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H: Is a probability distribution a measure on $( \mathbb{R}, {\cal R} )$?
Let $(\Omega, {\cal B}, P )$ be a probability space. Let $( \mathbb{R}, {\cal R})$ be the usual
measurable space of reals and its Borel $\sigma$- algebra. Let $X: \Omega \rightarrow \mathbb{R}$ be a random variable.
I am wondering if the following is correct: the probability distribution $Q$ of the random variable $X$ is a measure on $( \mathbb{R}, {\cal R})$ such that $Q( A ) = P( \{ w: X(w) \in A \} )$ for $A \in {\cal R}$.
AI: Yes, the distribution of a random variable $X:\Omega\to\mathbb{R}$ is indeed a probability measure (or probability distribution) on $(\mathbb{R},\mathcal{R})$ and is often called the pushforward measure of $P$ by $X$ and denoted by $P\circ X^{-1}$.
Note that it is well-defined, since $X$ is $(\mathcal{B},\mathcal{R})$-measurable and so $X^{-1}(A)\in\mathcal{B}$ for all $A\in\mathcal{R}$. Hence it makes sense to assign the probability of the sets $X^{-1}(A)$ for all $A\in\mathcal{R}$, and
$$
P_X(A):=P(X^{-1}(A)),\quad A\in\mathcal{R},
$$
defines a probability measure on $(\mathbb{R},\mathcal{R})$. This follows from basic properties of the pre-image and from the fact that $P$ is a probability measure (try showing this yourself). |
H: "Any"; universal or existential quantifier?
For any integers $m$ and $n$, if $7m+5n=147$, then $m$ is odd or $n$ is odd.
$$Q(m,n) \equiv 7m+5n=147$$
$$∀m∀n: Q(m,n) → \bigl(m \not\equiv 0 \!\!\pmod 2 \lor n \not\equiv 0 \!\! \pmod 2\bigr)$$
Am I right in assuming $\forall$ means "any" in this case? It doesn't seem to make sense to me ($\exists$ to me means "at least one, many, one, all but one, etc; anything less than all but more than none"), but Wikipedia states that $\forall$ can also mean "for any".
Is this correct?
AI: $\forall m$ is usually read for all m or for each m; for any m is a possible paraphrase in some contexts. $\forall x\big(\varphi(x)\big)$ means that for all possible values of $x$ in the domain of discourse, the statement $\varphi(x)$ about $x$ is true. If you think of $\varphi(x)$ as saying that $x$ has some particular property, then $\forall x\big(\varphi(x)\big)$ says that every element in the domain of discourse has that property.
Your understanding of $\exists$ isn’t quite correct, however; $\exists m\big(\varphi(m)\big)$ means precisely that there is at least one thing in the domain of discourse that has the property in question. It does not exclude the possibility that every element of the domain of discourse has it. For example, if we’re talking about integers, $$\exists n(n\text{ is even }\lor n\text{ is odd}\}$$ is a true statement, even though $$\forall n(n\text{ is even }\lor n\text{ is odd}\}$$ is also true. |
H: Monoidal Category - Equalizer
We have a category $\mathbb C$ with finite products and terminal object $1$. Further $\mathsf{Mon}(\mathbb C)$ is the category of monoids in $\mathbb C$ where a monoid is a triple $(M,m:M\times M \rightarrow M,e:1 \rightarrow M)$ which fulfills certian associativity and unit axioms. I have already proven that $\mathsf{Mon}(\mathbb C)$ has binary products. Assume $\mathbb C$ has equalizers. Now I have to prove that $\mathsf{Mon}(\mathbb C)$ has equalizers, too.
Further we have an arrow $f:A_1 \rightarrow A_2$ in $\mathsf{Mon}(\mathbb C)$ iff $f \circ m_1 = m_2 \circ (f \times f)$.
I would appreciate some good hints, how to start with this.
AI: Hint: Let's look at the example of monoids in set. Given $f,g \colon M_1 \to M_2$, the set where the maps coinside $\{m \in M \mid f(m) = g(m)\}$ [i. e. the equalizer in $\sf Set$] is a submonoid (as $f,g$ are morphisms). Together with the structure inherited from $M_1$ it is the equalizer in the category of monoids (here: semigroups with identity).
Let this example guide you in the general case, take the equalizer $E \to A_1$ in $\mathcal C$, restrict $m_1 \colon A_1^2 \to A_1$ to $E$ and show that the unit $e_1 \colon 1 \to A_1$ can be pulled back. (A monoid morphisms must also fulfill $f \colon e_1 = e_2$, right?) |
H: Finding eigenvalues.
I'm working on the following problem:
Define $T \in L(F^n)$ (T an operator) by
$T(x_1,...,x_n) = (x_1+...+x_n,...,x_1+...+x_n)$
Find all eigenvalues and eigenvectors of $T$.
I've found that the eigenvalues of $T$ are $\lambda = 0$ and $\lambda = n$. Is there an easy way to prove that these are the only eigenvalues of $T$? Determining and solving the characteristic polynomial is messy for arbitrary $n$.
AI: Try this: by direct computation,
$T^2 = nT, \tag{1}$
since every entry of $T^2$ is $n$.
So $m_T(x) = x^2 - nx$ is the minimal polynomial of $T$; every eigenvalue $\lambda$ of $T$ satisfies
$m_T(\lambda) = 0, \tag{2}$
so the only possibilities are $\lambda = 0$ and $\lambda = n$. |
H: How to prove this max absolute value equation?
How to prove this equation?
$$\max(|x_1-x_2|,|y_1-y_2|) = \frac{\left|x_1+y_1-x_2-y_2\right|+\left|x_1-y_1-(x_2-y_2)\right|}{2}$$
AI: Let $a = x_1-x_2$ and $b = y_1-y_2$ to simplify the problem to proving $$\max(|a|,|b|)=\frac{|a+b|+|a-b|}{2}$$
Now let $c=|a|, d=|b|$ and consider the possibilities of $a$ and $b$ having the same sign to show this is equivalent to showing $$\max(c,d)=\frac{|c+d|+|c-d|}{2} \text{ for }c,d \ge 0$$ while if $a$ and $b$ have opposite signs it is equivalent to showing $\max(c,d)=\frac{|c-d|+|c+d|}{2}$ for $ c,d \ge 0$, which is the same thing.
Then consider that if $c \ge d \ge 0$ this is equivalent to showing $c = \max(c,d)=\frac{c+d+c-d}{2}$ while if $d \ge c \ge 0$ this is equivalent to showing $d = \max(c,d)=\frac{c+d+d-c}{2}$. These last two are clearly true. |
H: Math question from the GMATprep
If $xy=1$ what is the value of: $2^{(x+y)^2}/2^{(x-y)^2}$
A 1
B 2
C 4
D 16
E 19
$(x+y)^2/(x-y)^2$ because $2$ just cancels out from numerator and denominator, right?
AI: This isn't as hard as you think.
xy = 1*1 or -1*-1
If xy = 1*1 then:
2^(x+y)^2/2^(x-y)^2 = (2^(2)^2)/(2^0^2) = 16/1 = 16
If xy = -1 *-1 then:
2^(x+y)^2/2^(x-y)^2 = (2^(-2)^2)/2^(-1 - (-1) )^2 = 16/1 = 16.
So, D is the answer. Let me know if you're confused |
H: Is it possible to extract some expressions from modulus brckets?
I have an expression:
$$\left| x-y-a\cdot \left( \frac{1}{x^2}+\frac{1}{y^2} \right) \right|$$
where $a>0$, $x>0$ and $y>0$. Can I make something with it to get:
$$\left| x-y \right| \cdot SomethingElse$$
AI: Of course:
$$\left| x-y-a\cdot \left( \frac{1}{x^2}+\frac{1}{y^2} \right) \right| = \left| x-y\right|\left| 1-\frac{a\cdot \left( \frac{1}{x^2}+\frac{1}{y^2} \right)}{x-y} \right|$$ |
H: Partial fraction of 1 over (x^2+1)^2
Its been years since I solved PF. Now I am having hard time solving this partial fraction
$$
F = \frac{1}{\left( x^2+1\right)^2}
$$
I proceeded with(Is this right ?)
$$ \frac{1}{\left( x^2+1\right)^2} = \frac{A}{\left( x+\iota\right)} + \frac{B}{\left( x+\iota\right)^2} + \frac{C}{\left( x-\iota\right)} + \frac{D}{\left( x-\iota\right)^2}$$
I don't get the solution that MATLAB gives:
>> [A,B,C] = residue([1],[1,0,2,0,1])
A =
-0.0000 - 0.2500i
-0.2500
-0.0000 + 0.2500i
-0.2500
B =
-0.0000 + 1.0000i
-0.0000 + 1.0000i
-0.0000 - 1.0000i
-0.0000 - 1.0000i
C =
[]
Kindly help me solve this PF.
Thanks in advance!
AI: $$F(x) = \dfrac{1}{\left( x^2+1\right)^2}=\dfrac{1}{\left( x+i\right)^2 \left( x-i\right)^2}=\dfrac{A}{x+i}+\dfrac{B}{\left( x+i\right)^2} +\dfrac{C}{x-i}+\dfrac{D}{\left( x-i\right)^2}= \dfrac{A(x+i)(x-i)^2+B(x-i)^2+C(x-i)(x+i)^2+D(x+i)^2}{\left( x+i\right)^2 \left( x-i\right)^2}$$
Then
$$1\equiv {A(x+i)(x-i)^2+B(x-i)^2+C(x-i)(x+i)^2+D(x+i)^2}$$
$$\left.\begin{matrix}{ x^3 \\ x^2\\x=i\\x=-i}\end{matrix}\right|
\begin{matrix}{A+C=0 \\ A(-2i+i)+B+C(-i+2i)+D=0\\-4D=1\\-4B=1}\end{matrix}$$
$$\begin{cases}
A+C=0 \\ -Ai+B+Ci+D=0\\D=-\dfrac{1}{4}\\B=-\dfrac{1}{4}
\end{cases}\quad\Rightarrow \quad\begin{cases}
A+C=0 \\ (-A+C)i=\dfrac{1}{2}\\D=-\dfrac{1}{4}\\B=-\dfrac{1}{4}
\end{cases}
\quad\Rightarrow \quad\begin{cases}
A+C=0 \\ -A+C=-\dfrac{i}{2}\\D=-\dfrac{1}{4}\\B=-\dfrac{1}{4}
\end{cases}$$
so
$$\begin{cases}
A=\dfrac{i}{4}, \\ C=-\dfrac{i}{4},\\D=-\dfrac{1}{4},\\B=-\dfrac{1}{4}.
\end{cases}$$ |
H: Finding Taylor approximation for $x^4e^{-x^3}$
I'm trying to find Taylor approximation for the function: $$x^4e^{-x^3}$$
I started taking the first, second, third, etc. derivatives but the expression for it seems to explode with terms. I was just wondering is there a trick for this one or do I just have to use brute force in order to discover the pattern? :)
thnx for any help
AI: Notice $$\exp(x) = \sum \frac{x^n}{n!} \implies \exp(-x^3) = \sum\frac{(-1)^nx^{3n}}{n!} \implies x^4 \exp(-x^3) = \sum\frac{(-1)^nx^{3n+4}}{n!} $$ |
H: Constructing invariant subspaces from scratch. An algorithm
So, basically i am trying to prove that there exist a basis w.r.t which there exists an upper triangular matrix in a complex field.
Most of the books which i read incorporate induction as a method which i find sadly non intuitive. Please have a look at the following :
Suppose in the complex field, I select an eigen vector $v_1$ such that $Tv_1 = kv_1$.
Now, i must select a vector $v_2$ such that:
(a) $v_1$ and $v_2$ are linearly independent.
(b) $Tv_2$ lies in span$(v_1,v_2)$.
Writing mathematically, $v_2$ should be such that
(i) $rv_1 + sv_2 = 0 \implies r = s = 0$.
(ii) $Tv_2 = a v_1 + b v_2 \implies (T - b I )v_2 = av_1$
What can I infer for $v_2$ from here?
Some points which may be helpful like:
$av_1$ lies in range $(T - b I)$
Now since the range of any linear mapping is a subspace,
$v_1$ also lies in the range $(T - b I)$.
Help will be really appreciated as I have spent a long time analyzing this. Thank you.
AI: Well, think about the linear transformation an upper triangular matrix
$$\begin{pmatrix}
\lambda_1 & \ast & \ast \\
0 & \lambda_2 & \ast \\
0 & 0 & \lambda_3 \end{pmatrix}$$
represents.
It takes $e_3$ to $\lambda_3 e_3$ -- that is, it fixes $\operatorname{span} \{e_3\}$. It takes $e_2$ to a linear combination of $e_2$ and $e_3$; since it also takes $e_3$ to a linear combination of $e_2$ and $e_3$ (specifically, a linear combination where the coefficient of $e_2$ is zero!), this means it also fixes $\operatorname{span} \{e_2, e_3\}$. And of course it fixes $\operatorname{span} \{e_1, e_2, e_3\}$, which in this case is not saying anything because that's the entire space.
So, a matrix in upper triangular form gives us
a line fixed by the transformation;
a plane fixed by the transformation and containing that line;
a three-space fixed by the transformation and containing that plane;
and so forth up to the dimension of $n$.
Consequently, to put a matrix in upper triangular form, we should reverse this process by finding such a line, then such a plane, etc. |
H: Count of 3-digit numbers with at least one digit as 9
Find the number of $3$ digit numbers (repetitions allowed) such that at least one of the digit is $9.$
I've posted my answer below. If there is a better way to solve this question, I would be glad to learn about that.
AI: Suppose that 'three-digit' means $abc$, where $a>0$.
Now, we first count that there are $900$ of these numbers.
Of numbers without a $9$, then it's $8\times 9 \times 9$, since the first digit can be any of 1-8, and the rest 0-8. This gives $648$ numbers without a 9.
One then finds that there are $252 = 900-648$ numbers that contain at least one nine (or any other specific non-zero digit). |
H: How to find the representation of Lie algebra
I read a book about the Lie algebra, but I really don't understand the calculation of $ad(X)$. For example, we have a Lie algebra of bases:
$$e_1=\left[\begin{array}{cc}1 & 0\\0 & -1\end{array}\right], e_2=\left[\begin{array}{cc}0 & 1\\0 & 0\end{array}\right], e_3=\left[\begin{array}{cc}0 & 0\\-1 & 0\end{array}\right]$$
I want to know the format of $ad(e_1)$, $ad(e_2)$ and $ad(e_3)$.
All the authors said that "it is clearly that ad(e1)=...,ad(e2)=... and ad(e3)=...", but how did they find it?
another example:
$$e_1=\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right], e_2=\left[\begin{array}{cc}0 & 1\\0 & 0\end{array}\right], e_3=\left[\begin{array}{cc}0 & 0\\1 & 0\end{array}\right], e_4=\left[\begin{array}{cc}0 & 0\\0 & 1\end{array}\right].$$
the adjoint representation is
$$ad(e_1)=\left[\begin{array}{cccc}0 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & -1 & 0\\0 & 0 & 0 & 0\end{array}\right].$$
Some one said that it has a relationship to the structure constant $C_{ijk}$, but how to write $ad(e)$ with $C_{ijk}$?
