text stringlengths 83 79.5k |
|---|
H: A bagel shop has plain muffins, cherry muffins, chocolate muffins, almond muffins, apple muffins, and broccoli muffins.
How many ways are there to choose:
1.) two dozen muffins with at least two of each kind?
2.) two dozen muffins with at least five chocolate muffins and at least three almond muffins?
3.) two dozen muffins with no more than two broccoli muffins?
4.) two dozen muffins with at least one plain muffin, at least two cherry muffins, at least three chocolate muffins, at least one almond muffin, at least two apple muffins and no more than three broccoli muffins?
Answers???
1.) $\left(\!\!{6\choose 12}\!\!\right)$ Since when order doesn't matter there is only one way to choose 12 muffins (2 for each flavor), there are 12 left to choose from out of the 6 flavors.
2.) $\left(\!\!{6\choose 16}\!\!\right)$ Same reasoning as #1.
3.) $\left(\!\!{5\choose 22}\!\!\right)$ We are choosing 2 broccoli and thus eliminating 1 flavor and then have only 22 choices left.
4.) $\left(\!\!{5\choose 12}\!\!\right)$ Same reasoning as the above but combined. There is only 1 way to choose at least one plain muffin, at least two cherry muffins, at least three chocolate muffins, at least one almond muffin, at least two apple muffins. So we have 15 choices left, but then we are choosing at most 3 broccoli and so we have 5 flavors left with 12 choices.
Do these sound reasonable?
AI: Your first two are fine. For the third one you’ve solved the wrong problem: you’ve calculated the number of ways of choosing two dozen muffins including exactly two broccoli muffins, when the problem asked for the number of ways of choosing two dozen muffins with at most two broccoli muffins. This simply adds another two terms to your answer:
$$\left(\!\!{5\choose 22}\!\!\right)+\left(\!\!{5\choose 23}\!\!\right)+\left(\!\!{5\choose 24}\!\!\right)$$
You’ve made a similar error in the fourth problem: you’ve counted the number of ways that include exactly $3$ broccoli muffins. You also need the ways that include $2,1$, or $0$ broccoli muffins.
(Broccoli muffins?) |
H: characterization of uniform ellipticity
Let $B$ be a $n\times n$ matrix over $\mathbb{R}$ and define $A:=BB^*$. I read in a paper that the following two statements are equivalent:
(1) the matrix $A$ is uniformly elliptic; i.e. for all vectors $y\in\mathbb{R}^n$ and some constant $N>0$, it holds that $$y^*Ay \geq \frac{1}{N^2} |y|^2.$$
(2) for the same constant $N$, the Frobenius norm of the inverse of $B$ is bounded above: $$\text{tr}(B^{-1}(B^{-1})^*) \leq N^2.$$
Can anyone see why this is true?
AI: I'll help you with one direction; the other implication is done likewise.
Let's take $\det B\ne 0$ (otherwise the $A$ is not elliptic and $B^{-1}$ is not defined). $\|\cdot\|$ stands for Euclidean norm and the induced matrix norm.
For direct implication we take $N^2=n\|B^{-1}\|^2$.
$$\frac{y^\ast Ay}{\|y\|^2}\ge \inf_{y\ne 0}\frac{\|B^\ast y\|^2}{\|y\|^2}=\inf_{z\ne 0}\frac{\|z\|^2}{\| (B^{-1})^\ast z\|^2}=\frac{1}{\|(B^{-1})^\ast\|^2}=\frac{n}{N^2}\ge \frac{1}{N^2}.$$
On the other hand, $\|(B^{-1})^\ast\|^2$ is the largest eigenvalue of $ B^{-1} (B^{-1})^\ast$, and the Frobenius norm $tr (B^{-1} (B^{-1})^\ast)$ is the sum of all eigenvalues of the same matrix, therefore $$tr (B^{-1} (B^{-1})^\ast)\le n \|(B^{-1})^\ast\|^2 =N^2.$$ |
H: Continuity of bounded variation functions
A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is a BV function if there exists $M<\infty$ for which $$\sum_{k=1}^N|f(x_k)-f(x_{k-1})|\leq M$$ for every sequence $x_0<x_1<\ldots<x_N$ and every $N$.
Any BV function has only jump discontinuities. That is, at any point $a$, the limit $\lim_{x\rightarrow a^-}f(x)$ and $\lim_{x\rightarrow a^+}f(x)$ exist, but they may or may not be equal.
I want to use this to show that a BV function $f$ is continuous except at countably many points.
Suppose the function is discontinuous at uncountably many points. How can I choose $x_0,x_1,\ldots,x_N$ to contradict the definition of BV?
AI: Any BV function can be written as a sum of two monotone functinos, so there is no loss in assuming that $f$ is increasing. Suppose there were more than countably many discontinuities, at points $x_{\alpha}$; each of these are jump discontinuities. Choose a rational $r_{\alpha}$ such that
$$f(x_{\alpha}-) < r_{\alpha} < f(x_{\alpha}+)$$
Do you see why this leads to a contradiction? |
H: Complex power series (or not quite so?)
I'm stuck with this problem. Any hints are appreciated.
It just says
$$
\mbox{"For what values of}\ z\ \mbox{is}\quad
\sum_{n = 0}^{\infty}\left(z \over 1+z\right)^{n}\quad \mbox{convergent ?}
$$
The thing is that it doesn't look like a power series, but I guess I should transform it into that form so I can get its radius of convergence... Is that correct ?.
How can I go with that ?.
Thanks!
AI: It's not a power series, but it is a geometric series; we know that in general,
$$\sum\limits_{n = 0}^{\infty} r^n$$ converges if and only if $|r| < 1$. So notice that your sum would fit this format with $$r = \frac z {1 + z}$$ |
H: Interval of convergence for $\sum_{n=1}^{\infty}9(-1)^nnx^n$
I need to find the interval and radius of convergence and I'm really confused with what I'm supposed to be doing. Here is the problem:
$\sum_{n=1}^{\infty}9(-1)^nnx^n$
I then used the ratio test to get $|9||x|\lt1$ and took a guess at my interval of convergence as $-\frac{1}{9}$ and $\frac{1}{9}$
Am I on the right track and if so what do I do from here?
AI: The $9$ is irrelevant, it is a constant. The ratio of the absolute values of the terms is $$\left|\frac{(n+1)x^{n+1}}{nx^{n}}\right|\to|x|$$ |
H: restriction of functions of several variables
Let $f: \Bbb R ^n \to \Bbb R $ will be differentiable function satisfying the condition
$$ \sum_{i=1}^{n} y_i \frac{ \partial f}{ \partial x_i } (y) \ge 0 $$ for every vector $ y=(y_1,y_2,...,y_n)$
Demonstrate that the function is bounded from below by $f(0)$
I don't any ideas.
Partial derivatives and differentiability have a few days.
Maybe we need a definition of the gradient, but still I could not take advantage of her.
Help me, please
AI: When beginning to prove things in multivariable calculus, it is important to recognise when you can reduce problems to one-dimensional ones (i.e. bring them within the scope of single variable calculus).
In this case, we can do this by defining the function:
$g: [0,1]\rightarrow \mathbb{R},\quad g(t)=f(yt)$
for an arbitrary fixed y.
Our aim is to show that $g(1)\geq g(0)$. Use the chain rule to differentiate $g$ and show this. |
H: What is the congruence class of $x^3\mod x^3+x+1$?
I have a given Polynom congruence with a Polynom $x^3+x+1$ ... so the set of the congruence classes is $\{0, 1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1\}$
But what would look this like?
$$x^3\mod x^3+x+1\equiv ?$$
I think the result must be one of the congruence classes of the set, mentioned above, but I cannot figure out which one.
AI: You're working over $\Bbb Z/(2)$. Note that $-1=1$ there, so from $x^3+x+1\equiv 0$ you get $x^3\equiv -x-1\equiv x+1$, that is, the class of $x^3$ is that of $1+x$. |
H: Trouble with factoring polonomial to the 3rd degree
I am having trouble factoring this problem:
$\displaystyle{-x^{3} + 6x^{2} - 11x + 6}$
I know the answer but i can't figure out how it is done with this.
I have tried by grouping and is doesn't seem to work. Can someone show me how to do this.
AI: $$\begin{align}
& -x^3+6x^2-11x+6 \\
=& -x^3+1+6x^2-11x+5 \\
=& (1-x^3)+6x^2-6x-5x+5 \\
=& (1-x)(1+x+x^2)-6x(1-x)+5(1-x) \\
=& (1-x)(1+x+x^2-6x+5) \\
=& (1-x)(x^2-5x+6) \\
=& (1-x)(x^2-2x-3x+6) \\
=& (1-x)(x(x-2)-3(x-2)) \\
=& (1-x)(x-2)(x-3)
\end{align}$$ |
H: The trace map in a finite field.
Let $p$ be a prime number, and consider the mapping called the trace
$$ Tr \quad : \quad \mathbb{F}_{p^n} \ \longrightarrow \ \mathbb{F}_{p^n} \quad : \quad x \ \longmapsto \ x + x^p + x^{p^2} + \cdots + x^{p^{n-1}}$$
My syllabus Abstract Algebra states the following:
Every element $x \in \mathbb{F}_{p^n}$, is mapped to $\mathbb{F}_p$.
The restricted mapping $Tr' \ : \ \mathbb{F}_{p^n} \ \rightarrow \mathbb{F}_p $ is surjective.
I failed to prove both these statements, and I ask you for some help.
Research effort for the first statement
We can see $\mathbb{F}_{p^n}$ as a vector space with the scalar field $\mathbb{F}_p$. I showed that $Tr$ is a linear mapping in this setting. I convinced myself that the mapping is not an ismorfism in general by looking at the case where $p=n=2$ and $x\neq 0 \neq y$. Assume that $Tr$ is multiplicative. then
$$Tr(xy) \ = \ Tr(x)Tr(y) \quad \iff \quad xy + x^2y^2 \ = \ (x+x^2)(y+y^2) \ = \ x^2y^2 +xy^2+x^2y +xy$$
and by substracting we see
$$x^2y+x^2y=0 \quad \iff \quad xy(x+y) = 0 \quad \iff \quad x = -y$$
And this is not generally true in a field with four elements... I had no inspiration to continue some other way.
Research effort for the second statement
I knew that for all elements $x \in \mathbb{F}_p, \ Tr(x) = \sum_{j=1}^n x^{p^j}=\sum_{j=1}^n x$. If this the mapping would be injective here, we would be done since the sum $x+ \cdots +x$ is an element of $\mathbb{F}_p$. If $p$ divides $n$ though, this won't work because every element $x \in \mathbb{F}_p$ would be mapped to 0.
I hope you can provide me hints to prove the statements. Thank you for your time.
AI: First: The elements of $F_{p^n}$ that are in $F_p$ are exactly the elements $x$ that satisfy $x^p=x$ (if you know Galois theory, this is the generator of the Galois group of $F_{p^n}$ over $F_p$). Can you prove that for any image of the trace map?
By the way, the trace map is not multiplicative!
Second: The trace is a polynomial of degree $p^{n-1}$ so it can have at most so many zeros. This means that there is an element that has non-zero trace. |
H: Angle between vectors?
Here's the problem from my homework:
If the vector $\vec{a}+\vec{b}$ is perpendicular to the vector $7\vec{a}-5\vec{b}$, and if the vector $\vec{a}-4\vec{b}$ is perpendicular to the vector $7\vec{a}-2\vec{b}$, what is the angle between vectors $\vec{a}$ and $\vec{b}$?
So, if I use the fact that $\vec{a} \perp \vec{b} \iff\vec{a} \cdot \vec{b}=0$ I get these two equations:
$ (\vec{a}+\vec{b})\cdot(7\vec{a}-5\vec{b})=7|\vec{a}|^2+2\vec{a}\cdot \vec{b}-5|\vec{b}|^2=0$
$(\vec{a}-4\vec{b}) \cdot (7\vec{a}-2\vec{b})=7|\vec{a}|^2-30\vec{a}\cdot \vec{b}+8|\vec{b}|^2=0$
Also, Iknow that $cos(\varphi)=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$, but I don't know what to do next?
AI: You correctly got:$$(\vec{a}+\vec{b})\cdot(7\vec{a}-5\vec{b})=7|\vec{a}|^2+2\vec{a}\cdot \vec{b}-5|\vec{b}|^2=0$$$$(\vec{a}-4\vec{b}) \cdot (7\vec{a}-2\vec{b})=7|\vec{a}|^2-30\vec{a}\cdot \vec{b}+8|\vec{b}|^2=0$$If you now rearrange these you get:$$2\vec{a}\cdot \vec{b}=5|\vec{b}|^2-7|\vec{a}|^2\tag{1}$$$$30\vec{a}\cdot \vec{b}=7|\vec{a}|^2+8|\vec{b}|^2\tag{2}$$Now just add equations (1) and (2) to obtain:$$32\vec{a}\cdot \vec{b}=13|\vec{b}|^2$$$$\therefore\vec{a}\cdot \vec{b}=\frac{13|\vec{b}|^2}{32}$$And we also know that:$$\vec{a}\cdot \vec{b}=|\vec{a}|\cdot|\vec{b}|\cdot\cos(\theta)$$$$\therefore\vec{a}\cdot \vec{b}=\frac{13|\vec{b}|^2}{32}=|\vec{a}|\cdot|\vec{b}|\cdot\cos(\theta)$$$$\therefore\cos(\theta)=\frac{13|\vec{b}|^2}{32|\vec{a}|\cdot|\vec{b}|}=\frac{13|\vec{b}|}{32|\vec{a}|}\tag{3}$$You can then use equations (1) and (2) to eliminate $\vec{a}\cdot \vec{b}$ which will give you a value for $\frac{|\vec{b}|}{|\vec{a}|}$ which can be used in equation (3) to obtain the final answer. |
H: How to notice that $3^2 + (6t)^2 + (6t^2)^2$ is a binomial expansion.
The other day during a seminar, in a calculation, a fellow student encountered this expression:
$$\sqrt{3^2 + (6t)^2 + (6t^2)^2}$$
He, without much thinking, immediately wrote down:
$$(6t^2+3)$$
What bothers me, is that I didn't see that. Although I suspected there might be some binomial expansion, I was too confused by the middle term to see it. Even after actually seeing the answer, it took me a good deal to realize what's happening.
My initial idea, by the way, was to substitute for $t^2$ and solve a quadratic equation. Horribly slow.
My questions is perhaps tackling a greater problem. I now see what happened, I can "fully" understand binomial expansions, yet I feel like I might end up encountering a similar expression and yet again not be able to see the result. My questions is thus:
How can I properly practice so I don't repeat this mistake?
This is perhaps too much of a soft-question, but I feel like many others have been in a position of understanding a problem, yet not feeling like they would be able to replicate it, and might actually have an answer that goes well beyond the obvious "just practice more".
Thanks for any input.
AI: This may actually be a matter of "just practice more." If you've done enough factoring, you'll recognize the coefficients $4$, $4$, and $1$ from $(x+2)^2 = x^2 + 4x + 4$. Taking out the $3$ is also a matter of practicing factorization - noticing that the terms all have a common factor.
Being able to do this mentally is, yet again, a matter of practice - this time of mental arithmetic. It may not be that your fellow student consciously set aside time at some point in his life toward practicing mental arithmetic.
More likely, he has experience doing things like that in math classes - when teachers write expressions on the board, he immediately starts thinking about solutions in his head ... |
H: How to prove $(\frac{n+1}{n})^n
Prove that $(\frac{n+1}{n})^n<n\quad $ for any n=3,4,5... by using induction.
For n=3 is true. and lets assume $(\frac{n+1}{n})^n<n$ is true. we must show that $(\frac{n+2}{n+1})^{n+1}<n+1$ is true. How can I continue?
AI: If $(\frac{n+1}{n})^n<n$ is true, then
$(\frac{n+1}{n})^n*\frac{n+1}{n}<n*\frac{n+1}{n}$ is also true which is
$(\frac{n+1}{n})^{n+1}<n+1$
since $\frac{n+1}{n}>\frac{n+2}{n+1}$
$(\frac{n+2}{n+1})^{n+1}<n+1$ is true. |
H: Exponential of a complex line
Is there an "elementary" way to prove that if $D$ is a one-dimensional vector space in $\mathbb{C}$ (considered here as a real vector space), then $\exp(D) \neq \mathbb{C}^{\ast}$ ?
AI: What counts as elementary?
The restriction of $\exp$ to any subset whose diameter is smaller than $2\pi$ is injective, so for every compact "interval" $I \subset D$ whose length is smaller than $2\pi$, $\exp\lvert_I$ is a homeomorphism. In particular, $\exp(I)$ has empty interior. $D$ is the union of countably many compact intervals of length $5$ (arbitrary choice $< 2\pi$), so $\exp(D)$ is the union of countably many closed subsets of $\mathbb{C}^\ast$ with empty interior. Hence $\exp(D)$ is of the first category in $\mathbb{C}^\ast$, hence its complement is dense.
$\exp$ is a local diffeomorphism, hence $\exp(D)$ is a one-dimensional (real) submanifold of $\mathbb{C}^\ast$. |
H: Cantor construction is continuous
I define a function $f:\mathbb{R}\to\mathbb{R}$ as follows:
$f(x)=0$ for $x\le 0$. $f(x)=1$ for $x\ge1$.
$f(x)=\dfrac12$ for $x\in\left[\dfrac13,\dfrac23\right]$.
$f(x)=\dfrac14$ for $x\in\left[\dfrac19,\dfrac29\right]$, $f(x)=\dfrac34$ for $x\in\left[\dfrac79,\dfrac89\right]$.
and so on.
So this function has been defined on $\mathbb{R}$, except for the Cantor set. How can we fill in the function on the Cantor set, so that we get a continuous function?
AI: Denote the funcion you constructed on $\mathbb{R}\setminus C$ by $f$. The question is how to define Cantor's function on $[0,1]$. You can use the formula
$$
F(x)=\sup_{t\in[0,x]\setminus C} f(t),\qquad x\in[0,1]
$$
The function $F$ is continuous thanks to the following lemma.
