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H: A linear algebra related detail in a proof of Index Theorem
Here is a clipping from Milnor's Morse Theory. Since this question about linear algebra, I will present my question below so that no prior knowledge of the materials in the book required to answer this question.
The only relevant thing here is the matrix of size $(\lambda+v)^2$ at the bottom. Here $\Big((E^{\tau}_0)_{**}(W_i,W_j)\Big)$ is a negative definite matrix of size $\lambda^2$ and for convenient, let's call this matrix $M$. The $(\lambda+v)^2$-matrix become
$$
H_c=
\begin{pmatrix}
M & cA\\\ cA^t & -4I+c^2B
\end{pmatrix}
$$
where $A$ and $B$ are some fixed matrix of size $\lambda \times \nu$ and $v^2$ respectively, $M$ is a negative definite matrix of size $\lambda^2$, and $c$ is an arbitrary real number.
Now the claim is that we can pick $c$ such that $H_c$ is a negative definite matrix. I.e., there is $c$ such that for any $x \neq 0$, $x^tH_cx <0$. I've tried this by writting $x = (a \quad b)^t$ where $a$ and $b$ are column vectors of lenght $\lambda$ and $\nu$ respectively and then do the block matrix multiplication
$$
\begin{pmatrix}
a & b
\end{pmatrix}^t H_c
\begin{pmatrix}
a \\ b
\end{pmatrix}
$$
and see if i can adjust $c$ to make the whole expression negative no matter how big or small $a$ and $b$ are. But i think it's messy and i want a clean or general way to get the result. I hope somebody could help me with this. Thank you.
AI: Negative definite matrices form an open subset of all symmetric matrices, since they are just the symmetric matrices whose eigenvalues are all negative and the eigenvalues vary continuously with the entries of a matrix. Now $H_0=\begin{pmatrix} M & 0 \\ 0 & -4I\end{pmatrix}$ is clearly negative definite since $M$ and $-4I$ are. So, by continuity, $H_c$ is also negative definite for all $c$ sufficiently close to $0$. |
H: Are real numbers enough to solve simpler exponential equations such as $2^x=5$, $(1/e)^x=3$, and $\pi^x=e$?
How to prove that solutions of simpler exponential equations (*) are real numbers?
In other words, how to prove that set of real numbers is enough to solve something like $2^x = 5,$ or $(\frac{1}{e})^x = 3$ or even $\pi^{x} = e$?
(*) Assumption is that base is not something like $e^{\sqrt{2}}, e^{\pi}, 2^{\frac{1}{2}},$ (#) so only "given" (computed in some way) indeed real? Also, inverse question: How to prove that this numbers (#) are real?
Graphically: how to prove that exponential function $f(x) = e^x$ for $x \in (- \infty, + \infty)$ "picks up" all the real values?
AI: First recall that the map
\begin{align*}
\exp &: \mathbb{R} \to (0, \infty)
\end{align*}
is a bijection (you can prove this by e.g., proving that $\exp$ is continuous, strictly increasing (this gives injective), unbounded, and obtains arbitrarily small positive values - then invoke the intermediate value theorem (this gives surjective)). In particular $\exp(x) = r$ has a unique real solution whenever $r \in \mathbb{R}_{> 0}$, which we call $\log(r)$.
By definition, whenever $\alpha \in \mathbb{R}_{>0}$ we have
$$\alpha^x = \exp (\log(\alpha) x)$$
which defines a bijection $\mathbb{R} \to (0, \infty)$ whenever $\log(\alpha) \neq 0$ - i.e., when $\alpha \neq 1$ (since then it is a composition of bijections).
This also gives a solution to your second question, we know that $\exp(x) \in \mathbb{R}$ whenever $x \in \mathbb{R}$. Since $\log(\alpha) \in \mathbb{R}$ exists whenever $\alpha \in \mathbb{R}_{>0}$ we have $\alpha^x = \exp(\log(\alpha)x) \in \mathbb{R}$ whenever $x \in \mathbb{R}$.
Note that this question is much more subtle when $\alpha < 0$ since we must choose our $\log$'s etc. But for sure $(-1)^{1/2} \notin \mathbb{R}$. Moreover, your equations will also have complex solutions - so we do not have uniqueness if we extend our view to $\mathbb{C}$. The moral being - $\log$ is not well behaved in $\mathbb{C}$ so your question turns out to be a subtle one if we allow $\mathbb{C}$ to be involved. |
H: Show that the group in the center of this short exact sequence is abelian
Suppose that $$1\to A\xrightarrow{f} B\xrightarrow{g}C\to 1$$ is an exact sequence of groups, where $A$ has order $85$ and $C$ has order $9$. Show that $B$ is abelian.
I have proved earlier that for a Sylow $3$-subgroup $S$ of $B$, we have that $g$ maps $S$ isomorphically to $C$. Further, I know that $|\text{Aut}(A)|=64$. I am interested in a solution that actually uses one or both of these facts in what I presume is the spirit of the problem.
I think I am able to sketch a solution in my head involving tracing out cases for the nature of the maps $f$ and $g$ and then showing that $C$ must be in the center of $B$ (which would allow us to conclude instantly), but I can't imagine that this is the morally correct perspective. As such, I ask that any solution try to use the aforesaid facts, if possible.
Any and all help would be appreciated.
AI: You have already said that you can pretend that $C$ lies in $B$, so that $B$ has a normal subgroup $A$, a subgroup $C$, and $A\cap C=1$ with $B=AC$. Notice that as $A$ is normal, $C$ acts on $A$ by conjugation.
As $A$ has automorphism group a $2$-group, and $C$ is a $3$-group, the only map from $C$ to $\mathrm{Aut}(A)$ is the trivial map. Thus the conjugation of $C$ on $A$ is trivial, i.e., $C$ centralizes $A$. Thus $A\leq C_B(A)$ and $C\leq C_B(A)$, and thus $C$ is central in $B$. Thus $CA=C\times A$, and $B$ is abelian. |
H: Hessian form of a real valued function on a submanifold of $\mathbb{R}^{n+m}$
I recently came across a result in a text in the field of differential geometry and I'm wondering why it is true. Let $M\subset\mathbb{R}^{n+m}$ be an $n$-dimensional submanifold and $w:M\longrightarrow\mathbb{R},\,w(x)=u(x)-\langle x,z\rangle$, where $u:M\longrightarrow\mathbb{R}$, $z\in\mathbb{R}^{n+m}$ and $\langle\cdot,\cdot\rangle$ denotes the canonical inner product on $\mathbb{R}^{n+m}$. The author states that we have $D_{M}^{2}w(x)=D_{M}^{2}u(x)-\langle II_{x}(\cdot,\cdot),z\rangle$, where $II_{x}$ is the second fundamental form at the point $x$ and $x$ being a critical point of $w$. How is the exact computation to get this result?
What I tried is to first compute the gradient w.r.t $M$ of $w$: $\nabla^{M}w(x)=\nabla^{M}u(x)-z^{tan}$, where $z^{tan}$ is the tangential component of $z$. If this is correct (is it?), I don't know how to get the Hessian of $w$ at $x$. Thanks in advance!
AI: The derivative operator on an embedded submanifold is just the projection of the ambient space's derivative operator, so:
$$
\nabla^M w(x) = \nabla^M (u - \langle{}\cdot, z\rangle{})(x) = \nabla^M u(x) - \text{proj}_{T_xM} z = \nabla^M u - z^{\text{tan}},
$$
like you wrote. This is because the derivative of $x \in \mathbb{R}^d \mapsto \langle{}x, z\rangle{}$ is $z$. The Hessian of a function $u : M \to \mathbb{R}$ is defined as the linear map $H(u)_x : T_x M \to T_x M$ defined by
$$
H(u)_x(v) = \nabla_v^M(\nabla^M u)|_x, \ \ \ \ \forall v \in T_x M.
$$
The $D^2_M u$ that you write above, considered as a bilinear form, is obtained by just setting $D^2_M u(v, y) = \langle{}H(u)_x(v), y\rangle{}$. We'll take this dot product at the end.
Well, we computed $\nabla^M w$ above, and so let's compute $\nabla^M_v \nabla^M w$:
$$
\nabla^M_v \nabla^M w|_x = \text{proj}_{T_x M}(\nabla_v^{\mathbb{R}^d} \nabla^M w) = \text{proj}_{T_x M} \nabla^{\mathbb{R}^d}_v (\nabla^M u - z^\text{tan}) = H(u)_x(v) - \text{proj}_{T_x M}\nabla^{\mathbb{R}^d}_v z^\text{tan}.
$$
Now, letting $\nu$ be a unit normal vector field with respect to which we're defining the second fundamental form, $z^\text{tan}$ is just
$$
z^\text{tan} = z - \langle{}\nu, z\rangle{}\nu
$$
and thus
$$
\nabla^{\mathbb{R}^d}_v z^\text{tan} = -\langle{} \nabla^{\mathbb{R}^d}_v \nu, z\rangle{}\nu - \langle{}\nu, z\rangle{}\nabla^{\mathbb{R}^d}_v \nu.
$$
Projecting this to $T_x M$ gives $\nabla^M_v$ (and note the first term vanishes):
$$
\nabla^M_v z^\text{tan} = - \langle{}\nu, z\rangle{}\text{proj}_{T_x M}(\nabla^{\mathbb{R}^d}_v \nu).
$$
Taking an inner product with an arbitrary vector $y \in T_x M$ gives
$$
H(w)_x(v)\cdot y = -\langle{}\nu, z\rangle{}\langle{}\nabla^{\mathbb{R}^d}_v \nu, y\rangle{} = \langle{}II_x(v,y) , \nu\rangle{} = \langle{}II_x(v, y), z\rangle.
$$
The second equality follows from Weingarten's formula (relating the second fundamental form to the shape operator) and the last from the fact that $II_x(v, y)$ is perpendicular to $M$, thus in dot producting against $z$ we just pick up a perpendicular component, i.e. $\langle{}\nu, z\rangle{}\nu$.
Putting it all together gives that $D^2_M w(x) = D^2_M u(x) - \langle{}II_x(\cdot, \cdot), z\rangle{}$. The minus here comes from the fact that way above, remember we're subtracting off $\text{proj}\nabla z^\text{tan}$. |
H: Formal way of proving that the ring $2 \Bbb Z_{10}$ has an additive inverse
I'm trying to find a formal way of proving that the ring $2 \Bbb Z_{10}$ has an additive inverse. I understand that 8+2 = 0 mod 10 and 6+4 = 0 mod 10, etc. Is there a way to formally prove this is true?
AI: Since $ \mathbb{Z}_{10} $ is a ring it is a group under an operation $+$ and every element has an inverse. Let $z \in \mathbb{Z}_{10} $, then there exists $-z$ such that $ z - z = 0 $. Since $ \mathbb{Z}_{10} $ is a ring then $ 2 (z - z) = 2 (0) = 0$ and from the distributive property of a ring it follows that $2z - 2z = 0$ for all $2z \in 2 \mathbb{Z}_{10} $ where $z \in \mathbb{Z}_{10}$ because of the second operation of the ring. |
H: How can a subset be undecidable?
A subset of a set can have an undecidable member relation. Though how can you determine if $A$ is actually a subset of $B$ if the member relation of $A$ is not decidable?
That feels contradictory because the definition of the subset relation uses the member relation: “If all the members of set $A$ are also members of set $B$, then $A$ is a subset of $B$”
AI: Just because an implication is decidable does not mean that its premise or consequent are decidable.
For example, we know that $\sf ZFC$ proves that the $V=L$ implies the Continuum Hypothesis. Consider the sets $A=\{x\mid x=0\land V=L\}$ and $B=\{x\mid x=0\land\sf CH\}$. We can prove that $A\subseteq B$, but we cannot prove that either sets is non-empty, starting from just $\sf ZFC$.
Similarly, if you want to understand sets of natural numbers, then considering $\varphi(x)$ and $\psi(y)$, both in the language of arithmetic, such that both sets defined by the formulas are undecidable, but $\forall x(\varphi(x)\to\psi(x))$ will give you an example of two sets which are undecidable, but one is included in the other. For example, the set of all machines that halt with input $0$ or $1$, and the set of all machines that halt with input $0$.
We can concoct simple examples by taking $\psi$ to define a very simple set, e.g. $\{2^n\mid n\in\Bbb N\}$, or the set of primes, or any other set with a nice simple recursive enumeration, and then considering any undecidable set $A$, we can take the $n$th element of our nice set in its enumeration, for $n\in A$. This has the nice benefit of giving us "an explicit feeling about uncountably many sets", whereas the first method only applied to countably many sets.
Of course, we can improve upon the first method by introducing oracles, but it is still useful since it lets us see how this may happen even when neither sets is decidable themselves. |
H: Is 3D space just a surface with an infinite number of handles?
I have recently come across the concept of a genus, and I was wondering, is 3D space mathematically equivalent to a surface with an infinite number of handles? I ask this because I asked a question a few weeks ago on whether a graph can be 'non-planar' in 3D, or more clearly, whether a graph exists, that when drawn in 3D still has an edge crossing which cannot be avoided. Clearly no such graph exists, but, if the answer to my above question is yes, then I can prove so by saying:
Since every graph has a genus, let us call it g, this means that the graph can be drawn, without edge-crossings, on every surface with g or more handles. This would effectively mean that every graph can be drawn, without edge-crossings, on a surface with an infinite number of handles, i.e., 3D space. Hence no graph exists which cannot be drawn without edge-crossings in 3D space.
The more I look at this question, the less confident I become of being correct, but I'm extremely curious.
Thanks :)
AI: One argument you can make along those lines, if you really wanted to, is that for any $n$, you can embed an $n$-holed torus in $\mathbb R^3$.
So if you want to draw a graph in $\mathbb R^3$, you can draw it on an $n$-holed torus (for sufficiently large $n$), embed the torus in $\mathbb R^3$, and get a drawing of the graph in $\mathbb R^3$ as a result.
What's missing is the converse: there are objects you can embed in $\mathbb R^3$ but not on any surface. But none of those objects are graphs. |
H: classify stable and unstable equilibrium points for differential equation $\frac{dx}{dt} = x(\lambda -x)(\lambda + x)$
I am doing exercise to find equilibrium points and classify them as stable/unstable for the following differential equation:
$\frac{dx}{dt} = x(\lambda -x)(\lambda + x)$
With $\frac{dx}{dt} = 0$, I can find the equilibrium points $x = 0, x = \lambda, x = -\lambda$.
However, I have no idea how to infer the sign of $\frac{dx}{dt}\Bigr|_{x=0}, \frac{dx}{dt}\Bigr|_{x=\lambda}, \frac{dx}{dt}\Bigr|_{x=-\lambda}$ in order to classify them as stable or unstable.
Could you please provide some directions so that I can continue to work on it?
AI: It is easy to see that the equilibria of
$\dot x = x(\lambda - x)(\lambda + x) \tag 1$
occur at
$x = -\lambda, 0, \lambda, \tag 2$
since these are the zeroes of the cubic polynomial
$x(\lambda - x)(\lambda + x) = x(\lambda^2 - x ^2) = \lambda^2x - x^3. \tag 3$
Note that we cannot "infer the sign" of $\dot x$ at any of these three $x$-values, since $\dot x =0$ has no sign; we can, however, find the sign of $d\dot x /dx$ at each of the critical values $0, \pm \lambda$. Indeed, (3) yields
$\dfrac{d\dot x}{dx}(x) = \lambda^2 - 3x^2; \tag 4$
we see that, for $\lambda \ne 0$,
$\dfrac{d\dot x}{dx}(\pm \lambda) = -2\lambda^2 < 0, \tag 5$
but
$\dfrac{d\dot x}{dx}(0) = \lambda^2 > 0; \tag 6$
thus $x = \pm \lambda$ are stable, attractors, and $x = 0$ is an unstable zero, a repellor.
In the event that $\lambda = 0$, (1) reduces to
$\dot x = -x^3, \tag 7$
and (4) becomes
$\dfrac{d\dot x}{dx}(x) = -3x^2; \tag 8$
the equation now has one critical point at $0$, where
$\dfrac{d\dot x}{dx}(0) = 0; \tag 9$
the stability thus cannot be decided via the derivative of $\dot x$; we can nevertheless infer that $0$ is stable since $\dot x > 0$ to its left, and $\dot x < 0$ to its right, so points on either side flow towards $0$ under the action of $\dot x$. |
H: Finding a joint distribution given a marginal and conditional distribution
My question is given a marginal distribution $p_X(x)$ and conditional distribution $p_{X|Y}(x|y)$, am I guaranteed to be able to find a joint distribution $p_{X,Y}(x,y)$.
In almost every form of this question I have seen asked or discussed, the questions starts given $p_X(x)$ and $p_{Y|X}(y|x)$. I understand in this case, one simply must multiply these two, i.e. $p_{X,Y}(x,y)=p_X(x)\cdot p_{Y|X}(y|x)$.
In my case, I am considering the following
$$p_X(x)=\int_Y p_Y(y)\cdot p_{X|Y}(x|y)dy=\int_Y p_{X,Y}(x,y)dy$$
This in my eyes boils down to, am I guaranteed to be able to find a $p_Y(y)$ that is a valid marginal distribution to give me the above equalities.
AI: This paper talks about compatibility of conditional distributions. In particular, there is a quote in Section 10 that is relevant. I am attaching a picture.
This other paper that is referenced in the above quote talks about uniqueness of $p_Y$ in the context that you are considering, but does not seem to address existence. |
H: What is this integral with the reciprocal of sine?
$$f(z) = \int\limits_{-\infty}^{\infty} \frac{x}{\sin(x z)} dx$$
What is the function $f(z)$? This doesn't converge for real $z$ but when $z$ is purely imaginary then $\sin$ becomes $\sinh$. So for example $f(i)=-\frac{i}{2}\pi^2$. But that's as far as I got.
AI: For $z \in \mathbb{C}$, let $f(z)$ be defined as follows:
\begin{equation}
f(z)=2\int\limits_{0}^{+\infty} \frac{x}{\sin(zx)}\,\mathrm{d}x
\end{equation}
Using the complex exponential definition of $\sin(zx)$:
\begin{equation}
f(z)=4i\int\limits_{0}^{+\infty} \frac{x}{e^{izx}-e^{-izx}}\,\mathrm{d}x
\end{equation}
\begin{equation}
f(z)=4i\int\limits_{0}^{+\infty} \frac{xe^{-izx}}{1-e^{-2izx}}\,\mathrm{d}x
\end{equation}
Using the geometric series for $e^{-2izx}$, you get the following:
\begin{equation}
f(z)=4i\int\limits_{0}^{+\infty} xe^{-izx}\sum_{k=0}^{+\infty}e^{-2izxk}\,\mathrm{d}x
\end{equation}
\begin{equation}
f(z)=4i\sum_{k=0}^{+\infty}\int\limits_{0}^{+\infty} xe^{-izx(1+2k)}\,\mathrm{d}x
\end{equation}
With the substitution $s=izx(1+2k)$, you get that:
\begin{equation}
f(z)=\frac{4i}{z^{2}}\sum_{k=0}^{+\infty}\frac{1}{(1+2k)^{2}}\int\limits_{0}^{+\infty} se^{-s}\,\mathrm{d}s
\end{equation}
The integral is just $\Gamma(2)$, thus:
\begin{equation}
f(z)=\frac{4i}{z^{2}}\sum_{k=0}^{+\infty}\frac{1}{(1+2k)^{2}}
\end{equation}
The last sum is known to converge to $\pi^{2}/8$, so you can conclude that:
\begin{equation}
\int\limits_{-\infty}^{+\infty} \frac{x}{\sin(zx)}\,\mathrm{d}x= \frac{i\pi^{2}}{2z^{2}}
\end{equation}
From the result, one can indeed see that $f(i)=-i\pi^{2}/2$. |
H: Show that $\int_{E_n} f d \mu \to \int f d \mu$, when $n \to \infty$
Given a measure space $(\Omega, \mathcal{F}, \mu)$ and $f : \Omega \to \mathbb{R}$ integrable. Consider
$$E_n = \left\{x \in \Omega : |f(x)| \geq \frac{1}{n} \right\}, n \geq 1.$$
I have shown that $\mu(E_n) < \infty$, for all $n \geq 1$. However, I can't proof what is being asked using this first result.
AI: We first suppose $f\geqslant 0$ a.e.
$$ \int_{E_n}fd\mu=\int_{\Omega} f\chi_{E_n}d\mu $$
But $f\chi_{E_n}\leqslant f\chi_{E_{n+1}}$ a.e because $f\geqslant 0$ a.e and $|f\chi_{E_n}|\leqslant f$ which is integrable. Using the monotone convergence theorem, we have
$$ \lim\limits_{n\rightarrow +\infty}\int_{E_n}fd\mu=\int_{\Omega}fd\mu $$
Now in the general case, let $f^+=\max(0,f)$ and $f^-=\max(0,-f)$, then $f^+,f^-\geqslant 0$ a.e, $f^+-f^-=f$ a.e and $|f^+|,|f^-|\leqslant|f|$ a.e so that $f^+$ and $f^-$ are integrable. Using what said above, we have
$$ \int_{E_n}fd\mu=\int_{E_n}f^+d\mu-\int_{E_n}f^-d\mu \underset{n\rightarrow +\infty}{\longrightarrow }\int_{\Omega}f^+d\mu-\int_{\Omega}f^-d\mu=\int_{\Omega}fd\mu $$ |
H: Suppose $U \sim Unif(0,1)$ and $Z \sim Unif(U,3+U)$. How can I find the pdf for $U + Z$?
Suppose $U \sim Unif(0,1)$ and $Z\mid U \sim Unif(U,3+U)$. I would like to find the pdf for $U + Z$, which in my process on $Z$ is a continuation of $U$. Is there a straightforward way to derive this?