AI: We recall that for any $X \in \mathfrak{sl}_2(\Bbb{C})$, $\text{ad}_{X}$ is the linear operator on the vector space $\mathfrak{sl}_2(\Bbb{C})$ that sends every $Y$ in $\mathfrak{sl}_2(\Bbb{C})$ to $\text{ad}_X(Y) = [X,Y]$. Every linear operator on a finite dimensional space can be represented by a matrix. In this case, we want a matrix in the basis $e_1,e_2,e_3$ as you defined. So if $X = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$ then we need to calculate $\text{ad}_X(e_1),\text{ad}_X(e_2)$ and $\text{ad}_X(e_3)$. We find:
$$\begin{eqnarray*} \text{ad}_X(e_1) &=& 0\\
\text{ad}_X(e_2) &=& 2e_2 \\
\text{ad}_X(e_3) &=& 2e_3 .\\ \end{eqnarray*}$$
Thus the matrix of $\text{ad}_X$ in the basis $\mathcal{B} = \{e_1,e_2,e_3\}$ is given by:
$$[\text{ad}_X]_{\mathcal{B}} = \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right).$$ |
H: Identical complex functions.
Uniqueness principle theorem :
If $f$ and $g$ are analytic functions on a domain $D$, and if $f(z)=g(z)$ for $z$ belonging to a set that has a non isolated point, then $f(z)=g(z)$ for all $z\in D$.
Now my question is this:
If $f$ and g are analytic on a domain $D$, and if $f(z)=g(z)$ for infinitely many $z\in D$, Is it always true that $f\equiv g$ on $D$?
I think this is not true. As we know that an analytic function on some disk has a Taylor expansion in that disk. And If we expand any given analytic function (by using Taylor series) in a smaller disk than the given domain. By doing this we will get another function whose domain will be smaller than that of the given function but they will be equal on infinite many values.
This is a vague idea. Please tell me if its right. If not then correct me by giving some appropriate example to support the above statement.
Thanks.
AI: The assertion is false, for example take any countable set like $\pi\mathbb Z$ and functions like
$$\sin(x), 2\sin(x)$$
they are unequal, but $\sin(x) = 2\sin(x) = 0 \quad\forall\ x\in \pi\mathbb Z$ |
H: Unit Ball with p-norm
I am having trouble understanding the definition of p-norm unit ball. What I know is that for infinity (maximum norm), then it will shape as a square. I need a "click" to understand this, can someone be so kind to explain this to me in simple words?
If the norm is the distance of the vector, then where does this vector located? From where to where is the length of the vector? Is it from the (0,0) point?
AI: If the center of the unit-ball is in the origin $(0,0)$, then each point on the unit-ball will have the same p-norm (i.e. 1). The unitball therefore describes all points that have "distance" 1 from the origin, where "distance" is measured by the p-norm.
The easiest unit balls to understand intuitively are the ones for the 2-norm and the 1-norm.
Example 1: The 2-norm is simply the length of the vector ($\sqrt{x_1^2 + x_2^2}$ for the 2-dimensional case). Therefore it makes sense that all points of the same length form a circle around the origin.
Example 2: The 1-norm ($|x_1| + |x_2|$) is another case that can be easily interpreted. Just imagine the special cases
$$ x_{horizontal} = (1,0), x_{vertical}=(0,1) $$
Their 1-norm is
$$ |x_{horizontal}| = | x_{vertical}| = 1 $$
Every point on the line between these two points will also have a 1-norm of 1 since you linearly decrease the $x_1$-component while you increase the $x_2$-component.
Example 3: The infinity norm is defined as $\|x\|_\infty=\max\{ |x_1|, \dots, |x_n| \}$. Therefore, $\|x\|_\infty=1$ for all $(x_1,x_2$) where either $|x_1|=1$ and$|x_2| \leq 1$ or $|x_1|\leq1$ and $|x_2| = 1$ - this is how the square is found! |
H: Multiple choice question about the dimension of a space of $10 \times 10$ complex matrices
Problem: Let $A \in M_{10}(\Bbb C)$ ,the vector space of $10 \times 10$ matrices with entries in $\Bbb C$.
Let $W_A$ be subspace of $M_{10}(C)$ spanned by $\{\,A^n : n\geq 0\,\}$
Then which of the following correct ?
1) $\dim(W_A) \leq 10$
2) $\dim(W_A) < 10$
3) $10 <\dim(W_A)<100$
4)$\dim(W_A)=100$
Solution:I think $\dim(M_{10}(\Bbb C))=100$
So $\dim(W_A) \leq 100$
elements of $W_A$ is linear combination of $\{\,A^n : n\geq 0\,\}$
After that I have no idea.
AI: Note, that this will depend on $A$, for example, if $A = \def\Id{\mathop{\rm Id}}\Id$, we have $W_A = \mathbb C\cdot \Id$, so $\dim W_A = 1$, if $A$ is not a multiple of $\Id$, we have $\{\Id, A\} \subseteq W_A$ is linear independent, hence $\dim W_A \ge 2$.
In general, if $\mu_A$ denotes the minimal polynomial of $A$, that is the uniquely determinend smallest degree normed member of
$$ I_A = \{p \in \mathbb C[X] \mid p(A) = 0\} $$
($I_A$ is an ideal in $\mathbb C[X]$ and $\mu_A$ is its normed generator). Then we have $\mu_A(A) = 0$, so $A^{\deg \mu_A}$ is a linear combination of $\{A^i\mid i < \deg\mu_A\}$ and this cannot happen for lower powers (since this would give a lower degree member of $I_A$). Hence $\dim W_A = \deg \mu_A$.
Addendum (after the "options" were added in the OP): Note that we know by Cayley-Hamilton, that $\chi_A(A) = 0$, where $\chi_A(X) = \det(X\Id - A)$ denotes $A$'s characteristic polynomial. So $\dim W_A \le \deg \chi_A = 10$. This tells us that options 3) and 4) are wrong. To show that 1) is the right answer, we will give a matrix such that $A^i$ for $i = 0,\ldots, 9$ are linear independent, hence $\dim W_A = 10$ can occur. Let $A = (a_{ij})$ where
$$ a_{ij} = \begin{cases} 1 & j = i+1\\ 0 & \text{otherwise} \end{cases}
$$
Then $A^k = (a_{k,ij})_{ij}$ where
$$
a_{k,ij} = \begin{cases} 1 & j = i+k\\ 0 &\text{otherwise} \end{cases}
$$
Now for any $\lambda_k \in \mathbb C$ we have
$$ \left(\sum_{k=0}^{9} \lambda_k A^k\right)_{ij} = \begin{cases} a_{j-i} & j-i \in [0,9]\\ 0 & \text{otherwise}\end{cases} $$
which gives linear independence. |
H: A question related to uniqueness principle theorem.
We know that the equation $ \sin^2z+ \cos^2z=1$ which holds $\forall z \in\Bbb R$, also holds $\forall z \in\Bbb C$.
This is obvious under the shadow of following theorem:
Uniqueness principle theorem :If $f$ and $g$ are analytic functions on a domain $D$, and if $f(z)=g(z)$ for $z$ belonging to a set that has a non isolated point, then $f(z)=g(z)$ for all $z\in D$.
Can we extend the domain of function $ |\sin z|^2+|\cos z|^2$ from $\Bbb R$ to $\Bbb C$? (i.e Can we say that $|\sin z|^2+|\cos z|^2=1\,\forall z \in\Bbb C$)?
I think this extension of the domain from the real line to the entire complex plane is not possible. Since function on left-hand side i.e.$ |\sin z|^2+|\cos z|^2$ doesn't look analytic. And to use above theorem we need both functions to be analytic. How will we show this?
Thanks.
AI: As you say correctly, the function $z \mapsto |\sin z|^2 + |\cos z|^2$ is not analytic, so the comparision principle cannot be used. Moreover, want you want to prove is wrong, as - for example
$$ \def\abs#1{\left|#1\right|} \abs{\sin i}^2 + \abs{\cos i}^2 = \frac{1+e^4}{2e^2} \ne 1. $$ |
H: Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.
I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x\equiv2\pmod3$$What $x$ value?
AI: HINT: Notice that your congruences are equivalent to the following ones:
$$\left\{\begin{align*}
x\equiv-1\pmod6\\
x\equiv-1\pmod5\\
x\equiv-1\pmod4\\
x\equiv-1\pmod3
\end{align*}\right.$$
In other words, $x+1$ is divisible by $6,5,4$, and $3$. What’s the smallest positive integer with that property? |
H: For what values of m are the roots of $x^2 +2x+3 = m(2x+1)$ real and positive
I am only able to show that to be real, $m <-1$ or $m\geq2$
Don't know how to finish solution
Answer is $2 \leq m < 3$
So far:
After expanding and factorising,
$x^2 + 2(1-m)x + (3-m) = 0 $
Roots are real if discriminant $\ge 0 $
i.e. $4(m-2)(m+1)>=0 $
Therefore $m ≤ -1 \text{ or } m \ge 2$ for roots to be real
However I don't know how to find m for roots to be positive
AI: For the equation to have positive and real roots, two extra conditions are needed
$$2m-2>0$$ $$3-m>0$$
$$\therefore 1<m<3$$
Taking the intersection of the intervals found, we will get $$\therefore 2 \leq m <3$$
:) |
H: What will happen to the limits of this integration?
This is a joint probability distribution function (PDF)
f(x,y) = $C(x^2 + y)$ and the limits are $x>0$ and $0 < y < (1-x^2)$
We have to calculate the marginal PDF of Y, i.e. f(y).
I get that we will integrate the joint function with the limits of x. But can't figure out what they will be. And what will happen to dx?
AI: First note that your joint PDF can be written as
$$
f_{X,Y}(x,y) = c\cdot(x^2+y)1_{x>0}(x)1_{0<y<(1-x^2)}(x,y)
$$
where $1_A(\cdot)$ designates the indicator function over the set $A$, namely,
$$
1_A(a) = \begin{cases}
1,\ \text{if}\ a\in A\\
0,\ \text{else}
\end{cases}.
$$
Now note that
$$
1_{x>0}(x)1_{0<y<(1-x^2)}(x,y) = 1_{0<x<\sqrt{1-y},0<y<1}(x,y)
$$
Thus, (for $0<y<1$, other wise $f_Y(y)=0$)
\begin{align}
f_Y(y) &= \int_{\mathbb{R}}c\cdot(x^2+y)1_{x>0}(x)1_{0<y<(1-x^2)}(x,y)dx\\
&=\int_0^{\sqrt{1-y}}c\cdot(x^2+y)dx
\end{align}
and from here it is easy to continue. |
H: Question regarding diameter of subsets of a metric space
The question is :
Find a condition on a metric space$(X,d)$ that ensures that there exist subsets $A$ and $B$ of $X$ with $A \subset B$ such that $diam(A)$ = $diam(B)$.
I know that if $X$ is a metric space and $A$ and $B$ are subsets of $X$ with $A \subset B$ then
$diam(A)$ <= $diam(B)$.
If i assume that the metric $d$ on $X$ is the discrete metric then then diameters of $A$ and $B$ will be the same.
Is this good enough ? or can there be some other condition on the metric space $X$ ?
AI: The condition you need is that $X$ has at least one point. If $A=\emptyset$ and $B$ is a one-point set, then $A\subset B$ and $diam(A)=diam(B)=0$.
A commenter has claimed that $diam(\emptyset)$ is undefined. I quote from Kuratowski's Topology Volume I, 1966 edition, p. 207:
III. Diameter. Continuity. Oscillation. The diameter, $\delta(X)$, of a set $X$ is the least upper bound of the distances of its points. If $\delta(X)$ is finite, the set X is said to be bounded.The following propositions are easily proved:$$\{\delta(X)=0\}\equiv\{X\;is\;empty\;or\;is\;composed\;of\;a\;single\;point\};\;\;\;(1)$$ |
H: how can I sort a matrix
I have coordinates of some points$(x,y)$, that I want to sort them,at the first by $y$
coordinates and second sort $x$ coordinates between those points that have equal $y$ coordinate.
Is there anyone know any function in matlab about it?
I examined it by $SORT$ function, but this function just sort each column of matrix.
Please help me.
Thanks.
AI: sortrows should be what you are looking for.
Though you may need to swap $x$ and $y$ first, or use the optional argument to tell it which column to base the sort first.
(BTW, the kind of ordering you are looking for is called "lexicographical order"; that may help you do more searches on the internet.) |
H: Determine relationship between vectors.
Using the product i would like to understand if it is possible to determine relationship between two vectors,let us consider the following problem:
Vectors $u$ and $v$ have length $1$. Which of the assumptions $(a)-(g)$ below imply that vectors $u$ and $v$ are: $$\begin{array}{c|c}
\text{i} & \text{prependicular} \\
\hline
\text{ii} & \text{parallel and pointing in the same direction} \\
\hline
\text{iii} & \text{parallel and pointing in the opposite direction} \\
\hline
\text{iv} & \text{parallel and no information about directions}
\end{array}$$ $$\text{(a) } u \cdot v = -1$$ $$\text{(b) } u \cdot v = 1$$ $$\text{(c) } |u \cdot v| = 1$$ $$\text{(d) } |u \cdot v| = 0$$ $$\text{(e) } u \cdot v = 0$$
Sure one thing we can easily say that if dot product of two vector is zero,they are perpendicular (orthogonal),parallel and pointing in the same direction,for example if we multiply vector by some scalar ,we get parallel and same direction vector,maybe product of them should be positive?if opposite direction,maybe their product is $-1$ ?about $(iv)$ maybe if absolute value of product is $1$
AI: $\mathbf u. \mathbf v=|\mathbf u||\mathbf v|\cos\theta$ where $\theta$ is the angle between vectors. So If they are parallel and pointing to the same direction then $\mathbf u. \mathbf v=1$, parallel and pointing to the opposite direction then $\mathbf u. \mathbf v=-1$, parallel and no information about the direction $|\mathbf u. \mathbf v|=1$. |
H: Problem with definition of regular surface in classical differential geometry
I am reading Do Carmo's differential geometry book and the definition of a regular surface in the second chapter is given to be this:
I have few doubts about this definition:
1) Why we need to find a neighbourhood of point $p$? Is it because we can't always define a map $X$ that will work for the whole surface and we are trying to find local maps for every point. And later in chapter 3, when author talks about this parametrization/map for a surface, he says surface parametrized at point $p$, what does "at point $p$" means, are we talking about local parametrizations?
2) $X$ is differentiable to infinite order....is this really necessary? What if map $X$ is differentiable to a large but finite order?
3) X is continuous by condition 1. But I don't understand the use of continuity as described in the book. In book it is given that we need condition 2 for one to oneness of the map so that we can have single tangent plane at each point (basically to avoid self-intersections). So why don't just make condition 2 to be $X$ being one to one function?
4) I couldn't understand why we need condition 3 very clearly. Could you please give a geometric intuition for this?
AI: 1) Because you usually can't map the whole surface to a planar set. Like the sphere, for example: every world map has to cut it somewhere.
2) Not really necessary. Much of the theory works for $C^k$ instead of $C^\infty$ as long as $k$ is large enough. But some constructions become more difficult and statements cumbersome, because one has to keep track of how many derivatives you took so far, and how many you can still take. So there's a nonzero cost for unclear benefit (especially at the textbook level).