Lemma. Let $F:[a,b]\to\mathbb{R}$ is non decreasing and $F([a,b])$ is dense in $[F(a),F(b)]$, then $F$ is continuous.
Proof. Since $F$ is non decreasing there exist finite one sided limits at any $x\in[a,b]$, denote them $F(x+)$ and $F(x-)$. Assume $F$ is discontinuous at $c\in[a,b]$, then $f(c-)<f(c+)$. In this case $F([a,b])\cap[F(c-),F(c+)]=\{F(c)\}$, so $F([a,b])$ is not dense in $[F(a),F(b)]$. Contradiction. |
H: Inner product on tangent space and metric tensor
In our class we talked about integrating on submanifolds and as a short side remark our teacher told us that by knowing the metric tensor, it is possible to define an inner product on a tangent space and then he said: $\alpha(t):=\phi(tb)$, where $t$ is a real number, phi is a chart and b a vector in a lower-dimensional $\mathbb{R}^n$.
And then we said that $||\dot{\alpha}(0)||_2^2=b^T (g_{ij})b$.
My problem is the following: I see that the tangent space contains all those curves, but I do not see the idea behind this vector $b$. Especially, why is it meaningful to say $\alpha(t):=\phi(tb)$.
Could anybody explain to me the idea behind this definition? Or maybe a good reference that uses this approach, may also be helpful. Unfortunately, I could not find something similar so far.
I'm somewhat sceptical about the circumstance in the given answer, how one can define b.
AI: If I am understanding this correctly, I believe he is trying to define an inner product on a tangent space to construct a metric that comes explicitly from the chart structure on the manifold.
One way to define the tangent space is as the space of vectors at your point which are the derivatives of curves going through your point. Specifically, suppose your manifold is $M$ and $\phi:U \to M$ is a chart with $U \subseteq \mathbb{R}^n$. Then the tangent space at $\phi(0)$ is exactly the set of vectors given by $(\phi \circ \alpha)'(0)$ where $\alpha$ is some curve in $U$ with $\alpha(0) = 0$. With this in mind we can understand the situation above.
He is saying for any vector $v$ in the tangent space $T_{\phi(0)}$ at $\phi(0)$, from the characterization above, there exists some curve (actually just a linear subspace) in $\mathbb{R}^n$ defined $\beta(t) = tb$ where $b$ is a vector in $U$, such that $v$ is given by $\alpha'(0)$ where $\alpha = \phi\circ \beta$ is the curve on $M$ defined by $\beta$. Then we can define the norm of $v$ by
$$
\langle v,v \rangle = ||\alpha'(0)||^2_2 := b^T(g_{ij})b
$$
By linearity, this lets us define $\langle -,-\rangle$ for any two vectors $a$ and $b$ in $T_{\phi(0)}$ by using the above formula applied to $a+b$. Thus we have constructed an inner product on $T_{\phi(0)}$ directly from the chart $\phi$. The metric tensor $g_{ij}$ is compatible with charts and varies continually so we get a well defined (independent of charts) and continuously varying inner product on the tangent spaces.
Edit: Here is one way to see that $b$ is uniquely defined. We get a basis of $T_{\phi(0)}$ given by the standard basis $e_1, \ldots, e_n$ of $\mathbb{R}^n$ and our parametrization $\phi$. Specifically, the vectors
$$
b_i := d\varphi_0(e_i) = \frac{d}{dt}(\varphi(e_i t))\big|_0.
$$
This is a basis of the tangent space because the tangent space by definition is the $d\varphi_0(\mathbb{R}^n)$, i.e., the pushforward of the tangent space of our parametrization by the differential of the parametrization. The RHS of this equation is one way of calculating the differential $d\varphi_0(e_i)$, the other being just calculating the Jacobian matrix of $\varphi$ and applying it to $e_i$. Now, since the $b_i$ form a basis of $T_{\varphi(0)}$, we have that for any $v \in T_{\varphi(0)}$, $v$ can be expressed uniquely as
$$
v = \sum a_i b_i
$$
Since $b_i$ form a basis. Thus, take $b$ to be the vector
$$
b = \sum a_i e_i.
$$
This maps to $v$ and is unique by construction since basis expressions are unique. Another way to see uniqueness is that $\varphi$ is a parametrization so $d\varphi_0$ is injective so if another vector say $b'$ mapped to $v$, we'd have $d\varphi_0(b) = d\varphi_0(b')$ so $b = b'$ by injectivity. |
H: Bi-implication theorem proving
While proving a theorem, i came across a situation like as follows
(P has a property) $\leftrightarrow $ $(x=y)$
(P has a property) $\leftrightarrow $ $(y=z)$
Now can i infer the following fact from the above two facts ?
(P has a property) $\leftrightarrow $ $(x=z)$
$\leftrightarrow $ stands for Bi-conditional(if and only if)
Thanks in advance :)
AI: The implication $\rightarrow$ is clearly correct, because if $P$ has the property, then $x = y = z$. The implication $\leftarrow$ doesn't hold in general though. For example, let the property of $P$ be "$x=y=z$ holds". Then in the case $x = z = 0$, $y = 1$ the statement $x = z$ is true but the property of $P$ is not satisfied. |
H: Why can't you solve this probability problem in this way?
Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?
This is how I tried to solve it.
Alice and Beatrix will visit together every $LCM(3,4) == 12$ days.
Alice and Claire will visit together every $LCM(3,5) == 15$ days.
Beatrix and Claire will visit together every $LCM(4,5) == 20$ days.
Thus the number times these occur in 365 days is $\left\lfloor \frac{365}{12} \right\rfloor\; +\; \left\lfloor \frac{365}{15} \right\rfloor\; +\; \left\lfloor \frac{365}{20} \right\rfloor\; == 72$
The answer seems to be $54$ though. Why is this?
AI: This solution uses the Principle of Inclusion and Exclusion.
It's very convenient that their names follow the first three letters of the alphabet, so I'll just call them A, B, and C $\ddot\smile$
Here's what you figured out:
A and B visit together every $12$ days
B and C visit together every $20$ days
A and C visit together every $15$ days
This is the expression you calculated, by adding the number of times each pair of friends visits per year: $$\left\lfloor \frac{365}{12}\right\rfloor + \left\lfloor \frac{365}{20}\right\rfloor + \left\lfloor \frac{365}{15}\right\rfloor = 72$$
However, this counts the times when all three of the friends meet three times each. The question asks for how many times exactly two friends visit.
How often to all three visit together? A, B, and C meet every $\text{lcm}(3,4,5) = 3 \cdot 4 \cdot 5 = 60$ days. Thus, the three friends meet exactly $$\left\lfloor \frac{365}{60}\right\rfloor = 6$$ times each year.
Since you've counted these six meetings three times each, we need to subtract $6 \cdot 3$ from our earlier answer, for a final total of $$\left\lfloor \frac{365}{12}\right\rfloor + \left\lfloor \frac{365}{20}\right\rfloor + \left\lfloor \frac{365}{15}\right\rfloor - 3 \left\lfloor \frac{365}{60}\right\rfloor = 72 - 18 = \boxed{54}$$ |
H: Generating Function for Binary String Question
The following is an assignment question I have been trying to work out for quite some time.
Let $C(x,y)=\sum_{n,k \geq 0} c_{n,k} x^{n} y^{k}$, where $c_{n,k}$ is the number of binary strings of length $n$ with $k$ blocks. Prove that
\begin{equation}
C(x,y)=\frac{1-x+yx}{1-x-yx}
\end{equation}
Then show that $[x^{n}] \frac{\delta}{\delta y} C(x,y) \left|_{y=1} \right.$ is the total number of blocks among all binary strings of length $n$.
I know that the first step is to find a recurrence relation for $c_{n,k}$. Consider a arbitrary binary string of length $n$ with $k$ blocks. The first digit is either zero or one; the difference is immaterial. The second digit is either the same or the opposite of the first digit.
In the first case, when the second digit is the same as the first, there are $k$ blocks in the $(n-1)$ digit long substring. The number of such substrings is $c_{n-1,k}$. In the second case, when the second digit is different than the first, one block has been found and so $(k-1)$ blocks remain in the $(n-1)$ digit long substring. The number of such substrings is $c_{n-1,k-1}$. This gives the following recurrence relation:
\begin{equation}
c_{n,k}=c_{n-1,k}+c_{n-1,k-1}.
\end{equation}
If $k>n$, then $c_{n,k}=0$ because there is no way to have more blocks of digits than digits in a binary string. If $k=n$, then $c_{n,k}=2$ because there are two binary strings with alternating digits.
Let $C(x,y)= \displaystyle\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c_{n,k} x^{n}y^{k}$. Find the bivariate generating function.
\begin{equation}
\begin{aligned}
\sum_{k=1}^{\infty} c_{n,k} y^{k} &= \sum_{k=1}^{\infty} c_{n-1,k} y^{k} + \sum_{k=1}^{\infty} c_{n-1,k-1} y^{k} \\
\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} c_{n,k} x^{n}y^{k} &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} c_{n-1,k} x^{n}y^{k} + \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} c_{n-1,k-1} x^{n}y^{k} \\
\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} c_{n,k} x^{n}y^{k} &= x\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} c_{n-1,k} x^{n-1}y^{k} + xy\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} c_{n-1,k-1} x^{n-1}y^{k-1} \\
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\sum_{n=0}^{\infty} \sum_{k=1}^{\infty} c_{n,k} x^{n}y^{k} - \sum_{k=1}^{\infty} c_{0,k}x^{0}y^{k} &= x\sum_{n=0}^{\infty} \sum_{k=1}^{\infty} c_{n,k} x^{n}y^{k} + xy\sum_{n=0}^{\infty} \sum_{k=1}^{\infty} c_{n,k-1} x^{n}y^{k-1} \\
\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c_{n,k} x^{n}y^{k} - \sum_{n=0}^{\infty} c_{n,0} x^{n} &= x \left( \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c_{n,k} x^{n}y^{k} - \sum_{n=0}^{\infty} c_{n,0} x^{n} \right) + xy\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c_{n,k} x^{n}y^{k} \\
\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c_{n,k} x^{n}y^{k} - 2 &= x \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c_{n,k} x^{n}y^{k} - 2x + xy\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} c_{n,k} x^{n}y^{k} \\
C(x,y)-1&= xC(x,y)-x+xyC(x,y) \\
(1-x-xy)C(x,y) &= 2-2x \\
C(x,y) &= \frac{2-2x}{1-x-xy}
\end{aligned}
\end{equation}
At this point, I am unsure how to proceed. Furthermore, I know that I have clearly made an error above but I cannot seem to find it.
Any help would be appreciated!
AI: The recurrence can be written
$$c_{n,k}=c_{n-1,k}+c_{n-1,k-1}+[n=k=0]-[n=1][k=0]+[n=k=1]\;,$$
where the square brackets are Iverson brackets; this makes the recurrence valid for all $n,k\ge 0$ on the assumption that $c_{n,k}=0$ if either $n$ or $k$ is negative. This is an easy way to build the initial conditions into the recurrence, and although I’ve not waded through the details, it looks as if this is probably where you went a little bit astray.
Multiply the recurrence through by $x^ny^k$ and sum over $n,k\ge 0$:
$$\begin{align*}
C(x,y)&=\sum_{n,k\ge 0}c_{n,k}x^ny^k\\\\
&=\sum_{n,k\ge 0}c_{n-1,k}x^ny^k+\sum_{n,k\ge 0}c_{n-1,k-1}x^ny^k+1-x+xy\\\\
&=xC(x,y)+xyC(x,y)+1-x+xy\;,
\end{align*}$$
so
$$C(x,y)=\frac{1-x+xy}{1-x-xy}\;.$$ |
H: Proving $\lim_{n \to\infty} \frac{1}{n^p}=0$ for $p > 0$?
I'm trying to prove 3.20a) from baby Rudin. We are dealing with sequences of real numbers.
Theorem.
$$\lim_{n \to {\infty}} \frac{1}{n^p} = 0; \hspace{30 pt}\mbox {$p > 0$}$$
Proof. Let $\epsilon > 0$. Because of the Archimedan property of real numbers, there exists an $N \in \Bbb{N}$ such that $n_0 \geq N$ implies $\frac{1}{n_0} < \epsilon$ and thus implies $n_0 > \frac{1}{\epsilon}$. Thus $\exists n \geq n_0 : n > (\frac{1}{\epsilon})^k$ where $k$ is any number. For the interesting case, pick $k = \frac{1}{p}$ where $p > 0$. Thus $n > (\frac{1}{\epsilon})^{\frac{1}{p}}$ implies $$\frac{1}{n} < \frac{1}{\left(\frac{1}{\epsilon}\right)^{\frac{1}{p}}} \Longrightarrow \frac{1}{n^p} < \epsilon$$
which further implies $d(n, N) < \epsilon$. QED.
AI: Another approach: you can show $x^p \rightarrow \infty $ as $x\rightarrow \infty$. Rewrite as $$\frac {x^{p+1}}{x}$$ , which is an indeterminate $\infty/\infty$ , and use L'Hopital, to get $$\frac {(p+1)x^{p}}{1} $$. Since $p$ is fixed and $p+1>1$, you can show this goes to $\infty$ , and then $1/x^p\rightarrow 0$ |
H: Convergence functions
Let X be a nonempty set. I define a convergence function on X to be a partial function from the set of all sequences in X, to X, that satisfies the five additional conditions:
Every constant sequence is assigned that constant.
If a sequence converges, so does any sequence obtained from altering, inserting, or deleting finitely many terms, and to the same value.
If a sequence converges, every subsequence converges to the same value.
If two sequences have the same convergence value, any weave of them also converges to the same value
This is the most complicated property. Let $(a_m)_n$ be a double sequence where every $(a_m)$ converges. And suppose the resulting sequence of convergence values also converges, say to $L$. Then there is a sequence, where the $n$-th element is in the $n$-th sequence, that converges to $L$.
The question I am trying to find out, as an independent scholar, is this: Does every convergence function on a set induce a topology such that every sequence that converges in the topological sense to a value $L$, also converges under $f$ to $L$? I have tried proving it, but I am completely stuck. Any help would be much appreciated.
AI: Yes.
A set $H$ will be closed iff it is closed under taking limit (of sequences within $H$). All you have to prove that any intersection and any finite union of closed sets is again closed.
Then define the open sets as complements of closed sets.
Note that the topology obtained this way is $T_1$: one point sets are closed by criterium 1.
Now if $x_n$ converges to $L$ according to our limit function $f$, and $U\ni L$ is any open set, then its complement, $U^\complement$ is closed, but cannot contain an infinite subsequence of $x_n$ as it doesn't contain $L$ but is closed under limit.
On the other hand, if $x_n$ converges to $L$ in the topology, then -eliminating constant subsequences- we can reduce it to the case when $x_n\ne L$ for all $n$.
Then consider $G:=\{x_n\,\mid\, n\in\Bbb N\}$. We claim that $G$ is not closed: its complement contains $L$ and if that were open, that would contain some $x_n$ because of the definition of limit.
So, by definition of the topology, $G$ is not closed under limit, moreover by the same reasoning, $L$ must be in the closure of $G$, i.e. there is at least a subsequence $(x_{n_k})_k$ that converge to $L$, according to $f$.
Now cut the original sequence $(x_n)$ into ${\bf A}_0:=(x_{n_k})_k$ and the rest, say $(x_{m_k})_k$ (So that $\{n_k\mid k\in\Bbb N\}\,\cup\,\{m_k\mid k\in\Bbb N\}\,=\,\Bbb N$, and repeat all with $(x_{m_k})_k$ which still converges to $L$ in topology. This yields a next subsequence ${\bf A}_1$ which already converges to $L$ according to $f$. Repeat it until all elements $x_n$ are used up (either for finite or infinite many times), then apply condition 4. or 5. |
H: Limit of convolution of measures is Cantor function
For positive integer $k$, let $\mu_k=\dfrac{1}{2}\left(\delta(x)+\delta\left(x-\dfrac{2}{3^k}\right)\right)$. Show that $$\lim_{k\rightarrow\infty}(\mu_1\ast\mu_2\ast\cdots\ast\mu_k)((-\infty,x))=C(x),$$ where $C$ is the Cantor function.
The definition of measure convolution that I find is
$$ (\mu_1\ast \mu_2)(I)=\int_\mathbb{R}\int_\mathbb{R}1_I(x_1+x_2)d\mu_1(x_1)d\mu_2(x_2)$$ for any interval $I$.
I'm not sure how to start on this. It's a convolution of $k$ measures...
AI: Here is a probabilistic proof. Recall that a convolution of measures corresponds to taking a sum of independent random variables. Recall that $\mu_k$ is the law of a random variable which takes on value $0$ with probability $\frac{1}{2}$ and $\frac{2}{3^k}$ with probability $\frac{1}{2}$. Therefore, if $X_k$ is sequence of $iid$ random variables taking on the value $2$ with probability $\frac{1}{2}$ and $0$ with probability $\frac{1}{2}$, $\mu_k$ is the law of $\frac{X_k}{3^k}$.
$\mu_1 * \dots * \mu_k$ is therefore the law of $\sum_{i=1}^k \frac{X_i}{3^i}$. It is not hard to see that this sum converges absolutely almost surely (and therefore in distribution) to $\sum_{i=1}^\infty \frac{X_i}{3^i}$ and that the limit random variable is uniformly distributed on the set of numbers which have only $0$ or $2$ in their tenary expansion. This characterizes the limit as the uniform measure on the Cantor set. The Cantor function is the distribution function of that measure. |
H: Question on Rudin sequences?
In baby Rudin, Rudin shows that $$\lim_{n \to \infty}\sqrt[n]{p} = 1.$$
In the proof of limit he tries to prove that the limit is $1$. So he takes $x_n = \sqrt[n]{p} - 1$.
