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H: How many solutions does the equation have?
(a) Sketch the graph of y=(x−p)^2 × (x−q) where p < q.
(b) How many solutions does the equation (x−p)^2 × (x−q)=k have when k > 0?
I have understood part(a) of the question and have got it correct. But I do not understand part(b) correctly. I thought the answer would be zero but the answer is one.
Can you explain what part(b) is asking for exactly and why the answer is one?
Thanks:)
AI: The roots of the equation $$y=f(x)=\color{red}{0}$$
are represented by the intersection of the graph with the $Ox$ axis whose equation is $y=\color{red}{0}$.
you see that $ p$ and $ q$ are roots.
The roots of the equation$$y=f(x)=\color{red}{k} \; with \; k>0$$
will be represented by the intersection of the curve and the horizontal line whose equation is $ y=\color{red}{k}$.
we see that kind of line intersects the curve at a single point. So, the equation $y=k (>0)$ has only one root.
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H: Limit $\lim_{x\to\infty} x(\arctan(a^2x)-\arctan(ax))$
I have a limit $\lim_{x\to\infty} x(\arctan(a^2x)-\arctan(ax))$ and I know the solution $\frac{a-1}{a^2}$, but I dont have any Idea, how to calculate this limit or at least how to start. Any idea?
AI: I assume that we are near $+\infty$.
If $ a=0 $, the limit is zero.
If $ a<0 $ the limit is$$+\infty(\frac{\pi}{2}-(-\frac{\pi}{2}))=+\infty$$
If $ a>0$, then we use the well-known identity, for $X>0 \;:$
$$\arctan(X)=\frac{\pi}{2}-\arctan(\frac 1X)$$
So, we want
$$\lim_{x\to+\infty}x\Bigl(\arctan(\frac{1}{ax})-\arctan(\frac{1}{a^2x})\Bigr)$$
we use the fact that, near $+\infty$
$$\arctan(\frac 1X)=\frac 1X(1+\epsilon(X))$$
thus
$$\arctan(\frac{1}{ax})-\arctan(\frac{1}{a^2x})=$$
$$\frac 1x(\frac 1a-\frac{1}{a^2})+\frac 1x\epsilon(x)$$
the limit is then
$$\frac 1a-\frac{1}{a^2}=\frac{a-1}{a^2}$$
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H: Calculate the curvature $k(t)$, for the curve $r(t)=\langle 1t^{-1},-5,3t \rangle$
I have that $k(t)=\frac{\mid r'(t)\times r''(t) \mid}{\mid r'(t)\mid^3}$.
So first, $r'(t)=\langle -\frac{1}{t^2},0,3 \rangle$.
$r''(t)=\langle \frac{2}{t^3},0,0 \rangle$.
$\mid r'(t)\mid = \sqrt{t^{-4}+9}$.
Then I did $r'(t) \times r''(t)$ to get $\langle 0,\frac{6}{t^{3}}, 0 \rangle$ and took the magnitude of this to get $\sqrt{\frac{36}{t^{6}}}$.
I then put that all over $\left(\frac{1}{t^4}+9\right)^{\frac{3}{2}}$
AI: What you have thus far looks correct. I'm not sure what your question is, but putting it all together yields:
$$k(t)=\frac{\big \lvert\frac{6}{t^3}\big \rvert}{{\left(t^{-4}+9\right)}^{\frac{3}{2}}}=\frac{6}{\left(1+9t^4\right)^{\frac{3}{2}}}$$
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H: Limit of $\frac{1}{r}\ln\left(1+r\sum\limits_{i=1}^n p_i \ln(x_i)+ \omicron(r)\right)$
Let be $n \in \mathbb{N}$ arbitrary but fixed, $\sum\limits_{i=1}^n p_i =1$ and $\forall ~ 1\leq i \leq n$ we assume: $x_i \in \mathbb{R}$.
What is the limit of
$\lim\limits_{r\to 0}~\frac{1}{r}\ln\left(1+r\sum\limits_{i=1}^n p_i \ln(x_i)+ \omicron(r)\right)$, where $\lim\limits_{r\to 0}\frac{\omicron(r)}{r}=0$?
I tried some manipulations and the theorem of L'Hospital but it only got worse...
AI: Let $$S(r)=r\sum_{i=1}^np_i\ln(x_i)$$
we have
$$\lim_{r\to 0}(S(r)+o(r))=0$$
thus
$$\color{red}{\lim_{r\to 0}\frac{\ln\Bigl(1+S(r)+o(r)\Bigr)}{S(r)+o(r)}=1}$$
therefore, the limit is $$\lim_{r\to0}\frac{S(r)+o(r)}{r}=\sum_{i=1}^np_i\ln(x_i)$$
because
$$\frac{\ln(1+S(r)+o(r))}{r}=$$
$$\color{red}{\frac{\ln(1+S(r)+o(r))}{S(r)+o(r)}}\frac{S(r)+o(r)}{r}.$$
and
$$\frac{S(r)}{r}+\frac{o(r)}{r}=$$
$$\sum_{i=1}^np_i\ln(x_i)+\frac{o(r)}{r}$$
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H: A type of set used in convergence in measure theory
This is not a specific problem, but a general question. Often when we're showing convergence of functions (particularly pointwise) or even of sets in certain cases, a set of the following form appears:
$$
\bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_{n}
$$
Often we will encounter situations where we want to show that something goes "wrong" on a set of this form, or a set of this form is measurable (which is obviously true if $E_{n}$ are measurable); I have a general question about a set of this form.
In what contexts does this show up? In my understanding, something is in this set if it "happens" infinitely often. This is a slightly probabilistic interpretation, but if some property holds for infinitely many sets, then something like this shows up. Is that a correct interpretation? I'd appreciate some elaboration.
Example:
If we have a sequence of functions $\{f_{n}\}$, where each $f_{n}$ is supported on a measurable set $E_{n}$, then
$\bigcap_{N} \bigcup_{n \geq N} E_{n} $
is precisely the set of points where the sequence $f_{n}$ is not eventually $0$. Thus, points $x$ at which arbitrarily many $f_{n}$ are non-zero (points in infinitely many supports $E_{n}$). Is this correct?
AI: What you've written is called $\limsup E_n$ and $x \in \limsup E_n$ iff $x \in E_n$ for infinitely many $n$. This understanding already shows your example is correct.
There is a similar notion that $\liminf E_n = \bigcup_{j \geq 1} \bigcap_{n \geq j} E_n$. $x \in \liminf E_n$ iff there is an $N$ such that $x \in E_n$ whenever $n \geq N$ (i.e. "$x \in E_n$ eventually").
These sets occur naturally when considering convergence as you noted. In fact, $\liminf E_n$ is defined to capture definitions like that of convergence which require some behaviour to occur "eventually". Then $\limsup E_n$ is just the complement of $\liminf E_n$.
For example, in the case of convergence we require that for fixed $\varepsilon > 0$, your sequence lies within distance $\varepsilon$ of the limit "eventually". More precisely, if $E_n^\varepsilon = \{x: |f_n(x) - f(x)| < \varepsilon\}$ then
$$\liminf E_n^\varepsilon = \{x: \exists N \text{ such that } \forall n \geq N, |f_n(x) - f(x)| < \varepsilon\}$$
so that we can write
$$\{x: f_n(x) \to f(x)\} = \bigcap_{k \geq 1} \liminf E_n^{k^{-1}}.$$
Conversely, if we want to find $x$ such that $f_n(x) \not \to f(x)$ then there is an $\varepsilon > 0$ such that $|f_n(x) - f(x)| > \varepsilon$ infinitely often which is the same as saying $x \in \limsup (E_n^\varepsilon)^c$. Hence we can write
$$\{x: f_n(x) \not \to f(x)\} = \bigcup_{k \geq 1} \limsup (E_n^{k^{-1}})^c.$$
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H: Find a ring homomorphism $\theta$ s.t. Ker $\theta = \mathbb{Z}_6 \times \{[0]\}$.
Find a ring homomorphism $\theta: \mathbb{Z}_6 \times \mathbb{Z}_{14} \to \mathbb{Z}_6 \times \mathbb{Z}_{14}$ for which Ker $\theta = \mathbb{Z}_6 \times \{[0]\}$.
Attempt: I know that Ker $\theta = \{(x, y) \in \mathbb{Z}_6 \times \mathbb{Z}_{14} | \theta (x, y) = (0, 0)\}$. So, I think of $\theta(x, y)=(0, 0)$ (the trivial homomorphism) as one example. Is this example correct?
I am also told that a correct answer is $\theta([x], [y])=([0], [y])$, but I don't understand how this can be true. For instance, $\theta(x, 1)=(0, 1) \notin Ker \theta$. Can someone please explain how this is a correct answer?
AI: Note that we have $(x, 1)\notin \Bbb Z_6\times \{[0]\}$. So in the answer you were given, having $(x, 1)\notin \ker\theta$ seems like it should be a good thing.
Note that if we set $\theta(x, y) = ([0], y)$, in order to get $([0], [0])$, the one and only requirement is that $y = [0]$. So that is the kernel: the subset of $\Bbb Z_6\times \Bbb Z_{14}$ where the second component is $0$. This is exactly $\Bbb Z_6\times \{[0]\}$, which is why the given answer is correct. (It's not the only correct answer, mind you, but it is the simplest one to describe.)
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H: How to quantify asymptotic growth?
Specifically, my research question is to find operator $A: (\mathbb{R}^+\rightarrow\mathbb{R}^+)\rightarrow\mathbb{S}$, where $\mathbb{S}$ is some totally ordered set, such that for $f, g: \mathbb{R}^+\rightarrow\mathbb{R}^+$ A(f) > A(g) iff $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty$ and A(f) = A(g) iff $\lim_{x \to \infty} \frac{f(x)}{g(x)}$ is a positive real less than infinity.
I'm wondering if anyone has knows sources regarding this question, or any guidance in general. I found a few properties of the operators so far, but have zero sources since I can't find anything similar.
Edit: in the paper so far I've been using Bachmann–Landau notation, but I used the limit definition since I think it is more clear.
AI: As a general comment, there shouldn't be a very precise description of the type of asymptotic growth that can occur for several reasons:
-the set of real valued functions is large and the variety of asymptotic growth is difficult to fathom, in particular there are theorems by du Bois-Reymond and Hardy that show that one always finds "growth orders" above a countable set of growth order, and between two countable sets as well.
-properties of comparability of real valued (or integer valued) functions like the maximal size of possible chains or the minimal size of cofinal chains may depend on axioms which are independant of ZFC
-there seems to be a sort of undiscernability of functions that asymptotically lie beyond a ray of fast growing functions called hyperexponential functions of finite strength
As stated, there are no such elements. It would imply that for any two such functions $f,g$, we have either $\lim_{+\infty} \frac{f}{g} \in \mathbb{R}$ or $\lim_{+\infty} \frac{g}{f}=+\infty$ or $\lim_{+\infty} \frac{f}{g}=+\infty$. Whereas in reality the limit may not exist.
If you restrict to a set $\mathbb{A}$ of functions for which this trichotomy holds, then you can just take $\mathbb{S}$ to be the quotient of $\mathbb{A}$ set by the relation $f \sim g$ iff $\lim_{+\infty} \frac{f}{g} \in \mathbb{R}$, and define the order $<$ as $[f]<[g]$ iff $\lim_{+\infty} \frac{g}{f}=+\infty$ (where $[\cdot]$ denotes the equivalence class). Of course this doesn't really tell you much about the nature of those orders of growth.
There are things in the litterature that look like that. In particular, Hardy fields are examples of such sets $\mathbb{S}$ where each function $f \in \mathbb{A}$ must be $C^k$ on some interval $[a_k,+\infty[$ depending on $f$ and $k$, and $\mathbb{A}$ must contain $g$ with $[g]=[f']$ for each $f \in \mathbb{A}$. And $\mathbb{S}$ must be stable under pointwise field operations (after taking equivalence classes).
So this is what you said plus ordered field operations plus derivation for $\mathbb{S}$.
Certain Hardy fields can be represented as fiels of formal series called transseries, and they can all be seen as fields of surreal numbers, but not in a canonical way. So in a sense in this context the field of surreal numbers works as a sort of universal domain of quantities that quantify asymptotic growth.
The problem of undiscernability is still witnessed the reliance on arbitrary choice (use of AC) to define the embeddings in general, but it still yields a way in which one can talk about certain Hardy fields in a formal way. For instance there is a conjecture (whose proof should appear not too late in the future) that any property of a "maximal Hardy field" (Hardy fields that cannot be extended as larger Hardy fields) stated in the first order language using field operations, the property "being a constant function", the exponential function and the derivation, is true if and only if it is true in the field of so-called logarithmic-exponential transseries. This is a model-theoretic conjecture by Aschenbrenner, van den Dries, and van der Hoeven.
You can find references on Hardy field and this type of question in the works of Hardyand Boschernitzan for instance. I suggest you look at the article On Numbers, Germs, and Transseries which will then direct you to many relevant references on this deep question. It should also help make sense of this messy answer.
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H: Grothendieck group "commutes" with direct sum
The Grothendieck completion group of a commutative monoid $M$ is the unique (up to isomorphism) pair $\langle \mathcal{G}(M), i_M\rangle$, where $\mathcal{G}(M)$ is an abelian group and $i_M\colon M\to\mathcal{G}(M)$ is a monoid homomorphism, satisfying the universal property: for every abelian group $G$ and monoid homomorphism $f\colon M\to G$ there exists a unique $\varphi\colon\mathcal{G}(M)\to G$ such that $f = \varphi\circ i_M$.
Let $M$ and $N$ be commutative monoids. It's easily seen that $M\oplus N$ is a commutative monoid with component-wise operation.
Question: Is it true that $\mathcal{G}(M\oplus N) \cong \mathcal{G}(M)\oplus\mathcal{G}(N)$ ?
The universal property applied to the monoid homomorphism $i_{M}\oplus i_{N}\colon M\oplus N\to\mathcal{G}(M)\oplus\mathcal{G}(N)$ gives a group homomorphism $\varphi\colon\mathcal{G}(M\oplus N)\to\mathcal{G}(M)\oplus\mathcal{G}(N)$ such that $i_{M}\oplus i_{N} = \varphi\circ i_{M\oplus N}$ and I was trying to prove that $\varphi$ is the desired isomorphism, without success.
Is the answer to the question affirmative? If so, is this the correct approach?
Any hints would be appreciated. Thanks in advance.
EDIT: Also, is it true if we replace $M\oplus N$ by $\bigoplus_{\alpha} M_{\alpha}$ ?
AI: Here's a sketch. You have to construct the inverse using universal properties as well. You have a composition $$M \hookrightarrow M\oplus N \stackrel{i_{M\oplus N}}{\longrightarrow} \mathcal{G}(M\oplus N)$$which induces a map $\mathcal{G}(M) \to \mathcal{G}(M\oplus N)$. Similarly, you get a map $\mathcal{G}(N) \to \mathcal{G}(M\oplus N)$. The universal property of the direct sum joins these two maps into a map $\psi\colon\mathcal{G}(M)\oplus \mathcal{G}(N) \to \mathcal{G}(M\oplus N)$. Now you have two maps $$\psi\circ \varphi\colon \mathcal{G}(M\oplus N) \to \mathcal{G}(M\oplus N)\quad\mbox{and}\quad\varphi\circ \psi\colon \mathcal{G}(M)\oplus \mathcal{G}(N) \to \mathcal{G}(M)\oplus \mathcal{G}(N).$$Using the uniqueness provided by the universal properties of $\mathcal{G}$ and $\oplus$, argue that these compositions equal the identity. It works the same for defining maps $$\bigoplus_{\alpha} \mathcal{G}(M_\alpha) \to \mathcal{G}\left(\bigoplus_{\alpha}M_\alpha\right) \quad\mbox{and}\quad \mathcal{G}\left(\bigoplus_\alpha M_\alpha\right) \to \bigoplus_\alpha \mathcal{G}(M_\alpha)$$and running the above argument through.
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H: I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$
I am trying to understand each step in order to get from $x^2-4$ to $(x-2)(x+2)$
I start from here and got this far...
$x^2-4 =$
$x*x-4 =$
$x*x+x-x-4 =$
$x*x+x-2+2-x-4 =$
$x*x+x-2+2-(x+4) =$
After this I try
$x(x-2)+2-(x+4) =$
and this clearly does not even equal the other factorings. I thought the $x$ could be factored out. I'm confused.
I know I can just insert $a^2-b^2$ into the difference of squares formula like so $(a-b)(a+b)$ but I am practicing factoring. I'm just curious to see each and every step of the factoring.
AI: \begin{align}
x^2-4&=x^2-2x+2x-4\\
&=x(x-2)+2(x-2)\\
&=(x+2)(x-2)
\end{align}
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H: 'Locally' Convex Function
I have a continously differentiable function $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ which I am trying to prove is globally convex. Computing the Hessian directly is very difficult as it is a somewhat complicated function of a matrix, other methods of proving global convexity have proved inconclusive. So far I am only able to show that it is 'locally convex' in the following sense:
For any $x\in\mathbb{R}^{n}$ there exists an $\varepsilon_{x}>0$ such that for $y\in\mathbb{R}^{n}$ where $\| y-x\|\leq\varepsilon_x$ it holds that
$$f(y)\geq f(x)+\nabla f(x)^{T}(y-x). $$
My question is a rather basic one, can we establish that local convexity of this kind implies global convexity? Are any extra conditions needed?
My intuition suggests that a continuously differentiable function on a convex set which is locally convex everywhere should be globally convex, but I have trouble constructing the argument. Any help is greatly appreciated!
AI: Here is proof if $f$ is assumed $C^2$. This is not necessary but simplifies the proof significantly.
It is sufficient to show the $f''(x) \ge 0$.
Pick some $x$, then there is a neighbourhood $U$ of $x$ such that
$f(x+h) -f(x) \ge f'(x) h$ for $x+h \in U$.
Since $f$ is $C^2$, Taylor gives (for $h$ sufficiently small) that
$f(x+h) = f(x) + f'(x)h + {1 \over 2} h^T f''(\xi_h)h$, where $\xi_h \in [x,x+h]$. This gives $h^T f''(\xi_h)h \ge 0$ for $h$ such that $x+h \in U$. If $h \neq 0$ then ${h^T \over \|h\|} f''(\xi_h) {h \over \|h\|} \ge 0$, of course.
Pick some unit vector $v$, and let $h = t v$ for small $t$, then
we have $ v^T f''(\xi_{tv}) v \ge 0$, and letting $t \to 0$
and using continuity of $f''$ we get $ v^T f''(x) v \ge 0$.
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H: Let $f:\Omega\to\textbf{R}^{m}$ be a function. Then $f$ is measurable if and only if $f^{-1}(B)$ is measurable for every open box $B$.
Let $\Omega$ be a measurable subset of $\textbf{R}^{n}$, and let $f:\Omega\to\textbf{R}^{m}$ be a function. Then $f$ is measurable if and only if $f^{-1}(B)$ is measurable for every open box $B$.
My solution
Let us prove the implication $(\Rightarrow)$ first.
If $f$ is measurable, the $f^{-1}(V)$ is measurable for every open subset $V\subseteq\textbf{R}^{m}$. In particular, since the open box $B$ is open, we conclude that $f^{-1}(B)$ is measurable for every open box $B$.
Let us prove the implication $(\Leftarrow)$ now.
