >0: \lim_{n \rightarrow +\infty } \int_{|t|>\delta}f_n(t)dt = 0
\end{cases}
Show that, for every $p>1$:
$$\lim_{n \rightarrow + \infty} ||f_n||_{L^p} = +\infty \quad \star$$
First I note that $1 = \int_{\mathbb{R}} f_n(t)dt = \int_{|t|<\delta}f_n(t)dt + \int_{|t|>\delta} f_n(t)dt$
and taking the limit follows that $$\lim_{n \rightarrow +\infty} \int_{|t|<\delta}f_n(t)dt=1$$
Now, in order to prove $\star$, I would like to show that for every $M>0$, there exists a $\bar{n} \in \mathbb{N}$ such that for every $n>\bar{n}$: $||f_n||_{L^p}^p>M^p$.
Therefore, since $f_n \geq 0$:
$$\int_\mathbb{R} f_n^p = \int_{|t|<\delta} f_n^p + \int_{|t|>\delta} f_n^p$$
Now I choose $\delta$ such that for $t \in \{t: |t|>\delta \}$ I have $f(t)<1$, therefore the second integral goes to $0$ as $n \rightarrow \infty$.
Also, since for $t \in \{t: |t|<\delta \}$ it holds that $f_n(t)>1$, and $f_n \geq 0$, it follows that $f_n^p>f_n$ for evry $p>1$.
So I can write $\int_{|t|<\delta}f_n^p > \int_{|t|<\delta}f_n$ and since the last term is $1$ as $n \rightarrow + \infty$, I have that $$\lim_{n \rightarrow \infty} \int_{|t|<\delta}f_n^p = 1 + \varepsilon_p$$ with $\varepsilon_p >0$
I don't know if it's okay up to now, but in any case I don't know how to conclude. Can anyone help me?
AI: By Holder's inequality. $\int_{|t|<\delta} f_n(x)dx \leq \int (f_n^{p}(x)dx)^{1/p} (2\delta)^{1/q}$ where $\frac 1 p +\frac 1 q=1$. The conclusion follows easily from this since $\delta$ is arbitrary. |
H: What is the motivation for sequences to be defined on natural numbers?
Possible duplicate: Sequences with Real Indices
I am trying to understand the motivation for various definitions in real analysis. Take for instance the definition of a sequence where it is defined as a function from natural numbers to that of real numbers. But why natural numbers? I cannot substantiate it with a reasonable argument and I am looking for one.
I have been able to substantiate certain definitions based on some of my own reasoning, while that might not have been the real reason why it was defined that way. Take for instance the definition of convergence of a sequence. I think one of the central definitions in analysis is that of convergence of a sequence. Here we say that a sequence is said to converge to a limit $L$ if any $\epsilon$-neighborhood of $L$ has all but finitely many terms of the sequence. I ask the usual set of questions to myself
Why do we need this definition?
Why was it defined this way?
I thought of some possible answers. There are questions about the operations of addition and subtraction, and the manipulation of those operations resulting in questions about re-arrangements of infinite series. The foundation to answering such questions lies in the answer to "does it even even make sense to represent the sum of infinite numbers by one number?" What is the logical basis to represent that infinite sum by one number? Then we get the answer that the sequence of partial sums converge and so we can represent them as one number. Then we get the questions "what is a sequence and what does it mean to say that the sequence converges?" Then we get the definitions for a sequence and the definition of convergence.
Definition of convergence makes sense to me. Let's assume we are thinking of two numbers $a$ and $b$. What are some of the rigorous ways to define the quality of two numbers being equal? A popular one is to say that $a \geq b$ and $b \geq a$. Another is the topologic definition where we say that $a$ always lies in any $\epsilon$-neighborhood of $b$ no matter how small $\epsilon$ is. Now, we can modify this definition to obtain the definition of convergence of a sequence where we replace "$a$" with "all but finite terms of the sequence". There you go. We got a definition for the convergence of a sequence.
The part I could not substantiate to myself was the need to define sequences as functions on natural numbers to real numbers. Why not define sequences on some other index set? For instance, why not define it on real numbers? More generally, we have countability properties defined but why even bother with natural numbers when what we want in the end is real numbers?
AI: When you have a sequence of real numbers, there is a first number in this sequence. Give it the index 1. Then there is a second number, give it the index 2. Continue. You might now ask: what is the real number with index $n$ in my sequence? For this, you define the function $f:\mathbb{N}\rightarrow\mathbb{R}$ which maps indices to the corresponding real numbers in your sequence.
When you now think a bit about it, your sequence is completely defined by the function $f$, and on the other hand $f$ is completely defined by the sequence. In other words, there is a bijection of sequences of real numbers and functions $\mathbb{N}\rightarrow\mathbb{R}$. This is the reason why we say that $f$ itself is the sequence. |
H: Finding curvature for $y = \sin ( -2x )$ at $x =\pi/4$
The answer is 4, but I got 1. I said $r(t)=<t,\sin(-2t)>$ and the $|r^\prime(\frac{\pi}{4})|$ is equal to 1. I also got 1 for $T^\prime(\frac{\pi}{4})$ but none of this information matters because I got the question wrong, can someone help me out please? thanks. Also, ideally, use the $\dfrac{|T^\prime|}{|r^\prime|}$ formula.
Perhaps, I shouldn't have translated it into a rectangular form? But I have only been taught how to compute curvature using a rectangular form.
AI: Your function can be parameterized as:
$${\displaystyle \mathbf{r}(t)\,=\,{\begin{pmatrix}t\\\sin(-2t)\end{pmatrix}}}$$
The curvature is given by :
$$\kappa = \frac{\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}$$
$${\displaystyle \mathbf{r}'(t)\,=\,{\begin{pmatrix}1\\-2\cos(-2t)\end{pmatrix}}}$$
$${\displaystyle \mathbf{r}''(t)\,=\,{\begin{pmatrix}0\\-4\sin(-2t)\end{pmatrix}}}$$
Setting $x \mapsto \frac{\pi}{4}$ follows:
$${\displaystyle \mathbf{r}'(\frac{\pi}{4})\,=\,{\begin{pmatrix}1\\0\end{pmatrix}}}$$
$${\displaystyle \mathbf{r}''(\frac{\pi}{4})\,=\,{\begin{pmatrix}0\\4\end{pmatrix}}}$$
Now use the formula:
$$\kappa = \frac{\|\mathbf{r}'(\frac{\pi}{4})\times\mathbf{r}''(\frac{\pi}{4})\|}{\|\mathbf{r}'(\frac{\pi}{4})\|^3}=\frac{\left\|\,{\begin{pmatrix}0\\0\\4\end{pmatrix}} \right\|}{\left(0^{2}+1^{2}\right)^{\frac{3}{2}}}=4$$ |
H: Split area under the curve into two halves
Draw a vertical line say $x=t$ on curve $y= \sqrt{x}$ between $x=0$ and $x=a$ such that area under the curve from $x=0$ to $x=t$ equals area under the curve from $x=t$ to $x=a$.
I am looking for a generalized approach for other functions such as $y = \ln(x)$
AI: Hint
Find $t \in (0,a)$ such that
$$\int_0^t \sqrt x dx = \frac{1}{2} \int_0^a \sqrt x dx.$$ |
H: Limit of sum of probability distributions
Suppose that $\lbrace X_i:i=1,2,...\rbrace$ is i.i.d. with density function $f$, finite $\sigma$ and mean $\mu <0$. Prove or disprove that
$$\mathbb{P}(\sum_{i=1}^n X_i \geq 0) \to 0 $$
as $n$ tends to infinity.
Note that
$$\mathbb{P}(\sum_{i=1}^n X_i \geq 0) = \mathbb{P}(\frac{\bar{X}_n-\mu}{\sigma/\sqrt{n}}\geq \frac{-\mu}{\sigma/\sqrt{n}})$$
My idea is to using CLT here, but don't know how to give a $\epsilon-\delta$ argument for the convergence.
AI: CLT is not the right approach. Use SLLN instead. Let $T_n =\frac 1 n \sum\limits_{k=1}^{n} X_k$. Then $X_n \to \mu $ almost surely. This implies convergence in distribution and hence $\lim \sup P(T_n \geq 0) \leq P(\mu \geq 0)=0$. |
H: What is the opposite side of convergence of function?
Assume we define $f(x)$ on the domain of $x\ge 0$. Given the following statement,
$$\lim_{x\rightarrow\infty} f(x) =0.$$
It is known that it is equivalent to
$\forall \epsilon>0, \exists X>0$, if $x\ge X$, $|f(x)| \le \epsilon$.
I want to know the negation statement for the one above, by using the $\epsilon$ language.
AI: $\exists \epsilon >0 , \forall X>0, \exists x\geq X, |f(x) | >\epsilon$. |
H: uniformly converge of a serie with given conditons
i have this question about series.
So i came along a question that started with :
$(f_{n})_{n}$:[-1,1]->$R^{+}$ is a row of continuous functions. The serie $\sum{f_{k}(x)}$ from k=0 to $\inf$ will converges and thefunction f:[-1,1]->$R^{+}$:x->$\sum(f_{k}(x))$ is continuous.
Now i wanted to prove that $\sum f_{k}$ uniformly converges.
I wanted to use the M-test from weierstrass. Therefore i only needed to prove that |$f_{k}(x)|$<= $f_{k}(x)$. For this i wanted to use that f is continous but i'm stuck and not sure if we can even say that it uniformly converge.
AI: Tha follows from Dini's theorem: since each $f_n$ take only non-negative values, the sequence $\left(\sum_{n=1}^Nf_n(x)\right)_{N\in\Bbb N}$ is monotonic increasing for each $x\in[-1,1]$. So, since each $\sum_{n=1}^Nf_n$ is continuous and since $f$ is continuous, the convergence is uniform. |
H: Writing $0$ with $6$ significant figures concisely
I have some calculations that result in exactly $0$, calculated to 6 significant figures. However, I do not want to write $0.00000$ for each of these calculations, as it makes it look messier. How can I write it more concisely, while still retaining the number of significant figures?
AI: If all the rest of your numbers will have the decimal point in them, consider writing it as simply $0$ and provide a reference note somewhere that $0$ is $0.00000$ throughout your paper/thesis/work. You've referred to this as being "exact" in the sense that if it's not exactly zero then it differs only in an $\varepsilon<10^{-6}$ so it's reasonable to use the "exact" value $0$.
This also has the minor advantage that your "exact" $0$s will stand out in tables and lists. |
H: What is the condition to split integral in Lebesgue theory?
Let $f:[0,1]^2\rightarrow \Bbb R$ be defined by
$$f(x,y)=1/y^2 \text{ if }0\leq x <y<1$$
$$f(x,y)=-1/x^2\text{ if }0\leq y <x \leq 1$$
$$f(x,y)=0 \text{ otherwise}$$
If can show that $$\int_{0}^{1}\int_{0}^{1}f(x,y)dxdy \neq \int_{0}^{1}\int_{0}^{1}f(x,y)dydx$$
In my course, it is written that the last equation does not contradict fubini's theorem since $\mid f\mid$ is not integrable on $[0,1]^2$. I don't know why and I can't compute the integral of $\mid f \mid$ on $[0,1]^2$. Any help would be appreciated
AI: By Tonelli's Theorem $\int |f| \geq \int_0^{1}\int_x^{1}\frac 1 {y^{2}} dydx=\int_0^{1}(\frac 1 x-1) dx =\infty$. |
H: Finding the upper bound of a linear functional
I have a linear functional $\phi_n:(C[0,1],\Vert{\cdot}\Vert_\infty) \to \mathbb{R}, n\in\mathbb{N}$ defined by $$\phi_n(x)=\int_0^1t^nx(t)dt$$
I have to calculate an upper bound for $\Vert\phi_n\Vert$ which is independent from $n$.
I thought about using the Cauchy–Schwarz inequality by saying that $$\vert\phi_n\vert=\vert\langle{t^n,x(t)\rangle}\vert\leq\Vert{t^n}\Vert\cdot\Vert{x(t)\Vert}$$ but I don't know how that would be independet from $n$.
I'm new to this area of mathematics so I don't really have a clue which other approach I should use to tackle this problem.
I would be really cool if someone could give me some advice or push me in the right direction.
Thanks!
AI: I'd say you're almost there: just realize that your domain is $[0, 1]$, a compact set, and that $x$ is continuous.
That means that there are no asymptotes in the output space/codomain for $x$ (which is $\mathbb{R}$) for any value your input set/domain. This means that the value $||x(t)||_\infty$, the supremum of $x(t)$ calculted over $t \in [0, 1]$, does exist and indeed is a finite value.
Finally, $\forall t \in [0, 1], \forall n \in \mathbb{N}, 0 \le t^n \le t \le 1$. So $\forall n \in \mathbb{N}, ||t^n||_\infty = 1$ (for the given domain).
Conclusion: By the previous, and the Cauchy-Schwartz inequality, $||\phi_n||_\infty \le ||x(t)||_\infty$
Ie, the trick was to bound the $t^n$ term to remove the $n$ term from the desired bound. |
H: Is countable infinity part of the natural numbers?
A hopefully relatively simple question: is countable infinity an element of the set of natural numbers? A related question here has been answered with the statement that (classic) infinity is not part of the natural numbers, but what about countable infinity?
Context for this question: A definition of a Reproducing Kernel Hilbert Space (RKHS) I have come across is as follows:
The reproducing kernel Hilbert space (RKHS) $\mathcal{H}$ is the closure of the linear span $\{f:f(x)=\sum_{i=1}^{m} a_ik(x,x_i), \space a_i \in \mathbb{R}, \space m \in \mathbb{N}, x_i \in \mathcal{X}\}$ equipped with inner products $\left< f,g \right>_{\mathcal{H}}=\sum_{ij}a_ib_jk(x_i,x_j)$ for $g(x)=\sum_{i=1}^{m} b_ik(x,x_i)$.
Many examples or RKHSs have infinitely many dimensions, so we would require $m=\infty \in \mathbb{N}$. Feel free to ignore this context if it is not required for your answer.
AI: No, you do not need $m = \infty$. Note that you are taking the closure of the linear span, which takes care of the elements requiring an infinite sum for their representation. |
H: Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively.
So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.
Any hints or explanations for this problem will be greatly appreciated !!
AI: Hint
You have
$$f(x)=(x^2-1)(ax+b)+(2x-5)$$
and
$$f(x)=(x^2-4)(cx+d)+(-3x+4)$$
From the second you get $f(1)=-3(c+d)+1$ and from the first you get $f(1)=-3$. Thus
$$\color{red}{c+d=\frac{4}{3}}$$
From the first you get $f(2)=3(2a+b)-1$ and from the second you get $f(2)=-2$. Thus
$$\color{red}{2a+b=\frac{-1}{3}}$$
You also get $$f(0)=-b-5=-4d+4 \implies\color{red}{4d-b=9}$$
NOTE You can make a simple observation about the coefficients of $x^3$ and $x^2$ in both the expressions for $f(x)$ to conclude that $a=c$ and $b=d$. With that the first two equations can suffice. |
H: How to properly write a solution to the following inequality?
Trying after many years to refresh my calculus and stack at very basic.
I'm following Thomas' Calculus and here is the problem:
Find a domain and a range for the following function:
$$f(t)=\frac{2}{t^2-16}$$
Domain I've found:
$D\in{\{t|(-\infty,-4)\cup(-4,4)\cup(4, \infty)\}}$, but I struggle with finding or more exactly write it in an understandable way steps to find a range.
To simplify range is a
$(-\infty,-\frac{1}{8}]\cup(0, \infty)$ but why? In an explanation which I had found said that
$t < -4 \Rightarrow t^2 - 16 > 0 \Rightarrow \frac{2}{t^2-16}>0$ but I feel that few steps here are skipped.
Can somebody write a detailed response for this part?
Thank you.
AI: $$t^2\ge0\implies t^2-16\in[-16,\infty).$$
As the expression can change sign we decompose before we can take the reciprocal
$$t^2-16\in[-16,0)\cup(0,\infty)\implies\frac2{t^2-16}\in\left(-\infty,\frac2{-16}\right]\cup(0,\infty).$$ |
H: What is $q$ in the factorized form $a(x+p)^2 + q$ of a quadratic equation?
I can't put the following equation into $a(x+p)^2 + q$ form:
$$-4x^2 - 2x + 3$$
I've gotten as far as: $-4(x + (\frac{1}{4})^2)$, but what would $q$ be? I got $\frac{1}{4}$ by the fact that $p = \frac{b}{2}.$
AI: Take out the factor of $-4$ from the first 2 terms to get $$-4\left(x^2 + \frac{1}{2}x \right) + 3 $$
Now note that $$x^2 + \frac{1}{2}x = \left(x + \frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2$$ Then put this all together to get $$-4x^2 - 2x + 3 = -4\left[\left(x + \frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2\right] + 3$$ which simplifies to $$-4x^2 - 2x + 3 = -4\left(x + \frac{1}{4}\right)^2 + \frac{13}{4}$$
and so $a = -4$, $p = \frac{1}{4}$ and $q = \frac{13}{4}$.
This is the method of completing the square.
The value of $q$ represents the $y$ coordinate of the turning point on the curve $y= -4x^2 - 2x + 3$
Alternative 1: you could take out the factor of $-4$ from all three terms, but this would lead to messier fractions.
Alternative 2: you could write $$-4x^2 -2x + 3 \equiv a(x+p)^2 + q $$ and expanding the RHS gives $$-4x^2 -2x + 3 \equiv ax^2 + 2apx + (ap^2 + q)$$ Then you can compare coefficients and this will give you some equations to solve for the values of $a$, $p$ and $q$ |
H: Fourier transform definition on $L^2$.
I'm trying to prove that $C^\infty_0(R^n)$ as the set of those smooth functions with compact support, is included in the set $X=\{f\in L^1(R^n) \ : \ \hat{f}\in L^1(R^n)\}$.
All because our professor said that one could define the Fourier transform on $L^2(R^n)$ by density of $X$ in $L^2$ and he claimed that this is a consequence of the density of $C^\infty_0$ in $L^2$ and of the inclusion $C^\infty_0\subseteq X\subseteq L^2$. Of course, $X\subset L^2$, but I can't see how $C^\infty_0\subset X$, since I only know that if $f\in C^\infty_0$, then $\hat{f}\in C^\infty\cap C_0$ (where $C_0$ is the set of all continuous functions which goes to 0 as $|x|\rightarrow+\infty$).
I tried to comb through some text, but I only saw authors using the density of $L^1\cap L^2$ in $L^2$ to define the Fourier transform on $L^2$...
AI: Well, you could show that for $f\in C_c(\mathbb{R})$, $\lvert\hat{f}(\xi)\rvert=O(\lvert\xi\rvert^{-1})$ as $\lvert\xi\rvert\to\infty$, and similarly for $C^1_c(\mathbb{R})$ you then get $O(\lvert\xi\rvert^{-2})$. This gives $\hat{f}\in L^1$.
Of course you would recognize it as part of the proof of Fourier transform of Schwartz is Schwartz, adapted to our case. |
H: If $E(X)=15$, $P(X\le11)=0.2$, and $P(X\ge19)=0.3$, what can be $V(X)$?
If $E(X)=15$, $P(X\le11)=0.2$, and $P(X\ge19)=0.3$, which of the following is impossible ?
$V(X)\le7$
$V(X)\le8$
$V(X)>8$
$V(X)>7$
I know $V(X)=E(X^2)-E(X)^2$.
Please, any other hints??
AI: Notice that you have $P(|X-E(X)|\geqslant 4)$ and try to use Chebyshev’s inequality:
$$\forall\alpha>0,\ P(|X-E(X)|\geqslant \alpha)\leqslant\dfrac{V(X)}{\alpha^2}$$ |
H: Finding perpendicular lines in $\mathbb R^4$
Let $$g=\begin{pmatrix}2\\-5\\-3\\-3\end{pmatrix}+\mathbb R\begin{pmatrix}1\\2\\3\\4\end{pmatrix}$$ and $$h=\begin{pmatrix}1\\-3\\0\\-1\end{pmatrix}+\mathbb R\begin{pmatrix}2\\3\\4\\5\end{pmatrix}.$$
Find all lines that are perpendicular to both $g$ and $h$.
Find the smallest affine subspace in $\mathbb R^4$ that contains both $g$ and $h$.
As for 1: One can easily see that the two lines are skew. Now, if $v_g$ and $v_h$ are the direction vectors of the lines I am first interested in a base of $U^\perp$ where $U=\langle v_g,v_h\rangle$. I got $$U^\perp=\left\langle\begin{pmatrix}2\\-3\\0\\1\end{pmatrix},\begin{pmatrix}-1\\1\\1\\-1\end{pmatrix}\right\rangle=:\langle v_1,v_2\rangle.$$ So now we should get two perpendecular lines $$l_1=p_1+\mathbb R v_1\quad\text{ and }\quad l_2=p_2+\mathbb R v_2$$ and need to find $p_1$ and $p_2$.
