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H: Product of $L^p$-convergent sequences are Cauchy
I'm working on showing:
If $\|f_n - f\|_p \to 0$ with $1≤p<\infty$ and $h_n \to h$ pointwise with $|h_n|<M$ for all $n$, then $\{f_nh_n\}$ is Cauchy in $L^p$
I've shown that $\|h_n-h\|_p \to 0$, but I'm having a hard time showing the product is Cauchy. I tried to manipulate the terms in the norm in additon to using Minkowski's inequality, but it hasn't worked thus far.
AI: Although $\|h_n - h\|_p$ may not converge to $0$, the dominated convergence theorem tells you that $\|f h_n - fh\|_p \to 0$. In particular $\{f h_n\}$ is Cauchy in the $L^p$ norm. Then apply the triangle inequality: for any $n$ and $m$ you have
\begin{align*} \|f_n h_n - f_m h_m\|_p &\le \|f_n h_n - f h_n\|_p + \|fh_n - fh_m\|_p + \|fh_m - f_m h_m\|_p \\
&\le K \|f_n - f\|_p + \|fh_n - fh_m\|_p + K \|f_m - f\|_p.
\end{align*}
All terms on the right tend to $0$ as $m,n \to \infty$.
On edit: you pointed out in the comments that you are actually trying to prove that $f_n h_n \to fh$ in $L^p$. You don't really need to go through Cauchy sequences to show this. Using the fact that $\|fh_n - fh\|_p \to 0$ you have $$\|f_n h_n - fh\| \le \|f_n h_n - f h_n\|_p + \|fh_n - fh\|_p \le K \|f_n - f\|_p + \|fh_n - fh\|_p \to 0.$$ |
H: What does $f:\mathbb R \rightarrow \mathbb R$ mean?
This is simply a basic notation question: what is the meaning of
$$f:\mathbb R \rightarrow \mathbb R$$
I imagine it's some sort of function to do with the set of real numbers, perhaps some sort of mapping. Until now I've only encountered functions of the form
$$f(x)=...$$
or $$f:x\mapsto...$$
Thanks in advance.
AI: It means that the function maps any real number (all real numbers are the domain) to one and only one real number (the codomain is a subset of $\mathbb R$).
For example the following notations are valid
$f:\mathbb R \rightarrow \mathbb R,\: f(x)=x^3$
$f:\mathbb R \rightarrow \mathbb R,\: f(x)=x^2$
$f:\mathbb R \rightarrow \mathbb R^+_0,\: f(x)=x^2$
$f:\mathbb R^+_0 \rightarrow \mathbb R^+_0,\: f(x)=\sqrt x$
and the following are wrong
$f:\mathbb R \rightarrow \mathbb R^+_0,\: f(x)=\sqrt x$
$f:\mathbb R \rightarrow \mathbb R,\: f(x)=\log x$
$f:\mathbb R \rightarrow \mathbb R,\: f(x)=\frac 1x$ |
H: Does the independence of the axiom of choice imply Gödel's incompleteness theorem?
I recently wrote this answer describing Gödel's completeness and incompleteness theorems, in which I came to the conclusion that a theory is (syntactically) complete if and only if all its models are elementarily equivalent, that is no formula in the theory can distinguish between two models of the theory.
The reason is that if for two models $\mathcal M,\mathcal M'$ with $\mathcal M\models\phi$ and $\mathcal M'\not\models\phi$, then neither $\phi$ nor $\neg \phi$ is provable by (semantic) completeness.
Since proving the independence of AC comes down to constructing a model of ZF which does not satisfy AC, is it correct to conclude that the independence of AC implies incompleteness of ZF?
This seems fishy to me because the incompleteness theorem requires some sort of nontrivial Gödel encoding, whereas the construction of the ZF+$\neg$AC uses a completely different technique.
AI: The answer depends on what you mean by "the incompleteness theorems". If all you mean is "$ZF$ is incomplete", then yes, the independence of $AC$ is enough to prove that $ZF$ is incomplete (though it's worth remembering that the consistency of $\neg AC$ was proved much later than Gödel's incompleteness theorems).
However, Gödel actually proved statements stronger than just "$ZF$ is incomplete". For example, the first incompleteness theorem tells you that (as long as $ZF$ is consistent) not only is $ZF$ incomplete, but you can't make it complete by adding any computably enumerable list of axioms to it. The second incompleteness theorem tells you specifically that (again, assuming that $ZF$ is consistent) one of the things $ZF$ can't prove is $Con(ZF)$. This is important because there are statements of interest in set theory (such as the consistency of large cardinals) that do imply $Con(ZF)$, and hence we know that $ZF$ can't prove that these statements are true (but remember, knowing that you can't prove $\sigma$ isn't the same thing as proving $\neg\sigma$!). |
H: How to compute the smallest face of a polytope containing a given point
Let $S\subseteq\mathbb{R}^d$ be a finite set and $P=conv(S)$ its convex hull so $P$ is a convex polytope. Let $p\in P$ be given. Note that if $F_1$ and $F_2$ are two faces of $P$ containing $p$, then $F_1\cap F_2$ is also a face of $P$ containing $p$. Thus, there exists a smallest face $F$ of $P$ containing $p$. Is there a polynomial time algorithm (in $n$ and $d$) that computes this face $F$? $F$ can be given as $conv(S_1)$ for some $S_1\subseteq S$.
I realized that if I can write $P=\{x\mid Ax\leq b\}$ then the rest follows. However, as I find out, computing this presentation of $P$ is called the Face Enumeration Problem and does not admit a polynomial time algorithm (unless I assume that $d$ is fixed).
AI: If $s \in S$, then $s \in F$ iff there is no $y \in \mathbb R^d$ with $y \cdot p \ge y \cdot x$ for all $x \in S$ and $y \cdot p > y \cdot s$. You can decide this using a polynomial-time algorithm for linear programming. |
H: All solutions of $f(w+z)=f(w)f(z)$, $f(1)=e$
Let $w=u+iv$ and $z=x+iy$ with $u,v,x,y\in\mathbb{R}$. Is the function
$$f:\, \mathbb{C}\to\mathbb{C},\, f(z)=e^x\cos y+ie^x\sin y$$
the only solution of the functional equation
$$f:\, \mathbb{C}\to\mathbb{C},\, f(w+z)=f(w)f(z),\, f(1)=e?$$
I was only able to prove that it is a solution:
$f(1)=e$ trivially and
$$\begin{align}f(w)f(z)&=(e^u\cos v+ie^u\sin v)(e^x\cos y+ie^x\sin y)\\&=e^{u+x}(\cos v\cos y-\sin v\sin y)+ie^{u+x}(\sin v\cos y+\cos v\sin y)\\&=e^{u+x}\cos (v+y)+ie^{u+x}\sin (v+y)\\&=f(w+z).\end{align}$$
AI: Without any additional assumption the answer is no, it's not the only solution. Consider for example $f(z)=e^{\overline{z}}$, or in your terms $f(z)=e^x\cos y-ie^x\sin y$. Another example could be $f(z)=e^{\Re(z)}$ (or $f(z)=e^x$), I guess you get the idea.
If you add an assumption for $f$ to be holomorphic (for example) then the solution is unique, see Find the holomorphic function satisfying an equation and agrees with $e^x$ in real line. |
H: $K(H)^{**}= B(H)$
Let $H$ be a Hilbert space. Is it true that we have an isometric linear isomorphism
$$B(H) \cong K(H)^{**}$$
where $K(H)^{**}$ is the bidual of the compact operators on $H$.
I think I proved this but I could not find a reference on the internet. Is the result true?
AI: Yes, this is completely standard and should appear in any operator theory book. One can easily show that $K(H)^*$ can be identified with $T(H)$ (the trace-class operators) via the duality
$$
\langle T,S\rangle=\operatorname{Tr}(TS).
$$
And similarly, $B(H)$ can be seen as the dual of $T(H)$ via the same duality.
To mention a few references:
Conway, A Course in Operator Theory: Section 1.19
Conway, A Course in Functional Analysis: Exercise IX.2.21
Farenick, Functional Analysis, Theorems 10.99 and 10.102
Reed-Simon, Methods of Mathematical Physics. Functional Analysis, Theorem VI.26
Davidson, C$^*$-Algebras by Example, Exercises I.18 and I.20
Murphy, C$^*$-Algebras and Operator Theory, Theorems 4.2.1 and 4.2.3
Takesaki, Theory of Operator Algebras, Theorem II.1.8 |
H: Strong Law of Large Numbers with randomly many summands
I'm wondering about a SLLN where the number of summands is also allowed to be a random variable.
More specifically assume that:
$\{X_i\}$ are iid with $\mathbb{E}[|X_1|] < \infty $
$N=\{N_n\}_{n\in\mathbb{N}}$ is a sequence of random positive integers
independent of $\{X_i\}$
$\lim_{n} N_n =\infty$ $a.s$
Can we conclude that $ \frac{1}{N_n} \sum_{i=1}^{N_n} X_i = \mathbb{E}[X_1] $ almost surely?
AI: This holds because in fact it is an easy consequence of the strong law of large numbers.
Let $\Omega$ be the underlying probability space. Define $Y_n = \frac{1}{n}\sum_{i = 1}^n X_i\,.$ By the strong law of large numbers, if we let $\Omega_1 = \{ Y_n \to E[X_1]\}$ then $P(\Omega_1) =1$. Similarly, define $\Omega_2 = \{N_n \to \infty\}$; then $P(\Omega_2) = 1$. Then $P(\Omega_1\cap \Omega_2) = 1$ and on the set $\Omega_1 \cap \Omega_2$ we have $Y_{N_n} \to E[X_1]$. Note that in fact you don't even need $N_n$ to be independent of $\{X_i\}$ just that $N_n \to \infty$. |
H: calculate $\oint_{|z|=1} \left(\frac{z}{z-a}\right)^n \, dz$
calculate $\oint_{|z|=1}\left(\frac{z}{z-a}\right)^{n}$
whereas a is different from 1, and n is integer.
My try:
\begin{align}
& \oint_{|z|=1}\left(\frac{z}{z-a}\right)^n \, dz\\[6pt]
& \oint_{|z|=1}\frac{z^n \, dz}{\sum_{k=0}^n z^k(-a)^{n-k}}\\[6pt]
& \oint_{|z|=1}\sum_{k=0}^n\frac{z^n \, dz}{z^k(-a)^{n-k}}\\[6pt]
& \oint_{|z|=1}\sum_{k=0}^n\frac{z^{n-k} \, dz}{(-a)^{n-k}}
\end{align}
I forgot here to write the binomial but it won't change the answer.
no singularities therefore it's $0.$
if $n$ is smaller than $0$:
\begin{align}
& \oint_{|z|=1}\left(\frac{z-a}{z}\right)^n \, dz\\[6pt]
& \oint_{|z|=1}\frac{\sum_{k=0}^n \binom n k z^k(-a)^{n-k} \, dz}{z^n}\\[6pt]
& \oint_{|z|=1}\sum_{k=0}^n
\binom n k \frac{z^{k}(-a)^{n-k}dz}{z^{n}}\\[6pt]
& \oint_{|z|=1}\sum_{k=0}^n \binom n k z^{k-n}(-a)^{n-k}\, dz
\end{align}
so if so we look $k-n=-1$
$k=n-1$
which is exactly $-na$
therefore the value of the integral is $-2n \pi i a$
AI: If $|a|>1$, and $n$ is a non-negative integer, then the function $f(z)=\left(\frac{z}{z-a}\right)^n$ is analytic on the disk $|z|\le 1$ and Cauchy Integral Theorem guarantees that
$$\oint_{|z|=1}\frac{z^n}{(z-a)^n}\,dz=0$$
METHODOLOGY $1$: For $|a|<1$
For $n\ge 0$, we can write
$$\left(\frac{z}{z-a}\right)^n=\left(1+\frac{a}{z-a}\right)^n=\sum_{k=0}^n \binom{n}{k}a^k(z-a)^{-k}$$
The coefficient on the $\frac{1}{z-a}$ term is $na$. So from the Residue Theorem, we have for $|a|<1$
$$\oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz=2\pi i na$$
For $n<0$, we can write
$$\left(\frac{z}{z-a}\right)^n=\left(1+\frac{a}{z}\right)^{|n|}=\sum_{k=0}^{|n|} \binom{|n|}{k}(-a)^k(z)^{-k}$$
The coefficient on the $\frac{1}{z}$ term is $-|n|a$. So from the Residue Theorem, we have
$$\oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz=-2\pi i |n|a=2\pi i n a$$
METHODOLOGY $2$:For $|a|<1$
If $n$ is a negative integer, then the function $f(z)=\left(\frac{z}{z-a}\right)^n$ has a pole of order $|n|$ at $z=0$. Hence, application of the residue theorem reveals
$$\begin{align}
\oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz&=\oint_{|z|=1}\left(\frac{z-a}{z}\right)^{|n|}\,dz\\\\
&=2\pi i \frac1{(|n|-1)!}\lim_{z\to 0}\frac{d^{|n|-1}}{dz^{|n|-1}}(z-a)^{|n|}\\\\
&=-2\pi i|n|a\\\\
&=2\pi i na
\end{align}$$
If $|a|<1$, and $n$ is a non-negative integer, then the function $f(z)=\left(\frac{z}{z-a}\right)^n$ has a pole of order $n$ at $z=a$ and application of the residue theorem reveals
$$\begin{align}
\oint_{|z|=1}\left(\frac{z}{z-a}\right)^ n\,dz&=2\pi i \frac1{(n-1)!}\lim_{z\to a}\frac{d^{n-1}}{dz^{n-1}}z^{|n|}\\\\
&=2\pi i n a
\end{align}$$ |
H: Can reflexive partial orders be defined from strict partial orders without using equality?
Consider a reflexive partial order $\leq$. We can define its strict counterpart without using equality, like so: $x \leq y$ $\land$ $\neg y \leq x$. But from a strict partial order $<$, can we define its reflexive counterpart without using any equality formulas in the definition?
AI: No, we cannot do this. (Below I'll write "$\mathcal{L}_{\mathsf{w/o=}}$" for the equality-free version of a logic $\mathcal{L}$.)
The key is that when we drop equality from our logical apparatus we get a notion of "near-isomorphism" which preserves all expressible structure: these are the maps which preserve and reflect all atomic formulas, and are surjective, but are possibly not injective.
To see where these come from, observe that when equality is built into our logical apparatus these are just the isomorphisms in the usual sense. Specifically, considering the atomic formula "$x=y$" we see that reflection of atomic formulas implies injectivity, and that gets us the usual notion of isomorphism. When equality is no longer part of our language, we have the possibility of non-injective maps of the above type. While these are absolutely not isomorphisms, they do suffice to demonstrate that the two structures in question are $\mathsf{FOL_{w/o=}}$-equivalent; this is a good exercise, and was previously used to answer another question of yours.
In fact, such maps establish far more than $\mathsf{FOL_{w/o=}}$-equivalence. For example, they give full $\mathcal{L}_{\infty,\infty\,\mathsf{w/o=}}$-equivalence. On the other hand they do not yield $\mathsf{SOL_{w/o=}}$-equivalence, since $\mathsf{SOL_{w/o=}}$ is just $\mathsf{SOL}$.
With this in mind it's easy to see that there are strict partial orders in which equality is not $\mathsf{FOL_{w/o=}}$-definable: consider the surjection from the discrete two-element strict partial order to the discrete one-element strict partial order.
On the other hand, the above is not the whole story. Certainly some strict partial orders let us define equality with an $\mathsf{FOL_{w/o=}}$ formula. For example, if the order is linear then $x=y$ iff $\neg x<y\wedge\neg y<x$. More generally (and Asaf Karagila mentions this above), equality is $\mathsf{FOL_{w/o=}}$-definable in every strict partial order which is weakly extensional (I don't actually know what the technical term is) in the sense that each element is determined by how it interacts with $<$: $$\{z: z<x\}=\{z: z<y\}\mbox{ and }\{z: x<z\}=\{z: y<z\}\quad\implies\quad x=y.$$
So as a natural follow-up question we can ask:
For which strict partial orders $\mathfrak{A}=(A;<)$ can we define $=$ with an $\mathsf{FOL_{w/o=}}$-formula?
And it turns out that these are exactly the weakly extensional ones in the sense above.
Suppose $\mathfrak{A}$ is not weakly extensional. Let $a,b\in\mathfrak{A}$ be distinct with $\{z: z<a\}=\{z: z<b\}$ and $\{z: a<z\}=\{z: b<z\}$. Now consider the induced suborder $\mathfrak{A}'=(A\setminus \{a\}; <\upharpoonright A\setminus\{a\})$ and the map $f: \mathfrak{A}\rightarrow\mathfrak{A}'$ which is the identity off $\{b\}$ and sends $b$ to $a$. It's easy to check that this $f$ preserves and reflects all atomic $\mathsf{FOL_{w/o=}}$-formulas in our language, but it doesn't reflect equality. |
H: Calculating expected value of $X$ with the density function $f(x)=16xe^{-4x}$
Suppose, $X$ be a random variable with probability density funciton,
$$
f(x) = \begin{cases}
16xe^{-4x}, & x \geq 0; \\
0, & \text{otherwise}
\end{cases}
$$
(source)
I tried to find the expected value of $X$, so I integrated $16x^2 e^{-4x}$ from $0$ to $\infty$.
After finding the indefinite integral with $u$-substitution:$-\frac{1}{2}e^{-4x}(8x^2+4x+1)+C$, I tried to compute the integral solution with the aforementioned boundaries and here is where I am not certain whether I calculated it right.
It might have gone wrong when I plugged in $\infty$ for $x$. I get $0-(-1/2)=1/2$. Is this correct or am I supposed to get something undefined and can one even get undefined expected value/mean?
When you subtract something from infinity, isn't it undefined? Same as multiplying $0$ with infinity?
AI: Yes your solution is fine, indeed we have
$$\int_0^\infty 16x^2e^{-4x}dx=\left[-\frac12e^{-4x}(8x^2+4x+1)\right]_0^\infty=0-\left(-\frac12\right)=\frac12$$
More in detail what we are solving is the following limit
$$\lim_{a\to \infty} \int_0^a 16x^2e^{-4x}dx=\lim_{a\to \infty}\left[-\frac12e^{-4x}(8x^2+4x+1)\right]_0^a=0-\left(-\frac12\right)=\frac12$$ |
H: Find partial derivatives for $\sqrt[3]{(x^3 + y^3)}$ at the point $(0,0)$.
For the function: $\sqrt[3]{(x^3 + y^3)}$ I'm requested to find both partial derivatives at the point $(0,0)$. I believe this is a problem that uses the limit definition of a partial derivative, however I always get 0 as an answer, which is incorrect. Can someone help me to sort this out?
AI: $\frac{\partial f}{\partial x}(0,0)=\lim_{t\to0} \frac{f((0,0)+t(1,0))-f(0,0)}{t}=\lim_{t\to0} \frac{\sqrt[3]{t^3}}{t}=\lim_{t\to0} \frac{t}{t}=1$. It is analogous for $\frac{\partial f}{\partial y}(0,0)$. |
H: Question about convexity: how do we prove that $\displaystyle \sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i}$?
Let $b_{1},b_{2},\ldots,b_{k}$ be nonnegative numbers and $p_{1} + p_{2} + \ldots + p_{k} = 1$ where each $p_{i}$ is positive. Then
\begin{align*}
\sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i}
\end{align*}
MY ATTEMPT
Since the logarithm function is strictly increasing, the proposed inequality is equivalent to
\begin{align*}
\ln\left(p_{1}b_{1} + p_{2}b_{2} + \ldots + p_{k}b_{k}\right) \geq p_{1}\ln(b_{1}) + p_{2}\ln(b_{2}) + \ldots + p_{k}\ln(b_{k})
\end{align*}
Once $f''(x) < 0$, where $f(x) = \ln(x)$, we conclude that $f$ is concave and the proposed inequality holds.
My question is: am I proving this result correctly? If this is the case, is there another way to prove it?
Any contribution is appreciated.
AI: Using $b_1,b_2,\cdots,b_k\ge 0$ with $p_1,p_2,\cdots,p_k$ as the respective weights, weighted AM-GM inequality gives $$\frac{\sum_{i=1}^k p_ib_i}{\sum_{i=1}^k p_i} \ge \left(\prod_{i=1}^k b_i^{p_i}\right)^{\dfrac{1}{\sum_{i=1}^k p_i}}$$ which gives the required inequality.
