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H: On $n\times n$ matrices $A$ with trace of the powers equal to $0$
Let $R$ be a commutative ring with identity and let $A \in M_n(R)$ be such that $$\mbox{tr}A = \mbox{tr}A^2 = \cdots = \mbox{tr}A^n = 0 .$$ I want to show that $n!A^n= 0$.
Any suggestion or reference would be helpful.
P.S.: When $R$ is a field of characteristic zero I can prove that $A^n=0$ but I have no idea for the general case.
AI: Let
$$\chi(\lambda) = \det(\lambda I_n - A ) = \lambda^n - e_1 \lambda^{n-1} + e_2 \lambda^{n-2} + \cdots + (-1)^n e_n$$
be the characteristic polynomial of $A$. By Cayley–Hamilton theorem (which is valid for any commutative ring), we have
$$\chi(A) = A^n - e_1 A^{n-1} + e_2 A^{n-2} + \cdots + (-1)^n e_n I_n = 0\tag{*1}$$
Let $p_k = \text{tr}A^k$ for $k = 1, \ldots, n$. The $e_k$ are elementary symmetric polynomials in the roots of $\chi(\lambda)$ and they are related to $p_k$ through the Newton identities (again valid for any commutative ring):
$$\begin{align}
e_1 = & p_1,\\
2 e_2 = & e_1 p_1 - p_2,\\
3 e_3 = & e_2 p_1 - e_1 p_2 + p_3,\\
4 e_4 = & e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4,\\
\vdots
\end{align}$$
As a result,
$$p_1 = p_2 = \cdots = p_n = 0\quad\implies\quad e_1 = 2 e_2 = 3 e_3 = \cdots = n e_n = 0$$
Multiply $(*1)$ by $n!$ then give us $\;n!A^n = 0\;$. |
H: Chain homotopy and compositions of morphisms.
Show that if $\alpha_1 \sim \beta_1$ and $\alpha_2 \sim \beta_2$ , then (whenever composition makes sense) $\alpha_1 \circ \alpha_2 \sim \beta_1 \circ \beta_2$.
I have two questions.
So are these morphisms ($\alpha_1$, $\alpha_2$, $\beta_1$, and $\beta_2$) maps from $\mathcal{A}$ to $\mathcal{C}$, where $\mathcal{A}$ and $\mathcal{C}$ are chain complexes with $$\mathcal{A} := ... \leftarrow A_{n-1} \leftarrow A_n \leftarrow ...$$ and $$\mathcal{C} := ... \leftarrow C_n \leftarrow C_{n+1} \leftarrow ...$$
Or do we have $\alpha_1 : A_1 \rightarrow C_1$, $\alpha_2: A_2 \rightarrow C_2$, $\beta_1: A_1 \rightarrow C_1$, and $\beta_2: A_2 \rightarrow C_2$. I'm assuming its the former, but I'm having problems with both cases.
If its the former, then we have, for every $n$, $\alpha_{1 (n)} : A_n \rightarrow C_n$ and $\alpha_{2(n)}: A_n \rightarrow C_n$. But then how can we compose them? Because for $a \in A_n$, we have $\alpha_{1(n)} \circ \alpha_{2 (n)}(a) = \alpha_{1(n)}(\alpha_{2(n)} (a))$. But $\alpha_{2(n)}(a) \in C_n$, so its not even in the domain of $\alpha_{1(n)}$.
If $\alpha_1: A_n \rightarrow C_n$ and $\alpha_1: A_n \rightarrow C_n$, then $\alpha_1 \circ \alpha_2 (a)$ still does not make any sense.
Thanks in advance
AI: I'm assuming that $\alpha_1,\alpha_2,\beta_1,\beta_2$ are chain complexes maps, in that case since $\alpha_1 \sim \beta_1$ e $\alpha_2 \sim \beta_2$ and so there must be $\mathcal A,\mathcal B,\mathcal C,\mathcal D$ four chain complexes such that $\alpha_1,\beta_1 \colon \mathcal A \to \mathcal B$ and $\alpha_2,\beta_2 \colon \mathcal C \to \mathcal D$.
In this way
$$\alpha_1 =\langle {\alpha_1}_n \colon A_n \to B_n\rangle_{n \in \mathbb Z}$$
$$\alpha_2 =\langle {\alpha_2}_n \colon C_n \to D_n\rangle_{n \in \mathbb Z}$$
$$\beta_1 = \langle {\beta_1}_n \colon A_n \to B_n \rangle_{n \in \mathbb Z}$$
$$\beta_2 = \langle {\beta_2}_n \colon C_n \to D_n\rangle_{n \in \mathbb Z}$$
Now in order to be possible that composites $\alpha_1 \circ \alpha_2$ and $\beta_1 \circ \beta_2$ exist it must be $\mathcal A= \mathcal D$:
in this way the composite maps are defined as
$${\alpha_1 \circ \alpha_2}_{n} = {\alpha_1}_n \circ {\alpha_2}_n \colon C_n \to B_n$$
$${\beta_1 \circ \beta_2}_{n} = {\beta_1}_n \circ {\beta_2}_n \colon C_n \to B_n$$ |
H: Exponentiation by squaring
I need to calculate $7^{2012} \mod {13}$ by hand using exponentiation by squaring, but I cant seem to figure it out.
I started with this but I don't know for sure if its correct or where it's going.
\begin{align}
(7^2)^{1006} &\equiv 49^{1006} \pmod {13} \\
(10^2)^{503} &\equiv 100^{503} \pmod {13}
\end{align}
AI: Looks good so far. Continuing, we find that:
\begin{align*}
100^{503} \pmod {13}
&\equiv 9^{503} \pmod {13} \\
&\equiv 81^{251} \cdot 9 \pmod {13} \\
&\equiv 3^{251} \cdot 9 \pmod {13} \\
&\equiv 9^{125} \cdot 3 \cdot 9 \pmod {13} \\
&\equiv 81^{62} \cdot 9 \cdot 1 \pmod {13} & \text{since $3 \cdot 9 = 27 = 2\cdot 13 + 1$}\\
&\equiv 3^{62} \cdot 9 \pmod {13} \\
&\equiv 9^{31} \cdot 9 \equiv 9^{32} \pmod {13} \\
&\equiv 81^{16} \pmod {13} \\
&\equiv 3^{16} \equiv 9^{8} \equiv 3^{4} \pmod {13} \\
&\equiv 9^2 \equiv 81 \equiv 3\pmod {13} \\
\end{align*} |
H: Show that a polynomial has at least one positive solution/root.
Let: $P(x) = a_nx^n + a_{n-1}x^{n-1} + ..... + a_1x + a_0$ where $a_0a_n < 0$
I have to prove that the Polynomial $P(x)$ has at least one positive root
how can I prove it? Any ideas?
AI: For $x$ very large, $P(x)\approx a_n x^n$ and so has a sign different from that of $P(0)=a_0$ since $a_0a_n<0$. By the intermediate value theorem, there is a positive root. |
H: Tensor Product and Direct Sum
Let $R$ be a commutative ring with identity and let $\{M_\alpha\}$ be a family of $R$-modules and $N$ another $R$-module. I've tried to show that
$$\left(\bigoplus_\alpha M_\alpha\right)\otimes N\simeq \bigoplus_\alpha (M_\alpha\otimes N).$$
I've tried to simply generalize the proof I've seem in the book for the case $\alpha=1,2$. For that, I've defined $\varphi : (\bigoplus_\alpha M_\alpha)\times N\to \bigoplus_\alpha (M_\alpha\otimes N)$ by
$$\varphi((m_\alpha), n)=(m_\alpha\otimes n),$$
this is well defined, since being $m_\alpha = 0$ unless for finite indices we have $m_\alpha\otimes n$ also well zero unless for finite $\alpha$. Also, $\varphi$ is clearly bilinear, so that the universal property grants that there is an $R$-module homomorphism $\phi : (\bigoplus_\alpha M_\alpha)\otimes N\to \bigoplus_\alpha (M_\alpha\otimes N)$ such that $\phi((m_\alpha)\otimes n)=(m_\alpha\otimes n)$.
Then, let $i_\alpha : M_\alpha \to \bigoplus_\alpha M_\alpha$ be the canonical injection, that is
$$(i_\alpha(m_\alpha))_\beta=\begin{cases}0 & \alpha\neq \beta, \\ m_\alpha & \alpha=\beta.\end{cases}$$
Then we define for each $\alpha$ the map $f_\alpha : M_\alpha\times N\to (\bigoplus_\alpha M_\alpha )\otimes N$ by $f_\alpha(m_\alpha,n)=i_\alpha(m_\alpha)\otimes n$, then $f_\alpha$ is bilinear and by the universal property induces $g_\alpha : M_\alpha\otimes N\to (\bigoplus_\alpha M_\alpha)\otimes N$ an $R$-module homomorphism such that $g_\alpha(m_\alpha\otimes n)=i_\alpha(m_\alpha)\otimes n$. Define then $\psi : \bigoplus_\alpha (M_\alpha\otimes N)\to(\bigoplus_\alpha M_\alpha)\otimes N$ by
$$\psi((m_\alpha\otimes n))=\sum_\alpha g_\alpha(m_\alpha\otimes n)$$
Now here's the point. I'm unsure $\psi$ is well-defined, since the sum can be infinite. My thought on that is: since just finitely many of the $m_\alpha\otimes n$ are nonzero and $g$ is an $R$-module homomorphism, this sum should become finite and we can continue the proof. Is this reasoning correct?
After that I would show that $\phi$ and $\psi$ are inverses. This sounds easy, because we have
$$\psi(\phi((m_\alpha)\otimes n)))=\psi((m_\alpha\otimes n))=\sum_\alpha g_\alpha(m_\alpha\otimes n)=\sum_\alpha i_\alpha(m_\alpha)\otimes n=(m_\alpha)\otimes n,$$
and similarly we have
$$\phi(\psi(m_\alpha\otimes n_\alpha))=\phi\left(\sum_\alpha i_\alpha(m_\alpha)\otimes n_\alpha\right)=\sum_\alpha \phi(i_\alpha(m_\alpha)\otimes n_\alpha)=(m_\alpha\otimes n_\alpha),$$
and so $\psi$ and $\phi$ are inverses $R$-module homomorphism and hence the isomorphism exists.
Is the proof allright? I'm mainly unsure with the reasoning about the sum.
Thanks very much in advance.
AI: Your reasoning is right and $g$ is well-defined (and hence a homomorphism as the $g_\alpha$ are). Given $(m_\alpha \otimes n) \in \bigoplus_\alpha M_\alpha \otimes N$, there are only finitely many $\alpha$ with $m_\alpha \otimes n \ne 0$, so for only finitely many $\alpha$ we have $g_\alpha (m_\alpha \otimes n) \ne 0$ (as $g_\alpha(0) = 0$). So the sum $\sum_\alpha g_\alpha(m_\alpha \otimes n)$ has only finitely many non-zero terms. |
H: prove or disprove about the statement for a limit of a sequence
Let $(a_n)$ be a sequence of numbers such that $\displaystyle\lim_{n\rightarrow\infty}a_{n+1}=L$. Does this imply that $\displaystyle\lim_{n\rightarrow\infty}a_{n}=L$?
AI: Yes, a limit of a sequence is determined by the "infinite tail" of the sequence.
Let $\varepsilon > 0$. Then there exists $N \in \mathbb{N}$ such that:
$$ n > N \Rightarrow |a_{n+1} - L| < \varepsilon $$. Take $N+1$, then for every $m > N+1 > 1$ we can write $m = n+1$ and then
$$ | a_m - L | = |a_{n+1} - L | < \varepsilon $$
since $m > N+1 \Rightarrow n+1 > N+1 \Rightarrow n > N$. |
H: Help solving the limit $\lim_{n \to \infty} \frac{2^{\ln n}}{n^3}$
I want to evaluate the following limit: $$\displaystyle\lim _{n\rightarrow\infty}\frac{2^{\ln n}}{n^3}.$$ Can someone help me how to solve it? I don't want the result but the solution.
AI: Note that $2^{\ln n} = n^{\ln 2}$ and so
$$
\lim _{n\rightarrow\infty}\frac{2^{\ln n}}{n^3}=
\lim _{n\rightarrow\infty}\frac{n^{\ln 2}}{n^3}=0
$$
because $\ln 2 < 3$. |
H: Is $\epsilon$ in every alphabet?
Given a $\Sigma$ an alphabet, is $\epsilon$ in it logically?
For example, if I have a function $ f : \Sigma \to \Sigma $, can I define it for example $ f(\sigma) = \epsilon$? even if my alphabet is for example only $\Sigma = \{a,b,c\}$?
AI: As Arthur Fischer wrote in the comments: $\epsilon$ is the empty string, the word of length zero. It is not contained in any alphabet (apart from some alphabet specifically defined to contain the empty string), but it is a string over any alphabet.
You should define $f:\Sigma^* \rightarrow \Sigma^*$, which is a function from the strings over the alphabet to the strings over the alphabet. If you were to leave out the stars, it'd be just from the alphabet to the alphabet, which I don't think is what you want.
(An analogue to this is regular set notation - usually, for some set $A$, $\emptyset \notin A$ unless explicitly defined otherwise, but $\emptyset \subseteq A$ for all $A$. Similarly for $\epsilon$ and $\Sigma$.) |
H: Expected Hamming distance
I choose two code words independently at random from $\mathbb F_2^n$ where each string has $n$ binary digits equally likely. $\mathbb F_2$ represens the binary digits.
The Hamming distance is between two vectors $x,y\in\mathbb F_2^n$ is the number of places where they differ: $d(x,y)=|\{j:x_j\not = y_j\}|$
How can I evaluate the expected distance then?
AI: In effect you’re choosing one word at random and asking for its expected distance from the zero word. That’s clearly $n/2$: every word at distance $k$ has a word at distance $n-k$ as its complement. |
H: Is $(f(x_2 )-f(x_1 ))/(x_2-x_1 )=f'(c)$ for $f$ is continuous derivative on $(a,b)$, $x_1,x_2,c∈(a,b)$
Suppose that $f$ is continuous derivative on $(a,b)$, and let $c∈(a,b)$. Prove or disprove: there exist points $x_1,x_2∈(a,b)$ such that
$(f(x_2 )-f(x_1 ))/(x_2-x_1 )=f'(c) $
This seem true, but I still have the feeling that it isn't true, but I can't give counter example that work for all $x_1,x_2$
AI: Although it is not explicitly stated, $x_1 \neq x_2$ should be required, otherwise $\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$ doesn't make sense.
Then a counterexample is given by $(a,b) = (-1,1)$, $c = 0$, and $f(x) = x^3$. |
H: Rewriting algebra
I'm working on calculations on polynomials and a paper gives the following algebra step:
$\frac{x^{15} - 1}{x^3 - 1} = x^{12} + x^9 + x^6 + x^3 + 1$
They do not explain how they get to this result, and I can not follow how they did this.
Could someone tell me what step i'm missing here?
AI: If we multiply both sides by $x^3 - 1$, we get $x^{15} - 1 = (x^3 - 1)(x^{12} + x^9 + x^6 + x^3 + 1)$. You can work out the right hand side to see that this is true. |
H: I'm not managing to prove that, $\left( \frac{1}{\sqrt{n}}\sum_{i=1}^na_i\right)^2\leq\sum_{i=1}^na_i^2$
I'm not managing to prove that, $$\left( \frac{1}{\sqrt{n}}\sum_{i=1}^na_i\right)^2\leq\sum_{i=1}^na_i^2$$I think it should be used something like domestic product, but do not know how to use.
AI: $$\left(\frac1{\sqrt n}\sum_{i=1}^na_i\right)^2\le\sum_{i=1}^na_i^2\iff\frac1n\left(\sum_{i=1}^na_i\right)^2\le\sum_{i=1}^na_i^2\iff$$
$$\iff\left(\sum_{i=1}^na_i\right)^2\le\sum_{i=1}^n 1^2\cdot\sum_{i=1}^na_i^2$$
and now apply directly Cauchy-Schwarz Inequality |
H: Uniform convergence with Lp functions
I have a convergence question:
Say we have a sequence of functions $\{u_{m}\}_{m=1}^{\infty}$ where $\{u_{m}\}_{1}^{\infty} \subset L^{p}(U)$ and where $U$ is bounded. Consider $u^{\epsilon} := \eta_{\epsilon}\ast u_{m}$ where $\eta_{\epsilon}$ is the usual mollifier.