We can parametrize $g$ via
$$
\vec{P}_{\lambda}=\left(\begin{array}{c}
2+\lambda\\
-5+2\lambda\\
-3+3\lambda\\
-3+4\lambda
\end{array}\right)
$$
and $h$ via
$$
\vec{G}_{\mu}=\left(\begin{array}{c}
1+2\mu\\
-3+3\mu\\
4\mu\\
-1+5\mu
\end{array}\right).
$$
So the connection of $g$ and $h$ has the direction vector
$$
v=\overrightarrow{P_{\lambda}G_{\mu}}=\left(\begin{array}{c}
-1+2\mu-\lambda\\
2+3\mu-2\lambda\\
3+4\mu-3\lambda\\
2+5\mu-4\lambda
\end{array}\right).
$$
The condition $v\perp g$ and $v\perp h$ yields
$$
\left\langle \left(\begin{array}{c}
-1+2\mu-\lambda\\
2+3\mu-2\lambda\\
3+4\mu-3\lambda\\
2+5\mu-4\lambda
\end{array}\right),\left(\begin{array}{c}
1\\
2\\
3\\
4
\end{array}\right)\right\rangle =0=\left\langle \left(\begin{array}{c}
-1+2\mu-\lambda\\
2+3\mu-2\lambda\\
3+4\mu-3\lambda\\
2+5\mu-4\lambda
\end{array}\right),\left(\begin{array}{c}
2\\
3\\
4\\
5
\end{array}\right)\right\rangle
$$
and thus,
$$
20+40\mu-30\lambda=0\,\,\,\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,\,\,\,26+54\mu-40\lambda=0.
$$
The solution of this system of linear equations is given by $\mu=1$
and $\lambda=2.$ With that, we find
\begin{align*}
l_{1} & =\vec{P}_{2}+\mathbb{R}\overrightarrow{P_{2}G_{1}}\\
& =\left(\begin{array}{c}
4\\
-1\\
3\\
5
\end{array}\right)+\mathbb{R}\left(\begin{array}{c}
-1\\
1\\
1\\
-1
\end{array}\right).
\end{align*}
Is this correct so far? But how do I get the second one?
As for 2: For the smallest subspace that contains both $g$ and $h$ I would take $g+v$ where $v$ is the direction vector between $g$ and $h$ as mentioned above. Does this make sense?
AI: Question 1
Your answer to the first question starting at We can parametrize $g$ via... looks good and you found the unique solution line.
The first part is wrong. You indeed computed well $U^\perp$. This space is of dimension $2$. That doesn't mean that there is two solutions. But just that the direction of the solutions belong to $U^\perp$.
Question 2
The smallest affine subspace is the one passing through $P_g$ and having for direction $Vect\{\vec{P_g P_h}, v_g,v_h\}$. This is an affine hyperplane.
The equation of such an affine hyperplane is
$$ax+by+cz+dt+e=0$$ and you need to find $a,b,c,d,e$. Which can be done by writing that $\vec{P_g P_h}, v_g,v_h$ belong to the associated vector hyperplane $ax+by+cz+dt=0$ while $P_g$ belongs to the affine hyperplane.
Leading (if I avoided computation mistakes...) to the equations
$$\begin{cases}
x &+2y &+3z &+4t & &= 0\\
2x &+3y &+4z &+5t & &=0\\
-x &+2y &+3z &+2t & &=0\\
2x &-5y &-3z &-3t &+ e &=0\\
\end{cases}$$
And finally to the affine hyperplane of equation
$$-x +3y-3z+t+11=0$$ |
H: Circle with points on coordinate axis.
If a circle has points $A$ and $B$ which lie on the coordinate axis and $AB$ is the diameter of the circle, would it always form a perfect circle where the midpoint of $AB$ is the centre of the circle?
The question states the line $y = -3x + 12$ meets the coordinate axis at $A$ and $B$. It asks you to find the equation of a circle that passes through points $A, B$and $O$ where $O$ is the origin.
The midpoint of $AB$ is $(2,6)$ so would this be the centre of the circle and if so, is it the same for all lines that meet the coordinate axis?
AI: Since $\angle AOB=\dfrac{\pi}{2}$, the circle must be centered at the midpoint of $AB$, which, as you say, would be the diameter of the circle. And the angle stays unchanged for random line intersecting both coordinate axis. |
H: Finding the joint probability density function of two independent random variables
Is there a way of determining the joint probability density function of two random variables? If we have two independent random variables, $X$ and $Y$ that both are uniform on [0,1], then how do one calculate the joint probability density function, knowing that the two PDFs are 1 each? It wouldn't simply be the product of the two PDFs, right?
AI: It would simply be the product of the two pdf's.
If $X$ and $Y$ have densities $f_X$ and $f_Y$ respectively then independence of $X$ and $Y$ is exactly the statement that $(X,Y)$ has density $g(x,y)=f_X(x)f_Y(y)$. |
H: Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $
I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[n]{\left(1+\dfrac{1}{t}\right)\left(2+\dfrac{1}{t}\right)...\left(n+\dfrac{1}{t}\right)}-\dfrac{1}{t}\\
&=\lim\limits_{t \to 0} \sqrt[n]{\left(\dfrac{t+1}{t}\right)\left(\dfrac{2t+1}{t}\right)...\left(\dfrac{nt+1}{t}\right)}-\dfrac{1}{t} \\
&=\lim\limits_{t \to 0} \dfrac{\sqrt[n]{(t+1)(2t+1)...(nt+1)}-1}{t}
\end{align*}
My idea is to use $\lim\limits_{x\to0}\dfrac{(ax+1)^{\beta}-1}{x} =a\beta .$ But after some steps (above), now I'm stuck.
Thanks for any helps.
AI: The idea is very good! The limit should be for $t\to0^+$, but since the limit for $t\to0$ exists, there's no real problem. However, you should use $t\to0^+$ for the sake of rigor.
The two-sided limit is the derivative at $0$ of the function
$$
f(t)=\sqrt[n]{(t+1)(2t+1)\dotsm(nt+1)}
$$
and in order to compute it, the logarithmic derivative is handy:
$$
\log f(t)=\dfrac{1}{n}\bigl(\log(t+1)+\log(2t+1)+\dots+\log(nt+1)\bigr)
$$
and therefore
$$
n\frac{f'(t)}{f(t)}=\frac{1}{t+1}+\frac{2}{2t+1}+\dots+\frac{n}{nt+1}
$$
which yields
$$
n\frac{f'(0)}{f(0)}=1+2+\dots+n=\frac{n(n+1)}{2}
$$
Since $f(0)=1$, we have
$$
f'(0)=\frac{n+1}{2}
$$ |
H: Velocity damping using acceleration yielding incorrect results
(I am using Python to demonstrate this question)
I have two functions, x and y. They should apply a damping factor of k to velocity v over the time t. The first function works fine, and you can see that I am using the power operator to achieve an accurate result. I am trying to rewrite the function to use acceleration instead, but the second function is not giving me the exact same results. Notably, the higher the k value, the less accurate the second function is.
import math
timestep = 0.01
def x(k, v, t):
for i in range(math.floor(t / timestep)):
v *= (1 - k) ** timestep
return v
def y(k, v, t):
for i in range(math.floor(t / timestep)):
a = - k * v
v += a * dt
return v
What changes can I make to the acceleration value a = - k * v to obtain the correct result? Or is it not possible to do this using acceleration? The idea is that a only depends on the variables k and v.
I understand that the function well never be 100% accurate because it is not continuous, but it should converge to the correct result as the timestep decreases, but this is not currently happening.
AI: Note: This "answer" is really two different answers, one of which applies to a particular definition of "damping factor", while the other applies to a different definition. Choose which part applies according to what kind of "damping factor" you really want to be using.
If your objective is to compute the differential equation
$$ \frac{dv}{dt} = -kv, \tag1$$
then your first function most definitely does not "work fine."
For $0 < k < 1$, after one unit of time the velocity returned by this function will be reduced by $kv$ units of speed, just as if the deceleration $-kv$ were applied uniformly over the entire first unit of time (instead of decaying as the velocity decays).
So at the end of one unit of time you have a velocity of $(1 - k)v$.
After two units of time you will have $(1-k)^2 v$, after three units of time you will have $(1-k)^3 v$, and so forth; in fact the velocity is decaying exponentially, but it is not decaying with damping factor $k$ according to Equation $(1)$ above.
Instead it is decaying with damping factor $(-\ln(1-k)).$
If you put $k = 1$ in your first function, the velocity goes instantly to zero at the first timestep. Is that correct behavior?
Try $k = 1.1.$ Running your first function in Python 3.8.3 with $k=1.1,$ $v = 100,$ and $t = 0.6$ I get (-7.762155952763054+23.88945958880576j) as a result. What do you get? Is your result good?
The exact solution for Equation $(1)$ is $v(t) = e^{-kt} v_0$ where $v_0$ is the initial velocity and $v(t)$ is the velocity after time $t.$
You could compute this result almost exactly in timesteps like so:
def f(k,v,t):
for i in range(math.floor(t / timestep)):
v *= math.exp(- k * timestep)
return v
Your second function is reasonably close to this for values of $kt$ that are not too large; its final velocity is about $\frac12\%$ too small for $k = 1,$ $t =1$ and this error compounds as you increase $t$.
It's an approximate solution using the Euler method, and its error will increase as you increase $k$, or the timestep, or the total time elapsed,
but the error is not nearly as much as the error of your first method.
If your objective in applying a "damping factor $k$" is that after one unit of time the velocity should be $1-k$ times the initial velocity,
then you can still approximate it via the Euler method as in your second function, but you need to use $\ln(1-k)$ instead of $(-k)$ as a factor.
That is, this definition of a damping factor will produce a velocity that is a solution to the differential equation
$$ \frac{dv}{dt} = v \ln(1-k), $$
assuming that everything is expressed in compatible units.
Because the the logarithm of a negative number is not real, this formula is valid only for $k < 1,$ but of course if you are using this definition of "damping factor" then you would naturally have the restriction that $k < 1$ anyway.
I observe that when you see "damping factor" or "damping coefficient" in the literature, the term is usually associated with damped harmonic motion, which does not seem to be the application here.
So it's up to you to determine which definition is correct for the particular application you require. |
H: Parabolic Bootstrapping
I started studying parabolic pdes. Often I come across an integral solution where the regularity is proven by a standard bootstrap argument or by standard parabolic results, but it is never explained how this works in detail.
For example, let $\Omega$ be an open, bounded subset of $\mathbb{R}^n$ with smooth boundary. Let $n>2$, $p \in (1, (n+2)/(n-2))$, and $T>0$. Consider the nonlinear heat equation with initial data $u_0 \geq 0$
$$\begin{aligned} u_t - \Delta u &= u^p & &\text{in }\Omega\times(0,T), \\ u &= 0 & &\text{on } \partial\Omega\times(0,T), \\ u(x,0) &= u_0(x) & &\text{for all } x \in \Omega.\end{aligned}$$
Now I have an integral solution $U$ in the sense that $U(x,t):\Omega\times(0,T) \rightarrow [0,\infty]$ is a nonnegative measurable function such that
$$ U(x,t) = \int_\Omega G(t,x,y) u_0(y) dy + \int_0^t \int_\Omega G(t-s,x,y) U^p(y,s) dy ds,$$
where $G(t-s,x,y)$ denotes the Green function of the heat equation with Dirichlet boundary condition. I also know that $U \in L^\infty_{loc}((0,T);L^\infty(\Omega))$. Now it says that one can show that $U \in C^{2,1}(\bar{\Omega}\times(0,T))$ with a standard bootstrap argument. Could someone explain how this works?
I have seen bootstrapping for elliptic pdes in the sense that we get higher regularity for a weak solution by applying Sobolev embeddings/schauder estimates over and over again. But it don't see how this would work in this case, because we have an integral solution and not a weak solution.
AI: This follows from linear theory. Set $f(x, t) = U^p(x, t)$.
Bounded weak solutions are Hölder continuous (see vor instance Porzio–Vespri), hence $f$ is. Then one can apply Theorem IV 5.2 (or 5.1?) of Linear and quasi-linear equations of parabolic type by Ladyženskaja, Solonnikov and Ural'ceva to obtain the desired result.
I am not too sure how helpful this answer is to you since looking at the proofs in these reference is not a very fun thing to do. Especially the second one is so general that is (in my opinion) quite hard to read. However, they are my go-to references for such questions.
Perhaps a better reference is Superlinear parabolic problems by Quittner and Souplet. They handle your equation in detail and somewhere (in the appendix, maybe) prove certain regularity results. |
H: Countably infinite Cartesian product of finite sets is infinite
Say that $A \neq \emptyset$, but $\emptyset \in A$. With the term "infinite set" I mean uncountable or countably infinite. I want to prove that
$A \times A \times \cdot \cdot \cdot A \times \cdot \cdot \cdot$
is an infinite set either if $A$ is finite or infinite. If A is countably infinite, I now how to proceed and I now that is true that the infinite Cartesian product is infinite. If A is finite, I simply would show that the cardinality of the above set is $|A| * |A| * \cdot \cdot \cdot$, so basically infinite. Am I right or am I missing something? Three questions (provided that my intuition is true):
if this result is true, does it hold even if $A$ has only two elements (as inferred by the assumptions)?
If $A$ is finite, the countably infinite Cartesian product is countably infinite or uncountable? I think it is countably infinite.
What about the case of $A$ uncountable? I think the product should be also uncountable.
Thanks.
AI: Your set:
$$
A\times A \times \ldots \times A \times \ldots
$$
can be written as $A^{\mathbb{N}}$ (the set of functions from $\mathbb{N}$ to $A$).
The cardinality of $A^{\mathbb{N}}$ is $|A|^{|\mathbb{N}|}$. This cardinality is infinite if and only if $|A|\geq 2$ (and in this case it is uncountably infinite).
So:
If $A=\emptyset$ then $A^{\mathbb{N}}=\emptyset$.
If $|A|=1$ then $|A^{\mathbb{N}}|=1$.
If $|A|\geq 2$ then $A^{\mathbb{N}}$ is uncountable.
The proof of (3) is a standard/famous exercise using Cantor's diagonal argument. Note that it is enough to prove (3) in the case that $|A|=2$. This is because if $|A|\geq 2$ then, letting $A_{0}$ be any $2$-element subset of $A$, observe that $A_0^{\mathbb{N}} \subseteq A^{\mathbb{N}}$.
For the case $|A|=2$ you may as well assume $A=\{0,1\}$. This puts $A^{\mathbb{N}}$ in 1-1 correspondence with the powerset of $\mathbb{N}$ (identify a subset of $\mathbb{N}$ with its characteristic function). The proof that the powerset of $\mathbb{N}$ is uncountable (using Cantor's diagonal argument) is worth trying on your own, if you haven't already. It's also easy to find in this network. |
H: How to study the monotonicity of $c_{n+1} = \sqrt{2+\frac{c_n}{3}}$ given that $c_1 =5$
I am studying the monotonicity of $c_{n+1} = \sqrt{2+\frac{c_n}{3}} = \sqrt{\frac{6+c_n}{3}}$, given than $c_1 = 5$
I proved by induction that $c_n > 0$.
My attempt
Lets make the hypothesis that:
$$ \frac{c_{n+1}}{c_n} = \sqrt{\frac{6+c_n}{3c_n^2}} < 1 \iff \frac{6+c_n}{3c_n^2} < 1 $$. Still that seems to lead nowhere
Any ideas?
AI: You can prove by induction that
$\frac{c_{n+1}}{c_n} < a $ for all $n \ge k$
This would be an absolutely valid argument.
You just start your induction with $n=k$ instead of with $n=1$ as one usually does.
In general, one can define a sequence which is not monotonous up to $n < k$
but is monotonous for $n \ge k$. So your example here is one such case, it seems.
FYI here are the first several values:
1 5
2 1.9148542155126762
3 1.6242797599256802
4 1.5941852422607272
5 1.591035851498506
6 1.59070590739861
7 1.590671337035468
8 1.5906677148329322
9 1.5906673353064673
10 1.5906672955405086
11 1.5906672913739177
12 1.5906672909373514
13 1.5906672908916089
14 1.5906672908868162
15 1.590667290886314
16 1.5906672908862614
17 1.5906672908862558
18 1.5906672908862554
19 1.5906672908862554
20 1.5906672908862554
21 1.5906672908862554
From here you can make all sorts of hypotheses and prove them formally by induction. |
H: Sum of two independent random variables (density)
I remember seeing the statement the sum of two independent random variables has a density if one of them has a density somewhere but forgot where it is. Is this statement true? If it is can someone provide me with a hint or a proof? Thanks!
AI: Let $X,Y$ be random variables, such that $X$ has density $f_X$ and let $\mu_Y$ be the distribution of $Y$. If $X,Y$ are independent, then $Z = X+Y$ has density given by:
$$f_Z(z) = (f_X * \mu_Y)(z) := \int_{\mathbb R} f_X(z-y)d\mu_Y(y)$$
Proof:
We need to check that for any $A \in \mathcal B(\mathbb R)$ we have $\mu_Z(A) = \int_{A}f_Z(z)dz$.
We get: $$ \mu_Z(A) = \mathbb P(X+Y \in A) = \mathbb E[1_A(X+Y)] = \mathbb E[\psi(X,Y)]$$ where $\psi(X,Y) = 1_{A}(X+Y)$.
So that $$ \mu_Z(A) = \int_{\mathbb R^2} 1_A(x+y) d(\mu_X \otimes \mu_Y)(x,y) = \int_{\mathbb R} \int_{\mathbb R} 1_A(x+y)d\mu_X(x)d\mu_Y(y)$$
Where use of Fubinii due to non-negativity and independence of $X,Y$ (which implies that the measure $\mu_{(X,Y)}$ is in the product form $\mu_X \otimes \mu_Y$). Now since $\mu_X$ has density $f_X$, we get :
$$ \mu_Z(A) = \int_{\mathbb R} \int_{\mathbb R} 1_A(x+y) f_X(x)dx d\mu_Y(y) = \int_{\mathbb R}\int_{\mathbb R}1_A(z)f_X(z-y)dzd\mu_Y(y) $$
Where at the last equality we substitution $z=x+y$ in the inner integral. Using Fubinii last time
$$ \mu_Z(A) = \int_A \int_{\mathbb R} f_X(z-y)d\mu_Y(y) dz = \int_A f_Z(z)dz$$
Since $A$ was arbitrary, by definition $f_Z$ is density of $Z$ (it is easy to see that it integrates to $1$ and is non-negative almost everywhere) |
H: 2 seemingly isomorphic groups
Take the following two groups:
$G_1$
$$\begin{array}{c|c|c|c|c}
\cdot & e
& a & b& c\\\hline
e & e & a & b & c \\\hline
a &a & e & c& b\\\hline
b & b & c & e & a \\\hline
c & c & b & a & e
\end{array}$$
$G_2$
$$\begin{array}{c|c|c|c|c}
\cdot & e
& a & b& c\\\hline
e & e & a & b & c \\\hline
a &a & e & c& b\\\hline
b & b & c & a& e \\\hline
c & c & b & e&a
\end{array}$$
In $G_1$ there are 3 normal subgroups, $\{e,a\},\{e,b\},\{e,c\}$
Each one leading to isomorphicly equivalent factor groups.
$$G_1 \cong \mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z} $$
$G_2$ has one normal subgroup $\{e,a\}$ which leads to
$$G_2 \cong \mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z} $$
Which seems to imply there is an isomorphism between them, but there clearly isn’t.
Where am I wrong?
AI: In$G_2$ we have $c=b^{-1}$ and $c^2=a$ so that $b^{-2}=a=b^2$, and hence $b^4=e$, but $b^2=a\neq e$. Hence $G_2$ has an element of order $4$, so that $G_2\cong C_4$, and hence is not isomorphic to $C_2\times C_2\cong G_1$. |
H: Write the polynomial equation given information about a graph
With the following information, I am to write the equation of the polynomial:
Degree 3, zeros at $x=-2$, $x=1$, $x=3$, y intercept: $0,-4$
I know that the answer is: $f(x)=\frac{-2}{3}(x+2)(x-1)(x-3)$
If you look at my post history you can see that I nearly always show what I've tried and where I've gotten stuck. In this case, I do not know where to start or how to approach this problem.
How can one take the given information and calculate the polynomial? Granular, baby steps preferred.
AI: If you know some zeroes of a polynomial, say $a$, $b$ and $c$, you can write
$$
P(x)=(x-a)(x-b)(x-c)Q(x)
$$
where $Q(x)$ is a new polynomial, in fact, if you substitute, for example, $a$, you get
$$
P(a)=(a-a)(b-a)(c-a)Q(a)=0\cdot(b-a)(c-a)Q(a)=0
$$
and the same happens if you substitute $b$ or $c$.
If you know the degree of the polynomial $P$, then you can foresee the degree of the polynomial $Q$, in fact 3 degree are already taken from $(x-a)(x-b)(x-c)$, so the degree of $Q$ is three unit lesser than the degree of $P$.
In the present case $P$ is known to have degree $3$, so $Q$ should have degree $0$, i.e. it is a constant polynomial, i.e. you have
$$
P(x)=(x-a)(x-b)(x-c)k
$$
You can obtain $k$ applying the remaining request, that the intercept is $y_0$,
$$
P(0)=y_0\quad\implies\quad(-a)(-b)(-c)k=y_0\quad\implies\quad k=-\frac{y_0}{abc}
$$
(obviously this is true if all of $a,b,c$ are different from $0$). |
H: Proving that $ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ converges by the comparison test
I would like to prove that the following series converges:$ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ by comparing it with a series that I already know converges. One such series could be the geometric series $ \sum_{k=1}^\infty \frac{2^{k}}{3^{k}} $. Since I know that $ \sum_{k=1}^\infty \frac{2^{k}}{3^{k}} $ converges, the only thing left to prove is that
$$ L = \lim_{k \to \infty}\frac{a_k}{b_k} < +\infty ,$$where $a_k = \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ and $ b_k = \frac{2^{k}}{3^{k}} $. However, when I try to prove this I get the following problem:
$$ \lim_{k \to \infty}\frac{3^{k}(k^{8}+2^{k})}{2^{k}(3^{k} - 2^{k})} $$
And I don't know how to solve this limit problem, and show that the above limit is less than $ +\infty$ so I'm kinda stuck and would appreciate any help!
AI: There are two well-known comparison tests. The first is the Limit Comparison Test: roughly, it says that if $a_k,b_k$ are positive sequences with $\lim a_k/b_k = L>0$, then $\sum a_k$ and $\sum b_k$ converge or diverge together. The second is the Direct Comparison Test; roughly, it says that if $0\leq a_k \leq c_k$ and $\sum c_k$ converges, so does $\sum a_k$; likewise, if $0\leq d_k\leq b_k$ and $\sum d_k$ diverges, so does $\sum b_k$. You can use either test here.