I have never noticed this before, but I could have tried to prove the limit for any $n \neq 1$ (and obviously it wouldn't hold). But in the process of proving that the limit is $1$, did he have some special insight that $\lim_{n \to \infty} = 1$? My question becomes clearer below: He now says that by the binomial theorem, $$0 < 1 + nx_n \leq (1 + x_n)^n = p$$
But how in the world, as an undergraduate, am I supposed to pull that out of my nearly-empty math toolbox? I never even knew such a technique existed. In fact, with Rudin, he uses many esoteric proofs (that all work out, anyway) and I have no idea what could have caused him to realize that is the way (in this case, using the binomial theorem) to prove such a sequence.
I noticed indeed that he just selected a sequence that is smaller than $p$ and show by the squeeze theorem of upper and lower limits that $x_n \to 0$. This is a red herring.
But here is what he specifically wrote:
$$0 < x_n < \frac{p - 1}{n}$$
But $p = (1 + x_n)^n$, so how do we know indeed that $\lim_{n\to\infty}\sqrt[n]{p} = 1$??
My main concern is, how can I learn the methods to prove sequences and series without being inefficient? Should I just look at many solutions to limits of sequences, go hard and kill myself at the problems anyway, or keep reading (which is very slow for me, at perhaps an hour a page)?
AI: The log trick is always helpful, when your base is non-negative:
$p^{1/n}=e^{(1/n)lnp}$. Now, $lnp$ is fixed, and $(1/n) \rightarrow 0$ , so you can find
an $M$ , so that for $n>M$ , $\frac {1}{n}<\frac{1}{\epsilon(lnp)}$ ....
EDIT: Another useful trick is that of sequential continuity, which works on the Reals:
the idea is : given a sequence $x_n $ with $x_n \rightarrow x$ , and $f$ is continuous, it follows that $f(x_n)\rightarrow f(x)$ . In this particular case, the function $x^a$ is continuous. You can then say, for $x^{1/n}:$ $Lim_{n\rightarrow \infty}x^{1/n}=e^{Lim_{n\rightarrow \infty}{(1/n)}=e^0=1}.$ You can also get a lot of mileage out of this one. |
H: exponential equation with a sum of exponents
I'm trying to solve the following exponential equation:
$e^{2x} - e^{x+3} - e^{x + 1} + e^4 = 0$
According to the the text I am using the answer should be $x = 1,3$ but I can't derive the appropriate quadratic $x^2 -4x + 3$ from the above equation using any of the methods I know. Can someone point me in the right direction? Is there a substitution I'm not seeing?
AI: $$e^{2x} - e^{x+3} - e^{x + 1} + e^4 = 0$$
$$\left(e^{x+1}\right)\left(e^{2x-\left(x+1\right)}-e^{\left(x+3\right)-\left(x+1\right)}-e^{\left(x+1\right)-\left(x+1\right)}+e^{4-\left(x+1\right)}\right)=0$$
$$\left(e^{x+1}\right)\left(e^{x-1}-e^{2}-e^{0}+e^{3-x}\right)=0$$
$$\left(e^{x+1}\right)\left(e^{x-1}-e^{2}-1+e^{3-x}\right)=0$$
$$\left(e^{x+1}\right)\left(e^{-x-1}\right)\left(e^x-e\right)\left(e^x-e^3\right)=0$$
$$e^0 \cdot \left(e^x-e\right)\left(e^x-e^3\right) =0$$
$$\left(e^x-e\right)\left(e^x-e^3\right)=0$$
Thus, either $e^x-e=0$ or $e^x-e^3 = 0$. These correspond to $x=1$ or $x=3$. |
H: Linear congruence proof, show congruence has exactly two incongruent solutions
Let p be an odd prime and k a positive integer. Show that the congruence $x^{2}$ $\equiv 1 \ mod p^{k}$ has exactly two incongruence solutions, namely, $x \equiv \pm 1\mod p^{k}$.
I'm not sure what to do after this:
$x^{2}$ $\equiv 1 \ mod p^{k}$ => $p^{k}$ | $x^{2}-1$=(x-1)(x+2)
AI: As you have already observed, $p^k\mid (x-1)(x+1)$. In particular $p\mid (x-1)(x+1)$, so either $p\mid x-1$ or $p\mid x+1$. In any case, we cannot have both since $p\not\mid 2$. This implies that $p^k\mid x-1$ or $p^k \mid x+1$. Why?
This works in general: $x^2=n\mod p^k$ has at most two incongruent solutions. |
H: Korean Math Olympiad 2000 (floor function, quadratic mod)
Let $p$ be a prime such that $p ≡ 1\ (\mathrm{mod}\ 4)$. Evaluate $\displaystyle\sum\limits_{k=1}^{p-1}\left({\left\lfloor\frac{2k^2}{p}\right\rfloor}-2{\left\lfloor\frac{k^2}{p}\right\rfloor}\right)$.
AI: Hint: Prove that
$$\lfloor2x\rfloor-2\lfloor x\rfloor =\begin{cases}0\\1\end{cases}$$
(Try writing $x=\lfloor x\rfloor+\delta$)
Then, the terms you are summing are either $0$ or $1$. Check how many times the condition for getting a $1$ happen during the sum using both facts: $p$ is prime and $p=4m+1$ for some positive integer $m$.
This should lead you to the sum $S(p)=2m=\frac{p-1}{2}$.
I am giving you a direction. But you should do some work as well. |
H: Addition in finite fields
For a question, I must write an explicit multiplication and addition chart for a finite field of order 8. I understand that I construct the field by taking an irreducible polynomial in $F_2[x]$ of degree 3, and creating the splitting field for that polynomial.
The polynomial I chose for the field of order 8 is $x^3 + x + 1$. Since every finite field's multiplicative group is cyclic, it's my understanding that I can write the 6 elements of this field that are not 0 and 1 and $a, a^2, ..., a^6$, where $a^7 = 1$. And if my thinking is correct about that, I know how multiplication works. But I'm lost on how to develop addition in this field. I know in this case that since 1 + 1 = 0, every element should be its own additive inverse, but beyond that I'm lost as to how, for example, I should come up with a proper answer for what $1 + a$ is.
That is, assuming I'm right about the first parts.
An answer that could help me understand how to do this in general would be very helpful, as I need to do this for a field of another order as well.
AI: Hint: $a$ being a root of your polynomial gives you the relation $a^3+a+1=0$ |
H: Showing that $E[X|X
I would like to show that:
$\hspace{2mm} E[X|X<x] \hspace{2mm} \leq \hspace{2mm} E[X] \hspace{2mm} $ for any $x$
X is a continuous R.V. and admits a pdf. I'm guessing this isn't too hard but I can't come up with a rigorous proof. Thanks so much.
AI: Hint: $$E[X] = E[X|X<x]P(X<x) + E[X|X\geq x]P(X\geq x)$$
Also:
$$E(X|X<x)< x \leq E(X|X\geq x)$$ |
H: I don't understand this PDE solution involving Fourier coefficients and orthonormal eigenfunctions.
$u_{xx} + u_{yy} = 0$ with $x \in (0,\pi)$ and $y \in (0, \pi)$
Initial Conditions:
$$
u(x,0) = x^2
$$
$$
u(x,\pi) = 0
$$
Boundary conditions:
$$
u_{x}(0,y) = 0 = u_{x}(\pi, y)
$$
I performed separation of variables, which yielded $\frac{Y''}{Y} = \frac{-X''}{X} = \lambda$
From this we derive two ODE's:
$$
X'' + \lambda X = 0
$$
$$
Y'' - \lambda Y = 0
$$
For the first ODE, we know the general solution is $X(x) = A\sin(\sqrt \lambda x) + B\cos(\sqrt \lambda x)$
Using the boundary conditions I differentiated my general solution which yielded $X'(x) = \sqrt \lambda A \cos ( \sqrt \lambda x ) - \lambda B \sin (\sqrt \lambda x)$ or with the first boundary condition when $x = 0$, yields simply $X'(0) = \sqrt \lambda A = 0$ which tells us A = 0, right?
Well I am just having real trouble understanding what is the logic behind the solution to this problem that my professor posted. Where do the orthonormal eigenfunctions come from? Why is there no talk of solving for the constants that come out of the general solution? Where do the solutions for Y come from? I would really appreciate if someone could work through this problem step by step because I am really lost as to where the thinking comes to take the steps taken.
Please help! Thank you in advance.
AI: Well I am just having real trouble understanding what is the logic
behind the solution to this problem that my professor posted. Where do
the orthonormal eigenfunctions come from?
You are finding them by solving the $X$ problem, i.e. by finding all values of the parameter $\lambda$ for which there exists nontrivial solutions to the $X$ ODE and the $X$ boundary conditions. Note that you have only applied the BC at $x=0$. You need to apply the BC at $x=\pi$ to complete your analysis. Also, note that you have assumed $\lambda>0$ since you got sines and cosines for your general solution. Based on where it sounds you are in your understanding, you need to consider the cases $\lambda=0$ and $\lambda<0$ as well.
Why is there no talk of
solving for the constants that come out of the general solution?
Well, you have to have the general solution (and $\lambda$ values) before you can do this. Once you find the eigenvalues $\lambda$ and eigenfunctions $X(x)$ (there will be an infinite sequence of each), then you can use the orthogonality of those eigenfunctions to determine the values of the constants in the general solution.
Where do the solutions for Y come from?
Once $\lambda$ is known, you simply solve the $Y$ ODE for those $\lambda$ values.
Edit: A little more to help you along (for the $\lambda>0$ case)... $X(x)=A\sin(\sqrt{\lambda}x)+B\cos(\sqrt{\lambda}x)\implies X'(x)=A\sqrt{\lambda}\cos(\sqrt{\lambda}x)-B\sqrt{\lambda}\sin(\sqrt{\lambda}x)$. Then $X'(0)=0\implies A\sqrt{\lambda}=0\implies A=0\text{ or }\sqrt{\lambda}=0\implies A=0$ since we are assuming $\lambda>0$.
On the other hand, $X'(\pi)=0\implies -B\sqrt{\lambda}\sin(\sqrt{\lambda}\pi)=0\implies B=0, \sqrt{\lambda}=0,\text{ or }\sin(\sqrt{\lambda}\pi)=0$. If $B=0$, we already had $A=0$ so $X(x)=0$, but we were seeking nontrivial solutions. Also, $\sqrt{\lambda}=0$ is impossible since we are assuming $\lambda>0$. That leaves us with $\sin(\sqrt{\lambda}\pi)=0\implies \sqrt{\lambda}=n\in\mathbb{Z}\implies \lambda=\lambda_n=n^2$, $n=1,2,\dots$ are the eigenvalues with associated eigenfunctions $X_n(x)=B_n\cos(\sqrt{\lambda_n}x)=B_n\cos(nx)$, $n=1,2,\dots$
Now that the $\lambda_n$ are known, solve the $Y$ problem... |
H: Finding vector and parametric equations provided only one point.
Normally to answer these questions I have a point and one or two vectors. However, for this one I only have a point. How can I concoct these equations provided there is limited information?
Find vector and parametric equations of the plane in $R^3$ that passes
through the origin and is orthogonal to $v$.
$$v = (3, 1, -6)$$
AI: In $\Re^3$, a plane can be characterized completely by a vector normal to the plane, and a point it goes through:
Imagine any plane. Note that every vector normal to it's surface is necessarily parallel to each other (even if you consider the normal pointing in the other direction, the lines all these normals live in are parallel). So, once you have a line you have your plane except for a translation in the direction of that normal, so you need to specify a point in the space that belongs to the plane to fully define it.
Now, in equations, a vector connecting to points in the plane must be perpendicular to $\vec{v}=(3,1,-6)$. Let's call this generic point in the plane $\vec{u}=(u_x,u_y,u_z)$ and for a second let $\vec{u}_0$ be the point in the plane that they are giving us (origin in this case), then
$$0=\vec{v}\cdot(\vec{u}-\vec{u}_0)=3u_x+u_y-6u_z$$
Please, note that what needs to be perpendicular to $\vec{v}$ is not a point in the plane $\vec{u}$ but a vector contained (or parallel) to the plane itself and therefore we need to use $(\vec{u}-\vec{u}_0)$ in the dot product above. Since the origin is contained in the plane in your case, then the distinction is hard to grasp but it is very important to understand. |
H: Composition relations and powers
Let $R$ be the relation on $\Bbb Z$ such that $xRy$ iff $x-y=c$
a.) Define $R^2$
b.) Define $R^i$ for abitrary $i\ge1$.
Well the problem I'm having with this is trying to figure out how get the power of a relation and how does $xR^2y$ for example. Well I know for starters that in the case with $R^2$ we have $R \cdot R$. Any input would be great.
AI: firstly think of a relation on a finite set $A$ with labelled elements $a_1, a_2, \dots, a_n$
the relation $R$ on $A$ is a subset of $A \times A$, so it can be represented by an $n \times n$ matrix $(r_{ij})$ whose elements take the value $0$ or $1$. $r_{ij}=1$ if $a_iRa_j$ and $0$ otherwise (c.f. the indicator function of a set).
the relation $R^2$ is defined by:
$$ a_iR^2a_k \equiv \exists j.a_iRa_j\land a_jRa_k
$$
thus $R^2$ can be computed by an operation very similar to matrix multiplication of $R$ by itself, except that we take the boolean union rather than the arithmetic sum, so that the matrix corresponding to $R^2$ has elements
$$r_{ij}^2 = \lor_k \;( r_{ik} \land r_{kj})$$
for $\mathbb{Z}$ it is easy to extend this idea to use an infinite matrix.
you can probably see that the matrix for $R$ on $\mathbb{Z}$ to be the identity matrix but shifted to the left by $c$ places. (when i wrote this i was assuming $c$ to be positive. of course if $c$ is negative we must shift to the right)
when multiplying this matrix by itself we find this shifts it another $c$ places to the left (or right for negative $c$, which means that:
$$ xR^2y \equiv x - y = 2c $$
if at first you have difficulty visualizing this, draw out the first few entries taking a small value of $c$ and do the "multiplication" - the underlying principle will soon become clear, and this should enable you to solve the second part of the question.
i would have chosen a superior notation if i had used the symbol $R_c$ for the original relation. with this notation you can investigate the product relation $R_cR_d$. you should be able to see that the relations thus defined form a (multiplicative) abelian group isomorphic to the additive group of $\mathbb{Z}$ |
H: Prove: The infinite series $x_n$ converges if and only if the infinite series $2^nx_{2^n}$ converges.
I am in desperate need for hints to get me in the right direction for this proof.
Let $(x_n)_{n\in \mathbb {N}}$ be a monotone decreasing null sequence. Prove that:
$$\sum_{n=1}^\infty x_n \text{ converges } \ \iff \sum_{n=0}^\infty 2^nx_{2^n} \text{ converges}$$
Thanks.
AI: Hint:This is standard cauchy condensation test. Please try to group terms in powers of 2. Thus $$\sum x_n = x_1 + (x_2 + x_3) + (x_4 + x_5 + x_6 + x_7) + \cdots$$ now use he fact that $x_n$ is decreasing and try to bound the above sum in terms of $\sum 2^{n}x_{2^{n}}$. |
H: Equivalence of definition for weak convergence
The Wikipedia page on convergence of measures says:
In the case $S=\mathbb{R}$ with its usual topology, if $F_n, F$ denote the cumulative distribution functions of the measures $P_n$, $P$ respectively, then $P_n$ converges weakly to $P$ if and only if $\lim_{n\rightarrow\infty} F_n(x) = F(x)$ for all points $x\in\mathbb{R}$ at which $F$ is continuous.
I'm interested in the proof of the following:
If $\lim_{n\rightarrow\infty} F_n(x) = F(x)$ for all points $x\in\mathbb{R}$ at which $F$ is continuous, then $$\lim_{n\rightarrow\infty}\int_\mathbb{R}gdF_n=\int_\mathbb{R}gdF$$ for any function $g\in C_0^\infty(\mathbb{R})$
AI: Integrate by parts; for each $n$ you have
$$\int_{-\infty}^\infty g(x)\,dF_n(x) = - \int_{-\infty}^\infty g'(x) F_n(x)\, dx$$
Now I think you can do the rest. |
H: Hunting Birds probability
A hunter locates 20 geese, 25 ducks 40 eagles, 10 ostriches, and 5 flamingos. He randomly selects 6 birds to target. What is the probability at least one of each species is targeted?
My reasoning $20 \choose 1$ $25 \choose 1$ $40 \choose 1$ $10 \choose 1$ $5 \choose 1$ $5\choose 1$ for the numerator because there are 5 ways to choose 1 species that will be targeted twice. For the denominator its $100 \choose 6$. However book says im wrong what did i forget?
AI: Looks like you forgot two things. You don't just have to choose which species is targeted twice, you have to choose which bird to target. Up to this point you've been counting individual birds, not species. So that last factor of $\binom51$ should be corrected to $\binom{95}1$. But now you have over counted by a factor of $2$, because the two birds of the same kind could have been chosen in either order.
Edited to answer follow-up questions.
Overcounting is not necessarily something to be avoided. Sometimes the best way to solve a counting problem is to overcount by a definite known factor (a factor of $2$ in your question) and then correct by dividing out the overcount factor.
For example, the number of $5$-card combinations from a $52$ card deck is, as you know given by the formula $\binom{52}5=\dfrac{52\cdot51\cdot50\cdot49\cdot48}{5!}$. Do you remember how that formula was derived? It was derived by first counting the number of $5$-card permutations, which is $52\cdot51\cdot50\cdot49\cdot48$, observing that we have overcounted the combinations by a factor of $5!$ (since there are $5!$ permutations for each combination, and so we divide the number of permutations by $5!$ to get the number of combinations.
In your problem, you're targeting two birds of the same kind, say two ducks. One duck was chosen first, and appears in the $\binom{25}1$ factor. The other duck was chosen second, and appears in the final $\binom51$ factor. Those two ducks could just as well have been chosen in the other order, so you're double counting.