Let us consider an open subset $V\subseteq\textbf{R}^{m}$. Thus we can express $V$ as a countable union of open boxes $(B_{j})_{j\in J}$. Thus we get
\begin{align*}
V = \bigcup_{j\in J}B_{j} \Rightarrow f^{-1}(V) = f^{-1}\left(\bigcup_{j\in J}B_{j}\right) = \bigcup_{j\in J}f^{-1}(B_{j})
\end{align*}
Since $f^{-1}(B_{j})$ is measurable and countable union of measurable sets is measurable, the result holds.
Could someone please verify if the wording of my proof is satisfactory?
AI: Yes, it is perfectly clear and correct! Well done!
Only the fact that every open subset is the countable union of open boxes might need a proof, if you haven't proven this fact before.
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H: Closed form of $\sum _{i=1}^n\:\frac{\left(m-i\right)!}{\left(n-i\right)!}$
I am interested in a closed form of the following sum:
$$
S_n := \sum _{i=1}^n\:\frac{\left(m-i\right)!}{\left(n-i\right)!}\;\; n,m \in \mathbb{N},\ n < m$$
Amongst other strategies I have also tried to compress the sum in order to gain some insight but failed in getting a useful result:
$$
S_n = (m-n)\left(\frac{m-(n-1)}{1}\left(\frac{m-(n-2)}{2}\left(\cdots\left(\frac{m-1}{n-1} + 1\right)\cdots +1\right) + 1\right)+1\right)$$
I thought one may be able to express $S_n$ in terms of $S_{(n-1)}$ but, as far as I can see it, that makes things more complicated.
This sum is used when analyzing the average runtime of a so called open addressing hash table.
AI: We have
$$\frac{(m-i)!}{(n-i)!}=\frac{(m-n)!(m-i)!}{(m-n)!(n-i)!}=(m-n)!\binom{m-i}{m-n}$$ so that $$\sum _{i=1}^n\:\frac{\left(m-i\right)!}{\left(n-i\right)!}=(m-n)!\sum_{k=m-n}^{m-1}\binom{k}{m-n}=(m-n)!\binom{m}{m-n+1}$$ by the hockey stick identity.
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H: SIR Model Specifics
I read on Wikipedia (under "Compartmental models in epidemiology") that the differential equations for the SIR Model was the following,
$$S'(t)=-\frac{\beta}{N}I(t)S(t)$$
$$I'(t)=\frac{\beta}{N}I(t)S(t)-\gamma I(t)$$
$$R'(t)=\gamma I(t)$$
It never states the range of $\gamma$ and $\beta$. However, based on the explanation of $\beta$, which is the transmission effectiveness of the disease and the average number of contacts between people per time multiplied, so theoretically if this average became arbitrarily large, $\beta \to \infty$? So it's range could be $\beta \in [0,\infty)$. Whereas, $\gamma=\frac{1}{D}$, where $D$ is how long someone is infected for which could mean $D \in (0,\infty)$ and as such $\gamma \in (0,\infty)$. However, I'm not entirely sure if these are correct (since I just assumed it based on the explanation), and I haven't found any other sources explaining this.
AI: Theoretically, yes, the ranges of $\beta$ and $D$ are as you stated (although $\beta = 0$ means the disease never spreads, which isn't really helpful). However, SIR is a model, and the purpose of models are to give a good approximation for things like diseases over a limited period of time. This means some parameter values may make the model very accurate, while other values may not even produce a practical model to work with.
Note that $\frac{\beta}{\gamma} = \beta D = R_0$, which is the basic reproduction number of a disease. Values of $R_0$ for common diseases seem to be around 0.5 to 20, depending on the disease. Hopefully this gives you some idea what realistic values of $\beta$ and $D$ are.
Also, this link may also provide more insight on the SIR model.
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H: Finding a fixed polynomial under the multiplicative inversion automorphism
Can anyone find a polynomial $f ∈ ℚ\left(X+\frac{1}{1-X} + \frac{X-1}{X}\right) ⊆ ℚ(X)$ that is fixed under the automorphism $(X ↦ \frac{1}{X})$? $f = X+\frac{1}{X}$ would be nice, but I don't know how to check if it's in the given (sub)field.
(End game / context: find $f ∈ \mathbb{Q}(X)$ s.t. $\mathbb{Q}(f) = \mathbb{Q}\left(X+\frac{1}{1-X} + \frac{X-1}{X}\right)^{(X ↦ \frac{1}{X})}$.)
AI: Letting
$$g(x)=x+\frac{1}{1-x} + \frac{x-1}{x}$$
it's easily verified that
$$g(x)+g\!\left({\small{\frac{1}{x}}}\right)=3$$
hence letting $h(x)=g(x)\bigl(3-g(x)\bigr)$, we get
$$h(x)=g(x)\,g\!\left({\small{\frac{1}{x}}}\right)$$
so $h$ satisfies the required conditions.
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H: Need help understanding two steps in the proof that limit point compactness implies sequential compactness
So I want to learn the proof that a compact metric space $(X,d)$ is also sequentially compact. The proof goes as follows:
(X,d) is compact, so it is also limit point compact. Let $\{x_k\}_{k=1}^\infty$ be a sequence in X and put $A = \{x_k : k\in \mathbf{N}\}$. Then A is either finite or infinite. If A is finite, we are done, so assume A is infinite. Then A has a limit point x. Since we are in a metric space, $B_{1/k}(x) \cap A$ is infinite. Choose $n_{k} > n_{k-1}$ s.t $x \in B_{1/k}(x) \cap A$. Then $\{x_{n_{k}}\}_{k=1}^\infty$ is a subsequence with $x_{n_{k}} \rightarrow x$, $k \rightarrow \infty$
I'm confused about the step where we intersect the ball of radius $1/k$ with $A$. Why is the intersection infinite? And why do we choose $n_k > n_{k-1}$? Is it just to say that the sequence is increasing $\forall k$?
AI: We know $x$ is a limit point of $A$. So by definition, every open neighborhood of $x$ must intersect $A$ at a point other than $x$. If there were only finitely many points in $B_\epsilon(x) \cap A$ for any $\epsilon>0$, where $B_\epsilon(x)$ is the open ball of radius $\epsilon$ centered at $x$, then take a point distinct from $x$ closest to $x$ and call this $P$. Then define the distance from $x$ to $P$ using your metric $d$ as $D= d(x,P)$. But then $B_{D/2}(x) \cap A$ is either empty or simply $x$ itself. But then $x$ is not a limit point of $A$, contradiction.
As for the next question, the choice of $n_k$ are merely so that you choose a point in $B_{1/k}(x) \cap A$ 'further down' in the sequence $\{x_n\}$ so that the point you choose will be even closer to $x$ (because the distance is at most $1/k$ for that choice of $k$). Then you are choosing points further and further down your sequence, i.e. forming a subsequence from your given sequence, where the points are getting closer and closer to $x$. Then because you can choose them arbitrarily close, you know that $x_{n_k} \to x$. Then you have found a convergent subsequence. [Not necessarily the only one, but one at least.]
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H: Does $(\frac{n}{4})^{\frac{n}{4}}$ have a higher asymptotic growth than $4^{n^{4}}$
I'm trying to determine how these 3 functions should be ordered in terms of asymptotic growth:
$$f(n) = \left(\frac{n}{4}\right)^{\frac{n}{4}}$$
$$g(n) = n^{\frac{n}{4}}$$
$$h(n) = 4^{n^{4}}$$
$f(n)$ seems to be somewhat similar to $n^{n}$ and that should grow faster than $4^{n}$, but I'm not sure how to prove it.
AI: Taking logs on both sides you have the comparison
$$\log g = \frac{n}{4}\log n$$
$$\log h = n^4 \log 4$$
Which is faster?
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H: Inequality on integrals of $L^1$ functions
Let $\lambda \geq 0$ and $(X,d,\mu)$ be a $\sigma-$finite measure space. Then for $f, g \in L^1(X,\mu)$
$$ \left| \int_X (|f|-\lambda)^{+} d\mu - \int_X (|g|-\lambda)^{+} d\mu \right| \leq \int_X ||f|-|g|| d\mu$$
holds (where $(x)^{+} = \text{max}(x,0)$).
I tried dividing $X$ space into sets where, $(|f|-\lambda)^{+}$ = $(|f|-\lambda)$ and so on, but I still did not manage to prove this inequality. Could you offer me some hints or help with the proof?
AI: Verify the the inequality:
$|x^{+}-y^{+}| \leq |x-y|$ for all real numbers $x$ and $y$.
If you bring the absolute value sign in LHS inside the integral this gives LHS $\leq \int |(|f)-\lambda| -(|g|-\lambda))|=\int ||f|-|g|| =$ RHS.
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H: Find inverse matrix
I have a matrix:
$$A = \begin{pmatrix}
1 & 1 & 1 & \cdots & 1\\
0 & 1 & 1 & \cdots & 1\\
0 & 0 & 1 & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & 1
\end{pmatrix}$$
I need to find $A^{-1}$. How could I do that, knowing that matrix A has size $n \times n$?
AI: HINT: Consider the matrix
$$B =
\begin{pmatrix}
1 & -1 & 0 &\cdots & 0\\
0 & 1 & -1 & \cdots & 0\\
0 & 0 & 1 & \ddots&0\\
\vdots & \vdots & \ddots & \ddots & -1\\
0 & 0 & 0 & \cdots &1
\end{pmatrix}$$
One easily verifies that
$$AB = I_n$$ and $$BA = I_n$$
Thus $B=A^{-1}$
Edit: How to come up with $B$? First of all you should compute some inverse matrices, maybe for the cases $n = 2$ and $n=3$. In order you may see a pattern and get an idea how a candidate $B$ for an inverse of $A$ should look like. Just verify $BA = I_n$ and $AB = I_n$ and you will end up with $A^{-1} = B$.
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H: Critical points of a multivariable function
Critical points of Z(x,y)=x^3+3xy^2-15x-12y
enter image description here
I get confuse on how you get x^4-5x^2+4=0
AI: They only substituted $y=\frac{2}{x}$ in the equation above and then they multiplied it by $x^2$ to get better-looking equation so it can be solved better.
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H: Find the arclength curve of $r(t)=i+3t^2j+t^3k$ for $0\leq t\leq \sqrt{12}$
I asked a question similar to this one, but I'm still confused on how to integrate this.
I have $r'(t)=\langle 0,6t,3t^2\rangle$. and so this gives you the integral from $0$ to $\sqrt{12}$ of $\sqrt{36t^2+9t^4}dt$. Step by step would be helpful, thanks!
AI: You've set up the calculation to do, which is good. From here it is just calculus. If you factor out the $t^2$ from the square root, we end up having to compute
$$\int_{0}^{\sqrt{12}}t\sqrt{36 + 9t^2} \mathrm{d}t.$$
Do you see a substitution you could perform?
|
H: Question regarding surjectivity of induced homormophism in an old version of Hatcher's proof of Prop. 4.13
So I am currently trying to understand the given proof of Hatcher's proof of proposition 4.13.
It's this particular part (in the middle of the screenshot) I don't understand:
The extended $f$ still induces a surjection on $\pi_k$ since the
composition $\pi_k(Z_k)\to \pi_k(Y_{k+1}) \to \pi_k(X)$ is
surjective.
How do we know that this composition is surjective? I understand the implication, but I don't understand why the composition is supposed to be surjective in the first place.
Does that follow from the induction hypothesis, that $\pi_k(Z_k) \to \pi_k(X)$ is surjective? If so, why does "factorizing through $\pi_k(Z_{k+1})$" not affect surjectivity?
I am sorry for not being able to express my confussion in any better way.
I hope someone can enlighten me.
Thank you very much!
AI: If $\pi_k(Z_k) \to \pi_k(X)$ is surjective, well, the composite $\pi_k(Z_k) \to \pi_k(Y_{k+1}) \to \pi_k(X)$ is that same map, so it is surjective. So yes, it follows from the inductive hypothesis.
|
H: Are perpendicular vectors always in different subspaces?
So I understand when two subspaces are considered perpendicular and what it means for vectors to be perpendicular/orthogonal.
The question I have is, if two vectors are perpendicular, do they always have to exist in orthogonal subspaces such as the nullspace and rowspace (I am using nullspace and rowspace as examples)? Can orthogonal vectors exist in the same subspace?
Finally, if $A^T = A$, then is the column space $\perp$ to nullspace and left nullspace?
AI: Given any two vectors $u, w$, you can for example consider $\text{span}(\{u,w \})$, which is a subspace that contains both $u$ and $w$. To answer your question more directly, take for example any two nonzero orthogonal vectors in $\mathbb{R}^3$. Then their span is a plane containing both vectors, but neither vector is contained in the line orthogonal to this plane.
For real matrices $A$, the row space is orthogonal to the null space. So if $A^T = A$, then yes, the column space is orthogonal to both the null space and left null space.
|
H: Establish the inequality for an analytic function.
So, I am self studying analysis. I came across this question. It asks us to establish the following inequality:
$\int_{-1}^{1} |f(x)|^2 dx \leq \frac{1}{2} \int_{0}^{2\pi} |f(e^{it})|^2 dt, $
for $f(z)$ analytic on the open unit disk and continuous on the closed unit disk.
My attempt: I have integrated $f(z) \overline{f(\overline{z})}$ along the upper semi-circle and have gotten the following:
$\int_{\gamma} f(z) \overline{f(\overline{z})}dz = \int_{-1}^{1} |f(x)|^2 dx + \int_{0}^{\pi} f(e^{it})\overline{f(e^{-it})}i e^{it}=0, $
by the residue theorem. Doing this again once more along the lower semi-circle with the proper parameterization for the circle portion gives me:
$\int_{\gamma} f(z) \overline{f(\overline{z})}dz = \int_{-1}^{1} |f(x)|^2 dx + \int_{2 \pi}^{\pi} f(e^{-it})\overline{f(e^{it})} (-i e^{-it})=0, $
Adding the two gives me:
$2 \int_{-1}^{1} |f(x)|^2 dx + \int_{0}^{\pi} f(e^{it})\overline{f(\overline{e^{it}})}i e^{it} + \int_{\pi}^{2 \pi} f(e^{-it})\overline{f(e^{it})} (i e^{-it}) = 0$
Taking modulus gives:
\begin{align*}
2 |\int_{-1}^{1} |f(x)|^2 dx| &\leq \int_{0}^{\pi} |f(e^{it})\overline{f(e^{-it})}| dt + \int_{\pi}^{2 \pi} |f(e^{-it})\overline{f(e^{it})}| dt \\
&\leq \int_{0}^{\pi} |f(e^{it})||f(e^{-it}))| dt + \int_{\pi}^{2 \pi} |f(e^{-it})||f(e^{it})| dt \\
&= \int_{0}^{2 \pi} |f(e^{-it})||f(e^{it})| dt
\end{align*}
But, how in the world do I finish it? If $|f(e^{-it})||f(e^{it})| = |f(e^{it})|^2.$ I would be done, but this doesn't seem true. Any hints would be great. Or maybe my approach is wrong?
AI: The idea is first to treat the case when $f(x) \in \mathbb R, -1 \le x \le 1$ as then you can just integrate $f^2$ on the two closed paths formed by the real diameter and the two half-circles (upper/lower); technically to apply Cauchy, one actually integrates on the smaller disc of radius $r$ and by continuity take $r \to 1$
So on the upper path $\int_{-r}^{r}f^2(x)dx + i\int_0^{\pi}f^2(re^{i\theta})e^{i\theta}d\theta=0$, hence
$\int_{-r}^{r}f^2(x)dx \le \int_0^{\pi}|f(re^{i\theta})|^2d\theta$ and with $r \to 1$ we get
$\int_{-1}^{1}f^2(x)dx \le \int_0^{\pi}|f(e^{i\theta})|^2d\theta$ and adding the lower path inequality we are done in this case.
But now let $f(z)=g(z)+ih(z)$ the usual decomposition in analytic functions that are real on the real axis (the Taylor coefficients of $g$ are the real parts of the Taylor coefficients of $f$ etc).
$\int_{-1}^{1}|f(x)|^2dx=\int_{-1}^{1}g^2(x)dx+\int_{-1}^{1}h^2(x)dx$ since on the real axis $g,h$ are real
Applying the previous case one gets:
$2\int_{-1}^{1}|f(x)|^2dx \le \int_0^{2\pi}|g(e^{i\theta})|^2d\theta+\int_0^{2\pi}|h(e^{i\theta})|^2d\theta$
But note that $g(\bar z)=\bar g(z), h(\bar z)=\bar h(z)$ so
$|f(z)|^2=f(z)\bar f(z)=|g(z)|^2+|h(z)|^2+i(h(z)g(\bar z)-h(\bar z)g(z))$
But on the unit circle $h(z)g(\bar z)-h(\bar z)g(z)=h(e^{i\theta})g(e^{-i\theta})-h(e^{-i\theta})g(e^{i\theta})$ is clearly an odd function in $\theta$ and since by periodicity
$\int_0^{2\pi}(h(e^{i\theta})g(e^{-i\theta})-h(e^{-i\theta})g(e^{i\theta}))d\theta=\int_{-\pi}^{\pi}(h(e^{i\theta})g(e^{-i\theta})-h(e^{-i\theta})g(e^{i\theta}))d\theta$ it clearly vanishes being odd, so the result follows since we get that
$\int_0^{2\pi}|f(e^{i\theta})|^2d\theta=\int_0^{2\pi}|g^2(e^{i\theta})|d\theta+\int_0^{2\pi}|h^2(e^{i\theta})|d\theta$
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H: Homeomorphism between $\mathbb{R}^2$ and $\mathbb{R}^2-B(0,1)$
Are $\mathbb{R}^2$ and $\mathbb{R}^2-B(0,1)$ homeomorphic? I'm trying to find a homeomorphism between the half of a hyperboloid of one sheet, $$C=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2-z^2=1,\textrm{ }z\geq 0\},$$ and the plane $\mathbb{R}^2$. I've already showed that $C$ is homeomorphic to $\mathbb{R}^2-B(0,1)$ by defining the function $f:\mathbb{R}^2-B(0,1)\to C$, $(x,y)\mapsto f(x,y)=(x,y,\sqrt{x^2+y^2-1})$.
AI: Putting my comment as an answer, since $\Bbb{R}^2$ is simply connected but $\Bbb{R}^2 - B(0,1)$ is not, and homeomorphism preserves simply connectedness, they are not homeomorphic.
|
H: Shilov Chapter 4 Problem 16
I am working on the captioned problem which is reproduced below.
And the hint for this problem is the following:
But I have no idea of how to use Chapter 3 Prob 12 for this problem. That problem is reproduced here:
I have no problem in finding out that three equations for unknown elements of A and B lead to equations for three minors of 2x3 matrix. But I have no clue as to how to proceed using Chap3 Prob 12.
Thank you.
AI: The equation
$$
\pmatrix{a&b\\ c&d}\pmatrix{x&y\\ z&w}=\pmatrix{P&-Q\\ R&-P}
$$
can be rewritten as
\begin{aligned}
bz-cy &= P,\\
(a-d)y-b(x-w) &= -Q,\\
c(x-w)-(a-d)z &= R.
