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H: Finding points of intersection between a line and a curve
I know this is easy but I am unsure about how to answer a multivariable-calculus question as a volunteer tutor, I left a couple of students with a method but a lot of others needed help today so I had to move on, I am hoping someone can confirm my method or reveal a correct method.
Question: Find the points of intersection between the parametrized line:
$$x = 1 + t$$
$$y = 3 + 2t$$
$$z = 1 + t$$
And the curve: $z = x^2 + 2y^2$
I noticed there are 4 equations and 4 unknowns. So was it correct to tell the students that it appears to be a substitution problem, and by simply plugging in the given values for $x, y, z$, we can solve for $t$. And it appears for $t$ there are two values for, which makes sense if there are two intersections. By plugging in the value of $t$ for the line one can surely find points. I assume the work can also be checked by using the equation of the curve for such points as well to determine they work in the equation of the curve.
I only have taken multivariable last term, so I am still getting comfortable with the idea here and I think I am the only tutor that can do it where I work, so verification is much appreciated.
AI: The method is fine: a point of intersection must satisfy the equation $$1+t=(1+t)^2+2(3+2t)^2\;.$$ However, the line and the curve do not intersect: one finds that
$$t=\frac{-25\pm\sqrt{-23}}{18}\;.$$
In this case one might also notice that on the line we have $z=x$ and $y=2x+1$, so that we can instead find points of intersection in terms of $x$ by solving $x=x^2+(2x+1)^2$.
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H: How can I solve this Initial Value Problem using the Euler method?
My Problem is this given Initial Value Problem: $$y^{\prime}=\frac{3x-2y}{x}\quad y(1)=0$$ I am looking for a way to solve this problem using the Euler method. I have a given Interval of $[1,2]$ and a given step size $h$ of $h=0.1$
My Approach was: I can see, this is a differential Equation of first-order. I have one given initial condition $y(1)=0$. So this must be a initial value problem of first order.
For the Euler Method, we need a step size of $h>0$. So our $0.1$ seems to be okay. Next thing, should be calculating: $$t_k=t_0+kh, \quad \quad k=0,1,2,\dots $$
And this is the point where i think i am stuck. I failed in calculating this and after reading the definitions i didn't succeed in making the connection from this method towards a solution for my given initial value problem, at all.
AI: We are given:
$$y'=\dfrac{3x-2y}{x}\quad y(1)=0$$
Interval is $x \in[1,2]$, and a given step size $h=0.1$
We have:
$h = \dfrac{b-a}{N} = .1 = \dfrac{2-1}{N} \rightarrow N = 10$
$x = a = 1$
$y(a) = y(1) = \alpha = 0 \rightarrow a = 1, \alpha = 0$
Set:
$x_0 = 1, x_i = 1 + 0.1 i, y_0 = 0$
Using Euler's, we have: $y_i = y_{i-1} + hf(x, y) = y_{i-1} + .1\left(\dfrac{3 x_{i-1} - 2 y_{i-1}}{x_{i-1}}\right)$
For $i= 1$, we have:
$x_0 = 1, y_0 = 0, y_1 = y_{0} + .1\left(\dfrac{3 x_{0} - 2 y_{0}}{x_{0}}\right) = 0 + .1 \dfrac{3 (1) - 2(0)}{1.1} = 0.3$
For $i= 2$, we have:
$x_1 = 1.1, y_1 = 0.3, y_2 = y_{1} + .1\left(\dfrac{3 x_{1} - 2 y_{1}}{x_{1}}\right) = 0.3 + .1 \dfrac{3 (1.1) - 2(.3)}{1.1} = 0.5454$
Continuing this way, we generate the table:
$~~~~~\text{Step} ~~|~~ x ~~~|~~ y $
$~~00 ~~| 1.0~~ | 0. $
$~~01 ~~| 1.1 ~~| 0.3 $
$~~02 ~~| 1.2 ~~| 0.545455 $
$~~03 ~~| 1.3 ~~| 0.754545 $
$~~04 ~~| 1.4 ~~| 0.938462 $
$~~05 ~~| 1.5 ~~| 1.1044 $
$~~06 ~~| 1.6 ~~| 1.25714 $
$~~07 ~~| 1.7~~ | 1.4 $
$~~08 ~~| 1.8~~ | 1.53529 $
$~~09 ~~| 1.9 ~~| 1.66471$
$~~10~~| 2.0~~ | 1.78947 $
The exact solution is given by:
$$y(x) = \dfrac{x^3-1}{x^2}$$
At $x=2$, we have: $y(2) = \dfrac{7}{4} = 1.75$
Compare that to Euler's method, which has $1.78947$.
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H: Probability of an event happening N or more times
I need to determine the probability of an event happening N or more times over M iterations. I know the probability of the event in question happening and its likelihood does not change over time.
For example, say I am rolling a six-sided die and I want to know what is the probability that over the course of 100 rolls I'll roll a 1 or a 2 at least 75 times?
Is there a nice and tidy formula for computing such probabilities?
Thanks
AI: Yes - it is called a binomial distribution. The probability of a trial having a probability of success of $p$ being successful $k$ times out of $n$ trials is
$$P(k) = \binom{n}{k} p^k (1-p)^{n-k}$$
For your numbers, the probability is
$$P(K \ge N) = \sum_{k=N}^M \binom{M}{k} p^k (1-p)^{M-k}$$
For your die example:
$$P(K \ge 75) = \sum_{k=75}^{100} \binom{100}{k} \left (\frac13\right)^k \left ( \frac{2}{3}\right )^{100-k} \approx 1.89 \cdot 10^{-17}$$
Here is the full distribution over all rolls, which explains why this probability is so small:
ADDENDUM
You may also have noticed that the distribution looks quite normal. In fact, for large $N$, it is well-known that binomial distributions (among others) approximate a normal distribution very well. In this case, $\mu = N p = 100/3$ and $\sigma^2 = N p (1-p) = 200/9$.
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H: Please explain me how can I show that the last limit does not exist?
I posted my answer with its question.
But how can I show that the last limit -on the second page- does not exist?
That is,
$\mathbf{\lim_{(h_1, h_2)\to (0,0)}\frac{\sqrt {|h_1.h_2|}}{\sqrt {h_1^2,+ h_2^2}}}$ does not exist. How?
11.2.3. Prove that $f(x,y)=\sqrt{|xy|}$ is not differentiable at $(0,0)$.
Solution.
By a theorem, we need to check that $f$ has first order partial derivatives and they are continuous at $(0,0)$.
$$\frac{\partial f}{\partial x}(0,0) = f_x(0,0) =\lim\limits_{h\to0} \frac{f((0,0)+h(1,0))-f(0,0)}h =
\lim\limits_{h\to0} \frac{f(h,0)-f(0,0)}h = \lim\limits_{h\to0} \frac{\sqrt{|h\cdot0|}}h=0$$
$$\frac{\partial f}{\partial y}(0,0) = f_y(0,0) =\lim\limits_{h\to0} \frac{f((0,0)+h(0,1))-f(0,0)}h =
\lim\limits_{h\to0} \frac{f(h,0)-f(0,0)}h = \lim\limits_{h\to0} \frac{\sqrt{|0\cdot h|}}h=0$$
Consequently, $\delta f(0,0)=(0,0)$, i.e., $f$ has first-order partial derivatives at zero.
Let's show the continuity of first order partial derivatives of $f$ at $(0,0)$.
$$\lim\limits_{(h_1,h_2)\to(0,0)} \frac{f((0,0)+(0,0)\cdot(h_1,h_2))-[f(0,0)+\partial f(0,0)(h_1,h_2))]}{\|(h_1,h_2)\|}=
\lim\limits_{(h_1,h_2)\to(0,0)} \frac{f(h_1,h_2)-f(0,0)-Df(0,0)(h_1,h_2)}{\|(h_1,h_2)\|} =
\lim\limits_{(h_1,h_2)\to(0,0)} \frac{f(h_1,h_2)}{\|(h_1,h_2)\|} =
\lim\limits_{(h_1,h_2)\to(0,0)} \frac{\sqrt{|h_1\cdot h_2|}}{\sqrt{h_1^2+h_2^2}} =
$$
AI: Try picking $h_1=h_2$, say. That is, approach the origin through the line $x=y$. What is the limit? Is it $0$? Now look at $h_1=2h_2$.
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H: What is the cardinality of $X$?
Let $X=\{x\in D^\mathfrak{c}:
0<|\{\xi<\mathfrak{c}:x(\xi)=1\}|\le\omega_1\}$, where $D=\{0,1\}$.
What is the cardinality of $X$? I think it is $\mathfrak c$, however I'm not sure. Also I don't know how to prove it. Thanks for your help.
AI: $|X|=\mathfrak{c}^{\omega_1}=(2^\omega)^{\omega_1}=2^{\omega_1}$. This can be $\mathfrak{c}$, if for instance, $2^\omega=2^{\omega_1}=\omega_2$, which is consistent with $\mathsf{ZFC}$, but under $\mathsf{CH}$ we have $2^{\omega_1}=2^{\mathfrak{c}}>\mathfrak{c}$.
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H: Sum of the first n Prime numbers
Let $P_i$ denote the i-th prime number. Is there any formula for expressing
$$S= \sum_{i=1}^m P_i.$$
We know that there are around $\frac{P_m}{\ln(P_m)}$ prime numbers less than or equal to $P_m$. So, we have:
$$S\le m\times P_m\le \frac{P_m^2}{\ln(P_m)}.$$
I want to know, if there is a better bound for $S$, in the litrature.
AI: Summation by parts gives
$$
\begin{align}
\sum_{p\le n}p
&=\sum_{k=1}^n(\pi(k)-\pi(k-1))\,k\\
&=n\,\pi(n)+\sum_{k=1}^{n-1}\pi(k)(k-(k+1))\\
&=n\,\pi(n)-\sum_{k=1}^{n-1}\pi(k)\tag{1}
\end{align}
$$
We have that $\pi(k)=\dfrac{k}{\log(k)}\left(1+O\left(\frac1{\log(k)}\right)\right)$ and so using the Euler-Maclaurin Sum Formula, we get that
$$
\sum_{k=1}^{n-1}\pi(k)=\frac12\frac{n^2}{\log(n)}+O\left(\frac{n^2}{\log(n)^2}\right)\tag{2}
$$
Therefore, we get
$$
\sum_{p\le n}p=\frac12\frac{n^2}{\log(n)}+O\left(\frac{n^2}{\log(n)^2}\right)\tag{3}
$$
Setting $n=P_m$ should give you a closer estimate.
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H: Field homomorphism respects arbitrary arithmetic expression
A field homomorphism $f:A \to B$ respects the fields' binary operations, in the sense that $f(x + y) = f(x) + f(y)$ and $f(xy) = f(x)f(y)$.
When you have an explicit expression like $expr = x^3 + 15x - 32/y$, it's easy to verify using the normal order of operations that the $f(expr) = f(x)^3 + 15f(x) - 32/f(y)$.
But how would you go about showing that $f$ respects arbitrary finite sequences of well-formed arithmetic operations, for example $f(a_1 S_1 a_2 S_2 a_3 S_3 a_4 S_4 \cdots S_{n-1} a_n )$, where each $a_k$ is a field element and $S_k$ is one of the field's binary or unary operations, or a parenthesis.
I've sort of taken this for granted for years.
Could you use induction on the number of symbols, or operations?
Or is there a much easier way?
AI: Don't think of the expression as a string of symbols. Instead, think of the expression as a finite composition of functions representing the binary and unary operations in the field. The string version of the expression is just convenient shorthand.
Now use induction on the number of function calls/operations, like Franklin.vp suggested.
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H: Interesting Differentiation Technique
@HansEngler Left the following response to this question regarding "bad math" that works,
Here's another classical freshman calculus example:
Find $\frac{d}{dx}x^x$.
Alice says "this is like $\frac{d}{dx}x^n = nx^{n-1}$, so the answer is $x x^{x-1} = x^x$."
Bob says "no, this is like $\frac{d}{dx}a^x = \log a \cdot a^x$, so the answer is $\log x \cdot x^x$."
Charlie says "if you're not sure, just add the two terms, so you'll get partial credit".
The answer $\frac{d}{dx}x^x = (1 + \log x)x^x $ turns out to be correct.
In this comment, @joriki asserts that this is not "bad math" but rather a legitimate technique,
You get the derivative of any expression with respect to $x$ as the sums of all the derivatives with respect to the individual instances of $x$ while holding other instances constant.
I had never previously seen such a technique so naturally I tested it on a few examples, including $\frac{d}{dx} \left( x^{ \sin x}\right)$, etc. and it provided the correct result. The following three questions arose,
$ \ \ $ 1. What is the proof of its is validity?
$ \ \ $ 2. Are there any examples where this technique outshines standard methods?
AI: Look at $y = f(u(x),v(x))$:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}$$
Now, note that $x^{\sin x}$ can be written as:
$$y=x^{\sin x} = f(x,\sin x), \ f(u,v) = u^v$$
So that:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}=
vu^{v-1} \cdot\frac{du}{dx} + \log u \cdot u^v \cdot\frac{dv}{dx}$$
In fact, this holds for any number of terms (since so does the chain rule), where each time we only look at the derivative by $x$ of only one of the terms, multiply those by the "internal" derivative, and sum them all up.
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H: Zorn's lemma exercise
Exercise: Let $L$ be a lattice in which every chain has an upper bound. Prove that $L$ has a unique maximal element.
I said:
p1) Suppose $L$ has two maximal elements. (Proof by contradiction is where I'm headed)
p2) Since $L$ is a lattice, then every two elements in the set have a least upper bound (lub) and a greatest lower bound (glb).
p3) Suppose there exists a $2$-element lattice $\{x_\alpha, x_\beta\}\in L$ such that $x_\alpha <x_\beta$ or $x_\beta<x_\alpha$ or they're equal. In this way we have also defined the requirement for a chain.
p4) Since $L$ has two maximal elements, $x_\alpha$ has to equal $x_\beta$.
p5) But that's impossible, because there would be no glb/lub defined in that lattice.
p6) The same can be said for three element subsets - if $x_\gamma\le x_\alpha$ and $x_\gamma\le x_\beta$, $x_\alpha$ and $x_\beta$ can't be equal - or else there is no glb/lub between $x_\alpha$ and $x_\beta$. If $x_\gamma$ is greater than both $x_\alpha$ and $x_\beta$, then $x_\gamma$ is maximal, contradictory to the premises.
Q) Thus $L$ has a unique maximal element.
I would gladly consider some criticism of my proof. Thanks!
AI: The biggest problem is that you’ve not shown that $L$ has any maximal element at all: you’ve only tried to show the other half of the result, i.e., that $L$ cannot have more than one maximal element. I realize that you may have omitted the existence part on the grounds that it’s a trivial application of Zorn’s lemma, so I’ll concentrate on the argument that you did present.
Your proof of uniqueness is horribly overcomplicated and not very clearly organized. The first point is fine, but you’ll make matters much clearer if you name the supposed maximal elements right at the start:
Suppose that $x$ and $y$ are maximal elements of $L$.
Your second point is premature, and your third is irrelevant. Your fourth is unclear, because you never said that $x_\alpha$ and $x_\beta$ are the maximal elements of the first point; you also haven’t clearly stated any argument to justify the conclusion that $x_\alpha=x_\beta$, though I think that you probably have the right general idea in mind. Now is actually the time to bring in your second point, but don’t clutter things up with the irrelevant part of it:
$L$ is a lattice, so $x$ and $y$ have a least upper bound $z$. Then $x\le z$, and since $x$ is maximal in $L$, it follows that $x=z$. Similarly, $y\le z$, and $y$ is maximal in $L$, so $y=z$. Thus, $x=y$, and $L$ has exactly one maximal element.
Note that it need not be cast as a proof by contradiction: the foregoing argument shows directly that any two maximal elements of $L$ are actually the same element.
I’ll comment briefly on the remaining points of your argument. The fifth is simply wrong: if $x_\alpha=x_\beta$, the least upper bound of $\{x_\alpha,x_\beta\}$ is simply $x_\alpha$. The sixth is partly wrong (for much the same reason) and wholly unnecessary.
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H: Two numbers are randomly selected from the set {0,1,2,3,4,5,6,7} without replacement
What is the probability that the sum of the two numbers is 7?
[what i did]
0+7
1+6
2+5
4+3
my ans: 4/8
Is this right or wrong? Does without replacement mean we can't use the same number twice? IE: 4+2 AND 2+4?
Thanks for the help
AI: Replacement does indeed mean that the same one cannot be used twice.
Order is not important, so we are using combinations in this case. Out of $_8 C_2 = 28$ possible sums, only 4 (7 + 0, 6 + 1, 5 + 2, 4 + 3) will yield 7. The probability is then $\frac{4}{28} \times 100\% = 14.3\%$.
