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H: Tetromino Proof
Prove that an 8 x 8 board cannot be covered by 15 L-tetrominos and one square tetromino (an L-tetromino is a plane figure shown below, constructed from four unit squares arranged in the form of L; a square tetromino is a plane figure shown below, constructed from four unit squares arranged in the form of a square)
AI: Do the following: color every cell on an odd row in white and color every cell on an even row in black. Suppose that the board can be covered with $15$ $L$-shaped tetrominos and one square-shaped tetromino. Notice now that since your $L$-shaped tetrominos will always cover an odd number of white cells and your square-shaped tetromino will alway cover an even number of white cells the total number of covered white cells will be always be odd. But this is not possible since the total number of white cells is $32$ which is an even number. Contradiction. |
H: Radon integrals and Radon charges
I'm reading chapter 6 (measure theory) of Pedersen's book Analysis now and I'm a bit puzzled in his passage from Radon integrals to Radon charges.
The book in 6.1.2 defines a Radon integral to be a linear complex valued positive functional (i.e which maps positive functions in positive real numbers)
defined on the space $C_c(X)$ of continuous (complex valued) functions with compact support on a locally compact Hausdorff space $X$.
In 6.5.5 the book gives three definitions of a Radon Charge:
A Radon charge is a linear combination of Radon integrals.
Let $\tau$ be the weak topology on $C_c(X)$ given by the family of semi norms
$f\mapsto |\int f|$ where $\int$ is a Radon integral and $f\in C_c(X)$.
A Radon charge is a continuous linear functional with respect to this topology.
A Radon charge is a linear functional $\Phi:C_c(X)\to\mathbb{C}$ such that
$$\sup\{|\Phi(g)|:g\in C_c(X),\, |g|\leq |f|\}$$
is finite for all $f\in C_c(X)$.
I can see the equivalence of 1 and 2. In theorem 6.5.6 it is proved that 3
holds for $\Phi$ iff $\Phi(f)=\int (f\cdot u)$ with $\int$ a Radon integral and
$u: X\to \mathbb{C}$ a Borel function of modulus one. I don't see why this theorem sheds any light on the equivalence of 3 with 1 or 2 above. Can someone help me?
AI: If you have
$$\Phi(f) = \int f\cdot u,$$
write $u = g^+ - g^- + ih^+ - ih^-$ with non-negative Borel functions $g^+,g^-,h^+,h^-$. Let $\Phi_1(f) = \int fg^+$, $\Phi_2(f) = \int fg^-$, $\Phi_3(f) = \int fh^+$, $\Phi_4(f) = \int fh^-$. Then the $\Phi_i$ are Radon integrals, and
$$\Phi = \Phi_1 - \Phi_2 + i\Phi_3 - i\Phi_4$$
is a linear combination of Radon integrals. |
H: Set where function has high values is small
Let $\mu$ be a probability measure on a set $A$, and let $f:A\rightarrow\mathbb{R}$ be a random variable. Given $\epsilon>0$, is it true that we can find $n$ such that $\mu\{|f(x)|>n\}<\epsilon$?
Intuitively it looks like it should be true (just choose $n$ large enough so that most of the values of $f$ is below $n$), but how can we prove it?
AI: Consider $A_n = \{ x : \lvert f(x)\rvert >n\}$. You have $A_n \supset A_{n+1}$ and
$$\bigcap_{n=0}^\infty A_n = \varnothing.$$
The continuity of the measure - since $\mu(A) < \infty$ - now shows $\lim\limits_{n\to\infty} \mu(A_n) = 0$. |
H: Predicate logic: "Everybody knows somebody who knows Alice"
I'm stuck on an undergraduate CS exercise: I am to translate "Everybody knows somebody who knows Alice" into predicate logic.
I'm having trouble bending my head around it (being a complete beginner), but this is what I'm thinking:
$x,y \in $ "the set of all people"
$A(y) = y$ knows Alice
$\forall x \exists !y A(y)$
However, this feels too cumbersome (if it's even valid), and I'm sure there must be a better, simpler way of doing it. Can anybody offer any suggestions?
AI: You need a binary predicate (two-place), let's say, $K(x, y)\,$ to denote "$\;x\,$ knows $\,y$", and then we can simply use the constant $\,a\,$ to denote Alice.
You used the uniqueness quantifier $\exists!$, which is not appropriate here.
What we want is to say "For all $x$, there is some $y$ such that $x$ knows $y$, and $y$ knows Alice."
$$\forall x \,\exists y\,\Big(K(x, y) \land K(y, a)\Big)$$ |
H: Integral of characteristic function is infinitely differentiable
Let $X$ be a set, $F$ a $\sigma$-field of subsets of $X$, and $\mu$ a probability measure on $X$. Given a random variable $f:X\rightarrow\mathbb{R}$, define $$\chi_f(t)=\int_Xe^{itf}d\mu$$ We can show that $\chi_f$ is continuous and $|\chi_f|\leq 1$.
Suppose $f$ is bounded. Show that all the derivatives $(d/dt)\chi_f,(d^2/dt^2)\chi_f,\ldots$ exist and are continuous.
We can write $$\chi_f(t)=\int_Xe^{itf}d\mu=\int_\mathbb{R}e^{itx}d\mu_f$$ But I'm not sure how to find its derivatives, since it is an integral with respect to $\mu_f$.
AI: I think it's nicer if you stay with
$$\chi_f(t) = \int_X e^{itf(x)}\,d\mu(x).$$
Differentiating the integrand produces
$$i f(x)e^{itf(x)},$$
which, since $f$ is bounded, is integrable, and dominated uniformly in $t$, hence by the dominated convergence theorem,
$$\chi_f'(t) = i\int_X f(x) e^{itf(x)}\,d\mu(x).$$
Iterate arbitrarily often. |
H: Proving sets A and B are countable
a) Let A and B be disjoint sets, which are both countable. Prove that $A$ U $B$ is also countable.
b) Use part (a) to show that the set of all irrational real numbers is not countable.
So for part a I understand that disjoint sets share no elements in common (other than the empty set). I tried to get a visual understanding and did: A = {1,2,3} and B = {4,5,6}.
I drew myself a diagram and realized that I can map each element in each set to a natural number, so there it has the same cardinality as the natural numbers, which implies it is countable. How can I do this as a formal proof? $A+B = N$?
AI: HINT: If $A$ and $B$ are countably infinite, there are bijections $a:A\to\Bbb N$ and $b:B\to\Bbb N$. Now modify these to get a map $a'$ that maps $A$ bijectively onto the even natural numbers and a map $b\,'$ that maps $B$ bijectively onto the odd natural numbers. Then combine $a'$ and $b\,'$ into a single bijection from $A\cup B$ to $\Bbb N$. |
H: Convergence of $|a_{n+1} - a_n| \le Cq^n$
Be $C\gt 0$, $0\le q\lt 1$ and $(a_n)_{n\ge 1}$ a sequence in $\mathbb R$ with
$$|a_{n+1} - a_n| \le Cq^n$$
Show that $(a_n)_{n\ge 1}$ converges.
AI: Given $m,n \in \mathbb{N}$ with $n>m$ we have
$$
|a_n-a_m|\le\sum_{i=m}^{n-1}|a_i-a_{i+1}|\le C\sum_{i=m}^{n-1}q^i=C\frac{q^m-q^{(n-1)-m+1}}{1-q}\le C\frac{q^m}{1-q}.
$$
It follows that
$$
\lim_{m,n\to \infty}|a_n-a_m|\le \lim_{m,n\to\infty}C\frac{q^m}{1-q}=0,
$$
i.e.
$$
\lim_{m,n\to \infty}|a_n-a_m|=0.
$$
Hence $(a_n)$ is a Cauchy sequence, consequently it converges. |
H: If $T$ is injective then there exists $\alpha>0$ such that $||Tx||\geq \alpha||x||$
Is this proof correct? I'm proving that if $T$ is a linear operator whose is injective then exist $\alpha>0$ such that $$||Tx||\geq\alpha||x||$$ for all $x$.
By contrapositive. Assume that for all $$\alpha>0$$ there is a $x$ such that $||Tx||<\alpha||x||$. In particular for all $n\in\mathbb{N}$ there is a $x_n$ such that
$$||T(x_n)||<\frac{1}{n}||x_n||.$$ Suppose (without lose of generally) that $||x_n||=1$ (in other case normalize the vector $x_n$).
Now since $x_n$ is a bounded sequence, by the Bolzano-Weirstrass there is a subsequence that converges. Let $$x_{n_k}\to x$$ then $x\neq 0$ because $||x_{n_k}||=1$. Now for any $\epsilon>0$ and $n$ large as necessary we have, $$||T(x)||\leq ||T(x-x_n)||+||T(x_n)||<\epsilon+\frac{1}{n}$$ therefore $||T(x)||<\epsilon\Rightarrow T(x)=0$ since $x\neq 0$, $T$ cannot be injective.
Any comment, sugest or new solution are welcome!!
AI: The result is not true. The typical example is the operator that maps the canonical basis $\{e_n\}$ as
$$
T:e_n\mapsto\,\frac1n\,e_n.
$$
This map is injective, but $\|Te_n\|=\frac1n$, so no $\alpha$ exists.
Note that if you follow your argument with this example in mind, you see that you are using Bolzano Weiertrass, which requires your sequence to be inside a compact set.
The finite-dimensional case:
If you want to do your proof in the finite-dimensional case then it works, but it can be done much more easily by noting that if $T$ is injective then it is invertible onto its image, then the inverse is a bounded map (every linear map is bounded in finite-dimension) and so you can take $\alpha=\|T^{-1}\|^{-1}$. |
H: The limit of integer valued random variables must be integer valued?
I saw something like this:
If $D_n$ are all integer valued random variables, and $D_n$ converges in distribution to $D$, then $D$ must also be integer valued.
I am a little bit suspicious about the statement. A trivial "counter example" might be $D_n \equiv n$ and $D \equiv \infty$. Since $\lim_{n\to\infty} D_n = D$, $D_n$ converges in distribution to $D$. But can we still say $D$ is integer valued?
AI: Let $P_n$ be the distribution of $D_n$, and $P$ the distribution of $D$.
The portmanteau theorem says that if $P_n\to P$ then
$$\liminf_n P_n(F)\leq P(F),$$ for any closed set $F\subset \mathbb{R}$.
Letting $F=\mathbb{Z}$, we see that any limit of integer-valued random variables is integer-valued, since
$$1=\liminf_n P_n(\mathbb{Z})\leq P(\mathbb{Z}).$$ |
H: Discrete math number problem
How would I justify the following statement.
Two integers are consecutive if and only if one is more than the other. Any product of four consecutive integers is one less than a perfect square.
I think this is true.
because for example
$2<3<4<5$
$2*4*5*3=120$
Which one less than 121 a perfect square.
So how would I justify it I did
Let n be a integer
$n(n+1)(n+2)(n+3)+1
=(m)^2$
But I am not sure how to proceed.
AI: $$
\begin{equation}
\begin{split}
n(n+1)(n+2)(n+3) &= n^4+6n^3+11n^2+6n +1-1 \\
& = n^2\left(n^2+6n+11+\frac{6}{n}+\frac{1}{n^2}\right) -1 \\
& = n^2\left(\left(n+\frac{1}{n}\right)^2+6\left(n+\frac{1}{n}\right)+11-2\right) -1 \\
& = n^2\left(\left(n+\frac{1}{n}\right)^2+6\left(n+\frac{1}{n}\right)+9\right) -1 \\
& = n^2\left(n+\frac{1}{n}+3\right)^2-1 \\
& = \left(n^2+3n+1\right)^2-1 \\
\end{split}
\end{equation}
$$ |
H: How to solve this limit 0/0 indetermination?
How can I solve this 0/0 indetermination, where the limit has t going towards 0?
$$\lim_{t\to 0} \frac{t^2}{\sin^2 t}$$
AI: Hint: Use the fact that
$$\lim_{t \to 0} \frac t {\sin{t}} = 1$$
This is easily proven either by L'Hospital's rule, or Taylor series (or by geometry). |
H: What mathematical structure models arithmetic with physical units?
In physics we deal with quantities which have a magnitude and a unit type, such as 4m, 9.8 m/s², and so forth. We might represent these as elements of $\Bbb R\times \Bbb Q^n$ (where there are $n$ different fundamental units), with $(4, \langle1,0,0\rangle)$ representing 4m, $(9.8, \langle1,0,-2\rangle)$ representing 9.8 m s⁻², and so forth, with the following rules for arithmetic:
$(m_1, u_1) + (m_2, u_2)$ is defined if and only if $u_1 = u_2$, in which case the sum is $(m_1 + m_2, u_1)$.
$(m_1, u_1) \cdot (m_2, u_2)$ is always defined, and is equal to $(m_1m_2, u_1+u_2)$, where $u_1+u_2$ is componentwise addition of rationals.
This structure has a multiplicative identity, namely $(1, 0)$, and a family of additive quasi-identities, namely $(0, u)$ for each $u$. We have $$(1, 0)\cdot (m, u) = (m,u)\cdot(1,0) = (m, u)$$ and $$(0, u) + (m, u) = (m,u)+(0,u) = (m, u)$$ for every $(m, u)$, but in the latter case the quasi-identity $(0,u)$ isn't a constant; it depends on the $u$ part of $(m,u)$.
Every element with $m\ne 0$ has a multiplicative inverse, and every element $(m, u)$ has an additive quasi-inverse $(-m, u)$ with $(m,u) + (-m, u) = (0, u)$, where $(0,u)$ is an additive quasi-identity. Multiplication distributes over addition. If either side of $$p\cdot(q+r) = (p\cdot q)+ (p\cdot r)$$ is defined, then so is the other, and they are equal.
All taken together this is very much like a field, except that the additive identity is peculiar. There is a $0_\text{m}$, a $0_\text{s}$ and a $0_\text{kg}$, represented as $(0, \langle1,0,0\rangle),
(0, \langle0,1,0\rangle),
$and $ (0, \langle0,0,1\rangle)$, and they can be multiplied but not added.
We might generalize this slightly, and define the same sort of structure over a ring $\langle G,+, \cdot\rangle$ and a group $\langle H,\star\rangle$: $G❄H$ is an algebraic structure whose elements are elements of $G\times H$, where $(g_1, h_1) + (g_2, h_2)$ is defined to be $(g_1+g_2, h_1)$ if and only if $h_1 = h_2$, and $(g_1, h_1) \times (g_2, h_2)$ is defined to be $(g_1\cdot g_2, h_1\star h_2)$ always. (Or we might relax the condition on $H$ and make it a monoid, or whatever.)
Does this thing have a name? Is it of any interest? Are there any interesting examples other than the one I started with?
