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H: Schwarz Inequality for Riemann-integrable
I want to show the Schwarz inequality,
$$
\left(\int_Qfg\right)^2\le\int_Qf^2\int_Qg^2,
$$
for Riemann-integrable functions $f:Q\subset\mathbb{R}^n\to\mathbb{R}$, where $Q$ is a rectangle.
But in the case where $\int_Qf^2 = 0$ I'm not seeing how to get $\int_Qfg=0$ from it. I have seen this question but I don't know what norm they're using and really didn't understand the inequality proposed in the first answer.
AI: Some notation: for any function $\phi:Q \to \Bbb{R}$, let's define the set $Z_{\phi} := \{x \in Q| \, \phi(x) \neq 0\}$. Here is a definition, and a few theorems:
Definition/Theorem
A set $Z \subset \Bbb{R}^n$ is said to have ($n$-dimensional Lebesgue) measure zero if for every $\epsilon > 0$, there is a countable collection $\{R_k\}_{k=1}^{\infty}$ of rectangles such that
\begin{align}
Z \subset \bigcup_{k=1}^{\infty}R_k \qquad \text{and} \qquad \sum_{k=1}^{\infty}\text{vol}(R_k) < \epsilon
\end{align}
Here, we define the volume of a rectangle in the obvious manner. One can prove it doesn't matter whether the rectangles in this definition are open or closed. It is also easy to prove that if $Z$ has measure zero, then for every subset $X\subset Z$, $X$ has measure zero.
Now, we have two theorems
Theorem $1$.
Let $Q \subset \Bbb{R}^n$ be a closed rectangle, and $\phi: Q \to \Bbb{R}$ be a Riemann-integrable function. If $Z_{\phi}$ has measure zero, then $\int_Q \phi = 0$.
Theorem $2$.
Let $Q \subset \Bbb{R}^n$ be a closed rectangle, and $\phi: Q \to \Bbb{R}$ be a Riemann-integrable, non-negative function. If $\int_Q \phi = 0$ then $Z_{\phi}$ has measure zero.
If I recall correctly, in Analysis on Manifolds, Munkres gives a pretty nice and quick proof of these facts using the equivalent characterization of Riemann-integrability (a bounded function on a rectangle is Riemann-integrable if and only if the set of discontinuities has measure zero)
Now, for your actual question. Suppose $\int_Q f^2 = 0$. Since $f^2$ is a non-negative, Riemann integrable function whose integral vanishes, theorem $2$ implies that set $Z_{f^2}$ has measure zero. Next, it is easily verified that $ Z_{fg} \subset Z_f = Z_{f^2}$; hence $Z_{fg}$ has measure zero. By theorem $1$, it follows that
\begin{align}
\int_Q fg &= 0.
\end{align}
Hence, we have an equality in the Cauchy-Schwarz inequality.
Edit:
Having taken a look at the answer in the link, I realized the issue is much simpler. It all boils down to the following simple lemma:
Let $a,b,c \in \Bbb{R}$, and consider the polynomial $p(x) = ax^2 + bx + c$. If for every $x \in \Bbb{R}$, we have $p(x) \geq 0$ (or for all $x \in \Bbb{R}$, $p(x) \leq 0$), then
\begin{align}
b^2 - 4ac \leq 0.
\end{align}
Note that if $a \neq 0$, this follows by a simple application of completeing the square: write $p(x) = a\left( x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}$, and from here, manipulate inequalities (there's like a few cases, but they're all easy to prove).
If $a = 0$, then we have $p(x) = bx + c$; but if $p(x)$ maintains a constant sign, then we must have $b=0$ (simply sketch the graph of $p(x)$ with $b \neq 0$ to convince yourself). Since $b=0$, and $a=0$, we of course have $b^2 - 4ac = 0 \leq 0$.
Now, we apply this simple lemma to our present situation. Consider the following polynomial in $\lambda$:
\begin{align}
p(\lambda) := \int_Q (\lambda f- g)^2 = \left(\int_Q f^2\right) \lambda^2 + \left(-2 \int_Q fg\right) \lambda + \int_Q g^2
\end{align}
Since $p(\lambda)$ was obtained by integrating a non-negative function, we clearly have that for every $\lambda \in \Bbb{R}$, we have $p(\lambda) \geq 0$.
Now, identify what $a,b,c$ are, and then you immediately find that $b^2 - 4 ac \leq 0$ implies the Cauchy-Schwarz inequality.
Finally, of course, if you wish to find out what the equality cases are, simply trace through the derivation above, and see when we can replace the $\leq$ with $=$. This I leave to you $\ddot{\smile}$.
|
H: Congruence mod p involving a product
Let $p$ be a prime, $p\equiv 3$ mod $4$. Numerically it appears that
$$
\prod_{n=1}^{p-1}\left(1+n^2\right)\equiv 4\mod p.
$$
How can one prove this? For $p\equiv 1$ mod $4$, the product is $0$ mod $p$ because $-1$ is a quadratic residue.
AI: Let's work over the finite field $\Bbb F_p$ with $p\equiv3\pmod 4$. One can adjoin
$i$ to this with $i^2=-1$ to get the finite field $k=\Bbb F_{p^2}$. Your product
is
$$\prod_{n=1}^{p-1}(n^2+1)=\prod_{n=1}^{p-1}(n-i)(n+i)=f(i)f(-i)$$
where
$$f(X)=\prod_{n=1}^{p-1}(X+n)=X^{p-1}-1$$
(we are working in characteristic $p$).
Then
$$f(i)=-1-1=-2.$$
Likewise, $f(-i)=-2$ and so the original product is $4$.
|
H: How are accumulation points related to topologies?
Is it possible to have two topologies, one strictly finer than the other, yet all of the accumulation points are the same?
AI: Since the closure of a set is the union of the set with the set of its accumulation points, if all accumulation points of all sets are the same, the closures of all sets are the same, and therefore the set of closed sets is the same. Because the open sets are the complements of the closed sets, then also the set of open sets is the same. But the set of open sets is the topology, therefore the topologies are the same.
|
H: Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$?
Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$.
Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?
Motivation: If we want to prove $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)\ge 0$, and $g(p, q, r, Q)$ is non-increasing with $Q$,
by using the known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a+b+c)^3 - abc$(see How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$),
sometimes, we may prove $g(a+b+c, ab+bc+ca, abc, \frac{4}{27}(a+b+c)^3 - abc)\ge 0$.
As a result, we deal with a symmetric inequality rather than the original cyclic inequality.
Is it known? Is it easily to prove?
Let us see some examples. Denote $p = a+b+c, q = ab+bc+ca, r = abc$ and $Q = a^2b+b^2c+c^2a$.
1) $ab^2 + bc^2 + ca^2 = (a+b+c)(a^2+b^2+c^2) - a^3-b^3-c^3 - (a^2b+b^2c+c^2a)$
2) From the known identity $a^3b+b^3c+c^3a +(ab+bc+ca)^2 = (a+b+c)(a^2b+b^2c+c^2a+abc)$, we have
$a^3b+b^3c+c^3a = p(Q + r) - q^2$.
3) $a^3b^2+b^3c^2+c^3a^2 = Qq - r(p^2-2q) - rq$.
4) $(a^2+b)(b^2+c)(c^2+a) = \cdots$
Any comments and solutions are welcome.
AI: Let $R$ be the ring generated by the symmetric polynomials and $f=a^2b+b^2c+c^2a$.
Certainly $f$ also contains $g=ab^2+bc^2+ca^2$ since $f+g$ is symmetric.
Let
$$u_{m,n}=a^mb^n+b^mc^n+c^ma^n.$$
To prove that $R$ contains all cyclic polynomials,
it suffices to prove that $u_{m,n}\in R$ for all $m$ and $n$. Indeed it
suffices to prove this when $m>n\ge1$.
Then,
$$(a+b+c)u_{m,n}=u_{m+1,n}+u_{m,n+1}+abc u_{m-1,n-1}\tag1$$
(when $n\ge1$)
and
$$(ab+bc+ca)u_{m,n}=u_{m+1,n+1}+abcu_{m-1,n}+abc u_{m,n-1}\tag2$$
(when $n\ge1$).
If we proceed by induction on $m+n$, (2) shows that if all $u_{m,n}\in R$
for $r\le k$ then so do all $u_{m,n}$ with $m+n=k+1$ save perhaps $u_{k,1}$.
But applying (1) for $(m,n)=(k-1,1)$ gives
$$u_{k,1}=(a+b+c)u_{k-1,1}-u_{k-1,2}-abcu_{k-2,0}.$$
As we have $u_{k-2,2}\in R$, then $u_{k,1}\in R$ too.
|
H: Prove that $|AUC| = |A|$, where $A$ is an uncountable set and $C$ is a countable set.
Let $A$ be an uncountable set and let $C$ be a countable set with $A \cap C = \{\}$. Show that $|A\cup C| = |A|$. I'm quite lost as to how to approach solving this problem. I know that CSB theorem is involved to conclude there is a bijection, but still confused. Any help is appreciated!
AI: We construct a bijection between $A\cup C$ and $A$.
Pick and remove an element from $A$ repeatedly, to obtain a sequence of distinct elements of $A$. (Note this requires the axiom of choice; I don’t think your statement can be proven without this axiom.) Call this sequence $B$, and everything else $D$. So $A=B\cup D$, where $B$ is countably infinite.
We can then construct a bijection from two countable sets to one (this is straightforward, and I’ll leave it as an exercise).
So suppose $f$ is a bijection from $B\cup C$ to $B$.
Then our bijection $g$ from $A\cup C$ to $A$ can be defined as $g(x) = x$ if $x \in D$, and $g(x)=f(x)$ if $x \in B\cup C$. This clearly works, and we are done.
|
H: What would a graph where $A\propto Y^2$ look like?
Consider a relation between $X$ and $Y$ such that the area under the curve of this was proportional to the square of $Y$, what would such a graph look like.
I was inspired of this question when I was going through my Physics textbook which mentioned such a thing when discussing Current, Voltage, Resistance and it's relationships.
I couldn't find anything online, so if you could point me to some resources and similar things it'd be very nice.
AI: $\int\limits_{x_0}^x y\,\mathrm{d}y=ky^2\Rightarrow y=2kyy'\Rightarrow y=0\hbox{ or }y'=\frac{1}{2k}\Rightarrow y=0\hbox{ or } y=\frac{x}{2k}+C$ in other words, it's an arbitrary non-vertical line.
|
H: Is there a closed formula for the sum of a geometric progression with binomial coefficients?
The title asks it all.
$$\sum_{i=0}^n{n\choose i}x^{i+1}=?$$
AI: The binomial theorem gives
$$(1+x)^n = \sum_{i=0}^n \binom n i x^i.$$
So your sum in question is simply $x(1+x)^n$.
|
H: Find Lebesgue Integral: $\lim_{n\rightarrow\infty}\int_0^2f_n(x)dx$
Consider a sequence of functions $f_n:[0,2]\rightarrow\mathbb{R}$ such that $f(0)=0$ and $f(x)=\frac{\sin(x^n)}{x^n}$ for all other $x$. Find $\lim_{n\rightarrow\infty}\int_0^2f_n(x)dx$.
My attempt:
$\sup_{x\in(0,2]}|f_n(x)-f(x)|=\sup_{x\in(0,2]}|\frac{\sin(x^n)}{x^n}-0|$, thus, we have uniform convergence. So, since the interval we are integrating over is finite and $f_n\rightarrow f$ uniformly, then by the Dominated Convergence Theorem, we may pass the limit into the integral. So, we then have $\lim_{n\rightarrow\infty}\int_0^2f_n(x)dx=\int_0^2\lim_{n\rightarrow\infty}f_n(x)dx=\int_0^20dx=0$.
I was wondering if I did this correctly, or if I made a mistake somewhere. In particular, are there any missing small details? I would really appreciate any insight! Thank you.
AI: Your argument is not correct and even the value of the limit is wrong. It is not true that $f_n(x) \to 0$ uniformly on $(0,2]$. In fact $f_n(x) \to 1$ for $0<x<1$ since $x^{n} \to 0$ and $\frac {\sin x} x \to 1$ as $ x \to 0$. For $x>1$ the limit is $0$ and you can apply DCT to show that $\int_1^{2} f_n(x)dx \to 0$. Now $\int_0^{1} f_n(x) dx \to 1$ by DCT since $|\sin t| \leq t$ for all $t \geq 0$. So the right answer is $1$.
|
H: An equivalent form of the definition of conditional probability
Problem: Suppose that the $\sigma$ algebra $\mathscr{G}$ is generated by a $\pi$ system $\mathscr{C}$. Please prove that $f\in\mathscr{G}$ is the conditional probability of the event $A$ with respect to $\mathscr{G}$ if and only if
$$\int_B f d\mathbb{P}=\mathbb{P}(A\cap B),\forall B\in\mathscr{C}.$$
I believe the necessity is obvious. For the sufficiency of the proof, notice that $\mathscr{C}$ is a $\pi$ system, the way I think of it is to take
$$\Lambda=\{B\in\mathscr{G}:\int_B f d\mathbb{P}=\mathbb{P}(A\cap B)\}.$$
Then $\mathscr{C}\subset\Lambda$ and if $\Lambda$ is proved to be a $\lambda$ system, then by monotone class theorem the problem is solved. Is easy to check if $B,C\in\Lambda$ and $C\subset B$, then $B-C\in\Lambda$. And is also easy to check if $A_n\in\Lambda,n=1,2,\cdots$ and $A_n\uparrow B$, then $A\in\Lambda$. But I don't know how to prove that the whole space $\Omega \in \Lambda$.
Please give me some ideas, thank you!
AI: It's not true. Suppose $\Omega = \{\omega_1, \omega_2\}$, $\mathscr C = \{\{\omega_1\}\}$, and $\mathscr G$ contains every subset of $\Omega$. Let $\mathbb P\{\omega_1\}=0$, $A = \{\omega_2\}$, and $f=2$. Then,
$$\int_{\{\omega_1\}}fd\mathbb P = 0 = \mathbb P(A \cap \{\omega_1\}).$$
But
$$\int_\Omega f d\mathbb P = 2 \neq 1 = \mathbb P(A \cap \Omega).$$
In practice, the generating set $\mathscr C$ is often an algebra (and therefore contains $\Omega$) or is easily seen to contain $\Omega$, in which case the result goes through using the proof you sketched.
|
H: Find $\int_0^{2\pi} \frac{x \cos x}{2 - \cos^2 x} dx$.
I have to find the integral:
$$\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx$$
I rewrote it as:
$$\int_0^{2\pi} \frac{x \cos x}{1 + \sin^ 2 x} dx$$
But nothing further. I plugged it in a calculator and the result was $0$. I can see that the following relation holds:
$$f(-x) = -f(x)$$
For
$$ f: [0, 2\pi] \rightarrow \mathbb{R} \hspace{2cm} f(x) = \frac{x \cos x}{1 + \sin^2 x}$$
so that means that the function is an odd function. So if the interval $[0, 2\pi]$ is a symmetric interval for $f(x)$ then the result would be $0$.
I can see that the interval $[0, 2\pi]$ is symmetric for $\sin x$ and for $\cos x$, so it is not far fetched to believe it is symmetric for $\dfrac{\cos x}{1 + \sin^2 x}$, but wouldn't multiplying it with $x$ interfere with that symmetry? I don't see why $[0, 2\pi]$ is symmetric for the function
$$f(x) = \frac{x\cos x}{1 + \sin^2 x}$$
How come that $x$ doesn't ruin the symmetry?
AI: $$I=\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx\tag 1$$
$$I=\int_0^{2\pi} \frac{(2\pi-x) \cos x}{2 - \cos^ 2 x} dx\tag 2$$
Adding (1) & (2) $$2I=\int_0^{2\pi} \frac{2\pi \cos x}{2 - \cos^ 2 x} dx$$
$$I=2\pi \int_0^{\pi} \frac{\cos x}{2-\cos^ 2 x} dx\tag 3$$
$$I=2\pi \int_0^{\pi} \frac{\cos (\pi-x)}{2-\cos^ 2(\pi- x)} dx$$
$$I=-2\pi \int_0^{\pi} \frac{\cos x}{2-\cos^ 2x} dx\tag 4$$
Adding (3) & (4), we get $$I=0$$
|
H: Suppose that $S\sim T$, $P\sim Q$, $S⋂P=\varnothing $ and $T⋂Q=\varnothing$. Prove that $(S∪P) \sim (T∪Q)$.
By definition of equivalent sets, a set $S$ is equivalent to set $T$ if and only if the function $f:S\to T$ is one-to-one and onto. A set $P$ is equivalent to set $Q$ if and only if the function $f:P\to Q$ is one-to-one and onto.
How do we prove $(S∪P)\sim (T∪Q)$?