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H: Why are the limits of integration of $\Gamma (z)$ the way they are?
$$\Gamma (z)=\int_0^\infty t^{z-1}e^{-t} dt$$ To make things easier, let's call this $G(z)$. If we take the derivative with respect to $t$ of both sides, we get $\frac {dG(z)}{dt}=\frac {d}{dt} [\int_0^\infty t^{z-1}e^{-t} dt]=0$ . The reason why I think this equals zero is because if we are just looking at $\int_0^\infty t^{z-1}e^{-t} dt$ , $z$ is fixed and this definite integral is equal to a constant (if it's convergent). If $\frac {dG(z)}{dt}=0$ , this implies that $G(z)=C\cdot t^0$ , so it looks like $G(z)$ is absolutely independent of the value of $t$ . Why then does $G$ have those specific limits of integration? Those limits, only affect $t$ , which $G$ is independent of. And I know that this is illogical because by my argument the limits of integration could even be equal to each other, which would obviously lead to a false solution. Thanks.
My background just Calculus $1$ so please don't use infinite series and such.
If you are interested in my motivation, the original motivation for this question is that I wanted to see if it is possible to have an elementary function $f(x)$ that that grows faster than $x!$ (please don't tell me the answer to this or give any hints, I wanna see if I can do this by myself). The motivation of THAT question was to see if it is even possible to have an elementary function of a finite number of terms that models $x!$ (again, please don't tell me the answer to that either.)
AI: $\frac{d}{dt} G(z)=0$ because, as a function of $t$, it is a constant. If $F(z)=z$, then $\frac{d}{dt}F(z)=0$, too. There is no free variable $t$ in the definition of $\Gamma$.
It is sometimes confusing notation to the uninitiated, but the $t$ inside the integral is really a placeholder, indicating what is changing - it only exists inside the integral.
Now, the step you are doing is a bit confusing, because you can't bring $d/dt$ from outside the interval, where $t$ is "unbound" and move it inside the integral, where $t$ is "bound." It simply can't be done, and there is no reason to think you can.
If you don't believe me, try it with other functions. It simply won't work. For example, $F(z)=\int_0^1 t^z dt = \frac{1}{1+z}$. Then, by your reason, $\frac{d}{dt} F = \int_0^1 zt^{z-1} dt = 1$. That doesn't make much sense, does it?
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H: How determinate the torsion subgroup and the normal maximal $\pi$-subgroup of $\mathbb{R}/\mathbb{Z}$?
Let $\mathbb{R}$ and $\mathbb{Z}$ the real numbers and the integers, respectively. Consider $G =\mathbb{R}/\mathbb{Z}$. Find
$O_{\pi}(G)$ - ie, the (unique) normal maximal $\pi$-subgroup of $G$;
$T(G)$ - ie, the torsion subgroup of $G$.
AI: If $G$ is a torsion abelian group it satisfies the $p$-group decomposition $G\cong\bigoplus G_p$, where each $G_p$ is the subgroup of all $p$-torsion elements (i.e. elts whose order is a power of $p$) of $G$, equivalently the union $\bigcup G[p^r]$ where $G[n]$ is defined as the subgroup of all $g\in G$ satisfying $g^n=e$.
If $G$ is a torsion abelian group and $\pi\subseteq{\cal P}$ a set of primes, a $\pi$-subgroup of $G$ is one in which every element has order which factors into primes only coming from $\pi$. Thus the maximal $\pi$-subgroup of any torsion abelian group $G$ is $\bigoplus_{p\in\pi}G_p$. Abelian automatically implies normal.
Note that $\bf R$ is an uncountable $\frak c$-dimensional $\bf Q$-vector space, so ${\bf R}\cong \bigoplus_{\frak c}{\bf Q}$. As $\bf Z$ is a subgroup of only one of the factors of $\bf Q$ in this decomposition, quotienting by $\bf Z$ only affects that factor, that is we have ${\bf R}/{\bf Z}\cong {\bf Q}/{\bf Z}\times\bigoplus_{\frak c}{\bf Q}$. All of the latter factors are nontorsion, so $T({\bf R/Z})\cong{\bf Q}/{\bf Z}$.
Finally, we have the Prüfer decomposition ${\bf Q}/{\bf Z}\cong \bigoplus_p {\bf Z}(p^\infty)$. The Prüfer $p$-groups are defined additively as ${\bf Z}[p^{-1}]/{\bf Z}$, or multiplicatively as the roots of unity in $\bf C$ with $p$-power order. This is in fact the $p$-torsion decopmosition of $G={\bf Q}/{\bf Z}$, so the maximal $\pi$-subgroup is $\bigoplus_{p\in\pi}{\bf Z}(p^\infty)$.
Tangent. An equivalent formulation is ${\bf Z}(p^\infty)\cong{\bf Q}_p/{\bf Z}_p$, where ${\bf Q}_p$ is the field of $p$-adic numbers with subring ${\bf Z}_p$ the $p$-adic integers. This allows the Prüfer decomposition to generalize to arbitrary global fields (quotienting a global field by its ring of integers decomposes as a direct sum of the local fields modulo their valuation subrings), in particular the function field case recovers the idea of partial fraction decomposition in algebraic language.
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H: Can different functions of a random variable be independent?
Let $X$ be a random variable. $f$ and $g$ are two non-trivial (e.g. non-constant) measurable functions defined on the range of $X$. Can $f(X)$ and $g(X)$ be independent? Thanks!
AI: $f$ and $g$ can be independent if they are simple functions that attain finitely many values. For example: take these functions on $[0,4]$
$$f(x)=
\begin{cases}
\frac 12 & \lfloor x \rfloor \text{ is even}\\
0 & \text{otherwise} \end{cases}\\
g(x)=
\begin{cases}
\frac12 & \lfloor x/2 \rfloor \text{ is even}\\
0 & \text{otherwise} \end{cases}
$$
We have $$P(g(x)=C|f(x)=D)=P(g(x) = C)$$
and
$$P(f(x)=C|g(x)=D)=P(f(x) = C)$$
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H: Extension of the chain rule to multiple variables?
Say there is a function $F(x,y,z)$, and $x$, $y$, and $z$ are all functions of a single variable $t$. That means that $F$ is by proxy a function of a single variable $t$. Is there an extension of the chain rule to multiple variables that would make it possible to solve this without plugging in the values of $x$, $y$, and $z$?
As an example, say we have $F(x,y,z)=x^2+y^2+z^2$, where $x=e^t$, $y=t^2$, and $z=t^3$. Is there something similar to the chain rule that would allow you to calculate $\frac{d}{dt}F(x(t),y(t),z(t))$? I know it's a trivial example and it would be easy to just substitute the values of $x$, $y$, and $z$ into $F(x,y,z)$, but I just chose this at random so that I could see how it is done through example.
AI: Yes. Look up the chain rule. It will tell you that
$$\frac d{dt} F(\vec x(t)) = \nabla F(\vec x(t))\cdot \frac{d\vec x(t)}{dt}.$$
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H: Why do we multiply in tree diagrams?
In probability, we always lay out the events through tree to see what depends on what. Then we were taught to "multiply" through that branch to get the probability of that event.
Why do we "multiply"? I've noticed we have the same sort of rule in calculus too (chain rule)
EDIT: I think my question was misinterpreted. I am asking just because we put a number (usually - if not almost fractional) beside a branch, why must we "multiply" through that number or that "thing" if I am being more abstract. Where "thing" could even mean a derivative operator $$\partial / \partial x$$
AI: The branch that goes from $R$ to $A$ has $1/3$ beside it. As you know, the $1/3$ means that if we are at $R$, we have probability $1/3$ of going to $A$. The $5/8$ and $3/8$ mean that if we have reached node $A$, we have probability $5/8$ of going to $D$, and probability $3/8$ of going to $E$. Let's stop here, though if you want to make the thing more tree-like, go ahead.
Now imagine say $1200$ people starting at $R$. They split along the $3$ branches coming out of $R$, with $1/3$ of them going to $A$, $1/6$ of them to $B$, and $1/2$ to $C$.
In particular, $400$ of them end up at $A$.
Now at $A$ they split, some going to $D$ and some to $E$, according to the probabilities written on the branches. So $5/8$ of the $400$ people end up at $D$. This is $250$ people.
Note that $250=(1200)(1/3)(5/8)=(1200)(5/24)$. So to determine what fraction of the $1200$ end up at $C$, we multiply $1/3$ by $5/8$.
When we are dealing with probabilities, the number who go to $A$ need not be exactly $400$, but the multiplication of probabilities takes place for essentially the same reason as in our people example.
diagram code:
\documentclass[11pt]{standalone}
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}[column sep=0.3in,row sep=0.5in]
{}& & & R \ar[-]{d}[swap]{\frac{1}{6}} \ar[-]{lld}[swap]{\frac{1}{3}}
\ar[-]{rrd}{\frac{1}{2}} & & \\
& A\ar[-]{ld}[swap]{\frac{5}{8}}\ar[-]{rd}{\frac{3}{8}} & & B &\hphantom{E} & C\\
D & & E
\end{tikzcd}
\end{document}
Remark: There is a strong analogy between the above situation, and the Chain Rule for differentiation. Let's stick to depth $2$, where we have $y=f(u)$ and $u=g(x)$. If we want the rate of change of $y$ with respect to $x$, we want to know the local scaling factor of $f(g(x)$ if we increase $x$ by a small amount $\Delta x$. What happens is that $u$ increases by about $\Delta u= (\Delta x)\frac{du}{dx}$, and $f(u)$ therefore increases by about $\Delta u \frac{dy}{du}$, that is, $(\Delta x)\frac{dy}{du}\frac{du}{dx}$.
So $\Delta x$ in a sense playes the same role as the $1200$ people did in the example above. In each case, we have a form of combined scalings. The question does a service in bringing out the analogy!
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H: verifying differential equation solution with sage
I solved the linear ODE system of equations:
\begin{equation}
x' = \begin{pmatrix}3&0&4\\0&2&0\\0&0&-3\end{pmatrix}x
\end{equation}
Skipping the details I got the following eigenpairs:
\begin{align}
\lambda_0=3, x_0=\begin{pmatrix}1\\0\\0\end{pmatrix}&\\
\lambda_1=2, x_1=\begin{pmatrix}0\\1\\0\end{pmatrix}&\\
\lambda_2=-3, x_2=\begin{pmatrix}\frac{2}{3}\\0\\1\end{pmatrix}&
\end{align}
Thus the general solution is:
\begin{equation}
x(t) = c_1e^{3t}\begin{pmatrix}1\\0\\0\end{pmatrix} + c_2e^{2t}\begin{pmatrix}0\\1\\0\end{pmatrix}+c_3e^{-3t}\begin{pmatrix}\frac{2}{3}\\0\\1\end{pmatrix}
\end{equation}
Here is what I got from sage:
sage: t = var('t')
sage: x = function('x', t)
sage: y = function('y', t)
sage: z = function('z', t)
sage: de1 = diff(x, t) -3*x -4*z == 0
sage: de2 = diff(y, t) -2*y == 0
sage: de3 = diff(z, t) +3*z == 0
sage: desolve_system([de1, de2, de3], [x, y, z])
[x(t) == 1/3*(3*x(0) + 2*z(0))*e^(3*t) - 2/3*e^(-3*t)*z(0),
y(t) == e^(2*t),
z(t) == e^(-3*t)*z(0)]
So my solution looks fine in terms of $y(t)$ and $z(t)$ but my $x(t)$ is different in the coefficient of $e^{3t}$
Why does sage give a different answer?
Specifically:
Did I solve the system incorrectly?
Did sage choose different eigenvectors?
Is sage only giving me a nicer scaling of the eigenvectors?
All help is greatly appreciated!
AI: Okay, we have:
$$x' = \begin{bmatrix}3&0&4\\0&2&0\\0&0&-3\end{bmatrix}x$$
The CP is: $-\lambda^3+2 \lambda^2+9 \lambda-18 \rightarrow \lambda_1 = -3 , \lambda_2 =3 , \lambda_3 = 2$.
We have three distinct and real eigenvalues, so we form $[A-\lambda I] v_i = 0$ to find the eigenvectors and get the following eigenvalue/eigenvector pairs:
For $\lambda_1 = -3$, the RREF gives us:
$c = 3, b= 0 , a = -(2/3) c = -2 \rightarrow v_1 = (-2,0,3)$ (note, c is a free variable)
For $\lambda_2 = 3$, the RREF gives us:
$c = 0, b= 0 , a = 1 \rightarrow v_2 = (1,0,0)$ (note, a is a free variable)
For $\lambda_3 = 2$, the RREF gives us:
$a = 0, c = 0, b = 1 \rightarrow v_3 = (0,1,0)$ (note, b is a free variable)
Summarizing these, we have:
$\lambda_1 = -3, ~v_1 = (-2, 0, 3)$
$\lambda_2 = 3, ~v_2 = (1, 0, 0)$
$\lambda_3 = 2, ~v_3 = (0, 1, 0)$
The solution is then given by:
$$x(t) = c_1 e^{\lambda_1 t}v_1 + c_2 e^{\lambda_2 t}v_2 + c_3 e^{\lambda_3 t} v_3 = c_1 e^{-3t}\begin{bmatrix}-2\\0\\3\end{bmatrix} + c_2e^{3t} \begin{bmatrix}1\\0\\0\end{bmatrix} + c_3e^{2t}\begin{bmatrix}0\\1\\0\end{bmatrix}$$
If we compare this to your solution, you note that for my first vector, I chose $c=3$, where you chose $1$, hence these are actually the same (actually, you lost a negative sign - please check that)!
If we write out the solution for this, we get:
$$x(t) = \begin{bmatrix}x(t)\\y(t)\\z(t)\end{bmatrix} = \begin{bmatrix} - 2c_1 e^{-3t}+c_2e^{3t} \\ c_3 e^{2t}\\ 3c_1e^{-3t}\end{bmatrix}$$
First question, why are the variables all negated in the Sage approach? That changes things. The eigenvalues are the same, but the eigenvectors are swapped around. Can you try it with the same signs as the matrix has?
It also looks like they are using some other algorithm to find these, but if you reduce all of their constants to a single constant, it is basically the same. It is odd that they chose $c_3 = 1$, but there is likely a reason for that.
Let me clarify what I am saying here. For their solution, we can equate their constants (make sure I did not make a silly algebra error here) to the ones we used as:
$c_1 = \dfrac{z(0)}{3}, c_2 = \dfrac{1}{3}(3 x(0) + 2 z(0)), c_3 = 1$.
These are totally arbitrary, but with that change, you can see we are talking the same thing. Whatever internal algorithm they are using chooses things in that form.
Update
I ran the script using the online version of Sage with the same variables and signs as the matrix $A$, and the result (I MathJax'ed it) is:
$x(t) = \dfrac{1}{3}(3 x(0) + 2 z(0))e^{-3t} - \frac{2}{3}e^{3t}z(0)$
$y(t) = e^{-2t}y(0)$
$z(t) = e^{3t}z(0)$
Which has the wrong sign for an eigenvalue (online version), maybe for all three as it is hard to tell!
I tried the online Maxima and here is the result, just like my eigenvalues/eigenvectors (note the sign and fraction on one of them).
So, why does Sage require users to change all of the signs of the matrix? Regardless, changing the signs does produce the correct eigenvalues and the constants as shown above and through the simple transformation, it is the same result I derived.
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H: What's the mean of all real numbers?
At first, I had thought the average must be zero, since for every positive number there's an equal magnitude negative number to cancel out the positive number's effect on the average, leaving only zero to set the average.
But you can make a similar argument about any number, for example using an arbitrary choice of 9, for every number x units greater than 9, there's another number x units less than 9, which would make 9 the mean. But since I could have chosen any number here instead of 9 that would mean that any and every number is the average of all real numbers.
So, what is the mean of all real numbers?
AI: As you proved for yourself, there can be be no well defined mean for the real numbers. Another way to think about it would be to think of the mean of a closed interval $[a,b]$ as:
$$\mu_{[a,b]}=\int_{a}^{b} \frac{x}{b-a} dx$$
To extend this to all the real numbers, you would have to take $a\to -\infty$, $b\to \infty$. You may be tempted to write the mean as:
$$\mu_\mathbb{R}=\lim_{M\to\infty}\int_{-M}^M \frac{x}{2M} dx =0$$, but since no one promises you that $a$ and $b$ go to $\infty$ at the same rate, you must write:
$$\mu_\mathbb{R}=\lim_{N\to\infty}\lim_{M\to\infty}\int_{-M}^N \frac{x}{N+M} dx$$
This is undefined, due to the fact that the value depends on your choice of rate (choose $N=2M$ or $N=M$ to see that).
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H: For a convex function, the average value lies between $f((a+b)/2)$ and $(f(a) + f(b))/2$
Suppose that $f\in C^2$, $f''(x)\geq 0$ $\,\,\,\forall x \in [a,b]$. I want to show that $$\frac{1}{2}(b-a)(f(a)+f(b))\leq \int_a^bf(t)\,dt\leq (b-a)f\left(\frac{a+b}{2}\right).$$If we divide by $b-a$, we see that the left term is less than the right term by definition of convexity, and it remains to show that the average value of the function lies between $f\left(\frac{a+b}{2}\right)$ and $\frac{1}{2}(f(a)+f(b))$.
The mean value theorem for integrals implies that the average value is attained at some point $c\in (a,b)$. But it's not clear to me why $f(c)$ should lie in between two other points. Perhaps there's another theorem about integration we should apply. Any ideas?