I did observe that this structure is also a bit like floating-point numbers, where the left component is the mantissa and the right component the exponent, except that floating-point numbers also have a normalization homomorphism that allows one to add $(m_1, e_1)$ and $(m_2, e_2)$ even when $e_1\ne e_2$, and to understand $(m_1, e_1)$ and $(b\cdot m_1, e_1 - 1)$ as different representations of the same thing.
AI: Tensor product of 1-dimensional graded vector spaces.
A number with units (as defined in the question) is an element of such a product and the grading carries the type of unit and its dimension. The product of spaces with different gradings gets a multiple grading, or you can see all the gradings as living in one large commutative group of all possible dimensions. The group does not have to be free, it can have relations between different dimensions, as are discovered from time to time in science.
There is also Arnol'd remark that dimensional analysis has been re-packaged as Toric Varieties, and internet postings by Terence Tao on units which he takes to be elements of a dual space.
Terry Tao: A mathematical formalisation of dimensional analysis |
H: How many triples of positive integers $(a,b,c)$ satisfy $a\le b\le c$ and $abc=1,000,000,000?$
How many triples of positive integers $(a,b,c)$ satisfy $a\le b\le c$ and $$abc=1,000,000,000?$$
I find combinatorial questions like this quite difficult.
I expressed $1,000,000,000$ as $2^95^9$.
I let $a=2^p5^s$, $b=2^q5^t$ and $c=2^r5^u$. So I need to find the number of non-negative integer solutions for $p+q+r=9$ and $s+t+u=9$, which is $\binom{11}{2}=55$.
Now I'm stuck. I would just say $55\times55$ but they're not* ordered pairs.
AI: You need to find the number of cases with $a \lt b \lt c$, $a=b\lt c$, $a \lt b =c$ and $a=b=c$, though you can join the second and third together into a set called "two equal", and we can call the first "none equal" and the last "three equal".
"Three equal" is easy: you want solutions to $3p=9$ and $3s=9$ and there only is $1 \times 1 = 1$ "three equal" solution.
"Two equal" requires the number of solutions to $2p+r=9$ and $2s+u=9$. This is $5 \times 5 = 25$. These will each appear once either as $(2^p5^s,2^p5^s,2^r5^u)$ or as $(2^r5^u,2^p5^s,2^p5^s)$ so we do not need to adjust for duplication. But $1$ is the "three equal" solution, leaving $25-1=24$ "two equal" solutions.
"None equal" is a little harder. You have the $55\times 55$ you found. But you need to subtract three times the $24$ "two equal" solutions [e.g. if you have a solution $(a,a,c)$ then it will appear also as $(a,c,a)$ and $(c,a,a)$] and one times the "three equal" solution. Even then five-sixths of these "none equal" solutions are in the wrong order. So you want $\frac{55\times 55-3\times 24 -1 \times 1}{6}=492$ "none equal" solutions.
And then your answer is $1+24+492=517$ ordered solutions as lnwvr has found. |
H: Solve for z(t) from the simultaneous equation using Laplace transform
Solve for $z(t)$ from the simultaneous equation using Laplace transform
$$
y' + 2y + 6 \int\limits_0^t z \mathrm{d}t = -2 u(t) \\
y' + z' + z = 0$$
subject to $y(0) = -5$ and $z(0) = 6$.
AI: First note that if $u(t)$ is the Heaveside step function, then
$$\int_0^t z(s)\,dt = u(t)\ast z(t)$$
where $\ast$ denotes convolution. So we can rewrite the system as
$$\left\{\begin{aligned} y^{\prime}+2y + 6u(t)\ast z(t) &= -2u(t)\\ y^{\prime} + z^{\prime} + z &= 0\end{aligned}\right.$$
Noting that $y(0) = -5$ and $z(0) = 6$, we take Laplace transforms of both equations to see that
$$\left\{\begin{aligned} (sY(s)+5) + 2Y(s) + 6\frac{Z(s)}{s} &= - \frac{2}{s}\\ (sY(s)+5) + (sZ(s)-6) + Z(s) &= 0\end{aligned}\right.$$
which simplifies to
$$\left\{\begin{aligned}(s^2+2s)Y(s)+6Z(s) &= -5s-2\\ sY(s) + (s+1)Z(s) &= 1\end{aligned}\right.$$
We now solve for $Z(s)$ by means of elimination; multiplying the second equation by $-(s+2)$ and then adding to the first leaves us with
$$(6-(s^2+3s+2))Z(s) = -4-6s\implies Z(s) = \frac{4+6s}{s^2+3s-4}$$
Note that $\dfrac{1}{s^2+3s-4} = \dfrac{1}{(s+4)(s-1)} = \dfrac{1}{5(s-1)} - \dfrac{1}{5(s+4)}$.
Therefore,
$$Z(s) = \frac{4}{5(s-1)} -\frac{4}{5(s+4)} +\frac{6s}{5(s-1)} - \frac{6s}{5(s+4)}$$
Can you take things from here with finding the inverse Laplace Transform? |
H: If a function has no critical points, then its zero set has no interior points
Let $f \in C^1$ in the open set $\Omega$ and have no critical points there. Let $E$ be the set where $f=0$. Show that $E$ has no interior points.
The back of the book says "If $f$ is constant on an open set then $Df=0$ there." I see this would contradict the fact that there are no critical points in $\Omega$. But I do not understand how "constant on an open set" relates to interior points.
AI: This problems appear in and old test of multivariable calculus.
Suppose the contrary. Let $a\in E$ an interior point. Then there is $\delta>0$ such that $$B_{\delta}(a)\subseteq E$$ and $$B_{\delta}(a)\subseteq \Omega$$ because is an open. Then for all $p\in B_{\delta}(a)$ we have $f(p)=0$. Let $U=B_{\delta}(a)$ a 'hood of $a$, then because $f\in C^1$ we have $$Df=\left(\frac{\partial f}{\partial x_1},\cdots,\frac{\partial f}{\partial x_p }\right)$$ but each partial derivate vanishes at $U$ (in at least one point say $p$). So $$Df(p)=0$$. A contradicction. |
H: How to show that an automorphism of $S_n$ is inner?
Given an automorphism $\phi:S_n\rightarrow S_n$ such that it maps all the transpositions on the transpositions, how do I show that this map is given by a conjugation with an element $s\in S_n$?
Thanks in advance!
AI: Let's suppose that we are looking at $\phi:\ S_N \to S_N$ with $N \geq 3$; for small $N=1,2$ this is trivial.
Let $\phi( (n,m) ) =: \sigma_{n,m}$. We can notice that:
$$\sigma_{n,m} := \phi((n,m)) = \phi((1,n)(1,m)(1,n)) = \sigma_{1,n} \sigma_{1,m}\sigma_{1,n}.$$
Now, because $\phi$ was supposed to be bijective, all the $\sigma_{n,m}$ are different for different pairs of $n,m$. In particular, if $\sigma_{1,n} = (a_n,b_n)$, then $\{a_n,b_n\} \cap \{a_m,b_m\} \neq \emptyset$ for any $n,m \neq 1$, $n \neq m$. A simple combinatorial argument shows the all the pairs $\{a_n,b_n\}$ have an element in common. Indeed, we may assume without loss of generality that $a_2 = a_3 =:a$ and $b_2 \neq b_3$. If $N =3$ we are done. Else for all $m \geq 3$ we have two options: either $a \in \{a_m,b_m\}$ (and we may wlog say that $a_m = a$) or $\{a_m,b_m\} = \{b_2,b_3\}$. If $N \geq 5$, the second case can never happen. If $N = 4$, the second case might a priori happen. Allowing $a_2 = a_3 = 2,\ b_3 = b_4 = 3,\ b_2 = a_4 = 4$, we find that $\phi$ would have to act as follows: $(12) \mapsto (24),\ (13) \mapsto (23),\ (14) \mapsto (34)$. But this quickly leads to a contradiction, because for instance $(12)(13)(14)$ is a $4$-cycle, while $(24)(23)(34)$ is a $3$-cycle.
Thus, wlog $a_n = a$ for all $n$ for some $a \in [N]$.
Consider the permutation $\pi$ which maps $1$ to $a$, and $n$ to $b_n$. Clearly, conjugation by $\pi$, say $\gamma_\pi$, sends $(1,n)$ to $$\gamma_\pi((1,n))= \pi(a_n,b_n) = (a,b_n) = \phi((1,n)).$$ Thus, $\phi$ and $\gamma_\pi$ agree on all transpositions of the form $(1,n)$. But they are both automorphisms, and these transpositions generate $S_N$, so they are equal everywhere, and we are done. |
H: Non-commutative indeterminates in polynomial rings.
Described below are some observations I have made while fiddling around with polynomials. In addition to the two questions below, I am looking for any sort of relevant information so I can read more.
Preliminary question:
Suppose I want to consider the polynomial ring $\mathbb{C}\left[x, y\right]$, except I don't want to assume that $x$ and $y$ are commutative. I believe that the resulting structure is still a ring, but I have never seen anything dealing with such a scenario. Is there a name for such a structure/assumption? Is it even a sound thing to do?
Main question:
Denote by $\mathbb{C}[x, y]^*$ the ring $\mathbb{C}[x, y]$ with the added restriction that $xy \ne yx$ in general. Let $R$ be a non-commutative ring and suppose I have a homomorphism
$$\varphi \colon \mathbb{C} \longrightarrow Z(R)$$
Then it seems I can use some "extended" form of the substitution principle to make the substitution $x = a$ and $y = b$ for any $a, b \in R$, for it appears to be the case that the map
$$\Phi \colon \mathbb{C}[x, y]^* \longrightarrow R$$
given by
$$\Phi(\sum_{i + j = 0}^{n} a_{ij}x^iy^j) = \sum_{i+j=0}^n\varphi(a_{ij})a^ib^j$$
is a homomorphism.
Is this in fact the case or am I missing something?
AI: The usual notation for non-commutative free $k$-algebras is $k\langle x,y\rangle$. And what you have verified is its universal property. All this is well-known, you can find it in Bourbaki's Algèbre for sure. Notice that $k\langle x,y\rangle$ is the tensor algebra over the free module of rank $2$, wheras $k[x,y]$ is the symmetric algebra over the free module of rank $2$. |
H: Conditional Probability with Independent Discrete Random Variables
Let X, Y be two independent Poisson random variables with lambda of X = 1, lambda of Y = 2. Find P(X = 40 | X + Y = 100).
I know P(X|Y) = P(X, Y) / P(Y), and since X and Y are independent P(X, Y) = P(X) P(Y). But P(X) and P(X + Y) are not independent, so how would I go about finding the joint mass function of X and X + Y? I already found the mass function of X + Y.
AI: I think you need a division $$P(X=40|X+Y=100)=\dfrac{P(X=40 \text{ and } X+Y=100)}{P(X+Y=100)}=\dfrac{P(X=40 \text{ and } Y=60)}{P(X+Y=100)}=\dfrac{P(X=40)P(Y=60)}{P(X+Y=100)}$$ |
H: Generating function - Partial fractions
I was Reading a book about generating functions, and it said me to do this:
$$\frac{1 - 2x + 2x^2}{(1-x)^2(1-2x)} = \frac{A}{(1-x^2)} + \frac{B}{1-x}+\frac{C}{1-2x}$$
It says:
Multiply both sides by $(1-x^2)$ so:
$$\frac{(1 - 2x + 2x^2)(1-x^2)}{(1-x)^2(1-2x)} = \frac{A(1-x^2)}{(1-x^2)} + \frac{B(1-x^2)}{1-x}+\frac{C(1-x^2)}{1-2x}$$
Then, it says to consider $x=1$, then $A = -1$. Well, I don't know why, since it will result in division by zero.
AI: I’m sure that it said to multiply both sides by $(1-x)^2$, not by $1-x^2$. When you do, you should have
$$\frac{1-2x+2x^2}{1-2x}=A+B(1-x)+\frac{C(1-x^2)}{1-2x}\;.$$
Now let $x=1$ and read off the value of $A$, since the last two terms on the righthand side are $0$.
Added: This is just a shortcut. What you’re really doing is finding the values of $A,B$, and $C$ that make the original equation an identity. After you put everything over the least common denominator, that equation is
$$\frac{1-2x+2x^2}{(1-x)^2(1-2x)}=\frac{A(1-2x)+B(1-x)(1-2x)+C(1-x)^2}{(1-x)^2(1-2x)}\;.\tag{1}$$
Two fractions with the same denominator are identically equal if and only if their numerators are identically equal:
$$1-2x+2x^2=A(1-2x)+B(1-x)(1-2x)+C(1-x)^2\;.\tag{2}$$
Substitute $x=1$ into $(1)$, and you immediately find that $-1=A$, or $A=-1$, and the suggested calculation is essentially the same thing.
Now it’s true that the fractions in $(1)$ are undefined at $x=1$ and $x=\frac12$, while the polynomials in $(2)$ are defined on all of $\Bbb R$. But making the polynomials in $(2)$ identically equal ensures that the fractions in $(1)$ will be equal wherever they are both defined, which is what we want, and making the polynomials identically easy is straightforward. If all else fails, we simply multiply out the righthand side of $(2)$ and equate coefficients to get a linear system of $3$ equations in the unknowns $A,B$, and $C$. Sometimes, however, as in this example, we can shortcut the process a bit by evaluating the polynomials at a point that knocks out all but one of the unknowns. |
H: Definitions of adjoints (functional analysis vs category thy)
If I have a linear operator $f$ on a Hilbert space, then I define the adjoint of $f$ to be $f^*$ where,
$(fx,y)=(x,f^*y)$ for all $x,y$.
I am confused because this definitions is very different to the definition of an adjoint used in category theory.
Is there some way to tie these two definitions together?
Thanks
Matthew
AI: The two notions of adjoint are related only by a symbolic similarity, enhanced by the once common but now apparently rare convention of writing $(X,Y)$ for the set of morphisms from $X$ to $Y$ (the category being understood or indicated by a subscript). Then if $L$ is left-adjoint to $R$, we have $(LX,A)\cong(X,RA)$, with obvious notational similarity to the Hilbert space definition of adjoint. |
H: Show and prove if $\sum_{j=1}^\infty{ \frac{(1+(1/j))^{2j}}{e^j}}$ converges or diverges
Show and prove if the following series converges or diverges
$$\sum_{j=1}^\infty{ \frac{(1+(1/j))^{2j}}{e^j}}$$
"I tried the comparison test, the root test, and the ratio test, but got messed up..."
Thanks!
AI: Apply the $\,n$-th root test:
$$\sqrt[n]{\frac{\left(\frac{n+1}n\right)^{2n}}{e^n}}=\frac{\left(\frac{n+1}n\right)^2}{e}\xrightarrow[n\to\infty]{}\frac1e<1\implies\;\text{the series converges}$$ |
H: Confusion with this definition of the derivative
This function is from my text: $$p(\theta) = \sqrt{13\theta}$$
It states that the derivative of the function $p(\theta)$ with respect to the variable $\theta$ is the function $p'$ whose value at $\theta$ is given by the following formula, provided that the limit exists:
$$\lim_{z \to \theta} \frac{p(z)-p(\theta)}{z-\theta}$$
Why is $z$ approaching $\theta$? I have always seen the definition of the derivative as the limit of a function as the increment $h \rightarrow 0$:
$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
Why can't we say: $\sqrt{13(\theta+z)}$ as $z \rightarrow 0$? I don't understand the limit derivative definition being applied to this problem by having $z \rightarrow \theta$.