When it's too hard to see what's going on by abstract thinking, here's a trick you can use: scale down the problem; create an analogous problem with the smallest numbers that make sense. E.g., $4$ ducks, $3$ geese; a hunter shoots $3$ of the $7$ birds; what's the probability that he gets at least one of each kind? The numbers are small enough so you can get the answer ($30/35$) by brute force. You can take the method you used on the big problem and apply it to the miniature problem, and you will know if you get the right answer or not. If your method produces the wrong answer on the miniature problem, you may be able to see why it's wrong. |
H: The Inverse Laplace Transform
What's the inverse Laplace transform of $\frac{s}{(s-5)^4}$? I'm thinking of adding zero to the top and dividing out to get rid of the top s.
AI: Hint:
Do the partial fraction expansion as:
$$\dfrac{s}{(s-5)^4} = \dfrac{5}{(s -5)^4} + \dfrac{1}{(s -5 )^3}$$
Now, use a Table of Laplace Transforms or use the ILT definition to find the inverse.
Spoiler
$\mathcal{L^{-1}}\left(\dfrac{5}{(s -5)^4} + \dfrac{1}{(s -5 )^3}\right) = \dfrac{6}{5}t^3~e^{5t} + \dfrac{1}{2}t^2~e^{5 t} = \dfrac{1}{6}~t^2~e^{5t}~\left(5t + 3\right)$ |
H: Calculus: finding integral
I need to compute the following two integrals: $$\int_0^{\infty}y^be^{-y/2} dy$$
$$\int_0^{\infty}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy$$ can I do this? Please show me the process.
How to use definition of gamma function to solve these?
AI: For the first integral, substitute $u=y/2$ with $2du=dy$ to get:
$$ 2^{b+1}\int_0^{\infty}u^be^{-u}du = 2^{b+1}\Gamma(b+1)$$
from the definition of the gamma function.
The second integral is a Gaussian integral, so I will not try to replicate what has been done so well before.
EDIT:
I see you want to use the definition of the gamma function.
For the second integral, let $u=y^2/2$ with $du=ydy$ to get:
$$\frac{1}{2\sqrt{\pi}}\int_0^\infty u^{-1/2}e^{-u}du=\frac{1}{2\sqrt{\pi}}\Gamma(1/2)$$
Since $\Gamma(1/2)=\sqrt{\pi}$, you get $1/2$ as a final answer. Of course finding $\Gamma(1/2)$ is another story.
Finding $\Gamma(1/2)$
The most logical way in your case is to go back, evaluate the second integral as a Gaussian integral, and then compare results. You can find online many good derivations of the Gaussian integral. I found this one well-explained.
But if you assume known the reflection property of the gamma function:
$$\Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin \pi z}$$
you can set $z = 1/2$ to get $\Gamma(1/2)^2=\pi$ or $\Gamma(1/2) = \sqrt{\pi}$. |
H: k critical graph cannot have k + 1 vertices
$k$-chromatic graph is called $k$-critical if removal of any vertex from graph makes it $k - 1$ vertex colorable.
Now i have to prove that if $G$ is a $k$ critical graph then it cannot have $k+1$ vertices.
I can see that the property is true as a triangle is 3 critical. Also we have a 5-cycle as 3 critical but no $4$ vertex graph is $3$ critical.
Can any one help me in what direction i should prove the theorem
AI: Suppose you have a k critical graph on k+1 vertices. If there exists a vertex of degree less than k-1, remove it. By assumption you can k-1 color the remaining graph. However if you add back in the vertex removed it can only be adjacent to k-2 colors, so we can complete it to give a k-1 coloring of the whole graph. So we can assume every vertex has degree k-1 or k.
Now choose a pair of degree k-1 not adjacent to one another, they must both be adjacent to all other vertices. (Not all vertices can have degree k, as $K_{k+1}$ is not k chromatic) Remove one, k-1 color the resulting graph then add back in the vertex giving it the same color as the other vertex in the pair. This is a valid k-1 coloring. |
H: How prove this $\int_{a}^{b}xf(x)dx\le\int_{a}^{b}xg(x)dx$
let $f(x),g(x)$ is continuous on $[a,b]$,and such
$$\int_{a}^{x}f(t)dt\ge\int_{a}^{x}g(t)dt,x\in[a,b)$$
and
$$\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt$$
show that:
$$\int_{a}^{b}xf(x)dx\le\int_{a}^{b}xg(x)dx$$
my try: we only prove this
$$\Longleftrightarrow \int_{a}^{b}x(f(x)-g(x))dx\le 0$$
if $[a,b]\subset (-\infty,0]$,then
$x<0$ this is true.But other case I can't.Thank you
AI: Let $F(x) = \int_{a}^{x}f(t)\,dt, G(x) = \int_{a}^{x}g(t)\,dt$ then we have $F(x) \geq G(x)$ for all $x \in [a, b)$ and $F(b) = G(b)$. Now we have $$\int_{a}^{b}xf(x)\,dx = bF(b) - aF(a) - \int_{a}^{b}F(x)\,dx = bF(b) - \int_{a}^{b}F(x)\,dx$$ and $$\int_{a}^{b}xg(x)\,dx = bG(b) - aG(a) - \int_{a}^{b}G(x)\,dx = bG(b) - \int_{a}^{b}G(x)\,dx$$ so that
$\displaystyle \begin{aligned}\int_{a}^{b}x\{f(x) - g(x)\}\,dx &= b\{F(b) - G(b)\} - \int_{a}^{b}\{F(x) - G(x)\}\,dx\\
&= -\int_{a}^{b}\{F(x) - G(x)\}\,dx \leq 0\end{aligned}$ |
H: separable space and open covers
If a topological space $X$ is separable, then every open cover of $X$ must be countable?
since $X$ is separable , then there exists a countable dense subset $S$. This implies, in every open cover any set must intersect with $S$.
AI: No, definitely not. The Mrówka space $\Psi$ is a separable space with an irreducible open cover of cardinality $2^\omega=\mathfrak{c}$. (Irreducible means that it has no proper subcover.) The Katětov extension of $\Bbb N$ is a separable space that has an irreducible open cover of cardinality $2^{2^\omega}=2^{\mathfrak{c}}$. |
H: Initial Value Problem with Laplace Transform
How do you solve the following with Laplace Transform?
$$
{\rm y}''\left(t\right) - 10\,{\rm y}'\left(t\right) + 25\,{\rm y}\left(t\right)
=
24\,t\,{\rm e}^{-2t}\,;
\qquad\qquad
{\rm y}\left(0\right) = -2\,,\quad {\rm y}'\left(0\right) = -10
$$
AI: Hints:
$\mathcal{L} (y''(t)) = s^2y(s) -s y(0) -y'(0)$
$\mathcal{L} (-10y'(t)) = -10(sy(s) -y(0))$
$\mathcal{L} (25y(t)) = 25y(s)$
$\mathcal{L} (24te^{-2t}) = \dfrac{24}{(s + 2)^2}$
Now, substitute the ICs, isolate $y(s)$ on the LHS, everything else on the RHS, do a partial fraction expansion and then find the Inverse Laplace Transform. You should end up with the following.
Spoiler
$y(t)=\dfrac{2}{343} e^{-2t} (84 t+e^{7 t} (84 t-367)+24)$ |
H: Finding the formula of a sequence.
I have a sequence $\{0,1,1,2,2,3,3,4,4,5,5,...\}$ and I am supposed to find a formula for the n-th term. I do not see a pattern between the terms except that each positive integer is repeated twice. I cannot see a relation between each term to determine whether it is arithmetic or geometric. Is there a strategy to find the formula of a sequence that is neither arithmetic or geometric?
AI: Hint: each term $a_n$ equals the previous term, plus $0$ if the previous term was even (i.e. $n$ of $a_n$ was even), or plus $1$ if the previous term was odd (i.e. if $n$ of $a_n$ was odd).
Edit: here's a perfectly legitimate formula. Note that I index $n$ at $0$. Adjust for your purposes if necessary.
$$a_n = \begin{cases}\frac{n}2, n \text{ even} \\ \frac{n+1}{2}, n \text{ odd} \end{cases}$$
Edit 2: if your book does not allow piecewise functions, then there's a handy trick: we use $(-1)^n$ to give us a $-1$ when $n$ is odd and $1$ when $n$ is even. From there, we can reconstruct our piecewise function:
$$a_n = \frac{n+x}{2} $$ Where we need $x = 0$ when $n$ is even, and $x=1$ when $n$ is odd. A sufficient $x$ will be $$x = \frac{1 + (-1)^n}{2}$$ Note that when $n$ is odd, we'll have $\frac{1+(-1)}{2} = \frac02$, and when $n$ is even, we'll have $\frac{1+1}{2} = \frac22 = 1$. Success!
So then we let $$a_n = \frac{n+\frac{1+(-1)^n}{2}}{2}$$ and we are done.
Unfortunately I cannot supply you with a good intuitition for how to 'see' that you need $x = \frac{1 + (-1)^n}{2}$. Of course, we want it to give either $0$ or $1$, and the way to create a $0$ a $1+(-1)$, but apart from that, this is just one of those things you have to play with to understand and remember. Using $(-1)^n$ is a very useful trick. |
H: How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola?
How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola, knowing the canonical form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\pm1$ where $a$ and $b$ are constants? Thanks !
AI: Taking the '+' sign, $$y^2-x^2=x+y+1\implies\left(y-\frac12\right)^2-\left(x+\frac12\right)^2=1^2$$
$$\implies\frac{\left(y-\frac12\right)^2}{1^2}-\frac{\left(x+\frac12\right)^2}{1^2}=1^2$$
Similarly for the '-' sign,
$$y^2-x^2=-(x+y+1)\implies\left(x-\frac12\right)^2-\left(y+\frac12\right)^2=1^2$$
Can you recognize $a,b$ here? |
H: Selecting Cards
A special deck of 51 cards constists of 25 pairs and 1 wild card. The deck is distrubuted evenly between 3 players (17 cards each). What is the probability that your hand has only two pairs that is the probability you have 13 single cards.
My reasoning was I have $51 \choose 17$ hands. There are $25 \choose 2$ ways to selcect two pairs. There are $51 \choose 4$ ways to select 4 cards. Im having trouble counting the single cards though. How many ways can I choose 13 single cards is where I am stuck
AI: Once two pairs have been selected, the remainder of the deck consists of $23$ pairs and the wild card.
Suppose that your hand does not contain the wild card; then you can fill it out by choosing $13$ of the $23$ remaining pairs, which can be done in $\binom{23}{13}$ ways, and then choosing one card from each of those pairs, which can be done in $2^{13}$ ways.
I’ll leave to you the calculation for the case in which your hand does include the wild card; it’s quite similar.
Matter are a bit different if the wild card is truly wild and can be counted as half of a pair with any other card in the deck. In that case you have hands with two true pairs and $13$ singletons not including the wild card, and you have hands with one true pair, the wild card, and, and $14$ singletons. But the methods of counting are pretty much the same. |
H: Public Key Cryptography
Assuming that a message has been sent via the RSA scheme with $p=37$, $q=73$, and $e=5$, what is the decoding of the received message "34?"
So far, I have $x^5 \pmod{37\times73} \equiv 34$. How do I invert a mod?
AI: We need the totient function of the modulus, hence we get:
$$\varphi(N) = \varphi(37 \times 73) = 36 \times 72 = 2592$$
First you need to find the decryption exponent.
$$d = e^{-1} \pmod {\varphi(n)} = 5^{-1} \pmod {2592} = 1037$$
Decryption:
$$\displaystyle m = c^{\large d} \pmod N \rightarrow 34^{1037} \pmod {2701} = 1415$$ |
H: How to prove a set is nowhere dense?
Im trying to solve a problem from chapter 8, Real Analysis, Carothers, 1ed, talking about compactness of metric space, :
I've finished the first problem actually.
For the second problem. I've proved the range of $f$ is closed with claim as following:
$f$ is continuous and $[0,1]$ is compact => range of f is compact in $[0,1]\times[0,1]$.
Since $[0,1]\times[0,1]$ is compact,
then we get range of $f$ is closed.
And then I got some problems:
The metric is unknown, which means that it is insufficient to say $[0,1]$ is compact and so does $[0,1]\times[0,1]$. How to solve it?
How to do next? I cannot use the hint left below.
AI: You don’t really need a metric on $[0,1]\times[0,1]$; it’s more convenient to use the product topology directly. It has a base consisting of all sets of the form $I\times J$, where $I$ and $J$ have one of the following forms:
$(a,b)$ for $0\le a<b\le 1$;
$[0,b)$ for $0<b\le 1$; or
$(a,1]$ for $0\le a<1$.
Now let $K$ be the range of $f$; you already know that $K$ is closed, since it’s a compact subset of the Hausdorff space $[0,1]\times[0,1]$. Thus, it is nowhere dense if and only if its interior is empty. Suppose, on the contrary, that it contains a non-empty open set. Then it contains a basic open set of the form $I\times J$ described above, and since it’s closed, it contains the closure of that basic open set; this is a rectangle $[a,b]\times[c,d]$ for some $a,b,c,d$ such that $0\le a<b\le 1$ and $0\le c<d\le 1$.
To use the hint, show that $R=[a,b]\times[c,d]$ must be the image of some subinterval $I$ of $[0,1]$. Further HINT: If not, there are $u,v,w\in[0,1]$ such that $u<v<w$, $f(u),f(w)\in R$, and $f(v)\notin R$; now consider the sets $$R\cap f\big[[0,v]\big]\qquad\text{and}\qquad R\cap f\big[[v,1]\big]\;.$$
Then let $g=f\upharpoonright I$, the restriction of $f$ to the subinterval $I$; $g$ is continuous and maps $I$ onto the rectangle $[a,b]\times[c,d]$, but this is impossible for the same reason that $f$ cannot map $[0,1]$ onto $[0,1]\times[0,1]$. |
H: Way to find the value of $ a_{2011}$
Let us consider the sequence $ a_1 , a_2 , a_3 ,\ldots $ where $ \frac{1}{a_{k+1}} = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_k}$ for $k >1$ and $a_1 = 2^{2009}$. How can I find the value of $ a_{2011}$?
AI: Observe that $ \frac{1}{a_{k+1}} = \frac{1}{a_k} + \frac{1}{a_k}$. This will give you a simple relation between $a_k$ and $a_{k+1}$. Unfold it upto $a_1$. You have your answer. |
H: How many ways the set D can be constructed?
The following relations hold for four non-empty sets $A,B,C,D$:
$ A \cup B \cup C \cup D = A \cap B $
$ B \cup C \cup D = B \cap C $
$ C \cup D = C $
If $A = \{1,2,3,4\} $ then in how many ways can the set $D$ be constructed ?
AI: (3) tells you that $D\subseteq C$; why?
(2) tells you that $B\cup C\subseteq B\cap C$. It’s always true that $B\cap C\subseteq B\cup C$, so $B\cup C=B\cap C$. This tells you that $B\subseteq B\cap C$, and it’s always true that $B\cap C\subseteq B$, so $B=B\cap C$, and therefore $B\subseteq C$. Use a similar argument to show that $C\subseteq B$, and conclude that $B=C$, so that $B\cup C=B\cap C=B=C$. Now (2) tells you further that $B\cup D=B$ and hence that $D\subseteq B=C$.
We can use the previous result to reduce (1) to $A\cup B\cup D=A\cap B$. Use the ideas of the previous paragraph to extract information from this. When you’ve done that, you’ll know how $D$ must relate to $A$. |
H: Limit $\lim (\frac{n!}{n^n})^{\frac{1}{n}}$
I need to calculate $$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$
My try:
When $n!$ is large we have $n!\approx(\frac{n}{e})^n\sqrt {2\pi n}$ (Stirling approximation)
$$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}=\lim_{n\rightarrow \infty}\frac{\left((\frac{n}{e})^n\sqrt {2\pi n}\right)^{\frac{1}{n}}}{n}$$
Simplifying we get,
$$\frac{1}{e}\lim_{n\rightarrow \infty} \left(\sqrt{2\pi n}\right)^{\frac{1}{n}}$$
I am stuck here. I don't know how to proceed further.
AI: Another way :
Let $$A= \lim_{n \to \infty}\left(\frac{n!}{n^n}\right)^{\frac1n}$$
$$\implies \ln A= \lim_{n \to \infty}\frac1n\sum_{1\le r\le n}\ln \frac rn$$
Now use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ |
H: problem with recurrence relation for series solution for ODE
I have
$$y''-xy'-y=0$$
and I'm trying to find the series solution around the ordinary point $x_0=1$. My last post I muscled through to the solution when the ordinary point was $x_0=0$, but this is proving to be tougher. Now I have obtained through power series analysis
$$y''-xy'-y=0=\sum_{k=0}^{\infty}[(k+2)(k+1)a_{k+2}-(k+1)a_{k+1}-(k+1)a_k](x-1)^k$$
which yields the recurrence relation
$$a_{k+2}=\frac{a_{k+1}+a_{k}}{k+2}$$
WHen I start plugging in sequential "k" values I'm not finding a very good pattern emerging.
$$a_2=\frac{a_0}{2!}+\frac{a_1}{2!}$$
$$a_3=\frac{a_0}{3!}+\frac{3a_1}{3!}$$
$$a_4=\frac{4a_0}{4!}+\frac{6a_1}{4!}$$
$$a_5=\frac{8a_0}{5!}+\frac{18a_1}{5!}$$
$$a_6=\frac{28a_0}{6!}+\frac{48a_1}{6!}$$
$$a_7=\frac{76a_0}{7!}+\frac{156a_1}{7!}$$
Outside just writing out term by term, substituting in the appropriate $a_k$, does this have a nice closed form? Or is is just the case that I have the answer with the $a_k$'s that I have
AI: I totally agree with JohnD' answer to your question. You have done the best you can achieve using power series. To try to clarify, let me go to the solution of your ODE which is
$$y=e^{\frac{x^2}{2}} \left(C_2+C_1\sqrt{\frac{\pi }{2}}
\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$
As you see, you have two arbitrary constants and then any polynomial expansion of the solution will depend on two constants : these are your $a_0$ and $a_1$ terms.
To me the nicest closed form is the recurrence relation you established; I suggest you do not go further. |
H: Drawing a simple graph with a certain number of vertices
I am supposed to see if it is possible to draw a graph with 8 vertices given the degree sequence: 1,1,2,3,5,5,6,7 First I tried the handshaking lemma and it holds.
