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H: What is the Cardinality of the euclidean space $\mathbf{R}^N$?
We know that the set of real numbers is not countable by Cantor's proposition and hence higher dimensional Euclidean space is not countable too. However I couldn't find any result about the cardinality of $\mathbf{R}^N$? Is it $\aleph_{n}$ or it has the same cardinality of $\mathbf{R}$ which is $\aleph_1$?
AI: The cardinality of $\mathbb N$ is $\aleph_0$. The cardinality of $\mathbb R$ is definitely not $\aleph_0$. It is usually denoted $\mathfrak c$ (continuum) and coincides with the cardinality $2^{\aleph_0}$ of the Cantor set. Since $\aleph_0\times n=\aleph_0\times\aleph_0=\aleph_0$ for any $n$ with $0<n<\aleph_0$, we see that the cardinality of $\mathbb R^n$ and even the cardinality of $\mathbb R^{\mathbb N}$ is $\mathfrak c$ again.
The point is that $(2^{\aleph_0})^n=2^{\aleph_0\times n}=2^{\aleph_0}$, and similarly for $(2^{\aleph_0})^{\aleph_0}$.
By the way, that the Euclidean spaces all have size continuum is due to Cantor himself. His proof consisted of interleaving the decimal expansions of two reals to produce another real, thus giving an injection of $(0,1)\times(0,1)$ into $(0,1)$. This, and the circumstances surrounding this result, is discussed in detail in this nice paper by Fernando Q. Gouvêa.
The technique of forcing can be used to show that the usual axioms of set theory do not suffice to decide whether $2^{\aleph_0}$ is $\aleph_1$, $\aleph_2$, some other $\aleph_n$, or something else. (Cantor's famous Continuum Hypothesis is the claim that $2^{\aleph_0}=\aleph_1$. Again, both that this is true and that this is false are statements unprovable from the standard axioms.) |
H: Prove that $x^4+x^3+x^2+x+1 \mid x^{4n}+x^{3n}+x^{2n}+x^n+1$
Problem: Prove that $x^4+x^3+x^2+x+1$ divides $x^{4n}+x^{3n}+x^{2n}+x^n+1$ for all positive $n$ that are not multiples of $5$.
I'd like to get some pointers about how to solve this. No full solutions, just a nudge in the right direction. I've been working on this for a bit now and I'm not getting anywhere. Thanks.
AI: $$X^4+X^3+X^2+X+1=\frac{X^5-1}{X-1}$$
$$X^{4n}+X^{3n}+X^{2n}+X^n+1=\frac{X^{5n}-1}{X^n-1}$$
Now, use the fact that $X^5-1|X^{5n}-1$ and that for $n$ not divisible by $5$ we have
$$gcd(X^5-1,X^n-1)=X^{gcd(5,n)}-1=X-1$$
Alternately
If $1,\omega_1, \omega_2, \omega_3, \omega_4$ are the fifth roots of unity, prove that
$$P(\omega_1)=P(\omega_2)=P(\omega_3)=P(\omega_4)=0$$
where $P(X)=X^{4n}+X^{3n}+X^{2n}+X^n+1$. Then, $P(X)$ must be divisible by $(X-\omega_1)(X-\omega_2)(X-\omega_3)(X-\omega_4)$. |
H: A square integrable martingale has orthogonal increments
I am really stuck with the following exercise:
$(\Omega,\mathcal{F},(\mathcal{F}_n)_{n\ge1},P)$ a filtered probability space. Let $(X_n )_{n\ge1}$ be a sequence of square-integrable random variables. Define for every $n\ge1$,
$S_n := X_1 +\dots + X_n$ . Suppose that $(X_n )_{n\geq1}$ is such that $(S_n )_{n\geq1}$ is a martingale. Show that if $i \neq j$, then $E[X_i X_j ] = 0.$
So, from the definition of a martingale, we know that
$E[S_{n+1}|\mathcal{F}_n]=S_n, \forall n\in\mathbb{N}$, but I don't know how to use this to prove the claim.
Thanks for your help!
AI: All the conditional expectations make sense since $X_n$ is square integrable for each $n$.
Assume that $i\lt j$. Then $X_iX_j$ has the same expectation as $\mathbb E[X_iX_j\mid\mathcal F_i]$, which is equal to $X_i\mathbb E[X_j\mid\mathcal F_i]$.
Since $\mathcal F_i\subset \mathcal F_{j-1}$, we can conclude by the towering property that $E[X_j\mid\mathcal F_i]=0$. |
H: How to determine a $\Theta$-class of a Function
I have 6 functions that I have to determine which of 4 given $\Theta$-classes or neither of them.
Example of a function I have been given:
\begin{align*}
\textit{$f_1$}(n) =&(17\textit{n}+1) \\
\end{align*}
The $\Theta$-classes I have been given:
\begin{align*}
\Theta&(1), \\
\Theta&(\log\textit{n}), \\
\Theta&(\textit{$n^a$}), \\
\Theta&(\textit{$a^n$}), \\
or& none
\end{align*}
How do I go about this subject and determine the classes?
So far I believe that $f_1$(n) has $\Theta(\textit{$n^a$})$,
But I'm not entirely sure.
AI: A function $f$ is $f = \Theta(g)$, that is bounded below and above by $g$ asymptotically, if $k_1,k_2,n_0 \in \mathbb Z_{\ge0}$ exist, such that $$k_1 g(n) \le f(n) \le k_2 g(n) \forall n \ge n_0.$$
For your first two functions you get the following result:
\begin{align*}
n \le &(17n+1) &\le 18 n &\text{ for } n \ge 1 &\text{ therefore }f_1(n) = \Theta(n^1)\\
n^2 \le &(n^2+10n+1) &\le 20 n^2 &\text{ for } n \ge 1 &\text{ therefore }f_2(n) = \Theta(n^2)\\
\end{align*}
Now for $f_3$ notice that $n^{1000}$ has a large exponent, but the term $1.001^n$ will dominate it around $n_0 = 1.664 \cdot 10^7$. So overall we have
\begin{align*}
1.001^n \le n^{1000} + 1.001^n + 1000 \le 3 \cdot 1.001^n& \text{ for } n \ge 10^8 &\text{ therefore }f_3(n) = \Theta(1.001^n)
\end{align*}
Note that the constants $k_1$ and $k_2$ don't need to be sharp for the actual analysis, as long as the inequalities hold.
Now you should be able to answer the other functions' complexity classes. |
H: Roots Calculation Question
How does one calculate the roots: $$ \sqrt {57-40\sqrt{2}} - \sqrt {57+40\sqrt{2}}$$ manually?
Also, how can one determine which of the pair is bigger:
$ \sqrt {3} +\sqrt {5} $ vs. $ \sqrt {2} +\sqrt {6} $
also, by hand?
AI: $a^2+2ab+b^2 = (a+b)^2$, $a+2 \sqrt{ab} +b=(\sqrt{a}+\sqrt{b})^2$
Therefore because $57=32+25, 40\sqrt{2}=2\sqrt{800}=2\sqrt{32×25}$, the expression can be written $(4\sqrt{2}-5) -(4\sqrt{2}+5)$.
So answer is -10.
$(\sqrt{3}+\sqrt{5})^2=8+2\sqrt{15} > (\sqrt{2}+\sqrt{6})^2=8+2\sqrt{12}$
So, $\sqrt{3}+\sqrt{5} > \sqrt{2}+\sqrt{6}$ |
H: Prove that $ \left(1-\frac{1}{n}\right)^n > \frac{1}{6} $ for $n\geq 2$
Prove that $ \left(1-\frac{1}{n}\right)^n > \frac{1}{6} $ for $n \in \mathbb{N}$, $ n\ge 2$
Indeed, the affirmation is true even if $n$ is not a natural ($ n\geq 2 $ ) and we can prove it using calculus. But, this is part of a question about limit of sequences (that is why I fixed $n$ a natural number), so I can't officially use derivative yet. Then, I was thinking to use induction.
For $n = 2$ it is clearly true. If we have $ (1 - \frac{1}{n})^n > \frac{1}{6}$ , then for $n>1$
$$ \left(1-\frac{1}{n+1}\right)^{n+1} > \left(1-\frac{1}{n}\right)^{n+1} > \frac{1}{6} \left(1-\frac{1}{n} \right) .$$ But it is not helping.
I would appreciate some help.
Thanks!
AI: Prove by induction that
$$\left(1 - \frac{1}{n+1}\right)^{n+1} > \left(1 - \frac1n\right)^n.$$
Bring that into the form
$$\left(1 + \frac{1}{n^2-1}\right)^n > 1 + \frac1n,$$
which is seen to be true by Bernoulli's inequality. |
H: Which one is BIG
Let we have two event :
Event A: $ab^3$
Event B: $a^3b$
The range of a and b are given as: $-1 < a <0 < |a|< b< 1$
It seems to me a big range to test that which event is big. How to do this quickly without putting lots of numbers to testify?
AI: A = $ab(b^2)$
B = $ab(a^2)$
Because $-1<a<0<b<1$, $ab<0$.
Because $|a|<b$, $a^2<b^2$.
Therefore, $ab^3<a^3b$. |
H: Distribution of distinct balls in identical boxes
how can I derive a formula for the number of distributions of $n$ different balls in $k$ identical boxes. Where $\mathbf{empty\ box}$ is allowed.
I know this is equivalent to finding the number of ways to partition a set of $n$ labelled (distinct) objects into $k\ \mathbf{non\ empty}$ unlabelled subsets, which is basically ${n\brace k}$ (stirling number of the $2^{nd}$ kind).
But the problem is, ${n\brace k}$ doesn't allow $\mathbf{empty}$ partitions, where as my problems does.
AI: I believe you'll find it's
$$
\sum_{m=1}^k {n\brace m}
$$
Note that I am taking ${n\brace m}=0$ when $m>n$.
Why? How many ways can you fill $m$ identical boxes with $n$ distinct balls such that each box has at least one ball? Now, clearly there must be some number of boxes filled, let that be $m$. $m$ can be anything from 1 to $k$, but if $k>n$, then obviously you can't fill all the boxes, and thus no combination requiring filling of all boxes matters. |
H: cubic equations which have exactly one real root
Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c=0$. So, $-a^3-a=c$.
so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$
i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$
Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$
Thus, quadratic has no real root and so is the cubic $x^3+x+c$
I would like to know if this justification is sufficient and if this can be generalized.
I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)
can this be generalized to any odd degree polynomial (at least for some special cases)
AI: You essentially split off the linear factor belnging to a real root and showed per determinant that the remaining quadratic has no real root.
This is fine but does not readily generalize to higher degrees.
Instead, it is probably easier to show that $f$ is injective: Assume $a,b$ are two real roots, i.e. $f(a)=f(b)=0$.
Then by Rolle, there exists $\xi$ between $a$ and $b$ (including the case $a=\xi=b$ if you also want to show that no multiple root - $a=b$ - exists) with $f'(\xi)=0$. But for $f(x)=x^3+x$ we have $f'(x)=3x^2+1\ge1$ for all $x$. |
H: How show that if Matrix A is not square, it cannot have an inverse.
How to show that if How A is not square, it cannot have an inverse.
Why is the the case and how can I prove it?
AI: Non-square matrices $m \times n$ matrices for which $m ≠ n$ do not have an inverse. However, in some cases such a matrix may have a left inverse or right inverse.
Let $A$ is $m \times n$ and the rank of $A$ is equal to $n$, then $A$ has a left inverse: an $n \times m$ matrix $B$ such that $BA = I$.
while, If $A$ has rank $m$, then it has a right inverse: an $n \times m$ matrix $B$ such that $AB = I$. |
H: A question about the dihedral group $D_n$
Let the dihedral group $D_n$ be given by elements $a$ of order $n$ and $b$ of order $2$, where $ba = a^{-1}b.$
(a) Show that $a^{-m}= a^{n-m}$ for all integers $m.$
Definition: Let $n$ greater than or equal to $3$ be an integer. The group of rigid motions of a regular n-gon is called the $nth$ dihedral group, denoted by $D_n.$
I'm having trouble showing (a). Any hints or thoughts will be much appreciated.
AI: Recall that $$a^{b+c} = a^ba^c$$
And recall that, since the order of $a$ is given to be $n$, we know that $a^n = e$.
Applying that here gives us:
$$a^{n-m}=a^na^{-m}=ea^{-m}=a^{-m}$$ |
H: Determine the least natural number $k$ such that $a(k)>1$
Let $a(n)$ be a sequence with $a(0)=1/2$ and $a(n+1)=a(n)+(a(n)^2)/2013$, $n$ natural number.
Determine the least natural number $k$ such that $a(k)>1$.
This problem is from Poland proposed to Romanian Masters of Mathematics.
Can you give me some hints? I don't want a complete solution.
Thank you!
AI: Motto: To iterate a quadratic function is basically impossible, to iterate a homographic function is trivial.
Consider any sequence $(x_n)$ defined by $x_{n+1}=x_n+\frac1cx_n^2$ for some $0\lt x_0\lt 1\lt c$. Then, for every $0\lt x\lt 1$,
$$
\frac{c+1}{c+1-x}\lt1+\frac{x}c\lt\frac{c}{c-x},
$$
hence, as long as $x_n\lt 1$,
$$
\frac{c+1}{c+1-x_n}x_n\lt x_{n+1}\lt\frac{c}{c-x_n}x_n,
$$
that is,
$$
\frac1{x_n}-\frac1c\lt \frac1{x_{n+1}}\lt\frac1{x_n}-\frac1{c+1}.
$$
Thus,
$$
\frac1{x_0}-\frac{n}c\lt\frac1{x_n}\lt\frac1{x_0}-\frac{n}{c+1},
$$
for every $n$ such that
$$
\frac1{x_0}-\frac{n-1}c\geqslant1,
$$
that is, such that $n\leqslant(c/x_0)-c+1$. Assume that $c$ is an integer and that $x_0\leqslant1/2$. Then $n=c$ and $n=c+1$ are admissible. For $n=c$, the lower bound of $1/x_n$ is $1/x_0-1$. For $n=c+1$ the upper bound is $1/x_0-1$, hence
$$
\frac1{x_{c+1}}\lt\frac1{x_0}-1\lt\frac1{x_c}.
$$
For every integer $c\geqslant2$ and $x_0\leqslant1/2$, the least $n$ such that $x_n\gt x_0/(1-x_0)$ is $n=c+1$. In particular, if $x_0=1/2$, the least $n$ such that $x_n\gt1$ is $n=c+1$. |
H: Prove that $\frac{d}{dx}\int_0^xf(x,y)dy = f(x,x)+\int_0^x\frac{\partial}{\partial x}f(x,y)dy$
Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$. Assume that $$\frac{d}{dx}\int_a^bf(x,y)dy=\int_a^b\frac{\partial}{\partial x}f(x,y)dy.$$
Use the above property and the chain rule to prove that
$$\frac{d}{dx}\int_0^xf(x,y)dy = f(x,x)+\int_0^x\frac{\partial}{\partial x}f(x,y)dy.$$
This doesn't seem like it should be that hard but I haven't been able to get it, I especially can't figure out how to take advantage of the chain rule. Here's my best shot:
Let $a<x$. Then we have
\begin{align}
\frac{d}{dx}\int_0^xf(x,y)dy &= \frac{d}{dx}\int_a^xf(x,y)dy +
\int_0^a\frac{\partial}{\partial x}f(x,y)dy.\\
&=\frac{d}{dx}(F(x,x)-F(x,a)) + \int_0^a\frac{\partial}{\partial x}f(x,y)dy.\\
&=f(x,x) - \frac{d}{dx}F(x,a)+\int_0^a\frac{\partial}{\partial x}f(x,y)dy.
\end{align}
Anyways I'm not sure where to go from here or if this is even correct, but it's the best I've been able to come up with.
AI: To compute $\frac{d}{dx} \int_{0}^{x} f(x,y) \, dy$, one way is to separate the variables by introducing :
$$ G(u,v) = \int_{0}^{u} f(v,y) \, dy $$
such that $G(x,x) = \int_{0}^{x} f(x,y) \, dy$. This way, you have :
$$ \frac{d}{dx} \int_{0}^{x} f(x,y) \, dy = \frac{d}{dx} G(x,x) = \frac{\partial G}{\partial u} (x,x) + \frac{\partial G}{\partial v} (x,x) \tag{$\star$}$$
The partial derivatives are both easy to compute :
$$ \frac{\partial G}{\partial u}(u,v) = f(v,u) $$
follows from the fundamental theorem of calculus (see http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
and
$$ \frac{\partial G}{\partial v}(u,v) = \int_{0}^{u} \frac{\partial f}{\partial x}(v,y) \, dy $$
follows from differentiation under the integral sign (see http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign).