AI: Looks like we have another uniform random variate, call it $V$ such that:
$$V \sim \text{Unif}(0,3)$$ and $$Z \sim U \star V \quad \text{(convolution)}$$ so that $$Z= U+V.$$
So now, $$U+Z = 2U + Z.$$
Let $$W=2U$$ so $$W \sim \text{Unif}(0,2).$$
$$U+Z = W+V \sim \text{Unif}(0,2) \star \text{Unif}(0,3).$$
The distribution will be a trapezoid:
$$ \left\{ \begin{array}{ll}
\frac{y}{6}, & 0 < y \le 2 \\
\frac{1}{3}, & 2<y\le 3 \\
\frac{5-y}{6}, & 3<y<5
\end{array} \right.$$ |
H: Prove that $u(x,t)\leq 6t+|x|^2$ for all $(x,t)\in U_T$. Here $U_T=U\times(0,T]$
let $U$ be the unit ball in $R^3$. Suppose $u$ solves the heat equation
$$
\left\{\begin{matrix}
u_t-\Delta = 0 \text{ in } U_T\\
u(x,t)=6t \text{ when } |x|=1\\
u(x,0)=g(x)
\end{matrix}\right.
$$
Suppose that $g\leq0$. Prove that $u(x,t)\leq 6t+|x|^2$ for all $(x,t)\in U_T$. Here $U_T=U\times(0,T]$
I know that for a heat equation, there is the maximum principle.
That is $$\max\limits_{\bar{U}_T}u=\max\limits_{\Gamma_T}u$$
Where $\Gamma_T=\bar{U}_T-U_T$
But I really don't see how to apply this theorem in the above context.
Appreciate your help
AI: Define
\begin{align}
U(x,t) = |x|^2+6t-u(x, t).
\end{align}
Observe that
\begin{align}
\partial_t U-\Delta U = 6-\partial_t u-(6-\Delta u) = 0
\end{align}
since $\Delta |x|^2= 2+2+2 = 6$. Note that $U(x, t) = |x|^2 = 1$ when $|x| =1$ and $U(x, 0) = |x|^2-u(x, 0)= |x|^2-g(x) \ge 0$.
Finally, by the minimum principle, we have that
\begin{align}
U(x, t) \ge 0 \ \ \implies \ \ |x|^2+6t\ge u(x, t).
\end{align} |
H: Is the volume of a cube the greatest among rectangular-faced shapes of the same perimeter?
My child's teacher raised a quesion in class for students who are interested to prove. The teacher says that the volume of a cube is the greatest among rectangular-faced shapes of the same perimeter and asks his students to prove this proposition.
I considered the relationship between the length of the sides of a cube and the lengths of the sides of rectangular-faced shapes in different situation. But when the calculations came down to polynomials, I couldn't proceed due to the uncertainty of the variables in the polynomials.
Can anyone please find a good way to prove the above proposition? Or is there already a proof? Thank you for your help!
AI: If you mean by "perimeter" the sum of the edges, then yes, the cube is the maximal rectangular parallelepiped among those with the same "perimeter".
Let the edges have lengths $(a,b,c)$.
Then the volume is $V=abc$ and the "perimeter" is $p=4(a+b+c).$
We can maximize volume while constraining the sum of the edges using Lagrange multipliers:
$$\begin{aligned}
L &= abc-\lambda \left(a+b +c-\frac{p}{4}\right)\\
0&=\frac{\partial L}{\partial a} = bc - \lambda\\
0&=\frac{\partial L}{\partial b} = ac - \lambda\\
0&=\frac{\partial L}{\partial c} = ab - \lambda\\
\end{aligned}$$
so that
$$bc=ac=ab$$ and
$$a=b=c.$$ |
H: Prove that $\mathrm{Ker}(T - \lambda I_V)^n = {0}$
I´m having trouble proving this statement, I already tried induction but I failed miserably.
Let $V$ a $K$-vectorial space and $T: V\to V$ endomorphism. Let $\lambda\in K$ that it's not a eigenvalue of $T$. Prove that $\mathrm{Ker}(T - \lambda I_V)^{n} = \{0\}$ for all $n\in\mathbb{N}$.
Does anyone know how to prove it?
Thanks beforehand
AI: Given, $V$ a $K$-vectorial space and $T: V\to V$ endomorphism. Whenever, $\mathrm{Ker}(T - \lambda I_V)^{n} \neq \{0\}$ for some $n\in\mathbb{N}$, it means that geometric multiplicity of $\lambda $ is $\ge 1 $ [because, $\mathrm{Ker}(T - \lambda I_V)^{n} \neq \{0\}$ for some $n\in\mathbb{N}$ $\implies$ $\mathrm{Ker}(T - \lambda I_V) \neq \{0\}$ $\implies $ $\dim(\mathrm{Ker}(T - \lambda I_V)) \ge 1 $, and Geometric multiplicity of $\lambda = \dim(\mathrm{Ker}(T - \lambda I_V)) $ ]
that means $\lambda $ is an eigenvalue of $A$.
So, whenever $\lambda $ is not an eigenvalue of $A$, $\mathrm{Ker}(T - \lambda I_V)^{n} = \{0\}$ for all $n\in\mathbb{N}$. |
H: Formula for unique orderings of N items from k bins sorted contiguously.
I'm looking for a formula that will tell me how many unique orderings I can get if I pick out $N$ items from $k$ bins that those items are divided into in continuous segments. For example, for $N=4$ and $k=2$, the $2$ bins are: $\{0,1\}$ and $\{2,3\}$. Each time I pick an item from the front of each bin (without replacement). So this gives me $6$ unique total orderings $\{0,2,1,3\}, \{2,0,1,3\}, \{0,2,1,3\}, \{0,2,3,1\}, \{2,0,3,1\}, \{0,1,2,3\}$ and $\{2,3,0,1\}$. Similarly for $N=8$ and $k=2$, the bins would be $\{0,1,2,3\}$ and $\{4,5,6,7\}$. With $k=1$, I should only get one single order and with $k=N$, I should get $N!$ orders.
I'm looking for a formula I can use to calculate the number of possible permutations for any given $N$ and $k$. I've used powers of $2$ here but I would like the formula to work with any value such as $N=7, k=4$ say. $k$ will only ever be a power of $2$.
AI: As JMoravitz indicated in his comment, all that matters is the sequence in which you pick the bins. The first time you pick the bin, you get the fits number in it. The second time you get the second number, and so on.
In your first example, the sequence $1122$ would mean pick the first bin twice, then then second bin twice, so it would result in the sequence $0123$. Similarly, $1221$ would give the sequence $0231$.
If there are $q$ bins with $a$ items each and $r$ bins with $b$ items each, where $aq+br=N$, then the number of possible sequences is $$\frac {N!}{(a!)^q(b!)^r}$$
In the example you gave of $7$ items in $4$ bins, $3$ of the bins have $2$ items and $1$ bin has one item, so the number of sequences is $$\frac{7!}{(2!)^3(1!)^1}=630$$ |
H: Analysis on $y = x$ line for Equilibria Points
I am having trouble analyzing what happens when I set the parameters to $(A, B) = (0,0)$ for the phase portrait in reference to this graph below. The first phase portrait graph solutions look undefined, but I wanted to know what it meant terms of the ab graph where $x=a$ and $y=b$. I get that it is on the $a = b$ line, but if we plug in $(a,b) = (3,3)$, then it looks like a source.
In this instance the TRACE is the $a = b$ line, but the phase portrait looks different for when $(a,b)$ is $(0,0)$ versus when $\{a,b > 0 \text{ and } a = b \text{ for } (a,b))\}$.
AI: We have the system
$$x' = ax + by \\ y'= -x - y$$
The critical point is $(x, y) = (0,0)$
For $a = b = 0$, the eigenvalues of the Jacobian at the critical point are $(-1, 0)$.
Note, the general eigenvalues are
$$\lambda_{1,2} = \dfrac{1}{2} \left(-\sqrt{a^2+2 a-4 b+1}+a-1\right),\dfrac{1}{2} \left(\sqrt{a^2+2 a-4 b+1}+a-1\right)$$
If one of the eigenvalues is zero and the other is negative, then the
origin is stable but not asymptotically stable.
We also have a $y'$nullcline as $y = -x$.
Drawing this phase portrait
If we solve the system $x' = 0, y' = -x-y$, we get
$$\begin{align}x(t) &= c_1 \\ y(t) &= c_2 e^{-t}-c_1 e^{-t} \left(e^t-1\right)\end{align}$$
For the case $a = b = 1$, we have
For the case $a, b > 0, a = 1, b = 5$, we have |
H: Ultrafilter with finite set
While I was working in some exercise about filters, a question came to my mind: let $X$ a set and $F\subseteq X$ a non empty finite set. How many ultrafilters $U$ there are such that $F\in U$? I think that there exist a unique ultrafilter that contains $F$ but I can't see why or how to prove but my intuition says that it is true. Am I wrong?
My work: take $F$ a non empty and finite subset of $X$. Suposse that there exist two different ultrafilters $U$ and $V$ such that $F\in U$ and $F\in V$. Since $U\neq V$, w.l.g., we can take $A\in U\setminus V$. Then $A\notin V$ but $V$ is an ultrafilter and therefore $X\setminus A\in V$. Moreover, $F\cap (X\setminus A)\neq\emptyset$ and $F\cap A\neq\emptyset$. But then, from here, what can I do? If my intuition is wrong, then, is there a bound over the number of ultrafilters that contains a fixed finite set? Thanks.
AI: Suppose that $F$ is a non-empty finite subset of $X$, and let $\mathscr{U}$ be an ultrafilter on $X$ such that $F\in\mathscr{U}$. Let $F=\{x_1,\ldots,x_n\}$, and for $k=1,\ldots,n$ let $A_k=X\setminus\{x_k\}$. Suppose that $\{x_k\}\notin\mathscr{U}$ for each $k\in\{1,\ldots,n\}$; then for each $k\in\{1,\ldots,n\}$ we must have $A_k\in\mathscr{U}$, and therefore $\bigcap_{k=1}^nA_k\in\mathscr{U}$. But $\bigcap_{k=1}^nA_k=X\setminus F$, which is certainly not in $\mathscr{U}$, since $F\in\mathscr{U}$. This contradiction shows that there must be some $k\in\{1,\ldots,n\}$ such that $\{x_k\}\in\mathscr{U}$, and in that case $\mathscr{U}$ is the fixed (or principal) ultrafilter over $x_k$:
$$\mathscr{U}=\{U\subseteq X:x_k\in U\}\;.$$
Conversely, if $x\in F$, then $\mathscr{U}_x=\{U\subseteq X:x\in U\}$ is an ultrafilter on $X$, and clearly $F\in\mathscr{U}$. Thus, the ultrafilters on $X$ that contain $F$ are precisely the fixed ultrafilters over the elements of $F$, and there are therefore $|F|$ of them. |
H: Proof with invertible matrices
Let $A,B,C$ be matrix of the same size, and suppose A is invertible, Prove that
$(A-B)C=BA^{-1}$ then $C(A-B)=A^{-1}B$
I tried to prove it as following.
$(A-B)C=BA^{-1}$ implies
$AC-BC=BA^{-1}$ so
$ACA-BCA=B$ taking $A^{-1}$
$CA-A^{-1}BCA=A^{-1}B$
Any hint will be appreciated
AI: Assume that $(A-B)C = BA^{-1}$. Then $A-B$ is invertible, and its inverse is $C+A^{-1}$. You can check:
$$ (A-B)(A^{-1}+C) = \mathrm{Id} + AC -BA^{-1} -BC = \mathrm{Id} + \bigg( (A-B)C - BA^{-1} \bigg) $$
But the assumption implies that the part in parentheses is zero, so this product is just the identity matrix. Inverse matrices always commute. By that I mean $X X^{-1} = X^{-1}X$ whenever $X$ is invertible. So you can re-arrange the product above in the opposite order:
$$ (A^{-1} + C)(A-B) = \mathrm{Id} $$
If you expand this out and re-arrange the terms, you'll get the expression you want:
$$ C(A-B) = A^{-1}B $$
How I came up with this solution:
The first thing I thought to do was a "change of variables". I thought let's give a name to $A-B$ and treat it as one single matrix. So I called it $X$:
$$ X := A-B $$
So then I re-wrote the original equation $(A-B)C = BA^{-1}$ using $X$ instead of $B$:
$$ XC = (A-X)A^{-1} = \mathrm{Id} - XA^{-1} $$
If you move the negative term to the other side and factor, you get
$$ X(C+A^{-1}) = \mathrm{Id} $$ |
H: Is there a closed form of $\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)!!}$?
This may be an impossible problem. But I imagine it's worth asking still.
What is the closed form of the sum:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)!!}$$
Perhaps there isn't a closed form. Double factorials are WAY out of my comfort zone so any tips would be appreciated. Thanks
AI: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^n \over \pars{4n + 1}!!}} =
\sum_{n = 0}^{\infty}{\pars{-1}^n \over \prod_{k = 0}^{2n}\pars{2k + 1}} =
\sum_{n = 0}^{\infty}{\pars{-1}^n \over
2^{2n + 1}\prod_{k = 0}^{2n}\pars{k + 1/2}}
\\[5mm] = &\
\sum_{n = 0}^{\infty}{\pars{-1}^n \over
2^{2n + 1}\pars{1/2}^{\overline{2n + 1}}} =
\sum_{n = 0}^{\infty}{\pars{-1}^n \over
2^{2n + 1}\bracks{\Gamma\pars{2n + 3/2}/\Gamma\pars{1/2}}}
\\[5mm] = &\
\sum_{n = 0}^{\infty}{\pars{-1}^n \over 2^{2n + 1}}\,{1 \over \pars{2n}!}\,
{\Gamma\pars{2n + 1}\Gamma\pars{1/2} \over \Gamma\pars{2n + 3/2}}
\\[5mm] = &\
{1 \over 2}\sum_{n = 0}^{\infty}{\pars{-1/4}^n \over \pars{2n}!}\,
\int_{0}^{1}t^{2n}\pars{1 - t}^{-1/2}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{\sum_{n = 0}^{\infty}{\pars{-t^{2}/4}^n \over \pars{2n}!}}\pars{1 - t}^{-1/2}\,\dd t
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}
\bracks{\sum_{n = 0}^{\infty}{\pars{\ic t/2}^{2n} \over \pars{2n}!}}
\pars{1 - t}^{-1/2}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}
\bracks{\sum_{n = 0}^{\infty}{\pars{\ic t/2}^{n} \over n!}\,{1 + \pars{-1}^{n} \over 2}}
\pars{1 - t}^{-1/2}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}
\bracks{\Re\sum_{n = 0}^{\infty}{\pars{\ic t/2}^{n} \over n!}}
\pars{1 - t}^{-1/2}\,\dd t =
{1 \over 2}\int_{0}^{1}{\cos\pars{t/2} \over \root{1 - t}}\,\dd t
\\[5mm] & =
\bbx{\root{\pi}\bracks{\mrm{C}\pars{1 \over \root{\pi}}\cos\pars{1 \over 2} +
\mrm{S}\pars{1 \over \root{\pi}}\sin\pars{1 \over 2}}}
\\[5mm] & \approx 0.9344
\end{align}
$\ds{\mrm{C}\ \mbox{and}\ \mrm{S}}$ are
FresnelC Function
and FresnelS Function, respectively.
In the last integral, the change
$\ds{\pars{\root{1 - t} = x \implies t = 1 - x^{2}}}$ yields inmediately the final solution. |
H: Proving an equality associated with symmetric positive definite matrices
Let $Q$ be an $n \times n$ symmetric positive definite matrix, $\vec{a}, \vec{b} \in \Bbb R^n$ be two random vectors. Prove that $a^TQ(ba^T-ab^T)Qb$ is non-positive.
Since $Q$ has $n$ independent eigenvectors with positive eigenvalues, I've tried expressing $\vec{a}$ and $\vec{b}$ as linear combinations of those eigenvectors, but it didn't work out. Really appreciate any help.
AI: By Cauchy-Schwarz $$|\langle \sqrt{Q}b,\sqrt{Q}a\rangle|^2=|a^TQb|^2\le \langle \sqrt{Q}b,\sqrt{Q}b\rangle \langle \sqrt{Q}a,\sqrt{Q}a\rangle=b^TQba^TQa$$ |
H: Finding modulus of a complex number
The question is as follows:
if $|a|=|b|=|c|=|b+c-a|=1$ where $a$,$b$,$c$ are distinct complex numbers , find $|b+c|$.
My attempt: By observation that $b=i$, $c=-i$, $a=1$ satisfy the following conditions
thus $|b+c| =|i-i|=0$
I realise that this is not a generalised method. Is there a better metod to solve this problem?
Any help will be appreciated.
AI: There are infinitely many solutions that satisfy that conditions.
Some of these will be in the form $b=-c$ and $|a|=1, a\in\Bbb C$. In this case, $|b+c|=0$ as you have discovered.
However, many more can be found. Define \begin{align}
a&=\cos\alpha+i\sin\alpha\\
b&=\cos\beta+i\sin\beta\\
c&=\cos\gamma+i\sin\gamma
\end{align}
In order to satisfy $|b+c-a|=1$, we can have any set of $\alpha,\beta,\gamma$ such that $$\sqrt{(\cos\beta+\cos\gamma-\cos\alpha)^2+(\sin\beta+\sin\gamma-\sin\alpha)^2}=1.$$
Sure, it may be possible to parameterize your solution, but I suspect that there is more to the problem that has been left out. This is probably as far as we can go. |
H: Prove that $2\cos^2(x^3+x) = 2^x + 2^{-x}$ has exactly one solution
I've been stuck on this question for some time now:
Show that there is exactly one value of $x$ which satisfies the equation $2\cos^2(x^3+x) = 2^x + 2^{-x}$.
Now this is obviously intuitively correct — I've modelled the equation with a function $f(x) = 2\cos^2(x^3+x) - 2^x - 2^{-x}$, and simply looking at the function reveals that the range of $2\cos^2(x^3+x)$ is $[0,2]$ and the range of $-2^x - 2^{-x}$ is $(-\infty, -2]$. That would imply that $f(x)$ ranges from $(-\infty, 0]$.
What I'm having trouble with is writing down a formal proof based on these intuitive ideas. I have identified $x=0$ as the sole solution; furthermore, I have tried to differentiate the function to find a maximum (a futile endeavor, since the equation becomes messy really quickly).
A hint would be appreciated.
AI: Because of AM-GM you have
$$2^x + 2^{-x}\geq 2\sqrt{2^x \cdot 2^{-x}}=2$$
where equality holds if and only if $2^x = 2^{-x} \Leftrightarrow x=0$.
Hence, we have
$$2\cos^2{(x^3+x)}\leq 2 < 2^x+2^{-x} \text{ for all } x \in \mathbb{R}\setminus\{0\}$$
Since $x=0$ satisfies the equation, it is the only solution. |
H: yes/ No :Is $f$ is uniformly continious .?
let $(X,T)$ be the subspace of $\mathbb{R}$ given by $X= [0,1] \cup [2,4] $. Define $f :(X, T) \rightarrow \mathbb{R}$ be given by
$f(x)= \begin{cases} 1 , \text {if x} \in [0,1] \\ 2 \ \text {if x} \in [2,4] \end{cases}$.
Is $f$ is uniformly continious ?
My attempt : I know that $f$ is continious since $f^{-1}( 2,4)= 2$ is open in $(X,T)$
Im confused that it is U.C or not ?
AI: A real-valued function $f$ on a subspace $X$ of $\Bbb R$ is uniformly continuous if for each $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. For your function this is the case, because for any $\epsilon>0$ you can take $\delta=1$: if $|x-y|<1$, then, as you can easily check, $f(x)=f(y)$ (why?), so $|f(x)-f(y)|=0<\epsilon$. |
H: The expected value of the Ito integral of functions in $\mathcal{V}$ is zero, $\mathbb{E}[\int_S^T f dB_t] = 0$ for $f\in\mathcal{V}$
In Oksendal's Stochastic Differential Equations, the set $\mathcal{V}(S,T)$ is defined to be all functions $f:[0,\infty)\times \Omega \to\mathbb{R}$ such that
$(t,\omega)\to f(t,\omega)$ is $\mathcal{B}([0,\infty))\otimes \mathcal{F}$-measurable
$f(t,\omega)$ is $\mathcal{F}_t-$adapted
$\mathbb{E}[\int_S^T f^2(t,\omega)dt]<\infty$
A function $\phi\in \mathcal{V}(S,T)$ is called elementary if $\phi$ has the form
$$\phi(t,\omega) = \sum_{j\geq 0}e_j(\omega)\chi_{[t_j, t_{j+1})}(t) $$
Then $e_j$ must be $\mathcal{F}_{t_j}$ measurable and thus independent from $B_{t_{j+1}}-B_{t_j}$ where $\{B_t\}$ is Brownian motion centered anywhere. The Ito integral is then defined for elementary functions to be $$\int_S^T \phi(t,\omega)dB_t(\omega) = \sum_{j\geq 0}e_j(\omega)[B_{t_{j+1}}-B_{t_{j}}](\omega)$$
Since $e_j$ is independent from $B_{t_{j+1}}-B_{t_{j}}$ then we have that
$$\begin{align*}\mathbb{E}\Big[\int_S^T \phi(t,\omega)dB_t(\omega)\Big] &= \sum_{j\geq 0}\mathbb{E}[e_j(\omega)[B_{t_{j+1}}-B_{t_{j}}](\omega)]\\
&= \sum_{j\geq 0}\mathbb{E}[e_j]\mathbb{E}[B_{t_{j+1}}-B_{t_{j}}] \\
&= \sum_{j\geq 0}\mathbb{E}[e_j]\cdot 0 = 0\end{align*}$$
Where we use that $B_{t_{j+1}}-B_{t_j}\sim\mathcal{N}(0, t_{j+1}-t_j)$.