So since drawing that graph is tedious I decided to remove the vertex with degree 7 since I know I can just add it back at the end and connect it to every point. So now the new degree sequence is 0,0,1,2,4,4,5
The handshaking lemma still holds for the degree sequence but Isn't it impossible to have a vertex of degree 5 since there aren't enough vertices in the degree sequence to have a degree of 5 for a vertex? I don't understand how such a graph could exist even though the handshaking lemma still holds. any thoughts/ suggestions?
AI: The key part missing is that if the handshaking lemma holds, then the sequence is the degree sequence of some multigraph. If you try and realise (i.e. draw) the original (or second) sequence as a simple graph, it can't be done, however it can be realised using parallel edges. |
H: Initial Value Problem with Repeated Eigenvalues
Given the matrix
$$
A=\left(\begin{array}{ccc}
1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)
$$
For $X'= AX.\quad$
$X\left(0\right)=\left(\begin{array}{r}1 \\ 0 \\ -2\end{array}\right)\,.\quad$
What is the solution ?.
AI: We find the eigenvalues and eigenvectors as:
$\lambda_{1,2,3} = 1$, which gives one eigenvector and two generalized eigenvectors as:
$v_1 = (0,1,0), v_2 = (1,0,0), v_3 = (0,0,1)$
We can write the general solution as:
$$X(t) = e^t\left(c_1v_1 + c_2(t v_1 + v_2) + c_3\left(\frac{t^2}{2!}v_1 + t v_2 + v_3\right)\right)$$
Using the initial condition, $X(0)$, we find:
$c_1 = 0, c_2 = 1, c_3 = -2$
Our final solution is then:
$x(t) = e^t(1 - 2t)$
$y(t) = e^t(t - t^2)$
$z(t) = e^t(-2)$
Note, there are many ways to do these types of problems from the matrix exponential, fundamental matrix, set of linear equations... |
H: Restriction of $f$ to $B(x_0,\delta)$ attains a maximum
Let $f: U\to \mathbb{R}$ $U\in \mathbb{R}$ be twice differentiable at $x_0\in U$. If $f'(x_0)=0$ and $f''(x_0)<0$ then the restriction of $f$ to $B(x_0,\delta)$ attains a maximum at $x_0.$
I know that by defintion $$\lim_{x\to x_0} =\frac{f'(x)-f'(x_0)}{x-x_0} = f''(x_0)<0$$ and since $f$ is differentiable at $x_0$ then it is also continuous at $x_0.$ How can I prove this question?
AI: Taylor's theorem gives $f(x) = f(x_0) + f'(x_0) (x-x_0) + { 1 \over 2} f''(x_0) (x-x_0)^2+ r_2(x-x_0)$, where $\lim_{h \to 0} { r_2(h) \over |h|^2 } = 0$. Choose $\delta>0$ small enough so that if $|x-x_0| < \delta$, then ${ r_2(x-x_0) \over |x-x_0|^2 } < -{ 1 \over 4} f''(x_0) $.
Since $f'(x_0) = 0$, if $|x-x_0| < \delta$ we have
$f(x) \le f(x_0) + { 1 \over 4} f''(x_0) (x-x_0)^2$, from which it follows that $x_0$ is a strict maximizer of $f$ on $B(x_0,\delta)$. |
H: Exponentials in complex numbers
If $\displaystyle z-\frac1z=i$, then find $\displaystyle z^{2014}+\frac{1}{z^{2014}}$.
The answer should be in terms of $1, -1,\;i\;or\;-i$. I am not able to understand how to simplify the given expression so that its $2014^{th}$ power can be found out easily.
AI: $(1)$
We have $\displaystyle z^2+\frac1{z^2}=1$
$\displaystyle\implies z^4-z^2+1=0$
Using $\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2),$
$z^6+1=(z^2+1)(z^4-z^2+1)=0\implies z^6=-1$
$(2A)$ We have $\displaystyle z^2=iz+1$
$\displaystyle \implies z^3=z\cdot z^2=z(iz+1)$
$\displaystyle=iz^2+z$
$\displaystyle=i(iz+1)+z=i+z(i^2+1)=i$
$(2B)$ As $\displaystyle z^2-iz-1=0, z^2-iz+i^2=0$
$\displaystyle\implies z^3+i^3=(z+i)(z^2-iz+i^2)=0\implies z^3=-i^3=i$
$\displaystyle\implies z^6=-1$
Now, $2014=335\cdot6+4$
Finally $\displaystyle z^4+\frac1{z^4}=\left(z^2+\frac1{z^2}\right)^2-2\cdot z^2\cdot\frac1{z^2}$ |
H: How to solve this gamma integral
Let we have the (p.d.f) of x which is:
$$f(x)=\frac {\Gamma{(n-1)/2)}}{\Gamma{(1/2)} \Gamma{(n-2)/2)}}x(1-x^2)^{(n/2)-2}$$
then to find the $E(x) = $$\int_{-\infty}^{\infty} x *(f(x) dx$ ; (Expectation of Mean)
Thus, we construct that
$$\int_{-\infty}^{\infty} (x) \frac{\Gamma{(n-1)/2)}}{\Gamma{(1/2)} \Gamma{(n-2)/2)}}(1-x^2)^{(n/2)-2}=??? $$
In my text book, it shows the result:
$$E(x) = 0 $$
Thanks
AI: *HINT*Whatever is your problem, did you notice that the derivative of (1 - r^2) looks like r ? On the other hand, what does happen is you split your integral (from -1 to 0 and from 0 to 1) ? Can you continue with this ? |
H: Find solutions to equation (ring/field theory, residue class)
I'm trying to solve this problem:
A residue class ring mod $n$ is a field if n is prime. Let $\mathbb{Z}_p$ be a residue ring, p prime. Let $a \in \mathbb{Z}_p$.
What are all solutions $x \in \mathbb{Z}_p$ to the following equation?
$$\widetilde{x^2} = \widetilde{a²}$$
Here's what I've tried so far (note: $a^{\#}$ is the inverse of a, n is identity element):
$\widetilde{x^2} = \widetilde{a²}$
$\iff \widetilde{x^2} \cdot (\widetilde{a^{\#}})^2 = \widetilde{a²} \cdot (\widetilde{a^{\#}})^2$
$\iff \widetilde{x^2} \cdot (\widetilde{a^{\#}})^2 = \widetilde{a} \cdot \widetilde{a} \cdot \widetilde{a^{\#}} \cdot \widetilde{a^{\#}}$
$\iff \widetilde{x} \cdot \widetilde{x} \cdot \widetilde{a^{\#}} \cdot \widetilde{a^{\#}} = \widetilde{a} \cdot n \cdot \widetilde{a^{\#}}$
$\iff \widetilde{(xxa^{\#}a^{\#})} = \widetilde{a \cdot a^{\#}}$
$\iff \widetilde{(xxa^{\#}a^{\#})} = \widetilde{1}$
$\iff \widetilde{x \cdot a^{\#}} \widetilde{x\cdot a^{\#}} = \widetilde{1}$
$\iff (\widetilde{x \cdot a^{\#} })^{2} = \widetilde{1}$
This is where I'm stuck. Can anyone help me solve this?
Edit: For all of those unfamiliar with the tilde notation: It indicates the residue class of whatever is below. Another way of stating the problem would (I think) be: Solve for x in the following equation:
$$x^2 mod \space n = a^2 mod \space n$$
AI: Let $K$ be a field.
Let $a \in K$.
Consider the equation $x^2 = a^2$.
Since $x^2 - a^2 = (x - a)(x + a) = 0, x = \pm a$. |
H: Verify that the unit closed ball in $c_0$ has not extremal points
I want to understand this proof of the fact that $c$ and $c_0$ aren't isometrically isomorphic, but I have very little experience in working with extremal points.
So how can I verify that the unit ball in $c_0$ didn't have one of these points? I tried to manipulate a generic sequence to obtain two difference sequences but the fact that I have to use a convex combination gives me problems.
And how can i prove that the extremal points are preserved via isometry?
Thanks in advance
AI: If $x=(x_j )_{j\in\mathbb{N}} $ is any point of unit ball $B$ in $c_0 $, then there exists $k\in\mathbb{N} $ such that $|x_k|<\frac{1}{2} $ and therefore $ x=\frac{1}{2} u+\frac{1}{2} v $ where $$u_j=\begin{cases} x_j \mbox{ for } j\neq k\\ x_k-\frac{1}{2} \mbox{ for } j=k\end{cases}$$
$$v_j=\begin{cases} x_j \mbox{ for } j\neq k\\ x_k +\frac{1}{2} \mbox{ for } j=k\end{cases}$$
It is easy to see that $u,v\in B $ and thus $B$ have no extremal points.
If $I$ is an isometry between the Banach spaces $X$ and $Y$ then by Mazur - Ulam Theorem it must be affine, but affine injections preserved extremal points. |
H: Number of solutions in a given interval
How many solutions does the equation $\cos^2x-\cos x-x=0$ have in the interval $\displaystyle \left[0,\frac\pi2\right]$?
Clearly, $x=0$ is a solution. Are there any other? I couldn't proceed after differentiating it to find the extremes.
AI: You don't need to take the derivative for this.
We know that $x=0$ is an obvious solution and $\frac{\pi}{2}$ is not a solution.
We also know that on the interval of $(0,\frac{\pi}{2})$, we have $0<\cos x<1$. Thus, we have the inequality $\cos^2 x<\cos x<\cos x+x$. Therefore, $\cos^2 x - \cos x-x<0$. This means that the only solution is $x=0$. Therefore, there is only $1$ solution. |
H: How prove this matrix equation $AXB=C$ has a solution
Qustion:
if Matrix equation $AY = C$ and $ZB = C$ has solution,
show that:
the equation of $AXB= C$ has solution
This problem is from this PDF(page 3) problem 2 :http://wenku.baidu.com/view/d6625d1cff00bed5b9f31dba.html
My idea:we have
$$\left(A\bigotimes I\right)\overline{Y}=\overline{C}$$
$$\left(I\bigotimes B^T\right)\overline{Z}=\overline{C}$$
we only prove
$$\left(A\bigotimes B^T\right)\overline{X}=\overline{C}$$ has solution.
since
$$\left(A\bigotimes I\right)\left(I\bigotimes B^T\right)=A\bigotimes B^T$$
then ony prove follow
$$I_{m}\left(A\bigotimes I\right)\bigcap I_{m}\left(I\bigotimes B^T\right)\subset I_{m}\left(A\bigotimes B^T\right)$$
My idea is wrong? If is true,and follow How works? Thank you
and maybe have other methods.Thank you very much!
AI: I will assume that your matrices are over a regular ring (maybe even over a field). Then each of these is regular. By this I mean for $A$ there is matrix $A^-$ such that $AA^-A=A$.
Then the first two equations you have imply that
$AA^-C=C$ and
$CB^-B=C$.
Now you may like to guess what a solution of $AXB=C$ may be. If not see bellow.
Answer $X=A^- CB^-$. |
H: Rank-one update of singular matrix
My problem is the following:
Given a column vector $\mathbf{e''}\in\mathbb{R}^{3}$ with $||\mathbf{e''}||=1$ and a $3\times 3$ matrix $B$ such that $rank\{B\}=3$, I would like to find a vector $\mathbf{v}\in\mathbb{R}^3$ garantying that $rank\{A(\mathbf{v})\}=3$, where $A(\mathbf{v})=(\mathbf{e''}\mathbf{e''}^T-I_{3\times 3})B+\mathbf{e''}\mathbf{v}^T$.
One approach would be to try to maximize $|det\{A(\mathbf{v})\}|$ with respect to $\mathbf{v}$, but I think this would involve non-linear optimization. As my problem seems to relate to rank-one updates, I would rather find a closed-form formula if it exists. I found this thread, which is quite close but only provides a solution to rank-one updates of non-singular matrices. This is not the case here as $rank\{(\mathbf{e''}\mathbf{e''}^T-I_{3\times 3})B\}\leq rank\{\mathbf{e''}\mathbf{e''}^T-I_{3\times 3}\}=2$.
Can someone think of a way to solve this ?
Thanks
AI: Write $A(v) = (w w^T -I)B+w v^T = ((w w^T -I)+w v^T B^{-1}) B = (w w^T -I + w (B^{-T}v)^T)B$.
Then if we choose $B^{-T}v = -w$, or $v = - B^T w$, we get $A(-B^Tw) = -B$. |
H: What is meaning of strict weak ordering in layman's term?
I gone through many pages using Google, but not understand exact meaning of Strict-weak Ordering term.
I have this requirement while sorting strings.
AI: This notion, which sounds somewhat like an oxymoron, is not very commonly used in mathematics, but it is in programming. The "strict" just means it is the irreflexive form "$<$" of the comparison rather than the reflexive "$\leq$". The "weak" means that the absence of both $a<b$ and $b<a$ do not imply that $a=b$. However as explained here, the relation that neither $a<b$ nor $b<a$ holds is required to be an equivalence relation. The strict weak ordering then induces a (strict) total ordering on the equivalence classes for this equivalence relation.
This notion is typically used for relations that are in basically total orderings, but defined using only partial information about the identity of items. For instance if $a<b$ between persons means that $a$ has a name that (strictly) precedes the name of $b$ alphabetically, then this defines a strict weak order, since different persons may have identical names; the relation of having identical names is an equivalence relation.
One can easily show that for a strict weak ordering "$<$", the relation $a\not<b$ is (reflexive and) transitive, so it is a pre-order, and the associated equivalence relation is the same as the one associated above to the strict weak ordering. In fact "$a\not<b$" is a total pre-order which induces the same total ordering (or maybe it is better to say the opposite ordering, in view of the negation) on its equivalence classes as the strict weak ordering does. I think I just explained that the notions of strict weak ordering and total pre-order are equivalent. The WP article also does a reasonable job explaining this. |
H: A doubt in finding the number of real roots of a given polynomial using derivative
I have learnt so far that a polynomial of degree $n$ has $n$ roots. To find out the number of real roots it has, we have to take its derivative, equate it to $0$ and then find the roots. Those roots are the extremes in the graph of the original polynomial. Using those extremes, we can find out how many times the graph of the original polynomial intersects the $X-axis$, so we know the number of real roots. Now, consider a polynomial of degree $3$. Suppose its derivative has no real roots. Does this $necessarily$ mean that the original polynomial has only one real root? Moreover, consider a polynomial of degree $4$. If its second derivative has no real roots, then (taking into account that the first derivative has one real root) will the original polynomial have two real roots? And does this continue for higher degrees?
AI: Yes, a polynomial of degree$~3$ (or of any odd degree) whose derivative has no real roots defines a strictly monotonic function $\Bbb R\to\Bbb R$, which being bijective has exactly one zero. But no, for a polynomial of degree$~4$ (or of any even degree) all the information you can give about its derivatives will not allow telling whether it has any roots: the answer to that question can be altered by adding a constant to the polynomial, which does not change any of its derivatives. |
H: What is the value of $ a_{2009}$
I have the following sequence :
$a_0 = 3 $
$ a_n = 2 + a_0 a_1 a_2\text{ ...}a_{n-1} $
How can I find the value of $a_{2009} $ ?
AI: Rewrite $a_{n+1} = a_n(a_n-2) + 2$ as $a_{n+1} - 1 = (a_n -1)^2$
To clarify further.
If $b_n = a_n -1$, then we have $b_{n+1} = (b_n)^2$, and so $b_{n} = (b_0)^{2^n}$. |
H: Represent boolean OR opperator in non-boolean math notation
I'm trying to represent the boolean opperation OR in a regular formula, I am familiar with the boolean algebra notation, I came up with this (A+B)/(A+B)
this works for all binary values except if both A=0 and B=0
is there a simple alternative that also works if all inputs are 0
AI: Notice that for AND, we could use the formula:
$$
A \land B \equiv A \times B
$$
Likewise for NOT, we could use the formula:
$$
\neg A \equiv 1 - A
$$
By using double negation and applying DeMorgan's Law, we can combine these together to get:
\begin{align*}
A \lor B &= \neg(\neg(A \lor B)) \\
&\equiv \neg(\neg A \land \neg B) \\
&\equiv 1-(1 - A)\times(1-B)\\
&\equiv \boxed{A + B - (A \times B)}\\
\end{align*} |
H: Nspire CAS spitting out a wrong answer?
Consider the integral:
$$\int \frac{8x+11}{(2x+3)(x+1)}$$
My Nspire CAS tells me that the answer to this is
$$\ln\left\lvert(x+1)^3 \cdot (2x+3)\right\lvert$$
This is not the correct answer according to my calculations and Wolfram Alpha
Any ideas what's going on?
AI: HINT:
Using Partial Fraction Decomposition
$$\frac{8x+11}{(2x+3)(x+1)}=\frac A{2x+3}+\frac B{x+1}$$
Do you know $\displaystyle \ln a+\ln b=\ln ab$ and consequently $\displaystyle c\cdot\ln a=\ln (a^c)$ (assuming if each logarithm is defined) ? |
H: Using the intermediate value theorem for derivatives to infer that a function is strictly monotonic
My textbook Elementary Classical Analysis claims that by Darboux's theorem (the intermediate value theorem for derivatives), if a function $f:\mathbb R\to\mathbb R$ has a nonzero derivative on $\mathbb R$, then is $f$ strictly monotonic (i.e., either $f'(x)>0$ on $\mathbb R$ or $f'(x)<0$ on $\mathbb R$).
The claim is definitely true if $f's$ domain were instead a closed interval, but since $\mathbb R$ is open, I don't understand why Marsden's claim should be true.
AI: Argue by contradiction. If $f$ is not strictly monotonic, then, $f′$ would take both a positive value and a negative value. There is a closed interval containing the points at which these values occur. So... |
H: I want to show $e^{-\alpha t}B(e^{2\alpha t})$ is a Gaussian process.
Let $B(t)$ be Brownian motion. Show that $e^{-\alpha t}B(e^{2\alpha t})$ is a
Gaussian process. Find its mean and covariance functions.
thanks .