Using these two expressions and $(\star)$, you get :
$$ \frac{d}{dx} G(x,x) = f(x,x) + \int_{0}^{x} \frac{\partial f}{\partial x}(x,y) \, dy $$ |
H: Finding the eigenvalues of a matrix
I was trying to prove the following theorem and got stuck at one point :
Theorem :
Let $A\in M_n(F)$, then the scalar $\lambda$ is an eigenvalue of $A$ if and only if
$$
det(A-\lambda*I_n) = 0.
$$
Proof:
I started off like this ,
A scalar $\lambda$ is an eigenvalue of $A$ if and only if there exists a non zero vector $v\in F^n$ such that $Av = \lambda v$.
$$
\Rightarrow (A-\lambda I_n)(v) = 0
$$
$$
\Leftrightarrow (A-\lambda I_n) \text{ is not invertible}
$$
$$
\Leftrightarrow det (A-\lambda I_n) = 0
$$
I figured that $(A - \lambda I_n)$ should not be invertible to get the det as $0$ , but i don't really understand why it can't be invertible ?
AI: A matrix $A$ is not invertible if and only if there exists $v\neq 0$ such that $Av=0$.
To be more precise: if $Av=0$ for some $v\neq 0$ then $A$ cannot be invertible since it is not injective: $Av=A0$. |
H: What are the probability that A is neither symmetric nor skew-symmetric?
A is a 3 × 3 matrix with entries from the set {–1, 0, 1}. Then the probability that A is neither symmetric nor skew-symmetric is:
My thoughts: There can be nine members on a $3*3$ matrix and there are three possibilities for each member. Therefore the total no. of matrices possible is $3^9$. Subtracting the no. of possibilities for skew symmetric and symmetric matrices from it and dividing by $3^9$ will give the required probability but, I'm not able to figure out the values to be subtracted. Is there a relationship between the total number of possibilities and the number of possibilities for symmetric or skew-symmetric matrix.
Please help and thanks in advance.
AI: Skew-symmetric matrices have $0$s on the diagonal, and opposing entries are determined by each other. There are then $3$ entries we have to choose for a skew-symmetric matrix, with $3$ options for each entry, so $3^3$ skew-symmetric matrices.
For symmetric matrices, we can choose any entries on the diagonal, and again, opposing entries are determined by each other. Thus, there are $6$ entries we have to choose for a symmetric matrix, with $3$ options for each entry, so $3^6$ symmetric matrices.
The only remaining question is: which matrix or matrices are both symmetric and skew-symmetric? If the number of such matrices is $n$, then there are $3^3+3^6-n$ matrices that are skew-symmetric or symmetric. |
H: Independence between two random variables and a function of the two random variables
Say we have two random variables, $X$ and $Z$ that are independent. Then let $W=a+bXZ$ be the random variable that is a function of both $X$ and $Z$, $a$ and $b$ are just scalar constants. Then are $X$ and $W$ independent? How about $Z$ and $W$?
This is just something I thought of, not sure how I would go about proving (or disproving) it.
AI: The random variables $W$ and $X$ are called independent when their joint probability $P(W,X)$ equals the product of their marginal probabilities, $P(W)P(X)$, or, in other words, the conditional probability is $P(W|X)\stackrel{\mathrm{def}}{=}P(W,X)/P(X)=P(W)$.
You can establish independence by computing $P(W|X)$ and also $P(W)$, and checking if they are equal. This wikipedia page might help you compute these distributions when some of them are normal. |
H: What are the eigenvalues of this symmetric matrix?
Let
$$A=\begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{pmatrix}.$$
I'm trying to find the eigenvalues of $A$, but when I calculate the characteristic polynomial, I get
$$p(\lambda)=-\lambda^3+6\lambda^2+15\lambda+2,$$
and I don't know how to solve $p(\lambda)=0$.
I'd appreciate any help. Thanks in advance.
AI: You made an error in your CP, it should be:
$$p(\lambda)=-\lambda^3+6\lambda^2+15\lambda+8 = -(\lambda-8)(\lambda+1)^2 = 0$$
So you have the two eigenvalues $\lambda_1 = 8$ (single root) and $\lambda_{2,3} = -1$ (double root).
Can you continue with the eigenvectors now? |
H: How to find the $S_\kappa$ elements in the product $ \sum_{\kappa=1}^{K} S_\kappa Q_\kappa(r, \theta,\phi) = 1$?
So I have one K-dimensional complex column vector $\textbf{S}$ and a set of K complex functions $Q_\kappa(r,\theta,\phi) \forall \kappa \in \{1,\dots,K\}$ defined in spherical coordinates.
I know all the K functions and the following equality holds:
$$ \sum_{\kappa=1}^{K} S_\kappa Q_\kappa(r, \theta,\phi) = 1$$
Is there a way to find the $S_\kappa$ elements of vector $\mathbf{S}$? Perhaps multiples ways? Or an algorithm to systematically find solutions?
P.S. Sorry about the tags, but I have no idea what tags I should put in this.
AI: Select random $(r,\theta,\phi)$ triples and note down corresponding $\mathbf Q$ vectors. When you have enough $Q$ points that they define a $(K-1)$-dimensional hyperplane, calculate its equation and normalize so the constant term is $1$. The coefficients of the $Q_\kappa$s are then your desired $\mathbf S$.
This only works if $\mathbf S$ is unique. If there are several possible $\mathbf S$s, then you can't ever be sure that they all are possible solutions, without knowing something about how the $Q$ function work. It might me that there's a single $(r,\theta,\phi)$ that will reject one of them, but you just haven't tried it yet. |
H: Unique smallest and largest topology ideas
What does it mean to be a smallest topology of a set $X$. I would guess that it would be a topology of $X$ which has least number of elements and similarly for largest topology it would have to be largest number of elements? Am I correct? This looks fairly straightforward unless I am missing something
For example:
If I am understanding this correctly if $X=\{a,b,c\}$ and $\mathcal{T_1}=\{\emptyset, X, \{a,b\}, \{a\}\}$ and
$\mathcal{T_2}=\{\emptyset, X, \{b,c\}, \{a\}\}$. Then the largest topology will be $\mathcal{T_l}=\{\emptyset, X, \{a,b\}, \{a\}, \{b\}, \{b,c\}\}$ because you can get $\mathcal{T_1}$ and $\mathcal{T_2}$ from it and the smallest will be $\mathcal{T_s}=\{\emptyset, X, \{a\}\}$
AI: We tend not to use the terms 'smallest' or 'largest' topology, but finest and coarsest (see wikipedia for a definition). Some people do use smallest and largest, but it's less common and in my opinion more ambiguous.
In this sense, the discrete topology (every subset is open) is the finest topology, and the coarsest is the indiscrete (trivial - only $X$ and $\emptyset$ are open) topology.
In terms of their cardinality, the indiscrete topology is the 'smallest' topology on $X$ and the discrete topology is the 'largest' (with the caveat that for infinite sets, the discrete topology has the same cardinality as some coarser topologies on the set, and so is not the unique largest topology).
With regard to your edit, you may be thinking about the infimum and supremum of topologies with respect to the partial order given by subset inclusion. To expand on this a little, the set of all topologies on a set $X$ form a partially ordered set with the partial order given by subset inclusion $\subseteq$. And so we can formally say that a topology $\tau$ on $X$ is less than or equal to another topology $\tau'$ if $\tau\subseteq\tau'$, equivalently if $\tau'$ is a finer topology than $\tau$. Let's denote this partially ordered set $(P_X,\subseteq)$ where $P_X$ is the set of all topologies on $X$.
$(P_X,\subseteq)$ has a unique maximal element given by the discrete topology, and a unique minimal element given by the indiscrete topology. We can also consider the collection of elements $A=\{\tau_\lambda\mid \lambda\in\Lambda\}$ and ask if there exists an element $\tau\in P_X$ such that $\tau_\lambda\subseteq\tau$ for all $\lambda\in\Lambda$, and for any other $\tau'\in P_X$ with this property $\tau\subseteq\tau'$. It turns out that such an element exists and is in fact unique (existence is not always guaranteed of supremums in general partially ordered sets, although uniqueness is). We call $\tau$ the supremum (sometimes called the least upper bound) of $A$ and write $\tau=\sup A$. It is the coarsest topology which contains all open sets from the $\tau_{\lambda}$. In your example, $\mathcal{T}_l=\sup\{\mathcal{T}_1,\mathcal{T}_2\}$.
We can similarly define the infimum (sometimes called the greatest lower bound) $\tau^*=\inf A$ which is the finest topology which is contained in all $\tau_{\lambda}$. In your example $\mathcal{T}_s=\inf\{\mathcal{T}_1,\mathcal{T}_2\}$.
We can find the infimum $\inf A$ by simply taking the intersection of all the topologies in $A$, that is $$\inf A=\bigcap A=\bigcap_{\lambda\in\Lambda}\tau_{\lambda}$$ and it's not hard to see that this is in $P_X$ and it is the infimum of $A$. It is not the case that the supremum is the union of the topologies in $A$ as the union is not in general a topology, however the union does form a subbase of the supremum. |
H: Mathematical Analysis: Riemann Integration
This is a question from my Mid-term test which was held last week:
Define $f:[0,1] \to \mathbb{R}$ by $f(x) = x,$ if $x$ is irrational and $f(x) =0$, otherwise
Is $f \in R[0,1]?$
My Proof:
Consider any Partition $P = \{\ x_{0} = 0, ..., x_{n} =1\}$ of $[0,1]$. Then, $L(f,P)=0$ but $U(f,P)$ $= \sum^{n}_{i=1} sup \{\ f(x): x\in [x_{i-1},x_i] \} (x_i -x_{i-1}) $ $ =\sum^{n}_{i=1} x_i(x_i -x_{i-1}) \geq \sum^{n}_{i=1} x_1(x_i -x_{i-1}) = x_{1} >0$
But the hint in the question requires one to show that $U(f,P) > \frac{1}{2}$ . Any advice please?
AI: Hint: Use AM-GM to prove
$$\sum_{i=1}^n x_i(x_i-x_{i-1}) = \sum_{i=1}^n x_i^2 - \sum_{i=1}^n x_ix_{i-1} \ge \sum_{i=1}^n x_i^2 - \frac12(x_1^2+x_n^2)-\sum_{i=2}^{n-1} x_i^2 >\frac12.$$ |
H: Finding A,B,C s.t $f'(a)+O(h^2)=\frac{Af(a)+Bf(a+2h)+Cf(a+3h)}{h}$
Find constants A,B,C s.t for differtiable three times function f, $f'(a)+O(h^2)=\frac{Af(a)+Bf(a+2h)+Cf(a+3h)}{h}$
I know that $f'(a)+O(h^2)=\frac{f(a+h)-f(a-h)}{2h}$ so I need to solve $$0.5f(a+h)-0.5f(a-h)=Af(a)+B(a+2h)+Cf(a+3h)$$but I don't know how. How can I solve this equalation?
AI: I use Martin's idea:
$$
f(a+2h)=f(a)+2hf'(a)+2h^2 f''(a)+O(h^3),\\ f(a+3h)=f(a)+3hf'(a)+\frac{9h^2}{2}f''(a)+O(h^3).
$$
Then
$$
9f(a+2h)-4f(a+3h)=5f(a)+6hf'(a)+O(h^3)
$$
and we get the $O(h^2)$ approximation. |
H: Clarification regarding the monotone convergence theorem
I have a simple question about the monontone convergence theorem. Let $f_n : \mathbb{R} \to [0,\infty]$ be a sequence of nonnegative, extended real valued, measureable functions on $\mathbb{R}$. Suppose the sequence is monotone increasing i.e. $0 \leq f_1 \leq f_2 \leq \ldots$, with the inequalities holding pointwise. Define $f$ to be the pointwise limit. The monotone convergence theorem asserts that $f$ is also measureable and $\int f = \lim_{n \to \infty} \int f_n $.
I'm wondering... when we say that $f$ is the pointwise limit of the $f_n$, does that include the possibility that at some point $x$ the sequence diverges to infinity? Say $f_n(x) = n$ for all $n$, for example. Then would we say $f_n(x) \to \infty$? I ask because the way you define convergence to infinity is a little different i.e. $x_n \to \infty$ when, for all $M < \infty$ there's and $N > 0$ so $n \geq N$ implies $x_n > M$. It seems a bit suspect that both kinds of convergence are implicitly being allowed in the statement. But, I suspect this is fine.
AI: Yes. In most of measure theory one allows functions to take values on the "extended real line", i.e. $[-\infty,\infty]$.
The infinities in the Monotone Convergence Theorem play no role, in the sense that if they appear on a null-set then you can forget about them. An if they appear in a set of positive measure, then both sides of the equality become infinity. |
H: A small bit on partial differentials and general solutions
Consider the equation:
$$\frac{\partial u}{\partial t} - \frac{\partial u}{\partial x} = 0$$
I want to find a general solution.
So I do the following:
$\alpha = ax + bt$, where $\frac{\partial \alpha}{\partial x} = a$, and $\frac{\partial \alpha}{\partial t} = b$
$\beta = cx + dt$, where $\frac{\partial \beta}{\partial x} = c$, and $\frac{\partial \beta}{\partial t} = d$
We do the chain rule:
$$\frac{\partial u}{\partial t} = \frac{\partial \alpha}{\partial t} \frac{\partial u}{\partial \alpha} + \frac{\partial \beta}{\partial t} \frac{\partial u}{\partial \beta} \text{ and } \frac{\partial u}{\partial x} = \frac{\partial \alpha}{\partial x} \frac{\partial u}{\partial \alpha} + \frac{\partial \beta}{\partial x} \frac{\partial u}{\partial \beta} \text{ which is } $$
$$ b \frac{\partial u}{\partial \alpha} + d \frac{\partial u}{\partial \beta} - \left( a \frac{\partial u}{\partial \alpha} + c \frac{\partial u}{\partial \beta}\right)=0 \Rightarrow b \frac{\partial u}{\partial \alpha} + d \frac{\partial u}{\partial \beta} - a \frac{\partial u}{\partial \alpha} - c \frac{\partial u}{\partial \beta}=0$$
and from there we get
$$(b -a) \frac{\partial u}{\partial \alpha} + (d-c) \frac{\partial u}{\partial \beta} =0$$
We want to get rid of a partial to get this to being a simpler equation, and a simple way to do that is to set $b=1, a=0, d=1, c=-1$. After that I get:
$$\frac{\partial u}{\partial \alpha} = 0 \text{ which integrates to } u(\alpha) = C$$
and any solution to the above equation is going to be some variant on $u=f(x-t)$. With the above values for a, b, c, d, I get $\alpha = t$ and $\beta = -x + t$. So u ends up as a function of t I would think in this case.
So far so good, but the part that the text says (and eludes me) is that it should be $u=f(\beta)$. SO your general solution here is $f(-x + t)$. It says that f is a function of $\beta$ only. Is that because using $\alpha$ you lose one variable? If I knew this I could always do the final step in these kinds of problems. I feel like I am just taking things as a given, not really understanding which (alpha or beta) I should be plugging back in. Anyhow, any help on that question would be appreciated. I did out the whole problem because I wanted to make sure my steps were right too.
AI: What you're doing only seems to make sense if you're planning to view $u$ as a function of both $\alpha$ and $\beta$. However, then you can't say "... which integrates to $u(\alpha)=C$", because "$u(\alpha)$" does not make sense: $u$ is a function of two variables, not one.
What's going on is that since $\partial u/\partial \alpha = 0$ everywhere, whenever you choose some $\beta$ it must be the case that $u(1,\beta)=u(2,\beta)=u(42+\pi,\beta)=\cdots=u(\alpha,\beta)$ for all $\alpha$. In other words, changing $\alpha$ cannot change the value of $u(\alpha,\beta)$, and there must be some $f$ such that $u(\alpha,\beta)=f(\beta)$.
You lose the $\alpha$ variable exactly because $\partial u/\partial \alpha=0$ says that it does not matter. |
H: How to evaluate the summation $S_b$
This question is from my notebook, not hw or else, only exercise to understand better. I tried by myself. However, since my trail are too trivial, I dont need to write here. i am confused a bit. I want to learn how to solve such question types. I guess I need to use residue. I am grateful to help me thank you.