If I was able to show that for every $m$ we have $||u_{m}^{\epsilon}-u_{m}||_{L^{1}(U)} \leq \epsilon C$ for some constant $C$, then clearly I have shown that $u_{m}^{\epsilon} \rightarrow u_{m}$ in $L^{1}(U)$ as $\epsilon \rightarrow 0$. My question is simply how is the convergence "uniform in m"?
My understanding of uniform convergence is using the definition:
If $f_{n}: S \rightarrow \mathbb{R}$ converges uniformly to $f: S \rightarrow \mathbb{R}$ iff $\alpha_{n} = \text{sup}_{x}|f_{n}(x)-f(x)|$ tends to 0.
I don't see how this definition coincides with the above.
Thanks!
AI: The convergence being "uniform in $m$" here means that given any $\delta > 0$, you can find an $\varepsilon_0 > 0$, such that for all $0 < \varepsilon \leqslant \varepsilon_0$, you have
$$\lVert u_m^\varepsilon - u_m\rVert_{L^1(U)} < \delta$$
for all $m$.
Or,
$$\sup_{m} \lVert u_m^\varepsilon - u_m\rVert_{L^1(U)} \xrightarrow{\varepsilon\to 0} 0.$$ |
H: Non-isomorphic $\mathbb{C}$-algebras
The question is as follows:
Show that the $\mathbb{C}$-algebras: $A=\mathbb{C}[x,y]/(x^2y-xy)$, $B=\mathbb{C}[x,y]/(x^2y+xy^2)$, $C=\mathbb{C}[x,y,z]/(xy, yz, zx)$, and $D=\mathbb{C}[x,y]/(x^2y+xy^2+x^4+y^4)$ are pairwise non-isomorphic.
Is there a particularly elegant way to approach this? I've tried playing around with decompositions, but I can't quite get things to work. Any help would be much appreciated! Thanks in advance.
AI: STRONG HINTS
Look at how the algebraic sets decompose. The variety corresponding to $A$ is the union of three lines intersection in two points. It looks like this $\rightarrow \coprod$, because $(x^2y-xy)=(x) \cap (y) \cap(x-1)$.
The second decomposes as $(x) \cap (x+y) \cap (y)$, so the algebraic set again consists of three lines, but this time they intersect in three points, forming a "fan" of three lines.
Similarly, $C$ is the union of the coordinate axes, so they intersect in one point.
And the fourth equation is irreducible, so $D$ is an integral domain. |
H: Connected and irreducible topological spaces
A topological space is called connected if any presentation of $X$ as $X = V_1 \uplus V_2$ by disjoint open subsets implies that one of them is trivial ($V_1 = X$ or $V_2 = X$). By taking complement one can replace the word "open" by "close".
A topological space is called irreducible if any presentation of $X$ as $X = X_1 \cup X_2$ by two closed subsets implies that one of them is trivial ($X_1 = X$ or $X_2 = X$).
Clearly every irreducible space is connected. The converse is not always true but:
Proposition:
Let $X$ be a connected topological space which has an open covering by irreducible subspaces. Then $X$ is irreducible.
I want to prove this proposition.
AI: HINT: It’s easily checked that a space $X$ is irreducible if and only if $U\cap V\ne\varnothing$ whenever $U$ and $V$ are non-empty open subsets of $X$. Let $X$ be connected, and let $\mathscr{U}$ be an open cover of $X$ by irreducible subspaces. Suppose that $V$ and $W$ are disjoint open subsets of $X$; you want to show that one of them is empty.
Show that each $U\in\mathscr{U}$ intersects at most one of $V$ and $W$.
Let
$$\begin{align*}
&\mathscr{U}_V=\{U\in\mathscr{U}:U\cap V\ne\varnothing\}\;,\\
&\mathscr{U}_W=\{U\in\mathscr{U}:U\cap W\ne\varnothing\}\;,\text{ and}\\
&\mathscr{U}_0=\mathscr{U}\setminus(\mathscr{U}_V\cup\mathscr{U}_W)\;.
\end{align*}$$
Consider the open sets $\bigcup\mathscr{U}_V,\bigcup\mathscr{U}_W$, and $\bigcup\mathscr{U}_0$, bearing in mind that $X$ is connected. |
H: Find a non-principal ideal (if one exists) in $\mathbb Z[x]$ and $\mathbb Q[x,y]$
I know that $\mathbb Z$ is not a field so this doesn't rule out non-principal ideals. I don't know how to find them though besides with guessing, which could take forever.
As for $\mathbb Q[x,y]$ I know $\mathbb Q$ is a field which would mean $\mathbb Q[x]$ is a principal ideal domain, but does this still apply for $\mathbb Q[x,y]$ ?
AI: Here is a general result:
If $D$ is a domain, then $D[X]$ is a PID iff $D$ is a field.
One direction is a classic result. For the other direction, take $a\in D$, consider the ideal $(a,X)$, and prove that it is principal iff $a$ is a unit.
This immediately answers both questions: $(2,X)$ is not principal in $\mathbb Z[X]$ and $(X,Y)$ is not principal in $\mathbb Q[X,Y]=\mathbb Q[X][Y]$. |
H: Is exponential of a concave function concave?
is this function:
$$\exp\Big(-||Ax||^2\Big)$$
concave in A??
I know that exponential of a convex function is convex, but is exponential of a concave function concave??
AI: Did's example
Let $f(x)=e^{-x^2}$.
Then $f'(x)=-2xe^{-x^2}$, so $f''(x)=4x^2e^{-x^2}-2e^{-x^2}=(4x^2-2)e^{-x^2}$.
$e^{-x^2}$ is always positive, but $4x^2-2$ can be positive or negative. |
H: Definition of infinite limit
Following is an exercise I solved, it is not homework so I don't know how to check my solution. Can you please check the solution for me? The exercise is this:
Modify the definition of convergent sequence to obtain a definition of convergence to $\infty$ and then use your definition to show that $x_n = \sqrt{n}$ converges to $\infty$. What does your definition say about the sequence $1,0,2,0,3,0,4,0,...$
My solution: The new definition is: For every $\varepsilon$ there exists $N$ such that for all $n > N$, $x_n > \varepsilon$. Then: Let $\varepsilon > 0$. Let $N$ be such that $N > \varepsilon^2$. Then $n>N$ implies $\sqrt{n} > \sqrt{N} > \varepsilon$. The sequence $1,0,2,0,3,0,4,0,...$ does not converge to $\infty$.
AI: What you have written down so far is correct. Can you also give a full proof that $1,0,2,0,3,0,\ldots$ does not converge to $\infty$? |
H: Differential Equation with Given Initial Condition
Given is the differential equation
$N(t)' = 2 * \sqrt{N(t)}$
We have to show that the constant function N(t) = 0 is a solution for the initial condition N(0) = 0, and that the function N(t) = $t^2$ is a solution again for the same initial condition.
So I would plug in N(t) or N'(t) respectively in the above formula - but how do I use the initial conditions?
AI: Hint: the equation is separable.
We can write:
$$\dfrac{1}{\sqrt{N}}dN = 2 dt$$
We can now integrate each side and then solve for the constant from the initial condition. |
H: Why is $2^n$ the maximum number of subsets of a set of size $n$?
There is a set with $n$ elements. Why is the maximum number of subsets that can be formed out of it $2^n$?
AI: We must show that $${n\choose 0}+{n\choose 1}+{n\choose 2}+\cdots +{n\choose n}=2^n$$ is the number of subsets of an $n$-element set $S$ where $n\geq0$.
Every subset of $S$ is a $k$-subset of $S$ where $k=0,1,2,...,n$. We know that ${n\choose k}$ equals the number of $k$-subsets of S. Thus by the Addition Principle $${n\choose 0}+{n\choose 1}+{n\choose 2}+\cdots +{n\choose n}$$ equals the number of subsets to the set $S$. We can count the same thing by observing that each element of the set $S$ has two choices, either they are in a subset or they are not in a subset. Let $S=\{x_1,x_2,x_3,...,x_n\}$. So, $x_1$ is either in a subset or it is not in a subset, $x_2$ is either in a subset or it is not in a subset,..., $x_n$ is either in a subset or it is not in a subset. Thus by the Multiplication Principle there are $2^n$ ways we can form a subset of the set $S$. Hence ${n\choose 0}+{n\choose 1}+{n\choose 2}+\cdots +{n\choose n}=2^n$.
Another approach is to consider the Binomial Theorem $$(x+y)^n=\sum_{k=0}^n {n\choose k}x^{n-k}y^k.$$ Letting $x=1$ and $y=1$ we obtain$$2^n=\sum_{k=0}^n{n\choose k}.$$ |
H: beta reduction bascis
Hi I get the basics of beta reduction e.g.
$$(\lambda var.body)arg $$
you just replace the occurrences of var with arg in body.
However what happens here?
$$(\lambda x.xx)(\lambda x.xx) \rightsquigarrow_\beta (\lambda x.xx)(\lambda x.xx)$$
Let's call them $A$ and $B$, so I replace all occurrences of $x$ in $A$ with $xx$ (from $b$) and throw away $B$'s lamda. Giving me
$$(\lambda x.xx)(\lambda x.xx) \rightsquigarrow_\beta \lambda x.xxxx$$
Instead of
$$ (\lambda x.xx)(\lambda x.xx)$$
Can anyone explain where I'm going wrong?
AI: $$ (\lambda x ~.~ xx)(\lambda z ~.~ zz) ~ \rightarrow_{\beta} ~ (xx)[x \mapsto (\lambda z ~.~ zz)] ~=~ (\lambda z ~.~ zz)(\lambda z ~.~ zz)
$$ |
H: Show that if $ f $ is a symmetric bilinear form such that $f(u,u)=0 $, $ \forall u \in V$ so $f=0$.
Let $V$ a vector space of demension $n$ over a field $\mathbb{K}$. Show that if $ f $ is a symmetric bilinear form such that $f(u,u)=0 $, $ \forall u \in V$ so $f=0$.
I could answer this question by using the matrix transformation, but I was wondering if there is a more elegant than mine.
AI: If you consider $f(x,y) = x_1y_2 + x_2y_1$, you have a nonzero symmetric bilinear form ($f((1,0),(0,1)) = 1$) that nevertheless has $f(u,u) = 0$ for all $u$ in characteristic $2$.
For all other characteristics, it's true, since you can then divide $2 f(u,v) = f(u+v,u+v) - f(u,u) - f(v,v)$ by $2$. |
H: Is there any number which $n!$ is lower than $2^n$ or same?
I interested in this question.
how many numbers meet this condition?
I think a few of them meet this but I want a proof for this.
also I'm not very pro in mathematics.
AI: Observe that $n!<2^n$ if $n<4$ and $n!>2^n$ for $n=4$
Let $m!>2^m$ for $m\ge 4$
Then $(m+1)!>(m+1)2^m>2^{m+1}$ if $m+1\ge 2\iff m\ge1$ which is true as $m\ge 4$ |
H: Difference between NFA and DFA
In very simple terms please, all resources I'm finding are talking about tuples and stuff and I just need a simple explanation that I can remember easily because I keep getting them mixed up.
AI: Each input to a DFA or NFA affects the state of the automaton: if it was in state $q$ immediately before the input, either it will be in some state $q'$ after the input, or the input will cause it to choke. (Note that $q'$ may be the same as $q$.) Suppose that we have an automaton in a state $q$. The difference in behavior between a DFA and an NFA is this:
If it’s a DFA, each possible input determines the resulting state $q'$ uniquely. Every input causes a state change, and the new state is completely determined by the input. Moreover, the automaton can change state only after reading an input.
If it’s an NFA, some inputs may allow a choice of resulting states, and some may cause the automaton to choke, because there is no new state corresponding to that input. Moreover, the automaton may be constructed so that it can change state to some new state $q'$ without reading any input at all.
As a consequence of this difference in behavior, DFA’s and NFA’s differ in another very important respect.
If you start a DFA in its initial state and input some word $w$, the state $q$ in which the DFA ends up is completely determined by $w$: inputting $w$ to the DFA will always cause it to end up in state $q$. This is what is meant by calling it deterministic.
If you start an NFA in its initial state and input some word $w$, there may be several possible states in which it can end up, since some of the inputs along the way may have allowed a choice of state changes. Consequently, you can’t predict from $w$ alone in exactly which state the automaton will finish; this is what is meant by calling it nondeterministic. (And it’s actually a little worse than I’ve indicated, since an NFA is also allowed to have more than one initial state.)
Finally, these differences affect how we determine what words are accepted (or recognized) by an automaton.
If it’s a DFA, we know that each word completely determines the final state of the automaton, and we say that the word is accepted if that state is an acceptor state.
If it’s an NFA, there might be several possible final states that could result from reading a given word; as long as at least one of them is an acceptor state, we say that the automaton accepts the word.
What I’ve described informally is the view of an NFA that makes it look most like a DFA and that I think best explains why it’s called nondeterministic. There is, however, another way of looking at NFAs: it’s also possible to think of an NFA as being in multiple states at once, as if it were making all possible choices at each input. If you think of it in those terms, you can say that it accepts a word provided that at least one of the states in which it ends up after reading that word is an acceptor state. This point of view is perhaps most useful for understanding the algorithm used to turn an NFA into an equivalent DFA. |
H: undetermined coefficients. What am I doing wrong?
I am having some trouble to solve the following differential equation for the undetermined coefficient:
$$
y''+2y'+y=xe^{-x}
$$
I have been watching some videos on youtube and done some reading but still do not fully understand what must be done to solve it. I was able to solve the left side by getting the common roots:
$$
(x+1)^{2} = r^{2}+r+1
$$
roots are -1, so
$$
y=(A+Bx)e^{-x}
$$
Now after I found the left side I started to determine $$g(x)$$
so $$g(x)=xe^{-x}$$
$$yp = e^{-x}Ax+e^{-x}B$$
derivative of yp is
$$y'p=-e^{-x}Ax+e^{-x}A-e^{-x}B$$
double derivative of yp is
$$y''p=e^{-x}Ax-2Ae^{-x}+Be^{-x}$$
I then subbed these into the original equation:
$$
(e^{-x}Ax-2Ae^{-x}+Be^{-x}) + 2(-e^{-x}Ax+e^{-x}A-e^{-x}B) + (e^{-x}Ax+e^{-x}B)=xe^{-1}
$$
Now after factoring everything out I got that $$A=\frac{1}{2}e^{-x}$$
Which would bring me to my final answer of $$y=Ae^{-x}+Bxe^{-x}+\frac{1}{2}e^{-x}$$
However this is wrong as the final answer is actually $$y=Ae^{-x}+Bxe^{-x}+\frac{1}{6}x^{3}e^{-x}$$
What is it that I am doing wrong? I am not sure what else to do :(
Thank you :)
AI: You have an inhomogeneous differential equation $y'' + 2y' + y = xe^{-x}$. To solve these we apply the following steps:
1) Find the general solution to the homogeneous equation $y'' + 2y' + y = 0$.
2) Find one solution $y = p(x)$ to the equation $y'' + 2y' + y = xe^{-x}$.
3) Then any solution to the equation $y'' + 2y' + y = xe^x$ is of the form $y = p(x) + h(x)$, where $h(x)$ is a solution to the homogeneous equation $y'' + 2y' + y = 0$.
You have correctly solved step 1), the general solution to the equation $y'' + 2y' + y = 0$ is $y(x) = Ae^{-x} + Bxe^{-x}$. Next, you want to find a solution to the equation $y'' + 2y' + y = xe^{-x}$. To do this, we often "guess" the form of the solution to be similar to the right hand side. In this case the right hand side is $e^{-x}$ multiplied by a polynomial. You could "guess" $p(x)$ to be of the form $p(x) = e^{-x} (ax^3 + bx^2)$. We are taking degree $3$ here, because we take two derivatives at most, and we end up with a linear term. So we probably won't need any higher degree terms. There is no need to look at linear and constant terms, because they satisfy the homogeneous equation.
Now we have to compute $p'' + 2p' + p$. We compute $p' = e^{-x} \left(3ax^2 + 2bx - (ax^3 + bx^2)\right)$ and $p'' = e^{-x} \left(ax^3 + bx^2- 2(3ax^2 + 2bx) + 6ax + b \right)$. So we get $p'' + 2p' + p = e^{-x} \left(x^3(a -2a + a) + x^2(b + 2(-b+3a) + (b - 6a)) + x(2 \cdot 2b - 4b + 6a) + 2b \right) = (6ax+2b) e^{-x}.$
This should equal $xe^{-x}$, and so we find $a = \frac{1}{6}, b=0$ and $p(x) = \frac{1}{6} x^3 e^{-x}$. Now we can finally write down the general solution to the inhomogeneous equation $y'' + 2y' + y = xe^{-x}$, which is $y(x) = \frac{1}{6} x^3 e^{-x} + Ae^{-x} + Bxe^{-x}$. |
H: Ratio of angles in a triangle, given lengths of triangle's sides.