Note: This is not much different from your approach because the special case of Jensen's inequality for the function $f(x)=\ln(x)$ can be proved by weighted AM-GM or vice-versa, the weighted AM-GM inequality can be proved by the Jensen's inequality for $f(x)=\ln(x)$ as you have done, since the inequality you want to prove is directly the weighted AM-GM inequality in disguise. Also, these two can be proven independently without using each other. |
H: Properties of a continuous submartingale which is a function of Brownian motion
I'm attempting to solve the following question but I'm really unsure of what my approach should be. I've some progress but I need a lot of help, as I'm not sure if I'm anywhere near the right track:
For part (a) intuitively, $$\textbf{(1)} \quad \mu > \sigma^2/2 \implies \lim S_t = \infty \\ \textbf{(2)} \quad \mu < \sigma^2/2 \implies \lim S_t = 0$$ almost surely.
Note that $t W_{1/t}$ is also Brownian motion so that it suffices to show that $$\lim_{t \rightarrow \infty} S_0 \exp(t \left(W_{1/t} + \mu - \frac{\sigma^2}{2}\right) = \infty \text{ or } 0$$
This holds if and only if $$\bigcup_{k = N}^\infty \bigcap_{0<s <1/k \\ s \in \mathbb{Q}} \left \{ S_0 \exp \left(\frac{W_s + \mu - \sigma^2/2}{s} \right) \ge M \right \} \text{ or } \bigcup_{k = N}^\infty \bigcap_{0<s <1/k \\ s \in \mathbb{Q}} \left \{ S_0 \exp \left(\frac{W_s + \mu - \sigma^2/2}{s} \right) \leq 1/M \right \}$$ for each $M \in \mathbb{N}$ for the two cases, respectively.
Both of these events are $\mathcal{F}_{1/N}$ measurable, where $\mathcal{F}_{1/N} = \sigma(W_t : t \leq 1/N)$ for every $N$ and thus $\mathcal{F}_0^+ = \bigcap_{\epsilon > 0} \mathcal{F}_\epsilon$ measurable. By Blumenthal's 0-1 law, their probabilities are thus 0 or 1 so we simply need to show that they are positive, but I have no idea how to do this, or if there is an easier way. Please help if you can!
AI: First, forget about the exponent and $S_0$. Try to show the following:
If $a \neq 0$ then $W_t + a t$ tends to $-\infty$ or $\infty$ almost surely.
The statement you want follows immediately from this above fact.
To show this fact, there are loads of ways. Here's an easy way to do it: try to show that there is a random $T < \infty$ a.s. so that for all $t \geq T$ we have $|W_t| \leq t^{2/3}$ (loads of different bounds here would work, by the way).
Why is this the case? Again, there are loads of ways to do this, but recall that the reflection principle says $$P(\max_{s \leq t} |W_s| \geq M) \leq 2 P(|W_t| \geq M)$$
and so $$\sum_{n \geq 1} P( \max_{t \in [n,n+1]} |W_t| \geq n^{2/3}) < \infty\,.$$
This shows that we eventually have $|W_t| \leq t^{2/3}$ by the Borel Cantelli lemma. |
H: Given $AB\cdot F=JG$, $AD^E=AHJ$, ..., find the value of $G^E$
In the problem above, I know that the digits are all between 1 and 9, inclusive, at least most of them are, but I can't figure out a way other than trial and error to get the answer. Can someone please help? Thanks!
AI: To solve these types of problems, you should try to determine and check its "weakest" point at each stage. In your case, this seems to be first where you have exponentials, especially where the ratio of the number of digits of the result to that of the number being exponentiated is closer to $1$ as it seriously limits what the value of the exponential can be. In particular, consider
$$AD^{E} = AHJ \tag{1}\label{eq1A}$$
You have $E \neq 1$. Also, since $10^3 = 1000$, any $2$ digit integer of a power more than $2$ must be $4$ or more digits long. This only leaves that
$$E = 2 \tag{2}\label{eq2A}$$
Next, you can see that only $E$ and $C$ are used in one equation, so tackle that next, i.e.,
$$\begin{equation}\begin{aligned}
C^{E} & = EC \\
C^2 & = 20 + C \\
C^2 - C - 20 & = 0 \\
(C - 5)(C + 4) & = 0 \\
C & = 5
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
The next one to check is the remaining exponential value, i.e.,
$$B^D = HB \tag{4}\label{eq4A}$$
Some digit $B$, which when taken to some power greater than $1$, has the same final digit. $B = 1$ is too small, and $B = 2$ is not allowed due to \eqref{eq2A}. With $B = 3$, you have $D = 5$ or $D = 9$ only, with $3^5 = 243$ already too large. With $B = 4$, you have $D \in \{3, 5, 7, 9\}$, with the only one which gives a $2$ digit result being $4^3 = 64$. $B = 5$ is not available. With $B$ being $6$ or higher, $B^D$ is at least a $3$ digit number, so the only possibility which works is
$$B = 4, D = 3, H = 6 \tag{5}\label{eq5A}$$
With most of the digits now determined, using \eqref{eq2A} and \eqref{eq5A} in \eqref{eq1A} gives
$$\begin{equation}\begin{aligned}
(10(A) + 3)^2 & = 100(A) + 60 + J \\
100(A) + 60(A) + 9 & = 100(A) + 60 + J \\
60(A) + 9 & = 60 + J
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
This gives
$$A = 1, J = 9 \tag{7}\label{eq7A}$$
You now have $7$ of the digits determined, with just $2$ left to assign in the remaining equation, i.e.,
$$AB \times F = JG \implies (14)F = 90 + G \tag{8}\label{eq8A}$$
The only multiple of $14$ between $90$ and $98$ is where $14(7) = 98$, giving
$$F = 7, G = 8 \tag{9}\label{eq9A}$$
This gives the expression value to determine to be
$$G^E = 8^2 = 64 \tag{10}\label{eq10A}$$ |
H: Show that a Hibert-Schmidt operator is compact.
Let $H$ be a Hilbert space and $u: H \to H$ a Hilbert-Schmidt operator, i.e. for an orthonormal basis $E$
$$\Vert u\Vert_2^2:=\sum_{x \in E} \Vert u(x) \Vert^2 < \infty$$
(this does not depend on the choice of basis).
I want to show that $u$ is compact. I looked this question up but all references I found proved this in the case that $H$ is separable, i.e. has a countable orthonormal basis. Is there an elementary proof that works for general Hilbert spaces?
AI: For
$$\sum_{x \in E} \Vert u(x) \Vert^2 < \infty$$ to hold, only countably many terms can be non-zero, i.e. $u$ is zero on all $H$ except a separable subspace $V$. So just look at the restriction
$$u|_V: V \to W\subseteq H$$
where $W = \text{im}(u) \cong V$ is also separable. $u|_V$ is compact by any of the proofs you know, so it sends the unit ball of $V$ to a relatively compact set in $W$, which is also relatively compact in $H$. $u$ sends the rest of the unit ball of $H$ to 0 so it also sends the unit ball to a relatively compact set. |
H: Half-open topology on $\mathbb R$ is separable, and $A \setminus \hat A$ is countable
This is part of Exercise 9 in Section 2.2 of Topology and Groupoids, by Brown.
For each $x \in \mathbb R, N \subseteq \mathbb R$ is a neighborhood of $x$ if and only if there are real numbers $x^{\prime}, x^{\prime \prime}$ such that $x \in [ x^{\prime}, x^{\prime \prime}) \subseteq N$. This is called the half-open topology on $\mathbb R$.
I am supposed to show that this topological space is separable, and that if $\hat A$ is the set of limit points of a set $A \subseteq \mathbb R$, then $A \setminus \hat A$ is countable.
My attempt:
The space is separable because the subset $\mathbb Q \subseteq \mathbb R$ is countable, and every neighborhood of every point must intersect $\mathbb Q$.
Showing that $A \setminus \hat A$ is countable seems harder. I thought about proving it by contradiction. Assume it were uncountable. So there are uncountably many $x \in A$ that have at least one neighborhood that contains no other points of $A$. So there are uncountably many intervals $[a, b)$ that contain only one point of $A$.
I was going to say that none of the intervals can intersect, but I'm not sure that's true.
I'm stuck. Any help is appreciated.
Edit:
If there are uncountably many such points, does that mean they have to form at least a finite interval? Then there couldn't be uncountably many intervals $[a, b)$ each only containing one of the points within a finite interval, right?
AI: You're right about the countable dense subset. (every set $[a,b)$ contains an open interval $(a,b)$ which contains a rational etc.)
If $a \in A\setminus \hat{A}$, then there is a basic subset $[a,f(a))$ with $f(a) \in \Bbb Q$ such that $[a,f(a)) \cap A = \{a\}$ (this is what not being a limit point entails, plus we use the density of $\Bbb Q$). Suppose that we have $a_1 < a_2$ in $ A\setminus \hat{A}$ and $f(a_1) = f(a_2)$. But then $a_2 \in [a_1, f(a_1)$ and this contradicts how $f(a_1)$ was chosen. So $f: A\setminus \hat{A} \to \Bbb Q$ is injective and so the domain of $f$ is at most countable. |
H: How to prove $\sqrt{2}$ is irrational in Type Theory?
Using Intuitionistic Type Theory, how would one go about proving $\sqrt{2}$ is irrational?
I read that we can not use law of excluded middle. (So does this mean we cannot use proof by contradiction).
So I assume we want to prove $\forall a,b\in \mathbb{N^2}: \lnot (a^2 = 2 b^2)$. In type theory this means we need to find an instance $P$ of the type corresponding to the theorem. The type would be something like $\prod \limits_{a:\mathbb{N}}\prod \limits_{b:\mathbb{N}} NotEqualsQ(Square(a),Times(Two,Square(b))$. I assume. Then we might have to define these function recursively?
I'm not sure how you would find an instance of this proof.
AI: Negation is defined $\neg A := A\to \bot$, the type of naturals $\mathbb N$ is defined inductively, and addition and multiplication are defined recursively.
To prove $\Pi_{a,b:\mathbb N}\neg(a^2=2b^2)$ we can introduce variables $a,b:\mathbb N$ and a proof that $a^2=2b^2$ like so:
$$a,b:\mathbb N, p : (a^2=2b^2) \vdash \bot$$
and the goal is to construct a term of type $\bot$. (Note: I suppose this could be called a "proof by contradiction" because we derive $\neg A$ from $A\vdash\bot$, but it does not require the excluded middle. A problematic "proof by contradiction" is when we derive $A$ from a proof of $\neg A\vdash \bot$.)
The idea is to construct $c,d:\mathbb N$ such that $(c^2=2d^2)$ and $\neg(2\div d)$, and then prove that $2\div d$, from which you obtain $\bot$. Constructing $c$ and $d$ can be done by splitting cases on whether $a$ and $b$ are even or odd, assuming you've proved something of the form $$(\forall k.\phi(2k)\times\phi(2k+1))\to(\forall n.\phi(n)).$$
For a more concrete justification, you could check out this proof in Coq. |
H: A graph theoretic definition of functions?
The graph(1) $G$ of the function $f:X\to Y$ is the set $\{(x,f(x)):x \in X\}$.
Unless I'm mistaken, this can be interpreted as a directed, bipartite graph(2) where every vertex in one of the partitions (domain) has out-degree $1$ and in-degree $0$. The other partition is the image.
CLARIFICATION: graph(2) refers to the graph-theoretic definition of graph, distinct from graph(1)
According to the post Difference between a function and a graph of a function?, the difference between a function $f$ and its graph $G$ is that the function's codomain cannot be recovered from the graph.
However, the domain of $f$ could be recovered from the disjoint union of $G$ with a second graph $H$ of isolated vertices which satisfies the condition -- the union of the set of vertices in $H$ and the set of vertices in the image partition of $G$ is the codomain.
Could you then say that $f$ is equivalent to $G\oplus H$ up to an isomorphism? Is there any benefit to viewing functions as a specific type of graph?
AI: As a visualizing tool, yes, some times a graph is better than a graph. Visualizing a function $\Bbb R\to\Bbb R$ as a collection of arrows from one number line to another (or just as a device which, when you press somewhere in the domain, the corresponding place in the codomain lights up) is, in my opinion, a better framework than a curve in the plane for understanding things like differentiation, the $\varepsilon$-$\delta$ definition of continuity, and function composition (one can, for instance, give an entirely clear and obvious intuitive argument for the chain rule in a matter of seconds, once derivatives and function chaining / composition are both understood).
It is also more easily generalized and therefore often used (at least in an abstract sense) in both analysis and topology.
However, while chaining functions is basically a trivial operation, and it neatly separates domain from codomain, which is nice some times, arithmetic operations are more difficult to visualize (I don't know intuitively how to combine my images of the graph-theory graphs for $x\mapsto x^2$ and $x\mapsto x$ into their sum $x\mapsto x^2+x$, but for the curves in the plane it is obvious). Additionally, it does have the drawback of not working too well on a printed page. Which I suspect is why it is a sadly underrepresented representation in calculus classes. |
H: Differentiation while transforming the expression by introducing polar coordinates
The expression that is to be transformed by polar coordinates is $\Big(\frac{\partial z}{\partial x}\Big)^2+\left(\frac{\partial z}{\partial y}\right)^2$.
So I defined $x=x(r,\phi)=r\cos\phi$ and $y=y(r,\phi)=r\sin\phi$, where $r>0$ and $\phi\in[0,2\pi)$, and apparently, the derivative of $z$ with respect to $r$ $\frac{\partial z}{\partial r}$ is equal to $\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial r}$, and the similiar goes for $\frac{\partial z}{\partial \phi}$.
So my question is how is this so (regarding the derivatives)? Doesn't $\frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial r}=2\frac{\partial z}{\partial r}$? What am I missing?
The solution later on is pretty much straightforward, but I'm absolutely stuck here, and while I believe the answer to my question is fairly simple, I would appreciate if someone would help me as I have shaky basis because I didn't have any classes. Again, apologies if the question is too simple.
AI: Your chain rule is correct, we do have
$$ \frac{\partial z}{ \partial r} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial r}. $$
But derivatives are not fractions, you cannot just cancel the $\partial x$'s and $\partial y$'s to get $2\frac{\partial z}{\partial r}$. Derivatives don't work like that. You have to work it through to get
$$ \frac{\partial z}{ \partial r} = \frac{\partial z}{\partial x} \cos \phi+\frac{\partial z}{\partial y} \sin\phi. $$
The same process yields
$$ \frac{\partial z}{ \partial \phi}= \frac{\partial z}{\partial x} (-r\sin \phi)+\frac{\partial z}{\partial y} (r\cos\phi).$$
You can then solve for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ and substitute into the expression you want. Hint: I would calculate $\left(\frac{\partial z}{\partial r}\right)^2 + \left(\frac{1}{r}\frac{\partial z}{\partial \phi}\right)^2$. |
H: The second largest eigenvalue
Let $A$ be an $n\times n$ positive semidefinite real matrix with the trace of $A$ is $2$. Suppose there exist unit vectors $y,z$ satisfying $|Ay|=|Az|=1$ and $y\cdot z=0$.
How is it showed that the second largest eigenvalue of $A$ is greater than or equal to $1$?
AI: It's false. Let $A=diag(\sqrt{2},0,2 - \sqrt{2})$. Let $y = (\frac 1 {\sqrt 2}, \frac 1 {\sqrt 2}, 0)$, $z = (\frac 1 {\sqrt 2}, - \frac 1 {\sqrt 2}, 0)$. The conditions are satisfied, while the second largest eigenvalue is $2 - \sqrt2 < 1$. |
H: Can you explain the behavior of $\frac{1}{x} \sin(\frac{1}{x})$ and $|\frac{1}{x} \sin(\frac{1}{x})|$ as $x$ approaches $0$
When I graph it out, I can see what the behavior of the graph is, but my intuition is failing to grasp why it acts this way.
As $x\to 0$, $f(x) = \frac{1}{x} \sin(\frac{1}{x})$ only oscillates. If $\sin(\frac{1}{x})$ is bounded by $1$ and $-1$, and $\frac{1}{x}$ diverges to $\infty$, then why doesn't the whole function diverge? Shouldn't the divergence overwhelm the bounded sin function?
Then if we apply the absolute value, as $x\to 0$, $f(x) = |\frac{1}{x} \sin(\frac{1}{x})|$ now does diverge to $\infty$, but I don't understand what changed. $\sin$ now oscillates between $0$ and $1$, but somehow it is now overpowered by the divergence of $\frac{1}{x}$, where it wasn't before.
AI: As to requested behaviour, then limit for both doesn't exist at $x \to 0$. Limit point 0 have both. Difference is that for $\left|\frac{1}{x} \sin(\frac{1}{x})\right|$ we have limit point $+ \infty$ and for $\frac{1}{x} \sin(\frac{1}{x})$ we have limit points $\pm \infty$
By the way, imho, main difference seems that $\left|\frac{1}{x} \sin(\frac{1}{x})\right|$ have no derivative at $x=\frac{1}{\pi n}$. |
H: What is the name of sets of nodes of a tree that cover/span/cut the entire width of the tree?
What are the following sets called?
They may be constructed by repeating the following recipe.
Choose a parent node
Remove the parent node's descendents
Repeat as desired while parent nodes remain.
Here is a tree in graphviz dot.
digraph G {a->b;
a->c;
a->d;
b->e;
b->f;
e->g;
e->h;
c->i;
c->j;
}
Here are four of many possible sets of nodes that cover the full width of the tree.
g,h,g,i,j,d
graph1
a
graph2
b,c,d
graph3
b,i,j,d
graph4
My best guess is covering set, spanning set, cutting set?
David
AI: If I understand correctly, these are the maximal antichains: sets of nodes in which no two are comparable, but such that any node outside the set is comparable with some node in the set. These make sense for arbitrary partial orderings, not just trees specifically.
(As an aside, note that it's easy to see that maximal antichains exist in every finite partial ordering; for infinite partial orderings, we need the axiom of choice.) |
H: Solving Exponential Equation $x^2 \cdot 2^{x+1} + 2^{|x-3|+2} = x^2 \cdot 2^{|x-3|+4} + 2^{x-1} $
The solution for the Exponential Equation considering $x\in\mathbb{R}$ is $\boxed{x\in[3, \infty)\cup \left\{ -\frac12;\frac12 \right\} }$
My try for solving $x^2 \cdot 2^{x+1} + 2^{|x-3|+2} = x^2 \cdot 2^{|x-3|+4} + 2^{x-1} $:
I initially considered that
$$x^2 \cdot b+c=x^2 \cdot d+e \Longrightarrow b=d \land c=e $$
Therefore, I would have:
$$2^{x+1}=2^{|x-3|+4} \Longrightarrow x+1=|x-3|+4 \Longrightarrow |x-3|=x-3 $$
$$\text{For } x-3=x-3 \Longrightarrow x\in(-\infty, \infty) $$
$$\text{For } -x+3=x-3 \Longrightarrow x=3 $$
$$\text{The interval would be } [3, \infty)$$
The same for
$$2^{|x-3|+2} =2^{x-1} \Longrightarrow |x-3|+2=x-1\Longrightarrow |x-3|=x-3 $$
$$\text{For } x-3=x-3 \Longrightarrow x\in(-\infty, \infty) $$
$$\text{For } -x+3=x-3 \Longrightarrow x=3 $$
$$\text{The interval would be } [3, \infty)$$
I don't know how to get $\left\{ -\frac12;\frac12 \right\}$ and I could't perform $$x=\pm \frac{\sqrt{e-c}}{\sqrt{b-d}} $$
This is the wrong approach? Any hints to solve it? Anything would be great.
AI: HINT
\begin{align*}
x^{2}\times 2^{x+1} + 2^{|x-3| + 2} = x^{2}\times 2^{|x-3|+4} + 2^{x-1} & \Longleftrightarrow 2x^{2}\times 2^{x} + 4\times 2^{|x-3|} = 16x^{2}\times 2^{|x-3|} + \frac{2^{x}}{2}\\\\
& \Longleftrightarrow 4x^{2}\times (2^{x} - 8\times 2^{|x-3|}) = 2^{x} - 8\times 2^{|x-3|}\\\\
& \Longleftrightarrow (4x^{2} - 1)(2^{x} - 8\times 2^{|x-3|}) = 0
\end{align*}
Can you take it from here? |
H: How does this statement follow algebraically from the last one? (Algebraic eqations)
I was reviewing some combinatorics notes and i found the following.