AI: Let $f : S \to T$ and $g : P \to Q$ be a one-to-one and onto functions. To show that $(S∪P)\sim (T∪Q)$ it suffices to give a one-to-one and onto function $h : S∪P \to T∪Q$. Define $h :S∪P \to T∪Q$ to be given by: $$h(x) =\begin{cases} f(x) &\text{if} &x \in S\\
g(x) &\text{if} &x \in P.\end{cases}$$
Now it is routine to check that $h$ is one-to-one and onto, using the facts that $f$ and $g$ are and also that $S⋂P=\varnothing = T \cap Q$.
|
H: Switching limit and infinite product
Dominated convergence theorem for an infinite product states that:
$$\lim_{n \to ∞} \prod_{k=1}^{∞}(a_{kn}+1)=\prod_{k=1}^{∞}\lim_{n \to ∞} (a_{kn}+1)$$
If
There exists a convergent sum $$\sum_{k=1}^{\infty}b_{k}$$ such that (for all k)$$b_{k}{\ge}|a_{nk}|$$
To prove this based on hint :
I have said that if : $$\lim_{n \to ∞} \prod_{k=1}^{∞}(a_{kn}+1)=\prod_{k=1}^{∞}\lim_{n \to ∞} (a_{kn}+1)$$ then $$\lim_{n \to ∞} \sum_{k=1}^{∞}\ln(a_{kn}+1)=\sum_{k=1}^{∞}\lim_{n \to ∞}\ln (a_{kn}+1)$$ This is true since log is continuous and log of a product is the sum of logs. Now we may use the familiar dominated convergence for infinite sum on this and say that this is true if there exists a convergent sum with terms $c_{k}$ such that(for all k)$$c_{k}{\ge}|\ln(a_{nk}+1)|$$ But I do not understand why it is enough to find a convergent series with terms $$b_{k}{\ge}|a_{nk}|$$I think we can somehow construct a convergent series with terms $ c_{k}{\ge}|\ln(a_{k}+1)|$ using a convergent series with terms $b_{k} {\ge} |a_{k}|$ and so than the above will be enough but I don’t know how exactly to do this.
AI: $\frac {\ln (1+t)} t \to 1$ as $ t \to 0$. Hence there exists a consatnat $c$ such that $|ln (1+t)| \leq c|t|$ for $|t|\ \leq \frac1 2$. Note that for $k$ sufficiently large we have $b_k <\frac 1 2$ and this makes $|a_{nk} |<\frac 1 2$ for all $n$ and $k$ sufficiently large. I hope the rest of the argument is clear.
|
H: Is this vector proof question wrong?
If a, b, c and d are not equal or 0 and (a.b)c=(b.c)a`,
show that a and b are parallel.
Since the dot product is a scalar, I can see that c and a are scalar multiples of each other, but what if b is prependicular to a and c? Wont the given equation still be satisfied whilst allowing a and b to be anything in the plane perpendicular to b? What am I doing wrong here?
Sorry for the poor formatting. I have no idea how it works
AI: What you are saying is correct. In $\mathbb R^{3}$ let $e_1,e_2,e_3$ be the standard basis. Take $a=e_1,b=e_2$ and $c=e_3$. Then the equation holds but $a$ and $c$ are linearly independent.
|
H: KKT $\min x_1^2+2x_2^2+x_1$
Given
$\min x_1^2+2x_2^2+x_1$
s.t : $x_1+x_2\leq a$
Prove:
a)for every $a\in\mathbb{R}$ the problem has a unique solution.
b)find the optimal solution as a function of $a$.
c)let f(a)be optimal value find explict formula for f(a) and prove it is convex.
So for a) i said the the target function is strictly convex and the constrain is convex hence we have a unique optimal solution.
for b)I found that $x_1=\frac{4a-1}{6},x_2=\frac{2a+1}{6}$ (not sure if it's right)
for c) I didn't understand what i'm suppose to do.
Hints/answers will be welcome.
AI: The function $f(a)$ is given by
$$
f(a) = \max_{x_1,x_2} x_1^2 + 2x_2^2 + x_1 + \mu(x_1+x_2-a).
$$
The Envelope Theorem is that you can differentiate the Lagrangian wrt the parameter of interest, then evaluate at the solution to see how the optimized solution varies with the parameter (https://en.wikipedia.org/wiki/Envelope_theorem). So
$$
f'(a) = -\mu^*
$$
Now, to prove convexity, you have to show that $f'(a)$ is increasing in $a$. That is equivalent to showing that the multiplier on the constraint, $\mu^*$, is increasing in $a$. If you increase $a$, you expand the choice set, and the objective must decreasing, so $\mu^*$ is decreasing in $a$. That implies $f''(a) = -D_a \mu^*(a) \ge 0$, and $f$ is convex.
Or you could plug your optimal solution back into the objective, and differentiate wrt $a$ and see what you get.
|
H: How do you find the limit of $8+(-1)^n 8$? And the intuitive way of thinking why it approaches the limit.
Given $\lim_{n\to \infty} 8+(-1)^n8$ does this converge towards some number or does it diverge? What is the intuitive way of thinking about the limit of this function?
AI: Notice that if $n$ is even, then $(-1)^n=1$ (because $n$ is even) so we have
$$
8+(-1)^n8=8+8=16
$$
But if $n$ is odd then $(-1)^n=-1$ because $n$ is odd and then we have
$$
8+(-1)^n8=8-8=0
$$
If the limit is to exist, then if $n$ is 'large', it must approach a single value. But the value 'bounces' back and forth between $0$ and $16$. Therefore, the limit does not converge, i.e. the limit diverges.
People new to the idea of convergence typically get the idea that 'diverge' means to get very large or very negative. But diverge just means that a sequence does not converge. To converge, you must get very 'close' to a single number. But there are 'plenty' of sequences which do not approach anything in particular. As another example, $\{(-1)^n\}$ simply alternates between $1$ and $-1$. Hence, this sequence also diverges. The sequence $\{(-1)^nn\}$ alternates between large even numbers and very negative odd numbers and hence also diverges (and does not always get 'large' or 'very' negative). You can also come up with sequences that do not exhibit this effect. For instance, $\{\sin n\}$ does not approach any particular value, hence diverges. As does the sequence $\{1,\dfrac{1}{1},2,\dfrac{1}{2},3,\dfrac{1}{3},4,\dfrac{1}{4},\ldots\}$. Come up with lots more on your own!
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H: Question About An Induction Problem: What Am I Supposed To Prove?
I've encountered this induction problem, and I'm not sure what I'm supposed to prove. I would like an explanation on a few things about the problem; I'm not looking for a hint/solution to the problem. The problem here is:
What I'm confused about is the following:
What is "cap"/hemisphere? Right now, I'm imagining something that looks like this:
What does choosing a hemisphere mean? And what does removing a hemisphere mean? Right now, I'm imagining something that looks like this:
What does "keeping the sphere covered mean"? Right now, I think it means that as long as the sphere has caps on it, it's "covered". But then I doubt this is correct, because the problem would be trivial- since I've chosen 4 caps to remain, it's covered. So I'm wondering what the right interpretation of this is.
Thanks in advance.
AI: I think the important part here is that the hemispheres don't actually stick out of the sphere; they're hemispheres of the actual sphere itself.
Here's a better way to phrase the question: A cap is a hemisphere of the sphere, and we have some set of caps such that their union is the whole sphere. Prove that there is a subset of four caps whose union is still the whole sphere.
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H: how to calculate （1987^718） mod 60
My attempt:
$1987=60×33+7$，$\phi（60）=16$，so $7^{16}≡1 \bmod60$,
$1987^{718} ≡ 7^{718} ≡ 7^{(16×44+14)} ≡ 7^{14} \bmod60$,
then I have no idea how to solve it.
What do you think about it? Could you please show me?
Regards
AI: Once you've reduced it to $7^{14} \pmod {60}$, note that $7^2 \equiv 49 \equiv -11$, and so
$$7^4 \equiv (7^2)^2 \equiv (-11)^2 \equiv 121 \equiv 1 \pmod {60}.$$
So $7^{14} \equiv (7^4)^3 \cdot 7^2 =\equiv 1^3 \cdot 49 \equiv 49 \pmod{60}$.
|
H: It is true that $1 \leq k \leq n$ then $k^2+n+1 \leq 2k(n+1)$?
Problem:
Suppose $k \in \mathbb{N}$ and $n \in \mathbb{N}$.
Suppose $1 \leq k \leq n$.
It is true that $k^2+n+1 \leq 2k(n+1)$?
How can I prove it?
My attempt:
I tried using induction on $n$ but actually I am not sure the statement is true.
AI: We want to prove $k^2 + n + 1 \leq 2k(n+1)$. Moving the $n+1$ to the right, this is the same as $k^2 \leq (n+1)(2k-1)$.
Since $k \geq 1$, we have $2k-1 = k+k-1 \geq 1+k-1 = k$. So
$$(n+1)(2k-1) \geq (n+1)k \geq (k+1)k > k^2,$$
as required.
|
H: Let S, T and P be three nonempty set. Prove that (a)S~S (b)If S~T, then T~S
(a) S~S means it is reflexive
(b) If S~T, then T~S means it is symmetry
Using the definition of equivalent sets,
set S is equivalent to T if and only if there exists a function f:S->T which is one-to-one and onto.
set T is equivalent to T if and only if there exists a function f:T->S which is one-to-one and onto.
How to prove
(a)S~S
(b)If S~T, then T~S
AI: I'll use $S eq T$ to mean $S$ is equivalent to $T.$ Ok, so then for part (a) of your problem, $S eq S$ means there is a function $f:S \to S$ which is one-to-one and onto. To finish that claim you only need one such bijection $f,$ and the first thing one would try is the identity mapping $f:S \to S$ defined by $f(s)=s$ for each $s.$
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H: Which is the canonical base for the Space of Complex matrices $A\in \Bbb{C}^{n\times n}$?
Which is the canonical base for the Space of Complex matrices $A \in \Bbb{C}^{n\times n}$?
I know in real matrices $\in \Bbb{R}^n$, the canonical base is trivial, just $n$ vectors with a $1$ in the $i$-th component for the $i$-th vector. But for complex matrices, it would be something like a tensor, because we need to generate a complex entry in each component of the vector. Nevertheless, I would like to know which is the formalization of this idea.
AI: I don't know if it is called “canonical”, but a natural basis of $\Bbb C^{n\times n}$ is $\{E_{ij}\mid1\leqslant i,j\leqslant n\}$ wher $E_{ij}$ is the $n\times n$ matrix whise entries are all $0$ except for the entry located at the $i$th line and the $j$th column, which is a $1$.
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H: Derivative of parametric equations
Consider the parametric equations $x=x(t), y=y(t)$.
I'm told that the derivative can be expressed as $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ provided $dx/dt\neq 0$.
However, in the derivation of this, it was assumed that we can express $y$ as a function of $x$, i.e. $y=f(x),$ for some $f$.
In my experience with parametric equations, it is not always possible to write $y=f(x)$, so how can we make this assumption?
AI: Consider the case of the cycloid
$$\begin{cases}x=t-\sin t,\\y=1-\cos t\end{cases}.$$
You cannot solve the transcendental equation $x=t-\sin t$ analytically for $t$, so that you cannot express $y=f(x)$ with a closed-form expression. Anyway,
$$\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{1-\cos t}{\sin t}$$ allows you to compute the slope at any point of the curve.
Addendum:
Notice that we can eliminate $t$ by
$$x=\arccos(1-y)-\sin(\arccos(1-y))$$
which gives the inverse of $f$ (but one has to consider several branches of the arc cosine).
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H: $ |T z|= |z| \,\,\forall z\in \mathbb{C} \implies ab=0 $
If $T: \mathbb{C} \to \mathbb{C}$ is given by $ T(z)= a z + b \bar z$ where $a,b \in \mathbb{C}$.
Then is it true that, $ |T z|= |z| \,\,\forall z\in \mathbb{C} \implies ab=0 $ ?
If yes, how can I show this?
Any help would be appreciated. Thanks in advance.
AI: Note
\begin{align*}
|Tz|^2 &= (Tz) \overline{T(z)}\\
&= (az+b \bar z) (\bar a \bar z + \bar b z)\\
&= |a|^2 z \bar z + |b|^2 z \bar z + a \bar b z^2 + \bar a b \bar z^2)\\
&= (|a|^2+|b|^2) |z| + (a \bar b z^2) + \overline{(a \bar b z^2)}\\
&= (|a|^2+|b|^2) |z| + 2 \text{Re}(a \bar b z^2).\\
\end{align*}
If $a \bar b \neq 0$, then we can vary $z$ while keeping $|z|$ constant, and $|Tz|^2$ will change due to the $\text{Re($a \bar b z^2$)}$ term, but $|z|^2$ will not, a contradiction. So $a \bar b = 0$, giving either $a=0$ or $\bar b = 0$. The result follows.
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H: Probability function question - Uniform distribution
if $X∼Uc(1,8)$ and $Y=X^2$. I need to calculate $P(Y≤7)$.
So Isn't it just the cumulative uniform distribution function but to calculate $x$ first? meaning $(x-a)/(b-a)$.
So I do $x-1/8-1$, but what's getting me confused is if I can do this -
x = $2.645$ (the square root of 7).
AI: $P(Y\leq 7)=P(-\sqrt 7 \leq X\leq \sqrt 7)=P(1 \leq X\leq \sqrt 7)$ since $X$ takes values in $(1,8)$. Hence $P(Y\leq 7)=\frac {\sqrt 7 -1} 7$.
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H: what's the meaning of ordinates of a Gaussian distribution?
In a Gaussian distribution, what's the meaning of the height (ordinate) at $x$?
according to [1], the funtion is called the probability distribution function of a Gaussian distribution, according to [2], it calculates the height of a Gaussian distribution.
Does this function mean the probability at $[-\infty，x]$, or the height (ordinate)?
TIA.
AI: The function you mentioned (better to write in with MathJax) is NOT the probability distribution, it is the pdf: probability Density function (the ordinate, as you said).
$$ \bbox[5px,border:2px solid black]
{
f(x|\mu\;\sigma^2)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2\sigma^2}(x-\mu)^2}
\qquad
}
$$
The probability distribution is its integral function
$$\mathbb{P}[X\leq x]=\int_{-\infty}^x f(t)dt$$
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H: Find the value of $p$ for which $f^{'\:}\left(\frac{1}{2}\right)\:=\:g^{'\:}\left(\frac{1}{2}\right)$
Given that $f\left(x\right)\:=\:tan^{-1}\left(2x\right)$ and $g\left(x\right)\:=\:p\:tan^{-1}\left(x\right),$ where $p$ is a constant. Find the value of $p$ for which $f^{'\:}\left(\frac{1}{2}\right)\:=\:g^{'\:}\left(\frac{1}{2}\right)$
AI: Well, let's find:
$$\frac{\text{d}\text{f}\left(x\right)}{\text{d}x}=\frac{\text{d}}{\text{d}x}\left(\arctan\left(2x\right)\right)=\frac{2}{1+4x^2}\tag1$$
$$\frac{\partial\text{g}\left(x\right)}{\partial x}=\frac{\partial}{\partial x}\left(\text{p}\arctan\left(x\right)\right)=\text{p}\cdot\frac{\partial}{\partial x}\left(\arctan\left(x\right)\right)=\frac{\text{p}}{1+x^2}\tag2$$
So, we have:
$$\frac{2}{1+4\left(\frac{1}{2}\right)^2}=\frac{\text{p}}{1+\left(\frac{1}{2}\right)^2}\space\Longleftrightarrow\space\text{p}=\frac{5}{4}\tag3$$
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H: Notation question: linear map $P(x_1,x_2,x_3,x_4,x_5)$
Consider the vector space $\mathbb{C}^5$ and the map $P:\mathbb{C}^5\mapsto\mathbb{C}^5$,
$P(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,0,0,0)$.
Am I supposed to read $x_1,x_2,x_3,x_4,x_5$ as 5 vectors or as the 5 entries of a single vector?
AI: They're five entries of a vector; $P$ has domain $\mathbb C^5$ so each $x_i$ is a complex number.
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H: Multivariable Calculus proof explanation help, Cauchy Sequences
Proof here. This is a proof from the Advanced Calculus book by Gerald.B Folland. I understand all the steps except where the author goes on to say that $|\textbf{x}_k|<|\textbf{x}_{K+1}|+1$ for all $k>K$. How did he get to this inequality?
AI: It's just using the definition of Cauchy convergence with $\epsilon=1$.
For all $k,k'> K$ you have $|x_k-x_{k'}|< \epsilon = 1$. In particular for $k' = K+1$ you have that for all $k> K$
$$|x_k-x_{K+1}|< 1\,.$$
Since $|a|-|b|\leq |a-b|$ for all $a,b$, in particular you have $$|x_k|-|x_{K+1}|\leq |x_k-x_{K+1}|< 1\,,$$
from which the statement follows.
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H: Dividing polynomial, multiple divisors and remainders
Problem:
If we divide $P(x)$ by $x-1$, then the remainder is $1$.
If we divide $P(x)$ by $x^2+1$, then the remainder is $2+x$.
What is the remainder when we divide $P(x)$ by $(x-1)(x^2+1)$?
My attempt:
I know $P(1)=1$ from first condition, I tried to see if I can extract more information from second condition, but I am clueless.
Any ideas?
AI: Hint: Use the division algorithm: $A(x)=B(x)Q(x)+R(x)$, $\deg(R)<\deg(B)$
Another Hint: if you have a good background about modular arithmetic, try to do it the same way as the Chinese Remainder Theorem problems.
Solution:
So let's try
$$P(x)=(x-1)Q_1(x)+1 \tag{1}$$
$$P(x)=(x^2+1)Q_2(x)+x+2 \tag{2}$$
Now let's work on this a little bit. Put $x=1$ in $(2)$ knowing that $P(1)=1$ from $(1)$, we get
$$1+2Q_2(1)+3 \iff Q_2(1)=-1$$
Note that $$Q_2(x)=(x-1)Q_4(x)+a \implies a=Q_2(1)=-1$$
$$\implies P(x)=(x^2+1)((x-1)Q_4(x)-1)+x+2$$
$$=(x^2+1)(x-1)Q_4(x)+(-x^2+x+1)$$
And we're done!