If I have a triangle $\,\triangle ABC,\,$ with sides of lengths $\,AB=6, \;BC=4, \;CA=5,\,$
then what can I know about the ratio of $\,\dfrac{\angle ACB}{\angle BAC}\,$?
AI: You can use the Law of Sines to find the ratio of the sines of your two angles:
$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \,=\, D \! $$
In your case, you'd have $$\dfrac 6{\sin(\angle ACB)} = \dfrac 4{\sin(\angle BAC)} \iff \dfrac {6}{4} = \dfrac{\sin(\angle ACB)}{\sin(\angle BAC)} = \frac 32$$
Alternatively, to compute the measures of your angles directly, use the Law of Cosines. $$c^2 = a^2 + b^2 - 2ab\cos\gamma\,\iff \cos \gamma = \dfrac{a^2 + b^2 - c^2}{2ab}$$ |
H: Question on Polynomial function
Suppose $f(x)$ is a polynomial in $x$ having integer coefficients and $13 < a < b < c$ are integers such that $f(13) = f(a) = 13$ and $f(b) = f(c) = 19$: Determine the possible values of $a,b,c$.
Could someone give me an idea what type of $f(x)$ I should assume? The term polynomial is too general and further, four values are given and I tried a cubic polynomial but was not successful.
AI: What you need is Bezout theorem: If $f$ is integral polynomial, and $A,B$ are integers, then: $ A-B $ divides $f(A) - f(B)$.
In your example, you find that $6 = 19 - 16$ has to be divisible by $b-a,b-13,c-a,c-13$. This leaves very few possibilities: for starters, $b,c$ are among the numbers $13+1,13+2,13+3,13+6$. For each choice, just take a look if you can find suitable $a$. This is clearly impossible for $b=14$, and also for $c=19$ (it would imply $a \geq 16$, but then there is no choice for $b$). Thus, you can only take $b=15,\ c=16$, and then indeed $a = 14$ is a legitimate choice.
Now, just try to find a polynomial though these points. |
H: Geometry GRE question
This is a GRE quesrtion, and I could not find the length to save my life, please help!
A circle with diameter PQ of length 10is internally tangent at P to a circle of radius 20. A sqare ABCD is constructed from A, B on the larger circle, CD tangent at Q for the smaller circle. and the smaller circle is outside ABCD.
What is the length of AB?
AI: Let $x=|AB|$, the side of the square, let $O$ be the centre of the large circle, and let $M$ be the midpoint of $\overline{AB}$. If you make a sketch, you’ll find that $\triangle AMO$ is a right triangle with legs $\frac{x}2$ and $x-|OQ|=x-10$ and hypotenuse $20$. Thus,
$$(x-10)^2+\frac14x^2=400\;,$$
which is just a quadratic in $x$. |
H: Functions from ordinals to ordinals
I'm trying to solve this problem, which appears in Schimmerling's "A Course in Set Theory."
Problem. Find two functions
$$f:\omega\rightarrow\omega\cdot2$$
and
$$g:\omega\cdot2\rightarrow\omega\cdot3$$
such that $\sup\{f(\omega)\}=\omega\cdot2$ and $\sup\{g(\omega\cdot2)\}=\omega\cdot3$, but if $h=g\circ f$, then $\sup\{h(\omega)\}<\omega\cdot3$.
Setting $f(n)=n\cdot2=n+n$ for all $n<\omega$, as well as
\begin{aligned}
g(m) & = m & \text{if $m<\omega$}\\
g(m) & = \omega+\omega +m & \text{otherwise}
\end{aligned}
it seems that $g\circ f(n)=g(n\cdot2)=n\cdot2$ (since $n\cdot2<\omega~~\forall n<\omega$). Thus it follows that $\sup\{g\circ f(\omega)\}=\omega\cdot2<\omega\cdot3$.
So far so good. But it turns out that the next exercise requires us to show the following.
Theorem. Let $\kappa<\lambda<\mu$ be three limit ordinals and
$$f:\kappa\rightarrow\lambda$$
and
$$g:\lambda\rightarrow\mu.$$
Suppose $\sup\{f(\kappa)\}=\lambda$ and $\sup\{g(\lambda)\}=\mu$, and that $g$ is nondecreasing, i.e., if $\alpha\le\beta<\lambda$ then $g(\alpha)\le g(\beta)$. Then letting $h=g\circ f$ we must have $\sup\{h(\kappa)\}=\mu$.
It now seems that my construction for the first problem is a counterexample to the theorem to be proved in the second. Could someone explain to me the loophole in my argument? Thanks!
AI: The main problem is that your first example doesn’t work: $f[\omega]$ isn’t cofinal in $\omega\cdot 2$. In fact $\sup f[\omega]=\omega$. Note that the theorem requires $g$ to be non-decreasing; this suggests that in the first problem you should look for functions $f$ and $g$ such that $g$ is not non-decreasing. Here’s one possibility: let $f(n)=\omega+n$, and then let
$$g(\alpha)=\begin{cases}
\omega\cdot 2+\alpha,&\text{if }\alpha<\omega\\
0,&\text{if }\alpha\ge\omega\;.
\end{cases}$$
If you want to make $g$ injective, let $g(\omega+n)=n$ for $n\in\omega$. |
H: On simple extension
Show that $\mathbb{R}$
is not a simple extension of $\mathbb{Q}$
as follow:
a. $\mathbb{Q}$
is countable.
b. Any simple extension of a countable field is countable.
c. $\mathbb{R}$
is not countable.
I 've done a. and c. Can anyone help me a hint to prove b.?
AI: Let $F$ be a countable field, then the collection of all polynomials of degree $\leq n$ is countable. Hence, $F[x]$ is countable, being the countable union of countable sets. Hence, $F[x] \times F[x]\setminus\{0\}$ is countable. There is a surjective function $F[x]\times F[x]\setminus\{0\} \to F(x)$ by
$$
(f(x), g(x)) \mapsto \frac{f(x)}{g(x)}
$$
Hence, $F(x)$ is countable. If $a$ is transcendental over $F$, then $F(x) \cong F(a)$, which is thus countable. |
H: Kleene Star operation on sets
I have the following question, and do not understand the Kleene star operation in the context of relations.
Let R be the relation $R=\{(0,1),(0,2),(1,4),(1,5),(2,3),(2,4),(2,5)\}^*$ on the set $A=\{0,1,2,3,4,5\}$. Find all minimal, maximal, smallest and largest elements, if possible, of set A with regards to relation R.
We have only defined the Kleene star operation in the context of formal languages. Namely $V^* = \bigcup_{i\in \mathbb{N}} A^n$, where $A^n$ is defined recursively as $V^0 = \{\epsilon\}$, V is the set of all strings of length 1 and $V^{i+1} = \{vw : w\in V^i \wedge v\in V\}$.
Does the Kleene star have it's own definition when used with sets? What does the above set R mean? I assume it somehow affects the maximal elements and largest element of the given set. Could someone please clarify this for me?
AI: My best guess is that the author is using this notation to denote the transitive closure of the relation $\{\langle 0,1\rangle,\langle 0,2\rangle,\langle 1,4\rangle,\langle 1,5\rangle,\langle 2,3\rangle,\langle 2,4\rangle,\langle 2,5\rangle\}$. That is, it represents repeated composition rather than repeated concatenation. On that interpretation you do get a relation for which the question makes good sense. |
H: Proving that a polynomial about the volume of a tetrahedron is irreducible
We know that the volume of a tetrahedron $ABCD$ can be represented as
$$144V^2=(a^2b^2d^2+b^2c^2e^2+c^2a^2f^2+b^2a^2e^2+c^2b^2f^2+a^2c^2d^2+c^2e^2f^2+a^2f^2d^2+b^2d^2e^2+c^2d^2f^2+a^2e^2d^2+b^2f^2e^2)-(a^2b^2c^2+a^2e^2f^2+b^2f^2d^2+c^2d^2e^2)$$
where $a=BC, b=CA, c=AB, d=AD, e=BD, f=CD$.
Let $F(a,b,c,d,e,f)$ be the right hand side of the above equation. Then, here is my question.
Question : Is $F(a,b,c,d,e,f)$ irreducible as a polynomial?
The answer must be YES, but I don't know how to prove that. Can anyone help?
AI: Notice there are no linear terms in $a$. So it is a polynomial of the form $Aa^2+B$. So, $B/A$ must be a square.
$A=b^2(d^2+e^2+f^2e^2+c^2)+(e^2f^2+c^2f^2+c^2d^2+f^2d^2+c^2d^2f^2+e^2d^2)$
and
$B=b^2(f^2d^2+c^2e^2+c^2f^2+d^2e^2)+(c^2d^2e^2+c^2e^2f^2)$
if I didn't miss some term.
Check that $A$ and $B$ don't have a common factor, which I haven't done. And then it is clear they are not squares themselves, since they have no terms linear in $b$. |
H: Properties of the $\text{lcm}(1,2 ,... n)$ function
I was thinking the other day about the following function - a sort of prime factorial:
$$f(n) = \text{lcm}(1,2,\cdots,n) $$
Does this function have a name? Does it have any interesting properties analagous to $n!$ (e.g. a version of Sterling's formula?)
EDIT A previous version of this question asserted that $f(n)$ is equal to the primorial. As pointed out below, that is demonstrably false.
AI: The function $\text{lcm}(1, \dots, n)$ is equal to $$e^{\psi(n)}$$
where $\psi$ is the second Chebyshev function. Many proofs of the Prime Number Theorem proceed by first estimating the Chebyshev function, so you'll find a wealth of information on the Wikipedia page. To make a long story short, the least common multiple of the integers from $1$ to $n$ is approximately $e^n$. |
H: If $u \in K$ is transcendental over $F$, then $F(u) \cong F(x)$
This isn't for homework, but I would just like a small hint please. The question asks
Let $K$ be an extension field of $F$. If $u \in K$ is transcendental over $F$, then $F(u) \cong F(x)$ (where $F(x)$ is the field of quotients of $F[x]$).
I should mention that the definition of $F(u)$ that I am working with is: $F(u)$ is the intersection of all subfields of $K$ containing $F$ and $u$. I have seen some definitions of $F(u)$ that have $F(u) = \left\{ f(u)/g(u) : f(x), g(x) \in F[x] \right\}$. This problem is essentially asking me to show that these two definitions are equivalent.
So I defined a function $\varphi : F(x) \to F(u)$ by $\varphi(f(x)/g(x)) = f(u)g(u)^{-1}$. I have shown that this function is a homomorphism of rings (routine), as well as the fact that it is injective (since $\ker \varphi = \{ 0 \}$), but I am having problems on showing surjectivity. I would really appreciate a small hint to put me back on track.
AI: You can avoid checking injectivity/surjectivity by noting that $\varphi$ is invertible, with inverse function $$\varphi^{-1}:F(u) \to F(x) \ : \ f(u) g(u)^{-1} \mapsto f(x)g(x)^{-1}.$$
This is well defined since there every element of $F(u)$ has a unique representation as $f(u)g(u)^{-1}$ (written in lowest terms). This is because two distinct such representations contradicts that assumption that $u$ is transcendental. |
H: On simple extension on Q
Prove that $\mathbb{Q}\left(\sqrt{5},\sqrt{7}\right)$ is a simple extension on $\mathbb{Q}$. More precisely, show that $\mathbb{Q}\left(\sqrt{5},\sqrt{7}\right)=\mathbb{Q}\left(\sqrt{5}+\sqrt{7}\right)$
AI: Clearly, $\mathbb{Q}(\sqrt{5}+\sqrt{7})\subset\mathbb{Q}(\sqrt{5},\sqrt{7})$ as $\sqrt{5}+\sqrt{7} \in \mathbb{Q}(\sqrt{5},\sqrt{7})$.
Now, $\sqrt{5}-\sqrt{7} \in \mathbb{Q}(\sqrt{5}+\sqrt{7}) \because \sqrt{5}-\sqrt{7}= \frac{-2}{\sqrt{5}+\sqrt{7}}$
$\therefore \sqrt{5}, \sqrt{7} \in \mathbb{Q}(\sqrt{5}+\sqrt{7})$.
$\therefore \mathbb{Q}(\sqrt{5},\sqrt{7})\subset \mathbb{Q}(\sqrt{5}+\sqrt{7})$.
$\therefore \mathbb{Q}(\sqrt{5},\sqrt{7})= \mathbb{Q}(\sqrt{5}+\sqrt{7})$. |
H: Probability of a three-pairs in 6-card poker
Suppose out of a deck of 52 regular playing cards, you are dealt a hand of 6 cards. What is the probability of you getting a three-pair hand? That is, three faces that each appear twice.
I've worked out $${\binom{13}{1}\binom{4}{2}\cdot\binom{12}{1}\binom{4}{2}\cdot\binom{11}{1}\binom{4}{2}\over\binom{52}{6}}\approx0.01821$$
as the answer, since you're picking three faces and any two suits. Is this correct, or did I miss something?
AI: It's a good thought, but you're overcounting. For example, your numbers would suggest that the following are different: (1) a pair of aces, then a pair of twos, then a pair of sixes and (2) a pair of sixes, then a pair of aces, then a pair of twos. Due to this overcounting, your answer is precisely $6$ times as large as it should be, since you're counting each ordered arrangement of the three pairs differently, and there are $3!=6$ ways to arrange $3$ distinct objects in order.
Rather, to determine the faces, we use instead $\binom{13}{3}$ to prevent such overcounting (since we're choosing $3$ of the $13$ possible face values). Then we use $\binom{4}{2}$ for each face (since we're choosing $2$ of the available $4$ suits for each face), so we have: $$\frac{\binom{13}3\cdot\binom{4}{2}^3}{\binom{52}{6}}$$ |
H: Convergence of iterative method
Assume that iterative method:
$x_{k+1}=F(x_{k})$
where $(k=0,\,1,\,2,\,...)$
converges to $\alpha$
which is root of $f(x)=0$
equation.
Prove that if $F(\alpha)=\alpha$, $
F'(\alpha)=F''(\alpha)=...=F^{(p-1)}(\alpha)=0
$ and $ F^{(p)}(\alpha)\neq0$
then convergence of that method is equal to $p$
, that means:
$\underset{n\rightarrow\infty}{\lim}\frac{|x_{n+1}-\alpha|}{|x_{n}-\alpha|^{p}}=C\neq0$
What's formula to calculate constant $C$?
AI: Start with the expression $$| x_{n+1} - \alpha | = | F(x_n) - F(\alpha) |.$$
Make a Taylor expansion of $F(x_n)$ around $\alpha$ (up to the order $p$). Notice that all the terms in the expansion excepted $F(\alpha)$ and the terms with order higher than $p$ do not vanish.
Now, plug the expansion in $| x_{n+1} - \alpha | = | F(x_n) - F(\alpha) |$ and you should only get one term on the right plus a reminder. Divide both sides by $|x_n - \alpha|^p$ and observe that the reminder term goes to $0$ when $n \rightarrow \infty$. You should now have your constant, which should look like $$ C = |F^{(p)}(\alpha)| / p! $$
Hope it helps! |
H: Given the index of an element in a triangular array, how do I find its row?
Consider a triangular array(numbers laid out in rows, where the r-th row contains r elements). Given the index i of an element in this array (assuming the numbers are laid out at indices 1, 2, 3, etc. starting from the top and moving down), which row does that number lie in?
AI: The last element in row $r$ has index $\sum_{k=1}^rk=\frac12r(r+1)$. Thus, an index $k$ is in row $r$ if and only if
$$\frac12r(r-1)<k\le\frac12r(r+1)\;,$$
or $(r-1)r<2k\le r(r+1)$. In other words, given $k$, you’re looking for the smallest positive integer $r$ such that $2k\le r(r+1)$ or, equivalently, such that $r^2+r-2k\ge 0$. Now $r^2+r-2k=0$ when
$$r=\frac{-1+\sqrt{1+8k}}2\;,$$
so you want
$$r=\left\lceil\frac{-1+\sqrt{1+8k}}2\right\rceil\;,$$
where $\lceil x\rceil$ is the smallest integer greater than or equal to $x$. |
H: A nonlinear first-order differential equation
How do we solve the following differential equation?
$$(y{}')^{2}+p(x)(1+y^2)^{3}=0$$
AI: Here is one approach.