Can someone please help me understand how in the world does that statement follow from the monstrosity above it?
In case it matters, the task is attempting to solve a recurrence relation for its closed form.
Here is the full solution: https://i.stack.imgur.com/Z7aRM.jpg
AI: Note $2^2 - 6 \cdot 2 + 8 = 0$, $4^2 - 6 \cdot 4 + 8 = 0$.
(Thanks @AlexeyBurdin for pointing out an oversight) |
H: Area under the graph of a convex function
Consider the following problem:
Suppose that $f$ is a twice differentiable real function such that
$f''(x)>0$ for all $x\in[a,b]$. Find all numbers $c\in[a,b]$ at which
the area between the graph $y=f(x)$, the tangent to the graph at
$(c,f(c))$, and the lines $x=a$, $x=b$, attains its minimum value.
The solution starts with the observation that the condition on $f''$ implies convexity of $f$, i.e. the graph of $f$ is always above any tangent to it, and therefore
$$A(c)=\int\limits_a^b (f(x)-f(c)-f'(c)\cdot(x-c))dx,$$
where $A(c)$ is the area the problem refers.
I'm really confused with this formula for the area and can't comprehend it even after drawing the picture. Can anybody explain it? Any help is appreciated.
AI: Convexity on differentiable functions is equivalent to $f(y)\geq f(x)+f′(x)(y−x)$ (the tangent line is below the function).
Then, the point is that you don't need to worry about the sign of the diference between $f$ and the tangent lines, since $f$ is always above the lines $x\mapsto f(c)+f′(c)(x−c)$ for every $c\in[a,b]$. Therefore, the distance between $f$ and the tangent line at $(c,f(c))$ is $f(x)−f(c)−f′(c)(x−c)$.
So,
$$ A(c) = \int_a^b f(x)−f(c)−f′(c)(x−c) dx = \int_a^b f(x) dx - (b-a)f(c) + f'(c)\frac{ (a-b)(a+b-2c) }{2}.$$ |
H: Revenue and quadratic formula - for every x increase in price there are y fewer sales
"When a shoe costs $\$80.00$, there are $300$ sales. Every $\$5.00$ increase in price will result in 10 fewer sales. Find the price that will maximize income."
I am able to solve the question just fine, but I am confused about the logic in setting up the quadratic formula.
The first step is to let $x$ be a $\$5$ increase in price, and you would plug in $x$ in the equation.
$$y = (80 + 5x)(300 - 10x)$$
The equation above is where I am confused. If $x$ is equal to one $\$5$ increase in price, then why wouldn't the equation be:
$$y = (80 + x)(300 - 2x)$$
AI: $x$ is not the price increase, $\$5$ is the price increase.
$x$ is just the number of price increases (like, say, number of five-dollar bills customer would have to pay extra) and also the corresponding number of sales decreases (each decrease is $10$ sales).
One price increase brings one sales decrease, so:
$$ y = (80 + 5\cdot 1)(300 - 10\cdot 1). $$
Two price increases bring two sales decreases:
$$ y = (80 + 5\cdot 2)(300 - 10\cdot 2). $$
And so on. |
H: Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $|\frac{1}{x-5}| \leq 1$ then condition $(1)$ and $(2)$ will not hold. So wouldn't the range be $x < 4$ and $5 \leq x$ ? I know this is not right since the answer should be $ 5\frac{1}{3} \leq x$ and $x < 4 \frac{2}{3}$ ... could someone provide a solution?
AI: What about the ratio test?
\begin{align*}
\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = \limsup_{n\to\infty}\frac{n3^{n}|x-5|^{n}}{(n+1)3^{n+1}|x-5|^{n+1}}\\\\
& = \limsup_{n\to\infty}\left(\frac{n}{n+1}\right)\frac{1}{3|x-5|} = \frac{1}{3|x-5|} < 1\\\\
& \Rightarrow |x - 5| > \frac{1}{3} \Rightarrow \left(x > \frac{16}{3}\right)\vee\left(x < \frac{14}{3}\right)
\end{align*}
Moreover, for $x = 16/3$, one has that
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}3^{-n}} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} < \infty
\end{align*}
due to the Leibniz test.
On the other hand, for $x = 14/3$, we have that
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}(-1)^{n}3^{-n}} = -\sum_{n=1}^{\infty}\frac{1}{n} = \infty
\end{align*}
since it is the harmonic series.
Hopefully this helps. |
H: Doubt on showing that when $p$ is prime, the only unipotent class is $p-1$?
I am trying to understand the solution of the following exercise:
A number $a$ is unipotent if $a\neq1$ and $a^2\equiv 1 \pmod{p}$.
Show that when $p$ is prime, the only unipotent class is $p-1$.
The answer is:
$(p-1)x\equiv1 \pmod{p}$ has a unique solution in $\Bbb{Z}/p\Bbb{Z}$. This solution is $x\equiv -1 \pmod{p}$. It is the same as $x=p-1$.
I may be doing some very silly mistake: We want to prove that $a=p-1$ is the only number such that $a^2\equiv 1 \pmod{p}$ but why does the above proof proves it? Aren't we proving only that $a$ is unipotent without - somehow - checking it for $(p-1),(p-2),(p-3),\dots$?
AI: Frankly I'm not quite sure what the answer's proof is trying to say, but the easiest way to see this result is to note that $\Bbb Z/p\Bbb Z$ is a field, and thus the polynomial $x^2-1$ can only have at most two roots (i.e., as many roots as its degree). As $x=\pm1$ both solve this polynomial, these are the only roots. As we're excluding $x=1$, this shows that there is a unique unipotent element modulo $p$.
That being said, in the case $p=2$ there are actually no unipotent elements since $1=-1\mod 2$ is the only element that squares to $1$ in $\Bbb Z/2\Bbb Z$. |
H: Calculating limits of a system
For the following system
$$x'=x(-x^2-y+4)=f(x,y)$$
$$y'=y(y^2+8x-1)=g(x,y)$$
I need to find the location of the critial points and determine each points type and stability.
I then need help finding calculating the limits
$$\lim\limits_{t \to \infty} x(t), \lim\limits_{t \to \infty} y(t)$$
if $x(0)=\frac{-5}{2}, y(0)=1$
I found the critical points and the type/stability but it seems like a large amount
$(0,0)$ eignevalues are $(4,-1)$ unstable saddle point
$(0,1)$ eigenvalues are $(2,3)$ unstable improper node
$(0,-1)$ eigenvalues are $(2,5)$ unstable improper node
$(2,0)$ eigenvalues are $(-8,15)$ Unstable saddle point
$(-2,0)$ eignevalues are $(-17,-8)$ Asymp stable improper node
$(-1, 3)$ eigenvalues are $(-2\sqrt 31 +8, 2\sqrt 31 +8)$ Unstable Saddle point
$(-3,-5)$ eigenvalues are $(-2\sqrt 259 +16, 2\sqrt 259 +16)$ unstable saddle point
$(2+i, 1-4i)$ eigenvalues are $(-29.22-18.774i, -6.78-5.226i)$ asymp stable spiral
$(2-i, 1+4i)$ eigenvalues are $(-29.22+18.774i, -6.78+5.226i)$ aspym stable spiral
In regards to the limits, I'm unsure of what $x(t)$ and $ y(t)$ values I am to use
AI: For the system
$$x'=x(-x^2-y+4) \\ y'=y(y^2+8x-1)$$
We find seven real critical points by setting $x' = y' = 0$ as
$$(x, y) = (-3,-5),(-2,0),(-1,3),(0,-1),(0,0),(0,1),(2,0)$$
We find the Jacobian as
$$J(x, y) = \dfrac{\partial f(x,y)}{\partial g(x,y)} = \begin{pmatrix} \dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y} \\\dfrac{\partial g}{\partial x} & \dfrac{\partial g}{\partial y}\end{pmatrix} = \begin{pmatrix}
-3 x^2-y+4 & -x \\
8 y & 8 x+3 y^2-1 \\
\end{pmatrix}$$
We find the eigenvalues of the Jacobian at each critical point and determine their stability (you can also include the type of point)
$$\begin{array} {|r|r|}\hline \mbox{Critical Point (x,y)} & \mbox{Eigenvalues}~~~(\lambda_1, ~\lambda_2) & \mbox{Stability} \\ \hline (-3,-5) & (2 \left(\sqrt{259}+8\right),2\left(8-\sqrt{259}))\right. & \mbox{Unstable} \\ \hline (-2,0) & (-17, -8) & \mbox{Stable} \\ \hline (-1,3) & (2 \left(\sqrt{31}+4\right),2 \left(4-\sqrt{31}\right)) & \mbox{Unstable} \\ \hline (0,-1) & (5, 2) & \mbox{Unstable} \\ \hline (0,0) & (4, -1) & \mbox{Unstable} \\ \hline (0,1) & (3, 2) & \mbox{Unstable} \\ \hline (2,0) & (15,-8) & \mbox{Unstable} \\ \hline \end{array}$$
From this information, we can plot the phase portrait
For the last question,what do you notice using the IC as a starting point as $t$ approaches infinity (see the red line). |
H: How to compute $\sum_{n=1}^\infty{\frac{n}{(2n+1)!}}$?
In a calculus book I am reading I have encountered the following problem:
$$\sum_{n=1}^\infty{\frac{n}{(2n+1)!}}$$
The hint is to use Taylor series expansion's for $e^x$. I tried to express the sum as the form $$e^x=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$
But I could not find a consistent method, I always end un with different sums of factorials that does not help me solve the problem
The official solution is $$\boxed{\frac{1}{2e}}$$
The excersise is in a chapter that mixes calculus with summation, so the solution will probably include both.
Any help or hint is highly appreaciated! Thanks in advance.
AI: METHODOLOGY $1$:
The Taylor series for the hyperbolic sine function $\sinh(x)$ is given by
$$\sinh(x)=\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}\tag1$$
If we divide both sides of $(1)$ by $x$, differentiate, and set $x=1$, awe find that
$$\underbrace{\cosh(1)-\sinh(1)}_{=e^{-1}}=2\sum_{n=1}^\infty\frac{n}{(2n+1)!}\tag2$$
Finally, dividing $(2)$ by $2$ yields the coveted result
$$\sum_{n=1}^\infty \frac{n}{(2n+1)!}=\frac1{2e}$$
as was to be shown!
METHODOLOGY $2$:
We begin with the Taylor series for $e^x$ at $x=-1$. Then, we see that
$$\begin{align}
\frac1e&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\\\\
&=\sum_{n=0}^\infty\left(\frac1{(2n)!}-\frac1{(2n+1)!}\right)\\\\
&=\sum_{n=0}^\infty \frac{(2n+1)!-(2n)!}{(2n)!(2n+1)!}\\\\
&=\sum_{n=1}^\infty \frac{2n}{(2n+1)!}\\\\
&=2\sum_{n=1}^\infty \frac{n}{(2n+1)!}
\end{align}$$
from which we arrive at the coveted result
$$\sum_{n=1}^\infty \frac{n}{(2n+1)!}=\frac1{2e}$$
as expected! |
H: Completeness of $(\mathcal M (2,\mathbb R),\lVert \cdot\rVert)$
Let $\mathcal M (2,\mathbb R)$ be the space of $2\times2$ matrices with inner product $ \langle A,B\rangle=\operatorname{tr}(AB^T) $
We define the normed space : $ \bbox[1px,border:1px solid green] { X:=(\mathcal M (2,\mathbb R),\lVert \cdot\rVert)}$ with norm $\bbox[1px,border:1px solid green] {\lVert A\rVert=\sqrt{\langle A,A\rangle } }$
I want to check the completeness (a.k.a Banach) of $X$ but how can we define the limit of a sequence in that space?
I.e Let $A_n$ be a sequence of matrices s.t $A_n=\bigg ($$ \left[
\begin{array}{cc}
x_1&y_1\\
z_1&w_1
\end{array}
\right] $$, $$ \left[
\begin{array}{cc}
x_2&y_2\\
z_2&w_2
\end{array}
\right] $$,...., $$ \left[
\begin{array}{cc}
x_n&y_n\\
z_n&w_n
\end{array}
\right] ,....$$\bigg) $.
Is my question misleading? let me know.
AI: The distance is given by the norm. So $A_n\to A$ means $\|A_n-A\|\to0$. Here the norm, if you make the computation, is
$$\tag1
\|A\|=\biggl({\sum_{k,j}a_{kj}^2}\biggr)^{1/2}.
$$
From $(1)$ it is easy to see that
$$
|a_{kj}|\leq\|A\|
$$
for any $k,j=1,2$. So, if $\{A_n\}$ is Cauchy, then for any $k,j$ you have that $$|(A_n)_{k,j}-(A_m)_{k,j}|\leq \|A_n-A_m\|.$$ Thus the sequence of $k,j$ entries is Cauchy. Now you can use the completeness of $\mathbb R$ to obtain the limit for each $k,j$, and you need to show that entrywise convergence implies norm-convergence; this last property follows easily from
$$\tag2
\|A\|\leq\max\{|a_{k,j}|:\ k,j=1,2\}.
$$ |
H: What is this topological space called?
Given any topological space $T=(X,\tau)$ and any map $f:X'\to X$ note we can define the space: $T_f=(X',\{f^{-1}[S]:S\in \tau\})$ what is this topological space called with respect to $T$ and $f$?
AI: The topology on $X'$ you've described is called the initial topology on $X'$ induced by $f\colon X'\to X$.
Its open sets are precisely (per construction) the preimages $f^{-1}(S)$ of open sets $S\in \tau$. By this construction you'll get the coarsest topology on $X'$ for which $f$ is continous.
Having induced this topology on $X'$ via $f$, you can induce the final topology (quotient topology) on the quotient $X'\big/{\sim}$ via $$p\colon X'\to X'\big/{\sim}$$ where $X'\big/{\sim}$ is the quotient given by $$a\sim b :\Leftrightarrow f(a)= f(b).$$ |
H: How to show that two reccurence relations are equal?
I encountered the following problem:
I've never seen this on tutorials or similar. How could i go about proving this? I tried using induction but its looking pretty grim.
AI: Induction works, with just one tricky bit. I’ll use $G$ to denote the function with the second-order recurrence and show by induction that it’s equal to the original $F$.
Let $G(0)=2$, $G(1)=4$, and $G(n)=4G(n-1)-3G(n-2)$ for $n\ge 2$. Clearly $G(0)=F(0)$ and $G(1)=F(1)$. Suppose that $n\ge 2$, and $G(k)=F(k)$ for $0\le k<n$. Then
$$\begin{align*}
G(n)&=4G(n-1)-3G(n-2)\\
&=4F(n-1)-3F(n-2)\\
&\overset{(*)}=4F(n-1)-\big(F(n-1)+2\big)\\
&=3F(n-1)-2\\
&=F(n)\,,
\end{align*}$$
and the result follows by induction. The one tricky step is starred. For that step I used the recurrence for $F$ with the index shifted down one: $F(n-1)=3F(n-2)-2$. The previous step is of course just the induction hypothesis. |
H: Proving the Marginal Distribution of a $\text{Gamma}(2,\lambda)$ Joint Distribution follows an Exponential
I derived the bivariate joint distribution with a transformation from two exponential distributions with $\lambda = 1.5$ for $X,Y$ and $U=X+Y$ and $V=X-Y$. With $X=(U+V)/2$ and $Y=(U-V)/2$ I derived the joint $f_{u,v}$ as follows:
Skipping a few steps:
$f_{u,v}=\frac{4}{9}e^{\frac{2}{3}(\frac{-(U+V)}{2}-\frac{U-V}{2})}\cdot\frac{1}{2}=$
$f_{u,v}=\frac{2}{9}e^{-\frac{2}{3}U}$ for $0<V<U$
So to find the marginal, I take the integral with respect to V. I should expect a $\text{Gamma}(2,1.5)$ distribution.
$f_u=\int_{0}^{U}\frac{2}{9}e^{-\frac{2}{3}U} dv$
$f_u=\frac{2}{9}Ue^{-\frac{2}{3}}$
The above should be $\frac{4}{9}$ not $\frac{2}{9}$. I have a feeling the bounds should be $0<V<2U$. I'm probably missing something really basic, I just am not able to show why that would be the bounds.
AI: Since they are exponentially distributed, the domain for each of $X$ and $Y$ is $[0,\infty)$.
Defining $U=X+Y$ means the support for $U$ is $[0;\infty)$.
Also when given $U$ the conditional support for $Y$ is $[0;U]$ (as it may not have a value greater than $U$), and $X=U-Y$ (of course).
Now $V=X-Y$, or $V=U-2Y$, so when given $U$, the conditional support for $V$ is $[U-2U;U-0]$, that is: $[-U;U]$
So the joint support for $\langle U,V\rangle$ is $\{\langle u,v\rangle:0\leq u, -u\leq v\leq u\}\\=\{\langle u,v\rangle: \lvert v\rvert\leq u\}$ |
H: $\sigma(n)$ is injective?
I was reading about the sum function of divisors $\sigma(n)$, and the question arose as to whether this is injective. Does anyone know the answer? in case of being injective some idea for the demonstration? Thanks in advance. And I apologize if it turns out to be an obvious question, I have a suspicion that it is injective but I'm not sure.
AI: Injective would mean that no two distinct numbers have the same sum of divisors. If there are two distinct numbers whose divisors have the same sum, those would serve as a counterexample to show the function is not injective.
There are many such counterexamples, the smallest is $\sigma(6)=\sigma(11)=12$ |
H: Possible mistake in the solution of Baby Rudin Ch. 6 Ex. 11
This is exercise 11, Chapter 6 in Baby Rudin:
Let $\alpha$ be a fixed increasing function on $[a, b]$. For $u \in \mathscr{R}(\alpha)$, define
$$ \lVert u \rVert_2 = \left\{ \int_a^b \lvert u \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}. $$
Suppose $f, g, h \in \mathscr{R}(\alpha)$, and prove the triangle inequality
$$ \lVert f-h \rVert_2 \leq \lVert f-g \rVert_2 + \lVert g-h \rVert_2 $$
as a consequence of the Schwarz inequality
where by "Schwarz inequality," Rudin means the following version of Holder's inequality:
$$\left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \leq \left\{ \int_a^b \lvert f \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2} \left\{ \int_a^b \lvert g \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}. $$
The solution to this question in the solutions manual is:
In the first display on the left-hand side, it should read $||f-h||_2^2$ instead of $||f-h||_2^1$. My question is about the only inequality in the proof. In particular, how is it true that $$\int_a^b |f-g||g-h| \ d\alpha \le \left\{ \int_a^b \lvert f-g \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2} \left\{ \int_a^b \lvert g-h \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}?$$ This does not looks like the proper usage of the "Schwarz inequality" or am I missing something? Can someone please propose a correct version of the solution?
The solution to the same question here seems to suffer from the same issue, if this is an issue at all, that is.
AI: Regarding your first question I agree, it seems it should be $\|g-h\|_{2}^{2}$ on the LHS.
For your second question, the Cauchy-Schwarz inequality is a special case of H$\ddot{\text{o}}$lder's inequality in the sense that it is the case that is applicable to inner products. So in the second display you have which explains this the intermediate steps are,
\begin{align}
\bigg|\int_{a}^{b}fgd\alpha\bigg|\leq\int_{a}^{b}|f||g|d\alpha=\|fg\|_{1}\leq\|f\|_{2}\|g\|_{2}=\bigg(\int_{a}^{b}|f|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g|^{2}d\alpha\bigg)^{1/2}
\end{align}
The other thing to note here is it seems you are misreading the solution. You are asking how,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha\leq\|f\|_{2}\|g\|_{2},
\end{align}
which is not what is being presented in the proof. In Rudin's proof he is using,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha\leq\|f-g\|_{2}\|g-h\|_{2}.