Another solution (but with the same algorithm) is using complex numbers:
$$P(x)=(x-1)Q_1(x)+1 \tag{$A(x)$}$$
$$P(x)=(x^2+1)Q_2(x)+x+2 \tag{$B(x)$}$$
$$B(i) \implies P(i)=2+i$$
$$A(i) \implies 2+i=(i-1)Q_1(i)+1 \iff Q_1(i)=\frac{i+1}{i-1}=-i$$
and we note that
$$Q_1(x)=(x^2+1)Q_3(x)+ax+b$$
$$x\mapsto i\implies -i=ai+b \implies a=-1,b=0$$
$$\implies Q_1(x)=(x^2+1)Q_3(x)-x$$
$$\implies P(x)=(x-1)((x^2+1)Q_3(x)-x)+1 = (x-1)(x^2+1)Q_3(x)+(-x^2+x+1)$$
Another Solution: Treat it as chinese remainder theorem but in polynomials, I'll solve it the same way I solve the normal chinese remainder theorem:
$$P(x) \equiv1 \pmod{x-1} \implies P(x)=Q(x)(x-1)+1$$
$$Q(x)(x-1)+1\equiv x+2 \pmod{x^2+1}$$
$$\implies Q(x)(x-1) \equiv x+1 \equiv x+1-(x^2+1)\equiv x(1-x) \pmod{x^2+1}$$
$$\implies Q(x)\equiv -x \pmod{x^2+1} \implies Q(x)=H(x)(x^2+1)-x$$
$$\implies P(x)=(x-1)(H(x)(x^2+1)-x)+1=(x-1)(x^2+1)H(x)+(-x^2+x+1)$$
and that's it. We're done.
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H: Length of a line in a regular hexagon
The picture belongs to a regular hexagon. I don't understand that if a line is drawn from $E$ to $C$, how its length equals to $|KL|$ and how it is parallel to $|KL|$?
Would you mind drawing its explanation?
AI: The two lines are not parallel or have the same length. We know the interior angle of a regular hexagon is $120^\circ$ and that the angles ECB and CEF are right ones.
The line EC has a length of $6\sqrt3$. This can be calculated by the cosine rule: $a^2=c^2+b^2 -2bc\cos(a)$ substituting: $a^2 = 72+72\cos(120)$ so $a = 6\sqrt3$.
We can find the length of KL by finding the hypotenuse of a right angled triangle with base $6\sqrt3$ nd side length 3. By Pythagoras' theorem this gives us $a^2=3^2+(6\sqrt3)^2$ -o $a^2 = 9+108 = 117$ so $a = 3\sqrt13$
We know the base of the triangle is $6\sqrt3$ because the lines CB and EF are parallel. We can find the height of this triangle by taking the difference of 4(the length LC) and 1(the length KE) So in conclusion, the lines are not parallel of of the same length. Line KL is $3\sqrt13$ units long while EC is $6\sqrt3$ units long.
I hope this was of some assistance.
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H: What does it mean for a point to be on the segment $[1,i]$?
I came up on a problem where I have to prove that there does not exist a point $a\in\mathbb{C}$ on the segment $[1,i]$ such that $i+1=3a^2$.
I am struggling to understand what is meant by $a$ being on the segment $[1,i]$. Does it mean that $a$ is on the segment from $(1,0)$ to $(0,i)$ on the complex plane? Or does it means something different?
AI: The line segment from $(1,0)$ to $(0,i)$ is the set of points $(t, i(1-t)), 0 \leq t \leq 1$. If this satisfies the given equation you get $1=2t-1$ and $1=2t(1-t)$. The first equation gives $t=2/3$ so the second equation cannot hold.
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H: Let $a > 1$ and $x > 0$. Prove that $a^x > 1$
What i'm proving is in the title. Essentially, I know that it holds for rational numbers. I want to prove it for all real numbers. The following is the definition for $a^x$ that I'll be using.
Let $\{r_n\}$ be any sequence of rationals that converge to $x$. Then, the exponential of the base $a$ is defined:
$$a^x = \lim_{n \to \infty} a^{r_n}$$
Proof Attempt:
Let $\{r_n\}$ be any sequence of rational numbers that converges to $x$.
This sequence is convergent, so it is bounded. By the Bolzano-Weierstrass Theorem, this has a convergent subsequence. We denote this subsequence by $\{s_n\}$.
This subsequence converges to $x$ as well. If it didn't, then there would exist an $\epsilon > 0$ such that for all $N>0$, $n>N$ and $|s_n-x| \geq \epsilon$. But since this subsequence is a part of the original sequence, which does converge to $x$, it follows that such an $\epsilon$ cannot exist.
If the subsequence is decreasing, then:
$$\forall n \in \mathbb{N}: s_n > 0$$
So, it follows that:
$$\forall n \in \mathbb{N}: a^{s_n} > 1$$
In the limit, it would follow that $a^x > 1$.
If the subsequence is increasing, then:
$$\exists N > 0: n > N \implies s_n > 0$$
$$\implies \exists N>0: n > N \implies a^{s_n} > 1$$
In the limit, it follows that $a^x > 1$. This proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?
AI: It doesn't work, because you merely proved that $a^{s_n}\ge1$, if we assume the map is continuous.
To fix it, we'll assume the following exponentiation fact:
$a^q$ is increasing for rational $q$, $a>1$
$a^q$ is greater than $1$ for for positive rational $q$, $a>1$
Consider any sequence $r_n$ that converges to $x$. Then we know that for $\varepsilon=\frac x2$, eventually every term in the sequence $a^{r_n}$ is greater than $1+\kappa$ for $\kappa=a^{\frac x2}-1>0$.
Thus, since eventually every term is greater than $1+\kappa$, the limit is at least $1+\kappa$ (note that it could be equal to $1+\kappa$, but this is still $>1$).
Of course, if you haven't already done so, you also need to show that this limit exists and is well-defined (independent of which sequence you choose).
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H: Gebrane's Hanoi
I need some help to solve a modification of the Tower of Hanoi problem found on a french forum.
The classic problem is described here.
This modification is called Gebrane's Hanoi problem (from the name of its inventor), the $n$ disks are numbered from top to bottom from $1$ to $n$ and are also placed on peg $A$, and the goal is to move all the disks to pegs $B$ and $C$ to form two towers: the tower on peg $B$ is formed with all the even-numbered disks and the one on peg $C$ is formed with all the odd ones (obviously).
The other rules are the same: each move consist in displacing an upper disk of a tower on top of another tower or on an empty peg, and no disk may be placed on a smaller one.
Let $HG(n)$ be the minimum number of moves to solve Gebrane's Hanoi problem.
Prove that: $$HG(n)=-\frac 1{21}\cos\left(\frac2 3n\pi\right)+\frac 1 7\sqrt 3\sin\left(\frac2 3n\pi\right)+\frac 5 7 2^n-\frac 2 3.$$
AI: Let's rename pegs $B$ and $C$ as $B'$ and $C'$ where $B'$ is the peg where
disc $n$ and all other discs with the same parity are destined to move, and
$C'$ is where the discs of opposite parity are to go.
Let $n\ge3$.
We need to move disc $n$ to $B'$. First we need to move discs $1$ to $n-1$ to $C'$;
this needs $2^{n-1}-1$ moves. After $2^{n-1}$ moves we have disc $n$ on $B'$
the remaining discs on $C'$ and $A$ empty.
Both discs $n-1$ and $n$ are where we want them. We need to move disc $n-2$
to $B'$. So we need to move discs $1$ to $n-3$ from $C'$ to $A$, which takes
$2^{n-3}-1$ moves, then an extra move to get $n-2$ to $B'$.
After $2^{n-1}+2^{n-3}$ moves, we have $n-2$ and $n$ on $B'$ and $n-1$ on $C'$,
and the remaining discs on $A$.
We are back to the original puzzle but with $n$ replaced by $n-3$.
Therefore
$$HG(n)=HG(n-3)+2^{n-1}+2^{n-3}=HG(n-3)+\frac58 2^{n}.$$
Now solve this recurrence
(the solution will have the form $a2^n+b+c\cos(2n\pi/3)+d\sin(2n\pi/3)$).
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H: Prove the following: If $\lim(x_n) = x$ and if $x > 0$, then there exists a natural number $M$ such that $x_n > 0$ for all $n\ge M$.
I recently encountered the following question:
Prove that if $\lim(x_n) = x$ and if $x > 0$, then there exists a natural number $M$ such that $x_n > 0$ for all $n\geq M$.
My solution manual suggests that I should choose $\epsilon=\frac{x}{2}$. I tried working out the same procedure with $\epsilon=\frac{x}{3}$. I could still obtain $x_n>0$ for $n\geq M$ ($x_n > \frac{2x}{3} $ to be precise).
Is it correct to choose $\epsilon=\frac{x}{3}$? To extend the principle, is it correct to choose $\epsilon=\frac{x}{a}$ where a is a natural number and $a<x $?
Can anyone please help?
Thanks in advance!
AI: Yes, it is correct. More generally, you can take $\varepsilon$ equal to any number smaller than or equal to $x$ (and greater than $0$). As far as I am concerned, the more natural choice is to take $\varepsilon=x$, but that's a matter of taste.
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H: Glueing together Riemann surfaces
In Miranda's , he wrote such a proposition:
proposition 1.6 of chapter 1: Let $X,Y$ be 2 Riemann surfaces. $U\subset X$, $V\subset Y$ are their open subsets. Suppose $\phi: U\rightarrow V$ is an isomorphism of Riemann surfaces, then there's a unique complex structure on $Z:=X\amalg Y/\phi$ such that the 2 natural embeddings $j_X: X\rightarrow Z$ and $j_Y: Y\rightarrow Z$ are holomorphic maps. In particular, if $Z$ itself is Hausdorff, then we obtain a Riemann surface.
My question is: for example, we can glue 2 torus (of genus 1) together by choosing $U,V$ as 2 disks small enough. Of course in this case $Z$ is a Hausdorff space. But what's the genus of it? What makes me puzzled is that topologically this identification space doesn't look like a closed surface! (it looks like the "tube" in the middle of the genus 2 torus has been closed!) Can anybody explain this space in detail to me? thanks in advance!
AI: "Of course in this case $Z$ is a Hausdorff space" - of course not. The point on boundary of $U$ in the first torus, and "corresponding" point on boundary of $V$ in the second torus (corresponding meaning they are limits of the identified sequences of points in $U$ and in $V$) will have no disjoint open neighborhoods in $Z$. (You can see the same thing when trying to glue two segments into a Y shape by identifying along a open subsegments; there are two points at the center of the Y, and they don't have disjoint neighborhoods.)
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H: Why is $\left( -a,a \right)$ equal to $|x| < a$
I have a problem understanding the equality of an open Interval as given $\left( -a,a \right) = \{x \in R | -a< x< a\}$ to say $|x| < a$..
Maybe someone can get me intuitive to understand that?
AI: Since the end points are not contained in the interval, we have $|x|<a$ instead of $|x|\le a$. A number $x$ is between $-a$ and $a$ ($-a<x<a$) if and only if its absolute value is smaller than $a$.
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H: Relation between areas in a trapezoyd
A trapezoid of ABCD vertices is inscribed in a circle, with radius R, being AB = R and CD = 2R and BC and AD being non-parallel sides. The bisectors of the internal trapezoidal angles, so that the bisector of  intercepts that of Dˆ at point Q, that of B intersects that of C N at point N and that of C ˆ intercepts that of Dˆ at point M. Knowing that the points M, N and Q are inside the ABCD trapezoid and that point P is the intersection of the Bˆ and Aˆ guidelines, determine the relationship between the areas of the MNPQ and ABCD polygons. Ans .: 1/9.
I tried a lot to do this, but everytime I found a segment with negative size
Please, can someone help me?
Thanks for antetion.
AI: The hint.
$CD$ is a diameter of the circle, which gives $AB=AD=BC=R$.
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H: For $f(x) = e^x + x^3 - x^2 + x$ find the limit $\lim\limits_{x\to \infty} \frac{f^{-1}(x)}{\ln x}$.
I have the function:
$$f : \mathbb{R} \rightarrow \mathbb{R} \hspace{2cm} f(x) = e^x + x^3 -x^2 + x$$
and I have to find the limit:
$$\lim\limits_{x \to \infty} \frac{f^{-1}(x)}{\ln x}$$
(In the first part of the problem, I had to show that the function is strictly increasing and invertible. I don't know if that's relevant to this, since I could show that the function is invertible, but I can't find the inverse.)
So this is what I tried:
I showed
$$\lim_{x \to \infty} f(x) = \infty$$
and so I concluded that $\lim\limits_{x \to \infty} f^{-1}(x) = \infty$. I'm not sure if this is correct, it might be wrong. But if would be right, then we could use l'Hospital, knowing that:
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$$
but after trying to use all of this on paper, I got nowhere. It just complicated thing a lot more.
So how should I solve this limit?
AI: Hint : $$\forall x\geq 1\quad 2e^x \geq e^x+x^3-x^2+x\geq e^x+1$$
So $$\ln(\frac{x}{2})\leq f^{-1}(x)\leq \ln(x-1)$$
So we deduce that your limit is a constant by the sandwich theorem .
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H: Showing that $\mathbb{C}^5/\ker(P)\simeq \mathbb{C}^2$
Consider the vector space $\mathbb{C}^5$ and the map
$P:\mathbb{C}^5\mapsto\mathbb{C}^5,\,P(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,0,0,0)$
According to my lecture notes, this implies: $\mathbb{C}^5/\ker(P)\simeq \mathbb{C}^2$. But what exactly does this mean? That $\mathbb{C}^5\ni x\mapsto \mathbb{C}^5/\ker(P)$ is isomorphic to $\mathbb{C}^2$?
AI: For a vector space $V$ and a subspace $U$ the quotient $V/U$ is defined as the set of equivalence classes of $v\sim w\iff v-w\in U$ equipped with the inherited operations. You can show that these are well-defined and $V/U$ is in fact a vector space itself. Elements of $V/U$ are commonly written as $[v]:=v+U$.
In your case $V=\Bbb C^5$ and $U=\ker P$ where $P$ is the projection
$$P\colon\Bbb C^5\to\Bbb C^5,\,(z_1,z_2,z_3,z_4,z_5)\mapsto(z_1,z_2,0,0,0)$$
The kernel of $P$ is readily identified with $\Bbb C^3$. The isomorphism theorem (or also called rank-nullity theorem in case of vector spaces) yields that ${\rm im}\,P=\Bbb C^2$ and the desired isomorphism $\Bbb C^5/\ker P\cong\Bbb C^2$.
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H: Khan Academy - Factoring sum of squares
I am working on Precalculus - Factoring sum of squares
The video starts with example:
$36a^8 + 2b^6 = 6a^{4^{2}} + \sqrt{2}b^{3^{2}}$
Can someone explain this to me? What's happened here?
Thanks.
AI: Actually, the video starts with
$$36a^8+2b^6=(6a^4)^2+(\sqrt{2}b^3)^2$$
which is of course true since $(ab)^n=a^nb^n$ and $(a^n)^k=a^{nk}$
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H: Prove that a planar graph with all vertices degree $3$ must has a face with at most $5$ edges.
I have some problems when I prove "For a planar graph $G$ and $\deg(v) = 3$ for any vertex $v$, there is a face with at most $5$ edges". I want to prove with contradiction.
Suppose that every face has more than $5$ edges.
Then $2e > 5r$ ($e$ is the number of edges and $r$ is the number of faces) because each edge occurs on the boundary of a face exactly twice.
Also, I can get $2e = 3v$. (for every edges has $2$ vertices)
So we have $3v > 5r$.
Then, with Euler's formular: $r = e - v + 1$, and $2e = 3r, 3v > 5r$ we can get $r > 20$.
But I cannot lead to a contradiction with the steps above.
Did I miss some points or my idea is not ok? Thanks.
AI: If $G = (V, E)$ is a planar graph with $|E|\geq g$ and no cycle of length $< g$, then
$$e \leq \frac{g}{g − 2}(v − 2)$$. This is a standard generalisation of the result $e \leq 3v-6$ which can be proved easily.
In this problem $g=6$, so we get $e \leq \frac{3}{2}(v-2)$, but since all vertices have degree $3$, number of edges is exactly equal to $\frac{3v}{2}$, hence a contradiction. Hence the graph must have a face with at most 5 edges.
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H: What is the "unit axis"?
As the highlighted text in the screenshot, what is the unit axis? And what mathematical properties the three coordinates X, Y, Z have?
AI: It means a vector of length 1.
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H: How to calculate the integral of $ \int\frac{1}{\cosh^{2}x}dx $
Ofcourse one can notice that
$ \left(\frac{\sinh\left(x\right)}{\cosh\left(x\right)}\right)'=\frac{\cosh^{2}\left(x\right)-\sinh^{2}\left(x\right)}{\cosh^{2}\left(x\right)}=\frac{1}{\cosh^{2}\left(x\right)} $
But im looking for a straight way to prove that
$ \int\frac{1}{\cosh^{2}x}dx=\frac{\sinh\left(x\right)}{\cosh\left(x\right)}+ C $
Here's what I tried:
$ \int\frac{1}{\cosh^{2}x}=\int\frac{1}{1+\sinh^{2}x}dx=\frac{\arctan\left(\sinh x\right)}{\cosh x} + C$
Now im not sure how to continue. Thanks in advance
AI: Based on your comment to another answer, you want to show that $\displaystyle \int \mathrm{sech}^2 x dx = \tanh x + c$.