AI: We have to prove that if $(t_1,\dots,t_d)$ are non-negative numbers and $(a_1,\dots,a_d)$ real numbers, then $\sum_{j=1}^da_jX_{t_j}$, with $X_{t}=e^{-\alpha t}B(e^{2\alpha t})$, is Gaussian.
For $d=2$, we write
$$a_1X_{t_1}+a_2X_{t_2}=a_1e^{-\alpha t_1}B(e^{\alpha t_1})+a_2e^{-\alpha t_2}\left(B(e^{2\alpha t_2})-B(e^{2\alpha t_1})\right)+a_2e^{-\alpha t_1}B(e^{2\alpha t_1}).$$
We then rearrange the terms and use the independence of increments of Brownian motion. From this computation, we can get a formula for the characteristic function of $(X_{t_1},X_{t_2})$. |
H: Integration of parabola
I have this homework question I am working on:
The base of a sand pile covers the region in the xy-plane that is bounded by the parabola $x^2 +y = 6$ and the line $y = x$: The height of the sand above the point $(x;y)$ is $x^2$: Express the volume of sand as (i) a double integral, (ii) a triple integral. Then (iii) find the volume.
I have drawn the $x^2 + y = 6$ and $y=x$ plane and found the intersection between the functions to be $(-3,-3)$ and $(2,2)$. So I now know what the base looks like. Now I am REALLY confused what the question means about the $x^2$ being the height. What point are they talking about?
Also, if it is a volume then doesn't it HAVE to be a triple integral? How can I possibly express it as a double integral?
AI: $$\int_{-3}^{2}\int_{y=x}^{y={6-x^2}}z dydx,~~~z=x^2$$ Or $$\int_{-3}^{2}\int_{y=x}^{y={6-x^2}}\int_{z=0}^{x^2}f(x,y,z) dzdydx,~~~f(x,y,z)=1$$ |
H: Calculation of a square root of a big number
How can I calculate the following number:
$$ \sqrt{444 \cdots (2n \text{ digits}) + 111 \cdots (n+1 \text{ digits}) - 666 \cdots (n \text{ digits})}.$$
My trying :
I have tried to calculate these data by observing the pattern of $ 7^2 , 67 ^2 ,667^2 , 6667^2 , \dots$. But there is no promising news to solve it by this process.
AI: Hint:
Use the fact that $111\dots 111 = \frac{10^n-1}{9}$ (where there are $n$ ones)
Then your number is
$$4\frac{10^{2n}-1}{9} + \frac{10^{n+1}-1}{9} - 6\frac{10^n-1}{9} = \frac{4\times 10^{2n}-4 +10^{n+1}-1-6\times 10^n+6}{9}$$
$$=\frac{4\times 10^{2n}+4\times 10^n+1}{9}$$
$$= \frac{(2\times 10^n+1)^2}{9}$$ |
H: Computing a probability of finding defects by random sampling
This is a problem from my semester end exams (which have got over). I suspect that the problem below is vague or open for misinterpretation. I would really like to know the actual answer to the problem if they are correctly asked, if not, I would like how can you also interpret it & the solution to that misinterpretation.
Note - The questions are as-is in the paper (no changes at all)
Question 1 -
An assembly line produces continues product that is 3% defective. If 100 pieces are randomly checked, find the probability that (i) exactly 3 are defective & (ii) none are defective.
My solution (as I recall)
(i) $\left(\binom{100}{3} \times 0.03^{3}\right) \times 0.97^{97} = 0.2275$
(ii) $\left(\binom{100}{0} \times 0.03^{0}\right) \times 0.97^{100} = 0.0476$
Question 2 -
A stack of construction material contains 8 good sample & 2 defective samples. A part is chosen at random and inspected by 4 different inspectors. Find the probability that two of the inspectors found defective sample?
My solution
$\left(\binom{8}{2} \times \binom{2}{2}\right) \left/ \binom{10}{4}\right. = 0.1333$
Does permutation matter in any of the above?!
Thanks!
AI: The probability $p$ of selecting one defective part is:
$$p=\frac{2}{10}=\frac{1}{5}=0.2$$
This means that two inspectors must pick a defective part and two inspectors must pick a non-defective part. Therefore,
$$P=\binom{2}{2}p^2\binom{8}{2}(1-p)^2=\binom{8}{2}p^2(1-p)^2$$ |
H: Definition of logarithm function derived from its useful properties
My understanding of the point of logarithms is that they turn multiplication into addition, and exponentiation into multiplication.
i.e.
$$ \ln cx = \ln c + \ln x $$
$$ \ln x^c = c \ln x $$
Let's call the above two statements about logarithms their "useful" properties.
The above two are somewhat the "point" of logarithms in so far as they were originally invented to simplify calculations: Take the logarithm, do all the calculations in the easier world of "log space" where multiplication is addition, then at the end take the inverse-logarithm (exponentiation) to convert the result back.
I'm aware of various definitions of logarithm, such as:
$$ \ln c = \lim_{h\to 0} \frac{c^h-1}{h} $$
and there are then proofs that this definition leads to what I have called the "useful" properties.
But are there any derivations that go the other way?
i.e. a definition of logarithm that starts with
"Define $\ln x$ as the function that turns multiplication into addition" etc.
and concludes with
"$\ln$ exists and it can be computed by calculating $\lim_{h\to 0} \frac{x^h - 1}{h} $
AI: Let us try with the simple functional equation $f(xy) = f(x) + f(y)$ satisfied by $f(x) = \log x$. We will show that any solution of this equation has to be of the form $k\log x$ where $k$ is a constant. We need to assume the differentiability of $f$. Treating $y$ as constant and differentiating with respect to $x$ we get $yf'(xy) = f'(x)$ and exchanging $x, y$ we get $xf'(xy) = f'(y)$ so that $xf'(x) = yf'(y) = xyf'(xy)$. Since $x, y$ are arbitrary it follows that $xf'(x) = k$ and then $f'(x) = k/x$.
Clearly if $k = 0$ then $f(x)$ is constant and from the function equation this constant must be $0$. If $k \neq 0$ then we can write $f(x) = k\int_{a}^{x}(1/t)\,dt$ where $a, x$ are of same sign. Putting this into the functional equation we will get $a = 1$ so that $f(x) = k\int_{1}^{x}(1/t)\,dt$
Also note that the functional equations can't determine $k$. Hence from the function equation you can't derive $$\ln x = \lim_{h \to 0}\frac{x^{h} - 1}{h}$$ There is another fundamental problem that we can't talk about the above limit unless we define $x^{h}$ for all $h$. This can only be done is a simple way if $\log$ and $\exp$ functions are already defined. |
H: Every projective f. g. module is f. p.
I want to show that if $P$ is a finitely generated (f.g.) projective module then $P$ is finitely presented (f.p.).
AI: As $P$ is f. g. we have an exact sequence $0\rightarrow Q\rightarrow A^n\rightarrow P\rightarrow 0$, $Q$ denoting the kernel of the map $A^n\rightarrow P$. As $P$ is projective, this exact sequence splits, $A^n\cong Q\oplus P$. The exact sequence $A^n\cong Q\oplus P\rightarrow A^n\rightarrow P\rightarrow 0$ shows $P$ to be f. p. (where $Q\oplus P\rightarrow A^n$ is the projection onto $Q$). |
H: How find this invertible matrix $C=\left[\begin{smallmatrix} A&B\\ B^T&0 \end{smallmatrix}\right]$
let matrix $A_{n\times n}$,and $\det(A)>0$, and the matrix $B_{n\times m}$,and such $rank(B)=m$,and let
$$C=\begin{bmatrix}
A&B\\
B^T&0
\end{bmatrix}$$
Find this Invertible matrix
$C^{-1}$
my try: I found this matrix Invertible matrix $C$,it must find $B^TAB$ Invertible matrix.But I can't
Thank you for your help
AI: Let
$ C^{-1} = \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} $ with appropriate sizes (i.e. $X$ is $n \times n$, $Y$ is $n \times m$, $Z$ is $m \times n$ and $W$ is $m \times m$). Then,
$C C^{-1} = \begin{bmatrix} A & B \\ B^T & 0 \end{bmatrix} \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} = \begin{bmatrix} I_n & 0 \\ 0 & I_m \end{bmatrix}$
Hence, we have the following equations:
$\begin{align}
AX + BZ &= I_n \\
AY + BW &= 0 \\
B^T X &= 0 \\
B^T Y &= I_m
\end{align}$
From the first and third equations:
$\begin{align}
X + A^{-1}BZ &= A^{-1}\\
B^T X + B^T A^{-1}BZ &= B^T A^{-1}\\
Z &= (B^T A^{-1}B)^{-1}B^T A^{-1}
\end{align}$
Now, if we put this in the second equation we get
$ X = A^{-1} - A^{-1}B (B^T A^{-1}B)^{-1}B^T A^{-1} $
From the second and forth equations:
$\begin{align}
Y + A^{-1}BW &= 0\\
B^T Y + B^T A^{-1}BW &= 0\\
B^T A^{-1}BW &= -I_m\\
W &= -(B^T A^{-1}B)^{-1}
\end{align}$
Now, if we put this in the second equation we get
$ Y = A^{-1}B(B^T A^{-1}B)^{-1} $
which is consistent with the forth equation.
Note that $A$ and $B^T A^{-1}B$ must be invertible. |
H: Question about tensor product of modules and ideals
Trying to prove some properties of tensor product with a given module, I came up with questions some of them I can't prove. Maybe it is also because Im not very used to work with tensor products and I think they can be tricky.
So let $A$ be a ring and let $M$ be an $A$-module. Let $I,J$ be any ideals of $A$. Now, is it true
$$JI \otimes_A M \subset J(I \otimes_A M)$$
maybe the above is an equality but I only need that...
Thank you!
AI: There is a canonical surjective homomorphism $JI \otimes_A M \to J(I \otimes_A M)$. It is injective when $M$ is flat (but not in general). |
H: Prove that f(x) can have any value between A and B
I need to prove the following statement:
Let f be a bounded and continuous function in the interval (a,b). if $\lim_{x\to a+}f(x)=A$ and $\lim_{x\to b-}f(x)=B$ and $A\neq B$ then f can get any value between and A and B in the interval (a,b).
What I did:
Let $ g(x) =
\begin{cases}
f(x), & \text{$a<x<b$} \\
A, & \text{x=a} \\
B, & \text{x=b} \\
\end{cases}$
. We can see that $g(x)$ is a continuous function in the the interval [a,b]. Therefore, I can use the Intermediate value theorem to prove that for any value between A and B, there exists $a<c<b$ such that g(c) is equal to that value, but in the interval (a,b) , $g(x)= f(x)$, so f(x) can get any value between A and B in the interval (a,b).
My Questions:
1)Can i define a function that is divided to 3 cases?
2)I didn't use the fact that f(x) is bounded, and that seems weird. what is wrong with my proof?
3) I am sure there must be a better way to prove it because the next statement i need to prove the same just for $A=\infty$ and $B=-\infty$(f(x) isn't bounded) and my trick won't work there, that's why I have no idea how to prove it.
AI: 1) Yes, you can, if $A$ and $B$ are finite. You have just extended $f$ to a continuous function on $[a,b]$
2) You have implicitly used it in assuming that $A$ and $B$ are finite. In fact, $|A|, |B|<+\infty$ if and only if $f$ is bounded (why?)
3) If $A=+\infty, B = -\infty$, then for any real $\alpha\in \mathbb{R}$, there is $a'>a, b'<$ such that $f(a') < \alpha < f(b')$. So you apply intermediate value theorem to $[a',b']$ |
H: Exact sequence in a category with zero morphisms
Let $C$ be a category with zero morphisms (equivalently, $\mathsf{Set}_*$-enriched), for example it could be a linear category. Then we can talk about kernels and cokernels of morphisms in $C$. I wonder if the following definition is already established and appears somewhere in the literature:
Definition: If $f : A \to B$ and $g : B \to C$ are morphisms, then $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ is called exact if $f$ is a kernel of $g$ and $g$ is a cokernel of $f$.
Here, "$0 \to A$" and "$C \to 0$" are just notation; I don't require that a zero object exists.
The definition is well-known when $C$ is abelian (and simplifies a bit in that special case).
AI: This appears as Definition 4.1.5 in [Borceux and Bourn, Mal'cev, protomodular, homological and semi-abelian categories]:
In a pointed category, a sequence of morphisms
$$1 \longrightarrow K \stackrel{k}{\longrightarrow} A \stackrel{q}{\longrightarrow} Q \longrightarrow 1$$
is a short exact sequence when $k = \ker q$ and $q = \operatorname{coker} k$.
Note that their definition of pointed category includes a zero object (denoted by $1$). |
H: Range of function
Given the function, $y=f(x)=\frac3{2-x^2}$, find its domain and range.
The domain is of course = $R - \{-\sqrt2,\sqrt2\}$.
However, the range I got was wrong(rather incomplete).
Rewriting the function for x in terms of y, I got $x=\pm \sqrt{\frac{2y-3}{y}} $
$\frac{2y-3}{y} \ge 0 \implies y\ge \frac32$
Therefore the $range(f) = [\frac32,\infty)$
However, the correct answer is $range(f)= [\frac32,\infty) \cup (-\infty,0)$
I dont understand why and how?
AI: Your mistake is that
$$\frac{2y-3}{y} \geq 0 \Leftrightarrow (y>0 \wedge y \geq \frac32) \vee (y<0 \wedge y \leq \frac32) \Leftrightarrow y \in [\frac32, \infty) \cup (-\infty, 0)$$
You fogot the $y<0$ case, in which the relation sign reverses when mulltiplying by $y$. |
H: Limit problem. Probability picking an item once out of M picks where p(pick) = 1/M+k.
What is the probability of picking an item one or more times if $p(pick) = \frac{1}{M} + \epsilon$ and we pick $M$ times?
$$p(pick~one~or~more) = 1-p(not~pick)^M = 1 - (1-(\frac{1}{M}+\epsilon))^M$$
I am interested in what happens when M goes to infinity.
$$\lim_{M \to \infty} (1-1/M-\epsilon)^M$$
If $\epsilon = 0$ then we can solve $$e^{\lim_{M \to \infty} M log(1-\frac{1}{M}) }$$
Which after subsitution $t = \frac{1}{M}$ can be handled with L'Hôptial's rule. But if that is not the case I have no idea. Computer caluclations for different $\epsilon$ show that $p(not~pick)^M$ goes to $0$.
AI: If $\epsilon>0$ then $\left(1-\frac1M-\epsilon\right)^M < (1-\epsilon)^M\to 0$.
And if $\epsilon=0$, $\lim \left(1-\frac1M\right)^M=\frac1e$ should be more or less well-known (and is close to some often used definition of $e$ in the first place). |
H: How to find the expresion such that its derivative must meet a certain condition
Suppose $a$ and $b$ are expressions in terms of the variable $x$.
We know:
$\begin{align}
a &= a \cdot\frac{b}{b} \\
&= \frac{ab}{b} \\
\end{align}
$
Is there a systematic way to find $b$ such that:
$\frac{db}{dx} = ab$
For example:
If $a=\sec(x)$, then $b=\tan(x)+sec(x)$ since $\frac{db}{dx} = sec(x)tan(x)+sec^2(x)=ab$
AI: Yes! In fact, this is how to solve first order linear differential equations:
$y'(x)+a(x)y(x)=f(x)$,
by multiplying both sides by b(x). Then if $b'(x)=a(x)b(x)$, we have
$\frac{d}{dx}(b(x)y(x))=f(x)b(x)$,
which can be solved for $y(x)$ by integrating.
To find the `integrating factor' $b(x)$, note that we can write the condition as
$\frac{d}{dx}\log(b(x))=a(x)$
which can be systematically solved by integrating. |
H: The arithmetic mean of $X$ when arithmetic mean of $X^2 = 29$.
Sorry if my question is a beginner because of my mathematical knowledge is low.
arithmetic mean is :
$$
\overline x=\dfrac{x_1+x_2+\cdots+x_n}n
$$
What method can solve it?
$$
\overline{X^2}=29\quad\Rightarrow\,\text{Not }\left(\overline X\right)^2 \\\,\\
\overline X=\boxed{?}\qquad\qquad\qquad\quad\,\,\,
$$
I got the square root but answer is not right.
For example :
$$
\overline X=\dfrac{1+2}2=\dfrac32=1.5 \\\,\\
\overline{X^2}=\dfrac{1+4}2=\dfrac52=2.5 \\\,\\
\sqrt{2.5}\approx1.581\ne1.5
$$
AI: An example of why it's impossible with the given information. Suppose $x_1 = \sqrt{29}$, $x_2 = \sqrt{29}$. Then $\overline{x^2} = 29$, and $\overline{x} = \sqrt{29}$.
Now suppose $x_1 = \sqrt{29}$, $x_2 = -\sqrt{29}$. Again $\overline{x^2} = 29$, but now $\overline{x} = 0$. So we have two sets with the same $\overline{x^2}$ but different $\overline{x}$. Therefore, without more data, it's impossible to get $\overline{x}$ from $\overline{x^2}$. |
H: A question about infinitary proofs and First Order Peano Arithmetic
In certain proof systems, infinite proofs are allowed; a common example is a version of Induction:
Given $\Sigma \vdash \phi(S^n 0)$ for all $n \in \Bbb N$, infer $\Sigma \vdash \forall x \phi(x)$.
What is the relation with the system described in S.C.Kleene, Mathematical Logic [1967], pag.206 - Axiom Schema 3 (Induction):
$A(0) \land \forall x(A(x) \rightarrow A(S(x))) \rightarrow A(x)$
In presence of (→ -E) may I "translate" the schema into a derived rule ?
If so, it does not look like an infinite rule.