Given:
$$(y')^{2}+p(x)(y^2+1)^{3}=0$$
We can solve for $y'$ by taking square roots, yielding:
$$y' = \pm ~ i ~ \sqrt{p(x)}~(y^2+1)^{3/2}$$
We can now separate and integrate:
$$\displaystyle \int \dfrac{1}{(y^2+1)^{3/2}}~ dy = \pm \int i ~ \sqrt{p(x)}~dx$$ |
H: General formula for roots for cubic equation
I have the following cubic equation with $ \beta \in [0,1]$ and $ \delta\in [0,1]$ are 2 parameters. Is it possible to use software to get the explicit expression of the solution.
\begin{equation*}
U^{3}\left( -\left( \frac{\delta^{2}}{2} + \beta\right)+\delta-\frac{1}{2} \right)+U^{2}\left(\frac{3}{2}-\delta-\left( \frac{\delta^{2}}{2} + \beta\right)\right)+U\left(\left( \frac{\delta^{2}}{2}+\beta -2 \delta\right)\right)+\left( \frac{\delta^{2}}{2} + \beta\right)=0
\end{equation*}
I want to solve this equation usin Maple or Mathematica, but I was not able to do it.
AI: You can find formula for cubic equation in Wikipedia: https://en.wikipedia.org/wiki/Cubic_equation
I will shortly describe the process how to solve such equation. This can be done in steps to minimize the use of long complicated expressions.
Step 1: write $aU^3 + bU^2 + cU + d = 0$.
Step 2: reduce to the case $T^3 + pT + q = 0$ (as explained in Wikipedia, divide the equation by $a \ne 0$ and then subsitute $U = T - \frac{b}{3a}$).
Step 3: use Cardano's formula. Taking the real cubic roots you get
$$ T_{\mathbb{R}}=u+v=\sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} $$
Step 4: to find the rest of the roots divide $T^3 + pT + q$ by $T-T_{mathbb{R}}$ and then solve the quadratic that you get.
In all the steps you intermediate parameters-variables, e.g.
$$ A = -\left( \frac{\delta^2}{2} + \beta \right) + \delta - \frac{1}{2} $$
etc. |
H: Finding the limit of $\lim_{v\to180}\frac{360\cos\left(\frac{v}{2}\right)}{180-v}$
I need to find this limit:
$\displaystyle\lim_{v\to180}\frac{360\cos\left(\dfrac{v}{2}\right)}{180-v}$, with $v$ in degrees.
I have tried to do this:
$\displaystyle\lim_{v\to180}\frac{360\cos\left(\dfrac{v}{2}\right)}{180-v}=^{L'H}\displaystyle\lim_{v\to180}\frac{-\frac{360}{2}\sin\left(\dfrac{v}{2}\right)}{-1}=180\sin\left(\dfrac{v}{2}\right)=180$.
This is wrong though, because its limit is actually $\pi$. I realise that $\pi$ radians = 180 degrees, but as far as i can tell nothing is linking degrees and radians in the formula. Could someone show me how to find this limit?
AI: Translate to radians in order to make sense of the usual differential stuff we know:
$$\lim_{x\to \pi}\frac{2\pi\cos\frac x2}{\pi -x}\stackrel{\text{l'Hôpital}}=\lim_{x\to\pi}\frac{-\pi\sin\frac x2}{-1}=\pi$$ |
H: Identify the singularity in this ..
$\frac {\sin^2z}{z^2} $
What kind of singularity is present in this ?
My take on this is that, the limit at $z = \infty$ , is $0\,\,$.So, limit is finite.Thus, there should be no essential singularity at $z = \infty$. But,the answer says there is an essential singularity at that point. Also, the series expansion gives no principal part.Thus, shouldn't there be a removable singularity at z=0 ?
Kindly help. I've got a test tomorrow. Confused. Thanks !
AI: Why is the limit $0$ as $z\to\infty$? Keep in mind that the sine function is not bounded on the complex plane as it is on the reals.
Let $f(z)=\frac{\sin^2(z)}{z^2}.$ Then $$\begin{align}f(z) &= \frac{1-\cos 2z}{2z^2}\\ &= \frac1{2z^2}\left(1-\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(2z)^{2n}\right)\\ &=\frac1{2z^2}\sum_{n=1}^\infty\frac{(-1)^{n-1}4^n}{(2n)!}z^{2n}\\ &= \sum_{n=1}^\infty\frac{2(-4)^{n-1}}{(2n)!}z^{2n-2}\\ &= \sum_{n=0}^\infty\frac{2(-4)^n}{(2n+2)!}z^{2n},\end{align}$$ so $f$ has a removable singularity at $z=0,$ since $\lim_{z\to 0}f(z)=1.$
Now, note that $f(\frac1z)$ readily has an essential singularity at $z=0,$ meaning that $f(z)$ has an essential singularity at $z=\infty.$ |
H: Find a minium value of a function
Given x,y,z are positive real numbers such that
$$
x^2+y^2+6z^2=4z(x+y).
$$
Find the minimum value of the following function
$$
P=\frac{x^3}{y(x+z)^2}+\frac{y^3}{x(y+z)^2}+\frac{\sqrt{x^2+y^2}}{z}
$$
AI: let $$a=\dfrac{x}{z},b=\dfrac{y}{z}$$
then $$a^2+b^2+6=4(a+b)$$
and we only find this minimum
$$\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}+\sqrt{a^2+b^2}$$
use Cauchy-Schwarz inequality,we have
$$\left(\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}\right)(ab(a+1)^2+ba(b+1)^2)\ge (a^2+b^2)^2$$
then we only find this follow minimum
$$\dfrac{(a^2+b^2)}{ab[(a+1)^2+(b+1)^2]}+\sqrt{a^2+b^2}$$
since
$$\dfrac{(a^2+b^2)}{ab[(a+1)^2+(b+1)^2]}\ge\dfrac{2}{a^2+b^2+2(a+b)+2} $$because use
$a^2+b^2\ge 2ab$
so we ony find
$$\dfrac{2}{a^2+b^2+2(a+b)+2}+\sqrt{a^2+b^2}=\dfrac{4}{3a^2+3b^2+10}+\sqrt{a^2+b^2}$$
let $x=\sqrt{a^2+b^2}\in[\sqrt{2},3\sqrt{2}]$
so
$$f(x)=\dfrac{4}{3x^2+10}+x$$
then follow is easy to find minimum. |
H: Can there exist an injective function from $\mathbb R$ to $(0,1)$?
I've been trying high and low to find an injective function from $\mathbb R$ to $(0,1)$, but to no avail. I've tried all sorts of polynomial functions, exponential functions, etc. but I've had no luck so far. Can anyone provide an example of such a function, or prove that such an injection is not possible?
AI: Firstly you can take $\mathbb{R}$ to $(0,\infty)$
using $f_1(x)= e^x$.
Then you can translate that by $2$ using $f_2(x)=x+2$ and obtaining $(2,\infty)$.
For last step shrink $(2,\infty)$ by $f_3(x)=\frac{1}{x}$to end up with$(0,\frac{1} {2})$.
All these were $1-1$ so the composition of all of them is injective.
Moral of the story try sucessive transformations, do not try to do at one step. By using injective function you are transforming to equivalent domains which may have been more easy to transform. The idea would be to compose and each time shrinking the domain. |
H: Finding a point in a 2D space
After sitting with this one for half a day, I'm not sure it is possible anymore, given my scenario.
The idea is to get a location on a map (coordinates) from some wifi data.
I have this wonderful system that on demand gives me my estimated position in a building. These positions are limited to a number of fixed points, meaning that if I am a few meters off, it will place me on the nearest fixed point. This system allows me to fetch the positions of all nearby fixed points in forms of coordinates (latitude, longitude), and wifi signal strength, which is how the position is estimated in the first place. That gives each of those fixed points two sets of data - Wifi strengths and coordinates. So for a given point, the data could be something like {A: -30dB, B: -67dB, C: -10dB}, {latitude 57.892305, longitude 54.234999}
The dB indicates the signal strength/distance from the wireless router. For now, I assume that this strength is linear (so that if the distance straight from A: -30dB and A: -40dB = 5 meters, then the distance between A: -30dB and 20dB = 5 meters).
All good so far.
Now what if I did not want an approximate fixed location, but an actual location, using these fixed points as reference? In other words, if I could get the signal strengths at any location, is there any way I could use the fixed points to calculate the coordinates of my current location?
I hope this is understandable, if not, please ask me to rephrase. :)
I tried drawing the idea to make it clearer: https://i.stack.imgur.com/Zna5b.png
AI: There is no really good answer to your question, because wifi signals are poor proxies for distance -- they are affected by walls, electronics, etc. That said, here is an optimization regime that will give you an answer.
Let the desired position be $(x,y)$, while the known locations are $(x_1,y_1),\ldots (x_n, y_n)$. Let the desired wifi strengths be $(a,b,c)$ while the known wifi strengths are $(a_1,b_1,c_1),\ldots, (a_n,b_n,c_n)$.
We first need a model for calculating wifi signal at a given spot knowing the data we know. Here is one way. Let $d_1'=d_1'(x,y)=\frac{1}{(x-x_1)^2+(y-y_1)^2}$, $d_2'=\frac{1}{(x-x_2)^2+(y-y_2)^2}, \ldots$, $d_n'=\frac{1}{(x-x_n)^2+(y-y_n)^2}$. We now normalize as $d_i=d_i(x,y)=\frac{d_i'}{d_1'+d_2'+\cdots+d_n'}$. The result is a set of positive real numbers $\{d_1,d_2,\ldots, d_n\}$ (depending on $x,y$), whose sum is $1$, and where larger numbers correspond to closer distance. With this, we define expected signal strengths $$a(x,y)=d_1a_1+d_2a_2+\cdots+d_na_n\\b(x,y)=d_1b_1+d_2b_2+\cdots+d_nb_n\\c(x,y)=d_1c_1+d_2c_2+\cdots+d_nc_n$$
This gives expected signal strength as a convex linear combination of the known signal strengths, with closer known nodes contributing a larger factor. Lastly, we want to incorporate our data to find $(x,y)$ as
$$F(x,y)=(a-a(x,y))^2+(b-b(x,y))^2+(c-c(x,y))^2$$
This is an optimization problem in $x,y$ that can be solved numerically, or analytically with partial derivatives. The coordinates $(x,y)$ that minimize $F$ are the answer you seek. |
H: Prove that if $ac + bc + c^2 < 0$ then equation (usual notation) has two roots
We have $a, b, c$ real parameters, $a ≠ 0$.
Prove that $ax^2 + bx + c = 0$ has two different roots ($b^2-4ac > 0$), if $ac + bc + c^2 < 0$
AI: If $P(x) = ax^2+bx+c$, from $c(a+b+c)<0$, we have $P(0)P(1) < 0$. So there is a root in $(0, 1)$. Further, note it cannot be a double root, as then the sign wouldn't have changed... |
H: Spectral radius of a compact operator and convergence
Let $T$ be a compact operator such that the spectral radius $\rho(T)<1$
does it then follow that $||T^n|| \rightarrow 0$ as $n \rightarrow \infty$?
AI: You need the spectral radius formula
$$
\rho(T) = \lim \|T^n\|^{1/n}
$$
Fix $\delta > 0$. Suppose $\rho(T) < 1$, then there is $1 > \epsilon > 0$, and $n_0 \in \mathbb{N}$ such that
$$
n \geq n_0 \Rightarrow \|T^n\|^{1/n} < (1-\epsilon)
$$
Or equivalently,
$$
\|T^n\| < (1-\epsilon)^n \quad\forall n\geq n_0
$$
Since $(1-\epsilon) < 1$, there is $n_1 \in \mathbb{N}$ such that
$$
(1-\epsilon)^n < \delta \quad\forall n \geq n_1
$$
Hence, for all $n\geq \max\{n_0,n_1\}$, it follows that
$$
\|T^n\| < \delta
$$
Hence, $\lim \|T^n\| = 0$ |
H: Centre manifold non-hyperbolic fixed points
I have the following dynamical system $\dot x=-x^3, \dot y=-y$.
I would like to prove that there are an infinite number of trajectories that become tangent to the line $y=0$ for $x\rightarrow 0$. Each of the trajectories is a centre manifold and they satisfy $y=h(x)$ where all possible forms of $h$ have zero taylor expansiion about origin.
First of all we know that the line $y=0$ is invariant. The only fixed point is $(0,0)$, the eigenvalues are $-1,0$ with Eigenvectors $(0,1), (1,0)$.
How can I proceed? Do I have to find some guess for $h(x)?$
AI: I think you can just solve for $h(x)$ directly.
$x$ will satisfy
$$\dfrac{1}{2}x^{-2} - C_1 = t$$
and $y$
$$ y = C_2e^{-t}. $$
Combine these expressions to get $y$ as a function of $x$. |
H: Clarification on $\frac{0^n}{0}$ when $n>0$
If this is a duplicate, I will gladly delete this if there is a duplicate but I've had difficulty finding one. I had, until recently, believed that we don't define $\frac 0 0$ as the limits coming from different directions vary widely and so no value works for all cases. I thought that $0^0$ wasn't defined as $0^1*0^{-1}=0^0=\frac 0 0$. I have been corrected on this but am now confused about such numbers as $\frac {0^2} 0$.
I had believed that as $\frac{0^2}{0} = 0^2*0^{-1} = 0^1 = 0$. Is $\frac {0^n} 0$ for $n>1$ equal to zero as I had previously believed? To be clear, I am mostly curious if I can state that $y=\frac{x^2}{x}$ is zero when $x=0$ or whether I have to specify that the limit as $x \rightarrow 0$ is $0$.
AI: There is quite a big distinction between numbers and limits. In the case of numbers $0^2=0$ and so, no matter how much algebra you do, $0^2 \div 0 = 0\div 0$. You already know that $0\div 0$ is undefined and then so tell is $0^n$ for all $n \ge 1$ because $0^n = 0$ for all $n \ge 1$.
Limits are really very different. Even though $x^2 \to 0$ as $x\to 0$, the limit
$$\lim_{x \to 0} \frac{x^2}{x}=0$$ is well-defined because $x^2$ tends to zero "faster" than $x$ does. It's all to do with how things behave on their way to the limit. It's not the actual destination that matter. Similarly
$$\lim_{x \to 0} \frac{x}{x^2} = \infty$$
because $x^2$ gets smaller faster than $x$ and so you have something much smaller on the bottom and something much bigger on the top. |
H: $\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$ if $(a,b)=1$
I hope to show that
$$\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$$
for $(a,b)=1$.
I made an effort to find a homomorphism but I failed.
Can you give a hint?
AI: Hint: $(a,b)=1$ if and only if there exist integer $m,n$ such that $an+bm=-1$.
Incidentally, the result should be that $$\Bbb Z[i]/(a+bi)\cong\Bbb Z_{a^2+b^2}.$$ |
H: Can you please check my limit proof
I proved the following:
If $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$
How I proved it: First I proved it for $x=0$. Then: Let $\varepsilon > 0$. If $x_n \to x$ then $(x_n - x) \to 0$. Since we have shown for $x=0$ we may assume that $x \neq 0$. Then
$$ x_n - x = (\sqrt{x_n} - \sqrt{x})(\sqrt{x_n} + \sqrt{x}) $$
implies
$$ {x_n - x \over \sqrt{x_n} + \sqrt{x}} = \sqrt{x_n} - \sqrt{x}$$
Since $\lim \sqrt{x_n} + \sqrt{x} = 2 \sqrt{x}$ and $x \neq 0$ it follows that
$$\lim \sqrt{x_n} - \sqrt{x} = \lim {x_n - x \over \sqrt{x_n} + \sqrt{x}} = 0$$
and therefore $\sqrt{x_n} \to \sqrt{x}$.
Is it correct?
AI: I'm afraid you used your conclusion to prove your conclusion, when you said: "Since $\lim\sqrt{x_n}+\sqrt x=2\sqrt x$...."
It's also worth noting that you must be very careful with your proof in the case that $x=0$. For example, consider the sequence $x_n=-\frac1n.$
It would be more accurate (in the context of real analysis) to say the following:
For any $x\ge 0$ and any sequence of non-negative numbers $x_n,$ if $x_n\to x,$ then $\sqrt{x_n}\to\sqrt x.$
I leave the case that $x=0$ to you. For the $x>0$ case, we can use the observation that $$|\sqrt{x_n}-\sqrt{x}|=\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}.$$ Show that for any $\epsilon>0,$ we can find $N$ such that for all $n\ge N$ we have $\sqrt{x_n}+\sqrt x\ge\frac54\sqrt x$ (hint: $t\mapsto\sqrt t$ is an increasing function on the non-negative reals) and $|x_n-x|<\frac54\epsilon\sqrt x.$
Hence, for any $\epsilon>0,$ we can find $N$ such that for all $n\ge N,$ we have $$|\sqrt{x_n}-\sqrt{x}|=\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}\le\cfrac{|x_n-x|}{\frac54\sqrt{x}}<\cfrac{\frac54\epsilon\sqrt{x}}{\frac54\sqrt{x}}=\epsilon.$$ |
H: Having trouble with a Lim (infinity-infinity)
so I am learning about Limits at the moment and I am having some trouble with this Lim..