Edit: Ending up answering my own question and keeping this up to avoid confusion when people look up this post Itō Integral has expectation zero when they are asking about a property in Oksendal.
AI: The key observation is in the assumption that $\phi\in\mathcal{V(S,T)}$ because this ensures that $\phi(t,\omega) = \sum_{j\geq} e_j(\omega)\chi_{[t_j, t_{j+1})}(t)$ where $e_j$ is $\mathcal{F}_{t_j}-$measurable and thus independent of $B_{t_{j+1}}-B_{t_j}$ which is a normally distributed random variable with mean $0$. Further because $B_{t_{j+1}}-B_{t_j}$ is independent of $\mathcal{F}_{t_j}$ we get this result. These are the two fact that caused me enough problems that I tried posting it here. |
H: Help with Baby Rudin Theorem 5.5 Proof
I have a question about the last sentence. I know $s \rightarrow y$ as $t \rightarrow x$; hence $u(t) \rightarrow 0$ as $t \rightarrow x$. But I'm confused that why the term $v(s)$ disappears. That is, why $v(s) \rightarrow 0$ as $t\rightarrow x$?
I know $s \rightarrow y$ as $t \rightarrow x$ and $v(s) \rightarrow 0$ as $s \rightarrow y$. But does that mean $v(s) \rightarrow 0$ as $ t \rightarrow x$?
I would appreciate if you could explain in details. Thank you!
AI: Look at the line after (5), there you have that $s$ is selected to be $f(t)$. We can do that, because the only restriction we have for $s$ is $s \in I$ and $f(t) \in I$ (because the range of $f$ is contained in $I$). We use that $s = f(t)$ to say that $g(s) = h(t)$ but we still have that $v(s) = v(f(t))$. Now as $t \rightarrow x$ we have that $f(t) \rightarrow f(x)$ by continuity of $f$ but $f(x) = y$ (look at the first line of the proof) then we have that $s \rightarrow y$ as $t \rightarrow x$ which means $v(s) \rightarrow 0$ as $t \rightarrow x$. |
H: Is there a tighter bound on $\prod_{i=1}^I x_i$?
If $x_i\in[0,1]$ for all $i=1,2,\cdots,I$, then we have a simple bound on $\prod_{i=1}^Ix_i$, i.e.,
$$\prod_{i=1}^I x_i \le \sum_{i=1}^I x_i.$$
I wonder if there is a tighter bound than the one above?
AI: The tightest you can get bounding by one element is $\bar{x} = \min\{x_i : i \in \{1, \ldots, I\}\}$, since
a multiplication by a number less than 1 will lower the product. Hence,
$$\prod_{i=1}^I x_i \leq \bar{x} $$
But of course, you can always bound multiplying by one element less
$$\prod_{i=1}^I x_i \leq \prod_{i=2}^I x_i \leq \cdots$$ |
H: Why can't you lose a chess game in which you can make $2$ legal moves at once?
So here is the Problem :-
Consider a normal chess game in an $8*8$ chessboard such that every player makes $2$ legal moves at once alternatively . Now imagine that you was asked to play with Magnus Carlsen .Then Prove that it's impossible for Magnus Carlsen to make you lose, or atleast can make you draw.
I was actually stumped when I first saw this . Also I tried thinking many normal chessgames and tried to understand what type of answer this question can take . From here I can say that a check on the $1st$ move made by any player is actually a checkmate . Other than that I have no idea, can anyone help ?
Edit :- I forgot to add another thing . It's given that I will be white and Magnus Carlsen will be black .
AI: While some of the details of the rules may still be ambiguous in boundary situations, it is clear that white can avoid a loss by opening with
♘b1-c3,♘c3-b1
or
♘g1-f3,♘f3-g1
More precisely, if either of these no-ops in fact leads to a position where black can force a win, then white could force a win by playing by black's strategy mirrored. |
H: How $|f(a) |= e^{-i\alpha}f(a) ?$
I have some confusion in maximum modulas theorem , my confusion marked in red box given below
How $|f(a) |= e^{-i\alpha}f(a) ?$
My thinking : here we have taken $\alpha = Arg (f(a))$
By definition of Arg we have
$arg(f(a))= -i \log \frac{f(a)}{|f(a)|}$
Now if u put the value $\alpha=arg(f(a))$
Then $|f(a)|= e^{-i \log \frac{f(a)}{|f(a)|}}f(a)$
Im confused that How $|f(a) |= e^{-i\alpha}f(a) ?$
AI: Note that for $z$ any complex number, $ z = |z|.e^{i\alpha} $, where $\alpha$ is the argument of $z$.
Then using $z = f(a), f(a) = |f(a)|.e^{i\alpha}$. Rearrange to get your answer. |
H: For which $n \in N$ is the following matrix invertible?
For which $n \in N$ is the following matrix invertible?
$$\left[\begin{array}{[c c c]}
10^{30}+5 & 10^{20}+4 & 10^{20}+6 \\
10^{4}+2 & 10^{8}+7 & 10^{10}+2n \\
10^{4}+8 & 10^{6}+4 & 10^{15}+9 \\
\end{array}\right]$$
My attempt:
For the matrix to be invertible, it must be non-singular. To compute the determinant, I split the large determinant into smaller determinants using the column addition property but it got too tedious to compute (and too lengthy to type out here :P)
The answer:
Replacing even numbers by zero and odd numbers by one, we have
$$|A| = \left| \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|$$
which is an odd number and hence $|A|$ can not be zero. Hence A is invertible for all $n \in N$.
I did not understand how all the odd numbers were simply replaced with 1 and the even numbers with 0. I would appreciate it if someone could provide a different answer or explain the given answer.
AI: Just expand the determinant using the first column. You see quickly that you get a sum of $3$ odd integers. Hence the determinant is an odd integer.
For example one term would be $[(10)^{30}+5] [((10)^{8}+7)((10)^{15}+9)-((10)^{10}+2n)((10)^{6}+4)]$ which is odd. |
H: Interpret convolutions with different arguments. For example, $f(t) * \delta(t-\alpha)=f(t-\alpha)$
I'm trying to learn signal processing and I don't know how to process this.
My textbook says that the pure time shift LTI system that goes $y(t) = x(t-t_0)$ has an impulse response $h(t) = \delta(t-t_0)$, which means that $f(t) * \delta(t-t_0) = f(t - t_0)$.
That's signal processing. It makes sense.
Mathematically, a convolution between signals $f(t)$ and $g(t)$ are defined as $f(t) * g(t) = \int_{-\infty}^{\infty} f(\tau)g(t-\tau) d\tau$. So, it requires that the arguments to $f$ and $g$ be the same. But it is often that the arguments are not the same, like in this case (i.e. $t$ vs. $t-t_0$).
How do I expand a convolution like $f(t) * \delta(t - t_0)$ into an integral, and prove this identity mathematically?
Here's what I tried.
Let $f_d (t) = f(t + t_0)$ so that $f(t) = f_d (t - t_0)$. Then, since $f_d (t - t_0) * \delta (t - t_0) = f_d (t - t_0)$, $f(t) * \delta (t - t_0) = f(t)$. (!!!)
Not the result I was expecting.
Fundamentally, I want to understand what is meant when engineers talk about convolutions that are not exactly of the form $(f * g)(t)$, but rather of $f(p(t)) * g(q(t)), p \ne q$.
My background is college calculus + intro differential equations.
AI: You do it by writing every function involved as a distinct function of $t$ only (as you did with $h$), and only then applying your formula for a convolution. Let me also write $f_{t_0} (t) := f(t-t_0)$. Then
\begin{align}
f(t) * \delta(t-t_0)=f(t)*h(t)
&= \int_{-\infty}^\infty f(\tau) h(t-\tau) d\tau
\\&= \int_{-\infty}^\infty f(\tau) \delta(t-\tau-t_0) d\tau
\\&= \int_{-\infty}^\infty f(\tau) \delta(t-(\tau+t_0)) d\tau
\\&= \int_{-\infty}^\infty f(\tau-t_0) \delta(t-\tau) d\tau
\\&= \int_{-\infty}^\infty f_{t_0}(\tau) \delta(t-\tau) d\tau
\\&=f_{t_0}(t)
\\&=f(t-t_0)\end{align}
I want to point out that there is a notation nightmare happening. (Nevermind that you introduce $x$ and $f$ to mean the same thing). The object on the left has $t$s that indicate that the whole expression is evaluated at some yet unspecified value $t$, but beware that if you just e.g. replaced $t=1$, $f(1)$ would be a number, maybe $6$, $\delta(1-t_0)$ only barely makes a little sense if at all, and who knows what $6*\delta(1-t_0)$ would be, probably not equal to $$\big(f(t) * \delta(t-t_0)\big)\big|_{t=1} = f(1-t_0).$$
This notation landmine explains why your attempt failed. If we now subscript by $z$ to mean translation by $z$, so that $f_z(t) = f(t-z), \ \delta_z(t) = \delta(t-z) $, etc. (in particular what you called $f_d$, I call $f_{-t_0}$ and $h = \delta_{t_0}$), then by applying the above correct result at the point marked ✅ for the function $f=g_{t_0}$,
$$ g(t-t_0)* \delta(t-t_0)=g_{t_0}(t) * \delta_{t_0}(t) \overset{✅}= g_{t_0}(t-t_0) = g(t-2t_0)$$
We see that in general, $g(t-t_0)* \delta(t-t_0) \neq g(t-t_0)$. This happened because you cannot treat the $t$ on the left as something that can be substituted with $t-t_0$.
Its unambiguous if one writes something like
$$ [f*\delta(\cdot-t_0)](t), [f*\delta(\cdot-t_0)](1), \text{etc}.$$
but there is a certain naturality to writing it in the way you did, which is OK if you "know the rules of the game". |
H: Finding $\iint_D \frac{10}{ \sqrt {x^2 + y^2}}dx\,dy$ for $D = \{(x,y) \mid 0 \leq x \leq 1, \sqrt{1-x^2} \leq y \leq x\}$
As I said title,
$$\iint_D \frac{10}{ \sqrt {x^2 + y^2}}\,dx\,dy$$ for $$D = \{(x,y) \mid 0 \leq x \leq 1, \sqrt{1-x^2} \leq y \leq x\}$$
I tried it using integration by substitution by $(x,y) = (r\cos\theta, r\sin\theta)$
Then Only just we left to find the $\iint_{D'}10 \,dr \,d\theta$ for the domain $D'$ whose variable $r$ and $\theta$
We can surely say $0\leq \theta \leq {\pi \over 4}$ But I couldn't find the range of the $r$.
The answer sheet said $1 \leq r \leq {1 \over \cos\theta}$. So Why does the range of the $r$ should that be?
Any help would be appreciated. Thanks.
AI: Putting polar coordinates to inequaility $\sqrt{1-x^2} \leqslant y$ gives exactly $1 \leqslant r $ and putting to $x \leqslant 1$ gives $ r \leqslant {1 \over cos\theta}$. By the way, restrictions on $\theta$ is obtained on same way and plot on $(r,\theta)$ is very nice. |
H: Why is $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] \le 4$
I am working on the following exercise:
Show that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $4$ over $\mathbb{Q}$ by showing that $1,\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent.
I have shown that $1,\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent, but if I am not mistaken this only shows that the degree of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ is at least $4$, i.e. $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] \ge 4$. I do not see how this should also prove that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] \le 4$. Could you explain that to me?
AI: That it has at most $4$ follows from the fact that $p(x)=(x^2-2)(x^3-3)$ has degree $4$ and splits over $\Bbb Q(\sqrt{2},\sqrt{3})$. So that's already obvious. (By standard results the minimal polynomial of this extension, whose degree equals $[\Bbb Q(\sqrt{2},\sqrt{3}): \Bbb Q]$, must thus be a divisor of $p$ and so $2$ or $4$). |
H: If $\omega$ vanishes in a surface, so do $d\omega$
It seens like an easy question, but I am newbie in differential forms.
Let $\omega$ a $C^\infty$ $r$-form in an open set $U\subset\Bbb{R}^n.$ If $\omega$ vanishes over the tangent vectors of a surface contained in $U$, so do $d\omega$.
First things first, I want to write down everything.
Let $U\subset\Bbb{R}^n$ open, and $\omega:U\rightarrow\mathcal{A}_r(\Bbb{R}^n)$ a $C^\infty$ differential form. So, I can write
$$\omega(x)(v_1,\dots,v_r)=\sum_{I}a_I(x)dx_I(v_1,\dots,v_r).$$
Here, the sum extends over all ascendent $r$-uples $I=\{i_1,\dots,i_r\}$ of integers less than $n$, $a_I:U\rightarrow\Bbb{R}$ is a $C^\infty$ function, and $dx_I=dx_{i_1}\wedge\dots\wedge dx_{i_r}$.
Let $M\subset U$ a surface, $\dim M=m$, and let $\varphi:V_0\subset\Bbb{R}^m\rightarrow V\subset M$ a parametrization in $M$, with $\varphi(u)=x\in V.$
I'm trying to right $\omega$ in $M$, but I'm confused, because the functions $a_I$ take vectors in $\Bbb{R}^n$, not in $\Bbb{R}^m$, and I want to write in the parametrization $\varphi.$
Ok, besides this problem, basically I need to prove something like:
$$\sum_{I}a_I(u)du_I(v_1,\dots,v_r)=0\implies \sum_{I}da_I(u)\wedge du_I(v_1,\dots,v_r).$$
AI: Let $S$ be that surface and $\iota : S\to U$ be the inclusion. The fact that $\omega$ vanishes over the tangent vectors of $S$ is the same as saying that $\iota^*\omega = 0$.
With that in mind, the proof is trivial: Since pullback $\iota^*$ commutes with exterior differentiation $d$,
$$ \iota^* (d\omega) = d (\iota^* \omega) = d(0) = 0.$$ |
H: If $n > 1$, there are no non-zero $*$-homomorphisms $M_n(\Bbb{C}) \to \Bbb{C}$
If $n > 1$, there are no non-zero $*$-homomorphisms $M_n(\Bbb{C}) \to \Bbb{C}$. A $*$-homomorphism is an algebra morphism $\varphi: M_n(\Bbb{C}) \to \Bbb{C}$ with $\varphi(\overline{A}^T) = \overline{\varphi(A)}$
I tried to show that every $*$-morphism must be zero: if $\varphi$ is such a $*$-morphism, then $E_{ij}^2 = 0$ for $i \neq j$ (here $E_{ij}$ is the matrix that is $1$ on position $(ij)$, $0$ elsewhere). Thus $\varphi(E_{ij}) = 0$ for $i \neq j$. Hence,
$$\varphi(A) = \sum_i a_{ii} \varphi(E_{ii})$$
so if I can show that $\varphi(E_{ii}) = 0$ I will be done. Unfortunately, I can't see why this should be true. I did not yet fully use the fact that it is a $*$-morphism. The only thing I can see is that $\varphi(A) \in \Bbb{R}$ for all $A$ since $\varphi(E_{ii})= \varphi(\overline{E_{ii}}^T) = \overline{\varphi(E_{ii}})$. If $\varphi$ is non-zero, then there is $A$ with $\varphi(A) \in \Bbb{R}\setminus \{0\}$. But then $\varphi(iA) = i\varphi(A) \notin \Bbb{R}$, a contradiction. Thus $\varphi=0$. Is this correct?
AI: We can prove that when $n\ge2$ and $\mathbb F$ is a field, the only mapping $\phi:M_n(\mathbb F)\to\mathbb F$ that is additive and multiplicative is zero. (So, we don't need $\phi$ to be an algebra morphism, not to say a $\ast$-homomorphism over $\mathbb C$.)
For any $A\in M_n(\mathbb F)$, we have $A=P_1DQ$ for some nonsingular matrices $P_1,Q$ and some diagonal matrix $D$. Let $P_2$ be any permutation matrix with a zero diagonal. Then $P_2^TD$ has a zero diagonal too. It follows that $A=(P_1P_2)(P_2^TD)Q$ can be written as $A=P(L+U)Q$ where $P,Q$ are nonsingular, $L$ is strictly lower triangular and $U$ is strictly upper triangular.
Since $L^n=0$, we have $\phi(L)^n=\phi(L^n)=0$. Therefore $\phi(L)=0$ and similarly $\phi(U)=0$. Hence $\phi(A)=\phi(P)\left(\phi(L)+\phi(U)\right)\phi(Q)=0$. |
H: Is $\operatorname{Aut}(D_{12})\simeq D_{12}$?
Let $D_{12}$ be the dihedral group of order 12. Then
$$|\operatorname{Aut}(D_{12})|=6\phi(6)=12=|D_{12}|,$$
and the standard method of proof for
$$\operatorname{Aut}(D_6)\simeq D_{6}\qquad\mbox{and}\qquad \operatorname{Aut}(D_8)\simeq D_{8}$$
seems to also work for
$$\operatorname{Aut}(D_{12})\simeq D_{12}.$$
But in this article (p.461, 14th line from the top), it says that $n=3$ and $n=4$ are the only numbers for which
$$\operatorname{Aut}(D_{2n})\simeq D_{2n}.$$
Could this be an error? Or am I missing something?
AI: The article is wrong. First, the automorphism group is definitely $D_{12}$: here is Magma code that proves it.
> G:=DihedralGroup(6);
> A:=AutomorphismGroup(G);
> A:=PermutationGroup(A);
> IdentifyGroup(A);
<12, 4>
> IdentifyGroup(G);
<12, 4>
Second, even his Theorem A states that $\mathrm{Aut}(D_{12})\cong \mathrm{Hol}(Z_6)$, and this is dihedral of order $12$. |
H: How to show that $\begin{pmatrix} 0 & -x \\\ 1/x & 0\end{pmatrix}$ is conjugate to a rotation?
Let $x >0$, and set $A=\begin{pmatrix} 0 & -x \\\ 1/x & 0\end{pmatrix}$.
Question: How to show that $A \in \operatorname{SL}_2(\mathbb R)$ is conjugate to an element of $\operatorname{SO}(2)$? That is, I am trying to show that there exist $C \in \operatorname{SL}_2(\mathbb R)$, and $Q \in \operatorname{SO}(2)$ such that $A=CQC^{-1}$.
Note that $A=\begin{pmatrix} x & 0 \\\ 0 & 1/x\end{pmatrix}R_{\pi/2}$, where
Let $R_{\pi/2}=\begin{pmatrix} 0 & -1 \\\ 1 & 0\end{pmatrix}$ is a rotation by $\pi/2$. However, I don't see how that representation helps us.
AI: $\pmatrix{1/\sqrt{x}\\ &\sqrt{x}}\pmatrix{0&-x\\ 1/x&0}\pmatrix{\sqrt{x}\\ &1/\sqrt{x}}=\pmatrix{0&-1\\ 1&0}=\pmatrix{\cos\frac\pi2&-\sin\frac\pi2\\ \sin\frac\pi2&\cos\frac\pi2}$.
Since all elements of $SO_2(\mathbb R)$ have unimodular eigenvalues, not every $A\in SL_2(\mathbb R)$ is similar to a special orthogonal matrix. In fact, $A\in SL_2(\mathbb R)$ is similar to some $Q\in SO_2(\mathbb R)$ if and only if $A$ is diagonalisable over $\mathbb C$ and its eigenvalues are unimodular. In this case, $Q$ is just the real Jordan form of $A$. |
H: Are all distributions with conjugate priors exponential families?
The Wikipedia page for conjugate priors lists several examples. Of the ones I'm immediately familiar with, all are exponential families. This leads me to wonder whether all families distributions that admit conjugate priors are exponential families.
More explicitly: suppose that $D$ is a family of distributions over $X$, parameterised by $\Theta$, that is, a probability measure $p(x;\theta)$ for each $\theta\in\Theta$. Suppose $D$ admits a conjugate prior. Does it follow that the family $D$ is an exponential family? If not, what is a simple counterexample, i.e. a family of distributions that admits a conjugate prior but is not an exponential family?
AI: what is a simple counterexample, i.e. a family of distributions that admits a conjugate prior but is not an exponential family?
Uniform model with Pareto prior |
H: Why is the second derivative of Cumulant Generating Function positive?
Wikipedia states (without reference) that the cumulant generating function of a random variable has the property
Its first derivative ranges monotonically in the open interval from the infimum to the supremum of the support of the probability distribution, and its second derivative is strictly positive everywhere it is defined, except for the degenerate distribution of a single point mass.
I think the strict positivitivity of second derivative boils down to the claim that
$$
\mathbb{E}\left(e^{x Z}\right) \mathbb{E}\left(Z^2 e^{x Z}\right)>\mathbb{E}\left(Z e^{x Z}\right)^2
$$
whenever the expectations exist, where $Z$ is any non-degenerate random variable. Is this correct?
Is there any good reference for the properties of Cumulant Generating Functions?
AI: Yes, the inequality is correct. A proof of the inequality is quite simple: Let $Q$ be the probability measure on the sample space of $Z$ defined by $\frac {dQ} {dP}=\frac {e^{zZ}} {Ee^{zZ}}$. Then the inequality says $(E_QZ)^{2} \leq E_Q Z^{2}$ and strict inequality holds except when $Z$ is a constant. |
H: Help in proving that $\int_{0}^{\infty} \frac{dx}{1+x^n}=\int_{0}^{1} \frac{dx}{(1-x^n)^{1/n}},n>1$
I wanted to prove that $$\int_{0}^{\infty} \frac{dx}{1+x^n}=\int_{0}^{1} \frac{dx}{(1-x^n)^{1/n}}, n>1.$$ I converted them to Gamma functions but I could not prove it. Please help me.