Thank you.
AI: You can use either of the definitions
$$p'(\theta)=\lim_{h \to 0}\frac{p(\theta+h)-p(\theta)}{h}, \tag{1}$$
which seems close to your approach, or else the one suggested in the text,
$$p'(\theta)=\lim_{z \to \theta}\frac{p(z)-p(\theta)}{z-\theta}. \tag{2}$$
For this problem, definition $(2)$ might look cleaner in applying it, since the technique (in either case) is to multiply top and bottom by the conjugate of the difference of radicals, and definition $(2)$ makes the radicals simpler to manipulate.
Nevertheless the definition $(1)$ has the advantage that, no matter what the algebra is, one is always looking to at some point factor out $h$ from something and cancel it, so that at that point making $h \to 0$ will give an answer using limit laws.
The reason these are both the same is that, if you believe $(1)$, then defining $h=z-\theta$ we see that $h \to 0$ is the same as $z \to \theta$, and also $\theta +h$ becomes replaced by $z$. Thus everything matches up in the two definitions. |
H: Proving $n^3$ is even iff $n$ is even
I am trying to prove the following statement:
Prove $n^3$ is even iff n is even.
Translated into symbols we have:
$n^3$ is even $\iff$ $n$ is even
Since it's a double implication, I started assuming n is even, then eventually concluded:
$$n \;\text{ is even }\;\implies \; n^3\;\text{ is even.}$$
However, since it's a double implication I have to conclude
$$n^3\;\text{ is even }\;\implies n \;\text{ is even.}$$
I assume $n^3$ is even. Then $n^3 = 2k$ for some integer $k$.
Then $n = (2k)^{1/3}$...
But I can't really seem to find a way to get a $2k$ equivalent expression for $n$...
Can you guide me?
AI: HINT: For the second implication, try proving the contrapositive of the implication. Suppose $n$ is not even (i.e., assume $n$ is odd), and prove that, then, $n^3$ is not even (i.e., $n^3$ is odd).
$$P \implies Q \equiv \lnot Q \implies \lnot P$$ |
H: Show the series $ \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} $ converges
Show the following series converges
$$ \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} . $$
I tried to use the comparison test and tried to compare it with the series of $\dfrac{2^j}{3^j}$ because this is the geometric series. However, $\dfrac {2^j}{3^j}$ is smaller than $\dfrac{2^j+j}{3^j-j}$, so by comparison test, the series $\sum_{j=1}^{\infty}\dfrac{2^j}{3^j}$ converges does NOT indicate that the series $\sum_{j=1}^{\infty}\dfrac{2^j+j}{3^j-j}$ converges... Thus I'm confused. Thanks!
Thanks!
AI: I think a simple way of doing this would be to use the fact that exponential functions grow must faster linear ones. So notice that $2^j>j$ for $j\geq1$ and $3^{j-1}>j$ for $j\geq 2$. But $\sum_{j=1}^\infty stuff$ converges only if $\sum_{j=2}^\infty stuff$ converges. So we will deal with the second summation. We can make the summation even bigger by adding $2^j$ in the top instead of $j$ and subtracting $3^{j-1}$ instead of just $j$.
$$
\sum_{j=2}^\infty\frac{2^j+2^j}{3^j-3^{j-1}}=\sum_{j=2}^\infty \frac{2^{j+1}}{2\cdot 3^{j-1}}=\sum_{j=2}^\infty\frac{2^j}{3^{j-1}}
$$
Now we have
$$
0\leq\sum_{j=2}^\infty \frac{2^j+j}{3^j-j} \leq \sum_{j=2}^\infty \frac{2^j}{3^{j-1}}=4
$$
(the sum on the right is simple to calculate as it is a geometric series). This shows your original sum converges.
EDIT: You asked how to show $3^{j-1}\geq j$ for $j \geq 2$. Well, you can just graph $3^{j-1}$ and $j$ and notice that it is obvious that $3^{j-1}$ is bigger than $j$ for all $j \geq 2$. If you wanted to show it with rigor, you could show this easily using induction. Notice when $j=2$, we have $3^1>2$. So it works when $j=2$. We show that this works for all $j$ by assuming it works for $j=2,3,4,5,\cdots k$ and show it works for $k+1$. Since it works for $j=k$, we know we have
$$
3^{k-1}>k
$$
Let's add $1$ to both sides
$$
3^{k-1}+1>k+1
$$
Now let's make the really simple observation that $1+1+1=3$. So notice that
$$
3^{k-1}+1+1+1>3^{k-1}+1>k+1
$$
Let's clean up the left hand side, we then have
$$
3^{k-1}+3=3^k>3^{k-1}+1>k+1
$$
So
$$
3^{k}>k+1
$$
But this is our original formula $3^{j-1}>j$ when $j=k+1$. What does this all mean? Well, we just showed that if this inequality is true for $j=k$ then it is true for $j=k+1$. Big deal! So what?
Ah, there is the magic. We showed that $3^{j-1}>j$ when $j=2$. But we just showed that if this is true for $j=k$ then it is true for $j=k+1$. So since $3^{j-1}>j$ is true when $j=2$, it is true when $j=3$ by what we showed above. But it is true for $j=3$, then it is true for $j=4$. AH HA! But it being true for $j=4$ means it is true for $j=5$. We can continue this forever. So it is true for all $j\geq 2$ as promised. This is what is called mathematical induction. Hopefully, this made sense to you. If not, stick to doing it graphically as above. Since you are probably learning calculus at the moment. Note that $3^{j-1}>j$ when $j=2$. Notice that the derivative of $3^{j-1}$ is always bigger than the derivative of $j$ for $j\geq 2$, so $3^{j-1}$ is bigger than $j$ at $j=2$ and always increase faster for $j>2$. This also shows this. |
H: Discrete Gaussian density function sum drawn from Gaussian distribution
Doing some analysis of my problem,
I have come up with the following equation
(I tried to search this problem but no luck)
$S_N = \sum \limits_{i=1}^{N} \exp{\left(-\frac{X_i^2}{2\sigma^2}\right)}$
where $X_i \ \ \text{~} \ \ {N(0, \sigma^2)}$
It is obvious that $S_N \rightarrow \infty$ as $N \rightarrow \infty$
The solution I was thinking of is (it may be wrong)
$\frac{N}{S_N} \simeq \sqrt{2}$
which I got from Matlab by generating 1000000 numbers, and they were all same for different standard deviation, and different number of samples.
Here is Matlab script I used,
N = 1000000;
sd = 3; % arbitrary standard deviation
x = sd*randn(1, N);
S = sum(exp(-x.^2/(2*sd^2)));
N/S
How can I solve this problem?
Can you recommend any approximation, or bound theorem to play this with?
Thanks!!
AI: I tried it, and I got 1.7314 which looks more like $\sqrt 3$ to me.
Anyway,
$$ E(S_N) = N E(\exp(-X^2/\sigma^2)) = N \frac1{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty \exp(-x^2/\sigma^2) \exp(-x^2/2\sigma^2) \, dx = N \frac1{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty \exp(-3x^2/2\sigma^2) \, dx = \frac N{\sqrt 3} .$$
Oops, I tried it again. And then I got $\sqrt 2$. But your code is different than the formula - in the code you are doing
$$ \sum_{n=1}^N \exp(-X_n^2/2\sigma^2) .$$
Anyway, similar calculation - you will get $N/\sqrt 2$ this time. |
H: Partial Fraction Decomposition in $ \frac{s^3-1}{(s^2+6)^2(s+12)^2} $.
What special considerations do you need to take when decomposing the following fraction and why?
I'm trying to decompose the following:
$$ \frac{s^3-1}{(s^2+6)^2(s+12)^2} $$
$$ \frac{s^3-1}{(s^2+6)^2(s+12)^2} = \frac{A}{(s^2+6)^2} + \frac{B}{(s+12)^2} $$
It obviously isn't correct. What to do?
AI: When doing Partial Fractions you need to consider all of the increasing powers and should have (see the handy table in the link):
$$ \dfrac{s^3-1}{(s^2+6)^2(s+12)^2} = \dfrac{As + B}{(s^2+6)} + \dfrac{Cs + D}{(s^2+6)^2}+ \dfrac{E}{(s+12)} + \dfrac{F}{(s+12)^2} $$
You should arrive at:
$A = \dfrac{758}{140625}$
$B = \dfrac{6841}{562500}$
$C = -\dfrac{67}{1875}$
$D = -\dfrac{167}{3750}$
$E = -\dfrac{758}{140625}$
$F = -\dfrac{1729}{22500}$ |
H: Determine the number of solutions for $x^2 = c$ such that $c \in \mathbb{F}$
I am working on the problem from number system class taught by the professor who shares his knowledge about the relation of abstract algebra and number systems. Here is the problem:
Assume $c$ and $\mathbb{F}$ are arbitrary. For $c \in \mathbb{F}$, how many distinct solutions can the equation $x^2 = c$ have? Be specific about what properties of $c$ lead to what number of solution and justify your reasoning using properties a field must have.
Thoughts for the Problem
I believe that there is a solution when $c = 0$ for any field. If $c = 1$, and we have $\mathbb{F}_2$, then we obtain exactly one solution. BUT if $\mathbb{F}$ is any arbitrary field and $c = 1$, then we have two solutions. This situation could occur the same way for any arbitrary field and $c \geq 2$, being an integer. The thing is: what if $c$ is either trascendental or irrational (like roots number)? Is it possible for the equation to have solutions like this? I am having a hard time, approaching this problem.
AI: Equivalently, we are talking about the solutions over $x \in \mathbb{F}$ to $x^2 - c= 0$, where $c\in \mathbb{F}$.
In any case, one of two things happens: either there is no solution, or there is a solution $d$ such that $d^2 = c$. In such a case, we may write
$$
x^2 - c = (x+d)(x-d)
$$
So that $x^2 = 0$ if and only if $x = d$ or $x = -d$. If (and only if) $d = -d$, our solution is unique. Otherwise, there are two solutions.
Note that
$d = -d \iff 2d = 0$. So, if $c$ has a unique square root, then either $c = 0$ or there is an element $d \in \mathbb{F}$ for which $2d = 0$.
I'm not sure what you mean with the statement "what if $c$ is transcendental or irrational". What about it? |
H: Show that the natural number $n$ with base ten representation ($r_{k}r_{k-1}$. . . $r_{1}r_{0}$)$_{10}$ is a multiple of $4$
Show that the natural number $a$ with base ten representation ($r_{k}$$r_{k-1}$. . . $r_{1}$$r_{0}$)$_{10}$ is a multiple of 4 if and only if the number ($r_{1}$$r_{0}$)$_{10}$, consisting of the rightmost 2 digits of $a$, is a multiple of 4,that is show 4|$a$ = ($r_{k}$$r_{k-1}$. . . $r_{1}$$r_{0}$)$_{10}$ $\iff$ 4|($r_{1}$$r_{0}$)$_{10}$.
I have no idea where to begin.
AI: HINT:
$$\begin{align*}
(r_kr_{k-1}\ldots r_1r_0)_{10}&=100(r_k\ldots r_2)_{10}+(r_1r_0)_{10}\\
&=4\cdot25(r_k\ldots r_2)_{10}+(r_1r_0)_{10}
\end{align*}$$ |
H: Differentiation under integral sign for exponential
This question arises from this question:
Suppose $P(x)$ is a polynomial. Why is it the case that $$\dfrac{d}{dy}\int_\mathbb{R}iP(x)e^{-x^2/2}e^{-ixy}dx=\int_\mathbb{R}xP(x)e^{-x^2/2}e^{-ixy}dx?$$ I'm thinking about using dominated convergence thm, but not sure how to apply it here.
AI: Hint: You can use Leibniz rule. |
H: Monotonic uniformly continuous function - Unique $f(t) = t$
Let $f:[0,1] \rightarrow [0,1]$ be such that $|f(x)-f(x')| <|x-x'|$ for all $x,x'\in [0,1]$ with $x \not= x'$. Show that there is a unique point $t\in [0,1]$ such that $f(t)=t$.
I noticed that $f$ is a Lipschitz function so that $f$ is a uniformly continuous function on $[0,1]$. By the intermediate value theorem, there exists at least one $t \in [0,1]$, such that $f(t)=t$.
To show uniqueness, how would like to show that $f$ is strictly monotonic on $[0,1]$. That's where I'm stuck... I tried using sequences : $|x_{n+1} > x_n|$ (or $<$). Is that the way to go?
AI: Suppose $t$ and $t'$ are distinct fixed points of $f$. Then $f(t) = t$ and $f(t') = t'$ so $|f(t) - f(t')| = |t - t'|$, contradicting $|f(x) - f(x')| < |x - x'|$ for all $x, x' \in [0, 1]$ with $x \neq x'$. |
H: $\mathbb{E}[e^{Xt}] = \mathbb{E}[\mathbb{E}[e^{Xt}\mid Y]] = \mathbb{E}[M_{X\mid Y}(t)]$?
$$\mathbb{E}[e^{Xt}] = \mathbb{E}[\mathbb{E}[e^{Xt}\mid Y]] = \mathbb{E}[M_{X\mid Y}(t)]$$
How do I get the above statement? I don't understand how in the 1st step $e^{Xt}=\mathbb{E}[x^{Xt}\mid Y]$ then in the 2nd $\mathbb{E}[e^{Xt}\mid Y] = M_{X\mid Y}(t)$?
This is from the part (b) of the below question:
And its provided solution: (see 1st 3 lines)
AI: The first step is what is called the law of Iterated Expecations. Simply put, if $X,Y$ are random variables then, $$\mathbb{E}_X[X] = \mathbb{E}_Y[\mathbb{E}_{X \mid Y}[X \mid Y]]$$ and by definition, $\mathbb{E}_{X \mid Y}[X \mid Y]$ is a function of $Y$ and again by definition $\mathbb{E}_{X \mid Y}[e^{xt}\mid Y]$ is the Moment generating function of the random variable defined by the PDF $f_{X\mid Y}(x\mid y)$. |
H: Irreducibility in $\mathbb{Z}_7[x]$
Is the polynomial $x^2+3$ irreducible in $\mathbb{Z}_7[x]$?
I am not sure how to even start the problem. I have been looking online trying to find help to teach myself how to do irreducibility and I am not having much luck. Can anyone offer any help as to where I should begin?