AI: we'll use the integral $$ \oint_{C_N} \frac{\pi\cot(\pi z)}{z^2 + b^2 } \ dz $$
where $C_N $ is the square contour of side $ 2N + 1 $
By residues theory :
$$ \oint_{C_N} \frac{\pi\cot(\pi z)}{z^2 + b^2 } \ dz = 2\pi i\left( \sum_{-N}^N \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , n \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , bi \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , -bi \right) \right)$$
the integral $ \to 0 $ as $ N \to \infty $
$$ \Rightarrow \sum_{-\infty }^{\infty} \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , n \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , bi \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , -bi \right) = 0 $$
$$ \Rightarrow \sum_{-\infty}^{-1} \frac{1}{n^2 + b^2} + \sum_{1}^{\infty} \frac{1}{n^2 + b^2 } + \frac{1}{b^2} + \lim_{z \to bi} \frac{\pi\cot(\pi z)}{z + bi} + \lim_{z \to -bi} \frac{\pi\cot(\pi z)}{z - bi} = 0$$
$$ \Rightarrow 2\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2 } + \frac{1}{b^2} - \frac{\pi\coth(b\pi)}{b} = 0 $$
$$ \Rightarrow \sum_{n=1}^{\infty} \frac{1}{n^2 + b^2 } = \frac{\pi\coth(b\pi)}{2b} - \frac{1}{2b^2} $$ |
H: What is the probability C was the one with the wrong answer?
Students A, B and C each independently answer a question on a test. The probability of getting the correct answer is 0.9 for A, 0.7 for B and 0.4 for C. If two of them get the correct answer, what is the probability C was the one with the wrong answer?
AI: We interpret "two got the right answer" as meaning "exactly two."
Let $A,B,C$ be the events that, respectively, A, B, C get the right answer, and let $A',B',C'$ be the events that A, B, C get the wrong answer. Let $T$ be the event that exactly two got the right answer. We want $\Pr(C'|T)$.
By the definition of conditional probability, we have
$$\Pr(C'|T)=\frac{\Pr(C'\cap T)}{\Pr(T)}.$$
We want to calculate the two probabilities on the right.
We have $\Pr(C'\cap T)=\Pr(A\cap B\cap C')=(0.9)(0.7)(0.6)$.
Also, $\Pr(T)=\Pr(A\cap B\cap C')+\Pr(A\cap B'\cap C)+\Pr(A'\cap B\cap C)$. |
H: Derivation of the quadratic equation
So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
AI: $ax^2+bx+c=0 (a\neq0)$
$x^2+\frac{b}{a}x+\frac{c}{a}=0$
$x^2+\frac{b}{a}x=-\frac{c}{a}$
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$
$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$
$(x+\frac{b}{2a})=\pm\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$
$\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ |
H: What is the Mirror/PingPong clamp mode algorithm?
I do programming as a hobby, and in a dynamic system various numerical values inevitably change. Those values can be greater than or less than the expected range, in which case they need to be "wrapped" according to a "mode". I am aware of three of these wrapping modes: Clamp, which simply limits a value to a maximum or minimum, Wrap, which brings a value back around like the hand on an analog clock, and Mirror AKA Ping-Pong which is sort of like Wrap except it bounces the value back and forth as it gets further out of range. I am able to produce the Clamp and Wrap effects, but the last kind is beyond my comprehension. Does anyone know what the function would be (assuming you can use operations such as floor()) that would produce such a Mirror or Ping-Pong effect for input number A and maximum number B if the minimum number is 0?
AI: The intuition is that "mirror" would work something like this pseudocode:
sub mirror(0,a,b) {
while( a < 0 || a > b ) {
if( a < 0 )
a = -a;
if( a > b )
a = b-(a-b);
}
return a;
}
If you already have an efficient implementation of "wrap" (say, using fmod or %), you can implement "mirror" without a loop as
sub mirror(0,a,b) {
a = wrap(0,a,2*b);
if( a < b )
return a;
else
return 2*b-a; // this equals b-(a-b)
} |
H: How find this inequality$\sqrt{\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)}+1$
let $x,y,z>0$,show that
$$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\dfrac{(x^2+yz)(y^2+xz)(z^2+xy)}{x^2y^2z^2}}$$
My try:
$$\Longleftrightarrow \left(\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\right)^3\ge \dfrac{8(x^2+yz)(y^2+xz)(z^2+xy)}{x^2y^2z^2}$$
let
$$a=\dfrac{x}{y},b=\dfrac{y}{z},c=\dfrac{z}{x}$$
$$\Longleftrightarrow \left(\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)(a+b+c)}+1\right)^3\ge 8\left(1+\dfrac{c}{a}\right)\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)$$
then I can't ,so I think this inequality maybe have other nice methods,Thank you
AI: As the inequality is homogeneous, we can normalise by $xyz=1$. Then we have:
$$\sqrt{\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)}+1\ge 2\sqrt[3]{\left(x^2+\frac{1}{x}\right)\left(y^2+\frac{1}{y}\right)\left(z^2+\frac{1}{z}\right)}$$
$$\sqrt{3+\sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)}+1\ge 2\sqrt[3]{2 + \sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)}$$
Let $\displaystyle a = \sum_{cyc} \left(x^3 + \frac{1}{x^3}\right)\ge 6$. Then the inequality is reduced to
$$f(a) = \sqrt{3+a} +1 - 2 \sqrt[3]{2+a} \ge 0$$
which is easy to do as $f(6)=0$ and $f'(a) > 0 $ for $a > 6$.
Addendum: alternate way to show $\sqrt{3+a} +1 \ge 2 \sqrt[3]{2+a}$ would be to cube, group terms and then square, to get the equivalent $(a+2)(a-6)^2 \ge 0$, which is obvious for $a \ge 6$. |
H: Catagorical Definition of Coproduct and Abelian Groups
I have the definition of a coproduct which is as follows:
A coproduct of $\{A_\alpha\}$ in $\mathcal{G}$, where $\mathcal{G}$ is a category and $\{A_\alpha\}$ a collection of objects, is an object $Q$ of $\mathcal{G}$ with $\pi_\alpha:A_\alpha\rightarrow Q$ s.t. $\forall{C}\in obj\{\mathcal{G}\}$ with morphisms $\phi_\alpha:A_\alpha\rightarrow C$ s.t $\exists! f:Q\rightarrow C$ s.t. $\phi_\alpha=f\circ\pi_\alpha$
Now I have been given the example with if our objects are abelian groups and morphisms are homomorphisms then this defines the direct sum. I was wondering how we go about proving this? I can kind of see why it does but how do we show for example that nothing else but the direct sum satisfies this definition?
Thanks for any help (sorry if my question is stupid or muddled)
AI: Let's prove that the coproduct is unique in a category. Assume that there are two, let's say $Q$ and $Q'$ for the collection $\{A_a\}$. Let $\pi_a:A_a\to Q$ and $\pi':A_a\to Q'$ be the morphisms you talked about. Then by the universal property you mentioned, we have that there are morphisms $\phi:Q'\to Q$ and $\psi:Q\to Q'$ such that $\pi_a=\phi\circ\pi_a'$ and $\pi_a'=\psi\circ\pi_a$. Now these morphisms $\phi$ and $\psi$ are unique. If we now take $\phi\circ\psi:Q\to Q$, this also satisfies $\phi\circ\psi\circ\pi_a=\phi\circ\pi_a'=\pi_a$, the same as the identity. Since the map of the universal property is unique, we have that $\phi\circ\psi=\mbox{id}$. The same argument shows that $\psi\circ\phi=\mbox{id}$.
So the coproduct is unique. Therefore, if you can show that the direct sum of abelian groups satisfies what you are looking for (it seems you already did this?), then it must be "the" coproduct (up to isomorphism). |
H: Who has a higher chance of drawing an easy question?
From a pool of N questions n are easy. Two people draw a question without returning it. Who has a higher chance of drawing an easy question?
AI: Assume $N\geq 2$
$P($First person draw an easy question$)=\frac{n}{N}$
$P($First person draw a hard question$)=1-\frac{n}{N}=\frac{N-n}{N}$
$P($Second person draw an easy question$)=\frac{n}{N}\cdot \frac{n-1}{N-1}+\frac{N-n}{N}\cdot \frac{n}{N-1}=\frac{Nn-n}{N(N-1)}=\frac{n}{N}$
So their chances are equal. |
H: Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.
AI: Try multiplying both sides by $11 \cdot 5 \cdot 6$, to get $$121 = 11 \cdot 6 + 11 \cdot 5 = 2 \cdot 5 \cdot 6 = 60$$ in $\mathbb{Z}/p\mathbb{Z}$. |
H: Finding the limit of a sequence? $\lim_{n\to\infty}\frac{3^{n-2}}{4^{m+2}}$
How would the limit of the following sequence be found.
$\displaystyle \frac{3^{n-2}}{4^{m+2}},\;$ as $\;n\rightarrow \infty$.
Would you use the squeeze theorem to find the limit of the sequence like this
$\displaystyle \frac{-3^m}{4^m}<\frac{3^{n-2}}{4^{m+2}}<\frac{3^m}{4^m}$
AI: We have $$0 \lt \dfrac{3^{n - 2}}{4^{m+2}} \lt \dfrac {3^n}{4^n} = \left(\frac 34\right)^n$$
The limit, as $n\to \infty$, of each end doing the "squeezing" is $0$, so the limit of the squeezed term of your sequence must be $0$ as well. |
H: Show that this is not a countable cover for the graph of a continuous function.
Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. Consider any rational number $q$. For any $\epsilon >0$, there is a $\delta > 0$ such that $|q - y| < \delta \implies |f(q) - f(y)| < \epsilon$.
I was using this to try and make a countable cover of the graph of $f$ using rectangles of vertical height $2 \epsilon$ and horizontal width $2 \delta$. There are countable number of rationals number and so there are countable number of such rectangles. In any interval that these rectangles might "miss", there is a rational number, and so a covering rectangle.
What is the flaw in this argument? It is possible for these rectangles to miss a part of the graph of $f$?
AI: Yes, it’s possible. Fix $\epsilon>0$. For each $q\in\Bbb Q$ you have a $\delta_q>0$ such that $|f(x)-f(q)|<\epsilon$ whenever $|x-q|<\delta_q$. Let $$R_q=(q-\delta_q,q+\delta_q)\times\big(f(q)-\epsilon,f(q)+\epsilon\big)\;;$$ the open rectangle $R_q$ covers $\{\langle x,f(x)\rangle:|x-q|<\delta_q\}$, so $\{R_q:q\in\Bbb Q\}$ covers the graph of $$f\upharpoonright\bigcup_{q\in\Bbb Q}(q-\delta_q,q+\delta_q)\;,$$ but the intervals $(q-\delta_q,q+\delta_q)$ need not cover $\Bbb R$, so the rectangles $R_q$ need not cover the graph of $f$.
To see why the intervals $(q-\delta_q,q+\delta_q)$ need not cover $\Bbb R$, let $\alpha\in\Bbb R$ be irrational, and suppose that $\delta_q<|q-\alpha|$ for each $q\in\Bbb Q$; then for each $q\in\Bbb Q$ we have $\alpha\notin(q-\delta_q,q+\delta_q)$. In fact it’s possible for
$$\Bbb R\setminus\bigcup_{q\in\Bbb Q}(q-\delta_q,q+\delta_q)$$
to contain a set homeomorphic to the middle-thirds Cantor set, in which case
$$\left\vert\Bbb R\setminus\bigcup_{q\in\Bbb Q}(q-\delta_q,q+\delta_q)\right\vert=|\Bbb R|=2^\omega=\mathfrak{c}\;,$$
and the rectangles $R_q$ fail to cover a lot of points of the graph of $f$. |
H: Limit of sequences: $ \lim (1 + \frac{r}{n} )^n = e^r $
Consider the sequence $x_n = \left (1+\frac{r}{n} \right )^n $ for $r \in \mathbb{Q} $. I need to prove that $ \lim x_n = e^r $
My attempt of proof (for r>0) is to find a subsequence of $x_n$ that converges to $e^r$, but it will be enough if $x_n$ is a Cauchy sequence. Since all the convergent sequences are Cauchy sequences, I was thinking to prove that $x_n$ is convergent and it will finish my problem. So if I prove that that $x_n$ is increasing and have an upper bound ($3^r$, for example), then it is done. But I'm having trouble with this last step.
But maybe it is easier to prove using, for example, that $\left(1+\frac{r}{n}\right)^n = \left(1+\frac{1}{n/r}\right)^n$ but I don't see how to proceed too.
I appreciate your help.
Thanks!
AI: I will use the following definitions. $\log$ is defined as
$$
\log(x) = \int_1^x \frac{1}{t} \, dt.
$$
for all $x > 0$. From this definition it can be shown that all the familiar properties of $\log$ holds. $e^x$ is defined as $e^x = \log^{-1}(x)$ for all $x \in \mathbb{R}$ (it can be proved that $\log$ is increasing so its one to one and maps onto $\mathbb{R}$, so the inverse exists). Then $e$ is defined as $e = e^1 = \log^{-1}(1)$.
Now I will prove the limit. By the fundamental theorem of calculus, $\log'(\frac{1}{r}) = r$. Using the definition of derivatives, we have
$$
\lim_{h \to 0} \frac{\log(\frac{1}{r} + h) - \log(\frac{1}{r})}{h} = r \\
\lim_{h \to 0} \frac{1}{h}\log(1 + rh) = r \\
\lim_{h \to 0} \log\left((1 + rh)^{\frac{1}{h}}\right) = r.
$$
Since $\log$ is continuous, we can write
$$
\log\left(\lim_{h \to 0} (1 + rh)^{\frac{1}{h}}\right) = r \\
\lim_{h \to 0} (1 + rh)^{\frac{1}{h}} = e^r.
$$
So we have $\lim_{n \to \infty} (1 + \frac{r}{n})^n = e^r \\$ where the expression inside the limit is a function. So it must also be true for the sequence with integer $n$.
EDIT:
Let $a_n = \left(1 + \frac{r}{n}\right)^n$. This sequence is increasing. To see this, note that $(1 + a)^b \geq 1 + ab$ (by binomial theorem). Apply this for $a = \frac{r}{n + 1}$ and $b = \frac{n + 1}{n}$ to get
$$
\left(1 + \frac{r}{n + 1}\right)^{\frac{n + 1}{n}} \geq 1 + \frac{r}{n + 1} \cdot \frac{n + 1}{n} = 1 + \frac{r}{n}.
$$
Taking both sides to the power of $n$ gives $\left(1 + \frac{r}{n + 1}\right)^{n + 1} \geq \left(1 + \frac{r}{n}\right)^n$ which shows that the sequence is increasing. To show that it is bounded, first use the binomial theorem to get
$$
\left(1 + \frac{r}{n}\right)^n = \sum_{i = 0}^n \binom{n}{i} \left(\frac{r}{n}\right)^i = \sum_{i = 0}^n \frac{n!}{i!(n - i)!} \cdot \frac{r^i}{n^i} \\
= \sum_{i = 0}^n \frac{r^i}{i!} \left(\frac{n(n - 1) \dotsb (n - i + 1)}{n^i}\right) \leq \sum_{i = 0}^n \frac{r^i}{i!}.
$$
since $n(n - 1) \dotsb (n - i + 1) \leq \underbrace{n \dotsb n}_{i} = n^i$. So
$$
\left(1 + \frac{r}{n}\right)^n \leq \sum_{i = 0}^n \frac{r^i}{i!} < \sum_{i = 0}^{\infty} \frac{r^i}{i!}.
$$
But the series on the right most side converges which we can check by the ratio test. So $a_n$ is bounded by that number. Thus $a_n$ converges since it is increasing and bounded. |
H: Prove that either$ P(A∩B)=0$ or else $P(A'∩B')=0$
Let $E$ and $F$ be independent with $E = A∪B$, and $F = A∩B$. Prove that either
$P(A∩B)=0$ or else $P(A'∩B')=0$.
AI: We are told that $E$ and $F$ are independent. By the definition of independence, we have
$$\Pr(E\cap F)=\Pr(E)\Pr(F).\tag{1}$$
But $E\cap F)=(A\cup B)\cap (A\cap B)$. The intersection of $A\cup B$ and $A\cap B$ is $A\cap B$, that is, $F$. Thus from (1) we obtain
$$\Pr(F)=\Pr(E)\Pr(F).\tag{2}$$
Now there are two cases to consider.