AI: The usual axiom scheme of induction includes each axiom of the form
$$
(\phi(0) \land (\forall n)[\phi(n)\to\phi(n+1)]) \to (\forall n)\phi(n)
$$
If we were to write this as a derived inference rule, it would be
$$
\phi(0)\quad (\forall n)[\phi(n)\to\phi(n+1)]\quad \vdash (\forall n)\phi(n)
$$
This is not the same as the infinitary $\omega$-rule, which for each formula $\phi$ includes the following rule:
$$
\phi(0)\quad \phi(1)\quad \phi(2)\quad \cdots\quad \vdash (\forall n)\phi(n)
$$
Note that for the derived rule from induction, there are exactly two hypotheses, while for the $\omega$-rule there are infinitely may hypotheses. A proof from the $\omega$-rule allows for completely separate subproofs of each $\phi(i)$, and so a proof that uses the $\omega$-rule will necessarily be infinite. |
H: Lower triangular matrices
Is the inverse of an invertible and lower triangular matrix still both lower triangular and invertible?
AI: Yes. The lower triangular matrix $L$ is invertible if and only if its determinant is nonzero, and since its determinant is the product of its diagonal entries, if and only if its diagonal entries are nonzero.
If we show the sum and product of lower triangular matrices is again lower triangular, then the proof goes quickly from there.
The sum of two lower triangular matrices is pretty evident, so let's jump into the product of two $n\times n$ lower triangular matrices $L$ and $M$. Consider an entry of the product above the main diagonal, i.e. row $i$ and column $j$ with $i \lt j$, obtained by multiplying the $i$th row of $L$ and the $j$th column of $M$:
$$ \sum_{k=1}^n L(i,k) M(k,j) $$
Notice that for $k \gt i$ the factor $L(i,k)=0$, while for $k \lt j$, the factor $M(k,j)=0$. Use the fact that $i \lt j$ and it turns out all the summands in the above expression are zero.
Now one way to finish the proof is by appealing to Cayley-Hamilton for an expression of $L^{-1}$ as a polynomial in $L$. From what we just showed, any polynomial in lower triangular $L$ is again lower triangular, and Cayley-Hamilton for invertible $L$ gives:
$$ L (L^{n-1} + \ldots ) = \pm \det(L) $$
$$ L^{-1} = \frac{1}{\pm \det(L)} (L^{n-1} + \ldots ) $$
showing the inverse of $L$ is indeed a polynomial in $L$. |
H: $\frac{\mathrm d}{\mathrm dt} \exp\{\lambda((q+pe^t)-1)\}$
$$\frac{\mathrm d}{\mathrm dt} \exp\{\lambda((q+pe^t)-1)\}$$
How do I do this? Do I use the chain rule?
$$= \exp\{\lambda((q+pe^t)-1)\} \frac{\mathrm d}{\mathrm dt} (\lambda((q+pe^t)-1)) \\
= \exp\{\lambda((q+pe^t)-1)\} \lambda pe^t$$
But the solution appears to be without the $e^t$ term? (see last line of image below)
We have $$\begin{align}M_X(t)&=\Bbb E[e^{Xt}]\\
&=\Bbb E[\Bbb E[e^{Xt}|Y]]\\
&=\Bbb E[(q+pe^t)^Y]\\
&=\sum_{y=0}^\infty(q+pe^t)^y\dfrac{\lambda^ye^{-\lambda}}{y!}\\
&=\exp\{\lambda((q+pe^t)-1)\}.
\end{align}$$Differentiating$$\dfrac{\mathrm dM_x(t)}{\mathrm dt}=\exp\{\lambda((q+pe^t)-1)\}\lambda p.$$
AI: How do I do this? Do I use the chain rule?
Yes, exactly, you use the chain rule. And your result is correct, the official solution has mistakenly forgotten the $e^t$ term from the outer $\exp$'s argument. |
H: Intuitive Understanding of Projective Modules
I was wondering if anyone could give me any sort of intuitive explanation of what a projective module is or a useful way to think about them. I know the definition(s) in terms of lifting, split exact sequence, direct summand but I have no intuitive understanding of what it means for a module to be projective?
Thanks for any help
AI: A finitely generated projective $A$ module $M$ can be thought as a locally free module, in the sense that each $M_{\mathfrak p}$ is free for a prime ideal $\mathfrak{p}$ (or that it´s free at each point in $Spec(A)$). This intuition come from Serre-Swan theorem, which states roughly that the sections functor is an equivalence between projective modules and complex vector bundles (furthermore vector bundles are locally trivial, and trivial bundles correspond to free modules). Note that a finitely generated projective module such that $M_{\mathfrak p}$ is free and have the same rank at each localization need not to be free (you cannot glue everything as in sheaf to get a free thing).
I think that it´s useful to cite too that finitely projective modules over local rings are free (this fact is know as Kaplansky theorem) and that over Von Neumann regular rings you can glue everything to get a free module. |
H: Conditional Expectation Probability
Walking in the street for 10 minutes, the number of people you cross has a
Poisson distribution with mean λ. Suppose that each has a cold with probability p. During
those 10 minutes, what is the expected number of walkers you cross who have a cold?
Attempt:
Let X~Poi(λ) where x= # of people you cross and y=# of people with cold
Find E[X|Y]= sum from n to infinite of x f(x, y) / p
AI: Continue with your notation $X \sim \operatorname{Poi}(\lambda)$.
Intuitive Answer: Note that on average you will enocounter $1/\lambda$ people in $10$ minutes (since $\Bbb{E}[X] = 1/\lambda$). Each of those you encounter has a cold with probability $p$. So, intuitively we should believe that the expected number you encounter with a cold is $(1/\lambda)p = p/\lambda$.
More Rigorous Argument: Let $Y$ be the number of people you encounter in $10$ minutes who have colds. Then, let's figure out $\Bbb{E}[Y|X]$. By the setup of the problem, given the event $X = n$, then $Y$ is a $\operatorname{Bin}(n,p)$ (why!?). Therefore $$\Bbb{E}[Y\mid X=n] = \Bbb{E}[\operatorname{Bin}(n,p)] = np$$
Hence $\Bbb{E}[Y\mid X] = Xp$ (why!?). Finally, since $\Bbb{E}[Y] = \Bbb{E}\big[\Bbb{E}[Y\mid X]\big]$ (why?!) you find
$$
\Bbb{E}[Y] = \Bbb{E}\big[\Bbb{E}[Y\mid X]\big] = \Bbb{E}[Xp] = p \Bbb{E}[X] = p(1/\lambda) = p/\lambda.
$$
So, once you fill in the details of the rigorous argument, your intuition should match the mathematics set up for this situation. |
H: If a function is smooth is 1 over the function also smooth
If $f(x):\mathbb{R}\rightarrow\mathbb{C}$ is $C^\infty$-smooth. Is $1/f(x)$ also $C^\infty$-smooth? $f(x)\neq0$
AI: If $f$ is differentiable and non-zero at some point $a$, then $1/f$ is differentiable at $a$, and $(1/f)'=-f'/f^2$. This is a "base case" of an induction argument for the following statement:
$1/f$ is $n$ times differentiable, and $(1/f)^{(n)}$ equals a polynomial in the functions $f,f',f'',\ldots,f^{(n-1)}$ divided by a power of $f$.
The induction step is just applying the quotient rule.
Have fun! |
H: The normal approximation of Poisson distribution
(I've read the related questions here but found no satisfying answer, as I would prefer a rigorous proof for this because this is a homework problem)
Prove: If $X_\alpha$ follows the Poisson distribution $\pi(\alpha)$, then
$$\lim_{\alpha\rightarrow\infty}P\{\frac{X_\alpha-\alpha}{\sqrt{\alpha}} \leq u \} = \Phi(u)$$
where $\Phi(u)$ is the cdf of normal distribution $N(0,1)$
Hint: use the Laplace transform $E(e^{-\lambda(X_\alpha-\alpha)/\sqrt{\alpha}})$, show that as $\alpha\rightarrow\infty$ it converges to $e^{\lambda^2/2}$
I did the transform but failed to sum the series(which is essentially doing nothing)
Here's what I got:
$$g(\lambda)=\sum_{n=0}^{\infty} \frac{e^{-\alpha}}{n!}\alpha^n e^{-\frac{\lambda(n-\alpha)}{\sqrt{\alpha}}}$$
and $\lim_{\alpha\rightarrow\infty} g(\lambda)=e^{-\lambda^2}$ is what I'm trying to arrive at. I tried L'Hospital only to find that the result is identical to the original ratio.
AI: Let $X_{\alpha}$ Poisson $\pi(\alpha)$, for $\alpha = 1, 2, \ldots$
The probability mass function of $X_{\alpha}$ is
$$f_{X_{\alpha}}(x)=\frac{{\alpha}^x\operatorname{e}^{-x}}{x!} \qquad \alpha = 1, 2, \ldots$$
The moment generating function of $X_{\alpha}$ is
$$
M_{X_{\alpha}}=\Bbb{E}\left(\operatorname{e}^{tX_{\alpha}}\right)=\operatorname{e}^{\alpha(\operatorname{e}^{t}-1)}\qquad t\in(-1,1)
$$
Now consider a “standardized” Poisson random variable $Z_{\alpha}=\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}$
which has limiting moment generating function
$$
\begin{align}
\lim_{\alpha\to\infty}M_{Z_{\alpha}}&=
\lim_{\alpha\to\infty}\Bbb{E}\left(\exp{(tZ_{\alpha})}\right)\\
&=\lim_{\alpha\to\infty}\Bbb{E}\left(\exp{\left(t\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}\right)}\right)\\
&=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\Bbb{E}\left(\exp{\left(\frac{tX_{\alpha}}{\sqrt{\alpha}}\right)}\right)\\
&=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\exp\left(\alpha(\operatorname{e}^{t/\sqrt\alpha}-1)\right)\\
&=\lim_{\alpha\to\infty}\exp\left(-t\sqrt\alpha +\alpha\left[t\alpha^{-1/2}+\frac{t^2\alpha^{-1}}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right]\right)\\
&=\lim_{\alpha\to\infty}\exp\left(\frac{t^2}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right)\\
&=\exp\left(\frac{t^2}{2}\right)
\end{align}
$$
by using the moment generating function of a Poisson random variable and expanding the
exponential function as a Taylor series. This can be recognized as the moment generating function of a standard normal random variable. This implies that the associated unstandardized random variable $X_{\alpha}$ has a limiting distribution that is normal with mean $\alpha$ and variance $\alpha$. |
H: How to find this limit:
The question is to find this limit without using H.R (Hopital Rule):
$$\lim_{x\to 1}\frac{x^4 -1}{x^3 -1}$$
So this will be $\frac{0}{0}$
Which is indetermined but using H.R we find it is
$\frac{4}{3}$
But I need another method to get the answer so how could I get it?
AI: You should decompose the two polynomial and the result will be very easy:
$$
\lim_{x\rightarrow 1} \frac{x^4-1}{x^3-1}=\lim_{x\rightarrow 1} \frac{(x-1)(x+1)(x^2+1)}{(x-1)(x^2+x+1)}=\lim_{x\rightarrow 1} \frac{(x+1)(x^2+1)}{(x^2+x+1)}=\frac{4}{3}
$$ |
H: Triple Integral with bounds in first octant
I am really confused on how to get my integrating function because I don't know, even after graphing, how the tetrahedron intersects the x-y-z axis.
I am supposed to find the triple integral for the volume of the tetrahedron cut from the
first octant by the plane $6x + 3y + 2z = 6$.
I have found the bounds of integration by isolating $x,y,z$ in the tetrahedron equation. So I know how the bounds will be in any order of integration that I want. I am now stuck on what the integrating function will be.
So for example, if I was integrating w.r.t dxdydz then, according to the bounds I found through a numerical method, should be:
$$\int_{z=0}^3\int_{y=0}^{\frac{6-2z}{3}} \int_{x=0}^{\frac {6-3y-2z}{6}} dx dy dz$$
Now these are the bounds I have found. How do I get the integrating equation? Would the integrating equation be x for the first definite integral?
so something like this?
$$x \rvert _{0}^{\frac{6-3y-2z}{6}}$$
Is that right?
So then to find the volume this would be the whole equation with the bounds.
{0,3} {0, (6-2z)/3 } {0, (6-3y-2z)/6} dx dy dz
As you can see in the graph above, it is bounded by the given plane so are my bounds correct? is the function a constant "1" ?
AI: You have a plane $P:~6x+3y+2z=6$ which cut the axes in the first octant as you see through the below plot. $P$, clearly, intersects $z=0$ in a line: $$z=0\to 6x+3y+2\times 0=6\to 6x+3y=4$$ I don't want to tell you that all triple integrals will start from this point that we did above, but we do for many certain triples. From the line above we can get the proper limits for $x$ and $y$:
$$y=0\to x=2/3\Longrightarrow x|_0^{2/3}\\ y=\frac{4-6x}{3}=4/3-2x\Longrightarrow y|_0^{\frac{4}3-2x}$$ Therefore we have: $$V=\large{\int_x\int_y\int_0^{\frac{6-6x-3y}{2}}}dzdydx$$ |
H: How to find the limit of an algebraic function
The question is to find this limit:
$$\lim_{x\to\infty}\frac{2x^\frac{5}{3}- x ^\frac{1}{3}+7}{x^\frac{8}{5} +3x + \sqrt{x}}$$
I need any hint to help since I tried so much and couldn't solve it.
AI: As $x$ tends to infinity, the dominant term of polynomial is the one with the largest power. Hence
$$\lim_{x\to\infty}\frac{2x^\frac{5}{3}- x ^\frac{1}{3}+7}{x^\frac{8}{5} +3x + \sqrt{x}}=\lim_{x\to\infty}\frac{2x^\frac53}{x^\frac85}\to\infty$$ |
H: Unable to understand combination of quantifiers and set notation
I know what universal and existential quantifiers are but following is confusing,may be its comibination of set notation and quantifers. What does the following statement means?
∀xP(x)
AI: For all $x$, the statement $P(x)$ is true.
It helps to separate things out: $\forall x$, $P(x)$.
Edit to answer your question: $P(x)$ is just a statement. It is just some property of the $x$s. Something we say about $x$. For example, $P(x)$ could be $x>0$ or "$x$ is blue", or $x \in \mathbb{Z}$.
So $P(x)$ is not a set, is is just a statement about the $x$s. However, we can use $P(x)$ to make a set. Suppose $P(x)$ means $x>0$. Then we could have $$A = \{x \in \mathbb{R} \mid x > 0\}$$ Which reads "A is the set of all real numbers $x$ such that $x$ is greater than zero." |
H: How to solve this integral: $\int_{-1}^{1} x^k (1-x^2)^{(n/2)-2} \, dx$
How to solve this integral step by step:
$$\int_{-1}^{1} (x^k) (1-x^2)^{(n/2)-2}dx=??? $$
In my text book, it shows the result like below:
$$\int_{-1}^{1} (x^k) (1-x^2)^{(n/2)-2}dx= \frac{(x^{(1+k)} ~_2F_1((1+k)/2, ~2-n/2, ~(3+k)/2, ~x^2))}{(1+k)} ; for ~x=-1~to~1~$$
$$=\frac{((1+(-1)^k) ~\Gamma(\frac{1+k}{2}) ~\Gamma(\frac{-1+n}{2})}{2Γ( \frac{1}{2}(-1+k+n))}$$
Note: $~_2F_1((1+k)/2, ~2-n/2, ~(3+k)/2, ~x^2)$ is a hypergeometric function, is there a simple way to find and solve that integral?
Thanks in advanced
AI: Following Daniel's comments:
Case $1$ --- $\;k\ge 0\;$ is odd: In this case, the integrand $\;x^k(1-x^2)^{\frac n2-2}\;$ is an odd continuous function in a symmetric interval and thus the integral equals zero.
Case $2$ --- $\;k=2m\ge 0\;$ is even: in this case our integral equals twice the integral over half the interval and we can substitute:
$$u:=x^2\implies dx=\frac{du}{2\sqrt u}\implies$$
$$\int\limits_{-1}^1 x^k(1-x^2)^{\frac n2-2}dx=2\int\limits_0^2 (x^2)^m(1-x^2)^{\frac n2-2}dx=$$
$$2\int\limits_0^1u^m(1-u)^{\frac n2-2}\frac{du}{2u^{1/2}}=\int\limits_0^1u^{m-1/2}(1-u)^{\frac n2-2}du=:B\left(\frac{k+1}2\,,\,\color{}{\frac{n-2}2}\right)$$
and, for example using the relation between gamma and beta functions, we have
$$B\left(\frac{k+1}2\,,\,\frac{n-2}2\right)=\frac{\Gamma\left(\frac{k+1}2\right)\Gamma\left(\frac{n-2}2\right)}{\Gamma\left(\frac{k+n-1}2\right)}$$ |
H: Conditional Probability/ Bayes' Theorem puzzle
I always believed that problems on conditional probability could be solved with common logic without using Bayes' theorem (because I cannot understand Bayes' theorem intuitively and I didn't bother because I knew another way).
But this problem gives me varying answers with general logic and Bayes' Theorem.
A test for a sickness is 99% reliable. That is, 99% of the healthy people test negative, and 99% of sick people test positive.Given that 1% of the entire population is actually sick, what are the chances that you are actually sick if you test positive?
Here is my solution without using Bayes' theorem.
$$P(being\ sick\ |\ test\ positive) = \frac{Number\ of\ people\ who\ are\ actually\ sick}{Number\ of\ people\ who\ test\ positive} $$
$$So,P(being\ sick\ |\ test\ positive)= \frac{1\%\ of\ \ the\ total\ population}{1\%\ of\ healthy\ people+99\%\ of\ sick\ people} $$
$$Let\ the\ total\ population\ be\ x. So,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
$$P=\frac{0.01x}{0.01*0.99x+0.99*0.01x}$$
$$P=\frac{1}{0.99+0.99}=\frac{1}{1.98}$$
But, when you use Bayes' theorem, you get P=1/2. Why is the denominator off by 0.02?
Either I'm wrong (but I really can't see why) or there is some kind of an approximation in Bayes' theorem (highly impossible). Can someone please tell me what are the other factors that Bayes' theorem considers but my solution doesn't? I would also be extremely grateful if someone could explain the theorem can be understood intuitively.
AI: Your numerator should be those people that get sick AND test positive, or $0.99\times 0.01x$.