\end{aligned}
That is, if
$$
A=\pmatrix{b&c&a-d\\ y&z&x-w},\tag{$\ast$}
$$
then the minors obtained by deleting respectively the third column, the second column and the first column of $A$ are given by $P,Q$ and $R$. So, if one can construct $A$, one can pick two arbitrary values for $d$ and $w$ and recover $a,b,c,x,y,z$ from $(\ast)$.
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H: Proving limits in terms of epsilon delta questions
$ f(x)= {x^3}-2x+1. $
We want to show that $ \lim_{x\to 2} ({x^3}-2x+1)= 5 $.
So here,
$ \lvert f(x)-f(2)\rvert = \lvert {x^3}-2x+1-5\rvert$ = $ \lvert {x^3}-2x-4\rvert $.
$ \lvert ({x^3}-2x)+(-4)\rvert \leq \lvert {x^3}-2x\rvert + \lvert -4\rvert = \lvert {x^3}-2x\rvert + 4 $, by triangle inequality.
I'm confused on what to do next for the $ \epsilon$ and $\delta$.
AI: As you have computed, we have
$$|f(x)-f(2)|=|x^3-2x-4|.\tag{0}$$
Now, we have
$$x^3-2x-4=(x-2)(x^2+2x+2)=(x-2)[(x+1)^2+1].\tag{1}$$
If $|x-2|<1$, then $1<x<3$ and we can bound
$$0<(x+1)^2+1\leq 16+1=17.\tag{2}$$
Therefore, for any $\epsilon>0$, we can choose $\delta=\min\{ 1, \epsilon/17\}>0$
such that if $|x-2|<\delta$, then
$$|f(x)-f(2)|=|x^3-2x-4|~~\mbox{ by }(0)\\
=|x-2|\cdot|(x+1)^2+1|~~\mbox{ by }(1)\\
\leq \frac{\epsilon}{17}\cdot 17 ~~\mbox{ by }(2)\\
=\epsilon.
$$
|
H: Calculate $ \int_0^{2\pi} \ln(2-2\cos(t)) \ln(2-2\cos(t+\theta)) dt$
I'm trying to evaluate
$$ \int_0^{2\pi} \ln(2-2\cos(t)) \ln(2-2\cos(t+\theta)) dt$$
but I'm not sure the best way to proceed. I've been trying to factor the inner terms in to rational functions of $e^{it}$ and use logarithm identities to split the integral in to the sum of simpler complex integrals, but to limited success.
AI: This integral has an expression in terms of special functions.
Using the fact that
$$-\sum_{n=1}^{\infty}\frac{\cos(nx)}{n}=\ln(2|\sin(x/2)|)$$
the above integral can be directly rewritten as
$$I=4 \int_{0}^{2\pi}\ln\Big(2\Big|\sin\frac{t}{2}\Big|\Big)\ln\Big(2\Big|\sin\frac{t+\theta}{2}\Big|\Big)dt=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{4}{mn}\int_{0}^{2\pi}dt~\cos(nt)\cos(mt+m\theta)$$
However it is true that
$$\int_{0}^{2\pi}dt~\cos(nt)\cos(mt+m\theta)=\pi\cos m\theta(\delta_{m+n,0}+\delta_{m-n,0})$$
and thus the desired quantity reduces to
$$I(\theta)=4\pi\sum_{n=1}^{\infty}\frac{\cos n\theta}{n^2}=2\pi\Big[\text{Li}_2(e^{i\theta})+\text{Li}_2(e^{-i\theta})\Big]$$
It seems that from this very simple form, a simpler one is unattainable, however, finding particular values of the integral in terms of known constants is not impossible. For example
$$I(0)=4\pi \zeta(2)=\frac{2\pi^3}{3}\\I(\pi)=-2\pi\zeta(2)=-\frac{\pi^3}{3}$$
EDIT:
It is actually not difficult to obtain a closed form for this sum as in here (thanks @Jay Lemmon for catching this).
Here's a way to show it different from the one cited on the link:
We know that
$$\sum_{n=1}^{\infty}\frac{\sin nx}{n}=\frac{\pi-x}{2}~~, x\in(0,2\pi)$$
Integrating this equation in $[0,\theta]$ we obtain
$$-\sum_{n=1}^{\infty}\frac{\cos n\theta}{n^2}+\frac{\pi^2}{6}=\frac{\pi \theta}{2}-\frac{\theta^2}{4}$$
which shows that the integral in question has a very simple form
$$I(\theta)=\pi\theta^2-2\pi^2\theta+\frac{2\pi^3}{3}$$
|
H: Minimum of $\sqrt {{x^2} + \frac{1}{{{x^2}}}} + \sqrt {{y^2} + \frac{1}{{{y^2}}}} + \sqrt {{z^2} + \frac{1}{{{z^2}}}} $?
I have been ponder around a difficult Vietnamese University entrance exam of the year 2003
that is to find the minimum of $\sqrt {{x^2} + \frac{1}{{{x^2}}}} + \sqrt {{y^2} + \frac{1}{{{y^2}}}} + \sqrt {{z^2} + \frac{1}{{{z^2}}}} $ given that $x + y + z \le 1$ ? and x,y,z are all greater of equal to zero
The answer for this problem $\sqrt {82}$, is there anyway we could find the minimum of this problem from the convex optimization point of view because the term $\sqrt {{x^2} + \frac{1}{{{x^2}}}}$ is convex on the interval [0,1]
Also is there any underlying hidden convex structure in this problem (least square, second order cone , quadratic...) ?
Thank you very much for your enthusiasm !
AI: As you point out, the function is convex on the standard simplex $\Delta^2$. Therefore, for all $(x,y,z) \in \Delta^2$, the function value at the average is less than or equal to the average of the function values:
$$f(m,m,m) \leq (f(x,y,z) + f(x,z,y) + f(y,x,z) + f(y,z,x) + f(z,x,y) + f(z,y,x))/6,$$
with $m=(x+y+z)/3$.
Due to symmetry:
$$f(x,y,z) = f(x,z,y) = f(y,x,z) = f(y,z,x) = f(z,x,y) = f(z,y,x),$$
and therefore:
$f(m,m,m) \leq f(x,y,z)$
for all $(x,y,z) \in \Delta^2$. The minimum is therefore attained at $x=y=z=1/3$.
|
H: Average value of the orders of all elliptic curves over the finite field of p-elements
Is true that the average value of the orders of all elliptic curves over $\mathbb F_p$ is $p+1$?
More precisely, fix a prime $p$ and let $\mathbb F_p$ be the field of $p$ elements. Consider the set $S=\{(a,b)\in\mathbb F_p\times\mathbb F_p \,:\, 4a^3+27b^2\neq 0\,\, \text{in}\,\,\mathbb F_p\}$, so that every element $(a,b)\in S$ defines an elliptic curve
$$E(a,b,p)=\{(x,y)\in\mathbb F_p\times\mathbb F_p \,:\,y^2 =x^3+ax+b\,\, \text{in}\,\, \mathbb F_p\}\cup\{\infty\}$$
Is it true that
$$\frac{1}{|S|}\sum_{(a,b)\in S}|E(a, b, p)| = p + 1$$
where $|S|$ and $|E(a, b, p)|$ denote the orders of these sets.
I have verified it through computation for some small primes and am just wondering if it is true in general.
AI: Yes. If $E$ defined by $y^2=x^3+ax+b$ has $p+1-t$ elements, then the curve $E_c$
defined by $y^2=x^3+ac^2x+bc^3$ has $p+1-t$ points if $c$ is a quadratic
residue modulo $p$ but $p+1+t$ points if $c$ is a quadratic non-residue.
(In the first cases $E_c$ is isomorphic to $E$, in the second case it's the
"quadratic twist" of $E$). Averaging out over the $E_c$ gives $p+1$ points.
Overall the set $S$ splits up into various sets of $E_c$s each one average
with $p+1$ points.
(I'm assuming here $p\ge5$)
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H: Example of non compact sets whose union and intersection is compact
Give an example of two non compact sets $A$ and $B$ such that $A \cup B$ is compact and $A \cap B$ is compact.
My attempt:
Let $A=\{1/n:n=1,2,...\}$ and $B=\{0\} \cup (1,2]$.
Then $A \cup B$ is closed and bounded so it is compact and $A \cap B$ is empty so it is also compact but $A$ and $B$ are not compact as they are not closed.
Is it correct?
AI: Your answer is just fine! Just remember, depending on what this is for, you may want to also show/explain why $A$ and $B$ are not compact (show that $A,B$ have a limit point that they do not contain so that they are not closed, hence not compact). You have justified that $A \cap B$ is compact. It is clear that $A \cup B$ is bounded. You may be expected to verify that $A \cup B$ is closed, which just require a little bit of routine extra work.
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H: The probability of choosing a 5 element subset from the set {1,2,...,20}, with 1 element from{1,4,6,8,9} and 1 from{11,13,17,19} but 0 from {2,3,5,7}
I need to make a 5 element subset using at least one of {11, 13, 17, 19} and at least one of {1, 4, 6, 8, 9} from the set of the first twenty integers but I cannot use any of {2, 3, 5, 7}. I tried doing
$$ \frac {\left({16 \choose 5} - {11 \choose 5}\right) \left({16 \choose 5} - {12 \choose 5}\right)}{{20 \choose 5}} $$
Is this correct?
AI: No, this is not correct. You should have some words explaining how you got your answer. $16 \choose 5$ is the number of ways to select five elements without $2,3,5,7$. It is a good start. You then subtract $11 \choose 5$ which is the number of ways to select five elements without $2,3,5,7$ nor $1,4,6,8,9$. Instead of multiplying, you should then subtract $12 \choose 5$, but now you have double subtracted the cases missing one of $1,4,6,8,9$ and one of $11,13,17,19$, so you have to add them back in. See the inclusion-exclusion principle.
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H: Find polynomial of degree n+1 for n+1 data points.
Generally, if we have n+1 data points, there is exactly one polynomial of degree at most n going through all the data points. What can we tell about existence of polynomial of degree n+1? How can we find polynomial of degree n+1 for a table:
x: 0 1 2 3
y: 1 0 -5 -20
AI: Suppose you have a set of points $\{(x_1,y_1,\ldots,(x_{n+1},y_{n+1})\}$. Construct a polynomial $p_n(x)$ of degree $n$ going through the points. Then you can find a polynomial of any degree $n+k$ going through that set of points. You could, as Arturo points out, add any $k$ points, distinct from your original set, to the set of points then find a polynomial of degree $p_{n+k}(x)$ going through those points using your method of choice. Then you have a polynomial of degree $n+k$ going through your original points.
If $k$ is particularly large, you could do this with the method above coupled with multiplying $p_n(x)$ by a polynomial which interpolates the points $\{(x_1,1),\ldots,(x_n,1)\}$ (so that it increments the degree but does not change the value of $p_n(x)$ at your $x_i$'s).
As another method, you find a polynomial of degree $k$, say $g(x)$, and then form
$$
\tilde{p}(x)= p_n(x) + g(x) \prod_i (x-x_i)
$$
so that $\tilde{p}(x)$ takes the values $y_i$ at each $x_i$ but is of degree $\deg p_n(x) + \deg g(x)= n+k$
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H: Prove that $L \geq M$.
In the above proof, it is taken that for both of the sequences $\exists N$ s.t $|a_n - L|<\epsilon $ and $|b_n - M|<\epsilon$. Is it correct to take the same $N$ for both sequences?
After that they have assumed the inequality which is true for $n\geq N_0$ is also true for $n\geq N$. Is that correct ?
AI: They omitted some details, but it can be done. Let's say $|a_n - L| < \epsilon$ for $n \geq N_1$ and $|b_n - M| < \epsilon$ for $n \geq N_2$. Let $N = \max\{N_0,N_1,N_2\}$ and this $N$ would work for the whole proof (we just need the existence of such an $N$).
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H: It's possible to construct a disjoint sequence of open balls with center elements of a sequence in a metric space
Let $(X,d)$ a complete metric space, if $(x_n)_{n \in \mathbb{N}}$ is a non constant sequence such that $x_n \to x$. I try to construct a sequence of disjoint open balls with center in $x_n$ for all $n \in \mathbb{N}$.
I have this idea, like $x_n \to x$, we know that $(d(x_n,x))_{n \in \mathbb{N}}$ is decreasing and $d(x_n,x) \to 0$, using this, I pose the sequence of open balls $(B(x_n,r_n))_{n \in \mathbb{N}}$ where $r_n=\dfrac{1}{2}\min\{d(x_n,x_{n-1}),d(x_n,x_{n+1})\}$. With this, I proved that $B(x_n,r_n)\cap B(x_{n+1},r_{n+1})=\emptyset$ for all $n \in \mathbb{N}$, but I can't prove that $B(x_n,r_n)\cap B(x_m,r_m)$ for all $m,n \in \mathbb{N}$ with $m \not = n$. Is this true?, Can you help me with any ideas?
AI: As highlighted in my comment, this is not possible if $x_n = x$ for some $n \in \Bbb{N}$, so we shall assume that's not the case.
Your example fails for $X = \Bbb{R}$, with $x_0 = x_2 = 4$, $x_1 = \frac{1}{2}$, and $x_n = \frac{1}{n}$ for $n \geq 3$. Then $x_n \to 0$, but $B(x_1,r_1)$ contains all of $x_n$ for $n \geq 3$.
The construction is actually simple (if you do not require explicit formulae for each $r_n$). For each $k \in \Bbb{N}$, since $\lim_{k \to \infty} x_k \not\to x_n$, $x_n \in \{x_k\}_{k=1}^\infty$ is isolated so there exists $r_n > 0$ such that $x_k \notin B(x_n,r_n)$ for all $k \neq n$. Choose such an $r_n$ for each $n$, and you have a sequence of disjoint open balls with each centering around $x_n$.
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H: Show that $ d(x, A) = 0 $ if, and only if, $ x \in \overline{A} $
Let $ X $ be a metric space with distance $d$ and let $ A \subset X $ not empty.
(a) Show that $ d(x, A) = 0 $ if, and only if, $ x \in \overline{A} $
(b) Show that if $ A $ is compact, $ d (x, A) = d(x, a) $ for some $ a \in A $
(c) Define the $ \epsilon $ - neighborhood of $ A $ in $ X $ as the set $$U(A, \epsilon) = \{x : d(x, A) < \epsilon \}$$ Show that $ U (A, \epsilon) $ matches the union of the open balls $ B_{d}(a, \epsilon) $ for $ a \in A$.
(d) Suppose $ A $ is compact and let $ U $ be an open containing $ A $. Show that some $ \epsilon $ - neighborhood of $ A $ is contained in $ U $.
(e) Show that statement (d) is not true if we put $ A $ closed but not compact.
Definition: Let $ (X, d) $ be a metric space and $ A $ a non-empty subset of $ X $. For each $ x \in X $, we define the distance from $ x $ to $ A $ by the equation $$ d(x, A) = \inf \{d(x, a): a \in A \} $$
For part (a), if $ A $ has the property of the supreme, then I want to use the following inequality: $$ \rho(x, y) \leq d(x, y) \leq \sqrt{n} \rho(x, y) $$ where $ d $ is the euclidean distance and $ \rho $ is the distance of the supreme. However this seems to be a particular case of exercise. Is this reasoning correct? For the other paragraphs I have not concluded anything.
AI: (a) See here
(b) We first show that, for any non-empty set $E\subset X$, $f_E(x):=d(x,E)$ is continous. Given any $x,y,z\in X$, we have the inequality
$$d(y,z)\le d(x,y)+d(x,z)$$
Taking the inf over $z\in E$, we see that
$$
d(y,E)\le d(x,y)+d(x,E)
$$
Which implies
$$ d(y,E)-d(x,E)\le d(x,y)$$
By symmetry, we also have
$$ d(x,E)-d(y,E)\le d(x,y)$$
Therefore,
$$|d(x,E)-d(y,E)|\le d(x,y)$$
The last inequality shows that $f_E$ is continuous (in fact Lipschitz).
In particular, given $x\in X$, $f_{\{x\}}$ is continuous, and so attains a minimum on the set $A$, say at $a\in A$. But then $d(x,a)=d(x,A)$.
(c) By definition of $d(x,A)$,
$$x\in U(A,\varepsilon)\iff d(x,A)<\varepsilon \iff \exists a\in A(d(x,a)<\varepsilon) \iff x\in \bigcup_{a\in A}B_\varepsilon (a)$$
(d) Since $U$ is open, $U^c$ is closed. By (a), $d(x,U^c)>0$ for every $x\in A$. Let $a\in A$ minimize the function $f_{U^c}$ on $A$. This is possible by compactness and continuity. Put $\varepsilon:=d(a,U^c)$. If $x\in A$ and $d(x,y)<\varepsilon$, then $y\not\in U^c$, i.e. $y\in U$. So, $B_\varepsilon (x)\subset U$. This holds for every $x\in A$, and so we are done by (c).
(e) Look at $A=\{1/n: n\in\mathbb{Z}^+\}$ and $U=(0,\infty)$.
|
H: How to evaluate $\int_{|z|=2} \frac{1}{z^2+z+1} dz$
I found the poles of $\displaystyle f(z) = \frac{1}{z^2+z+1}$ given by the solutions for $z^2 + z + 1$, $ \displaystyle z_1 = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\displaystyle z_2 = -\frac{1}{2}-i\frac{\sqrt{3}}{2}$.
But when I use the Residue's Theorem, I get an indetermination. For example, for $z_1$ $$\displaystyle Res\left(\frac{1}{z^2+z+1}, -\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)= \lim_{z \to -\frac{1}{2}+i\frac{\sqrt{3}}{2}} \left((z - \left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right)\left(\frac{1}{z^2+z+1}\right) $$$$= (0)(\frac{1}{0})$$
Is this the way I should treat this problem or should I try with something else?
AI: Hint:
Just observe that
\begin{align*}
(z - (-\frac{1}{2}+i\frac{\sqrt{3}}{2}))(\frac{1}{z^2+z+1})&=(z-z_1)\times\frac 1{(z-z_1)(z-z_2)}\\
&=\frac 1{z-z_2}\qquad\text{for all }z\neq z_1
\end{align*}
So, the limit you are looking for is
$$\lim_{z\to z_1}\left[(z - z_1)\times\frac{1}{(z-z_1)(z-z_2)}\right]=\lim_{z\to z_1}\frac{1}{z-z_2}=\frac{1}{z_1-z_2}=\frac{1}{i\sqrt 3}=-\frac 1{\sqrt 3}i$$
|
H: How to solve this implicit equation? (use trial and error)
I want to solve this implicit equation and find $f$ When $Re$ is constant: $$\frac{1}{\sqrt{f}}=2\log({Re.\sqrt{f})}- 0.8$$
I tried to make the equation simple By using: $\sqrt{f}=t>0,Re=a$:
$$\frac{1}{t}=2\log({at})-0.8$$
$$0.8t=2t\log(at)-1$$
I can not find $t$ in this eqution. Should I draw graphs of $\frac{1}{t},2\log({at})-0.8$ and find intersection points for given "a" as question said with "trial and error method". But in general I don't like use this method for math questions. So is it possible to solve equation in other way ?