If $\Pr(F)=0$, then Equation (2) holds, and we have $\Pr(F)=\Pr(A\cap B)=0$.
If $\Pr(F)\ne 0$, from (2), by cancelling $\Pr(F)$, we get $\Pr(E)=1$.
Thus $\Pr(E')=1-\Pr(E)=0$. But $E'=(A\cup B)'=A'\cap B'$. It follows that $\Pr(A'\cap B')=0$.
So we have shown that if (1) holds, then either $\Pr(A\cap B)=0$ or $\Pr(A'\cap B')=0$. That is what needed to be shown.
Remark: We cannot interpret the "or else" as exclusive or. For if $\Pr(A)=0$ and $\Pr(B)=1$. then $E$ and $F$ are independent but $\Pr(A\cap B)=0$ and $\Pr(A'\cap B')=0$. |
H: Proving that $\frac{6\tan\phi}{\tan^2\phi-9}=\tan A$
In a triange $ABC$, points $D$ and $E$ are taken on side $BC$ such that $BD=DE=EC$. If $\angle ADE=\angle AED=\phi$, how can we prove that $$\frac{6\tan\phi}{\tan^2\phi-9}=\tan A$$
AI: The key here is recognizing the identity we wish to prove. Note that
$$\tan{A} = -\frac{\frac{2}{3} \tan{\phi}}{1-(\frac13 \tan{\phi})^2}$$
This looks like a double-angle formula for tangent. Now note that, because of symmetry, $B=C$. Thus $A = \pi-2 B$ and $\tan{A} = -\tan{2 B}$. Therefore, the formula implies that
$$\tan{B} = \frac13 \tan{\phi}$$
This is what we should need to prove. We do this by noting two items: 1) The triangle $\Delta ABE$ is isosceles with side $AD=AE=d$ given by
$$d \cos{\phi} = \frac{a}{6}$$
2) The law of sines in either side triangle, say $\Delta ABD$ produces the equation
$$\frac{\sin{(\phi-B)}}{\sin{B}} = \frac{a}{3 d}$$
or
$$\frac{\sin{\phi}}{\tan{B}} - \cos{\phi} = 2 \cos{\phi} $$
The above assertion - and consequently, the original relation - is proven. |
H: Is conjugate of holomorphic function holomorphic?
If $f(z)$ is holomorphic, does it follow that $g(z)=\overline{f(z)}$ is holomorphic?
I'm looking at $$\lim_{z\rightarrow a}\dfrac{g(z)-g(a)}{z-a} = \lim_{z\rightarrow a}\dfrac{\overline{f(z)-f(a)}}{z-a}$$
Can we pull the limit out to get $\overline{\lim_{z\rightarrow a}\dfrac{f(z)-f(a)}{z-a}}$?
AI: No. When you took the conjugation out, it forces you to conjugate $z-a$ in the denominator.
An easy (and canonical) way to see that the conjugate of a holomorphic function is not holomorphic is to consider $z\mapsto \overline z$. This is easily confirmed by looking at the Cauchy-Riemann equations. |
H: Order Topology on $\mathbb{Z_+}$
Why is the order topology on $\mathbb{Z_+}$ a discrete one? I understand that the discrete topology will have all subsets of $\mathbb{Z_+}$ which means that all subsets of $\mathbb{Z_+}$ are open which is not necessarily true.
AI: Why do you say not every subset is open? Every subset is open!
All the singletons $\{ n \}$ are basic open sets in the order topology, because
$$\begin{alignat}{2}
\{ 1 \} &\; = \{ x \in \mathbb{Z}_+ : x < 2 \} &\ &= (-\infty, 2)\\
\{ n+1 \} &\; = \{ x \in \mathbb{Z}_+ : n < x < n+2 \} &&= (n, n+2)
\end{alignat}$$
But open sets needn't all be basic open sets: in particular, any unions of basic open sets is open.
Any subset $X \subseteq \mathbb{Z}_+$ can be written as the union of singletons:
$$X = \bigcup_{n \in X} \{ n \}$$
...and so all subsets are open. |
H: Prove : If $I = (p(x))$ is a prime ideal in $F[x]$ then $p(x)$ is irreducible.
I have to show :
If $I = (p(x))$ is a prime ideal in $F[x]$, where F is a field, then $p(x)$ is irreducible.
In the book I use, there is the proof of the converse which uses Euclid's Lemma.
I tried to suppose $I=(p(x))$ is an ideal, then every polynomial in I is divisible by $p(x)$. Then for $a(x)b(x) \in F[x]$ we have $a(x)b(x)=p(x)q(x)$ for $q(x) \in F[x]$ with $\delta a<\delta p$ and $\delta b<\delta p$.
From there, I don't see how to deduce $p(x)$ is irreducible ie there is no factorisation over F $p(x)=c(x)d(x) \in F[x]$ with $\delta c < \delta p$ and $\delta d < \delta p$
We could express $p(x)$ in function of $a(x),b(x)$ and $g(x)$ but it does not help.
Any hints ? :)
Thanks
AI: Suppose $p(x)=c(x)d(x)$, i.e. $p(x)$ is reducible. Since $c(x)d(x)=p(x)\in(p(x))$, either $c(x)\in(p(x))$ or $d(x)\in(p(x))$. Without loss the former, so $p(x)|c(x)$, and also $c(x)|p(x)$. Hence $c,p$ are associates, so $d$ is a unit, which contradicts the reducibility assumption. |
H: Is the following sentence a tautology: $(p\Rightarrow q)\vee(r \Rightarrow p)\vee(r\Rightarrow s)\vee(r\Rightarrow q)$?
If both $p$ and $q$ are false then ($p\Rightarrow q$) is true.
If either $p$ or $q$ is true then one of ($r\Rightarrow p$) or ($r\Rightarrow q$) is true.
If both $p$ and $q$ are true then all are true.
Since they are OR'ed, one way of an other is true.
IT IS A TAUTOLOGY..
My prof dint accept my answer, he said need little more consideration.
AI: Your professor is mistaken, and your argument is correct, though it could be simplified slightly without any major change, since you could combine the second and third cases:
If $p$ and $q$ are both false, then $p\to q$ is true; otherwise at least one of $p$ and $q$ is true, and therefore at least one of $r\to p$ and $r\to q$ is true. In all cases, therefore, at least one of the implications is true and therefore their disjunction is true.
However, it can be simplified even more by looking just at $p$: if $p$ is true, then $r\to p$ is true, and if $p$ is false, then $p\to q$ is true. |
H: Show that $ x_n = \left(1 + \frac{r}{n} \right)^n $ has an upper bound
I asked this question but maybe my doubt was not enough clear. So I will ask something more specific: Show the sequence $x_n = \left(1 + \frac{r}{n}\right)^n$ for $ r \in \mathbb{Q}, r>0$ has an upper bound.
I tried to show it as we do for $a_n = \left(1 + \frac{1}{n} \right)^n $ (using the expansion of $a_n$ but I had no sucess.
I would like some hint!
Thanks!
AI: If you know that $\log(x)\leq x-1$ which is clear from the symmetry with the inverse function $\text e^x\geq x+1$ it follows that
$$
0\leq \log(x_n)=n\ \log\left(1+\frac{r}{n}\right)\leq r
$$
showing that $x_n$ is bounded. In fact this shows that $1\leq x_n\leq \text{e}^r$. Also it should be noted here that $\log(x)$ is an increasing function thus preserving inequalities. |
H: How was this derivative simplified further?
Here is the steps on how the derivative gets simplified:
We are given the point (0,0) for it to be evaluated at.
\begin{align*}\tan(3x+y) &= 3x \\
(3+y^{\prime}) \sec^2(3x+y) &= 3 \\
y^{\prime} &= \frac{3-3\sec^2(3x+y)}{\sec^2(3x+y)} \\
&= \frac{-3\tan^2(3x+y)}{\tan^2(3x+y)+1} \\
\qquad&=-3\sin^2(3x+y)
\end{align*}
and this is where I get confused! How does it further get simplified to this:}
\begin{align*}& = \frac{-27x^2}{9x^2+1}\end{align*}
Is the point used to simplify it further? Is it a trig identity or formula that they are using?
Thanks so much.
AI: $$
\tan(3x+y) = 3x
$$
Then
$$
\frac{-3\tan^2(3x+y)}{\tan^2(3x+y)+1}=\frac{-3 (3x)^2}{(3x^2)+1}== \frac{-27x^2}{9x^2+1}
$$ |
H: Cauchy's integral formula problem
Let $D$ be an open disk in $\mathbb{C}$ and let $\overline{D}$ be the closure of $D$. Suppose $f:\overline{D}\to\mathbb{C}$ is continuous on $\overline{D}$ and analytic on $D$, prove that
$$f(w)=\frac{1}{2\pi i}\displaystyle\int_{\partial{D}}\frac{f(z)}{z-w}\mathrm{d}z$$
for any $w\in D$.
Here $\partial{D}$ is the boundary of $D$. In this case $f$ is not analytic on $\overline{D}$ so we can not apply Cauchy's integral formula. Any hints? Thanks!
AI: For ease of notation, let's assume the centre of the disk is $0$. Then for $0 < r < 1$, the function
$$f_r(z) = f (r\cdot z)$$
is holomorphic in a neighbourhood of $\overline{D}$. Thus
$$f_r(w) = \frac{1}{2\pi i} \int_{\partial D} \frac{f_r(z)}{z-w}\,dz.$$
Since $f_r \to f$ uniformly on $\overline{D}$ for $r\to 1$, we can interchange limit and integral to obtain
$$f(w) = \lim_{r\to 1} f_r(w) = \lim_{r\to 1} \frac{1}{2\pi i} \int_{\partial D} \frac{f_r(z)}{z-w}\,dz = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z)}{z-w}\,dz.$$ |
H: Constructing regular grammar
I'm trying to make a regular grammar for this language:
$$ L = \{ a^ncb^m(cc)^p : n\ge 1, m\le 1, p\ge 0\} $$
Where the alphabet is $ \Sigma = \{a,b,c\}$
It seemed like right-linear. This may be disastrously wrong, but here's how I started:
$S \to aA$
$A \to aA | cB $
$B \to bC | C $
$C \to cD |$ $\lambda$
$D \to cC$
EDIT:
Corrected for the errors pointed out.
AI: You’ve a couple of problems right off the bat: your grammar can generate at most one $a$, and it can generate the word $a$, which isn’t in $L$. You can solve both of these problems by replacing $A\to cB\mid\lambda$ with $A\to aA\mid cB$: with those and the initial production $S\to aA$ you can generate all strings of the form $a^ncB$ with $n\ge 1$.
You’re in trouble again with the $B$ productions: $B\to bB$ allows you to generate as long a string of $b$’s as you like, when you’re supposed to generate at most one $b$. You needed something like this for the $a$’s; you don’t want it for the $b$’s. You don’t need the $\lambda$ production, either: you can take care of that at the $C$ stage. What you need here is $B\to bC\mid C$: generate at most one $b$ and shift to generating $cc$’s. $C\to ccC\mid\lambda$ is then fine, provided that your definition of regular grammar is what many refer to as an extended regular grammar. This is a right linear grammar, but note that right linearity is a property of the grammar, not of the language.
If you’re required to use a regular grammar in the strict sense of the term, you need to replace $C\to ccC$ with regular productions, e.g., $C\to cD\mid\lambda$ and $D\to cC$. (Note that you do not want a $\lambda$ production for $D$, since you don’t want to stop with an odd number of final $c$’s.) |
H: Computing complex integral with absolute value
What is the value of $\int_{|z|=1}|z-1||dz|$?
By definition, if $z$ is parametrized by $z=e^{i\theta}$, the integral is $$\int_0^{2\pi}|e^{i\theta}-1|\cdot|ie^{i\theta}|d\theta = \int_0^{2\pi}|e^{2i\theta}-e^{i\theta}|d\theta$$ It has an absolute value, so I don't know how to integrate it.
AI: The differential $|dz|$ is an infinitely small increment of arc length. In fact
$$
|dz| = |d(e^{i\theta})|=|ie^{i\theta}\,d\theta| = d\theta.
$$
As pointed out by njguliyev,
$$
|e^{i\theta}-1| = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} = \sqrt{2-2\cos\theta}.
$$
So you have
$$
\int_0^{2\pi} \sqrt{2-2\cos\theta} \, d\theta.
$$
The tangent half-angle substitution $t = \tan\dfrac\theta2$ leads via trigonometric identities to
$$
\cos\theta=\frac{1-t^2}{1+t^2}\text{ and }d\theta=\frac{2\,dt}{1+t^2},
$$
and as $\theta$ goes from $0$ to $2\pi$, then $t$ goes first from $0$ to $\infty$ and then from $-\infty$ to $0$.
The integral becomes
$$
\int_{-\infty}^\infty \sqrt{1-\frac{1-t^2}{1+t^2}} \, \frac{2\,dt}{1+t^2}
= \int_{-\infty}^\infty \frac{2\sqrt{2}|t|\,dt}{\sqrt{1+t^2}^3}
= 2\int_0^\infty\frac{2\sqrt{2}t\,dt}{\sqrt{1+t^2}^3}.
$$
The factor $|t|$ comes from the fact that $\sqrt{t^2}=|t|$ and vanishes when the bounds exclude negative values of $t$.
Then let $u=1+t^2$ so that $du=2t\,dt$ and we get
$$
2\int_1^\infty \frac{2\sqrt{2}\,du}{u^{3/2}}.
$$
Note that as $t$ goes from $0$ to $\infty$, $u$ goes from $1$ (not $0$) to $\infty$. |
H: Calculating Variance to the power of a variable
Let X be standard uniform random variable. That is, X has the density $f_x(x) = 1$ for 0 < x < 1 and 0 elsewhere. Suppose that we toss a fair coin (independently of the value of X) and set $Y = X$ if the coin shows heads, and $Y = 1$ if the coin shows tails.
Calculate $Var(Y^p)$ for any p>0
Calculate the limit of $Var(Y^p)$ as p $\to \infty$ Can you think of a way to figure out the limit without having to do the calculation in the first part?
So I'm not exactly sure how to start this problem, as far as I know there is no real rule for variances to take out the power of p factor. I imagine I would need to try something akin to
$Var(Y^p) = E(Y^{2p}) - (E(Y^p))^2 $. However I'm not sure how to calculate $E(Y^p)$. I know $E(Y)$ would simply be $E(Y) = (\frac12)(\frac12) + (\frac12)(1) $ As there is a 1/2 chance of heads and if it is heads the distribution $f_x$ has an expected value 1/2, similarly if it is tails it has 1/2 chance of being 1. Would $E(Y^p)=(\frac12)(\frac12)^p + (\frac12)(1)$ make any sense?
For part (b) the limit is simply 0 or infinity depending whether the variance is a fraction or greater than 1. I'm not sure what another method would be.
AI: Your approach is fine but you have to modify $\mathbb E(X^p)$ which is calculated as follows:
$$
\mathbb E(X^p)=\int_0^1 x^pdx=\frac{1}{p+1}
$$
Therefore:
$$
\mathbb E(Y^p)=\frac{1}{2}+\frac{1}{2}\frac{1}{p+1}\\
\mathbb E(Y^{2p})=\frac{1}{2}+\frac{1}{2}\frac{1}{2p+1}
$$
and finally
$$
\mathbb{Var}(Y^p)=\frac{1}{2}+\frac{1}{2}\frac{1}{2p+1}-\left(\frac{1}{2}+\frac{1}{2}\frac{1}{p+1}\right)^2
$$
and we can say:
$$
\lim_{p\to\infty}\mathbb{Var}(Y^p)=\frac{1}{4}.
$$ |
H: Volume of a solid involving integration by parts.
Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve $y=e^x$ and the line $x=\ln 5$, rotated about the line $x= \ln 5$
Here is my work:
$$V= 2\pi \int _{a}^{b}(\text{shell radius})(\text {shell height})dx$$
$$V= 2\pi \int _{0}^{\ln 5}(x)(e^x)dx$$
$$u=x\,; du=dx\,\,\,\,\,dv=e^xdx\,;v=e^x$$
Using the integration by parts formula: $\int udv = uv-\int vdu$
$$2\pi [ xe^x|_{0}^{\ln 5} - \int_{0}^{\ln 5}e^xdx]$$
$$2\pi [ xe^x|_{0}^{\ln 5} - e^x|_{0}^{\ln 5}]$$
$$2\pi [(\ln 5)5 - (5-1)]$$
$$2\pi [5\ln 5 - 4]$$
This was the answer I got but the answer on the online homework says the correct answer is $2\pi(4-\ln5)$. Are these two answers equivalent or did I do something wrong? Any help is appreciated!