3) Continuous and one-to-one is not the same as being a homeomorphism. (There are some situations when they are, but that is something one has to think through.) Stating the definition as it is allows us to say: locally, as far as topology is concerned, the surface is exactly like a plane. That's a nice thing to have. All local topological properties of the plane are immediately available to us, because they are preserved by homeomorphisms.
4) This is easier to illustrate with an example of a curve. The curve $y= x^{2/3} $ is visibly non-smooth - it has a cusp at $(0,0)$. But it admits a parametrization $(x,y)=(t^3,t^2)$ which is infinitely differentiable. There seems to be a disconnect between the geometric roughness of the curve and the smoothness of its parametrization. Imposing regularity ensures that we don't have this problem; by some form of the implicit function theorem, a surface with a smooth regular parametrization is indeed a smooth-looking geometric object.
A related surface example: $z=x^{2/3}$ with parametrization $(u,v)\mapsto (u^3,v,u^2)$. The differential map has rank $1$ at some points, and this is what allows the cusp to form.
Ultimately, the proof of the pudding is in the eating. Complicated as it is, this definition succeeds at matching a formal mathematical concept to an informal geometric notion of smooth surface. This is something you get convinced of when you actually use it. |
H: When a pole lies outside the circle of integration, what does Cauchy integral formula state?
I have the following complex line integral:
$$ \int_{|z| = 2} \frac{z}{z - 3} $$
My prof said it is $0$, but did not explain. He just said that the point $3+0i$ lies
outside the circle.
But the Cauchy integral theorem does not mention anything about it.
Can anybody give me the proof and also mention does this happen even if the
point lies outside the circle/loop and the function is not analytic inside it.
AI: The reason the answer is zero is because the function
$f(z)=\frac{z}{z-3}$
is analytic inside the circle. It is generally true that the integral around a simple closed curve of a function that is analytic in the interior is 0. So you are not using the Cauchy integral formula in this case.
If you are trying to calculate
$\int\frac{f(z)}{z-a}$
around some curve $\gamma$ where $f$ is not analytic Cauchy integral formula does not say anything about the integral no matter where $a$ is located. |
H: Find the Fourier series S(t) of the period 2$\pi$
Find the Fourier series S(t) of the period $2\pi$ function
$f(t)=\begin{cases}
-1& \text{if −$\pi$ < t < 0;}\\
\;\;1& \text{if $\:$0 < t < $\pi$;}\\
\;\;0&\text{if $t = −\pi, 0, or \;\pi$ }
\end{cases}$
Use MATHEMATICA to graph partial sums $S_N(t)$ of the Fourier series for f(x) with
N = 3, 6, 12, 24. What do you notice? Depending on the correctness of your $S_N(t)$, your graphs should portray what is known as Gibbs’s phenomenon.
For Fourier sine coefficients $b_n$ of the square wave, I get:
$b_n =\frac{2}{\pi} \int\limits_0^\pi$sin nx dx = $\frac{2}{\pi}[\frac{-cos\; nx}{n}] x = 0$ to $\pi$
$\frac{cos n\pi}{n}=\begin{cases}
0& \text{if n is even}\\
\frac{4}{n\pi}& \text{if n is odd}
\end{cases}$
Thus for n = 1, 2, 3, 4, 5...
I get:
{$\frac{4}{\pi},0,\frac{4}{3\pi}, 0, ... $}
I have difficulty in partial graphing sums. I need some help.
AI: You are not summing the coefficients, but rather the sine series having these coefficients. Thus you want the graphs of
$$\frac{4}{\pi} \sin{x} + \frac{4}{3 \pi} \sin{3 x} + \cdots + \frac{4}{(2 N+1) \pi} \sin{(2 N+1) x}$$
over $x \in [-\pi,\pi]$ for various values of $N$. |
H: Harmonic function (PDE) - Orthogonal matrix
Let $u\in C^2(\mathbb{R}^n)$ be harmonic in $\mathbb{R}^n$, i.e.
$$
\Delta u:=\sum\limits_{k=1}^{n}\frac{\partial^2 u}{\partial x_k^2}=0\mbox{ in }\mathbb{R}^n.
$$
Let $Q\in\mathbb{R}^{n^2}$ be an orthogonal matrix (i.e. $QQ^T=E_n$). Show that then the function $v\in C^2(\mathbb{R}^n)$ with $v(y):=u(x)$ with $y=Qx$ is harmonic in $\mathbb{R}^n$, too.
Hello, it is to show that $\Delta v=0$. So I already started with
$$
y=\begin{pmatrix}y_1\\\vdots\\y_n\end{pmatrix}=\underbrace{\begin{pmatrix}q_{11}\ldots q_{1n}\\\vdots\\q_{n1}\ldots q_{nn}\end{pmatrix}}_{=Q}\cdot\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}=\begin{pmatrix}\sum\limits_{i=1}^{n}q_{1i}x_i\\\vdots\\\sum\limits_{i=1}^{n}q_{ni}x_i\end{pmatrix},
$$
getting then
$$
\Delta v(y)=\sum\limits_{k=1}^{n}\frac{\partial^2 v(y)}{\partial\left(\sum\limits_{i=1}^{n}q_{ki}x_i\right)^2}.
$$
Now I would like to know how I can continue at this point.
Thank you very much!
AI: let $y_l=\sum_{i=1}q_{li}x_i$ then
$\frac{\partial v}{\partial x_i}=\sum_{l=1}^{n}\frac{\partial u}{\partial y_l}.q_{li}$
and
$\frac{\partial^2 v}{\partial x_i\partial x_k}=\sum_{l=1}^{n}\sum_{j=1}^{n}\frac{\partial ^2u}{\partial y_l\partial y_j}.q_{li}.q_{ji}$
then
$\Delta v=\sum_{l=1}^{n}\sum_{j=1}^{n}\frac{\partial ^2u}{\partial y_l\partial y_j} (\sum_{i=1}^{n}q_{li}.q_{ji})=\Delta u$ |
H: How does my professor go from this logarithm to the following one of a different base?
I don't understand on the second last line how my professor goes from $2^{log_3(n)} = n^{log_3(2)}$ how is that relation formed?
AI: Note that $\log_b a = \frac{\log b}{\log a}$, so
$$2^{\log_3 n} = 2^{\frac{\log n}{\log 3}} = \exp\left(\log 2 \cdot \frac{\log n}{\log 3}\right) = \exp \left(\log n \cdot\frac{\log 2}{\log 3}\right) = n^{\frac{\log 2}{\log 3}} = n^{\log_3 2}.$$ |
H: Just a question regarding continuous differentiability
$ f: [0,1] \to [0,1] $ be a MONOTONE & CONTINUOUS function. Does it always imply that: $ f(x) $ is continuously differentiable??
AI: No: Take, for example, $f(x) = \frac{1}{2} x$ for $x < \frac{1}{2}$, and $f(x) = \frac{3}{2}x - \frac{1}{2}$ for $x \ge \frac{1}{2}$.
Or any function with a corner, or something more complicated like the Devil's Staircase. |
H: Second order implicit derivative
The equation
$\begin{equation}
x^3 \ln x + y^3 \ln y = 2z^3 \ln z
\end{equation}$
defines $z$ as a differentiable function of $x$ and $y$ in a neighbourhood of the point $(x,y,z) = (e, e ,e)$. Calculate $z_1 (e,e)$ and $z_{11}(e, e)$.
Attempt at solution
Define $F(x,y,z) = x^3 \ln x + y^3 \ln y - 2z^3 \ln z = 0$. Then according to the implicit function theorem
$\begin{align}
z_1 &= -\frac{F_1}{F_3} \\
&= \frac{3x^2 \ln x + x^2}{6z^2 \ln z + 2z^2}
\end{align}$
so $z_1(e,e) = \frac{1}{2}$. I have solved $z_{11}$ but to be honest, using the quotent rule was a pain so I was wondering if there exists a more efficient way. Brute forcing my way through the problem I noticed that
$\begin{align}
z_{11} &= -\frac{F_{11}}{F_3} \cdot z_{1}
\end{align}$ but I'm at a loss how I could obtain that result without all the ... rigmarole. Any help would be greatly appreciated.
AI: Just a thought: avoid the quotient rule by multiplying out your $z_1$ to get
$$6z_1z^2\ln z +2z_1 z^2 = 3x^2 \ln x +x^2.$$
Now do the derivatives wrt $x$ of each term, where you apply the chain rule, at two points the desired $z_{11}$ appears, and at some other points you get a $(z_1)^2$.
Of course that's still a bit messy since one has to plug in the previous $z_1$ result into the places where $(z_1)^2$ appears...
I once tried to write a general formula for the second derivative in the two variable case, but it got so unwieldy that I thought it better to do concrete problems by the above method. [The formula had so many mixed partials and second pure partials to put in it, that it looked like more trouble than it was worth to try to apply the formula rather than do the particular problem directly.] |
H: Intersection of two congruent spirals
Let $S_1$ and $S_2$ be two congruent circular spirals in $\mathbb{R}^3$,
both with their axes passing through the origin. They are congruent in that their
radii are equal, as are their winding frequencies; but aside from being constrained
to have their axes through the origin, one is an arbitrary rigid reconfiguration of the other.
My question is: Do they always intersect,
and if so, where? It appears that they always intersect in two points, as in the example
below (origin: green), but I am not seeing a proof. Any help would be appreciated.
AI: No, they don't have to intersect and generally will not.
Each helix lives on a circular cylinder, so if they intersect at all, it must be somewhere on the two orthogonal ellipses where the cylinders intersect. But each helix meets those ellipses only in finitely many points, so if you have a configuration where the helices do intersect, turning one of them by an infinitesimal amount will make it miss the point of intersection without thereby hitting any of the other possible intersection points.
Intuitively, if the winding period is much larger than the radii, then one certainly won't expect any intersection. |
H: Checking of continuity
While compactness & connectedness are preserved under continuous maps, this question comes to my mind:
$f : \mathbb R \to \mathbb R$ is strictly monotone increasing function such that {$ f(x) : x \in \mathbb R$} is dense in $\mathbb R$ , then prove or disprove, that: $f$ is continuous on $\mathbb R$
AI: A strictly monotonic function can only have jump discontinuities (try proving this, or read a proof at this question), and a jump discontinuity would imply that the image set isn't dense; so $f$ must be continuous. |
H: Is ∑ with negative value solvable?
Is it possible to have a negative value in sigma?
e.g.
$y = \Sigma_{k=0}^{k=-2} k \times 10$
Will this give the result $(0 \times 10) + (-1 \times 10) + (-2 \times 10) = -30 $?
Or will it be $\infty$ because $k$ will be increased with $1$ until it equals $-2$ (which is never).
Or something else?
AI: There are several things to remark here:
First, $$\sum_{k=0}^{k=-2}\times 10$$ is not actually the correct notation.
You seem to mean $$\sum_{k=0}^{k=-2}k\times 10$$ which is correct notation (though usually the $k=$ part is not included in the top), but would be $0$. The reason is that the notation $$\sum_{k=i}^{k=j}k$$ means "take the sum of $k$ for each $k$ which is $\geq i$ and $\leq j$". In the case at hand, there are no such $k$, and by convention, this means the sum is $0$.
To get the desired result, you can do $$\sum_{k=-2}^{k=0}k\times 10$$ or, with the more common notation, $$\sum_{k=-2}^0k\times 10$$ |
H: An exercise in Rudin's RCA
Would you please give me some help on the following problem?
Suppose $1 \leq p \leq \infty$, and $q$ is the exponent conjugate to $p$. Suppose $\mu$ is a positive $\sigma$-finite measure and $g$ is a measurable function such that $fg\in L^1(\mu)$ for every $f\in L^p(\mu)$. Prove that then $g\in L^q(\mu)$.
If $p=\infty$. It is obvious that $g\in L^1(\mu)$, so it suffices to consider the case $1\leq p \lt \infty$. I first tried to think of a counterexample by constructing some $g\notin L^q(\mu)$ with $fg\notin L^1$. But since the $\sigma$-finiteness of $\mu$ is given, I think that the proof should rely on the duality of $L^p$ and $L^q$. I guess solving the problem amounts to show that the map $\Psi:L^p \to \mathbb{C}$ defined by $\Psi(f)=\int fg$ is bounded, but I am stuck here.
AI: I'll suggest a brief sketch: Start by assuming that $f \ge 0$.
Choose sets $E_k$ with $\mu(E_k) < \infty$, $E_k \subseteq E_{k + 1}$, and $X = \bigcup E_k$ by $\sigma$-finiteness. Take $G_n = \{x : |g(x)| < n\}$, and let $$g_n = \chi_{G_n \cap E_k}\cdot g$$ be an indicator function. Note $g_n \in L^q$.
Define (bounded) linear functionals $\Lambda_{n, k}$ on $L^p$ by
$$\Lambda_{n, k}(f) = \int g_n f d\mu$$
If these have a common bound, we're done. If not, use the Banach-Steinhaus theorem to find $f$ for which $$\sup_{n, k} |\Lambda_{n, k} f| = \infty$$ Use the monotone convergence theorem to conclude that $gf \notin L^1$, a contradiction. |
H: cohomology of Eilenberg-Maclane space
In line 5, Page 394 of Allen Hatcher's book Algebraic Topology, it is claimed that $H^n(K(G,n);G)=Hom(H_n(K(G,n),\mathbb{Z});G)$ for any abelian group $G$. How to get it?
I have tried but cannot continue: by universal coefficient theorem, we need $Ext(H_{n-1}(K(G,n);\mathbb{Z}),G)=0$. This holds if $H_{n-1}(K(G,n);\mathbb{Z})$ is free.
Thanks a lot.
AI: I think the group $H_{n-1}(K(G,n);\mathbb{Z})$ is actually zero. This follows from Hurewicz theorem. |
H: find maximum length of sum of two vector
let us consider following problem
so let us introduce vector with $2$ coordinates ,namely
$v=(v_1,v_2)$ with length $12$,which means that
$v_1^2+v_2^2=144$
and vector $s=(s_1,s_2)$ with length $10$,so it means
$s_1^2+s_2^2=100$
now let us consider sum of two vector
$v+s=(v_1+s_1,v_2+s_2)$
if we take square of each member ,we get
$v_1^2+2*v_1*s_1+s_1^2+v_2^2+2*v_2*s_2+s_2^2$
so it means that
$244+2*v_1*s_1+2*v_2*s_2$
now for vector $v$, possible solutions is $(8,6)$ or $(6,8)$ ,for vector $s$ we may write $144$ what we could do?please help me
AI: You are making it too hard.
If the two vectors are $\mathbf{u}$ and $\mathbf{v}$, what does the triangle inequality say about the maximum length of $\mathbf{u}+\mathbf{v}$?
Once you have an upper bound, you need to show that it is tight. It won't be hard. |
H: Calculation of $\frac {1-q^{n+1}}{1-q} \times \frac {1-q^{n+2}}{1-q}$?
I am trying to solve this equation:
$$\sum^{n}_{i=0} q^i \times \frac {1-q^{n+2}}{1-q} = \frac {1-q^{n+1}}{1-q} \times \frac {1-q^{n+2}}{1-q} = \frac {1-q^{n}q^1}{1-q} \times \frac {1-q^{n}q^2}{1-q} = ...?$$
where
$$\sum^{n}_{i=0} q^i = \frac {1-q^{n+1}}{1-q}$$
However, I am stuck at the last part. Can I simplify this even more. How to do that?