AI: I think the inequality needs to be reversed.
$\displaystyle \frac{b-a}{2}f\Big(\frac{b+a}{2}\Big)-\int_a^{\frac{b+a}{2}}f(t)\,dt=\int_a^{\frac{b+a}{2}}\int_t^{\frac{b+a}{2}}f'(s)\,ds\,dt\le\int_{\frac{b+a}{2}}^{b}\int_{\frac{b+a}{2}}^{t}f'(s)\,ds\,dt$ (since $f'(s)$ is bigger in the latter region and the integration is over a domain of the same size) $\displaystyle=\int_{\frac{b+a}{2}}^{b}f(t)\,dt-\frac{b-a}{2}f\Big(\frac{b+a}{2}\Big),$ hence $\displaystyle (b-a)f(\frac{b+a}{2})\le\int_a^b f(t)\,dt$.
Other side:
$\displaystyle \int_a^{\frac{b+a}{2}}f(t)\,dt-\frac{b-a}{2}f(a)=\int_a^{\frac{b+a}{2}}\int_a^{t}f'(s)\,ds\,dt\le \int_{\frac{b+a}{2}}^{b}\int_{t}^{b}f'(s)\,ds\,dt=\frac{b-a}{2}f(b)-\int_{\frac{b+a}{2}}^b f(t)\,dt\,,$ hence $\displaystyle \int_a^b f(t)\,dt\le\frac{b-a}{2}(f(b)+f(a))\,.$
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H: Trigonometry Functions/equations
Given that $\sin A=\dfrac{8}{17},\;$ find the possible values of $\cos A$ and $\cot A$.
Can someone please explain this question? I' new to the topic, and I'm very unsure as to how its done.thanks
AI: Well, $\sin^2A+\cos^2A=1$; this is the Pythagorean theorem.
So you know two possible values of $\cos A$. Hence you can deduce the possible values of $\cot A$.
Answer: $\cos A=\pm\sqrt{1-(8/17)^2}=\pm15/17$. Hence $\cot A=\pm15/8$.
Hope this helps.
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H: Fourteen indviduals are to be divided into comittee of three, four and seven members
How many ways can the individuals be grouped?
$${14\choose3}\cdot{11\choose4}\cdot{7\choose7}$$
Is this the right way to approach this?
AI: Yes, your answer is fine. It reduces to the multinomial coefficient
$$\binom{14}{3,4,7}=\frac{14!}{3!\,4!\,7!}$$
if you write it out and perform the most obvious cancellations:
$$\binom{14}3\binom{11}4\binom77=\frac{14!}{3!\,\color{green}{11!}}\cdot\frac{\color{green}{11!}}{4!\,\color{magenta}{7!}}\cdot\frac{\color{magenta}{7!}}{7!\,0!}=\frac{14!}{3!\,4!\,7!}\;.$$
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H: Question on induction-1 is the least positive integer
Question on induction
prove: 1 is the least positive integer.
proof:
Let $A=\left\{x\geq 1\left|x\in Z^+\right.\right\}$, and then $1\in A$, if positive integer $n\in A$, then $n\geq 1$,
Since $1\leq n<n+1$, so $n+1\in A$. By induction, we have $A=Z^+$. So for all positive integer that it is greater than or equal to $1$, and $1$ is the least positive integer.
question:
There is something different from the induction pattern
$\text{when } n=1,\text{...}n=k,\text{...}\text{and } n=k+1,\text{...}$
So, can you explain me a bit more about the proof and the corresponding of the pattern I listed?
I saw this as one first example in the first section of first chapter of one text of mathematical analysis. So I do not know the definition of "integer".
AI: Here's a slight rewording of the given proof in the induction format that you're used to:
Claim: For all $n \in \Bbb{Z}^+,~$ $n\ge 1$.
Proof: We proceed by induction on $n \in \Bbb{Z}^+$.
Base Case: For $n=1$, we have $1 \ge 1$, which works.
Induction Hypothesis: Assume that the claim is true for $n=k$, where $k$ is a positive integer. That is, assume that $k\ge 1$.
It remains to prove the claim true for $n=k+1$. Since $k \in \Bbb{Z}^+$, we know by the definition of the positive integers that $k+1\in \Bbb{Z}^+$. Recall that $k+1>k$. But by the induction hypothesis, we know that $k\ge1$. Hence, $k+1\ge1$, so the claim is true for $n=k+1$. This completes the induction.
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H: Given that $\cos x =-3/4$ and $90^\circ
Given that $\;\cos x =-\frac{3}{4}\,$ and $\,90^\circ<x<180^\circ,\,$ find $\,\tan x\,$ and $\,\csc x.$
This question is quite unusual from the rest of the questions in the chapter, can someone please explain how this question is solved? I tried Pythagorean Theorem, but no luck. Is it possible to teach me how to use the circle diagram?
AI: The cosine of an angle corresponds to the $x$-coordinate in the unit circle, and the sine of an angle cooresponds to its $y$-coordinate on the unit circle.
Note that $\,\cos = -\frac 34 < 0\,$ if and only if the angle $x$ terminates in either the second or third quadrant, where the angle $x$ is measured with respect to the positive $x$-axis.
Since we are given that $\,90 \lt x \lt 180,\,$ we know that the angle $x$ terminates in the second quadrant. So $\sin x > 0$.
So, $\,\tan x = \dfrac{\sin x}{\cos x} <0,\;$ and $\,\csc x = \dfrac{1}{\sin x}>0$.
Now, we know that by the Pythagorean Theorem as it relates to trigonometry identities, $${\bf \sin^2 x + \cos^2 x = 1} $$ $$ \begin{align} \iff \sin^2 x & = 1 -\cos^2 x \\ \\ \implies \sin x & = \pm \sqrt{1 -\cos^2 x} \\ \\ & = \pm \sqrt{1 - \left(\dfrac {-3}{4}\right)^2} \\ \\ & = \pm \sqrt{\frac 7{16}} \\ \\ & = \pm \frac{\sqrt 7}{4}\end{align}$$ Since we know that in the second quadrant, $\sin x > 0$, we take the positive root: $$\sin x = \frac{\sqrt 7}{4}$$
and so you have all you need to compute $$\tan x=\dfrac{\sin x}{\cos x} = \dfrac{\sqrt 7/4}{-3/4} = -\left(\dfrac{\sqrt 7}{3}\right)$$ $$\csc x=\dfrac1{\sin x}= \dfrac{1}{\sqrt 7/4} = \dfrac 4{\sqrt 7} $$
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H: Limit of ${x^{x^x}}$ as $x\to 0^+$
Can you please explain why
\begin{align*}
\lim_{x\to 0^+}{x^{x^x}} &= 0
\end{align*}
AI: Ley $y=x^{x}$. Then taking log on both sides, $\ln y=x\ln x$. Taking $x\to0^+$, we can use L'Hopital on $\displaystyle{\frac{\ln x}{\frac{1}{x}}}$, as both the numerator and denominator approach $\infty$. We find the limit to be $0$. Hence, $y\to 1$.
Now take $z=x^{x^x}$. It is $0$ to the power something that approaches $1$. Hence, it is $0$.
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H: Given that $\csc A=-\frac{5}{2}$ and $\sin A< 0$, find $\cot A$
Trig question, Given that $\csc A=-\frac{5}{2}$ and $\sin A< 0$, find $\cot A$
Ive done this so far, although my answer is abit off, where the correct answer is $-\frac{1}{2}\cdot\sqrt{21}$ while my answer is missing the negative. Can you please spot the mistake?
AI: HINT:
If $\csc A=-\frac52<0, -\sin A=\frac1{\csc A}=-\frac25$
So, $\sin A<0$ is redundant information
We can not determine the sign of $\cot A$ from here
We can just use $$\cot A=\pm\sqrt{\csc^2A-1}$$
As Jonas Meyer has pointed out $\cos A>0$
So, $\tan A=\frac{\sin A}{\cos A}<0$ as $\sin A<0$
$\implies \cot A<0,\cot A=-\sqrt{\csc^2A-1}$
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H: Evaluating $\mathbf{\lim_{(h_1, h_2)\to (0,0)}\frac{\sqrt {|h_1.h_2|}}{\sqrt {h_1^2+ h_2^2}}}$
I was looking at Peter Tamaroff's answer to this question. The question was to show that the following limit did not exist: $$\mathbf{\lim_{(h_1, h_2)\to (0,0)}\frac{\sqrt {|h_1.h_2|}}{\sqrt {h_1^2+ h_2^2}}}$$
In the answer it was explained that since if you let $h_1=h_2$ you would get a different limit than if you let $h_1=2h_2$ , the limit of the original does not exist. Why are you allowed to choose different "speeds" at which the variables approach $0$? Doesn't $\mathbf{\lim_{(h_1, h_2)\to (0,0)}}$ imply that both variables approach at the same "speed" so the only valid solution is $h_1=h_2$? Thanks.
P.S. I know this might be terribly wrong because I've never done multivariable limits, so do you also have a recommendation for an online source for me to learn this?
AI: No, the notation $\lim_{(x,y) \rightarrow (a,b)}$ means you are allowed to approach the point $(a,b)$ in EVERY possible way. In 2-dimensions (in the plane) $lim_{(t,t):t \rightarrow (a,a)}$ will mean approaching the point $(a,a)$ at equal speed and $lim_{(t,2t):t \rightarrow (a,2a)}$ will mean approaching $(a,2a)$ along a line of slope $2$.
You should think that approaching a point along any one dimensional continuous curve (like a line, a parabola etc) in two or higher dimensions is not very far from working in one dimensions (because everything is constrained to move in one dimension). This is called parametrization.
In other words, a two dimensional space is inherently two dimensional and hence cannot (and shouldn't) be continuously parametrized by one dimension.
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H: Viewing $R/I$ as $R/I$-module
While doing exercise 4 of Chapter 6 in Atiyah & Macdonald's Introduction to Commutative Algebra, I got stuck at this step:
I have shown that $R/I$ is a Noetherian $R$-module. Here $R$ is a commutative ring with $1$ and $I$ is some ideal of $R$. How can I (no pun intended) conclude
from here that $R/I$ is a Noetherian $R/I$-module?
Well, in the exercise $I$ is actually the annihilator of $R$-module $M$, but the argument above probably works for all ideals $I$.
Thanks!
AI: If you already know that $R/I$ is a Noetherian $R$ - module this solves your problem. Namely because of the following. What is an $R$ - submodule of $R/I$? Well it's just an ideal of $R/I$! So to say that $R/I$ satisfies the ACC on $R$ - submodules is the same as saying it satisfies the ACC on ideals, or that $R/I$ is a Noetherian ring.
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H: Basic probability confusion
A machine has six switches. The probability that any particular switch works properly is $0.98$. Assuming independent operation of the switches, calculate the probability that at least one switch fails to work properly.
Why isn't it this: P(not working) = $1 - 0.98 = 0.02$
There are six switches, so $1/6$ of picking on the switches. We want to find AT LEAST one. Therefore we have $6 * 1/6 * 0.02 = 0.02$
EDIT: I want to know why I am wrong instead of finding the correct answer
AI: OK, I think I now understand how your thinking went, and therefore where it went wrong:
You are failing to take into account that after testing the first switch, you would not test it again. Also, if the first switch already failed, you don't need to test it again to see that it failed.
Also I don't see the need to test them in random order, you just test them one by one.
So you test the first switch. With a probability of $0.02$, it fails. Then you are ready (you know it failed). Otherwise (that is, with probability $0.98$) you proceed to the second switch. Again, when you test it it will fail with probability $0.02$ and succeed with probability $0.98$. But there's only a probability of $0.98$ that you'll even get to test it, so the probability that you find it to fail is $0.98\cdot 0.2$ and the probability that you find it to succeed is $0.98\cdot 0.98$. In the latter case, you proceed to the third switch, and so on. Unless you've found the last switch to also succeed, one of the switch failed. But the probability that the last switch succeeded is $0.98^6$ because each switch contributed a factor of $0.98$. Since the probabilities add up to $1$, this means that the probability that at least one failed is $1-0.98^6$.
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H: Convergence of Cantor series
Show that for every nonnegative real number $x$, there is a series (called Cantor series) with $$x=\sum_{k=1}^\infty\frac{c_k}{k!}$$ where $c_k$ are integers such that $0 \leq c_{k+1} \leq k$ for all $k \in \mathbb{N}\setminus\{0\}$
Note: By defining $c_1$ to be the largest integer $\leq x$, $0 \leq x - c_1 < 1$. Thus we can assume that $x \in [0,1)$.
Any hints or a solution? Thanks.
AI: Hint: Continue as you started. For each $k\ge 2$ let $c_k \in [0, k-1] \cap \mathbb N$ largest with $x_k := \sum_{l=1}^k \frac{c_l}{l!} \le x$. Proof (by induction) that $x - x_k \le \frac 1k$.
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H: Topologies coinciding at a point or a set.
Consider a set equiped with two topologies. What does it mean to say that the two topologies coincide at a point in the set? Is it meaningful to talk about this concept in general. Is it meaningful in the context of metric spaces?
p.s. I know what the relative topology is for any subset of a topological space and this concept is a different one, as is apparent from this problem.
AI: I would guess that it means the every open neighborhood of $x$ in $\tau_1$ contains an open neighborhood of $x$ in $\tau_2$, and visa versa. This essentially means that "continuity at $x$" is the same in each topology.
It's possible, though, that the stronger statement is intended: the neighborhoods of $x$ are the same in each topology.
There's a lovely way to define the "local topology" at a point. Let $(X,\tau)$ be a topology and $x\in X$. Define a new topology, $\tau_x$ which has as a basis the singletons $\{y\}$ where $y\neq x$ and $U\in\tau$ such that $x\in U$. So the open sets are any set not containing $x$, or any set contain a $\tau$-neighborhood of $x$. It is not hard to see that this is a topology.
In this case, my first definition would be equivalent to saying, for topologies $\tau$ and $\rho$, they coincide at $x$ if and only if $\tau_x=\rho_x$.
Now, in the complete lattice of topologies on $X$, $$\bigcap_{x\in X}\tau_x = \tau$$ So we can recover $\tau$ if we know each $\tau_x$. Then, for a topological space, $Y$, we can define a function $f:X\to Y$ to be $\tau$-continuous at $x$ if it is continuous on $(X,\tau_x)$, and it turns out, due to the lattice meet property above, the function $f$ is continuous on $(X,\tau)$ if and only if it is continuous at each $(X,\tau_x)$.
There's also a simple "local range topology", $\tau^x$ defined as just the $\tau$-open sets containing $x$, and the empty set. Then you can say that $f:Y\to X$ is continuous "to" $x$ if it is continuous when using the topology $\tau^x$. Again, we can recover $\tau$ if we know ecah $\tau^x$. Indeed, $\tau = \bigcup \tau^x$.
The stronger definition above would happen exactly when $\tau^x=\rho^x$. It's true that if $\tau^x=\rho^x\implies\tau_x=\rho_x$, though so maybe this stronger definition is reasonable.
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H: Ratio Test and L = 1
Consider the series $$\sum\limits_{n=1}^\infty\frac{(2n)!}{a^n(n!)^2}$$ with $a > 0$. Determine if the series converges for:
i) a > 4
ii) 0 < a < 4
iii) a = 4
For i) and ii), I will use the ratio test, $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{4}{a}$$
So, if $a > 4$, the series converge and, if $0 < a < 4$, the series do not.
iii) If $a = 4$, I am a bit stuck. Can someone please give me some tips?
Thank you
AI: Your coefficients are $${2n\choose n}\frac{1}{4^n}\ge \frac{1}{n+1}$$
where ${2n\choose n}$ are central binomial coefficients, and I used the bound ${2n\choose n}\ge \frac{4^n}{n+1}$. Hence by comparison with the harmonic series, this diverges.
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H: How can I use the Heun's method to solve this first order Initial Value Problem?
My Problem is this given Initial Value Problem: $$y^{\prime}=\frac{3x-2y}{x}\quad y(1)=0$$ I am looking for a way to solve this problem using Heun's method. I have a given Interval of $[1,2]$ and a given step size of $h=0.1$
The example is already solved with a numerical solution. But i want to know more. See here. After discussing the solution by Eulers Method with a friend, he told me about Heun's method. But we failed to apply it to our example. How would Heun's method be applied to this problem?