AI: It can be solved in terms of the Lambert $W$ function: $$\frac{1}{2t}+\log\frac{1}{2t}=b:=\log\frac{a}{2}-0.4$$ or, for $x=1/(2t)$, we have $xe^x=e^b$, thus $x=W(e^b)$.
|
H: Prove that $\lim_{y \to 0^+}\frac{\sin(yt)}{y}$ Uniformly converges to $t, t \in [0,R], R > 0$
I need to show that
$$
\lim_{y \to 0^+}\frac{\sin(yt)}{y}
$$
Uniformly converges to $t \in [0,R], R > 0$
The answers show that:
$$
\forall t \in [0,R]: 0 < yt < \delta \Rightarrow |t-\frac{\sin(yt)}{yt}| < \varepsilon
$$
Therefore the converges is uniform.
I dont understand why? The definition of uniformly converges from what i remember says:
$$
\forall \varepsilon > 0, \exists \delta > 0, \forall x,y \in I, |x-y| < \delta \Rightarrow |f(x)-f(y)| < \varepsilon
$$
How what they did prove this?
AI: What you wrote is the definition of uniform continuity of the function, without any reference to uniform convergence. Here the issue
is of uniform convergence of a family of functions, parameterised by $t$.
The family $f_t(y)=\frac{\sin(ty)}{ty}$ converges uniformly in $[0,R]$ to $t$ as $y\to 0^+$ if and only if $\lim_{y\to 0^+}\sup_{t\in [0,R]}|f_t(y)-t|=0$, and that follows from the answers' claim.
|
H: Surface integral of piecewise volume boundary?
How should I go about solving the surface integral
$$\iint_{\partial V} 2(x^2+y^2) \, dS,$$ where $V$ is the region bounded by the paraboloid $$z=\frac12-x^2-y^2$$ and the cone $$z^2=x^2+y^2?$$
I have done surface integrals before, but I am unsure how to proceed with finding the boundary of the solid $V.$
AI: I assume the surface in question lies above the plane $z = 0;$ however, if not, you can easily adapt this solution to the case that the surface contains both the upper and lower nappe of the cone.
Graphing the two surfaces shows that the surface for which we would like to compute the surface area resembles an ice cream cone: it consists of the cap $\mathcal C$ of the paraboloid $z = \frac 1 2 - x^2 - y^2$ and part of the upper nappe $\mathcal N$ of the cone $z^2 = x^2 + y^2.$ Ultimately, we will seek to compute $$\iint_{\partial V} 2(x^2 + y^2) \, dS = \iint_\mathcal C 2(x^2 + y^2) \, dS + \iint_\mathcal N 2(x^2 + y^2) \, dS.$$
Observe that $\mathcal C$ and $\mathcal N$ intersect if and only if $z^2 = x^2 + y^2$ and $z = \frac 1 2 - x^2 - y^2$ if and only if $z = \frac 1 2 - z^2$ if and only if $z^2 + z - \frac 1 2 = 0$ if and only if $z = \frac{\sqrt 3 - 1}{2}.$ Consequently, we have that $$\mathcal C = \biggl \{(x, y, z) \,|\, z = \frac 1 2 - x^2 - y^2 \text{ and } \frac{\sqrt 3 - 1}{2} \leq z \leq \frac 1 2 \biggr \} \text{ and}$$ $$\mathcal N = \biggl \{(x, y, z) \,|\, z^2 = x^2 + y^2 \text{ and } 0 \leq z \leq \frac{\sqrt 3 - 1}{2} \biggr \}. \phantom{\text{ and butt }}$$ Geometrically, the cap is a "deformation" of disk in the $xy$-plane, so we may parametrize $\mathcal C$ by polar coordinates $F(r, \theta) = \bigl \langle r \cos \theta, r \sin \theta, \frac 1 2 - r^2 \bigr \rangle$ for $0 \leq r \leq \sqrt{1 - \frac{\sqrt 3}{2}}$ and $0 \leq \theta \leq 2 \pi.$
Likewise, the upper nappe $\mathcal N$ of the cone can be parametrized most easily by polar coordinates $G(r, \theta) = \langle r \cos \theta, r \sin \theta, r \rangle$ for $0 \leq r \leq \frac{\sqrt 3 - 1}{2}$ and $0 \leq \theta \leq 2 \pi.$
Can you finish the solution from here? Use the definition of the surface integrals $$\iint_\mathcal C 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{\sqrt{1 - \sqrt 3 /2}} 2r^2 ||F_r(r, \theta) \times F_\theta(r, \theta)|| \cdot r \, dr \, d \theta \text{ and}$$ $$\iint_\mathcal N 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{(\sqrt 3 - 1)/2} 2r^2 ||G_r(r, \theta) \times G_\theta(r, \theta)|| \cdot r \, dr \, d \theta. \phantom{\text{ and }}$$
|
H: Implicit differentiation of a function with 3 variables
The problem says: If the equation $x^2 +y^2 +z^2 = G(ax+by +cz)$ defines $z=f(x,y)$, $f$ and $G$ being differentiable, $a$, $b$ and $c$ constants, find $\partial z/\partial x$.
Is this correct?:
$$
\frac{\partial z}{\partial x} = -\frac{Gx}{Gz} = - \frac{2ax}{2cz}.
$$
I'm kind of confused with the constant as they are not in $x^2 +y^2 +z^2$.
AI: The problem suggests $x^2 + y^2 + z^2 =\text{const}$ and $G(ax+by+cz)=\text{const}$ define two surfaces respectively. Their intersections, defined by $x^2 + y^2 + z^2 = G(ax+by+cz)$, is a line $z=f(x,y)$. Here, we consider $x,y$ to vary independently, meaning $\partial y/\partial x = 0$.
Back to the problem, we define the line equation as
$$
x^2 + y^2 + z^2 - G(ax+by+cz) = 0.
$$
And differentiating with respect to $x$ gives
$$
\begin{align}
& 2x + 0 + 2 z \frac{\partial z}{\partial x} - \frac{\partial G(ax+by+cz)}{\partial x} = 0 \\
\Rightarrow \quad &
2x + 2 z \frac{\partial z}{\partial x} - G'(\gamma) \left.\left(a+c\frac{\partial z}{\partial x}\right)\right|_{\gamma=ax+by+cz} =0 \\
\Rightarrow \quad &
\frac{\partial z}{\partial x} = \left.\frac{aG'(\gamma)-2x}{-cG'(\gamma)+2z} \right|_{\gamma=ax+by+cz}.
\end{align}
$$
|
H: Weil divisors associated to Cartier divisors
Let $X=\{x_3^2=x_1^2+x_2^2\}\in \mathbb{P}^3$, let $L_1=Z(x_2,x_1+x_3)$, and $L_2=Z(x_2,x_1-x_3)$.
I don't quite understand how to get that $\operatorname{div}(x_2)$ is associated to $[L_1]+[L_2]$ and $\operatorname{div}(x_1+x_3)$ is associated to $2[L_1]$.
I think this means $\operatorname{ord}_{L_1}(x_2)=\operatorname{ord}_{L_2}(x_2)=1$ and $\operatorname{ord}_{L_1}(x_1+x_3)=2$. But I think $\operatorname{ord}_{L_1}(x_2)=\operatorname{ord}_{L_2}(x_2)=2$ as well, since we just need to calculate the length of $((k[x_1,x_2,x_3]/(x^2_1+x^2_2-x^2_3))_{(x_2,x_1+x_3)})/(x_2))$, and since $\overline{(x_2)} \subset \overline{(x_2,x_1+x_3)} \subset (k[x_1,x_2,x_3]/(x^2_1+x^2_2-x^2_3))_{(x_2,x_1+x_3)})$, $\operatorname{ord}_{L_1}(x_2)$ should be 2, why is this not true?
This comes from Gathmann's notes on Algebraic Geometry, page 179.
AI: I'll write $A$ for the ring $$\left(\frac{k[x_1, x_2, x_3]}{(x_1^2 + x_2^2 -x_3^2)}\right)_{(x_2, x_1+ x_3)}.$$ There seems to be an error in your calculation of the length of $A/x_2$. In particular, we have that the ideals $(x_2)$ and $(x_2, x_1+ x_3)$ of $A$ are actually equal. This is because $x_1-x_3$ is invertible in $A$, so we have that $$x_1 + x_3 = -\frac{x_2^2}{x_1 - x_3}$$ which is already in the ideal $(x_2)$.
|
H: Is the function with prescribed Fourier coefficients bounded a.e.?
Consider a function $F\in L^2(0,1)$ whose Fourier coefficients are:
\begin{align*}
\widehat{F}(n)=
\begin{cases}\frac{1}{n},\quad &n=1,2,3,\dots,\\
0, \quad &n\le 0.
\end{cases}
\end{align*}
$\textbf{Question}:$ Is $F$ a bounded function on $(0,1)$, with the possible exception of a null-set?
It resembles a little bit the Fourier series of the function $f(x)=x$, but in the case of $F$ the Fourier coefficients don't oscillate. Moreover, if we consider a formal series expansion $$
F(\theta)=\sum_{n=1}^\infty\frac{e^{in\theta}}{n},
$$
then it clearly diverges for $\theta=0$. It doesn't disqualify my question though, since $F$ doesn't have to be continuous.
I would appreciate any comments/hints.
AI: Hint: The idea here is that $-\log(1 - z) = \sum_{n=1}^{\infty} {z^n \over n}$, so that $F(\theta) = -\log(1 - e^{i\theta})$ which is unbounded as $\theta$ goes to zero.
|
H: a countable open interval cover of the irrationals must also cover the rationals?
let $A:=[0,1]\setminus \Bbb Q$. Is it necessary for a countable open interval cover $(I_k)_{k=1}^\infty$ of $A$ also covers $[0,1]\cap \Bbb Q$, therefore, $A$?
I think this is true, due to the fact that $\Bbb Q$ and $\Bbb R \setminus \Bbb Q$ are dense and each $I_k$ is open. But I have not found the details.
AI: $(-1,\frac 1 2)$ and $(\frac1 2, 2)$ cover $A$ but not $\mathbb Q \cap [0,1]$ since $\frac 1 2$ is not in either one of these intervals..
|
H: Deriving a polynomial with certain properties...
What is an algebraic expression for a polynomial, $q$, with the following properties:
1. $q$ has real coefficients.
2. The only real zeros of $q$ are $-2$ with multiplicity $3$ and $1/4$ with multiplicity 2.
3. A complex zero of $q$ is $i$ with multiplicity $1$.
4. $q(0)=-5$.
Out of all polynomials with the above properties, there is exactly one with the smallest degree. What is the polynomial?
So we have $(x+2)^3$ and $(x-1/4)^2$ the pops up in $q$ but not sure what to do next. Thanks for the help.
AI: Let us go point by point :
$q$ has real coefficients.
We can write $q(x) =(x+2)^3(x-1/4)^2p(x)$ where $p(x)$ is a product of polynomials with degree 2, up to a constant factor, and no real roots.
Thus $i$ is a root of $p$ with multiplicity $1$. By point 1, $-i$ is also a root with multiplicity $1$ : $p(x)=(x-i)(x+i)m(x)$ where $m$ is a product of polynomials with degree 2, no real roots and $m(i) \ne 0$. $(x-i)(x+i)=x^2+1$. Finally we have $q(x) = (x+2)^3(x-1/4)^2 (x^2+1) m(x)$.
$q(0)=5$, for this we just needs to adjust $m$.
The smallest degree polynomial $q$ satisfying all this is if we take $m(x) = c$ for some $c \in \mathbb{R}$. You just need to find its value by solving $q(0)=5$.
|
H: An identity of Arithmetic Functions
Problem: Show that for all positive integers $n$,
$$ \sum_{a=1, (a,n)=1}^{n} (a-1, n) = d(n)\phi(n)$$
where $(a, b)$ stands for $\text{gcd}(a, b)$ and $d, \phi$ are the divisor and Euler's totient function, i.e., number of numbers co-prime to n and less than n = $\phi(n)$.
I find this one really fascinating because of $a-1$. This problem is from Niven and Zuckermann 'Introduction to the Theory of Numbers'.
My approach is to show that the L.H.S. is a multiplicative function. because it is easy to compute it for powers of primes.
Let $d_1d_2=n$ where $(d_1, d_2)=1$
I want to show that
$ \sum_{a=1, (a,n)=1}^{n} (a-1, n) = (\sum_{a=1, (a,d_1)=1}^{d_1} (a-1, d_1))( \sum_{a=1, (a,d_2)=1}^{d_2} (a-1, d_2)) $ but I am not able to proceed other than showing that some terms are cancelling. THe main problem is that there are $x$ such that $(x, n)=1$ but $x > d_1, d_2$.
Please help. Any hints are appreciated.
AI: For a divisor $d|n$, the number of $a>1$ such that $gcd(a-1,n)=d$ is equal to $|\{1 \leq q \leq \dfrac{(n-1)}{d}|(qd,n)=d\}|=|\{1 \leq q \leq \dfrac{(n-1)}{d}|(q,n/d)=1\}|=\varphi(n/d)$.
Now there's a multiplicative function.
Also, it is known that if $f$ is multiplicative, then $\sum_{d|n}f(d)$ is also muliplicative, so having an expression of this form is useful.
|
H: Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$.
I can show that $x^2+y^2=\frac12(1+\log2)$ is the equation of the circle of largest area inscribed in $y=\pm e^{-x^2}$:
The minimum distance $r$ (which will be the radius of the circle) between the origin and $y=e^{-x^2}$ can be found by finding the critical numbers of the derivative of the distance function.
\begin{align}
r&=\sqrt{x^2+(e^{-x^2})^2}\\
\frac{dr}{dx}&=\frac{2x-4xe^{-2x^2}}{2\sqrt{x^2+e^{-2x^2}}}\\
0&=2x(1-2e^{-2x^2})\\
x&=0\quad\text{(obviously inadmissible, or)}\\
x&=\sqrt{\frac12\log2}\\
\\
r&=\sqrt{\frac12\log2+e^{-\log2}}\\
r^2&=\frac12\log2+\frac12\\
\implies\quad x^2+y^2&=\frac12\left(1+\log2\right)
\end{align}
as desired.
The relationship between a circle and an ellipse is like that of a square and a rectangle: given a set perimeter, the more square the rectangle, the larger the area. But since this question is based on the curve $y=e^{-x^2}$ and not on any set perimeter, it does not seem like the same conclusion can be drawn. I would need something stronger (perhaps based on concavity?) to show whether $x^2+y^2=\frac12(1+\log2)$ is the largest ellipse or not.
AI: We can assume by symmetry and without loss of generality that the ellipse can be parametrized by $$(x,y) = (a \cos \theta, b \sin \theta), \quad a, b > 0, \quad \theta \in [0,2\pi).$$ We require tangency to the curve $y = e^{-x^2}$ as well as a single point of intersection in the first quadrant. That is to say, $$b \sin \theta = e^{-(a \cos \theta)^2}$$ has a unique solution for $\theta \in (0, \pi/2)$, and at this point, $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{b}{a} \cot \theta = -2x(\theta)e^{-x(\theta)^2} = -2a (\cos \theta )e^{-(a \cos \theta)^2}.$$ Consequently, $$-\frac{b}{a} \cot \theta = -2ab \cos \theta \sin \theta,$$ or $$\sin \theta = \frac{1}{a \sqrt{2}}.$$ Note if $a < 1/\sqrt{2}$, no such angle exists. The ellipse is "too narrow"--this is due to the fact that the point of tangency is at $(x,y) = (0,1)$. At the point of tangency, we also have $$\cos \theta = \sqrt{1 - (2a^2)^{-1}}$$ so that we now have $$\frac{b}{a \sqrt{2}} = e^{1/2 - a^2}$$ or $$b = a e^{1/2 - a^2} \sqrt{2}.$$ Finally, we seek to maximize the area of this family of ellipses parametrized by $a$. Since the area is proportional to $ab$, we need to maximize $$f(a) = a^2 e^{1/2-a^2}.$$ Computing the derivative with respect to $a$ and solving for critical values, we get $$0 = \frac{df}{da} = 2(a-1)a(a+1)e^{1/2-a^2},$$ hence $$a = 1$$ is the unique solution, with $b = \sqrt{2/e}$ and the ellipse has equation $$\frac{x^2}{1} + \frac{y^2}{2/e} = 1.$$ The area of this ellipse is simply $$\pi a b = \pi \sqrt{\frac{2}{e}}.$$ This is obviously not a circle.
For your enjoyment, I have included an animation of the family of ellipses $$\frac{x^2}{a^2} + \frac{y^2}{2a^2 e^{1-2a^2}} = 1,$$ for $a \in [1/\sqrt{2},2]$:
|
H: Prove $\left( \frac{ n }{ n-1 } \right)^{ n-1 } = \left( \frac{ n-1 }{ n } \right)^{ 1-n }$
I have seen this equation:
$$\left( \frac{ n }{ n-1 } \right)^{ n-1 } = \left( \frac{ n-1 }{ n } \right)^{ 1-n }$$
As you can see the numerator switched with the denominator and I wonder how. I know power laws and yet I can not quite figure out what happened here.
AI: $\left(\frac{n}{n-1}\right)^{n-1}=\frac{(n)^{n-1}}{(n-1)^{n-1}}=\frac{(n-1)^{-(n-1)}}{(n)^{-(n-1)}}=\frac{(n-1)^{1-n}}{(n)^{1-n}}=\left(\frac{n-1}{n}\right)^{1-n}$
|
H: How to find Pythagorean triples with one side only?
I was using the formula to find triples, but I can only find two of them.
the pythagorean triple associate with 102 are
102 136 170,
102 280 298,
102 864 870,
102 2600 2602,
$a = m^2 - n^2$
, $b = 2mn$
, $c = m^2 + n^2$
let $a = 102
= (m+n)(m-n)$
since m and n are odds
(m+n)(m-n) would be a multiply of 4, but 102 isn't a multiply of 4
There are no solution when a = 102
let b = 102, mn = 51
case 1 :
m= 51, n=1
we get 102 2600 2602
case 2 :
m= 17, n=3
we get 102 280 298
AI: This is a common mistake. It's not the case that the formula you give finds all Pythagorean triples. Rather, the formula finds all primitive Pythagorean triples—triples whose greatest common divisor equals $1$. Furthermore, in that formula the $m$ and the $n$ should have opposite parity and be relatively prime.
Since $(m,n)=(51,1)$ and $(m,n)=(17,3)$ are the only relevant factorizations of $\frac{102}2$, and neither of them has integers with opposite parities, we conclude that there are no primitive Pythagorean triples at all with $102$ as a leg.
But we can also look for primitive Pythagorean triples with a leg that is a divisor of $102$, and scale it up appropriately. The divisors of $102$ are $1,2,3,6,17,34,51,102$, and doing this process on each of these divisors individually yields four primitive Pythagorean triples:
$$
(3,4,5), \quad (17,144,145),\quad (51,140,149), \quad (51,1300,1301).
$$
Multiplying these through by $34,6,2,2$ respectively gives the four triples listed in your answer.
The fact that the formula for primitive triples, when the primitivity is ignored, sometimes does produce some Pythagorean triples makes this mistake even easier to make. The moral of the story: we need to appreciate the exact wording of a theorem—including all its hypotheses and its precise conclusion.
|
H: Elements of the subgroup for Galois group over $\mathbb{Q}$ (Cyclotomic extension)
Well... this question looks like silly though, I have a curious about the below.
Let $K= \mathbb{Q}(\omega)$ for $\omega=e^{2\pi i \over n}$
My book said $G(K/\mathbb{Q}) = \{\sigma_i \vert \sigma_i : \omega \to \omega^{i}$ for $(i,n)=1$$ \} \simeq \mathbb{Z}_n^*$
The notation $\mathbb{Z}_n^*$ is a multiplicative group for $\mathbb{Z}_n$
But It doesn't say the elements of the subgroup of the $G(K/Q)$, So I couldn't help only guessing about that.