I really appreciate your answers!!
AI: Before question was edited:
This looks like the multiplication of two Geometric Progressions:
$$(1+q^1+q^2.....+q^{(n)} ).(1+q^1+q^2.....+q^{(n+1)} )$$ |
H: Find Greatest Common Divisor and Least Common Multiple
Find GCD (320, 112) and LCM[320, 112].
Solve the equation 320x + 112y = a in the following situations:
(i) a = 32
(ii) a = 10.
Using Euclids Algorithm to find the GCD I have the following:
320 = 112*2 + 96
112 = 96*1 + 16
96 = 16*6 + 0
GCD(320, 112) = 16
(edit) Okay, after working a bit I remembered how to find LCM with prime factors
320 = 2^6 * 5
112 = 2^4 * 7
So, LCM[320, 112] = (2^6)(5)(7) = 2240
Moving on, I need to know how to apply this information to solve the equations.
AI: (i) In the first case :
$$320x + 112y = 32$$
Which is an equation of a line and has infinite solutions for $(x,y)$
(ii)Similar explanation. |
H: Calculate Var(XY)
I was practicing and came across this, which I couldn't solve.
Calculate Var(XY) where X ~ Uniform(0,1) and Y ~ Normal(0,1)
AI: Let $W=XY$. We want to calculate $E(W^2)-(E(W))^2$.
On the (unstated) assumption that $X$ and $Y$ are independent, all we need is $E(X^2)$ and $E(Y^2)$, since $E(Y)=0$.
The calculation of $E(X^2)$ is an easy integration. And since $\text{Var}(Y)=E(Y^2)-(E(Y))^2$, we have $E(Y^2)=1$. |
H: Show that there exists a $k>o$ such that solutions of this system of differential equations never cross the line $y = kx$.
For the system:
$\frac{d x}{d t} = -y$
$\frac{d y}{d t} = x(1-x) - Ay$
Where $A \geq 2$.
I want to show that there exists a $k > 0$ such that $(x(t), y(t))$ cannot cross through the line $\{ y = k x, x > 0 \}$ when we start below this line.
I started with showing that for the line $y = k x$ we have that $\frac{d y}{d t} = k \frac{d x}{ d t} = -k y$, and now I wanted to show that if we start below this line, we have a lower $\frac{dy}{dt}$ so we can never cross the line.
But I have been unsuccesful.
Any ideas? Thanks.
AI: I think the following consideration can answer your question.
So, the $\lbrace y = kx \rbrace$ is a boundary for domain of interest. Let's define the normal field at $y = kx$ such that it points "in" $\lbrace y \leqslant kx \rbrace$; for $k > 0$ constant vector field $(k, -1)$ is such. Then the idea is following: if we compare the vector field on boundary with normal field on the same boundary, we can define where trajectories go in our out of domain. So, the sign of dotproduct between normal field and vector field shows the direction of crossing, "in" or "out: if it's positive, trajectory goes in, if it's negative it goes out. You're interested in situation such that trajectories don't go out of domain, so there's a positive sign of dotproduct along boundary. So, let's check that for our vector field:
$$ \vec{n} \equiv (k, -1), \; \vec{v}\vert_{\lbrace y = kx \rbrace} = (-kx, x(1-x) - kAx), $$
$$ (\vec{n}, \vec{v}) = x \lbrack x + Ak - k^2 -1 \rbrack $$
Here it goes: if you want trajectories not to leave through upper part of boundary, then $Ak - k^2 -1 \geqslant 0$ would be necessary and sufficient for that. In other cases you'll have a segment where trajectories can leave through boundary. |
H: Inverse Laplace Transform Problem
I have this problem $\frac{1}{(s^2+1)^3}$. I have to find its Inverse Laplace Tranformation.
I already try using partial fraction but it didn't work because I found it will back to the problem form.
Any other way for solutions?
AI: You may use the residue theorem. The ILT is
$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{(s^2+1)^3}$$
where $c \gt 0$. Consider a contour integral in the complex $s$ plane:
$$\frac{1}{i 2 \pi} \oint_C ds \frac{e^{s t}}{(s^2+1)^3}$$
where $C$ includes the line $[c-i R,c+i R]$ and a semicircle of radius $R$ that closes to the left when $t \gt 0$ and to the right when $t \lt 0$. We take the limit as $R \to \infty$. When $t \gt 0$, the contour integral - and therefore the ILT, is then equal to the sum of the residues of the integrand at the poles $z=\pm i$. In this case, each pole is a triple pole, so that the residues are given by
$$\sum_{\pm} \frac12 \left [\frac{d^2}{ds^2} \frac{e^{s t}}{(s \pm i)^3} \right ]_{z=\pm i} $$
The derivative term produces
$$\frac{d^2}{ds^2} \frac{e^{s t}}{(s \pm i)^3} = \frac{(s\pm i)^2 t^2 - 6 (s\pm i) t + 12}{(s \pm i)^5} e^{s t}$$
so that performing the above sum produces, as the ILT
$$\frac12 \frac{-4 t^2 - i 12 t + 12}{32 i} e^{i t} + \frac12 \frac{-4 t^2 + i 12 t+ 12}{-32 i} e^{-i t} = -\frac18 (t^2-3) \sin{t} - \frac{3}{8} t \, \cos{t}$$
When $t \lt 0$, on the other hand, there are no poles within $C$, so the ILT is zero there, as is expected. |
H: combined reliability
I thought I knew the answer to this question, but further reflection is showing some holes in my knowledge; my college math is twenty-five years old and google isn't helping today.
Let's say you have a computer with five disk drives. If any one of the drives crashes, the computer is down. For this model of drives, 99% of them run a year without crashing. So the probability that this computer will be up the entire year is
.99 number of drives
or
.99 5 ≈ .95
That's correct, right?
Now the obverse observation is that each drive has a 1% chance of failure over the course of a year. How does one express the total chance of failure for the whole computer? The answer should be 5%, because that's what's left over from the 95% given above.
The naive approach would be to do the same thing again:
.015
but that gives 0.0000000001 which obviously isn't right. What am I missing here?
AI: $0.01^5$ is the probability that ALL 5 drives crash.
To see why, consider that an individual drive failing is unaffected by what happens with the other drives, and the probability of any one of them failing is 0.01. So the probability of all of them failing is the product of each probability.
To calculate the probability of failure, you do exactly as you have done, calculate 1 - Probability of not failing. |
H: On a sum involving fractional part of an integer
I was interested in estimating the sum of the form
$$
\sum_{j=1}^{N} \{ \sqrt{j} \}.
$$
I was wondering if there is a reference or maybe some one could
help me figure out what to do.
Thanks!
$\{ \alpha\}$ denotes the fractional part of the real number $\alpha$.
AI: The desired sum is equal to $$\left(\sum_{j=1}^N \sqrt{j}\right) - \left(\sum_{j=1}^N \lfloor \sqrt{j}\rfloor\right)$$
The second sum is computed here as $(N+1)a - \frac{a^3}{3} - \frac{a^2}{2} - \frac{a}{6}$, where $a=\lfloor \sqrt{N+1}\rfloor$.
The first sum may be estimated by an integral as follows
$$\frac{2}{3}N^{3/2}=\int_0^N\sqrt{x}dx < \left(\sum_{j=1}^N \sqrt{j}\right) < \int_1^{N+1} \sqrt{x}dx=\frac{2}{3}((N+1)^{3/2}-1)$$
The difference between the upper and lower estimates is $O(\sqrt{N})$, which isn't too bad I think. With care the integrals can be refined, sharpening the estimates. |
H: Example function $f:(0,1) \to (0,1)$ such that $f^{-1}(y)$ is uncountable for all $y$.
Intuitively the interval $(0,1)$ can be an uncountable union of disjoint uncountable sets, but I'm wondering if there is a nice function to show this easily. Namely, what is a simple example of $f:(0,1) \to (0,1)$ such that $f^{-1}(y)$ is uncountable for every $y$?
AI: $f(0.x_1x_2x_3x_4x_5\ldots)=0.x_1x_3x_5\ldots$ |
H: How would one prove that $\# (E \cup G) = \#(E) + \#(G)$?
For a set $E$ and $n \in \mathbb{N}$ we say $\# (E) = n$ if there is a bijection from $I_n$ to $E$, where
$$I_n \overset{\text{def}}= \{k\in \mathbb{N}: k \leq n\}$$
Suppose that $\# (E) = n$, $\#(G) = k$ and $E \cap G = \varnothing$. Show that
$$\#(E \cup G) = \#(E) + \#(G)$$
My thoughts
It may be tempting to use $E \cup G = E + G - (E \cap G)$, set $\#$ by both sides and evaluate both sides to get the statement I want to prove, but is this right? I thought of the bijection as some sort of homomorphism. It's something like this
$$\#(E \cup G) = \#(E) + \#(G) - \#(E \cap G)$$
$$\#(E \cup G) = \#(E) + \#(G)$$
Another thing: Do I need to apply the ordering on the set to show the given equation, or is that unnecessary? I thought of this because of the given set.
I'm quite stuck on proof because I don't really know how it goes. Any advices or comments?
AI: There is a bijection $f : E \to I_{\#E}$, and a bijection $g : G \to I_{\#G}$. Let us extend $f$ to $E \cup G$ by definining
$$f(x) = \left\{ \begin{array}{lr} f(x) & : x \in E \\ 0 & : x \in G\end{array}\right.$$
Since $E \cap G$ is empty, $f$ is well-defined. Do the same for $g$.
Now consider $h = f + g$ defined pointwise; can you show that $h$ is a bijection with $I_{\#E + \#G}$? |
H: Divisibility of a prime number
I need help with the following:
Show that:
If $p$ is prime such that $p$ divides $a^n$
Then $p^n$ divides $a^n$
I know that if $p$ is a prime and divides a square number $a$ then $p$ also divides $a$ but I'm not sure how to apply this to the given problem.
AI: Notice that if $p$ is a prime and $p$ divides $a^n$, then $p$ divides $a$. Why this is follows from the fact that if $p$ is a prime and does not divide $a$ (but divides $a^n$), then $p$ cannot be a prime. So it's a contradiction, and $p$ must divide $a$. Let $q$ denote the product of all other primes that divide $a$. Then we can write
$$
a=pq
$$
and hence
$$
a^n=(pq)^n=p^nq^n.
$$
So $p^n$ divides $a^n$. |
H: P p-sylow with $ P ⊂ Z(G) $
For me this problem is hard .
If $ P⊂ Z(G)$ is a $p$-sylow of $G$ then there is a $N$ normal subgroup of $G$ such that $P ∩ N = 1$ and $G = PN$.
I try use the Schur-Zassenhaus: All normal subgroup cop rime has a complement and all conjugates are conjugate, but i stuck in this problem please help thanks.
AI: By applying Schur-Zassenhaus to $P$, you can find a subgroup $N$ such that $P \cap N = 1$ and $G = PN$.
Now $P$ is central, so $P$ normalizes $N$. Thus $N$ is normal. |
H: Ask a question about the definition of trace norm.
Suppose $X\in \mathbb{R}^{M\times N}$
$\|X\|_*=\mathrm{trace}(\sqrt{X^*X})=\sum_i^{\min{M,N}}\sigma_i$
where $\sigma_i$ is the singular values of $X$.
I know that $\mathrm{trace}({X^*X})=\sum_i^{\min{M,N}}\sigma_i^2$. Thus, the following equation holds.
$\sqrt{\mathrm{trace}({X^*X})}=\sqrt{\sum_i^{\min{M,N}}\sigma_i^2}$.
But I wonder how to prove $\mathrm{trace}(\sqrt{X^*X})=\sum_i^{\min{M,N}}\sigma_i$
AI: Let $X = U \Sigma V^*$, where $U$ and $V$ are unitary and $\Sigma$ is nonnegative real diagonal, be an SVD of $X$. Then
$$X^*X = V \Sigma^2 V^*,$$
so
$$\sqrt{X^*X} = V \sqrt{\Sigma^2} V^* = V \Sigma V^*,$$
Notice that this is a similarity relation, so
$$\operatorname{tr} \sqrt{X^*X} = \operatorname{tr} (V \Sigma V^*) = \operatorname{tr} \Sigma.$$ |
H: If $p$ is prime and any integer $k>1$, then $p^{\frac 1k} $ is irrational. Prove this by assuming $p^{\frac 1k}$ rational
I've tried setting $p^{\dfrac 1k}= \dfrac a b$, and then raising $p^{\dfrac 1k}$ to the $k^\text{th}$ power, but I'm stuck.
AI: let $p^{\frac1k}$ be rational then $p^{\frac1k}=\frac ab$ with $(a,b)=1$ or $p=\frac {a^{k}}{b^{k}}$,
since p is a prime hence integer we have $b=1$ and $p=a^k$, and if $k>1$ then p will be divisible by more then two numbers, hence it will not be prime, a contradiction, therefore $p^{\frac1k}$ is irrational for $k>1$ and p be a prime. |
H: Verify the identiy, (cos(x+h) - sin x)/h = cos x * ((cos h - 1)/h)- sin x * (sin h /h)
Verify the identity:
$$\frac{(\cos(x+h) - \cos x)}{h} = \cos x \left(\frac{\cos h - 1}{h}\right)- \sin x \left(\frac{\sin h }{h}\right)$$
=(Cosxcosh - sin x sin h -cos x)/h.
I can't think of where to go from here.
Thanks
AI: If I understood correctly what you have written:
$$\frac{(\cos(x+h) - \cos(x))}{h}\stackrel{?}{=} \cos x \left(\frac{\cos h - 1}{h}\right)- \sin x \left(\frac{\sin h }{h}\right)$$
Or
$$\frac{\cos(x+h) - \cos(x)}{h}\stackrel{?}{=}\frac{\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h} $$
Or
$$
\begin{eqnarray}
\cos(x+h) - \cos(x)&=&\cos(x)(\cos(h)-1)-\sin(x)\sin(h)=\\
&=&\cos(x)\cos(h)-\cos(x)-\sin(x)\sin(h)
\end{eqnarray}
$$
$$\cos(x)\cos(h)-\sin(x)\sin(h)=\cos(x+h)$$
and so you have
$$\cos(x+h) - \cos(x)=\cos(x+h) - \cos(x)$$ |
H: Prove that $x$ is any positive real number greater than $0$, $x>0$, then exists $N$ in the natural numbers such that $\frac{1}{N^3}
Prove that $x$ is any positive real number greater than $0$, $x>0$, then exists $N$ in the natural numbers such that $\frac{1}{N^3}<x$
My steps:
Well I begin with $N\in\mathbb{N}$ and $x\in\mathbb{R}$ such that $N$ could be any integer number begining in 1; and $x$ could be any real number.
Then if $\frac{1}{N^3}<x$ and $N$ always be and integer number... and $x$ could be real, I will analyze a few number:
With $N=1$ and $x=1$, then $\frac{1}{1^3}<1$ its totally false. Well if $N=2$ and $x=2$, then $\frac{1}{2^3}<2$ it's true. Then I will check that $\frac{1}{N^3}<x$ is true if and only if $N>0$, and $x>\frac{1}{N^3}$
I am in the correct way???
AI: Your doing things in the wrong order, cf. comments above.