For the LCT, you've correctly identified a good candidate for $b_k$. To compute the limit, try dividing by $6^k$ and then using $\lim_{k\to\infty} k^n/ r^k = 0$ if $r>1$ and $n\in \mathbb{R}$:
$$
\lim_{k\to \infty}\frac{3^k(2^k+k^8)}{2^k(3^k-2^k)} = \lim_{k\to \infty}\frac{1+k^8/2^k}{1-(2/3)^k}= \frac{1+0}{1-0}=1
$$Since $\sum b_k$ converges, so does $\sum a_k$.
For the DCT, let's throw away some stuff to make a good comparison. Observe that $k^8 < 10^{20}\cdot 2^k$ for $k\geq 1$. Likewise, $ (5/2)^k /3<3^k - 2^k $ for $k\geq 1$. Thus
$$
0 < \frac{2^k +k^8}{3^k-2^k} < \frac{(1+10^{20}) 2^k}{3^k-2^k} < \frac{(1+10^{20})2^k}{(5/2)^k / 3} = 3(1+10^{20}) \left(\frac{4}{5}\right)^k
$$These are the terms of a convergent geometric series. So the original series converges. |
H: question about sequences convergent to $e$
Let $(x_n)$ be an increasing divergent sequence of positive reals. Then $(1+\frac{1}{x_n})^{x_n}\to e$, it is a standard theory, it is proven by comparing $x_n$ with $[x_n]$ and $[x_n]+1$ and we use a known fact that $(1+\frac{1}{n})^n$ is increasing and $e$ is its limit.
Is it true that $(1+\frac{1}{x_n})^{x_n}<e$ for all $n$?
Is $(1+\frac{1}{x_n})^{x_n}$ increasing?
As mentioned, I know this for natural numbers, but couldn't find a more general result for positive reals.
Thanks in advance.
AI: Let $f: (0,\infty) \to \mathbb R$ be given by $f(t) = (1 + \frac{1}{t})^t = \exp(t \ln(1+\frac{1}{t}))$
To show it is increasing function of $t$, it is equivalent to showing that function $g: (0,\infty) \to \mathbb R$ given by $g(t) = t\ln(1+\frac{1}{t})$ is increasing.
We have $g'(t) = \ln(1+\frac{1}{t}) + t\cdot\frac{-1}{t^2}\frac{t}{t+1} = \ln(1+\frac{1}{t}) - \frac{1}{t+1}$
So that
$g''(t) = \frac{-1}{(t+1)t} + \frac{1}{(t+1)(t+1)} = \frac{1}{t+1}(\frac{1}{t+1} - \frac{1}{t})$
We immediatelly see that $g''$ is negative on the whole $(0,\infty)$, which means that $g'$ is decreasing on the whole $(0,\infty)$. But $\lim_{t \to 0^+} g'(t) = + \infty$ and $\lim_{t \to \infty} g'(t) = 0$, so that by continuity, $g'$ is positive on the whole $(0,\infty)$, which means $g$ is increasing on the whole $(0,\infty)$. So as long as your sequence $(x_n)_{n \in \mathbb N} \in \mathbb R^{\mathbb N}$ is increasing, then $(1+\frac{1}{x_n})^{x_n}$ is increasing, too. The bound $(1+\frac{1}{x_n})^{x_n} < e$ is then clear. |
H: Definition of geodesic not as critical point of length $L_\gamma$ [*]
Context of this question:
This question follows from a post Decomposition of a function and chain rule. and discusses on something different. Using calculation of variation we can find critical points of a function of a vaiable curve $\gamma$ with its end points fixed at $a,b$ and therefore defining geodesic on a manifold. (The rest of the paragraph is unnecessary reading for the question; it's mainly for the purpose of arranging my several posts on a topic.) Along the geodesic, exponential maps on a manifold project a tangent vector at a point $p$ (locally approximately linearly) to another point, as discussed here What is exponential map in differential geometry (a related but different concept of exponential maps of Lie group is discussed here Relations between two definitions of Lie algebra). Geodesics have as such properties like the 'closed curve' {$\exp_p(v),\forall v$ of the same norm and belonging to $T_pM$} is perpendicular to all the geodesics passing through $p$, and is the shortest curve connecting $a,b$ (i.e. it's also a critical point for length). So we can say the 'closed curve' is very much resembles a circle, and a geodesic a radius or a straight line (We can perhaps even say that with geodesic and exponential maps we 'maps' projective geometry on to a manifold, similar to what we do when, with homeomorphism in definition of a manifold, we 'map' Euclidean space to a manifold). With the fact that geodesics is the shortest curve, we can define a metric (a measure of distance, NOT Riemannian metric which is an inner product and 2-tensor, as discussed here: Calculation of inner product for Riemannian metrics.) on a manifold. The metric is homeomorphic to the original metric of the manifold, as discussed here Comparison of metrics on a manifold..
My question is as follows:
A critical point of 'energy' (as Spivak calls it) $E(\gamma)=\int_a^b \langle \frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle dt$--where $\frac{d\gamma}{dt}$ is tangent vector along $\gamma$ at the point of $\gamma(t)$--is called geodesic. (I guess he uses the name 'energy' for in physics square of velocity is proportional to energy.)
Why we define critical point for energy, instead of critical point for length $L(\gamma)=\int_a^b\sqrt{\langle \frac{d\gamma}{dt},\frac{d\gamma}{dt}\rangle} dt$, to be geodesic?
AI: The quantity
$$
{\displaystyle E(\gamma )~{\stackrel {\text{def}}{=}}~{\frac {1}{2}}\int _{a}^{b}\left\|\gamma '(t)\right\|^{2}~\mathrm {d} {t}}{\displaystyle E(\gamma )~{\stackrel {\text{def}}{=}}~{\frac {1}{2}}\int _{a}^{b}\left\|\gamma '(t)\right\|^{2}~\mathrm {d} {t}}
$$
is sometimes called the energy or action of the curve; this name is justified because the geodesic equations (derive from) the same Euler–Lagrange equations of motion for this action.
So a a unit speed can include geodesics with orbits minimizing energy as well as lengths ... the concepts extending to Hamiltonian and Relativity formulations. |
H: Is there a convention to interpret equalities of functions as series?
I am wondering about whether there is a default or standard interpretation of statements such as $$\sum_{n=1}^\infty f_n(x) = f(x)$$ or equivalently
$$\sum_{n=1}^\infty f_n = f$$
In some cases these statements can mean 'uniformly convergent to $f$' or just 'pointwise convergent to $f$'. But sometimes I come across these equalities without the uniform or pointwise qualification, and thus in these situations I don't know whether as a default to interpret them as meaning pointwise or uniform convergence.
For example, when I first learnt about power series, we had not yet met the notions of uniform convergence (or pointwise). We simply defined $f(x) = \sum_{n=1}^\infty a_nx^n$. In hindsight, this equality really is equivalent to asserting the pointwise convergence of the series to $f$ over the radius of convergence. (Although it also turns out to be uniformly convergent within the radius)
Another example comes from the second answer in this question: When can a sum and integral be interchanged?, from the user Jonas Teuwen. In particular, he states that $f = \sum_n f_n$ in his answer. How should these equalities be interpreted? Is there a default, e.g. just assume it means pointwise, or is it entirely context dependent?
[Note: my current understanding is that when we deal with infinite series of functions, writing it as an equality is really a shorthand for some first order logic statement. I.e. it is completely analogous to the fact that stating $\lim_{n \rightarrow \infty} a_n = l$ in the case of real sequences really means $\forall \epsilon >0 \exists N \forall n>N (|a_n - l|< \epsilon)$. In this way I think of the equality symbol as just shorthand for a more verbose expression when it comes to series of functions, rather than meaning equality of mathematical objects so to say. In this sense, I don't know how to interpret the statements abut equality of series without any context.]
AI: By default, it means that the convergence is pointwise. If it's meant to be uniform convergence, then it's either mentioned, or you are working in a function space with a topology where convergence means uniform convergence.
Like the normed space $(C^0(K),\Vert\cdot\Vert_\infty)$ of continuous functions $K\longrightarrow\mathbb R$ on a compact set $K$ with the norm $\Vert f\Vert_\infty:=\sup_{x\in K}\vert f(x)\vert$. Convergence in this normed space is equivalent to uniform convergence.
Or you could consider the topological space $(\mathcal H(\mathbb C),\tau)$ of holomorphic functions on $\mathbb C$ with the topology of compact convergence $\tau$. There, $\sum_n f_n=f$ would usually be interpreted to mean compact convergence.
So essentially, if it's explicitly mentioned that you're working in a function space whose topology implies a specific kind of convergence, then that kind of convergence is probably implied. Otherwise, the default is pointwise. |
H: Confusion regarding usage of Lambert function
I stumbled upon an equation that goes like:
$$e^{\pi x} - \frac{x}{k} = -1$$
I learnt that Lambert function is useful when dealing with such equations where it can take the form $f(x) = xe^x$.
So, the equation essentially becomes:
$$ x = \frac{1}{\pi} \ln\Big(\frac{x}{k} - 1\Big)$$
Is there any way that I can make $\ln\Big(\frac{x}{k} - 1\Big)$ as some $e^{f(x)}$ so that I can use the Lambert function?
Any other way of solving the equation is always welcomed :) Thanks.
AI: We begin with the expression
$$e^{\pi x}-x/k=-1$$
Upon rearranging, we find that
$$\begin{align}
\pi k&=e^{-\pi x}(\pi x-\pi k)\\\\
&=e^{-\pi(x-k+k)}(\pi x-\pi k)\\\\
-\pi ke^{\pi k}&=e^{-\pi (x-k)}(-\pi(x-k))
\end{align}$$
Can you finish now? |
H: Notation: setting multiple variables to zero
I would like to set multiple variables to zero in an latex algorithm environment. Is there a more pleasing way than to state simply
$a=b=c=d=e=f=0$ ?
This looks cluttered. Maybe someone with more knowledge in math notation knows a better (shorter) way to express this. Thanks!
AI: If you have an index set $I$ such that $I = \{a, b, c, d, e, f\}$, then you can simply say
$$\forall i \in I, i = 0$$ |
H: Confusion in solutions of logarithm.
An expression $log_2 x^2=2$ can be written as $2log_2 x=2$ but leads to loss of a root. I am having difficulty is recognizing the expressions in which above property is applicable. Another expression $log_2^3 x=log_2 x^3$ can be written as $log_2^3 x=3log_2 x$ and doesn't leads to any loss. My teacher has finished logarithm and it is very difficult to ask him as he rarely replies to queries(also the teaching is online).
I thought that it would be related to the power of $x$ if it is either odd or even i.e. we apply property if the power is odd but not when it is even. And also if the base is a constant or variable i.e. property applicable to logs with variable bases and not in constant bases.
I want to know how to recognize in questions that this property doesn't lead to any loss. Kindly help.
AI: You seem to be assuming the logarithm power law holds for any real numbers. This isn't generally true. What we can say is that
$$\log_a{(x^n)}=n\log_a{(x)}$$
holds for all $a,x\gt0$ (with $a\ne1$) and $n\in\mathbb{R}$. But you can also instead use $|x|\ge0$ to somewhat extend this identity to the entire non-zero reals. Namely
$$\log_a{(|x|^n)}=n\log_a{(|x|)}$$
holds for all $x\in\mathbb{R}\setminus\{0\}$. For even $n\in\mathbb{Z}$ this becomes
$$\log_a{(x^n)}=n\log_a{(|x|)}$$ |
H: Probability density of $R=\sqrt{X^2+Y^2}$ when $(X,Y)$ is distributed in a disk
I am struggling with the following question:
Let $\left(X, Y\right)$ be a pair of random variables with joint density function $\mathrm{g}\left(x,y\right) = \frac{1}{2}xy$ if $\left(x,y\right)\in D$, $0$ else. Here, $D$ denotes the first quarter of the disk of radius $2$ centered at $\left(0,0\right)$.
First the problem asks if $X$ and $Y$ are independent. This is easy, we can simply integrate.
My issue is with the second question:
Find the density of the random variable $R = \,\sqrt{\, X^{2} + Y^{2}\, }\,$ and then that of $R^{2}$.
I imagine this is related to a change of variables as it looks like polar coordinates but I don't understand what I am supposed to do here.
AI: Hint
The cumulative distribution function of $\ R\ $ is given by
\begin{align}
P(R\le \rho)&=\iint_{D\cap \cal{D}_\rho}\frac{xy}{2}dxdy\\
&=\frac{1}{2}\int_0^\rho\int_0^\frac{\pi}{2}r^3\cos\theta\sin\theta\, d\theta dr\ ,
\end{align}
where $\ \cal{D}_\rho=\left\{(x,y)\left|\sqrt{x^2+y^2}\le \rho\right.\right\}\ $, and its density function can be obtained by differentiating this. And $\ P(R^2\le r)=$$P\left(R\le \sqrt{r}\right)\ $. |
H: Gram matrix of $A$ is equal to $A^T \bar{ A }$
$A$ is an $m \times n$ matrix, $( \vec{ a_1 }, \vec{ a_2 }, \ldots, \vec{ a_n } )$ and $A'$, the Gram matrix of $A$, is a matrix having $\vec{ a_i } \cdot \vec{ a_j }$ as $(i,j)$ element.
How can we get $A'= A^T \bar{ A }$ ? ($A^t$ is the transpose of $A$ and $\bar{ A }$ is the complex conjugate of $A$.)
AI: Under the assumption that the dot product between two vectors $\vec{v}$ and $\vec{w}$ is $\vec{v}^T \overline {\vec{w}}$, then it follows from the fact that given any two matrices $A$ and $B$, the $(i, j)$th element is the $i$th row of $A$ dotted with the $j$th column of $B$.
Note that if the dot product is simply $\vec{v}^T \vec{w}$, this property is not true (unless we're dealing with real matrices, in which case the complex conjugate is meaningless). Take
$$A = \begin{bmatrix} 1 & 0 \\ i & 1\end{bmatrix}, ~~A^T \overline{A} = \begin{bmatrix} 1 & 0 \\ i & 1\end{bmatrix}$$
for example, which doesn't satisfy your property. |
H: Let $R$ be a $\mathbb Z$ graded ring and let $f \in R_1$ where $f \ne 0$. Show that $R[f^{-1}]$ is a $\mathbb Z$ graded ring.
I've defined $R[f^{-1}]_i = \{ \frac{r}{f^k} \mid r \in R_{i+k}\}$ and I've shown that each $R[f^{-1}]_i$ is an abelian group.
I've shown that $R[f^{-1}]_i R[f^{-1}]_j \subset R[f^{-1}]_{i+j}$.
I just need to show that $R[f^{-1}]$ is a direct sum of the $R[f^{-1}]_i$.
If $\frac{r}{f^k} \in R[f^{-1}]$, then $r \in R$ can be written uniquely $r=r_a + \cdots + r_{a+n}$ where $a \in \mathbb Z$ and $n \in \mathbb N$.
So, $\frac{r}{f^k} = \frac{r_a + \cdots + r_{a+n}}{f^k} = \frac{r_a}{f^k}+\cdots + \frac{r_{a+n}}{f^k} \in R[f^{-1}]_{a-k} + \cdots + R[f^{-1}]_{a+n-k}$.
So, $R[f^{-1}]$ is a sum of the $R[f^{-1}]_i$.
How can we show $R[f^{-1}]$ is a direct sum of the $R[f^{-1}]_i$?
AI: Hint : multiply any relation $\sum_i \frac{r_i}{f^{k_i}} = 0$ in $R[f^{-1}]$ by a sufficiently large power of $f$ to get a relation in $R$. |
H: Weak Convergence in $W^{1,2}(D)$ implies strong convergence in $L^{2}(D)$
Let $D$ is some bounded domain in $\mathbb{R^n}$. At the bottom of page 20/top of page 21 in this book, they observe that a sequence of functions $\{u_{\epsilon}\} \in W^{1,2}(D)$ is uniformly bounded so that $ u_{\epsilon} \rightarrow u$ weakly in $W^{1,2}(D)$ and strongly in $L^2(D)$. I understand the weak convergence in $W^{1,2}(D)$ follows from a theorem functional analysis (In reflexive Banach spaces, every bounded sequence has a weakly convergent subsequence). What fact does the strong convergence in $L^2(D)$ rely on?
AI: That follows from Rellich–Kondrachov_theorem, which says that
$$ W^{1, p} (\Omega) \to L^{q} (\Omega)$$
is a compact embedding when $1 \le q < p^* = \frac{np}{n-p}$.
Since when $p=2$, $p^* = 2n/(n-2)>2$. The boundedness of $\Omega$ implies that
$$ L^q (\Omega ) \to L^2(\Omega)$$
is a bounded embedding when $2\le q< p^*$ (see here). Thus the composition is also compact. Since compact operator sends weakly convergent sequence to strong convergent sequence (see here), $u_\epsilon \to u$ strongly in $L^2$.
Remark: When $n=2$ it is still true. The boundedness of $\Omega$ again implies
$$ W^{1, 2} (\Omega) \to W^{1, p}(\Omega)$$
for all $1\le p<2$. Then the Rellich-Kondrachov theorem implies that
$$ W^{1, p} (\Omega) \to L^{p^*} (\Omega)$$
is compact. Since we want $2p/(2-p) = p^*=2$, we choose $p = 1$. Indeed by choosing $p\to 2$, one can show similarly that $W^{1, 2}(\Omega)$ compactly embed into $L^q$ for all $q\ge 1$, when $n=2$.
Lastly, when $n=1$, $W^{1, 2}$ compactly embed into $C^{0, \alpha}$ for $\alpha <1/2$, and in particular it embeds compactly into $L^2$. |
H: In a metric space is a dense subset of a dense subspace dense in the space itself?
In a metric space is a dense subset of a dense subspace dense in the space itself?
I think that must be false, but I couldn't think of any examples of the contrary.
AI: It is true. Suppose $(X,d)$ is a metric space and we have inclusions $A \subseteq B \subseteq X$ where all the sets get the induced topology/metric. Suppose $A$ is dense in $B$ and $B$ is dense in $X$.
We show that $\operatorname{cl}_X(A) = X$, which will show denseness of $A$ in $X$.
Let $x \in X$. Let $\epsilon > 0$. Then since $B$ is dense in $X$ there is $b \in B$ with $d(b,x) < \epsilon/2$. Since $A$ is dense in $B$, there is $a \in A$ with $d(a,b) < \epsilon/2$. But then $d(a,x) \leq d(a,b) + d(b,x) < \epsilon$ so we have shown that $x \in \operatorname{cl}_X(A)$. |
H: Convergent or Divergent Series? $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt[n+1]{10}}$
I need to find whether this series is convergent or divergent:
$$
\sum_{n = 1}^{\infty}\frac{\left(-1\right)^{ n + 1}}
{\,\sqrt[n + 1]{\, 10\, }\, }
$$
(1) Alternating series test does not provide any additional information since $\lim_{n \to \infty} \frac{1}{\sqrt[n+1]{10}} = 1$ and not $0$. A ratio test with its larger series $\sum\limits_{n=1}^{\infty} \frac{1}{\sqrt[n+1]{10}}$ results in $1$ meaning its inconclusive. Can someone guide me in the right direction?
AI: The general term of the series is not converging to $0$: as you mentioned it, its absolute value converges to one.
Therefore the series diverges. |
H: How to prove or intuitively understand that $\operatorname{P}(\max X_i > \varepsilon) = \operatorname{P}(\bigcup {X_i > \varepsilon})$
I am trying to at least get a feeling of how this equality works. I have a basic understanding of probability theory. However I could not wrap my head around this equality. Any explanation or a proof would help me.
If I understand this it might improve my understanding of the relationship between a random variable and seeing it as an event. Is that union of random variables a common thing to see? Don't we have union of events but joint dist. of random variables mostly?
Let $ X_1, \dots , X_n $ be $n$ random variables (can be independent).
$$\operatorname{P}\left( \max_{1\le i \le n} X_i > \varepsilon\right)= \operatorname{P}\left(\bigcup_i^n \{ X_i > \varepsilon\} \right)$$
I repeat what I read but it doesn't make sense to me even though maybe it is sensible. The probability of a maximum of a set random variable being bigger than epsilon is equal to the probability of the union of the set of random variables that are bigger than epsilon.
AI: If you know anything about measure theory, it may help some to expand out the part of the probability notation that people usually abbreviate. The LHS is
$$P\left ( \left \{ \omega \in \Omega : \max_{i=1,\dots,n} X_i(\omega)>\epsilon \right \} \right ).$$
The RHS is
$$P \left ( \bigcup_{i=1}^n \left \{ \omega \in \Omega : X_i(\omega) > \epsilon \right \} \right ).$$
Now if $\omega$ is such that $\max X_i(\omega)>\epsilon$, then there is some $i^*$ (depending on $\omega$) such that $X_{i^*}(\omega)>\epsilon$, in which case $\omega$ is in the $i^*$th set in the union. On the flip side, if $\omega$ is in the union, then there is some $i^*$ such that $X_{i^*}(\omega)>\epsilon$, and then the max is also bigger than $\epsilon$.