AI: $$I=\int_{0}^{\infty} \frac{1}{1+x^n} dx$$
Let $x=1/y\implies dx=-dy/y^2$, then
$$I=\int_{0}^{\infty} \frac{y^{n-2} dy}{1+y^n}$$
Next, use $1+y^n=z^n \implies y^{n-1}dy=z^{n-1} dz$ to get
$$I=\int_{1}^{\infty} \frac{dz}{z (z^n-1)^{1/n}}.$$
Lastly, let us take $z=1/u \implies dz=-du/u^2$, to get
$$I=\int_{0}^{1} \frac{du}{(1-u^n)^{1/n}}.$$ |
H: How to define a morphism from the Spec of the completion of $O_{Y,y}$ to $Y$?
Let $Y$ be a noetherian scheme and $y \in Y$. We denote by $\hat{O}_{Y,y}$ the completion of the local ring $O_{Y,y}$. I want to define a morphism
$$
\operatorname{Spec} \hat{O}_{Y,y} \to Y
$$
which sends the closed point $\hat{m}_y$ to $y$ and the map of the stalks is the canonical injection $O_{Y,y} \hookrightarrow \hat{O}_{Y,y}$.
How do I define this morphism? Thank you!
AI: By taking affine neighbourhood of $y$, we may assume $Y=\operatorname{Spec}B$, and define it as $B\to O_{Y,y}\to\hat{O}_{Y,y}$ where the first map is localization $O_{Y,y}=B_{m_y}$ and the second map is the completion. |
H: Use trapezoidal rule to find $\lim _{n\rightarrow \infty }\frac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}$
When $f\left( x\right) =\log x$ and section $\left[ 1,2\right]$
$$\lim _{n\rightarrow \infty }n\left[ \int _{a}^{b}f\left( x\right) dx-\dfrac{b-a}{n} \left\{\dfrac{f\left( a\right) +f\left( b\right) }{2}+\sum ^{n-1}_{k=1}f\left( a+\dfrac{\left( b-a\right) k}{n}\right) \right\} \right] =0$$
Use this to find the value of the limit:
$$\lim _{n\rightarrow \infty }\dfrac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}-①$$
I have assigned $f(x)$ and section $\left[ 1,2\right]$.
$$\lim _{n\rightarrow \infty }n\left[ \dfrac{1}{2}\log 2-\dfrac{1}{n}\left\{\dfrac{1}{2}\log 2+\sum ^{n-1}_{k=1}\log \left( 1+\dfrac{k}{n}\right) \right\} \right]-②$$
I can't find common points of $①$ and $②$. So I don't know what to do next.
AI: This answer is incorrect.
I'm leaving it here (instead of deleting it, got the advice here) just so that other solvers don't make the same mistake.
As rightly pointed out in the comments, This lemma should be used to obtain the correct answer of $1/\sqrt 2$
$$\begin{gather}
L = \lim_{n \to \infty} \left( \frac{4}{e} \right)^n \cdot \left( \frac{n}{n+1} \cdot \frac{n}{n+2} ... \frac{n}{n+n}\right) \\
\log L = \lim_{n \to \infty} n \log \left( \frac 4e \right) + \sum_{r=1}^n\log\left(\frac{n}{n+r}\right) \\
\log L = \lim_{n \to \infty} n \left( \log \left( \frac 4e \right) - \frac 1n\sum_{r=1}^{n}\log\left(1+\frac rn\right)\right)
\end{gather}$$
the term on the right is a Riemann sum, and it can be derived by using the trapezoidal rule to get the area for infinitesimal parts and summing them up. I'll leave this part up to you but the solution of the riemann sum is $\log\left( \frac 4e\right)$. This makes the limit
$$\log L = \lim_{n \to \infty} n \left( \log \left( \frac 4e \right) - \log \left( \frac 4e \right) \right) \\
\log L = 0 \\
\boxed{L = 1}$$ |
H: For What positive values of x ; the below mentioned series is convergent and divergent?
$$ \sum \frac{1}{x^{n}+x^{-n}}$$
My attempt
$$
\begin{aligned}
&\begin{aligned}
\therefore & u_{n+1}=\frac{x^{n+1}}{x^{2 n+2}+1} \\
\therefore & \frac{u_{n+1}}{u_{n}}=\frac{x^{n+1}}{x^{2 n+2}+1} \cdot \frac{x^{2 n}+1}{x^{n}} \\
& \frac{u_{n+1}}{u_{n}}=x \frac{1+\left(\frac{1}{x}\right)^{2 n}}{x^{2}+\left(\frac{1}{x}\right)^{2 n}}
\end{aligned}\\
& i f x>1 \quad \therefore \quad \frac{1}{x}<1 \quad \therefore \text { when } n \rightarrow \infty,\left(\frac{1}{x}\right)^{2 n} \rightarrow 0
\end{aligned}\\
$$
What to do next?
AI: In your attempt you have shown that $\lim \sup \frac {u_{n+1}} {u_n} \leq x \frac {1+0} {x^{2}+0}=\frac 1 x <1$. Hence Ratio Test tells you that the series is convergent for $x>1$. But there is a bettter approach as shown below:
For $x>1$ it is dominated by $\sum \frac 1 {x^{n}}$ which is convergent.
For $0<x<1$ it is dominated by $\sum x^{n}$ which is convergent.
For $x=1$ it is divergent. |
H: How to evaluate $\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2}$?
Note: Similar questions have been asked here and here, but this is quite different.
I am trying to evaluate
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} \quad (1)$$
I re-wrote the fraction as $$ \frac{2n+1}{2n(n+1)^2} = \frac1{2(n+1)} \cdot \frac{2n+1}{n(n+1)}= \frac1{2(n+1)} \left( \frac1n + \frac1{n+1} \right) = \frac1{2} \left( \frac1{n(n+1)} + \frac1{(n+1)^2} \right) = \frac12 \left( \left( \frac1n -\frac1{(n+1)} \right) + \frac1{(n+1)^2} \right)$$
Hence
$$(1) = \frac12 \sum_{n=1}^{\infty}\: \left( \frac1n -\frac1{(n+1)} \right) + \frac12 \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac12\lim_{n \to \infty}1-\frac1{n+1}+\frac{\pi^2}{12} = \frac{\pi^2}{12} + \frac12$$
Therefore
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} = \frac{\pi^2}{12} + \frac12$$
I am unsure about $$ \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac{\pi^2}{12} $$
We know the basic p-series $$ \sum_{n=1}^{\infty}\:\frac1{n^2} = \frac{\pi^2}{6} $$
Is this solution correct?
AI: As has been already pointed out in the comments,
$$\sum_{n = 1}^\infty \frac{1}{(n+1)^2} = \frac{1}{4} + \frac 19 + ... = \frac{\pi^2}{6}-1$$
Therefore, your solution would be
$$\frac{1}{2}\left( \sum_{n=1}^\infty \frac 1n - \frac {1}{n+1} + \sum_{n=1}^\infty\frac{1}{(n+1)^2}\right) = \frac 12 \left( 1 + \frac{\pi^2}{6} - 1\right) = \boxed{\frac{\pi^2}{12}}$$ |
H: Can we prove we know all the ways to prove things?
The things like induction and contradiction, they're all ways we prove things. Is that set of ways to prove things complete? Does the self referential nature of this question make it unprovable with something related to Gödel's incompleteness theorem? Has it been proven to be unprovable if this is the case? Is there some proof that shows we have found all the ways to prove things?
AI: We don't and we can't. Ways of proving things are just axioms. According to the Gödel's First Incompleteness Theorem, if your axiomatic system includes Peano Arithmetic, there are true statements that you can't prove, moreover, if you add more axioms (or proving ways), there will be other true statements that are unprovable. |
H: Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$
How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3}
\sec\theta\ \tan\theta d\theta $$
$$=\int \tan^2\theta \sec^4\theta(1-3\cos^2\theta)^{4/3} \sec\theta\ \tan\theta d\theta $$
$$=\int \tan^3\theta \sec^5\theta(1-3\cos^2\theta)^{4/3}\ d\theta $$
$$=\int\dfrac{ \sin^3\theta}{ \cos^8\theta}(1-3\cos^2\theta)^{4/3}\ d\theta $$
I can't see if this substitution will work or not. This has become so complicated.
Please help me solve this integral.
AI: I agree with the other answers. My response is long-winded so...
Often when attacking indefinite integrals, you will immediately suspect that a substitution [i.e. $u = g(x)$] is needed, but won't be sure which substitution to try.
I have to ask the OP:
Why did you think that $x = \sec \theta$ was the right substitution? Had you recently been exposed to problems that seemed similar where $x = \sec \theta$ was the right substitution?
The point of my response/rant is to develop the OP's intuition. Since the integral contains $(x^3 - 3x)^{(4/3)},$ my first guess as to the right substitution to try would be $u = (x^3 - 3x).$ This would convert this portion of the integral to $u^{(4/3)}.$
The idea is that (as a first guess for the right substitution), I would be hoping that (except for the $u^{(4/3)}$ factor), the remainder of the integral would be a polynomial in $u$, where each term has an integer exponent.
As I say, the point of my response is simply to expand the OP's intuition (and perspective). |
H: What type of polygon will fit this description?
A particle moving with constant speed turns left by an angle of 74° after travelling every 1m distance. It returns back to the starting point in 18s and we are required to find out the speed of the particle.
My attempt: I know that the particle will return back to its initial position if the line joining it to the orgin rotates by a multiple of 360°,therefore after it changes it's direction 180 times,it will have rotated an angle of 13320° which is 37 times 360.Therfore,the time taken for one rotation will be 18/180= 0.1 second and hence the speed is distance divided by time which is 1/0.1=10m/s.How will the motion of the particle look like?After solving the problem independently, I tried to look up the solution where it was mentioned that the motion will describe a regular polygon which seems incorrect.
AI: Because $74$ is not of the form $\frac{n-2}{n}*180$ for some integer $n\geq 3$, this will not trace a convex regular polygon. It will however trace a star polygon. Let $n$ be the number of edges of this star polygon. Then we have that $74*n$ is divisible by $360$ and $n$ is the smallest such number. After some calculations, you get that $n = 180$ and thus that the speed $180/18=10$ meters per second. Below is a picture of the polygon. |
H: Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?
AI: A polynomial $f(x)$ is a multiple of $x^2-x-1$ (the characteristic polynomial of the Fibonacci sequence) iff $f(\varphi)=f(\bar{\varphi})=0$, with $\varphi=\frac{1+\sqrt{5}}{2}$, $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$ being algebraic conjugates. Since $\varphi^2=\varphi+1$ we have by induction $\varphi^k=F_{k}\varphi+F_{k-1}$ and the same holds for $\bar{\varphi}$. It follows that $ax^9+bx^8+1$ is a multiple of $x^2-x-1$ iff
$$\begin{eqnarray*} 0 &=& a\varphi^9+b\varphi^8+1 = a(34\varphi+21)+b(21\varphi+13)+1\\&=&(34a+21b)\varphi+(21a+13b+1)\end{eqnarray*}$$
and
$$ 0 = (34a+21b)\bar{\varphi}+(21a+13b+1). $$
It follows that we must have $21a+13b=-1$ and $34a+21b=0$, so $\color{red}{a=21,b=-34}$ works.
You may also invoke the more general identity
$$ (x^2-x-1)\sum_{k=0}^{n}(-1)^{k+1}F_k x^k = (-1)^{n+1}F_n x^{n+2}+(-1)^n F_{n+1} x^{n+1}-x.$$ |
H: Number of solutions of $2011^x$ $+$ $2012^x$ $+$ $2013^x$ $=$ $2014^x$
The problem is : To find number of real solutions of $ \ \ 2011^x$ $+$ $2012^x$ $+$ $2013^x$ $=$ $2014^x$
My attempt : I first tried to see if the equation has zero solutions ;i.e ,the LHS of the equation was even and so was the RHS so I couldn't prove it had zero solutions.
Next I tried finding the derivative of $ \ \ 2011^x$ $+$ $2012^x$ $+$ $2013^x$ $-$ $2014^x \ $ to get an idea of the graph and it turned out to be :
$$2011^x \ln 2011 +2012^x \ln 2012 +2013^x \ln 2013 - 2014^x \ln 2014 $$
Now I am stuck ,how do I proceed
AI: Divide by $2014^x$ to get $$a^x+b^x+c^x=1$$ The derivative is $(\ln a) a^x+(\ln b)b^x+(\ln c)c^x<0$ yet at $x=0$, the LHS $=3>1$ and at $x\to\infty$, LHS$\to0$. So only one solution. |
H: Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative?
I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hence,
$$ (1) = \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = - \sum _{n=1}^{\infty } \frac1{2^{n+1}} = - \sum _{n=2}^{\infty } \frac1{2^{n}} = -\left( \sum _{n=0}^{\infty } \left(\frac12\right)^n -1-\frac12\right) = -2+\frac32 =-\frac12 $$
Therefore,
$$ \boxed{\sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} =-\frac12} $$
I can't spot any algebraic mistakes, thus I assume the sum is correct. But I don't understand the result. How can a real valued sum converge to a negative number?
AI: I reckon your original sum is
$$\sum _{n=1}^{\infty }\left(\frac{1}{2^{2n-1}}-\frac{1}{2^{2n}}\right).$$
Of course, this is a GP. |
H: Throwing dice: Compute the probability of strictly increasing values in the first $n$ throws
Let a die be thrown $m$ times. What is the probability of strictly increasing values in the first $n< m$ throws? In other words, compute the probability $P(X=n)$ with $X\,\widehat{=}$"The first $n$ face values are strictly increasing".
The problem here is that I can't seem to find a rule which is common to how the probabilities of $P(X=1),P(X=2),\ldots$ seem to form in order to arrive at a general rule. How do I approach this?
AI: Consider the set of all rolls of $n$ dice, each die having $N$ sides. There are $N^n$ such sequences. How many of these sequences are strictly increasing?
Every possible sequence of $n$ strictly increasing die rolls can be obtained by selecting $n$ distinct values, without replacement, from the set of the die's $N$ faces, and then sorting them into a strictly increasing order. There are $N \choose n$ distinct ways to do this.
Thus, the probability of a sequence of $n$ rolls being strictly increasing is
$$
p = \frac{1}{N^n} { N \choose n}.
$$
For $N = 6$, these values work out to be:
$n = 1$: $p = 1$
$n = 2$: $p = 5/12$
$n = 3$: $p = 5/54$
$n = 4$: $p = 5/432$
$n = 5$: $p = 1/1296$
$n = 6$: $p = 1/46656$ |
H: How to prove the statement about the rank of a block matrix?
Let $A$ and $B$ be real matrices with the same number of rows. Prove that:
$$\mbox{rank} \begin{bmatrix} A & B\\ 2A & -5B\end{bmatrix} = \mbox{rank}(A) + \mbox{rank}(B)$$
I have no idea how to approach the problem. Could you give me a hint?
AI: Subtract the double of the first block row from the second one, we get
$$
\pmatrix{A&B\\ 0&-7B}.
$$
Then divide the second block row by $-7$ to obtain
$$
\pmatrix{A&B\\ 0&B}.
$$
Finally, subtract the second block row from the first block row to obtain
$$
M=\pmatrix{A&0\\ 0&B}.
$$
If you can prove that $\operatorname{rank}(M)=\operatorname{rank}(A)+\operatorname{rank}(B)$, then you are done, because the rank of a matrix is unaffected by elementary row operations. |
H: dealing infinities in the book frank jones lebesgue integration on euclidean spaces
I am currently reading the book "Frank Jones : Lebesgue Integeartion on Euclidean Spaces". The writing is not completely rigorous. for example measure can be $\infty$ but he doesn't tell what to do in such cases.
what does the following statement mean when one of $\lambda(A_i) = \infty.$ (I mean how does infinite series defined when one of them is $\infty$)
$$\lambda\Big( \bigcup_{i=1}^{\infty} A_i \Big) = \sum_{i=1}^{\infty} \lambda(A_i).$$
It is particularly confusing when argument depends on measure of set not being $\infty$
let $A_1 \subset A_2 \subset A_3 \dots$ then $\displaystyle \lambda \Big( \bigcup_{n=1}^{\infty} A_n \Big) = \lim_{N \rightarrow \infty} \lambda(A_N)$
we can write union of $A_i$ as disjoint union as follows $ \displaystyle\bigcup_{n=1}^{\infty} A_n = A_{1} \cup \bigcup_{n=1}^{\infty} (A_{n+1} \sim A_{n}) $ then we have that
$\displaystyle \lambda(\bigcup_{n=1}^{\infty} A_i) = \lambda(A_{1}) + \sum_{n=1}^{\infty} \lambda(A_{n+1} \sim A_{n}) $ he then says that
$ = \displaystyle \lim_{N \rightarrow \infty} \Big[ \lambda(A_{1}) + \sum_{n=1}^{N} \lambda(A_{n+1} \sim A_{n}) \Big]$
$ = \displaystyle \lim_{N \rightarrow \infty} \lambda \Big( A_{1} \cup \bigcup_{n=1}^{N} (A_{n+1} \sim A_{n}) \Big)$
$ = \displaystyle \lim_{N \rightarrow \infty} \lambda(A_N)$
then following is the excercise
Let $A_1 \supset A_2 \supset A_3 \dots$ if $ \lambda(A_1) < \infty$ then we have that $\displaystyle \lambda \Big( \bigcap_{n=1}^{\infty} A_n \Big) = \lim_{n \rightarrow \infty} \lambda(A_i))$
now we know that $\displaystyle A_{K} = \Big( \bigcap_{n=1}^{\infty} A_n \Big) \cup \Big(\bigcup_{n=k}^{\infty}(A_{n} \sim A_{n+1}) \Big) $
so we have that
$\lambda(A_{K}) = \lambda(A) + \sum_{n=k}^{\infty} \lambda(A_{n} \sim A_{n+1})$
and $\displaystyle \lim_{K \rightarrow \infty} A_{K} = \lim_{K \rightarrow \infty} \Big[ \lambda(A) + \sum_{n=k}^{\infty} \lambda(A_{n} \sim A_{n+1}) \Big] = \lambda(A)$
But I didn't use the condition that $\lambda(A_1) < \infty.$ ?? what is wrong with the arugment. and I am really confused handling with limit of sequence of numbers which contains $\infty.$ Book doesn't talk about what it means for limit when sequence contains $\infty$ or handle it separately.
AI: For your last example, the assumption that $\lambda(A_j)<\infty$ for some $j$ is crucial.
Let $\lambda$ be Lebesgue measure and $A_k$ be the
interval $(k,\infty)$. Then each $\lambda(A_k)=\infty$
but $\bigcap_k A_k=\emptyset$. The usual error here is to assume that "$\infty-\infty$" equals $0$. |
H: The meaning of $\frac{d^2 \sin x}{d (\cos x)^2}$
Just wondering if there is a meaning in the the following $$\frac{d^2 \sin x}{d (\cos x)^2}$$
I know that the Leibinz notation is some kind of symbolic representation of derivative.
But is there any formalism behind this?
Actually let me tell the story from the beginning: I would calculate the derivative:
$$\frac{d^n}{dx^n}(1-x^2)^{n+1/2}$$
for $x=\cos kt$
Would like to find a formal way of doing this.
AI: By definition differential for $f$ function is linear function $g(h)=A\cdot h$, for constant $A$, which represent so called principal part of the change in a function. Formally for $x_0$ point this means $$f(x_0+h)-f(x_0)=g(h)+o(h)=A\cdot h +o(h),h \to 0$$
Proved, that existence of differential is equivalence to function differentiability and $A=f^{'}(x_0)$.
For linear function $g$ is accepted notation $df(x_0) = f^{'}(x_0)dx$, where $dx$ is differential for identity function: $dx(h)=h$. So exact definition should be written:
$$df(x_0)(h) = f^{'}(x_0)dx(h)=f^{'}(x_0) h$$
In brought example $d (\cos x)^2 = -2\cos x \sin x dx$
High order differential, omitting some details, is defined for simple variable as
$$d^n y=y^{(n)}dx^n$$
So for brought example
$d^2 \sin x=-\sin x dx^2$
Note: When I finished first part of question, I found, that you added second one starting with word "Actually". For composition situation is little more difficult and, for example, for composition $z=z[y(x)]$ second differential is
$$d^2z=z^{''}_{yy}dy^2+z^{'}_{y}d^2y$$
Let me write answer to your example little later.
Addition.
As promised:
$$
\begin{array}{}\frac{dy}{dx} = -(n+1)x\left(1-x^2 \right)^{\frac{n-1}{2}} \\
\frac{d^2y}{dx^2} = (n^2-1)x^2\left(1-x^2 \right)^{\frac{n-3}{2}}-(n-1)\left(1-x^2 \right)^{\frac{n-1}{2}} \\
\frac{d^3y}{dx^3} = -(n+1)n^2x^3\left(1-x^2 \right)^{\frac{n-5}{2}}+4nx^3\left(1-x^2 \right)^{\frac{n-5}{2}}-\\ -3(n+1)x^3\left(1-x^2 \right)^{\frac{n-5}{2}}+3nx(n+1)\left(1-x^2 \right)^{\frac{n-3}{2}}-\\ -3x(n+1)\left(1-x^2 \right)^{\frac{n-3}{2}}
\end{array}$$
Sorry, yet, do not see general form.
Meanwhile, if you put $x=\cos kt $, then, whole expression, obviously, become more easy $y(t)=\left(\sin kt\right)^{(n+1)}$, but how it will simplify work for $\frac{d^ny}{dx^n}$, I do not see yet ,
If you would like have $\frac{d^ny}{dt^n}$ then What is the $n$th derivative of $\sin(x)$? help. |
H: RMM 2015 /P1: Does there exist an infinite sequence of positive integers $a_1, a_2, a_3, . . .$
Does there exist an infinite sequence of positive integers $a_1, a_2, a_3, . . .$ such that $a_m$ and $a_n$ are coprime if and only if $|m - n| = 1$?
My Progress:This is a very beautiful problem ! I think I have got a construction , but I am not able to have/define the explicit formula for the nth term.