Since $x\ne0$ we can consider $\frac1x$ and since $x>0$ we also have that $\frac1x>0$.
By the Archimedean property of $\mathbb R$, there exists some $N\in\mathbb N$ with $\frac1x<N$.
Since $N\ge 1$, we have $N^3\ge N>\frac1x$ and by taking reciprocals (all numbers involved are positive!), $\frac1{N^3}<x$. |
H: Linearly independent vectors and matrix
If $\{v_{1},v_{2},\cdots,v_{n}\}$ is $n$ linearly independent vectors
in $\mathbb{R}^{n}$, what would be necessary and sufficient condition of $A$ ($n\times n$ matrix) $A$ so that the vectors $Av_{1}$,
$Av_{2}$, $\cdots$, $Av_{n}$ are linearly independent?
AI: Hint:
The vectors $Av_1, \ldots ,Av_n$ are linearly independent if, and only if, $\det \left([Av_1 | \ldots |Av_n]_{n\times n}\right)\neq 0$.
Now note that $[Av_1 | \ldots |Av_n]_{n\times n}=A[v_1|\ldots |v_n]_{n\times n}$. |
H: Semigroup isomorphism
Does there exist an isomorphism between the semigroups $S(4)$ and $\mathbb{Z}_{256}$?
$S(4)$ is the set of all maps from the set $X$ to itself and $X =\{1, 2, 3, 4\}$,
$S(4)$ is a semigroup under the composition of mappings, and $\mathbb{Z}_{256}=\{0, 1, 2, \dots , 255\}$ is the semigroup under multiplication modulo 256.
AI: The semigroup $S(4)$ contains exactly $24$ invertible elements (which form the symmetric group $S_4$).
But $256^Z$ contains $128$ invertible elements (all odds). So they can't be isomorphic.
Alternatively: $256^Z$ has an absorbing element $0$ with $0x=x0=0$ for all $x$. No such map exists in $S(4)$. |
H: Approximation to Lattice points inside circle
In the Wikipedia here, there is an approximation that $N(r)=\pi r^2+E(r)$ and then it says that Gauss managed to prove that $E(r)\leq2 \sqrt2 \pi r$ can anyone prove the original formula and tell me how Gauss got his approximation.
AI: Around each lattice point consider the axe-parallel square of side length $1$ and the lattice point as center.
The area of the circle is $\pi r^2$ is approximately $N(r)$ (the sum of square areas for interior points) and the difference comes only from lattice points whose associated square is neither fully inside nor fully outside the circle. But then such squares are fully outside the circle with radius $r-\frac12\sqrt 2$ and fully inside the circle with radius $r+\frac12\sqrt 2$. In other words, if $r>\frac12\sqrt 2$ then
$$ \pi\left(r-\frac12\sqrt 2\right)^2\le N(r)\le \pi\left(r+\frac12\sqrt 2\right)^2.$$
This already gives $E(r)\le \pi r\sqrt 2+\frac12\pi$, which is better than needed.
For $r\le \frac12\sqrt 2$, note that $N(r)=1$ and this may differ from $\pi r^2$ by more than $2\pi r$, so the claim does not even hold for all $r$! |
H: How to calculate expected value?
Expected numbers of 'yes' for following data
A = [1,2,3,5]
B = [5,6,7,8]
A 'yes' is when a^b > b^a where a is value randomly chosen from A and b is value randomly chosen from B . Also numbers a and b are discarded from A and B after randomly chosen.
Please explain how to solve it.
AI: Hint: There are only $16$ possibilities, so you can go through them. All with $1$ from $A$ will be no, all with $2$ or $3$ will be yes, and $5$ will be yes unless it is paired with $5$. Presumably you are picking four disjoint pairs and looking for the count of the pairs which produce yes. |
H: Understanding the theorem about the almost disjoint sets partial order
I have read the Kunen's set theory. But I get a struck to understand the proof of a theorem.
Let $\mathcal{A}\subset\mathcal{P}(\omega)$ be a family of pairwise almost disjoint sets, then we define $(\Bbb{P}_\mathcal{A},\le)$ as $\Bbb{P}_\mathcal{A}=\{\langle s,F\rangle : s\in [\omega]^{<\omega},\,F\in[\mathcal{A}]^{<\omega}\}$ and $\langle s,F\rangle \le \langle s',F'\rangle$ iff $s\subset s'$, $F\subset F'$ and $x\cap s'\subset s$ for each $x\in F$.
Lemma 2.8. In $\Bbb{P}_\mathcal{A}$, $\langle s,F\rangle$ and $\langle s',F'\rangle$ are compatible iff $x\cap s'\subset s$ for all $x\in F$ and $x\cap s\subset s'$ for all $x\in F'$.
Lemma 2.10. If $G$ is a filter in $\Bbb{P}_\mathcal{A}$ and $\langle s,F\rangle \in G$ then $x\cap d_G\subset s$ for each $x\in F$
(Where $d_G:=\bigcup\{s\mid\exists F:\langle s,F\rangle \in G\}$.)
Proof. If $\langle s',F'\rangle\in G$ then $\langle s',F'\rangle$ and $\langle s,F\rangle$ are compatible, so by Lemma 2.8, $x\cap s'\subset s$ for all $x\in F$.
I don't understand the proof of Lemma 2.10. Especially, I don't understand $\langle s',F'\rangle\in G$ implies the compatibility of $\langle s,F\rangle$ and $\langle s',F'\rangle$. In this book there is no proofs of this statement (as I know) and I don't know how to prove this statement. Thanks for any help.
AI: Given $x \in F$, note that for any $\langle s^\prime , F^\prime \rangle \in G$ we have that $\langle s^\prime , F^\prime \rangle , \langle s , F \rangle$ are compatible -- since $G$ is a filter and $\langle s , F \rangle$ also belongs to $G$ -- so by Lemma 2.8 it follows that $x \cap s^\prime \subseteq s$.
It then follows that $x \cap d_G = x \cap \bigcup_{\langle s^\prime , F^\prime \rangle \in G} s^\prime = \bigcup_{\langle s^\prime , F^\prime \rangle \in G} ( x \cap s^\prime ) \subseteq s$.
Recall that given a poset $\langle \mathcal{P} , \preceq \rangle$, a (nonempty) subset $F \subseteq \mathcal{P}$ is a filter if
whenever $p,q \in \mathcal{P}$ and $p \preceq q$, then $p \in F$ implies $q \in F$;
whenever $p,q \in F$, then $p,q$ there is an $r \in F$ with $r \preceq p , q$.
This second condition implies that every two elements of a filter are compatible as elements of the poset. |
H: Conditional probability, Bayes' rule and chain rule
I'm reading the following paper A Rational Account of the Perceptual Magnet Effect and I'm puzzled by equation (3) on page 2:
$$p(T|S,c) \propto p(S|T)p(T|c)$$
where $T$, $S$ and $c$ are random variables. Apparently it should be trivial, but I tried to play with conditional probability, Bayes' rule and chain rule without getting anything like this. Am I missing something very obvious?
AI: It doesn't seem to be explicitly written out, but it appears as if they assume $S$ and $c$ to be independent conditional on $T$ such that $p(S|T, c)=p(S|T)$.
$$
p(T|S, c)=\frac{p(T, S, c)}{p(S, c)}=\frac{p(S|T, c)p(T, c)}{p(S, c)}=\frac{p(S|T, c)p(T| c)p(c)}{p(S, c)}\propto p(S|T, c)p(T| c)=p(S|T)p(T| c)
$$
If they are not, then the last equality does not hold. |
H: How do I caclulate the probability of hash collisions?
I have a 10Gb file and the entire file is overwritten with random data every day.
Afterwards, I divide the data into blocks and hash each block to generate a fingerprint.
I am trying to choose an appropriate1 hash size to make it unlikely that that will be a collision (a false match in the hash for a particular block as compared to the previous iteration: a match is OK if the data is randomly the same but not if it's showing the same hash for different data).
How can I calculate the probability of a false match in 10 years given the block size and hash length?
1 appropriate means: as small as possible without making the first collision more than 50% likely in 10 years
AI: Assuming that the hash function behaves like a random oracle, then the probability that any given block hashes to the same value as the previous version of the same block is 2-n, for a hash function output size of n bits. If you have k blocks in your file, then, over ten years, you are computing k·3653 block hashes (or k·3652 or k·3654, depending on leap years), so the probability of not hitting a collision in 10 years will be equal to (1-2-n)k·3653. It is simpler to count the average time between two occurrences of a collision: it will be 2n/k days.
For instance, if using 1 kB blocks in a 10 GB file, then k = 107 (ten million blocks). If n = 128, then the average time between two collisions will be close to 1040 years, i.e. a lot of time. Said otherwise, if you want collisions to occur once every 3653 days on average, then you need a hash function with an output of about 35 bits, because 235 is roughly equal to 3653·107. More bits lower collision rate, of course.
All of the above assumes that you are fighting against random collisions. If some evil attacker tries to force collisions to happen, and can choose the contents of some blocks, then you need the hash function to be collision resistant up to the computational power that the attacker could muster; in that case, use a hash function with an output of at least 160 bits, corresponding to a theoretical resistance of 280 (the traditional limit of attacker's power). Note that some theoretical weaknesses have been found in SHA-1 (the usual 160-bit hash function) so you would prefer using SHA-256, which offers a 256-bit output. If size is constrained, then you may want to truncate the SHA-256 output, but don't reduce it below 160 bits.
An interesting point is that when fighting an attacker, then how many hashes you compute is no longer important; what matters is how many the attacker may be able to compute, and he may compute billions per second, way more than your millions per day. |
H: Prove that a module is projective or not
Let $R=\left(\begin{array}{cc}\mathbb{Q}&\mathbb{Q}\\0&\mathbb{Q}\end{array}\right), J=\left(\begin{array}{cc}0&\mathbb{Q}\\0&0\end{array}\right)$. Prove that $R/J$ is not a projective $R$-module.
I'm really misunderstand about $R/J$, about it's equivalent class. I can understand $\mathbb{Z}/ 2\mathbb{Z}=\mathbb{Z}_2$, but in general I can't imagine $A/B$.
Let $R=\left(\begin{array}{cc}\mathbb{R}&\mathbb{R}\\\mathbb{R}&\mathbb{R}\end{array}\right), M=\left(\begin{array}{cc}\mathbb{R}&\mathbb{R}\\0&0\end{array}\right)$. Prove that $M$ is a right projective module $R$-module but not free.
Prove that with universal property really hard with me because of maps (homomorphisms) constructive.
Thanks for regarding!
AI: Hints:
1) One characterization of projective modules you could use is that in the exact sequence
$$
0\to A\to B\to P\to 0
$$
with $P$ projective, the sequence splits. That is, $B\cong A\oplus P$ as $R$ modules.
In your case: $R\cong J\oplus (R/J)$. But for $J$ to be a direct summand of $R$, it would have to contain a nonzero idempotent... does it?
2) Here, $R$ is a four dimensional $\Bbb R$-algebra. $M$ is only a two dimensional $\Bbb R$ vector space. Could $M$ be the direct sum of copies of $R$? On the other hand, it is relatively easy to see another module $N$ such that $M\oplus N=R$, showing that $M$ is projective. In case you don't have this fact yet, it would also be a good thing to prove: a module is projective iff it is a direct summand of a free module. This follows immediately from the lemma I gave in the hint for 1).
In general, any quotient $A/B$ can be thought of as "like $A$, except anything that I see from $B$ is considered zero." For example, $\begin{bmatrix} a&b\\0&c\end{bmatrix}=\begin{bmatrix} a&0\\0&c\end{bmatrix}+\begin{bmatrix} 0&b\\0&0\end{bmatrix}=\begin{bmatrix} a&0\\0&c\end{bmatrix}$ mod $J$.
So in your case, $R/J\cong \begin{bmatrix}\Bbb Q&0\\0&\Bbb Q\end{bmatrix}\cong \Bbb Q\oplus \Bbb Q$.
Actually, it is true in general for rings with identity that the Jacobson radical $J$ (which is what $J$ is in the case above) is never a right or left summand of $R$ (unless it's zero.) The reason is that the Jacobson radical is a superfluous submodule of $R_R$ and $_RR$. By definition, that is that $B+J=R$ implies that $B=R$. |
H: Why is it sufficient for a normal to only be orthogonal to 2 vectors on a plane instead of 3?
The following is an excerpt from my textbook:
It is clear geometrically that there is a unique plane containing any 3 points $A,B$ and $C$ that are not all on a line. In determining the equation of this plane, the problem is finding a normal vector for the plane. Since any nonzero vector which is orthogonal to both $\overrightarrow{AB}$ and $\overrightarrow{AC}$ will do as a normal, $\dots$
Why don't we need to consider the vector $\overrightarrow{BC}$ too? Or is it sufficient to just choose any two of the 3 possible vectors that we can create given points $A,B$ and $C$? How do I prove this?
AI: It is given that 3 points give a unique plane.
Similarly 2 vectors having a common point also give a unique plane.
Hence you can use $\overrightarrow{BA}$,$\overrightarrow{BC}$ or $\overrightarrow{CA}$,$\overrightarrow{CB}$ as well |
H: Question concerning finite rings
Let $R$ be a finite ring. Is it possible that $R$ has an element $a\in R$ such that $a$ is a left divisor of zero and $a$ is not right divisor of zero?
Thanks.
AI: Let $S$ be a finite semigroup of left zeroes ($ab=a$ for all $a,b\in S$), $F$ a finite field, $FS$ the semigroup ring of $S$ over $F$. Then $a(b-c)=0$, so every $a$ is a left divisor of zero but not a right divisor.
(Thanks @rschwieb for a remark) |
H: Finding centre of ellipse using a tangent line?
I need to determine the centre coordinates (a, b) of the ellipse given by the equation:
$$\dfrac{(x-a)^2}{9} + \dfrac{(y-b)^2}{16} = 1$$
A tangent with the equation $y = 1 - x$ passes by the point (0, 1) on the ellipse's circumference.
I'm guessing I have to find the implicit derivative first, but I'm not quite sure how to derive the part of the right. According to my calculator, the implicit differentiation is:
$\frac{-16(x-a)}{9(y-b)}$
But I'd really like to try and do this by hand. I'm just really not sure of the steps I need to take to solve this.
Thanks for any help.
AI: Hints: Follow, understand and prove the following
Since the point $\;(0,1)\;$ is on the ellipse then
$$\frac{a^2}9+\frac{(1-b)^2}{16}=1$$
Now differentiate implicitly:
$$\frac29(x-a)dx+\frac18(y-b)dy=0\implies \frac{dy}{dx}=-\frac{\frac29(x-a)}{\frac18(y-b)}=-\frac{16}9\frac{x-a}{y-b}$$
But we know that
$$-1=\left.\frac{dy}{dx}\right|_{x=0}=-\frac{16}9\frac{-a}{1-b}$$
Well, now solve the two variable equations you got above... |
H: What is the simplification of sum of $i(i+1)$?
I am trying to simplify $$\frac{\sum_{i=1}^n i(i+1)}{n(n-1)}$$.
I am not sure how to simplify ${\sum_{i=1}^n i(i+1)}$ part.
How can I simplify it?