I find the easiest way is to use complex numbers. Start with the circular trigonometric version $\displaystyle \int \sec^2 x dx = \tan x + c$, which I assume you can assume or you know how to prove. Then transform to the hyperbolic trigonometric version with the substitution $x=iy$:
Note that $\cos iy = \cosh y \implies \sec iy = \mathrm{sech}\ y$ and $\tan iy = i\tanh y$.
$\displaystyle \int \sec^2 iy d(iy) = \tan iy + c$
$\displaystyle i\int \mathrm{sech}^2 y dy = i\tanh y + c$
$\displaystyle \int \mathrm{sech}^2 y dy = \tanh y + z$
where $z$ is an arbitrary complex constant. If you're dealing with a purely real domain, $z$ will be real as well.
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H: How to solve such equations: $z^4 = -4$, $z \in \mathbb{C}$?
I have the following task:
Compute in each case all $z\in\mathbb{C}$ such that $z^4 = -4$ and $z^3 = 5i$.
I do not know how to solve such eqautions fast. Do you have any idea of how to solve such equations?
And one further question:
The multiplication of two complex numbers is defined as
$$wz := (u+iv)(x+iy)=(ux-vy)+i(vx+uy)$$
But what if w or z would be defined in the following ways:
$$w = (u-iv)$$ or $$w = (-u+iv)$$
Would the formula change?
Thank you for helping me out guys
AI: You have to keep in mind that taking roots are multi-valued in the complex plane. By the fundamental theorem of algebra every non-constant polynomial of degree $n$ has exactly $n$ in the complex plane when counted with multiplicity. This applies also for polynomials of the form $p(z)=z^n-c$ for some complex number $c\in\Bbb C^*$.
Of particular interest are the so-called roots of unity which are the roots of polynomials of the form $p(z)=z^n-1$. Solutions to the equation $z^n-1=0$ are called the $n^\text{th}$ roots of unity. Using polar coordinates we find that these are of the form $\zeta_n^k=\exp\left(k\frac{2\pi}n\right)$ for $k=0,...,n-1$.
Now, given an equation of the form $z^n-c=0$ we shall write $c$ as $c=r\exp(i\theta)$ where $r=|c|$ and $\theta=\arg c\in[0,2\pi)$. Then we can conclude that $z=r^\frac1n\exp \left(i\frac\theta n\right)$. But wait; there is more. In fact, every $z_k=r^\frac1n\exp \left(i\frac\theta n\right)\zeta_n^k$, $k=0,\dots,n-1$, is a solution too! We can express the general solution(s) therefore in the following manner
$$z_k=r^\frac1n\exp\left(i\frac{\theta+2\pi k}n\right),\,k=0,\dots,n-1$$
Can you take it from here?
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H: Minimising variance of the expected return for two correlated investments
For two investments with returns $X$ and $Y$, we can allocate the cash in proportions $\alpha$ to $X$ and $1-\alpha$ to $Y$, making the total variance of our investment $$\text{Var}(\alpha X+(1-\alpha)Y).$$ In a textbook I'm told that the value of $\alpha$ such that this quantity is minimised (i.e. the value of $\alpha$ to minimise the risk of the distribution of capital) is $$\alpha=\frac{\sigma_Y^2-\sigma_{XY}}{\sigma_X^2+\sigma_Y^2-2\sigma_{XY}},$$ where $\sigma_X^2$ and $\sigma_Y^2$ are the variances of $X$ and $Y$, and $\sigma_{XY}=\text{Cov}(X,Y)$. (This is page 187 of 'An Introduction to Statistical Learning' by Gareth James etc.)
I'm struggling to yield the answer when I try to prove this myself.
My Attempt
I start off by reducing the variance of the expected return to an easily differentiable function:
\begin{align}
\text{Var}(R) & = \text{Var}(\alpha X+(1-\alpha) Y) \\
& = \text{Var}(\alpha X)+\text{Var}((1-\alpha)Y)-\text{Cov}(\alpha X,(1-\alpha)Y) \\
& = \alpha^2\text{Var}(X)+(1-\alpha)^2\text{Var}(Y)-\alpha(1-\alpha)\text{Cov}(X,Y) \\
& = \alpha^2\sigma_X^2+(1-\alpha)^2\sigma_Y^2-\alpha(1-\alpha)\sigma_{XY} \\
& = \alpha^2\sigma_X^2+\sigma_Y^2-2\alpha\sigma_Y^2+\alpha^2\sigma_Y^2-\alpha\sigma_{XY} +\alpha^2\sigma_{XY}\\
& = \alpha^2(\sigma_X^2+\sigma_Y^2+\sigma_{XY})-\alpha(2\sigma_Y^2+\sigma_{XY})+\sigma_Y^2 \\
\end{align}
Now we can take the derivative and set it equal to zero to minimise the variance of the return:
\begin{align}
\frac{\partial\text{Var}(R)}{\partial \alpha} & = 2\alpha(\sigma_X^2+\sigma_Y^2+\sigma_{XY})-2\sigma_Y^2-\sigma_{XY} \\
0 & =2\alpha(\sigma_X^2+\sigma_Y^2+\sigma_{XY})-2\sigma_Y^2-\sigma_{XY} \\
\implies \alpha & = \frac{2\sigma_Y^2+\sigma_{XY}}{2(\sigma_X^2+\sigma_Y^2+\sigma_{XY})}
\end{align}
My issue
My final result is not the same as the one given in the text. Have I misused some identities for variances and covariances? I don't think I've made any algebraic mistakes - I've gone over it a number of times.
What am I doing wrong here?
AI: $\newcommand{\Var}{\operatorname{Var}}\newcommand{Cov}{\operatorname{Cov}}$The variance was not expanded correctly. Recall that $\Var(X+Y)=\Var(X)+\Var(Y)\color{blue}{+2\Cov(X,Y)}$. Thus $$\Var(\alpha X+(1-\alpha)Y)=\alpha^2\Var(X)+(1-\alpha)^2\Var(Y)\color{blue}{+2\alpha(1-\alpha)\Cov(X,Y)}.$$
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H: Is this result of the derivative of the composition of two functions correct?
I have the following functions:
$f(x,y)=xy,\; g(t)=(e^t,\cos(t))$
And I want to calculate the derivative matrix of $f$ after $g.$
I calculate the jacobian matrix of $f$, which was a matrix, like
$$
D_f=\begin{bmatrix}
y & x
\end{bmatrix}
$$
then on the components of $g$
$$
D_f(e^t,\cos(t))=\begin{bmatrix}
\cos(t) & e^t
\end{bmatrix}
$$
then calculate the matrix of $g$
$$
D_g(e^t,\cos(t))=\begin{bmatrix}
e^t \\
-\sin(t)
\end{bmatrix}
$$
then to calculate the matrix of the derivatives of $f$ after $g$, I multiplied the matrices
$$
D_f(e^t,\cos(t))=\begin{bmatrix}
\cos(t) & e^t
\end{bmatrix}
$$
times
$$
D_g(e^t,\cos(t))=\begin{bmatrix}
e^t \\
-\sin(t)
\end{bmatrix}$$
$= \cos(t)e^t-e^t\sin(t).$
I have no way of checking if this is correct. Is this correct?
AI: Yes, this is correct. You can easily check this yourself. The composition is given by
\begin{align}
f(g(t))=f(e^{t},\cos (t))=e^{t}\cos(t).
\end{align}
If we differentiate this function we get
\begin{align}
(f(g(t))'=\frac{d}{dt}(e^t\cos(t))=e^t\cos (t)-e^t\sin(t)
\end{align}
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H: Showing that if $\phi:\Bbb{Z}\oplus\Bbb{Z}\to\Bbb{Z}\oplus\Bbb{Z}$ is an epimorphism of abelian groups, then it is an isomorphism.
I am a mathematician working in analysis and my knowledge in algebra is rusty.
Is there a direct argument showing that if
$\phi:\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$
is an epimorphism of abelian groups, then it is an isomorphism?
My argument goes as follows: Since $\phi$ is a surjection onto a free abelian group $\mathbb{Z}\oplus\mathbb{Z}$, the group $\mathbb{Z}\oplus\mathbb{Z}$ is isomorphisc to
$\mathbb{Z}\oplus\mathbb{Z}\oplus\ker\phi$. $\ker\phi$ is also a free abelian group (as a subgroup of free abelian) and hence has to be zero as otherwise bases of the two free abelian groups:
$\mathbb{Z}\oplus\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{Z}\oplus\ker\phi$
would have different cardinality.
Adding proofs of all results that I am using here would make the proof somewhat lengthy. Is there a more direct argument?
AI: Here's a different perspective, at least.
Let $M$ be the $2 \times 2$ matrix with $\mathbb Z$ coefficients such that
$$\phi(v)=Mv
$$
where $v \in \mathbb Z \oplus \mathbb Z$ is regarded as a column matrix.
If $\det(M)=0$, the map $\phi$ is not surjective.
If $|\det(M)| > 1$, the map $\phi$ is not surjective.
If $|\det(M)|=1$, the map $\phi$ is both injective and surjective.
All of these can be proved rather simply, using methods of solving 2 linear equations in 2 unknowns.
But frankly, your outline seems fine to me, and my outline may have as many details to fill in as yours, although not bringing in much in the way of "deep" mathematics. At least it's a different perspective
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H: How would I cut one washer into two equal-area washers?
I'm trying to mill an asymmetric graphite crucible in the shape of a hollow cone (imagine an ice cream cone with the end bitten off). I require identical "horizontal resistance" at the wide top, as well as the narrow bottom (and throughout) - hence - the walls of my cone need to be fatter at the top, and they taper towards the bottom. Resistance goes up with length, and down with area - my length (around the circular path) gets longer at the top, so I need more area to keep it constant. (This is a hobby project, not a job or homework)
Viewed from above, the start of the cone is a disc with a hole in it - e.g. the shape of a big washer with a big hole.
Viewed from below, again, the end (that you bit off) is the shape of a washer, but this time, the hole is much smaller.
To calculate the approximate* resistance that electricity** will encounter when travelling from one side to the other of each washer-shaped area, I decided to draw an imaginary line on this washer such that it equally divides the total area.
*= it enters sideways from a square join-point equal to the outer-inner radius
**= it is 3v, so no skin effects
Another way to imagine the same problem: if you start with one wide washer with a small hole, how do you calculate the circumference of the line that divides that washer into two equal-area washers? (e.g if you were to cut along this mid-line, you'd get one big washer of the original circumference with a big hole in the middle, and a second smaller washer with the original small hole in the middle, and they would both weigh the same).
I've spent an hour in google with no luck so far (I'm not sure of the terms to search on, but I'm pretty sure Pythagoras isn't helping, and it's dominating the results).
An approach I attempted which didn't work out was as follows:
I reasoned that if the outer radius is A and the hole radius is C, then the midpoint radius can be B. From there, I reasoned that C is likely to be a radio of A to B somehow, so I defined C as B - n * (A-B), and ended up trying to solve for n, which gave me n=(sqrt(A^2-2B^2)-B)/(A-B) which broke immediately of course when I tried an example because: sqrt(-3).
I'm guessing this is probably harder math than I realized?
Right now, I'm obsessing over how to divide washer-shapes equally. The real problem (of which I'm fairly sure that answer is part) is how to find the formula to draw concentric expanding washer shapes such that their "midpoint circumferential length" multiplied by their total area is a constant.
Any help/tips/clues are most welcome. Go easy on me - the stuff I'm trying to do now was drummed into me 40 years ago, plus I know "programmer math" not "symbol math", so to say I'm rusty/unprepared is an understatement...
AI: Interesting geometry problem!
Let’s suppose we have a “washer” (or, a circle with a smaller circle removed from the middle) with outer radius $R$ and inner radius $r$, like this:
We want to solve for $r^*$, the radius of the circle dividing this washer into two smaller equal-area (or equal-mass) washers.
First, note that the area of an arbitrary washer with inner radius $r$ and outer radius $R$ is given by
$$\pi R^2-\pi r^2$$
since the washer is the result of removing a circle with area $\pi r^2$ removed from a circle with area $\pi R^2$. Now, we would like to calculate $r^*$ such that the areas of the two “sub-washers” produced is the same, or
$$\pi R^2-\pi {r^*}^2=\pi {r^*}^2-\pi r^2$$
Solving this equation for $r^*$ gives us the solution
$$r^*=\sqrt{\frac{R^2+r^2}{2}}$$
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H: Help understanding step in the proof of Taylor's theorem with remainder
Let $k\in\mathbb{N}$, $x_{0},x\in A\subseteq\mathbb{R}$ and $f:A\to\mathbb{R}$ satisfy that $f^{(j)}$ exists and is continuous on the closed interval between $x_{0}$ and $x$ and differentiable on the open interval between $x_{0}$ and $x$ for all $j\leq k$. Then there exists a $c$ between $x_{0}$ and $x$, such that:
\begin{equation}
f(x)=\sum_{j=0}^{k}\frac{f^{(j)}(x_{0})}{j!}(x-x_{0})^{j}+\frac{f^{(k+1)}(c)}{(k+1)!}(x-x_{0})^{k+1}.\quad\quad (*)
\end{equation}
My question: How does one realise that $(*)$ has exactly one solution $M\in\mathbb{R}$ when $x\neq x_{0}$:
\begin{equation}
f(x)=\sum_{j=0}^{k}\frac{f^{(j)}(x_{0})}{j!}(x-x_{0})^{j}+M(x-x_{0})^{k+1}
\end{equation}
Is it because $(*)$ is a linear function or? Thanks in advance.
AI: If $x \ne x_0$,we have
\begin{equation}
\frac{f(x)-\sum_{j=0}^{k}\frac{f^{(j)}(x_{0})}{j!}(x-x_{0})^{j}}{(x-x_0)^{k+1}}=M
\end{equation}
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H: How many functions f : [n] → [n] are there for which there exists exactly one i ∈ [n] satisfying f(i) = i?
As the title asks, I'm looking for the number of functions $f : [n] \rightarrow [n]$ for which there exists exactly one $i \in [n]$ satisfying $f(i) = i$.
I know the number of functions $f : [n] \rightarrow [n]$ is $n^n.$ I'm looking for how I would make that more specific as to only find the functions where there exists exactly one $i \in [n]$ satisfying $f(i) = i$.
EDIT: I think I may have come up with a solution, and verification would be greatly appreciated as my textbook offers no solutions. My thinking is this:
Because there's exactly one $f(i) = i$, any function satisfying our criteria must have all but one $f(i) \neq i$. So, for each $i \in [n]$, we have $(n-1)$ options for $f(i)$. This is true for all but one $i \in [n]$, where the number of options for $f(i)$ is $1$. So, my proposed solution is $(n-1)^{(n-1)}$. This provides us with with the number of functions $f : [n] \rightarrow [n]$ where every $i \in [n]$ satisfies $f(i) \neq i$. Then there is the single case of $f(i) = i$, which does not change the number of possible solutions from $(n-1)^{n-1}$.
EDIT 2: Solved. My logic above was ignoring the fact that there are $n$ choices for the element $i$ where $f(i) = i$. My solution above finds the number of functions $f : [n] \rightarrow [n]$ where $(n-1)$ elements satisfy $f(j) = j$, and one fixed $i \in n$ always satisfies $f(i) = i$. By multiplying this number, $(n-1)^{n-1}$, by the number of choices for the fixed element $i$, we get $n \times (n-1)^{n-1}$, which is the answer.
AI: HINT since this seems like homework.
Pick a specific $i$ for which $f(i) = i$. Now for each of the remaining $f(j)$ how many choices are there?
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H: $f(2-x)=f(2+x)$ and $f'(1/2)=0=f'(1)$. Find minimum number of roots of $f''(x)=0$.
Let $f$ be a non constant twice differentiable function satisfying $f(2-x)=f(2+x)$ and $f'(1/2)=0=f'(1)$. Find the minimum number of roots of $f''(x)=0$ in the interval $(0,4)$.
Answer: $4$
I managed to rewrite the given equation as $f(x)=f(4-x)$. Using this, I obtained
$$f'(x)+f'(4-x)=0$$
And
$$f''(x)=f''(4-x)$$
So it suffices to show that $f''(x)=0$ has at least $2$ roots in the interval $(0,2)$ but I'm unsure what's gonna happen at $x=2$ here. Also, I have $f'(7/2)=0=f'(3)$ from the given info and the first equation I obtained, but I don't know how to use this.
Any help would be great.
AI: Your function is symmetric about the line $x=2$. One property of a symmetric continuous function is that at the line of symmetry, the slope must be zero (can you see why?)
Hence, we have
$f'(1/2) = f'(1) = f'(2) = 0$
Now, using Rolle's theorem, there exists at least one point each in the intervals $(1/2,1)$ and $ (1,2)$ such that $f''(x) = 0$. For the absolute minimum, we can have 2 zeros. Now since that would be repeated for the mirror image, the answer should be $4$
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H: Negative arc length value. True or not?
I saw a post about this and someone said the arc length is an integral of a positive function, so it is positive.
But by solving this exercise I found the arclength as a negative value.
The arc length for $ y= ln (1-x^2)$ from x=0 to x=1/2.
The result I found is $L= -ln(2)-ln(3/2)+1/2 = -0.5986 $. I calculated it by myself and then I use an online program to calculate it and it gives the same number.
So, is it true? Or can't an arc length have a negative value?