Here is the construction, Let $a_1=2\cdot 3$, $a_2=5\cdot 7$, $a_3=2\cdot 11$, $a_4=3\cdot 5 \cdot 13$, $a_5=2\cdot 7\cdot 17$ , $a_6=3\cdot 5 \cdot 11 \cdot 19$ , $a_7=2\cdot7\cdot13\cdot23$, $a_8=3\cdot5\cdot11\cdot17\cdot29$ , $a_9=2\cdot7\cdot13\cdot 19 \cdot 31$ and so on .
I am trying to find some patterns, but I am not able to observe anything.
So what I am doing is, for the construction of $a_n$ term, I look at $a_{n-1}$ ,then I start from $a_1$ and then try to put a factor $p$ of $a_1$ in $a_n$ such that gcd ($a_{n-1},p$)=$1$ . Similarly for $a_2$, $a_3$, and so on. At the end I add another prime which was not used in any of the $a_i$'s. Also we have to make sure that no ${a_i} \mid a_j$ for $i<j$
Also note that I am only using primes .
Sorry, if something is not clear. Hope one can provide me some hints and guide.
Thanks in advance.
AI: By using the canonical indexing of primes, it is sufficient to show that there exists a sequence $\{A_n\}_{n\geq 1}$ of finite subsets of $\mathbb{N}\setminus\{0\}$ such that $A_{n+1}$ belongs to the complement of $A_n$ but has a non-trivial intersection with every member of the family $A_1,A_2,\ldots,A_{n-1}$. Your sequence is associated to
$$ \{1,2\},\{3,4\},\{1,5\},\{2,3,6\},\{1,4,7\},\{2,3,5,8\},\{1,4,6,9\},\{2,3,5,7,10\},\ldots$$
and I can see a pattern here: starting with $A_5=\{1,4,7\}$, $A_n$ is given by
$$ (A_{n-2}\setminus\{\max A_{n-2}\})\cup\{\max A_{n-2}-1\}\cup\{n+2\}. $$
Decrease by one the maximum element of $A_{n-2}$, then append $n+2$.
By induction it should not be difficult to prove that this actually works. I'll start the proof:
$A_n\cap A_{n+1}=\emptyset$. This is blatantly true for any $n\leq 6$, hence we may assume $n>6$. Since $\max A_{n+1}=n+3>n+2=\max A_n$, $\max A_{n+1}$ is not an element of $A_n$. The set $A_{n+1}\setminus\{\max A_{n+1}\}$ equals $A_{n-1}$ with the maximum element ($n+1$) being replaced by $n$. $A_n\cap A_{n-1}=\emptyset$ by inductive hypothesis, hence the proof of $A_n\cap A_{n+1}=\emptyset$ boils down to the proof of $n\not\in A_n$, which follows from $\max(A_n\setminus\{\max A_n\})=n-1$.
$A_n$ has a non-trivial intersection with $A_1,A_2,\ldots,A_{n-2}$. By direct inspection we may assume $n>6$ as well. By definition $A_n$ has non-trivial intersections with $A_{n-2},A_{n-4},\ldots,A_2$, so it is sufficient to prove that $A_n$ has non-trivial intersections with $A_{n-3},A_{n-5},\ldots,A_1$. In the previous point we have shown $\max(A_n\setminus\{\max A_n\})=n-1=\max A_{n-3}$, so $A_n\cap A_{n-3}\neq\emptyset$. In a similar way we may show that if we remove the two greatest elements from $A_n$, the maximum becomes the maximum of $A_{n-5}$, so $A_n\cap A_{n-5}\neq \emptyset$ etcetera.
This is basically the reverse approach of the one taken by Eigen von Eitzen here (his sets end with $2n-1,2n$, our sets start with $1,4$ or $2,3$). We gain a pleasant bit of regularity if we pick $A_3$ as $\{2,5\}$ instead of $\{1,5\}$:
$$ \{1,2\},\{3,4\},\{2,5\},\{1,3,6\},\{2,4,7\},\{1,3,5,8\},\{2,4,6,9\},\{1,3,5,7,10\},\ldots$$ |
H: $71\nmid a$ and $x^7\equiv a\pmod {71}$ has an integer solution. How many positive integer solution there is for this equation that lower than 71?
I need a little help with the following question:
If $a$ is an integer such as $71\nmid a$ and $x^7\equiv a\pmod {71}$ has an integer solution. How many positive integer solution there is for this equation that lower than 71?
All I understand is that $71$ is prime.
And that because $71\nmid a$ then $\gcd (a,71) = 1$ so we know there is $s,t\in \mathbb Z$ such as $sa+71t=1$ then $71t=1-sa$ then $71 \mid 1-sa$ then $sa \equiv 1\pmod {71}$.
If $c$ is a solution then $c^7\equiv a\pmod {71}$ then $71\mid c^7-a.$
Thanks in advance
AI: The multiplicative group mod a prime $p$ is cyclic. A generator of this group is called a primitive root mod $p$. If $b$ is a primitive root, it means for each $x \in \{1,\ldots, p-1\}$ we can write $x \equiv b^j \mod p$ for some unique $j \in \{1,\ldots, p-1\}$. $b$ has order $p-1 \mod p$. It turns out, for example, that $11$ is a primitive root mod $71$, but you don't need to know that for this question, you only need to know the theorem that a primitive root exists for each prime.
If $d$ is any divisor of $p-1$ and $x^d \equiv 1 \mod p$ where $x \equiv b^j \mod p$, that says $b^{jd} \equiv 1 \mod p$, so $jd$ must be divisible by $p-1$. So, the number of such $x$ is the same as the number of $j \in \{1, \ldots, p-1\}$ such that $p-1 \mid jd$. So in your case, how many members $j$ of $\{1,\ldots, 70\}$ are there such that $7j$ is divisible by $70$? |
H: Contrapositive of a statement involving integers
If we need to write a contrapositive and negation for c and d are integers, then cd is an integer. So does it mean that c and d can be fractions or natural numbers for negation? I understand that contrapositive is making it to negative and back to positive but I can't make head or tails about integers part. I am new to this so please advise.
AI: The "contrapositive" of "if A then B" is "if not B then not A".
Here, your statement is "if c and d are integers then cd is an integer" so the contrapositive is "if cd is not an integer then c and d are not both integers".
That can be written more simply by using the fact that "not A and B" is the same as "not A or not B".
If cd is not an integer then either c is not an integer or d is not an integer. |
H: Show $(\Bbb R\setminus\{−1\},*)$ is an Abelian group, where $a*b:=ab+a+b$. Solve $3 * x * x = 15$ in that group.
I have stumbled across the following problem and cannot understand how to solve it. Especially part b. I understand that part a must show that the rules of a Abelian group must apply for it to be an Abelian group to solve the problem, I'm just unsure how to state this.
We consider $(\Bbb R\setminus \{−1\}, *)$, where
$a * b := ab + a + b, a, b \in \Bbb R\setminus \{−1\}$
a. Show that $(\Bbb R\setminus \{−1\}, *)$ is an Abelian group.
b. Solve $3 * x * x = 15$ in the Abelian group $(\Bbb R\setminus \{−1\}, *)$, where $*$ is defined above.
Any help with this would be greatly appreciated. Once I have a firm understanding of this I will be able to solve other Abelian group problems.
AI: Let $A = \mathbb{R} \backslash \{-1\}$, and $a,b,c\in A$.
For part (a), we need to check the group axioms:
Closure: Supposing to the contrary that $-1 = a*b=ab+a+b$, then $0=ab+a+b+1=(a+1)(b+1)$, or either $a=-1$ or $b=-1$ (so $A$ is closed).
Association: $(a*b)*c = (ab+a+b)*c = (ab+a+b)c+ab+a+b+c = abc+ac+bc+ab+a+b+c = abc+ab+ac+a+b+c = a(bc+b+c)+a+(bc+b+c) = a*(bc+b+c) = a*(b*c)$ (which follows because standard addition and multiplication of reals is associative).
Identity: $a*0 = a0 + a + 0 = 0 + a + 0 = a$
Inverse: $a*(\frac{-a}{1+a}) = a\frac{-a}{1+a} + a + \frac{-a}{1+a} = \frac{-a^2}{1+a} + \frac{a^2+a}{1+a} + \frac{-a}{1+a}=0$
Commutativity follows due to the commutativity of addition and multiplication of reals: $a*b = ab+a+b = ba+b+a = b*a$.
To solve part (b), we've shown that $(A,*)$ is an Abelian group, so it doesn't matter what grouping we use to evaluate $3*x*x$ (as $*$ is associative).
Then $3*x*x = 3*(x*x) = 3*(xx+x+x) = 3*(x^2+2x)=3(x^2+2x)+3+x^2+2x$
And now it's a polynomial: $4x^2+8x+3 = 15$, or $4x^2+8x-12 = 0$
which has roots at $x = -3$ and $x = 1$. |
H: If $f$ is a open and continuous function then $f$ is injective.
I think I managed to prove:
"If $f:\mathbb{R} \rightarrow \mathbb{R} $ is continuous and open,( i.e., any open $\mathcal{A}\subset\mathbb{R}$ then $f(\mathcal{A})$ is open,) then $f$ is injective."
but I have some doubts especially in the end of the proof. I started with:
Suppose that exists $x\neq y$ such that $f(x)=f(y)$. The restriction of $f$ to $[x,y]$ can not be constant otherwise it would contradict the hypothesis of being an open function. Then exists $z\in(x,y)$ such that $f(z)\neq f(x)$. Suppose that $f(z)>f(x)$ and then prove that $\sup_{w\in (x,y)}\text{$f$}\in f((x,y))$, which is sufficient to contradict the fact of $f$ is an open function.
Consider two sequences $a_{n}$ and $b_{n}$ such that:
$x<a_{n+1}<a_{n}<z<b_{n}<b_{n+1}<y$ and that $a_{n}\rightarrow x$ and $b_{n}\rightarrow y$.
Define $\mathcal{M} _{n}=\max_{w\in [a_{n},b_{n}]}\text{$f$}$, which exists because $f$ is continuous and $[a_{n},b_{n}]$ is compact.
Take $\epsilon$=$\frac{f(z)-f(x)}{2}$ then exists $\delta_{i}>0$, for i=1,2, such that:
$|w-x|<\delta_{1}$ $\Rightarrow$ $|f(w)-f(x)|<\epsilon$
$|w-y|<\delta_{2}$ $\Rightarrow$ $|f(w)-f(x)|<\epsilon$
Take $\delta$=min{$\delta_{1},\delta_{2}$}. As $a_{n}\rightarrow x$ and $b_{n}\rightarrow y$ there exists $p_{i}\in\mathbb{N}$ for i=1,2 such that
$n\geqslant p_{1}$ then $|a_{n}-x|<\delta$
$n\geqslant p_{2}$ then $|b_{n}-y|<\delta$
Again take $p$=max{$p_{1},p_{2}$} and as $(x,y)=(x,a_{p})\cup[a_{p},b_{p}]\cup(b_{p},y)$ we can conclude:
$\forall w\in(x,a_{p})\subset(x,x+\delta)$ $\Rightarrow$ $f(w)<f(x)+\epsilon<f(z)$
$\forall w\in(b_{p},y)\subset(y-\delta,y)$ $\Rightarrow$ $f(w)<f(x)+\epsilon<f(z)$
So exists maximum of $f$ in $(x,y)$ which is attained in $[a_{p},b_{p}]$ and is $\mathcal{M}_{p}$.
Any suggestions would be deeply appreciated.
AI: Here is a proof of the claim that uses the Extreme Value Theorem. I will not spell out complete details because it might be worthwhile to compare the strategy to your proof attempt, especially regarding how much of your proof overlaps with the proof of EVT.
We assume for the sake of a contradiction that there are distinct $a,b$ such that $f(a)=f(b)$. Let $I$ be the open interval $(a,b)$. So $f(I)$ is open by assumption. By the Extreme Value Theorem, $f$ achieves a maximum and minimum value on the closed interval $[a,b]$. Since $f(a)=f(b)$, it follows that at least one of the maximum value or the minimum value is attained in $I$. If the maximum value is attained in $I$, then $f(I)$ contains its supremum, hence is not open. If the minimum value is attained in $I$, then $f(I)$ contains its infimum, hence is not open. So in every case we get a contradiction. |
H: How can I describe this domain?
I solved an equation on a 3D-domain which is the unitary sphere where a torus has been removed.
How can I describe it with proper mathematical words?
AI: If the solid sphere is given by $S(x,y,z)\le0$ and the solid torus is given by $T(x,y,z)\le0$, then your region is given by $S(x,y,z)\le0, T(x,y,z)\ge0$.
A parametric representation of the boundary of your region depends on the size and location of the torus and is bound to be complicated because there are many cases to consider. See this paper for instance. |
H: Chain rule proof confusion
Here is a common informal proof for the chain rule:
If $S(a)=f(g(x))|_{x=a}$, then $S'(a)$ is given by
\begin{align}
\lim_{x \to a}\frac{S(x)-S(a)}{x-a}&=\lim_{x\to a}\frac{f(g(x))-f(g(a))}{x-a} \\
&=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}\cdot\frac{g(x)-g(a)}{x-a} \\
&=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}\cdot\lim_{x \to a}\frac{g(x)-g(a)}{x-a} \\
&=f'(g(a))g'(a)
\end{align}
The reason this proof is considered informal is because it does not take into account the case where $g(x)=g(a)$ as $x \to a$. This is not, however, where my confusion lies. I am confused as to how $f'(g(a))$ equals $$\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}$$
As far as I understand, $f'(g(a))$ means apply the function $f'$ to $g(a)$. The function $f'$, evaluated at a particular point $p$, is defined by
$$
\lim_{x \to p}\frac{f(x)-f(p)}{x-p}
$$
Thus, $f'(g(a))$ is defined as
$$
\lim_{x \to g(a)}\frac{f(x)-f(g(a))}{x-g(a)}
$$
And so I don't see how these two expressions can be equal to each other. What am I missing?
AI: Since $g$ is differentiable in $a$ and $f$ is differentiable in $g(a)$, then they are also continuous in those points; therefore $g(x) \to g(a)$ as $x\to a$ and you can use substitution $y=g(x)$ in the limit $\lim_{x\to a} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)}$ to get:
$$\lim_{x\to a} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)} \stackrel{y=g(x)}{=} \lim_{y\to g(a)} \frac{f(y) - f(g(a))}{y - g(a)} = f^\prime (g(a))\; .$$ |
H: $Q(n)-Q(n-1) = T(n)$ Prove that $Q(n)$ degree is $k+1$
I was given this problem and I've been thinking a lot of time and still I have nothing.
$Q:ℕ↦ℕ$
$Q(n)-Q(n-1) = T(n)$
$T(n)$ degree is $k$
Prove that $Q(n)$ degree is $k+1$
any idea?
Thank you
AI: Let $Q(n)$ be a $r^{th}$ degree polynomial.
Thus $Q(n)=a_0n^r+a_1n^{r-1}+a_2n^{r-2}+\ldots+a_{r-1}n+a_r$
which means that $Q(n-1)=a_0(n-1)^r+a_1(n-1)^{r-1}+a_2(n-1)^{r-2}+\ldots+a_{r-1}(n-1)+a_r$
Now in the expansion of $(n-1)^r$, the term with highest power, that is $n^r$ always have a coefficient of $1$. Hence $(n-1)^r$ can be written as $n^r-P_r(n)$ where $P_r(n)$ is some polynomial of degree $r-1$
Thus $a_0(n-1)^r=a_0n^r-a_0P_r(n)$, $\space a_1(n-1)^{r-1}=a_1n^{r-1}-a_1P_{r-1}(n)$ and so on
$\therefore Q(n-1)=Q(n)-T(n)$ where $T(n)=a_0P_r+a_1P_{r-1}+\ldots+a_{r-1}P_1$
Now it is evident that $T(n)$ would be of degree $(r-1)$ and also $Q(n)-Q(n-1)=T(n)$
We have shown here that if the degree of $Q(n)$ is $r$, then the degree of $T(n)=Q(n)-Q(n-1)$ is $(r-1)$.
The degree of $T(n)$ given here is $k$
$\therefore $ The degree of $Q(n)$ is $(k+1)$ |
H: Relating actions of intersections of subgroups of a finite group.
Suppose that we have a finite group $G$ with three subgroups $A,B,C$. I am interested in relating the action of $A \cap B \cap C$ on $B \cap C$ with the action of $A \cap B$ on $B$ (both via right multiplication). In particular, I am interested in relating transversals (a collection of representatives for the orbit space) for these actions.
I'm not sure exactly what we might be able to say in this situation. I think the following may be true:
There exists a transversal $T \subseteq B$ for the coset space $B / (A \cap B)$ such that $T \cap C$ is a transversal for $(B \cap C) / (A \cap B \cap C)$.
and also perhaps the stronger statement
Let $T' \subseteq B \cap C$ be a transversal for the coset space $(B \cap C) / (A \cap B \cap C)$. Then there exists a transversal $T' \subseteq T \subseteq B$ for the coset space $B / (A \cap B)$ such that $T \cap C = T'$.
Now I'm not sure if we could go further and say something like:
If $T \subseteq B$ is a transversal for the coset space $B / (A \cap B)$, then $T \cap C$ is a transversal for $(B \cap C) / (A \cap B \cap C)$.
or perhaps
If $T \subseteq B$ is a transversal for the coset space $B / (A \cap B)$, then $T \cap C$ is a transversal for $(B \cap C) / (A \cap B \cap C)$, and every such transversal is obtained in this way.
If not, is there something we can say generally about this set up? Perhaps something like:
AI: We assume that $A,C\leq B$. Let $B$ be three copies of $C_2$, generated by $x,y,z$. Let $A=\langle x,y\rangle$. Then $T=\{1,z\}$ is a transversal to $A$ in $B$. Now let $C=\langle y,z\rangle$. Then $A\cap C=\langle y\rangle$, and $T=T\cap C$ is indeed a transversal to $A\cap C$ in $C$.
But $T=\{1,xz\}$ is also a transversal to $A$ in $B$. And it has a different intersection with $C$, namely $\{1\}$.
So, in conclusion, your statement does not hold, that $T\cap C$ is a transversal to $A\cap C$ in $B$. Thus extend $T'$ to $T$ in any way.
Edit: You can however always choose such a transversal. Simply choose a transversal $T$ such that $T\cap C$ is a transversal to $A\cap B\cap C$ in $B\cap C$. Choose a transversal $T'$ to $A\cap B\cap C$ in $B\cap C$. If $x$ and $y$ are different coset representatives, then $xy^{-1}\not\in A\cap B\cap C$. If $xy^{-1}\in A\cap B$ though, then $xy^{-1}\in A\cap B\cap C$ (as $x,y\in C$), a contradiction. Thus they label different cosets of $A\cap B$ in $B$. |
H: Problem with the power of functions' set and number of discontinuity points
I am considering functions $f:\mathbb{R}\rightarrow \mathbb{R}$ with property that $\forall_{r\in\mathbb{R}}$ exists a limit $\lim_{x\rightarrow r}f(x)$ (it doesn't have to be equal to $f(r)$). I have two problems.
The first with showing that such function $f$ has countable number of discontinuity points. I have found only that monotonic functions have such property that it doesn't help me a lot.
Second problem is connected with evaluating the power of set of such functions. We know of course that the power of set of the continuous functions is equal to continuum which is a hint, but still it doesn't help me much.
I appreciate any help, because I am thinking about it from several days.
AI: Whatever you indicate, it can be either a function with removable discontinuity or a continuous function.
Whenever $f$ have removable discontinuity, then $f$ can't be a monotone function,
For example, $f(x) =
\begin{cases}
2, & \text{if x=1} \\
x, & \text{otherwise}
\end{cases} $
And, how many are there such type functions?
Then it is uncountable, you can see that easily.
As, we can creat, for each $a\in\mathbb{R} $, $f(x) =
\begin{cases}
a, & \text{if x=a+1} \\
x, & \text{otherwise}
\end{cases} $
And, so, clearly cardinality of power set of such functions is equal to the cardinality of $\mathbb{R}^{\mathbb{R}} $. |
H: How to find the minimum of $f(\mathbf{x}) = \| \mathbf{A}\mathbf{x}-\mathbf{b} \|^2_\mathbf{p} + \| \mathbf{x}-\mathbf{c} \|^2_\mathbf{q}$
Let $\mathbf{A} \in \mathbb{R}^{M \times N}$, $\mathbf{x} \in \mathbb{R}^{N}$, $\mathbf{b} \in \mathbb{R}^{M}$, $\mathbf{c} \in \mathbb{R}^{N}$, $\mathbf{p} \in {\mathbb{R}^+}^{M}$ and $\mathbf{q} \in {\mathbb{R}^+}^{N}$. Given fixed $\mathbf{A}$, $\mathbf{b}$, $\mathbf{c}$, $\mathbf{p}$ and $\mathbf{q}$, how to solve the following least-squares optimization problem? Does it have a closed-form solution?
$$
\mathbf{x}^* = \arg\min_{\mathbf{x}} \| \mathbf{A}\mathbf{x}-\mathbf{b} \|^2_\mathbf{p} + \| \mathbf{x}-\mathbf{c} \|^2_\mathbf{q},
$$
where $\|\mathbf{x}\|^2_\boldsymbol{\tau} \triangleq \sum_j (x_j^2 / \tau_j)$.
AI: Welcome to MSE! Usually we encourage the OP to post whatever he/she/they have tried to solve the problem, so try to keep that in mind as you continue your (hopefully long-lasting and meaningful) journey on MSE.