AI: You may know the wellknown results $$ \sum_{i=1}^n i=\frac{n(n+1)}{2}$$
and
$$ \sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6},$$
from which you can combine your sum. |
H: Unsolved uniform continuity
Prove or disprove that: the function $ f : \mathbb R \to \mathbb R $ defined by $ f(x) := x^{1/3} \log (1+|x|)$ is uniform continuous!
I have tried all possible (I THINK) to prove it to be uniform continuous but failed. I think it would be NOT uniform continuous!Now, here I am still unable to find suitable sequences having their distance going to zero, BUT the distance between their functional images converging to some non-zero!
AI: Hint: Prove that its derivative is bounded. |
H: Relationship Little '$\mathcal{o}$' and Big '$\mathcal{O}$'
I'm learning about asymptotic analysis and, as a starting point, big and little o definitions.
On the Wikipedia page, http://en.wikipedia.org/wiki/Big_O_notation
further down under the heading for little-$\mathcal{o}$ notation it states
"In this way little-$\mathcal{o}$ notation makes a stronger statement than the corresponding big-$\mathcal{O}$ notation: every function that is little-$\mathcal{o}$ of $g$ is also big-$\mathcal{O}$ of $g$, but not every function that is big-$\mathcal{O}$ g is also little-$\mathcal{o}$ of $g$ (for instance $g$ itself is not, unless it is identically zero near ∞"
This makes sense to me, given the definitions. However, I have an example that seems to be to the contary. In the limit $x\rightarrow0$ the little order of the sin function could be characterized as
$$
\sin(x)=\mathcal{o}(1)
$$
i.e., it is always a lot less than 1. According to the arument, this should also be the big-$\mathcal{o}$ of $\sin$ in the same limit. However, in the limit the big $\mathcal{O}$ is
$$
\sin(x)=\mathcal{O}(x)
$$
and which as at odds with the relationship between litttle-$\mathcal{o}$ and big-$\mathcal{O}$ in the wiki article.
Where have I went wrong? Is it because saying $\sin(x)=\mathcal{o}(1)$ hasn't characterised the order in terms of a dominant function, but rather picked a constant larger than zero?
Cheers
AI: There is no problem since $O(x)\subset o(1)$ hence $\sin x=O(x)$ is more precise than $\sin x=o(1)$. (But recall that $f(x)=O(g(x))$ is only a shorthand for $f\in O(g)$, likewise for $o$.) |
H: Polynomials in nature
What polynomials occur in "nature"? I am interested in polynomials of degree three and higher. I am aware of Stefan Boltzmann Law and Chemical Equilibrium Examples.
Edit:
There are some formulas under beam deflection category.
I am particularly interested in full polynomials, as opposed to pure power laws. For example $y=at^2+v_0 t + y_0$ for a projectile uses a complete polynomial, while Boltzman law is just a power law.
AI: Power Laws are probably what you are looking for and are very, very prevalent in physics, economics, linguistics and well, nature. They occur for phenomena that exhibit scale invariance. All forms of exponents occur. The M-sigma law has a power of 4 for example. |
H: Semigroups isomorphism
Does there exist an isomorphism between the semigroups $S(4)$ and $\mathbf Z_{256}$.
$S(4)$ is the set of all maps from the set $X$ to itself and $X = \{1, 2, 3, 4\}$. $S(4)$ is a semigroup under the composition of mappings and $\mathbf Z_{256} = {0, 1, 2, … , 255}$ is the semigroup under multiplication of integers modulo 256.
AI: Note that in $\mathbb Z_{256}$ we only have two elements $x$ so that
$$x^2=x$$
Indeed, if $x$ is odd, it is invertible, otherwise $x-1$ is invertible $\mod{256}$.
In $S(4)$ there are many functions $f$ so that $f \circ f =f$, for example, all functions with only one element in the image.
So the answer is no.
Second solution The invertible elements in $S(4)$ are the permutations, thus $S(4)$ has $4!=24$ invertible elements. The invertible elements in $\mathbf Z_{256}$ are the numbers relatively prime to $256$ (i.e. odd numbers). Thus $\mathbf Z_{256}$ has 128 invertible elements. |
H: Need help with detail on proof regarding intermediate values of a derivative.
This is Theorem 5.12 in Rudin's Principles of Mathematical Analysis.
Suppose $f$ is a real differentiable function on $[a,b]$ and suppose $f^{'}(a) < \lambda < f^{'}(b)$. Then there is a point $x \in (a,b)$ such that $f^{'}(x)=\lambda$.
Proof:
Put $g(t)=f(t)-\lambda \cdot t$. Then $g^{'}(a)<0$, so that $g(t_1) <g(a)$ for some $t_1 \in (a,b)$.
Can someone help me see why $t_1$ exists?
AI: $$g'(a)=\lim_{h\to0}\frac{g(a+h)-g(a)}{h}<0$$
implies that $g(a+h)-g(a)<0$ for sufficiently small $h>0$. |
H: Apollonius circle, its radius and center
I've got the following set:
$\{|z-a|=k|z-b|\}$, where z is a complex number, a an b are fixed, and $k>0$,$k \ne 1$.
I need to prove that this is a circle (called Apollonius circle).
I also have to prove that this circle's radius is equal to $k|a-b||1-k^2|^{-1}$ and it's centre is $(a-k^2b)(1-k^2)^{-1}$.
I don't know what to do. I've tried to work with the analytic circle equation ($|z-c|^2=r^2$), substituting given radius and center, but it didn't work. I also tried to square both sides of the first given equation ($|z-a|=k|z-b|$), which usually works with complex number, but also didn't get any result... Can somebody show me how to solve this problem? I will be very grateful.
AI: The algebra is messy but straightforward. Let $z=x+i y$, $a=a_r+i a_i$ and $b=b_r+i b_i$. Square both sides of the defining equation to get
$$(x-a_r)^2+(y-a_i)^2 = k^2 (x-b_r)^2 + k^2 (y-b_i)^2$$
Rearrange and expand to get
$$(1-k^2) x^2 + (1-k^2) y^2 - 2 (a_r-k^2 b_r) x - 2 (a_i - k^2 b_r) y + |a|^2-k^2 |b|^2 = 0$$
Now complete the square. This is where it gets messy.
$$\begin{align}(1-k^2) \left ( x-\frac{a_r-k^2 b_r}{1-k^2}\right )^2 + (1-k^2) \left ( y-\frac{a_i-k^2 b_i}{1-k^2}\right )^2 \\= \frac{(a_r-k^2 b_r)^2+(a_i-k^2 b_i)^2}{1-k^2} - (|a|^2-k^2 |b|^2)\\ = \frac{|a|^2 + k^4 |b|^2 - 2 k^2 (a_r b_r+a_i b_i) - (|a|^2-k^2 |b|^2)(1-k^2)}{1-k^2}\\ = \frac{k^2 (|a|^2+|b|^2) - 2 k^2 (a_r b_r+a_i b_i)}{1-k^2}\\ = \frac{k^2}{1-k^2} |a-b|^2\end{align}$$
From here, I hope it is clear that the above reduces to, in complex notation:
$$\left |z-\frac{a-k^2 b}{1-k^2}\right| = \frac{k}{1-k^2} |a-b|$$
So, a circle of center $$\frac{a-k^2 b}{1-k^2}$$ and radius $$\frac{k}{1-k^2} |a-b|$$ |
H: Show that two logical expressions are equivalent without using the truth table.
I would like to show that the expression (S => P) AND (NOT S => Q) and the expression (S AND P) OR (NOT S AND Q) are equivalent.
I am not interested in any solution using truth tables or exhaustive trying of truth assignments, I find these approaches rather unilluminating.
AI: The following facts will get you most of the way there: $$A\Rightarrow B\equiv(\neg A)\vee B$$ $$A\wedge(B\vee C)\equiv(A\wedge B)\vee(A\wedge C)$$ $$A\vee(B\wedge C)\equiv(A\vee B)\wedge(A\vee C)$$ |
H: Existence of some linear mappinng's problem
Asume that $X$ is a linear finite dimensional vector space over $\mathbb R$ and $W$ is its subspace.
Let $f$, $h$ be linear mappings from $X$ to $X$ such that $f(X)=W$ and $h(X) \subset W$.
Is there a linear mapping $g:X \rightarrow X$ such that
$$
h=f\circ g
$$
AI: Hint: Does there exist a linear $s \colon W \to X$ with $f\circ s = \operatorname{id}_W$? How would that $s$ help you answer the question? |
H: Equivalent definition of locally compact when $X$ is Hausdorff. How did they get $\overline{V} \cap C$ is empty?
The theorem in Munkres' Topology is
Theorem 29.2. Let $X$ be a Hausdorff space. Then $X$ is locally compact if and only if given $x \in X$, and given any neighborhood $U$ of $x$, there is a neighbordhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subset U$.
One direction is clear, namely the new formulation implies $X$ is locally compact. The converse argument goes: Let $U$ be a neighbordhood of $x$. Take the one-point compactification $Y$ of $X$, and let $C$ be the set $Y - U$. Then $C$ is closed in $Y$, so that $C$ is a compact subspace of $Y$. Then there exist disjoint open sets $V$ and $W$ containing $x$ and $C$, respectively. Then the closure $\overline{V}$ of $V$ in $Y$ is compact; furthermore, $\overline{V}$ is disjoint from $C$, so that $\overline{V} \subset U$, as desired. (mostly quoted from Munkres').
How did they get that disjointness (bolded)?
AI: Thanks to Daniel Fischer's comment. If $U$, $V$ are two disjoint open sets, then $\overline{U} \cap V = \varnothing$. Proof. If $x \in \overline{U} \cap V$, then every open set around $x$ contains points of $U$ and $V$ contradicting the openness and disjointness of the two sets. |
H: Conditional Probability using Venn Diagrams
In class we have been told (for now) to use Venn diagrams to solve probability questions, we were also set this question for homework. Instead of answers, a tip, or a nudge in the right direction would be help me the most.
Two events $A$ and $B$ are such that $P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{2}$. Calculate $P(A'\cap B)$ when:
i) $P(A \cap B) = \frac{1}{8}$.
ii) $A$ and $B$ are mutually exclusive
iii) $A$ is a subset of $B$
Here's my working for part i)
$P(A'\cap B) = P(B) - P(A\cap B)$
$\therefore P(A'\cap B) = \frac{3}{8}$
Part 2, and three are more elusive, though. I know mutually exclsuive means they can't both happen at the same time. Does this mean it's $\frac{2}{3}$, or $\frac{1}{2}$?
AI: It is the probability of the intersection of $A'$ and $B$, so it cannot be bigger than $P(B)$
In fact for (ii) it is $\frac12$
For (iii): $A \subset B \implies A \cap B = A$ |
H: Evaluation of the integral.
$$I\left(n,\epsilon\right)=\int_{-{\rm i}\infty}^{+{\rm i}\infty} \frac{{\rm e}^{\epsilon z}}{\left(z+\epsilon\right)^n}\,{\rm d}z$$
The integration is taken along the imaginary axis,
an integer $n\gt 0$ and
$\epsilon \in \left[-1, +1\right]$
How to evaluate $I\left(n, \epsilon\right)$?
I saw this question in a textbook, when I am studying. This question seems so interesting to me. Thus, I am asking here. I never have seen such question up to today. And I have No idea to solve this. Only I guess, I need to use residue theorem. Therefore, Please can someone explain and solve this clearly? I am willing to learn:) thank you so much:)
AI: Let's take the case $\epsilon=-1$. Then consider the following contour integral:
$$\oint_C dz \frac{e^{-z}}{(z-1)^n}$$
where $C$ is the semicircle of radius $R$ oriented positively, with center at the origin and diameter along the imaginary axis, closing to the right (i.e., enclosing the positive real axis). Then we may split this contour integral into two pieces:
$$-\int_{-i R}^{i R} dz \frac{e^{-z}}{(z-1)^n} + i R \int_{-\pi/2}^{\pi/2} d\theta \, e^{i \theta} \frac{e^{-R \cos{\theta}} e^{-i R \sin{\theta}}}{(R e^{i \theta}-1)^n}$$
Now, we may show that the magnitude of the second integral vanishes as $R \to \infty$ when $n \gt 0$; this magnitude is bounded by:
$$2 R \int_0^{\pi/2} d\theta \, \frac{e^{-R \cos{\theta}}}{R^n} = \frac{2}{R^{n-1}} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2}{R^{n-1}} \int_0^{\pi/2} d\theta \,e^{-2 R \theta/\pi} \le \frac{\pi}{R^n}$$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=1$, which is
$$\frac{1}{(n-1)!} \left [\frac{d^{n-1}}{dz^{n-1}} e^{-z} \right ]_{z=1} = \frac{(-1)^{n-1}}{(n-1)!} \frac{1}{e}$$
Therefore, we can say that
$$\int_{-i \infty}^{i \infty} dz \frac{e^{-z}}{(z-1)^n} = i 2 \pi \frac{(-1)^{n}}{(n-1)!} \frac{1}{e}$$
Similarly, for $\epsilon = +1$, you can find using the above method (and being careful to orient the contour positively) that
$$\int_{-i \infty}^{i \infty} dz \frac{e^{z}}{(z+1)^n} = i 2 \pi \frac{1}{(n-1)!} \frac{1}{e}$$ |
H: $1_{\limsup A_n} = \limsup 1_{A_n} $
Do you have some hints on how to prove the following relation? I think it should be quite straightforward, but I cannot see it.
$$
1_{\limsup A_n} = \limsup 1_{A_n} $$
AI: Each side is a way of writing the function
$$ x\mapsto\begin{cases}1 & \text{if }x\in A_n\text{ for infinitely many }n \\ 0 & \text{otherwise} \end{cases} $$
This should be easy to prove by unfolding the definitions of $1_A$ and $\limsup$ (for sequences for sets and numbers, respectively). |
H: A difficulty solving a limit
Could anyone help me with this limit?
$$\lim_{x \to \infty} (\log_2^2 x - \sqrt{x})$$
All my usual method fail with this one, Wolfram Alpha looks at me funny, and I know the answers is $\ - \infty$, but have not idea whatsoever as to how to come to this conclusion.
AI: Hint:
$$
\log_2^2 x -\sqrt{x}=\sqrt{x}\left(\frac{\log_2^2 x}{\sqrt{x}}-1\right)
$$ |
H: Using change of variables, solve the integral and show the domain obtained by the change.
I need to solve the following integral using change of variables:
$$\int\int_D\frac{\sqrt[3]{y-x}}{1+x+y}dA$$
where D is the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$.
I tried to change the variables to $u=y-x$ and $v=1+x+y$, but then I couldn't solve the integral.
Any tips on how to solve it (or the full solution) would be highly appreciated!
Thanks in advance!
EDIT:
I should've been more detailed in what I did. I tried the change $u=y-x$ and $v=1+x+y$, found the new domain and calculated the Jacobian. The integral then is:
$$\int_1^2\int_{1-v}^{v-1}\frac{\sqrt[3]{u}}{v}\frac{1}{2}dudv$$
That is what I can't solve.
I can solve it in u, but I don't even know if it's right:
$$\frac{3}{8}\int_1^2\frac{{(v-1)}^{4/3}-{(1-v)}^{4/3}}{v}dv$$
AI: Your triangle is bounded above by the line $y=1-x,$ meaning $y\le 1-x,$ which we can rewrite as $x+y\le1.$ This suggests using $v=x+y$ as your other variable, rather than your choice of $v$.