If you don't know anything about measure theory, suffice it to say that you should read "$\{ X_i>\epsilon \}$" as "the event that the random variable $X_i>\epsilon$", and "$\bigcup_{i=1}^n \{ X_i>\epsilon \}$" as "the event that at least one of the events "$X_i>\epsilon$" occurs". That is, $\{ X>\epsilon \}$ is not a set containing a random variable, it is an event, i.e. a set containing elements of the probability space. |
H: Prove that $ |A| = \lim_{t\rightarrow \infty}| A \cap (-t,t)|$ for all $A \subset \mathbb{R}$
Problem taken from the books sheldon Axler Measure , integration Real analysis
Prove that $ |A| = \lim_{t\rightarrow \infty}| A \cap (-t,t)|$ for all $A \subset \mathbb{R}$
My attempt : $\lim_{t\rightarrow \infty}| A \cap (-t,t)|=|A \cap (-\infty,\infty)|=\min| A|$
Im newly learning measure theory
AI: Take a sequence $t_n\nearrow \infty$ (i.e., $t_n>t_{n-1}$) and let $A_n:=A\cap(-t_n,t_n)$. By monotonicity, $\lim_{n\to\infty}|A_n|\le |A|$ ($\because A_n\subseteq A$ for each $n\ge 1$ and $|A_n|$ is nondecreasing). Thus, it remains to show that
$$
\lim_{n\to\infty}|A_n|\ge |A|.\tag{1}\label{1}
$$
Let $B_n:=A_n\setminus A_{n-1}$, where $A_0\equiv \emptyset$. Using Problem 2A.8 in Axler's book, we have
\begin{align}
|A_n|&=\left|\bigcup_{i=1}^{n} B_{i}\right|=\left|\bigcup_{i=1}^{n} B_{i}\cap(-t_{n-1},t_{n-1})\right|+\left|\bigcup_{i=1}^{n} B_{i}\cap (\mathbb{R}\setminus(-t_{n-1},t_{n-1}))\right| \\
&=\left|\bigcup_{i=1}^{n-1} B_{i}\right|+\left|B_n\right|=\cdots= \sum_{i=1}^n |B_i|.
\end{align}
Consequently, $\sum_{n\ge 1}|B_i|=\lim_{n\to\infty}|A_n|$. If the sum on the RHS is infinite, the inequality $\eqref{1}$ is trivially satisfied. Otherwise, since $|\cdot|$ is subadditive, for each $m\ge 1$,
$$
|A|\le |A_m|+\sum_{n> m}|B_i|,
$$
and the second term on the RHS converges to $0$ as $m\to\infty$. |
H: Adjunction isomorphisms imply full and faithful
Let $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$ be two functors such that $\alpha:1_{\mathcal{B}}\cong F\circ G$ and $\beta:G\circ F\cong 1_{\mathcal{C}}$. I want to show that $F$ is full and faithful.
It can be deduced that $F\circ G$ is full and faithful. From this, we can conclude that $G$ is faithful. I am not able to prove that $F$ is full.
Let $C,C'\in\mathcal{C}$ and $b:F(C)\rightarrow F(C')$. We have to find a morphism in $\mathcal{C}(C,C')$ whose image under $F$ is equal to $b$. I thought about using the natural isomorphism $\alpha$:
$$\alpha_{F(C')}\circ b=F(G(b))\circ\alpha_{F(C)}.$$
But this seems like a dead-end. I imagine the proof is rather trivial but I am not able to see it. Any hints?
Edit:
Solved: I think the candidate is $\beta_{C'}\circ G(b)\circ\beta^{-1}_C$.
AI: Since $\mathcal{D}(d, d’)\xrightarrow{G}\mathcal{C}(Gd, Gd’)\xrightarrow{F} \mathcal{D}(FGd, FGd’)$ is an isomorphism, the second map is surjective. Now by $G\circ F\cong \mathrm{id}$ we know $G$ is essentially surjective, so for any $c, c’\in \mathcal{C}$ the above observation can be used to prove that $\mathcal{C}(c, c’)\xrightarrow{F} \mathcal{D}(Fc, Fc’)$ is surjective. |
H: Is this an equivalent way of stating Cauchy's convergence test for series?
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of elements from $\mathbb{R}^p$. Then Cauchy's test states that $\sum\limits_{i=0}^\infty x_i$ converges if and only if for every $\epsilon >0$ there is a natural number $N$ such that $||x_{n+1}+x_{n+2}...+x_{n+p}||<\epsilon$ holds for all $n>N$ and all $p\in \mathbb{N}$.
I was wondering if this is equivalent to saying that $\sum\limits_{i=0}^\infty x_i$ converges if and only if $\lim\limits_{n\to \infty}(x_{n+1}+x_{n+2}...+x_{n+p})=0$, $\forall p\in \mathbb{N}$. To me, they look equivalent, but since I am new to series, I just want to make sure.
AI: Let $S_{np} = \sum_{k=n+1}^{n+p} x_k \to \mathbf{0}$ as $n \to \infty$. By definition for any $\epsilon > 0$ there exists $N_p$ such that $n > N_p$ implies $\|S_{np} - \mathbf{0}\| =\|S_{np}\| < \epsilon$.
Since $| \|S_{np} \| - 0| = \|S_{np}\|$ this implies $\|S_{np}\| \to 0$ if and only if $S_{np} \to \mathbf{0}$ for any fixed $p \in \mathbb{N}$.
The Cauchy criterion states that for every $\epsilon > 0$ there exists $N$ (which depends at most on $\epsilon$) such that if $n > N$ we have $\|S_{np}\| < \epsilon$ for all $p \in \mathbb{N}$. This is equivalent to saying that $S_{np} \to \mathbf{0}$ as $n \to \infty$ uniformly with respect to $p$.
By the same argument we have $\|S_{np}\| \to 0$ uniformly with respect to $p$ if and only if $S_{np} \to \mathbf{0}$ uniformly with respect to $p$.
The equivalent condition is $\lim_{n\to \infty} S_{np} = \mathbf{0}$ uniformly for $p \in \mathbb{N}$. |
H: Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)[0,1]
(C)$[-1,\frac{1}{\sqrt{2}}]$
(D)$[\frac{1}{\sqrt{2}},1]$
My approach is as follow
${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) - {\sin ^{ - 1}}x = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)}^2}} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {1 - \left( {\frac{{{x^2} + 1 - {x^2} + 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{2 - 1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
Not able to proceed from here
AI: Substitute $x = \sin\theta$ and let's restrict our attention to $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ since that covers the whole range of $\sin\theta$:
$$\sin^{-1}\left(\frac{\sin\theta+\cos\theta}{\sqrt{2}}\right) = \frac{\pi}{4}+\sin^{-1}(\sin\theta)$$
$$\implies \sin^{-1}\left(\sin\left(\theta+\frac{\pi}{4}\right)\right) = \frac{\pi}{4}+\theta$$
However, the equality is only true when $$\left|\theta+\frac{\pi}{4}\right|\leq \frac{\pi}{2} \implies -\frac{3\pi}{4}\leq \theta \leq \frac{\pi}{4}$$
Taking the intersection of this set with our original domain, we have that
$$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{4} \implies -1 \leq x \leq \frac{1}{\sqrt{2}}$$ |
H: When trying to prove some $M$ is an $A$-module, with $A$ being an algebra, how do I deal with the set of generators of $A$?
Suppose I want to find out if $M$ is an $A$-module, with $A$ being an algebra with a set of generators $S$ and $R$ a set of equations satisfied by the generators of $A$.
For the sake of simplicity, suppose there are only two elements in the generating set $S$, say $x, y$, and assume further that $R=\{x^2=y\}$. I am now interested in showing that $M$ is an $A$-module.
I have defined multiplication only between the generators $x, y$ of $A$ and the elements $m$ of $M$. Our teacher said this does not suffice to give $M$ an $A$-module structure. He said that we need to also make sure that $x^2*m=y*m$ for all $m\in M$, after defining the multiplication. Why is this required?
AI: Recall that an abelian group $M$ is a left $R$-module if the scalar product
$$ R \times M \to M, (r,m) \mapsto rm $$
satisfies the usual axioms.
We need this scalar product to be well-defined! |
H: Calculating the $p$-value
How do I calculate the $p$ value of the following?
Students' height is approximately normal with
s.d = $4$ inches,
sample = $10$,
mean height = $68$ inches.
Calculate the $p$ value corresponding to the following null hypotheses.
$H_o$ = Avg. height is $70$ inches
$H_1$ = Avg. height is not $70$ inches
My approach:
$$\frac{68-70}{4\sqrt{10}} = -1.5811$$
Since the sample size is less than $30$, I thought I could use the $t$-table with $9$ degrees of freedom. The critical level was not stated, so I used $0.95 = 1.833$. I know this is a two-sided test from the null hypothesis so I would have to multiply it by $2$. However this is incorrect.
The answer is $$P(|N(0,\frac{4}{\sqrt{10}}| >2)= 2(1-0.9431) = 0.1138$$
Where is the number $0.9431$ coming from?
AI: The reason why a $t$-test is not used here is because, although the sample size is small, you are given the population standard deviation and are told that the data are approximately normally distributed. Consequently, the test statistic has the form $$Z \mid H_0 = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}}.$$ The standard deviation is not being estimated from the sample.
Sometimes, it can be difficult to tell from the language of the question whether the standard deviation is being estimated. This is one such case; the main clues are that you are not told it is a "sample standard deviation," and that it is stated before the sample size and the sample mean are given; specifically, it is provided in the same sentence as the statement that height data is approximately normal.
The value $0.9431$ comes from the value of the test statistic above: with $\sigma = 4$, $n = 10$, $\bar X = 68$, and $\mu_0 = 70$, we obtain $$Z = -\sqrt{\frac{5}{2}} \approx -1.58114.$$ Then the $p$-value of this two-sided test is $$2 \Pr[Z < -1.58114] = 2 (1 - \Phi(1.58114)),$$ where $\Phi(1.58114) \approx 0.943077$ is the cumulative distribution function of the standard normal distribution, for the probability that a standard normal random variable is less than or equal to $1.58114$. |
H: Order Preserving Bijection
Consider the posets $(\mathbb{Z}^+,\leq)$ and $(\mathbb{Z}^-,\leq)$. Is the bijection $f: \mathbb{Z}^+ \rightarrow \mathbb{Z}^-$ not order preserving?
I am new to set theory and I don't have an idea on how to show whether $f$ is order preserving. Any help will be highly appreciated.
AI: HINT: There is no order-preserving bijection $f:\Bbb Z^+\to\Bbb Z^-$. Suppose that $f(1)=a$, where $a$ is some negative integer. If $f$ is a bijection, there must be some positive integer $n$ such that $f(n)=a-1$. And $a-1<a$, so what must be true of $n$ if $f$ is to be order-preserving? |
H: Equal number of $n$th roots of unity within character values $f_1(a), \dots, f_m(a)$. (Apostol exercise 6.12 Intro to ANT)
This problem is taken from Exercise 6.12 from Apostol's "Introduction to Analytic Number Theory".
Verbatim, the problem states
Let $f_1, \dots, f_m$ be the characters of a finite group $G$ of order $m$, and let $a$ be an element of $G$ of order $n$. Theorem 6.7 shows that each number $f_r(a)$ is an $n$th root of unity. Prove that every $n$th root of unity occurs equally often among the numbers $f_1(a), f_2(a), \dots, f_m(a)$.
Theorem 6.7 above simply states that every $f_r(a)$ is an $n$th root of unity.
The case of $n = 1$ is trivial, I'll assume $n > 1$ throughout the rest of this question. Apostol gives as a hint that evaluation of the sum
$$\sum_{r=1}^m \sum_{k=1}^n f_r(a^k) e^{-2\pi ik/n}$$
in "two different ways" shows how many $f_r(a) = e^{2\pi i/n}$. The sum above evaluates to $m$, since
$$S = \sum_{r=1}^m \sum_{k=1}^n f_r(a^k) e^{-2\pi ik/n} = \sum_{k=1}^n e^{-2\pi ik/n} \sum_{r=1}^m f_r(a^k), $$
and, by Theorem 6.13,
$$\sum_{r=1}^m f_r(a^k) = \begin{cases}
m & k = n \\
0 & \text{otherwise}.
\end{cases}$$
Therefore, $S$ vanishes for $k = 1, 2, \dots, n-1$, which implies $S = me^{-2\pi i} = m$.
The "second way" to evaluate $S$ is to notice that the sum
$$\sum_{k=1}^n [f_r(a)e^{-2\pi i/n}]^k$$
equals $n$ only when $f_r(a) = e^{2\pi i/n}$, otherwise, $f_r(a)e^{-2\pi i/n}$ is some non-one $n$th root of unity $\omega$ and the sum above equals
$$\sum_{k=1}^n \omega^k = 0$$
by properties of roots of unity. Thus, since $S = m$, if $\alpha$ represents the number of $f_r(a) = e^{2\pi i/n}$, then $m = \alpha n$, or $\alpha =m/n$.
However, what would lead one to consider the sum $S$ in the first place? Where did it come from? It seems almost as if $S$ appeared out of "thin air". Is there another way to prove this statement without the evaluation of $S$?
Thanks in advance!
AI: Here is a proof, using only facts from Chapter 6 of Apostol, and using its notation. Let $a\in G$ have order $n$.
The characters form a group under multiplication. Therefore, if $f_1,\dots,f_r$ are the characters, then multiplying by $f_i$ simply permutes the characters. In particular, for any fixed $a\in G$, it permutes the values in the set
$$X=\{f_j(a)\mid 1\leq j\leq r\}.$$
This is the set we wish to prove is simply each $n$th root of unity appearing $r/n$ times. Suppose that $f_i(a)$ is a primitive $n$th root of unity. Then multiplying by $f_i$ has the effect of multiplying all elements in the set $X$ by $f_i(a)$, a primitive $n$th root of unity. The only way such a multiplication can preserve $X$ is if $X$ is uniform.
Thus we must check that there is such an element $f_i$. If not, then all of the $f_j(a)$ (since it is closed under multiplication as the $f_i$ form a group) must be all $m$th roots of unity for some $m$ strictly dividing $n$ (as they are already $n$th roots of unity).
But now, $a^{n/m}$ is a non-trivial element, and $f_i(a^{n/m})=1$ for all $1\leq i\leq r$. By Theorem 6.10 from Apostol (column orthogonality relation), $a^{n/m}$ is the identity, a contradiction. |
H: Prove that the circumference of an ellipse is given by this infinite series
Prove that the circumference of an ellipse is given by :
$$2\pi a\left[1-\sum_{n=1}^{\infty}\left(\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\right)^{2}\frac{e^{2n}}{2n-1}\right]$$
The parametric of an ellipse is :
$$x=a\cos(\theta)$$
$$y=b\sin(\theta)$$
The Circumference of parametric curve can be computed via:
$$\int_{\alpha}^{\beta}\ \sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}d\theta$$
Assuming the curve is not self-intersecting over the given interval.
Using this we see that the circumference of an ellipse is:
$$4\int_{0}^{\frac{\pi}{2}}\ \sqrt{a^{2}-\left(a^{2}-b^{2}\right)\cos^{2}\left(\theta\right)}d\theta$$
Assuming $a>b$,then $e=\frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a}$ follows the integral is :
$$4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^{2}\cos^{2}\left(\theta\right)}=4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^{2}\sin^{2}\left(\theta\right)}$$
It's well-known that :$$\sqrt{1-x}=-\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^{n}}{4^{n}\left(2n-1\right)}$$
Which is convergent for $\left|x\right|<1$.
Since $0<e^{2}\sin^{2}\left(\theta\right)<1$,hence the integral maybe written as:
$$4a\left[\int_{0}^{\frac{\pi}{2}}d\theta+\int_{0}^{\frac{\pi}{2}}-\sum_{n=1}^{\infty}\binom{2n}{n}\frac{e^{2n}\sin^{2n}\left(\theta\right)}{4^{n}\left(2n-1\right)}d\theta\right]$$
Fubini/Tonelli theorems implies that the integral is indeed:
$$4a\left[\frac{\pi}{2}-\sum_{n=1}^{\infty}\binom{2n}{n}\frac{e^{2n}}{4^{n}\left(2n-1\right)}\int_{0}^{\frac{\pi}{2}}\sin^{2n}\left(\theta\right)d\theta\right]$$
From :
$$\int_{0}^{\frac{\pi}{2}}\sin^{2n}\left(\theta\right)d\theta=\frac{\pi}{2}\prod_{k=1}^{n}\frac{2k-1}{2k}\tag{$n \in \mathbb N^+$}$$
Then integral transforms to:
$$2\pi a\left[1-\sum_{n=1}^{\infty}\binom{2n}{n}\frac{e^{2n}}{4^{n}\left(2n-1\right)}\prod_{k=1}^{n}\frac{2k-1}{2k}\right]$$$$2\pi a\left[1-\sum_{n=1}^{\infty}\frac{\left(2n-1\right)!!}{\left(n!\right)^{2}4^{n}}\frac{e^{2n}}{2n-1}\right]$$
But how to finish ?
AI: Note that
$$\begin{align}
\binom{2n}{n}\frac{e^{2n}}{4^n(2n-1)}\prod_{k=1}^n\frac{(2k-1)}{2k}=\frac{(2n)!}{(n!)^2}\frac{e^{2n}}{4^n(2n-1)}\frac{(2n-1)!!}{(2n)!!}\tag1
\end{align}$$
We also have the identities
$$\begin{align}
(2n)!!&=(2n)(2n-2)(2n-4)\cdots (2)\\\\
&=2^nn!\tag3
\end{align}$$
and
$$\begin{align}
(2n-1)!!&=(2n-1)(2n-3)\cdot 1\\\\
&=\frac{(2n)(2n-1)(2n-2)(2n-3)\cdot 1}{(2n)(2n-2)\cdots (2)}\\\\
&=\frac{(2n)!}{2^n(n!)}\tag3
\end{align}$$
Using $(2)$ and $(3)$ in the right-hand side of $(1)$, we find that
$$\begin{align}
\binom{2n}{n}\frac{e^{2n}}{4^n(2n-1)}\prod_{k=1}^n\frac{(2k-1)}{2k}=\left(\frac{(2n-1)!!}{(2n)!}\right)^2\frac{e^{2n}}{4^n(2n-1)}
\end{align}$$
as was to be shown! |
H: Is $X$ a Borel subset of $\beta X$?
Consider a Tychonoff space $X$ and $\beta X$ its Stone-$\check{\rm C}$ech compactification. I'm currently studying the existence of certain types of regular Borel measures on $X$. Since it's much simpler to obtain regular Borel meaures for a compact, I'd like to obtain them for $\beta X$, and then consider the restriction to $X$. To do so, I'd like to know whether $X$ is a Borel subset of $\beta X$.
AI: It need not be. I’ve not read the paper, but K.C. Goswami has a paper Density topology on R is not a Borel subset of its Stone-Čech compactification [PDF]. It uses results from Ichiro Amemiya, Susumu Okada, Yoshiaki Okazaki, Pre-Radon measures on topological spaces [PDF], which I have also not read. |
H: Mathematical proof regarding circular permutations
The question here is in how many ways can $n$ people stand in order to form a ring.
Now I understand the concept behind it. I have understood it like this:
$n$ people can be arranged in $n!$ ways. Now in each ring, if the ring is broken from a unique point and is straightened to form a row, it would give rise to a unique permutation of $n$ people.
Since each ring consists of $n$ people, it would give rise to $n$ different permutations which in turn means that there are $n$ distinct permutations corresponding to any single ring.
$\therefore $ The number of ways in which a ring can be formed by $n$ people is $\dfrac{n!}{n}=(n-1)!$
Now what I feel is that this explanation is not sufficiently rigorous mathematically. Please help in providing a proper mathematical explanation for it.
I'm sorry if the question is very basic but please help.
Thanks
AI: I would say this (not very different from your justification): suppose you enumerate the $n$ people from a specific person, whatever its position on the ring. Two different arrangements then correspond exactly to two different permutations of the $n-1$ remaining people. Hence there are $(n-1)!$ such arrangements. |
H: Let $X$ be a set and ${\rm Sym}(X)$, the symmetric group on $X$. If $x,y\in X$, is there guaranteed to be an $f\in{\rm Sym}(X)$ such that $f(x)=y$?
Let $X$ be a set and $\operatorname{Sym}(X)$ be the symmetric group on $X$. If $x, y \in X$, is there guaranteed to be an $f \in \operatorname{Sym}(X)$ such that $f(x) = y$?
I think the answer is yes, but I'm having difficulty justifying it logically.
AI: Simply take the involution (transposition) $\;(x\,y)\;$, meaning: the bijective map
$$f:X\to X\;,\;\;f(a):=\begin{cases}y,&a=x\\{}\\x,&a=y\\{}\\a,&a\neq x,y\;\end{cases}\;,\;\;\;\;x\neq y$$
If $\;x=y\;$ then the identity map fulfills what you want. |
H: Isomorphism equivalence relation
I am reading through Real Analysis by Fomin and Kolmogorov, and the book makes the statement that:
"Isomorphism between partially ordered sets is an equivalence relation as defined in Sec. 1.4, being obviously reflexive, symmetric, and transitive".
So I have a couple questions regarding this:
When equivalence relations are expounded upon earlier in the book, it is between members of a given set, but here we are speaking of a mapping between two distinct sets. I don't understand how to translate this into an equivalence relation when it is a mapping between two sets.
How then does the isomorphism end up being symmetric? I understand the reflexivity and transitivity, as they are included in the definition of a set having a partial ordering, but how does the symmetry come into play?
Any help would be appreciated! Thanks!
AI: They’re talking about a relation on the class of all partial orders, specifically, the relation of being isomorphic as partial orders. I’ll write $\langle P,\le\rangle\equiv\langle Q,\preceq\rangle$ to mean that the partial orders $\langle P,\le\rangle$ and $\langle Q,\preceq\rangle$ are order-isomorphic.
All that they’re saying is that this relation is reflexive, symmetric, and transitive.
Any partial order $\langle P,\le\rangle$ is isomorphic to itself: $\langle P,\le\rangle\equiv\langle P,\leq\rangle$.
A partial order $\langle P,\le\rangle$ is isomorphic to a partial order $\langle Q,\preceq\rangle$ iff $\langle Q,\preceq\rangle$ is isomorphic to $\langle P,\le\rangle$: $\langle P,\le\rangle\equiv\langle Q,\preceq\rangle$ iff $\langle Q,\preceq\rangle\equiv\langle P,\leq\rangle$.