AI: We are given:
$$y'=\dfrac{3x-2y}{x}\quad y(1)=0$$
Interval is $x \in[1,2]$, and a given step size $h=0.1$
For Heun's (Improved Euler's), we have:
$h = \dfrac{b-a}{N} = .1 = \dfrac{2-1}{N} \rightarrow N = 10$
$x = a = 1, y(a) = y(1) = \alpha = 0 \rightarrow a = 1, \alpha = 0$
Set: $x_0 = 1, x_i = 1 + 0.1 i, y_0 = 0$
Using Heun's (Improved Euler's), we have:
$y_{i+1} = y_{i} + \dfrac{h}{4}\left[f(x_i, y_i)+3f(x_i + \dfrac{2}{3}h, y_i+\dfrac{2}{3}hf(x_i,y_i))\right]$, so
$y_{i+1} = y_{i} + \dfrac{1}{40}\left[\dfrac{3 x_{i-1} - 2 y_{i-1}}{x_{i-1}} + 3\left(\dfrac{3 (x_i + \dfrac{2}{3}(.1)) - 2 (y_i+\dfrac{2}{3}(.1)\dfrac{3 x_{i-1} - 2 y_{i-1}}{x_{i-1}}}{x_i + \dfrac{2}{3}(.1)}\right) \right]$
For $i= 1$, we have:
$x_0 = 1, y_0 = 0, y_1 = 0 + \dfrac{1}{40}\left[3 + 3\dfrac{3(1+(2/3)(.1)) - 2(2(.1))}{1 + (2/3)(.1)} \right] = 0.271875$
Continuing this way, we generate the table:
$~~~~~\text{Step} ~~|~~ x ~~~|~~ y $
$~~00 ~~| 1.0 ~~| ~~ 0.00000 $
$~~01 ~~| 1.1 ~~| ~~ 0.271875 $
$~~02 ~~| 1.2 ~~| ~~ 0.503084 $
$~~03 ~~| 1.3 ~~| ~~ 0.705482 $
$~~04 ~~| 1.4 ~~| ~~ 0.886908 $
$~~05 ~~| 1.5 ~~| ~~ 1.05271 $
$~~06 ~~| 1.6 ~~| ~~ 1.20664 $
$~~07 ~~| 1.7 ~~| ~~ 1.35138 $
$~~08 ~~| 1.8 ~~| ~~ 1.48891 $
$~~09 ~~| 1.9 ~~| ~~ 1.6207 $
$~~10 ~~| 2.0 ~~| ~~ 1.74786$
The exact solution is given by:
$$y(x) = \dfrac{x^3-1}{x^2}$$
At $x=2$, we have: $y(2) = \dfrac{7}{4} = 1.75$
Compare that to Heun's (Improved Euler's) method, which has $1.74786$.
Look at how much better this estimate is over regular Euler's from earlier.
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H: Holomorphic function with zero derivative is constant on an open connected set
I was wondering about this fact, as I do not know how to prove it correctly. I tried with Cauchy-Riemann, but since they are PDE's I found it hard to show that this is the only thing that can cause this zero derivative
AI: By integration: Let $\Omega \subset \mathbb{C}$ be a connected open set, $f : \Omega \to \mathbb{C}$ be a holomorphic function and $x_0 \in \Omega$. For every $x \in \Omega$, there exists a path $\gamma : [0,1] \to \Omega$ such that $\gamma(0)=x_0$ and $\gamma(1)=x$.
$$0= \int_{\gamma} f'(z)dz= \int_0^1f'(\gamma(t)) \gamma'(t)dt= \int_0^1(f \circ \gamma)'(t)dt= f(x)-f(x_0)$$
Therefore, $f(x)=f(x_0)$ for every $x \in \Omega$, ie. $f$ is constant.
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H: Random vector with uniform distribution.
Let $(X,Y)$ be a random vector with uniform distribution at $0 \leq x \leq 1$, $x \leq y \leq x+h$ with $0<h<1$.
Find $E(X)$ and $E(XY)$.
What i did:
(1) Find densities:
$f_X(x) = \left\{
\begin{array}{lr}
1 & : 0 \leq x \leq 1 \\
0 & : otherwise
\end{array}
\right.$
$f_Y(y) = \left\{
\begin{array}{lr}
1/h & : x \leq y \leq x+h \\
0 & : otherwise
\end{array}
\right.$
(2) Find expected values
$E(X) = \displaystyle\int_0^1xf_X(x)dx = \displaystyle\int_0^1xdx =1/2$
$E(XY) = \displaystyle\int_{-\infty}^{+\infty}\displaystyle\int_{-\infty}^{+\infty}xf_{X,Y}(x,y)dxdy$, this mean i should find $f_{XY}$ or there's another way to do it?. Because here i don't know how can i get $f_{XY}$, my first attempt was consider $f_{XY}= f_{X}f_{Y}$ but it's only true if $X$ and $Y$ are independent.
AI: Outline of Solution: We give two approaches, with the second significantly simpler than the first.
We have a constant density over a certain parallelogram. The corners are $(0,0)$, $(1,1)$, $(1,2)$, and $(0,1)$.
The density function of $X$ is right, it is uniform. The density function of $Y$ is not right. Even the shape is not right. The density function of $Y$ can only involve the variable $y$, you have to "integrate $x$ out. A look at the picture shows that $Y$ lives on the interval $[0,2]$, and that the densities near $y=0$ and $y=2$ are close to $0$. You can get the density of $Y$ either by integration or more easily by geometry. (First find the cdf. It will be a different formula for $y\lt 1$ than for $y\gt 1$).
To find the expectation of $XY$, we luckily don't need the distribution of $XY$. Note that our parallelogram has area $1$, so we just need to integrate $xy$ over the parallelogram.
Another way: Let $Z=Y-X$. Then $(X,Z)$ is uniform over the unit square. So to find the density of $Y$, you need the density of a sum of two uniforms. This may be a problem you have seen before.
To find the expectation of $XY$, we need the expectation of $X(X+Z)$. The expectation of $X^2$ is easy, you know the density of $X$. The expectation of $XZ$ is also easy. Add.
|
H: If $\tan A=2$, find the possible values of $\csc A$
If $\tan A=2$, find the possible values of $\csc A$.
Can someone please show me thorough steps as to how to do this question?
AI: The $\cot(A) = 1/2$, then use $\csc^{2}(A) = 1+ \cot^{2}(A)$ to get $\csc(A)= \pm \sqrt{\dfrac{5}{4}}$. Depending on increasing level of trig. sophistication the answer is $\sqrt{\dfrac{5}{4}}$ or $\pm \sqrt{\dfrac{5}{4}}$.
|
H: a basic problem about lagrange multipliers
Find the max and min values of the function $f(x,y)=x^2+y^2$ under the restriction $g(x,y)=\frac{x^2}{2}+y^2-1=0
$
Note that we can use Lagrange Multipliers Theorem , since $grad(g(p))\ne 0$ $\forall p$ in the ellipse.
Then we put the system of equations $$
\left( {2x,2y} \right) = \nabla f\left( {x,y} \right) = \lambda \nabla g\left( {x,y} \right) = \left( {\lambda x,2\lambda y} \right)
$$
Then if $x,y \ne 0 $ I have two values for $ \lambda =1,2$ But I can't find $x,y$.
Maybe I'm doing something wrong. Please help me.
AI: You want to solve the system
$$
\frac{x^2}{2}+y^2-1=0\\
2x=\lambda x\\
2y=2\lambda y
$$
for $(x,y,\lambda)$. Clearly any triple with both $x$ and $y$ non-zero is not a solution because it would force $\lambda$ to be equal to both $1$ and $2$. Similarly, any triples with both $x$ and $y$ equal to zero are not solutions because such triples cannot satisfy the first equation.
If $x=0$ and $y\neq 0$, your system of equations reduces to
$$
y^2=1,\quad 2y=2\lambda y,
$$
with solutions $(0,\pm1,1)$. These two points correspond to values of your target function $x^2+y^2$ of $1$.
If $y=0$ and $x\neq 0$, the system of equations simplifies to
$$
x^2=2,\quad 2x=\lambda x,
$$
with solutions $(\pm\sqrt{2},0,1)$; the value of the target function in this case is $2$.
The maximum of the function $(x,y)\mapsto x^2+y^2$ under the constraint $x^2/2+y^2=1$ is thus equal to $2$, and the minimum is equal to $1$.
|
H: Prove that $\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$
The question is:
Prove that:
$$ \cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B} $$
I have tried expanding it as $\dfrac{\cos(A+B)}{\sin(A+B)}$ and $\dfrac{1}{\tan(A+B)}$.
AI: Witout using the identity for $\tan(A+B)$:
$$\cot(A+B)=\frac{\cos(A+B)}{\sin(A+B)}=\frac{\cos A\cos B-\sin A\sin B}{\sin A\cos B+\cos A\sin B}=\frac{\sin A\sin B(\cot A\cot B-1)}{\sin A\sin B(\cot B+\cot A)}$$
|
H: a problem using Lagrange multipliers
Prove that $
\frac{{n!}}
{{n^{\frac{n}
{2}} }}
$ is the max of the function $
f\left( x \right) = \prod\limits_{i = 1}^n {x_i }
$ under the restriction $
g\left( x \right) = \sum\limits_{i = 1}^n {\frac{{x_i ^2 }}
{{i^2 }} - 1 = 0}
$.
Well Using LaGrange multiplier I have the system $
\nabla f = \lambda \nabla g
$ then for each $k$ we have that $
\prod\limits_{i \ne k} {x_i } = \frac{{2\lambda x_k }}
{{k^2 }}
$ Thus $
\prod\limits_{i = 1}^n {x_i } = \frac{{2\lambda x_k ^2 }}
{{k^2 }}
$
Considering the following Sum, we conclude that:$$
n\left( {\prod\limits_{i = 1}^n {x_i } } \right) = \sum\limits_{k = 1}^n {\left( {\prod\limits_{i = 1}^n {x_i } } \right)} = \sum\limits_{k = 1}^n {\frac{{2\lambda x_k ^2 }}
{{k^2 }}} = \lambda
$$
I don't know how to continue :S
AI: Put $x_i^2=\frac{i^2}{2\lambda}\prod_{j}x_j$ in the constraint; you should obtain $1=\sum_{i=1}^n \frac{x^2_i}{i^2}=\sum_{i=1}^n \frac{\prod_{j}x_j}{2\lambda},$ i.e.
$$\lambda=\frac{n}{2}\prod_{j}x_j.$$
Substitute the value of the Lagrange multiplier $\lambda$ back in $x_i^2$ getting
$$x^2_i=\frac{i^2}{n},$$
for all $i=1,\dots,n$. Selecting the positive root to deduce the $x_i$'s, the maximum of $f$ is computed at
$$x=(\frac{1}{\sqrt{n}},\dots,\frac{n}{\sqrt{n}})$$
and is equal to
$$f(x)=\frac{1}{\sqrt{n}}\cdots\frac{n}{\sqrt{n}}=\frac{1\cdots n}{(\sqrt{n})^n}=\frac{n!}{n^{\frac{n}{2}}}.$$
|
H: Unknown symbol '#' in set
I am reading a text on Complexity theory. There is a set whose notation I cannot understand:
"Let $\sum$ = {0,1,#}"
From the context, and given that the book is used computer science courses, it seems like $\sum$ is the set which contains all combinations of bit strings up to length '#'.
In the text, a subset of $\sum$ is being used to define a set of palindromes in a Turing machine.
I would like to know more about this # symbol.
Another notation that I don't understand is in the following expression:
$x \in \{0,1\}^{n/4}$
What is the purpose of the $n/4$?
AI: $\Sigma = \{0,1,\#\}$ means the set of these 3 symbols, which one usually calls alphabet in complexity theory. And $x\in \{0,1\}^{n/4}$ means that $x$ is a word of length $n/4$ over the alphabet $\{0,1\}$.
These are exactly the same definitions as in mathematics, as 'word' means exactly the same as finite sequence which is a function from $\frac{n}{4} \rightarrow \{0,1\}$!
|
H: "Nice" application of the fundamental group
I'm looking for an example of a topological result which is easy to prove using the fundamental group, but hard or impossible elementarily.
First I thought about something like $\mathbb{R}^2\not\cong\mathbb{R}^n$ for $n>2$, but this should be possible to show without using the fundamental group but just simple connectedness.
Do you have any ideas? Thanks!
Edit: What about Brouwer's fixed point theorem; does it have an easy proof without using the fundamental group? If not, I think it would fit.
AI: Brouwer's fixed point theorem can be proved quite elementarily using Sperner's Lemma.
The fundamental group can be used quite effectively to show certain topological spaces are not homotopic (and thus also not homeomorphic). For instance, none of the following spaces are homotopic: The torus, $\mathbb R^2$, projective plane.
With a little bit of the theory of covering spaces, one gets a very short and elegant proof that a free group on two generators contains subgroups that are free groups on any finite number of generators. This result is not terribly hard, but certainly not elementary in group theory.
There is a nice proof of the fundamental theorem of algebra using the fundamental group (essentially using degree theory).
|
H: Black and white balls in the urn. What is the probability of...?
There are $100$ balls in the urn - $50$ black and $50$ white. A person randomly chooses a batch of $50$ balls and replaces it with a batch of inversely colored balls (i.e. for example a batch of $34$ black and $16$ white balls would be replaced by a batch of $16$ black and $34$ white balls).
It is obvious that after replacement took place there could be anything from all black to all white balls. But what is the probability that one can still find at least $40$ balls of either color?
Honestly I'm not even sure how to figure out the probability of urn still having 50-50 balls, and for my original question apparently I have to sum the probabilities of $60-40$, $59-41$, $58-42$, $\ldots$, $41-59$, $40-60$ ball combinations.
AI: If the person chooses $k$ black and $50-k$ white balls, then after inverted replacement, we have $100-2k$ black and $2k$ white balls.
You will find at least $40$ balls, iff $2k\geq 40$ or $100-2k\geq 40$, in other words, iff $20\leq k\leq 30$.
There are $\binom{100}{50}$ possible batches. Now count the batches with exactly $k$ black balls. You can choose the $k$ black balls in $\binom{50}{k}$ ways and the $50-k$ white balls in $\binom{50}{50-k}=\binom{50}{k}$, ways. So the result should be:
$$\sum\limits_{k=20}^{30}\frac{\binom{50}{k}^2}{\binom{100}{50}}\approx 97\%$$
|
H: Expected number of shots for a game to end
A basketball player plays a shooting game. He gets +1 point if he scores a basket and -2 points if he misses. He starts with 0 points. The game ends when the player reaches +10 or -10. What is the expected number of shots taken for a game to end, given a player scores a basket with probability p.
AI: For $-11\le n\le 10$, let $e_n$ denote the expected number of rounds until a score $\le -10$ or $\ge 10$ is reached when starting with a score of $n$.
Clearly, $e_{-11}=e_{-10}=e_{10}=0$ while for $-10<n<10$ we have $e_n=1+pe_{n+1}+(1-p)e_{n-2}$. This gives you $19$ linear equations in $19$ unknowns.
|
H: I solved the question. But I am asking a little bit. $\det\big(D(f\circ g)(a)\big)=?$
$\color{red}{\text{b)}}$ If $n=m$ and $f:B_r\big(g(a)\big)\to\mathbb{R}^n$ is differentiable at $g(a)$, prove that $$\det\big(D(f\circ g)(a)\big)=\det\big(Df(g(a))\big)\det\big(Dg(a)\big)$$ Solution: $$D(f\circ g)(a)=Df(g(a))Dg(a)\quad\text{by the chain rule.}$$
After here, how can I show its determinant?
AI: Let $f:B_r(g(a))\subset \mathbb R^n\rightarrow \mathbb R^n$ and $g:\Omega\subseteq \mathbb R^n\rightarrow \mathbb R^n$ both of class $\mathcal C^1$. With $\Omega$ we denote an open subset of $\mathbb R^n$. We select the point
$$a=(x_1,\dots,x_n)$$
in $\Omega$, with $g(a)=(y_1\dots,y_n)$. By the chain rule
$$ (D(f\circ g)(a))_{ij}:=\frac{\partial f_i }{\partial y_k}\frac{\partial y_k}{\partial x_j}= \sum_{k=1}^n(D(f)(g(a))_{ik}D(g)(a)_{kj},$$
i.e.
$$D(f\circ g)(a)=D(f)(g(a))D(g)(a),~~(*)$$
as you correctly remarked. Eq. $(*)$ is an equality of $n\times n$ matrices. As
$$\operatorname{det}(AB)=\operatorname{det}(A)\operatorname{det}(B),~~ (**)$$
for all $n\times n$ matrices $A$ and $B$, then you arrive at the statement. You can prove (**) using the Cauhy-Binet formula shown in Cauchy-Binet formula for determinants.
|
H: Parameter integrals - show continuity
One little question concerning the proof of
Let $I=[a,b], D=[c,d]$ compact ranges and $f\in C^0(I\times D)$ with values in $\mathbb{R}$ or $\mathbb{C}$. Then
$$
F(t):=\int_a^b f(x,t)\, dx
$$
is continious for all $t\in D$.
Proof: $I\times D$ is compact in $\mathbb{R}^2$. Therefore $f$ is uniformly continious on $I\times D$, that means:
For any $\varepsilon > 0$ there is a $\delta >0$ in such a way that
$$
\lvert f(x_1,t_1)-f(x_2,t_2)\rvert\leq\frac{\varepsilon}{b-a}
$$
for all $(x_i,t_i)\in I\times D$ with $\lvert (x_1,t_1)-x_2,t_2)\rvert\leq\delta$.
My question is: Where does the denominator $(b-a)$ come from? I cannot explain the denominator.
AI: \begin{gather}
\begin{split}
&|F(t_1) - F(t_2)| \\
&= |\int_a^b f(x,t_1) dx - \int_a^b f(x,t_2) dx |\\
&\leq \int_a^b|f(x,t_1) - f(x,t_2)| dx\\
\end{split}
\end{gather}
If $|f(x,t_1) - f(x,t_2)| < \epsilon $, then $|F(t_1) - F(t_2)| < \epsilon (b-a)$.