Here is the my guess.
Let the subfield $E(\leq K)$(or the fixed field for $G(K/E)$) $s.t.$ $\vert G(K/E) \vert = [K;E] =d$ and $d \vert \phi(n)$
Then $G(K/E) = \{\sigma_j \vert \sigma : \omega \to \omega^j$ for $(j,n)=1\}$ for $G(K/E)\leq G(K/\mathbb{Q}) $
(Here the $j(\in Z_n^*)$ is $\vert j \vert \vert d$, $\vert j \vert$ means the order of the $j$)
Is my guess right? If my guess incorrect, how do I revise that? Any answer or comment would be appreciated.
AI: No, this is not a valid way to describe $G(K/E)$. The reason is that when $\Bbb{Z}_n^*$ is not cyclic we will not get all the subgroups by listing the elements of orders $\mid d$.
The smallest counterexample is when $n=8$. In that case we have
$$
\Bbb{Z}_8^*=\{1,3,5,7\}.
$$
Note that all the elements of $\Bbb{Z}_8^*$ have order one or two.
The field $K=\Bbb{Q}(e^{\pi i/4})=\Bbb{Q}(i,\sqrt2)$ has three distinct subfields that are degree two extensions of the rationals. Namely
$$
E_1=\Bbb{Q}(i),\quad E_2=\Bbb{Q}(\sqrt2),\quad E_3=\Bbb{Q}(\sqrt{-2}).
$$
The corresponding Galois groups $G(K/E_i)$ are (leaving that as an exercise for now)
$$
\begin{aligned}
G(K/E_1)&=\{\sigma_1,\sigma_5\},\\
G(K/E_2)&=\{\sigma_1,\sigma_7\},\\
G(K/E_3)&=\{\sigma_1,\sigma_3\}.
\end{aligned}
$$
Observe that $d=2$ for all these intermediate fields.
Knowing $[E:\Bbb{Q}]=d$ is insufficient information to describe $E$, and hence insufficient to describe $G(K/E)$.
The lattice of subgroups and intermediate fields becomes more complicated roughly as a function of the number of prime factors of $n$. The Chinese Remainder Theorem lets us describe $\Bbb{Z}_n^*$ as a direct product of groups
$\Bbb{Z}_{p_i^{\ell_i}}^*$ with $p_i^{\ell_i}$ ranging over the prime powers appearing in the factorization of $n$. With the exception of $p_i=2$ these are always cyclic, but enumerating all the subgroups, while possible, is not very simple.
How to revise your guess? The general facts are that any intermediate field $E$ is of the form $E=\Bbb{Q}(\rho)$ for some element $\rho$. The trivial thing is then that
$$
G(K/E)=\{\sigma\in G(K/\Bbb{Q})\mid \sigma(\rho)=\rho\}.
$$
I doubt this is very useful for you in that it amounts to regurgitating the basic idea of Galois correspondence. The intermediate fields can be described in terms of Gaussian periods. Wait for Will Jagy to show up with sources to more information :-)
|
H: Calculate the integrals for $f(x) = x^2 + 1$
Let $ f \colon \ \mathbb{R} \to \mathbb{R}$ so that $ f(x) = x^2 + 1$. Then compute $$15\int_0^1 f^7(x) \,dx - 14\int_0^1 f^6(x) \,dx$$
There is the direct calculation option for me: using Newton binomial, exploiting $(x^2+1)^7$ and $(x^2+1)^6$ and then integral all terms.
Is there any other elegent solution?
AI: By integration by parts we have \begin{align*} I & = \int_{0}^{1} (x^2 + 1)^7\ dx \\ & = x(x^2 + 1)^7 \ \bigg |_{0}^{1} - 14 \int_{0}^{1} \left [(x^2 + 1)^7 - (x^2 + 1)^6 \right ]\ dx \end{align*}
Can you now finish?
So the required integral evaluates to $2^7.$
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H: Prove that $\operatorname{rank}(L_{A})=\operatorname{rank}(L_{A^\top})$
I want to prove $\operatorname{rank}(L_{A})=\operatorname{rank}(L_{A^\top})$ and $A\in M_{m\times n}(\Bbb F)$ by using two facts. 1. If $W$ is a subspace of $V$ which is a finite dimensional vector space, then $\dim(W)+\dim(W^{\circ})=\dim(V)$. 2. Suppose that $W$ and $V$ are finite dimensional vector spaces and that $T:V \to W$ is linear. Then $N(T^{\top})=R(T)^{\circ}$. Notation : $S^{\circ}=\{f\in V^{\ast}|f(S)=0\}$means the annihilator of $S$ and $T^{\top}:W^{\ast}\to V^{\ast}$ is defined by $T^{\top}(g)=gT$ for all $g\in W^{\ast}$.
My attempt of the proof:
$$\operatorname{rank}(L_A)=\dim(R(L_A))=m-\dim(R(L_A)^{\circ}) $$ by fact 1. $$=m-(N(L_A^{\top}))$$by fact 2. $$=\dim(R(L_A^{\top}))$$ by dimension theorem. I have no idea to keep going. Thanks for someone helping me.
AI: Let me call $\textsf T$ your $L_A : \mathbb F^n \to \mathbb F^m$. Let $e_1,\dots,e_n$ be the standard basis for $\mathbb F^n$; $e_1^*,\dots,e_n^*$ the corresponding dual basis, and consider the unique linear map $\phi : (\mathbb F^n)^* \to \mathbb F^n$ such that $\phi(e_i^*) = e_i$. Now, since $\phi$ is an isomorphism (why?) the dimensions of $R(\textsf T^t)$ and $\phi \big( R(\textsf T^t) \big)$ coincide. Finally, check that $\phi \big( R(\textsf T^t) \big)$ is precisely the image of the linear map $x \mapsto A^tx$. As a hint, what is the matrix representation of $\textsf T^t$?
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H: How to show the following cubic graph is Hamiltonian?
Suppose we have a complete graph $K_{2n}$, where $n$ is an integer, and we number vertices counterclockwise as $1,2,\cdots,2n.$ Then we give weight to the edge connecting $u,v$ as $|u-v|$.
Now, delete all the edges with weight less than $n-1$, that is, keep all the edges with weight $n$ and $n-1$, we get a cubic graph $G$.
I want to show that $G$ is Hamiltonian. I got a complicated construction. I am wondering if there is some theorem that can get this result quick and clean.
AI: There's no good general results for when a cubic graph is Hamiltonian that apply here; it's easiest to solve such problems by finding a Hamiltonian cycle. But the construction can be made simple.
If $n$ is even, then we get a Hamiltonian cycle by going from $i$ to $i + (n+1)$, starting at any vertex and of course wrapping around modulo $2n$.
If $n$ is odd, this instead finds two disjoint $n$-cycles; each one visits every other vertex around $K_{2n}$. Deleting the edges $(1,n+2)$ and $(2,n+1)$ in those cycles gives two paths of length $n-1$: one starting at $1$ and ending at $n+2$, and one starting at $2$ and ending at $n+1$. We can use the edges $(2,n+2)$ and $(1,n+1)$ to join these two paths into a Hamiltonian cycle.
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H: Question on $L^p$ convergence.
Suppose $q \in [1, \infty)$, I'm trying to find a sequence of functions $g_n \in L^p$ (for all $p \in [1,\infty]$) such that $g_n$ converges in $L^p$ when $p \in [1, q]$ but doesn't converge into $L^p$ when $p \in (q, \infty]$.
I tried the sequence of functions $g_n(x) = \displaystyle{\frac{x^{1/p}}{n}}$, over the measure space $([0,1], \mathcal{B}([0,1]), \lambda)$, where $\mathcal{B}([0,1])$ are the Boreal sets over the interval $[0,1]$. If $p \leqslant q$ this clearly converges to $0 \in L^p$, however this is still true for $p > q$.
Any help would be greatly appreciated.
AI: On the measure space $([0,1], \mathcal{B}([0,1]),\lambda)$, take $$g_n = n^{1/q} \mathbf{1}_{[0,(n\log(n))^{-1}]}. $$
Then
$$\int g_n^p \, d \lambda = \frac{n^{p/q-1}}{\log(n)} \to 0\tag{1}$$
if and only if $p \leq q$. Therefore $g_n \to 0$ in $L^p$ for every $p \leq q$. Moreover, $g_n$ does not converge in $L^p$ for any $p>q$. Indeed, if it did converge, the limit would necessarily be $0$ since convergence in $L^p$ for any $p>q$ implies convergence in $L^q$ (the measure $\lambda$ being finite). This is impossible due to $(1)$.
Bonus: if you remove the term $\log(n)$ from the definition of $g_n$, you get a sequence converging in $L^p$ for every $p<q$ but not in $L^p$ for $p\geq q$.
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H: Question about Vector Calculus
Mine is probably a very trivial question. I didn't get good training on vector calculus and I am not able to find the correct materials to learn to understand the following equation.
Given
$F = \int \frac{D^2}{8\pi\epsilon} dV$
where $\vec{D}$ is a vector, the book gives:
$\delta F=\int\frac{\vec{D}\cdot\delta\vec{ D}}{8\pi\epsilon}-\int\frac{D^2}{8\pi\epsilon^2}dV$
Can you explain why the 1st term is not $\int\frac{\vec{D}\cdot\delta\vec{ D}}{4\pi\epsilon}$? I would like to understand more about vector differential (is this what it call?). Can you give a suggestion on what this is called so that I can google the right keyword to learn more? Thanks a lot!
snapshot:
AI: First of all, I have to say you're very brave for reading Landau and Lifshitz's Volume 8 about electrodynamics of continuous media $\ddot{\smile}$.
Anyway, you're right, it should be $4 \pi \epsilon$ in the denominator of the first term, and in my version of the book, that's how it is stated (so perhaps, yours is an older edition with typos?). So, we have:
\begin{align}
\delta\mathcal{F} &= \int\dfrac{\mathbf{D} \cdot \delta \mathbf{D}}{4 \pi \epsilon} \, dV - \int \dfrac{D^2}{8 \pi \epsilon^2} \delta \epsilon \, dV
\end{align}
Anyway, this $\delta$ operation being done on the various quantities is called the "variation".
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H: is a matrix $A*A^{t}$ or $A^{t}* A $ Symmetric?
if $\mathbb{K}$ is a field and $A\in M_{m\times n}(K)$ proof or give a counterexample that $A\cdot A^t $ and $A^t\cdot A $ are Symmetric matrix
AI: Yeah they are symmetric
If A is an m × n matrix and A^t is its transpose, then the result of matrix multiplication with these two matrices gives two square matrices: AA^t is m × m and
A^tA is n × n. Furthermore, these products are symmetric matrices. Indeed, the matrix product A.A^t has entries that are the inner product of a row of A with a column of A^t. But the columns of A^t are the rows of A, so the entry corresponds to the inner product of two rows of A. If pi j is the entry of the product, it is obtained from rows i and j in A. The entry P(ij) is also obtained from these rows, thus p(ij)= p(ji), and the product matrix p(i,j) is symmetric. Similarly, the product A^t*A is a symmetric matrix.
A quick proof of the symmetry of A AT results from the fact that it is its own transpose:
(A * A^t)^t = (A^t)^t * A^t = A * A^t
refer this for more info https://en.wikipedia.org/wiki/Transpose
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H: Solving an equation having trignometric functions on RHS and an number in LHS
If $$\cos\frac\pi{2n}+\sin\frac\pi{2n}=\frac{\sqrt n}2, n\in\Bbb N,$$ find $n$.
I haven't been able to get an answer to this problem but I know that since the range of $$\sin x+\cos x$$ is $1,√2$
So the value of $n$ must lie between $4,8$.Ofcourse I can just check for all values but is there a way to do it analytically?
AI: This probably steers away from the intention of the problem, but you can square both sides of the equation and take it from there.
\begin{align}
\cos\frac\pi{2n}+\sin\frac\pi{2n}&=\frac{\sqrt{n}}2\\
1+\sin\frac\pi n&=\frac n4\\
\sin\frac\pi n&=\frac n4-1
\end{align}
For $n\in\Bbb N$, we have LHS in range $(0,1]$. This limits the possibilities of $n$ to $(4,8]$. On this interval, $\sin\frac\pi n-\frac n4+1$ is strictly decreasing; therefore, there can only be one solution. By inspection, $n=6$ is the only solution.
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H: Vector spaces uniqueness proof: If $z_1$ and $z_2$ are two such elements, then $z_1 + z_2 = z_1$ and $z_1 + z_2 = z_2$; thus, $z_1 = z_2$.
I am currently studying Introduction to Hilbert Spaces with Applications, by Debnath and Mikusinski. Chapter 1.2 Vector Spaces says the following:
(a) $x + y = y + x$;
(b) $(x + y) + z = x + (y + z)$;
(c) For every $x, y \in E$ there exists a $z \in E$ such that $x + z = y$;
From (c) it follows that for every $x \in E$ there exists a $z_x \in E$ such that $x + z_x = x$. We will show that there exists exactly one element $z \in E$ such that $x + z = x$ for all $x \in E$. That element is denoted by $0$ and called the zero vector.
Let $x, y \in E$. By (c), there exists a $w \in E$ such that $x + w = y$. If $x + z_x = x$ for some $z_x \in E$, then by (a) and (b),
$$y + z_x = (x + w) + z_x = (x + z_x) + w = x + w = y.$$
This shows that, if $x + z = x$ for some $x \in E$, then $y + z = y$ for any other element $y \in E$. We still need to show that such an element is unique. Indeed, if $z_1$ and $z_2$ are two such elements, then $z_1 + z_2 = z_1$ and $z_1 + z_2 = z_2$. Thus, $z_1 = z_2$.
I find the uniqueness "proof" suspicious. The authors just claim that $z_1 + z_2 = z_1$ and $z_1 + z_2 = z_2$, and that therefore $z_1 = z_2$, without really "proving" anything. Is this actually a valid proof? I would greatly appreciate it if people would please take the time to clarify this.
AI: Note that they have established that $x + z_1 = x$ and that therefore for all $y \in E$ we have $y + z_1 = y$. Thus they can choose their $y$ to be $z_2$ and thus get $z_2 + z_1 = z_2$. The same can be repeated for $z_2$.
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H: Proof of an inequality (maybe using mean value theorem)
Suppose that $f:[a,b]\rightarrow \mathbb{R}$ is a twice differentiable function. The inequality is given by
$$|f(x+h)-f(x)-f'(x)h|\leq \frac{h^2}{2}|f''(\eta)|$$
where $x,x+h\in [a,b]$. The aim is to prove that there is a $\eta \in[a,b]$ such that above inequality holds.
I'm guessing that the proof can be derived using some kind of mean value theorem. But I haven't figure it out yet.
AI: If $f$ is twice differentiable on $[a, b]$ then for $x, x+h \in [a, b]$ we can compute its Taylor approximation with Peano remainder
$$
f(x+h) = f(x) + h f'(x) + \frac{h^2}{2} f''(\xi),
$$
for some $\xi \in [x, x+h] \subset [a, b]$. We shuffle things around, take the absolute value and get
$$
\lvert f(x+h) - f(x) - h f'(x) \rvert = \left \lvert \frac{h^2}{2} f''(\xi) \right \rvert \tag{I} \label{i}
$$
Now let
$$
\eta = \underset{\xi \in [a, b]}{\arg \max}\, \lvert f''(\xi) \rvert.
$$
By definition $\eta \in [a, b]$ and
$$
\left \lvert \frac{h^2}{2} f''(\xi) \right \rvert \leq \left \lvert \frac{h^2}{2} f''(\eta) \right \rvert, \quad \forall \xi \in [a, b]
$$
which we insert into \eqref{i} to derive
$$
\lvert f(x+h) - f(x) - h f'(x) \rvert \leq \left \lvert \frac{h^2}{2} f''(\eta) \right \rvert
$$
for all $x, x+h \in [a, b]$.
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H: Do we need to rigorously prove why a set is non-empty with non-negative integers?
For proofs using Well-Ordering Principle(WOP), can we prove the set is a non-empty set of nonnegative integers simply by "stating"?
Eg of what I mean by "stating":
Given any integer $n$ and any positive integer $d$, there exist integers $q$ and $r$ such that
$n = dq + r$ and $0 ≤ r < d$.
I would start off by saying let set $S={x\underline{\in}N| x=n-dk}$ and then simply "state" that since $x\underline{\in}N$,its a non-empty set of non-negative integers and then proceed to use WOP.
Meanwhile, some solutions I see prove that its a non-empty set of nonnegative integers rigorously.
Eg:
Are there any guidelines on when to prove rigorously or when to "state"? Or is it simply up to the reader to accept it?
AI: When you're given a set, you cannot just state it is non-empty by picking an element from the set. In fact, to be able to say $x \in N,$ you need to justify why you can do that.
Many times, it is trivial to show that the set is non-empty. There also are times, when the given assertion trivially holds even for the empty set you can split off the proof into cases when the given set is empty or non-empty. In the latter case, you can pick an element from the set.
So for example, you have two sets $A$ and $B$ and the assertions says $A\subseteq B.$ Then you take two cases.
Case 1: $A = \emptyset.$ Then $A \subseteq B$ is trivial.
Case 2: $A \neq \emptyset.$ Then you pick $x \in A$ and use any known information to conclude $x \in B.$
You cannot say "Let $x \in A$," since $A$ maybe the empty set.
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H: Uniform convergence of $\tan(x)^n$
I am considering the uniform convergence of $(\tan(x))^n$ in the interval$[0,π/4)$. This function sequence is pointwise convergent everywhere in the given interval and converges to $0$.
Is this function uniform convergent in the given interval? In my opinion, it is uniform convergent because the supremum of $|f_n(x)-f(x)|$ as $n$ tends to $\infty$ is $0$. Then by the $M_n$ test, we can say that the given function sequence is uniformly convergent.
Had the interval been $[0,π/4]$, the function sequence would have pointwise converged to a discontinuous function and hence been not uniform convergent.
AI: It is not uniformly convergent. Let $x_n=\arctan (1-\frac 1 n)$. Then $0<x_n <\frac {\pi} 4$ and $(\tan x_n)^{n}=(1-\frac 1 n )^{n} \to\frac 1 e$ so $\sup_x (tanx )^{n}$ does not tend to $0$.
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H: Show that basis $\mathscr{A}$ equals the intersection of all topologies on $X$ that contain $\mathscr{A}$
Attempt:
Let $\mathcal{T}$ be the topologie generated by $\mathscr{A}$. If $\{ \mathcal{T}_{\alpha} \}$ is a collection of all topologies on $X$ that contain $\mathscr{A}$, we prove that
$$ \mathcal{T} = \bigcap_{\alpha; \mathscr{A} \subset \mathcal{T}_{\alpha}} \mathcal{T}_{\alpha} $$
First of all, we realize that intersection of topologies on X is still a topology on $X$ and it is the smallest such topology thus we always have $\bigcap \mathcal{T}_{\alpha} \subset \mathcal{T} $.
Next, Take any $U \in \mathcal{T}$. IF we can show that $U$ is in every $\mathcal{T}_{\alpha}$ then we are done.