\end{align}
EDIT: To clarify the last inequality, since $f,g,h$ are all Riemann integrable with respect to $d\alpha$ then $f-g$ and $g-h$ are also Riemann integrable. So consider,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha=\|(f-g)(g-h)\|_{1}\leq\|f-g\|_{2}\|g-h\|_{2}=\bigg(\int_{a}^{b}|f-g|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g-h|^{2}d\alpha\bigg)^{1/2}
\end{align}
EDIT2: As Paramanand pointed out, since $f-g$ and $g-h$ are Riemann integrable replace $f$ by $|f-g|$ and $g$ by $|g-h|$ in the given Cauchy-Schwarz inequality to get,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha=\bigg|\int_{a}^{b}|f-g||g-h|d\alpha\bigg|\leq\bigg(\int_{a}^{b}|f-g|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g-h|^{2}d\alpha\bigg)^{1/2}
\end{align} |
H: Calculate: $\int_0^\infty [x]e^{-x} \, dx$ where $[x]:=\max \{k\in\mathbb{Z}:k\leq x\}$
Calculate: $\int_0^\infty [x]e^{-x} \, dx$ where $[x]:=\max \{k\in\mathbb{Z}:k\leq x\}$
Solution: $$\int_0^\infty[x]e^{-x} \, dx = \sum_{k=0}^\infty \int_k^{k+1}[x]e^{-x} \, dx = \sum_{k=0}^\infty k\int_k^{k+1}e^{-x} \, dx = \sum_{k=0}^\infty k(-e^{-1}+1)e^{-k}$$
It is right? How can I solve the series? Thank you.
AI: To evaluate the final sum, rewrite it as a telescoping sum as follows: $$\sum_{k=0}^{\infty} ke^{-k}(1-e^{-1}) = \sum_{k=0}^{\infty} ke^{-k} -ke^{-(k+1)} $$
Writing out the first couple of terms: $$ 0-0 + \frac{1}{e} - \frac{1}{e^2}+\frac{2}{e^2}-\frac{2}{e^3}+\frac{3}{e^3}-\frac{3}{e^4} = \frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+... $$
This is an infinite geometric series and $\frac{1}{e}$ is within the radius of convergence, that is $\big|\frac{1}{e}\big|\le 1$. It is, however, missing the first term, $1$.
$$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r} $$
Setting $r=\frac{1}{e}$ and subtracting $1$, the sum is now $$\frac{1}{1-\frac{1}{e}}-1=\frac{e}{e-1}-\frac{e-1}{e-1}=\frac{1}{e-1} $$ |
H: Let $F$ be a field of characteristic $p$. Do the morphism $x^p-a$ always has a root in $F$ for every $a \in F$?
Do you have an example in which $x^p - a$ does not have a solution in $F$?
AI: Take the field $\Bbb F_p(t)$ of rational functions in $t$ with coefficients in $\Bbb F_p$, then $x^p-t$ does not have a root.
Indeed, the Frobenius endomorphism $x\mapsto x^p$ on $\Bbb F_p(t)$ sends $f(t)\mapsto f(t^p)$ for any $f(t)\in\Bbb F_p(t)$ (this is because $x^p=x$ for $x\in\Bbb F_p$). Therefore, the image of the Frobenius endomorphism is $\Bbb F_p(t^p)$, which does not contain $t$, showing that there is no $f(t)$ such that $f(t)^p=t$. |
H: Determining whether the function is exponential?
Here's a screenshot of the problem:
Because the $x$-value is an exponent, then this must be an exponential function. By definition, an exponential function is where the independent variable (the $x$-value) is the exponent.
To write the function in the form $K(x) = ab^x$, I first converted the radical into its exponent form and moved the denominator over to the numerator to get
$K(x) = (3^x)(3^{-\frac{1}{2}})(6^{-x})$
I then simplified it further, following the exponent rules of $a^ma^n = a^{m + n}$
$K(x) = (3^{x - \frac{1}{2}})(6^{-x})$
However, as shown, my answers were incorrect. What did I do wrong?
AI: Because the x-value is an exponent, then this must be an exponential function. By definition, an exponential function is where the independent variable (the x-value) is the exponent.
While you're not wrong in this respect, this only really holds if the function is in its simplest form. Consider, for instance,
$$f(x) = \frac{3^x}{3^x}$$
Sure, the variable $x$ is only in exponents here. However, if you simplify, you realize immediately $f(x) = 1$, which is obviously not exponential.
So we have to be very careful about how we define things as a result of things like these. We define $f$ to be an exponential function if there exist constants $a,b$ such that $f(x) = ab^x$. To show a function is exponential, then, you need to somehow manipulate it into that form and determine the constants $a,b$ necessary.
To reiterate: this is the real definition of an exponential function. (Well, to an extent; there are modifications to the definition you can make, but this is the most relevant one for your case.) The notion of "$x$ appears in the exponent" is an intuition, but clearly doesn't always hold.
So, with this mind, we need to figure out what constants $a,b$ you need for
$$f(x) = \frac{3^x}{\sqrt 3 \cdot 6^x}$$
Well, first note that, through exponent rules,
$$f(x) = \frac{1}{\sqrt{3}} \cdot \frac{3^x}{6^x} = \frac{1}{\sqrt 3} \cdot \left( \frac 3 6 \right)^x = \frac{1}{\sqrt 3} \cdot \left( \frac 1 2 \right)^x$$
With this manipulation, it's clear what your constants $a,b$ are, and thus you can conclude that $f$ is exponential.
Again, be very careful about what $a,b$ can be: this is where you went wrong. They have to be constants, not functions themselves. $3^{x-0.5}$ is not a constant. Basically, you cannot have your constants depend on $x$. |
H: Linear program for flow into a node should exit all on one edge
Suppose I'm given a problem where I want to route some flow from a set of sources to a set of sinks in a directed graph; however, as opposed to the standard flow constraints, I also want to constrain some nodes in the following way: all flow into the node must leave along only one outgoing edge. In other words, the normal flow constraint on the non-source/non-sink nodes is the following: $$\sum_{u \in V} f_{(u, v)} - \sum_{w \in V} f_{(v, w)} = 0.$$
However, for vertex $v$, I instead want: $\sum_{u \in V} f_{u, v} - f_{(v, w)}= 0$ for exactly one outgoing edge $(v, w)$. (All other outgoing edges have zero flow.) How do I write a set of linear constraints to ensure this fact in an ILP? There must be a fairly standard way to do this but I am having trouble formulating it/finding it via searching.
AI: Let $M$ be a big number.
$$\sum_{u \in V} f_{u,v} - \sum_{w \in V} f_{v,w}=0$$
We can introduce some indicator variable, $x_{v,w}= \begin{cases} 1 &, f_{v,w}>0 \\ 0 &, f_{v,w}=0 \end{cases}$
We impose the following constraints:
$$Mx_{v,w}\ge f_{v,w}$$
$$x_{v,w} \le Mf_{v,w}$$
$$\sum_{w \in V} x_{v,w}\le 1$$ |
H: Proof-verification: Any prime $p>3$ can be expressed in the form $(6n+1)$ or $(6n+5)$
I have already seen answers to this question, but I would like to get my own proof verified.
Proposition: Any prime $p>3$ can be expressed in the form $(6n+1)$ or $(6n+5)$.
Definition: A prime $p$ is a natural number greater than $1$ which is divisible only by $1$ and itself. Thus, no prime $p>3$ is divisible by $3$.
Proof: $$p=3n'+r\;\mathrm{where}\;0<r<3$$
$r\neq 0,3$ because stating otherwise would imply divisibility by 3. The above statement is equivalent to saying: $$(p=3n'+1)\,\mathrm{or}\,(p=3n'+2)$$ Further, $(3n'+2)\equiv5\mod3\iff(3n'+5)\equiv2\mod3$. Thus, $$p=(3n'+1)\;\mathrm{or}\;p=(3n'+5)$$Substituting $n'=2n$, we get: $$p=(6n+1)\;\mathrm{or}\;p=(6n+5)$$ Which completes the proof.
I'm sceptical of the step where I used congruence to switch $2$ and $5$. I would like to know the flaws in my proof, and measures that I can take to rectify them.
AI: Regarding your proof, you need to consider the case where $n'$ is odd: when $n' = 2k+1$ for a positive integer $k$. When $n'$ is odd, what happens to the parity of $p$ in both cases? |
H: Seeking Starting Point for $f(a) = \int_0^1 \frac{\ln\left(ax^2 + 1\right)}{x + 1}\:dx$
I would like to solve the following parameterised definite integral.
$$
f(a) = \int_0^1 \frac{\ln\left(ax^2 + 1\right)}{x + 1}\:dx
$$
Where $a \in \mathbb{R}^+$
I have tried a few different methods that haven't resulted in anything useful. I was hoping to ask for a starting point on this question.
Note - if possible, I'm trying to resolve this integral using Real Analysis only.
As per a recommendation provided, I will employ Feynman's Trick (coupled with the Dominated Convergence Theorem and Leibniz's Integral Rule) and differentiate. Firstly though, we observe that $f(0) = 0$. We now proceed by differentiating under the curve with respect to $a$:
\begin{align}
f'(a)& = \frac{d}{da}\int_0^1 \frac{\ln\left(ax^2 + 1\right)}{x + 1}\:dx = \int_0^1 \frac{\frac{\partial }{\partial a}\big[\ln\left(ax^2 + 1\right)\big]}{x + 1}\:dx \\
& = \int_0^1 \frac{x^2}{ax^2 + 1} \cdot \frac{1}{1 + x}\:dx = \int_0^1 \frac{x^2}{\left(ax^2 + 1\right)\left(x + 1\right)}\:dx
\end{align}
We now apply a Partial Fraction Decomposition on the integrand:
\begin{align}
f'(a) &= \int_0^1 \frac{1}{a + 1}\left[\frac{x}{ax^2 + 1} - \frac{1}{ax^2 + 1} + \frac{1}{x + 1}\right]\:dx\\
& = \frac{1}{a + 1}\left[\frac{1}{2a}\ln\left(ax^2 + 1\right) - \frac{1}{\sqrt{a}}\arctan\left( \sqrt{a}x\right) + \ln\left(x + 1\right) \right]_0^1 \\
&= \frac{1}{a + 1}\left[\frac{\ln\left(a + 1\right)}{2a} - \frac{\arctan\left( \sqrt{a}\right) }{\sqrt{a}} + \ln(2) \right]
\end{align}
From here, we resolve $f(a)$ by integrating with respect to $a$:
\begin{align}
f(a) &= \int \frac{1}{a + 1}\left[\frac{\ln\left(a + 1\right)}{2a} - \frac{\arctan\left( \sqrt{a}\right) }{\sqrt{a}} + \ln(2) \right]\:da \\
&= \frac{1}{2}\int \frac{\ln(a + 1)}{a\left(a + 1\right)}\:da - \int \frac{\arctan\left(\sqrt{a}\right)}{\sqrt{a}\left(a + 1\right)} + \ln(2)\int \frac{1}{a + 1}\:da \\
&= \frac{1}{2}A - B + \ln(2)D
\end{align}
We now resolve each individually (note I will omit the constant of integration till the end). Starting with the easiest $D$:
$$
D = \int \frac{1}{a + 1}\:da = \ln(a + 1)
$$
We now resolve $A$ by applying a Partial Fraction Decomposition:
\begin{align}
A &= \int \ln(a + 1) \left[ \frac{1}{a} - \frac{1}{a + 1} \right]\:da = \int \frac{\ln(a + 1)}{a}\:da - \int \frac{\ln(a + 1)}{a + 1}\:da \\
&= -\operatorname{Li}_2(-a) - \frac{1}{2}\ln^2(a + 1)
\end{align}
We now resolve $B$. Here we make the substitution $a = b^2$, $b > 0$
\begin{align}
B = \int \frac{\arctan(b)}{b\left(b^2 + 1\right)} \cdot 2b \:db = 2\int \frac{\arctan(b)}{b^2 + 1}\:db = 2 \cdot \frac{1}{2}\arctan^2(b) = \arctan^2\left(\sqrt{a}\right)
\end{align}
Thus, we now may form $f(a)$
\begin{align}
f(a) &= \frac{1}{2}A - B + \ln(2)D \\
&=\frac{1}{2}\left[-\operatorname{Li}_2(-a) - \frac{1}{2}\ln^2(a + 1) \right] - \arctan^2\left(\sqrt{a}\right) + \ln(2)\ln(a + 1) + C
\end{align}
Where $C$ is the constant of integration. We resolve $C$ using $f(0) = 0$:
\begin{align}
f(0) = 0 = \frac{1}{2}\left[-\operatorname{Li}_2(0) - \frac{1}{2}\ln^2(1) \right] - \arctan^2\left(\sqrt{0}\right) + \ln(2)\ln(0 + 1) + C = 0 + C \rightarrow C = 0
\end{align}
And so,
$$
\int_0^1 \frac{\ln\left(ax^2 + 1\right)}{x + 1}\:dx = \frac{1}{2}\left[-\operatorname{Li}_2(-a) - \frac{1}{2}\ln^2(a + 1) \right] - \arctan^2\left(\sqrt{a}\right) + \ln(2)\ln(a + 1)
$$
AI: Feynman's trick seems to be a way
$$f(a) = \int_0^1 \frac{\log\left(ax^2 + 1\right)}{x + 1}\,dx$$
$$f'(a) = \int_0^1 \frac{x^2}{(x+1) \left(a x^2+1\right)}\,dx=\frac 1 {a+1} \int_0^1\left(\frac{x-1}{a x^2+1}+\frac{1}{x+1}\right)\,dx$$Assuming $-1\leq \Im\left(\sqrt{a}\right)\leq 1\lor \Re\left(\sqrt{a}\right)>0$
$$f'(a)=\frac{\log (2)}{a+1}+\frac{\log (a+1)}{2 a(a+1)}-\frac{\tan ^{-1}\left(\sqrt{a}\right)}{\sqrt{a} (a+1)}$$
$$\int f'(a)\,da=\log (2) \log (a+1)-\frac{1}{2}\text{Li}_2(-a)-\frac{1}{4} \log ^2(a+1)-\arctan ^{2}\left(\sqrt{a}\right)$$ |
H: Spivak's Calculus Assumptions 2
Spivak's Calculus is known for being the best Calculus book when it comes to rigour, but I think I've found an assumption he makes during his very first proof!
The axioms are:
P1 (Associativity of Addition): (a + b) + c = a + (b + c)
P2 (Additive Identity): a + 0 = a
P3 (Additive Inverse): a + (-a) = 0
P4 (Commutativity): a + b = b + a
The Proof:
(1) If a + x = a,
(2) Then (-a) + (a + x) = (-a) + a *(adding (-a) to both sides)
(3) Then ((-a) + a) + x = 0 *(using P1 and P3)
(4) Then 0 + x = 0 *(using P3)
(5) Then x = 0 *(using P4 and P2)
Conclusion: If a + x = a, Then x = 0
My problem is in the very first step from (1) to (2). Where is the justification for adding (-a) to both sides? Seems to me if we are going to do such formal proofs with such basic axioms for such simple theorems, such an assumption is quite severe? May as well assume the conclusion! To clarify, mathematically the assumption is:
If a = b, Then a + c = b + c
You can only add numbers to both sides of the equation if you establish this truth (either as a theorem or an axiom). I don't think it's possible to prove it with P1-P4.
So should it be included as one of Spivak's axioms?
Am I missing something?
AI: You are right, Spivak hasn't defined what the symbol $=$ means and well a plethora of other problems if you want to be absolutely formal. But I believe you can make the exact same objection about any step of the proof. What does it mean mathematically for something to follow? Why if (2) then (3)? The thing is, Spivack is using another structure implicitly which is first order logic, that provides a way to structure mathematical thought unambiguously. In https://en.wikipedia.org/wiki/First-order_logic#Equality_and_its_axioms you can see that you can define the symbol, but well in the same page you can see it still not as simple as just making a list of more and more axioms. You can see that it follows from substitution for formulas. I would still consider Spivack rigorous in the same way Euclid's Element is rigorous, the things that are implicitly assumed are for the purposes of the book "universal", even though by modern standards of formality they clearly aren't. Though you would need to read on logic and set theory to have the modern perspective and it is no simple task. |
H: Quetion about this irrationality proof.
I found this proof on mathoverflow. It's about the irrationality of $(\arcsin 1/4)/\pi$.
My question is: Assuming that we don't know the value of $(\arcsin 1/2)$, and only knew that $\sin(\arcsin 1/2) = 1/2$, then this proof wouldn't work just the same to establish the irrationality of $(\arcsin 1/2)/\pi = 1/6?$
AI: No. It would not work the same way. With $\theta=\arcsin(1/2)$ the argument would still lead to the conclusion that $2i\sin\theta$ is an algebraic integer. However, this time $$2i\sin\theta=2i\cdot\frac12=i$$ actually is an algebraic integer, a zero of the monic polynomial $x^2+1$ with integer coefficients. So we did not reach that same contradiction. |
H: Does $\lim_{n \to \infty}\sum_{k=1}^n \left[\zeta\left(2k-1-\frac{1}{2n}\right) + \zeta(2k)\right]$ equal the Euler-Mascheroni constant?
Let $\zeta(s)$ be the Riemann zeta function and $\gamma$ be the Euler-Mascheroni constant. Is the following formula for the Euler-Mascheroni constant true?
$$
\lim_{n \to \infty}\sum_{k=1}^n \left[\zeta\left(2k-1-\frac{1}{2n}\right) + \zeta(2k)\right]
= \gamma
$$
Related question:
AI: Yes, and the method is the same. Writing $f(n)\approx g(n)$ for $f(n)=g(n)+\mathcal{O}(n^{-1})$ as $n\to\infty$,
\begin{align*}
\sum_{k=1}^n\left[\zeta\left(2k-1-\frac{1}{2n}\right)+\zeta(2k)\right]
&\approx-2n+\gamma+\zeta(2)+\sum_{k=2}^n\sum_{m\geqslant 1}(m^{1+1/(2n)}+1)m^{-2k}
\\&=-2+\gamma+\zeta(2)+\sum_{k=2}^n\sum_{m>1}(m^{1+1/(2n)}+1)m^{-2k}
\\&\approx-2+\gamma+\zeta(2)+\sum_{m,k>1}(m^{1+1/(2n)}+1)m^{-2k}
\\&=-2+\gamma+\zeta(2)+\sum_{m>1}\frac{m^{1+1/(2n)}+1}{m^2(m^2-1)}
\\&=-2+\gamma+\zeta(2)+\sum_{m>1}\frac{1}{m^2(m-1)}+\sum_{m>1}\frac{m^{1/(2n)}-1}{m(m^2-1)}
\\&\approx-2+\gamma+\zeta(2)+\sum_{m>1}\frac{1}{m^2(m-1)},
\end{align*}
and the last sum is equal to $$\sum_{m>1}\left(\frac{1}{m-1}-\frac{1}{m}\right)-\sum_{m>1}\frac{1}{m^2}=1-\big(\zeta(2)-1\big)=2-\zeta(2).$$ |
H: Homeomorphism between circle and ellipse
I want to show that the circle represented by $x^2+y^2=1$ is homeomorphic to the ellipse $x^2/4+y^2=1$
They are both subsets of $R^2$ of course.
I just don’t know what the functions would look like
What is the function taking the circle to the ellipse and vice versa and how to show that it is continuous
AI: A homeomorphism is a continuous bijective map that has a continous inverse. Thus, you do not only need to prove continuity but additionally that your inverse is continuous, too.
As mentioned in the comments, the map $$\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} \frac{x}{2} \\ y \end{pmatrix} $$ does the job.
It's clearly bijective and continous. It's hopefully obvious what the inverse is and you can easily check that it is indeed continous, therefore it's a homeomorphism. |
H: Kernel of matrix $\begin{pmatrix}x&y&z\end{pmatrix}$ over $k[x,y,z]$
Let $R=k[x,y,z]$ and let $M=\langle f_1,f_2,f_3\rangle$ where
$$
f_1=\begin{pmatrix}y\\-x\\0\end{pmatrix},\;f_2=\begin{pmatrix}z\\0\\-x\end{pmatrix},\;f_3=\begin{pmatrix}0\\z\\-y\end{pmatrix}.
$$
Show $M=\ker A$ where $A=\begin{pmatrix}x&y&z\end{pmatrix}$.
What I have so far:
Let $f\in M$. So $f=af_1+bf_2+cf_3$ for some $a,b,c\in R$. Observe
$$
\begin{align*}
Af&=aAf_1+bAf_2+cAf_3\\
&=a\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}y\\-x\\0\end{pmatrix}+b\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}z\\0\\-x\end{pmatrix}+c\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}0\\z\\-y\end{pmatrix}\\
&=a(yx-xy+0z)+b(zx+0y-xz)+c(0x+zy-yz)\\
&=0.