$\lim_{x\to ∞}=\sqrt{x}(\sqrt{16x^3+4x-1}-\sqrt{16x^3+3}) $
I just can't get the right answer, which is $ \frac 12 $ ,
I know how to do the -> $\lim_{x\to ∞}=(\sqrt{16x^3+4x-1}-\sqrt{16x^3+3}) $
but I can't do the exact lim with square root from x in front... I just don't know what to do with it, tried everything and looked everywhere but I can't even find one example with this particular setup...
help ? anyone?
AI: Distribute the $\sqrt x$: $$\sqrt a(\sqrt b - \sqrt c) = \sqrt a\sqrt b - \sqrt a \sqrt c = \sqrt {ab} - \sqrt{ac}$$
In terms of your limit:
$$\lim_{x\to ∞}=\sqrt{x}(\sqrt{16x^3+4x-1}-\sqrt{16x^3+3}) = \lim_{x\to \infty} (\sqrt{x(16x^3+4x-1)}-\sqrt{x(16x^3+3)})$$
Now evaluate the limit as you know how to do: by multiplying the function by $\dfrac{\text{conjugate}}{\text{conjugate}}$ |
H: Leisure reading for an undergraduate student
I am a freshman at a local university. I never really had much passion for math, but I always did well in math exams . I attribute this lack of passion to rote learning/emphasis on methods/formulas than philosophy behind those methods that is so common in Indian education system.
My question is what are the journals/fiction can I read for fun that can inspire passion and excitement. I perused through math journals like American Mathematical Monthly in my uni's library, but I found them to be way too complicated for me.
AI: There's the classic "Godel, Escher, Bach" by Douglas Hofstadter. I won't try to give you a description or synopsis (as you can find countless others on the internet), but it's worthwhile to say the least. Long, and at times difficult, but worthwhile.
There are recent graphic novels about the lives of Bertrand Russell and Richard Feynman (respectively "Logicomix" and "Feynmann") that are more biographies than mathematics, but still very good reads.
There's also no shortage of recreational mathematics written by the late Martin Gardner. Gardner used to write a column called "Mathematical Games" in Scientific American. These can be found compiled into numerous anthologies.
There's one called "The Mathematical Experience" that gives insight into how mathematicians think, and what research mathematics actually consists of.
Finally, there's the short work of fiction "Flatland" which will take no more than an afternoon to read, but is another classic.
EDIT: I can also recommend the Orthogonal series by Greg Egan. These are science-fiction stories set in a universe with physics different than ours; among other things, light travels at varying speeds depending on its frequency. Again, these are sci-fi, and not mathematics, but Egan does a very good job of detailing and explaining the physics he creates, and sees-through their consequences in a manner that is very satisfying to a scientific mind. |
H: Complex number - how to find the angle between the imaginary axis and real axis?
Assume I have complex number $z = a + ib$.
$z$ can be represented by a polar representation as $r(\cos \theta+i\sin \theta)$,
when $r$ is the absolute value of $z$, $\sqrt{a^2 + b^2}$.
But how can I find $\theta$?
AI: Consider the following Argand-diagram
The y-axis is the imaginary axis and the x-axis is the real one. The complex number in question is
$$x + yi$$
To figure out $\theta$, consider the right-triangle formed by the two-coordinates on the plane (illustrated in red). Let $\theta$ be the angle formed with the real axis.
$$\tan\theta = \frac{y}{x}$$
$$\implies \boxed{\tan^{-1}\left(\frac{y}{x}\right)}$$
The hypotenuse of the triangle will be
$$\sqrt{x^2 + y^2}$$
Therefore,
$$\sin\theta = \frac{y}{\sqrt{x^2 + y^2}}$$ |
H: Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem)
Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(2\cos\theta \sin\theta) - \sin^2\theta\\
= \cos^2\theta - \sin^2\theta + i(2\cos\theta \sin\theta)
$
so $\cos2\theta = \cos^2\theta - \sin^2\theta$
Is this correct?
AI: Yes,
$$\cos^2\theta - \sin^2\theta = (\cos\theta)(\cos\theta) - (\sin\theta)(\sin\theta)$$
$$=\cos(\theta + \theta) = \cos(2\theta)$$ |
H: Please explain to me why $ \int_b p(a|b) db \neq p(a) $
I have one question that bugs me. How is it that:
$
\int_a p(a|b) da = \int_a \frac{p(a,b)}{p(b)} da = 1
$
but
$
\int_b p(a|b) db = \int_b \frac{p(a,b)}{p(b)} db \neq p(a)
$
I don't understand this because $ p(a,b) = p(b,a) $ and therefore
$
p(a|b) = \frac{p(a,b)}{p(b)} \\
\Rightarrow p(a,b) = p(a|b) p(b) = p(b|a) p(a) = p(b,a)
$
and if I'm not mistaken:
$
p(a|b)p(b) = p(b|a)p(a) \\
\frac{p(a,b)}{p(b)} p(b) = \frac{p(a,b)}{p(a)} p(a) \\
p(a,b) = p(a,b)
$
But how is it, that this is not correct:
$
\int_b p(a|b) db = \int_b \frac{p(a,b)}{p(b)} db = \int_b p(a,b) \frac{1}{p(b)} db = \int_b p(a,b) db \int_b \frac{1}{p(b)} db \\
\int_b p(a,b) db \int_b \frac{1}{p(b)} db = \int_b p(a,b) db *1 = \int_b p(a,b) =p(a)
$
which should hold because of the marginalization rule:
$
p(a) =\int_b p(a,b)
$
AI: What is true is that $\int_bp(a,b)db = p(a)$. Roughly speaking the probability that $a$ and $b$ occur, summed over all $b$, is just the probability that $a$ occurs.
I don't really understand what you are asserting to be true and what you are asserting to be false and what you are asking. But there are several mistakes in your post. Firstly, the equality $p(a|b) p(b) = p(b|a) p(b)$ is wrong, in general, unless $p(a)=p(b)$ or one of $p(a)$, $p(b)$ is 0. So is $\int_b p(a,b) \frac{1}{p(b)} db = \int_b p(a,b) db \int_b \frac{1}{p(b)} db$. And so is $\int1/p(b)db=1$, as you can easily convince yourself by considering a finite probability space. I am not sure: does that resolve your difficulties? |
H: Simple Derivative
I am wanting trying to remember how to solve a derivative of this nature:
$$ \frac{dM}{dt} = rM(t)$$
$$ dM = rM(t)dt $$
Solving when t = 0 equals 1 we can get the solution as
$$ M(t)= M(0)e^{rt} $$
AI: divide both side by $M(t)$,we get $dM/M(t)=rdt$ now know that integral of $1/t$ is $ln(t)$,can you follow from this?
EDit:
if we integrate we get $ln(M(t))=rt+c$ or $M(t)=e^{r*t+c}$
now let us represent $e^{r*t+c}=e^{r*t}*e^{c}$;if we denote $e^c=C$,we get
M(t)=$e^{r*t+c}=C*e^{r*t}$
from there $M(0)=C=1$ |
H: Limit of $\frac{\sqrt{mx^2}}{\sqrt{\sin(m+1)x^2}}$
I know it's an easy question, but I could use your help here, anyway.
Could you tell me how to prove that $ \lim _{x \rightarrow 0}\frac{\sqrt{mx^2}}{\sqrt{\sin(m+1)x^2}} = \sqrt{\frac{m}{m+1}}$? I know I should use the fact that $\lim _{x \rightarrow 0} \frac{\sin \phi}{\phi}=1$. I've tried writing $\sin(mx^2 + x^2) = \sin(mx^2)\cos (x^2) + \cos(mx^2)\sin (x^2)$, but nothing comes to mind.
I would really appreciate your help. I know it should be obvious, but I don't see it.
AI: $$\left(\frac{\sqrt{mx^2}}{\sqrt{\sin(m+1)x^2}}\right)^2= \frac{mx^2}{\sin(m+1)x^2}\\=
\dfrac{mx^2}{\sin(mx^2)\cos (x^2) + \cos(mx^2)\sin (x^2)} \\
=\dfrac{1}{\frac{\sin(mx^2)\cos (x^2)}{mx^2}+\frac{\cos(mx^2)\sin (x^2)}{mx^2}}\underset{x\to{0}}\to\dfrac{1}{1+\frac{1}{m}}=\dfrac{m}{m+1}.$$ |
H: Is this a poset?
Is $(S, R)$ a poset where $S$ is the set of all people in the world and $(a, b) \in R$, where $a$ and $b$ are people if $a$ is no shorter than $b$?
My attempt:
$a$ is no shorter than $a$. This is not reflexive because $a$ can be taller than $a$. Contradiction
True?
AI: Are you shorter than yourself? I kind of doubt it. Thus, $\langle\text{Don Larynx},\text{Don Larynx}\rangle\in R$. The same goes for anyone else, so $R$ is reflexive.
You may be having trouble with no taller than: $a$ is no taller than $b$ if the height of $a$ is $\le$ the height of $b$. From this it’s also not hard to check that $R$ is transitive.
The crucial property is antisymmetry. What happens if there are two different people whose heights are equal? |
H: How to prove sin(sin(sin...(sin(1))) converges to 0 without using continuity?
Using continuity I was able to show the sequence $x_0 = 1$, $x_{n+1} = sin(x_n)$ converges to 0, but I was wondering if there was a way to prove it using only properties and theorems related to sequences and series, without using continuity.
So far, I know the sequence is monotonically decreasing and bounded below by 0, so it must converge to its infimum. From here I'm not exactly sure how to show 0 is the infimum of this set of numbers.
Alternatively, I could check convergence to $0$ by comparison, but no sequences come to mind that are greater than the given sequence for all $n \geq N$ and converge to $0$.
I've already seen the answers at the following:
Compute $ \lim\limits_{n \to \infty }\sin \sin \dots\sin n$
Prove that $\sin(\sin...(\sin(x))..)$ converges asymptotically to zero
AI: We know that for all $x \in (0,1]$, we have
$$
0 < \sin x < x
$$
From there, you can show that $x_0,x_1,\dots$ is a strictly monotonically decreasing sequence. Is this enough? That is, can we forgo continuity? No. As a counterexample, consider the seuqence $x_k = f(x_{k-1}); x_0 =1$ with
$$
f(x) =
\begin{cases}
\sin(x) + 0.1 & x^*<0\leq 1\\
\sin(x) & 0 < x \leq x^*
\end{cases}
$$
Where $x^*$ is the positive solution to $= x^* = \sin(x^*) + 0.1$.
We note that $x_k$ is a strictly monotonically decreasing sequence converging to $x^*$ (and therefore bounded below by $0$), and we even have $f(x)<x$ for every point in $(0,1]$. However, because of the lack of continuity, this is not enough to "force" $\{x_k\}$ to converge to $0$. |
H: System of equations with multiplication
I want to find all $x,y,z\in\mathbb{R}$ such that $(x+1)yz=12, (y+1)zx=4, (z+1)xy=4$.
I can multiply all three equations to get $(x+1)(y+1)(z+1)x^2y^2z^2=192$.
I can divide the first equation by the second to get $\dfrac{(x+1)y}{(y+1)x}=4$, which simplifies to $3xy+4x-y=0$.
None of these seems to help. How can I solve the equations?
AI: Your equations are
$$\begin{align*}I&\;\;xyz+yz&=12\\
II&\;\;xyz+xz&=4\\
III&\;\;xyz+xy&=4\end{align*}$$
Well, now a little algebra:
$$\begin{align*}I-II:&\;\;(x-y)z=8\\
I-III:&\;\;(x-z)y=8\\
II-III:&\;\;(z-y)x=0\end{align*}$$
Since $\;x\neq 0\;$ (why?), we get $\;y=z\;$ and thus $\;xz=xy=y^2+8\;$ ...take it from here. |
H: Removing logs from equation
I have a simple question that I need clarification on:
If
$$\log(a) = \log(b) + c$$
is it true that
$$a = b + \exp(c)$$
Is this correct or am I missing something really basic that I cant remember from maths class?
AI: You’re missing something, namely, one of the laws of exponents: if $\log a=\log b+c$, where these are natural logs, then
$$a=e^{\log a}=e^{\log b+c}=e^{\log b}\cdot e^c=be^c\;.$$
(If they are common logs, with base $10$ rather than $e$, replace every $e$ by $10$.) |
H: Prove using Jensen's Inequality
Let $\alpha_1, \alpha_2, . . . , \alpha_n$ be the interior angles of a convex (but not necessarily regular) n-gon. Prove, that for all integers $n\geq3$:
$$\cos \alpha_1 + \cos \alpha_2 + \cdots + \cos \alpha_n + n \cos\left(\dfrac{2\pi}{n}\right) \leq0$$
The prof said that I need to use Jensen's Inequality, and may be something else.. But I don't see it!
AI: If all the interior angles are acute, then noting that $\cos \theta$ is concave for $\theta \in (0, {\pi \over 2})$, we can use Jensen's Inequality.
$$\frac1n \sum \cos \alpha_i \le \cos \left(\frac1n \sum \alpha_i \right) = \cos \left( \frac{\pi(n-2)}{n} \right) = \cos(\pi - \frac{2\pi}{n}) = -\cos(\frac{2\pi}n)$$
Note that this is more restrictive than just being convex, and in fact your inequality does not hold in general if one of the angles can be larger than $\pi \over 2$. |
H: Third-degree cosine inequality for obtuse triangle
Suppose $\triangle ABC$ is an obtuse triangle with side lengths $a=BC, b=CA, c=AB$. I want to show that $$a^3\cos A+b^3\cos B+c^3\cos C<abc.$$
My idea is to use the cosine rule. I have $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, etc. Plugging into the inequality I get
$$a^4b^2+a^4c^2-a^6+b^4a^2+b^4c^2-b^6+c^4a^2+c^4b^2-c^6<2a^2b^2c^2.$$
How can I show this?
AI: Using $\sin2x,\cos2x$ formula
$$a^3\cos A=(2R\sin A)^3\cos A= 2R^3(2\sin^2A)(2\sin A\cos A)$$
$$=2R^3(1-\cos2A)\sin2A=R^3(2\sin2A-2\sin2A\cos2A)=R^3(2\sin2A-\sin4A)$$
Using this, $\sum \sin2A=4\prod \sin A$
Now, $\sin4A+\sin4B+\sin4C=2\sin(2A+2B)\cos(2A-2B)+2\sin2C\cos2C$
Now, $\cos2C=\cos\{2\pi-2(A+B)\}=\cos2(A+B)$ and $\sin2(A+B)=\sin(2\pi-2C0=-\sin2C$
$\implies\sin4A+\sin4B+\sin4C=-2\sin2C\cos(2A-2B)+2\sin2C\cos2(A+B)$
$=-\sin2C\{\cos(2A-2B)-\cos(2A+2B)\}=-\sin2C\cdot2\sin2A\sin2B$
$=-2(2\sin C\cos C)(2\sin A\cos A)(2\sin B\cos B)$
For an obtuse triangle, only one angle is between $(\frac\pi2,\pi)$ so exactly one of the cosine ratio $<0$ and all the sine ratios are $>0$
$\implies\sin4A+\sin4B+\sin4C>0$
$\implies\sum a^3\cos A <2R^3(\sum\sin2A)=2R^3(4\sin A\sin B\sin C)=\prod(2R\sin A)$ |
H: Find asymptotes $e^{-x}-e^{-2x}$
I should find the asymptotes for $e^{-x}-e^{-2x}$. So I will take limits.
My attempt goes like this:
1) when x → $+\infty$ then $e^{-x}$ → $0$ and so does $e^{-2x}$. Thus: $0$. Quite easy.
2) when x → $-\infty$ then $e^{-x}$ → $\infty$ and so does $e^{-2x}$. So we have $\infty - \infty$ but the right $\infty$ is "bigger" since $e^{-2x}$ dominates over $e^{-x}$ for x close to $-\infty$.