Since we put $$u=y-x\\v=x+y,$$ then we have $$y=\frac{u+v}2\\x=\frac{v-u}2.$$ It can be shown, then, that the Jacobian is $-\frac12,$ so that $$\iint_D\frac{\sqrt[3]{y-x}}{1+x+y}\,dA=-\frac12\iint_D\frac{\sqrt[3]{u}}{1+v}\,du\,dv.$$ It remains only to find the new limits of integration that describe $D$.
Since $x$ is bounded below by $0,$ then $$0\le\frac{v-u}2\\0\le v-u\\u\le v$$ Since $y$ is bounded below by $0,$ then $$0\le\frac{u+v}2\\0\le u+v\\-v\le u.$$ Hence, we have $-v\le u\le v$ and $v\le 1$ (by previous discussion). Also, we have $0\le v$ (why?), and so the integral becomes $$-\frac12\int_0^1\int_{-v}^v\frac{\sqrt[3]{u}}{1+v}\,du\,dv=-\frac12\int_0^1\frac1{1+v}\left(\int_{-v}^v\sqrt[3]{u}\,du\right)\,dv.$$
Can you take it from there? |
H: Solving for x in a equation involving natural logarithms
How would I solve for x in this equation here:
$$\ln(x)+\ln(1/x+1)=3$$
I realize that the answer is $e^3-1$, but I am not sure as to how to get it.
Any input is appreciated.
AI: $$ \ln x + \ln \left( \frac{1}{x} + 1 \right) = \ln \left( x(1/x + 1) \right) = 3 $$
$$ \Rightarrow \ln(x+1)= 3 \Rightarrow x + 1 = e^3 \Rightarrow x = e^3 - 1 $$ |
H: limit 0 times infinity, rewrite to find the limit
I need some help with: $\lim_{x\to 0+} x^3\cdot e^{1/x}$. How to start?
I've tried substitution $(y=1/x)$ without any luck.
I would prefer not to use L'Hopitals rule and apologizes for a bad title line.
AI: You can carry on with your substitution. In the case $y = 1/x$, then as $x \to 0^+$, $y \to \infty$, and you want to look at the limit of
$$
\lim_{y\to\infty} (1/y^3) e^y = \lim_{y\to\infty} \frac{e^y}{y^3}.
$$
If you know, for instance, that the exponential grows faster than any polynomial, you can avoid L'Hopital's rule. |
H: Adding Logarithms
Studying for my midterm.
Solve the following algebraically:
$$\log_2x+\log_2(x+4)=5$$
So I know that $\log_b(mn)=\log_b(n)+\log_b(m)$ therefore:
$$5=\log_2(x(x+4))$$
$$\text{or}$$
$$5=\log_2(x^2+4x)$$
$$5=2^{x^2+4x}$$
Now I don't know how to solve for x at this point. I'm stuck. Please let me know if I've done anything wrong in my calculations. The answer key says that $x=4$.
Thanks.
AI: You rewrote the expression incorrectly. It should be $x^2 + 4x = 2^5$. |
H: Monic polynomials with integer coefficients
We have $\Pi_{j=1}^n (z-z_j)$ a polynomial with integer coefficients. Is also $\Pi_{j=1}^n (z-z_j^k)$ for k=1,2,3,... a polynomial with integer coefficients?
AI: Yes.
The coefficients of the latter are symmetric polynomials in the $z_j$, hence are (integer!) polynomials in the elementary symmetric polynomials in the $z_j$, that is in the coefficients of the former polynomial. |
H: Characterization of $\mathrm{Im} f=\ker f$
Let $E$ a finite dimensional space over a field $\mathbb F$ and $f\in\mathcal{L}(E)$. The question is to prove the equivalence of:
$\mathrm{Im} f=\ker f$
$f^2=0$ and there's $h \in\mathcal{L}(E)$ such that $hf+fh=\mathrm{id}$
I proved that $2.\Rightarrow 1.$ and $1.\Rightarrow f^2=0$ but I didn't know how complete the proof. Any help will be appreciated.
AI: Since $E$ is finite-dimensional we might pick a basis $v_1,\ldots ,v_m$ of $\ker f$ and, as $\ker f\subseteq \operatorname{im} f$, pick $w_1,\ldots,w_m$ with $f(w_i)=v_i$. Then $v_1,\ldots ,v_m,w_1,\ldots,w_m$ are linearly independant: Assume $$a_1v_1+\ldots +a_mv_m+b_1w_1+\ldots+b_mw_m=0. $$
Then after applying $f$ we obtain $b_1v_1+\ldots+b_mv_m=0$, hence $b_1=\ldots=b_m=0$ and then from $a_1v_1+\ldots +a_mv_m=0$ also $a_1=\ldots =a_m=0$. From $\dim\ker f+\dim\operatorname{im}f=\dim E$, we knwo that $\dim E=2m$, so this linearly independant system is a basis of $E$ (this is the only point where $\dim E<\infty$ is needed!).
Now define $h$ by letting $h(v_i)=w_i$ and $h(w_i)=0$. Then $hf(v_i)+fh(v_i)=0+v_i$ and $hf(w_i)+fh(w_i)=w_i+0$ and hence $hf+fh=\operatorname{id}_E$. |
H: Showing $\lor$ in terms of $\to$ and $\lnot$
I was looking at a question a user posted yesterday (link below). And one of the answers mentioned one could express $∨$ in terms of $\to$ and $\lnot$.
In terms of $\land$ he stated it can be expressed like such:
\begin{align*}
p \land q &\equiv \neg\neg p \land \neg\neg q \\
&\equiv \neg(\neg (p) \lor (\neg q)) \\
&\equiv \neg(p \to \neg q) \\
\end{align*}
I've been thinking about how something like that could be done for a while now and its starting to bug me. Out of curiosity was that the right procedure and if so how could it be shown for $\lor$.
link: Proving that $\{\to, \lnot\}$ is logically complete
AI: $p\lor q=\lnot(\lnot p\land\lnot q)=\dots$ |
H: Proving that $n^2-l$, $n^2$ and $n^2+l$ can't all be perfect squares
I tried to write a proof and used the argument that if $n^2$ is a perfect square, $n^2-l$ and $n^2+l$ can't both be perfect squares. However, I can't find a proof for this statement. Can you help me with this?
What I have tried:
Suppose that $n^2-l$, $n^2$ and $n^2+l$ are all perfect squares. Then this must hold
$$n^2-l = \sum_{i=0}^{m-a}(2i+1)$$
$$n^2 = \sum_{i=0}^{m}(2i+1)$$
$$n^2+l = \sum_{i=0}^{m+b}(2i+1)$$
From first two I can obtain that
$$l=\sum_{i=m-a+1}^{m}(2i+1)$$
And from last two:
$$l=\sum_{i=m+1}^{m+b}(2i+1)$$
While it seems that these both can't hold, I am not able to show an obvious contradiction there.
I also got a suggestion to start with this:
$$n^2+l=x^2$$
$$n^2-l=y^2$$
I tried to count them together but $2n^2=x^2+y^2$ also doesn't seem a straightforward contradiction to me.
AI: What you're trying to prove is false. A counterexample is given by $n=5$ and $l = 24$. |
H: Number of ways to fill a grid with balls
Find the number of ways to fill a $3\times 3$ grid (with corners defined as $a,b,c,d$) if you have 3 black and 6 white marbles.
Note: This question was asked in an e-litmus exam and is not an assignment question.
AI: area of the grid =$9$
No. of ways of arranging 3 black marbles or 6 white marbles = $9 \choose 3$ = $84$. |
H: Find number no of ways to fill a grid with balls[Part 2]
Find the number of ways to fill a 3*3 grid (with corners indistinguishable) if you have 3 black and 6 white marbles.
Approach till now[Incorrect]:
No. of ways of arranging 3 black marbles or 6 white marbles = $(^9C_3)$ = 84.
Since there are 4 different ways to view the grid hence the answer is $84\over4$=$21$
AI: In the marked-corner case, this amounts to choosing spots for the three black balls, so there are $\binom93=84$ ways to do it.
Now, the answer to the unmarked-corner case depends on what is allowed.
Case 1: Let's suppose that if one arrangement is a rotation of another, then they are considered to be the same (my initial interpretation). Note that only $4$ of the $84$ arrangements have any rotational symmetry--in particular, those in which one black ball is in the middle slot, and the other two black balls are on opposite sides or corners. Each of those $4$ can be viewed in only two "different-looking" ways without markings, while the other $80$ can all be viewed in $4$ "different-looking" ways without markings. Consequently, if we "unmark" the corners, then there are $$\frac{80}4+\frac{4}{2}=22$$ ways to arrange the marbles, and not $21$.
Case 2: Let's suppose that reflections are also permitted, as well as rotations. The $4$ arrangements mentioned above still each have only two "different-looking" ways that they can be viewed. Let's say that there are $m$ arrangements with neither reflection symmetry nor rotational symmetry. Since the $4$ mentioned above have both kinds of symmetry, then there are $80-m$ arrangements with reflection symmetry, but no rotational symmetry. Those with neither type of symmetry (for example, with black balls at top-left, center, and bottom) can be viewed in $8$ "different-looking" ways, and those with reflection symmetry only (for example, all three black balls in the top row) can be viewed in $4$ "different-looking" ways. Thus, there are $$\frac42+\frac{80-m}4+\frac{m}8=\frac{176-m}8$$ ways to arrange the marbles in this case, so it remains to determine the number $m$.
Note that if no corner has a black ball, then the arrangement necessarily has at least one kind of symmetry. Thus, we need only consider those arrangements (on the marked grid) with a black ball in at least one corner, having no symmetry. Further note that if all three black balls are in corners, then we have reflection symmetry, so we need not consider arrangements with black balls in more than two corners. Let us consider the distinct possibilities:
There are black balls in two opposite corners, such as $a$ and $c$ in the original problem ($2$ ways to do this). To avoid any kind of symmetry, we must place the final black ball in one of the edge slots ($4$ ways to do this). Hence, there are $8$ non-symmetric arrangements with black balls in two opposite corners.
There are black balls in two non-opposite corners ($4$ ways to do this). To avoid any kind of symmetry, we must place the final black ball in one of the edge slots adjacent to one of the previously-placed black balls, but not adjacent to the other ($2$ ways to do this). Hence, there are $8$ non-symmetric arrangements with black balls in two non-opposite corners.
There is a black ball in exactly one corner ($4$ ways to do this), and no black ball in either adjacent edge slot. Note that if both remaining black balls are in edge slots, then we have reflection symmetry along a diagonal, so one of the remaining black balls is in the center, and one must be in one of the other two edge slots. Hence, there are $8$ non-symmetric arrangements with exactly one black ball in a corner, and no black ball in either adjacent edge slot.
There is a black ball in exactly one corner ($4$ ways to do this), and a black ball in at least one of the two adjacent edge slots. To avoid reflection symmetry along a diagonal, we can't have black balls in both adjacent edge slots, so there are $2$ ways to place the adjacent black ball. There are then three available slots in which to place the final black ball, and none of them yield any symmetry. Hence, there are $24$ non-symmetric arrangements with exactly one black ball in a corner, and a black ball in at least one of the two adjacent edge slots.
There are no other non-symmetric arrangements, and the above types are distinct, so $m=8+8+8+24=48,$ and so there are $$\frac{176-m}8=\frac{128}8=16$$ ways to arrange the marbles in the unmarked grid, if both rotations and reflections are allowed. |
H: Speed of object towards a point not in the object's trajectory?
Trying to study for my mid-term, but I'm having slight difficulties understanding what I'm supposed to do in this one problem:
A batter starts running towards first base at a constant speed of 6 m/s. The distance between each adjacent plate is 27.5 m. After running for 20 m, how fast is he approaching second base? At the same moment, how fast is he running away from third base? (see image below)
This is what I have so far:
Let $d$ be the distance the batter has run thus far
The distance between the batter and first base is 7.5 m
The distance between the batter and second base is $\sqrt {27.5^2 + (27.5-d)^2}\ $, or approx. 28.5044 m when $d = 20$
The distance between the batter and third base is $\sqrt {27.5^2 + d^2}\ $, or approx. 34.0037 m when $d = 20$
No need to hand feed me the answer, I'd just like a bit of insight on how to solve the problem.
AI: The first question is easy: the batter is running straight towards the first base, so he is approaching the first base with a speed of $6$ m/s. To answer the second question, try to find the function $f(t)$ of time that gives the distance to the third base. Then find the derivative at the point $t=\frac{20}{6}$, when the batter has been running for 20 metres. |
H: Interesting trigonometric equation
Find all $x\in \mathbb{R}$ such that
$\arccos x = 2\arctan \sqrt{\frac{1-x}{1+x}}$
Now, here is my approach, please state anything that is not correct/fully motivated.
$\cos 2\theta =2\cos ^2\theta -1$
$\cos ^2\theta =\frac{\cos ^2\theta}{1}=\frac{\cos ^2\theta}{\cos ^2\theta+\sin^2\theta}=\frac{1}{1+\tan ^2\theta }$
$\Rightarrow \cos 2\theta = \frac{2}{1+\tan ^2\theta}-1$
$\cos \arccos x = x = \cos(2\arctan \sqrt{\frac{1-x}{1+x}} ) = \frac{2}{1+\frac{1-x}{1+x}}-1=1+x-1=x. $
Hence, we have shown that $\cos LHS=\cos RHS$ and thus the solutions are given by the intersections of the domains of the functions.
$f(x)=\arccos x \ \ \ \ \ D_{f} = \left \{ x\in \mathbb{R}: -1\leq x\leq 1 \right \}$
$g(x)=2\arctan \sqrt{\frac{1-x}{1+x}} \ \ \ \ \ D_{f} = \left \{ x\in \mathbb{R}: x\neq 1 \wedge x\leq 1 \wedge x> -1 \right \}$
Solutions are thus given by:
$D_{f}\cap D_{g}=\left \{ -1< x< 1 \right \}$
Follow up question: Is it enough to show that $\cos LHS = \cos RHS$ and proceed as I've done?
AI: Hint
By calculating the derivative
$$\left(2\arctan \sqrt{\frac{1-x}{1+x}}\right)'=-\frac{1}{\sqrt{1-x^2}}=(\arccos x)'$$
and verifying that the two expressions have the same value at $0$ (for example) we conclude that the equality hold for all $x\in(-1,1]$ |
H: Continuity of a Function with complex analysis
Problem:
Let $f$ be defined $$f(z)=\frac{{{\rm Re}(z^{2})}^2}{\left \| z^2 \right \|}$$ if $z\neq 0$ and $f(0)=0$.
Prove that $f$ is continuous at $0$.
Does any one have any idea on how to solve it?
AI: Recall that $|\Re z|\leqslant |z|$ so...? |
H: Rate of Change of Cylindrical Roll's Volume as it Unrolls
This is purely a "for-fun" question. I was minding my own business in the washroom this morning when I began to unroll some toilet paper from the roll, and in typical Breaking Bad fashion (sorry if you don't understand the reference) I had a serious moment where I wondered how the rate of change of the volume of the roll of paper changes as it is unrolled. Obviously the circumference of the cylinder keeps decreasing and as a result the ROC changes as well. I thought about it a bit but got stumped when I started writing things out (my math skills are a bit rusty).
I believe the beginning part of this solution Accounting for changing radius of a paper roll to always unroll the same amount of paper is the correct starting point for this question.