AI: The radius of the shell at $x$ isn’t $x$: that’s the distance to the $y$-axis, and you’re not revolving the region about the $y$-axis. The distance from the shell at $x$ to the axis at $\ln 5$ is $\ln 5-x$. The rest of your integral is fine, so you should have
$$2\pi\int_0^{\ln 5}(\ln 5-x)e^x\,dx\;.$$
Note that all you have to do now is compute
$$2\pi\ln 5\int_0^{\ln 5}e^x\,dx$$
and combine it properly with the erroneous value that you already calculated. |
H: Hard Integral $\frac{1}{(1+x^2+y^2+z^2)^2}$
Prove that $\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz = \pi^2$
I tried substitution, trigonometric substitution, and partial fraction decomposition, but I can't solve this problem, I only know that
$\frac{1}{(1+x^2+y^2+z^2)^2}$ is a even function :( then
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz = $$
$$ 2 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{0}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx \, dy \, dz $$
AI: You can use spherical coordinates. Then the integral changes to the following:
$$
\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty} \frac{1}{(1+x^2+y^2+z^2)^2}\, dx dy dz\\ = \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{\pi}\displaystyle\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2\sin\theta d\rho d\theta d\phi \\=2\pi.2\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2 d\rho
$$
Now it is enough to calculate the following:
$$
\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2 d\rho.
$$
which you can find by $\rho=\tan y$ and then:
$$
\int_{0}^{\infty} \frac{1}{(1+\rho^2)^2}\,\rho^2 d\rho=\int_{0}^{\pi/2} \sin^2 y dy=\frac{\pi}{4}
$$ |
H: Limit as $x$ approaches $0$ for a function.
Unfortunately this question does not make the slightest sense to me. I know how to find limits but this is confusing. Any help will be appreciated...
$$\lim_{x\to 0}\frac{\sin(x^m)}{(\sin x)^p}$$
where $m$ and $p$ are positive integers. (Hint: the answer will depend on
$m$ and $p$.)
How will I find that? Thanks!
AI: Hint:
$$\lim_{x\to 0}\frac{\sin(x^m)}{(\sin{x})^p}=\lim_{x\to 0}\frac{\sin(x^m)x^p x ^m}{x^p x^m(\sin{x})^p}=\lim_{x\to 0}\left(\dfrac{\sin(x^m)}{x ^m}\cdot\dfrac{x^p}{(\sin{x})^p}\cdot\dfrac{x^m}{x^p}\right)$$ |
H: Solving a second order homogenous ODE with non-real complex roots
I'm working on solving this problem:
$$ y''+ 48y = 0 $$
For a typical homogenous ODE with real roots we let $y=e^{rx}$ and solve for the roots $r_1$ and $r_2$:
$$
y=e^{rx}$$ $$
y'=re^{rx} $$ $$
y''=r^2e^{rx} $$
$$ r^2e^{rx} + 48e^{rx} = 0 $$
$$ e^{rx} (r^2+48) = 0 $$
Since $e^{rx}$ cannot equal zero we set $r^2+48$ equal to zero then solve. However, this results in a imaginary number. What's a young math student to do in this situation?
AI: You can also just write off the top:
$$r^2 + 48 = 0 \rightarrow r_{1,2} = \pm i \sqrt{48} = \pm i 4 \sqrt{3}$$
So, we have:
$y_1 = e ^{+4\sqrt{3} i~x}$
$y_2 = e ^{-4\sqrt{3} i~x}$
We know: $ e^{it} = \cos t + i \sin t$.
So, we get:
$$y(x) = y_1(x) + y_2(x) = c_1 \cos( 4 \sqrt{3}x) + c_2 \sin( 4 \sqrt{3}x)$$ |
H: Is this a correct way to express $\left|f(x)\right| \leq \left|x\right|^9$?
If $\left|f(x)\right| \leq \left|x\right|^9$, then, is it correct to say that
$f(x) \leq x^9$ and $f(x) \geq -x^9$ ?
If it is not, could someone explain why? Thank you.
AI: To expand on what Mark Bennet said, your formulation is almost correct: it's not $-x^9\leq f(x)\leq x^9$, but it is $-|x|^9\leq f(x)\leq |x|^9$.
To get rid of the absolute values entirely takes some care, but it can be done. The expression is the same as the piecewise expression
$$
\left\{
\begin{array}{lr}
-x^9\leq f(x)\leq x^9 &: x>0 \\
-x^9\geq f(x)\geq x^9 &: x<0 \\
\end{array}
\right.
$$
which can be phrased using OR: Either $-x^9\leq f(x)\leq x^9>0~$ OR $~-x^9\geq f(x)\geq x^9 <0$. There's no really good reason to write something this silly, but if you need it in a particular form for reasons, this is one possibility. |
H: Question about calculating $2^{32101}\bmod 143$?
I am trying to calculate $2^{32101}\bmod 143$ with using paper and a calculator.
What bothers me is that $32101 = 47 \times 683$ and $143=11 \times 13$, so they are not prime numbers. This means i can't use Fermat's little theorem. I tried also to solve it by square and multiply algorithm, but i am stuck there too, so maybe it doesn't work too.
I have been thinking of using the Chinese reminder theorem, but i don't know how to apply it here, because of the prima facorization of $32101$...
Does anyone have an idea how to calculate suche large numbers when they are not prime?
I would be glad if someone could help me. Thank you in advance!
AI: Well, you already thought about using Fermat's Little Theorem and Chinese Remainder theorem, which combined can do this.
The basic idea is that you can compute
$2^{32101} \pmod{11}$ and $2^{32101} \pmod{13}$ using Fermat's Little Theorem, and then use the results to compute $2^{32101} \pmod{143}$ using Chinese Remainder Theorem. |
H: How do I prove by induction?
For example if i wanted to prove:
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start.
Thanks.
AI: Prove for the base case, n=1:
$$1^{2} = \frac{1(1+1)(2+1)}{6} = \frac{2\cdot3}{6}=1$$
The "sum" of just $1^2$ is indeed 1. Base case proven.
Now for the induction step, proving that it holds for n+1:
Observe that for $1^{2} + 2^{2} +... + n^{2}$, the sum including $n+1$ equals the original summation plus the $n+1$ term: $1^{2} + 2^{2} + ... + n^{2} + (n+1)^{2}$, i.e. equals $ \frac{n(n+1)(2n+1)}{6} + (n+1)^{2}$.
So all we need to do is show that $\frac{n(n+1)(2n+1)}{6} + (n+1)^{2} = \frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$.
(Edit: To be clear: $\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$ is $\frac{n(n+1)(2n+1)}{6}$ with $n+1$ substituted in for $n$.)
Let's do that now:
$$\frac{n(n+1)(2n+1)}{6} + (n+1)^{2} = \frac{n(n+1)(2n+1)}{6} + \frac{6(n+1)^{2}}{6}$$
$$ = \frac{n(n+1)(2n+1)+6(n+1)^2}{6}$$
$$ = \frac{n(n+1)(2n+1)+6(n^{2}+2n+1)}{6}$$
$$ = \frac{n(2n^{2}+3n+1)+6n^{2}+12n+6)}{6}$$
$$ = \frac{2n^{3}+3n^{2}+n+6n^{2}+12n+6}{6}$$
$$ = \frac{2n^{3}+9n^{2}+13n+6}{6}$$
Having simplified this, we'll now show that it is equal to $\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$:
$$\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} = \frac{(n+1)(n+2)(2n+3)}{6}$$
$$ = \frac{(n^{2}+3n+2)(2n+3)}{6}$$
$$ = \frac{2n^{3}+6n^{2}+4n+3n^{2}+9n+6}{6}$$
$$ = \frac{2n^{3}+9n^{2}+13n+6}{6}$$
And we're done! |
H: Partial differentiation with respect to multiple variables?
Let's say you have $y(v,t)$ where $v(x(s,t),t) = y(s,t)$.
I know that
$\frac{\partial y}{\partial t} = \frac{\partial v}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial v}{\partial t} $
However, I see in my textbook that the second partial derivative of $y$ with respect to $t$ is
$\frac{\partial^2 y}{\partial t^2} = \frac{\partial^2 v}{\partial x^2}(\frac{\partial x}{\partial t})^2 + 2(\frac{\partial^2 v}{\partial x \space \partial t}) + \frac{\partial^2 v}{\partial t^2} + \frac{\partial v}{\partial x}\frac{\partial^2 x}{\partial t^2}$
and I don't understand how to get there. At first I thought that in the first term the denominator of $\frac{\partial^2 v}{\partial x^2}$ was to the second power (yes, I am aware that that simply means the second partial derivative with rspect to x and has nothing to do with the operation of multiplying the same thing twice) because the $(\frac{\partial x}{\partial t})$ was also squared, and then I attempted to apply the normal power rule (fg)' = f'g + g'f, but I am really confused about this and any help would be really appreciated. This is from page 3 of H.F. Weinberger's Introduction to PDEs and I am just trying to relearn partial differentiation for my PDE class. Thanks
AI: Let $y(s,t) = v \big( x(s,t), t \big)$. For the partial derivative $\frac{\partial y}{\partial t}$, I prefer to write :
$$ \frac{\partial y}{\partial t} = \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) + \frac{\partial v}{\partial y}\big( x(s,t),t \big) \tag{$\star$}$$
This way, we clearly see what depends on $t$. To compute $\frac{\partial^{2} y}{\partial t^{2}}$, we have to differentiate the two terms in $(\star)$ with respect to $t$.
For the first one, we have :
$$
\begin{eqnarray*}
\frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) \Big] & = & \Big[\frac{\partial}{\partial t} \frac{\partial v}{\partial x}\big( x(s,t),t \big) \Big] \frac{\partial x}{\partial t}(s,t) + \frac{\partial v}{\partial x}\big( x(s,t),t \big) \Big[ \frac{\partial}{\partial t} \frac{\partial x}{\partial t}(s,t) \Big] \\
& = & \Big[ \frac{\partial^{2} v}{\partial^{2} x}\big( x(s,t),t) \frac{\partial x}{\partial t}(s,t) + \frac{\partial^{2} v}{\partial x \partial y}\big( x(s,t),t \big) \Big] \frac{\partial x}{\partial t}(s,t) + \\
& &\frac{\partial v}{\partial x}\big( x(s,t),t \big) \Big[\frac{\partial^{2} x}{\partial^{2} t}(s,t) \Big] \\
& = & \frac{\partial^{2} v}{\partial x^{2}}\big( x(s,t),t \big) \left( \frac{\partial x}{\partial t}(s,t) \right)^{2} + \frac{\partial^{2} v}{\partial x \partial y}\big( x(s,t),t \big)\frac{\partial x}{\partial t}(s,t) + \\ & & \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial^{2} x}{\partial t^{2}}(s,t) \\
\end{eqnarray*}
$$
Which we can summarize :
$$ \frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) \Big] = \frac{\partial^{2} v}{\partial x^{2}} \left( \frac{\partial x}{\partial t} \right)^{2} + \frac{\partial^{2} v}{\partial x \partial y} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial x} \frac{\partial^{2} x}{\partial t^{2}} \tag{1}$$
And for the second term, we have as well :
$$\frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial y}\big( x(s,t),t \big) \Big] = \frac{\partial^{2} v}{\partial y \partial x}\big( x(s,t),t \big) \frac{\partial x}{\partial t}(s,t) + \frac{\partial^{2} v}{\partial y^{2}}\big( x(s,t),t \big) $$
Which also writes :
$$ \frac{\partial}{\partial t} \Big[ \frac{\partial v}{\partial y}\big( x(s,t),t \big) \Big] = \frac{\partial^{2} v}{\partial y \partial x} \frac{\partial x}{\partial t} + \frac{\partial^{2} v}{\partial y^{2}} \tag{2}$$
Since $v$ is $\mathcal{C}^{2}$, from Schwartz lemma, we have :
$$ \frac{\partial^{2} v}{\partial x \partial y} = \frac{\partial^{2} v}{\partial y \partial x} $$
Hence, $(1)$ and $(2)$ give :
$$ \frac{\partial^{2} y}{\partial t^{2}} = \frac{\partial^{2} v}{\partial x^{2}} \left( \frac{\partial x}{\partial t} \right)^{2} + 2 \frac{\partial^{2} v}{\partial x \partial y} \frac{\partial x}{\partial t} + \frac{\partial^{2} v}{\partial y^{2}} + \frac{\partial v}{\partial x}\frac{\partial^{2} x}{\partial t^{2}}$$ |
H: What function space do $u(x)=\sin(x)$ and v$(x)=e^{-2x}$ belong to?
I'm trying to follow notes, a question is presented in the middle of the notes that isn't answered.
$u(x)=0$ when $-1\leq x<0$
$u(x)=x^2$ when $0\leq x \leq 1$
The $u(x)$ function belongs to $C^1[-1,1]$. What space do $u(x) = \sin x$ and $v(x) = e^{-2x}$ belong to?
AI: If the question is what's the largest $n \in \{0,1,2,\ldots\}\cup\{\infty\}$ such that $u \in C^n\lbrack-1,1\rbrack$ where $u(x) = \sin x$ or $u(x) = e^{-2x}$, then the answer (in both cases) is $n = \infty$. In fact we know that $u,v:\mathbb{R}\to\mathbb{R}$ given by $u(x) = \sin x$ and $v(x) = e^{-2x}$ are both infinitely differentiable on $\mathbb{R}$, which is a stronger assertion. |
H: Regular expression for a language
I made a regular expression to match this language but I'm not sure it's right. Perhaps someone can show me where it deviates.
The language:
$L = {a^{n} c b^{m} (cc)^{p} : n \geq 1, m \leq 1, p\geq 0}$
The expression:
$r = a a^{*} c (\lambda + b) (c c)^{*}$
AI: Yes, that’s exactly right. If you allow the abbreviation $a^+$ for $aa^*$, you can use that instead. |
H: Find the equation of the tangent line to the curve at the given point. $y = 1+2x-x^3$ at $(1,2)$
I have the equation $y = 1+2x-x^3$ and the point $(1,2)$.
When I work it out I come up with the derivative of $2-3x^2$.
When I apply the point I come up with a slope of $-1$ and a tangent line of $y=4-x$.
Can someone work it out and confirm my answer or show me where I am going wrong?
AI: Try using the point-slope formula to obtain the equation of the line, given $m = -1$ and $(x_0, y_0) = (1,2)$, which is given by:
$$y - y_0 = m(x - x_0):$$
$$ y - 2 = -(x - 1) \iff y = -x + 3$$ |
H: Language made by a regular expression
I created a language from this regular expression but I'm not sure about it, especially where I wanted to use the $w$ to express a sequence of terminals.
The expression:
$r = a a ^{*} (b + bb + bbb) (a + b) (a + b) ^ {*}$
The alphabet:
$\Sigma = \{a,b\}$
The language:
$L = \{ a a ^{n} b b^{m} w^{p} : n \geq 0, m \leq 2, p \geq 1, $w is any sequence of a's and b's including $\lambda$ $ \}$
AI: Your description of $L$ is not wrong, but it’s a bit clumsy. For instance, instead of writing $aa^n$ and requiring $n\ge 0$, you can write just $a^n$ with the requirement $n\ge 1$. You can also combine $bb^m$ into $b^m$ by changing the requirement on $m$. More important, there’s no need for an exponent on $w$, since $w$ is already an arbitrary string of $a$’s and $b$’s. If you make these changes, you get
$$L=\left\{a^nb^mw\in\{a,b\}^*:n\ge 1,1\le m\le 3,\text{ and }|w|\ge 1\right\}\;.$$
In other words, the language consists of those words that start with a string of one or more $a$’s, followed by one to three $b$’s, followed by any non-empty string of $a$’s and $b$’s. |
H: Existence of a certain 2-coloring of a graph
There are n studio apartments in a building. Some of the apartments are connected with each other by direct phone line. Prove that it is possible to assign to each apartment a female or a male in such way that each person has direct connection with at least as many people of the opposite sex as he/she has with people the same sex.
AI: Assign the apartments in a way that maximizes the number of direct lines connecting a male to a female. Consider any tenant, assumed w.l.o.g. to be male. If he were connected to more men than women, then replacing him with a woman would increase the number of male-female connections, contradicting the assumed maximality.
By a standard "compactness" argument [*], the result can be generalized to locally finite graphs, i.e., the number of apartments may be infinite, but each one is connected to only finitely many others.