\end{align*}
$$
Therefore $f\in\ker A$. For the reverse direction, let $g\in\ker A$. How can I write $g$ as a linear $R$-combination of the $f_1,f_2,f_3$? This shouldn't be very complicated, for some reason I just can't see it right now.
AI: Let $g=(p,q,r)^t\in \ker A$ so that
$$
xp+yq+zr=0. \label{eq1}\tag{$*$}
$$
Since $k[x,y,z]=k[x,y][z]$, we may uniquely decompose $p,q$ as
\begin{align*}
p &= p_1 + z p_2, \\
q &= q_1 + z q_2
\end{align*}
with $p_1,q_1\in k[x,y]$ and $p_2, q_2\in k[x,y,z]$.
Evaluating at $z=0$, the equation \eqref{eq1} becomes $xp_1+yq_1 = 0$ in $k[x,y]$, so that $p_1=ay$, $q_1=-ax$ for some $a\in k[x,y]$.
We conclude that
$$
g=\begin{pmatrix}p\\q\\r\end{pmatrix}
=\begin{pmatrix}ay+zp_2\\-ax+zq_2\\r\end{pmatrix}
= a f_1 + \begin{pmatrix}zp_2\\zq_2\\r\end{pmatrix}.
$$
Since $af_1\in \ker A$, we know that $(zp_2, zq_2, r)^t$ is still an element of the kernel, so we have
$$
xzp_2 + yzq_2 + zr = 0,
$$
which yields
$$
r = -xp_2 - yq_2.
$$
Finally,
$$
g= a f_1 + \begin{pmatrix}zp_2\\zq_2\\-xp_2-yq_2\end{pmatrix}
= a f_1 + p_2 f_2 + q_2 f_3 \in M.
$$ |
H: If $z \in \mathbb C$ such that $|z|+|z-2019|=2019$ then $z \in \mathbb R$
Let $z$ be a complex number such that $|z|+|z-2019|=2019$. Note that
$$|z+(2019-z)|=2019=|z|+|z-2019|=|z|+|2019-z|$$
This equality occurs when $0,z,2019-z$ are collinear.
But, how to show that z is a real number from that?
Note. By using the definition of modulus, i can show that $\text{Im}(z)=0$ from the equation $|z|+|z-2019| = 2019$.
But, i wonder if i can get the same result with the previous way. Thanks.
AI: $|a+b|=|a|+|b|$ iff $a =t b$ for some $t \geq 0$ (or $b =t a$ for some $t \geq 0$). Here we get $z=t(z-2019)$ (which implies $t \neq 1$). So $z= \frac{(2019) t} {t-1}$ which is real. |
H: Are there elements of sets that are no sets in Zermelo Fraenkel Set theory?
I've seen the ZFC formalised in a lecture where the lecturer introduced, part by part, propositional logic, 1. order logic, and then zermelo-fraenkel-set theory. The lecturer didn't introduce any notion of identity ("=") in the part about 1. order logic, and defined identity in the part about set theory.
There, two sets where defined equal when from something being an element of the first set followed that it must be an element of the 2nd set as well, and vice versa (the definition captures what I read would be the "axiom of extensionality", later on).
However, in the further proceeding of the lecturer (here: https://youtu.be/AAJB9l-HAZs?t=4456), the lecturer used the "=" sign, and the notion of identity, not only for sets, but also for elements of sets.
Did he, in this moment, assume that every element is a set? And does this mean (for the further use of the ZFC) that I can only use ZFC to describe "collections" of entities that are as well a set?
AI: In ZFC, everything is a set, although we do use shorthands like $15$ and $G$ when we want to refer to mathematical objects that "we know" aren't sets. The idea is that all mathematical objects can be coded up as sets, so we forget that there are any other mathematical objects than sets.
For example $0$ can be implemented as $\emptyset$, $1$ can be implemented as $\{\emptyset\}$, and in general the number $n$ can be implemented as the set of all smaller numbers. (This is just one possible choice of implementation, but it turns out to be convenient.) Given an encoding of the natural numbers into set theory, we can then construct e.g. the reals using your favourite quotients of appropriate products. This allows ZFC to talk about real numbers (although it does so in an extremely unnatural way).
The general point is that by removing all the objects of maths other than sets, we get a simpler theory which we can talk more easily about.
For set theories which contain non-sets, look up "urelements" a.k.a. "atoms". |
H: Two players until one player wins three games in a row. Each player will win with probability $\frac{1}2$. How many games will they play?
QUESTION: Suppose two equally strong tennis players play against each other until one player wins three games in a row. The results of each game are independent, and each player will win with probability $\frac{1}2$. What is the expected value of the number of games they will play?
MY APPROACH: I have tried to set up some kind of recurrence relation here but could not succeed.. Observe that there can be at most a winning streak of $2$. A winning streak of $3$ means that the game ends.. If we assume that the number of $1$ game winning streak is $x$ and the number of $2$ game winning streak is $y$ then $x+y+1$ obviously yields the desired answer..
Note: A $1$ game winning streak simply means that they win alternately.. Since each one of them has a $50\%$ chance of winning therefore we can do this..
Now, somehow, we have to find the value of $x$ and $y$.. But here I am stuck.. With so less information, neither can I set up a recurrence relation, nor do I see any way to compute two variables..
Any help will be much appreciated.. :)
AI: Recurrence works fine.
All we care about is the length of the current win streak, we don't even care who has been winning. Accordingly, let $E_i$ denote the expected number of games it will take if one player currently has a winning streak of length $i$. The answer we seek is $E_0$.
We get: $$E_2=\frac 12\times 1+\frac 12\times (1+E_1)=1+\frac 12\times E_1$$
Similarly $$E_1=1+\frac 12\times (E_1+E_2)$$ and $$E_0=1+ E_1$$
This is easily solved and yields $$\boxed {E_0=7}$$ |
H: Convergence of reciprocal of a linear function
For what values of $m$ does the integral $\int_m^\infty \frac{1}{4x-16}\;dx$ converge?
I began by taking the following limit $$\lim\limits_{k\rightarrow\infty}\frac{1}{4}\int_m^k\frac{1}{x-4}\;dx=\lim\limits_{k\rightarrow\infty}\left(\frac{1}{4}\ln{|x-4|}\right)\bigg|_m^k$$
which, as far as I can tell, diverges no matter the $m$.
The next thought was that maybe the question was playing fast and loose with the term "value" and I could try letting $a\rightarrow\infty$ since the symmetry of the function suggests that the integral could be $0$.
However, splitting up the integral $$\int_{-\infty}^\infty\frac{1}{4x-16}\;dx=\int_{-\infty}^4\frac{1}{4x-16}\;dx+\int_4^\infty\frac{1}{4x-16}\;dx$$
just leads to divergence as well. Is there something I'm missing here? Are there truly values of $m$ for which we have convergence?
AI: For $m \ge 4$ the integral $\int_m^{\infty}\frac{1}{x-4}\;dx$ is divergent.
If $m<4$ the integral $\int_m^{\infty}\frac{1}{x-4}\;dx =\int_m^{4}\frac{1}{x-4}\;dx+\int_4^{\infty}\frac{1}{x-4}\;dx$ is divergent.
Conclusion: there is no $m$ such that $\int_m^\infty \frac{1}{4x-16}\;dx$ converges. |
H: Given joint pdf, how can I find $\operatorname{Var}(X)$ without using the marginal distribution of $X$?
$(X,Y)$ has joint pdf $\frac{1}{y}$ for $0<x<y<1$.
I would usually use the marginal pdf to get the expected value. But the question doesn't let me use the marginal distribution of $X$. I think this means that I am not allowed to use the marginal pdf... Is there any other way to do this?
AI: The variance is $\int\int x^{2} f(x,y)dxdy -(\int\int x f(x,y)dxdy)^{2}$. This can be written as $(\int_0^{1} \int_o^{y}x^{2} \frac 1 y dxdy-\int_0^{1} \int_o^{y}x \frac 1 y dxdy)^{2}$. I will leave the rest to you. |
H: Number of subsets with $m$ elements of a set with $n$ elements is $\frac{n!}{m!}$
I wonder if my proof of the statement in the title is correct:
Let $n,m\in\mathbb{N}: m<n$.
Let $X_n$ be a set with $n$ elements.
Let $A_m\subset X_n$ with $m$ elements.
Let $B_m$ be a set of all possible sets $A_m$.
$A_0=\emptyset\implies B_0=\{\emptyset\}\implies |B_0|=1$
$B_{m+1}=\{A_m\cup x\quad \forall x\in A_m',\forall A_m\in B_m, \forall m\}$
$|A_m'|=|X_n|-|A_m|=n-m$
$|B_{m+1}|=|B_m|\times|x\in A_m'| = |B_m|\times|A_m'| = |B_m|\times(n-m)$
$|B_m|=1\cdot n\cdot (n-1)\cdot \dots \cdot (n-m)=\frac{n!}{m!}$
I suppose the idea is correct but my notation may be misleading or incorrect in some places.
AI: No, the number of $m$-set elements from a set of $n$ elements (so indeed $m \le n$) is exactly $\binom{n}{m}$, almost by (the semantic/combinatorial) definition of the binomial coefficient.
If we pick a subset in order, we cannot re-use elements, so we have $n \times (n-1) \times \ldots$ ($m$ terms) choices, which is $\frac{n!}{(n-m)!}$ but then all $m!$ sequences that lead to the same set (which have no order) are identified, and so we divide by $m!$ again and so we get $$\frac{n!}{(n-m)!m!}$$ which is the algebraic definition of $\binom{n}{m}$. |
H: Prove that $10^n + 1 \equiv 0 \ \mod \ 1 \ldots 1, n \geqslant 2$ has no solutions.
Prove that $$10^n + 1 \equiv 0 \ \mod \ 1 \ldots 1 = \dfrac{10^k-1}{9}, k \geqslant 3, n \geqslant 2$$ has no solutions. Where the number of units is greater than or equal to $3$.
AI: Let $m=\frac{10^k-1}{9}$ so that $10^k=9m+1\equiv 1\pmod m$.
Let $n=qk+r$ with $0\leq r<k$.
Then
$$10^n=(10^k)^q10^r\equiv 10^r\pmod m.$$
If $10^r+1\equiv 0\pmod m$, then $m|(10^r+1)$ hence for $k\geq 3$ we get the contradiction
$$10^{k-1}+1<m\leq 10^r+1\leq 10^{k-1}+1.$$ |
H: Linear independence in $\mathbb{Z}_2$ and $\mathbb{R}$
Let $v_1,...,v_k$ be vectors whose entries are all in $\{ 0, 1 \}$. These vectors can be considered both as elements of $\Bbb{Z}_2^n = \{0,1\}^n$ (with modulo 2 addition), and as elements of $\Bbb{R}^n$ (with regular addition).
Question: Is it true that $v_1,...,v_k$ are linearly independent with respect to $\mathbb{Z}_2$ (modulo 2 addition) $\Leftrightarrow v_1,...,v_k$ are linearly independent with respect to $\mathbb{R}$?
If not, is either direction true?
AI: Since these vectors have entries in $\{ 0, 1 \}$, it follows that any linear dependence of such vectors over $\Bbb{R}$ is actually (up to a constant multiple) a linear dependence over $\Bbb{Q}$, which can easily be turned into a linear dependence over $\Bbb{Z}$ by clearing denominators, and then by reducing mod $2$ can be turned into a linear dependence over $\Bbb{Z}_2$. (If this linear dependence over $\Bbb{Z}_2$ is initially trivial, that means the $\Bbb{Z}$-coefficients of the original linear dependence all had a factor of $2$ in them to begin with, which may be cancelled out; if the resulting linear dependence over $\Bbb{Z}_2$ is still trivial, we repeat as necessary until not all the $\Bbb{Z}$-coefficients are even, at which point we get the desired nontrivial linear dependence relation over $\Bbb{Z}_2$.)
But the other direction isn't true: the vectors $[1, 0, 0, 1], [1, 1, 0, 0], [0, 1, 0, 1]$ are linearly dependent over $\Bbb{Z}_2$ (their sum is $[0, 0, 0, 0]$), but linearly independent over $\Bbb{R}$. |
H: Comparison of Product-Like Topologies
Let $X$ be a Banach space, let $\{w_n\}_{n=1}^{\infty}$ be a sequence of (strictly) positive real numbers, and consider the two associated topological spaces $\prod_{n=1}^{\infty} X$ and $$
X_1:=\left\{
x \in \prod_{n=1}^{\infty} X:\, \sum_{n=1}^{\infty} w_n \|x\|_X < \infty
\right\}
$$
where $\|\cdot\|_X$ is the norm on $X$ and $X_1$ is equipped with the metric
$$
d_1((x_n),(y_n)):= \sum_{n=1}^{\infty} w_n \|x_n - y_n\|
.
$$
Since $\prod_{n=1}^{\infty} X$ is a countable product then its topology is given by the metric
$$
d_{\prod}((x_n),(y_n)):= \sum_{n=1}^{\infty} \frac{1}{2^n}\frac{\|x_n-y_n\|}{1 +\|x_n-y_n\|}.
$$
It seems to me that there should be a constant $C>0$ such that $d_{\prod}\leq C d_1$ but not ther other way around. I.e.: The metric on $\prod_{n=1}^{\infty} X$ restricted to the subset $X_1$ should be strictly coarser than the topology on $X_1$ defined by the metric $d_1$. But is this true?
AI: There need not be any $C$ such that $d_{\prod} \leq C d_1$. Take $X=\mathbb R$ and suppose $w_n=\frac1 {4^{n}}$. Let $k$ be such that $2^{k} >2C$. Let $(x_n)=(0,0,...)$ and $y_n=1$ when $n=k$, $y_n=0$ when $n \neq k$. Then the inequality becomes $\frac 1{2^{k+1}} \leq C \frac 1 {4^{k}}$ or $2^{k} \leq 2C$.
There may be no such inequality in the opposite direction either. |
H: What is the probability that player A rolls a larger number if player B is allowed to re-roll (20-sided die)?
The problem statement is:
2 players roll a 20-sided die. What is the probability that player A rolls a larger number if player B is allowed to re-roll a single time?
The question is a bit ambiguous, but I am going to operate on the following 2 assumptions:
(a) Player B doesn't know what player A rolls when deciding whether or not to reroll.
(b) If player B re-rolls, his first roll is discarded. In other words, when comparing player A's roll to player B's roll, only the last roll of player B is taken into account.
(c) Player B doesn't want player A to win, so it will play optimally.
I solved this problem, but it seems my solution does not match the answer given , which is $\frac{1}{4}$. Below is my solution process.
I know the following:
(1) Probability that A rolls a larger number if player B isn't allowed to re-roll. The probability that they roll the same number is $\frac{20}{400}$. The probability that player A rolls a larger number is thus $\frac{190}{400} = \frac{19}{40}$.
(2) How does player B decide if they should toss again? It's obvious to me that he should toss again if the first toss is $\leq 10$. If he tosses $> 10$, he should not toss again. So with probability $0.5$, he will get an expected value of $15.5$, and with probability $0.5$, he will toss again and get expected value $10.5.$
His expected outcome toss when considering that he can re-roll is thus
$$
E[B] = 0.5 \cdot 15.5 + 0.5 \cdot 10.5 = 13
$$
2.5 higher than the case where he isn't allowed to re-roll. Seems reasonable...
I found the threshold of $b = 10$ (where $b$ is the largest value on the first toss at which player B decides to do a second toss) by intuition, but we could have formulated an optimization problem
$$
\arg \max_b \frac{20-b}{20} \frac{20 + b + 1}{2} + \frac{b}{20} 10.5
$$
and solved for $b$ that maximizes $E[B]$.
Then I define disjoint events to be $B$ deciding to retoss (denote as $RR$) and $B$ deciding not retoss (denote as $NR$). Then we can write
$$
P(A > B) = P(A > B | RR) P(RR) + P(A > B | NR) P(NR)
$$
Previously we saw that $P(RR) = P(NR) = 0.5$.
For $P(A > B | RR)$, where player $B$ retosses, I believe the probability that I computed in (1) is the same as the conditional probability $P(A > B | RR)$, i.e., $P(A > B | RR) = \frac{19}{40}$. I think this is true because the die tosses are IID and memoryless. So that when $B$ retosses, we can treat this case as simply both $A$ and $B$ tossing a single time.
For $P(A > B | NR)$, when we condition on $NR$, i.e., $B$ stopping on the first toss, then this means that $B$ rolled a $11, 12, \ldots, 20$. There are $20 \cdot 10$ possibly outcomes for $(A,B)$ conditioned on $NR$. $9 + 8 + \ldots 1 = 45$ of these outcomes are such that $A > B$. So $P(A > B | NR) = \frac{45}{200} = \frac{9}{40}$
So $P(A > B) = \frac{19}{80} + \frac{9}{80} = \frac{28}{80} = \frac{7}{20}$ for the case where $B$ is allowed to re-toss. This is only $\frac{1}{8}$ less than the case where $B$ isn't allowed to re-toss. This seems reasonable.
I don't think I made a mistake in my solution, but it doesn't match $\frac{1}{4}$.
AI: What you have done looks right (and I've checked the calculations and get the same answer). In particular, if B doesn't know the result of A's roll, it is correct to reroll on 10 or below, and keep the original roll on 11 or above, since if B keeps a roll of $r$ the chance of B winning is $r/20$, whereas if B rerolls the chance of winning is $21/40$.
The value of $1/4$ cannot possibly be correct. Even if we give B every possible advantage, by letting them pick the higher of two rolls rather than having to choose before seeing the second (and assuming that A has to get strictly higher to win), A wins more than $1/4$ of the time. This is because if all three rolls are different, A wins with probability $1/3$, and all three rolls are different with probability $\frac{19}{20}\times\frac{18}{20}$, so A's chance of winning must be greater than $\frac{19}{20}\times\frac{18}{20}\times\frac13=0.285$. (In fact the exact value with these assumptions would be $\frac{247}{800}$.) |
H: Subset of a normal subgroup is a subgroup?
I am studying Dummit & Foote abstract algebra. This question is what I have a problem.
Let $G$ be a finite group, $H$ be a subgroup of $G$, $N$ be a normal subgroup of $G$. If $|H|$ and $|G:N|$ are relatively prime, prove that $H$ is a subgroup of $N$.
I founded that $H$ is a subset of $N$. But I can't find how to prove about subgroup. Does normality implies that subset becomes a subgroup? If not, how can I prove it?
AI: Your hypothesis includes that H is a subgroup of G, to prove that it is a subgroup of N, you only need to prove that it is a subset of N which you just did. |
H: Distinct subrings of cardinality $p$ of the field $\mathbb F_{p^2} $
This particular question was asked in masters entrance of a university for which I am preparing.
Question Let $p$ be a prime number. How many distinct subrings ( with unity) of cardinality $p$ does the field $\mathbb F_{p^2}$ have?
$0$
$1$
$p$
$p^{2} $
Unfortunately, I am unable to even start this question as such kind of questions were not done during Abstract Algebra course. So, I am unable to show any attempt.
Also, this question was asked before here:
Number of subrings of a Field
But I had some questions in proof of Answer, so I asked user who answered to supply the proof. But as he didn't supplied the proof, so I have no option other than asking it here.
Question in the answer of User who answered the question in link are :
1.why subring will always be a vector space over prime field?
2.Can you please prove it. Also, how does dimension of $F_{p^2}$ over prime field must be 2.
Also if someone wants to answer with a different approach, that also would be really appreciated.
But kindly give a proof of a result which you are assuming.
AI: A subring contains $1$, so it contains $2$, $3, \ldots, p-1$ so all of the subfield $\Bbb F_p$. As this subfield is a ring and has $p$ elements, that's all the subring can be. So the second answer $1$ is correct: the unique subring of $p$ elements is $\Bbb F_p$, the prime subfield. |
H: Is $R(T)$ a closed subspace of $\ell^2$?
Let $T$ be the diagonal operator on the Hilbert space $\ell^2$ such that
$$T=\text{diag}(0,\frac{1}{2},\frac{1}{3},\ldots).$$
Clearly the range of $T$ denoted $R(T)$ is not $\ell^2$ since $e_1=(1,0,0...)\notin R(T)$.
Is $R(T)$ a closed subspace of $\ell^2$?