If a partial order $\langle P,\le\rangle$ is isomorphic to a partial order $\langle Q,\preceq\rangle$, and $\langle Q,\preceq\rangle$ is isomorphic to a partial order $\langle S,\sqsubseteq\rangle$, then $\langle P,\le\rangle$ is isomorphic to $\langle S,\sqsubseteq\rangle$: if $\langle P,\le\rangle\equiv\langle Q,\preceq\rangle$, and $\langle Q,\preceq\rangle\equiv\langle S,\sqsubseteq\rangle$, then $\langle P,\le\rangle\equiv\langle S,\sqsubseteq\rangle$ |
H: Are algebraic varieties strictly more general than (differentiable) manifolds?
I have read that every non-singular algebraic variety is a smooth manifold. However, I was wondering if every smooth manifold can be expressed as a non-singular algebraic variety, or even just a general algebraic variety; such that algebraic varieties are a strict generalization of manifolds.
If not, does not restricting algebraic varieties to be defined in terms of polynomial equations allow for a generalization?
(Any references on the relation of varieties and manifolds, and any practical use of such a relation, would be appreciated, as I'm new to algebraic geometry. Thanks!)
AI: The classes of varieties and manifolds have nontrivial intersection, but neither one contains the other. A complex projective variety can have singularities which would disqualify it from being a complex manifold. If it is nonsingular, then using the Jacobian criterion we see that we get a complex manifold. For the converse, if we can holomorphically embed our manifold in $\Bbb{P}^n$ as a closed subset then by Chow's Lemma it can be given the structure of an algebraic variety.
Of course, there are complex manifolds that are not algebraic. For instance the Hopf Surfaces $(\Bbb{C}^2\setminus \{0\})/\Gamma$ with $\Gamma$ a discrete group acting freely admit no Kähler metric and cannot be projective.
Actually, Serre's GAGA paper gives a way to translate between analytic geometry and algebraic geometry in the complex case. However, the suitable "analytic" notion of algebraic variety is an "analytic space" which is like a complex manifold but allowing for singularities.
Once we stop working over $\Bbb{C}$, we notice that algebraic varieties make sense over any field $k$ whereas the notion of manifold pretty much needs one to use $\Bbb{R}$ or $\Bbb{C}$. (Apparently there are analogues of manifolds over $\Bbb{Q}_p$ but I don't know much about that.) So, the notion of variety is much more general in that being cut out by polynomial equations makes sense over any ring let alone a field; varieties are more "algebraically general" whereas manifolds are more general in that over $\Bbb{R}$ and $\Bbb{C}$ there are many functions that are smooth/holomorphic but not polynomial. |
H: Let $f$ be a bounded linear functional on a Hilbert space. Show the function $\|x\|^2+f(x)$ achieves a minimum.
Let $f$ be a bounded linear functional on a Hilbert space. Show the function $\|x\|^2+f(x)$ achieves a minimum, describe the point and the minimum value.
The only way I know of solving something of the sort is to introduce some constant to create a real function so we can take derivatives, but I do not see how to incorporate that trick here.
AI: Presumably this is a real Hilbert space since you are talking about a $\min$.
Let $g(x)= \|x\|^2 + \langle c , x \rangle$.
Note that $g(x) = \|{1 \over 2} c + x\|^2 - \|{1 \over 2} c\|^2$ and hence
$g(x) \ge g(-{1 \over 2} c) = - \|{1 \over 2} c\|^2$ for all $x$. |
H: Are "forgetful" and "mindful" Turing machines equivalent?
Premise:
Define a "mindful" Turing machine (MTM) to be a Turing machine (TM) with a log that records the configuration of the head (i.e. current state, symbol being read, next state, symbol to write, and shift) at each step while the machine is running. In addition, an MTM may employ transitions of the form "if [log lists configuration] then [transition]." That is, and MTM "remembers" its past actions and can "choose" its next action accordingly. Call such transitions "unscoped conditional transitions" or "conditional transitions."
Define a "numerate" Turing machine to be an TM with a log which may employ transitions of the form "if [log lists configuration $n$ states ago] then [select this transition]," but not unscoped conditional transitions. That is, an NTM can "remember" its past actions within a particular timeframe and "choose" its next action accordingly. Call such transitions "scoped conditional transitions."
A "forgetful" Turing machine (FTM) is a TM which is neither mindful nor numerate.
A "wise" Turing machine (WTM) is a TM which is both mindful and numerate.
Define two TMs to be equivalent iff for all given starting configurations, the tapes of the two machines are identical upon halting (provided that both machines halt).
Observations:
The following are clear, and an explanation is provided for each:
FTM$\subseteq$MTM$\subseteq$WTM - every MTM which lacks conditional transitions is trivially equivalent to an FTM (since it never reads its log). Every WTM which lacks scoped conditional transitions is trivially equivalent to an MTM.
FTM$\subseteq$NTM$\subseteq$WTM - every NTM which lacks conditional transitions is trivially equivalent to an FTM (since it never reads its log). Every WTM which lacks unscoped conditional transitions is trivially equivalent to an NTM.
Questions:
Are FTMs and MTMs equivalent - i.e. for every MTM, is there an equivalent FTM? If so is there a method for converting MTMs to FTMs? If not, what can an MTM do that an FTM cannot?
Are FTMs and NTMs equivalent - i.e. for every MTM, is there an equivalent FTM? If so is there a method for converting NTMs to FTMs? If not, what can an NTM do that an FTM cannot?
What is the relation between MTMs and NTMs? It is clear that there are steps performed by MTMs that are not performed by NTMs (and vice-versa), but the outputs may be the same regardless of the steps taken to produce them.
AI: These are all equivalent. Basically, a normal Turing machine can record its configuration history elsewhere on the tape; or, if you're using a multi-tape Turing machine, on one of the "auxiliary tapes; or, you can work in a larger-than-normal symbol set which allows you to "double-use" cells (e.g. using a symbol set $\Sigma\times\Gamma$ you can think of a configuration of symbols as a configuration of $\Sigma$-symbols and a separate configuration of $\Gamma$-symbols). That last one is probably the slickest, and a good example of how Cartesian products can be used to clever effect; it also plays an important role in more limited models of computation like finite-state automata (think about intersections of sets accepted by automata).
To get a feel for this, I think it's a good idea to master the proof that two-tape Turing machines are equivalent to one-tape Turing machines; it uses the same basic ideas and is somewhat more streamlined. |
H: Complete projection matrix
I am trying to solve some exercise, which seems to be beyond me.
Let $P$ be a projection. "A projection $P$ is a linear map, that has the following property"1: $P*P=P$, where as $P \in \mathbb{R}^{n\times n}$.
Furthermore, $P$ is defined as
$$P =
\begin{Bmatrix}
\frac{3}{2} & -1 & -1 \\
\alpha & 0 & \beta \\
\frac{1}{4} & -\frac{1}{2} & \frac{1}{2} \\
\end{Bmatrix}
$$
Find $\alpha$ and $\beta$, so that $P$ is a projection.
What I tried:
I thought, by multiplying $P * P = P$, I should receive enough "equations", which allow me to find $\alpha \text{ and } \beta$.
So I tried and got the following result:
$$P * P =
\begin{Bmatrix}
\frac{3}{2}^2-\alpha-\frac{1}{4} & -\frac{3}{2}+\frac{1}{2} & -\frac{3}{2}-\beta-\frac{1}{2} \\
\vdots & \ddots & \vdots \\
\end{Bmatrix}
$$
Which got me following equations:
$$
\begin{align}
\frac{3}{2} & = \frac{3}{2}^2-\alpha-\frac{1}{4} \Rightarrow \alpha = 0.5 \\
-1 & = \frac{3}{2} - \beta - \frac{1}{2} \Rightarrow \beta = 1\\
\end{align}
$$
However, inserting $\alpha = 0.5$ and $\beta = 1$ into $P$ and calculating $P * P$ yields $P * P \neq P$.
Where did I go wrong? Is this even a right way of solving this? If not, can somebody lead me to the right direction?
1 Translated from the exercise I am trying to solve.
AI: You just missed a negative sign on the last equation:
$-1 = -\frac{3}{2} - \beta - \frac{1}{2}$
$\beta = -1$ |
H: Weaker Choice of the Real Numbers
In set theory, a set $A$ is a projective set if for any other sets $B, C$ and for any function $f:A\rightarrow B$ and surjective function $g:C\rightarrow B$, there exists a function $h:A\rightarrow C$ such that $g \circ h = f$. The Axiom of Choice is the statement that all sets are projective sets, while the weaker Axiom of Countable Choice is the statement that $\mathbb{N}$ is a projective set.
Suppose that the Axiom of Choice in classical set theory is replaced with the weaker axiom that the set of real numbers $\mathbb{R}$ is a projective set. Let us for the sake of this question call this new axiom the Axiom of Real Choice, for lack of a better term.
Is the Axiom of Countable Choice provable from the Axiom of Real Choice? If so, then:
Which results in mathematics (such as in real analysis, group theory, etc) that are proven using the full Axiom of Choice and cannot be proven using the Axiom of Countable Choice are now provable using the Axiom of Real Choice?
Which results in mathematics that are proven using the full Axiom of Choice and cannot be proven using the Axiom of Countable Choice still cannot be proved using the Axiom of Real Choice?
Are there any commonly used axioms that are inconsistent with the full Axiom of Choice that are consistent with the Axiom of Real Choice?
AI: Yes. If you have a countable family of sets, $\{X_n\mid n\in\Bbb N\}$, extend it to a family of size $\Bbb R$.
Yes. "Every set of size $\Bbb R$ admits a choice function".
The Axiom of Choice, and therefore all of its equivalents: e.g. every vector space has a basis; every commutative unital ring has a maximal ideal; etc. Including every equivalent of the Boolean Prime Ideal theorem (e.g. every commutative unital ring has a prime ideal).
This is not particularly clear, but likely to be false.
This is Form 181 in the Howard–Rubin dictionary of choice principles. But the biggest problem with this principle is that it will follow from the conjunction of:
$|\Bbb R|=\aleph_1$ and
$\sf AC_{\aleph_1}$.
And while it's a somewhat nontrivial conjunction, it is also not particularly powerful or interesting (for example, it is easy to arrange a model where this conjunction holds, but there are sets which cannot be linearly ordered). You can find more information in the book, or in the online graphing website: https://cgraph.inters.co/. |
H: Greatest Common Divisor Problem: Prove that $\gcd(\frac{a^3+b^3}{a+b}, a+b) = \gcd(a+b, 3ab)$
I've been stuck in this problem for some time now. Currently what I have accomplished is, using the propriety $\gcd(a,b) = \gcd(b,a \bmod(b))$ to get in the equation
$$ \gcd(a+b, \frac{a^3+b^3}{a+b}\bmod(a+b)) $$
but I don't know where to go anymore or even if I'm in the correct path. Any tips or solutions would be great appreciated.
AI: hint: $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2 -3ab)$ |
H: Corestriction of a full and faithful functor
Let $F:\mathcal{C}\rightarrow\mathcal{D}$ be a full and faithful functor. Consider the corestriction $F:\mathcal{C}\rightarrow F(\mathcal{C})$ of $F$ to its image. Note that for $D\in F(\mathcal{C})$, we have $D=F(C)$ for some $C\in\mathcal{C}$: i.e. $D\cong F(C)$. This implies that the corestriction of $F$ to its image is an equivalence of categories.
Now, I don't understand the following:
Replacing $F(\mathcal{C})$ by the full subcategory
$\mathcal{C}'\subset\mathcal{D}$ generated by all the objects
$D\in\mathcal{D}$ isomorphic to an object of the form $F(C)$, one
still has an equivalence $F:\mathcal{C}\rightarrow\mathcal{C}'.$
I don't follow this characterization of $\mathcal{C}'$: specifically, I am not sure what a "subcategory generated by all the objects etc." means. I know that
$$\mathcal{C}'(D,D'):=\mathcal{D}(D,D')$$
for all $D,D'\in\text{Ob}(\mathcal{C}')$–but what are the object of $\mathcal{C}'$?
Can someone please give an explicit description of the full subcategory $\mathcal{C}'$?
Edit:
Is $\text{Ob}(\mathcal{C}')$ merely $\{D\in\text{Ob}(\mathcal{D})\ |\ (\exists C\in\mathcal{C})( D\cong F(C)\}$? Then, why say "generated"?
AI: If $\mathcal{D}$ is a category then a subcategory $\mathcal{C}'$ of $\mathcal{D}$ is called full if for any two objects $X$ and $Y$ of $\mathcal{C}'$ we have
$$
\mathcal{C}'(X, Y) = \mathcal{D}(X, Y) \,.
$$
A full subcategory is uniquely determined by its class of objects.
Given any class of objects $\mathcal{O} \subseteq \operatorname{Ob}(\mathcal{D})$ we can therefore talk about the full subcategory of $\mathcal{D}$ generated by $\mathcal{O}$.
This is the unique subcategory $\mathcal{C}'$ of $\mathcal{D}$ with $\operatorname{Ob}(\mathcal{C'}) = \mathcal{O}$ and $\mathcal{C}'(X,Y) = \mathcal{D}(X,Y)$ for all objects $X, Y \in \mathcal{O}$.
In the given example we take
$$
\mathcal{O}
=
\{
D \in \operatorname{Ob}(\mathcal{D})
\mid
\text{there exists $C \in \operatorname{Ob}(\mathcal{C})$ with $D \cong F(C)$}
\} \,.
$$
The resulting full subcategory $\mathcal{C}'$ of $\mathcal{D}$ is known as the essential image of $F$.
Regarding your edit:
I don’t think there is a particular reason why the verb “generated” is used here.
One could also use a different verb which describes the situation.
(I personally would talk about the full subcategory of $\mathcal{D}$ whose class of objects is given by $\mathcal{O}$.)
On a side note:
The image $F(\mathcal{C})$ won’t necessarily be a subcategory of $\mathcal{D}$.
So in general you won’t get an equivalence of categories between $\mathcal{C}$ and $F(\mathcal{C})$.
(But if $F(\mathcal{C})$ is a subcategory of $\mathcal{D}$ then the corestriction of $F$ will indeed be such an equivalence of categories.) |
H: Is a function increasing if the derivative is positive except at one point of an interval?
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable on $(a,b)$. Suppose $f' > 0$ on $(a,b)$ except at a point $c \in (a,b)$ (that is, $f'(c) \leq 0$).
Is $f$ increasing on $(a,b)$?
Must $f'(c)$ be zero, or can it be negative?
Clearly $f$ is increasing on $(a,c) \cup (c,b)$ but I'm not sure about how the value at $c$ compares with the values at other points.
And I think $f'(c)$ must be zero: If $f'(c) < 0$ then for small positive $h$ we have $\frac{f(c+h) - f(c)}{h}$ is also negative (by definition of the derivative as a limit of this ratio), so $f(c+h) - f(c) < 0$. Since $f'$ is positive on $(c,c+h)$, the Mean Value Theorem implies that $f(c+h) - f(c) = f'(d)h$ for some $d \in (c, c+h)$, and $f'(d)h$ is a product of two positive numbers, hence positive. So $f(c+h) - f(c) > 0$, a contradiction.
AI: Let $f\colon (a,b)\to \Bbb R$ be continuous, and $f'(x)>0$ for $x\in(a,b)\setminus\{c\}$. We do not even need to assume that $f'(c)$ exists.
Then $f$ is strictly increasing:
Suppose $a<x_1<x_2<b$. Then $f(x_1)<f(x_2)$ follows from the Mean Value Theorem if $x_2\le c$ or if $x_1\ge c$. If $x_1<c<x_2$, just go in two steps via $c$.
Now suppose additionally that $f'(c)=$ exists. Then directly from the increasing property we get $f'(c)\ge0$. |
H: Finding a polynomial $f(x)$ of degree 5 such that $f(x)$ is divisible by $x^3$ and $f(x)+2$ is divisible by $(x+1)^3.$
There is some polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)$ is divisible by $x^3$.
$f(x)+2$ is divisible by $(x+1)^3.$
Find that polynomial.
I know that because $f(x)$ is divisible by $x^3$ our polynomial is in the form of $ax^5+bx^4+cx^3.$ However, I'm not very sure how our second condition comes into use. Any help?
AI: If $f(x)+2=ax^5+bx^4+cx^3+2=(x^3+3x^2+3x+1)(dx^2+ex+f)$,
then multiplying out and equating coefficients yields
$a=d, b=e+3d, c=f+3e+3d, 0=d+3e+3f, 0=e+3f,$ and $ 2=f$.
This system is easy to solve for $a, b, c, d, e, $ and $f, $
and then the answer is $f(x)=ax^5+bx^4+cx^3$. |
H: How to solve for a specific gradient in an implicit relationship, when no points are known?
The problem
Consider the following relation:
$$x^2-3xy+y^2=7$$
I'm struggling with what is essentially the following task:
Find all coordinates of all points where the gradient of the tangent of the curve is ${2\over3}$.
Using implicit differentiation, I arrive at the following derivative of $y$ wrt $x$:
$${\text dy\over\text dx}=\frac{3y-2x}{2y-3x}$$
This is somewhat of a mindbender for me, since it looks like the derivative itself depends on both $x$ and $y$. How can this be? How does one work with this?
Even more daunting is the task of solving $\frac{\text dy}{\text dx}=\frac{2}{3}$. The equation cannot be solved as we have a single, two-variable equation: $$2(2y-3x)=3(3y-2x)$$
Sadder yet is that by my logic, there is an infinite number of solutions ($xy$ pairs) that satisfy this. Unfortunately, graphing this relation gives the solution set $(x,0)$ where $x\inℝ$.
The correct answer is $(\sqrt7,0)\cup(-\sqrt7,0)$.
My questions
How do I work with a derivative which depends on both $x$ and $y$, and conceptually, how is that even possible?
Why is it that graphing the solutions of $2(2y-3x)=3(3y-2x)$ did not work and gave erroneous solutions?
How is the answer $(\sqrt7,0)\cup(-\sqrt7,0)$ obtained?
Thank you very much :)
AI: Note: As per comment everywhere where you read ellipse you should read hyperbola.
In your case the curve is an ellipse. If you go to a value $x$, what is the derivative of the function? Well, it is not clearly defined. First you need to find $y$. You can have 0, 1, or 2 $y$ values corresponding to a single $x$. So where do you calculate the derivative? You need to choose $y$ to be on the ellipsis.
Think for example that you have a circle of radius $1$, centered at the origin. And you want to calculate the derivative at $x=0.5$. You notice that you have two intersections. At $y=\sqrt 3/2$ you have a negative derivative, but at $y=-\sqrt 3/2$ the derivative is positive.
So you have an equation for the derivative that involve $x$ and $y$. But you also know that $(x,y)$ is on the given curve. That's the second equation.
What you need to do is to write $y(x)$ from the derivative equation, plug it into the equation for the curve, find $x$, then find $y(x)$.
Let me know how these steps work for you. |
H: Hartog's lemma: does a proof for $\mathbb N$ generalise?
I'm reading some online notes taken from Imre Leader's Cambridge lectures on logic and set theory. I find the notes very clear on the whole, but one particular proof - the proof of Hartog's Lemma on page 21 - strikes me as odd.
Hartog's lemma: For any set $X$, there exists an ordinal that does not inject into $X$.
In the notes, Hartog's lemma is first proved in the special case where $X = \mathbb N$, which is merely the statement that there exists an uncountable ordinal. To prove this special case, one considers the set $B$ consisting of distinct ordinals defined on subsets of $\mathbb N$. One then constructs the ordinal $\omega_1 = \sup B$, the least upper bound on the ordinals in $B$. (This ordinal $\omega_1$ is constructed by thinking of the ordinals within $B$ as being nested inside one another and patching them together, as described in detail on pages 19 and 20 of the notes.) One then argues that $\omega_1$ must be an uncountable ordinal. For if $\omega_1$ is countable, then the ordinal $\omega_1^+$, defined as $\omega_1^+ = \omega_1 \cup \{ x \}$ where $y < x$ for all $y \in \omega_1$, is also countable, and is greater than $\omega_1$, contradicting the fact that $\omega_1$ is by definition an upper bound on countable ordinals.
What confuses me is when the notes claim that the above proof for $X = \mathbb N$ generalises immediately to arbitrary sets $X$.
To me, this seems like invalid reasoning. Here are some examples where the proof seems to fall over:
If $X$ is a finite set and $B$ is the set of distinct ordinals defined on subsets of $X$, then $\omega_1 = \sup B$, as a set, is $X$ itself. The construction has failed to produce an ordinal bigger than $X$.
If $X$ is an infinite set other than $\mathbb N$, we can follow the same construction, producing an ordinal $\omega_1 = \sup B$ as before. We then want to argue that if $\omega_1$ has the same cardinality as $X$, then $\omega_1^+ = \omega_1 \cup \{ x \}$ has the same cardinality as $X$ too. But it isn't obvious to me that an arbitrary infinite set plus one extra element has the same cardinality as the infinite set without this extra element. (Except for when the infinite set is countable.)
What am I missing?
AI: If $|X|=n$, the construction produces $\{k\in\omega:k\le n\}=n+1>n$.
More generally, let $\alpha$ be the set of ordinals that inject into $X$. Then $\alpha$ is a transitive set of ordinals, so $\alpha$ is an ordinal. Of course $\alpha\notin\alpha$, so $\alpha$ does not inject into $X$.
Alternatively, if $X$ is infinite, you can come closer to imitating the earlier argument by observing that if $\alpha$ injected into $X$, then the ordinal $\alpha+1$ would also inject into $X$ (since for infinite $\alpha$ there is an easy bijection between $\alpha$ and $\alpha+1$), and clearly $\alpha+1\notin\alpha$. It appears that at this point Leader had not proved that a transitive set of ordinals is an ordinal, so he may well have done it this way (or expected his audience to fill in this detail). |
H: Show that the transformations are linear
I'm a little confused with this question:
Let $W=V \bigoplus U$. where U and V are subspaces of W. Let $P_{1}$ and $P_{2}$ The transformations of W in W such that $w=u+v$ of W (u $\in$ U and v $\in$ V) associate, respectively, u and v, that is, $P_{1}(w)=u$ and $P_{2}(w)=v$. Show that $P_{1}$ and $P_{2}$ are linear.