Now, each $\mathcal{T}_{\alpha}$ is topology of $X$ which contains $\mathscr{A}$ but we dont know how the open sets look here so we may not conclude that $U$ is open with respect to these topology... BUT since $\mathcal{T}$ is generated by basis $\mathscr{A}$, then we can find some $A \in \mathscr{A} $ such that $U \subset A $ So $U \in \mathscr{A} \subset \mathcal{T}_{\alpha} $ for all $\alpha$ and ${\bf we \; are \; done}$. QED
If $\mathscr{A}$ is a subbasis, then the topology $\mathcal{T}$ generated by it is is instead the collection of all ${\bf unions}$ of finite intersections of elements of $\mathscr{A}$
. Just above we have $\bigcap \mathcal{T}_{\alpha} \subset \mathcal{T}$
so we may just show the reverse inclusion. In topology $\mathcal{T}$, the open sets are finite intersection of elements of $\mathscr{A}$ so they all lie in $\mathscr{A}$ and thus in all topologies $\mathscr{T}_{\alpha}$ and so we are basically done.
IS this correct?
AI: If $\mathcal{A}$ is a base then often the topology generated by that base is the set
$$\mathcal{T}_{\mathcal{A}}:=\{\bigcup \mathcal{A}': \mathcal{A}' \subseteq \mathcal{A}\}$$
the set of all unions of subcollections of $\mathcal{A}$.
So if this is your definition, the proof for
$$\mathcal{T}_{\mathcal{A}} = \bigcap\{\mathcal{T}: \mathcal{A} \subseteq \mathcal{T}, \mathcal{T} \text{ a topology on } X\}$$
goes slightly differently: let $O$ be in $\mathcal{T}_A$ so that $O=\bigcup \mathcal{A'}$ for some $\mathcal{A'} \subseteq \mathcal{A}$. Let $\mathcal{T}$ be a topology on $X$ that contains $\mathcal{A}$. Then $\mathcal{T}$ also contains $\mathcal{A}'$ and so $O \in \mathcal{T}$ as topologies are closed under unions. As $\mathcal{T}$ was arbitrary, $O$ is in the right hand intersection, and one inclusion has been shown.
Now, $\mathcal{T}_A$ is itself a topology (because we know $\mathcal{A}$ is a base) on $X$ that contains $\mathcal{A}$, as $A = \bigcup \{A\}$ for all $A \in \mathcal{A}$, and so is one of the topologies we're intersecting on the right so that the right to left inclusion is trivial.
For a subbase $\mathcal{A}$ you can make some minor adaptations to this proof, or note that a topology contains $\mathcal{A}$ iff it contains all finite intersections of members of $\mathcal{A}$ (as all topologies are closed under finite intersections) and if we define those collections as $\mathcal{B}$ then $\mathcal{T}_{\mathcal{A}} = \mathcal{T}_{\mathcal{B}}$ which equals the intersection of all topologies that contain $\mathcal{B}$ by the previous case (where we had a base) and so also the intersection of the (same, as said !) topologies that contain $\mathcal{A}$ and we're done by a reduction to the previous case.
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H: Asymptotic rate of decrease of error function
The complementary error function is defined as
$$
\text{erfc}(x) = 1 - \frac{2}{\sqrt{\pi}}\int_0^{x} e^{-t^2} dt
$$
and is related to the Gaussian (Normal) distribution. Is there an approximation of the form $\exp(g(x))$ that converges to $\phi(x)$ asymptotically? i.e. can we find $g(x)$ such that
$$
\underset{x \rightarrow \infty}{\lim} \frac{\exp(g(x))}{\text{erfc}(x)} = 1
$$
and that $\exp(g(x))$ "approximates" $\text{erfc}(x)$ in some sense when $x$ is large but not infinite?
AI: There is one asymptotic expansion in the Handbook of Mathematical functions and also listed in your link; it is not quite in the form you wanted but close.
If
$$\text{erfc}(z) = \frac{2}{\sqrt{\pi}} \int_z^\infty e^{-t^2}~ dt = 1 - \text{erf}(z)$$
then
$$\text{erfc}(z)\sim \frac{1}{\sqrt{\pi}}e^{-z^2} \cdot
\sum_{k=0}^\infty
(-1)^k~\frac{(2k)!}{2^{2k}k! } \cdot \frac{1}{z^{2k+1}}
$$
as $z\to\infty$, $\lvert \arg z \rvert < 3\pi/4. $
This can be adapted to the required form; for example, the first term is,
$$\text{erfc}(z) = \frac{1}{\sqrt \pi} e^{-z^2}\cdot \frac{1}{z} = e^{-z^2-\log(z\sqrt{\pi})}$$
which gives $g(z) = -z^2-\log(z\sqrt{\pi})$.
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H: Calculate $\sum_{n=1}^\infty\frac{n^x}{n!}$
I want to evaluate function defined by following sum: $$\sum_{n=1}^\infty\frac{n^x}{n!}$$ I was thinking about writing Taylor series expansion for it. However my try resulted in sum that looks even harder to calculate:
$$\sum_{n=1}^{\infty}\frac{\ln^k(n)}{n!}$$Thanks for all the help in solving this problem.
AI: For $x=0$, you recognize
$$\sum_{n=1}^\infty\frac1{n!}=e-1.$$
For $x=1$,
$$\sum_{n=1}^\infty\frac n{n!}=\sum_{n=1}^\infty\frac1{(n-1)!}=e.$$
For $x=2$,
$$\sum_{n=1}^\infty\frac{n^2}{n!}=\sum_{n=1}^\infty\frac{n(n-1)+n}{n!}=\sum_{n=2}^\infty\frac1{(n-2)!}+\sum_{n=1}^\infty\frac1{(n-1)!}=2e.$$
For larger powers $x$, the sum will depend on the decomposition of $n^x$ as a sum of falling factorials $(n)_x$, which is given by the Stirling numbers of the second kind, https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Definition.
Hence the sums are given by $e$ times the sum of the first Stirling numbers, which are the Bell numbers.
There does not seem to be an easy generalization to negative nor fractional powers. Asymptotic expressions for the Bell numbers are available.
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H: How to show that $\Bbb Q$ is incomplete under this special metric?
Fix a prime $p$ and a positive rational number $r$ can be written uniquely in a form $r=p^e\frac{b}{c}$, with $e\in \Bbb Z$ and $p,b,c \in \Bbb N$ being pairwise coprime.
Define $|\cdot|_p$ as
$|r|_p=0$, if $r=0$,
$|r|_p=|-r|_p$, if $r<0$,
$|r|=p^{-e}$, if $r=p^eb/c$.
The special metric is defined as $d(r_1,r_2)=|r_1-r_2|_p$.
I meet this problem in exercises, I want to construct a Cauchy sequence such that the difference of two items is of the form $p^e$, and show its limit is irrational. But I am confused with how to show a limit is irrational.
Are my thinkings right? I just want some hints (solution better).
AI: This is the $p$-adic metric on $\Bbb Q$.
Textbooks describe how if $p$ is odd and $a$ is a quadratic residue modulo $p$
then $x^2=a$ is soluble in the $p$-adic numbers by using a method called
"Hensel lifting". This involves constructing a sequence of integers converging
to a square root of $a$ modulo $p$, so this sequences of integers is a non-convergent
Cauchy sequence within $\Bbb Q$.
But we can proceed more naively using fixed point iteration. Let $p=7$
and $a=2$. Take $x_0=3$, so that $x_0^2\equiv2\pmod 7$. Define a sequence
$(x_n)$ recursively by
$$x_{n+1}=\frac12\left(x_n+\frac2{x_n}\right).$$
It's easy to show that $x_n\to\sqrt2$ inside $\Bbb R$ but we want this to be
true in our metric, more precisely $|x_n^2-2|_7\to0$.
In fact $|x_n^2-2|_7=7^{-n-1}$ and $|x_{n+1}-x_n|_7=7^{-n}$ (exercises for reader).
From this one deduces that $(x_n)$ is Cauchy, and if $x_n\to a$ then $x_n^2\to a^2$,
but $x_n^2\to2$.
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H: Two disjoint sets of positive measure everywhere.
Background
This question was asked a few minutes ago and then deleted after another user exhibited what he believed to be a duplicate but I fail to see the link between the two.
Here is the statement of the deleted question, $m$ being the Lebesgue measure :
Can we find $A,B \subset \mathbb R$ such that $m(A \cap I)$ and $m(B
\cap I)$ are positive for each open interval $I$ and $A \cap B $ is the empty set ?
And here is the other question, which happens to have a positive answer.
I copy it on this page as well :
Starting from a countable basis of $\mathbb R$, I am asked to construct a Borel set such that $0<m(E∩I)<m(I)$ for every non empty segment $I$.
Can someone explain how a positive answer to the second question answers the first ?
AI: Take $A=E$ and $B=E^{c}$. If $I$ is any open interval $m(A\cap I) >0$ so $A\cap I$ is non-empty. This proves that $A$ is dense. Similarly $B$ is also dense.
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H: Proof for arithmetic progression with different indices
How do I show that $\left(a_n\right)_{n\geq1}$ is an arithmetic progression if and only if $a_i+a_j-a_k = a_{\left(i+j-k\right)}$.
I tried by using the different definitions for arithmetic progressions and equating them, but I have no clue how to solve this. Any ideas would be greatly appreciated!
AI: Suppose $(a_n)_{n \geq 1}$ is an AP. Then $a_n = a_1 + (n - 1)d$ for some $d$, so:
\begin{align*}
a_i + a_j - a_k &= (a_1 + (i - 1)d) + (a_1 + (j - 1)d) - (a_1 + (k - 1)d) \\
&= a_1 + (i + j - k - 1)d \\
&= a_{i+j-k}
\end{align*}
Now suppose $a_i + a_j - a_k = a_{i+j-k} \; \forall i,j,k \geq 1$. Then:
$$
a_{i} + a_{i+2} - a_{i+1} = a_{i + (i+2) - (i+1)} = a_{i+1} \implies a_{i+2} - a_{i+1} = a_{i+1} - a_i
$$
Thus, $(a_n)_{n \geq 1}$ is an AP.
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H: Homotopic functions-proof
Let $f(x)$ and $g(x)$ be continuous mappings from topological space $X$ to $S^n$, such that $f(x)$ is different from $-g(x)$ for all $x$. Prove that $f$ and $g$ are homotopic.
So, my idea was to start off with the definition of homotopy:
Let $F: X \times I\to S^n$
$F(x,0)=f(x)$, $F(x,1)=g(x)$, $F(x,t)=tg(x)+(1-t)f(x)$.
But how to prove that $F(x,t)\in S^n$?
AI: Defining $F(x,t)=tg(x)+(1-t)f(x)$ is a nice try. However, $tg(x)+(1-t)f(x)\not\in S^n$ in general, and we only know that $tg(x)+(1-t)f(x)\in\mathbb{R}^{n+1}$.
Therefore, we have to divide it by its norm. That is, we define
$$F(x,t)=\frac{tg(x)+(1-t)f(x)}{\|tg(x)+(1-t)f(x)\|}.\tag{0}$$
Now, we have to worry if $\|tg(x)+(1-t)f(x)\|=0$, which would make the map not well-defined. That's where the assumption comes in.
If $\|tg(x)+(1-t)f(x)\|=0$ for some $(x,t)\in X\times I$, then we have $tg(x)+(1-t)f(x)=0$, which gives
$$tg(x)=-(1-t)f(x).\tag{1}$$
Taking the norm of both sides, we get
$$t\|g(x)\|=(1-t)\|f(x)\|\tag{2}$$
since $t, 1-t\geq 0$.
Using the facts that $\|g(x)\|=\|f(x)\|=1$ (since $g, f:X\to S^n$), we obtain from $(2)$ that $t=1-t$. Putting this back to $(1)$, we obtain $g(x)=-f(x)$, which contradicts to the assumption.
Therefore, $F$ defined in $(0)$ is well-defined, and $F:X\times I\to S^n$. And $F$ is the required homotopy between $f$ and $g$, which I invite you to check.
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H: Can a real 2 by 2 matrix have one eigenvalue with geometric multiplicity 2?
Given the a real matrix $A=\begin{bmatrix} a & b \\ c& d\end{bmatrix}$, we assume that it has only one real eigenvalue $\lambda$. I am wondering if it is possible that the eigenvalue $\lambda$ has geometric multiplicity 2, but it seems like it is not possible.
Let $v=\begin{bmatrix} v_1 \\ v_2\end{bmatrix}$. When I solve the usual equation $(\lambda I-A)v=0$, because of the dimension of course, I only obtain one condition for the eigenvector, namely $v_1=\frac{(\lambda-d)}{c}v_2$, which would indicate that there is only one eigenvector and would not be possible to have 2 linearly independent eigenvectors of the repeated eigenvalue $\lambda$. Perhaps this is a very trivial observation for real $2\times 2$ matrices? Am I missing something very silly?
For posterity after the comments: ... with $c\neq0$ indeed is not possible.
AI: Your construction of the eigenvector assumes $c$ is non-zero. A matrix with full geometric multiplicity is diagonalizable, so the only such matrix is $\lambda I$.
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H: Verification of power series solution to differential equation.
Verify that $$y=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^n$$ is the solution of the differential equation $(x+1)y+y’=0$. So we differentiate $y$ to get $$y’=\sum_{n=1}^{\infty} (-1)^{n+1}x^{n-1}$$ and substitute into the differential equation but I can’t seem to get zero. Please help.
AI: If we rewrite the differential equation:
$$
y' = - y (x+1) \implies \frac{y'}{y} = - (x+1) \implies \left(\ln {y}\right)' = -(x+1)
$$
Hence integrating we obtain
$$
y(x) = k \operatorname e ^{-\frac{(x+1)^2}{2}}
$$
but as pointed out by @Lutz Lehmann , the series is the Taylor series of $\ln(1+x)$:
$$
\ln(1+x) = \sum_{n = 1 }^ {\infty}\frac{ (-1)^{n+1}}{n}x^n
$$
They are different.
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H: finding solution of differential equation $y''-9y=(x^2-2)\sin(4x)$
Using method of undetermined coefficient finding solution of $y''-9y=(x^2-2)\sin(4x)$
What i try: First we will find characteristic solution
$r^2-9=0\Longrightarrow r=\pm 3$
So our characteristic solution
$y_{c}(x)=C_{1}e^{3x}+C_{2}e^{-3x}$
But i did not understand How can i find perticular solution.
How do i solve it help me please.
AI: But i did not understand How can i find perticular solution.
The exercise explicitly asks you to use the method of undetermined coefficients: perhaps you could explain what you don't understand about this method? There are plenty of examples (on this site, other websites, youtube) available to see how the method works.
Note that you could save yourself a bit of time and work though.
Using method of undetermined coefficient finding solution of $y''-9y=\color{blue}{(x^2-2)\sin(4x)}$
Based on the right-hand side, you would generally suggest a particular solution $y_p$ of the form:
$$y_p = \left(Ax^2+Bx+C\right)\sin(4x)+\left(Dx^2+Ex+F\right)\cos(4x) \tag{$*$}$$
and this leads to a system of 6 equations in the 6 unknowns ("undetermined coefficients") $A,\ldots,F$.
However, $\color{blue}{(x^2-2)\sin(4x)}$ is an odd function so $y''-9y$ has to be odd which implies that $y$ has to be odd. Therefore only the coefficients in $(*)$ corresponding to odd functions will be non-zero and you immediately have $B=D=F=0$ as those correspond to even functions. The system is reduced to a 3-by-3.
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H: Quadratic variation of $ X_t = t W_t $
Let $W = \{W_t : t \geq 0\}$ be the standard Brownian motion and consider
$$X_t=t W_t$$
Show that the quadratic variation of $X_t$ is $\frac{t^3}{3}$
I know this question has been answered here but I would like to do it by definition, ie, i would like to show that
$$\lim_{n\to\infty}\sum_{i=1}^{n}(\Delta X_{t_i})^2\stackrel{L^2}{=}\frac{t^3}{3}, $$
$\text{ with } \ \Delta X_{t_i}=X_{t_i}-X_{t_{i-1}} \text{ and } t_{i}=\frac{t}{n}i$
My Attempt: I have shown that $\Delta X_{t_i} \sim \mathcal{N}(0,(\frac{t}{n})^3(i^2+i-1))$ and I was thinking about applying the LLN but I cannot do it since the variance has a term which is not constant in $i$.
Any ideas?
AI: The limit is to be understood in mean-square or $L^2$ sense. Use the independence of the increments and properties of the Gaussian distribution in order to show that $E[(\sum_i(\Delta X_i)^2-t^3/3)^2]$ converges to zero.
Edit (additional details):
First show that $E[(\sum_i(\Delta X_i)^2]\to t^3/3$. Using your result that $\Delta X_i\sim\mathcal{N}(0,\sigma_i^2)$ with $\sigma_i^2=(t/n)^3(i^2+i-1)$, we obtain as $n\to\infty$
$$ E\left[\sum_{i=1}^n(\Delta X_i)^2\right]=\sum_{i=1}^n E\left[(\Delta X_i)^2\right] = \sum_{i=1}^n \sigma_i^2= \frac{t^3}{n^3}\frac{n^3+3n^2-n}{3}=\frac{t^3}{3}+O(1/n)\to \frac{t^3}{3}.$$
Therefore, because of independence, as $n\to\infty$
$$E\left[\left(\sum_{i=1}^n(\Delta X_i)^2-t^3/3\right)^2\right]=\text{Var}\left[\sum_{i=1}^n(\Delta X_i)^2\right]+O(1/n^2)=\sum_{i=1}^n\text{Var}\left[(\Delta X_i)^2\right]+O(1/n^2)\\=2\sum_{i=1}^n\sigma_i^4+O(1/n^2)=\frac{2t^6}{n^6}\frac{n^5+5n^4+5n^3-5n^2-n}{5}+O(1/n^2)=O(1/n)\to 0.$$
This shows that the $L^2$ limit of the sum of quadratic increments is equal to $t^3/3$.
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H: Well-ordered sets
I would like to ask how many well-ordering of set $\mathbb N$ exist. And they shouldn't be isomorphic to each other. I found out that the set of well ordering of set $A$ is the subset of $A \times A$. And that the set can be always well-ordered, because it is enough to do an injection between these set and ordinal numbers. But how should I find out how many well-ordering of the set $\mathbb N$ exist? :))
AI: First, $\mathbb{N}$ comes with a natural well-order, which is the usual $\leq$.
Let $(\mathbb{N}, R)$ be a well-ordered set. Then we have an associated ordinal $O(\mathbb{N}, R) \cong (\mathbb{N},R)$. Thus the associated ordinals must be countable.
Conversely, given a countable ordinal, we can lift an ordinal structure on $\mathbb{N}$.
Thus, the amount of (non-isomorphic) well-orders on $\mathbb{N}$ is the amount of countable infinite ordinals, which is $\aleph_1$ (the cardinality of the first uncountable ordinal).
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H: Finding orthogonal projection on Hilbert space
Let $H = L^2(−1, 1)$ and $L \subset H$ be the set of all continuous functions such that $f(0) = 0$. Find the orthogonal projection $P : H → \bar{L}$.