The result easily generalizes to any number of sexes. To put it in standard graph-theoretic terminology:
Theorem. Given a locally finite graph $G$ and a nonempty finite set $S$, we can find a function $f:V(G)\to S$ such that, for any $x\in V(G)$ and any $s\in S$, we have $$|\{y\in N(x):f(y)=s\}|\ge|\{y\in N(x):f(y)=f(x)\}|.$$
[*] In case anyone is interested, here is (one version of) the "compactness argument" used to get from the finite to the locally finite case. Let $F$ be the set of all functions $f:V(G)\to S$. Viewed as the Tychonoff product of copies of $S$ endowed with the discrete topology, $F$ is a compact space. For each vertex $x$, let $K_x$ be the set of all functions $f\in F$ such that the displayed inequality holds at $x$ (for all $s\in S$). Note that $K_x$ is a closed set. (The assumption of local finiteness is used here.)
I claim the the family $\{K_x:x\in V(G)\}$ has the finite intersection property, i.e., that $\bigcap_{x\in X}K_x\ne\emptyset$ for each finite set $X\subseteq V(G)$. This follows from the finite case of the theorem, applied to the finite graph obtained by restricting $G$ to the finite set consisting of the vertices in $X$ and their neighbors. (Local finiteness is used again here.) Hence, by compactness, we have $\bigcap_{x\in V(G)}K_x\ne\emptyset$, Q.E.D. |
H: How can we directly see that the number of random walks starting and ending at the origin is ${n\choose n/2}^2$?
In an infinite two-dimensional square-shaped grid, we define four directions, north, south, east, west. We thus have $4^n$ random walks of length $n$. If we end where we started, for every north step we have a south one, and similarly for east and west. Supposing $k$ north-south and $n/2 -k$ east-west steps, we find that the number of possible walks is $$\sum_{k=0}^{n/2}\frac{n!}{k!^2(n/2-k!)!^2}={n\choose n/2}\sum_{k=0}^{n/2}{n/2\choose k}^2={n\choose n/2}^2$$ a beautiful result that hints a cleaner interpretation. How can we see this directly from the grid?
AI: Since $n$ must be even in order for the walk to be closed, I’ll write $n=2m$. Let $n_N,n_S,n_E$, and $n_W$ be the number of steps north, south, east, and west, respectively. Then $n_N=n_S$ and $n_E=n_W$, so $n_N+n_E=n_S+n_W=m$ and $n_N+n_W=n_S+n_E=m$.
Note also that if we choose $n_N,n_S,n_E$, and $n_W$ so that $n_N+n_E=n_S+n_W=m$ and $n_N+n_W=n_S+n_E=m$, then automatically $n_N=n_S$ and $n_E=n_W$.
Now imagine charting an $n$-step walk by starting with a strip of $n$ squares and labelling each square N, S, E, or W for steps north, south, east, and west, respectively. However, we’ll do it in a slightly odd way. First choose any $m$ squares and mark them $\nearrow$. These will be the $m$ steps that go either north or east. Then choose any $m$ squares and mark them $\nwarrow$; these will be the $m$ steps that go either north or west. Clearly this procedure can be carried out in $\binom{2m}m^2$ ways. Now go through the strip and change the markings according the following rules:
a square marked with both $\nearrow$ and $\nwarrow$ is marked N.
a square marked only with $\nearrow$ is marked E.
a square marked only with $\nwarrow$ is marked W.
an unmarked square is marked S.
It follows from the remark in the second paragraph that we’ve laid out a chart for a path that returns to the origin, and it’s not hard to see that every $n$-step path that returns to the origin has a chart that can be produced in this way. There are $\binom{2m}m^2$ charts produced in this way, so there are $\binom{2m}m^2$ $n$-step paths that return to the origin. |
H: How does $\zeta(i\pi)$ converge?
$$\zeta(i\pi) = \sum_{r=1}^{\infty}r^{-i\pi} = \sum_{r=1}^{\infty}e^{-i\pi \ln(r)} = \sum_{r=1}^{\infty}\operatorname{cis}(-\pi\ln(r))$$
Did I mess up somewhere in the steps above? I can't see how the last expression would converge.
AI: Your mistake is thinking that $\zeta(i \pi)$ is defined by that sum.
That summation only defines $\zeta$ for complex numbers with real part bigger than one. This definition is extended to the rest of the complex plane by the process of analytic continuation.
The basic theoretical (but not practical) idea is that once you know $\zeta(z)$ for such complex numbers, you could then, say, find its Taylor expansion around the point $1.01 + 3i$. This Taylor expansion has radius of convergence larger than 0.01, and so it lets you define $\zeta(z)$ for even more numbers. And you keep going until you get everywhere in the plane.
And happily, $\zeta(z)$ is a sufficiently nice function that you can do this consistently. More annoying functions would have branch cuts and "monodromy" and stuff.
Of course, this theoretical description is only really useful for explaining that it can be done, not how to actually compute it in practice or study it fruitfully in theory. That uses tools like the functional equation for $\zeta$, or the construction of other series that sum to $\zeta$. Your wolfram alpha link gives a few of those. |
H: Prove by induction that $1^3 + \dots + n^3 = (1 + \dots + n)^2$
I'm suppose to prove by induction:
$1^3 + \dots + n^3 = (1 + \dots + n)^2$
This is my attempt; I'm stuck on the problem of factoring dots.
AI: Your attempt looks OK as far as it goes (except for a missing superscript $2$ at one point, but that's not causing any further errors).
To prove that something is equal to $1^3+2^3+3^3+4^3+\cdots+n^3$, one must show that at each step the amount that gets added is the next cube.
So how much has to be added to $(1+2+\cdots+n)^2$ to get $(1+2+\cdots+n+[n+1])^2$? We need to show that that is the next cube, $(n+1)^3$. The amount that gets added is
$$
\begin{align}
& \phantom{={}} (1+2+\cdots+n+[n+1])^2 - (1+2+\cdots+n)^2 \\[6pt]
& = (A + [n+1])^2 - (A)^2 \\[6pt]
& = A^2 + 2A[n+1]+[n+1]^2 - A^2 \\[6pt]
& = 2A[n+1] + [n+1]^2.
\end{align}
$$
Now here it helps to know that
$$
A = 1 + 2 + 3 + \cdots+n = \frac{n(n+1)}{2}.
$$
Thus the difference above is
$$
2A[n+1]+[n+1]^2 = n(n+1)[n+1]+[n+1]^2
$$
and this simplifies to
$$
n^3+3n^2+3n+1
$$
and finally to
$$
(n+1)^3.
$$
Next, there's the problem of organizing that into a presentable proof by induction. |
H: Is this set of integer sequences countable?
I'm faced with a set of strictly increasing functions $\Bbb N\to \Bbb N$, i.e. positive integer valued sequences. The only thing I know about them is that they are pairwise eventually disjoint, by which I mean that given $U$, $V$ in this set, $U(\Bbb N)\cap V(\Bbb N)$ is finite.
Is this family countable? Uncountable? Does it depend?
Obviously the set of all positive integer sequences is uncountable, so this feels like a bit of a long shot, but I'm hoping the pairwise eventually disjoint condition might put a cap on the cardinality.
One thing I've been able to do, using a sort of diagonal argument, is to show that given such a set, if it's countable, I can find a new sequence that's eventually pairwise disjoint to every element in the set, and yet has no number assigned to it. But this new sequence won't actually be in the set, so this doesn't produce a contradiction like a diagonal argument is supposed to.
If the set is countable, then I can "decapitate" each sequence by a finite amount to obtain a set of pairwise disjoint sequences, which I'd really like to be able to do.
AI: You can have as many as $2^\omega=\mathfrak{c}$ such sequences. First, we can identify strictly increasing sequences with their ranges. Two subsets of $\Bbb N$ are almost disjoint if their intersection is finite. There are families of $2^\omega$ pairwise almost disjoint infinite subsets of $\Bbb N$: see my answer to this question and the answers to the question that it duplicated for a variety of constructions. |
H: Conditional Probability of census burea
The question seems very straight forward , but when i attempted it the answer is wrong.
I just took the probability of all three groups and multiplied by themselves the intersections and addition them.
this is the question.
In early 2001, the United States Census Bureau started releasing the results of the latest census. Among many other pieces of information, the bureau recoded the race or ethnicity of the residents of every county in every state. From these results the bureau calculated a 'diversity index,' which measures the probability that two people chosen at random are of different races or ethnicities. The census determined that in a county in Wisconsin, 83% of its residents are white, 10% are black, and 7% are asian. Calculate the diversity index for this county.
Diversity Index =
AI: What you're required to find is the following:$P(\text{Picking 2 people from different ethnicities})$. Let $C$ be the event that we pick 2 people and they are from 2 different ethnicities. $$P(\text{Picking 2 people from different ethnicities})=1-P(\text{2 chosen people belong to the same ethnicity})$$
The simpler question now is, given that you pick 2 people, what is the probability that they are from the same ethnicity; given that there are 3 different ethnicities spanning the whole Sample Space? |
H: Can someone explain these matrix operations to me?
So I have a row reduced echelon form matrix, that is $\begin{bmatrix}
I & F\\
0 & 0
\end{bmatrix}$.
The transpose of this matrix is thus $\begin{bmatrix}
I & 0\\
F^T & 0
\end{bmatrix}$.
This is all well and good and makes sense, but why does their product equal to $\begin{bmatrix}
I & F\\
F^T & 0
\end{bmatrix}$?
Do you multiply each block matrix like vectors? I'm confused.
AI: Yes, if the sizes of the corresponding blocks match, then we have
$$\pmatrix{A&B\\C&D}\cdot\pmatrix{X&Y\\U&V}\ = \ \pmatrix{AX+BU & AY+BV\\
CX+DU & CY+DV}\,,$$
which is a consequence of the matrix multiplication. |
H: Summation and Time Complexity
So I'm studying for my data structures midterm that's this monday, and my professor gave out a sample midterm with the answers, but I'm having a hard time understanding one of the questions.
Here's a screen cap:
Could someone give me a walk through of the math that's happening in each step? And possibly a conceptual explanation in computer science terms?
AI: You see no matter what values i,j take, k goes from 1 to n. Thus cont will run n times runs of the first two loops. Consider i,j, fix i, j goes from 1 to i. Thus the total time complexity
$$T(n)=n\cdot\left((1)+(1+2)+...+(1+2+...+n)\right)$$ |
H: Why are the roots of the polynomial $z^N = a^N$ equal to $z_k = a \ e^{j\frac{2 \pi k}{N}}$?
I am trying to understand equation 3.28 from this image in my book.
I get everything that the author is saying, except for when he finds the roots, (zeros), of $z^N = a^N$. Of course, there are going to be $N$ roots. He says the roots are given by $z_k = a \ e^{j \frac{2 \pi k}{N}}$.
What I do not understand is why they HAVE to be evenly spaced around a circle here? Where does it say that they have to be evenly spaced here? Why arent they all squished to one part of the circle, or why cant they be randomly spaced within the circle? I get that yes, those are solutions, but I can think of an infinitely more number of solutions that are all on the circle, but not evenly spaced...
Would appreciate someome elucidating this for me. Thank you.
AI: To find the solutions of $z^n=a^n$ you write the equation in polar coordinates. If $z=re^{j\theta}$, then we have
$$r^ne^{jn\theta}=a^ne^{j0}.$$
The magnitudes must be equal so we get $r^n=a^n$, which implies $r=a$. We also have $e^{jn\theta}=e^{i0}(=1).$ This is satisfied only if $n\theta=2\pi k$ for some integer $k$. So the solutions to the equation are $$z=ae^{j\frac{2\pi k}{n}}$$ where $k$ is an integer. Now you can check that you only get $n$ different complex numbers that correspond to $k=0\ldots n-1$, any other value of $k$ will just be a repeated solution.
Edit: We are not forcing the solutions to be evenly spread around the circle, we are just trying to find all the possible solutions. A priori we have a solution for each integer $k$. What I claim is that we have a lot of repeated solutions. For example, $k=0$ and $k=n$ give you the same solution
$$ae^{j\frac{2\pi 0}{n}}=a=ae^{j\frac{2\pi n}{n}}.$$
In fact, it is enough to consider $k=1\ldots n$ because any other $k$ would give you a solution which is already represented in ${1,\ldots,n}$. One way you can think about it is that you start with the solution given by $k=1$ and then increasing $k$ by one corresponds to rotating your solution by a $\frac{2\pi}{n}$ angle. Thus, when your reach $k=n+1$ you have completed a full turn in the circle and you start going through the same solutions. |
H: Invertibility of a square matrix with zero diagonal elements and positive non-diagonal elements
$M$ is square and
$$M(i,j)=0, i=j$$
$$M(i,j)>0, i\ne j$$
Is $M$ full-rank or invertible?
Actually the $M$ I am studying has much stronger properties but I guess the simple conditions above might be enough to make $M$ non-singular. The stronger properties of $M$ are:
All elements in $M$ are non-negative integers between $0$ and $N$;
The sum of each row is equal to $N$.
Edit
$N$ is not the dimension of $M$. It's just a constant positive integer.
I did search for this problem, but it seemed there was no much work on such matrices. It looks easy, but I don't know how to prove it and I couldn't find a counter-example either.
AI: The matrix $M$ doesn't have to invertible. I give a simple counterexample here.
$$M=\begin{pmatrix}
0&2&1&1\\
2&0&1&1\\
1&1&0&2\\
1&1&2&0
\end{pmatrix}$$
You will see $r_1+r_2-r_3-r_4=0$ which indicates they are linear dependent. |
H: 8-digit sequences invoving exactly 6 different digits
How many 8-digit sequences are there involving exactly six different digits? How can I approach this problem?
AI: HINT:
Count the ways to choose a set of $6$ digits to use.
Given a set of $6$ digits, either you use one digit $3$ times, or you use two digits twice each. Treat those cases separately.
In each case you must first choose the digit(s) to be duplicated; in how many ways can that be done? Then you’re solving a problem of the same general type as counting the number of distinguishable arrangements of the letters in the name MISSISSIPPI; I expect that you’ve done some problems of that type.
Properly combine the results of the previous steps. |
H: Suppose $F: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous functions. If $f$ and $g$ are measurable, then
Suppose $F: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous functions. If $f$ and $g$ are measurable, then $h(x) = F( f(x), g(x) ) $ is also measurable.
Proof:
For all $a$, $ \{ x : h(x) > a \} = \{ x : (f(x),g(x)) \in G_a \} $ where $G_a := \{ (u,v) : F(u,v) > a \} = F^{-1}((a, \infty))$. Since $F$ is continuous, then given any $(a, \infty)$, we must have $F^{-1}((a, \infty))$ open set. Since open sets are union of open rectangles on $\mathbb{R}^2$, then it must be the case that
$$ G_a = \bigcup_{n=1}^{\infty} (a_n,b_n) \times (c_n,d_n) $$ Therefore, Can we say the next step follows from the above? Im having trouble proving this holds. Can someone help me?
$$ \{ x : h(x) > a \} = \bigcup_{n=1}^{\infty} \{ x : f(x) \in (a_n,b_n) \} \cap \{ x : g(x) \in (c_n, d_n) \} $$
And the above is measurable since it is union of measurable sets, thence $F$ must be measurable.
AI: \begin{align*}
h(x) > a &\iff
(f(x),g(x)) \in G_a
\\
&\iff
(f(x),g(x)) \in (a_n,b_n) \times (c_n, d_n) \text{ for some } n
\\
&\iff f(x) \in (a_n,b_n) \text{ and } g(x) \in (c_n, d_n) \text{ for some } n
\\
\end{align*}
So
\begin{align*}
\{x : h(x) > a\} = \bigcup_n \{x : f(x) \in (a_n,b_n)\} \cap \{x : g(x) \in (c_n, d_n)\}
\end{align*} |
H: Conditional Probability injection-molding
I thought about this questions for a while now. I understand the fact that the foreman forgets to shut up the injection is 0.48. then the questions confuses me when they to tell us the probability of the defective molding will be produced 0.04 to 0.19 in early morning. this fact really threw me off. I am now confused. I am not even sure how to approach this problem.
A foreman for an injection-molding firm admits that on 48% of his shifts, he forgets to shut off the injection machine on his line. This causes the machine to overheat, increasing the probability that a defective molding will be produced during the early morning run from 4% to 19%. If a molding is randomly selected from the early morning run of a random day, what is the probability that it is defective?