AI: Your analysis is correct, but absolutely not formal enough. The second case can for example be done as follows:
$$e^{-x}-e^{-2x} = e^{-x}(1-e^{-x})$$
Taking the limit $x\rightarrow-\infty$ you find a limit of the form $\infty\cdot(1-\infty)$, thus giving you $-\infty$. |
H: Power series help please!
$$ \frac{2}{3-x} $$
I need to find a power series representation for this. I figured i'd pull the two out, but I can't figure out what to do with the three.
AI: Recall that for $|x|<1$:
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
So you'd like to bring your function in a similar form. You can do that by manipulating as follows:
$$\frac{2}{3-x} = \frac{2}{3}\frac{1}{1-\frac{x}{3}} = \frac{2}{3}\sum_{n=0}^\infty \left(\frac{x}{3}\right)^n$$ |
H: solution of $y^2 - x = 15$ and $x^2 -xy = 2009$
Find all the integer solutions to the equations:
\begin{eqnarray}
y^2 - x &=& 15 \\
x^2 -xy &=& 2009
\end{eqnarray}
Not sure how to solve this :/, tried the usual algebra way (solving for something and substetuting) but didn't really work out
$x+15$ must be a square number, so $x$ has to be $1,10,21,34,49,\ldots$
Using $49$ it works and gives me a solution, but is that the only one? And if, how do I show it.
AI: x² -xy = 2009
x(x-y)=1*2009 =2009 *1 =-2009 (-1)=(-1)(-2009)
x(x-y)=7*287 =... like that
x(x-y)=49*41=...like above
find x,y then compare with other equation |
H: How can I express the non-intersecting sections of multiple sets with a single set operation?
I don't have a lot of experience with set theory, as I suspect this question will make clear!
As the title says, I'm interested in expressing the non-intersecting sections of three sets using a single set operation.
What I mean is, how would you express the gray shaded regions in the following Venn Diagram?
My best guess is that this has to do with complements... something like
((A ∪ B) ∪ (A ∪ C) ∪ (B ∪ C))C
...maybe? But I am concerned that this complement would also include the universe, U, which I don't want to be part of the selection. I also suspect that my notation is wrong, so apologies if that's the case.
Thanks for any help!
AI: So you want the part $A$ without those parts in $B$ and $C$; hence you want $$A \backslash (B \cup C)$$
Likewise you want $B$ without either $A$ or $C$, hence you also-want $$B \backslash(A \cup C)$$; and similarly you want $C$ without $A$ or $B$; hence also desire $$C \backslash(A \cup B)$$
Now overall you want any of the above, hence, you union to obtain
$$\boxed{A \backslash (B \cup C) \;\;\cup\;\; B \backslash(A \cup C) \;\;\cup\;\; C \backslash(A \cup B)}$$
Hope that helps!
Note:: $X \backslash Y$ means the set $X$ without the elements in the set $Y$; read "X less Y". |
H: Solving recursion formula with sum
I am trying to solve the following recurrence, but i am stuck...
$$t(n)=n + \sum_{j=1}^n t(n-j)$$
I really appreciate your help,
Tarcísio.
AI: Notice that as $j$ runs from $1$ to $n$ in the summation, $n-j$ runs from $n-1$ down to $0$, so we can rewrite the recurrence as
$$t(n)=n+\sum_{k=0}^{n-1}t(k)\;.\tag{1}$$
It never hurts to gather some numerical data:
$$\begin{align*}
t(0)&=0+t(0)\\
t(1)&=1+t(0)\\
t(2)&=3+2t(0)\\
t(3)&=7+4t(0)\\
t(4)&=15+8t(0)\\
t(5)&=31+16t(0)
\end{align*}$$
At this point there’s a pretty obvious pattern: it appears that
$$t(n)=2^n-1+2^{n-1}t(0)=2^{n-1}\big(t(0)+2\big)-1\tag{2}$$
for $n>0$. Proving $(2)$ is just a matter of verifying that it gives the correct value for $t=1$ and satisfies the recurrence $(1)$, i.e., that
$$2^n-1+2^{n-1}t(0)=n+\sum_{k=1}^{n-1}\left(2^k-1+2^{k-1}t(0)\right)+t(0)\;,$$
which is a fairly straightforward calculation.
Added: For a completely different approach, note that
$$t(n)-t(n-1)=n+\sum_{k=0}^{n-2}t(k)=1+\left((n-1)+\sum_{k=0}^{n-2}t(k)\right)=1+t(n-1)$$
and hence $$t(n)=1+2t(n-1)$$ for $n\ge 2$. This is a simple recurrence that can be solved in a variety of ways. |
H: Why factorials when divided by factorials less than the number have a remainder 0?
Lets take the example, if we take the expression $\frac{X!}{y_1!\cdot y_2!\cdots y_n!} $as long as summation $S=y_1+y_2+...y_n$ is less than or equals $X$, the remainder is always $0$. Thats How the permutation of $X$ things where there is $y_1$ things same , $y_2$ things same works. My question is, why does this happen, what is the mathematical explanation behind this?
when its like $\frac{100!}{49!\cdot49!}$ that still works? Here the first $49$ consecutive digits already divided, but how the second consecutive $1..49$ also divides by $50...100$?
AI: Not a full answer, but an outline of how I convinced myself of this fact:
Prime factorization is the key word. The key result is that if $y_1 + y_2 + \cdots y_n \leq x$ then for any prime $p$ the power of $p$ in the factorization of $x!$ is at least as high as in $y_1!y_2!\cdots y_n!$.
Once you've convinced yourself of this, you can (pretend to) write down a prime factorization of the numerator and denominator and see that the denominator can be simplified away completely. Thus you end up with just a product of primes (whatever's left of the numerator after the simplification), which is trivially an integer. |
H: Theoretical impossibility? Deviation from normality with a sample greater than 300?
Huge thanks in advance!
I've been lead to believe that the following is a theoretical impossibility: a population larger than 300 records without an approximation of a normal distribution. The dataset I used is a set of amounts of financial transactions (specifically, donations to a charity).
Using R, I performed multiple Shapiro-Wilks tests using multiple random samples (sized 5000) drawn from about 100000 gift amounts and received a p value of 0. This is supposed to indicate that that the sample deviates from normality.
Are the following claims in fact inconsistent (as they seem to be):
My sample of gift amounts is larger than 30
My sample of gift amounts approximates a normal distribution
According to the Shapiro-Wilks test, my data set deviates from a normal distribution
AI: I strongly suspect the reason that the Shapiro-Wilks test is telling you that the sample is not coming from a normal distribution is because the underlying base distribution is not normal. Basically if the underlying distribution is not normal then approximating it with a normal distribution isn't going to work because even if you get a lot of samples its still not a normal distribution.
EDIT: I just saw your comment about textbooks and using the normal approximation after a certain sample size. These textbooks will be referring to the sample distribution being normally distributed after a certain number of samples. This is not the same as finding the underlying or base distribution though and might have led to some confusion.
For some intuition building: The reason why approximating the sample distribution with the normal distribution tends to work out frequently is because very many distributions found in nature are either normal or satisfy the requirements for the CLT to be true. The most basic case is the times we have a normal distribution as the underlying distribution, in this case the sample distribution will be normal for any sample size. For example we might have a different non-normal base distribution say the height of certain species of trees. You might see that there are some that are double the height of others and even less that are 3 times the height and so on. Now the interesting thing is that say the underlying distribution is not normal but satisfies the requirements for the CLT to hold so then the sample distribution is normal for a large enough sample size. I think this is what the textbooks are mostly getting at when they say for sample of N > 30 you can start to use the normal distribution as an approximation for the sample distribution. This is all well and good if the distribution you are approximating is indeed normal or satisfies the requirements for the CLT. However the financial world has many non-normal distributions in it and further has situations where the variance and mean are poorly defined. For something like donations it is entirely plausible that someone makes a donation that is 10^n times bigger than someone else (with n being a large number). This type of difference is not something that you would see in say a normally distributed sample of say the height of trees in nature, there is no tree that will be 1000000 times taller than the average and yet in the financial world these types of things certainly can, and do, happen. The impact of extreme events on the mean and variance of an individual sample results in having to have a larger sample. If the mean of a sample is overwhelmingly defined by the maximum donation made in that sample then I think we are starting to get into the realm of the extreme value theorem and that I suspect is what has happened here. In this case we would be essentially looking for the maximum value found in each sample and this is no longer a normal distribution.
As for finding the underlying base distribution, the last time I looked at something similar to this I found that the underlying distribution was actually an extreme value distribution. I would suggest as a starting point looking at generalized extreme value distributions for this.
I would check using some Q-Q plots what the underlying distribution is. It might be a bit qualitative for people's liking on here but you can start to see if a distribution is not a normal distribution this way. |
H: Equation $\sqrt{x}+\sqrt{y}=\sqrt{2013}$ in rationals
Can we find all rational numbers $x,y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{2013}$?
Certainly possible answers are $(2013,0)$ and $(0,2013)$.
If we square the equation, we get $x+y+2\sqrt{xy}=2013$, so $\sqrt{xy}$ must be rational.
AI: if $x\neq 0$ then $x,\sqrt{xy},y$ make a geometric progression, so we can write $y=xq^2$ and $q$ is rational.
$x+y+2\sqrt{xy} = 2013$
$x + xq^2+2xq = x(q+1)^2=2013$
$x= \frac{2013}{(q+1)^2}, y=xq^2, q\in \mathbb Q^+\cup\{0\}$
if $x=0$ then $y = 2013$. |
H: Separable set, the real [0,1] interval and measure
I am having a hard time understanding exactly what "separable" means, and I am trying to relate that to the measure of the real [0,1] segment (which is 1, right?).
My confusion started when studying measure for Lebesgue integration. The set of rationals in [0,1] is given measure 0, so that we can integrate the indicator function of the rationals on [0,1] (which has value 0 - so the complement, the the integral of the indicator of the non-rationals on [0,1] must be 1?).
So far so good, it's not too hard to accept that the rationals are countable, so they "count for nothing" on [0,1], which has "many more" elements. I saw a rigorous presentation where an explicit cover of the rationals in [0,1] is built and its measure brought to zero when taking a limit. No problem.
However, I started thinking that maybe I could similarly cover the non-rationals, and bring that cover's size to 0 too. Which must fail somewhere, because that set (of non-rationals in [0,1]) must have size 1. Then I met the concept of "separability", and I started thinking: if the reals are separated by the rationals (the rationals are "dense" in $\mathbb{R}$, I can't find "two reals that are adjacent", and therefore I would be able to find a cover who limit size is 0?
Obviously, I am making several mistakes along the way, and I am not having my topology straight. What does "separable" really entail for the reals in [0,1]? How/why will the cover size for the non-rationals in [0,1] converge to 1 rather than 0?
AI: Being separable means that there is a dense countable subset. This means that there is a subset we can write as $\{x_n\mid n\in\Bbb N\}$, such that whenever $x$ is a point in the space, there is a sequence $x_{n_k}\to x$.
That what it means being separable. Indeed $\Bbb Q\cap[0,1]$ is a dense countable subset of $[0,1]$, so it witnesses the separability.
And indeed as you said, one can enumerate the countable subset, then for every $\varepsilon$ take the interval of length $\frac1{2^n}\cdot\varepsilon$ around $x_n$, and that would amount to a cover of the countable set of measure at most $\varepsilon$.
Why would that fail if we take the irrational numbers instead?
There are uncountably many irrational numbers.
This means that we can't enumerate them and use the trick of $\frac1{2^n}$. At some point we would have to resort to infinite powers of $2$, which is not a well-defined notion in the real numbers.
Measure is volume, whereas density is an order related property (in the real numbers, anyway, as the topology is induced by the order). So while the rational numbers are dense, their volume is zero. The irrational numbers, on the other hand have a volume of $1$ (in the unit interval, of course).
This is exactly the meaning of the fact that the irrational have a full measure.
Finally, given $\varepsilon$, we construct the $\varepsilon$-cover of the countable set, we may want to claim that it actually covers all the interval. But that would be false. Depending on the enumeration (the choice of how we indexed the $x_n$'s to begin with) we can construct irrational numbers which are not covered.
The process is similar to the construction of the Cantor set. Take the complements of all the intervals, and show that not finitely many of them have an empty intersection (which is simple enough to see). Then by the fact that these are all compact sets, the intersection is non-empty and must contain an irrational number not covered. |
H: about completion of a metric space
Is the following proposition true? If yes, how would you prove this?
Proposition :
Let $(X,d)$ a metric space and let $(\widetilde{X},\widetilde{d})$ be a completion of $(X,d)$
then if : $Y\subset \widetilde{X}$ and $\overline{Y}=\widetilde{X} \Rightarrow X=Y.$
AI: It’s clearly false. Let $X=\Bbb Q$ with the usual metric; then $\widetilde{X}=\Bbb R$ with the usual metric. Let $Y=\Bbb Q\cup\{\pi\}$; clearly $\operatorname{cl}Y=\Bbb R$, but $Y\ne\Bbb Q$. |
H: List of preferences for four ladies and four gentleman where no one obtains his or her first pick.
How can I determine a list of preferences for four ladies and four gentlemen where no one obtains his or her first choice in a stable matching determined by the Gale-Shapely algorithm, regardless of who proposes.
I understand how to use the Gale-Shapely algorithm, but not sure how do this. Have tried using brute force but have not gotten very far. I can't seem to make it so the 4th boy does not receive his 1st choice when they are proposing.
AI: can you confirm this:
women are 1-4; men a-d; x can be anyone not already in that preference.
[ \begin{array}{lcr}
\mbox{1} & a&b&c&d \\
\mbox{2} & d & c&a&b \\
\mbox{3} & b & d&x&x \\
\mbox{4} & b & a&x&x \\
\ \\
\mbox{a} & 2 & 4&3&1 \\
\mbox{b} & 2 & 1&3&4 \\
\mbox{c} & 1 & 2&x&x \\
\mbox{d} & 1 & 3&4&2 \end{array}]
Men Propose:
step 1: 1 accepts c, rejects d. 2 accepts a, rejects b.
step 2: 1 accepts b, rejects c. 3 accepts d.
step 3: 2 accepts c, rejects a. 4 accepts a.
Women Propose:
step 1: a accepts 1. b accepts 3, rejects 4. d accepts 2.
step 2: a accepts 4, rejects 1.
step 3: b accepts 1, rejects 3.
step 4: d accepts 3, rejects 2. c accepts 2. |
H: Examples of rare, meager and nonmeager sets in $\mathbb{R}$
Kreyszig Functional Analysis book presents the following definition.
I'm trying to get some examples.
(a) The cantor set $K$ is rare in $\mathbb{R}$ because it's closed and has empty interior so that $$\operatorname{int}\left(\overline{K}\right)=\operatorname{int}(K)=\varnothing.$$
(b) The set $\mathbb{Q}$ of rationals numbers is meager in $\mathbb{R}$ because it's enumerable so that we can write $$\mathbb{Q}=\{r_1,r_2,...,r_n,...\}=\bigcup_{i=1}^\infty{\{r_i}\},$$ where $\{r_i\}$ is rare for all $i\in\mathbb{N}$.
(c) I think that $(0,1)$ (like any other non-degenerate interval) is a nonmeager set in $\mathbb{R}$. My question is: how to prove it?.
Thanks.
AI: Your first two examples are spot on.
Actually, so is the third.
The proof of this relies on the Baire Category Theorem, mentioned in the comments above.
Baire Category Theorem: If a metric space $X\neq\varnothing$ is complete, it is nonmeager in itself.
To see this, recall that any non-degenerate open interval $(a,b)$ is homeomorphic to the real line $\mathbb{R}$. As $\mathbb{R}$ is a complete metric space, it follows that $(a,b)$ is completely metrizable (of course, the usual metric is not complete on $(a,b)$, but some other metric — one derived from a homeomorphism from $\mathbb{R}$ — makes it complete). Therefore $(a,b)$ is nonmeagre-in-itself by the Baire Category Theorem.
If $(a,b) = \bigcup_n Z_n$ where each $Z_n$ is nowhere dense in $\mathbb{R}$, it would follow that each $Z_n$ is also nowhere dense in the subspace $(a,b)$, which would then contradict the fact that $(a,b)$ is nonmeagre-in-itself.
A quick comment about your second example: It is good that you note that the singleton sets in $\mathbb{R}$ are nowhere dense. For instance, $\mathbb{N}$ is countable, but not meagre-in-itself, because $\mathbb{N}$ is discrete (all subsets are open) and so the singleton sets are not nowhere dense in $\mathbb{N}$. (Of course, $\mathbb{N}$ is meagre in $\mathbb{R}$.) |
H: Order of Automorphism group of a group of prime order.
An easy question I am having a difficult time finding a straightforward answer to.