Well i don't even know exactly how to start, but here is my trial:
I defined
$$P_{1}: W \rightarrow U, P_{1}(w)=u.\\
P_{2}: W \rightarrow V, P_{2}(w)=v.$$
So what I'd have to do was just prove that $P_{1}(w_{1}+w_{2})= P_{1}(w_{1})+P_{1}(w_{1})$ and $P_{1}({\lambda}w)={\lambda}P_{1}(w)$. The same for $P_{2}$. But since i have no information about the elements of the given spaces, i don't know how to proceed or even if my trial is right.
If you guys could give any tip of how to solve this it would be great. Thank you
AI: Let $w_1,w_2\in W$. Then there exist unique $u_1,u_2\in U$ and $v_1,v_2\in V$ such that $w_1=u_1+v_1$ and $w_2=u_2+v_2$.
Also, for $w_1+w_2\in W$, there exist unique $u\in U, v\in V$ such that $w_1+w_2=u+v$, which means $P_1(w_1+w_2)=u$ and $P_2(w_1+w_2)=v$ by definition of the maps.
But we also know that $w_1+w_2=u_1+v_1+u_2+v_2=(u_1+u_2)+(v_1+v_2)$, with $(u_1+u_2)\in U$ and $(v_1+v_2)\in V$. Due to the uniqueness of the decomposition of $w_1+w_2$ as a sum of vectors in $U$ and $V$, we realize that $u=u_1+u_2$ and $v=v_1+v_2$.
Thus we get $P_1(w_1+w_2)=u=u_1+u_2=P_1(w_1)+P_2(w_2)$. Similarly for $P_2(w_1+w_2)$.
I'll leave it to you to prove that $P_1(\lambda w)=\lambda P_1(w)$ and $P_2(\lambda w)=\lambda P_2(w)$. |
H: is the integral of any polynomial of the form $a+bt+ct^2......+dt^n$ from zero to t always not zero.
For instance,
I am looking at a question in linear algebra,
Define $T : P_{3}\to P_{4}$ such that
$T(p)=\int_{0}^{t}p(x)dx$
find the Nul(T)
This ends up being = {0}. Since this is true I was wondering how no polynomial is from zero to t where t is the dimension of the polynomial is zero
AI: hint
$$T(p)=0\implies $$
$$\frac{dT(p)}{dt}=0\implies$$
$$\frac{d}{dt}\int_0^tp(x)dx=p(t)=0$$ |
H: Smallest closed ideal containing an element in a $C^*$-algebra
Let $A$ be a $C^*$-algebra and $a \in A$. I want to describe the smallest closed ideal containing $a$. If the algebra is unital, I think this ideal will be $\overline{AaA}$. But can we describe this ideal when $A$ is non-unital? Maybe something like
$$\overline{AaA + \Bbb{Z}a} $$
can work?
AI: If there is no unit then the ideal is $$\overline{ AaA +Aa+aA+span(a)}.$$ |
H: Question about whether function is odd or even
If x is an odd number, would the function $2020^x + x^{2020} $ be odd or even?
I'm currently having trouble figuring out the answer to this question. Currently I believe the result would be odd, since an even number raised to an odd power results in an even number, while an odd number raised to an even power results in an odd number. The sum of an even and an odd is odd, therefore the function will produce odd outputs when x is an odd input.
However, when I try and test the outputs using a function, the result is undefined, so I cannot tell for sure if I'm correct. Any clarification on my answer and insight would be appreciated.
Edit- I
AI: If you think of the expression
$$
2020^x + x^{2020}
$$
as telling you the value of a function $f$ at a variable point $x$ then that function is an odd function if and only if for every value of $x$
$$
f(x) = f(-x).
$$
In this example, the function is not an odd function.
That has nothing to do with the value of the function when $x$ is an odd integer.
It happens to be true that when $x$ is a positive odd integer, $2020^x$ is even and $x^{2020}$ is odd, so their sum is an odd number.
When $x$ is a negative odd integer, $2020^x$ is not an integer and $x^{2020}$ is odd, so their sum is not an integer and so is neither even nor odd. |
H: Finding the limit of $\frac{N_n}{\ln(n)}$ where $N_n$ is the number of digits of $n$
I came across this question in an entrance exam of a local college where we are asked to evaluate the limit : $$\lim\limits_{n\to\infty} \frac{N_n}{\ln(n)}$$
Where $N_n$ denotes the number of digits of $n$, with the latter being a non zero positive integer.
It seems I am lacking some sort of relationship between $N_n$ and $n$ (i.e equality/inequality). I could spot that $n \ge N_n$ but it doesn't seem useful in this case.
Appreciate any help!
AI: HINT
Following the suggestion given in the comments, we have that
$$\frac{\log_{10}n}{\ln(n)}\le \frac{N_n}{\ln(n)}\le \frac{1+\log_{10}n}{\ln(n)}$$
then we can conclude by squeeze theorem. |
H: Explicit description of $\mathcal{O}_{\Bbb{P}^1}(-1)$ as a line bundle
I understand the construction of $\mathcal{O}_{\Bbb{P}^1}(-1)$ as a sheaf on $\Bbb{P}_\Bbb{C}^1$, but I'm trying to understand how exactly does this define a line bundle and why people call this the "tautological line bundle".
Following the suggestion of "tautological", my first idea was to define the map:
\begin{align*}
\pi:\Bbb{A}^2&\to\Bbb{P}^1\\
(x_0,x_1)&\mapsto (x_0:x_1)
\end{align*}
whose fibers are clearly lines.
Now, I can neither see how to define the trivialization maps nor how this relates to the sheaf $\mathcal{O}_{\Bbb{P}^1}(-1)$, so probably I'm on the wrong path.
I'm having a hard time trying to come up with different ideas, because I don't even know how to find a variety $X$ so that $\pi:X\to\Bbb{P}^1$ is the line bundle I'm looking for.
AI: Maybe it helps to think about it if you don't choose a basis right away. Let $V$ be a two-dimensional vector space and $\mathbb{P}(V)$ the variety of lines through the origin in $V$. Then $\mathbb{P}(V)$ certainly has a trivial two-dimensional vector bundle $V \times \mathbb{P}(V)$. Inside this bundle is the more interesting bundle
$$
\{ (v, \lambda) \in V \times \mathbb{P}(V): v \text{ is on the line } \lambda\}.
$$
This is why it's called tautological: a point of $\mathbb{P}(V)$ is a line and this is the bundle that attaches that line to it.
Of course you can't compute like this; you need coordinates. In coordinates, what you want is
$$
\{(x_0, y_0), [z_0: w_0] \in \mathbb{C}^2 \times \mathbb{P}^1 : \exists a \in \mathbb{C} \text{ such that } az_0 = x_0 \text{ and } aw_0 = y_0 \}.
$$
together with its normal projection to $\mathbb{P}^1$. |
H: Justify the equation $\sum_{n=0}^{\infty} \frac{(-1)^n}{m+nk} = \int_{0}^1 \frac{t^{m-1}}{1+t^k} dt$, $m > 0$.
I am trying to solve this problem from my exam prep under real analytic function section.
Justify the equation $\sum_{n=0}^{\infty} \frac{(-1)^n}{m+nk} = \int_{0}^1 \frac{t^{m-1}}{1+t^k} dt$, $m > 0$.
I am not sure what $k$ is as the question doesn't really specify.
Any help will be appreciated.
AI: HINT:
$$\frac1{m+nk} =\int_0^1 t^{m+nk-1}\,dt $$ |
H: How do I compute my limits of integration for a density function?
$ f(x,y) = 6x^2y \ $ if $0 \leq x \leq 1 , 0\leq y \leq 1$ and $0$ in other case.
How to compute $ P(X+Y>1) $ ?
$$ P(X+Y>1) = P(X>1-Y) = \int_{0}^{1} \int_{y}^{1-y} 6x^2y \ dx \ dy $$
Is this correct?
AI: No. It is $\int_0^{1}\int_{1-y}^{1} 6x^{2}y dx dy$. |
H: Prove the inequality $1 - \tanh(xy) \leq \cosh(x)^{-y}$
Using some tricks in statistical mechanics I came across the inequality.
$$
1 - \tanh(xy) \leq \cosh(x)^{-y}
$$
for all $x,y >0$.
Do you have a proof (or counterexample)?
AI: Let $\alpha$ be a positive real number greater than or equal to $1$.
Since $f_\alpha(z)=z^\alpha:\left[0,+\infty\right[\to\mathbb{R}$ is a convex function, it results that
$f_\alpha\left(\frac{z_1+z_2}{2}\right)\le\frac{f_\alpha(z_1)+f_\alpha(z_2)}{2}$ for any $z_1,z_2\in\left[0,+\infty\right[.$
Let $x,y$ be any two positive real numbers with $y\ge1$, we get that
$f_y\left(\frac{e^{x}+e^{-x}}{2}\right)\le\frac{f_y(e^{x})+f_y(e^{-x})}{2}\;\;$ that is
$\left(\frac{e^{x}+e^{-x}}{2}\right)^y\le\frac{e^{xy}+e^{-xy}}{2}$
$\frac{2}{e^{xy}+e^{-xy}}\le\left(\frac{e^{x}+e^{-x}}{2}\right)^{-y}$
$1-\frac{e^{xy}-e^{-xy}}{e^{xy}+e^{-xy}}=\frac{2e^{-xy}}{e^{xy}+e^{-xy}}<\frac{2}{e^{xy}+e^{-xy}}\le\left(\frac{e^{x}+e^{-x}}{2}\right)^{-y}$
$1-\tanh(xy)=1-\frac{e^{xy}-e^{-xy}}{e^{xy}+e^{-xy}}<\left(\frac{e^{x}+e^{-x}}{2}\right)^{-y}=\left(\cosh x\right)^{-y}.$
Hence,
$1-\tanh(xy)<\left(\cosh x\right)^{-y}\;$ for any $x,y\in\mathbb{R}^+$ with $y\ge1$. |
H: How to write recursive relationship as a summation of matrices/vectors
I have a recursive relationship, where for a given iteration $k$,
\begin{align*}
k=0, f(k) &= v_1\\
k=1, f(k) &= v_1 - 2Av_2\\
k=2, f(k) &= v_1 - 3Av_2 + 3ABv_1 - ABAv_2\\
k = 3, f(k) &= v_1 - 4Av_2 + 6ABV_1 - 4ABAv_2 + ABABv_1\\
\end{align*}
where $v_1$ is a $m_1 \times 1$ vector, $v_2$ is a $m_2 \times 1$ vector, $A$ is a $m_1 \times m_2$ matrix, and $B$ is a $m_2 \times m_1$ matrix.
I want to write $f(k)$ as a summation. I can see that the coefficients are the binomial expansion coefficients. Moreover, the coefficients change from 1 to -1. Therefore, I have
$$f(k) = \sum_{i=0}^k \binom{k}{i} (-1)^{i}...$$
to start with. However, I am stumped by the $A$, $AB$, $ABA$, $ABAB$, terms. How can I express those in my summation?
AI: I think the best you can do is $(AB)^{\lfloor i/2 \rfloor} A^{i \text{ mod } 2}$. Though a "definition by example", as you've given above, will likely suffice for the reader to understand what you are defining. You could also define a function such as $m_i = \begin{cases} (AB)^{i/2} & i \text{ even}\\ (AB)^{(i-1)/2}A & i \text{ odd} \end{cases}$. Then you could define your term using this $m_i$.
I hope this helps ^_^ |
H: Let $S_1$ and $S_2$ be the symmetric closures of $R_1$ and $R_2$, respectively. Prove that $S_1 \subseteq S_2$.
This is an exercise from Velleman's "How To Prove It".
Suppose $R_1$ and $R_2$ are relations on $A$ and $R_1 \subseteq R_2$. Let $S_1$ and $S_2$ be the symmetric closures of $R_1$ and $R_2$, respectively. Prove that $S_1 \subseteq S_2$.Let $T_1$ and $T_2$ be the transitive closures of $R_1$ and $R_2$, respectively. Prove that $T_1 \subseteq T_2$.
The symmetric closure $S$ of $R$ is the smallest set (under the subset partial order) such that $R \subseteq S$ and $S$ is symmetric. The definition for transitive closure is similar.
I am stuck on solving both of these problems.
For the first one, if I could somehow show that an arbitrary element of $S_1$ is contained in $R_1$, it would be easy to show that $S_1 \subseteq S_2$, since $R_1 \subseteq R_2$ and $R_2 \subseteq S_2$.
Any hints would help. Thanks in advance!
AI: Contains = is a superset of.
The symmetric (resp. transitive) closure of $R$ is the least symmetric (transitive) relation containing $R$.
The symmetric (transitive) closure of $R_2$ contains $R_2$ and thus contains $R_1$; since the symmetric (transitive) closure of $R_1$ is the least symmetric (transitive) relation containing $R_1$, it must be contained by the closure of $R_2$ by definition of "least".
Edit: going in more detail. The case for "transitive" is analogous.
Consider a relation $R \subseteq A^2$. We define the symmetric closure of R to be $S = \{(x, y) \in A^2 : $ for every symmetric relation $S' \subseteq A^2$ such that $R \subseteq S'$, $(x, y) \in S'\}$.
Note that the symmetric closure of $R$ is symmetric and contains $R$.
Note also that whenever we have $S' \subseteq A^2$ satisfying the same properties (that is, whenever we have symmetric $S'$ such that $R \subseteq S'$), we will always have $S \subseteq S'$. Therefore, $S$ may be described as "the least symmetric relation containing $R$".
Back to the problem at hand. Note therefore that we have $R_2 \subseteq S_2$ and $S_2$ symmetric since it is given that $S_2$ is the symmetric closure of $R_2$. Therefore, since it is given that $R_1 \subseteq R_2$, we have $R_1 \subseteq S_2$. That is, $S_2$ is a symmetric relation containing $R_1$. But it is given that $S_1$ is the symmetric closure of $R_1$; therefore, we have $S_1 \subseteq S_2$.
Interestingly, it turns out that the symmetric closure can be given a simpler description. The symmetric closure of $R$ can be written as $R \cup \{(x, y) \in A^2: (y,x) \in R\}$. It can easily be shown that this definition is equivalent to the above.
Sadly, the transitive closure doesn't have a simpler description. It does, however, have a more "concrete" (or, to be precise, more predicative) definition. The transitive closure can also be defined as $\{(x, y) \in A^2 : $ there is some $i > 1$ and a sequence $x_1, ..., x_i$ such that $x = x_1$, $y = x_i$, and for all $j$ s.t. $1 \leq j < i$, $(x_j, x_{j + 1}) \in R\}$. This definition too is equivalent to the above (of transitive closure).
However, it's easiest to approach the problem using the "least relation" definition. |
H: Show if $\int_U g$ exists, so does $\int_U f$ (extended integral question)
Let $U \subseteq \mathbb{R}^n$ be open(not necessarily bounded), let $f,g: U \rightarrow \mathbb{R}$ be continuous, and suppose that $|f(x)| \leq g(x)$ for all $ x \in U$. Show if $\int_U g$ exists, so does $\int_U f$
Definition: Let $f:S \rightarrow \mathbb{R}$ be a (not necessarily bounded) function on a set $S \subseteq \mathbb{R}^n$ (not necessarily bounded), and suppose $f \geq 0$, define $\int_S f=\textbf{sup} \{\int_D f: D \subseteq S$ $\textbf{is compact and rectifiable} \}$, provided this supremum exists. For general $f$ (which may be negative at some point), define $\int_S f=\int_S f^+-\int_S f^-$, provided both these integrals exist. (Re call: $f^+=\textbf{max}\{f,0 \}, f^-=\textbf{max \{-f,0\}}$). $\int_S f$ is the extended integral over S.
Lemma: Let $ u \subseteq \mathbb{R}^n$ be open. There exists a sequence $\{C_N:N \in \mathbb{N} \}$ of compact, rectifiable sets such that $C_N \subseteq C_{N+1}$ for each $N \in \mathbb{N}$, and $u=\bigcup_{N=1}^{\infty} C_N$.
Theorem: Let $u \subseteq \mathbb{R}^n$ be open, let $f: u \rightarrow \mathbb{R}$ be continuous, and let $\{C_N:N \in \mathbb{N} \}$ be a sequence of compact rectifiable sets. f is integrable over $u$ if and only if $\{\int_{C_N} |f| \}^{\infty}_{N=1}$ is a bounded sequence of regular integrals.
Corollary: Let $fLu \rightarrow \mathbb{R}$ be a continuosu function on an open set $u$. Then $f$ is integrable over $u$ iff $|f|$ is integrable over $u$.
Theorem: let $u \subseteq \mathbb{R}^n$ be open and bounded, and let $f: u \rightarrow \mathbb{R}$ be bounded and continuous. Then the extended integral of $u$ exists. If the regular integral of $f$ also exists, the two are equal.
Question: The above are the tools I have for this one. Yet I am not sure how to use them. We know one already exists. Should I say that the other one is bounded and lead all the way to integrability and hence exists?
AI: Hint:
Since $f$ is continuous, it is integrable over each compact rectifiable set $C_N$ and, since $|f| \leqslant |g|$, we have for all $N \in \mathbb{N}$
$$\tag{*}\int_{C_N} |f| \leqslant \int_{C_N}|g| \leqslant \sup_K\int_{C_K}|g|$$
Apply the first theorem (both directions) -- to conclude that the RHS of (*) is finite and finally that $f$ is integrable over $U$. |
H: Idempotent linear operators are projections
I'm working on this problem:
Suppose $V$ is a Hilbert space and $P : V \to V$ is a linear map such that $P^2= P$ and $\Vert Pf \Vert \leq \Vert f \Vert$ for every $f \in V$. Prove that there exists a closed subspace $U$
of $V$ such that $P = P_U$.
I did do some searching online and I know that $P$ is an idempotent matrix and I know that $U$ is supposed to be $\text{range}P$, which I can show is closed. However, I'm having a hard time showing $P_{\text{range}P} = P$. I know that I can decompose $f = g + h$, where $g \in \text{range}P$ and $h$ is in the orthogonal complement of $\text{range}P$. I'm not sure how to proceed.
Most of the online searches require that $P$ be self-adjoint, but this question doesn't have this condition.
Can I get some help?
AI: Let $U$ be the image of $P$ and let $W$ be the image of $1-P$. It suffices to show that $U$ and $W$ are orthogonal to each other, since then $$v=Pv+(1-P)v$$ must be the orthogonal decomposition of $v$ with respect to $U$ and $W$ and so $P$ is the projection onto $U$.
So, suppose $U$ and $W$ are not orthogonal; say $u\in U$ and $w\in W$ with $\langle u,w\rangle\neq 0$. The idea is to now consider linear combinations of $u$ and $w$: if $v=au+w$ for some scalar $a$, then $$\|v\|^2=|a|^2\|u\|^2+\|w\|^2+2\operatorname{Re}(a\langle u,w\rangle)$$ and $$\|Pv\|^2=|a|^2\|u\|^2$$ (here we use the fact that $P(1-P)=0$ so $Pw=0$ since $w\in W$). To get a contradiction, you now just have to choose $a$ such that $2\operatorname{Re}(a\langle u,w\rangle)<-\|w\|^2$, which is possible since $\langle u,w\rangle\neq 0$. |
H: A multiplication problem: $OUI \times OUI = OOUUI + OOUUI$
Here's a $6$th-grade exam problem:
$$OUI \times OUI = OOUUI + OOUUI$$
$O, U$ and $I$ are digits.
e.g $365 \times 365 = 33665 + 33665$ (not the case)
Thus, because the digits are into account I believe there's no algebraic equation. Also, $6$th graders don't even know algebra anyway.
With some extremely clever insight, someone could end up with the conclusion that $I=2$, pretty fast, and that it has to be $212 > OUI > 102$, giving only $10$ possibilities to test out.
There's a catch though; They don't have calculators. Can you tell me what kind of thinking or if there's an analytical solution?
AI: Someone who has done puzzles like this will get $I=2$ quickly. We need $I^2$ to end with the same digit as $2I$. Next would be to see $O=1$. If $O=2$ it conflicts with $I$ and if $O$ is greater $O^2 \gt 2\cdot O$. We are now down to eight possibilities for $U$. Multiplying by hand to check if the tens digit of $UI^2$ is the same as the tens digit of $2\cdot UI$ is not too much work. It is only a two digit multiply and you don't have to do the hundreds column. |
H: First Variation of $L_2$ with Linear operator
Let $\Omega \subset \mathbb{R}^2$ be open and bounded, $P:C^1(\Omega) \to L_2(\mathbb{R})$ be a linear and bounded operator. I want to calculate the first variation of $\|Pu - f\|^2_{L_2}$, $u \in C^1(\Omega), f \in L_2(\Omega)$.
${d \over dt}\|P(u+tv) - f\|^2_{L_2} = {d \over dt}\int_{\Omega} |P(u+tv) - f|^2 = {d \over dt}\int_{\Omega} (P(u+tv) - f)^2$.
I thought that the most straightforward way is to use chain rule but I don't see this leading to the result which should be $\langle v,2P^T(P(u)-f)\rangle_{\Omega}$.
Any help appreciated.
AI: First, $P$ is linear so $P(u+tv)=P(u)+tP(v)$ and so
$$ \int_{\Omega} \big(P(u+tv)-f\big)^2=\int_{\Omega} t^2\,P(v)^2+2t\,P(v)(P(u)-f)+(P(u)-f)^2 $$
whose derivative with respect to $t$ at $t=0$ is
$$ 2\int_{\Omega} P(v)(P(u)-f)=2\langle P(v),P(u)-f \rangle_{L^2}=\langle v,2P^T(P(u)-f)\rangle_{\Omega}. $$
I assume this is brunt of what you want. |
H: Prove $D \in \mathcal{L}(\mathcal{P}(\mathbf{R}),\mathcal{P}(\mathbf{R})) : \text{deg}(D(p)) = \text{deg}(p) - 1$ is surjective
Suppose $D \in \mathcal{L}(\mathcal{P}(\mathbf{R}),\mathcal{P}(\mathbf{R}))$ is such that $\deg(D(p)) = \deg(p) - 1$ for every nonconstant polynomial $p \in \mathcal{P}(\mathbf{R})$. Prove that $D$ is surjective.