So if we want to show $|F(t_1) - F(t_2)| < \epsilon $ we should take $|f(x,t_1) - f(x,t_2)| < \epsilon (b-a) $\
So the denominator $b-a$ comes.
|
H: Diameter of finite set of points is equal to diameter of its convex hull
Let $M\subset \mathbb{R}^2$ be a finite set of points, $\operatorname{C}(M)$ the convex hull of M and
$$\operatorname{diam}(M) = \sup_{x,y\in M}\|x-y\|_2$$
be the diameter of $M$
What I want to show now is, that it holds
$$\operatorname{diam}(M) = \operatorname{diam}(\operatorname{C}(M))$$
Because $$M\subseteq\operatorname{C}(M)$$ we obtain $$\operatorname{diam}(M) \le\operatorname{diam}(\operatorname{C}(M))$$ but how to proof that $$\operatorname{diam}(M) \ge \operatorname{diam}(\operatorname{C}(M))$$
I suppose it should be possible to construct a contradiction assuming $\operatorname{diam}(M) <\operatorname{diam}(\operatorname{C}(M))$ but i do not see how at this moment.
AI: Hint:
Prove this for a triangle and then use the fact that for every point of $C(M)$ there is a triangle that contains it, there are many ways to go from there.
I hope this helps ;-)
|
H: Use the compound angle identity to prove $\sin(90^\circ +A)=\cos A$
Use the compound angle results to prove
$$\sin(90+A)=\cos A.$$
I know this is rather easy question, but I cant wrap my head around it. So can you please show me how to do it. thanks
AI: I'm not sure what you call "compound angle results" to, but we can prove the claim either by
$$\sin(90+\alpha)=\sin(180-(90+\alpha))=\sin(90-\alpha)=\cos\alpha$$
or by
$$\sin(90+\alpha)=\underbrace{\sin90}_{=1}\cos\alpha+\sin\alpha\underbrace{\cos90}_{=0}=\cos\alpha$$
|
H: proving a sum of binomial coefficients
How can i prove that $\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=2^{2n-1}$
I tried using induction and pascal's identity but it didn't help me.
AI: We know:$$ (1+x)^{2n}=\sum_{j=0}^{2n}{2n \choose j}x^j$$
Now put $x=1$ and and $x=-1$ in turn and add,
$\displaystyle 2^{2n}=(1+1)^{2n}+(1-1)^{2n}=\sum_{j=0}^{2n}{2n \choose j}+\sum_{j=0}^{2n}{2n \choose j}(-1)^j=\sum_{j=0}^{n}{2n \choose 2j}+\sum_{j=0}^{n-1}{2n \choose 2j+1}+\sum_{j=0}^{n}{2n \choose 2j}-\sum_{j=0}^{n-1}{2n \choose 2j+1}=2\sum_{j=0}^{n}{2n \choose 2j}$
$\displaystyle \Rightarrow 2^{2n-1}=\sum_{j=0}^{n}{2n \choose 2j}$
|
H: Can I change the order of two terms when factoring: $x^2(x^2-4-3x)$ to $x^2(x^2-3x-4)$?
I'm doing homework and I'm stuck on this assignment:
$$x^4 - 4x^2 - 3x^3$$
I figured this would equal
$$x^2(x^2-4-3x)$$
Now I know if I would change the order to
$$x^2(x^2-3x-4)$$
I can factorise it again, but my question is if it's allowed changing the order like that in algebra.
AI: Yes, the law of commutativity (together with the law of associativity) allows you to rearrange summands (or factors) in arbitrary order. (However, because these laws hold for addition, not for subtraction, You better view the minus sign as part of the coefficient for this, i.e. $x^2-4-3x=x^2+(-4)+(-3x) = x^2+(-3x)+(-4) = x^2-3x-4$, just to prevent you from falling for $x-5=5-x$ or the like)
|
H: Solving the multiple integral $\int \int_D \sqrt{x^2+y^2+3}dx dy$, D:{$1\le x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3}$}
$\iint_D \sqrt{x^2+y^2+3}\, dx dy$, $D=\left\{(x,y) \in \mathbb{R}^2 \mid 1 \leq x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3} \right\}$.
So I've started by drawing two circles, One with $R_1 =1$, Other with $R_2=2$
Then i drawed the lines $y_1=\sqrt3x$ and $y_2 = \frac x{\sqrt 3}$.
Using the polar system i'll need to calculate:
$\int^{\pi/3}_{\pi/6}dθ\int_1^2\sqrt{r^2+3}*r*dr$
Using substiution on: $\int_1^2\sqrt{r^2+3}*r*dr$ i get: $\frac {7^{3/2}}{3}-\frac {4^{3/2}}{3}$.
Then i need to multiply by $\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
And the final answer is: $\frac {7^{3/2}-4^{3/2}}{18}\pi$.
The answer should be: $\frac {5(7\sqrt 7 -8)}{18}$, I'm pretty close but it seems like i'm missing something, I tried solving this one a couple of times and still got to the same answer.
I need help, What am i missing?
AI: I think you've done almost everything right, but $\;1\le r\le\frac6{\sqrt{10}}=\frac{3\sqrt2}{\sqrt 5}\;$ ! , so
$$\int\limits_1^2r\sqrt{r^2+3}\,dr=\left.\frac12\frac23(r^2+3)^{3/2}\right|_1^2=\frac13\left(7^{3/2}-8\right)$$
and together with
$$\int\limits_{\pi/6}^{\pi/3}d\theta=\frac{\pi}6\;,\;\;\text{we get:}$$
$$\frac{\pi}{18}(7\sqrt 7-8)$$
I can't understand where does that five you wrote there come from...
|
H: Simple groups of order 300
I am aware of No simple group of order $300$.
It is said that "there would be 6 Sylow 5-groups, one of which will have an index of 6", but why does one have index 6? If we write $|G|=p^k m$ where $p\nmid m$, then by Lagrange's theorem any $p$-Sylowgroup has index $m$, which in this case is 12?
AI: What you said is said in that link is a serious blunder: as all the Sylow $\,5-$subgroups obviously have the same order then they all have the same index.
What is true is that the number of different Sylow $\,5-$subgroups equals the index of the normalizer of any of them, so we have a subgroup $\,N\le G\,$ with $\,[G:N]=6\;$.
This already leads to a contradiction if we suppose $\,G\;$ is simple, as then the regular action of $\,G\,$ on $\,N\,$ gives us an injective homomorphism $\,G\to S_6\,$...but this is impossible (why?) .
|
H: Does $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge?
Sorry for the oversimplified question, but does the series $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge? The ratio test of it gives the result of "1".
Thanks a lot.
AI: If the ratio test gives $1$ then we are in a situation where we can't conclude wether the series is convergent or divergent.
There are several ways to prove that your given series is divergent and one way is: let the partial sum be
$$S_n=\sum_{k=1}^n \frac{1}{1+k}$$
Then we have
$$S_{2n}-S_n=\sum_{k=n+1}^{2n}\frac{1}{1+k}\geq n\times\frac{1}{1+2n}\geq\frac{1}{3}\tag{1}$$
If the series were convergent, then the sequence $(S_{2n}-S_n)$ would have been convergent to $0$. This is not the case, as shown by the inequality $(1)$.
|
H: DFT of binomial coefficients
I want to compute the DFT of a vector whos entries are binomial coefficients, i.e.
$$(v)_k = \binom{N-1}{k},$$
where $k$ runs from 0 to $N-1$.
I use the DFT matrix
$$F_{k,j} := \frac{1}{\sqrt{N}}e^{-\frac{2\pi i k j}{N}}.$$
So I want to compute
$$(Fv)_\ell = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} \binom{N-1}{k}e^{-\frac{2\pi i \ell k}{N}}.$$
I don't see how to evaluate this sum. I noticed the symmetry of the binomial coefficients and the elements of the DFT matrix, but I'm not sure how to exploit it.
Any suggestions?
AI: You could try to apply the binomial formula with well chosen numbers, like $1$ and $\exp(-2\pi i l/N)$.
EDIT:
The binomial formula is this one:
$$(a+b)^m=\sum_{k=0}^m\binom{m}{k}a^k\; b^{m-k}\, .$$
|
H: Chain rule question
Use chain rule to calculate $\frac{\partial f}{\partial u}$ and $\frac{\partial f}{\partial v}$ when:
$f(x,y) = e^{xy}, x(u,v) = u^2 + v^2, y(u,v) = u^2 - v^2$
So I said:
$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial u} = e^{xy} \cdot ( 2uy + 2ux)$.
But the answer sheet does not agree with me.
What did I do wrong?
AI: First of all, you are computing
$$\frac{\partial f}{\partial u}(u,v),$$
(this notation is a bit sloppy; I should introduce the function $g$, s.t. $g(u,v)=(x(u,v),y(u,v))$ and compute the partial derivatives of the composition $f\circ g$)
so your expression must be a function of $u$ and $v$ (and not a mix of $u$, $v$, $x$ and $y$). Moreover
$$\frac{\partial y}{\partial u}(u,v)=2u.$$
In summary
$$\frac{\partial f}{\partial u}(u,v)=ye^{xy}2u+xe^{xy}2u=2ue^{xy}(x+y)=
2ue^{(u^2+v^2)(u^2-v^2)}(2u^2)=4u^3e^{u^4-v^4}.$$
I hope it helps!
|
H: Set contains a proper infinite subset
Consider that we define an "infinite" set as one which contains some proper subset that is in bijection with itself. Now suppose that $B$ is a set and $A \subset B$ is an infinite proper subset of $B$. Can we prove using just that definition and elementary facts about functions that $B$ must also be infinite?
AI: You are assuming $A$ to be infinite according to your definition, right?
Let $C$ be a proper subset, and $f : A \to C$ a bijection.
Consider the proper subset of $B$ given by $D = C \cup (B \setminus A)$.
Now consider the bijection $g : B \to D$ which is the identity on $B \setminus A$, and $f$ when restricted to $A$.
|
H: Proving a Lambert series identity from Ramanujan's Collected Papers
While studying Ramanujan's Collected Papers I came across an identity $$q(1 + q + q^{3} + q^{6} + \cdots)^{8} = \frac{1^{3}q}{1 - q^{2}} + \frac{2^{3}q^{2}}{1 - q^{4}} + \frac{3^{3}q^{3}}{1 - q^{6}} + \cdots$$ which I am unable to establish (although the identity looks simple). I tried to use the the following identities:
$\displaystyle Q(q) = 1 + 240\left(\frac{1^{3}q}{1 - q} + \frac{2^{3}q^{2}}{1 - q^{2}} + \frac{3^{3}q^{3}}{1 - q^{3}} + \cdots\right) = \left(\frac{2K}{\pi}\right)^{4}(1 + 14k^{2} + k^{4})$
$\displaystyle Q(-q) = \left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2} + 16k^{4})$
$\displaystyle Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})$
where we have $q = e^{-\pi K(k')/K(k)}$ and was trying to put the RHS (of the identity to be established) as a combination of the above $Q$ functions but was not able to do so. In this way I thought to transform the RHS in terms of $K, k$ and then reach the LHS which is $q\psi^{8}(q)$ where Ramanujan's $\psi(q)$ is related to Jacobi's theta function via $\vartheta_{2}(q) = 2q^{1/4}\psi(q^{2})$. I was able to write the RHS as
$\displaystyle \sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{2n}} = \frac{1}{2}\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}} + \frac{1}{2}\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 + q^{n}} = S_{1} + S_{2}$
where $S_{1}$ can be expresssed in terms of $Q(q)$, but somehow could not express $S_{2}$ in terms of $Q$ functions. Maybe I am on the wrong path. If there is any simpler way to prove the identity please let me know.
AI: Paramanand Singh already found the answer. Nevertheless I'll try to post an answer here.
To avoid issues with branch cuts of $K$, let us replace
$$\frac{2K}{\pi} = \vartheta_3^2(q);\quad
k = \frac{\vartheta_2^2(q)}{\vartheta_3^2(q)}$$
where
$$\vartheta_3(q) = \sum_{n\in\mathbb{Z}} q^{n^2}$$
is another Jacobi thetanull function.
Thus
$$Q(q)-Q(q^2) = 15\,\vartheta_2^4(q)\,\vartheta_3^4(q)$$
Now use the doubling formula:
$$\vartheta_2^2(q) = 2\,\vartheta_2(q^2)\,\vartheta_3(q^2)$$
Therefore
$$Q(q)-Q(q^2) = \frac{15}{16}\vartheta_2^8(q^{1/2})
= 240\,q\,\psi^8(q)$$
Consequently, your LHS equals
$$q\,\psi^8(q) = \frac{1}{240}\left(Q(q)-Q(q^2)\right)
=\sum_{n=1}^\infty\left(\frac{n^3\,q^n}{1-q^n}
-\frac{n^3\,q^{2n}}{1-q^{2n}}\right)$$
$$= \sum_{n=1}^\infty n^3\,\frac{q^n\,(1+q^n)-q^{2n}}{1-q^{2n}}
= \sum_{n=1}^\infty \frac{n^3\,q^n}{1-q^{2n}}$$
which is the RHS, QED.
(Edit: typo corrections.)
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H: prove：$\left|x+\frac{1}{x}\right|\geq 2$
prove：
$\left|x+\frac{1}{x}\right|\geq 2$
Can I just use
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$
and
$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?
AI: Hint: Prove that $$\left(x+\frac{1}{x}\right)^2 \geq 4$$
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H: Descartes' theorem: Find the midpoint of the $4$th circle
I'm thinking about Descartes' theorem "Wikipedia".
I understood how to find the radius with algebra. Now I'm trying to use ruler and compass to find the midpoint of the 4th circle.
I thought about using the midpoints of $k_1-k_3$ to get a triangle and the centroid as midpoint, but it's not the right approach. Any ideas, what point I have to find to get the midpoint?
And further more: Is it possible using ruler and compass to get the midpoint of a $4$th circle touching the first three, if $k_1-k_3$ are not touching?
Any toughts about this are welcome.
AI: A convenient transformation, when you are dealing with circles, is inversion.
Inversion takes circles and straight lines to circles and straight lines. Moreover, if a circle passes through the center of inversion then it turns into a line. Here we have circles tangent to each other. So, if we use as center the tangency point of two of the circles, they both will turn into lines. Even more, since inversion preserves angles, these two lines will be parallel.
Take the tangency point, $P$, of two of the given circles, $c_1$ and $c_2$, and apply inversion with center $P$ and radius, say $1$.
(Actually it might be better to use a radius, if possible, such that the circles get intersected at two points. That way you know that those two points are not going to move during inversion.)
The two circles $c_1$ and $c_2$ are going to become parallel lines, $L_1$ and $L_2$, so the center of the circle you are looking for lies in the middle line, $L_3$. Notice that $L_1$ and $L_2$ are easy to construct since they pass through the intersection points of a circle with center $P$ and radius $1$ and each circle $c_1$ and $c_2$. In the worst case you can always invert two or three points of the circle, as needed, and the inverse line or circle will pass through those inverted points.
The other circle $c_3$ becomes a circle $C_3$ in between the two lines because tangency (i.e. angles) is preserved. So, the tangency points of the inversion $C_4$, of the circle to be constructed to that circle $C_3$ are the intersection of the middle line $L_3$ with that third circle, $C_3$.
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H: Using simple linear algebra for encryption?
e.g.
the character $a = 97$ (it's computer decimal format, commonly known)
and then using a pattern/key like $y = 31 x + 5$ to get $3012$ (substitute $97$ into $x, y$ is now the encrypted code).
How easy/hard would it be for someone to crack a code like this? 3012 3197 3135 (without them knowing the pattern)
If someone had these encrypted codes, are there any methods they could use besides trial and error to figure out the pattern and then the code?
AI: What you have described is nothing more than a single-alphabet substitution cipher. This sort of cipher has been around for centuries, with variations on the type of mapping. It is very easy/simple to break if given enough ciphertext encoded with the method.
To break this cipher, first analyze the frequency of the resulting numbers. (e.g. 3012 occurs what percent of the time in the message?) Now, compare this to a frequency table of letters in the English alphabet. This will give you a starting place. You can also look at the frequency of pairs of letters. By that time, you should have a pretty good guess for the substitutions.
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H: An exercise in Fulton's Algebraic Curves
I'm doing Fulton's Algebraic Curves [Ex 1.34]
Let $R$ be a UFD. $(a)$ Show that a monic polynomial of degree two or three in $R[X]$ is irreducible if and only if it has no roots in $R$. $(b)$ The polynomial $X^2 - a \in R[X]$ is irreducible if and only if $a$ is not a square in $R$.
But I think the require of $R$ a UFD is not necessary. I think we only need $R$ to be an integral domain (because it requires that the polynomial must be monic). Is my thought correct?
Thank you!
AI: We define the notion of irreducible elements only in integral domains, so, yes, $R$ should be an integral domain (though commutativity is not really used).