My thoughts on this:
We say $g=P_{\bar{L}}(f)$ iff $f-g \perp \bar{L}$, for $g \in \bar{L}$ and $\forall f \in H$. So we need to find the set describing $g$. In particular, $f-g$ would need to be perpendicular to $g$, that is:
$\int_{-1}^{1} (f-g)\bar{g} \text{d}x = \int_{-1}^{1} f \bar{g} \text{d}x - \int_{-1}^{1} |g(x)|^2 \text{d}x$ $\forall f$ and some $g \in \bar{L}$.
But then I am not sure how to proceed further.
I would appreciate any guidance/hints.
AI: Actually $\overline {L}=H$ so the projection is the identity map!
A well know result in measure theory shows that continuous functions are dense in $H$. Given any continuous function $g$ define $h(x)=xg(\epsilon)$ for $0 \leq x \leq \epsilon$ and $h(x)=g(x)$ for $x \geq \epsilon$. Use a similar construction for $x<0$. Note that $h \in L$. Then $\int |g(x)-h(x)|^{2} dx \to 0$ as $ \epsilon \to 0$. This proves that any $f \in h$ can approximated in the norm of $H$ by an element of $L$.
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H: Proving that $ \sup E[X_n] \geq E[X]$
Consider the sequence of random variables $\{X_n\}_{n\geq1}$ such that $X_n$ are non-negative and $X_n \rightarrow X$ almost surely, with $\sup E[X_n]<\infty$.
Prove $E[X]\leq \sup E[X_n].$
My attempt
$$E[X]=E[\lim X_n]=\lim E[X_n]\leq \sup E[X_n]$$
Is this correct? I would appreciate any suggestion about.
AI: You did not say in what sense $X_n \to X$. Also $E(\lim X_n)=\lim EX_n$ is false in general.
Assuming that $X_n \to X$ almost surely we get $EX =E\lim \inf X_n \leq \lim \inf EX_n$ by Fatou's Lemma and $\lim \inf EX_n \leq \sup_n EX_n$.
For Fatou's Lemma see https://en.wikipedia.org/wiki/Fatou%27s_lemma
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H: if M is a maximal ideal of R such that every element in 1 + M is a unit, then R is a local ring: Counter-example?
I think I have a counter-example to the theorem
If $M$ is a maximal ideal in $R$ such that for all $x \in M$ , $x+1$ is a unit, then $R$ is a local ring with maximal ideal $M$. $R$ has a unique maximal ideal $M$.
A proof is supplied here, but I feel I have a counter-example. Can someone tell me why my counter-example is wrong?
Consider $R = \mathbb Z/6 \mathbb Z$. This has the ideal $M = \{0, 2, 4\}$ as a maximal ideal.
$1 + M = \{1, 3, 5\}$ has every element a unit in $\mathbb Z / 6 \mathbb Z$. Thus, from the theorem, we are to conclude that $(M, \mathbb Z / 6 \mathbb Z)$ is a local ring.
But the ring $\mathbb Z / 6 \mathbb Z$ is not local! It has another maximal ideal $J = \{0, 3 \}$.
What am I missing?
AI: $3$ is not a unit of $\Bbb Z/6\Bbb Z$, as you've pointed out in the following paragraph.
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H: Evans and Murthy: if $\sum_{i=0}^r a_iA^ib=0 , a_i > 0 , i = 0,1,\dots,r$ then $x$ can be expressed as a linear combination of $A^ib$
In the article of Evans and Murthy (1977) the following lemma is given:
If $A,b$ satisfy the relationship $$\sum_{i=0}^r a_iA^ib=0 \quad \quad a_i > 0 \quad i = 0,1,\dots,r$$ then any vector $x$ in the subspace $V \leq \mathbb{R}^n$ spanned by $A^ib$, $i=0,1,\dots,r$ can be expressed as a linear combination of $A^ib$, $i=0,1,\dots,r$ with positive coefficients. That is, for any $x \in V$ there exist (nonunique) $c_i > 0, i=0,1,\dots,r$ such that $$x = \sum^r_{i=0} c_iA^ib$$
There is no proof given in the article for this lemma, but I would like to find out the proof for why this holds.
I think that, since $a_i >0$, $A^ib \leq 0$ for some $i$. But I don't get any further... Can someone help me with a hint or something?
AI: To illustrate, let $y = -A^1 b + A^2b$. This representation is not valid since the coefficient for $A^1 b$ is negative. Since we have $$a_1 (A^1 b) + \cdots + a_r (A^r b) = 0$$, we can write $$ A^1 b = -\frac{a_2}{a_1} (A^2 b) - \cdots -\frac{a_r}{a_1} (A^r b) $$; the coefficient of $A^j b$ in RHS are all negatives. So we can write $y$ as
\begin{align*} y &= - \left( -\frac{a_2}{a_1} (A^2 b) - \cdots -\frac{a_r}{a_1} (A^r b) \right) + A^2 b \\
& = \left( \frac{a_2}{a_1} (A^2 b) + \cdots +\frac{a_r}{a_1} (A^r b)\right) + A^2 b
\end{align*}
whose coefficients are now all positive.
This special case is just illustrative for general case; all negative coefficients can be flipped into positives via put the following equality to negatively coefficiented terms:
$$A^k b = -\frac{1}{a_k}\left( a_1 A^1b + \cdots + a_{k-1}A^{k-1} b +
a_{k+1}A^{k+1} b +\cdots + a_r A^r b\right)$$
This is from the fact that $A^i b$ are linealy dependent with positive coefficients.
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H: How many equivalence classes does the following equivalence relation (over permutations) have?
Let $n\in\mathbb{N}$. Consider the set of all permutations over $n$.
Two permutations $\pi_1 = i_1..i_n$ and $\pi_2 = j_1..j_n$ are not equivalent if either $i_n i_1$ appears in $\pi_2$ (that is, $\exists k: j_k=i_n \land j_{k+1}=i_1$) or $j_n j_1$ appears in $\pi_1$. How many equivalence classes there are?
AI: Your question cannot be answered because it assumed a false premise. The operation you described is not an equivalence relation, because it violates transitivity:
$$(1, 2, 3, 4) = (4, 3, 2, 1) = (3, 4, 1, 2)$$
$$(1, 2, 3, 4) \neq (3, 4, 1, 2)$$
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H: Can someone help me solve this quartic equation?
$$x^4+5x^3-18x^2-10x+4=0$$
I cannot solve this quartic equation - is there any way to solve it apart from the quartic equation? It has no integer roots, and a hint given on the worksheet says to first divide through by $x^2$. Can somebody help?
AI: Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $x^2$ as $x\ne0$
The left hand side becomes
$$x^2+\left(\dfrac2x\right)^2+5\left(x-\dfrac2x\right)-18 =\left(x-\dfrac2x\right)^2+4+5\left(x-\dfrac2x\right)-18$$
So, we have a Quadratic Equation in $x-\dfrac2x$
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H: Obtain value of variable through inverse function
I am trying to implement a controller for a strict feedback system and I am reading a paper regarding the procedure. My problem has to do only with some math involved so this is why I am posting here. Suppose I have a function defined as:
$$ S_1(ε_1) = \frac{e^{2ε_1}}{e^{2ε_1}+1} $$
My goal is to compute the value of $ε_1$ in order to be able to proceed with the design procedure. I have an equation which holds for $S_1(ε_1)$:
$$ x_1(t) = p_1(t)S_1(ε_1)+a_0(t) $$
where $p_1(t)$ is a known performance function, $x_1(t)$ is the state of the system which is known and $a_0(t)$ is some reference trajectory also known. By working out the math of this equation, we come up with:
$$ S_1(ε_1) = \frac{x_1(t)-a_0(t)}{p_1(t)} \Rightarrow ε_1(t) = S_1^{-1}(\frac{x_1(t)-a_0(t)}{p_1(t)}) $$
So now, do I have to find the inverse function of
$$ S_1(ε_1) = \frac{e^{2ε_1}}{e^{2ε_1}+1} $$
and replace $ε_1$ at the inverse function with the expression $ \frac{x_1(t)-a_0(t)}{p_1(t)} $ or do I misunderstand something ?
AI: Let $S(x):= \frac{e^{2x}}{e^{2x}+1}$. Then we have $0<S(x)<1$ for all $x \in \mathbb R.$
Try to show that $S : \mathbb R \to (0,1)$ is bijective. To determine $S^{-1}$, let $y \in (0,1)$ and consider the equation $S(x)=y.$
Elementary computations give
$$x= \frac{1}{2} \ln (\frac{y}{1-y}).$$
Thus
$$S^{-1}(y)=\frac{1}{2} \ln (\frac{y}{1-y}).$$
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H: If $x \not= 0$ and $\lambda x = 0$, then $\lambda = 0$.
I am trying to use the definition of vector spaces to prove that, if $x \not= 0$ and $\lambda x = 0$, then $\lambda = 0$.
Expressing the proposition differently, it seems to me that it is saying the following: If $x \not= 0$ and $\lambda x = 0$, then there exists some $\lambda \in E$ such that $\lambda x = 0$.
Let $x \not= 0$ and $\lambda x = 0$, where $x, \lambda \in E$.
Therefore, by the definition of vector spaces, there exists some $w \in E$ such that $x + \lambda = w$.
Since $x, w \in E$, let $w = x$ (that is, we select $w$ to be equal to $x$).
$\therefore x + \lambda = w \Rightarrow \lambda = w - x = 0 \ \ \ Q.E.D.$
I would greatly appreciate it if people would please take the time to clarify whether or not this is correct.
AI: The sentence “If $x\ne0$ and $\lambda x=0$, then there exists some $\lambda\in E$ such that $\lambda x=0$.” makes no sense; what is the first $\lambda$?
And the problem of proving that $x\ne0$ and $\lambda x=0$ implies that $\lambda=0$ contains the implicit assumption that $x$ is a vector and that $\lambda$ is a scalar. But then $x+\lambda$ is meaningless.
You can prove what you want to prove as follows: if $\lambda\ne0$, then\begin{align}0&=\lambda^{-1}0\\&=\lambda^{-1}(\lambda x)\\&=\left(\lambda^{-1}\lambda\right)x\\&=1x\\&=x.\end{align}But we are assuming that $x\ne0$.
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H: prove that $ \frac{\sigma(1)}{1}+\frac{\sigma(2)}{2}+\dots+\frac{\sigma(n)}{n} \leq 2 n $
[HMMT 2004] For every positive integer $n$, prove that
$
\frac{\sigma(1)}{1}+\frac{\sigma(2)}{2}+\dots+\frac{\sigma(n)}{n} \leq 2 n
$
If $d$ is a divisor of $i,$ then so is $\frac{i}{d},$ and $\frac{i / d}{i}=\frac{1}{d} .$
Summing over all divisors $d$ of $i$ (which is $\sigma(i)$ ), we see that $\frac{\sigma(i)}{i}$ is the sum of all the reciprocals of the divisors of $i ;$
that is,
$
\frac{\sigma(i)}{i}=\sum_{d | i} \frac{1}{d}
$ .....
But how they concluded that $
\frac{\sigma(i)}{i}=\sum_{d | i} \frac{1}{d}
$ ??? ***
i am getting trouble in understanding this question in recent few days, what i till now understand is that
if
$d$ is divisor of $i$ then $i/d$ is also divisor of $i$ , so
$\frac{i / d}{i}=\frac{1}{d} .$ but they are not all divisors of i so how we get *** ,i also tried to take some examples but still can't get it,
can anyone explain this ..
thankyou
AI: Why do you say that “they are not all divisors of $i$”? You have$$\sigma(i)=\sum_{d\mid i}d$$and therefore$$\frac{\sigma(i)}i=\sum_{d\mid i}\frac di.$$But the numbers of the form $\frac di$, with $d\mid i$ are precisely the numbers of the form $\frac1{d'}$, with $d'\mid i$. That is, the are the inverses of the divisors of $i$.
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H: Proving $xRy\iff xy^{-1}\in\ker(f)$ is equivalent relation, where $f:(G,.) \to (H,.)$ is homomorphism of groups $G$ and $H$.
I know for sure it is but I failed to prove how. I tried to use the homomorphism definition where $f(x*y^{-1}) = f(x)*f(y^{-1})$ but I didn't see a pattern.
AI: Note that $ {\rm Ker}(f) = \{x\in G|\, f(x) = 1_H\}.$
$\bullet$ Reflexive property: Since $f$ is a homomorphism, $f(1_G) = 1_H.$ So, for all $x\in G,$ we have
$$ f(xx^{-1}) = f(1_G) = 1_H,$$
so $xx^{-1}\in {\rm Ker}(f).$ This implies $xRx.$
$\bullet$ Symmetric property: For all $x, y\in G,$ if $xRy,$ then $xy^{-1}\in {\rm Ker}(f).$ So $f(x) = f(y).$ Then, we have
$$ f(yx^{-1}) = f(y)[f(x)]^{-1} = f(x)[f(x)]^{-1} = 1_H,$$
so, $yx^{-1}\in {\rm Ker}(f).$ This implies $yRx.$
$\bullet$ Transitive property: For all $x, y, z\in G,$ if $xRy$ and $yRz,$ then $xy^{-1}\in {\rm Ker}(f)$ and $yz^{-1}\in {\rm Ker}(f).$ So $f(x) = f(y) = f(z).$ Then, we have
$$ f(xz^{-1}) = f(x)[f(z)]^{-1} = f(z)[f(z)]^{-1} = 1_H.$$
so, $xz^{-1}\in {\rm Ker}(f).$ This implies $xRz.$
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H: Counter example with series $\sum_{k=2}^{\infty} \frac{1}{k}$ to show that $ \sum_{k=2}^{\infty} \frac{2^k}{\lfloor \frac{k}{2} \rfloor!} $ diverges?
I am inspecting the convergence of different series and I ran across the following series:
$$ \sum_{k=2}^{\infty} \frac{2^k}{\lfloor \frac{k}{2} \rfloor!} $$.
I know that the following series shoud diverge, but unfortunately the D'Alembert's criterion gave me the answer zero. Furthermore the $\lim_{n} u_{n}$ seems to not go to zero . So I would like to give the counter example with the harmonic series:
$$ \sum_{k=2}^{\infty} \frac{1}{k}$$
to show that the first series diverge. Can someone show me how to do that counter example or have I totally missed the point and failed to understand the principles of series?
AI: $[\frac k 2] \geq (9!) 9^{[\frac k 2]-9}$ (since $9,10,11,..., [\frac k 2]$ are all $\geq 9$). Hence the series is dominated by a constant times the series $\sum \frac {2^{k}} {3^{3([\frac k 2]-9)}}$. Now observe that $3([\frac k 2]-9) \geq k$ for all $k$ sufficiently large. Comparing the given series with $\sum (\frac 2 3)^{k}$ we get convergence.
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H: How to show that two Euler-Lagrange-equations have the same solution?
Let $\phi:\mathbb R^n \to \mathbb R$ be a twice continuously differentiable function and let
$$L_{\phi}(t,x,\dot x) =\nabla\phi(x)^T\dot x = \sum_i\frac{\partial \phi}{\partial x_i}(x_1,..,x_n)\dot x_i.$$
Let $L:[a,b] \times \mathbb R^n \times \mathbb R^n \to \mathbb R$ another arbitrary $C^1$-Lagrange function. How can one show that the Euler-Lagrange-Equations $L$ and $L+L_{\phi}$ have the same solutions?
AI: First, notice$$\begin{align}\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L_\phi&=\sum_i\left(\frac{\partial^2\phi}{\partial x_i\partial x_j}\dot{x_i}-\frac{d}{dt}\left(\frac{\partial\phi}{\partial x_i}\delta_{ij}\right)\right)\\&=\sum_i\frac{\partial^2\phi}{\partial x_i\partial x_j}\dot{x_i}-\frac{d}{dt}\left(\frac{\partial\phi}{\partial x_j}\right)\\&=0.\end{align}$$So$$\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)(L+L_\phi)-\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L=\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L_\phi=0.$$So$$\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)L=0\iff\left(\frac{\partial}{\partial x_j}-\frac{d}{dt}\frac{\partial}{\partial\dot{x_j}}\right)(L+L_\phi)=0,$$i.e. the two systems of ELEs are equivalent.
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H: Infinite series of sequences
Let S be the set of sequences whose series converge absolutely. We define 2 norms on S: $$\| \{ a_n \}_{n=0}^{ \infty } \|_1 = \sum_{n=0}^\infty | a_n |$$ and, $$\| \{ a_n \}_{n=0}^\infty \|_{\sup} = \sup \{ |a_n|_{n=0}^\infty \} $$
Note: S is the set of sequences such that $\| a \|_1 < \infty.$ (The sup-norm is sometimes called the infinity-norm.)
Define a linear operator $\Sigma : S \to \mathbb{R}$ by: $$\Sigma \big( \{ a_n \}_{n=0}^\infty \big) = \sum_{n=0}^\infty a_n$$
Question 1: find the operator norm of $\Sigma$ using $\| . \|_1$.
Question 2: show that the operator norm of $\Sigma$ using $\| .\|_{\sup}$ is unbounded.
Can anyone help me answer this question or give me hints. Also sorry about my latex code I am new at it and very bad but I tried my best!
AI: HINTS:
$1)$ Prove an easy inequality of that norm and then find an example where the equality holds.
$2)$ Think harmonically :)
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H: What exactly does it mean for a vector to have a direction?
Vectors are defined as having magnitude and direction. If I understand it correctly, their magnitude is their length, meaning they have the properties of a line segment. What does it mean for a vector to have a direction? Let me be more specific:
Let $\vec v=\begin{bmatrix}2\\3\end{bmatrix}$
The angle of this vector is $\arctan(\frac y x)=\arctan(\frac3 2)$ from the positive $x$ axis and we know that the direction of this vector is "pointing to the upper right". What does this mean geometrically? What defines direction for a vector? (I am new to the topic so go easy on me.)
AI: That is the correct definition in physics. If you incorporate the correct definition of a vector in maths, you can easily understand the "direction" concept alternatively.
In physics the direction associated with a vector means you first fix a coordinate system and see in which direction the effect of the vector is as a whole. For example take velocity v.
So say $v=a\hat{i}$. This means that the velocity is entirely in the direction of x axis of your chosen coordinate system. That effect of a vector is 0 in the direction perpendicular to the direction in which it's effect is as a whole. Now you could represent the same vector for velocity in another basis. But logically changing basis should not change the vector ( the velocity of any object has to be in the same direction in which ever coordinate system you represent it). So in the new coordinate system the entire effect of the vector ( v here) would still be in the x direction ( of your previous coordinate system) with the only change that you are representing your vector in a different coordinate system.
Mathematically a vector is an element of a vector space and there are vectors with which you cannot associate any sense direction. Like a wavefunction is a vector but you cannot define a direction with it. A matrix is a vector but you cannot associate a direction with it in the normal sense.
I should mention that only in our 3 dimensional space or Euclidean space can we sensibly denote a sense of direction to a vector and not generally in other spaces. Because our sense of direction itself doesn't exist for other spaces. You will need to first define what a direction means in a vector space and only then can a vector in that space be given a sense of direction. That direction might be completely different in sense than the directions we are used to understand in our world. So you see how algebra helps us do things in higher dimensions where geometry couldn't.
I hope that helped
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H: probability of non smokers equals probability of smokers
Suppose that $10$ % students are smokers.In a random sample of $10$ students, the probability that number of nonsmokers equals the number of smokers is
What i try: If $10$% students are smokers. Then $90$ % students are non smokers.
Let $p$ be the probability of smokers and $q$ be the probability of nonsmokers.