AI: It is not closed. The sequences $(0,\frac 1 {1^{3/2}},\frac 1 {2^{3/2}},..., \frac 1 {N^{3/2}},0,0,...)$, $N=1,2,...$, are all in the range of $T$ but their limit $(0,\frac 1 {1^{3/2}},\frac 1 {3^{3/2}},,...)$ is not in the range of $T$: If you write this as $T(a_n)$ you would get $\sum a_n^{2}=\infty$, a contradiction. |
H: Congruent numbers have congruent squarefree parts?
The problem is:
If $a\equiv b\pmod{p},$ then $Squarefree(a)\equiv Squarefree(b) \pmod{p}.$ Is this true?
I encountered such problem in homework I'm doing, where I need to check if the squarefree value of a polynomial is congruent to some numbers modulo $5$. Now I'm wondering if it would be enough to check it only for values $0,\ldots,4$.
I think the statement is true, I just don't know how to prove it.
Edit: by a squarefree part od $a$ I mean $Squarefree(a)=d,$ where $a=b^2d$ and $b^2$ is the largest perfect square divisor of $a$.
AI: This is not true. For example,
$$3^3\equiv2^3\pmod{19}$$ but $$3\not\equiv2\pmod{19}$$ where $squarefree(3^3)=3$ and $squarefree(2^3)=2$. |
H: $a,b\in\mathbb R^2$, is it ok to define $[a,b]$ to be the segment connecting $a$ and $b$?
How to denote higher dimensional segments?
For example, $a,b\in\mathbb R^2$, it is not necessarily that $a\leq b$. Is it ok to denote $[a,b]$ as the segment connecting $a$ and $b$?
AI: As long as you say that that's what you mean the first time you use it, then it's completely fine. It doesn't conflict with other common notation, and it's somewhat similar to the meaning we are used to. |
H: Probability of getting a specific sequence of length $4$ in $10$ coin tosses
This is more of a thinking question maybe, hope that's ok.
Suppose I toss a coin $10$ times. What is the probability that within these 10 tosses I get the sequence THHT.
My attempt:
If I have 10 coin tosses THHT can happen in only 1 way, so similar to seating arrangements I treated it as one outcome.
Then I have THHT _ _ _ _ _ _ six more spots to fill, each of which have 2 possible outcomes, heads or tails.
THHT can take any of the 7 spots so we have $ 7*2^6 $ ways this can happen.
Then because there is six spots, we could have the sequence THHT twice.
THHT THHT _ _ there is twelve ways this can happen times 4 because of the double T's.
Now I have $$\frac{7*2^6-48}{2^{10}}$$ $2^{10}$ are all the possible coin toss combinations.
This gives me $$\frac{400}{1024}$$
The correct answer is $$\frac{393}{1024}$$.
This is a question to a homework that I did a few months ago.
I am quite enthusiastic about these types of problems (counting, probability, combinatorics) and that's why I did it differently than how we learned in the lectures (inclusion, exclusion).
My answer was wrong and I couldn't find out why during the semester, I miscounted something.
I was told (by TA) my answer is wrong because I didn't do inclusion-exclusion.
I know how to do it inclusion exclusion way, after all it's just following a few repetitive steps.
But I enjoy finding the correct way to count $$\frac{\text{possibilities this can happen}}{ \text{all possibilities}}$$
However after spending all morning tinkering, I can't find out why.
I think my solution could have been correct as it was very close, but I counted 7 variations extra.
Because 7 is the number of "seats" I can have THHT take, I feel like this isn't a coincidence.
Thank you very much for any insight you can share in what I missed, or if indeed, me being close to the answer was pure coincidence (the irony) and this can only be solved with inclusion and exclusion mechanics.
AI: You started with THHT_ _ _ _ _ _ and similar in $7 \times 2^6$ different ways
But you have double counted:
THHTHHT_ _ _ and _ THHTHHT_ _ and _ _ THHTHHT_ and _ _ _ THHTHHT in $4\times 2^3$ ways
THHT_ _ THHT and _ _ THHTTHHT and _ THHTTHHT_ and _ THHT_THHT and THHT_THHT_ and THHTTHHT_ _ in $6\times 2^2$ ways
If you subtract these, then you oversubtract THHTHHTHHT. That string appears three times in the initial count, twice in the first double count and once in the second double count, and you only want it counted once overall
So I think the total count is
$$7 \times 2^6 - 4\times 2^3-6\times 2^2 + 1=393$$ |
H: Congruences: Solving $ax \equiv c \pmod m$,
This is from the Joseph Silverman book on Number Theory
Solving $ax \equiv c \pmod m$,
https://i.stack.imgur.com/H3QcK.png
I understand upto Step X in the above image. But how does he get from Step X to Step Y?
AI: Note that we will have infinitely many solutions such that $au+mv=g$ , where $g=$gcd$(a,m)$ .
Now, let $u_0$ and $v_0$ be one such solution which satisfies $au+mv=g$ .
Given that $g\mid c \implies \frac {c}{g} \in \Bbb Z$ .
Since $au_0+mv_0 =g$ , we get that $\frac {c}{g} ( au_0+mv_0)=c $
$\implies \frac {c}{g} au_0 + \frac {c}{g} mv_0 = c $.
$\implies \frac {c}{g} au_0 - c = -[\frac {c}{g} mv_0]$
Now, since $\frac {c}{g}$ is an integer , we get that $\frac {c}{g} au_0 \equiv c \pmod m$
So if we take $x_0= \frac {c}{g}u_0 $, we get a solution that satisfies $ax\equiv c \pmod m $
Do tell me if there is any mistake or something is not clear . |
H: $(a \cdot b)\circ c = (a \circ c) \cdot (b \circ c)$ and $(a \circ b)\cdot c = (a \cdot c) \circ (b \cdot c)$
Do $(a \cdot b)\circ c = (a \circ c) \cdot (b \circ c)$ and $(a \circ b)\cdot c = (a \cdot c) \circ (b \cdot c)$ lead to a contradiction?
Where $\cdot$ and $\circ$ are both binary operations who form groups.
(I try to use the minimal amount of group structure )
Attempt 1
“Direct Attack” using just the two statements to prove a contradiction.
$$(a \cdot b) \circ c = (a \circ c) \cdot (b \circ c)$$
$$=(a \cdot (b \circ c)) \circ (c \cdot (b\circ c))$$
I continued doing this but it didn’t help.
Attempt 2
Using some group axioms.
the identity axioms don’t seem to help, my biggest suspicion is the axiom of inverses.
Can any of you help?
Note: I’m looking for answers with minimal added structure.
AI: Let $1$ denote the identity with respect to $\cdot$. Notice that
$$1\circ x=(1\cdot 1)\circ x=(1\circ x)\cdot (1\circ x),$$
so $1\circ x$ is an idempotent. The only idempotent in a group is the identity, so $1\circ x=1$ for all $x$, but this is a clear contradiction.
What this shows is that you only need to know that one operation distributes over another to obtain a contradiction to them both being groups. Indeed, this is why the additive identity in a ring is a zero for multiplication.
Edit: As Mark Kamsma pointed out, if the group has a single element then of course everything does work. So you have to exclude this literally trivial case. This is also why one of the axioms of a field is that $0\neq 1$, to stop the trivial ring being a field. |
H: A question on application of derivatives
If the area of the circle increases at a uniform rate, show that the rate of increase of the circumference varies inversely as the radius.
My approach-
If area is increasing at a uniform rate then the area function should be $A(r)=pr+q$ where $p$ and $q$ are certain constants and $r$ is the radius of the circle.
We can say that-
$$A(r)=\int_0^r C(r)dr$$where $C(r)$ is the circumference as a function of $r$.
Now differentiating both sides with respect to $r$-
$$A'(r)=\frac{d}{dr}(\int_0^r C(r)dr)$$
Now, using Newton-Leibniz formula-
$$p=C(r)$$
Circumference turns out to be a constant quantity! Where did I go wrong?
AI: I think you assumed that the area is linear with $r$- but the relationship between $A$ and $r$ is fixed: $A= \pi r^2$. What the question means is a constant $A'(t)$ ($A$ is linear with time). If you use the above $A(r)$, you'll get the desired result. |
H: Definition essential supremum
Consider the following fragment from Axler's "Measure, Integration & Real analysis":
Is the definition of $\Vert f \Vert_\infty$ unaffected when we change any of the strict inequalities into non-strict inequalities?
AI: If $\mu \{x:|f(x)| >t\}=0$ then $\mu \{x:|f(x)| \geq t+\epsilon\}=0$ for any $\epsilon >0$ and it follow from this that the infima are the same whether you have greater than or greater than or equal to in the definition. |
H: Calculating $ \lim_{n \to \infty} \left(\frac{n-3}{n}\right)^n $
While calculating the following limit:
$$ \lim_{n \to \infty} \left(\frac{n-3}{n}\right)^n $$
I have used the following procedure:
\begin{align}
\lim_{n \to \infty} \left(\frac{n-3}{n}\right)^n =
\lim_{n \to \infty} \left(\frac{\frac{1}{n}\cdot(n-3)}{\frac{1}{n}\cdot n}\right)^n =
\lim_{n \to \infty} \left(1-\frac{3}{n}\right)^n = \\=
\lim_{n \to \infty} \left(1-\frac{3}{\infty}\right)^\infty =
\lim_{n \to \infty} \left(1-\frac{3}{\infty}\right)^\infty =
\lim_{n \to \infty} (1)^\infty = 1\\
\end{align}
I'm aware that the solution is $ e^{-3} $ but I'd like to know that rules I'm breaking in my process so the answer is wrong. I'm suspicious of the last two steps. I think that assuming that $ 3/\infty $ tends to 0 as $ n $ approaches $ \infty $ is fine and the result would approach 1 without the power. But in this case, the power makes the result approaches $ 0.05 $ instead.
AI: From this step
$$\ldots=\lim_{n \to \infty} \left(1-\frac{3}{n}\right)^n =\ldots$$
we can't "plug" $\infty$ to solve since $1^\infty$ is an indeterminate form.
We can use
$$\left(1-\frac{3}{n}\right)^n=\left[\left(1+\frac{(-3)}{n}\right)^{\frac n{(-3)}}\right]^{-3}$$
and conclude by the standard limit
$$\lim_{x \to \pm\infty} \left(1+\frac{1}{x}\right)^x=e $$ |
H: Extension of function beyond boundary of a closed set is continuous
Let $(X,\mathcal T)$ be a topological space and $A\subset X$ be a closed subset. Assume $g\in C(A,\mathbb C)$ is such that $g=0$ on $\partial A$. Define the extension $\tilde g$ by $\tilde g = g$ on $A$ and $\tilde g = 0$ on $A^c$. Prove that $\tilde g$ is continuous.
This is Exercise 4.15 in Folland's Real Analysis. I've worked out a proof (see below). I prove that $\tilde g$ is continuous at each $x\in X$ by considering cases : $x\in \text{int} A$, $x\in \partial A$ and $x\in A^c$. I want to know if this is the best possible proof (I find it quite tedious).
Let $x\in \text{int} A$ and $\epsilon >0$. Since $g\in C(A,\mathbb C)$, there is some $U\in \mathcal T$ such that $x\in A\cap U$ and $g(A\cap U) \subset B(g(x),\epsilon)$. Then $x\in \text{int}(A)\cap U \in \mathcal T$ and $\tilde g(\text{int}(A)\cap U) = g(\text{int}(A)\cap U) \subset B(g(x),\epsilon)= B(\tilde g(x),\epsilon)$.
Let $x\in \partial A$ and $\epsilon >0$. Since $g\in C(A,\mathbb C)$, there is some $U\in \mathcal T$ such that $x\in A\cap U$ and $g(A\cap U) \subset B(0,\epsilon)$. Note that $\tilde g(A^c\cap U)\subset \{0\}$, thus $\tilde g(U) \subset B(0,\epsilon)=B(\tilde g(x),\epsilon)$.
Let $x\in A^c$ and $\epsilon >0$. Note that $A^c\in \mathcal T$ and $\tilde g(A^c)\subset \{0\} \subset B(\tilde g(x),\epsilon)$.
AI: You have that $$\tilde{g}(x)= \begin{cases}g(x) & x \in A\\
0 & x \in \overline{A^\complement}
\end{cases}$$
where both cases overlap on $A \cap \overline{A^\complement} = \partial A$, where both are $0$ by assumption.
So the pasting lemma for the case of two closed subsets (a proof of which in general topology terms can be found on this site) implies that $\tilde{g}$ is continuous on $X$ iff $g$ is continuous on $A$ (the constantly $0$ map being trivially continuous anyway). |
H: Why is the Conway 'Look and Say' sequences constant defined by this polynom?
In his work on 'Look and Say' sequences,for instance beginning with $1$.
$$1//
11//
21//
1211//
111221//
312212$$
If $L_n$ is the length of the $n-th$ sequences, then it follows from Conway work that :
$$\lim_{n\to\infty} \
\frac{L_{n+1}}{L_n} =\lambda=1.303577269034... $$
where $\lambda$ is the unique real, stricly positive root of
\begin{align}
x^{71} - x^{69} - 2x^{68} - x^{67} + 2x^{66} + 2x^{65} + x^{64} - x^{63} \\
- x^{62} - x^{61} - x^{60} - x^{59} + 2x^{58} + 5x^{57} + 3x^{56} - 2x^{55} - 10x^{54} \\
- 3x^{53}- 2x^{52} + 6x^{51} + 6x^{50} + x^{49} + 9x^{48} - 3x^{47} - 7x^{46} - 8x^{45} \\
- 8x^{44} + 10x^{43} + 6x^{42} + 8x^{41} - 5x^{40} - 12x^{39} + 7x^{38} - 7x^{37} + 7x^{36} \\
+ x^{35} - 3x^{34} + 10x^{33} + x^{32} - 6x^{31} - 2x^{30} - 10x^{29} - 3x^{28} + 2x^{27} \\
+ 9x^{26} - 3x^{25} + 14x^{24} - 8x^{23} - 7x^{21} + 9x^{20} -3x^{19} - 4x^{18} \\
- 10x^{17} - 7x^{16} + 12x^{15} + 7x^{14} + 2x^{13} - 12x^{12} - 4x^{11} - 2x^{10} + 5x^9 \\
+ x^7 - 7x^6 + 7x^5 - 4x^4 + 12x^3 - 6x^2 + 3x - 6
\end{align}
My question is: why that polynom? How did Conway manage to get it? Is it an approximation of the experimental values of $\lambda$ he got?
If there exists any paper, I would appreciate to read it. Thanks for your help.
AI: Have a look here for the derivation: http://www.njohnston.ca/2010/10/a-derivation-of-conways-degree-71-look-and-say-polynomial/
The gist of it is every term after the 8th term can be constructed from some of 92 strings. Then its a matter of counting how the sequence length increases and then computing this limit. |
H: Find the linear function $f$ whose composition $f\circ f\circ f\circ\dots$ ($6$ times) is equal to $2x-1$ for all $x$
I am not able to start the solution, so can get a hint for this one?
Thanks
AI: If $f(x)=ax+b$ then $f(f(...(f(x)))))=a^{6}x+b(1+a+a^{2}+a^{3}+a^{4}+a^{5})$. Equate this to $2x-1$ and compare coefficients to find the values of $a$ and $b$. |
H: How can I prove $\left|\frac{e^{it_p x_j}-1}{t_p}\right| \leq 2|x|$?
Background:
I am trying to understand why, if $X$ is a random variable with $\mathbb{E}\big[|X|\big]< +\infty$ and $\mu=\mathbb{P}^{X}$, then $$\frac{\partial}{\partial x_j}\overline{\mu}(u)=i\int e^{i \langle u,x \rangle} x_j \mu(dx).$$
My problem:
Suppose $t_p \to 0$ and $x \in \mathbb{R}^n$. How can I prove that $$\left|\frac{e^{it_p x_j}-1}{t_p}\right| \leq 2|x|,$$ where $x_j$ is the $j$-th coordinate of $x$?
AI: In fact you can get $|x|$ as a bound. Use the fact that $|e^{i\theta} -1| \leq |\theta|$ for all real numbers $\theta$.
[$\int_0^{1} e^{it\theta} dt=\frac 1 {i \theta} (e^{i\theta}-1)$ and LHS is bounded in modulus by $\int_0^{1}dt=1$]. |
H: Use of Hôpital's rules to calculate also the sequences
Hoping that my question is clear, I would like to understand because the L'Hôpital's rules are used in several questions on Math.SE (an answer for example) to calculate the sequences,
$$(a_n)_{n\in\Bbb N}, \quad \text {or} \quad \{a_n\}, \quad n\in\Bbb N$$
During my university period I had been instructed that Hôpital's theorems cannot be applied.
AI: In this answer, L'Hôpital's rule is actually used to compute
$$
\lim_{\substack{x \to \infty \\ x \in \Bbb R}} \frac{ \ln(1-\frac{3}{x})}{1/x} = \lim_{\substack{x \to \infty \\ x \in \Bbb R}} \frac{\frac{3}{x^2}}{\frac{-1}{x^2} (1-\frac{3}{x})} = -3 \, ,
$$
and that implies
$$
\lim_{\substack{n \to \infty \\ n \in \Bbb N}} \frac{ \ln(1-\frac{3}{n})}{1/n} = -3 \, .
$$
Only in that answer $n$ is used both as the (integer) index of the sequence and as the real-valued argument of a function.
Generally, if your sequence is $a_n = f(n)$ with a function $f: [1, \infty) \to \Bbb R$, and if you can show that $\lim_{x \to \infty} f(x) = A$ (using L'Hôpital's rule or any other method), then $\lim_{n \to \infty} a_n = A$ follows. |
H: Is it possible to reach a monochromatic configuration only using 2x2 and 5x5 flips?
The following problem has been troubling me for quite a while now:
"The cells of a $10\times 10$ grid are either coloured blue or green. In a move you are allowed to select any $2\times 2$ or $5\times 5$ grid and reverse the colour of each cell in that "sub-grid". Is it possible to make all cells blue given any starting configuration?"
My intuition says tells me that this is possible. I first tried to find a sequence of moves to switch the colour of one cell and thus use this as many times as needed but this was easier said than done and I could not find this algorithm. I could however prove that this was impossible only using $2\times 2$ flips using the sum of the cells modulo $2$ being an invariant (assigning 1 to green and 0 to blue). Hints would be appreciated.
AI: Think of the cell values as $0$ or $1$, and using mod-2 arithmetic, an operation like "change, in a 2x2 square, all greens to blue and blues to green" becomes "add $1$ to each item in the $2 \times 2$ square."
For $i, j = 1, \ldots 9$, let's call $T(i,j)$ the $10 \times 10$ matrix that's all zeroes except for locations $(i,j), (i+1,j), (i, j+1), (i+1,j+1)$, i.e., all locations except for a $2 \times 2$ block whose upper left corner is at location $(i, j)$. Similarly, let $F(i, j)$ be the $10 \times 10$ matrix that's all zeroes except for having $1$s in an $5 \times 5$ block whose upper left corner is at $(i, j)$. Here $i, j = 1, \ldots, 6$), because you can't fit a $5 \times 5$ matrix of $1$s starting anywhere after the 6th row or columns.
Now the problem becomes (everything is 10 x 10 from now on):
Given a matrix $M$, is there a collection of $T$ and $F$ matrices such that
$$
M + T_1 + T_2 + \ldots + T_k + F_1 +
\ldots + F_p = 0?
$$
which is equivalent, $\bmod 2$, to
$$
M = T_1 + T_2 + \ldots + T_k + F_1 +
\ldots + F_p = 0.
$$
In other words, do the matrices $T(i,j)$ and $F(i, j)$ span $M_10(\Bbb F_2)$?
By writing out each $10 \times 10$ matrix as a column vector (just stack the columns on top of each other!), you get $81$ column vectors $t_{ij}$ for the $T$ matrices, and another $36$ vectors $f_{ij}$ for the $F$ matrices. The question is whether these $117$ matrices actually span. Fortunately, that's easy to answer: you take the $100 \times 117$ mod-2 matrix, and row reduce it to see whether it has at least $100$ independent columns. (I said it was easy; I didn't say it'd be fast!)
My guess, from a little fiddling, is that they do not span. I could write some matlab code to check, but...it's time to move on.
====
OK, I wrote the code, in analogy with the other answer mentioned in the comments, and found that each of the $T$ and $F$ matrices has the property that $M \cdot X = 0 \bmod 2$, where $X$ is this matrix
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0
1 1 0 1 1 1 1 0 1 1
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0
1 1 0 1 1 1 1 0 1 1
To put it differently: every $2 \times 2$ and $5 \times 5$ submatrix of $X$ sums to $0 \bmod 2$. So in summing up $T$ and $F$ matrices, you will never get the matrix $E$ with a $1$ in the lower left corner and zeroes everywhere else, because the dot-product of $E$ with $X$ is $1$ rather than $0$.