Anyway, to your question. Note that $\|\mathbf{x}\|_\mathbf{\tau} = \|D_\tau \mathbf{x}\|$, where $\|\cdot\|$ denotes the standard Euclidean norm in the appropriate dimension and $D_{\tau} = \text{diag}\left(\tau_1^{-1/2}, \tau_2^{-1/2}, \cdots, \tau_n^{-1/2}\right)$. So it suffices to find
$$\mathbf{x}^* = \arg \min_{\mathbf{x} \in \mathbb{R}^n} \left[\|A \mathbf{x} - \mathbf{b}\|^2 + \|D\mathbf{x} - \mathbf{c}\|^2 \right] ~~~~~~~~~~~~~~~ (*)$$
for arbitrary constants $A$, $D$, $\mathbf{b}$, and $\mathbf{c}$. (To solve your problem, just take $A$ in $(*)$ to be $D_{\mathbf{p}} A$, $\mathbf{b}$ in $(*)$ to be $D_{\mathbf{p}} \mathbf{b}$, take $D$ in $(*)$ to be $D_\mathbf{q}$, and take $\mathbf{c}$ in $(*)$ to be $D_{\mathbf{q}} \mathbf{c}$.) We can rewrite the objective function as
$$\mathbf{x}^T A^T A \mathbf{x} - 2 \mathbf{b}^T A \mathbf{x} + \mathbf{b}^T \mathbf{b} + \mathbf{x}^T D^T D \mathbf{x} - 2 \mathbf{c}^T D \mathbf{x} + \mathbf{c}^T \mathbf{c} \\ = \mathbf{x}^T (A^T A + D^T D) \mathbf{x} - 2 (\mathbf{b}^T A + c^T D)\mathbf{x} + \mathbf{b}^T \mathbf{b} + \mathbf{c}^T \mathbf{c}$$
Since the last two terms are constants, the problem is equivalent to minimizing
$$\mathbf{x}^T (A^T A + D^T D) \mathbf{x} - 2(\mathbf{b}^T A + \mathbf{c}^TD) \mathbf{x}$$
Note that $A^T A + D^T D$ is a real symmetric positive definite matrix and can hence be orthogonally diagonalized (see Spectral Theorem), say as $Q^T E Q$, with $Q$ an orthogonal matrix and $E$ a positive real diagonal matrix. We can then make a change of variables $\mathbf{y} \equiv Q \mathbf{x}$ to get the new objective function
$$\mathbf{y}^T E \mathbf{y} - 2 \mathbf{d}^T \mathbf{y} = (\mathbf{y}^T - 2\mathbf{d}^T E^{-1}) E \mathbf{y}$$
where $\mathbf{d}^T = (b^T A + c^T D) Q^{-1} = (b^T A + c^T D) Q^T$. Since the change of variables from $\mathbf{x}$ to $\mathbf{y}$ was a bijection from $\mathbb{R}^n \to \mathbb{R}^n$, to solve the original problem we can simply find $\mathbf{y} \in \mathbb{R}^n$ maximizing this modified objective function. But doing so is easy, because if
$$\mathbf{y} = \begin{bmatrix} y_1 & \cdots & y_n \end{bmatrix}^T, ~~ 2 \mathbf{d}^T E^{-1} = \begin{bmatrix} f_1 & \cdots & f_n \end{bmatrix}, ~~ E = \text{diag}(e_1, \cdots, e_n)$$
then the objective function is simply
$$\sum_{i = 1}^n e_i (y_i^2 - y_i f_i)$$
whose minimum (recall all the $e_i$ are positive) occurs whenever each $y_i = f_i / 2$ (we just minimize each individual quadratic term). To recover the answer to $(*)$, we simply remember that $\mathbf{x} = Q^{-1} \mathbf{y} = Q^{T} \mathbf{y}$ to obtain the final answer of
$$\mathbf{x} = Q^T (\mathbf{d}^T E^{-1})^{T} = Q^T [E^{-1} Q (A^T \mathbf{b} + D^T \mathbf{c})] = Q^T E^{-1} Q (A^T \mathbf{b} + D^T \mathbf{c}) \\
= (A^T A + D^T D)^{-1} (A^T \mathbf{b} + D^T \mathbf{c})$$
You can check that this agrees with the one variable least-squares solution by taking $D = 0$ and $\mathbf{c} = \mathbf{0}$. $\square$
Hence, the answer to the original question is
\begin{align*}
\mathbf{x}^* &= \left[(D_\mathbf{p} A)^T D_\mathbf{p} A + D_\mathbf{q}^T D_\mathbf{q} \right]^{-1} \left[ (D_\mathbf{p} A)^T D_\mathbf{p} \mathbf{b} + D_\mathbf{q} D_\mathbf{q} \mathbf{c}\right] \\ &= \boxed{(A^T D_\mathbf{p}^2 A + D_\mathbf{q}^2)^{-1} (A^T D_\mathbf{p}^2 \mathbf{b} + D_\mathbf{q}^2 \mathbf{c})}
\end{align*} |
H: What is wrong with my approach in converting a complex equation into polar form?
The equation I wanted to convert was $|z^2-1|=1$. This is a very easy example but I have no idea where I made my mistake.
Putting $z=re^{i\theta}$, we have $z^2=r^2e^{2i\theta}$. Square both sides on the equation above and use the fact that $|z|^2=z\bar{z},$ we have
$$1=|r^2e^{2i\theta}-1|^2=(r^2e^{2i\theta}-1)(\overline{r^2e^{2i\theta}-1})=(r^2e^{2i\theta}-1)({r^2e^{-2i\theta}-1})=r^4-r^2e^{-2i\theta}-r^2e^{2i\theta}+1.$$
So this gives, (either $r=0$ or) $r^2=e^{-2i\theta}+e^{2i\theta}=2\cos(2\theta).$
However, naturally I though this should corresponds to ($r=0$ or) $r=2\cos(\theta).$ (which can be obtained via $z=x+iy$)
This should be relatively easy but I have no idea why two methods dont agree.
Many thanks in advance!
AI: Alternatively, notice that ${|z^2 - 1|=1}$ would imply that ${z^2-1}$ lies on the complex unit circle. So ${\Re(z^2 - 1)=\cos(\theta)}$, and ${\Im(z^2-1)=\sin(\theta)}$ (since ${\cos(\theta) + i\sin(\theta)}$ describes the complex unit circle). You also know from simple algebraic manipulation that
$${z^2-1 = [\Re(z)^2 - \Im(z)^2 - 1] + [2\Re(z)\Im(z)]i}$$
And so you get a system of equations
$${\begin{array}{cc}\Re(z)^2 - \Im(z)^2 - 1 &=\cos(\theta)\\2\Re(z)\Im(z)&=\sin(\theta)\end{array}}$$
Which, for a given value of ${\theta}$ - is enough to deduce the value(s) of ${z \in \mathbb{C}}$. |
H: How to find a probability that sum of geometric variables is less than a number
Let $X_i, i=1, \ldots, n$ be Geometric i.i.d random variables, which represent the number of fails, with parameter $p$.
Calculate or estimate from above and below:
$$
P(\sum_{i=1}^n X_i\leq A), \quad A \in N.
$$
I know that sum of the geometric random variables is the negative binomial, but I would not know all the parameters for the negative binomial r.v.
AI: You probably mean $A \in \mathbb{N}$, i.e. $A$ is a natural number. You should probably use CLT here, as $n$ is large, $X_i$ are iid with $\mu <\infty, \sigma^2<\infty$
$$
P(S_n<A) = P(\frac{S_n-n \mu}{\sigma \sqrt{n}}<\frac{A-n \mu}{\sigma \sqrt{n}}) \approx \Phi(z)
$$
Here $z = \frac{A-n \mu}{\sigma \sqrt{n}}$, and $\mu, \sigma$ are mean and sd of each $X_i$, which you can easily find. |
H: Let $E_1 \subset E_2$ both be compact and $m(E_1) = a, m(E_2) = b$. Prove there exists a compact set $E$ st $m(E) = c$ where $a < c < b$.
Exercise 1.27 (Stein & Shakarchi): Suppose $E_1$ and $E_2$ are a pair of compact sets in $\mathbb{R}^d$ with $E_1 \subset E_2$ and let $a = m(E_1)$ and $b = m(E_2)$. Prove that for any $c$ with $a < c < b$, there is a compact set $E$ with $E_1 \subset E \subset E_2$ and $m(E) = c$.
Hint: As an example, if $d = 1$ and $E$ is a measurable subset of $[0, 1]$, consider $m(E ∩ [0, t])$ as a function of t.
Here is my intuition for tackling this exercise
Consider a countable sequence of closed cubes centered at the origin which increase in measure (essentially I imagine that the cubes fill $\mathbb{R}^d$ as they grow infinitesimally in terms of side length). Now, as the cubes grow we should be able to create a measurable set $E$ such that $E_1 \subset E \subset E_2$ where $m(E) = c$ and $a < c< b$, by taking the intersection of some limit of these closed cubes with $E_2$. Now, it's clear by construction that this set would be compact, as a closed subsets of a compact set are themselves compact.
However, I am having trouble justifying that this $E$ does exist. I'm thinking that it should because $E_1 \subset E_2$ with measures $m(E_1) = a$ and $m(E_2) = b$ remind me of $[a,b]$ some connected interval in $\mathbb{R}$. Thus, with the hint in mind, we should be able to create a continuous function and use the Intermediate Value Theorem.
With all this in mind, I'm having trouble formalizing this intuition, it's that classic tip of the tongue feeling! Any hints or correction of my intuition would be most welcome!
AI: Let $E_1, E_2$ be two compact sets in $\mathbb R^d$. Let $f : \mathbb R \to \mathbb R$ be defined by
$$ f(t) = m(E_1 \cup (E_2 \cap \{ x_1+ \cdots + x_d \le t\})),$$
where $x_1, \cdots, x_d$ are the standard coordinates of $\mathbb R^d$. Then it is easy to see that
$$ \lim_{t\to -\infty} f(t) = m(E_1), \ \ \lim_{t\to +\infty} f(t) = m(E_2).$$
Also, for any $t>s$,
$$ f(t) - f(s) \le m (E_2 \cap \{s \le x_1+ \cdots + x_d \le t\})) \le C(t-s)$$
for some $C$ depending on $n, E_2$. Thus $f$ is continuous and so your result follows from the intermediate value theorem: for any $ c$ so that $m(E_1)< c<m(E_2)$, there is $t\in \mathbb R$ so that $f(t) = c$. Let
$$ E = E_1 \cup ((E_2 \cap \{s \le x_1+ \cdots + x_d \le t\}).$$
Then $E$ is compact, $E_1 \subset E\subset E_2$ which has $m(E)=f(t) = c$. |
H: Help with proving/disproving an inequality
$\textbf{Question:}$Let $x_1,x_2,x_3,x_4 \in \mathbb{R}$ such that $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1) =16 $. Is it then true that $x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4- x_1x_2x_3x_4 \le 5$, with equality $\iff x_1=x_2=x_3=x_4=\pm 1$?
Rough calculations seem to suggest that this is indeed the case, but I am unable to prove it.
For some context, this question is actually related to USAMO $2014$ P$1$. The original question was that: given a polynomial $P(x)=x^4+ax^3+bx^2+cx+d,$ and $ b-d \ge 5$, where all $4$ roots $x_1,x_2,x_3,x_4$ of $P(x)$ are real, find the smallest value of the expression $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$.
Indeed, I managed to prove that this expression is at least $16$. But to show that the minimum value of $16$ is actually attainable, I have to find a construction of some polynomial $P(x)$ satisfying the conditions of the question. While it is indeed obvious to see that setting $ x_1=x_2=x_3=x_4=1$ or $ x_1=x_2=x_3=x_4=- 1$ both works ( just expand $(x-1)^4$ and $(x+1)^4$ respectively), are there other non-trivial values that would work too?
In particular, using $\textbf{Vieta's formula}$ gives us that $b=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4$ and $d=x_1x_2x_3x_4$, so the complicated looking expression in the first paragraph is actually just equivalent to $b-d$.
AI: Since $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2,$$ we obtain:
$$16=\prod_{k=1}^4(1+x_k^2)=((1-x_1x_2)^2+(x_1+x_2)^2)((1-x_3x_4)^2+(x_3+x_4)^2)=$$
$$=((x_1+x_2)(x_3+x_4)-(1-x_1x_2)(1-x_3x_4))^2+$$
$$+((1-x_1x_2)(x_3+x_4)+(1-x_3x_4)(x_1+x_2))^2\geq$$
$$\geq((x_1+x_2)(x_3+x_4)-(1-x_1x_2)(1-x_3x_4))^2=$$
$$=(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4- x_1x_2x_3x_4 -1)^2.$$
Can you end it now?
The equality occurs for $$(1-x_1x_2)(x_3+x_4)+(1-x_3x_4)(x_1+x_2)=0,$$ which is
$$x_1+x_2+x_3+x_4=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4$$ and for $$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4- x_1x_2x_3x_4 -1=4.$$
We have for example also the following case of the equality occurring:
$$(x_1,x_2,x_3,x_4)=\left(5,\frac{-5+\sqrt5}{2},\frac{-5-\sqrt5}{2},0\right).$$ |
H: Why is the definite integral from $a$ to $b$ is negative of integral from $b$ to $a$ graphically?
I was studying properties of definite Integrals, and I came across this property which was easy to prove.
$\int_{b}^{a}f(x) \,\mathrm{d}x = -\int_{a}^{b}f(x) \,\mathrm{d}x$
However, I had a problem with the graphical approach. If we look at the graph of $x^3-x$
Now the definite integral represents algebraic area, but in $[-1,0]$, the graph is above $x-axis$, So I thought the area should be positive either going from $a$ to $b$ or from $b$ to $a$.
But obviously, this is not true, and I guessed that it is because of the $dx$ in multiplication but I couldn't find out the proper explaination anywhere, so posting it here.
AI: The claim: The area should be positive either going from a to b or from b to a
$\!>$ is incorrect
Consider the conversion of addition to integration. In this, one assumes an element $\delta x=x_{final}-x_{initial}$.
So while going from $b$ towards $a$, the result will surely be negative of what it would have been for $a$ to $b$.
This was formulated by Rienmann as well, and called Riemann Sums.
In the above graph, $\xi_i$ are sample points in region bounded by $x_{i-1}<x<x_i$. The integration can now be simplified into summation of rectangular regions. Given by : $${A = \lim\limits_{\left| P \right| \to 0} \sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} }$$
where ${\left\| P \right\| }={ \max \left\{ {\Delta {x_1},\Delta {x_2}, \ldots ,\Delta {x_n}} \right\}}$ and ${\Delta {x_i} = {x_i} – {x_{i – 1}}}$ |
H: Integral of Series over Domain of Convergence
I'm looking for a little help regarding integration of series where the domain of integration gets very close to the edge of the series' domain of convergence. My particular case is the logistic function and it's Maclaurin series (via the geometric series expansion around $x=0$),
$$
f(x) = \frac{1}{1+e^{-x}} = \sum_{n=0}^{\infty} (-1)^n e^{-nx}.
$$
It's easy to check with the Alternating Series Test that this series converges strictly for $x > 0$. What I'm trying to wrap my head around is the following. This particular integral of $f$ is pretty straightforward:
$$
\int _{0}^{1} f(x) \, dx = \ln (1 + e^x) \Big\vert _0 ^1 = \ln (2).
$$
However, if we consider the integral of the series,
$$
\int _{0}^{1} \sum_{n=0}^{\infty} (-1)^n e^{-nx} \, dx,
$$
it's not so clear to me anymore how we may justify integrating all the way over to $x=0$ if the series does not actually converge for $x=0$. I've read other questions on the site pointing to the fact that the Dominated Convergence theorem allows you (especially in cases like these with alternating series, where Tonelli/Fubini can't help much) to exchange the limit with the integration sign, but (unless I'm missing something) this is only true provided that the sequence of partial sums converges pointwise to $f$, which to my understanding is not the case for $x=0$.
So, how is it that you go about justifying a procedure like this? A naive integration yields
$$
\sum _{n=1}^{\infty} \frac{(-1)^{n+1}}{n} e^{-nx},
$$
which actually does converge at $x=0$, and correctly evaluates to $\ln (2)$. However, I'm not thoroughly convinced that integrating the series from $0$ to $1$ is legal. Any and all help is appreciated :)
EDIT: A comment kindly pointed out that the integral does NOT in fact equal $\ln(2)$, but rather $\ln(1+e) - \ln(2)$. Still, my doubt about the validity of the integration remains. If it's valid on integrate all the way to $0$, why is it so? If it's not, then why?
AI: Indeed your made a very nice observation that is often neglected by practitioners of arcane art of integral and series.
This type of technical issue is often overcome by realizing the given expression as the limit of perturbed expressions with additional parameters. (In this regard, we might possibly borrow the physics jargon 'regularization' for this technique) Abel's Theorem is an archetypal example of this approach.
1. Let us consider OP's example in detail. One obvious resolution is to cut-off the domain of integration around the origin. So let $\epsilon \in (0, 1)$ and consider
$$ \int_{\epsilon}^{1} \frac{1}{e^x + 1} \, \mathrm{d}x. $$
Then Fubini-Tonelli Theorem is now applicable since
$$ \sum_{n=1}^{\infty} \int_{\epsilon}^{1} \left| (-1)^{n-1}e^{-nx} \right| \, \mathrm{d}x < \infty, $$
and so,
\begin{align*}
\int_{\epsilon}^{1} \frac{1}{e^x + 1} \, \mathrm{d}x
&= \sum_{n=1}^{\infty}(-1)^{n-1} \int_{\epsilon}^{1} e^{-nx} \, \mathrm{d}x \\
&= \sum_{n=1}^{\infty}(-1)^{n-1} \frac{e^{-n\epsilon} - e^{-n}}{n} \\
&= \log(1+e^{-\epsilon}) - \log(1 + e^{-1}).
\end{align*}
Now letting $\epsilon \to 0^+$ shows that the original integral is equal to $\log 2 - \log(1+e^{-1})$. So the inapplicability of Fubini-Tonelli Theorem to the original integral can be overcome by this cut-off.
2. Of course, this cut-off technique is not the only way of perturbing the integral. For instance, we may introduce a new parameter $r$ taking values in $(0, \infty)$ and then perturb the integral to introduce
$$ I(r) := \int_{0}^{1} \frac{1}{e^x + r} \, \mathrm{d}x. $$
Then it is routine to prove that $I(r) \to I(1)$ as $r \to 1$. Moreover, if $r \in (0, 1)$, then we may utilize Fubini-Tonelli Theorem to compute
$$ I(r)
= \sum_{n=1}^{\infty}(-1)^{n-1} r^{n-1} \int_{\epsilon}^{1} e^{-nx} \, \mathrm{d}x
= \frac{\log(1+r) - \log(1 + r e^{-1})}{r}. $$
Then letting $r \uparrow 1$ yields the same answer as before. |
H: Why does $\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n)} = \ln(2)$?
While proving some results on series I encountered that, one of those result implies that
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n)}$$
is convergent and it has sum equal to sum of alternating harmonic series. (And we know that alternating harmonic series converges to $\ln2$.)
However I am not able to find the sum of series $\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n)}$ directly (without that result). Is there is any way to show that sum equals to $\ln2$?
AI: $$\frac{1}{(2n-1)(2n)}=\frac{1}{2n-1}-\frac{1}{2n}$$
So the given sum is nothing but
$$S=(1-1/2)+(1/3-1/4)+(1/5-1/6)+(1/7-1/8)+.....+...$$
Now note that $$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5+....$$
So $S$ nothing but $x=1$ case of the logarithmic series written above.
Therefor $s=\ln 2$ |
H: Indefinite integral of $\sin^8(x)$
Suppose we have the following function:
$$\sin^8(x)$$
We have to find its anti-derivative
To find the indefinite integral of $\sin^4(x)$, I converted everything to $\cos(2x)$ and $\cos(4x)$ and then integrated. However this method wont be suitable to find the indefinite integral $\sin^8(x)$ since we have to expand a lot. Is there any other way I can evaluate it easily, and more efficiently?
AI: I copied and pasted this answer, since I actually wrote this answer to a different question which you can find here: Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? ).
Define ${S_n = \int\sin^{2n}(x)dx}$. Then
$${S_{n}=\int \sin^2(x)\sin^{2n-2}(x)dx=S_{n-1}-\int \cos^2(x)\sin^{2n-2}(x)dx}$$
On the rightmost integral, using integration by parts yields
$${\int\cos^2(x) \sin^{2n-2}(x)dx=\frac{\cos(x)\sin^{2n-1}(x)}{2n-1}+\frac{1}{2n-1}\int \sin^{2n}(x)dx}$$
So overall
$${\Rightarrow S_n = S_{n-1}-\frac{\cos(x)\sin^{2n-1}(x)}{2n-1} - \frac{1}{2n-1}S_n}$$
And so
$${\left(\frac{2n}{2n-1}\right)S_n = S_{n-1} - \frac{\cos(x)\sin^{2n-1}(x)}{2n-1}}$$
$${\Rightarrow S_n = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$
Now you have a recursion relation that will help you compute the integral for higher even powers of ${\sin(x)}$:
$${S_{n} = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$ |
H: Separability of a $\sigma$-algebra generated by an algebra
Let $X:(Ω,\mathcal{F})$ be a measurable set and $\mathcal{F}=\sigma(C)$ where $C$ is an algebra .
Now We define $B_\mathcal{w}=\underset{\mathcal{w}∈A,A∈C}{\cap}A$, which suggests the intersection of all sets in $C$ that contain $\mathcal{w}$.
Similarly, we define $\mathcal{F}_\mathcal{w}=\underset{\mathcal{w}∈A,A∈\mathcal{F}}{\cap}A$,which suggests the intersection of all sets in $\mathcal{F}$ that contain $\mathcal{w}$.
If there exists another element $x$ which does not equal to $\mathcal{w} $ and $x$ belongs to $B_\mathcal{w}$. Now I think its trivial that $x$ belongs to $\mathcal{F}_\mathcal{w}$ but I couldn't give a precise proof.
Moreover, could we replace the condition "$C$ is an algebra" by "$C$ is an arbitrary set class which generates $\mathcal{F}$" and get the same conclusion?