The quotient $\frac{G}{C_G(P)}$ is isomorphic to a subgroup of $Aut(P)$ where $P$ is a group of prime order.
Is $|Aut(P)|=p-1$?
AI: First of all, let's try to demonstrate that every group of order $p$ in cyclic.
Then it's easy to prove that the only automorphisms of $P$ are defined by $\phi(g)=g^k$, where $g$ is a generator of $P$ and $(k,p)=1$.
It's now easy to check that $Aut P \simeq \left(\mathbb{Z}_p\right)^*\simeq \mathbb{Z}_{p-1}$ |
H: Find the number $c > 0$ such that the region bounded by the curves $y = x$, $y = -2x$, $x=c$ has area 6.
My intuition leads me to believe that I need to separate this into two regions and use a definite integral to compute the area. I am having troubles associating $x=c$ and each of the given functions.
AI: Considering the below plot, you're obviously called to do the following equation having a definite integral:
$$\int_0^c(x-(-2x))dx=6$$ |
H: I want to prove that these definitions of expected value hold
Let $(\Omega,\mathcal B,P)$ be a probability space. I have two (related) questions. Assuming that $g:\mathbb{R}\to\mathbb{R}$ is Borel measurable, and understanding that
$$E(g(X)) = \int_{\Omega}g(X(\omega))dP(\omega),$$
how do I prove that these equalities hold for the two following circumstances?
First, $X$ is discrete, with range $\{x_{i}:i\in \mathbb N\}$. Then
$$E(g(X))=\sum_{i=0}^{\infty}g(x_{i})P(X=x_{i}),\ \ \text{ provided } \sum_{i=0}^{\infty}|g(x_{i})|P(X=x_{i})<\infty.$$
Second, $X $ is absolutely continuous with density $f.$ Then
$$E(g(X))=\int g(x)f(x)dx\ \ \text{ provided }\int |g(x)|f(x)dx<\infty.$$
and my try:I think about number one I can write $$E(g(X))=\int_{\mathbb R}g(X)dP$$ As regards $X$ is discrete I think $$E(g(X))=\sum_{i=1}^{\infty}g(x_{i})\int dP=\sum_{i=1}^{\infty}g(x_{i})P(X=x_{i}) $$
about number two i just now $X$ is absolutely continuous hence$$P_{X}(E)=\int_{E}f(x)dx \qquad \forall E \in \mathcal B_{\mathbb R}$$
thanks for any help.
AI: In the discrete case $g(X)$ is a discrete RV and you can write $g(X)$ as the nearly-simple function $$g(X) = \sum_i g(x_i)1_{\{X =x_i\}}$$ If $\sum_i |g(x_i)|< \infty$ approximate $g(X)$ with $Y_n := \sum_{i=1}^n g(x_i)1_{\{X=x_i\}}$. Notice that for every $n$, $|Y_n| \leq \sum_{i}|g(x_i)|$ and $Y_n \to g(X)$. This will allow you to apply DCT to
$$E[Y_n] = \sum_{i=1}^n g(x_i)\int_{\Omega} 1_{\{X=x_i\}} dP = \sum_i^n g(x_i)P(X=x_i).$$
A similar argument will work in the non-discrete case. Approximate $g$ by simple functions $\varphi_n = \sum_{i=1}^{N_n} \lambda^n_i 1_{A_i}$ (here $A_i$ are measurable sets). Then $\varphi_n(X) \to g(X)$ and
$$
E[\varphi_n(X)] = \sum_{i=1}^{N_n} \lambda^n_i \int_{\Omega} 1_{A_i}(X) dP = \sum_{i=1}^{N_n} \lambda^n_i \int_{X \in A_i} dP = \sum_{i=1}^{N_n}\lambda_i^n P(X \in A_i) \\= \sum_{i=1}^{N_n} \lambda_i^n \int_{A_i}f(x)dx = \int \varphi_n(x) f(x)dx
$$
Once again use the bound on $\int|g(X)|$ to use the DCT and finish off the proof! |
H: Under what conditions is this true? A finitely generated ideal minus a generator equals an ideal containing that.
Let $(r_1, \dots, r_{s-1}) \subset I \subset J = (r_1, \dots, r_s)$ be ideals in a commutative ring $R$. In other words $r_s \notin I$. Also, let $\{r_1, \dots, r_s\}$ be a minimal generating set for $J$. Then does $I = (r_1,\dots, r_{s-1})$?
My attempt. Consider $J$ as an $R$-module of dimension $s$. $\{r_1, \dots, r_s\}$ is a basis as it spans $J$. If $r_i R \cap r_j R \neq 0$, then $r_i = r_j x$ for some non-unit $x \in R$ or $r_i R= r_j u R = r_j R$. Thus $r_i R \subset r_j R$ and the set isn't minimal, therefore $r_i R \cap r_j R = 0, \ \forall i\neq j$. Therefore $J = r_1 R \oplus \dots \oplus r_s R$.
AI: No. Look at $(x^2 + y) \subset (x^2+y, y) = (x^2,y) \subset (x,y)$ in the polynomial ring ${\mathbb C}[x,y]$.
Note that $(x,y)$ is the radical ideal of $(x^2,y)$, so that looking at the corresponding varieties you have $V(x^2 + y) \supset V(x^2+y,y) = V(x^2,y) = V(x,y)$, which is what inspired this particular counterexample. |
H: Understanding the last few lines in a proof by Royden in Real Analysis.
The book states:
$\textbf{Proposition 2}$: Let $C$ be a countable subset of the open interval $(a, b)$. Then there is an increasing function on $(a, b)$ that is continuous only at points in $(a, b) \setminus C$
$\textbf{Proof}$ If $C$ is finite the proof is clear. Assume $C$ is countably infinite. Let $\{q_n\}_{n = 1}^\infty$ be an enumeration of $C$. Define the function $f$ on $(a, b)$ by setting $$f(x) = \sum_{\{n | q_n \leq x\}} \frac{1}{2^n} \text{ for all } a < x < b.$$
Since a geometric series with a ratio less than 1 converges, $f$ is properly defined. Moreover, $\begin{align}\text{if } a < u < v < b \text{, then } f(v) - f(u) = \sum_{\{n | u < q_n \leq v\}} \frac{1}{2^n}.\end{align}$
Thus $f$ is increasing. Let $x_0 = q_k$ belong to $C$. Then, by (1),
$$f(x_0) - f(x) \geq \frac{1}{2^k} \text{ for all } x < x_0.$$
Therefore $f$ fails to be continuous at $x_0$. Now let $x_0$ belong to $(a, b) \setminus C$. Let $n$ be a natural number. There is an open interval $I$ containing $x_0$ for which $q_n$ does not belong to $I$ for $1 \leq k \leq n$. We infer from (1) that $|f(x) - f(x_0)| < 1/2^n$ for all $x \in I$. Therefore $f$ is continuous at $x_0$. Q.E.D.
Ok, I have read the last few sentences, about showing $f$ is continuous on the irrational numbers, a hundred times and I still cannot see why what Royden has written shows that $f$ is continuous on the irrational numbers. Could someone explain it to me in a more expounded upon way? He is too terse here for me to decipher.
AI: Starting at $x_0$, there is some neighborhood of $x_0$ that doesn't contain $\{q_1,\dots,q_n\}$. We might have to make this neighborhood very small, if one of these $q_k$'s is close to $x_0$, but eventually we can avoid all of them (since there are only finitely many).
Another way of thinking about this is that $$\inf_{1\leq k\leq n}|x_0 - q_k| = \min_{1\leq k\leq n}|x_0 - q_k|$$ since the $\inf$ is taken over finitely many things, and that $$\min_{1\leq k\leq n}|x_0 - q_k| > 0$$
since $x_0 \in (a,b) \backslash C$.
Now, suppose $x_0 > q_k$ for some $k \leq n$. Then, for every other $x$ in this interval, we have $x > q_k$ as well, since $q_k$ isn't in this interval (i.e. $q_k$ is either to the left or right of all the points in this interval). So, when we go to check the value of $f$ at $x$ vs. the value of $f$ at $x_0$, the contribution to the sum from $\{q_1,\dots,q_n\}$ will be the same. The contribution from other $q_k$'s for $k>n$ might be different, but this difference will be at most
$$
\sum_{k>n}\frac{1}{2^k} < 1/2^n.
$$ |
H: Finding a matrix where the column space is a subset of the null space
Let $A_{3x3}$ be a matrix $ \ne 0$ such that the column space of $A$ is a subset of the null space of $A$. I need to find $A$.
Here's my process so far:
let $v_1, v_2, v_3$ be the column vectors of $A$
$Col(A)=c_1 v_1 + c_2 v_2 + c_3 v_3$ is a subset of $Null(A)$
let $b$ be the column space of A such that $Ax=b$, then assume $b$ is also in the null space of $A$ so $Ax=0=A(Ab)$
How do I go about finding a particular matrix that satisfies this? Am I even heading on the right direction?
AI: Take $A=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Then $Ax=0$ iff $x_3 = 0$. It is easy to see that the null space (also known as the kernel, or $\ker A$) is given by $\ker A = \{ (x_1,x_2, 0)^T \}_{x_1,x_2 \in \mathbb{R}}$.
Then $y = Ax$ for some $x$ iff $y = (x_3,0,0)^T$ for some $x_3 \in \mathbb{R}$, so the range space is is given by ${\cal R} A = \{ (x_1,0, 0)^T \}_{x_1 \in \mathbb{R}}$.
It should be clear that ${\cal R} A \subset \ker A$. |
H: Integration problem in real analysis
Let $f: [a,b]\rightarrow \mathbb{R}$ be a function, and let $m$=inf{$f(x)|x\in [a,b]$}, $M$=sup{$f(x)|x\in[a,b]$}.
Prove that for any partition $P$ and sample set $S$, $m(b-a)\leq RS(f,P,S)\leq M(b-a)$.
My attempt: $RS(f,P,S)=f(s_1)\Delta x_1 + ... + f(s_n)\Delta x_n$ ; $m(\Delta x_1 + ... + \Delta x_n)\leq f(s_1)\Delta x_1 + ... + f(s_n)\Delta x_n$.
Then what do I do?
AI: Hint:
$$ m\leq f(x_i^{*}) \leq M,\quad x_i^*\in [x_{i-1}, x_i]\implies m\Delta x_i\leq f(x_i^*)\Delta x_i\leq M\Delta x_i. $$
Can you finish it now? |
H: Show that a progression converges and assess the limit.
$a_{n} = \sqrt{9n^2+2n+1} -3n$
I tried to simplify the term to show that the limit is $\frac{1}{3}$ with the binominal formula for example $\sqrt{9n^2+6n+1-4n} = \sqrt{(3n+1)^2-4n}$ or playing around with the $9n^2$ but because of the $-3n$ after the radical, it never becomes easier as it was... Running out of ideas now and would aprreciate a new idea how to solve this.
AI: Hint: $a_{n} = \sqrt{9n^2+2n+1} -3n = \frac{(\sqrt{9n^2+2n+1} -3n)(\sqrt{9n^2+2n+1} +3n)}{\sqrt{9n^2+2n+1} + 3n}=\frac{(2n+1)}{\sqrt{9n^2+2n+1} + 3n} = \frac{(2+\frac{1}{n})}{\sqrt{9+\frac{2}{n}+\frac{1}{n^2}} + 3}$ |
H: Help with limit of radical function
$$\lim_{x \to \infty} \frac{\sqrt{4x^{4}+3}}{5x^2+3}$$
$$= \lim_{x \to \infty} \frac{(4x^{4}+3)^{1/2}}{5x^2+3}$$
$$= \lim_{x \to \infty} \frac{(\frac{4x^{4}}{x^{1/2}} +\frac{3}{x^{1/2}})^{1/2} }{5+\frac{3}{x^2}}$$
$$= \lim_{x \to \infty} \frac{2x+\frac{\sqrt{3}}{x}}{5+\frac{3}{x^2}}$$
The answer is $\frac{2}{5}$. Why is this the answer? I guess what confuses me is this part: $(4x^2)^{1/2}= 2x^1=2x.$ Is that not so? How are they getting $2x^0=2$ from that?
AI: When you divided by $x^2$ in the numerator and denominator and moved it inside the radical, you should have been dividing by $x^4$, not $x^{1/2}$. Consequently, you should have $$\lim_{x\to\infty} \frac{\sqrt{4+3x^{-4}}}{5+3x^{-2}}$$ Which does, indeed, evaluate to $2/5$ |
H: semisimplicity of Lie algebra
Let $L$ be a lie algebra. Then if $L$ is semisimple, we have $L = L_1 \oplus \cdots\oplus L_n$ for some simple ideals $L_i$. But we can also consider the adjoint representation. In this representation, each $L_i$ will be an irreducible submodule. So we get that the representation is completely reducible.
I don't get why the converse is false. I don't want a counterexample, but rather, an explanation. Since the submodules in the adjoint representation correspond to ideals, the converse should be true, no? If $L$ (when considered as a module with the adjoint representation) is a sum of irreducible modules, then it seems that each of these irreducible submodules will be simple ideals.
AI: Part of the definition of a simple ideal is that this ideal is nonabelian as a Lie algebra. If $L$ has a one-dimensional ideal $I$, then $I$ is an irreducible submodule of the adjoint representation, but $I$ is abelian and therefore not a simple ideal of $L$.
We say that a Lie algebra $L$ is reductive if its adjoint representation is completely reducible. One can show that $L$ is reductive if and only if it is the direct sum of a semisimple Lie algebra and an abelian Lie algebra. |
H: Open and Closed Quotient Maps.
a) Find a subset $A$ of $\Bbb{R}$ such that the quotient map $p: \Bbb{R} \rightarrow \Bbb{R}/A$ is not open.
If we let $A= \Bbb{Q}$, then we can see that $(0,1)$ is open in $\Bbb{R}$. But if we pick a rational number $p/q \in (0,1)$ and draw a ball around it in $\Bbb{R}/\Bbb{Q}$, we will see that $0$ and $1$ are not contained in the ball...they they are equal to $p/q$ since they are rational. This is a contradiction, right?
b) Find a subset $B$ of $\Bbb{R}$ such that the quotient map $q : \Bbb{R} \rightarrow \Bbb{R}/B$ is not closed.
Let $A= \{a,b\}$ for $a,b \in \Bbb{R}$ such that $a < b$. Then $[a,b]$ is closed in $\Bbb{R}$, since we cannot have a ball around the point $a$ without including points outside of the interval. However, in $\Bbb{R}/\{a,b\}$, we can just take the ball with radius $\frac{b-a}{2}$. Since $b=a$ in $\Bbb{R}/\{a,b\}$, we don't have to include points outside that interval.
c) Using on of these examples, show that a quotient space of a Hausdorff space need not be T1.
We can look at $\Bbb{R}/\Bbb{Q}$. If we take a rational number $q \in \Bbb{Q}$ and an irrational number $a \in \Bbb{R}$, we cannot draw a ball around $a$ without including $q$ since the rational numbers are dense in $\Bbb{R}$ and all rational numbers are equal in $\Bbb{R}/\Bbb{Q}$.
Are my answers correct? If not, can you give me any hints?
Thanks in advance
AI: It’s true that the quotient map from $\Bbb R$ to $\Bbb R/\Bbb Q$ is not open, but your argument doesn’t really make sense: $\Bbb R/\Bbb Q$ isn’t a metric space, so it’s not clear what you mean by a ball in $\Bbb R/\Bbb Q$, and $0$ and $1$ aren’t points of $\Bbb R/\Bbb Q$. Let $q$ be the quotient map; then $$q^{-1}\big[q[(0,1)]\big]=(0,1)\cup\Bbb Q\;,$$ which is not open in $\Bbb R$, so $q[(0,1)]$ is not open in $\Bbb R/\Bbb Q$, and $q$ is therefore not an open map.
The same quotient space works for (b). This time consider the closed set $F=[0,1]$; $$q^{-1}\big[q[F]\big]=[0,1]\cup\Bbb Q\;,$$ which is not closed in $\Bbb R$, so $q[F]$ is not closed in $\Bbb R/\Bbb Q$, and $q$ is not a closed map.
Your example for (b) does not work: if $f:\Bbb R\to\Bbb R/\{a,b\}$ is the quotient map, then $f$ is closed. Let $F$ be any closed set in $\Bbb R$. If $F\cap\{a,b\}=\varnothing$, then $f^{-1}\big[f[F]\big]=F$, so $f[F]$ is closed in $\Bbb R/\{a,b\}$. Otherwise, $f^{-1}\big[f[F]\big]=F\cup\{a,b\}$, which is closed in $\Bbb R$, so again $f[F]$ is closed in $\Bbb R/\{a,b\}$, and $f$ is a closed map.