Here's how I found that matrix (with amazingly ugly matlab code)
function grid_puzzle2()
twos = zeros(10, 10, 9, 9); % room for all 81 10x10 mats with 2x2 blocks of ones.
fives = zeros(10, 10, 6, 6);
for i = 1:9
for j = 1:9
twos(i:i+1,j:j+1, i, j) = 1; % fill in the blocks
end
end
% restructure the 4-index matrix to a 2-index one, where each column
% is the result of reading out the columns of one 10x10 matrix.
% Yeah, putting the indices in the right order for this takes practice.
t2 = reshape(twos, [100, 81]);
for i = 1:6
for j = 1:6
fives(i:i+4,j:j+4, i, j) = 1;
end
end
t5 = reshape(fives, [100, 36]);
T = [t2, t5]; % a 100 x 117 matrix
S = rref(T'); % the row-reduced version of the transpose of $T$
U = ones(117, 1); % a list of 117 "1"s
x = S\U ; % "best possible" solution of Sx = U
These computations are all done over the reals, so the vector $x$ ends up with entries that are all (by chance) multiples of $0.25$. So I computed
t = round(4*x);
s = mod(t, 2)
and got the matrix that I pasted in above.
Now WHY is solving $Sx = u$ the right thing to do? I wanted a matrix whose dot product with each $2 \times 2$ and $5 \times 5$ matrix was the same. If I'd picked $u = 0$, which is the obvious choice, the natural solution would be $x = 0$, which would be of no use. So picking $u$ to be all $2$s is a better choice. I used all $1$s because it was easier to express in Matlab, and then fiddled a little at the end. |
H: Proving a limit using the $\epsilon$ - $\delta$ definition of limit.
Given $\lim _{x\to a}\left(f\left(x\right)\right)=\infty$ and $\lim _{x\to a}\left(g\left(x\right)\right)=c$ where $c \in R$, prove $\lim _{x\to a}\left[f\left(x\right)+g\left(x\right)\right]=\infty$.
My attempt:
Let for every $M>0$ exists $\delta_1$ which satisfies $0 < |x-a| < \delta_1 \implies f(x) > M$.
Let for every $\epsilon > 0$ exists $\delta_2$ which satisfies $0 < |x-a| < \delta_2 \implies |g(x) - c| < \epsilon$. Or, I can write it as $0 < |x-a| < \delta_2 \implies c - \epsilon < g(x) < c + \epsilon$.
Let for every $N > 0$ exists $\delta$ which satisfies $0 < |x-a| < \delta \implies f(x) + g(x) > N$.
Using $\delta =$ min{$\delta_1, \delta_2$} so $f(x) > M$ and $g(x) > c - \epsilon$, I get $f(x) + g(x) > M + c - \epsilon$.
And I'm stuck. I've seen a solution somewhere which divides the final equations for $c = 0, c > 0,$ and $c < 0$ but I don't get the idea why do I have to solve it in cases. I don't know how to construct such $\delta$ that satisfies $N$.
I have taken a look at a similar question, proof of limit using epsilon-delta definition, but I'm not quite enlightened with the answer yet.
AI: We can set $N=M + c - \epsilon$ at any value and since
$$f(x)+g(x)>M + c - \epsilon=N$$
the proof is complete. |
H: Proof of Cauchy-Schwarz in $\mathbb{R}^n$ using Law of Cosines
Let us consider the Cauchy-Schwarz Inequality for vectors in $\mathbb{R}^n$:
Let $u, v \in \mathbb{R}^n$. Then $|u \cdot v| \leq |u||v|$.
Wikipedia provides a proof for this setting by enforcing a condition on the discriminant of a polynomial.
What I don't understand is why this specific case of Cauchy-Schwarz doesn't immediately follow from the law of cosines $u \cdot v = |u| |v| \cos{\alpha}$, where $\alpha$ is the angle between $u$ and $v$.
We know that the span of $u$ and $v$ is at most two-dimensional, and so if we restrict ourselves to $\mathrm{Span}(u,v)$ we are in standard Euclidean geometry, the angle $\alpha$ is well-defined, and the law of cosines holds.
I would be very grateful if someone could point out an error in the argument I've made (a quick proof of Cauchy-Schwarz for vectors in $\mathbb{R}^n$ as an immediate consequence of the law of cosines, without any recourse to discriminants of polynomial equations); or, if I haven't made an error, a perspective on why Wikipedia and other references (like Hubbard and Hubbard, Vector Calculus, Theorem 1.4.5) prefer what seems to me a more complicated and less intuitive proof.
AI: Your reduction to two dimensions is valid, but in practice it requires you to also show $u\cdot v=|u||v|\cos\alpha$.holds in $2$ dimensions, and that $\cos\alpha\in[-1,\,1]$ for any angle in the plane, i.e. $\alpha\in\Bbb R$ (give or take your favourite modulo-$2\pi$ restriction convention). I suspect you already know how to prove these results, but they do take a little work.
I think Wikipedia just wanted to show you there's a purely non-trigonometric way to do it, which isn't surprising given that CS can be stated as an inequality in real variables, no matter how little the reader knows about vector spaces.
It's also worth noting that algebraic methods, while often written with a real vector space in mind, are very easily tweaked to deal with complex vector spaces instead. Is it obvious one can define an angle, equal to a real number of radians, between two vectors spanning a complex plane? |
H: Question regarding stochastic independence
Let $X,Y,Z$ be random variables and pairwise independent, i.e. $X$ independent from $Y$, $Y$ indepedent from $Z$, and $X$ independent from $Z$. I am interested in a rigorous argument, why $ (X,Y) $ is independent from $Z$?
AI: Counterexample: Let $\Omega$ be the sample space consisting of all the permutations of $(1,2,3)$ as well as the three vectors $(1,1,1), (2,2,2)$ and $(3,3,3)$ (total: $3!+3$ elements in $\Omega$).
One vector is chosen at random from $\Omega$, denoted by $(X,Y,Z)$.
Clearly, each of these is uniform over $\{1,2,3\}$ and they are pairwise independent, but given $(X,Y)$ the value of $Z$ is uniquely determined, so $(X,Y)$ and $Z$ are not independent. |
H: Rearranging basic algebraic expression
I am starting to relearn algebra as I am starting to use more linear algebra at work. I apologise for how simple this question is going to sound.
The formula that I am working with is:
$$
x_{1}+2\left(\frac{-1}{5}y_{1}+\frac{3}{5}y_{2}\right)=y_{2}
$$
and I want to rearrange it to find
$$x_{1}$$
I know that the result I am looking for is:
$$
x_1=\frac{2}{5}y_1-\frac{1}{5}y_2
$$
But I just do not understand the steps to get there. If anyone could give me a walkthrough, I would really appreciate it.
Cheers,
AI: The main principle needed to solve the problem is the distributivity of multiplication over addition.
In the equation, we have 2 times the sum of two things, the distributive law states that it is equal to 2 times one thing plus 2 times the other.
$$\begin{align} x_1 + 2(\frac{-1}{5} y_1 + \frac{3}{5} y_2= y_2)\\
x_1 + 2\frac{-1}{5}y_1+ 2\frac{3}{5} y_2=y_2&&\text{Using the distributive law}\\
x_1 - \frac{2}{5} y_1 + \frac{6}{5}y_2 = y_2&&\text{Simplifying }
\end{align}$$
$$\begin{align}
x_1= \frac{2}{5} y_1 - \frac{6}{5} y_2+ y_2&&\text{Isolating } x_1\\\
x_1 = \frac{2}{5} y_1 -\frac{1}{5} y_2 && \text{Simplifying} \end{align}$$
And we’re done! |
H: If $\varphi$ is a linear functional, show that $\{\varphi\ge0\}^\circ=\{\varphi>0\}$
Let $E$ be a normed $\mathbb R$-vector space, $\varphi\in E'\setminus\{0\}$ and $H:=\{\varphi\ge0\}$.
I would like to show that $H^\circ=\{\varphi>0\}$ and $\partial H=\{\varphi=0\}$.
Should be trivial enough. It's clear that $H$ is closed and hence $\partial H=H\setminus H^\circ$. So, we only need to determine $H^\circ$.
Now, $\{\varphi>0\}=\varphi^{-1}((0,\infty))$ is open and contained in $H$. So, we only need to show that if $x\in\{\varphi=0\}$, it cannot be an interior point of $H$. I think the correct approach is to take an arbitrary $\varepsilon>0$ and show that $B_\varepsilon(x)$ contains a point not in $H$. But how do we do that?
AI: If $\phi(x)$ is $0$ then $\phi[B(x, \delta)]$ is open and contains $0$ and so $B(x, \delta)$ contains points with $\phi(x) < 0$ for all $\delta>0$. |
H: Logic expressions for an English verse.
Question:
This is an assignment question that I am having trouble with. The question goes like this.
What I did:
I made the predicate logic expression as follows.
$$\forall x\ LOVES(x, MYBABY)\ \wedge\ \forall x\ (x=ME \Longleftrightarrow LOVES(MYBABY, x))$$
Please tell me if this is correct. Also, I can't seem to find any logic on how to end up with $ME=MYBABY$, so please help with that as well. Sorry if this question seemed too naïve or seemed like I am getting an assignment done from you. Thanks for the attention.
AI: Your translation is technically correct, but the instructions don't list $\Longleftrightarrow$ as a permissible connective, so try to rewrite it with the connectives given.
Hint for the second question: The first part says that everybody loves $mybaby$. This implies in particular that $mybaby$ loves $mybaby$. Now think about how to use the rest to derive that $me = mybaby$. |
H: (Dummit and Foote) Group of order 105 with $n_3 = 1$ must be abelian
I was working on this problem: Let $G$ be a group of order $105 = 3\times 5\times 7$. Assume it has a unique normal Sylow 3-subgroup. Then prove that $G$ is abelian.
I worked out the following from Sylow's theorem:
$n_5 = 1$ or $n_5 = 21$
$n_7 = 1$ or $n_7 = 15$
and by showing that a homomorphism from $G$ into $\operatorname{Aut}(P_q)$ must be trivial if $q-1$ is coprime to $|G|$:
Since $n_3 = 1$ the Sylow 3-group lies in the center.
if $n_5 = 1$ the Sylow 5-group lies in the center.
and counting elements of order $q$:
$n_5 = 21$ would mean the Sylow 5-subgroups contribute 84 elements of order 5.
$n_7 = 15$ would mean the Sylow 7-subgroups contribute 90 elements of order 7.
This implies we cannot have both, one of them must be a unique normal subgroup.
Is this correct so far? How can I continue from here and finish the proof? Is there a way to avoid splitting into two different cases?
AI: In the hope that this will be the definitive answer to understanding groups of order $105$, I will talk about the ways to solve this.
The question assumes that the Sylow $3$-subgroup is normal in $G$. The condition on the Sylow $3$-subgroup here is necessary. There are two groups of order $105$, both with a normal Sylow $5$- and $7$-subgroup, but one is cyclic and the other is $C_5\times F_{21}$, where $F_{21}$ is a non-abelian group, the normalizer of a Sylow $7$-subgroup of $A_7$.
The fastest way to proceed is to notice that $P_3$, the Sylow $3$-subgroup, is not only normal but central. To see this, you can recall that $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(H)$, which has order $2$ in this case. The from scratch method is to notice that $C_3$ has only two non-identity elements, so for any element $g\in G$, $g^2$ must act trivially on $P_3$. But $|G|$ is odd, so every element is a square, and $P_3$ is central.
At this point, there are two ways to proceed. The first is to notice that $G/P_3$ has order $35=5\times 7$, and groups of order $35$ are cyclic. If $G/Z(G)$ is cyclic then $G$ is abelian, and we are done. (Clearly $G$ is therefore in fact cyclic.)
The alternative proof is to note that $P_3\leq C_G(P_5)$ and $P_3\leq C_G(P_7)$. Thus $|C_G(P_5)|\geq 15$, and $|C_G(P_7)|\geq 21$. (Recall that $C_G(P_q)\leq N_G(P_q)$ and $n_q$, the number of Sylow $q$-subgroups, is equal to $|G:N_G(P_q)|$.) From Sylow's theorem ($n_q\equiv 1\bmod q$) we see that $n_5=n_7=1$, as needed.
If you don't want to do this you can count elements, although it's more subtle than most such arguments. Let's do this without the assumption that $n_3=1$, to obtain the full classification.
The number $n_5$ of Sylow $5$-subgroups is either $1$ or $21=3\times 7$. We want to prove the former, so assume the latter. Then there are $21\times 4=82$ elements of order $5$, and since $C_G(P_5)=P_5$, there are no elements of order $5n$ for any $n>1$. This leaves exactly $105-82=23$ elements of order not $5$, and these must have order $1$, $3$, $7$ or $21$. If $n_7\neq 1$ then $n_7=15$, but this is impossible as there are only $23$ elements left. So $n_7=1$, removing six elements of order $7$. There are seventeen elements left, so $n_3\leq 8$ (as each Sylow $3$-subgroup requires two elements of order $3$). Thus $n_3=1$ or $n_3=7$. If $n_3=7$ then that removes fourteen elements of order $3$, and the identity, so there are two elements remaining, which must have order $21$. But in any cyclic group of order $21$ there are twelve elements of order $21$, which is too many.
Thus $n_3=1$, and the Sylow $3$- and $7$-subgroups are both normal. Thus $P_3P_7$ is normal in $G$, has index $5$, and therefore contains every element of order dividing $21$. So where are the two remaining elements? This yields a contradiction, so $n_5=1$.
If $n_7\neq 1$ then $n_7=15$, as it must be $1$ modulo $7$. Again, you can obtain a contradiction as before, because $C_G(P_5)$ contains $P_7$ but $C_G(P_7)$ does not contain $P_5$. Let's try to count elements, and see what goes wrong. This yields $15\times 6=90$ elements of order $7$. There are five elements in $P_5$, leaving ten elements. Thus $n_3\leq 5$, so $n_3=1$. Thus we have a subgroup $P_3P_5$ of order $15$. This contains ten more elements (as we have already counted $P_5$), and so we have exactly the right number of elements, $105$.
If $15$ were a prime, then this would be fine. Then $7\mid (15-1)$ and there would be a map from $C_7$ into $\mathrm{Aut}(C_{15})$, which would have order $14$. But $15$ is not a prime, so we can obtain a contradiction using centralizers, as above, but element counting will not work in this case. The group $P_3P_5$ has normal subgroups $P_3$ and $P_5$, on which $P_7$ cannot act. Thus $P_3P_5$ is actually central, and $G/(P_3P_5)$ is cyclic, so $G$ is abelian. Alternatively, $P_3$ is central, so $P_3$ centralizes $P_7$. But $n_7=15$, so $P_7$ does not centralize $P_3$. This is a clear contradiction.
Thus $n_7=1$ as well. The subgroup $P_5P_7$ is a normal, cyclic, subgroup of order $35$. Since there is no map from $P_3$ to $\mathrm{Aut}(P_5)$, this is actually central. The subgroup $P_7P_3$, of order $21$, complements this, so $G\cong P_5\times P_7P_3$. If $n_3=1$, equivalently $P_3$ centralizes $P_7$, then you end up with an abelian (cyclic) group of order $21$. If $n_3=7$, equivalently $P_3$ acts non-trivially on $P_7$, then $P_3P_7$ is a Frobenius group of order $21$. This is the normalizer in $A_7$ of a Sylow $7$-subgroup. |
H: Formula for sequence of 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4 and so on.
I have been trying to figure out a formula for the sequence: $0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, ...$ and so on . It is not geometric and it is not arithmetic, I tried to apply these formulas, but of them are failing leading me to believe it is sum of arithmetic and geometric, can this be possible? How to get the formula for the sequence?
AI: If you want the sum up to term $n$, the number of triplets up to term $n$ is $\lfloor \frac {n-1}3 \rfloor$, so multiply the sum of the numbers up to that by $3$. Then add the last one or two numbers, if any. |
H: How do I solve $(\cos2x+1)^2=1/2$?
$x$ belongs to $$[0,\pi/2[$$
$$(\cos(2x+1))^2 = \frac{1}{2}$$
I tried to find $x$ using
$$2x+1=\frac{\pi}{4} +\frac{n\pi}{2}
\qquad\text{or}\qquad
2x+1=\frac{3\pi}{4} + n\pi$$
but I didn't find my answer in MCQ which is
$$\frac{3\pi-4}{8}
\qquad\text{and}\qquad
\frac{5\pi-4}{8}.$$
I just need a hint on how to solve such equation.
AI: Your inital step is right, indeed we have that
$$(\cos(2x+1))^2 =\frac12 \iff \cos(2x+1)=\pm\frac{\sqrt 2}2 \iff 2x+1=\frac \pi 4+k\frac \pi 2$$
then
$$x=\frac \pi 8+k\frac \pi 4-\frac12$$
and since $x\in [0,\pi/2[$ we obtain
$$x_1=\frac 3 8\pi-\frac12, x_2=\frac 5 8\pi-\frac12$$ |
H: Let $x_1,x_2,x_3,x_4$ denote the four roots of the equation $x^4 + kx^2 + 90x - 2009 = 0$. If $x_1x_2 = 49$, then find the value of $k$.
Let $x_1,x_2,x_3,x_4$ denote the four roots of the equation $$x^4 + kx^2 + 90x - 2009 = 0.$$ If $$x_1x_2 = 49,$$ then find the value of $k$.
What I tried :- From Vieta's Formula for quartic equations I get that
$$ x_1 + x_2 + x_3 + x_4 = 0$$
$$x_1x_2 + x_1x_3 + x_2x_3 + x_2x_4 + x_3x_4 + x_4x_1 = k$$
$$x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_1 + x_4x_1x_2 = (-90)$$
$$ x_1x_2x_3x_4 = -2009$$
Since $x_1x_2 = 49$, we have $x_3x_4 = (-41)$.
Hence in equation $3$ I get
$$x_1x_2(x_3 + x_4) + x_3x_4(x_1 + x_2) = (-90)$$
$$\Rightarrow 49(x_3 + x_4) - 41(x_1 + x_2) = (-90)$$
As $(x_1 + x_2 + x_3 + x_4) = 0$, $(x_1 + x_2) = -(x_3 + x_4)$, so $$49(-x_1 - x_2) - 41(x_1 + x_2) = (-90)$$
$$\Rightarrow (-90)(x_1 + x_2) = (-90)$$
$$\Rightarrow (x_1 + x_2) = 1,$$ so $(x_3 + x_4) = (-1)$.
From here I don't know how to proceed, can anyone help?
AI: Your Vieta for $k$ is wrong and should be $k=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4$ and this can be rewritten in terms of information you already have. Give that a go. |
H: Examples of finite-dimensional, semisimple, non-separable $k$-algebras?
1. Motivation
Let $k$ be a field.
Apparently, any separable $k$-algebra is finite-dimensional and semisimple.
Using Maschke's theorem for Hopf algebras, one can prove that for Hopf algebras the following stronger statement holds:
A $k$-Hopf algebra is separable if and only if it is finite-dimensional and semisimple.
I believe, if you replace "$k$-Hopf algebra" by "$k$-algebra", this proposition is wrong.
2. Question
What are examples of finite-dimensional, semisimple, non-separable algebras?
Apparently, if $k$ is a perfect field any finite-dimensional, semisimple $k$-algebra is separable. Hence, any example has to be over an imperfect field.
AI: For any prime $p$, the $\mathbb{F}_p(T)$-algebra (which is simple since it is a field) $\mathbb{F}_p(T)(\sqrt[p]{T})$ is such an example. |
H: proof that $\log(\det(XX^H))$ is not concave
Is there an elegant way to proof that $\log(\det(XX^H))$ is not concave, with respect to the complex-valued matrix elements of $X$?
AI: If $f:X\mapsto \log\det (X X^H)$ were concave you would have $f(0)\ge(f(X)+f(-X))/2$. This is violated by any $X$ of full rank, for which $\det XX^H > 0$. |
H: Poisson's equation Evan's book
I do not understand the line - ' naive differentiation through the integral near the sigularity is unjustified'. I am wondering if anyone could help explain that?