AI: Say $\mathscr{E}\subset \mathcal{P}(\Omega)$ separates points if for every $x,y\in \Omega$, if $x\neq y$ then there is some $E\in\mathscr{E} $ such that exactly one of $x$ or $y$ belongs to $E$.
$\mathscr{E}$ separates points iff $\sigma(\mathscr{E})$ separates points.
$\Rightarrow)$ Trivial.
$\Leftarrow)$ Fix $x,y\in\Omega$ with $x\neq y$. Suppose $\mathscr{E}$ does not separate $x$ and $y$. Let
$$
\mathfrak{X}:=\{A\subset \Omega: \{x,y\}\subset A\vee \{x,y\}\cap A=\emptyset\}
$$
You can check that $\mathfrak{X}$ is a $\sigma$-algebra. By our assumption, $\mathscr{E}\subset\mathfrak{X}$, so $\sigma(\mathscr{E})\subset \mathfrak{X}$, so $\sigma(\mathscr{E})$ does not separate $x$ and $y$.
Remark: $\mathscr{E}$ was an arbitrary collection of subsets of $\Omega$, not necessarily an algebra. |
H: Prove that $1<\int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx<\sqrt{\frac{\pi}{2}}$using integration.
Prove that $$1<\int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx<\sqrt{\frac{\pi}{2}}$$ using integration.
My Attempt
I tried using the Jordan's inequality
$$\frac{2}{\pi}x\leq\sin x<1$$
Taking square root throughout
$$\sqrt{\frac{2x}{\pi}}\leq \sqrt{\sin x}<1$$
On integrating throughout
$$1<\frac{\pi}{3}\leq \int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx<\frac{\pi}{2}$$
But I am not getting $\sqrt{\frac{\pi}{2}}$ as required.
AI: We can do better: using this post
$$
\int _0^{\pi/2} \sqrt{\sin(x)} \,dx = \sqrt{\frac{2}{\pi}}\Gamma\left(\frac{3}{4}\right)^2 \approx 1.19
$$
If you want a numerical answer,
$$
\int _0^{\pi/2}\sqrt{\sin(x)}\,dx > \int _0^{\pi/2} \sin(x)\,dx = 1
$$
On the other hand, by Cauchy-Bunyakovsky-Schwarz
$$
\left(\int _0^{\pi/2}\sqrt{\sin(x)}\,dx\right)^2\leq \left(\int _0^{\pi/2} 1\,dx \right)\cdot \left(\int _0^{\pi/2} \sin(x)\,dx \right) = \frac{\pi}{2}
$$ |
H: How do you find the area of the 'floors' in a torus?
Take a Torus such that 'r', the radius of the circle of the torus's cross-section (see fig. 1), is $300$ m.
Now, given that there are two 'floors' in said Torus and the sum of their areas is $864,000 \; m^2$, find R. (see fig. 2) [Floors highlighted.]
This is what I tried:
To find the length of the 2nd floor, draw a line from the centre to F (Fig. 2). This is 300 m . We then get a right angle with hypotenuse 300 m, height 200 m. The base is easily calculated to be 223.60 (2d.p.) by The Pythagorean Theorem. Double that to get the length of the 2nd floor as the line segment from the centre is perpendicular (known) to the chord and thus bisects it.
Length of Floor-1 = $600 \; m$
Length of Floor-2 = $447.21 \; m $(2d.p.)
First consider the Triangle T-1,
$\text{Height} = 200 \ m \quad \; and \quad \text{Base} = 300 - 223.60 = 76.40 \; (2 d.p.) $
So, $$ \text{Hyp.} = 214.1 \quad (1 \, d.p.) $$
Also, note that the angle between the base and hypotenuse is $69.1$ DEG.
Now, consider T-2,
The largest angle is $180 - 69.1 = 110.9$ DEG, and the unlabelled short length is 214 m.
Applying Law of Cosines we get:
$$c = \sqrt{R^2 + 159.64R + 45796} $$
Finally consider T-3,
Since the height of the triangle remains 200 m, Applying Pyhtaogra's Theorem:
$$ d = \sqrt{R^2 + 159.64R + 5796} \qquad (1)$$
We can now finally calculate the areas of the floors (as rings) in terms of 'R'.
For Floor-1:
$$A_1 = \pi [ \ (R+600)^2 - R^2 \ ] $$
Which simplifies to:
$$ A_1 = \pi [ \ 1200R + 3.6 \times 10^5 \ ] \qquad (2) $$
Likewise for Floor-2,
$$A_2 = \pi[ \ (d+447.21)^2 - d^2 \ ] \implies A_2 = \pi[\ 894.42d + 1.999 \times 10^5 \ ] \qquad (3)$$
Total Area then is $A_1 + A_2 = \pi[ \ 894.42d + 1200R + 5.60 \times 10^5 \ ]$
Which becomes:
$$A_{Net} = \pi[\ \sqrt{R^2 + 159.64R + 5796} + 1200R + 5.6 \times 10^5 \ ] \qquad (4) $$
Since $A_{Net}$ is given to be $864,000 \ m^2$:
$$ \pi[\ \sqrt{R^2 + 159.64R + 5796} + 1200R + 5.6 \times 10^5 \ ] = 864,000 $$
I then proceed to solve it as follows:
$$\sqrt{R^2 + 159.64R + 5796} = 1200 R - \big[ {864000\over \pi} {-5.6 \times 10^5} \big] $$
Squaring both sides,
$$ R^2 + 159.64R + 5796 = 1.44 \times 10^6 R^2 + 6.84 \times 10^8 R + 8.12 \times 10^{10} $$
This reduces to the following quadratic:
$$ 1.44 \times 10^6 R^2 + 6.84 \times 10^8 R + 8.12 \times 10^{10} = 0 \qquad (5)$$
(You might've noticed that I've taken some liberties with rounding. I'm just working towards getting an answer that's good enough, hopefully that shouldn't mess up the calculations too much)
Solving this quadratic -- and here's the problem! -- leads to negative results, which is obviously bad.
$R_1 = -233 {1\over 3}$ & $R_2 = -247 {2\over 3}$
So what exactly am I doing wrong? Is the problem with the working? Or is there something wrong with the method? I tried solving this with another method, which was basically thinking of the torus as it's cartesian (or polar) equation and then solving for the sum of the areas of intersection (area 'within the torus') of two parallel planes 200m apart where one cuts through the centre of the torus, horizontally. It didn't go well though...
Hyelp!
[Feel free to edit the question, working, pictures & tags of the question to make it more readable]
AI: The surface of the upper floor is $$\pi\left((R+r)^2-(R-r)^2\right)
=4\pi Rr$$ and of the lower floor $$\pi\left((R+\frac{\sqrt{5}}{3}r)^2-(R-\frac{\sqrt{5}}{3}r)^2\right)=\frac{4\sqrt{5}}{3}\pi Rr.$$
Since $$4\pi Rr+\frac{4\sqrt{5}}{3}\pi Rr=864000$$ and $r$ is known, we find $$R=\frac{864000}{4\pi\left(1+\frac{\sqrt{5}}{3}\right)r}\approx 131.31024<r$$
so no such torus can exist, thus you can get arbitrary non-consistent equations (and even $R<0$ too) assuming it exists (like "false implies anything").
Edit:
Let all the points mentioned above lie on a vertical plane passing through the center of symmetry $O$ (if the floors are horizontal). Let $O,C,A,B$ be on the upper floor and $O_1,F,E,G$ be on the lower one. $OA=R$ and $AC=AF=AD=AG=AB=r$, $AD\perp BC$, $E\in AD$, $AE=\frac 23 AD$ thus (by Pythagorean theorem) $EF=r\sqrt{1-\frac 49}=r\frac{\sqrt{5}}{3}$. $OAEO_1$ is a rectangle, hence $O_1E=OA=R$ thus
$$O_1G=O_1E+EG=O_1E+EF=R+r\frac{\sqrt{5}}{3},\\
O_1F=O_1E-EF=R-r\frac{\sqrt{5}}{3}.$$ |
H: How can I prove that each component of a 2-regular graph is a cycle?
I tried to look at the similar questions asked before, however, they all assume that if a graph is simply one vertex then that also constitutes a cycle which is an assumption I'm not allowed to make. I did a lot of cases and my question statement turned out to be true in each case, however I don't know how to show that each and every component is a cycle?
Any help is appreciated
AI: Start with a vertex $v_1$. It has degree $2$, so let $v_2$ be one of the vertices adjacent to $v_1$. The degree of $v_2$ is $2$, so it is adjacent to $v_1$ and one other vertex; call that other vertex $v_3$. Keep going, constructing a path $v_1,v_2,v_3,\ldots\;$. The graph is finite, so at some point you’ll run out of unused vertices and have a path $v_1,v_2,\ldots,v_n$ such that the vertices adjacent to $v_n$ are $v_{n-1}$ and one of the other vertices in the path. That vertex cannot be any of the vertices $v_2,v_3,\ldots,v_{n-2}$: if $2\le k\le n-2$, $v_k$ is adjacent to $v_{k-1}$ and $v_{k+1}$, and $\deg(v_k)=2$, so $v_k$ isn’t adjacent to any other vertex. The only possibility is $v_1$: $v_n$ must be adjacent to $v_1$, and we have a cycle.
The regularity hypothesis clearly implies that there are no edges from this cycle to any other vertex, so this cycle must be a component of the graph. If you set up the proof as an induction on the number of components in the graph, this argument covers both the base case of a connected graph and the induction step, since it reduces the number of potentially non-cyclic components by $1$. |
H: Find the eccentricity of the conic $4x^2+y^2+ax+by+c=0$, if it tangent to the $x$ axis at the origin and passes through $(-1,2)$
Solving this would require three equations
(1) Tangent to x axis at origin
Substituting zeroes in all $x$ and $y$ gives $c=0$
(2) Passes through (-1,2)
$$4(1)+4-a+2b=0$$
$$-a+2b=-8$$
How do I find the third equation?
AI: As you mentioned, one of the points on the ellipse is (0,0). Equation of the ellipse is
$\frac{(x+\frac{a}{8})^2}{1^2}+\frac{(y+\frac{b}{2})^2}{2^2} = {(\frac{a}{8})}^2 + {(\frac{b}{4})}^2$
For ellipse equation $\frac{(x \pm h)^2}{A^2}+\frac{(y \pm k)^2}{B^2} = 1$, eccentricity of ellipse
= $\frac{\sqrt{B^2-A^2}}{B}$ (where $B \ge A$)
= $\frac{\sqrt3}{2}$ (where $A = t, B = 2t$)
Here, $t = \sqrt{{(\frac{a}{8})}^2 + {(\frac{b}{4})}^2}$
Alternatively:
Based on the fact that it is tangent to x-axis at origin, the center of the ellipse will have to be on y-axis (as it is not a slanting ellipse). That means $a = 0, c = 0, b = -4$. So equation of ellipse is $x^2 + \frac{{(y-2)}^2} {4} = 1$ with major axis (along $y$ axis) $= 4$, minor axis $= 2$.
So, eccentricity = $\frac{\sqrt3}{2}$ |
H: Transitive models of $V≠L$ within L
Suppose $V=L$. Can there be transitive models of $ZFC+V≠L$?
Let $M$ be a transitive model of ZFC. If $x\in M$, then $x\in L_\alpha$ for some $\alpha$ because $V=L$, but it's not evident to me that $\alpha\in M$.
Such an $M$ would necessarily have to be a set, since the only inner model is $L$ itself.
AI: Yes, transitive models inside $L$ can be very non-$L$-ish.
Specifically, recall Shoenfield absoluteness. Since "$T$ has a countable transitive model" is $\Sigma^1_2$,$^*$ by applying Downwards Lowenheim-Skolem in $V$ we have that whenever $T$ is a theory with a transitive set model in $V$ then $T$ has a countable transitive model in $L$.
So, for example, if $\mathsf{ZFC+\neg CH}$ + "There is a proper class of supercompacts" has a transitive model, then it has a constructible transitive model, despite the fact that of course that theory is highly incompatible with the axiom of constructibility for both large cardinal and combinatorial reasons.
It may help, visualization-wise, to consider e.g. $L_{\omega^2}(\mathbb{R}^L)$. This is a transitive set contained in $L$ of height $\omega^2$ but containing all constructible reals - basically, it's "short and wide" in a way which makes it very different from any level of $L$. Now that's sort of a bad example since it's "informationally" equivalent to the better-behaved $L_{\omega_1^L}$ - each $L_\alpha$ for $\alpha<\omega_1^L$ is represented by a real in $L_{\omega^2}(\mathbb{R}^L)$ and can be "decoded" in a definable way - but it's a good first taste of how the particular shape of the levels of $L$, rather than their mere constructibility, constrains their behavior.
EDIT: There's another theorem which is relevant here. It gets a weaker conclusion than Shoenfield, but is quite different and interesting:
(Barwise) Every countable model of $\mathsf{ZF}$ has an end extension which is a model of $\mathsf{ZFC+V=L}$.
See here. Of course, that end extension will probably be ill-founded - that's why Barwise's theorem doesn't prove $\mathsf{V=L}$ outright. Incidentally, the picture at that blogpost is quite nice on its own - note the added width, in addition to height, per the comment about the shape of levels of $L$ in the previous section. Barwise's theorem does not give us a top extension.
Barwise's theorem lets us transfer consistency results: if $\mathsf{ZFC}$ + "There is a transitive model of $T$" is consistent then so is $\mathsf{ZFC+V=L}$ + "There is a transitive model of $T$." More generally, note that end extensions preserve internal transitivity satisfaction: if $M\models\mathsf{ZF}$, $A$ and $T$ are in $A$, $M$ thinks $T$ is a theory and $A$ is a transitive set satisfying $T$, and $N$ is an end extension of $M$ (perhaps one satisfying $\mathsf{ZFC+V=L}$!), then $N$ also thinks that $A$ is a transitive model of the theory $T$.
$^*$OK, that's not strictly true: rather, it's $\Sigma^1_2$ relative to $T$. So really all we can conclude is that every $\{\in\}$-theory which is in $L$ which has a transitive model in $V$ also has one in $L$.
A good example of how this can play out is to consider the following. Let $T_0=\mathsf{ZFC}$ + "$0^\sharp$ exists," and let $T_1$ be $T_0$ + axioms correctly stating each bit of $0^\sharp$. Now per the above (under reasonable hypotheses) $T_0$ has a transitive model in $L$. On the other hand, $T_1$ definitely won't: a transitive model of $T_1$ has to compute $0^\sharp$ correctly, which $L$ can't. But this is fine, since $T_1$ itself computes $0^\sharp$: $T_1\not\in L$ so we can't apply Shoenfield. |
H: How to find parameters of negative binomial random variable?
Consider $X_i \sim NB(r_i, p)$, where Let $X_i$ be i.i.d. for $1, \ldots, N$. Let $Y_i\sim Geometric (p)$.
Then,
$X_1\sim NB(r_1,p)$ satisfies $X_1 = Y_1 + \cdots +Y_{r_1}$,
$X_2\sim NB(r_2,p)$ satisfies $X_2= Y_{r_1+1} + \cdots + Y_{r_1+r_2}$,
$\ldots$
$X_N\sim NB(r_N,p)$ satisfies $X_N= Y_{r_{N-1}+1} + \cdots + Y_{r_1+\ldots+r_N}$.
Denote $Z_1=\max_{1\leq j\leq r_1}(Y_j), \ldots, Z_N=\max_{r_{N-1}+1\leq j\leq r_1+\ldots +r_N}$.
Question:
Is it possible to find parameter $r$ of Negative Binomial random variable $S=\sum_{i=1}^N Z_i \sim NB(r,p)$?
AI: So $Z_i$ are the max of $r_i$ independent geometric random variables with parameter $p$ (with the convention that $0$ is in the support). Thus $P(Z_i \leq n)=(1-(1-p)^{n+1})^{r_i}$.
This is not a geometric CDF, thus you should not expect the sum of the $Z_i$ to be negative binomial distributed. |
H: Existence of an analytic function in a unit disc
Let $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$.Then, is it true that there exists a holomorphic function $f:\mathbb{D} \to\mathbb{D}$ such that $f(\frac{1}{2})=-\frac{1}{2}$ and $f'(\frac{1}{4})=1$?
Can we use Schwarz's lemma to solve it?
AI: This is a double application of Schwarz-Pick; let $f(1/4)=w$
first $1=|f'(1/4)| \le \frac{1-|w|^2}{1-|1/4|^2}$, so one gets $|w| \le 1/4$
then: $|\frac{f(1/2)-f(1/4)}{1-\overline {f(1/2)} f(1/4)}| \le |\frac{(1/2)-(1/4)}{1-\overline {(1/2)} (1/4)}| =\frac {2}{7}$, so:
$|\frac{2w+1}{2+w}| \le \frac {2}{7}$ with $|w| \le 1/4$
But (see below for a direct proof) $|\frac{2w+1}{2+w}| \ge |\frac{2(-1/4)+1}{2+(-1/4)}|=2/7$ when $|w| \le 1/4$ with equality iff $w=-1/4$ hence we get $f(1/4)=-1/4$ and $f$ is a disc automorphism as we have equality in Schwarz-PIck, while it is obvious that $f(z)=-z$ is the only such, but $f'(1/4)=-1$ so contradiction and no such function exists!
(for example $|\frac{2w+1}{2+w}| =|2-\frac{3}{2+w}| \ge 2-|\frac{3}{2+w}|$, while $-|\frac{3}{2+w}| \ge -|\frac{3}{2+(-1/4)}|=-12/7$ since $|2+w| \ge |2+(-1/4)|, |w| \le 1/4$) |
H: Can an integer that is $3\pmod 7$ be expressed as a sum of two cubes?
I was going back through my past exams in a Discrete Mathematics Course and came across this problem which I failed to solve- Does there exist $x$ and $y$ s.t $x^3+y^3 \equiv 3\pmod 7$? Give a convincing proof of your assertion.
I went through some examples and couldn't find any such $x$ and $y$. My attempt was that of considering parity. For an integer which is $3\pmod 7$, it can be either even or odd. If it is even then either $x$ and $y$ must be even or $x$ and $y$ must be odd. Considering the even case, we get that $8a^3+8b^3 \equiv 3\pmod 7$. But This expression is always even and I wasn't really sure where to go from here. I feel like considering parity would be the right approach but I did similar things considering odds and other cases and couldn't gain any traction.
Any help would be appreciated.
AI: Hint: Use Fermat little theorem:
If $7\nmid x$ then $$ x^6\equiv 1 \pmod 7 $$ and so $$ x^3\equiv \pm 1 \pmod 7 $$ |
H: Why isn't Universal enveloping algebra graded?
Given a Lie algebra $L$, define $U(L) = T(L)$ mod $I(L)$ where $T(L)$ is the tensor algebra of $L$ and $I(L)$ is the two sided ideal of $T(L)$ generated by all elements of the form $xy-yx-[x,y]$ where $x,y \in L$. Can somebody explain to me why the generators of $I(L)$ are not homogeneous for the grading of $T(L)$? It seems to me that since every generator is in $L$ this shouldn't be a problem...
AI: You wrote down a generator of $I(L)$: $xy-yx-z$
where $z=[x,y]$. This is inhomogeneous: the $xy$ and $yx$ have degree $2$ but $z\in L$
has degree $1$. |
H: How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$?
This is INMO 2000 Problem 2.
Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}
My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)=z(z-3) \implies y=\frac{z(z-3)}{3x}$ and $x=\frac{z(z-3)}{3y} $.
Note that since $z$,$x$,$y$ is an integer, we must have $3\mid z$.
So, let $z=3k$.
So we have $y=\frac{3k(3k-3)}{3x}=\frac{k(3k-3)}{x}$ and $x=\frac{z(z-3)}{3y}=\frac{k(3k-3)}{y}$ .
Then I am not able to proceed.
Hope one can give me some hints and guide me.
Thanks in advance.
AI: We have $(1-z)(x^2-xy+y^2)=1-z^2.$
If $z=1$, so $x+y=0$ and we obtain $(t,-t,1)$, where $t$ is an integer.
Let $z\neq1$.
Thus, $$x^2-xy+y^2=z+1$$ and $$x+y=1-z,$$ which gives $$(1-z)^2-3xy=z+1$$ or
$$3xy=z^2-3z.$$
Thus, $z$ is divisible by $3$ and $$(1-z)^2-\frac{4}{3}(z^2-3z)\geq0$$ or
$$z^2-6z-3\leq0$$ or
$$3-\sqrt{12}\leq z\leq 3+\sqrt{12},$$
which gives $$0\leq z\leq 6$$
Can you end it now? |
H: how many $\sqrt{2}^{\sqrt{2}^{\sqrt{2}\ldots }}$
Use $y=x^{\frac{1}{x}}$ graph and think the following calculate.
$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}\ldots }}$$
I want to know how many $\sqrt{2}^{\sqrt{2}^{\sqrt{2}\ldots }}$ will be.
When $\sqrt{2}^{\sqrt{2}^{\sqrt{2} }}$ =$2$
I can't solve this calculate with using graph.
AI: Let $a_1=\sqrt2$ and $a_n=\sqrt2^{a_{n-1}}$.
First of all, the $a_n$ is bounded above by $2$. This is clearly true when $n=1$. For $n>1$, suppose $a_{n-1}<2$. Then $a_n=\sqrt2^{a_{n-1}}<\sqrt2^2=2$. By induction, $a_n<2$ for all $n$.
Next, I claim this sequence is increasing. This is clear when $n=1$. When $n>1$, suppose $a_n>a_{n-1}$. Then $a_{n+1}=\sqrt2^{a_n}>\sqrt2^{a_{n-1}}=a_n$. By induction $a_n<a_{n+1}$ for all $n$.