In some sense, your universe is those people who test positive, which is $0.01x$. That includes healthy people with false positives, and sick people with true positives. In this problem those two populations are of equal size, hence $\frac{1}{2}$. |
H: What's the difference between isomorphism and homeomorphism?
I think that they are similar (or same), but I am not sure.
Can anyone explain the difference between isomorphism and homeomorphism?
AI: Homomorphism - an algebraical term for a function preserving some algebraic operations. For a group homomorphism $\phi$ we have $\phi(ab)=\phi(a)\phi(b)$ and $\phi(1)=1$, for a ring homomorphism we have additionally $\phi(a+b) = \phi(a) + \phi(b)$ and for a vector-space homomorphism also $\phi(r\cdot a)=r\cdot\phi(a)$, where $r$ is a scalar and $a$ is a vector.
Isomorphism (in a narrow/algebraic sense) - a homomorphism which is 1-1 and onto. In other words: a homomorphism which has an inverse.
However, homEomorphism is a topological term - it is a continuous function, having a continuous inverse.
In the category theory one defines a notion of a morphism (specific for each category) and then an isomorphism is defined as a morphism having an inverse, which is also a morphism.
With such an approach, morphisms in the category of groups are group homomorphisms and isomorphisms in this category are just group isomorphisms. Similarly for rings, vector spaces etc.
In the category of topological spaces, morphisms are continuous functions, and isomorphisms are homeomorphisms.
Extra remark: A fundamental difference between algebra and topology is that in algebra any morphism (homomorphism) which is 1-1 and onto is an isomorphism - i.e., its inverse is a morphism. In topology it is not true: there are continuous and bijective functions whose inverses are not continuous. That's (one of the reasons) why we like compact Hausdorff topological spaces: for them inverses are always continuous, just like in algebra inverses of homomorphisms are homomorphisms. |
H: Cauchy integral formula for derivatives intuition
A nice way to remember a formula is to connect it to something you already know. For example, to remember the Cauchy integral formula, I remember that $f(z_0)=\frac{1}{2\pi i}\int_\Gamma\frac{f(z)}{z-z_0}\,dz$ because if you write out the Laurent series for $\frac{f(z)}{z-z_0}$ you're likely to get something like:
$$ \frac{f(z)}{z-z_0}=\frac{a_{-1}}{z-z_0}+a_0 +a_1(z-z_0)+\dots \,\,\,\,(1)$$
If you take the integral of the right hand side of (1), you're going to be left with $a_{-1} 2\pi i$, but we were really looking for the behavior of $f(z)$ at $z_0$, and since we used an integral to find the behavior we got $2\pi i$ as an artifact, so we're just going to divide our answer by $2\pi i$ and be left with $f(z_0)$ a.k.a $Res(f;z_o).$
My questions:
is there a similar way to think about $f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\Gamma\frac{f(z)}{(z-z_0)^{n+1}}\,dz$? I've tried writing out a Laurent series for $\frac{f(z)}{(z-z_0)^{n+1}}\,dz$, but it didn't get me anywhere.
if $f(z)=\frac{g(z)}{(z-z_0)^m}$, is $g(z_0)=a_{-1}$ ? (where $a_{-1}$ is from its Laurent series)
Obviously there is a strong connection between the singularities of a function, its Laurent series, residue theory and the Cauchy integral formulas. After a semester of introductory complex analysis, I'm struggling to place everything in a coherent narrative. Any help would be greatly appreciated.
Edit: Based on the answers I've formed a new narrative to remember the formulas. I thought I'd share it in case someone finds it helpful.
Let's say we have a function:
$$ \frac{ f(z) }{ (z-z_0)^{n+1} } $$
We want to study the behavior of $f^{(n)}(z)$ at $z=z_0$. We know that a Taylor coefficient is on the form $a_n=\frac{f^{(n)}(z_0)}{n!}$, so we can find $f^{(n)}(z_0)$ by manipulating the Taylor series.
So expand to Taylor form:
$$ \frac{ f(z) }{ (z-z_0)^{n+1} }=\frac{c_0}{(z-z_0)^{n+1}}+\frac{c_{1}}{(z-z_0)^{n}} +\dots+\frac{c_{k}}{(z-z_0)^{1}}+c_{k+1}+c_{k+2}(z-z_0)+\dots $$
We see that we can recover $c_k=\frac{f^{(k)}(z_0)}{n!}$ by integrating the equation above. But since we used an integral to isolate $c_k$, we get $2\pi i$ as an artifact, and since we used the Taylor series we also got $\frac{1}{n!}$ as an artifact.
We can think of the forumla $f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\Gamma\frac{f(z)}{(z-z_0)^{n+1}}$ a formula for recovering $f^{(n)}(z_0)$ from a Taylor series: Use $\int_\Gamma\frac{f(z)}{(z-z_0)^{(n+1)}}$ to isolate the Taylor coefficient and remove the extraneous information by multiplying by $\frac{n!}{2\pi i}$.
When $n=0$ it's just the regular Cauchy integral formula, of course; $\frac{0!}{2\pi i}=\frac{1}{2\pi i}$.
AI: Expand $f$ into its Taylor series,
$$f(z) = \sum_{k=0}^\infty c_k(z-z_0)^k.$$
Then you have
$$\frac{f(z)}{(z-z_0)^{n+1}} = \sum_{k=0}^\infty c_k (z-z_0)^{k-n-1}.$$
When you integrate termwise, the only term with a nonzero integral is the one with the exponent $k-n-1 = -1$, that is, $k = n$, so the integral is $2\pi i c_n$. But $$c_k = \frac{f^{(k)}(z_0)}{k!}$$ for all $k$. |
H: Please help me with this Integral
Calculate the integral (complex):
$$\oint_{D(0,1)}\overline ze^z \mathrm dz$$
While $D(0,1)$ is the unit circle.
AI: Collecting the hints from the comments in an answer:
On the unit circle, we have $\overline{z} = \dfrac{1}{z}$. Hence the integral can also be written
$$\int_{\lvert z\rvert = 1} \frac{e^z}{z}\,dz,$$
which by the Cauchy integral formula evaluates to
$$2\pi i e^{0} = 2\pi i.$$ |
H: Show that $|\sin\frac{1}{n^2}|<\frac{1}{n^2}$, $n=0, 1, 2, \dots$
As part of showing that
$$
\sum_{n=1}^\infty \left|\sin\left(\frac{1}{n^2}\right)\right|
$$
converges, I ended up with trying to show that
$$
\left|\sin\left(\frac{1}{n^2}\right)\right|<\frac{1}{n^2}, \quad n=1, 2, 3,\dots
$$
since I know that the sum of the right hand side converges. But I can't show this. I've tried searching but I haven't been able to find anything.
What I've tried is that firstly, the absolute values are not needed since $\sin x>0$ if $0<x<1$. I rearranged a little bit:
$$
\sin\left(\frac{1}{n^2}\right)-\frac{1}{n^2}<0
$$
and the derivative is
$$
\frac{2}{n^3}\left(1-\cos\left(\frac{1}{n^2}\right)\right)> 0
$$
so my idea of showing that it is decreasing and negative for the first $n$ wouldn't work.
How can I show this? Help is appreciated.
Edit: Maybe I should add that I'm not completely sure it is true, but I tried it numerically and it seems like it.
AI: Consider $f(x)=\sin x -x\Rightarrow f'(x)=\cos x-1 < 0 \text { for x > 0 } \Rightarrow f(x)\text{ is strictly decreasing function}$
But then, $f(0)=0$ which would imply $f(x)< 0 $ for $x > 0$ i.e., $\sin x < x$ for $x >0$
Thus, $\sin \big(\frac{1}{n^2}\big)< \frac{1}{n^2}$
Does this imply $|\sin\big(\frac{1}{n^2}\big)|< \frac{1}{n^2}$ |
H: Prove that there is no homomorphism from $\mathbb{Z}_{8} \oplus \mathbb{Z}_{2}$ ont0 $\mathbb{Z}_{4} \oplus \mathbb{Z}_{4}$
Prove that there is no homomorphism from $\mathbb{Z}_{8} \oplus \mathbb{Z}_{2}$ ont0 $\mathbb{Z}_{4} \oplus \mathbb{Z}_{4}$.
My idea for the proof : Let $\phi$ be such homomorphism. Since Ker $\phi$ is a subgroup of $\mathbb{Z}_{8} \oplus \mathbb{Z}_{2}$ , then i must find all possible subgroups of it and then prove that $(\mathbb{Z}_{8} \oplus \mathbb{Z}_{2})/$Ker$\:\phi $ is not isomorphic with $\mathbb{Z}_{4} \oplus \mathbb{Z}_{4}$ where Ker $\phi$ can be any of the subgroup. To prove it isnt isomorphic is by finding elements in $\mathbb{Z}_{8} \oplus \mathbb{Z}_{2}$ that has order such that no elements in $\mathbb{Z}_{4} \oplus \mathbb{Z}_{4}$ have that order.
Is there any better way..?
AI: A surjective function between two finite sets of the same cardinality is also injective, so your homomorphism would be an isomorphism. This is not possible, because the first group has an element of order $8$ and the second one doesn't. |
H: if the matrix such $B-A,A$ is Positive-semidefinite,then $\sqrt{B}-\sqrt{A}$ is Positive-semidefinite
Question:
let the matrix $A,B$ such $B-A,A$ is Positive-semidefinite
show that:
$\sqrt{B}-\sqrt{A}$ is Positive-semidefinite
maybe The general is true?
question 2:
(2)$\sqrt[k]{B}-\sqrt[k]{A}$ is Positive-semidefinite
This problem is very nice,because we are all know this
if $$x\ge y\ge 0$$,then we have
$$\sqrt{x}\ge \sqrt{y}$$
But in matrix,then this is also true,But I can't prove it.Thank you
AI: Two proofs: (thanks for Julien for helping me fix the first proof)
First proof:
First assume $A,B>0$. Suppose $B \geq A$.
$$B-A \geq 0 \\ A^{-1/2}BA^{-1/2} - I \geq 0, $$
since $M\geq 0 \Rightarrow N^{*}MN \geq 0$ for $N$ invertible. This means all eigenvalues of $A^{-1/2}BA^{-1/2}$ are strictly greater than 1. Therefore for all $\|x\|=1$
$$\langle A^{-1/2}BA^{-1/2}x,x \rangle \geq 1 \\ \langle BA^{-1/2}x, A^{-1/2}x \rangle \geq 1 \\ \langle B^{1/2}A^{-1/2}x, B^{1/2}A^{-1/2}x \rangle = \|B^{1/2}A^{-1/2} x\|^2\geq1 \\ \|B^{1/2}A^{-1/2} x\|\geq1$$
so all eigenvalues of $B^{1/2}A^{-1/2}$ are strictly greater than 1 in modulus. But by a similarity transformation, these are also the eigenvalues of $A^{-1/4}B^{1/2}A^{-1/4}$, and those are all positive since $A^{-1/4}B^{1/2}A^{-1/4}$ is positive (being of the form $N^* M N$ for $M>0$ and $N$ invertible). So
$$A^{-1/4}B^{1/2}A^{-1/4}-I \geq0\\B^{1/2}-A^{1/2}\geq0,$$
finishing the case $A,B>0$. To extend to $A,B\geq 0$, let $A_\epsilon:=A+\epsilon I$, and similarly for $B_\epsilon$. Then $B\geq A \Rightarrow B_\epsilon \geq A_\epsilon \Rightarrow (B_\epsilon)^{1/2}\geq (A_{\epsilon})^{1/2} \Rightarrow B^{1/2}\geq A^{1/2}$, if you believe that $\epsilon \mapsto M_\epsilon^{1/2}-N_{\epsilon}^{1/2}$ is right-continuous in the eigenvalues at zero (follows from $\epsilon \mapsto M_\epsilon^{1/2}$ being right-continuous in the eigenvalues).
Second Proof (from Lax's book Linear Algebra and its Applications): This relies on the fact that if $A(t)$ is a matrix-valued function whose derivative is everywhere positive definite, then $t_2 > t_1 \Rightarrow A(t_2) > A(t_1)$ (unstrict inequality corresponds to positive semidefiniteness). Also it uses the lemma that if $A$ is positive definite and $AB+BA$ is positive definite, then $B$ is positive definite. And it uses that positive semidefinite matrices are a convex subset of $\mathbb{C}^{n^2}$ (all these proofs can be found in Lax's book in the section on matrix inequalities).
Let $B\geq A \geq 0$. Define a function $M(t) = A + t(B-A)$. It is positive on $[0,1]$ by convexity (as a matter of fact, it is positive for all $t\geq0$, I believe). Further, it has a positive semidefinite matrix as its derivative. Also $\sqrt{M(t)}$ is positive semidefinite. Define $R(t) = \sqrt{M(t)}$. Then $R^2 = M$, so
$$R\dot{R} + \dot{R} R = M'(t) \geq 0.$$
$R$ is positive and $R\dot{R} + \dot{R} R$ is positive, so $\dot{R}$ is positive. Therefore $R$ is a non-decreasing function of $t$ and $R(1) \geq R(0)$. QED |
H: Average value of double integral
I am trying to work on this homework problem but I am lost. The only thing I know to do here is divide the double integral by 8 or multiply by 1/8 since area = l*w and 4*2=8.
But now, please help me understand how I can convert the C to x and y and get the bounds for integration. I want to understand it so if you can explain it then I would be happy.
AI: HINT: The image isn’t entirely clear, but it appears that the formula is $$C(x,y)=100-15d^2\;.$$ Use the Pythagorean theorem: $d^2=x^2+y^2$, so $$C(x,y)=100-15x^2-15y^2\;.$$ |
H: The question related to a regular surface.
Prove that an equation of the form $f(x,y,z)=c$ determines a regular surface if $f$, defined on some open subset $S$ of $\mathbb{R}^3$, is smooth and $\nabla f\neq 0$ everywhere in $S$.
I know these following definitions:
(1) a smooth surface $S$ is regular if for any surface patch $\sigma$ of $S$, we have $\vert\vert \sigma_u \times \sigma_v\vert\vert \not=0$
(2) a topological surface $S$ in $\Bbb R^3$ issmooth if all coordinate charts are smooth.
(3) $f:U\to \Bbb R^m$ is smooth if all partial derivatives of $f$ of any order exist on $U$.
Please help me solving the question. Thank you.
AI: Hint: use the implicit function theorem.
Added: the implicit function theorem has many (seemingly) different statements, but the gist of it is that the level sets of a regular (i.e., one which admits a full-rank differential) $C^1$ (or $C^\infty$) map are locally graphs of a $C^1$ (or $C^\infty$, respectively) function. This function is precisely the patch (which I assume is a different name for a local coordinate system) you require, and if I may presume, I urge you to verify that you are comfortable with this 'language'.
More precisely, in this context, the implicit function theorem affords us for each $(x_0,y_0,z_0)\in S$ such that $f(x_0,y_0,z_0)=c$ and (w.l.o.g.) $\frac{\partial f}{\partial x}(x_0,y_0,z_0)\neq 0$, a neighborhood $U$ of $(y_0,z_0)$ and $V$ of $x_0$ such that $\frac{\partial f}{\partial x}$ doesn't vanish in $V$, $V\times U\subset S$, and a $C^1$ function $g:U\to V$ such that for all $(x,y,z)\in V\times U$ one has $f(x,y,z)=c$ if and only if $x=g(y,z)$.
Moreover, we have
$$\frac{\partial g}{\partial y}(y,z) = -\left(\frac{\partial f}{\partial x}(g(y,z),y,z)\right)^{-1}\frac{\partial f}{\partial y}(g(y,z),y,z),$$
and similarly for $z$, which helps us to easily verify that $g$ is also $C^\infty$, whenever $f$ is.
This is precisely what we need to note that $\varphi:U\to S$ defined $\varphi(y,z) = (g(y,z),y,z)$ is a patch for the surface $f=c$ in a neighborhood $V\times U$ of $(x_0,y_0,z_0)$, and that
$$\varphi_y(y,z)\times\varphi_z(y,z) = \left(g_y(y,z),1,0\right)\times\left(g_z(y,z),0,1\right) =\\ = (1,-g_y(y,z),-g_z(y,z)) =\\ = \left(\frac{\partial f}{\partial x}(g(y,z),y,z)\right)^{-1}\nabla f(g(y,z),y,z)\neq \overline{0}$$ |
H: Throwing balls onto cliffs
A man throws a ball from the ground towards the top of a cliff. Suppose that as soon as the ball reaches its maximum height, the ball lands on top of the cliff.
Let time t=0 be the moment the ball is released from the man's hand. For simplicity's sake, let the ball reach a maximum height of 4 at t=2. Assuming "gravity" is the only force acting on the object once released, the ball's movement can be described by following function:
$$y(t)=\begin {cases}-(t-2)^2+4 &\text{if } 0<t<2\\ 4 &\text{if }t\ge2\end{cases}$$
This seems all good; y is a continuous function, and
$$y'(t)\begin {cases}-2(t-2) &\text{if } 0<t<2\\ 0 &\text{if }t\ge2\end{cases}$$
is continuous. But
$$y''(t)\begin {cases}-2 &\text{if } 0<t<2\\ 0 &\text{if }t\ge2\end{cases}$$
is clearly not continuous. This means that the ball would have an infinite change in its acceleration. So this movement is not possible. Are you not able to throw a ball onto a cliff so that it reaches its max height when it reaches the top of the cliff?
AI: As in your drawing, you can think of the ball as coming to rest on the top of the cliff. To the first approximation, there is no problem with an infinite change in the acceleration. It lands on the cliff and stops. At another approximation, you need to supply upward force on the ball to stop it from falling. That force has to come from slightly compressing the ground the ball rests on and the bottom of the ball, so the ball will descend a very small distance from its peak to establish these forces.
Similar issues occur if you think of a car starting to turn instantaneously. On one level, you can go from zero transverse acceleration to some amount instantaneously. On the next level, you need to rotate the steering wheel, transmit that to the tires, get them to deform, and so on to initiate the turn. None of this happens instantaneously, but it is close enough. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.