Apart from that, the proof goes through:
(a) Assume $f$ with $\deg f>1$ has a root $a$ in $R$. Using polynomial division (thanks to $X-a$ being monic) we have $f(X)=(X-a)g(X)+b$, find $b=f(a)=0$ and $\deg g>0$, hence $f(X)=(X-a)g(X)$ is the product of two nonunits, i.e. reducible.
On the other hand if monic $f$ can be written as product of two nonunits, $f=gh$, then the product of the leading terms equals the leading term $1$ of $f$. Hence after moving a unit constant factor around, both $g$ and $h$ can be assumed monic. As $g,h$ are nonunits they are not simply the constant $1$, so they both have positive degree and $\deg g+\deg h\le 3$ implies that at least one of $g,h$ is linear, i.e. of the form $X-a$ with $a\in R$, which implies that $f(a)=0$.
(b) Using (a), a root of $X^2-a$ is precisely the same as a witness that $a$ is a square.
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H: prove that $f(t)$ is orthogonal to $f'(t)$ for som all $t \in I$
Suppose that $I$ is nonempty open interval and that $f: I \to \Bbb R^m$ is differentiable on $I$
If $f(I) \subseteq \partial B_r(0) $ for some fixed $r>0$, prove that $f(t)$ is orthogonal to $f'(t)$ for som all $t \in I$
I guess I need to use the chain rule. Sorry, I cannot say any idea about the question.
But what is orthogonal?
How can I link between orhogonal and the chain rule of partial derivatives ?
I am glad to show me the proof step by step. Thanks for helping:)
AI: Edit I am just adding some details to the approach suggested by @David Mitra in the above comments.
Hint: by definition,
$$f(t)=(x(t),y(t))$$
with $x^2(t)+y^2(t)=r^2$ for all $t\in I$. The trick, as pointed out by @David Mitra is to note that
$$\|f(t)\|^2=x^2(t)+y^2(t)=r^2,$$
i.e. $\|f(t)\|$ is constant for all $t\in I$. Try now to compute the derivative
$$\frac{d}{dt}\|f(t)\|^2 $$
using the definition of $\|f(t)\|^2$. You should arrive at
$$\frac{d}{dt}\|f(t)\|^2=2\langle f(t),f'(t)\rangle,$$
where $f'(t)=(x'(t),y'(t))$. This is sufficient to finish the problem (why?).
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H: Why the sum of the squares of the roots of the $n$th Hermite polynomial is equal to $n(n-1)/2$?
How to prove that the sum of the squares of the roots of the $n$th Hermite polynomial is $\frac{n(n-1)}{2}$?
I tried with Vieta formulas, but it's hard. I appreciate a proof or reference to it.
An idea is to use the definition of sum of Hermite polynomials, but do not know.
AI: Let us write
\begin{align}H_n(x)&=A_n(x-x_1)\ldots(x-x_n)=\\
&=A_n\left(x^n-e_1(x_1,\ldots,x_n)x^{n-1}+e_2(x_1,\ldots,x_n)x^{n-2}+\mathrm{poly}_{n-3}(x)\right),\tag{1}
\end{align}
where $e_k(x_1,\ldots,x_n)$ denote elementary symmetric polynomials:
\begin{align}
& e_1(x_1,\ldots,x_n)=\sum_{k=1}^nx_k,\\
& e_2(x_1,\ldots,x_n)=\sum_{1\leq i<j\leq n}^n x_ix_j.
\end{align}
We want to find
$$\sum_{k=1}^{n}x_k^2=e_1^2(x_1,\ldots,x_n)-2e_2(x_1,\ldots,x_n),\tag{2}$$
and therefore it will suffice to know the coefficients of $x^n$, $x^{n-1}$ and $x^{n-2}$ in $H_n(x)$. But they can be determined from the series representations of Hermite polynomials:
$$H_n(x)=2^n\left(x^n-\frac{n(n-2)}{4}x^{n-2}+\mathrm{poly}_{n-4}(x)\right).\tag{3}$$
Together with (1) and (2), this gives the result:
$$\sum_{k=1}^{n}x_k^2=\frac{n(n-1)}{2}.$$
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H: Can we check whether a Cantor set is self-similar or not?
Given a Cantor set $C$ on the real line, do we have some ways to determine whether it is self-similar or not? In particular, how can we check that $C$ is not self-similar?
Edited:
Definition: Let $\{f_i\}_i$ be a family of contraction maps, i.e. $|f_i(x)−f_i(y)|=r_i|x−y|$ where $0<r_i<1$. $C$ is self-similar if there are such maps so that $E=\bigcup_i f_i(E)$.
To make things concrete, we may consider the following central Cantor set as an example.
Suppose $r_k = \frac{1}{k+2}$ for $k \geq 1$. Let $I_e = [0,1], I_0 = [0, r_1], I_1 = [1-r_1 , 1]$. For each $k \geq 1$, $w \in \{0,1\}^k$, let $I_w$ be a subinterval at level $k$. We take $I_{w1} , I_{w2}$ to be two subintervals of $I_w$ placed on the left and right with lengths $$|I_{w1}| = |I_{w2}| = |I_w| \cdot r_{k+1} . $$
Then $C= \bigcap_{k=1}^\infty \bigcup_{w \in \{0,1 \}^k} I_w$ is a Cantor set. Is it not self-similar?
AI: First of all, self-similar sets are not necessarily topological Cantor sets. For example, closed intervals are self-similar, as it is easy to check. However, if we assume (using your notation) that the pieces $f_i(E)$ are pairwise disjoint, then $E$ is a Cantor set. This is known as the strong separation condition. When dealing with self-similar sets, a weaker separation that is often used and assumed is the open set condition: there exists a nonempty open set $U$ such that $f_i(U)\subset U$ and $f_i(U)$ are pairwise disjoint. Closed intervals are self-similar sets under the open set condition, so the open set condition is not enough to guarantee a self-similar set is a Cantor set.
In general, there isn't a simple way to determine whether a given compact set (Cantor or not) is self-similar. However, self-similar sets are more regular than arbitrary compact sets. For example:
Different concepts of fractal dimension such as Hausdorff and box-counting (Minkowski) dimension coincide for arbitrary self-similar sets.
If a self-similar set $E$ is not a point, then it has strictly positive Hausdorff dimension. This can be seen as follows: after reordering, let $f_1, f_2$ be two of the maps generating $E$ with distinct fixed points (if all the maps have the same fixed point, then $E$ equals that fixed point). Let $F$ be the self-similar set corresponding to the maps $f_1^n, f_2^n$, where $n$ is large enough that $f_1^n(I)$ and $f_2^n(I)$ are disjoint intervals, where $I$ is the closed interval joining the fixed points of $f_1, f_2$. Then $F$ is a self-similar set with the strong separation condition, for which there is a well-known formula for the Hausdorff dimension which is always strictly positive (this can also be checked directly for $F$), and $F\subset E$.
Under the open set condition (but not in general) self-similar sets have positive and finite Hausdorff measure in their dimension.
The particular set $C$ defined in the question has Hausdorff dimension $0$. This can be seen by using the natural covers by sets $I_w, w\in\{0,1\}^k$, and is essentially a consequence of the fact that $r_k\to 0$, so that the relative size of intervals of generation $k+1$ inside intervals of generation $k$ tends to $0$. Since $C$ is not a point, by 2. above, $C$ cannot be self-similar.
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H: Möbius transform calculation, over an annulus
I started learning about Möbius transformations in my Complex Analysis textbook. This question appeared as an exercise (no solutions are provided, sadly):
Let's say you have a Möbius transform that maps the annulus $r<|z|<1$ to a region bounded by two circles (for the sake of example take $|z|=1$ and $|z-1/4|=1/4$). Is this enough information to figure out $r$?
What I've tried: Trying to figure out what different compositions of Möbius transforms will affect in here. I was thinking taking the inverse somehow would be a good idea but I couldn't find my way around the details of how I'd do this.
I'm stumped, so an explanation of how this is possible would be great. Thanks!
AI: Yes, $r$ is determined by the other doubly-connected region $\Omega$ (even if we do not restrict ourselves to Möbius transformations or to domains bounded by circles). A high-tech explanation involves the concept of conformal modulus, see Conformal maps of doubly connected regions to annuli.
Here is a low-tech approach which still does not rely on solving for the Möbius map. Let $L$ be a line of symmetry of $\Omega$; in your example it's the real line. Its image under a Möbius transformation onto $r<|z|<1$ is a line (or circle) $L'$. Since $L$ is invariant under inversion with respect to either boundary component of $\Omega$, $L'$ must be invariant under inversion in the circle $|z|=1$ and under inversion in $|z|=r$. Since the composition of two latter inversions is the scaling map $z\mapsto r^2z$, it follows that $L'$ is a line through the origin $z=0$. Composition with a rotation ensures that $L'$ is the real line.
Now recall that the cross-ratio is preserved by Möbius transformations. The line $L$ crosses $\partial \Omega$ at four points: in your example they are $-1,0,1/2,1$. The line $L'$ crosses the boundary of $r<|z|<1$ at the points $-1,-r,r,1$. Therefore, the cross-ratios of these quadruples are equal:
$$(-1,0;1/2,1)=\frac{-1-1/2}{0-1/2}\cdot\frac{0-1}{-1-1}= \frac32 \tag1$$
must be equal to $$(-1,-r;r,1)=\frac{-1-r}{-r-r}\cdot\frac{-r-1}{-1-1}=\frac{(1+r)^2}{4r}\tag2$$
Hence $r=2-\sqrt{3}$.
Added: One can do without the knowledge of cross-ratio. The Möbius map sending $0,1/2,1$ to $-r,r,1$ is described by the equation
$$\frac{z-1/2}{0-1/2}\cdot\frac{0-1}{z-1} = \frac{w-r}{-r-r}\cdot\frac{-r-1}{w-1}\tag3$$
Since we also want $w=-1$ at $z=-1$, equation (3) yields the equality of (1) and (2). Incidentally, (3) also gives a formula for our Möbius map, should we want to have it.
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H: number of binary sets - combinatorics
Just ran into this question:
let $f(n,m)$ be the number of binary strings where there are at most $n$ 1's and at most $m$ 0's.
the empty string also counts as a string.
show that $f(n,m)=\binom{n+m+2}{n+1}-1$.
thanks in advance,
Yaron.
AI: This is a pretty direct proof. One would really hope for a combinatorial proof.
The obvious expression is $f(n,m)=\sum_{i=0}^n\sum_{j=0}^m \binom{i+j}{i}$.
But $$\sum_{j=0}^m\binom{i+j}{i}=\binom{i+m+1}{i+1}=\binom{i+m+1}{m}$$
So $$f(n,m)=\sum_{i=0}^n \binom{i+m+1}{m} = -1 + \sum_{i=0}^{n+1}\binom{i+m}{m} = \binom{m+n+1+1}{m+1}-1$$
Which is the result you wanted.
This is using twice the identity $\sum_{k=0}^A \binom{B+k}{B} = \binom{B+A+1}{B+1}$, the first time with $A=m$ and $B=i$, the second time with $A=n+1$ and $B=i$.
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H: The variational derivative
I'm reading about variational derivative from the following pdf-file:
http://www.cs.iastate.edu/~cs577/handouts/variational-problems.pdf
On the very first page, the author takes the partial derivative of $J(y_1, ..., y_n)$ w.r.t $y_k$ on formula $(1)$ and gets the formula $(2)$ from it. Could someone explain how did he get the result? What I get if I take the partial derivative is (Note that $y_i = y_i(x_i)$):
$$\frac{\partial}{\partial y_k} [J(y_1, ..., y_n)] = \frac{\partial}{\partial y_k} \left[ \sum_{i=0}^{n} F\left( x_i, y_i, \frac{y_{i+1}-y_i}{\Delta x}\right)\Delta x\right] = \frac{\partial}{\partial y_k} \left[ \cdots + F\left( x_{k-2}, y_{k-2}, \frac{y_{k-1}-y_{k-2}}{\Delta x}\right)\Delta x + F\left( x_{k-1},
y_{k-1}, \frac{y_{k}-y_{k-1}}{\Delta x}\right)\Delta x + F\left( x_k, y_k, \frac{y_{k+1}-y_k}{\Delta x}\right)\Delta x + F\left( x_{k+1}, y_{k+1}, \frac{y_{k+2}-y_{k+1}}{\Delta x}\right)\Delta x + \cdots\right] = \frac{\partial}{\partial y_k} \left[ F\left( x_{k-1},
y_{k-1}, \frac{y_{k}-y_{k-1}}{\Delta x}\right)\Delta x + F\left( x_k, y_k, \frac{y_{k+1}-y_k}{\Delta x}\right)\Delta x \right] = F_{y_k}\left( x_{k-1},
y_{k-1}, \frac{y_{k}-y_{k-1}}{\Delta x}\right)\Delta x + F_{y_k}\left( x_k, y_k, \frac{y_{k+1}-y_k}{\Delta x}\right)\Delta x$$
Did I forget to use chain rule or something?...Thank you for your help :)
AI: For each $k$, $y_k$ appears in $J(y_1,\dots,y_n)$ only twice, i.e. in the terms
$$F_{k-1}:=F(x_{k-1},y_{k-1},\frac{y_k-y_{k-1}}{\Delta x})\Delta x ~~(*)$$
and
$$F_{k}=F(x_{k},y_{k},\frac{y_{k+1}-y_{k}}{\Delta x})\Delta x.~~(**)$$
I think there is a typo in formula $(2)$ in your reference: the last term should have the variables
$$(x_{k},y_{k},\frac{y_{k+1}-y_{k}}{\Delta x}) $$
as it corresponds to the choice $i=k$ in $J(y_1,\dots,y_n)$. Let us introduce the notation
$$F=F(x,y,g(y)),$$
where $x$ is any of the $x_k$'s, $y$ any of the $y_k$'s and $g(y_k):=\frac{y_{k+1}-y_{k}}{\Delta x}$. In the paper you are reading the author uses the notation
$$g(y)=y'.$$
in formula $(2)$.
Note that $g(y_{k-1})=\frac{y_k-y_{k-1}}{\Delta x}$, by definition.
Le us compute the partial derivative $\frac{\partial J}{\partial y_k}$ using the chain rule on the terms $(*)$ and $(**)$. We arrive at
$$\frac{\partial J}{\partial y_k}=\frac{\partial F_{k-1}}{\partial g(y_{k-1})}\frac{\partial g(y_{k-1})}{\partial y_k}\Delta x+\left(
\frac{\partial F_{k}}{\partial y_{k}}\frac{d y_{k}}{d y_k}\Delta x+
\frac{\partial F_{k}}{\partial g(y_{k})}\frac{\partial g(y_{k})}{\partial y_k}\Delta x\right).$$
($F_{k-1}$ does not depend on $y_k$). Now
$$\frac{d y_{k}}{d y_k}=1, $$
$$ \frac{\partial F_{k-1}}{\partial g(y_{k-1})}\frac{\partial g(y_{k-1})}{\partial y_k}\Delta x=\frac{\partial F_{k-1}}{\partial g(y_{k-1})} \frac{1}{\Delta x}\Delta x=\frac{\partial F_{k-1}}{\partial g(y_{k-1})},$$
$$\frac{\partial F_{k}}{\partial g(y_{k})}\frac{\partial g(y_{k})}{\partial y_k}\Delta x=
-\frac{\partial F_{k}}{\partial g(y_{k})} \frac{1}{\Delta x}\Delta x=-\frac{\partial F_{k}}{\partial g(y_{k})}. $$
In summary
$$\frac{\partial J}{\partial y_k}=\frac{\partial F_{k-1}}{\partial g(y_{k-1})}+\left(
\frac{\partial F_{k}}{\partial y_{k}}\Delta x-
\frac{\partial F_{k}}{\partial g(y_{k})}\right)=(\text{replicating the order of the factors in formula (2) contained in your reference})=\frac{\partial F_{k}}{\partial y_{k}}\Delta x+\frac{\partial F_{k-1}}{\partial g(y_{k-1})}-\frac{\partial F_{k}}{\partial g(y_{k})}. $$
In the (a bit confusing) notation of the paper
$$\frac{\partial J}{\partial y_k}=F_{y}(x_{k},y_{k},\frac{y_{k+1}-y_{k}}{\Delta x}) \Delta x+
F_{y'}(x_{k-1},y_{k-1},\frac{y_k-y_{k-1}}{\Delta x})-
F_{y'}(x_{k},y_{k},\frac{y_{k+1}-y_{k}}{\Delta x}) $$
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H: Proving a language is not regular using pumping lemma
I had an exam today and the professor gave us the following problem:
Let $L = \{a^nb^m : n|2m \}$. Prove that $L$ is not regular.
Ok this sounds easy. Here is my solution: Assume opposite -- $L$ is regular. Then by the pumping lemma there exist decomposition $xyz$ of string $s \in L$ such that
$|y| \ge 1$, $|xy| \le p$, where $p$ is the pumping lemma length and $xy^iz \in L$ for all $i \ge 0$.
Setting $ s = а^pb^{2p}$, clearly $s$ is in $L$. Then $s = xyz$, and from the condition $|xy| \le p$ it follows that $y$-part consist only of $a's$.