Thus, $\lim_{m\to\infty}a_n$ exists. Call it $x$. Then $x=\lim_{n\to\infty}a_{n+1}=\sqrt2^{\lim_{n\to\infty}a_n}=\sqrt2^x$. Since $x\le 2$, the only solution to this is $x=2$.
P.S. The solution $x=2$ is unique since for $x\le 2$, we have $(x^{1/x})'=-x^{1/x-2}(\ln(x)-1)<0$.
P.P.S. The value of $\lim_{n\to\infty}a_n$ changes by what value you choose $a_1$ to be (with the same recursive definition). What we have seen above is that for $a_1=\sqrt2$, we have $\lim_{n\to\infty}a_n=2$. However, clearly, if $a_1=4$, $\lim_{n\to\infty}a_n=4$. In fact, it can be shown that for $a_1<4$ the limit is $2$ and for $a_1=4$ the limit is $4$. For $a_1>4$ the limit diverges. |
H: Creating a graph within specific vertex/degree parameters
Suppose that T is a tree with four vertices of degree 3, six vertices of degree 4, one vertex of degree 5, and 8 vertices of degree 6. No other vertices of T have degree 3 or more. How many leaf vertices does T have?
So this seemed simple enough at first; eventually it became abundantly clear that it was not. I assumed that there would be multiple configurations all yielding the same amount of leaves, but I kept adding in extra vertices that disqualified my attempted graphs. Presumably, one would start with the one 5-degree vertex and then go on from there, but it seems like the only way to find a tree that satisfies the listed condition is hidden behind a wall of tedious trial and error.
Is there a property or formula I'm unaware of that'd leave a framework for making such a graph?
AI: HINT: You can remove any vertex of degree $2$ from a tree without changing the number of leaves, so you might as well assume that there are no vertices of degree $2$. Then you have $19$ internal vertices whose degrees sum to $89$. Suppose that the tree has $\ell$ leaves and $e$ edges. Then $\ell+19=e+1$, since the graph is a tree. Can you find a second equation relating $\ell$ and $e$? If you get completely stuck, there’s a further hint in the spoiler-protected block below; mouse over to see it.
Handshake lemma. |
H: If $A$ is a positive definite real symmetric matrix and $\lambda$ its eigenvalue then $A^r v = \lambda^r v$ for all real $r>0$ .
Suppose $A$ is a positive definite real symmetric matrix with eigen value $\lambda$ and $v$ an eigenvector corresponding to $\lambda$. How to prove that
$$A^rv= \lambda^r v$$
holds for all real $r>0$.
So far I have manage to prove that $A^n v = \lambda^n v$ holds for all natural number $n>0$ by induction on $n$. From this point I hope I have to prove that it holds for all positive rational number $r>0$ then go on to prove for the real case. But the problem is how to proceed from natural number to rational number then to real number.
Let alone the equality how did they even define rational power and real power of matrices. Can anyone help me.
AI: Let $\lambda_1,\cdots,\lambda_n>0$ be eigenvalues of $A$. Then there is an orthonormal matrix $S$ (namely $S^TS=1$) such that
$$ S^TAS=\text{diag}(\lambda_1,\cdots,\lambda_n). $$
So
$$ S^TA^rS=\text{diag}(\lambda_1^r,\cdots,\lambda_n^r) $$
which gives
$$ A^rS=S\text{diag}(\lambda_1^r,\cdots,\lambda_n^r). $$
Let $S=(v_1,\cdots,v_n)$ and then
$$ A^rv_i=\lambda_i^rv_i. $$ |
H: When is it possible to "move" an exponent out of a radical?
It seems when a value in a radical is positive it's valid to "move" the exponent out of the radical.
Consider the function $\sqrt{x^2}$. When $x\geq0$ then
$\sqrt{x^2} = (\sqrt{x})^2$
For example, when $x = 5$
$\sqrt{5^2} = (\sqrt{5})^2$
$\sqrt{25} = \sqrt{5} \cdot \sqrt{5}$
$5 = 5$
However, when $x<0$ it's no longer valid to "move" the exponent out.
For example, when $x=-5$
$\sqrt{(-5)^2} \not= (\sqrt{-5})^2$
$\sqrt{25} \not= \sqrt{-5} \cdot \sqrt{-5}$
$5 \not= -5$
Also, it seems if a value inside a radical can be rendered positive, then an exponent can be "moved" out.
For example,
$\sqrt{(x^2)^3} = (\sqrt{x^2})^3 = |x|^3$
Thus, is it correct to say that if a value inside a radical is positive or can be rendered positive, an exponent can be "moved" out from a radical?
I'm guessing if this is the case, then it has something to do with the product rule of radicals which requires the values inside the radicals to be positive beofre being "combined":
if $a \ge 0$ and $b \ge 0$, then $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$
AI: Yes, it is correct, and it follows from the properties of powers:
$$(x^{\alpha})^{\beta}=(x^{\beta})^{\alpha}$$
whenever all the quantities involved are well defined, e.g. for positive $x$ and real $\alpha,\beta$. |
H: What are the set theoretic properties of vector spaces?
I am reading introductory Quantum Mechanics, where it says-
For a classical system, the space of states is a set (the set of
possible states), and the logic of classical physics is Boolean. The space of states of a quantum system is not a mathematical set; it is a vector space.
Further, in the footnote, it says-
To be a little more precise, we will not focus on the set-theoretic
properties of state spaces, even though they may, of course, be regarded as sets.
What does "set-theoretic
properties of state spaces" mean?
AI: Although I think @Michael's comments may be onto something, I have my own guess as to what Susskind means. A vector space is an ordered $4$-tuple $\mathcal{V}:=(V,\,\Bbb K,\,+,\,\cdot)$, with $+$ telling us how to add elements of $V$, and $\cdot$ telling us how to multiply them by elements of the field $\Bbb K$. I won't repeat all the vector space axioms you probably already know. The important point is that the vector space is not just $V$ itself. But we cheat the set theory a little:
elements of $V$ will often, in an abuse of notation/terminology, be called elements of $\mathcal{V}$, which in turn may just be called $V$;
we don't care about the details of how your favourite set theory defines an ordered $n$-tuple. |
H: Multiplication group for $\mathbb Z_n$ modulo $n$
By definition:
Let $\mathbb Z^+_n = \{[0],[1],[2],\ldots,[n−1]\}$
$\mathbb Z^+_4 = \{[0],[1],[2],[3]\},$
but
how $\mathbb Z^*_{12}$ is $\{[1],[5],[7],[11]\}$ ?
how $\mathbb Z^*_{7}$ is $\{[1],[2],[3],[4],[5],[6]\}$ ?
AI: You should distinguish between the additive groups $Z_4$ and
$Z_{12}$, which has $\{[0],[1],[2],[3],[4],...,[11]\}$,
and the multiplicative groups $Z_4^*$, which has $\{[1],[3]\}$, and $Z_{12}^*$, which has $\{[1],[5],[7],[11]\}$.
Those marked with $^*$ have the elements with multiplicative inverses.
Addendum to answer additional question in comment from OP:
The elements with multiplicative inverses in the ring $Z_{12}$ are those
whose representatives are relatively prime to $12$, i.e., not multiples of $2$ or $3$. |
H: Calculate the sum of the series $\sum_{n\geq 0}^{}a_{n}x^n$ with $(a_n)$ periodic
Let $(a_{n})_{n\geq0}$ be a periodic sequence of period $T$ satisfying for all $n\geq0$: $a_{n+T}=a_{n}.$
Calculate the sum of the series of coefficients $a_{n}$ with
$$S(x)=\sum_{n=0}^{\infty}a_{n}x^n$$ for $|x|<R $, where $ R $ is the radius of convergence.
My try:
If $|x|<R$, so $$S(x)=\sum_{n=0}^{\infty}a_{n}x^{n}=\sum_{n=0}^{T-1}a_{n}x^n+\sum_{n=T}^{\infty}a_{n}x^n$$
then by changing the summation index. $$S(x)=\sum_{n=0}^{T-1}a_{n}x^n+\sum_{n=0}^{\infty}a_{n+T}x^{n+T}$$
then using the periodicity
$$S(x)=\sum_{n=0}^{T-1}a_{n}x^n+\sum_{n=0}^{\infty}a_{n}x^{n+T}=\sum_{n=0}^{T-1}a_{n}x^n+x^{T}\sum_{n=0}^{\infty}a_{n}x^n$$
and finally $$S(x)=\sum_{n=0}^{T-1}a_{n}x^n+x^{T}S(x) \implies S(x)=\frac{\sum_{n=0}^{T-1}a_{n}x^n}{1-x^{T}}$$
Is it right or wrong? We are waiting for your answer. Thank you for your help
AI: This looks alright! It's technically not a rigorous proof, as you don't use limits or determine the radius of convergence (which should be about $1$). But if you know the ratio test, you can say that for $x < 1$ the ratio test says that it converges, and so it must converge to this! |
H: If $\int_{-2}^{3} [2f(x)+2]\,dx = 18$, and $\int_1^{-2} f(x)\,dx=-8$, then $\int_1^{3} f(x)\,dx$ is equal to what?
If $\int_{-2}^{3} [2f(x)+2]\,dx = 18$, and $\int_1^{-2} f(x)\,dx=-8$, then $\int_1^{3} f(x)\,dx$ is equal to what?
So I've tried the regular sum rules and tried plugging in everything I can for any case of substitution, but I can't find out a correct answer. I thought 8/3=f(x) from the second integral, but that doesn't hold up for the first integral, because f(x) there is equal to 4. No idea where to go.
AI: If $\int_{-2}^{3} [2f(x)+2]\,dx = 18-> \int_{-2}^{3} f(x)dx=4 $, then $\int_1^{3} f(x)\,dx$= $\int_{1}^{-2} f(x)dx$ + $\int_{-2}^{3} f(x) dx$=$-8+4=-4$ |
H: Converse of Deduction Theorem
I have a basic question about natural deduction and deduction theorem. I learn from my textbook that the deduction theorem
$$\textit{If }\ \Gamma,A\vdash B,\ \textit{ then }\ \Gamma\vdash A\rightarrow B.$$
corresponds to the introduction rule of $\rightarrow$ in natural deduction. This is trivially in fact. But what about the converse?
$$\textit{If }\ \Gamma\vdash A\rightarrow B,\ \textit{ then }\ \Gamma,A\vdash B.$$
It also holds. But what does it correspond to in natural deduction? Or how is it expressed in natural deduction?
It seems to me that the elimination rule of $\rightarrow$ does similar work, but it's different from the converse of the deduction theorem.
Thanks!
AI: Yes, it is just a quick consequence of the elimination rule.
$\to $ Elimination (in its most common phrasing) says that if $\Gamma\vdash A\to B$ and $\Gamma\vdash A$ then $\Gamma\vdash B.$
So, if $\Gamma\vdash A\to B,$ then $\Gamma, A\vdash A\to B.$ And of course $\Gamma,A\vdash A,$ so applying the elimination rule gives $\Gamma,A\vdash B.$ |
H: Which open sets are invariant under rotations?
Let $Q \in \operatorname{SO}(2)$, and let $U \subseteq \mathbb R^2$ be an open, bounded, connected subset. Suppose that $QU = U$.
Is it true that $Q$ must be a disk, or the interior of a regular polygon (if $Q$ is a rotation by $2\pi/n$ then a regular $n$-gon would be invariant).
If not, can we characterize all such possible invariant $U$'s?
Edit:
As mentioned by Brian M. Scott, open annuli and their regular polygonal counterparts are also possible.
AI: Since $Q$ is a rotation, it either has finite order $n$ or is an irrational rotation. If $Q$ is an irrational rotation and $Q(U)=U$, then $U$ is invariant under all rotations, and every connected component of $U$ is either a disk, an annulus or the whole plane.
If $Q$ has finite order, I don't think there can be a reasonnable classification. Indeed, in that case $X:=\mathbb C^*/\langle Q \rangle$ is homeomorphic to $\mathbb C^*$, and you can take any open set in $X$ and lift it to $\mathbb C$ to get a $Q$-invariant open set.
Edit: more details on the irrational case.
Observe that if $z_0 \in U$ and $U$ is $Q$-invariant (where $Q$ is an irrational rotation), then the circle $|z|=|z_0|$ is in $U$. Let $A:=\{ r \geq 0: S_r \subset U\}$, where $S_r$ is the circle of center $0$ and radius $U$.
Using the previous observation, you can check that $A$ is an open subset of $\mathbb R^+$, so it's a union of intervals (open except possibly at $r=0$).
Again with the observation, you can see that $U=\{ r e^{it}, (r,t) \in A \times [0,2\pi)\}$, hence a union of (at most one) disk and annuli, or the whole plane if $A=\mathbb R^+$. |
H: Do algebraic fields in mathematics have any connection to the fields in physics?
Basic question, but is there some sort of connection between the two, or are they just separate definitions.
AI: They are different things. Algebraically fields are just a number system. Good examples are the rational, real, and complex numbers.
In physics fields are typically vector functions over some domain, often real valued. An example would be a function that returned air pressure given some point on Earth. They're only related in so much as fields play an important role in defining vector spaces. |
H: Two different definitions for a model in first order logic?
So, if $T$ is a theory in a first order language $\mathcal L$, I thought a model for $T$ is
a set $M$ with interpretations for all the constant, function and relation symbols of $\mathcal L$, in which all statements in $T$ are true.
But recently, I had someone calling a model
a function $\beta$ that
assigns a truth value to each $\mathcal L$-statement,
assigns "true" to all statements in $T$, and
is compatible w.r.t. the usual inference rules, e.g. $\beta(\phi\land\psi)$ gives "true" if and only if $\beta(\phi)$ and $\beta(\psi)$ both give "true".
Question:
Is the latter also called a model of $T$?
Are these definitions equivalent, and if so, are they equivalent in an obvious way?
Some thoughts
So clearly, each "set-model" $M$ given such a "function-model" $\beta$ by defining
$$(*)\quad \beta(\phi)=\text{"true"}\;\Longleftrightarrow \; M\Vdash \phi$$
Also, if $T$ is inconsistent by proving $\phi\land \neg\phi$, then we cannot choose a value for $\beta(\phi)$, and so no such "function-model" exists.
The other direction seems non-obvious.
I can imagine that the equivalence follows from Gödel's completeness theorem, but I am uncertain.
Is it true that for every choice of such a "function-model" $\beta$ there is a "set-model" $M$ so that $(*)$ holds?
AI: No, that second definition is totally unsatisfactory (for first-order logic anyways - it's the correct definition for propositional logic).
First, let me rephrase it slightly. By thinking about $\beta^{-1}(\{\top\})$ instead of $\beta$ itself, we see that it's just a definition of a maximal consistent theory containing $T$. This is a bit easier to think about in my opinion, so I'll do so. Maximal consistent sets (even with additional bells and whistles - see below) are definitely not the same things as a models, for two big reasons:
Until we prove the completeness theorem, we don't know that a maximal consistent $S$ has a model.
More fundamentally, maximal consistent theories do not have unique models up to isomorphism (unless they're the theories of finite structures), per the compactness theorem. Put another way, isomorphism is finer than elementary equivalence (and indeed inasmuch as model theory studies complete theories - which is most of the time, really - it's leaning on the richness which can occur within a single elementary equivalence class).
Now to be fair, if you really really want to avoid talking about sets you could look at expansions of the language instead and get something kinda-reasonable. Specifically, a "syntax-only model" of a theory $T$ in a language $\mathcal{L}$ could be defined as a pair $(S, \hat{\mathcal{L}})$ where:
$\hat{\mathcal{L}}\supseteq \mathcal{L}$,
$S$ is a maximal consistent $\hat{\mathcal{L}}$-theory containing $T$, and
$S$ has the witness property: whenever $\exists x_1,...,x_n\varphi(x_1,...,x_n)$ is a sentence in $S$, there are closed $\hat{\mathcal{L}}$-terms $t_1,...,t_n$ such that $S\vdash\varphi(t_1,...,t_n)$. (I've said a bit about the importance of the witness property here which may be relevant.)
Such a syntax-only model has a naturally associated structure in the usual sense which (has a reduct which) is indeed a model of $T$. Moreover, every consistent theory has a syntax-only model; this is really what Henkin's argument shows. So we can get something via a syntax-only approach (but we have to add the witness property, which isn't stated in your second definition - or at best is unclearly implicit in its final buletpoint).
However, not all models of $T$ will occur this way. In particular, any model of $T$ with cardinality $>\vert T\vert+\aleph_0$ will not be so represented. So again, lots of model-theoretic topics vanish if we make this shift (or at least require tedious circumlocutions). |
H: A problem to show that a certain field extension is not normal
Problem: Let $\alpha$ be a real number such that $\alpha^4=5$. Then show that $\mathbb Q(\alpha +i\alpha) $ over $\mathbb Q$ is not a normal extension.(where $i^2=1$)
My approach: I could show that the minimal polynomial of $(\alpha+i\alpha)$ over $\mathbb Q$ is $x^4+20$ and in the splitting field it splits as $$x^4+20=(x-(\alpha+i\alpha))(x+(\alpha+i\alpha))(x-(\alpha-i\alpha))(x+(\alpha-i\alpha))$$$$=(x-(\alpha+i\alpha))(x+(\alpha+i\alpha))(x^2+2i\alpha^2)$$
As $(\alpha+i\alpha)\in \mathbb Q(\alpha+i\alpha)$ we have $-(\alpha+i\alpha)\in \mathbb Q(\alpha+i\alpha)$.
If I can show that $(x^2+2i\alpha^2)$ is irreducible over $\mathbb Q(\alpha+i\alpha)$ then I am done, that is $(\alpha-i\alpha)\not\in\mathbb Q(\alpha+i\alpha)$ which is equivalent to showing $i\not\in Q(\alpha+i\alpha)$ $\left( \because \frac{(\alpha+i\alpha)}{(\alpha-i\alpha)}=-i\right)$
But I could not figure that out. I am stuck here. I don't know if I have to use Galois' Correspondence here. Help me out. Thanks in advance!
Other approaches are welcome as well.
AI: If $i \in \mathbb Q(\alpha + i\alpha)$, then $1 + i \in \mathbb Q(\alpha + i\alpha)$, and hence $\alpha = \frac{\alpha + i \alpha}{1 + i} \in \mathbb Q(\alpha + i\alpha)$ too.
So $\mathbb Q(\alpha + i \alpha) = \mathbb Q(\alpha, i)$.
However, $$[\mathbb Q(\alpha, i) : \mathbb Q] = [\mathbb Q(\alpha, i) : \mathbb Q(\alpha)] \times [\mathbb Q(\alpha) : \mathbb Q] = 2 \times 4 = 8,$$
while
$$ [\mathbb Q(\alpha + i \alpha) : \mathbb Q] = {\rm deg}(X^4 + 20) = 4,$$
which is a contradiction. |
H: When is a limit of measure preserving maps also measure preserving?
Take some compact space, for simplicity take the interval $[0,1]$.
Let $f_n:[0,1] \to [0,1]$ be measure preserving, i.e. $\mu (f_n^{-1}(A))=\mu(A)$ for all lebesgue measurable $A \subset [0,1]$.
The question is under what kinds of convergence $f_n \to f$ do we also ensure that $f$ is measure preserving?
I think I can show that for $f_n \to f$ in $L^2$ this holds. I'm now wondering about weak convergence $f_n \rightharpoonup f$ in $L^2$? Anyone know the answer (and maybe a proof/counterexample)?
AI: Let $f_n(x) = \{nx\}$ where $\{x\}$ denotes the fractional part of $x$. Then $f_n \rightharpoonup \tfrac12$ in $L^2$. |
H: Showing divergence of $\sum_{n=2}^{\infty} \frac{1}{n\ln n + \sqrt{\ln^3n}}$
$$\sum\limits_{n=2}^{\infty} \frac{1}{n\ln n + \sqrt{\ln^3n}}$$
As $n \to \infty$, we see that $n \ln n \gg (\ln n)^{3/2}$. Hence
$$\sum\limits_{n=2}^{\infty} \frac{1}{n\ln n + \sqrt{\ln^3n}} \sim^{\infty} \sum\limits_{n=2}^{\infty} \frac{1}{n \ln n}, 0 \leq a_n$$ By the integral test, the last sum is diverging making the original sum divergent. Is this correct?
AI: Yes your argument is correct by limit comparison test indeed
$$\frac{\frac{1}{n\ln n + \sqrt{\ln^3n}}}{\frac{1}{n\ln n}}=\frac{n\ln n}{n\ln n + \sqrt{\ln^3n}}\to 1$$
and $\sum \frac{1}{n \ln n}$ diverges (also by condensation test). |
H: Why $I^2$ a vector space where $I$ is the space of differentiable functions vanishing at a point $x$
I am reading the following wikipedia article on tangent spaces, in particular, this subsection on the definition via the cotangent space. Here is a paraphrasing of the first two sentences of the first paragraph at the given link.
"Consider the ideal $I$ that consists of all smooth functions $f$ from the manifold $M$ to $\mathbb{R}$ that vanish at $x\in M$. Then $I$ and $I^2$ are real vector spaces."
I am bothered by the claim that $I^2$ is a real vector space. It is my understanding that $I$ is the set of all smooth functions from $M$ to $\mathbb{R}$ that vanish at $x$ and that $I^2$ is the set of all functions that can be expressed as the product of two elements of $I$. (If I have gotten anything wrong so far please let me know)
It is not clear to me that $I^2$ is closed under vector sum, as this would require that for any functions $f_1,f_2,f_3,f_4$ in $I$, there must exist functions $g_1,g_2$ such that $f_1f_2+f_3f_4 = g_1g_2$. It is not clear to me that this is the case.
Thanks for the help.
AI: There is usually an abuse of notation when defining sets like $I^2$. In this case, $I^2$ is defined as the space of the functions that look like
$$
\sum_{i} f_ig_i
$$
where $f_i, g_i \in I$. This is now easily seen to be a real vector space. |
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