For simplicity, assume that the paper is rolled onto itself - no cardboard piece in the centre and the innermost layer's circumference approaches 0.
Also, assume that the roll is being unravelled at 1 circumference per second ($2\Pi r$/s) of the outermost (initial) layer.
Lastly, assume that there are 100 layers (100 circles as defined in the mentioned question) and each is 1 unit-of-your-liking.
Feel free to adjust any of these parameters to your liking!
AI: If the roll is rotating at a constant angular rate, you will remove one layer per second all the way down. As you started with $100$ layers, it will take $100$ seconds to unroll. The remaining volume is proportional to the square of the remaining time. If the starting volume is $V$, at time $t$ (seconds) there is $V\frac {(100-t)^2}{100^2}$ of the initial volume remaining.
Added: if you want the paper removed at a constant rate, the angular velocity goes up as $\frac 1r$. This is because the linear velocity is constant and $v=r\omega$. Then we have $\frac {dr}{dt}=-k\frac 1r, \frac 12r^2=C-kt, r=\sqrt{2(C-kt)}$. With $r(0)=r_0, r(100)=0$ we have $C-100k=0, C=\frac 12r_0^2,\\ r=r_0\sqrt{1-\frac t{100}}$ |
H: Probability question: given $P(A|B)$ and $P(B)$ how do I find $P(A)$?
I have a probability distribution for some quantity $A$ given a fixed $B$, i.e. $P(A|B)$. I also have a prior distribution $P(B)$ for $B$. I'm trying to find the distribution $P(A)$.
I had thought about using Bayes' theorem which implies that:
$P(A) = \frac{P(B)P(A|B)}{P(B|A)}$
But I don't have any information about $P(B|A)$, or at least I don't think I do...
I suspect I'm missing something obvious here, can anyone help out?
Thanks!
EDIT
Based on the answers given I fear I might have tried to frame the question in too general a manner. The specific forms I'm working with are as follows:
$P(A|B) = A\exp{\left[-\frac{A^2 + B^2}{2}\right]} I_0\left(AB\right)$ (a slightly modified Rice distribution)
$P(B) \propto B^{-n}$ (generic power-law with $n>0$)
Does this change the situation at all? thanks!
AI: I understand from your question that you have the distributions $P(B)$ and $P(A|B)$, which I assume are functions.
Then the joint distribution is $P(A, B)=P(A|B)P(B)$. Now to find the distribution $P(A)$ just sum $P(A, B)$ over $B$. Or if continuous case use integration over density functions. Refer http://en.wikipedia.org/wiki/Marginal_distribution for more information.
$$P(A=a)=\sum_{b}P(a,b)=\sum_{b}P(A=a|B=b)P(B=b).$$
I guess this is what you want.
Edit (to match the edit of the question): You have continuous case. So use integration as below. I guess $[0 +\infty)$ is sufficient as Rice is in non negative domain.
$$P(A=a)=\int_{-\infty}^{+\infty}P(a,b)db$$ |
H: What am I doing wrong when trying to find a determinant of this 4x4
I have to find the determinant of this 4x4 matrix:
$
\begin{bmatrix}
5 & -7 & 2 & 2 \\
0 & 3 & 0 & -4 \\
-5 & -8 & 0 & 3 \\
0 & -5 & 0 & -6 \\
\end{bmatrix}
$
Here is my working which seems wrong according to the solutions. What am i doing wrong?:
And here is the solution:
AI: What you are doing wrong is precisely what the solution said you were doing wrong. The $2$ was alright, since that's the same as $2\cdot(-1)^{1+3},$ but the $-5$ was not, since $$-5\cdot(-1)^{2+1}=-5\cdot-1=5.$$ Keep in mind that we have an alternating sign factor as we move along a row/column, and that the starting sign depends on the row/column that we're in. |
H: Example of a commutative ring with identity with two ideals whose product is not equal to their intersection
I need a specific example of a commutative ring with identity, and two ideals in the ring whose product is not equal to their intersection.
I know that for two such ideals I and J, IJ = I ∩ J if I + J = R. I have seen some results about what conditions guarantee a ring will have ideals whose intersections are not equal to their products, but I didn't really understand the terminology of those results well enough to come up with a specific example.
AI: Hint: Consider the ring $R=\mathbb{Z}$ and the ideal $I=2\mathbb{Z}$. |
H: Show uniform convergence (Gamma function)
In order to exchange limit and integral of
$\displaystyle\lim_{n\to\infty}\int_{0}^{n}x^{s-1}\left(1-\frac{x}{n}\right)^n dx$,
I need to show uniform convergence of $f_n(x)=x^{s-1}(1-\frac{x}{n})^n$. Am i right?
It converges pointwise to $f(x)=x^{s-1}e^{-x}$, so $|f(x)-f_n(x)|$ should be smaller than $\epsilon$ for $\epsilon>0$
I know the definition of uniform convergence, but i don't know how to show it!
Any hints are welcome. Thank you.
AI: I need to show uniform convergence of $f_n(x)=x^{s-1}(1-\frac{x}{n})^n$. Am i right?
No. Using the uniform convergence to exchange a limit and an integral works well on a fixed, bounded interval. However, here, the bound of the integral is increasing, so you are actually working on the whole positive half-line. To see this, we can rewrite the limit as:
$$\lim_{n \to +\infty} \int_0^{+\infty} x^{s-1}\left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x) dx,$$
where $1_{[0, n]}$ is the function whose value is $1$ on $[0, n]$ and $0$ elsewhere. On an unbounded interval, some mass can escape at infinity even if functions get closer and closer uniformly, so the uniform convergence is not enough.
If you know them, my advice would be to use the monotone convergence theorem or the dominated convergence theorem (the later is more convenient here). You will then have to prove that, for all integer $n \ge 1$ and all real $x \ge 0$:
$$0
\le \left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x)
\le e^{-x}. \ \ \ \ (1)$$
[if you choose to use the monotone convergence theorem, you also need to prove that the sequence of functions we are working with is non-decreasing. It is true, but not very fun to show. The inequalities above are easier.]
If you can't use either of these theorems, you can succeed by a clever use of $\varepsilon$'s, especially is you manage to prove the uniform convergence on any subset $[0, M]$ with $M > 0$. But that is a bit cumbersome.
~~~~~
EDIT: So, if you want to use the last method, here are some additional hints. Let $M > 0$. We can assume without loss of generality that $n \ge M$. We will divide the integral in two parts:
$$\int_0^n x^{s-1}\left(1-\frac{x}{n}\right)^n dx = \int_0^M x^{s-1}\left(1-\frac{x}{n}\right)^n dx + \int_M^{+ \infty} x^{s-1} \left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x) dx.$$
We have two pieces. We can control the first piece if we know that the sequence of functions converges uniformly on all compact sets:
$$\lim_{n \to +\infty} \int_0^M x^{s-1}\left(1-\frac{x}{n}\right)^n dx = \int_0^M x^{s-1} e^{-x} dx.$$
We have also a upper bound on the second piece (provided we have proved $(1)$):
$$0 \le \int_M^{+ \infty} x^{s-1} \left(1-\frac{x}{n}\right)^n 1_{[0, n]} (x) dx \le \int_M^{+ \infty} x^{s-1} e^{-x} dx.$$
But the function $x \mapsto x^{s-1} e^{-x}$ is integrable, so that:
$$\lim_{M \to +\infty} \int_M^{+ \infty} x^{s-1} e^{-x} dx = 0.$$
~~~~~
Let's wrap it up. Choose any $\varepsilon > 0$. Then, you can find a $M>0$ such that:
$$\int_M^{+ \infty} x^{s-1} e^{-x} dx < \varepsilon.$$
In addition, if $n$ is large enough, then:
$$\left| \int_0^M x^{s-1}\left(1-\frac{x}{n}\right)^n dx - \int_0^M x^{s-1} e^{-x} dx \right| < \varepsilon,$$
whence:
$$\left| \int_0^n x^{s-1}\left(1-\frac{x}{n}\right)^n dx - \int_0^{+ \infty} x^{s-1} e^{-x} dx \right| < 2\varepsilon.$$ |
H: Elementary number theory, sums of two squares
Suppose x can be written as a sum of two squares of integers, and y can be written as a sum of two squares of integers. Show that xy can also be written as a sum of two squares of integers.
AI: $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2 +(ad+bc)^2.$$
Remark: let $u=a+ib$ and $v=c+id$ be complex numbers, with $a,b,c,d$ real. Then $uv=(a+ib)(c+id)=(ac-bd)+i(ad+bc)$. The identity of the answer just says that the norm of the product $uv$ is equal to the product of the norms. Everything is connected to everything else! |
H: Is $0^\infty$ indeterminate?
Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is $0^\infty$ indeterminate? Or is it only 1 raised to the infinity that is?
AI: No, it is zero.
Consider the function $f(x,y) = x^y$ and consider any sequences $\{(x_0, y_0), (x_1, y_1), \ldots\}$ with $x_i \to 0$ and $y_i \to \infty$. It is easy to see that $f(x_n,y_n)$ converges to zero: let $\epsilon > 0$. For some $N$, $|x_i| < \epsilon$ and $y_i > 1$ for all $i \geq N$, so $|f(x_i,y_i)| < \epsilon$ for all $i\geq N$.
More generally, as $x\to c$ and $y\to \infty$, $x^y$ converges to 0 for $|c|<1$, diverges to infinity for $c>1$, oscillates without converging for $c \leq -1$, and is indeterminate when $c=1$. |
H: Derive the equation of the locus of all points (exact question in description)
Derive the equation of the locus of all points such that the line joining a point in the locus to the point (6,2) and the line joining the same point to the point (2,6) are perpendicular.
I need help with this proof for my calculus class
I have a test coming up an I'm having trouble with proofs like this. This is on my study guide. I was able to complete a problem where u had to derive an equation whose points are equidistant from 2 points. I just want to learn how to do itn no need to give me answers
AI: Hint:
Let us call $A=(6,2)$ and $B=(2,6)$. If $P$ is a point satisfying the given condition, then
$APB$ is a rectangular triangle.
Consider the middle point between $A$ and $B$, $C=\frac{A+B}{2}$.
Draw a picture.
Edit: (Notation: $||X-Y||$: distance between $X$ and $Y$)
We have,
$$
4||P-C||^2=(P-A+P-B)\cdot(P-A+P-B)=||P-A||^2+||P-B||^2=||A-B||^2
$$
Then we obtain a circunsference with centre $C$ and ratio $\frac{||A-B||}{2}$.
$$
(x-4)^2+(y-4)^2=16
$$ |
H: Equivalence class help
I have a question that goes as follows:
Let d be a positive integer. Define the relation Rho on the integers Z as follows: for all m,n element of the integers.
m rho n if and only if d|(m-n)
Prove that rho is an equivalence relation. Then list its equivalence classes.
Now the first d that comes to mind is 1, so I proved it was an equivalence relation as follows:
Reflexive: m rho m <=> d|m-m
Symmetric: m rho n <=> d|m-n => d|n-m with n-m = -(m-n) <=> n rho m
Transitive: k is an element of integers:
m rho n and n rho k => d|m-n & d|n-k => d|(m-n) + (n-k) => d|m-k => m rho k
I am unsure if this is a sufficient proof or if my logic holds. However I can't think of any equivalence classes for this as d can vary. If d was 7 for example I would think equivalence classes would be 1 = {...,-13,-6,1,8,15,...}, 2 = {...,-12,-5,2,9,15,...} etc...
Does this relation have equivalence classes and I am missing something or?
AI: There is a different relation for each $d$. What is being asked is "for all $d$, is the corresponding relation an equivalence relation?"
As for your proof, it is correct, but you may want to be clearer with some of the steps, depending on how familiar the intended audience is with divisibility. |
H: Finding the permutation that shows two permutations are conjugates method?
Problem:
Given $\sigma=(12)(34)$ and $\gamma=(56)(13)$ find $\tau\in S_6$ with
$\tau^{-1}\sigma\tau=\gamma$
Attempt: I'm kind of new to this but from what I understanding find $\tau$ that satisfies this will show that $\sigma$~$\gamma$ right? This means that they are conjugates of each other. I started off by writing out what the permutations are in $S_6$ but I was not seeing anything. I also rewrote what we are trying to prove as $\sigma\tau$=$\tau\gamma$ by left multiplying by $\tau$.
Question: My main question is whether or not there's a method to solving these types of problems and if there is how can it be applied to this one? Any step in the right direction is appreciated, thank you.
AI: Hint: for any cycle $(i_1\,i_2\,\ldots\,i_k)$, we have
$$
\sigma (i_1\,i_2\,\ldots\,i_k)\sigma^{-1}=(\sigma(i_1)\,\sigma(i_2)\,\ldots\,\sigma(i_k))
$$
for all $\sigma\in S_n$. |
H: Closed under intersections
I read this definition: "A collection C of subsets of E is said to be closed under intersections if A ∩ B belongs to C whenever A and B belong to C."
How could the intersection of ANY A and B belonging to C ever NOT belong to C?? Whats the point of this definition?
AI: Let $E=\{1, 2, 3\}$ and suppose the collection $C$ of subsets was $C=\{\{1, 2\}, \{2, 3\}\}$. Then this collection of two subsets is not closed under intersection, since $\{1, 2\}\cap\{2, 3\}=\{2\}$, which is not in $C$. |
H: Proving supremum for non-empty, bounded subsets of Q iff supremum in R is rational
Let E be a nonempty bounded subset of ℚ. Prove that E has a supremum in ℚ if and only if its supremum in ℝ is rational and that in this case, the two are equal.
This seems intuitive enough, and I know that not all nonempty, bounded subsets of R have a supremum in Q, but how do I prove this?
I can proceed by proving that if E's sup in R is irrational, E does not have sup in Q - but this I have only though one example, not in a general case. How do I say this generally?
I also have to prove the reverse, which is that if sup in R is rational, then E has supremum in Q. Again, I get this intuitively, but how do I prove it?
Help appreciated!
AI: I find it easier to prove directly.
Suppose $E$ has a supremum in $\mathbb{Q}$. Let $\alpha = \sup_\mathbb{Q} E$. Then $\alpha \in \mathbb{Q}$, $\alpha$ is an upper bound for $E$, and if $\beta \in \mathbb{Q}$ is an upper bound for $E$, then $\alpha \le \beta$.
Now suppose $\gamma \in \mathbb{R}$ is an upper bound for $E$, and let $q_n \in \mathbb{Q}$ be such that $q_n \downarrow \gamma$. Since $\gamma$ is an upper bound, and $q_n \ge \gamma$, $q_n$ is an upper bound for $E$, and hence $q_n \ge \alpha$. Hence we have $\gamma \ge \alpha$. Hence $\alpha$ is the supremum of $E$ in $\mathbb{R}$.
Now suppose $E$ has a supremum in $\mathbb{R}$, and the supremum is rational. That is, $\alpha = \sup_\mathbb{R} E \in \mathbb{Q}$. Then $\alpha$ is an upper bound for $E$. Suppose $\beta \in \mathbb{Q}$ is an upper bound for $E$, then since $\beta \in \mathbb{R}$, we must have $\alpha \le \beta$. Since $\alpha \in \mathbb{Q}$, it follows that $\alpha = \sup_\mathbb{Q} E$. |
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