Probability =
AI: Hint:
Suppose there was $1$ early morning moulding a day for $10000$ days. How many defective mouldings might you expect? Fill in the question marks beloe
on ????? days you would expect the machine to overheat, and so you would expect ???? defective mouldings from those days
on ????? days you would expect the machine not to overheat, and so you would expect ???? defective mouldings from those days
so in total you would expect ???? defective mouldings out of $10000$, so the probability of a defective moulding is ???? |
H: Question about conformal maps.
By definition, a diffeomorphism $\sigma:(M,g)\to (N,h)$ is called conformal if $\sigma^*h=ug$. Another definition I've seen in other contexts is that conformal maps are ones that preserve angles. Now I've considered maps $\sigma$ from $M$ to $N$ that preserve angles (the angle between vectors being defined by $\frac{g(x,y)}{\sqrt{g(x,x)g(y,y)}}$), and I've found that if $\sigma$ preserves angles, then for $\sigma^*h$ to be equal to $ug$ it must be the case that $\frac{\sigma^*h(x_p,x_p)}{g(x_p,x_p)}$ is independent of the vectors. This isn't obvious to me if it is true, and if it isn't true then I'm wondering why conformal maps are defined as they are, as it would seem like an unnecessarily strong definition.
AI: Here's a physics-style (or 18th-century math style) explanation. It can be brought up to modern standards of mathematical rigor, but I think it's clearer without that. Imagine an infinitesimal triangle. Being infinitesimal, it has essentially straight edges. It gets mapped to another infinitesimal triangle with the same angles. But triangles with the same angles are similar, so the edges are enlarged or shrunk by the same factor. |
H: Find the formula for a derivative
Suppose: $$n(x)=\frac{1}{x}+1$$
Find $n'(x)$
The book indicates the answer is: $$\frac{-1}{x^2}$$ However I am not sure how it got that conclusion. Would anyone mind walking me though the problem step by step?
AI: (Hints)
You can write $\frac1x + 1 = x^{-1} + 1$
The derivative of $x^{-1} + 1$ is the same as the derivative of $x^{-1}$ plus the derivative of $1$
The derivative of $x^n$ is $nx^{n-1}$ for any $n $
What is the derivative of a constant? |
H: Equation of a tangent line: $f(x)=(x-4)(x^2-5)$, at $(2,2)$
Find an equation of the tangent line to the graph of f at the given point:
AI: What you want to use is the product rule for finding the derivatives. $\frac{d}{dx} uv = u'v + v'u$
So $u = (x-4), v = (x^2-5)$, find u' and v' and substitute into the equation to find the derivative.
Then substitute in $x = 2$ to find the slope of the line at $x = 2$, and then to find the equation of the line (since you already have the point) just substitute it into point slope form:
$y - y_1 = m(x - x_1)$
You should get $y - 2 = -9(x - 2)$ |
H: Seeking a combinatorial proof of the identity$1+3+\cdots+(2n-1)=n^2$
I would appreciate if somebody could help me with the following problem
Q: Seeking a combinatorial proof
$$1+3+\cdots+(2n-1)=n^2$$
AI: Consider a bag with balls numbered from $1$ to $n$. Number of ways of choosing $2$ balls with replacement is $n^2$.
We can also count the same in a different way. The pair of balls can be represented as $(i,j)$. Let us now look at the number of ways such that $\max\{i,j\} = k$, where $k \in \{1,2,\ldots,n\}$. If $C_k$ denotes the number of ways such that $\max\{i,j\} = k$, we then have
$$C_k = 2k-1$$
This is because if $\max\{i,j\} = k$, then either $j<i=k$ or $i<j=k$ or $i=j=k$.
Number of ways such that $j<i=k$ is $k-1$.
Number of ways such that $i<j=k$ is $k-1$.
Number of ways such that $i=j=k$ is $1$.
Hence, we have $C_k=2k-1$. Hence, the total number of ways of choosing a pair of balls from $n$ balls with replacement is
$$\sum_{k=1}^n C_k$$ |
H: Minimum size of directed path an orientation of a graph with a given independent set size
The problem I am working on is as follows:
Let $G$ be a graph with $n$ vertices and at most $r>=2$ independent vertices (no two are adjacent). Prove that if D is an orientation of G which does not contain a directed cycle then $D$ contains a directed path of length at least ⌈$n/r$⌉$-1$.
I have a theorem proven in class that says every directed graph has a path cover with the number of paths at most the size of the maximum independent set of vertices, and since there are always r paths that cover n vertices, if the paths have large differences in size one of them should easily be bigger than ⌈$n/r$⌉$-1$, and if they're as equal as possible then the largest of them (or if they're all equal, any one of them) will have size $n/r$ or $(n+k)/r$ for some positive integer $k$, and $(n+k)/r > n/r >= $⌈$n/r$⌉$-1$. But if this works (and I'm doubtful it does for exactly this reason), then why does D need to contain no directed cycles?
AI: I agree with you that the problem is oddly stated. The assumption that $D$ is acyclic is superfluous, as is the condition $r\ge2$, and there is no need to mention the undirected graph $G$. I would have stated it this way: "If $D$ is a directed graph with $n$ vertices and at most $r$ independent vertices, then $D$ contains a directed path of length $\lceil n/r\rceil-1$."
I find your discussion a little confusing. What do you mean by the "size" of a path? the number of vertices or the number of edges? I would explain it this way:
By the (Gallai-Milgram) theorem about path covers, the vertices of $D$ can be covered by a set of $r$ paths. The longest of those paths has $m$ vertices where $m\ge n/r$. Since $m$ is an integer, this means that $m\ge\lceil n/r\rceil$. Since the path has $m$ vertices, its length is $m-1\ge\lceil n/r\rceil-1$. |
H: How is this matrix called (two diagonals)?
I need to write an algorithm for solving this matrix but I wanted to first make a search online and that's why I need its name.
AI: You can rearrange terms, to get the following system
$$
\left [ \begin{array}{ccccccc}
d_1 & a_{2n+1} \\
a_1 & d_{2n+1} \\
& & d_2 & a_{2n} \\
& & a_2 & d_{2n} \\
& & & & \ddots \\
& & & & & d_n & a_{n+2} \\
& & & & & a_n & d_{n+2} \\
& & & & & & & d_{n+1}
\end{array}\right ] \left [ \begin{array}{c}
x_1 \\ x_{2n+1} \\ x_2 \\ x_{2n} \\ \vdots \\ x_n \\ x_{n+2} \\ x_{n+1}
\end{array}\right ] = \left [ \begin{array}{c}
b_1 \\ b_{2n+1} \\ b_2 \\ b_{2n} \\ \vdots \\ b_n \\ b_{n+2} \\ b_{n+1}
\end{array}\right ]
$$
which is equivalent to solving $n$ $2\times 2$ linear systems and $1$ algebraic equation of $1^{st}$ order. |
H: Prove that if $(v_1,\ldots,v_n)$ spans $V$, then so does the list $(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$
Prove that if $(v_1,\ldots,v_n)$ spans $V$, then so does the list $(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$
Proof:
Suppose that $V = \text{span } (v_1,\ldots,v_n).$ Then for any $v \in V$, there exist $a_1,\ldots,a_n \in \mathbb{F}$ such that
\begin{align}
v &= a_1v_1+\cdots +a_nv_n
\\ &= a_1v_1-a_1v_2 + a_1v_2+a_2v_2+a_3v_3+\cdots+ a_nv_n
\\ &= a_1v_1-a_1v_2+a_1v_2-a_1v_3+a_2v_2-a_2v_3+a_1v_3+a_2v_3+a_3v_3+a_4v_4+\cdots+a_nv_n
\\ &= a_1(v_1-v_2)+(a_1+a_2)(v_2-v_3)+(a_1+a_2+a_3)v_3+a_4v_4+\cdots+a_nv_n
\\ &=\sum_{i=1}^{n-1} \left[\left(\sum_{k=1}^{i}a_k\right)(v_i-v_{i+1})\right] + a_nv_n \in \text{span}(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).
\end{align}
Now let $u \in \text{span}(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$ Then there exist $b_1,\ldots,b_n \in \mathbb{F}$ such that
\begin{align}
u &= b_1(v_1-v_2)+ b_2(v_2-v_3)+\cdots +b_{n-1}(v_{n-1}-v_n) + b_nv_n
\\ &= b_1v_1 + (b_2-b_1)v_2+(b_3-b_2)v_3+\cdots+(b_n-b_{n-1})v_n \in \text{span}(v_1,\ldots,v_n)
\end{align}
completing the proof. $\hspace{125mm} \Box$
AI: Just for fun you can prove this more general result.
Let $V$ be an $n-$dimensional vector space, and let $\vec{v}\in V$. Assume that $\left( \vec{\beta}_1,\ldots ,\vec{\beta}_n\right)$ is a basis for $V$. If $\vec{v}=c_1\vec{\beta}_1 +\cdots+ c_n \vec{\beta}_n$, then for $c_j \ne 0$, we can replace one of the $\beta_j$'s with $v$ so that $\left( \vec{\beta}_1,\ldots , \vec{\beta}_{j-1},\vec{v},\vec{\beta}_{j+1},\ldots, \vec{\beta}_{n}\right)$ is a basis for $V$.
So you can use it $n$ times for each element in your list. |
H: Am I missing something with this question about nested compact sets?
On my homework, I have a question that states:
Let $X$ be a compact space, and suppose $\{ F_n \}$ is a countable collection of nonempty closed subsets of $X$ that are nested. Show that $\cap_n F_n$ is nonempty.
It seems trivial since the finite intersection, $\cap_{k=1}^n F_k$, is just $F_n$, which is nonempty.
AI: That's not what the notation means; by definition,
$$\bigcap_n F_n = F_1 \cap F_2 \cap F_3 \cap ...$$
and in general, might not be equal to any of the $F_n$'s.
The notation used here is slightly ambiguous, and it's much better to write something like
$$\bigcap_{n \in \Bbb{N}}$$ or $$\bigcap_{n = 1}^{\infty}$$ |
H: Characteristic polynomial of a linear operator T
Suppose that $V$ is a finite dimensional vector space with $B_1$ and $B_2$ as its ordered basis and let $T$ be a linear operator on $V$.
Then the matrices $[T]_{B_1}$ and $[T]_{B_2}$ are similar.
Why are they similar?
Also, because they are similar, it would follow that their characteristic polynomials are the same, which would further imply that, the characteristic polynomial of $T$ is independent of the choice of ordered basis, right ?
AI: For any two bases, there is a matrix $C$ such that if a vector has coordinate-vector $a$ in the first basis, then it has coordinate vector $Ca$ in the second. The best way to figure out what this matrix is is to consider the two bases as creating two entirely distinct vector spaces, $U$ for the first, and $W$ for the second.
To get the domains to match up properly, you have to have $C:U\to W$. Remembering that basis elements in $U$ and $W$ are just ones and zeros, you should be able to figure out what $C$ and $C^{-1}$ do to all vectors. It turns out that this $C$ is the right one such that $[T]_{B_2}=C[T]_{B_1}C^{-1}$.
And yes, everything you said in the last paragraph is right. |
H: Can we ever get an irrational number by dividing two rational numbers?
If we try to divide any two random arbitrarily long rational numbers like
103850.2387209375029375092730958297836958623986868349693868398659825528365...
and
127.123123123...
Is it guaranteed that the result is also a rational number?
AI: The quotient of two rationals is always a rational.
For if $\alpha = \frac{a}{b}$ and $\beta = \frac{c}{d}$ with $a, b, c, d$ integers with none of $b, c, d$ being zero, then
$$\frac{\alpha}{\beta} = \frac{ad}{bc}$$
is a quotient of integers, and so is rational. |
H: Right-linear grammar from regular expression
I made a right-linear grammar that from this regular expression:
The alphabet is:
$Σ = \{a, b, c\} $
Regular expression:
$r = cc^*(ba)^*bb$
My solution, it seems a little too short like I'm leaving something out. Maybe someone can see where I went wrong on the right-linearity:
$ S \to cA $
$ A \to b a A | B | cA $
$ B \to bb $
AI: You have a problem with the $A$ productions: you can have the derivation
$$S\Rightarrow cA\Rightarrow cbaA\Rightarrow cbacA\Rightarrow cbacB\Rightarrow cbacbb\;,$$
which you clearly don’t want. You need to make sure that once you stop generating $c$’s and generate something else, you never return to generating $c$’s. Why not let $S$ do the work of generating $c$’s, and confine $A$ to generating $ba$ pairs or stopping: $S\to cS\mid cA$, and $A\to baA\mid bb$? We now have
$$\begin{align*}
&S\to cS\mid cA\\
&A\to baA\mid bb\;.
\end{align*}$$
Any derivation must begin with some number $n$ of applications of $S\to cS$, where $n$ can be $0$, followed by an application of $S\to cA$; at that point we have
$$S\Rightarrow^* c^ncA\;.$$
Now we can only apply $A\to baA$ $m$ times for some $m\ge 0$ and finish off with $A\to bb$, and the result is
$$S\Rightarrow^* c^ncA\Rightarrow^* c^nc(ba)^mbb$$
for any $m,n\ge 0$. This is exactly what your regular expression requires, since $cc^*=c^*c$. |
H: How to show a set is nowhere dense
I want to show that $A=\{(x_1,x_2,x_3,0,0,...,0): x_i\in \mathbb{R}\}$ is nowhere dense in $\mathbb{R}^{n}$ with usual topology.
I know that I have to show $(\overline{A})^{0}=\emptyset$. But I couldn't find $\overline{A}$. Please help.
AI: $\overline{A} = A$ because of $(x_n) \in A, x_n \to x$, then each component of $x_n$ converges to the corresponding component of $x$. Hence,
$$
x(j) = 0 \quad\forall j\geq 4
$$
which implies $x\in A$
If $x\in A$, and $\epsilon > 0$, then the element $y\in \mathbb{R}^n$ given by
$$
y = (x_1, x_2, x_3, \epsilon/2, 0, 0, \ldots)
$$
satisfies $\|x-y\| < \epsilon$, but $y\notin A$. Hence, $B(x,\epsilon)$ is not contained in $A$ for any $\epsilon > 0$. |
H: Probability that exactly $n$ trials are required is $\binom{n-1}{k-1}p^{k}(1-p)^{n-k}$
Independent trials that result in a success with probability $p$ are
successively performed until a total of $k$ successes is obtained.
Show that the probability that exactly $n$ trials are required is
$$\binom{n-1}{k-1}p^{k}(1-p)^{n-k}$$
Let $A$ be the event that $k$ successes occur in $n$ trials. Then $P(A) = \binom{n}{k}p^k(1-p)^{n-k}$. Now, let $B$ be the event that exactly $n$ trials are required to obtain $k$ successes (as in the problem statement above). Then $P(B) = \binom{n-1}{k-1}p^{k}(1-p)^{n-k}$.
Since $P(B) \cdot \frac{n}{k} = P(A)$, I was thinking maybe that could lead to a proof.
How $A$ is different from $B$? For example, is there a formal way to state this, such as $B = \overline{A} \cap \dots$.)
I'm not looking for the answer, just some hints please.
AI: To see how $A$ is different from $B$:
Suppose that $k=3$ and $n=5$. If S is a success and F is a failure, SSSFF, SSFSF, SFSSF, and FSSSF all count towards $A$ but not towards $B$. The strings that count towards $A$ are those that have $3$ S’s and $2$ F’s; the ones that count towards $B$ are those that have $3$ S’s and $2$ F’s and end with an S. |
H: Doubt in the definition of graded rings
My doubt is quite simple, I didn't understand what is the operation $\cdot$ in this definition:
Thanks
AI: Given a ring $R$ and any two subsets $X\subseteq R$ and $Y\subseteq R$, we can define the subset
$$X\cdot Y=\{x\cdot y\mid x\in X,y\in Y\}$$
where $x\cdot y$ is just the ordinary product of two elements of a ring. So the statement that
$$S_d\cdot S_e\subseteq S_{d+e}$$
just says that the product of any element of $S_d$ with any element of $S_e$ is an element of $S_{d+e}$. |
H: Suppose $A$ is a subset of $\mathbb{R}^n$ with $m_{*}(A) = 0$, then show that $m_{*}(A \times R^m) = 0$ in $\mathbb{R}^{n+m}$.
Suppose $A$ is a subset of $\mathbb{R}^n$ with $m_{*}(A) = 0$, then show that $m_{*}(A \times R^m) = 0$ in $\mathbb{R}^{n+m}$.