AI: If $F$ is a field, and $P(x)$ is a polynomial of degree $2$ or $3$ over $F$, then $P(x)$ is irreducible over $F$ if and only if the equation $P(x)=0$ has no roots in $F$. |
H: Two idempotent matrix
Let $A,G$ be two $n\times n$ matrix satisfying:
$$A^2=A, GAG=G, im(G)\subset im(A).$$ Prove that $G^2=G$.
I do not know how to prove it.
AI: $im(G) \subset im(A)$,so for any vector $u$,there is a vector $v$ that
$$Gu=Av$$.
So we get:
$$Gu=GAGu=GAAv=GA^2v=GAv=GGu=G^2u$$
.
QED. |
H: Where is the error? Determining why a proof is incorrect.
Why is the following proof incorrect?
I have tried to find out why for a few hours...
aproof http://forosdelecuador.com/proof.png
The answer is:
We aren't really proving the conclusion of each case is valid for the other case.
Consider the case assumptions for each case to define the real boundaries for x.
AI: From case 1 and case 2, you are using the following logic:
If $x<6$ or $0 < x$, then $0<x<6$.
What you need to do is to deduce from case 1:
If $x-3 \ge 0$ then $3 \le x < 6$.
Similarly for case 2. Then you should use the logic:
If $3 \le x<6$ or $0< x < 3$, then $0<x<6$. |
H: $a_n = n ( \sqrt{1+ \frac{1}{n}} - 1 )$
i have to show convergence and the limit.
$a_n = n ( \sqrt{1+ \frac{1}{n}} - 1 )$
So far, i tryed to used $a^2 - b^2$, and got
$ \frac {1}{\sqrt{1+\frac{1}{n}}+1} $
how can i go on?
AI: $a_n = n \left( \sqrt{1+ \frac{1}{n}} - 1 \right)= \frac {1}{\sqrt{1+\frac{1}{n}}+1} \to \frac{1}{2}$ for $n\to\infty$ because $\frac{1}{n}\to 0$. |
H: Compositum of finite field extensions
If $L_{1}$ and $L_{2}$ are field extensions of $F$ that are contained in a common field, show that $L_{1}L_{2}$ is a finite extension of $F$ if and only if both $L_{1}$
and $L_{2}$ are finite extensions of $F$. (Patrick Morandi, Field and Galois Theory, Exercise $14$ page $14$.)
Can anyone tell me a hint to prove it?
AI: For a hint, you might start out by looking at $L_1(\lambda)$ where $\lambda\in L_2$, and showing that this field is finite over $F$. From there it shouldn’t be too hard.
EDIT, in response to request for further help:
First step, $L_1(\lambda)$ is finite over $L_1$ because $\lambda$ is root of an $F$-polynomial, and thus of an $L_1$-polynomial. Second, this shows that if $\{a_a,a_2,\cdots,a_n\}$ is an $F$-basis of $L_2$, then you have the finite sequence of finite extensions
$$
L_1\subset L_1(a_1)\subset L_1(a_1,a_2)\subset\cdots\subset L_1L_2\,.
$$ |
H: How to find $ \sum\limits_{n=0}^{\infty}\frac{1}{3-8n-16n^2}$?
I have to show, that this series converges and determine its limit.
$$\sum\limits_{n=0}^{\infty}\frac{1}{3-8n-16n^2}$$
So far, my idea was to transform it into a telescoping series, but I don't really know how to do it.
AI: Factorise and use partial fractions.
$\dfrac{1}{3-8n-16n^2}=\dfrac{1}{4}\left(\dfrac{1}{4n+3} - \dfrac{1}{4n-1}\right)=\dfrac{1}{4}\left(\dfrac{1}{4(n+1)-1} - \dfrac{1}{4n-1}\right)$
So $\displaystyle\sum\limits_{n=0}^{\infty}\frac{1}{3-8n-16n^2}=\frac{1}{4}$ |
H: $\left<2,x\right>$ is a maximal ideal of $\Bbb Z[x]$
I want to show that $\left<2,x\right>$ is a maximal ideal of $\Bbb Z[x]$.
My game plan is to use the 3rd isomorphism theorem to somehow get that $Z[x]/\left<2,x\right>$ isomorphic to $Z_2$ (since every this would mean that $Z[x]/\left<2,x\right>$ is a field and hence $\left<2,x\right>$ would be a maximal ideal) but I'm not entirely sure how to arrive at that conclusion.
My first thoughts are that since $\left<x\right>$ is an ideal of $Z[x]$ and $\left<x\right>\subset\left<2,x\right>$ we have that $Z[x]/\left<2,x\right> \cong Z[x]/\left<x\right>/\left<2,x\right>/\left<x\right>$ by the 3rd isomorphism theorem, but I'm not sure if this is the right direction. Any hints or tips would be great.
AI: This seems like a possible direction. As a next step, you can say that:
$$ \mathbb{Z}[x]/(2,x) \simeq \left( \mathbb{Z}[x]/(x)\right) /\left((2,x)/(x) \right) \simeq \mathbb{Z}/(2) = \mathbb{Z}_2,$$
and you're done. The second isomorphism needs some justification, perhaps. This would involve either waiving your hands around it, or doing it by hand.
You can also refer to the second isomorphism theorem. Because $ \mathbb{Z}[x] = \mathbb{Z} + (2,x)$, we have that:
$$ \mathbb{Z}[x]/(2,x) = \left( \mathbb{Z} + (2,x) \right) /(2,x) \simeq \mathbb{Z}/\left( \mathbb{Z} \cap (2,x) \right) \simeq \mathbb{Z}/(2) = \mathbb{Z}_2.$$ |
H: Definition of rank
In Hatcher P146, the rank of a finitely generated abelian group is defined to be the number of $\mathbb{Z}$ summands when the group is expressed as a direct sum of cyclic groups.
$\mathbb{Q}$ $1$: What does "$\mathbb{Z}$ summands" really mean? For example, $H_2(K \times S^1) = \mathbb{Z \oplus Z_2}$. So the number of $\mathbb{Z}$ summands is 1? In other words, $\mathbb{Z}_2$ does not count?
$\mathbb{Q}$ $2$: Is the rank of $H_n$ equal to the number of $n$-cells? Why?
AI: For question $1$, you are correct. The rank is $1$. For question $2$, you can say that the rank of $H_n$ is less than or equal to the number of $n$-cells. This is because in cellular homology, $H_n$ is a quotient of a subgroup of $C_n\cong \mathbb Z^{\text{number of $n$-cells}}$. Combine this with the observation that taking quotients and subgroups reduce rank. |
H: Continuity of the sum of functions.
I know that if $f(x)$ and $g(x)$ are continuous, then $h(x)=f(x)+g(x)$ is continuous. What about the converse? Suppose I knew that a function $j(x)$ is continuous and knew that it is composed of a sum of functions. Furthermore, suppose that I can create different expressions whose sum still yield me $j(x)$. For example, let $j(x)=x^4-9x+2x^2+4$, then the possible expressions whose sum still yield me $j(x)$ would be:
$$j(x)=k(x)+l(x)\;\;where\;\;k(x)=x^4, l(x)=-9x+2x^2+4$$
$$j(x)=m(x)+n(x)\;\;where\;\;m(x)=x^4-9x, n(x)=2x^2+4$$
And so on. Would each component of every possible expression whose sum yield me the continuous function be continuous? (Using the framework above, would $k(x), l(x),m(x),n(x)$ be continuous given that $j(x)$ is continuous?).
AI: Let $f(x)$ be ANY function, then $j(x) = 0 = f(x) + (-f(x))$ is continuous. |
H: How to prove this equality using boolean algebra?
I have approximately no idea on how to solve the following problem, so any help would very much be appreciated:
$$x' y'+ x y = (x y' + x' y)'$$
I can't figure out how to prove the equalities using boolean algebra
Thank you for your help!
AI: Here's one way. First, use DeMorgan's law:
$$
(xy' + x'y)' = (xy')'(x'y)'
= (x' + y)(x + y')
$$
From there, you can expand the product using the distributive property, noting that $x'x$, which means "$x$ and not $x$", is $0$ (i.e. false).
$$
(x' + y)(x + y') = (x'x) + xy + x'y' + (yy') = xy + x'y'
$$ |
H: Number of monic polynomials of degree $2$
I know that there are $p^2$ monic polynomials of degree $2$ over the field $\mathbb{Z}_{p}
$ but I want to prove it precisely. Help me a hint to prove that. Thanks a lot.
AI: A degree 2 polynomial has the form $ax^2 + bx +c$. It is monic when $a=1$. How many choices of $b$ are there? How many choices of $c$ are there? |
H: Pumping Lemma - Clarification of Usage
I'd like to make sure my understanding of the Pumping Lemma is correct.
Consider $L=\{ 0^n1^m2^{n-m}:\, n \ge m \ge 0\}$
I'm going to give 2 solutions to prove that $L$ is not regular. One using "pumping down" and the other using a "cheap trick". I'm not sure whether either solution is correct, so any comments would help clarify my understanding.
Solution 1: Pumping Down
Let p be the pumping length. Let $S= 0^{p+1}1^p2 = xy^iz$ for $i \ge 0$
Now, $ |xy| \le p$ implies $y$ contains only $0's$
So, for $i=0,$ $S=xy^0z=xz$ and since $xy$ had only one more $0$ than the number of $1's, \, x$ will not have more 0's than the number of 1's. Hence, $S \notin L$ giving the required contradiction.
Solution 2: Cheap Trick
As before, let p be the pumping length. But this time, $S= 0^p1^p2^0 = 0^p1^p$ which is generally the example proven in most textbooks. However, I'm not sure this is a valid approach, so some clarification would be helpful.
AI: Here is a very careful application of the pumping lemma for regular languages, showing all of the relevant details. It uses pumping down with your second choice of word to be pumped.
Suppose that $L$ is regular; then it has a pumping length $p$. Let $w=0^p1^p2^0$; the pumping lemma for regular languages says that we can decompose $w$ as $w=xyz$, where $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for each $k\ge 0$. Clearly $xy$ is contained in the initial $0^p$ of $w$, so there are $r\ge 0$ and $s>0$ such that $x=0^r$ and $y=0^s$, and therefore $z=0^{p-r-s}1^p$. Then
$$xy^0z=xz=0^r0^{p-r-s}1^p=0^{p-s}1^p\in L\;,$$
but $p-s<p$, so $xy^0z\notin L$. This contradiction shows that $L$ is not regular after all. |
H: verify, that this sequence is cauchy-sequence
Be $ 0 < q < 1, (a_n)_{n \in \mathbb{N}} $ a sequence in $\mathbb{R}$ and $ n_0 \in \mathbb{N} $
One has: $ |a_{n+1} -a_n| \leq q |a_n-a_{n-1}|$
for all $n \geq n_o$. Show, that the sequence $(a_n)_{n \in \mathbb{N}}$ convergates.
I tryed to verify, that its a Cauchy-sequence, but i didn't gone very far..
AI: Hint 1:
$|a_{n+1} - a_n| \leq q^{n-1}|a_2 - a_1|$ (why?)
Hint 2:
For $m\leq n\in \mathbb{N}$ $|a_{n+1} - a_{m}| \leq \sum_{k=m}^n|a_{k+1}-a_k|$ by the triangle inequality
Hint 3:
For any $\epsilon>0$, we would like to show that there is an $N$ such that $|a_{n+1} - a_{m}|<\epsilon$ whenever $N<m\leq n$.
Now, for a particular choice of $N$, we have
$$
\begin{align}
|a_{n+1} - a_{m}| &\leq \sum_{k=m}^n|a_{k+1}-a_k|\\
&\leq \sum_{k=m}^nq^{k-1}|a_2-a_1|
\\
&\leq \sum_{k=N+1}^\infty |a_2-a_1| q^{k-1}
= |a_2-a_1| q^{N} \sum_{k=0}^\infty q^k\\
&= |a_2-a_1| q^{N} \cdot \frac{1}{1-q}
\end{align}
$$
You may choose an $N$ so that this is less than $\epsilon$ in order to complete your proof. |
H: How do I determine the maximum value for a quadratic equation on an interval?
I need to determine the maximum value for y = ax^2 + bx + c, where I know the coefficients and the upper and lower x values.
Say the input values are:
a = 5
b = 1
c = 2
x lower limit = -5
x upper limit = 5
Given these input, how do I determine the the maximum value for the quadratic equation above?
My goal is to implement a function in a computer programming language that has a signature such as funcMax(int a, int b, int c, int xUpper, int xLower). The part that is confusing me is that I have two x values for this signature but one x in the formula above.
But computer programming aside for now, mathematically, how do I get the max value for the above equation, using the above input (with two x values)?
AI: For this quadratic function, the global maximum happens either at the two ends or at a local maximum.
The first two cases are simple, you have to only compare the $y$ evaluated at the upper and lower limits of $x$ respectively.
And there is a local maximum only if the $a$ value is negative, shown by the second derivative test. When this is the case, the maximum value is given by taking $x = -\frac b{2a}$. (You can verify this by completing square or first derivative test.) Then you have to check if this $x$ is within your domain given by the upper and lower limits.
If I write your required function out:
$$\begin{align*}&\text{funcMax}(a,b,c,x_{Upper}, x_{Lower}) \\
=& \begin{cases}
\max(ax_{Upper}^2 + bx_{Upper}+c, ax_{Lower}^2 + bx_{Lower}+c,\frac{b^2}{4a}-\frac{b^2}{2a}+c) & \text{if }a\ne0\text{ and }x_{Upper}\le-\frac{b}{2a}\le x_{Lower}\\
\max(ax_{Upper}^2 + bx_{Upper}+c, ax_{Lower}^2 + bx_{Lower}+c) & \text{otherwise}
\end{cases}\end{align*}$$ |
H: The Moore Plane's topology
In the definition of the Moore plane $X=L{_1}\cup L{_2}$, where $L{_1}$ is the line $y=0$ and
$L{_2}=X\setminus L{_1}$ , I have a problem. In the Engelsking's book, for each $x\in L{_1}$
neghbourhood of $x$, is the form $U(x,1/i)\cup \{ x \}$ where $U(x,1/i)$ be the set of $X$ inside the circle centered $x$ and radius $1/i$ for i=1,2,..... So, I wonder that whether or not radius is greater than $1$?If it is, how can cover all $X$ with small radius? thanks
AI: You don’t need to cover $X$ with basic open nbhds of points of $L_1$: you also have the basic open nbhds of points of $L_2$, which are ordinary Euclidean balls small enough to stay within $L_2$. Specifically, the following collection is a base for $X$:
$$\left\{\{\langle x,0\rangle\}\cup B\left(\left\langle x,\frac1k\right\rangle,\frac1k\right):k\in\Bbb Z^+\right\}\cup\left\{B(\langle x,y\rangle,\epsilon):y>0\text{ and }0<\epsilon\le y\right\}\;,$$
where for $B(p,r)$ is the usual Euclidean open ball of radius $r$ centred at $p$.