AI: Your mistake was in taking the square root of
$$1+\left(\frac{dy}{dx}\right)^2 = \left(\frac{x^2+1}{x^2-1}\right)^2$$
This is $\dfrac{x^2+1}{1-x^2}$, not $\dfrac{x^2+1}{x^2-1}$, because $x^2-1$ is negative for $x\in[0,\frac12]$.
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H: $\forall x \in X: (y < x \implies y \in E) \implies x \in E$
Let $(X, \leq)$ be well-ordered and non-empty.
If $E \subseteq X$ satisfies
(i) $\min E \in X$
(ii) $\forall x \in X: ((y < x \implies y \in E) \implies x \in E)$
Then $E=X$.
Proof: Assume $E \neq X$. Then we can put $x:= \min X \setminus E$. If $y < x$, then $y \notin X\setminus E$, i.e. $y \in E$. By (ii) $x \in E$, a contradiction. Thus $E= X$.$\quad \square$
Note now that in this proof $(i)$ was not used. But $(ii)$ vacuously holds if $E=\emptyset$, while the theorem fails if $E=\emptyset$. Where does the proof implicitely use $(i)$ then? Why is $(i)$ important?
AI: Property (i) is superfluous, it follows from (ii). In fact (i) does not really make sense for $E = \emptyset$ and if $E \neq \emptyset$ then it is immediate from the fact that $E \subseteq X$. Property (ii) already guarantees $E \neq \emptyset$.
To see this, let $x = \min X$. Then we vacuously have for all $y < x$ that $y \in E$. So we must have $x \in E$.
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H: Calculate probability based on depth
Task: Calculate probability based on depth
$$f(x)=\begin{cases} \frac{1}{2} & x \in [1,3]\\\ 0 & x \not\in [1,3]\end{cases} $$
$$P([1,2])=?$$
My take on it:
$$ \int_3^1 f(x)dx= [ \frac{x}{2} ]^{3}_{1} = \frac{3}{2} - \frac{1}{2} = 1$$
Is this correct?
AI: That's not correct. You have to calculate $P(X \in [1,2]) = P(1 \leq X \leq 2) = \int_{1}^{2} f(x) \, dx = [x/2]_{1}^{2} = 1-1/2 = 1/2.$
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H: Convert English statements to logical forms
Everyone has a roommate who dislikes everyone
$ R(x,y)=$ $x$ and $y$ are roommates
$L(x,y)=$ $x$ likes $y$
so it becomes
$$\forall x \exists y (R(x,y) \land \lnot L(y,x))$$
Is this correct ? Thank you
AI: No, what you have written is that everyone has a roommate who dislikes them. You used $x$ twice, once for the person who has a roommate and once for the disliked person. The second "everyone" is everyone in the world, not just one of the roommates.
The correct answer is
$$\forall x \exists y \forall z (R(x,y) \land \lnot L(y,z))$$
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H: Floor function of a product
I'm reading a book about proofs and I'm currently stuck in this problem.
Prove that for all real numbers $x$ and $y$ we have that:
$$\lfloor x\rfloor \lfloor y\rfloor \leq \lfloor xy\rfloor \leq \lfloor x\rfloor \lfloor y \rfloor + \lfloor x \rfloor + \lfloor y \rfloor$$
I though I could do it by cases, considering when the number is a pure integer and when is an integer plus some real number. But by doing this I end up with having a lot of cases to show. Is there any better, simpler and clever approach? Thank you!
PS: I end up with a lot of cases because there is a point where I will have to consider "subcases" of cases specifically when the integer part is multiplying with the positive "rest" less than one
AI: I shall assume that $x, y \ge 0$. We also have $\lfloor x\rfloor, \lfloor y\rfloor \ge 0.$
Write $x = \lfloor x\rfloor + \{x\}$ and $y = \lfloor y\rfloor + \{y\}.$ Clearly, $0 \le \{x\}, \{y\} < 1$.
Note that
$$xy =\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor\{y\} + \lfloor y\rfloor\{x\} + \{x\}\{y\}. \quad (*)$$
Using the fact that all the rightmost three terms on the RHS of $(*)$ are nonnegative, we see that
$$xy \ge \lfloor x\rfloor\lfloor y\rfloor.$$
Note that the RHS is an integer which is lesser than $xy$. By definition of floor, $\lfloor xy\rfloor$ must be the greatest such integer. Thus, we have
$$\lfloor xy\rfloor \ge \lfloor x\rfloor\lfloor y\rfloor,$$
giving us the left inequality.
Using $(*)$ and the fact that $\{x\}, \{y\} < 1$, we see that
$$xy \le \lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y\rfloor + \{x\}\{y\}.$$
Since $\lfloor .\rfloor$ is an increasing function, we see that
$$\lfloor xy\rfloor \le \lfloor\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \{x\}\{y\}\rfloor. \quad (**)$$
Note that $\lfloor n + z\rfloor = n + \lfloor z\rfloor$ for any $n \in \Bbb Z$ and $z \in \Bbb R$. Using this, the RHS of $(**)$ can be written as
\begin{align}
&\lfloor\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \{x\}\{y\}\rfloor\\
=&\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \lfloor\{x\}\{y\}\rfloor\\
=& \lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor, \quad (\because 0 \le \{x\}\{y\} < 1)
\end{align}
giving us the right inequality.
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H: Does the complex field with dictionary order have the least-upper-bound property?
If any clarification is necessary, here is a little definition of what "dictionary" order encompasses. In Rudin's book it was stated in the following way.
Let $z=a+bi$ and $w=c+di$ where $z,w$ are arbitrary complex numbers and $a,b,c,d$ are reals.
$z<w$ if $a<c$ or if $a=c$ and $b<d$.
It is not hard to prove that this definition of ordering does indeed turn the complex field into an ordered set, but there is another question that is posed in the book and it is about the least upper bound property of this set. More specifically, I would like to know if the complex number set with order defined in this way does actually have the least upper bound property.
Intuitively speaking, I think that it should, because even the fact that the complex number set becomes an ordered set in this setting relies pretty much only on the fact that reals are an ordered set. I do not know, however, how to construct a fairly decent proof of this, or even how to approach this question a bit more formally.
Any help would be much appreciated.
AI: The complex numbers $\mathbb C$ with dictionary order do not have the l.u.b. property.
Consider the following subset of $\mathbb C$: $\{a+bi|a\le0\}$.
In the dictionary order, any $x+yi$ with $x>0$ is an upper bound of this subset,
and any $x+yi$ with $x\le0$ is not an upper bound of this subset.
But you can't say there is any least upper bound $x+yi$ with $x>0$,
because $\dfrac x2+yi$ is a smaller upper bound.
|
H: get 2 numbers and return the second number with the sign of the first number
I know it's more about programming, but I can use arithmetic operators only so I think it fits the math community.
AI: If $x$ is not zero, its sign is $|x|/x$. Thus you'd just need to return $|xy|/x$, where $x$ is the first number, and $y$ is the second number (assuming we don't care about the second number's sign).
E.g. $x=5$ and $y=-3$ gives $|5(-3)|/5 = 3$.
If one of them is zero, I guess it would depend on the application. Do you have a positive/negative zero?
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H: How to re-write this recurrence in linear form?
I have the following recurrence
$$T(n) = \frac{1 - T(n - 1)}{T(n - 2)}$$
and $T(0) = a, T(1) = b$. I re-wrote it as such
$$T(n)T(n - 2) = 1 - T(n - 1)$$
I am now trying to re-write this in linear form by making some sort of substitution. What could I do to make this recurrence linear?
AI: This isn’t exactly an answer to your question, but I think it’s relevant. I doubt there will be a nice way to express $T$ linearly, because it turns out that $T$ is actually periodic. This becomes clear after calculating the first couple of terms:
$$\color{green}{T(2)=\frac{1-b}{a}}$$
$$\color{green}{T(3)=\frac{a+b-1}{ab}}$$
$$T(4)=\frac{1-a}{b}$$
$$T(5)=\frac{a+b-1}{ab}$$
$$\color{green}{T(6)=\frac{1-b}{a}}$$
$$\color{green}{T(7)=\frac{a+b-1}{ab}}$$
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H: What is meant by this indicator function?
In Hastie et al. (2009) p.509, it is written that $N_k=\sum_{i=1}^N I(C(i) = k)$. To my understanding, $I$ should be an indicator function.
However $C(i)$ is defined as "Each observation is assigned to one
and only one cluster. These assignments can be characterized by a many to-one mapping, or encoder $k = C(i)$, that assigns the $i$th observation to
the $k$th cluster".
How is the first expression an indicator function then? I don't see any true/false condition within the $I$.
AI: $N_k=\sum_{i=1}^N I(C(i) = k)$ is counting the number of observations assigned to the $k$th cluster. It is the number of times $C(i) = k$ is true (meaning that observation $i$ is in cluster $k$) for a particular cluster $k$, iterating over all of the observations $i = 1, \dots, N$.
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H: Expected number of red balls in urn of two-coloured balls with color substitution.
There are $m$ black and $n$ red balls in an urn. One randomly picks a ball from the urn. If the picked ball is the black one, person changes it with the red ball and returns to the urn. If the picked ball is red, he does nothing and just returns the ball to the urn.
The questions are as follows:
Find the expected value of number of red balls in the urn after $k$ iterations.
Find the variance of number of red balls in the urn after $k$ iterations.
Also, there was given a note that answer are compact expressions, i.e. they do not contain summation signs.
I tried solving this problem in rather straightforward way, made a few steps and realised it would be really complicated to go that way. So, let $\xi$ be our random variable of number of red balls after $k$ iterations, and let $N = n + m$.
If there was no black balls picked, namely, $k$ red balls are taken out in a row, the probability is simple to write:
$$P(\xi = n) = \left( \frac{n}{N} \right)^k$$
If there is one black ball picked. Let $j$ be a position where the black ball appeared. This means that at from $1$-st to $(j - 1)$-th positions there were sampled only red ball with probability $\frac{n}{N}$, and from $(j + 1)$-th to $k$-th there were again red balls but with probability $\frac{n + 1}{N}$ altogether:
$$P(\xi = n + 1) = \sum\limits_{j = 1}^k \left( \frac{n}{N} \right)^{j - 1} \frac{m}{N} \left( \frac{n + 1}{N} \right)^{k - j - 1}.$$
And I stucked already at this point as have no idea how to write this sum compactly, moreover, if in this manner star looking at $\xi = n + 2, \ldots, \xi = n + k$ there will be to many inner summations to handle.
I also tried to think about possible standard way of handling that kind of problems with indicators, but also failed creating one.
So, I think there should be a smarter solution. Would appreciate any help with this problem!
AI: For the Expected Value: Just use indicator values.
Of course all of the red ones are still red at the end, so define $X_1, \cdots, X_m$ to be the indicator variables for the $m$ balls that start out black. That is $X_1=1$ if black ball $\#1$ is red at the end and it $=0$ otherwise. Then, of course, All the $X_i$ have the same expectation so $$E=n+m\times E[X_1]$$
What is $E[X_1]$? Well, in order to stay black the ball must be missed in each of the $k$ selections. The probability of being missed in any given selection is $\frac {m+n-1}{m+n}$ so the probability it is missed in all $k$ selections is $\left(\frac {m+n-1}{m+n}\right)^k$. Thus $$E[X_i]=1-\left(\frac {m+n-1}{m+n}\right)^k$$ which implies that $$E=n+m\times \left(1-\left(\frac {m+n-1}{m+n}\right)^k\right)$$
Sanity checks: if $k=0$ then $E=n$ as it should. As $k\to \infty$ we have $E=n+m$, again as it should. And if $n=m=k=1$ then $E=1.5$ which is easily verified. More broadly, if $k=1$ and $m,n$ are arbitrary then this gives the correct answer of $n+\frac m{m+n}$.
For the Variance we can proceed along the same lines. Letting $$Z=X_1+\cdots +X_m+n$$ we see that $E=E[Z]$ and that the variance is given by $$E[Z^2]-E^2$$
To compute $E[Z^2]$ we remark that $$Z^2=X_1^2+\cdots +X_m^2+n^2+2n\sum X_i+2 \sum_{i<j}X_iX_j$$ $$=n^2+(2n+1)\sum X_i+2\sum_{i<j}X_iX_j$$
Where we have used the fact that $X_i^2=X_i$ for all $i$. The only tricky terms to compute are those of the form $E[X_iX_j]$ for $i\neq j$. To do that, we need to compute the probability that both the $i^{th}$ and the $j^{th}$ black balls are selected at least once. I think it is easier to compute the complement. The probability that neither are selected at all is just $$\left( \frac {m+n-2}{m+n}\right)^k$$ and the probability that one is selected and the other is not is $$2\times \left( \frac {m+n-1}{m+n}\right)^k\times \left(1-\left(\frac {m+n-2}{m+n-1}\right)^k\right)$$
The rest is just a matter of putting the pieces together. The algebra should be messy, but not all that difficult.
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H: A question on Lebesgue integral
Let $f\in L^1(\mathbb{R}^n)$. I want to prove that when $m(B-B’)\to 0$
$$\left|\int_{B’}f(x)dx-\int_{B}f(x)dx\right|\to 0.$$
In Riemann integral, if $f$ is absolutely integrable on $[a,b]$, then $|f|$ in bounded, namely $|f(x)|\le M$, on $[a,b]$, and hence we have
$$\left|\int_{B’}f(x)dx-\int_{B}f(x)dx\right|= \left|\int_{B’-B}f(x)dx\right|\le \int_{B’-B}|f(x)|dx\le M\cdot m(B’-B),$$ where $B\subset B’$. If $m(B-B’)\to 0$, then we will have $$\left|\int_{B’}f(x)dx-\int_{B}f(x)dx\right|\to 0.$$
But for Lebesgue integral, we know $|f|$ is not necessarily bounded and it’s only finite almost everywhere. In this case we cannot find the supremum; in fact, we even cannot necessarily find the essential supremum. For example,
$$f(x):=\left\{\begin{array}{l}
\dfrac{1}{\sqrt{|x|}},~~x\in[-1,0)\cup(0,1],\\
+\infty,~~x=0,\\
\dfrac{1}{x^2},~~x\in(-\infty,-1)\cup(1,+\infty).
\end{array}\right.$$
We can easily check that $f\in L^1(\mathbb{R})$ and of course $|f|$ is integrable on $[-1,1]$. But now we can see that $|f|$ is not bounded and even the essential supremum is $+\infty$. Now we cannot prove that in the same way.
How to do that? The claim seems to be correct but I don’t know how to prove it. Any help is appreciated. Thanks!
AI: HINT: for a simple function $s:=\sum_{k=1}^n c_k \mathbf{1}_{A_k}$ you have that
$$
\int_{C}s \mathop{}\!d\lambda =\int_{\mathbb{R}^n}\mathbf{1}_{C}s \mathop{}\!d \lambda = \int_{\mathbb{R}^n}\left(\sum_{k=1}^n c_k \mathbf{1}_{C}\mathbf{1}_{A_k}\right)\mathop{}\!d \lambda \\
=\int_{\mathbb{R}^n}\left(\sum_{k=1}^nc_k \mathbf{1}_{C \cap A_k}\right)=\sum_{k=1}^n c_k \lambda (A_k\cap C)
$$
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H: Show that $F(x)=\frac{x}{(1-x)^2}-\frac{2x}{(2-x)^2}=\sum_{n=0}^{\infty}n(1-2^{-n})x^n$
I've been working on a recent exercise question where I was asked to show that:
$$F(x)=\frac{x}{(1-x)^2}-\frac{2x}{(2-x)^2}=\sum_{n=0}^{\infty}n(1-2^{-n})x^n$$
Now I cansee that the infinite sum is a power series, which leads me to believe that I might argue that the Taylor series of a function is the power series for that function and then simply show that the taylor series for F(x) can be written as the sum seen on the far right. Now I've been able to show that the first term of the taylor series evaluated at x=0 is equivalent to the factor of the $x^1$ part of the infinite sum, but I'm confused as to how to go on from here.
I thought perhaps I might try to do some proof by induction, but that would require a generalized expression for the derivative of F(x) which doesn't exactly seem trivial.
I'd really appreciate some help as this has been puzzling me for hours.
AI: If $|x|<1$, then$$\sum_{n=0}^\infty nx^n=\frac x{(1-x)^2}.$$Therefore, if $|x|<2$,$$\sum_{n=0}^\infty n\left(\frac x2\right)^n=\frac{x/2}{(1-x/2)^2}=2\frac x{(2-x)^2}.$$In other words,$$\sum_{n=0}^\infty n2^{-n}x^n=\frac{2x}{(1-x)^2}.$$Can you take it from here?
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H: Is the theorem 5.7 given in Apostol's Mathematical Analysis correct?
I came across this theorem in Apostol's Mathematical Analysis and the proof makes sense and seems to be correct.
https://drive.google.com/file/d/1-sRgtgGlQ5N8h9Vq5eosftjcEbOavOv0/view?usp=drivesdk
However, I also encountered this counterexample in the book 'Counterexamples in Analysis' by Gelbaum and Olmstead. I think I am missing something here and these two might be different, but I am not able to say how.
https://drive.google.com/file/d/1-sNtOdB06WBH5_unbgdJSdpUEugK2IeS/view?usp=drivesdk
AI: This is a somewhat subtle point.
The theorem in the first link is saying that there is a neighborhood of $c$ such that if $x$ is in that neighborhood then $f(x)-f(c)$ has the same sign as $x-c$. This does not mean that $f$ is increasing on a neighborhood of $c$, which would amount to saying that there exists a neighborhood of $c$ such that if $x,y$ are both in that neighborhood then $f(x)-f(y)$ has the same sign as $x-y$.