Here is my problem: I say -- let $y = a$, choose $i = p+1$, then it should $xy^{p+1}z= a^{2p+1}b^p \in L$ -- a contradiction, so $L$ is not regular.
Is it my proof correct?
Many thanks to all,
Ivan
AI: You've made assumptions about the length of $x,y,z$ if you know exactly what $xy^{p+1}z$ is.
You've shown that $x=a^k, y=a^\ell, z=a^{p-k-\ell}b^{2p}$ where $p\geq k+\ell$, $\ell>0$. Then can $xy^2z$ be in $L$?
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H: Interpolating a linear transformation
I'm experimenting with some rudimentary ideas for data encryption (I've never formally taken a cryptology class). An idea that I had for an encryption was to use matrices. So I treat a data set as vector and apply a matrix (non-singular of course) to the vector.
So my question is two-fold:
First of all, how would someone go about finding the encryption matrix from the image and the pre-image? Like given $\vec{v}=A\vec{u}$, where only $\vec{v}$ and $\vec{u}$ are known, could one find $A$?
Secondly, how easily could such a code be broken programmatically? I presume it would be fairly difficult because the hacker would not only need to know the matrix, but would also need to know the matrix itself (or there could even be a set of matrices alternating between vectors).
AI: If matrix A is $n\times n$, you need at least $n$ linealry independent pairs of $(\vec u, \vec v)$ to reconstruct $A$.
I hope this replies to both of your questions.
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H: Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
example1: $|x-1|+|x-2|\geq 1$
my solution：(substitution)
$x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$
square,
$t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$
so proved.
question1 : Is my proof right? Alternatives?
one reference answer:
$1-|x-1|\leq |1-(x-1)|=|1-x+1|=|x-2|$
question2 : prove:
$|x-1|+|x-2|+|x-3|\geq 2$
So I guess:( I think there is a name about this, what's that? wiki item?)
$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
How to prove this? This is question3. I doubt whether the two methods I used above may suit for this general case.
Of course, welcome any interesting answers and good comments.
AI: Actually, $|x-n| + |x-1| = |n-x| + |x-1| \geq |n-1|=n-1$. Don't need the other terms.
Your first proof is correct. You have to take some care moving from $U^2\geq V^2$ to $U\geq V$, but in this case, $U=|t-1|$ is non-negative.
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H: Finding an inner product
Question:
Given 2 vector spaces $U=sp(1,1), W=sp(2,0)$.
How do I find an inner product in $\Bbb R^2$ s.t. $U=W^{+}$ (orthogonal)
I would love an explanation for the algorithm really, more than this specific question.
Thanks.
AI: Given two non-collinear vectors $u$ and $v$ you can get an inner product to make them orthogonal in the following way.
Find first a linear transformation that sends $u$ to $(1,0)$ and $v$ to $(0,1)$. If you already have their coordinates in this basis (the standard basis) then the transformation is the inverse $A$ of the matrix $[uv]$ formed by putting both vectors as columns. So you get $Au=(1,0)$ and $Av=(0,1)$. Now the standard inner product makes $(1,0)$ and $(0,1)$ orthogonal. So, if you define
$$<x,y>:= (Ax)\cdot(Ay),$$
where $\cdot$ denotes the standard inner product. You get an inner product such that $<u,v>=0$.
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H: For $f,g~(f0$ let $\{h\in\mathcal C[0,1]:t-c
For $f,g~(f<g),t\in\mathcal C[0,1],c>0$ let $\{h\in\mathcal C[0,1]:t-c<h<t+c\}$$=\{h\in\mathcal C[0,1]:f<h<g\}.$ I want to show that $t-c=f,~t+c=g.$
$$t-c<t<t+c\text{ and } \\f<\dfrac{f+g}{2}<g.\\\text{Then }t-c<\dfrac{f+g}{2}<t+c\text{ and } f<t<g.$$
I don't know how to contradict the following cases:
For some $y\in[0,1].$
Let $t(y)-c>f(y)$
Let $t(y)-c<f(y)$
Let $t(y)+c>g(y)$
Let $t(y)+c<g(y)$
This problem can more clearly be written as:
For $f_1,g_1,f_2,g_2\in\mathcal C[0,1]$ $$\{h\in\mathcal C[0,1]:f_1<h<g_1\}=\{h\in\mathcal C[0,1]:f_2<h<g_2\}\implies f_1=f_2,g_1=g_2.$$
AI: Let $A=\{h\in\mathcal C[0,1]|\;t-c<h<t+c\}$ and $B=\{h\in\mathcal C[0,1]|\;f<h<g\}$.
For every $\epsilon\in(0,c)$ we have $t-\epsilon,t+\epsilon\in A$ by definition of $A$. Since $A=B$, this means that $t-\epsilon,t+\epsilon\in B$ for all $\epsilon\in(0,c)$, which by definition of $B$ means that $$f<t-\epsilon<t+\epsilon<g.$$ This implies that $$f\leq t-c<t+c\leq g.$$ To complete the proof, we have to show that $f\geq t-c$ and $g\leq t+c$. To do this, note that (since $f<g$) for every $\epsilon\in(0,1)$ we have $$f<(1-\epsilon)f+\epsilon g<g,$$ therefore $(1-\epsilon)f+\epsilon g\in B$. But this means that $(1-\epsilon)f+\epsilon g\in A$ for all $\epsilon\in(0,1)$, which means that $$t-c<(1-\epsilon)f+\epsilon g<t+c.$$ Now, let $\epsilon\to 0$. This gives us $$t-c\leq f\leq t+c.$$ Similarly, letting $\epsilon\to 1$ gives us $$t-c\leq g\leq t+c.$$ This completes the proof.
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H: Are there any algorithms or methods to compute Landau function $g(n)$?
The details about Landau function can be found in A000793.
Maybe there is some methods in A000793, but I don't understand what it says. If possible, can someone illustrate the method?
AI: I considered Michael Somos' pari script and added some comments :
{ a(n) =
local(m, t, j, u);
if (n < 2,
n >= 0, \\the result 0 or 1
\\else
m = ceil( n/exp(1) );
t = ceil( (n/m)^m );
j=1; \\init of result
for (i=2, t,
u = factor(i); \\returns [p1,a1],[p2,a2],...,[pm,am] if i=p1^a1 p2^a2...pm^am
u = sum( k=1, matsize(u)[1], u[k, 1]^u[k, 2]); \\sum k=1 to m of pk^ak
if (u <= n, j=i) \\if the sum is not larger than n memorize i
);
j \\the result
)
}
Note that pari/gp is free and rather convenient for tests !
This requires quite some factorizations. The referenced paper by Deleglise, Jean-Louis Nicolas and Paul Zimmermann 'Landau's function for one million billions' should be more interesting for large values.
|
H: convergence test : $\int_{0}^\infty \mathrm 1/(x\ln(x)^2)\,\mathrm dx $
I have to check if $\int_{0}^\infty \mathrm 1/(x\ln(x)^2)\,\mathrm dx $ is convergent or divergent.
My approach was to integrate the function , hence : $\int_{0}^\infty \mathrm 1/(x\ln(x)^2)\,\mathrm dx=-\lim_{x \to \infty} 1/\ln(x)+ \lim_{x \to 0} 1/\ln(x)=0 $
Still my book says that it is divergent. Maybe the $\infty$ sign of the integral means to check for $+\infty$ and $-\infty $ or i just overlooked something. Any help would be appreciated.
AI: Look at the following improper integral:
$$
\int_1^2f(x)dx
$$
Certainly, $\lim_{x\to 1^+}(x-1)f(x)=+\infty$ so the Comparison test admits the series is divergent.
|
H: Inverse image presheaf
Let $f:X\rightarrow Y$ be a continuous map of topological spaces, and $\mathscr{G}$ a sheaf on $Y$. So far I failed to come up with a simple example where the presheaf $f^{-1}\mathscr{G}$ on $X$ obtained via the direct limit
$$f^{-1}\mathscr{G}(U):=\lim_{f(U)\subset V}\mathscr{G}(V)$$
is not a sheaf. If anyone could give me an example (the simpler the better), it would be greatly appreciated.
AI: Take $X = $ two points with the discrete topology and $Y = $ a point. Let $\mathscr{G}$ be a set with more than two elements, sitting on $Y$. What you get is the corresponding constant pre-sheaf on $X$, which fails to satisfy the gluing axiom.
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H: prove that $\operatorname{Ker}T=\operatorname{Im}(S)^\perp$ for some transformation
Question
Let $V$ be an inner product space of finite dimension. Given linear transformations $T,S\colon V \to V$ such that $\langle T(v),w\rangle=\langle v,S(w)\rangle$, for all $v,w \in V$. Show that $\operatorname{Ker}T=(\operatorname{Im}S)^{\perp}$
Thought of a proof:
Let $w \in V$, $k \in \operatorname{Ker}T$. Therefore
$\langle T(k),w\rangle=\langle k,S(w)\rangle$
and $\langle 0,w\rangle=\langle k,S(w)\rangle$ so $0=\langle k,S(w)\rangle$.
Therefore $k$ is orthogonal to every $w\in V$ and so $k \in (\operatorname{Im}S)^\perp$.
Is this direction true?
AI: Yes, this is correct. Note that one usually denotes the orthogonal by $(\mbox{Im } S)^\perp$. So you have proved that $\mbox{Ker T}\subseteq (\mbox{Im } S)^\perp$. Note also $S$, as it is defined, is called the adjoint of $T$ and is denoted by $S=T^*$.
For the other direction, you could use dimension, noting that both subspaces have the same dimension.
For a proof working in an infinite-dimensional Hilbert space, assuming $T$ is bounded, you could use that $V=\mbox{Ker } T\oplus (\mbox{Ker } T)^\perp$. Then take $x=k\oplus l$ decomposed according to this orthogonal decomposition and assume that $x$ lies in $(\mbox{Im } S)^\perp$. Then $0=(x,Sy)=(Tx,y)=(Tk+Tl,y)=(Tl,y)$ for every $y\in V$. This implies that $Tl=0$, so $l$ is in $\mbox{Ker } T\cap (\mbox{Ker } T)^\perp$. Hence $l=0$ and $x=k+l=k$ belongs to $\mbox{Ker } T$.
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H: $\mathbb A^n$ is not a complete quasiprojective variety
Suppose that $\mathbb K$ is an algebrically closed field. In the course of Algebraic Geometry I've attended we gave the following definition (in what follows every space is to be intended on the field $\mathbb K$):
Definition. A set $X \subset \mathbb P^N$ is said to be complete iff for every quasiprojective variety $Y$ the projection $\pi_2 \colon X \times Y \to Y$ is a closed map.
I would like to prove (as an exercise) that the affine space $\mathbb A^n$ is not complete.
If $n=1$ this is very easy: we take $Y=\mathbb A^1$ and we consider the closed subset $Z:=V(x_1x_2-1) \subset \mathbb A^{1}\times \mathbb A^{1}$; the projection onto the second factor $\pi_2(Z) \simeq \mathbb A^{1} \setminus \{0\}$ which is not Zariski closed.
Is there an easy trick to adapt this argument to higher dimensional case? How would you prove it?
Thanks.
AI: Here's a generalization of your idea in three dimensions. We let $X = \Bbb{A}^2$ and $Y = \Bbb{A}^1$. Consider $V(zy -1) \subseteq X \times Y = \Bbb{A}^3$ and think of $X$ as being the $y - z$ plane and $Y$ as the $x$ - axis. Then projection onto $Y$ is isomorphic to $\Bbb{A}^1 - \{0\}$ which is not Zariski closed (because $k$ - algebraically closed guarantees that $\Bbb{A}^1 - \{0\}$ is not a finite set of points). Now generalize to higher dimensions.
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H: Volume of revolution $y = x^2 + 2$
$$y = x^2 + 2, 0 \leq x \leq 2$$
I know what this looks like, I was suppose to find it for a revolution around the x axis for which I used the disk method and around the y axis which I used shell so I never had to change variables.
I got two different numbers after about an hour of calculations, that is not possible right? Just to make a logical check if I take my calculator and spin it in any way it has the same volume whether I spin it left or right.
Is it possible that the shape has different volumes depending on the axis?
AI: It is absolutely possible you'd get two different volumes if you rotated it about 2 different axis... think of extreme cases like y=.001x . If you revolved that around the x axis from [0,2], you're going to get a much different volume than if you rotated it around the y axis from [0,2].
|
H: question about functions (asymptotic)
This is right?
$f=\Omega(g)\Rightarrow2^f=\Omega(2^g)$?
If not I'd like to get a Counter-example.
Thank you!
AI: No.
Let $f(x) = x$ and $g(x) = 2x$. Then $f(x) = \Omega(g(x))$ since $f(x) = \frac{1}{2} g(x)$ but for any $c>0$ it will not be true that $2^{f(x)} = 2^x \geq c 2^{2x} = c 2^{g(x)}$ whenever $x \geq -\log_2 c$. Therefore it is not true that $2^{f(x)} = \Omega(2^{g(x)})$ as $x \to \infty$.
|
H: Find the value of $\space\large i^{i^i}$?
Is $\large i^{i^i}$ real ? How to find it?
Thank You!
AI: $i^i=e^{i\log i}$
Now on principal branch,using $i=e^{i\pi/2}\implies \log i=i\pi/2$ gives $i^i=e^{-\pi/2}$
Therefore, $i^{i^i}=i^{e^{-\pi/2}}=e^{e^{-\pi/2}\log i}=e^{i(\pi e^{-\pi/2})/2}=\cos\left(\pi \frac{e^{-\pi/2}}{2}\right)+i\sin\left(\pi \frac{e^{-\pi/2}}{2}\right)$
and hence its imaginary part is $\neq 0$ as $ \frac{e^{-\pi/2}}{2}$ is not an integer.
|
H: Is the point $(3, 2, −1, 4, 1)$ in the open ball $B^{5}$ $((1, 2, −4, 2, 3), 3)?$
This is question 11 from the text here.
Is the point $(3, 2, −1, 4, 1)$ in the open ball $B^5 ((1, 2, −4, 2, 3),
3)$?
In my attempt to solve this question, I took the point, subtracted each corresponding component from the ball's center, and then summed the squares of these values:
$$(3-1)^2+(2-2)^2+(-1-(-4))^2+(4-2)^2+(1-3)^2=21$$
Because $21>r^2$, that is $21>9$, I concluded that the point is not in the open ball. However, the correct answer is yes, the point is inside the open ball. (Answers are posted here)
For reference (because it might help explain my confusion) this is the definition of an open ball from the book:
The set of all points $(x_1, x_2,...,x_n )$ in $\mathbb{R}^n$ which satisfy the inequality
$(x_1 − p_1 )^2 + (x_2 − p_2)^2 +...+ (x_n − p_n )^2 < r^2$
(1.1.15)
is called an open n-dimensional ball with radius r and center p, which we denote $B^n(\textbf{p},r)$.
Can someone help me figure out where I am going wrong? Thanks!
Update:
I contacted the author of this textbook, Dan Sloughter, and he promptly corrected the answer guide. Also, I was using an outdated version of the book/answers. In case anyone needs them in the future, the updated links are:
http://www.synechism.org/wp/the-calculus-of-functions-of-several-variables/
and
http://dananne.org/dw/doku.php?id=cfsv:cfsv
AI: Your definition of the Euclidean metric, your work, and your conclusion are all correct. Based on that metric, the given point is not in the open ball $B^5 ((1, 2, −4, 2, 3), 3)$.
I think you can trust your work.
I suggest you post a comment (as requested by the "author" of the answers) stating that there is an error in the answer to $(11)$.
Additional suggestion: In that same comment, post a link to the URL for you question here at math.se, to show both the correct answer, and how your correct solution is obtained. That way, the solution can be corrected by its author, while at the same time and until the correction, help alert anyone else who is perplexed by that same question.
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H: Induction proof, help please?
I have a problem that I need to prove using induction. Prove that a surjective function has at least as many members in its domain as it does in its codomain.
Do I begin by using the axiom of choice?
Thanks.
AI: If you need to use induction then it is probably that your function is between finite sets. In that case you don't need the axiom of choice. For the general case, you do need the axiom of choice. But you can't really use induction either. Since there seem to be a dispute on why induction is not a valid method for the infinite case, let me spell out the reason.
The reason you can't use induction is that in order to pass from a set of cardinality $\kappa$ to a set of cardinality $\kappa^+$, and in order to use the induction hypothesis (and that's where induction really works, otherwise using the word induction won't make it a proof by induction), you will have to start with a set of cardinality $\kappa^+$ and remove some part of it. Removing a part of cardinality less than $\kappa^+$ won't change the cardinality and therefore you cannot use the induction hypothesis; removing a part of cardinality $\kappa^+$ doesn't guarantee that you end up with a set of cardinality less than $\kappa^+$ in order to use the induction hypothesis; and if you do remove almost the entire set, then in order to deduce from the induction hypothesis on $\kappa$ to the case of $\kappa^+$ you have to actually assume that the case of $\kappa^+$ was correct to begin with -- thus begging the question.