I have attempted an answer, however, I think it is incorrect:
We can redefine this as a linear map between two finite dimentional vector spaces:
$$
D \in \mathcal{L}(\mathcal{P}_m(\mathbf{R}),\mathcal{P}_{m-1}(\mathbf{R}))
$$ for $m > 0$.
Let $(1, x, x^2 \ldots, x^{m-1})$ be a basis for $\mathcal{P}_{m-1}$. We can extend this to a basis of $\mathcal{P}_m$ because $\mathcal{P}_{m-1} \subset \mathcal{P}_m$: $$(1, x, x^2 \ldots, x^{m-1}, x^m)\text{.}$$
Then define $D$:
\begin{align}
D(x^i) &= x^i, i = 0, \ldots, m - 1 \\
D(x^m) &= 0
\end{align}
Clearly then, $\text{range}(D) = \mathcal{P}_{m-1}$, as $(1, x, x^2 \ldots, x^{m-1})$ is a basis for $\text{range}(D)$. Hence $D$ is surjective.
The reason I think this answer is incorrect, is because I have chosen my own definition of $D$, not proved it for an arbitrary $D$. However, for similar questions, I often see the answers choose a specific mapping, and I struggle to know when that is acceptable and when it isn't.
AI: I'm not sure if this is the most efficient solution for this problem, but I'll give it a try.
We can first show that the subspace $\mathbb{R}_{m}[x]$ of polynomials of degree at most $m$ is contained in the image of $D$, $\operatorname{im } D$, for any $m \geq 0$. This will imply that the image of $D$ has polynomials of all degrees, so it should indeed be all of $\mathbb{R}[x]$. For that purpose, the following result will come in handy:
$\textbf{Lemma}$. Let $p_{0}, \ldots, p_{m}$ be $m + 1$ polynomials such that $\deg p_{i} = i$ for $i = 0, \ldots, m$. Then $p_{0}, \ldots, p_{m}$ is a basis for $\mathbb{R}_{m}[x]$.
$\textit{Proof.}$ I can expand on this if you wish.
Let $m \geq 0$ be arbitrary and consider the $m + 1$ nonconstant polynomials $x, \ldots, x^{m+1}$. Now let's take a look at their values under $D$:
$$ D(x), \ldots, D(x^{m+1}) .$$
By the hypothesis of the problem, we know that these polynomials have degrees from $0$ to $m$, so they form a basis for $\mathbb{R}_{m}[x]$. In particular:
$$ \mathbb{R}_{m}[x] = \operatorname{span}(D(x), \ldots, D(x^{m+1})) $$
Notice that $D(x), \ldots, D(x^{m + 1})$ are polynomials in $\operatorname{im }D$, which is a subspace of $\mathbb{R}[x]$. A fundamental property of $\operatorname{span}(D(x), \ldots, D(x^{m+1}))$ is that it is the smallest subspace containing $D(x), \ldots, D(x^{m+1})$. We can therefore deduce that
$$ \mathbb{R}_{m}[x] = \operatorname{span}(D(x), \ldots, D(x^{m+1})) \subseteq \operatorname{im } D .$$
Now consider an arbitrary polynomial $p(x) = a_{0} + a_{1}x + \ldots + a_{m}x^{m}$. Then $p \in \mathbb{R}_{m}[x]$, so $p$ must be in the image of $D$ as well. Since $p$ was an arbitrary, we can conclude that $\mathbb{R}[x] \subseteq \operatorname{im }D$, so $ \mathbb{R}[x] = \operatorname{im }D $ and $D$ is surjective. |
H: Advantages of Each Coordinate System
I am currently learning about the spherical coordinate system in class, but I do not know its advantages or even if it is advantageous in using this coordinate system over another.
I am very comfortable in using the rectangular coordinate system and the cylindrical coordinate system (polar coordinate system but just in 3D), as the rectangular coordinate system is just the cartesian coordinate system with another dimension and the cylindrical coordinate system is just the polar coordinate system with an additional dimension.
However, the concept of spherical coordinates come out of nowhere (that I know of) and I am unable to see its advantages. For example, if wanting to calculate an integral in the first octant, you can just restrict to $x>0$, $y>0$, and $z>0$ for the rectangular coordinate system. And for cylindrical coordinates, you can restrict $z>0$, $0<\theta<\frac{\pi}{2}$, and $r$ its corresponding boundary conditions.
My question is are there ever cases when using spherical coordinates are more intuitive than using cylindrical or rectangular coordinates?
AI: In one word - for sphere.
In more words - in spherical coordinates we obtain parallelepiped from sphere:
$$\begin{array}{}
x = r \sin \phi \cos \theta; \\
y = r \sin \phi \sin \theta; \\
z = r \cos \phi
\end{array}
$$
$$\{(x,y,z): x^2+y^2+z^2 \leqslant R^2\} \to \{(r,\theta,\phi): 0 \leqslant r \leqslant R, \theta \in [0, \pi], \phi \in [0, 2 \pi)\}$$ |
H: Is $\oint_{\left | z \right |=2} \frac{e^{\frac{1}{z}}}{z(z^{2}+1))}dz$ equal to zero?
I based my analysis on the fact that the only residue that's outside the curve is the reside in $\infty$ that's equal to zero, so all the other resides inside the curves must add to zero too.
Am I correct?
AI: The substitution $w=1/z$ gives
$$\int_{|z|=2}\frac{e^{1/z}}{z(z^2+1)}\,dz=\int_{|w|=1/2}\frac{we^w}{1+w^2}\,dw=0$$
by Cauchy's theorem. This is basically the same as your argument that there
are no singularities outside the circle. |
H: Correct way to integrate $\int x(x^2-16)dx$
Evaluate:
$$\int x(x^2-16)dx$$
I have noticed that this integral can be solved using two different methods, but I am not sure which one is the correct one.
Way 1: Using $u$-subtitution
Let $u=x^2-16, du = 2xdx$
Then, we have $$\int x(x^2-16)dx$$ $$= \frac{1}{2}\int udu$$ $$= \frac{1}{2}(\frac{1}{2}u^2)+C$$
$$= \frac{1}{4}(x^2-16)^2+C$$
Way 2:
$$\int x(x^2-16)dx$$ $$= \int(x^3-16x)dx$$ $$= \frac{1}{4}x^4-\frac{16}{2}x^2+C$$ $$= \frac{1}{4}x^4-8x^2+C$$
AI: Both methods are correct. Notice that $$\frac14(x^2-16)^2=\frac14(x^4-32x^2+256)=\frac14x^4-8x^2+64.$$
The $C$ in your first way is simply $64$ less than the $C$ in your second way. |
H: Finding $dy/dx$ from $ y^2 = \sin^4{2x} + \cos^4{2x} $ using implicit differentiation
Problem:
Find $\frac{dy}{dx}$ by implicit differentation for the following:
$$ y^2 = \sin^4{2x} + \cos^4{2x} $$
Answer:
\begin{align*}
2y \frac{dy}{dx} &= 4(2) \sin^3(2x) \cos(2x) - 4(2) \cos^3(2x) \sin(2x) \tag1\\
y \frac{dy}{dx} &= 4 ( \sin^3(2x) \cos(2x) - \cos^3(2x) \sin(2x) ) \tag2\\
y \frac{dy}{dx} &= 4 ( \sin(2x)\cos(2x)) ( \sin^2(2x) \cos(2x) - \cos^2(2x) \sin(2x) ) \tag3\\
\\
y \frac{dy}{dx} &= 2 ( \sin(4x)) ( \sin(2x) \sin(2x) \cos(2x) - \cos(2x) \cos(2x) \sin(2x) ) \tag4\\
y \frac{dy}{dx} &= 2 ( \sin(4x)) ( \sin(2x) \left( \frac{ \sin(4x)}{2 }\right) - \cos(2x) \left( \frac{ \sin(4x)}{2 } \right) ) \tag5\\
y \frac{dy}{dx} &= ( \sin(4x)) ( \sin(2x) \left( \sin(4x)\right) - \cos(2x) \sin(4x) ) \tag6\\
y \frac{dy}{dx} &= ( \sin^2(4x)) ( \sin(2x) - \cos(2x) ) \tag7
\end{align*}
Hence my answer is:
$$ \frac{dy}{dx} = \frac{ ( \sin^2(4x)) ( \sin(2x) - \cos(2x) ) } { y } \tag{$\star$}$$
The book's answer is: $\frac{-\sin{8x}}{y} $
Where did I go wrong?
AI: Your mistake is at the $3^{rd}$ step, where you are not extracting the common factor correctly.
\begin{align*}
2y \frac{dy}{dx} &= 4(2) \sin^3(2x) \cos(2x) - 4(2) \cos^3(2x) \sin(2x) \\
y \frac{dy}{dx} &= 4 ( \sin^3(2x) \cos(2x) - \cos^3(2x) \sin(2x) ) \\
y \frac{dy}{dx} &= 4 ( \sin(2x)\cos(2x)) ( \sin^2(2x) - \cos^2(2x) ) \\
y \frac{dy}{dx} &= -2\sin(4x) \cos(4x) \\
\frac{dy}{dx} &= \frac{-\sin(8x)}{y} \\
\end{align*} |
H: Proving that a certain definition of neighborhoods forms a neighborhood topology
This is Exercise 2 from Section 2.1 on page 21 of Topology and Groupoids, by Brown.
Exercise:
Let $\leq$ be an order relation on the set $X$. Let $x \in X$ and $N \subseteq X$. We say that $N$ is a neighborhood of $x$ if there is an
open interval $I$ of $X$ such that $x \in I \subseteq N$.
Prove that these neighbourhoods of points of $X$ form a neighbourhood
topology on $X$. This topology is called the order topology on $X$.
What is the order topology on $\mathbb{R}$?
More information:
The axioms for a neighborhood topology are given as follows (page 20 same book):
If $N$ is a neighborhood of $x$, then $x \in N$.
If $N$ is a subset of $X$ containing a neighborhood of $x$, then $N$ is a neighborhood of $x$.
The intersection of two neighborhoods of $x$ is again a neighborhood of $x$.
Any neighborhood $N$ of $x$ contains a neighborhood $M$ of $x$ such that $N$ is a neighborhood of each point of $M$.
My attempt:
To prove that neighborhoods as defined here form a neighborhood topology, we need to verify that the neighborhoods in this exercise satisfy the four axioms given earlier.
Assume that $N$ is a neighborhood of $x$. Since there is an open interval $I$ of $X$ such that $x \in I \subseteq N$, we see that $x \in N$, which means the first axiom is satisfied.
Assume that $M$ is a neighborhood of $x$, and $M \subseteq N$. There is an open interval $I$ of $X$ such that $x \in I \subseteq M \subseteq N$. This implies that $x \in I \subseteq N$, which shows that $N$ is a neighborhood of $x$, so the second axiom is satisfied.
Assume that $M$ and $N$ are neighborhoods of $x$. Then there are open intervals $I_N$ and $I_M$ such that $x \in I_M \subseteq M$ and $x \in I_N \subseteq N$. Let $I = I_M \cap I_N$. Clearly $x \in I \subseteq M \cap N$. If we can show that $I$ is an open interval, then it follows that $M \cap N$ is a neighborhood of $x$, so the third axiom is satisfied.
Assume that $N$ is a neighborhood of $x$. There is an open interval $I$ of $X$ such that $x \in I \subseteq N$. Let $M = I$. Clearly $M$ is a neighborhood of $x$, because trivially, $x \in I \subseteq M$. We want to show that $N$ is a neighborhood of each point of $M$. Let $y \in I$, and suppose $I$ is written in the form $I = (a, b)$. Let $\delta = \min \{y - a, b - y \}$. Then $(y - \delta, y + \delta) \subseteq I \subseteq N$, which shows that $N$ is a neighborhood of $y$. Since $y$ was an abritrary selection from $I = M$, we conclude that $N$ is a neighborhood of each point of $M$.
As for the question "What is the order topology on $\mathbb{R}$?", I think this is called the "usual topology" on $\mathbb{R}$.
Questions and comments:
I have searched for "order topology" on StackExchange and Google but found nothing that addresses this exercise. I know almost nothing about topology beyond the neighborhood axioms mentioned above.
I think 1. and 2. are probably okay. I think 3. is fine as long as the intersection of two open intervals is an open interval, but I don't know how to prove this. I haven't even been given a definition of "open interval" in this abstract setting, so I'm not quite sure what I'm working with.
For 4., my concern is that I'm using more properties than $X$ and $\leq$ necessarily have. I understand that in topology in general, I cannot assume properties of the real numbers that I'm used to taking for granted, but I'm not sure if I'm doing that here, and if so, how to fix it.
Thanks for any help.
AI: Since there is no generally accepted standard topology on arbitrary partially ordered sets, I suspect that by order relation he means a linear (or total) order on $X$. In that case an open interval in $X$ is simply a set of the form
$$(a,b)=\{x\in X:a<x<b\}\;,$$
where $a,b\in X$ and $a<b$. These sets form a base for a topology on any linearly ordered set, and the topology is indeed called the order topology; a space endowed with a linear order and topologized with the associated order topology is a linearly ordered topological space, or LOTS for short. The order topology on $\Bbb R$ is indeed the usual Euclidean topology, which is also induced by the metric $d(x,y)=|x-y|$.
Let $(a,b)$ and $(c,d)$ be open intervals in $X$. Then
$$(a,b)\cap(c,d)=\begin{cases}
\varnothing,&\text{if }b\le c\text{ or }d\le a\\
(c,b),&\text{if }a\le c<b\le d\\
(c,d),&\text{if }a\le c<d<b\\
(a,b),&\text{if }c<a<b\le d\\
(a,d),&\text{if }c<a<d<b\;,
\end{cases}$$
and you can verify that these are the only possibilities. Thus, the intersection of two open intervals is an open interval, provided that it is non-empty. (Technically, it’s an open interval anyway, since $(a,b)=\varnothing$ if $a\ge b$.)
Added: I have now seen Brown’s definitions, and it turns out that he is indeed talking about what I would call a linear order. However, his definition of open interval includes the open rays $(\leftarrow,a)$ and $(a,\to)$ for $a\in X$ as well as $\varnothing$ and $X$ itself. That adds a few possibilities, like $(a,b)\cap(c,\to)$, but they are easily analyzed along the same lines. For instance, $(a,b)\cap(c,\to)$ is $\varnothing$ if $b\le c$, $(c,b)$ if $a\le c<b$, and $(a,b)$ if $c<a$. It does not change the generated topology, since $X$ and the open rays are unions of open intervals in the stricter sense.
Your answer to (4) doesn’t quite work, because there is not reason to think that arithmetic operations are meaningful in an arbitrary linearly ordered space $X$. You’re also working too hard: if $y\in M(= I)$, then $y\in I\subseteq N$, so $N$ is a nbhd of $y$.
Otherwise your answers are fine, apart from the acknowledged gap in (3). |
H: What are the unit elements in $\Bbb{Z}[i]$?
What are the unit elements in $\Bbb{Z}[i]$, where $\Bbb{Z}[i]$ is defined to be the set of Gaussian integers ?
My progress: My mentor gave this problem and he said to use determinants and scalefactor.
Then I was able to proceed and got the unit elements in $\Bbb{Z}[i]=1,-1,i,-i$ , which is correct . However, I wonder if there is some other way to proceed .
Any solution or hint is appreciated .
Thanks in advance.
AI: If $a+bi$ is a unit ($a,b$ are integers) then for some integers $c,d$ we have $$(a+bi)(c+di)=1.$$ Then the norms of the LHS and RHS should be the same. To get the norm of the LHS, we multiply by $a-bi$ and $c-di$, we get $(a^2+b^2)(c^2+d^2)$. So $(a^2+b^2)(c^2+d^2)=1$. Since both factors are non-negative integers, both are equal to 1. So $|a+bi|=1$. Conversely if $|a+bi|=1$ then $a^2+b^2=(a+bi)(a-bi)=1$ so $a+bi$ is a unit. For integers $a,b$
$a^2+b^2=1$ is possible iff $a=\pm1, b=0$ or $a=0, b=\pm1$. Thus there are four units: $1,-1, i, -i$. |
H: Uncountable sets - Why is the following proof false?
Let $S$ be any subset of the natural numbers. Then the sum
$$
\sum_{n \in S} \frac{1}{2^n}
$$
converges against a unique value for each subset $S$.
This sum yields a computable number because it is possible to compute it digit by digit. Therefore, these sums map every subset of $\mathbb{N}$ to a unique computable number.
This is a contradiction because the set of all subsets of $\mathbb{N}$ is uncountable but the computable numbers are countable.
Where is the error here?
AI: Given a subset $S \subseteq \mathbb{N}$, there may not be an algorithm to compute whether an arbitrary $n \in \mathbb{N}$ is in $S$. Therefore, you can't actually compute it digit by digit. In fact, your argument shows that there can't always be an algorithm for deciding this problem.
Edit:
Say that a function $f : \mathbb{N} \to \mathbb{N}$ is "computable"' iff there is a Turing machine which takes as input a binary number and always halts with a binary number on the tape. Note that this is equivalent to many other definitions of "computable" - for example, that $f$ is general recursive, computable in $\lambda$-calculus, that $f$ can be coded in Haskell (or most any programming language), etc. In some literature, the term used is "recursive".
Consider some $S \subseteq \mathbb{N}$. $S$ is said to be "decidable" if there is a computable function $f$ such that for all $n$, $f(n) = 0$ if $n \notin S$ and $f(n) = 1$ if $n \in S$. We say that such an $f$ is the "characteristic function" of $S$. In some literature, the term used is "recursive set".
There are some sets $S \subseteq \mathbb{N}$ which are not decidable. This doesn't mean that some particular human can't decide whether some $n$ is in $S$ or not; it means that no "algorithm" (Turing machine) can take as input a number $n$ and output whether or not $n \in S$.
We say that a function $f : \mathbb{N} \to \mathbb{Z}$ is computable if there are computable functions $g, h : \mathbb{N} \to \mathbb{N}$ such that for all $n$, $f(n) = g(n) - h(n)$.
Similarly, we say that a function $f : \mathbb{N} \to \mathbb{Q}$ is computable if there are computable functions $g, h : \mathbb{N} \to \mathbb{Z}$ such that for all $n$, $f(n) = g(n) / h(n)$.
Finally, we say that a real number $x$ is computable if there is some computable $f : \mathbb{N} \to \mathbb{Q}$ such that for every $n$, $|f(n) - x| \leq 1/(n + 1)$. We say $f$ computes $x$ in this case.
Not every real number is computable. In particular, it can be shown that $x_S = \sum\limits_{n \in S} \frac{1}{3^n}$ is computable iff $S$ is decidable. For if $S$ is decidable, let $g$ be its characteristic function and define $f(n) = \sum\limits_{i = 0}^n \frac{g(n)}{3^n}$; then $f$ computes $x$. And if $x_S$ is computable, let $g$ be a function that computes $x$. Then by computing $g(3^{n + 2})$, we get close enough to $x_S$ to determine its base-3 expansion up to the $n$th place after the "decimal" point, so we can compute whether this place has a zero (in which case $n \notin S$) or a 1 (in which case $n \in S$).
Since the set of all Turing machines is countably infinite, so too is the set of decidable subsets of $\mathbb{N}$. But the collection of all subsets of $\mathbb{N}$ (that is, the power set of $\mathbb{N}$) is well known not to be countable. Therefore, there must exist some $S$ which is not decidable. In this case, $x_S$ is not a computable number. There is no algorithm at all that can list its digits one at a time. It's not a matter of humans being too stupid to come up with one; it's simply impossible to do so.
A specific example of such an $S$ can be given as follows: suppose given an enumeration of all Turing machines. Let $S = \{n \in \mathbb{N}: $ the $n$th Turing machine halts on the input of $0\}$. The fact that $S$ cannot be dedicable is a corollary of (and equivalent to) the famous "Halting Problem". |
H: $\pi(n)$ is always more than the sum of the prime indices of the factors of composite $n \geq 12$
Let $n=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \geq 12$ be any composite integer.
Then it seems that this is true: $$\pi(n) > \sum_{i=1}^{k}{\pi(p_i)a_i}\ .$$
You get equality instead iff $n$ is prime.
I also assume that if it is true, it's a known result. Can anyone point me towards a resource discussing it if so? Alternatively, if I've made a mistake and/or this is a trivial result, please point out how.
Also, if true, I think Bertrand's Postulate immediately follows:
For prime $p$, it gives $\pi(2p)>\pi(p)+1$, implying at least one prime between $p$ and $2p$.
By the same token, $\pi(3p)>\pi(p)+2$, and $\pi(p^2)>2 \pi(p)$.
(See my answer below for additional thoughts.)
AI: Not quite complete answer:
Suppose $km=n$ with $6\le k\le m$. A result of Rosser–Schoenfeld says that $\pi(x) < 1.25506x/\log x$ for $x>1$, so
$$
\pi(k)+\pi(m) \le 2\pi(m) < 2.51012m/\log m < 5.02024m/\log n,
$$
since $m\ge\sqrt n$. On the other hand, another result of Rosser–Schoenfeld says that $\pi(x) > x/\log x$ for $x>17$. The fact that $n\ge17$ and $k\ge6$ now forces
$$
\pi(n) > n/\log n = km>\log n > 5.02024m/\log n > \pi(k)+\pi(m).
$$
On the other hand, a result of Ramanujan says that $\pi(2x) \ge \pi(x) + 2$ and $\pi(3x) \ge \pi(x) + 3$ when $x\ge6$. Therefore when $m\ge6$,
\begin{align*}
\pi(2m) &> \pi(m) + 1 = \pi(m) + \pi(2) \\
\pi(4m) \ge \pi(3m) &> \pi(m) + 2 = \pi(2m) + \pi(3) \\
\pi(5m) &\ge \pi(4m) \ge \pi(2m)+2 > (\pi(m)+1)+2 = \pi(m) + \pi(5).