Note that your description of basic open nbhds at points of $L_1$ isn’t actually correct: the ball $U$ is tangent to $L_1$ at $x$ and therefore does not have its centre at $x$. |
H: Complete function space proof
An exercise in my textbook is as follows.
Let $T>0$ and $L\geq 0$. Consider $C[0, T]$, the space of all continuous real valued funcitons on $[0,T]$, with the metric $\rho$ defined by $$\rho(x, y)=\sup_{0<t\leq T}e^{-Lt}|x(t)-y(t)|.$$ Verify that $(C[0, T], \rho)$ is a complete metric space.
The proof of completeness given in the book goes like this.
Note that $e^{-LT}\leq e^{-Lt}\leq 1$ for $t\in [0, T]$. Therefore, for any $x, y\in C[0, T]$ we have the following inequality: $$e^{-LT}\sup_{0<t\leq T}|x(t)-y(t)|\leq\sup_{0<t\leq T}e^{-Lt}|x(t)-y(t)|\leq \sup_{0<t\leq T}|x(t)-y(t)|$$ This implies that a sequence in convergent in $(C[0,T],\rho)$ if and only if it is convergent in $(C[0,T],d)$ where $d$ is the uniform metric on $[0,T]$. That is, $$d(x, y)=\sup_{0<t\leq T}|x(t)-y(t)|.$$ Therefore, $(C[0, T], \rho)$ is complete since $(C[0, T], d)$ is complete.
My question is from what theorem the completeness of $(C[0, T], \rho)$ and $(C[0, T], d)$ is equivalent? Also from what theorem $(C[0, T], d)$ is complete? Thank you!
AI: Let $X$ be a space with two equivalent metric $d_1, d_2$. Then $\{a_n \}$ is a Cauchy sequence with respect to $d_1$ if and only if it is a Cauchy sequence with respect to $d_2$. From here you can show easily that $(X, d_1)$ is complete if and only if $(X, d_2)$ is. |
H: Solving $\log(x+2) - \log(x) = 3$
I have work through the whole problem, but I cannot get passed the last step.
The original equation was: $\log(x+2) - \log(x) = 3$
I worked it out to this: $\frac{x+2}{x} = 1000$.
I know the answer is $\frac{2}{999}$ but I don't know how to get there. It's probably really simple, but I am just drawing a blank! Any help would be just great, thanks!
AI: You are almost there!. While: $$\frac{x+2}{x}=1000\to x+2=1000x\to 1000x-x=2\to 999x=2\to x=\frac{2}{999}$$ |
H: Finding Fourier transform of initial condition
Consider the equation $$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2} + a\frac{\partial u}{\partial x}$$ for a function $u(x,t)$ with initial value $$u(x,0)=f(x).$$ Let $\hat{u}(y,t)$ and $\hat{f}(y)$ denote the Fourier transform in the $x$ variable of $u$ and $f$. I want to solve for $\hat{u}$ in terms of $\hat{f}$.
Taking the Fourier transform and using the formula for Fourier transform of derivatives, I get $$\frac{\partial}{\partial t}\hat{u}(y,t)=(iy)^2\hat{u}(y,t)+aiy\hat{u}(y,t)=-y^2\hat{u}(y,t)+aiy\hat{u}(y,t)$$
I don't understand how to get $\hat{u}$ in terms of $\hat{f}$. There is no $\hat{f}$ in the equation.
AI: $ \hat u(y,0) = \hat f(y) $. Solve the ODE as an equation in $t$, which happens to depend upon a parameter $y$. That is, treat $y$ as a constant while you solve the ODE. |
H: Show there are distinct $\xi,\eta$ s.t. $\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$
Let $f:[0,1]\to\mathbb{R}$ be continuous. Suppose $f$ is differentiable on $(0,1)$ and $f(0)=0,f(1)=1$. Show that for any positive real numbers $a,b$, there are distinct points $\xi,\eta\in(0,1)$ s.t.
$$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$$
The word distinct makes this problem much harder.
AI: Since $a,b>0$ and $f(0)=0,f(1)=1$, $\exists \nu\in(0,1)$ s.t. $f(\nu)=\dfrac{a}{a+b}$. Hence we have $$\frac{f(\nu)-f(0)}{\nu-0}=f'(\xi)\Longrightarrow \frac{a}{a+b}=\nu f'(\xi)\Longrightarrow\frac{a}{f'(\xi)}=(a+b)\nu$$where $\xi\in(0,\nu)$.
Similarly, we have$$\frac{f(1)-f(\nu)}{1-\nu}=f'(\eta)\Longrightarrow \frac{b}{a+b}=(1-\nu) f'(\eta)\Longrightarrow\frac{b}{f'(\eta)}=(a+b)(1-\nu)$$where $\eta\in(\nu,1)$.
Therefore $$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=(a+b)\nu+(a+b)(1-\nu)=a+b$$and it's also obvious that $\xi\ne\eta$. |
H: T- invariant subspace and continuous functions
Let $C(\mathbb R)$ be the vector space of all continuous functions over $\mathbb R$. For $f,g \in C(\mathbb R)$, define the linear map $T_f: C(\mathbb R) \rightarrow C(\mathbb R)$ by $T_f(g)=g \circ f$.
Then, $(T_f(g))(x)=g(f(x))$ for all $x\in \mathbb R$. For every such $f$, find a nontrivial $T_f$-invariant subspace.
I know that a subspace $W$ of $V$ is $T$-invariant if $T(W) \subseteq W$, and nontrivial just means it's not the zero subspace or the entire space.
So would I just let $W$ be a subspace of $C(\mathbb R)$ and use the linear transformation properties to prove that $T(W) \subseteq W$?
AI: hint: constant functions are quite continuous. |
H: If $a_n\leq b_n$, then $\lim \sup a_n \leq \lim b_n$
Hello I want to make sure my work is correct!
Suppose that $\{a_n\}$ and $\{b_n\}$ are sequences such that $a_n \leq b_n$ for all $n $ and $b_n \to b$. Prove that $\lim \sup a_n \leq b.$
Let $\epsilon > 0$ be given. Choose an $N$ s.t. $|b_n-b|<\epsilon$ for $n\geq N$ where $b = \lim \sup b_n$.
$a = \lim \sup a_n$.
Proof by contradiction: If $a \geq b$, then, since $a_n\leq b$,
$b\leq a\leq b_n$, meaning $|b_n-a|<|b_n-b|<\epsilon $, implying $b_n \to a$. This is a contradiction,
therefore, $a\leq b$.
AI: No. It may happen that $b_n \to a$. That is not a problem.
What you can do is as follows : You know that $a_n \leq b_n$ for all $n \in \mathbb{N}$. Hence, for any $k \in \mathbb{N}$
$$
u_k := \sup\{ a_n : n\geq k \} \leq \sup\{ b_n : n\geq k\} =: v_k
$$
Now, $v_k \to b$, and hence
$$
\limsup a_n = \lim u_k \leq \lim v_k = b
$$ |
H: convergate and limit of a series, using geometric series
I have to show, that this series convergates and determine its limit.
$\sum\limits_{n=1}^{\infty}\frac{1}{5^{2n+3}}$
Well, i tried to transform it into a Geometric series, but i dont really know how to do it.
AI: Hint: $$\sum\limits_{n=1}^{\infty}\frac{1}{5^{2n+3}}=\frac{1}{5^3}\sum\limits_{n=1}^{\infty}\frac{1}{5^{2n}}=\frac{1}{125}\sum\limits_{n=1}^{\infty}\frac{1}{25^{n}}$$ |
H: Problem related to continuous complex mapping.
We are given with a map $g:\bar D\to \Bbb C $, which is continuous on $\bar D$ and analytic on $D$. Where $D$ is a bounded domain and $\bar D=D\cup\partial D$.
1) I want to show that: $\partial(g(D))\subseteq g(\partial D).$
And further, I need two examples:
a) First, to show that the above inclusion can be strict, that is: $\partial(g(D))\not= g(\partial D).$
b) Second example, I need to show that conclusion in (1) is not true if $D$ is not bounded.
So basically we have to show that the boundary of the open set $g(D)$ is contained in image of boundary of $D$ (and sometimes strictly contained). I think that we will use open mapping theorem. But how this theorem will help us here that is not clear.
AI: You know by the open mapping theorem that $g(D)$ is open. (I am assuming $g$ is non-constant.) But you also know $\bar D$ is compact (closed and bounded), so $g(\bar D)$ is also compact, and hence closed. Hence the closure of $g(D)$ is contained in $g(\bar D)$. You should be able to argue at this point that $\partial(g(D)) = g(\bar D) \setminus g(D)$. |
H: Exact Solution for Logarithmic Equation?
I am faced with this equation, and I don't really know where to begin:
$$x^2e^2 - 2e^x = 0.$$
I usually start these types of problems by factoring out a common term, but I don't see any in this particular example.
AI: Here is a the crucial step
$$ e^2x^2-2e^x =0 \implies \left( x-\frac{\sqrt{2}}{e}e^{x/2} \right)\left( x+\frac{\sqrt{2}}{e}e^{x/2} \right)=0. $$
$$ \implies \left( x-\frac{\sqrt{2}}{e}e^{x/2} \right)=0 \quad \rm{or}\quad \left( x+\frac{\sqrt{2}}{e}e^{x/2} \right)=0. $$
Now, follow the steps
i) make the change of variables $\frac{x}{2}=u$
ii) Use the result and follow the links to see the derivation if you want. |
H: On the image of $\mathbb{R}$ under an entire $f$ satisfying $f(n^{\frac{1}{n}})\in\mathbb{R}$.
Suppose $f$ is entire on $\mathbb{C}$ and $f(n^{\frac{1}{n}})\in\mathbb{R}$. Show then that $f(\mathbb{R})\subset\mathbb{R}$.
Any suggestions on how to even begin this problem? Thanks!
AI: Consider the function $g(z) = \overline{f(\bar z)} - f(z)$. Show first that $g$ is entire. Then show that $g(z) = 0$. Hint: there is nothing special about $n^{1/n}$ except that it is a real sequence that converges. |
H: Showing that $\lim\limits _{n\to\infty}{n \choose \left\lceil \frac{n}{2}\right\rceil }\cdot2^{-n}=0$ but the matching series does not converge
I want to show that: $$\lim\limits _{n\to\infty}{n \choose \left\lceil \frac{n}{2}\right\rceil }\cdot2^{-n}=0 $$
And also $${\displaystyle \sum_{n=1}^{\infty}{n \choose \left\lceil \frac{n}{2}\right\rceil }\cdot2^{-n}} $$
Does not converge. I'm not particularly good with this sort of approximation so I would really appreciate some help.
AI: These are the central binomial coefficients. You can see the wiki page for the estimates $$\frac{2^n}{n+1}\le {n\choose \lceil \frac{n}{2}\rceil} \le \frac{2^n}{\sqrt{1.5n+1}}$$
Multiply through by $2^{-n}$ and you get your terms. The upper bound gives the first part; the lower bound gives the second part by comparison with the harmonic series. |
H: Series, conditional or absolute? +more
This is for a presentation, so I not just want to solve it, but also be able to talk a bit about it
$$\sum \limits_{n=2}^\infty (-1)^{n+1}\frac{n-1}{n^2}$$
1)Show that the series converges, is it absolute or conditional?
First off we quickly see that the series is alternating and if w use the alternating series test:
$$\lim_{x \to \infty}(\frac{n-1}{n^2})=\lim_{x \to \infty}(\frac{1}{n}-\frac{1}{n^2})=0$$
And see that the series converges. But to determine weather it's absolute or conditional, we compare it to the harmonic series
$$\lim_{x \to \infty} \frac{\frac{n-1}{n^2}}{\frac{1}{n}}=\lim_{x \to \infty}\frac{n-1}{n}=\lim_{x \to \infty}1-\frac{1}{n}=1$$
Since this is different from what we found in our earlier test, the series is conditional convergent.
2)What can you tell about $|S_9-S|$?.
This is where it gets a bit unclear to me, what can I tell about this? I calculated that $S_9=\frac{8}{81}$ but I dont know what to do with it.
All in all: Is there anything Im missing on the first part and what am I supposed to possibly tell about the other?
AI: You are being asked to use Leibnitz' estimate for alternating series. That is, if you have a series that passes the alternating series test, the error in every partial sum is at most the very next term of the series. That is, $$|S_9-S|\le \frac{10-1}{10^2}=0.09$$ |
H: How prove this $\sqrt[5]{1782+\sqrt[3]{35+15\sqrt{6}}+\cdots}$ is positive integer numbers.
Prove that
$$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}\in N$$
This problem from this
My try: let
$$x=\sqrt[3]{35+15\sqrt{6}}+\sqrt[3]{35-15\sqrt{6}}$$
then
$$x^3=70+3\sqrt[3]{35^2-(15\sqrt{6})^2}\cdot x=70-3\sqrt[3]{125}x$$
then we have
$$(x^3-70)^3=-27\cdot 125 x$$
$$\Longrightarrow x^6-15x^4-140x^3+225x^2+1050x+4900=0$$
Then I can't ,and I think this is interesting problem, Thank you
AI: Denotes $t=\sqrt[3]{15 \sqrt{6}+35}+\sqrt[3]{35-15 \sqrt{6}},$ then $t^3+15 t-70=0.$
$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}$
$=\sqrt[5]{1782+405t}-t$
Since $(t+2)^5-(1782+405t)=(t+5)^2 \left(t^3+15 t-70\right)=0$, we get $\sqrt[5]{1782+405t}-t=2.$ |
H: Countably many worlds and Universal Sentence.
This is a naive, kind of informal argument.
Suppose we have a language with just one predicate $P$ and constants $a_{1}, a_{2}, a_{3}$ and so on. Suppose also that we have countably many worlds $1, 2, 3$ and so on.
Without loss of generality, we assume that $Pa_{1}$ holds in all the worlds.
That is, the set of worlds in which $Pa_{1}$ holds is $N$.
Now, $Pa_{1}\&Pa_{2}$ will hold in fewer or at most the same set of worlds.
For purpose of argument ,we take that it holds in fewer worlds. So the
Set $S_{2}$ of worlds in which $Pa_{1}\&Pa_{2}$ holds is a strict subset of $N$.
We continue this process, assuming that strict inclusion holds at every stage.
So we have a nested series of subsets $S_{i}$ of $N$, such that
$Pa_{1}\&Pa_{2}\&\dotsb\&Pa_{i}$ holds in $S_{i}$.
Clearly, the set of worlds in which the universal sentence ($\forall nPn$?) AnPn holds
Is the intersection of all the nested sets $S_{i}$ which is the null set.
That is , there is no world in which the universal sentence is true.
Alternately, we have to give up the assumption of strict inclusion of
The successive sets $S_i$, so that
There is some $k$, such that for all $i$ greater than $k$, $S_i =S_k$, which means
That $Pa_i$ for $i > k$ is true in all worlds in which $Pa_1\&Pa_2\&\dotsb\&P_k$ is true.