The counterexample given in the second link shows that in general being differentiable on $(a,b)$ with $f'(c)> 0$ doesn't imply that $f$ is increasing on a neighborhood of $c$.
Note that being continuously differentiable with $f'(c)>0$ does imply that $f$ is increasing on a neighborhood of $c$, but the counterexample in the second link is not continuously differentiable.
To visualize this, it can help to graph $g(x)=x+2x^2$ and $h(x)=x-2x^2$; the graph of $f$ bounces back and forth between the graph of $g$ and the graph of $h$ faster and faster as you approach $x=0$. But now think about the graphs of $g$ and $h$ themselves: both $g$ and $h$ have the same sign as $x$ as long as $|x|<1/2$. Therefore the same is true of $f$.
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H: If $A$ is finite and $f: A \rightarrow A$, $f$ is injective iff $\text{ran}f = A$: How Simple is My Proof Allowed to Be?
Synopsis
This exercise seems so obvious that I don't know how to put it into words. A simple diagram shows this to be true. But I've tried to write down a proof anyways, and I was wondering if any of you would be willing to check that its acceptable. In other words, if you were a teacher, would you accept my proof?
Exercise
Assume that $A$ is finite and $f: A \rightarrow A $. Show that $f$ is one-to-one iff $\text{ran}f = A$.
Proof
Suppose that $f: A \rightarrow A$ is one-to-one. Then every element in $A$ must be mapped to some other unique element in $A$. Since $A$ is finite, this implies that $\text{ran}f = A$. Now suppose $\text{ran}f = A$. Then by the pigeonhole principle, in order for $f$ to be a function, $f$ must be injective.
AI: The first one's fine, but for the second one, being more precise is a better option.
Assume for the other way that $\text{ran} ~ f =A$. As $A$ is finite, we can let $|A|=n$. Suppose there is a $x \in A$ such that there are $x_1,x_2 \in A$, $f(x_1)=f(x_2)=x$. Then there are $n-2$ elements in the domain, while there are $n-1$ in the range, and as the image of of an element is unique, by the pigeonhole principle, there must be at least one element in $A$ which does not possess a pre-image, contradicting our assumption that $\text{ran} f = A$
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H: Tensoring with projective module preserves injectivity.
Let $A$ be a ring, $P$ a projective left $A$-module and $E,F$ two
right $A$-modules. If $u:E\rightarrow F$ is an injective homomorphism,
the homomorphism $$u\otimes 1_P:E\otimes_A P\rightarrow F\otimes_A P$$
is injective.
Attempt:
Since $P$ is projective, there exists a free $A$-module $L$ with submodules $R,Q$ such that $\phi:R\oplus Q\rightarrow L,\,(r,q)\mapsto r+q,$ is an $A$-module isomorphism and there exists an $A$-module isomorphism $f:P\rightarrow R$. Thus the mapping
$$h:P\oplus Q\rightarrow L,\,(p,q)\mapsto f(p)+q,$$
is an $A$-module isomorphism. Furthermore, the mapping
$$g:E\otimes_A(P\oplus Q)\rightarrow(E\otimes_AP)\oplus(E\otimes_A Q)$$
such that $g(x\otimes(p,q))=(x\otimes p,x\otimes q)$, for $x\in E$ and $(p,q)\in P\oplus Q$, is a $\mathbf{Z}$-module isomorphism. Similarly
$$g':F\otimes_A(P\oplus Q)\rightarrow(F\otimes_A P)\oplus(F\otimes_A Q)$$
such that $g'(y\otimes(p,q))=(y\otimes p,y\otimes q)$, for $y\in F$ and $(p,q)\in P\oplus Q$, is a $\mathbf{Z}$-module isomorphism.
On the other hand $u\otimes 1_P:E\otimes_A P\rightarrow F\otimes_A P$ and $u\otimes 1_Q:E\otimes_A Q\rightarrow F\otimes_A Q$ are $\mathbf{Z}$-module homomorphisms. Hence
$$(u\otimes 1_P)\oplus(u\otimes 1_Q):(E\otimes_A P)\oplus(E\otimes_A Q)\rightarrow(F\otimes_A P)\oplus(F\otimes_A Q)$$
is a $\mathbf{Z}$-linear mapping. It follows that $\left[(u\otimes 1_P)\oplus(u\otimes 1_Q)\right]\circ g=g'\circ(u\otimes1_{P\oplus Q})$.
I don't know how much of this is useful. What should be the strategy here?
AI: Recall that every projective $A$-module is flat, hence by definition of flatness, we have that $E \otimes_A P \to F \otimes_A P$ is an injective group homomorphism. We will prove it in two steps.
First, we prove that for any collection $\{M_i\}_{i \in I}$ of (left) $A$-modules, we have that $M = \oplus_{i \in I} M_i$ is flat if and only if $M_i$ is flat for all indices $i.$
Proof. Given any (right) $A$-module $N,$ we have a natural isomorphism $N \otimes_A (\oplus_{i \in I} M_i) \cong \oplus_{i \in I} (N \otimes_A M_i).$ Consequently, if $\varphi : E \to F$ is an injective map of (right) $A$-modules, then the following diagram commutes. $$\require{AMScd} \begin{CD} E \otimes_A (\oplus_{i \in I} M_i) @>\varphi \otimes_A 1_M>> F \otimes_A (\oplus_{i \in I} M_i) \\ @VVV @VVV \\ \oplus_{i \in I} (E \otimes_A M_i) @>\oplus_{i \in I}(\varphi \otimes_A 1_{M_i})>> \oplus_{i \in I} (F \otimes_A M_i) \end{CD}$$ Considering that the vertical maps are isomorphisms, it follows that $M$ is flat if and only if $\varphi \otimes_A 1_M$ is injective if and only if $\oplus_{i \in I}(\varphi \otimes_A 1_{M_i})$ is injective if and only if $\varphi \otimes_A 1_{M_i}$ is injective for all indices $i$ if and only if $M_i$ is flat for all indices $i.$
We may now prove that every projective $A$-module is flat.
Proof. Certainly, $A$ is a flat $A$-module. By the previous claim, we have that $\oplus_{i \in I} A$ is flat. Consequently, every free $A$-module is flat. Considering that projective $A$-modules are direct summands of free $A$-modules, it follows that projective modules are flat.
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H: I do not understand an inequliaty that is used to prove the ratio test
I do understand:
$\exists q\lt1:\forall k\in\Bbb N_{0}:\lvert\frac{a_{k+1}}{a_{k}}\rvert \leq q$ is equivalent to
$\lvert a_{k+1}\rvert \leq q \cdot \lvert a_{k}\rvert$
Now the statement is that $
\sum \limits_{k=0}^{\infty}a_{k}$ converges absolutely.
The proof is what I do not understand:
"Inductively you can show:
$\forall k \in\Bbb N_{0}: \lvert a_{k}\rvert \leq q^{k}\cdot\lvert a_{0}\rvert$ "
How do we go from $$\lvert a_{k+1}\rvert \leq q \cdot \lvert a_{k}\rvert$$ to $$\lvert a_{k}\rvert \leq q^{k}\cdot\lvert a_{0}\rvert$$
And how did $q^{k}$ happen?
AI: "Inductively"
You have proven that for ever $k$ that $|a_{k+1}| \le q\cdot|a_{k}|$.
So that would mean $|a_1| \le q\cdot |a_0|$. But $|a_2| \le q \cdot|a_1|$.
And $|a_1| \le q\cdot |a_0|$ so $|a_2| \le q\cdot|a_1| \le q\cdot (q\cdot |a_1|) = q^2\cdot |a_0|$
And $|a_3| \le q\cdot |a_2| \le q\cdot (q^2|a_0|)=q^3\cdot |a_0|$.
And so on.
If you have $|a_{k}| \le q^{k}|a_0|$ then $|a_{k+1}| \le q\cdot |a_k|\le q(q^k\cdot |a_0|)= q^{k+1}|a_0|$.
That's a proof by induction.
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H: Constructing a model of ZFC in ZFC+Con(ZFC).
In ZFC+Con(ZFC), one can prove ZFC is consistent, so has a model. But how can I construct a model of ZFC in ZFC+Con(ZFC) formally?
I think the Con(ZFC) only says "the ZFC is consistent", not a technical statement, so if we formally construct a model of ZFC in ZFC+Con(ZFC), then there might be only axioms of ZFC in the way of constructing the model, so it can be also constructed in ZFC.
It contradicts the Goedel's second incompleteness theorem. Therefore, we cannot formally construct a model of ZFC even in ZFC+Con(ZFC). Is this argument right?
Thank you.
AI: No, that argument isn't right. $\mathsf{ZFC}$ proves the completeness theorem, and so in particular does indeed prove "$Con(\mathsf{ZFC})$ implies $\mathsf{ZFC}$ has a model."
Looking at the proof of the completeness theorem will demystify this: the proof actually constructs a model of the consistent theory in question, and the point is that that construction can be performed in $\mathsf{ZFC}$.
What this won't do is construct a transitive model of $\mathsf{ZFC}$. Indeed, the theory $$\mathsf{ZFC}+Con(\mathsf{ZFC})+\mbox{ "$\mathsf{ZFC}$ has no transitive model"}$$ is consistent (as long as $\mathsf{ZFC}+Con(\mathsf{ZFC})$ is). To see this, let $M\models\mathsf{ZFC}+Con(\mathsf{ZFC})$ and suppose $M\models$ "$\mathsf{ZFC}$ has a transitive model" (otherwise we're done). Let $\alpha$ be the least $M$-ordinal such that $M\models$ "$(L_\alpha)^M\models\mathsf{ZFC}$." Then $L_\alpha^M\models$ "$\mathsf{ZFC}$ has no transitive model" since otherwise (the $L$ of) that model would give us a level of $M$'s $L$-hierarchy below $\alpha$ which ($M$ thinks) is a transitive model of $\mathsf{ZFC}$.
So the models that the completeness theorem lets us construct from mere consistency are not particularly nice. But, they are models!
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H: Prove that $ \lim_{n\to\infty}\frac{\left(n!\right)^{2}2^{2n}}{\left(2n\right)!\sqrt{n}}=\sqrt{\pi} $
Prove that $ \lim_{n\to\infty}\frac{\left(n!\right)^{2}2^{2n}}{\left(2n\right)!\sqrt{n}}=\sqrt{\pi} $
What i want to do is to reach the form $ \sqrt{2\left(\prod_{k=1}^{n}\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)\right)} $
and then I want to use Wallis product, which will bring me to the result I seek. But I cant see how to reach this form. Any ideas would help.
AI: We have
\begin{align*}
\frac{{n!^2 2^{2n} }}{{(2n)!\sqrt n }} & = \frac{{2^n n!2^n n!}}{{(2n)!\sqrt n }} = \frac{{(2 \cdot 4 \cdots (2n))(2 \cdot 4 \cdots (2n))}}{{1 \cdot 2 \cdots (2n)\sqrt n }} \\ & = \frac{{2 \cdot 4 \cdots (2n)}}{{1 \cdot 3 \cdots (2n - 1)\sqrt n }} = \sqrt {\frac{{2 \cdot 2}}{{1 \cdot 3}} \cdots \frac{{(2n)(2n)}}{{(2n - 1)(2n + 1)}}\frac{{2n + 1}}{n}} .
\end{align*}
Now note that
$$
\frac{{2 \cdot 2}}{{1 \cdot 3}} \cdots \frac{{(2n)(2n)}}{{(2n - 1)(2n + 1)}} \to \frac{\pi }{2},\quad \frac{{2n + 1}}{n} \to 2.
$$
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H: Combinatorics: Why/how does this solution for counting three digit numbers work?
My homework book lists three different solutions for the exercise below, I used the most obvious one, (counting all possible configurations with one 2 and two 2's), but the book also gives this other (much shorter) solution. Sadly I do not understand what they are doing or why it works. Concretely I would like to know what they mean with: "the number of 3 digits numbers with all digits at least 2/3".
Exercise:
What is the probability that a random number of 3 (possibly identical) digits has a 2 as the
lowest digit? Note that N(S) = 1000 because numbers like 000 and 010 count too.
Shortest solution:
Consider the number of 3 digits numbers with all digits at least 2 (includes for example
367) and cancel all numbers with digits at least 3: 8^3 − 7^2 = 169. Thus: P(A) = 169/1000 = 16.9%.
AI: What's happening is you are taking the set of all strings with digits at least $2$, which is a prerequisite, and removing the subset that doesn't contain any $2$'s. You need at least one $2$ to be in the desired set, and need digits to be at least $2$. So you take the ones with all digits at least $2$ and remove the ones that don't have a $2$ in them.
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H: Is the inverse of an element in a group different than the element (except for $e$)? Also, are all the subgroup of a cyclic group cyclic?
Suppose the cyclic group $G=\{e,a,a^2,a^3,a^4,a^5,a^6,a^7,a^8,a^9,a^{10},a^{11}\}$ under some operation, say *.
$$G=\langle a\rangle$$
Now, each element must have an inverse, s.t. $b*b^{-1}=e$, and the inverse $b^{-1}$ is unique and different than $b$ except for $b=e$.
Here, the inverse of $a$ is $a^{11}$ and so on. However, the inverse of $a^6$ is $a^6$ (since $a^6*a^6=a^{12}=e$), which is my first issue here since $a^6\neq e$.
The second issue is that there is a theorem that says:
"Every subgroup of a cyclic group is cyclic."
Source: Contemporary Abstract Algebra - CH.4, Theorem 4.3: Fundamental Theorem of Cyclic Groups.
We now define $H=\{e,a^3,a^9\}$, which is a subset of G and also a subgroup (under *) [closed, associative, has an identity and all elements have inverse]. But H is not cyclic [needs to have $a^6$ to be so].
I would like to know why my example is incorrect, or that I did not quite understood the subject.
AI: It is not required that $b^{-1}\neq b$. In fact, $(a^6)^{-1}=a^6$. Your subset is not a subgroup because $(a^3)^2=a^6$, which is not in the set.
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H: Why the open-ball topology functor from $\boldsymbol{Met_c}$ to $\boldsymbol{Top_m}$ is not an isomorphism? Are they isomorphic at all?
Let $\boldsymbol{Met_c}$ denote the category of metric spaces whose morphisms are all continous maps. And let $\boldsymbol{Top_m}$ denote the category of metrizable topological spaces whose morphisms are all continous maps. According to the book Abstract and Concrete Categories (The Joy of Cats), the functor from $\boldsymbol{Met_c}$ to $\boldsymbol{Top_m}$ that associate with each metric space its induced topological space is an equivalence but not an isomorphism. Why it is not an isomorphism? Are they isomorphic at all?
AI: Different metrics on the same set can yield identical topologies (consider scaling a given metric). So the functor in question is not injective.
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H: The probability that a killed random walk on $[-N,N]$ escapes before dying
Let $X_t$ be a continuous time simple random walk in $\mathbb{Z}$ starting at $0$, let $\tau^*$ be an exponential r.v of parameter $1$. What is the probability
$$
\mathbb{P}(\tau ^* \ge \tau_{N})?
$$
Where $\tau_N = \inf \{t \in \mathbb{R}: X_t \in \{-N,N\} \}$.
My first attempt was to use the optional stopping theorem, but it didn't seem to be enough to extract the probability I am interested in.
EDIT: If an analytical representation is not possible, is there asymptotics as $N \to \infty.$
AI: Let $p_n$ denote the probability of escape before being killed when started from $n \in \mathbb{Z}$. We are interested in the value of $p_0$. By the memoryless property of the exponential distribution, conditioning on the first step we get
\begin{equation}
p_0 = \frac 13 p_1 + \frac 13 p_{-1}.
\end{equation}
More generally, for $-N + 1 \leq i \leq N-1$ we have
\begin{equation}
p_i = \frac 13 p_{i+1} + \frac 13 p_{i-1}
\end{equation}
with boundary conditions $p_N = p_{-N} = 1$. Solving this difference equation we get
\begin{equation}
p_n = \frac{\psi^n + \psi^{-n}}{\psi^N + \psi^{-N}} = \frac{\cosh(n \log \psi)}{\cosh(N \log \psi)}.
\end{equation}
where $\psi = \frac{3 + \sqrt{5}}{2}$. In particular, for $n=0$ we obtain
\begin{equation}
p_0 = \frac{1}{\cosh(N \log \psi)}.
\end{equation}
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H: How to end showing that a statement is true by using induction?
I am practicing how to show true statements using induction.
Since there are eventually different ways to show that a statement is true, I wanted to ask the following: When can you say that a true statement has been shown?
Example:
Prove that for any natural number n $\ge$ 2 we have $2^n > n+1$
Showing the base case is easy:
$$ 2^2 > 2+1 $$
But now we have the induction step and I wanted to ask whether it is okay to stop where I stopped.
$$ 2^{n+1} > (n+1)+1 $$
$$ 2^n * 2 > n+2 $$
$$ 2^n > \frac{n+2}{2}$$
$$ 2^n > \frac{n+2}{2}$$
$$ 2^n > \frac{n}{2} +1 $$
Can I know say that it is true because $n \in \mathbb{N}$ and $n \ge 2$?
Or do I have to transform the equation in another way? I sometimes do not know when something has been shown and when the transformation is not enough to conclude that the statement is true.
AI: Your proof is upside-down: you should start with something which is true and end up with what you want to prove
So instead say something like
Given $2^n > n+1$ and $n \gt \frac n2$
you have: $2^n > \frac n2 +1$
and multiplying both sides by two: $2^{n+1}> n+2$
and minor rearrangement: $2^{n+1} > (n+1)+1$
and with the base case you can say by induction ...