The result seems intuitive, but I'm not sure how to prove it.
$m_{*}(A) = 0$ implies that for every $\epsilon$, there exists a rectangle $R$ in $\mathbb{R}^n$ such that $R$ covers $A$ and $m(R) < \epsilon$. Then $R \times \mathbb{R}^m$ covers $A \times \mathbb{R}^{m}$ and $m(R \times \mathbb{R}^m) \leq \epsilon * \infty = \infty$ .
I know that $0 * \infty = 0$, but how do I get to that point when every $\epsilon$ only gets me to $\infty$?
edit: I realize it doesn't mean there's a single rectangle that covers $A$, but a sequence of rectangles.
AI: Let $B_k$ be the box around the origin in $\mathbb{R}^m$ each of whose sides is of length 2k. Write
$$
A\times \mathbb{R}^m = \cup_{k=1}^{\infty} A\times B_k
$$
Then,
$$
m_{\ast}(A\times \mathbb{R}^m) \leq \sum_{k=1}^{\infty} m_{\ast}(A\times B_k)
$$
So it suffices to prove that $m(A\times B_k) = 0$ for all $k$. Now apply your argument with $A\times B_k$ instead of $A\times \mathbb{R}^m$ |
H: Why distribution in PDE is so like the distribution in Probability?
I just learnt distribution in PDE class.
Campared with distribution in probability, they have:
Same name
Similar concept for Delta function
Same converge mode(converge as distribution)
...
AI: Probability distributions are measures, and a measure is a distribution in the PDE sense. |
H: How to get average of Arun's weight?
This is an aptitude problem I found in a website, question no 9:
In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
How can I solve this? Actually I am getting more confused at different probable weights. What is it meaning?
AI: If all are correct, then Arun's weight can not be greater than 68 Kg and can not be less than 66 Kg. Thus average is 67 Kg. |
H: How prove $p$ have contains infinite prime numbers
give the natural Numbers $a,b,c$,and such $a<b<c$, if exsit prime number $p$, such
$$p+a,p+b,p+c$$ are Composite numbers,
show that:such this contidtion $p$ have contains infinite prime numbers.
My idea: for example $a=2,b=3,c=4$.
(1)then we take $p=23$ is prime number.and
the
$$p+a=25,p+b=26,p+c=27$$ are composite numbers
(2)and an other prime number $p=31$,
then
$$p+a=33,p+b=34,p+c=35$$ are composite numbers too.
(3) and take $p=47$,is prime number.and
$$p+a=49,p+b=50,p+c=51$$ are composite numbers too
and so on $\cdots$
For other any example,such
$$a=3,b=5,c=7$$
(1)then we take $p=3$,then
$$p+a=6,p+b=8,p+c=10$$ are composite numbers
(2) take $p=5$, then
$$p+a=8,p+b=10,p+c=12$$ are composite numbers
and so on $\cdots$
Now How prove my question,I think this is interesting problem,But I don't prove it,Thank you
AI: Recall that there are arbitrarily long gaps between consecutive primes. So for example let $N\gt c$. Then there are $N-1$ consecutive composites between $N!+2$ and $N!+N$. Let $p$ be the largest prime $\le N!+1$.
We can "start" this construction after any prime $q$, thereby getting infinitely many $p$ with the required property. |
H: Prove $\mu(V)=\mu(E) $, when $V = \cap_{k=1}^{\infty} {V_k} $
Prove $\mu(V)=\mu(E)$, when $V = \cap_{k=1}^{\infty} {V_k}$
$E \subset {V_k}$,For each k , $\mu(V_k) < \mu(E) + \epsilon\cdot 2^{-k}$
AI: Since $V=\bigcap_{i=1}^{\infty} V_i\subseteq V_k$ for any positive integer $k$, we have $$\mu(V)\leq \mu(V_k)<\mu(E)+\frac{\varepsilon}{2^k}.$$ Since the strict inequality $$\mu(V)<\mu(E)+\frac{\varepsilon}{2^k}$$ is true for all $k$, its weak version must be true in the limit: $$\mu(V)\leq\mu(E)+\lim_{k\to\infty}\frac{\varepsilon}{2^k}=\mu(E).$$
As for the other direction, $E$ is contained in $V_k$ for all $k$, so it must be contained in their intersection as well: $E\subseteq V$. By the monotonicity of measures, we have $\mu(E)\leq \mu(V)$.
Putting the two weak inequalities derived above together, we conclude that $\mu(V)=\mu(E)$.
In the proof above, I used the following
$\textbf{Lemma}\quad$ Suppose $a\in\mathbb{R}$ and $(b_k)_{k\in\mathbb{Z}_+}$ is a sequence of real numbers such that
(i) $a<b_k$ for all positive integers $k$; and
(ii) $\lim_{k\to\infty}b_k$ exists and is equal to $b$.
Then, $a\leq b$.
$\textit{Proof}\quad$ Suppose (i) and (ii) are true, but $a>b$. I will derive a contradiction. If $a>b$, let $\varepsilon=a-b>0$. Since $(b_k)$ converges to $b$ by (ii), there exists a positive integer $K$ such that for any positive integer $k>K$, the following must be true: $$|\,b_k-b\,|<\varepsilon.$$ But then, $$b_k-b\leq |\,b_k-b\,|<\varepsilon=a-b,$$ that is, $b_k<a$, which contradicts (i). Hence, supposing $a>b$ has led to a contradiction, so that if (i) and (ii) are true, then $a\leq b$ must necessarily be true as well. |
H: Basic examples of monoids?
What are some (simple/elementary) examples of noncommutative monoids with no additional structure? I'm having a hard time thinking of examples of "pure" monoids that aren't monoids simply because they are groups...
I've read this and this and some of this, but would like more examples that presuppose little to no algebra.
AI: Free monoids are pretty intuitive at heart: strings of letters under a concatenation operation.
Also, the set of functions $X\to X$ on a set (no stipulation of being bijective). More generally the set of endomorphisms of an arbitrary object of an arbitrary category caries a monoid structure under composition, so in particular functions on a set and linear maps on a vector space (categories in general constitute an abstract algebra background, but sets and vector spaces not as much).
If $G$ is an arbitrary group (which is admittedly an essentially algebraic concept), in your case noncommutative, one can adjoin a zero element (in the literature's lingo, an absorption element) and easily make the set $G\cup\{0\}$ into a monoid that is not a group. (For fun, this process can be continued indefinitely to create bigger and bigger monoids.) This idea is pretty straightforward. |
H: Why is n mod 0 undefined?
I tried to find out what $n$ mod $0$ is, for some $n\in \mathbb{Z}$. Apparently it is an undefined operation - why?
AI: I might say it depends on how you define what it means to mod out by a number.
A typical first way of thinking about mods is to say that $a \equiv b \pmod d$ if $a = b + dk$ for some integer $k$. In this sense, there is nothing wrong with saying $a \equiv b \pmod 0$, although this says nothing more than $a = b$.
A different first way of thinking about mods is to say that $a \equiv b \pmod d$ if when you divide $a$ by $d$, you get remainder $b$ (or something very similar). In this sense, as it does not make sense to divide by $0$, we are at a loss.
A typical higher idea is to consider $\mathbb{Z}$ as a group, and for the 'mod by $d$' operation to mean the equivalence classes induced by taking cosets of the subgroup generated by $d$, which I'll denote by $\langle d \rangle$. In this sense, the subgroup $\langle 0 \rangle$ is the trivial subgroup, so modding out by $0$ falls more along the lines of the first way of thinking I mentioned above : it's well-defined, but not really useful. |
H: Problems in the Ramanujan Class Invariant $G_n$?
In page 290 of his second notebook , Srinivas Ramanujan defines 2 functions $G_n$ and $g_n$. And then proceeds to give a table of $G_n$.
But looking at the papers of Bruce C. Berndt, Heng Huat Chan, and LiangCheng Zhang -(http://www.math.uiuc.edu/~berndt/articles/toronto.pdf). The Values of $G_n$ of several n differ from Ramanujan's Table. Then what is the relation between these two values?
AI: Let us define Ramanujan function $\chi(q)$ by $$\chi(q) = (1 + q)(1 + q^{3})(1 + q^{5})\cdots = (-q;q^{2})_{\infty}$$ then the definition of $G'_{n}$ given in the above image in question is $$G_{n}^{'-1/24} = 2^{-1/4}e^{\pi n/24}\chi(e^{-\pi n})$$ or what is the same as $$G'_{n} = 2^{6}e^{-\pi n}\chi^{-24}(e^{-\pi n})$$
But in Berndt's book as well as any standard reference (and in particular Ramanujan's classic paper "Modular Equations and Approximations to $\pi$") the definition of $G_{n}$ is given by $$G_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}\chi(e^{-\pi\sqrt{n}})$$
The relation between now accepted definition of $G_{n}$ and old $G'_{n}$ (in this image) is clearly seen to be $$G_{n^{2}} = G_{n}^{'-1/24}$$
Similar is the case for definition of $g_{n}$. |
H: What is meant by linearity of a dot product?
I would like to know what is meant by linearity of a dot product.
Thank you
AI: It means that the dot product satisfies two properties:
If $u$, $v$ and $w$ are vectors such that $\cdot$ and $+$ make sense, $$u \cdot (v + w) = u \cdot v + u \cdot w$$ and vice-versa:
$$(u + v) \cdot w = u \cdot w + v \cdot w$$
If $u$ and $v$ are vectors such that $\cdot$ makes sense, and $c$ is a scalar, $$(cu) \cdot v = c (u \cdot v)$$ and vice-versa: $$u \cdot (cv) = c (u \cdot v)$$ |
H: Simple proof about XOR, (possibly a duplicate question)
Prove Whether this statement is True or False:
Other than solving by Truth Table
If $A \oplus b = A \oplus C$ then $A=C$
I saw this question online and I've been thinking about it for days now and I'm pretty sure this question may have been asked already but I didn't it so far,
Thanks
AI: $\;\oplus\;$ ('exclusive or') is just the opposite of boolean equivalence, which I usually write as $\;\equiv\;$. So $\;\oplus\;$ is just $\;\not\equiv\;$.
Since $\;\equiv\;$ is associative (which you may or not be allowed to use, if not you'll have to prove it first), it is allowed to leave out parentheses in the middle lines below, and calculate
\begin{align}
& (A \not\equiv b) \equiv (A \not\equiv C) \\
\equiv & \;\;\;\;\;\text{"negate both sides of $\;\equiv\;$"} \\
& A \equiv b \equiv A \equiv C \\
\equiv & \;\;\;\;\;\text{"symmetry of (middle) $\;\equiv\;$"} \\
& A \equiv A \equiv b \equiv C \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& b \equiv C \\
\end{align}
Note that if you're allowed to use that $\;\equiv\;$ and $\;\not\equiv\;$ are mutually associative, then the proof becomes even shorter. |
H: Show that $\det(A-\lambda B)$ is a nonconstant polynomial if $B$ is invertible
Let $A$ and $B$ be arbitrary complex square matrices. If $B$ is invertible, show that $$p(\lambda)=\det(A-\lambda B)$$ is a nonconstant polynomial in $\lambda$.
AI: If $B$ is invertible, the leading term is ${\sf det}(B)(-\lambda)^n$ which is nonzero. |
H: Extending a $C^1$ function up to the boundary
Let $U \subset \mathbb{R}^N$ be an open bounded set. let $f \colon U \to \mathbb{R}$ be a $C^1$ function.
I know that it is not always possible to extend $f$ continuously up to the boundary of $U$ since it could be unbounded.
Now, it sounds reasonable that this is possible if I have that the gradient of $f$ is bounded in $U$ since this should imply that the function can't blow up whenever I approach the boundary. Am I right? Do you know how to prove this result rigorously? If it turns out to be false, could you please explain me where does my intuition fail?
Also, if the first part turns out to be true, is there any chance to prove that some regularity still holds on the boundary? What I mean is: if I assume that the boundary is a smooth differential manifold, can I take derivatives on it?
Thank you!
AI: If the gradient of $f$ is bounded, then $f$ is Lipschitz in the intrinsic metric of $U$, which is defined as infimal length of curves connecting two given points within $U$. Without extra assumptions on the boundary, this is not enough to conclude that $f$ extends continuously. For example, if $U\subset \mathbb C$ is the unit disk minus the segment $[-1,0]$, then $f(z)=z\arg z$ has bounded gradient, but has no continuous extension to $\partial U$. (It approaches different values on two "sides" of the radial slit.)
Smoothness of $\partial U$ rules out the situations as above, and allows you to extend $f$ continuously. You can't expect smoothness, however. For example, let $U$ be the unit disk, and $f(z)=|z-1|$. Then $f\in C^\infty(U)$, but its extension to the boundary is not differentiable at $1$. |
H: Question on using the integral of conditional probability to get unconditional probability
I'm trying to solve this problem in my homework assignment and I get different result from the answer. I know the answer is right, but at the same time I also don't see where I did wrong in my solution. So here's the problem:
Let X and Y have a joint uniform distribution on the triangle with vertices (0,0), (3,0), (0,3).
Find:
(i) E(X|Y) and E(Y|X)
(ii) Var(X|Y) and Var(Y|X)
(iii) EX and Var(X)
I've correctly completed the first and second sub-problem, but for the third one, I struggle to get it right. I'm trying to use the fact that
EX = E(E(X|Y))
to derive EX. And the way I approach it is:
EX = $$ \int {E(X|Y)*f(Y)} dY$$ where f(Y) is the pdf of Y being a certain value
From the sub-problem 1, I have:
E(X|Y) = (3-Y)/2 and f(Y) = 3-Y
Hence, I tried to solve integration:
$$\int{\frac{(3-Y)^2}{2}} dY $$ where 0<=Y<=3
and I get result 4.5 whereas the correct answer is 1.
Can someone please show where I did wrong? I'd much appreciate your help!
Thanks in advance!
AI: The joint density is $\frac{1}{4.5}$ over the triangle. That has not been used in calculating the pdf of $Y$, which is $\frac{3-y}{4.5}$ on $[0,3]$ and $0$ elsewhere. |
H: finding $f$ function
For which of the sets $\mathbb{X}:= \mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$, there exist a function $f:\mathcal{P}(\mathbb{X})\rightarrow\mathbb{X}$ that for all $A\in\mathcal{P}(\mathbb{X})\setminus \lbrace\phi\rbrace$, $f(A)\in A?$ I want to find $f$ function? please help me.
AI: A function $f:\mathcal{P}(A)-\{\varnothing\} \to A$ satisfying the condition $f(A)\in A$ is called a choice function on $A$.
In the case of $\Bbb{N}$, $\Bbb{Z}$, $\Bbb{Q}$, you can construct the choice function explicitly. For example, you can define $f:\mathcal{P}(\Bbb{N})\to\Bbb{N}$ as like:
$$
f(A):=\min A
$$
In the case of $\Bbb{Z}$ you can construct the bijection $g:\Bbb{Z}\to\Bbb{N}$. If we define $h(A)=g^{-1}(f(g_*(A)))$ (where $g_*(A)$ is image of $A$ under $g$) then $h$ satisfies $f(A)\in A$ for all nonempty $A$. Similarly, you can construct the bijection between $\Bbb{Q}$ and $\Bbb{N}$ and you can also define the function satisfy the given condition.
In the case $\Bbb{R}$ and $\Bbb{C}$, you can prove the existence of these functions. However, there is no explicit construction for the function satisfy the given condition. In fact, the existence of these function requires the axiom of choice and it is necessary to prove the existence of choice function on $\Bbb{R}$. It is known that '$\Bbb{R}$ has no choice function' consistent with ZF. |
H: Operation of Sine and Cosine
Use condition $\displaystyle\sin\theta-\cos\theta=\sqrt{2}$,
Please find the value of $\displaystyle\frac1{\sin^{10}\theta}+\frac1{\cos^{10}\theta}$.
AI: If we set $a=\sin\theta,b=\cos\theta$
we have $a-b=\sqrt2$ and
as $a^2+b^2=1,(a-b)^2=2\iff -2ab=1$
$\displaystyle (a^2-b^2)^2=(a^2+b^2)^2-4a^2b^2=1^2-(2ab)^2=0$
$\displaystyle\implies a^2-b^2=0$ and we have $a^2+b^2=1$
Can you solve for $a^2,b^2?$
We need to find $\displaystyle\frac1{a^{10}}+\frac1{b^{10}}=\frac1{(a^2)^5}+\frac1{(b^2)^5}$ |
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