Now my question is ( assuming improbably that there is no flaw in the argument above)
Why cannot this argument go through formally? Is it something to do with
the Lowenheim-Skolem theorem ?
AI: There is a flaw in the argument. Consider the following sets $S_1,S_2, \dots$.
$S_1$ is all of $\mathbb{N}$.
$S_2$ is all of $\mathbb{N}$ except $1$.
$S_3$ is all of $\mathbb{N}$ except $1$ and $3$.
$S_4$ is all of $\mathbb{N}$ except $1$, $3$, and $5$.
And so on.
We have strict inclusion at all stages.
The intersection of the $S_i$ is the set of even numbers. It is very much non-empty. |
H: A bounded sequence has a convergent subsequence
Let ${a_n}$ be a bounded sequence of real numbers. Prove that ${a_n}$ has a subsequence that converges to lim sup $a_n$.
Thanks!
AI: $\limsup a_n$ is, by definition, the largest limit point of the sequence $a_n$. When you work with Real numbers (or, more generally, when you work with 1st-countable spaces), for every limit point, $a$ , there is a sequence of points that converge to $a$. |
H: Proof of an infinite sum involving cosines
I have a homework problem that requires proving the Poisson formula from the Jensen-Poisson formula. I have all steps of the problem complete except for proving the identity (for $r < 1$) $$1 + \sum_{n=1}^\infty 2r^n \cos (n(\phi - \theta)) = \frac{1-r^2}{1+r^2 - 2r \cos (\phi-\theta)},$$
on which I'm not sure how to get started.
AI: Use the identity
$$ \cos(t) = \frac{e^{it}+e^{-it}}{2} $$
and the geometric series
$$ \sum_{k=0}^{\infty} t^k = \frac{1}{1-t}. $$ |
H: Markov chains: simple random walk $S_n$
In the case of a simple random walk $\{S_n, n \ge 0\}$ what is $S_n$. I see this for
$P\{S_n=i |\ |S_n|=i_{n-1},...,|S_1|=i_1 \} = \frac{p^i}{p^i+q^i}$
What does this mean? is it: the probability of going to state i after n steps (not sure about that) is $\frac{p^i}{p^i+q^i}$ , which is independent of n.
AI: A random walk, in the context of Markov chains, is often defined as $S_n = \sum_{k=1}^n X_k$ where $X_i$'s are usually independent identically distributed random variables. My understanding of your given statement is the probability of the summation $S_n$ reaching value $i$ given all its previous history. |
H: How to prove that if $-1
I am trying to prove an equivalence.
I have already proved that:
$$x^2 + x < 0 \implies -1 < x < 0 $$
using a sub-proof by cases, in which I used the fact that when $xy < 0$, $x$ and $y$ have opposite signs. So I just factored the expression and went from there.
However, I can't seem to find an appropriate way of proving the second implication?
$$-1 < x < 0 \implies x^2 + x < 0$$
Answer:
I solved it by determining that x + 1 > 0, then since x < 0, (x)( x + 1 ) < 0 ,
then $x^2$ + x < 0
AI: Hint: $(x+1)>0$, and multiplying an inequality by a positive number doesn't change the inequality arrows. |
H: Integral of exponential with second degree exponent
I want to compute the integral $$\int_\mathbb{R}e^{-t\left(y-\dfrac{(at+x)i}{2t}\right)^2}dy$$
I know that $\int_\mathbb{R}e^{-ty^2}dy=\sqrt{\pi/t}$, but here there is an extra imaginary factor. What can I do?
AI: Integrate around a contour. Start at $-R$. Go in a straight line to $R$. Then go to $R + bi$ (let me use $b$ for the imaginary bit), then go to $-R+bi$, then finally go back to $-R$. Show that the integral of the bits from $R$ to $R+bi$ and $-R+bi$ to $-R$ converge to zero as $R \to +\infty$. You know the integral around the contour is zero. The integral along one of the horizontal lines is the integral you want, and the integral along the other horizontal line is the integral you know how to do. And so they must equal each other. |
H: Integration of exponential with square
It is known that $\int_\mathbb{R}e^{-tx^2}dx=\sqrt{\pi/t}$. What about $\int_\mathbb{R}e^{-t(x+ai)^2}dx$ for $a\in\mathbb{R}$? Is it still also $\sqrt{\pi/t}$?
I can't simply change the variable $y=x+ai$, because then the domain of integration would also change.
AI: Yes it is. Consider the following contour integral in the complex plane:
$$\oint_C dz \: e^{-t z^2}$$
where $C$ is a rectangle having vertices at $z=-R$, $z=R$, $z=R + i a$, and $z=-R + i a$. The contour integral is then
$$\int_{-R}^R dx \, e^{-t x^2} + i \int_0^a dy \, e^{-t (R+i y)^2} \\ + \int_R^{-R} dx \, e^{-t (x+i a)^2} + i \int_a^0 dy \, e^{-t (-R+i y)^2}$$
It should be clear that the second and fourth integrals vanish as $R \to \infty$. Further, by Cauchy's integral theorem, the contour integral is zero as there are no poles inside $C$. Therefore
$$\int_{-\infty}^{\infty} dx \, e^{-t (x+i a)^2} = \int_{-\infty}^{\infty} dx \, e^{-t x^2}$$
as was to be shown. |
H: How do you compute the Fourier Transform of this Unit-Impulse Function?
I have been given this problem from a textbook (not homework, trying to study for an exam. The goal is to find the Fourier transform of this function.
$\sum_{k=0}^\infty a^k*\delta(t-kT), |a|<1$
Can anyone give me a hint or point me in the right direction of how to compute the Fourier Transform? Thanks!
AI: The FT of each individual impulse is
$$\int_{-\infty}^{\infty} dt \, \delta(t-k T) \, e^{i \omega t} = e^{i k \omega T}$$
so that the FT of the sum is a geometric series:
$$\sum_{k=0}^{\infty} \left ( a \, e^{i \omega T}\right )^k = \frac{1}{1-a \, e^{i \omega T}}$$ |
H: Why are projective modules cohomologically trivial?
Let $G$ be a finite group, $H\subset G$ a subgroup, $k$ a commutative ring, $M$ a $kG$-module, $n\in\mathbb{Z}$, and $\hat{H}\,^n(H,M)$ the $n$th Tate cohomology group as defined in this question, where we consider $M$ a $kH$-module via $kH\hookrightarrow kG$.
We say that $M$ is cohomologically trivial if $\hat{H}\,^n(H,M)=0$ for all $n\in\mathbb{Z}$ and all subgroups $H\subset G$. Using Shapiro's lemma and the double coset formula, I can see that free $kG$-modules are cohomologically trivial. I'd like to use this to conclude that projective $kG$-modules are also cohomologically trivial.
If $N$ is projective, then we can find $N'$ such that $N\oplus N'=M$ is free, and hence cohomologically trivial. Then, the injection of $kG$-modules $N\hookrightarrow M$ gives a map of cohomology groups
$$\hat{H}\,^n(H,N)\to\hat{H}\,^n(H,M)=0$$
but unless this map is injective, we can't conclude that $N$ is cohomologically trivial. Is this map injective, and if not, how can I show that projective modules are cohomologically trivial? Perhaps I need to use the fact that $N$ is a direct summand of $M$, and not just a submodule.
AI: The inclusion of $N$ into $M$ has a section, that is, a map $M\to N$ whose composition with $N\to M$ is the identity of $N$. This and functoriality does what you want. |
H: Stationary points
I've a question related to stationary point of inflection. A function
$$ f(x)=ax^5+bx^3+cx $$ has stationary points at $ (-2, 64), (2,-64)~and~(0,0). $ find the values of a, b and c.
I found the first derivative and equate it to 0. I'm getting same equation when I equate $f'(x)$ to $0$ for $f(2)$ and $f(-2)$. For $f(0)$ I found $c=0$ that is correct.
Answer provided tells me $a=3$ and $b=-20$.
Please help how I can find this value of $a$ and b .
Thanks,
Arif
AI: You have found $c=0$, so we won't deal with that part.
We are told that $f(2)=-64$. From this we conclude that
$$(a)2^5+(b)2^3=-64.$$
The point $(-2,64)$ gives no additional information.
We have $f'(2)=0$. Since $f'(x)=5ax^4+3bx^2$, we have
$$(5a)2^4 +(3b)2^2=0.$$
We have $2$ linear equations in $2$ unknowns. Solve. The solving is easier if we note that the first equation is equivalent to $4a+b=-8$ and the second is equivalent to $20a+3b=0$. |
H: The residue at $\infty$
I am stuck on the following problem :
$\,\,\,\,$*Problem*$\quad$The residue of an entire function at $\infty$ is $0$.
Solution: True. This follows from the definition of the residue at $\infty$ together with the Cauchy-Goursat Theorem. Another way to see this is to take the Taylor expansion for $f$ at $0$ $\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n$ and replace $z$ with $1/z$ to get the Laurent expansion for $f(1/z)$ at $0$: $\displaystyle f(z)=\sum_{n=-\infty}^0a_{-n}z^n.$ Then we have that the residue at $\infty$ is given by the negative coefficient of the $1/z$ term of
$$
\dfrac1{z^2}f\left(\dfrac1z\right)=\sum_{n=-\infty}^{-2}a_{-n-2}z^n,
$$
which is clearly $0$.
I am having trouble to understand the last few lines (in italic) in the given solution. Can someone give lucid explanation? Thanks and regards to all.
AI: The thing is that functions do not have residues, but rather differentials have residues. This is something which can be quite confusing in a first complex analysis class. The "residue of a function" is not invariant under a change of local parameter, but the residue of a differential is. For this reason, what is usually called the "residue at $0$ of $f(z)$" is actually the residue at $0$ of $f(z)dz$.
When you change the coordinate from $z$ to $w=1/z$, the differential $dz$ is transformed into $-dw/w^2$, which explains the change of sign and the extra factor. Thus,
$$f(z)dz = \frac{-1}{w^2} f(1/w) dw.$$
The "residue of $f$ at $\infty$" is the residue at $0$ of $\frac{-1}{w^2} f(1/w) dw$. |
H: Understanding dimension of matrix
Consider the vector space $M_{n\times n}(\mathbb F)$, the set for all $n\times n$ matrices. The basis for this vector space has dimension $n^2$. I read somewhere that the basis has $n$ matrices that have nonzero diagonal entries. I am trying to understand this intuitively, can anyone help clarify why there are $n$ of these elements?
Thanks.
AI: There is not just one single basis for $M_{n\times n}(\mathbb F)$, nor any distinguished one. There are infinitely many such bases. The one usually referred to as the standard basis consists of the matrices $E_{i,j}$ where $1\le i,j\le n$ and $E_{i,j}$ has $1$ at position $(i,j)$ and $0$ elsewhere. For this particular basis it is then obvious that precisely $n$ of these matrices have non-zero diagonal entries, namely the matrices $E_{i,i}$.
But, as said, there are plenty of other bases. In particular, there are bases in which every basis element has non-zero diagonal entries. One thing that can be said in general is that any basis must contain, for every position $(i,j)$, at least one matrix with non-zero $(i,j)$ entry. |
H: closed form for $\int_0^{\infty}\log^n\left(\frac{e^x}{e^x-1}\right)dx$
How can I find a closed form for
$$\int_0^{\infty}\log^n\left(\frac{e^x}{e^x-1}\right)dx, n\in\mathbb{N}$$
AI: Substitute $u = -\log{(1-e^{-x})}$, then after a little algebra, you will find that $dx = -du/(e^u-1)$, and the integral becomes
$$\int_0^{\infty} du \frac{u^n e^{-u}}{1-e^{-u}}$$
This is a well known integral that has value
$$\Gamma(n+1) \zeta(n+1)$$
This may be shown by Taylor expanding the denominator to get
$$\sum_{k=0}^{\infty} \int_0^{\infty} du \, u^n \, e^{-(k+1) u} = n!\sum_{k=0}^{\infty}\frac{1}{(k+1)^{n+1}}$$
It should be noted in passing that restriction of $n$ to integers is not necessary and $n$ may be a real, or even a complex number so long as the integral converges. |
H: How to find $x$ in $\frac{x}{2}-\frac{1}{4{x^2}}=1$
I have the following equation: $\frac{x}{2}-\frac{1}{4{x^2}}=1$ and trying to find $x$ value.
I wasn't able to proceed after having common denominator $\frac{2{x^3}-1}{4x^2}=1$
AI: You now have a cubic as (cross multiply by $4x^2$ and then subtract it from both sides):
$$2x^3 - 4x^2 -1 =0$$
Do you know how to solve a cubic?
You get one real and two complex solutions.
Spoiler (Hover if you dare)
$x = 2.11208493554430$ |
H: Primitive Root question
Question:
Show that if $m$ is a positive integer and $a$ is an integer relatively prime to $m$ such that $ord_{m}a = m-1$, then $m$ is prime.
So if you could give me guidance and explanations of answers that would be great.
The textbook answer key this question is from does it by a proof by contradiction assuming m is not prime.
But the steps don't really logically follow, so I would appreciate other methods or clear explanation of steps. Thanks.
AI: By Euler's Theorem, we have $a^{\varphi(m)}\equiv 1\pmod{m}$.
It follows that $a$ has order $\le \varphi(m)$.
We will show that if $m$ is composite, then $\varphi(m)\le m-2$, so $a$ cannot have order $m-1$.
If $m$ is composite, then $m$ has a factor $d$ with $1\lt d \lt m$. Then $0$ and $d$ are not relatively prime to $m$. So there are at least $2$ integers in the interval $[0,m-1]$ which are not relatively prime to $m$. It follows that $\varphi(m)\le m-2$. |
H: Let $L/K$ be a Galois extension with $Gal(L/K)=A_{4}$. Prove that there is no intermediate subfield $M$ of $L/K$ such that $[M:K]=2$.
Let $L/K$ be a Galois extension with $Gal(L/K)=A_{4}$. Prove that there is no intermediate subfield $M$ of $L/K$ such that $[M:K]=2$.
Please tell me a hint. Thanks a lot.
AI: Such an intermediate extension corresponds to a subgroup of what index in $A_4$? Would this conflict with any properties that $A_4$ has, something about lack of solvability subgroups of index $2$ maybe. |
H: Getting better at proving continuity
How can I get better at distinguishing between continuity and uniform continuity?
If I were given an exam question asking me to prove if the function is uniformly continuous or continuous then I won't be able to answer it correctly. If it were asked explicitly, then I might not have as much difficulty, for example, prove $x^2$ is continuous on $[0,\infty)$. But if it asked, is the function $x^2$ uniformly continuous on $\mathbb{R}$ then I might have just said yes and proved continuity instead even though it isn't uniformly continuous.