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H: finding the values for $x_1+x_2+x_3=5$ by restriction
Let $x_1+x_2+x_3=5$ and $1 \leq x_1 \leq 4$ , $0 \leq x_2 \leq 4$ , $0 \leq x_3 \leq 4$ .
How many $x_1$, $x_2$, and $x_3$ are there?
Firstly i found the all cases such that $C (4+3-1,4)=15$ but i stuck in the rest.I could not make inclusion exclusion part. Can you help me? The correct answer is $14$.
AI: You can use a generating function. The answer is the coefficient of $x^5$ in $$(x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4)^2,$$ which is the coefficient of $x^4$ in $$(1+x+x^2+x^3)(1+x+x^2+x^3+x^4)^2.$$ Expanding yields $$1+3x+6x^2+10x^3 + 14x^4 +\cdots,$$ so the answer is $14.$
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H: Weierstrass' M-test in reverse
Weierstrass' M-test says that the series of functions on some set $X$:
$$\sum_{n=1}^\infty f_n(x)$$
if $\forall n \in \mathbb{N}, \exists M_n$, \forall x\in X where $M_n \geq |f_n(x)|$, so the majorant series $\sum_{n=1}^\infty M_n$ converges, then the original series converges uniformly and absolutely for any $x\in X$.
But is the reverse true?
That is:
If I can prove that for some specific sequence of functions, any sequence of $M_n$ with $M_n\geq |f_n(x)|$, will have a divergent series $\sum_{n=1}^\infty M_n$.
Does that prove that the original sum is not uniformly convergent? can you give any counterexamples?
AI: No, that will not prove that the original sum is not uniformly convergent. For instance, let $f_n(x)=\frac1n\chi_{[n,n+1)}$. Then $\sum_{n=1}^\infty f_n$ converges uniformly to $f\colon\Bbb R\longrightarrow\Bbb R$ defined by$$f(x)=\begin{cases}0&\text{ if }x<1\\1&\text{ if }x\in[1,2)\\\frac12&\text{ if }x\in[2,3)\\\vdots\end{cases}$$But $\sup f_n=\frac1n$ and $\sum_{n=1}^\infty\frac1n$ diverges. So, if, for each $n$, $M_n\geqslant\sup f_n$, the series $\sum_{n=1}^\infty M_n$ diverges too.
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H: When the quotient $\frac {13n^2+n}{2n+2}$ is an integer?
Find the nonzero values of integer $n$ for which the quotient $\frac {13n^2+n}{2n+2}$ is an integer?
My Attempt
I assumed $(2n+2 )| (13n^2+n)$ implies existence of integer $k$ such that $$13n^2+n=k(2n+2)$$
$\implies\ 13n^2+(1-2k)n-2k=0$
$\implies\ n=\frac{(2k-1)\pm\sqrt{(1-2k)^2+104}}{26}$
But this is taking me nowhere.
AI: For the sake of future readers, @MikeDaas's method ought to be an answer. It can be simplified to$$2n+2|2(13n^2+n)+(12-13n)(2n+2)=24\implies n+1|12,$$i.e. the nonzero options are $$n\in\{-13,\,-7,\,-5,\,-4,\,-3,\,-2,\,1,\,2,\,3,\,5,\,11\}.$$Now we check which of these actually satisfy $2n+2|13n^2+n$; @Ravi found the positive solutions to be $2,\,3,\,11$, but the negative ones are $-13,\,-5,\,-4,\,-2$.
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H: What would the notation for this binary string look like?
Every block of 1's of length $\ge 4$ cannot be followed by a block of 0's of length $\ge 4$, and any block of 1s of length 1, 2 or 3 must be followed by a block of 0s whose length is congruent to 1 mod 4.
The answer I came up with is this:
$$\{0\}^*\{(1111)(1)^*\{0, 00, 000\}, \{1, 11, 111\}0(0000)^*\}^*$$
The problem is, it cannot produce some strings like $11111$. What would the proper notation look like.
AI: It appears to me that you’re just missing the possibility of terminating with a string of ones. You can fix that by adding one more piece:
$$\{0\}^*\{(1111)(1)^*\{0,00,000\},\{1,11,111\}0(0000)^*\}^*\{(1111)(1)^*,\lambda\}$$
Here $\lambda$ is the empty word.
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H: Dose convergence in integral and measure imply convergence in L
Suppose a sequence of function $f_n \mapsto f$ converges in measure and in integral. by integral I mean $\int f_n \,d\mu \mapsto \int f \,d\mu$. Does that mean they converge in L1. It seems not entire true to me e.g. consider $f$ with compact support whose positive value cancels with the negatives and moves right $+ 1$ for each $n.$ Any hints?
AI: Consider $$f_n(x)=\begin{cases} n &\text{ if }x\in \big[0,\frac{1}{n}\big],\\ -n &\text{ if }x\in \big[-\frac{1}{n},0\big]\end{cases}$$ Then $f_n\xrightarrow{\text{ almost everywhere }\implies \text{ in measure}} 0$. Also, $0=\int_\Bbb R f_n\longrightarrow \int_\Bbb R0=0$, but $\int_\Bbb R|f_n-0|=2$.
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H: Proving $A - B \subset A - (B - A)$
I believe I have been able to prove that for sets $A$ and $B$, $A - (B - A) \supset A - B$, but my proof is not particularly elegant. I was hoping someone knew of a more clever or straightforward way to show this. My proof is:
Let $x \in A - B$. Then $x \in A$ and $x \not \in B$. So $x \not \in \{y \mid y \in B, \; y \not \in A \}$, so $x \not \in B - A$. Since $x \in A$ and $x \not \in B-A$, $x \in A - (B - A)$, so $A - B \subset A - (B - A)$.
AI: Let $x$ be an arbitrary element. In the table below, let $0$ denote that $x$ is not in the set while $1$ denote that $x$ is in the set.
$$
\begin{matrix}
A \ & B \ & A-B \ & B-A \ & A-(B-A) \\
0 \ & 0 \ & 0 \ & 0 \ & 0 \\
0 \ & 1 \ & 0 \ & 1 \ & 0 \\
1 \ & 0 \ & 1 \ & 0 \ & 1 \\
1 \ & 1 \ & 0 \ & 0 \ & 1
\end{matrix}
$$
Note that corresponding to every $1$ (in fact the only one $1$) in the column headed $A-B$, we have a $1$ in the column headed $A-(B-A)$, and there is a $1$ in the column headed $A-(B-A)$ for which the corresponding entry in the column headed $A-B$ is a $0$.
This shows that, for any sets $A$ and $B$, we always have
$$
A - B \subset A-(B-A),
$$
but there are examples when
$$
A - (B-A) \not\subset A-B.
$$
For example, let us consider the sets
$$
A = \{ p \} = B.
$$
Then $A-B = \emptyset = B-A$, but
$$ A-(B-A) = A - \emptyset = A \neq \emptyset. $$
In the table above, the discrepency between the columns headed $A-B$ and $A-(B-A)$ occurs in the very last line where our arbitrary element is in both the sets $A$ and $B$. And, this precisely is the hint we have used in constructing our counter-example above.
Hope this helps.
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H: If $\operatorname{ker} T \neq \{0\}$, there is $S: V \rightarrow V$, $S\neq 0$, with $T \circ S = 0$.
Let $V$ be a vector space and $T$ be linear transformation $T: V \rightarrow V$ . If $\operatorname{ker} T \neq \{0\}$, prove that there's a linear transformation $S: V \rightarrow V$ such that $S$ is not the zero transformation but $T \circ S = 0$.
is there any sentence or a any direction that anyone can give me ? I have no clue what to do(I could prove that The second side will prove the $\operatorname{ker} T$ is not zero but i dont have any idea to prove the first side...}
AI: Hint 1: $T \circ S = 0$ if and only if for every $x \in V$, $S(x) \in \ker T$.
Hint 2: For any vector spaces $U,V$ for which $U \neq \{0\}$ and $V \neq \{0\}$, there exists a linear transformation $S: U \to V$ with $S \neq 0$.
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H: Why traditional equation solving method gives different answer for some problems?
This question arose when I tried solving one of the math problems below,
Question: The average monthly income of Rakesh and Suresh is Rs. 5050. The average monthly income of Suresh and Ramesh is Rs. 6250 and the average monthly income of Rakesh and Ramesh is Rs. 5200. What is the monthly income of Rakesh?
Answer:
Rakesh +Suresh (total income) = 5050 x 2 = 10100.... (i)
Suresh +Ramesh (total income) = 6250 x 2 = 12500.... (ii)
Rakesh+ Ramesh (total income) = 5200 x 2 = 10400.... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500.... (iv)
Subtracting (ii) from (iv), we get P = 4000.
So, Rakesh's monthly income = Rs. 4000
When I tried solving the above problem with this method shown below, it gives different Answer Why?
My Answer: Let Rakesh = x, Suresh = y , Ramesh = z. Then the above statements can be written as,
x+y = 5050 --1st equation
y+z = 6250 --2nd equation
z+x = 5200 --3rd equation
Solving the 1st and 2nd equations we get x-z = -1200 Which can be
written x= z-1200 Which is substituted in the 3rd equation, So that z
+ (z-1200) = 5200 ==> 2z = 6400 ==> z = 3200 Now substitute z = 3200 in 3rd equation, we get x = 2000 which is Rakesh's Monthly Income.
Why I get the wrong answer 2000 instead of the right answer 4000 which follows a different method?.
AI: The system of equations should be the following:
$$x+y=10100$$
$$y+z=12500$$
$$z+x=10400$$
You forgot to multiply the incomes by $2$ because they're the averages.
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H: If $u(x,y)$ is harmonic, how do I show that $u(z(x,y),w(x,y))$ is harmonic?
Suppose that $u(x,y)$ is an harmonic function. Consider the transformation:
\begin{align}
z &= c + x\cos\theta + y\sin\theta, \\
w &= d - x\sin\theta + y\cos\theta,
\end{align}
where $d$, $c$, and $\theta$ are constants
Show that $u(z,w)$ is also an harmonic function ($u_{zz}+u_{ww}=0$).
It seems pretty obvious, but I am really struggling to find a way to show that $u(z,w)$ solves the equation.
AI: One way of doing this is -
Solve for $x,y$ in terms of $z,w$
As $\frac{\partial f}{\partial z} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}$, write $u_{zz} + u_{ww} $ in terms of $x,y$.
You should then have your answer.
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H: show that $T$ and $T^*$ have different eigenspaces
I have the matrix $[T]_{E}^{E}$= $\left(\begin{matrix}
1 & 1 \\
0 & 1
\end{matrix}\right)$
$V=\mathbb{R^2}$ with the standard inner product, I need to show that $T$ and $T^*$ have different eigenspaces. so for $T$ it was easy I found the eigenvalue $\lambda=1$ and its eigenspace is $SP{(1,0)}$
but now I am trying to find the eigenspace of $T^*$ for $\lambda=1$ and I don't understand how I should go from $T$ to $T^*$
AI: Notice that
$$
[T^*]_E^E = \begin{pmatrix}
1 &0 \\ 1 & 1
\end{pmatrix}.
$$
The corresponding eigenspace of $T^*$ is $\mathrm{span}\{(0,1)\}$, which is not the eigenspace of $T$
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H: Find $\iint_{U} \frac{x}{y+x^{2}}$
Could you help me with the following please:
Evaluate
$$\iint_{U} \frac{x}{y+x^{2}}dxdy$$
where $U$ is limited by $x = 1$, $y=x^{2}$, $y=4-x^{2}$. Suggestion consider $x=\sqrt{v-u}$ and find $y$ as a function of $u$ and $v$ and apply change of variable.
I have plotted the region and searched how to find $y$, to obtain the transformation and to be able to calculate the Jacobian, if you could please help me find it $y$ or if you could give me some advice, thank you.
AI: $$\int_{1}^{\sqrt 2}dx \int_{x^2}^{4-x^2}\frac{x}{y+x^{2}}dy$$
imho is more easy direct integrating.
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H: Permutations/Combinations with Categories
I have 3 categories. Each of these categories has three options.
Category 1 options: A, B, C
Category 2 options: D, E, F
Category 3 options: G, H, I
So a permutation of these may look like this: A E H
I am trying to find the number of permutations with these categories.
This looks like a special kind of permutations problem and I am not sure how to approach it, can anyone help please?
AI: There are $3$ ways to choose the letter from each category, and then $3!$ ways of permuting them, giving a total of $$3^3\times 3!=162$$ ways.
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H: Is the pushforward of flow equal to its time derivate?
Let $X \in \Gamma(M)$ with $\Phi^{X}_{t}(p)$ as the one parameter group at a point $p \in M$. I believe that in general the equality $$T_p\Phi^{X}_{t_0}(X_p)=X_{\Phi^{X}_{t_0}(p)}=(\frac{\partial}{\partial t}\Phi^{X}_{t}(p))(t_0)$$ (i.e. $(\Phi^{X}_{t_0})_*(X_p)=X_{\Phi^{X}_{t_0}(p)}$) does not hold. I'm looking for a simple counterexample. Thanks in advanced.
Adendum: I want to give a bit of context to the question. While studying the Lie derivative, I wondered whether the vector field (which generates the flow) would be invariant under the pushforward of the one parameter group.
As shown below this is not the case. However I do wonder under which conditions this should indeed hold. Say, for complete flows.
AI: EDIT: Yes, your formula is correct. What's going on here is the computation of the Lie derivative $\mathscr L_XX$, which is always $0$. Of course, for a general vector field $Y$, $\mathscr L_XY$ will no longer be $0$.
My original counterexample is now a non-counterexample:
Take $X = x^2\dfrac{\partial}{\partial x}$ on $\Bbb R-\{0\}$. Then
$\Phi_t(x) = \dfrac x{1-tx}$ for $x\ne 0$ and for $|t|<1/|x|$. Then $\dfrac d{dx}\Phi_t(x) = \dfrac 1{(1-tx)^2}$. In your notation, $T_p\Phi_t = \dfrac1{(1-tp)^2}$. Applying this to $X_p = p^2\dfrac{\partial}{\partial x}$, we get $\dfrac1{(1-tp)^2}\cdot p^2 \dfrac{\partial}{\partial x} = \left(\dfrac p{1-tp}\right)^2 \dfrac{\partial}{\partial x} = X_{\Phi_t(p)}$, as predicted.
Here now is a proof of your formula: If we consider the curve
$$\alpha(t) = \Phi_{t+t_0}(p),$$
then $\alpha'(0) = X_{\Phi_{t_0}(p)}$. On the other hand, setting $\beta(t) = \Phi_t(p)$, then of course $\alpha(t) = \Phi_{t_0}\circ\beta(t)$, so by the chain rule
$$\alpha'(0) = T_p\Phi_{t_0}\beta'(0) = T_p\Phi_{t_0}X_p.$$
This is your equality.
(By the way, this hideous $T_p$ notation, rather than $Df$ or $f_*$, is very annoying. :))
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H: What is the derivative of the following integral?
Consider the vector $(a_M, a_S) \in R^2_{+}$. Vector components are drawn from a joint log-normal distribution $G(a_M, a_S)$ with mean $(\mu_M,\mu_S)$ and covariance matrix $\Sigma$.
Now, I have the following expression
$$N_M(W_M)=\int^{\infty}_0 \int^{a_MW_M}_0 a_M \ g(a_M, a_S) da_M da_S$$
where the second integral sign refers to the second $d$, i.e. we are integrating over all $a_S$ that are lower than the product $a_MW_M$.
And I am interested in its derivative, and Eric gave the answer below, equal to $$\int^{\infty}_0 a^2_M \ g(a_M, a_MW_M) da_M$$
Now, I am interested in the ratio between the two expression, meaning
$$\frac{\int^{\infty}_0 a^2_M \ g(a_M, a_MW_M) da_M}{\int^{\infty}_0 \int^{a_MW_M}_0 a_M \ g(a_M, a_S) da_M da_S}$$
Is there a way to simplify it?
AI: Assuming "$\mathrm{d}a_M \,\mathrm{d}a_S$" is intended as a product measure so that the intended computation using iterated integral ordering of the measure factors (so that the upper bound on the second integral is the dummy variable bound in the first integral instead of a different free variable) is
$$\frac{\mathrm{d}}{\mathrm{d}W_M} \int_0^\infty \; \int_0^{a_M \cdot W_M} \; a_M g(a_M, a_S) \,\mathrm{d}a_S \,\mathrm{d}a_M \text{,} $$and assuming the undefined function, $g$, is sufficiently well behaved that the order of integration can be reversed (for example, the hypotheses of the Fubini-Tonelli theorem, of Fubini's theorem, or of other theorems are satisfied), we apply the (first half of the) fundamental theorem of calculus with the chain rule to obtain
$$ {} = \int_0^\infty a_M^2 g(a_M, a_M W_M) \,\mathrm{d}a_M \text{.} $$
Of course, the weird notation may not be what I guessed above, so, to get responsive answers, you may need to augment your Question with the definition of the notation you used.