Your answer to (c) is fine. |
H: How can I compute $\int\frac1{e^{ct}} dt $
$$ \displaystyle \int \dfrac{1}{e^{ct}} dt $$
How do I do this, I've tried substitution. Rewrite the above integral as $\displaystyle \int e^{-ct} dt $, then we have $u=-ct$, but at this point derivatives and integrals are confusing me. If we want $[e^u]'$, we just have $[e^u]u'$, but for integrals I've forgotten how to tackle this.
AI: Let $u=-ct$ then $du = -c\,dt$ or alternatively: $\,\,-\frac{1}{c}du = dt\,\,$ thus your integral becomes
$$\int e^{-ct} dt = -\int e^{u}\frac{1}{c}du = -\frac{1}{c}\int e^{u}du.$$
Can you take it from here? |
H: Basic induction proof that all natural numbers can be written in the form $2a + 3b$
The theorem given is:
If $n$ is a natural number then $n$ can be written in the form $2a + 3b$ for some integers $a$ and $b$.
How would I prove this by induction? I've had a go at proving this but I don't know if my technique is sound.
The base case would be when n = 1 = 2(-1) + 3(1) (if we take the natural numbers as excluding 0). Then if I assume n = 2a + 3b is true, n+1=2a+3b+1. Therefore n+1=2a+3b+2(-1)+3(1) which can be written as n+1=2(a-1)+3(b+1) which should conclude the proof.
Is this a proper proof or is there some other way of doing it? How would I prove the theorem if I took the natural numbers to include 0 (i.e. could I still use 1=2(-1)+3(1) when it would no longer be the base case)?
Thanks for your help.
AI: Your proof is perfectly good. You can use whatever integer $b$ you like as the base case, to prove some proposition $P(n)$ is true for all integers $n\ge b$. $0$ and $1$ are both very common base cases. You can also use induction in the other direction (e.g., for negative numbers) to prove that every integer below $b$ satisfies the proposition.
Formally, induction is usually defined in the upwards direction, and usually to start at $0$ (or $1$, depending which text you use), but extending it to do other things is quite straightforward. The downward induction can be recast as upwards: rather than induction downward in $n$, do induction upward in $-n$. Same thing. |
H: The rule $\phi \leadsto \phi\land\psi$ is not sound
I have been asked to show that this rule is not sound: $$\frac{\varphi}{\varphi\wedge\psi}\wedge I'$$
Any help with this would be greatly appreciated.
AI: Work in propositional logic with variable letters $A$ and $B$. Let $\phi = A \lor \lnot A$ and let $\psi = B$. Now $\phi$ is true in every valuation, but $\phi \land \psi$ is not true in every valuation. Therefore, the inference rule is not sound. |
H: Proving there is a max,with two limits given.
So I need help with this exercise. If $f$ is a positive and continuous function with
$$\lim_{x \to -\infty} f(x) = \lim_{x \to +\infty} f(x) = 0 $$
Prove that $f(x)$ has a maximum. Thanks in advance :)
AI: Well, go back to the definition of a limit that $\forall\; \varepsilon>0$, there exists $b$ such that $|f(x)-L|<\varepsilon\; \forall\; x>b$. Similarly, there exists $a$ such that $|f(x)-L|<\varepsilon\; \forall\; x<a$. So fix $\varepsilon > 0$. Now, $f(x)$ is bounded by let's say $M>0$ on $[a, b]$ by compactness and the continuity of $f$. But it's also bounded by $\varepsilon$ everywhere else. Hence $f$ is bounded above by $\text{max}(M, \varepsilon)$. Now, continuity demands that $f$ reaches its supremum (it exists by boundedness). That is $f$ has a maximum.
EDIT: Note that I assumed $f$ was continuous on all of $\mathbb{R}$. Of course, counterexamples exist otherwise. |
H: Model and countermodel to $\exists x.\forall y. x
Can someone please help me with this question. I have been struggling with it for ages and can't quite seem to work it out:
Let $<$ be a binary relation symbol that we will write infix. Let $\phi = \exists x.\forall y.x<y$. Define structures $A,B$ such that $A \models \phi$ and $B \not\models \phi$.
I would be really grateful of any help.
AI: HINT: The symbol chosen to represent the relation is a big hint. Suppose that $<$ really is, as the symbol suggests, some kind of order relation. Then $\varphi$ just says that there is a smallest element with respect to $<$. Can you think of sets $A$ and $B$ and relations on $A$ and $B$ for which we normally use the symbol $<$ or $\le$ such that $A$ does have a smallest element with respect to the, and $B$ does not? |
H: Find a matrix with the null space equal to the column space of that matrix
I know this question has been asked and answered here: Can a matrix have a null space that is equal to its column space?. However, I'm not clear on the mechanism used to find an actual matrix so I figured I would make a new question. If this is inappropriate please flag or delete.
What I have done is let $A$ be a 4 x 4 matrix where $Null(A)=Col(A)$ (null space of A=Col Space of A). Therefore, the span of the columns of A = the null space of A.
I know that the RREF of A will have the bottom two rows zeroed out. But, what I don't know is where to go from here in terms of finding a matrix that satisfies the above condition.
Any help provided would be much appreciated. I'm taking an online, distance linear algebra course for credit and the materials provided are very minimal.
AI: You need $A \ne 0$ such that $A^2 = 0$ and $\operatorname{rank} A = n/2$.
So, for any nonsingular $S$, you can define $A = S^{-1} J S$, where $J$ is a Jordan matrix of the form
$$J = \bigoplus_{k=1}^{n/2} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$$ |
H: Variable Endpoint of definite integral.
My book asks to find the number $c$ such that the region bounded by the curves $y = x^2$, and $y = c$ has area 36. I understand the a definite integral will solve the question but am having a hard time applying the right interval. Right now I have the proper integral with constants moved out: $$2c - \int_0^{x} x^2 dx = 36$$
The resulting $2c - \frac{x^3}{3} = 36$ is confusing. I assume that using the correct interval will change the expression to be in terms of c. Any direction would be greatly appreciated.
AI: You need to find where $y = x^2 = c$:
$y = x^2 = c \implies x = \pm \sqrt c$. Use those points as your interval endpoints.
And you need to integrate the constant. Recall that $\int c\,dx = cx + \text{constant}$.
$$\int_{-\sqrt c}^{\sqrt c} (c - x^2) \,dx = 36.$$
Integrate, then evaluate at the bounds of integration, and set equal to $36$. Then you can solve for $c$ to complete the task.
$$\int_{-\sqrt c}^{\sqrt c} (c - x^2) \,dx = 36 $$
$$\iff cx - \frac{x^3}{3}\Big|_{-\sqrt c}^{\sqrt c} = 36.$$
Since $y = x^2$ and $y = c$ are both symmetric about the y-axis, and the interval of integration is centered at $x = 0$, it suffices to compute:
$$2\int_{0}^{\sqrt c} (c - x^2) \,dx = 36 $$
$$\iff 2\Big(cx - \frac{x^3}{3}\Big)\Big|_0^{\sqrt c} = 2c\cdot \sqrt c - \frac 23(\sqrt c)^3 = 36$$ Simplify and solve for $c$. |
H: ordered topology on integers and local compactness
Hi I am trying to make sure my logic is sound,
Let's suppose that we declared the discrete topology on $\mathbb{Z}$. Let us consider the set $\{1,2,3\}$. This set is open. However, this set is also closed because it has no limit points (the open set $\{3\}$ contains 3 but does not intersect the set $\{1,2,3\}$, the case is similar for $\{1\}$ and $\{2\}$) , also, under the discrete topology the $\mathbb{Z}$ is hausdorff so any finite point set has to be closed.
Is what I said correct?
Also, consider $\mathbb{Z}$ in the order topology. I would guess that order topology on $\mathbb{Z}$ would equal the discrete topology on $\mathbb{Z}$. However, I considered the order topology for the sake that I was reading the Munkres Topology book (p.183 Example 3) where it stated that "Every simply ordered set $X$ having the least upper bound property is locally compact. Given a basis element for $X$ it is contained in a closed interval in $X$, which is compact.
I guess what I am asking, if we define the order topology on $\mathbb{Z}$ (because it is simply ordered) , it surely is locally compact because any closed subset of it, is also open, correct?
thank you.
AI: The order topology on $\Bbb Z$ is indeed the discrete topology, and it is locally compact. It’s locally compact because for each $n\in\Bbb Z$, $\{n\}$ is a compact open nbhd of $n$ that is contained in every open nbhd of $n$. |
H: Problems with definition of almost surely
Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space. In probability theory one says that and event $F\in\mathcal F$ happens $\mathbb P$-almost surely, if $\mathbb P(E)=1.$ Intuitively, as a beginner one thinks that this means that there exists an event $N\in\mathcal F$ with $\mathbb P(N)=0$ and such that the event $E\setminus N$ happens surely? Is this intuition generally true? Counterexample?
AI: In many probability spaces there are events $E \subsetneq \Omega$ and $\varnothing \neq N \subset \Omega$ such that $P(E)=1$ and $P(N)=0$.
Take for example the following scenario:
Suppose you flip a fair coin infinitely many times. The sample space $\Omega$ will be all sequences which look like $(x_1, x_2, x_3, …)$ where $x_i = 0$ if you flip a tail and $x_i= 1$ if you flip a heads. For instance, if the first two flips were tails and the third a heads, you have a sequence starting with $(0,0,1,x_4, …)$. Now, let's ask: What is the probability that the coin flipped all heads? The event we want to look at is $N = \{(1,1,1,1,1,1,…)\}$ and
$$P(N) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdots = \lim_{n \to \infty}\left(\frac{1}{2}\right)^n = 0.$$
This intuitively makes sense because how likely are you to actually flip all heads with a fair coin and infinitely many flips? But, you can use this same strategy and see that the probability of any one fixed sequence of coin flips must be $0$. But again, this intuitively makes sense because how likely are you to flip exactly the same faces of the coin in exactly the same order as the fixed sequence for infinitely many flips?
Now, if I ask what is the probability that you flip a heads on your first flip, lets call this event $H$, the answer is sensibly $P(H)=\frac{1}{2}$. So how can we merge a positive probability of events like $H$ when we have that each single outcome has probability $0$? This is the challenge that measure theory has come to answer. If $P$ is that measure (the measure of probability) and now you ask what is the probability of the event $E$ that you don't flip all heads? Well, the probability of flipping all heads is $0$, so $P(E) =1$. But $E \neq \Omega$ since $E = \Omega \backslash \{(1, 1, 1, 1, …)\}$. Therefore, can we say that if you flip a coin infinitely many times, then you will surely have an outcome that lands in $E$? No! It could so happen that you do flip all heads. Although, since the probability of that happening is $0$ (and just based on our intuition of how likely that is), you will almost surely flip a sequence that lands in $E$. |
H: Monotonicity of $2^x-x-1$
I'm trying to prove that the function $g(x) := 2^x-x-1$ is strictly increasing on $(1, \infty)$. Using its derivative I was able to conclude that it is increasing on that interval. So, the problem is now reduced to showing that $g$ is injective.
So far I haven't been able to prove this. It might be simple, but I cannot see it. Any help would be appreciated.
AI: $g'(x)=2^x\cdot ln2-1>0$ for every $x\in(1,\infty)$ and because $g'(x)>0$ and not $g'(x)\geq 0$ we have that $g$ is strictly increasing. |
H: Help with limit of radical expression
$$\lim_{x \to \infty} (\sqrt{x^2-49}-\sqrt{x^2-16} ) $$
I multiplied by the conjugate radical expression:
$$=(\sqrt{x^2-49}-\sqrt{x^2-16}) \times (\sqrt{x^2-49}+\sqrt{x^2-16}) $$
$$= x^2-49-(x^2-16)=x^2-49-x^2+16=-33$$
$$\lim_{x \to \infty}f(x) = -33$$
This is wrong. The correct answer is $0$. What is wrong in my process? Is it possible to solve this limit without multiplying by the conjugate? Thanks.
AI: $\lim_{x \to \infty} (\sqrt{x^2-49}-\sqrt{x^2-16} )=\lim_{x \to \infty}\frac{(\sqrt{x^2-49}-\sqrt{x^2-16} )(\sqrt{x^2-49}+\sqrt{x^2-16} )}{(\sqrt{x^2-49}+\sqrt{x^2-16} )}=\lim_{x \to \infty}\frac{-33}{(\sqrt{x^2-49}+\sqrt{x^2-16} )}=0$
The denominator goes to $+\infty$ |
H: Find CFG for $a^i b^j c^k$ with $!(i=j=k)$
I need to show that the following language is context free:
It contains all words $a^i b^j c^k$ with the condition that there're not exactly as many a's and b's and c's.
I would write it like this:
$$\{a^i b^j c^k\; : \;\;!(i=j=k) \;\}$$
Here's what I've come up with:
I split it up into four cases. In three cases two are equal. In the fourth case they're all distinct.
$i=j \neq k: \{a^i b^j c^k : \;i=j \lt k \;\} \cup \{a^i b^j c^k : \;i=j \gt k \;\}$
$i=k \neq j: \{a^i b^j c^k : \;i=k \lt j \;\} \cup \{a^i b^j c^k : \;i=k \gt j \;\}$
$j=k \neq i: \{a^i b^j c^k : \;j=k \lt i \;\} \cup \{a^i b^j c^k : \;j=k \gt i \;\}$
$i \neq j \neq k: \{a^i b^j c^k : i \lt j \lt k \;\} \cup \{a^i b^j c^k : j \lt i \lt k \;\} \cup \{a^i b^j c^k : k \lt i \lt j \;\} \cup \{a^i b^j c^k : k \lt j \lt i \;\}$
I'm not quite sure about the last cause, though.
I do understand how to create a CFG for conditions like $k=i+j$, $(i=j) \lor (j=k)$.
But I can't figure out how to create a CFG for these sub-languages.
Any help would be appreciated!
AI: First note that $ \neg(i=j$ and $j=k)$ is $(i \neq j $ or $j \neq k ) $ (Demorgan's Law)
$S \to S_1|S_2
\\ S_1 \to XC
\\X \to aA|bB|aXb
\\S_2\to AY\\
Y\to bYc|bB|cC\\
A\to aA|\epsilon\\
B\to bB|\epsilon\\
C\to cC|\epsilon$ |
H: Flatness, Hilbert polynomial and reduced schemes.
Let $f:X \to S$ be a projective morphism of schemes and $F$ a coherent $O_X$-module. We have that if $F$ is $S$-flat then the Hilbert polynomial $P(F_s)$ is locally costant as a function of $s \in S$. (for reference http://carlossicoli.free.fr/H/Huybrechts_D.,_Lehn_M.-The_Geometry_of_Moduli_Spaces_of_Sheaves,_Second_Edition_(Cambridge_Mathematical_Library)(2010).pdf, proposition 2.1.2). The converse is true if $S$ is reduced. Can you give me an example with $S$ not reduced, $P(F_s)$ locally costant but $F$ not $S$-flat?
AI: Let $X=S=\textrm{Spec }D$, where $D=\mathbb C[t]/t^2$, and let $f$ be the identity on $S$.
Let $F$ be the $\mathcal O_X$-module $\tilde M$, where $M=D/tD\cong \mathbb C$.
Aside. There is a criterion, which I learnt on Hartshorne's notes/book on Deformation Theory, saying that a $D$-module $M$ is flat (over $D$) if and only if the natural map ("multiplication by $t$") $M/tM\to M$ is injective. In other words, it is enough to check that the sequence $$0\to tD\to D\to \mathbb C\to 0$$ stays exact after taking $M\otimes_D-$.
Back to our $F$: it is not flat over $S$. Or, $M$ is not flat over $D$. Indeed, tensoring the inclusion $tD\to D$ with $M$ we get $M\otimes_DtD\to M\otimes_DD$, which is the zero homomorphism $\mathbb C\to\mathbb C$.
Added. (Example of a nonflat family).
Let $S=\mathbb A^1$, and $X\subset\mathbb A^1\times \mathbb P^3$ the family given by $$X_s=\{x_2=x_3=0\}\cup\{x_1=x_3-sx_0=0\}\subset \mathbb P^3.$$ The first component is always a line. If $s\neq 0$, the second component is another line, disjoint from the first. Hence $HP_{X_s}=2T+2$ in this case. But $X_0=\{x_1x_2=x_3=0\}$ is a reducible plane conic. Hence $HP_{X_0}=2T+1$. Conclusion: $F=\mathcal O_X$ is not $S$-flat, i.e. $X\to S$ is not flat. |
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