For continuity I know I must prove in the following matter: $$\forall \epsilon > 0, \forall x \in X, \exists \delta > 0 : |x - y| < \delta \implies |f(x) - f(y)| < \epsilon.$$
For uniform continuity:
$$\forall \epsilon > 0, \exists \delta > 0, \forall x \in X : |x - y| < \delta \implies |f(x) - f(y)| < \epsilon$$
But I still don't understand how to distinguish a function from being uniformly continuous or continuous if it doesn't tell me explicitly.
AI: The following properties might help you build some intuitions.
(1) Any continuous function defined on a closed and bounded interval $[a, b]$ must be uniformly continuous.
(2) If $f$ is a continuous function on $(a, b)$. then $f$ is uniformly continuous if and only if $f$ can be extended to a continuous function on $[a, b]$.
(3) If $f$ has the properties that $|f'|$ is bounded, then it is uniformly continuous. (on any interval)
(4) More generally, if $f$ satisfies $|f(x) - f(y)| \leq C |x-y|^\alpha$ for all $x, y$, for some $\alpha, C>0$, then $f$ is uniformly continuous.
(5) Let $ f$ be defined and continuous on $(a, b)$. If $f$ is uniformly continuous on $(a, c]$ and $[c, b)$ respectively, then it is uniformly continuous on $(a, b)$. |
H: Polynomial ring with uncountable indeterminates
In Rotman's Advanced Modern Algebra second edition (2010), on page 883 (or on page 905 in its first edition (2002)), in the proof of the existence of localization of a commutative ring $R$ on its multiplicative subset $S$, he writes:
"Let $X=(x_{s})_{s\in S}$ be an indexed set with $x_{s}\mapsto s$ a bijection $X \rightarrow S$, and let $R[X]$ be the polynomial ring over R with indeterminates $X$."
However in his definition of formal power series over $R$ he comments:
"To determine when two formal power series are equal, let us recognize that a sequence $\sigma$ is really a function $\sigma:\mathbb{N} \rightarrow R$, where $\mathbb{N}$ is the set of natural numbers, with $\sigma(i) = s_{i}$ for all $i \geq 0$."
So I want to ask if $S$ is uncountable, then is it still legitimate that he defines $R[X]$ in this way?
AI: If $X$ is an arbitrary set, then $R[X]$ is defined to be the free commutative $R$-algebra over $X$. It is also known as the polynomial ring (better would be polynomial algebra) with variables $X$ and coefficients $R$. You can construct it as the symmetric algebra over the free module over $X$, or equivalently as the monoid algebra over the free commutative monoid over $X$. Explicitly, it consists of maps $\sigma : \mathbb{N}^{(X)} \to R$ with finite support, where we think of this as the polynomial $\sum_{\alpha \in \mathbb{N}^{(X)}} \sigma(\alpha) \cdot X^{\alpha}$. The formal power series ring $R[[X]]$ is the $(X)$-adic completion of $R[X]$. Its elements are maps $\sigma : \mathbb{N}^{(X)} \to R$ such that for every $d \in \mathbb{N}$ the set $\{\alpha \in \mathbb{N}^{(X)} : \sum_x \alpha_x = d,~ \sigma(\alpha) \neq 0\}$ (monomials of degree $d$ appearing in the power series) is finite. Of course, nothing has to be assumed countable here, and your quote refers to the single variable case $X=\{x\}$. |
H: Find the four digit number?
Find a four digit number which is an exact square such that the first two digits are the same and also its last two digits are also the same.
AI: HINT:
So, we have $$1000a+100a+10b+b=11(100a+b)$$
$\implies 100a+b$ must be divisible by $11\implies 11|(a+b)$ as $100\equiv1\pmod{99}$
As $0\le a,b\le 9, 0\le a+b\le 18\implies a+b=11$
$$\implies11(100a+b)=11(100a+11-a)=11^2(9a+1)$$
So, $9a+1$ must be perfect square |
H: Evaluating the integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$
How can I evaluate the following integral?
$$
\int_0^{\pi/2}
\log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\,
\frac{1}{\cos\left(x\right)}\,{\rm d}x\,,
\qquad\left\vert\,a\,\right\vert \le 1$$
I tried differentiating under the integral with respect to the parameter $a$, and I also tried expanding the log term in a Taylor series and then switching the order of integration and summation. I ran into difficulties with both approaches.
AI: Use the expansion for $|z| < 1$
$$\log{\left ( \frac{1+z}{1-z}\right )} = 2 \sum_{k=0}^{\infty} \frac{z^{2 k+1}}{2 k+1}$$
Then the integral is equal to
$$2 \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \int_0^{\pi/2} dx \, \cos^{2 k}{x}$$
It is straightforward to show that
$$\int_0^{\pi/2} dx \, \cos^{2 k}{x} = \frac{1}{2^{2 k}} \binom{2 k}{k} \frac{\pi}{2}$$
Thus the integral $I(a)$ is
$$I(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k+1}}{2 k+1} \frac{1}{2^{2 k}} \binom{2 k}{k}$$
We may evaluate this sum by considering
$$I'(a) = \pi \sum_{k=0}^{\infty} \frac{a^{2 k}}{2^{2 k}} \binom{2 k}{k} = \pi \left (1-a^2\right)^{-1/2}$$
Integrating with respect to $a$ and noting that $I(0)=0$, we find that
$$I(a) = \pi \arcsin{a}$$ |
H: if matrix such $AA^T=A^2$ then $A$ is symmetric?
let matrix $A_{n\times n}$ is real matrix,such $AA^T=A^2$, The transpose of matrix $A$ is written $A^T$,
show that :
the matrix $A$ is Symmetric matrices
maybe this problem have more methos,because it is know that if matrix $A$ is symmetric,then we have
$AA^T=A^2$,But for my problem,I can't prove it.Thank you
AI: Although not an exact duplicate, the method for solving An equivalent condition for a real matrix to be skew-symmetric applies. Briefly speaking, for real square matrices, $\langle X,Y\rangle = \operatorname{trace}(Y^TX)$ defines an inner product and symmetric matrices are orthogonal to skew-symmetric matrices. Now, write $A=H+K$, where $H=\frac12(A+A^T)$ is the symmetric part and $K=\frac12(A-A^T)$ is the skew-symmetric part. Then $AA^T=A^2$ implies that $(H+K)K=0$ and in turn $HK=-K^2=K^TK$. Taking trace on both sides, we get $\langle K,H\rangle=\langle K,K\rangle$. Since $\langle K,H\rangle=0$, it follows that $\langle K,K\rangle=0$ and $K=0$, i.e. $A$ is symmetric. |
H: How to solve gamma function integral for 4!
I have been trying to test my knowledge of the gamma function by calculating 1! and 4!. I got the right result for 1! but I cannot get 4! analytically.
To be more specific, I think I am not integrating the gamma function correctly for 4!
I have $4! = \int_0^\infty e^{-t}t^{z-1}dt$ with $z=4$. How can you do it?
Do I have to integrate by parts?
Thanks
AI: Using integration by parts, we see that the gamma function satisfies the functional equation $\Gamma(z+1)=z\Gamma(z)$ and so $\Gamma(n+1)=n!$. So you can do it for $\Gamma(5)$.
ATTENTION $4!=\Gamma(5)$.
$$\small
\Gamma(z+1)=\int_0^\infty\operatorname{e}^{-t}t^{z}\operatorname{d}t=\underbrace{\left[-\operatorname{e}^{-t}t^{z}\right]_0^\infty}_0-\int_0^\infty\left(-\operatorname{e}^{-t}\right)\left(zt^{z-1}\right)\operatorname{d}t=z\underbrace{\int_0^\infty\operatorname{e}^{-t}t^{z-1}\operatorname{d}t}_{\Gamma(z)}=z\Gamma(z)
$$ |
H: Is there a way to express matematically that B is closer to A than C is to A?
Is there a way to express mathematically that B is closer to A than C is to A?
Eg. 3 is closer to 1 than 4 is to 1
or -1 is closer to 3 than 10 is to 3
AI: If you accept the possibility that the distance could be the same:
$|A-B|\le |C-A|$ otherwise, $|A-B|\lt |C-A|$ |
H: Positive integral solutions of $3^x+4^y=5^z$
Are there more integral solutions for $3^x+4^y=5^z$, than $x=y=z=2$ ?
If not, how do I show that? I could show that for $3^x+4^x=5^x$, but I'm stuck at the general case? Any ideas, maybe graphs, logarithms or infinite descent?
AI: I will prove that the only positive integral solution to $3^x+4^y=5^z$ is $x=y=z=2$.
Proof. Looking at the equation mod $4$, we see $3^x\equiv 1\pmod{4}$, or equivalently, $(-1)^x\equiv 1\pmod{4}$. This implies $x=2x_1$ for some integer $x_1$. Also, looking at the equation mod 3, we see $5^z\equiv 1 \pmod{3}$, or equivalently $(-1)^{z}\equiv 1\pmod{3}$. This implies $z=2z_1$ for some integer $z_1$. Thus,
$$
2^{2y}=4^{y}=(5^{z_1})^2-(3^{x_1})^2=(5^{z_1}+3^{x_1})(5^{z_1}-3^{x_1})
$$
Hence, $5^{z_1}+3^{x_1}=2^{s}$ and $5^{z_1}-3^{x_1}=2^{t}$, with $s>t$ and $s+t=2y$. Solving for $5^{z_1}$ and $3^{x_1}$, we get
$$
5^{z_1}=2^{t-1}(2^{s-t}+1) \ \ \textrm{ and } \ \ 3^{x_1}=2^{t-1}(2^{s-t}-1)
$$
Since the left side of both equalities is odd, $t$ must be equal to $1$. Let $u=s-t$. Then, the equation $3^{x_1}=2^{t-1}(2^{s-t}-1)$ becomes $3^{x_1}=2^{u}-1$. Looking at this equation mod $3$, we get
$0\equiv (-1)^{u}-1\pmod{3}$, and so $u$ is even, say $u=2u_1$ for some positive integer $u_1$.
Thus,
$$
3^{x_1}=(2^{u_1})^{2}-1=(2^{u_1}+1)(2^{u_1}-1)
$$
Hence, $2^{u_1}+1=3^{\alpha}$ and $2^{u_1}-1=3^{\beta}$ for some $\alpha>\beta$. But this gives, $3^{\alpha}-3^{\beta}=2$, and hence $\alpha=1$ and $\beta=0$. Consequently, $u_1=1$, and so $u=2$. This gives us the unique solution $x=y=z=2$. |
H: Prove or disapprove a propositions
Let p,q and r be three propositions. Prove or disapprove
$(p\to q) \land (q \iff r) \land (p \lor \lnot (\lnot q \lor \lnot r) \equiv p \land q \land r$
so, the way i do is
LHS = $(\lnot p\lor q) \land (\lnot q \lor r) \land (\lnot r \lor q) \land (p \lor q) \land (p \lor r)$
so what should I do next? Or am I completely wrong?
AI: The statement is false. Suppose q and r are true but p is false. The left side will be true and the right side will be false. |
H: Evaluate $\int_0^4 |\sqrt{x} - 1|~dx$
I am working on a problem set involving indefinite integrals. Currently stuck in this question:
$\int_0^4 |\sqrt{x} - 1|~dx$
I tried the following substitution:
Let $u=\sqrt{x}$ so, $du=\dfrac{dx}{2\sqrt{x}}$when $x= 4 \Rightarrow u=2$ and $x=0 \Rightarrow u=0$
This is where I am stuck. It seems that my substitution fails me.
Also, since it is an absolute value, I tried to express it in piecewise function. However, I can't possibly make sense out of it.
Can anyone help me?
PS. I worked out the previous item before this, however, I do not know if I did it correctly:
$\int_0^3 |x^2-4|~dx = \int_0^2 4-x^2~dx +\int_2^3 x^2-4~dx= 3$
Is this correct? Thanks
AI: Observe that between [0,1), $\sqrt x -1 \lt 0$. Hence |$\sqrt x -1$|=-$(\sqrt x -1)$.
Therefore $$\int_0^4 |\sqrt{x} - 1|~dx=-\int_0^1 (\sqrt{x} - 1)~dx+\int_1^4(\sqrt{x} - 1)~dx$$ |
H: About stabilizer in group action
Let $X$ be a finite set and $x$ is an element of $X$. Let $G_x$, the stabilizer subgroup, be the subset of $S_X$ consisting of permutations that fix $x$.
The question is
Is stabilizer always a normal subgroup?
I know that $G_x=\{g\in G\mid gx=x, x \in X\}$ and I have proved that stabilizer is a subgroup of $G$. But I got stuck at the point proving stabilizer is always normal.
I have found that 'if the group action is transitive, then the stabilizer is normal'.
So, I came to a conclusion that stabilizers are not always normal. But I cannot understand this. Is there some good explanation on this (stabilizer is not always normal) with examples?
Thank you.
AI: Take the action of $S_3$ on itself by conjugation and $x=(1\ 2)$. Then the stabilizer of $x$ is $\{e, x\}$ and this is not a normal subgroup of $S_3$. |
H: Convert 4-sat to 3-sat
I want to know in general how can I convert $4-SAT$ to 3-SAT.
And I have a specific case that if you can help me optimize it to 3-SAT it will be greate.
I want to do this so I be able to use sat solvers programs.
$(C \lor A \lor D) \land (C \lor B \lor D) \land (\lnot C \lor A \lor \lnot D) \land (\lnot C \lor B \lor \lnot D) \land (C \lor \lnot A \lor \lnot B \lor \lnot D) \land (\lnot C \lor \lnot A \lor \lnot B \lor D)$
AI: Every clause is of the form $(x_1 \vee x_2 \vee x_3 \vee x_4)$, where $x_i$, $i \in [4]$ is a literal. Replace every such clause with two clauses $(x_1 \vee x_2 \vee z) \wedge (\neg z \vee x_3 \vee x_4)$, where $z$ is a fresh variable. You can then verify that if some setting of $x_i$'s satisfies the original clause, you can find a setting of $z$ such that the two clauses are satisfied.
However, note that if you want to use a SAT solver, none of this is needed. A standard SAT solver can handle clauses of any length, and different clauses can be of different length. |
H: Names for certain numbers.
I am wondering if there is names for numbers with the following characteristics:
Numbers that end with 0.
Numbers divisible by 5.
If there are names for numbers with similar characteristics, I would be happy to learn about them as well. :)
Update:
10, 20, 30, ... is called?
5, 10, 15, 20, 25, 30, ... is called?
AI: Respectively:
(1) Positive multiples of $10$.
(2) Positive multiples of $5$.
If they satisfy both these properties, then they're positive multiples of $10$.