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H: LC differential equation
I want to solve the differential equation for the charge of an LC circuit:
$$d^{2} Q / d t^{2}+(1 / L C) Q=0$$
I create the characteristic equation:
$$\mu^{2}+\frac{1}{\mathrm{LC}}=0 \Rightarrow \mu^{2}=-\frac{1}{\mathrm{LC}} \Rightarrow \mu = \pm i \sqrt{\frac{1}{\mathrm{LC}}}= \pm i w_0\\$$
As a result, $$ q(t)=c_{1} e^{i\sqrt{\frac{1}{LC}}t}+c_2 e^{-i\sqrt{\frac{1}{LC}}t} \\
\Rightarrow q(t)=\left(c_{1}+c_{2}\right) \cos \left(w_{0} t\right)+i \cdot\left(c_{1}-c_{2}\right) \sin \left(w_{0} t\right)\\$$
But the charge is a real quantity, so: $$ q(t) \in R \Rightarrow c_{1}=c_{2} $$ Finally: $$ q(t)=2 c_{1} \cos \left(w_{0} t\right) \Rightarrow q(t)=C \cos \left(w_{0} t\right)
$$
My question is, how does the phase $\varphi$ occur in the equation?$$Q(t)=Q_{m} \cos \left(\omega_{0} t+\varphi\right)$$
AI: The fact that $q(t)$ is real does not imply that $c_1=c_2$. $c_1$ and $c_2$ are complex numbers such that $C_1=c_1+c_2$ and $C_2=i(c_1-c_2)$ are real numbers and
$$q(t)=C_1\cos \left(w_{0} t\right)+C_2 \sin \left(w_{0} t\right).$$
Note that
$$q(t)=Q_{m} \cos \left(\omega_{0} t+\varphi\right)=(Q_m\cos(\phi))\cos(\omega_{0} t)+(-Q_m\sin(\phi))\sin(\omega_{0} t)$$
and therefore
$$\begin{cases}
C_1=Q_m\cos(\phi)\\
C_2=-Q_m\sin(\phi)
\end{cases}$$
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H: Checking Optimization function whether its convex or not
The optimization function is defined as
$\frac{1}{2}.x^T.A.x$ where $A=\begin{pmatrix}
1 & 0.5 \\
0.5 & 1
\end{pmatrix}$
How to check if this is a convex or not? I know about the second derivative test and it gives
$A$ which should be greater than $0$ in order to be convex but what are we really checking here? how is $A$ compared against $0$?
AI: You have to check wheather $A$ (the hessian of the objetctive function $\frac{1}{2}x^T A x$) is positive semidefinit or not.
Here, $A$ is diagonally dominant and symmetric, which implies directly, that A is positive semidefinit, which implies, that your objective function $\frac{1}{2}x^T A x$ is convex.
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H: Discrepancy in calculating the volume formed by revolving the region above $y=x^3$ under $y=1$ and between $x=0$ and $x=1$ about the $x$-axis
Here's the problem:
Find the volume formed by revolving the region above $y=x^3$ under $y=1$ and between $x=0$ and $x=1$ about the $x$-axis
Now here's the integral I came up with:
$\int_{0}^{1}\pi\left(x^{3}-1\right)^{2}dx=2.02$, since the radius is $x^3-1$.
But when I plug it in Wolfram and others, I get $0.45$. What's wrong?
AI: You slices are not disks: they are annuli (disks with holes). The outer radius of the annulus at $x$ is $1$, and the inner radius is $x^3$, so you should be computing
$$\pi\int_0^1\left(1^2-(x^3)^2\right)\,dx=\pi\int_0^1(1-x^6)\,dx=\pi\left[x-\frac{x^7}7\right]_0^1=\frac{6\pi}7\;.$$
The volume that you computed in Wolfram is that generated by the region between the $x$-axis and the curve $y=x^3$ from $x=0$ to $x=1$, not the volume generated by the region between $y=x^3$ and $y=1$.
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H: how to solve this inhomogeneous second order differential equation
I want to solve this inhomogenenous differential equation of second order:
$$ x''+2x'+x= \sqrt{t+1}e^{-t} $$
with initial conditions $ x(0)= \pi $ and $ x'(0)= \sqrt{2} - \pi $
The solution is $ y= y_p+y_h $, so the particular solution added with the homogenous solution.
Because $ D:= 2^2-4 =0 $
the solution to the homogeneous part is $$ y(x)=(C_1+C_2x)e^{-x} $$
How do I solve the particular solution?
AI: The fact that $e^{-t}$ is a solution of the homogeneous equation might suggest the substitution $x(t) = e^{-t} u(t)$, which (after a bit of simplification) turns the differential equation into
$$ u''(t) = \sqrt{t+1} \ .$$
Now integrate twice.
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H: Extension of a map holomorphic on the unit disk to map holomorphic on the complex plane
I want to prove a statement regarding the holomorphic extension of a function holomorphic on the unit disk.
Let $D \subset \mathbb{C}$ be the (open) unit disk and $f: \bar{D} \to
> \mathbb{C}$ be a continous map, such that $f$ restricted to $D$ is
holomorphic and $|f(z)|=1$ for all $z \in \partial D$. Show that $f$
can be extended to a function holomorphic on $\mathbb{C}$ up to
finitely many isolated singularities, i.e. there are finitely many
points $z_1,..., z_n \in \mathbb{C}$ and a holomorphic function $g:
\mathbb{C} \setminus \{z_1,...,z_n \} \to \mathbb{C}$ such that $f=g$
on $D$.
Let $C_0=\partial D$. Then $C_0$ is the unit circle. Consider $f|_D: D \to \mathbb{C}$. Then $f|_D$ is holomorphic because $f$ is holomorphic. Let $K \subset C_0$ be a circular arc. Because $|f(z)|=1 $ for all $z \in \partial D$ it follows $f(K) \subset C_0$. Let $\sigma_0$ be the inversion with respect to $C_0$. Consider
\begin{align*}
g: D \cup K \cup \sigma_0(D), \ F(z)=\begin{cases}
f(z) & , \ z \in D \\
\sigma_0(f(\sigma_0(z))) & , \ z \in \sigma_0(G) \\
\sigma_0(f(\sigma_0(z))) & , \ z \in K
\end{cases}
\end{align*}
Then $f(z)=\sigma_0(f(\sigma_0(z)))$ for all $z \in K$. Because $f|_D$ can be extended to a function that is conitnous on $\bar{D}$ it follows from the Schwarz reflection principle that the function $g$ is holomorphic. The function $g$ has a singularity at $a \in \mathbb{C}$ if and only if $a\in \sigma_0(G)$ and $f(\sigma_0(a))=0$. Define $S:=\{z \in \sigma_0(D) \ | \ f(\sigma_0(z))=0 \}$. Then $S$ is the set of singularities of $g$. Suppose that $S$ contains infinitely many elements. Because $\sigma_0$ is bijective it follows that there are infinitely many elements $z \in D$ such that $f(z)=0$. Thus $f$ has infinitely many zeros. For every zero of order $m \geq 1$ there is a neighborhood and a holomorphic function $h$ such that $f(z)=(h(z))^m$ in that neighborhood.
But I do not see how to proceed.
AI: $f$ has only finitely many zeros because otherwise these zeros would have an an accumulation point in the closed unit disk.
But the zeros cannot accumulate in the interior of the disk because of the identity theorem. And they can also not accumulate at a point of the boundary because $|f(z)| = 1$ on the boundary.
If $z_1, \ldots, z_n$ is the (possibly empty) list of zeros of $f$ then $g$ can be defined in $\Bbb C \setminus \{ \sigma_0(z_1), \ldots, \sigma_0(z_n) \} $ as
$$
g(z) = \begin{cases}
f(z) & \text{ if $|z| \le 1$ } \\
\sigma_0(f(\sigma_0(z))) & \text{ if $|z| > 1$, $f(\sigma_0(z)) \ne 0$ } \\
\end{cases}
$$
As you said, $g$ is holomorphic in $\Bbb C$ save for poles at $\sigma_0(z_1), \ldots, \sigma_0(z_n)$.
In fact $g$ is a rational function because the singularity at $z = \infty$ is a pole or removable (depending on whether $f(0)$ is zero or not).
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H: Finite Abelian group and subgroup
Suppose $G$ is a finite abelian group and $H$ be a proper subgroup. Let $a$ be an element in $G$ not in $H$. Does there always exists an $m>0$ s.t. $a^m \in H$? If it is there what is the proof?
AI: This holds for any group $G$ and subgroup $H$ where $a \notin H$ has finite order: choose $m$ to be the order of $a$ so that $a^m = e \in H$.
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H: $2^X$ separable $\implies$ $X$ separable
Let $X$ be a $T_1$ space and let $2^X$ have the Vietoris topology. I know from an article that $2^X$ is separable only if $X$ is separable, but the article omits the proof as it is apparently obvious.
Given $D$ is a countable dense subset of $2^X$, how can I derive a countable dense subset $C$ of $X$?
I have no intuition for what $C$ should be. Candidates that I have tried end up being either not dense or not countable.
AI: For each $F\in D$ pick a point $x_F\in F$, and let $C=\{x_F:F\in D\}$. If $U$ is any non-empty open set in $X$, $\{H\in 2^X:H\subseteq U\}$ is open in $2^X$, so it contains some $F\in D$, and clearly $x_F\in U$.
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H: Evaluating volume of a sphere with triple integral in cylindrical coordinates
I need to evaluate $\iiint_V \frac{1}{x}\: dV$ where $V$ is the inside of a sphere given by $x^2+y^2+z^2=x$.
I write the equation as a sphere with centre in $\left(\frac{1}{2},0,0\right)$ and radius of $\frac{1}{2}$. Now I know that my sphere and sphere in $(0,0,0)$ with radius $\frac{1}{2}$ would have the same volume, but I'm confused about what do do with $\frac{1}{x}$.
I tried something like this:
$$x=\frac{1}{2}+\frac{1}{2} R\cos\phi$$
$$y=\frac{1}{2} R\cos\phi$$
$$z=\frac{1}{2z}$$ and using Jacobian of $|J|= \frac{1}{2R}$
Then I tried writing my coordinates as
$$x=x$$
$$y=R\cos\phi$$
$$z=R\sin\phi$$
I didn't mention my limits because I think the problem is with my choice of coordinates. Any hints on what the easiest substitution would be?
AI: Using the following spherical coordinates
$$\begin{cases} x = \rho \cos \phi \\ y = \rho\sin\phi\sin\theta \\ z = \rho \sin\phi\cos\theta \\ \end{cases} \implies J = \rho^2\sin\phi$$
we get that the boundary is given by
$$x^2+y^2+z^2 = x \implies \rho^2 = \rho \cos\phi \implies \rho = \cos\phi$$
which makes the integral
$$\iiint_E\frac{1}{x}\:dV = \int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^{\cos\phi}\rho \tan\phi \:d\rho \:d\phi \:d\theta$$
$$ = \pi \int_0^{\frac{\pi}{2}} \sin\phi\cos\phi\:d\phi = \frac{\pi}{2}$$
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H: Question about product of generating functions in a proof that for positive integer $n$, $\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1$.
I was recently looking at a copy of the Mathematics Magazine from 2004 and was reading Q944 (here). It asks this:
Show that for positive integer $n$, $$\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1.$$
The solution is here. Basically, if we let $S_n$ be the sum, we find that we can write $$S_n=\sum_{k=0}^na_{n-k}b_k,$$ where $a_k=(-1)^k\binom nk$ and $b_k=\binom{n+k}n$. Then we can find generating functions for $a_k$ and $b_k$. In particular, we find that $$\sum_{k=0}^na_kx^k=(1-x)^n$$ and $$\sum_{k=0}^\infty b_kx^k=\frac1{(1-x)^{n+1}}.$$
So far, all of this makes sense to me. Now for the final step of the solution, we note that $$\sum_{n=0}^\infty S_nx^n=\sum_{n=0}^\infty\left(\sum_{k=0}^na_{n-k}b_k\right)x^n=(1-x)^n\cdot\frac1{(1-x)^{n+1}}.$$
This last step doesn't really make a lot of sense to me. After all, aren't the generating functions for $a_k$ and $b_k$ dependent on $n$? And so, for example, don't we get that $a_1$ means different things depending on what $n$ is?
Sorry if I'm not being clear here--I'm having a bit of trouble formulating exactly what my confusion is. But, basically, if somebody could explain the last step in a bit more detail, that'd be fantastic.
AI: I’d carry out the last step a bit differently. What the first part shows is that $S_n$ is the coefficient of $x^n$ in the product
$$(1-x)^n\cdot\frac1{(1-x)^{n+1}}\;,\tag{1}$$
something that is often written
$$S_n=[x^n]\left((1-x)^n\cdot\frac1{(1-x)^{n+1}}\right)$$
with the $[x^n]$ operator. Clearly, then,
$$\begin{align*}
S_n&=[x^n]\left(\frac{(1-x)^n}{(1-x)^{n+1}}\right)\\
&=[x^n]\left(\frac1{1-x}\right)\\
&=[x^n]\sum_{k\ge 0}x^k\\
&=1\;.
\end{align*}$$
The $n$ in $(1)$ really does depend on which $S_n$ we’re computing, but $(1)$ simplifies to $\frac1{1-x}$ for all $n$, so in the end we really are looking at one power series.
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H: Contour Integral of $\frac{1}{1+e^z}$
How do I calculate $\displaystyle\int_{C_1(1)}(1+e^z)^{-1}\text{d}z$?
I have tried parametrizing $C_1(1)$ by $z=1+e^{i\theta}$ with $\theta \in [0,2\pi]$, but this does not help much as I would have a double exponential.
Any ideas on how to progress are welcome.
AI: That integral is equal to $0$, since the disk $D_\pi(1)$ is simply connected, the image of $C_1(1)$ is contained in it and the function$$\begin{array}{ccc}D_\pi(1)&\longrightarrow&\Bbb C\\z&\mapsto&\dfrac1{1+e^z}\end{array}$$is analytic.
I chose the disk $D_\pi(1)$ because, for each $z$ in that disk, $1+e^z\ne0$.
|
H: Series with limit
The problem is the following. Let $(a_{n})_{n\geq0}$ such that $\lim_{n\to\infty}a_{n}=\alpha\in\mathbb{C}$ and let $(b_{n})_{n\geq0}$ a succession of positive real numbers. We know that the series
$$
\sum_{n=0}^{\infty}b_{n}z^{n}
$$
converges for every $z\in\mathbb{C}$. Prove that, considering now $z\in\mathbb{R}$,
$$
\lim_{z\to+\infty}\frac{\sum_{n=0}^{\infty}a_{n}b_{n}z^{n}}{\sum_{n=0}^{\infty}b_{n}z^{n}}=\alpha.
$$
If one could switch the limit and the series, then it would be done, but it doesn't seem so easy. How can be proved the limit otherwise?
AI: Suppose $\alpha = 0.$ Let $\epsilon>0.$ Then there exists $N$ such that $|a_n|<\epsilon$ for $n>N.$ The quotient of interest can be written
$$\frac{\sum_{n=0}^N a_nb_nz^n + \sum_{n=N+1}^\infty a_nb_nz^n}{\sum_{n=0}^N b_nz^n + \sum_{n=N+1}^\infty b_nz^n}.$$
For $z>0,$ the absolute value of this is bounded above by
$$\frac{\sum_{n=0}^N |a_n|b_nz^n + \epsilon\sum_{n=N+1}^\infty b_nz^n}{\sum_{n=0}^N b_nz^n + \sum_{n=N+1}^\infty b_nz^n}.$$
Now divide top and bottom by $\sum_{n=N+1}^\infty b_nz^n$ and take the $\limsup$ as $z\to \infty.$ We see this $\limsup$ is no more than
$$\frac{0+\epsilon}{0+1} = \epsilon.$$
This takes care of the $\alpha=0$ case, and if you think about it, gives all other cases as well.
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H: Use of Bernoulli equation $x^2y' +2xy=y^3$
I am not sure what am I doing wrong here, I have $$x^2y'+2xy=y^3$$
I should apply the Bernoulli equation, so I mark $$z=\frac{1}{y^2}$$ $$z' = -\frac{2y'}{y^3}$$
$$-\frac{z'}{2} = \frac{y'}{y^3}$$
Now I just substitute $z$ and I get $$-\frac{x^2z'}{2}+2xz=1$$
How I should proceed from here? I am not recognize any pattern
AI: From that step you can write your equation in the form $$\frac{dz}{dx}+Pz=Q$$ where $P$ and $Q$ depend on $x$...
You get $$\frac{dz}{dx}-\frac{4}{x}z=-2x^{-2}$$
Your $P=-\frac{4}{x}$,
$Q=-2x^{-2}$
Now, $$ze^{\int-\frac{4}{x}dx}=\int{e^{\int-\frac{4}{x}dx}(-2x^{-2})dx}$$
So that $$zx^{-4}=-2\int{x^{-6}dx}$$.
|
H: quantiles of a monotonic function of two independent R.V.s
Consider two independent random Variables $X, Y$, and $X>=0, Y>=0$. $f(X, Y)$ is monotonic to $X, Y$ respectively.
Suppose that we know the 0%, 25%, 50%, 75%, 100% quantiles for both $X, Y$ and we have access to $f(\cdot, \cdot)$.
Is it possible to find some quantiles of the new R.V. $Z = f(X, Y)$?
Quantile is invariant to monotonic transformation but is there any nice property when we consider the scenario above?
I appreciate a lot for any hint. :)
AI: Well, if you know $100\%$ quantiles, i.e. you know $c$ and $d$ such that $X \le c$ and $Y \le d$ with probability $1$, then $Z = f(X,Y) \le f(c,d)$ with probability $1$. This says $f(c,d)$ is a $100\%$ quantile for $Z$. (I say a rather than the, because in general quantiles are not unique).
Similarly for $0\%$.
Not for any of the others, though: they would require knowing the actual distributions, at least in some interval.
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