So $$p=\frac{1}{10}$$ and $$q=\frac{9}{10}$$
I did not understand How $p=q$ which is asked in question.
Please,Help me , Thanks
AI: You can use binomial pmf
$$\mathbb{P}[X=5]=\binom{10}{5}(\frac{1}{10})^5(\frac{9}{10})^5$$
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H: Find zero of function $\sum_{n=0}^{\infty}\frac{cos(x(n+1))}{n!}$
I'm interested in finding the smallest positive zero of function
$$\sum_{n=0}^{\infty}\frac{\cos(x(n+1))}{n!},\qquad x\in\mathbb R$$
It is approximately equal to $0.832$. I've calculated Taylor series expansion of this sum, which is:
$$\sum_{k=0}^{\infty}e \cdot\frac{(-1)^k \ B_{2k}\ x^{2k}}{(2k)!}$$
Where $B_n$is n-th Bell number.
But I don't know, whether it helps.
Thanks for all the help in solving this problem.
AI: Your sum is equal to
$$\Re\bigg(\sum_{n=0}^\infty \frac{e^{ix(n+1)}}{n!}\bigg)$$
or
$$\Re\exp\big(ix+e^{ix}\big)=\Re\exp\big(\cos x+i\sin x+ix\big)$$
Using Euler’s formula, this is equal to
$$e^{\cos x}\cos(x+\sin x)$$
To find when this is equal to zero, you must compute the zeroes of the function
$$\cos(x+\sin x)$$
or the values of $x$ for which
$$x+\sin x = \pi(n+1/2)$$
for some $n\in\mathbb Z$. I do not suspect an elementary solution will exist, but you can use this explicit form to calculate some pretty good approximate solutions.
The smallest of the zeroes will occur when
$$x+\sin x = \pi/2$$
which is at approximately $x\approx 0.832$. See @Raymond’s answer for a more accurate approximation and a representation in terms of the Dottie number.
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H: Reverse the order of integration.
Reverse the order of integration:
$$\int_{0}^{1}\int_{2\sqrt{x}}^{2\sqrt{x}+1} f(x,y)dydx$$.
This is my solution:
$$\int_{0}^{1}\int_{0}^{\frac{y^2}{4}}f(x,y)dxdy+\int_{1}^{2}\int_{\frac{(y-1)^2}{4}}^{\frac{y^2}{4}}f(x,y)dxdy+\int_{2}^{3}\int_{1-\frac{(y-1)^2}{4}}^{0}f(x,y)dxdy$$
But it dosen't work. What am I doing wrong? I think my third integral is bad. Could someone help me?
AI: It helps to graph the region over which you are doing the integral. In the original case, you fix a value of $x$, and integrate over the values of $y$ between the two parabolas $y = 2\sqrt{x}$ and $y = 2\sqrt{x} +1$, and then varying $x$ from 0 to 1
Now, instead, let us fix a value of $y$. Then, we have the following cases
$$0 \leq y \leq 1$$
$$0 \leq x \leq \frac{y^2}{4}$$
This you've got right
Then,
$$1 \leq y \leq 2$$
$$\frac{(y-1)^2}{4} \leq x \leq \frac{y^2}{4}$$
This is also okay
Then,
$$2\leq y \leq 3$$
$$\frac{(y-1)^2}{4} \leq x \leq 1$$
This is where the mistake was done. The region would be bounded on the left by the upper parabola, and is not from 0 to 1 like in your attempt
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H: Automorphism of commutative groups.
For every group G there is a natural group homomorphism G → Aut(G)
whose image is the group Inn(G) of inner automorphisms and whose
kernel is the center of G. Thus, if G has trivial center it can be
embedded into its own automorphism group
The inner automorphism group of a group G, Inn(G), is trivial (i.e.,
consists only of the identity element) if and only if G is abelian.
Consider group C4 and Homomorpism f: C4->Aut(C4). Inn(C4)={e}. Ker(f)=Center(C4)={all elements of C4}.
But |Aut(C4)|=()=(4)=2, where () is Euler's function. Aut(C4) is isomorphic to C2. There is quotient map from C4 to C2. |Img(f)| = 2
Can you please suggest where is my mistake?
AI: There are two automorphisms of $C_4$, true. But one of them (given by $x\mapsto -x$) isn't an inner / conjugation automorphism, so it is not in the image of $f$. The kernel of $f$ is the center of $C_4$, which is all of it, and the image of $f$ is trivial (it is isomorphic to $C_4/Z(C_4)$, as it should be).
The fact that $\operatorname{Aut}(C_4)$ is (isomorphic to) a quotient of $C_4$ is incidental and not really relevant.
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H: Direct sum and determinant
Consider the vector space $V = V_1 \oplus V_2$, where $V_1,V_2$ are subspaces. Let $f_i\in\mathcal{L}(V_i)$ be a linear map for $i=1,2$, and define $f\in\mathcal{L}(V)$ by $f(v)= f_1(v_1)+f_2(v_2)$, where $v=v_1+v_2$ and $v_i\in V_i$, for $i=1,2.$
Prove that $\det(f) = \det(f_1)\cdot\det(f_2)$.
I have no idea where to start, any ideas are welcome.
AI: I think you are talking about finite dimensional vector spaces $V_1$ and $V_2$ (so that one can talk about determinant). Choose $(e_1,\ldots,e_n)$ and $(f_1,\ldots, f_m)$ be basis of $V_1$ and $V_2$. Then $(e_1,\ldots,e_n,f_1,\ldots,f_m)$ is a basis of $V=V_1 \oplus V_2$.
Take $f_i \in \mathrm{End}(V_i)$, and define $f = f_1 \oplus f_2$. Then $f \in \mathrm{End}(V)$. If $M_i$ is the matrix of $f_i$ in the basis $(e_j)$ (or $(f_k)$), then the matrix of $f$ in the basis $(e_1,\ldots,e_n,f_1,\ldots,f_m)$ is $M=\begin{pmatrix} M_1 & 0 \\ 0 & M_2\end{pmatrix}$.
Thus, $\det f = \det M = \det M_1 \times \det M_2 = \det f_1 \times \det f_2$.
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H: Show that the (conditional) probability of a triple occupancy of some cells equals $1/4$.
Seven balls are distributed randomly in seven cells. If exactly two cells are empty, show that the (conditional) probability of a triple occupancy of some cells equals $1/4$.
Let $H$ be the event that exactly two cells are empty. Let $A$ be the event that there is a triple occupancy.
I need to compute $P(A|H)$.
2 cells are empty, so a triple occupancy must occur in the remaining 5 cells. Thus, there are five ways to do it. Once the place for a triple occupancy is chosen, the rest 4 cells must contain 1 ball per each.
the number of ways to distribute 7 balls into 5 cells without any cell being empty is ${6 \choose 4}$.
Therefore, my answer is that $P(A|H) = 1/3$, and this is wrong. Can you point out where my logic is wrong?
AI: The problem is that you haven't considered the order of the balls. It's true that there are $5$ patterns (i.e. just counting how many balls are in each cell) with a triple and $10$ patterns without a triple, but each of the patterns with a triple occurs in $\binom 73\times 4!$ ways, since there are this many options for which three balls go in the triple and order the rest. Similarly each pattern without a triple occurs in $\binom 72\times \binom 52\times 3!$ ways.
If we distribute each ball at random, it is the ways with balls labelled which are equally likely, not the patterns.
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H: How would I solve this, considering I have no values whatsoever?
I think that SQ is straight and so have tried to use Pythagoras, which leaves me with $a + b + c = ac/b$, but I don't see any values. How could I find values?
AI: Notice the shorter sides of the rectangle are equal and hence $a+b=c$. Using Pythagoras, we have $$(b+c)^2+(a+b)^2=(a+c)^2 \\ (2c-a)^2+c^2=(a+c)^2 \\ 4c^2=6ac \implies \frac ac=\frac 23$$ And also $$\frac ac+\frac bc = 1\implies \frac bc =\frac 13$$ This gives $$a:b:c=2:1:3$$
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H: $f=g \implies \nabla f = \nabla g$?
I want to disprove the following statement: $$f=g \text{ on } S \implies \nabla f = \nabla g \text{ on } S$$ where $S$ is some smooth closed surface and $f,g$ are smooth.
I don't understand why this shouldn't be the case, assuming $S$ is more than a single point. In the simplest case, where $f,g:\mathbb{R}\to\mathbb{R}$ and S is some closed interval this is obviously true ($f'=g'$, right?).
I've tried looking for a relatively simple counterexample (something like $f:\mathbb{R}^2\to\mathbb{R}$) i can plot on geogebra and really see whats going on geometrically, but don't seem to be getting anywhere.
AI: Let $f(x,y):=x$ and $g(x,y):=-x$ in the $(x,y)$-plane, and let $S$ be the $y$-axis. Then $f\restriction S=g\restriction S$, but $\nabla f\restriction S=-\nabla g\restriction S$.
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H: Upper and lower darboux sum
I have recently been going through some questions on Riemann integration.I got stuck in one of those questions which happen to be a multiple-select question. It says that:
Let $f$ be a continuously differentiable real-valued function on $[a,b]$ such that $\left\lvert f'(x)\right\rvert\le k$ for all $x$ in $[a,b]$. Then for any partition $P$ of $[a,b]$, which of the following is/are correct?
$\left\lvert L(f,P)\right\rvert\le k(b-a) \le \left\lvert U(f,P)\right\rvert$
$U(f,P)-L(f,P) \le k(b-a)$
$U(f,P)-L(f,P) \le k \left\lVert P\right\rVert$ where $\left\lVert P\right\rVert$ is the norm of partition $P$
$U(f,P)-L(f,P )\le k\left\lVert P\right\rVert(b-a)$
This is what I have done so far:
Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ therefore by mean-value theorem there exists a $c$ in $(a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.
And then by the hypothesis we obtain that $\left\lvert f(b)-f(a)\right\rvert\le k(b-a)$.
Now how should I proceed?
AI: Put $$P=(x_i)_{i=0,n}$$
then
$$U(f,P)-L(f,P)=\sum_{i=0}^{n-1}(M_i-m_i)(x_{i+1}-x_i)$$
where
$$M_i=\sup_{[x_i,x_{i+1}]}f \; \; and $$
$$m_i=\inf_{[x_i,x_{i+1}] } f .$$
$f $ is continuous at $ [x_i,x_{i+1}] $, thus
$$M_i=f(b_i), \; and \; m_i=f(a_i)$$
with $(a_i,b_i)\in [x_i,x_{i+1}]^2$.
Wlog, we assume that $a_i<b_i$.
$f $ is continuous at $[a_i,b_i]$ and differentiable at $ (a_i,b_i) $, then by MVT,
$$M_i-m_i=f(b_i)-f(a_i)=(b_i-a_i)f'(c_i)$$
with $b_i-a_i\le x_{i+1}-x_i\le ||P||$.
therefore
$$M_i-m_i\le ||P||k$$
and finally
$$U(f,P)-L(f,P)\le k||P||(b-a)$$
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H: The problem of rebalancing an investement portfolio
I bought $30$ of stock $A$ at a price of 1 USD each. And 2 of sock $B$ at a price of $15$ USD each. Now 50% of my portfolio is stock $A$ ($\frac{30 \times 1}{60}$) and 50% of it is stock $B$ ($\frac{15 \times 2}{60}$). I want to keep this balance, so that each stock is 50% of my portfolio.
A month later prices changed, now $A$ costs 3 USD and $B$ costs $12$ USD. I want to rebalance my portfolio so that each stock contributes 50% of the overall value again. Under some budget $b$, how many of each stock do I have to buy to rebalance my portfolio?
This is a real life problem of mine. I am interested in a general solution: given budget $b$, stock quantities $s_1, s_2, \dots, s_n$, stock prices $c_1, c_2, \dots, c_n$, find how many of each stock to buy $s_1', s_2', \dots, s_n'$ such that each stock contributes a fixed percentage $p_1, p_2, \dots, p_n$.
I tried writing it down as a system of linear equations, but couldn't obtain a general solution from there. Important caveat is that the new obtained percentage does not have to be exact, it should be just close enough. You can't always obtain exactly 50%/50% or any other ratio.
Here's my attempt at a system of equations:
$$c_1(s_1 + s_1') + c_2(s_2 + s_2') + \dots + c_n(s_n + s_n') = S$$
$$c_i(s_i + s_i') = p_i S, \ i \in {1, 2, \dots, n}$$
$$c_1 s_1' + c_2 s_2' + \dots + c_n s_n' \leq b$$
Where $S$ is the total value.
I feel like there should be a simple and beautiful solution, but so far I have only arrived to equations where I need to guess some appropriate values. E.g for the simple problem statement above I have arrived to:
$$3 s_2' - s_1' = 16$$
From here, by substituting random values until it works, I arrive to two (of the many) solutions that restore the balance: $(s_1 = 2, s_2 = 6)$ and $(s_1 = 8, s_2 = 8)$. The problem with that I can't automate picking values until it works, and I can't run such computations every time I need to rebalance a real portfolio of 10 or more stocks.
AI: Given your budget $b$ and the current value of the portfolio $v$, after rebalancing the total portfolio will have value $b+v$. Each stock $i$ should represent $p_i(b+v)$ of the portfolio. Buy or sell enough of it to get to this level.
In your example, $A$ is currently $90$ and $B$ is currently $24$, so $v=114$. If your budget is $36$ the new portfolio value will be $150$, so each stock should represent $75$. You need to sell $15$ worth of $A$ and buy $51$ worth of $B$ to get to balance. Note that the net cost of $51-15=36$ matches your budget.
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H: De Morgan’s law: Wikipedia proof, cannot follow part 1, step 3.
I would like to prove De Morgan’s laws and have tried to follow the Wikipedia proof. However, I am stuck in part 1 of this proof, line 3:
1: Let $x \in (A \cap B)^c $. Then, $ x \notin A \cap B $.
2: Because $ A \cap B = \{y | y \in A \wedge y \in B \}$, it must be the case that $ x \notin A$ or $x \notin B$.
3: If $x \notin A$, then $x \in A^c$, so $x \in A^c \cup B^c$
Why is the part after the comma (“so, ...”) correct and where is it coming from?
Thank you.
AI: If $x$ belongs to a set, then it belongs to every larger set. So, if $x\in A^\complement$, then it also belongs to $A^\complement\cup B^\complement$.
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H: Describe all martingales that only take values in $\{−1, 0, 1\}$.
Describe all martingales that only take values in $\{−1, 0, 1\}=:\Omega$.
In the first instance i would try to find a filtration of
$$P(\Omega)=\{\emptyset,\{0\},\{1\},\{-1\},\{1,-1\},\{1,0\},\{-1,0\},\Omega\}.$$
Candidates are:
$P(\emptyset),P(\{0\}),P(\{1\})\subseteq P(\{0,1\})\subseteq P(\Omega)$
$P(\emptyset),P(\{0\}),P(\{-1\})\subseteq P(\{0,-1\})\subseteq P(\Omega)$
$P(\emptyset),P(\{0\}),P(\{1\})\subseteq P(\{0,1\})\subseteq P(\Omega)$
$P(\emptyset),P(\{1\}),P(\{-1\})\subseteq P(\{1,-1\})\subseteq P(\Omega)$
I'm not really sure how to proceed. Any assistance or thoughts would be much appreciated.
AI: Any martingale taking values in $\lbrace -1,0,1\rbrace$ has the form $X_n = E[X_\infty|\mathcal{F}_n]$ for some $\lbrace -1,0,1\rbrace$-valued random variable $X_\infty$ and some filtration $(\mathcal{F}_n)$. This is because any such martingale is bounded hence by the martingale convergence theorem it converges a.s. and in $L^1$ to some $X_\infty$ and $X_n = E[X_\infty|\mathcal{F}_n]$. ($X_\infty$ necessarily takes values in $\lbrace -1,0,1\rbrace$ by a.s. convergence). Furthermore, since there are finitely many sub-$\sigma$-fields in $\mathcal{P}(\Omega)$, the filtration has to be stationary, i.e. there exists $\mathcal{F} \subset \Omega$ such that $\mathcal{F}_n = \mathcal{F}$ for $n$ large enough.
In conclusion, every martingale taking values in $\lbrace -1,0,1\rbrace$ is stationary, in the sense that there exists a (nonrandom) integer $N \in \mathbb{N}$ such that
$$X_n = E[X_\infty|\mathcal{F}_n], \quad \forall n \leqslant N,$$
and
$$X_n = X_\infty, \quad \forall n >N,$$
where $X_\infty$ is some $\lbrace -1,0,1\rbrace$-valued random variable and $\mathcal{F}_1, \ldots, \mathcal{F}_N$ is a (finite) filtration.
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H: Row sum of square matrix $\mathbf{A}^T\mathbf{A}$ relation to row/column sum of $\mathbf{A}$
I have a matrix $\mathbf{A} \in \mathbb{R}^{m \times n}$ with positive entries. Its column sum is $k$ and its row sum is $1$, i.e. $\sum col(\mathbf{A}) = k$ and $\sum row(\mathbf{A}) = 1$.
Now looking at $\mathbf{A}^T\mathbf{A}$, I notice that it has a row (and column) sum of $k$, i.e. $\sum row(\mathbf{A}^T\mathbf{A}) = \sum col(\mathbf{A}^T\mathbf{A}) = k$. Is there a reason for this?
Could this be because $\sum row(\mathbf{A}^T\mathbf{A}) = \sum col(\mathbf{A}^T\mathbf{A}) = \sum col(\mathbf{A})*\sum row(\mathbf{A})$?
Thank you
AI: Let $u$ be the column vector of $n$ $1$'s and $v$ the column vector of $m$ $1$'s.
The column vector of row sums of $A$ is $A u$ and the row vector of column sums is
$v^T A$. You seem to be saying that $v^T A = k u^T$ and $A u = v$. Then
$A^T A u = A^T v = (v^T A)^T = k u^T$ and $u^T A^T A = (A u)^T A = v^T A = k u$, so the row and column sums of $A^T A$ are all $k$.
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H: Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$
Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following equation for $x$:
$$1-x=(1-u)^2$$ $$x=u(2-u)$$
and replacing $u$ by and $x$, and $=$ by $\to$ , which leads to
$$\frac{1}{(1-x)^2}=1+x(2-x)+x^2(2-x)^2+x^3(2-x)^3+... \tag{2}$$
Succumbing to the same 'substitution'approach with $\frac{x}{1+x^2}$, I put
$$\frac{1}{1-x}=\frac{u}{1+u^2}$$ $$x=1-\frac{1}{u}-u$$
so with $x \to 1-\frac{1}{x}-x$
I should have:
$$\frac{x}{1+x^2}=1+\left(1-\frac{1}{x}-x\right)+\left(1-\frac{1}{x}-x\right)^2+\left(1-\frac{1}{x}-x\right)^3+...\tag{3}$$
but this does not seem correct (at least according to my Desmos graph for $|x|<1$).
Can someone please explain what my conceptual errors are?
AI: Your series $(1)$ is valid for $x\in(-1,1)$.
So series $(2)$ is valid for $x(2-x)\in(-1,1)$, i.e. for $x\in(1-\sqrt 2,1)\cup(1,1+\sqrt 2)$.
But series $(3)$ is only valid if $1-\frac{1}{x}-x\in(-1,1)$. And it turns out that this is not true for any $x\in\Bbb R$.
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