AI: If you are given a function of the form $$u(x) = \int_{{\mathbf R}^n} g(x,y) \, dy$$ it may be tempting to calculate its partial derivatives as
$$\frac {\partial^k}{\partial x_i^k} (x) = \int_{{\mathbf R}^n} \frac {\partial^k}{\partial x_i^k} g(x,y) \, dy.$$
Evans refers to this as naive because unless you have a very specific justification for interchanging the integral with the partial derivative (the exchange of an integral with a limiting process typically requires some type of dominated convergence argument) you cannot expect the result to be valid. |
H: Prove that this continuous function takes the value $0$ for every $x$
Let $f : [0,1] \to \mathbb{R}$ be continuous, with $f(0) = f(1) = 0$. Suppose that for every $x ∈ (0,1)$ there exists $δ > 0$ such that both $x + δ$ and $x − δ$ belong to $(0,1)$ and $f(x) = (f(x −δ)+f(x+δ))/2$. Prove that $f(x) = 0$ for every $x ∈ [0,1]$.
I notice that $f(x) = (f(x −δ)+f(x+δ))/2$ is equivalent to
$$f(x+δ)-f(x)=f(x)-f(x-δ),$$ which gives the equality between derivatives when $δ$ tends to $0$.
So I think I should show that the derivative is constant in all $(0,1)$, and being $f(0)=f(1)=0$ the derivative is $0$, so $f(x)$ is constant, and is $0$ in all $[0,1]$.
But how do I move from that discrete equality to the continuous equality?
AI: Since $f$ is a continuous function and $[0,1]$ is compact, $f$ must have a maximum value $M$ and a minimum value $m$. If $M=m=0$, we are done.
Suppose that $M \neq 0$. Then, since $f(0)=f(1)=0$, we must have $M>0$. Since $f$ is continuous, we know that $f^{-1}(\{M\})$ is a closed subset of $[0,1]$, and so, it is compact. Let $x_M= \min f^{-1}(\{M\})$ (such $x_M$ exists because $f^{-1}(\{M\})$ is compact). Since $x_M \in f^{-1}(\{M\})$, we have that $f(x_M)=M$. It is easy to see that $x_M \in (0,1)$.
But there must exist $\delta >0$ such that $$M=f(x_M) = (f(x_M −δ)+f(x_M+δ))/2$$ But, $f(x_M −δ)\leq M$ and $f(x_M+δ)\leq M$. So we must have $f(x_M −δ)= f(x_M+δ)=M$. Ccontradiction since $x_M= \min f^{-1}(\{M\})$. So, $M=0$.
In a similar way, we prove that $m=0$.
Suppose that $m \neq 0$. Then, since $f(0)=f(1)=0$, we must have $m<0$. Since $f$ is continuous, we know that $f^{-1}(\{m\})$ is a closed subset of $[0,1]$, and so, it is compact. Let $x_m= \min f^{-1}(\{m\})$ (such $x_m$ exists because $f^{-1}(\{m\})$ is compact). Since $x_m \in f^{-1}(\{m\})$, we have that $f(x_m)=m$. It is easy to see that $x_m \in (0,1)$.
But there must exist $\delta >0$ such that $$m=f(x_m) = (f(x_m −δ)+f(x_m+δ))/2$$ But, $f(x_m −δ)\geq m$ and $f(x_m+δ)\geq m$. So we must have $f(x_m −δ)= f(x_m+δ)=m$. Contradiction since $x_m= \min f^{-1}(\{m\})$. So, $m=0$. |
H: Are finitely generated modules over a commutative ring always a direct sum of cyclic submodules?
Let's first motivate my question by looking at a finitely generated $k$-algebra $A$ over a field $k$.
Then $A$ in general does not have the form $k[a_1,a_2,\ldots,a_n]$ where $\{a_1,a_2,\ldots,a_n\}$ is a generating set for $A$. For example consider the two-dimensional irreducible representation $V$ of the quarternion group $Q_8$, then the ring of invariants is finitely generated by Hilbert's finiteness theorem, but the algebra of invariants, which is a subalgebra of a polynomial algebra in two variables holds the form $$ \mathbb{C}[V]^{Q_8} = \dfrac{\mathbb{C}[f,g,h]}{(h^2-f^2g+4g^3)},$$ where $f$ and $g$ are invariant polynomials of degree 4, and $h$ is of degree 6. The reason is that the generating polynomials are not algebraically independent.
Now consider a commutative ring $R$, and $M$ a finitely generating $R$-module, and $\{m_1,m_2,\ldots,m_n\}$ a generating set for $M$, I want to know whether it is true that $M$ holds the form $$ M = \bigoplus_{i=1}^n Rm_i$$
I think this is not true, but this is true if and only if $M$ is a finitely generated $\textit{free}$ module over $R$.
Can someone enlighten me?
AI: You're right, it is not generally true. Rings for which finitely generated modules (and also simply just all modules) are direct sums of cyclic modules have pretty good coverage in literature.
Start with
Behboodi, M., and G. Behboodi Eskandari. "On rings over which every finitely generated module is a direct sum of cyclic modules." Hacettepe Journal of Mathematics and Statistics 45.5 (2016): 1335-1342.
and follow the citations.
I think this is not true, but this is true if and only if is a finitely generated free module over .
Being free is sufficient for being that form, but it will not usually be necessary. For example, if the commutative ring has a nontrivial ideal, $R/I$ is clearly cyclic, and is certainly non-free (it has nonzero annihilator.) (Simplified thanks to metalspringpro) |
H: Uniform Boundedness and the Arzela-Ascoli Theorem in a Riemannian Manifold
Let $(M, g)$ be a complete Riemannian Manifold and let $(x_n)$ be a sequence of curves in $\Omega = \{x \in C^1([0, 1], \ M): \ x(0) = p, \quad x(1) = q, \quad \dot{x}(0) = v, \quad \dot{x}(1) = w \}$ such that $g(\dot{x}_n, \dot{x}_n)$ is uniformly bounded. A paper I am currently reading makes the claims that:
There exists a compact subset $K$ of $M$ containing the image of all the $x_n$
Arzela-Ascoli theorem gives the existence of a subsequence of $(x_n)$ which converges uniformly to some continuous curve satisfying $x(0) = p$ and $x(1) = q$.
I've come up with some reasoning as to why this may be true (which I've put below), but I'm not entirely confident because I'm fairly new to functional analysis (especially in the context of manifolds). Any help is appreciated.
1) This feels fairly intuitive, but I'm not certain that my reasoning holds. If the uniform bound is given by $g(\dot{x}_n, \dot{x}_n) < R$, then the sequence of lengths of the curves—say $L(x_n)$—also satisfies $L(x_n) < R$. Since $M$ is complete, we can construct a ball of radius $R$ around $x_n(0) = p$ which must contain the image of each $x_n$ (otherwise, we contradict the bound). The compact set $K$ can then be taken as the closure of this ball.
2) Now assuming 1 holds, we have the uniform boundedness of $(x_n)$ because their images are contained in a compact subset of a metric space. I suppose that equicontinuity is implied by the fact that these curves are $C^1$ together with the uniform boundededness—but I'm not sure how to show that in the context of manifolds. Assuming Arzela-Ascoli works as usual in this context, we'd then get a subsequence of $(x_n)$ which converges uniformly to some continuous curve $x$ (with respect to the topology induced by the metric, I suppose). The fact that $(x_n) \subset \Omega$ together with uniform convergence should imply that $x$ satisfies the relevant boundary conditions.
AI: You are correct about (1), except that you should get $L(x_n) \le R$ since
$$ L(x_n) = \int_0^1 \| x_n'(t)\|\, \mathrm d t = \int_0^1 \sqrt{g(x_n'(t), x_n'(t))} \, \mathrm d t.$$
The closed ball of radius $\sqrt R$ is closed, hence compact since $(M, g)$ is complete.
For (2), the equicontinuity follows from the bound on $g(x_n', x_n')$. Now $\{x_n\}$ is a sequence of continuous maps $x_n : [0,1] \to K$. Using the bound and the definition of the distance on $(M, g)$, we have
$$ d( x_n(t), x_n(s)) \le L( x_n) = \int_t^s \| x_n'(\tau)\| \, \mathrm d\tau \le \sqrt R |t-s|. $$
In particular, for all $\epsilon >0$, choose $\delta = \epsilon /\sqrt R$. Then whenever $t, s\in [0,1]$ and $|t-s|<\delta$, we have $d( x_n(t), x_n(s)) <\epsilon$ for all $n$. Thus the family is equicontinuous.
Lastly, note that Arzela-Ascoli theorem can be generalized to compact domain, see here. I suppose the proof is similar to the case of functions. Another way (which is an overkill) is to apply Nash embedding theorem, so that $(M, g)$ is isometrically embedded in $\mathbb R^N$ for some $N$. Then one can think of $x_n : [0,1] \to \mathbb R^N$ and apply the simpler version on each coordinates $1, 2, \cdots, N$. |
H: Convergence Range of $\sum\limits_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}$
$$\sum\limits_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}$$
By the ratio test we need $\lim_{n \to \infty} | \frac{a_{n+1}}{a_n}| < 1 \rightarrow\lim_{n \to \infty} x^2 < 1$. Hence we get the range $-1 < x < 1$.
My question is if we sub $x = -1$ we get the following series:
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^{2n-1}}{2n-1}$$
EDIT NOTE: My mistake here is not realizing that $(-1)^{2n-1} = -1$, and not an alternating series.
By the alternating series test we get $(1) \ b_n > b_{n+1}$ and $\lim_{n \to \infty} \frac{1}{2n-1} =0$, so wouldn't we also have to allow $-1 \leq x < 1$?
AI: The series $\sum\limits_{n=1}^{\infty} \frac{(-1)^{2n-1}}{2n-1}=\sum\limits_{n=1}^{\infty} \frac{-1}{2n-1}$ is harmonic and diverges therefore the convergence range is $-1 < x < 1$. |
H: $\text{rank}(A - I) = 0$
If $A$ is a square matrix with real entries, does $\text{rank}(A-I)=0$ tell me that $A=I$? Since the zero matrix is the only matrix of rank zero? ($I$ is the corresponding identity matrix).
AI: Yes that's correct, indeed we have that $\forall x$, $x$ is in the null space of $(A-I)$ that is
$$(A-I)x=0 \iff Ax=x \iff A=I$$ |
H: General real solution of a system of differential equations
Learning for an extra resit of a university exam I was trying to find my mistakes in the resit. However, I can't come to the given solution for the following question(the 4th answer is mine and the 3rd should be correct):
The question
When I calculated it myself with the following calculations, I calculated the eigenvalues to be $5+3i$ and $5-3i$
Calculation 1
Then using the following calculations I determined the corresponding eigenvectors to be $[1+3i;5]$ and $[1-3i;5]$.
Calculation 2
I verified this using Matlab and some websites such as Symbolab.
Then I used the following calculations to come to my final answer:
Calculation 3
This option is not given in the list of possible answers. The values for the last answer correspond to my findings but do not qualify due to the missing $3$ in front of the t's in the (co)sines. Can anybody find my mistake(s) and help me come to the final answer?
The teacher says she will not share the calculations and that the eigenvector should probably be multiplied with a (complex)constant first but I cannot find which.
AI: As far as I can see, your calculations do not contain an error. However, the checked answer "None of the other answers are correct" is wrong, as the third variant is indeed correct. At the stage
$$
A-\lambda I=\pmatrix{1-3i&-2\\5&-1-3i}
$$
you decided on forming the kernel vector from the components of the second row and got $\pmatrix{1+3i\\5}$. You could the same way form it from the first row to find that the kernel is also generated by $\pmatrix{2\\1-3i}$ which then using the same procedures for the remaining steps gives the third answer variant. You get indeed the first kernel vector from the second by multiplication with the complex scalar $\dfrac{1+3i}2$. |
H: An approximate solution of a second-order differential equation
In the book, Mathematical Methods for Students of Physics and Related Fields, Second Edition by Sadri Hassani, Page 667, the author has stated that, for the following differential equation
$y''(x) - x^2 y(x) \approx 0$,
where $x \to \infty$, one can easily obtain an approximate solution of the form $e^{\pm x^2/2}$.
Is there any approach to obtain this solution, besides solving the exact differential equation $y''(x) - x^2 y(x) = 0$ by the Frobenius method, and then taking the limit of the solutions (Hermite polynomials) as $x \to \infty$?
AI: In general the equation $y''-qy=0$ with $0<q(x)\to\infty$ for $x\to\infty$ has the WKB approximation (see wikipedia, this is a standard example) of basis solutions
$$
y(x)=q(x)^{-\frac14}\exp\left(\pm\int\sqrt{q(x)}dx\right)
$$
where the first factor is the second order term in the expansion. So indeed, in first order you would get $y(x)=\exp(\pm\frac{x^2}2)$, while in second order, there would be additionally a factor $\frac1{\sqrt{x}}$.
The approach is to set $y=\exp(S)$ with an expansion $S=S_0+S_1+S_2+...$ where $S_0\gg S_1\gg S_2\gg...$ for $x\to\infty$. This scale also translates to the derivatives. Set for shortness $S'=s$ then isolating the components of equal scale in $s'+s^2-q=0$ gives
$$
s_0^2=q\\
s_0'+2s_0s_1=0\\\vdots
$$
which implies $s_0=\pm\sqrt q$ and $s_1=-\frac{s_0'}{2s_0}\implies S_1=-\frac12\ln|s_0|=-\frac14\ln(q)$. |
H: Stationary solution to a inhomogeneous differential equation.
Use the Transfer function to determine the stationary solution.
Given:
$$y'''(t) +4y''(t)+9y'+10y(t) = 2e^{-3t}\cos(t)$$
I have determined the transfer function to
$$H(s)=\frac{2}{s^{3}+4s^{2}+9s+10}$$
I have attempted to use the theory below to determine the stationary solution but without any success. I would appreciate some help to figure this out. What I have tried does not give me the correct solution.
Correct solution:
$$y_{p}(t)=-0.04615384615 \cdot e^{-3t} \cdot \cos(t) +0.1692307692 \cdot e^{-3t} \cdot \sin(t)$$
AI: The right side of your equation is the real part of $2e^{(-3+i)t}$. The transfer function does not contain parts of the right side, it is just $H(s)=(s^3+4s^2+9s+10)^{-1}$, so that your solution is obtained as the real part of $2H(-3+i)e^{(-3+i)t}$. What remains is to compute the complex numbers in that expression, giving something like $\frac{2e^{(-3+i)t}}{-3+11i}$, then make the denominator real and read off the real part. |
H: Show that $S \leq \sup _{x \geq(1+\varepsilon) M_{n}} x^{-1 / r} \cdot x^{1 / r-1} \sum_{k=1}^{n} E|X_{k}|^{r} I(|X_{k}|>x^{1 / r}) $
Let $1<r<2$ and let $\left\{X_{n}, n \geq 1\right\}$ be a sequence of pairwise independent random tariables
with $E X_{n}=0$ and $E\left|X_{n}\right|^{r}<\infty$ for all $n \geq 1 .$
Set $M_{n}=\sum_{k=1}^{n} E\left|X_{k}\right|^{r}$, $$S =\sup _{x \geq(1+\varepsilon) M_{n}} x^{-1 / r}\left|\sum_{k=1}^{n} E X_{k} I\left(\left|X_{k}\right| \leq x^{1 / r}\right)\right| $$ and let $\varepsilon > 0$
Show that $$S \leq \sup _{x \geq(1+\varepsilon) M_{n}} x^{-1 / r} \cdot x^{1 / r-1} \sum_{k=1}^{n} E|X_{k}|^{r} I(|X_{k}|>x^{1 / r}) $$
My work so far :
since $EX_n = 0$ then $\forall a $ :
\begin{multline*} EX_n = 0 \Longrightarrow EX_nI(|X_n|>a) + EX_nI(|X_n|\leq a) = 0 \Longrightarrow |EX_nI(|X_n|>a)| = |EX_nI(|X_n|\leq a)| \\ \end{multline*}
using this and Jensen's inequality we get :
\begin{align}
S=\sup _{x \geq(1+\varepsilon) M_{n}} x^{-1 / r}\left|\sum_{k=1}^{n} E X_{k} I\left(\left|X_{k}\right|>x^{1 / r}\right)\right| \\
\leq \sup _{x \geq(1+\varepsilon) M_{n}} x^{-1 / r} \sum_{k=1}^{n} E\left|X_{k}\right| I\left(\left|X_{k}\right|>x^{1 / r}\right) \\
\end{align}
however I'm stuck here.
AI: You are almost done. Just note that
$\frac{|X_k|}{x^{1/r}} 1_{|X_k| > x^{1/r}} \leq \frac{|X_k|^r}{x} 1_{|X_k| > x^{1/r}} $ since, either $|X_k| < x^{1/r}$, in which case both sides are 0, or otherwise, $\frac{|X_k|}{x^{1/r}} \geq 1$, and if $a \geq 1$, then $a^r \geq a$ for $r \geq 1$. |
H: Convexity bound in Lieb and Loss.
I'm currently reading through Lieb and Loss's Analysis text. At the end of the proof of theorem 1.9 the authors prove the inequality
$$ \left( |a|+|b|\right)^{p}\leq(1-\lambda)^{1-p}|a|^{p}+\lambda^{1-p}|b|^{p} $$
where $a,b\in \mathbb{C} , p>1$ and $ 0<\lambda<1$. They cite the convexity of the map $|t|^p$ on $0<t<1$. I'm having trouble seeing this and I couldn't get the coefficients to work out so that I could use convexity. On the other hand, I managed to prove this estimate by setting
$$f(x)= \left(1-x \right)^{1-p}|a|^{p}+x^{1-p}|b|^{p}, 0<x<1$$ and using the second derivative to show that the minimum of $f$ occurs at $$\frac{|b|}{|b|+|a|}$$ and equals $$\left( |a|+|b|\right)^{p}.$$ I imagine a solution using the convexity of $|t|^p$ would be much cleaner.
I'd appreciate if anyone could illuminate this.
AI: If $\lambda\in (0,1)$:
$$
(|a| + |b|)^p = \left[ (1-\lambda) \left(\frac{|a|}{1-\lambda}\right) +
\lambda \left(\frac{|b|}{\lambda}\right)\right]^p
\leq
(1-\lambda)\left(\frac{|a|}{1-\lambda}\right)^p
+\lambda \left(\frac{|b|}{\lambda}\right)^p.
$$ |
H: Find a limit of radicals
How to find a limit where $x\to\infty$ of the following expression? What is the steps?
My first guess is to apply this rule to numerator: $(a+b)(a-b) = a^2 - b^2$
AI: No, that won't help. As $x \to \infty$, you simply look at the dominating powers within each sum. For example, in the numerator, $x^2+1$ looks like $x^2$. Then $\sqrt{x^2+1} \sim \sqrt{x^2} = x$. $x+\sqrt{x}$ then approximates $x$ in this limit, so the numerator behaves as $x$.
In the denominator, $x^3+x$ becomes $x^3$, so that $\sqrt[4]{x^3+x}$ approximates $x^{3/4}$. Then the denominator behaves as $-x$ because $x$ dominates $x^{3/4}$.
The limit is then $x/(-x) = -1$. |
H: Solutions to the equation $xk = x^k$
The equation, $xk = x^k$ (where $x$ and $k$ are both integers).
Are there any solutions other than $\{ (1,1), (2,2) \}$ ?
AI: Well, $x^k=kx$ implies $x(x^{k-1}-k)=0$. Thus, $x=0$ works for all $k>0$ and $k=1$ works for all $x$. Aside from that, there are only solutions when $k$ has a $(k-1)$-th root, which I believe only occurs when $k=1$ or $k=2$. |
H: Prove that if f g and 2 are bounded, then f is also bounded
Let $f$ and $g$ be two functions from $\Bbb{R}$ to $\Bbb{R}$.
Prove that if $f-g$ and $2fg$ are bounded, then $f$ is also bounded.
So from what I understand is that $|f| \le M$ and $|g| \le N$. I know I can get rid of the absolute values by squaring both sides, but then what would I do from there?
AI: If $f-g$ is bounded then $(f-g)^2=f^2+g^2-2fg$ is also bounded. You know that if $f$ and $g$ are bounded functions then $f+g$ is also bounded. So we get $f^2+g^2=(f-g)^2+2fg$ is bounded, and $f^2\le f^2+g^2$ so we get desired result. |
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