\end{align*}
In other words, we have shown that $\pi(k) + \pi(m) < \pi(km)$ for $k\ge2$ and $m\ge6$.
This should be very close to proving the entire statement by induction on the number of prime factors (counting multiplicity). |
H: How to prove these four propostions about Borel-Cantelli Lemma in Shriyaev's book 'Probability'?
The original problems are shown in picture linked below. Thank you.
Problem $\mathbf{2.10.19}$. (On the second Borel-Cantelli lemma.) Prove the following variants of the second Borel-Cantelli lemma: given an arbitrary sequence of (not necessarily independent) events $A_1, A_2, \ldots$, one can claim that:
(a) If
$$
\sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad
\liminf_n \frac{\sum_{i,k=1}^n {\sf P}(A_iA_k)}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} = 1,
$$
then (Erdös and Rényi [$37$]) ${\sf P}(A_n\text{ i.o.}) = 1$.
(b) If
$$
\sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad
\liminf_n \frac{\sum_{i,k=1}^n {\sf P}(A_iA_k)}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} = L,
$$
then (Kochen and Stone[$64$], Spitser [$125$]) $L \geq 1$ and ${\sf P}(A_n\text{ i.o.}) = 1/L$.
(c) If
$$
\sum_{n=1}^\infty {\sf P}(A_n) = \infty \quad \text{and} \quad
\liminf_n \frac{\sum_{1\leq i<k\leq n} [{\sf P}(A_iA_k)-{\sf P}(A_i){\sf P}(A_k)]}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2} \leq 0,
$$
then (Ortega and Wschebor [$92$]) ${\sf P}(A_n\text{ i.o.}) = 1$.
(d) If $\sum_{n=1}^\infty {\sf P}(A_n) = \infty$ and
$$
\alpha_H = \liminf_n \frac{\sum_{1\leq i<k\leq n} [{\sf P}(A_iA_k)-H{\sf P}(A_i){\sf P}(A_k)]}{\left[\sum_{k=1}^n {\sf P}(A_k)\right]^2},
$$
where $H$ is an arbitrary constant, then (Petrov [$95$]) ${\sf P}(A_n\text{ i.o.}) \geq \frac{1}{H+2\alpha_H}$ and $H+2\alpha_H \geq 1$.
Original at https://i.stack.imgur.com/oxcKz.jpg
AI: This is the Kochen-Stone Lemma. I will state this result and a short proof for you. But first a little technical result.
Lemma: If $0\neq f\in L_2$ and $\mathbb{E}[f]\geq0$, then for any $0<\lambda<1$
$$\begin{align}
\mathbb{P}\big[f>\lambda \mathbb{E}[f]\big]\geq (1-\lambda)^2 \frac{\big(\mathbb{E}[f]\big)^2}{\mathbb{E}[|f|^2]}\tag{1}\label{anty-cheby}.
\end{align}
$$
Here is a short proof:
By Hölder's inequality
$$
\mathbb{E}[f]=\int_{\{f\leq \lambda\mathbb{E}[f]\}}f \,d\Pr+ \int_{\{ f>\lambda\mathbb{E}[f]\}} f\,d\mathbb{P} \leq \lambda\mathbb{E}[f] + \Big(\|f\|_2\sqrt{\Pr[f>\lambda\mathbb{E}[f]]}\Big).
$$
Here is the result that we will used to get the version of Corel Cantelly closer to what you described in your problem.
Lemma(Kochen-Stone) Let $\{A_n\}\subset\mathscr{F}$. If $\sum_n\mathbb{P}[A_n]=\infty$, then
$$\begin{align}
\mathbb{P}\big[\bigcap_{n\geq1}\bigcup_{k\geq n}A_k\big]\geq\limsup_n\frac{\Big(\sum^n_{k=1}\mathbb{P}[A_k]\Big)^2}{\sum^n_{k=1}\sum^n_{m=1}\mathbb{P}[A_k\cap A_m]}\tag{2}\label{ko-sto}
\end{align}
$$
Here is a Sketch of the proof:
Without loss of generality, we assume that $\mathbb{P}[A_n]>0$ for all $n$. Let $f_n=\sum^n_{k=1}\mathbb{1}_{A_k}$, $f=\sum_{n\geq1}\mathbb{1}_{A_n}$, and for any $0<\lambda<1$, define $B_{n,\lambda}=\big\{f_n>\lambda\mathbb{P}[f_n]\big\}$.
Observe that
$$
A=\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k=\{f=\infty\}\supset\bigcap_{n\geq 1}\bigcup_{k\geq n}B_{k,\lambda}=B_\lambda;
$$
then, by $\eqref{anty-cheby}$, we obtain
$$
\mathbb{P}[A]\geq\mathbb{P}[B_\lambda]\geq\limsup_{n\rightarrow\infty}\mathbb{P}[B_{n,\lambda}]\geq(1-\lambda)^2\limsup_n\frac{\big(\mathbb{E}[f_n]\big)^2}{\mathbb{E}[f^2_n]}.
$$
Letting $\lambda\rightarrow1$ gives $\eqref{ko-sto}$.
Using Kochen-Stone's Lemma one can prove the following version of the reverse Borel-Cantelli Lemma
Theorem (reverse Borel-Cantelli) Suppose $\{A_n\}\subset\mathscr{F}$ is such that for any $i\neq j$, $\mathbb{P}[A_i\cap A_j]\leq\mathbb{P}[A_i]\mathbb{P}[A_j]$. If $\sum_n\mathbb{P}[A_n]=\infty$, then $\mathbb{P}\Big[\bigcap_{n\geq1}\bigcup_{k\geq n}A_k\Big]=1$.
Here is a short proof:
Denote by $A=\bigcap_{n\geq 1}\bigcup_{k\geq n}A_k$. Let $a_n=\sum^n_{k=1}\mathbb{P}[A_k]$,, $b_n=\sum_{i\neq j}\mathbb{P}[A_i]\mathbb{P}[ A_j]$, and $c_n=\sum^n_{k=1}\mathbb{P}^2[A_k]$. By Kochen--Stone's lemma we have
$$
\mathbb{P}[A]\geq\limsup_n\frac{c_n+b_n}{a_n+b_n}
$$
From $a^2_n=c_n+b_n\leq a_n+b_n$, and $a_n\nearrow\infty$, it follows that $b_n\nearrow\infty$ and $\lim_n\tfrac{c_n}{b_n}=0=\lim_n\frac{a_n}{b_n}$. Therefore, $\mathbb{P}[A]=1$.
Reference:
https://projecteuclid.org/euclid.ijm/1256059668 |
H: What is the correct choice of the contour in the case of undamped forced harmonic oscillator?
I am interested in finding the Green's function (GF) for the undamped forced harmonic oscillator equation: $$\Big(\frac{d^2}{dx^2}+\omega_0^2\Big)x(t)=f(t).$$ In order to find the GF, start by define it: $$\Big(\frac{d^2}{dx^2}+\omega_0^2\Big)G(t-t')=\delta(t-t').$$ First denoted $\tau\equiv t-t'$ and consider the Fourier transform $$G(\tau)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\tilde{G}(\omega)e^{i\omega \tau}d\omega,~ \delta(\tau)=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}e^{i\omega\tau}d\omega$$ which redialy gives $$\tilde{G}(\omega)=\frac{\sqrt{2\pi}}{\omega^2-\omega_0^2}\Rightarrow G(\tau)=\int\limits_{-\infty}^{\infty}\frac{e^{i\omega\tau}}{(\omega+\omega_0)(\omega-\omega_0)}d\omega$$ i.e., the integral has simple poles at $\omega=\pm\omega_0$
For $\tau>0$, there are three ways in which a closed semicircular contour can be chosen to enclose the poles.
$1.$ Both the poles $-\omega_0$ and $+\omega_0$ can be included in the contour, by choosing two small semicircular indentations in the lower half-plane around $-\omega_0$ and $+\omega_0$. In this case, the result is $G(\tau)=\frac{1}{\omega_0}\sin(\omega_0\tau)$.
$2.$ Both the poles $-\omega_0$ and $+\omega_0$ can be excluded from the contour, by choosing two small semicircular indentations in the upper half-plane around $-\omega_0$ and $+\omega_0$. In this case, the result is $G(\tau)=0$.
$3.$ The pole $-\omega_0$ is included from the contour while $+\omega_0$ is excluded. In this case, the result is $G(\tau)=\frac{i\pi}{\omega_0}e^{-i\omega_0\tau}$.
$4.$ The pole $+\omega_0$ is included from the contour while $-\omega_0$ is excluded. In this case, the result is $G(\tau)=\frac{i\pi}{\omega_0}e^{+i\omega_0\tau}$.
Which one is the correct choice of contour to find $G(\tau)$ and why?
AI: The issue is that you assume that the small semicircular integral is $0$. If you integrate over an arc segment with radius $r$ and angle $\alpha$ around a simple pole $z_0$, $$\lim_{r\to 0}\int_{C(r,\alpha)}f(z) dz=\alpha i \mathrm{Res}(f,z_0)$$
See for example this answer.
Then if you go around the pole on a semicircle counterclockwise your integral is $\pi i \mathrm{Res}(f,z_0)$, and if you go clockwise is $-\pi i \mathrm{Res}(f,z_0)$. So it does not matter which trajectory you choose, as long you do it correctly.
EDIT
Since there was a question in the comment, I've decided to add a few explanations to this answer. For simplicity of notation, we want to integrate $f(x)$ from $-\infty$ to $\infty$, with two simple poles at $\pm\omega_0$. So we create a contour in the complex plane, made up on a large semicircle $\Gamma$, with radius $R\to\infty$, where we know that the integral of $f(z)$ vanishes. Then on the real line we avoid the poles by making small semicircles, of radius $\epsilon\to 0$ around $\pm\omega_0$. We call these $\gamma_{+,-}^{u,d}$. The $+$ or $-$ sign identify the pole, and $u$ means we avid the pole going above the line, $d$ we go below.
$$\lim_{R\to\infty, \epsilon\to 0}\left(\int_{-R}^{-\omega_0-\epsilon}f(x)dx+\int_{-\omega_0+\epsilon}^{\omega_0-\epsilon}f(x)dx+\int_{\omega_0+\epsilon}^Rf(x)dx\\+\int_\Gamma f(z)dz+\int_{\gamma_+^{u,d}} f(z)dz+\int_{\gamma_-^{u,d}} f(z)dz\right)=2\pi i\sum_{z_i}\mathrm{Res}(f,z_i)$$
Here you have the choice of how you avoid the poles (up or down), but that will change the sum on the right. The first three integrals converge to $\int_{-\infty}^\infty f(x)dx$, the fourth is zero. So
$$\int_{-\infty}^\infty f(x)dx=2\pi i\sum_{z_i}\mathrm{Res}(f,z_i)-\int_{\gamma_+^{u,d}} f(z)dz-\int_{\gamma_-^{u,d}} f(z)dz$$
If you make point $\omega_0$ to be inside the contour, you add it to the sum. But you need to subtract the integral over $\gamma_+^d$, which is $\pi i \mathrm{Res}(f,\omega_0)$. So the net contribution of that pole is $\pi i \mathrm{Res}(f,\omega_0)$. If you avoid the pole going on the upward trajectory, you will not add it to the sum, but the integral on $\gamma_+^d$ is $-\pi i \mathrm{Res}(f,\omega_0)$, so the net contribution is once again $\pi i \mathrm{Res}(f,\omega_0)$. You can do the same for the $-\omega_0$ pole. So in this case
$$\int_{-\infty}^\infty f(x)dx=\pi i\left(\mathrm{Res}(f,-\omega_0)+\mathrm{Res}(f,\omega_0)\right)$$
This is independent of the choice of your contour. |
H: confusion about uniform probability distribution expectation value
I read in my statistics book about Discrete Uniform Probability Distribution and the Expected Value
A discrete random variable $X$ with $k$ possible outcomes $x_1, x_2, ...,x_k$ is said to follow a discrete uniform distribution if the probability mass function
(PMF) of $X$ is given by
$$P(X = x_i) = \frac{1}{k}, \forall i = 1,2,...,k $$
If the outcomes are the natural numbers $x_i = i (i = 1, 2, . . . , k)$, the mean and variance of $X$ are obtained as
$$E(X) = \frac{k+1}{2}, $$
$$Var(X) = \frac{1}{12}(k^2 - 1) $$
I have a question on $E(X)$. It seems I find it hard to understand the formula for $E(X)$ for coin toss example. in a fair coin toss we have $k = 2$ (heads or tails, here $0$ and $1$ respectivley) and $ E(X)= \frac{2+1}{2} = 1.5$
However, when you calculate the $E(X)$ as
$ E(X) = $ probability weighted sum of $k$ it is
$$ E(X) = 0\times0.5 + 1\times0.5 = 0.5 $$
So why is this difference? What am I missing?
AI: The way you define the random variable $X$ is not consistent between the two approaches, which is why you have different results. In the discrete uniform case, the support is $X \in \{1, 2\}$, with probability mass function $$\Pr[X = 1] = 1/2, \\ \Pr[X = 2] = 1/2.$$ Hence the expectation is $$\operatorname{E}[X] = 1 \cdot \Pr[X = 1] + 2 \Pr[X = 2] = \frac{1}{2} + \frac{2}{2} = \frac{3}{2}.$$ When you try to represent the outcomes of a fair coin toss with $X$, you instead wrote $X \in \{0, 1\}$, hence $$\Pr[X = 0] = 1/2, \\ \Pr[X = 1] = 1/2.$$ Obviously, you will not get the same result. |
H: The intersection assumption in the definition of direct sums
Let $W_1,W_2,\ldots,W_k$ be subspaces of a vector space $V$. We define the sum of these spaces to be the set
$$\{v_1+v_2+\cdots+v_k:v_i\in W_i\text{ for }1\leq i\leq k\},$$
which we denote by $W_1+W_2+\cdots+W_k$ or simply by $\sum_{i=1}^k W_i.$ If, in addition, $W_j\cap\sum_{i\not=j}W_i=\{\mathbf 0\}$ for each $j$, then we call $V$ the direct sum of $W_1,W_2,\ldots,W_k$ and write $V=W_1\oplus W_2\oplus\cdots\oplus W_k$. What I'm concerned about is whether we can prove that the intersection of all subspaces in the direct sum is the zero space, that is,
$$\bigcap_{i=1}^k W_i=\{\mathbf 0\}.$$
I think this is weaker than the statement that each subspace intersects the sum of the other subspaces only in the zero vector but don't know how to justify it. Please give me some help. Thank you.
AI: $\{\mathbf 0
\}\subseteq W_j\cap W_l\subseteq W_j\cap\sum_{i\not=j}W_i=\{\mathbf 0\}$ for all $l\not=j$. Thus also $\bigcap_{i=1}^k W_i=\{\mathbf 0\}$.
Moreover, let $V=\mathbb{R}^2$, $W_1=\mathbb{R}(1,0)$, $W_2=\mathbb{R}(0,1)$, $W_3=\mathbb{R}(1,1)$. The $W_1\cap W_2\cap W_3=\{(0,0)\}$, but $W_3\cap (W_1+W_2)=W_3\not=\{(0,0)\}$. |
H: Why does the negative of the direction of steepest ascent result in the direction of steepest descent?
The gradient is a vector composed of partial derivatives and is a vector that gives the direction of the steepest ascent.
In a YouTube video that I was watching, it talked about how the gradient descent of the function $f(x,y)$ is $-\nabla f(x,y)$. It does not seem that very intuitive to me that the negative the gradient, or the direction of the steepest ascent, results in the direction of the steepest descent.
I can think of counter-examples to this statement, but I am unsure if I am understanding this part of the video correctly.
Video link
AI: This is because $\nabla(-f)=-\nabla f$ is the direction of steepest ascent of the function $-f$, which is therefore the direction of steepest descent of $f$. |
H: Represent $f(x) - f(y)$ as an integral
Description
I've come across the following transition in a textbook of Convex Optimisation. I couldn't figure out what's going on so that I'd appreciate if anyone hits me with any hint!
Problem
Suppose $x, y \in \mathbb{R}^n$ and $f$ be a $\beta$-smooth convex function on $\mathbb{R}^n$;
Then the transition of interest goes as
$$f(x) - f(y) = \int^1_0 \nabla f(y + t(x - y))^{T} (x - y) dt $$
Additional Comment
This transition appears in the proof of the convergence rate of Gradient Descent for a smooth convex objective function $f$.
AI: This is just the fundamental theorem of line integrals, which says for (sufficiently smooth) functions $f$,
$$f(x) - f(y) = \int_{C} \overrightarrow{\nabla f(r)} \cdot \vec{dr}$$
where $x$ and $y$ are the endpoints of $C$. Think of it as a generalization of the fundamental theorem calculus to higher dimensions, which roughly says that the difference in a function evaluated at two points $x$ and $y$ can be calculated as the integral of its derivative over the range $[x, y]$. Here, the derivative is the gradient (since we have multiple variables now).
For your problem, $\beta$-smoothness is more than sufficient to use the theorem above, so just let $r(t) = y + t(x - y)$ be a parameterization of a straight line $L$ from $x$ to $y$ such that $r(0) = x$ and $r(1) = y$. Then from the fundamental theorem for line integrals it follows that
\begin{align*}
f(x) - f(y) &= \int_{L} \overrightarrow{\nabla f}(r) \cdot \overrightarrow{dr} = \int_0^1 \overrightarrow{\nabla f (r(t))} \cdot \overrightarrow{r'(t)} dt \\ &= \int_{0}^1 \overrightarrow{\nabla f (y + t(x - y))} \cdot \overrightarrow{(x - y)} dt
\end{align*} |
H: Simplify the radical $\sqrt{x-\sqrt{x+\sqrt{x-...}}}$
I need help simplifying the radical $$y=\sqrt{x-\sqrt{x+\sqrt{x-...}}}$$
The above expression can be rewritten as $$y=\sqrt{x-\sqrt{x+y}}$$
Squaring on both sides, I get $$y^2=x-\sqrt{x+y}$$
Rearranging terms and squaring again yields $$x^2+y^4-2xy^2=x+y$$
At this point, deriving an expression for $y$, completely independent of $x$ does not seem possible. This is the only approach to solving radicals which I'm aware of. Any hints to simplify this expression further/simplify it with a different approach will be appreciated.
EDIT: Solving the above quartic expression for $y$ on Wolfram Alpha, I got 4 possible solutions
AI: Consider the final relation you have obtained as a quadratic equation in $x$,i.e:
$$x^2-(2y^2+1)x+y^4-y=0$$
Solving the above gives you
$$x=y^2+y+1 \text{ or } x=y^2-y$$
Individually solve these the quadratics in $y$ to obtain the four solutions you got from Wolfram Alpha. |
H: When to use ds vs dy and dx for line integrals
Are $\int_{C}^{} f(x,y) ds $ and $\int_{C}^{} F(x,y) = \int_{C}^{} P(x,y) dx + \int_{C}^{} Q(x,y) dy$ equivalent to each other for some curve C? If not, when are they not interchangeable? And if so, when is there an advantage to using one over another?
Edit: it seems that the difference is when the function being integrated has $x$ and $y$ being multiplied or divided, i.e, inseparable into two separate functions. Is this correct?
AI: Strictly speaking there is no real diffetence between ds, dx, and dy. Just that we integrate over a different variable.
However, s is commonly used as a path length parameter. And x and y are commonly used to integrate parallel to the x-axis respectively the y-axis.
In your case the first integral assumes that we can write x and y as functions of s. That is, we have x(s) and y(s) that parametrize the curve C.
The sevond integral integrates along the first parameter of P, which is x, which represents the coordinate along the x axis. It means that we consider y a function of x, that is y(x) so that it follows the curve C.
As an example consider $(x(s), y(s))=(\cos s,\sin s)$ and $(x,y(x))=(x,\sqrt{x^2+y^2})$ respectively $(x(y), y)=(\sqrt{x^2+y^2},y)$. These are 3 different parametrizations of the first quarter of the unit circle. |
H: Evaluate the given limit $\lim_{n\to \infty} \frac{e^n}{(1+\frac 1n)^{n^2}}$
The procedure I am about to write is wrong, and I know why it is wrong yet it was the only one I could think of, so I will put it up anyway
$$\lim_{n\to \infty} \frac{e^n}{\left(1+\frac 1n\right)^{n^2}}$$
For the denominator
$$\begin{aligned}\lim_{n\to \infty} \left(1+\frac 1n\right)^{n^2} &=e^{\lim_{n\to \infty} (1+\frac 1n-1)n^2} \\
&=e^{\lim_{n\to \infty} n}\end{aligned}$$
And the same for the numerator, so their division should give $1$
As I said, I know this is wrong. The correct answer is $\sqrt e$. What is the right process?
AI: Note that
$$\frac{e^n}{(1+\frac 1n)^{n^2}}=\left(\frac{e}{(1+\frac 1n)^{n}}\right)^n$$
is an indeterminate form $1^\infty$ therefore we can't conclude that it is equal to one.
By $x=\frac1n \to 0^+$ we obtain
$$\large{\frac{e^n}{(1+\frac 1n)^{n^2}}=e^{\log\left(\frac{e^n}{(1+\frac 1n)^{n^2}}\right)}=e^{\frac{\frac1n-\log \left(1+\frac 1n\right) }{\frac1{n^2}}}}=e^{\frac{x-\log (1+x)}{x^2}}$$
then we can proceed by l'Hopital or Taylor's series to find
$$\frac{x-\log (1+x)}{x^2} \to \frac12$$
Assuming the limit exists, we can also use the result shown here:
Are all limits solvable without L'Hôpital Rule or Series Expansion |
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