I don't think there are any other names for them, although they each form an arithmetic progression, which is a sequence of natural numbers. If you want to get really fancy, I suppose you could say, respectively, that they're all the numbers greater than zero that are:
(1) zero modulo $10$
(2) zero modulo $5$. |
H: $\iint_V |y-x^{2}| \operatorname{d}x \operatorname{d}y$ with $V = [-1,1] \times [0,2]$
it's especially difficult because i don't understand how to integrate absolute value terms.
I only know that if you function, say $x^{2}-1$, is below the $x$-axis i need to integrate $1-x^2$ between the interval $[-1,1]$.
But in this case i got two variables which quite confuse me....
Can someone give me tips/hints/help with this problem?
Thanks!
AI: For every $x$ in $[-1,1]$,
$$I(x)=\int_0^2|y-x^2|\mathrm dy=\int_0^{x^2}(x^2-y)\mathrm dy+\int_{x^2}^2(y-x^2)\mathrm dy$$
hence
$$
I(x)=\left[x^2y-\tfrac12y^2\right]_0^{x^2}+\left[\tfrac12y^2-x^2y\right]_{x^2}^2=\ldots$$ |
H: A question about weighted forward unilateral shift operators
We define
$$
B(x_{1}, x_{2},...)=(0, \frac{x_{1}}{2}, \frac{x_{2}}{3},...,x_{n})\in l^{2}(N),
$$
How could be shown that that $B$ is a quasinilpotent?
AI: 1) By induction show that
$$
B^n(x_1,x_2,\ldots)=\left(0,0,\ldots,0,\frac{x_1}{2\cdot\ldots\cdot (n+1)},\ldots,\frac{x_k}{(k+1)\cdot\ldots\cdot(n+k)},\ldots\right)
$$
2) Then derive that
$$
\Vert B^n\Vert\leq\frac{1}{(n+1)!}
$$
3) Using Stirling's formula show that spectral radius
$$
r(B)=\lim\limits_{n\to\infty}\Vert B^n\Vert^{1/n}=0
$$
4) Recall that spectrum $\sigma(A)$ of any bounded operator $A$ satisfies
$$
\varnothing\neq\sigma(A)\subset\{\lambda\in\mathbb{C}:|\lambda|\leq r(A)\}
$$
hence $\sigma(B)=\{0\}$. This means by definition that $B$ is quasinilpotent. |
H: Open sets in product topology
For any two topological spaces $X$ and $Y$, consider $X \times Y$. Is it always true that open sets in $X \times Y$ are of the forms $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$?
I think is no. Consider $\mathbb{R}^2$. Note that open ball is an open set in $\mathbb{R}^2$ but it cannot be obtained from the product of two open intervals.
AI: Nice done. The forms $U\times V$ is a base for the topology. That means that every open set in $X\times Y$ is a union of elements of the form $U\times V$. |
H: Hard contest type trigonometry proof
Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
AI: note
$$\cos{(x+y)}=\cos{[(x+y+z)-z]}=\cos{(x+y+z)}\cos{z}+\sin{(x+y+z)}\sin{z}$$
and
$$\cos{(y+z)}=\cos{(x+y+z)}\cos{x}+\sin{(x+y+z)}\sin{x}$$
$$\cos{(z+x)}=\cos{(x+y+z)}\cos{y}+\sin{(x+y+z)}\sin{y}$$
add this three
\begin{align*}
&\cos{(x+y)}+\cos{(y+z)}+\cos{(x+z)}\\
&=(\cos{x}+\cos{y}+\cos{z})\cos{(x+y+z)}+(\sin{x}+\sin{y}+\sin{z})\sin{(x+y+z)}\\
&=p\cos^2{(x+y+z)}+p\sin^2{(x+y+z)}\\
&=p
\end{align*} |
H: Is something similar to Robin's theorem known for possible exceptions to Lagarias' inequality?
Robin's theorem says that if
$$\sigma(n)<e^\gamma n\log\log n$$
holds for all $n>5040$, where $\sigma(n)$ is the sum of divisors of $n$, then the Riemann hypothesis is true, but if there are any counterexamples, then they are colossally abundant numbers, and there are infinitely many counterexamples.
Lagarias' theorem says
$$\sigma(n)\leq H_n+e^{H_n}\log(H_n)$$
holding for all natural numbers $n$ is equivalent to the Riemann hypothesis, where $H_n$ is the $n$th harmonic number, the sum of the reciprocals of the first $n$ positive integers. It looks like Lagarias' inequality is sharper than Robin's, so any counterexamples to Robin's inequality must also be counterexamples to Lagarias'. Is known to be impossible for integers that aren't colossally abundant numbers or of some similar more general type to be exceptions to Lagarias' inequality?
AI: Lagarias wrote, ""Robin showed that, if the Riemann Hypothesis is false, then there will necessarily exist a counterexample to the inequality (1.2) that is a colossally abundant number; the same property can be established for counterexamples to (1.1). (There could potentially exist other counterexamples as well)." (1.2) is your first display, (1.1) is your second display. |
H: How to find out if a point lie in rectangle?
I have a rectangle in $2D$ space which is determined by $2$ points (each in opposite vertice) $p_1(x,y)$ and $p_2(x,y)$ . How can I find out numerically if a other point $p(x,y)$ is lying inside plane of the rectangle?
AI: Suppose
Left-Top vertex is $P_1(x_1,y_1)$, and
Right-Bottom vertex is $P_2(x_2,y_2)$
(in general, $2$ opposite vertices).
Point $P(x,y)$ is lying inside the rectangle, if
$$
\min\{x_1,x_2\} < x < \max\{x_1,x_2\}
$$
$$
and
$$
$$
\min\{y_1,y_2\} < y < \max\{y_1,y_2\}.
$$
If $x_1<x_2$, and $y_1<y_2$ (Left-Bottom and Right-Top vertices), then
$$
x_1<x<x_2
$$
$$
and
$$
$$
y_1<y<y_2.
$$
If edges of rectangle aren't parallel to coordinate axes, then $2$ points can't define unique rectangle (see image):
point $P$ belongs to blue rectangle, but not belongs to red rectangle.
Update (case of non-parallel to coordinate axes edges)
Points $P_1$ and $P_2$ determines diagonal of possible rectangle.
Other vertices are on the circle (circumference) with diameter $P_1P_2$:
So, if $P$ is inside the circle with diameter $P_1P_2$, then it is possible to build such rectangle, that $P$ belongs to rectangle.
Denote
radius of the circle: $R = \dfrac{1}{2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$;
center of the circle: $x_o = \frac{x_1+x_2}{2}$, $y_o = \frac{y_1+y_2}{2}$,
Condition of possibility to build such rectangle:
$$
(x-x_o)^2 + (y-y_o)^2 < R^2.
$$ |
H: If a subgroup of a symmetric group has an odd permutation, then it has a subgroup of index 2.
I want to show that for $n\geq 2$ and $H\leq S_n$ if $H$ contains an odd permutation then it necessarily has a subgroup of index 2. I am not sure how to start, if it has an odd permutation then it might not necessarily contain a transposition, perhaps it is possible to consider the group generated by odd permutation, but it can have many possible orders.
AI: Consider the restriction of $\operatorname{sign}$ to $H$. Since $H$ has an odd permutation, this restriction is still surjective. Hence, $\ker( \operatorname{sign}\mid H)$ has index $2$ in $H$. |
H: Is it possible to subtract a matrix from both side?
I have this equation $AX + B = I$ and I want to find Matrix $X$.
$$(A^{-1})AX + B = (A^{-1})I$$
$$X + B = (A^{-1})I$$
My question is, is it legal to do $X + B - B = (A^{-1})I - B$?
AI: If two matrices are equal, then their differences with $B$ must be equal. How could it be that when you subtract $B$ from the same matrix (equal matrices have equal entries) you get two different answers?
That said, if two matrices are equal, then their products with $A^{-1}$ on the left are equal, too, assuming the inverse and products are defined. But they may not be: not every matrix is invertible, and not every matrix product is defined, because of they might have the wrong dimensions.
However, as Riccardo.Alestra pointed out, your equal matrices are $AX+B$ and $I$, so when you multiply them by $A^{-1}$ on the left, you should have $A^{-1}\left(AX+B\right)=A^{-1}I$ which simplifies differently than what you wrote in your question. |
H: Homeomorphism between $D^n$ and $[0,1]^n$
I know that $D^n=\{x\in \mathbb{R}^n:\|x\|\leq 1\}$ is homeomorphic to $[0,1]^n$, but how to write down homeomorphism? How to find explicit formula?
Thanks in advance.
AI: Let $\| v \|_{p}$ denotes the $p$-norm of $v \in \Bbb{R}^{n}$. In particular,
$$ \| v \|_{2} = [ v_{1}^{2} + \cdots + v_{n}^{2} ]^{1/2} \quad \text{and} \quad \| v \|_{\infty} = \max \{ |v_{1}|, \cdots, |v_{n}| \}. $$
Then the following map
$$ F : D^{n} \to [-1, 1]^{n} : v \mapsto \frac{\|v\|_{2}}{\|v\|_{\infty}} v $$
gives the homeomorphism with the inverse
$$ G : [-1, 1]^{n} \to D^{n} : w \mapsto \frac{\|w\|_{\infty}}{\|w\|_{2}} w. $$
(Of course, we set $F(0) = 0 = G(0)$.)
The idea is simple:
$$ D^{n} = \{ \| v \|_{2} \leq 1 \} \quad \text{and} \quad [-1, 1]^{n} = \{ \| v\|_{\infty} \leq 1 \}. $$
So the map $F$ rescales the vector so that $\| F(v) \|_{\infty} = \| v \|_{2}$. Checking that both $F$ and $G$ are indeed continuous is not theoretically hard, though it may be somewhat cumbersome.
The following graph may help you what is actually going on in $n = 3$. The sphere (above) is mapped into the cube (below) by the mapping $F$. (The seams between the faces of the cube are software artifacts.) |
H: Finding -1 to irrational powers
I've been curious about $A=(-1)^r$ when $r$ takes different values. I know that if $r$ was an even number, $A$ will become $1$ and if it was odd, it will become $-1$. Now I want to know what will $A$ be if $r$ were an irrational number like $\sqrt{2}$.Here's what i have done: According to Euler's identity we have $e^{\pi i}=-1$ and so $e^{\pi i\sqrt{2}}=(-1)^\sqrt{2}$ and $e^{i(\sqrt{2}\pi)}=(-1)^\sqrt{2}$. We also know that $e^{i\theta}=\sin\theta+i\cos{\theta}$. According to what i just showed, $(-1)^\sqrt{2}=e^{i(\sqrt{2}\pi)}=\sin{\pi\sqrt{2}}+i\cos{\pi\sqrt{2}}$ and from this we can find the exact value of $(-1)^\sqrt{2}$. Is what I have done correct? Is $(-1)^\sqrt{2}$ even a number?
AI: I would prefer not to consider $(-1)^r$ as definable in any sense. Because for me, $(-1)^{p/q}$ is $\bigl(\sqrt[q]{-1}\bigr)^p$ or $\sqrt[q]{(-1)^p}$, and I get:
If $p,q$ are both odd, then both expressions are equal to $-1$, since $(-1)^q=-1$ therefore $\sqrt[q]{-1}=-1$.
If $p$ is even, then both are clearly equal to $1$.
If $q$ is even, then neither is well defined for me (it can be, but in a different way).
The problem here that the numbers $p/q$ with $q$ odd are dense in $\mathbb R_+$, therefore this exponential is quite badly discontinuous.
You can as well think that $(-1)^{p/q}$ is any (or all) of the roots of the polynomial $X^q=(-1)^p$, getting a similar thing as in the previous case, just with all the complex roots considered as well.
(All of these considerations strongly mimic the problem of $0^0$, which is considered to be $1$ by many people in combinatorics, however, we all know that it is not as simple as that.)
In the end, you can define $(-1)^r$ to be whatever you want, as long as you define it properly. However, as a reviewer of a paper, I would very likely reject such notation. |
H: Parseval equation for a Fourier series
Consider $f(x):=\lvert x\rvert, x\in [-\pi,\pi]$. Then the Fourier series is
$$
f(x)=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos((2n-1)x)}{(2n-1)^2}.
$$
Now my task is to write down the related Parseval equation.
The general Parseval equation is
$$
\frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)=\frac{1}{\pi}\int_{-\pi}^{\pi}\lvert f(x)\rvert^2\, dx,
$$
so here it is
$$
\frac{\pi^2}{2}+\frac{16}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\ dx.
$$
Is that the hole task or what?!
AI: You got
$$0=f(0)=\frac\pi2-\frac4\pi\sum_{n=1}^\infty\frac1{(2n-1)^2}$$
and as commented by Avitus you have a numerical value for the above series. Now check that
$$\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\implies\ldots$$ |
H: Argument over an induction proof
My friend gave me a problem.
Define a sequence $\langle a_n \rangle$ by the recurrence relation :$$ a_{n+2} - 6a_{n+1} + 8a_n = 0 $$ and
$a_1 = 4, a_2 = 8 $. Find the general term $a_n$ in closed form.
By inspection, I got the general term as $ a_n = 2^{n+1} $. Then I proved it using induction. However, he says that the proof isn't elegant!
He questions the source of my induction hypothesis: "how did you get that in the first place?".
He says that he is fine with the induction proof, but he isn't satisfied. He asks as to how I know that $a_n = 2^{n+1} $ is the only solution.
We had a heated argument over this matter. Now, what's wrong with what I did?!?
Btw, my proof:
Hypothesis: $P(n)$ = " $a_n = 2^{n+1} $ is the $n$*th* term of the given sequence."
Assumption: Assume that $ P(i),\; \forall\; i \in \{1,2,...k-1\} $ to be true.
To prove $P(k)$:
We know that $a_k = 6a_{k-1} - 8a_{k-2}$. Also, from our assumption, we have $a_{k-1} = 2^k$ and $a_{k-2} = 2^{k-1}$. Plugging that in the relation gives us $a_k = 2^{k+1}$.
Conclusion:
$P(k)$ is true if $ P(i),\; \forall\; i \in \{1,2,...k-1\} $. $P(1)$ is true. So $P(2)$ is true, and so on.
Therefore, $P(n)$ is true $\forall\; n \in \mathbb{N}$.
Hence proved.
AI: Except for minor formulation quibbles, your proof is a valid proof that $2^{n+1}$ is the only solution to the recurrence.
The minor quibble is that instead of making your induction hypothesis be "$a_n=2^{n+1}$ is the $n$th term of the given sequence", it would be more rigorous to make it "If $(a_k)_{k\ge 1}$ is any sequence that satisfies the recurrence, then $a_n=2^{n+1}$". Then what you have is a proof by induction on $n$ that every sequence that satisfies your recurrence is equal to the particular sequence $2^{n+1}$.
Strictly speaking you also need to prove that $2^{n+1}$ is a solution at all (because your induction proof assumes that you already have a sequence that solves the recurrence). But that proof is as simple as plugging $2^{n+1}$ into the recurrence relation and simplifying. No induction is needed there. |
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