Moreover, if you are not intending to use the axiom of choice then the claim is consistently false; that is to say it is perfectly possible that there is a set which can be mapped onto a strictly larger set than itself when the axiom of choice fails. So you have to use the axiom of choice in order to prove this claim, and if you do -- then you might as well use it directly. As in Peter's answer on this page.
In any case, for the issue of finite sets, here is some elaborated hint:
Suppose $f\colon A\to B$ is a surjection, we prove by induction that if $A$ has $n$ members then $B$ has no more than $n$ members.
For $n=1$ this is trivial because $A$ is a singleton, so $B$ must be a singleton as well. Suppose this is true for $n$, and let $A$ be a set of size $n+1$.
Pick some $a\in A$, and consider $A'=A\setminus\{a\}$. If $f\upharpoonright A'$ is still a surjection onto $B$, then we can use the induction hypothesis to conclude that $B$ has at most $n$ members and we are done.
In the other case, $f\upharpoonright A'$ is not surjective. Conclude that there is some $b\in B$ such that $f\upharpoonright A'\colon A'\to B\setminus\{b\}$ is a surjection, therefore $B\setminus\{b\}$ has at most $n$ members and therefore $B$ has at most $n+1$ members as wanted.
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H: Example of a restriction of sheaf not being a sheaf
Let $\mathscr{F}$ be a sheaf on $X$, and $Y\subset X$ a subset. Define a presheaf $\mathscr{F}|_Y$ on $Y$ via the direct limit
$$\mathscr{F}|_Y(V):=\lim_{V\subset U}\mathscr{F}(U),$$
where $V$ is an open subset of $Y$, and $U$ is an open subset of $X$. Clearly $\mathscr{F}|_Y$ is a sheaf if $Y$ is an open subset of $X$. But I can't think of an example where $\mathscr{F}|_Y$ is not a sheaf. Can anyone think of an example?
AI: $\DeclareMathOperator{\sh}{Sh}$
Edit: The question originally asked about $F|_Y(V)=\varinjlim_{U\subset V}F(U)$, and my answer reflects this. It now looks like the OP is asking out the
inverse image functor.
I claim that $F|_Y$ is the wrong thing to look at. In general, if $f:Y\to X$ is a continuous map between topological spaces (eg. the an embedding $Y\subset X$) then there is a natural functor $f_*:\sh(Y)\to\sh(X)$ given by
$$
(f_* F)(U) = F(f^{-1}(U))
$$
In your case ($f:Y\to X$ is the embedding $Y\subset X$) one has $(f_* F)(V) = F(V\cap Y)$. Now, this functor isn't in the direction that you're looking for, but $f_*$ has a left adjoint $f^*:\sh(X)\to\sh(Y)$, which is characterized by the universal property
$$
\hom_{\sh(Y)}(f^* F,G) = \hom_{\sh(X)}(F, f_* G)
$$
for $F\in\sh(X)$, $G\in\sh(Y)$. (Note: in algebraic geometry one often writes $f^{-1}$ instead of $f^*$, but I'm just talking about sheaves of sets here.) Anyways, $f^* F$ is defined as follows:
$$
(f^* F)(V) = \varinjlim_{U\supset f(V)} F(U)
$$
So in your case ($Y\subset X$) one has
$$
(f^* F)(V) = \varinjlim_{U\supset V} F(U)
$$
which is very different than $\varinjlim_{U\subset V} F(U)$. For example, if $V$ has empty interior, than your definition yields $F|_Y(V) = *$ for all $V$.
Edit: Georges pointed out that I was overly hasty: my definition of $f^* F$ yields a presheaf, the sheafification of which is the "actual" $f^* F$. Anyways, here is an example of an inclusion $f:Y\to X$ and a sheaf $F$ on $X$ for which $f^{pre}F:V\mapsto\varinjlim_{U\supset f(V)} F(U)$ is not a sheaf.
Let $Y=\mathbb{R}$ with the discrete topology, $X=\mathbb{R}$ with the usual topology, and $f:Y\to X$ be the identity map. Then it is easy to see that if $F\in\sh(X)$ is the "sheaf of continuous functions," then $f^{pre} F$ is not a sheaf.
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H: Vector spaces and finite dimensions related problem.
Please can you help me whit this problem.
For $1.$ I did it as it's classical.
What I am having trouble with are the other questions. Hints would be good but if you can explain that would be great.
Let $E, E'$ be two vector spaces over $\mathbb{K}$, $\phi$ is a linear map from $E$ to $E'$ and $F$ (resp $F'$) a subspace of $E$ (resp of $E'$).
$1.$
$a)$ Show that $\phi(F)$ (resp $\phi^{-1}(F')$) is a subspace of $E'$ (resp of $E$).
$b)$ deduce that $Ker\phi$ (resp $\text{Im}\phi$) is a subspace of $E$ (resp of $E'$).
$2.$ we suppose that the dimension of $E$ is finite. Let $f, g$ be two endomorphisms of $E$ and $V$ be a subspace of $E$.
$(i)$ Show that $$\dim f^{-1}(V)= \dim(\text{Ker}f)+\dim(V\cap \text{Im}f)$$
$(ii)$ Deduce that $$\dim (\text{Ker}f\circ g)\leq \dim(\text{Ker}g)+\dim(\text{Ker}f)$$
$(iii)$ Show that $$\text{Ker}(g\circ f)= f^{-1}(\text{Ker}g\cap \text{Im}f)$$
$3.$ We suppose that $f\circ g=g\circ f$.
Show that
$$f(\text{Ker}g)\subseteq \text{Ker}g \text{ and } f(\text{Im}g)\subseteq \text{Im}g$$
$4.$ We suppose that the dimension of $E$ is finite.
Show that $$\dim(\text{Im}f\cap \text{Ker}g)=\text{rank}f-\text{rank}(g\circ f)$$
AI: Since 1 is done, I will try the rest.
For 2 (i), try to use the rank-nullity theorem.
For 2(ii), apply the formula to $g^{-1}(\ker f)$ and use $\dim_{K}(M+N)=\dim(M)+\dim(N)-\dim(M\cap N)$.
For 2(iii), $v$ lies in the kernel iff $g\circ f(v)=0$. So $f(v)$ lies in the kernel of $g$ and in the image.
For 3, use the definition of the kernel, image and commutativity.
For 4, use $\mbox{rank}(g\circ f)+\dim(\ker g|_{f(V)})=\mbox{rank}(f)$. (rank-nullity theorem again)
Hope these helps!
|
H: Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.
Evaluate the integral $$\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\, \mathrm dx.$$
How can i evaluate this one? Didn't find any clever substitute and integration by parts doesn't lead anywhere (I think).
Any guidelines please?
AI: As $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$$
If $$\begin{eqnarray}I &=& \int^{\frac{\pi}{2}}_0 \frac{\sin^nx}{\sin^nx+\cos^nx} \,dx\\
&=& \int^{\frac{\pi}{2}}_0 \frac{\sin^n\left(\frac\pi2-x\right)}{\sin^n\left(\frac\pi2-x\right)+\cos^n\left(\frac\pi2-x\right)}\, dx\\
&=& \int^{\frac{\pi}{2}}_0 \frac{\cos^nx}{\cos^nx+\sin^nx}\, dx
\end{eqnarray}$$
$$\implies I+I=\int_0^{\frac\pi2}dx$$ assuming $\sin^nx+\cos^nx\ne0$ which is true as $0\le x\le \frac\pi2 $
Generalization :
$$\text{If }J=\int_a^b\frac{g(x)}{g(x)+g(a+b-x)}dx, J=\int_a^b\frac{g(a+b-x)}{g(x)+g(a+b-x)}dx$$
$$\implies J+J=\int_a^b dx$$ provided $g(x)+g(a+b-x)\ne0$
If $a=0,b=\frac\pi2$ and $g(x)=h(\sin x),$
$g(\frac\pi2+0-x)=h(\sin(\frac\pi2+0-x))=h(\cos x)$
So, $J$ becomes $$\int_0^{\frac\pi2}\frac{h(\sin x)}{h(\sin x)+h(\cos x)}dx$$
|
H: Hypothetical probability question
I found this question on an online University of Washington course assignment related to Bayesian Probabilities:
You've lost contact with your safari leader and now you find yourself confronted by a legion of 100 marmosets. Your training tells you that you must give a present to the one king in order avoid being clawed. Fortunately, you have some raisinets and are a master of conditional probabilities. A lilliputian wearing a tophat approaches you directly. The king always wears a hat, and the custom has caught-on with 5% of the others. What is the probability that this is the king?
Before you get a chance to decide, a moustachioed marmoset pushes the first one out of the way. He too is wearing a hat. Good thing you memorized the 'M' section of the encyclopedia: the king is seen with a moustache half the time, whereas only 1% of ordinary monkeys have one. What is the probability that he is the king? Assume that moustaches and hats are independent of each other given the identity of the marmoset.
My probability skills are lacking. Would the first question be 1/6? Given that 5 lilliputians wear tophats and include one more for the king.
For the second question, I was having difficulty with the fact that there are 100 monkeys and only 1% have a moustache. Does that mean this has to be the king?
AI: Note: It is not the case that $6\%$ of marmosets wear hats. From the sample of $100$ marmosets, all we know is that $1$ of them (the king) definitely wears a hat and that $5\%$ of the remaining $99$ ordinary marmosets wear a hat (not $5\%$ of all $100$ marmosets).
Let $K \equiv$ the marmoset is a king, $H \equiv$ the marmoset wears a hat, and $M \equiv$ the marmoset has a mustache. We know that:
$Pr(K)=1/100 \implies Pr(\overline{K})=99/100 $
$Pr(H \mid K)=1$
$Pr(H \mid \overline{K})=5/100=1/20$
$Pr(M \mid K) = 1/2$
$Pr(M \mid \overline{K}) = 1/100$
Hence, for the first question, we have:
$$ \begin{align*}
Pr(K \mid H)&=\dfrac{Pr(K,H)}{Pr(H)}\\
&=\dfrac{Pr(K)Pr(H \mid K)}{Pr(K)Pr(H \mid K) + Pr(\overline{K})Pr(H \mid \overline{K})}\\
&=\dfrac{\dfrac{1}{100} \cdot 1}{\dfrac{1}{100} \cdot 1 + \dfrac{99}{100} \cdot \dfrac{1}{20}} \\
&=\dfrac{20}{20 + 99} \\
&=\dfrac{20}{119} \\
\end{align*} $$
For the second question, we exploit the fact that mustaches and hats are independent of each other given the identity of the marmoset:
$$ \begin{align*}
Pr(K \mid H,M)&=\dfrac{Pr(K,H,M)}{Pr(H,M)}\\
&=\dfrac{Pr(K)Pr(H \mid K)Pr(M \mid K,H)}{Pr(K)Pr(H \mid K)Pr(M \mid K,H) + Pr(\overline{K})Pr(H \mid \overline{K})Pr(M \mid \overline{K},H)}\\
&=\dfrac{Pr(K)Pr(H \mid K)Pr(M \mid K)}{Pr(K)Pr(H \mid K)Pr(M \mid K) + Pr(\overline{K})Pr(H \mid \overline{K})Pr(M \mid \overline{K})}\\
&=\dfrac{\dfrac{1}{100} \cdot 1 \cdot \dfrac{1}{2}}{\dfrac{1}{100} \cdot 1 \cdot \dfrac{1}{2} + \dfrac{99}{100} \cdot \dfrac{1}{20} \cdot \dfrac{1}{100}}\\
&=\dfrac{1000}{1000 + 99}\\
&=\dfrac{1000}{1099}\\
\end{align*} $$
|
H: Calculating this Riemann sum limit
Calculate the limit $$\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}$$
How exactly do we calculate this limit of the Riemann sum? I am never able to find what is the partition. I know that our $f(x)$ is $\sin(x)$.
AI: Rewrite the sum as
$$\frac{1}{n} \sum_{k=1}^n \left ( \frac{k}{n} - \frac{1}{n^2}\right ) \sin{\left ( \frac{k}{n}\right)}$$
As $n \to \infty$, the $1/n^2$ term vanishes and we are left with
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{n} \sin{\left ( \frac{k}{n}\right)}$$
which is the Riemann sum for the integral
$$\int_0^1 dx \, x \, \sin{x}$$
NB in general
$$\int_a^b dx \, f(x) = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left (a + \frac{k}{n} (b-a) \right)$$
when the integral on the left exists.
ADDENDUM
I was asked to expand upon the claim that $1/n^2$ vanishes. If we use this term, we see that its contribution is
$$\frac{1}{n^3} \sum_{k=1}^n \sin{\left ( \frac{k}{n}\right)}$$
which, in absolute value, is less than $(1/n^3) (n) = 1/n^2$, which obviously vanishes as $n \to \infty$.
|
H: How many distinct pairwise difference multisets exists?
For all $K$-point integer-valued sets whose elements are between $1$ and $N$ and also distinct, how many distinct pairwise difference multiset exists?
Note that the difference is evaluated modulu N.
for example: for $(N,K)=(5,3)$ we have:
$\color{green}{\{1,2,3\} \rightarrow \{1,1,2,3,4,4\}}$
$\color{red}{\{1,2,4\} \rightarrow \{1,2,2,3,3,4\}}$
$\color{green}{\{1,2,5\} \rightarrow \{1,1,2,3,4,4\}}$
$\color{red}{\{1,3,4\} \rightarrow \{1,2,2,3,3,4\}}$
$\color{red}{\{1,3,5\} \rightarrow \{1,2,2,3,3,4\}}$
$\color{green}{\{1,4,5\} \rightarrow \{1,1,2,3,4,4\}}$
Note that other possible 3-point sets are shifted or reversed version of shown sets and lead to same pairwise difference multisets. So we can assume that the first element of set is $1$, without loss of anything. So for $(N,K)=(5,3)$ we have only 2 distinct pairwise difference multisets.
I'm looking for a formula which computes all distinct possible pairwise difference multisets for any pair $(N,K)$ ?
As I didn't know what is the most preferable site for this question, I posted in on these sites too: mathoverflow, Theoretical Computer Science
AI: I used the following Python code to tabulate some numbers:
import sys, itertools
for n in range(1, 25):
sys.stdout.write("{:3}:".format(n))
for k in range(1, n + 1):
r = set()
for c in itertools.combinations(range(n), k):
d = [0]*n
for i, c1 in enumerate(c):
for c2 in c[(i + 1):]:
d[c2 - c1] += 1
d[n + c1 - c2] += 1
d = tuple(d)
r.add(d)
sys.stdout.write(" {:4}".format(len(r)))
sys.stdout.write("\n")
Demo run of that code is also available at http://ideone.com/R0z3dg. Since you requested an explanation in a comment, here is how this works: The outer two loops iterate over reasonable values for $N$ and $K$, i.e. all the (non-empty) cells of the table. The choice of $0<K\leq N$ is somewhat arbitrary; I might as well have included $0$ or excluded $N$, since both of these will always lead to $1$. c then iterates over all possible combinations of $K$ elements taken from the set range(n) $=\{0,1,\dots,(N-1)\}$. These are your point sets. The combinations will be already sorted. The next two loops choose c1 and c2 from that combination such that c1 < c2. For each such pair, you have two differences. The difference c2 - c1 will already be in the range $[0, N)$. The other difference will be negative, so you add $N$ as part of your modulo semantics. For both of these differences, you increment the associated count in the multiset. So d[i] will indicate the number of times the difference i (mod $N$) has already been seen in the current combination c. Once this count is complete (i.e. the loops over c1 and c2 are done), you turn the mutable list into an immutable tuple which can be used as an element of the set r. The result for a given $N$ and $K$ is simply the size of that set r once all combinations c have been processed.
The resulting table looks like this:
N↓ K→ 1 2 3 4 5 6 7 8 9 10 …
1: 1
2: 1 1
3: 1 1 1
4: 1 2 1 1
5: 1 2 2 1 1
6: 1 3 3 3 1 1
7: 1 3 4 4 3 1 1
8: 1 4 5 7 5 4 1 1
9: 1 4 7 10 10 7 4 1 1
10: 1 5 8 16 13 16 8 5 1 1
11: 1 5 10 20 26 26 20 10 5 1 1
12: 1 6 12 28 35 35 35 28 12 6 1 1
13: 1 6 14 34 57 74 74 57 34 14 6 1 1
14: 1 7 16 47 73 120 85 120 73 47 16 7 1 …
15: 1 7 19 56 106 160 222 222 160 106 56 19 7 …
16: 1 8 21 70 137 246 327 254 327 246 137 70 21 …
17: 1 8 24 84 196 376 576 698 698 576 376 196 84 …
18: 1 9 27 104 237 493 762 1121 701 1121 762 493 237 …
19: 1 9 30 120 324 735 1311 1962 2338 2338 1962 1311 735 …
20: 1 10 33 143 384 932 1778 2649 3651 2377 3651 2649 1778 …
21: 1 10 37 165 507 1257 2609 4662 6006 7866 7866 6006 4662 …
22: 1 11 40 195 601 1717 3576 6830 9860 13268 7944 13268 9860 …
23: 1 11 44 220 759 2244 5302 10395 17105 23958 28216 28216 …
24: 1 12 48 253 881 2635 6574 12844 21499 32342 42389 25220 …
25: 1 12 52 286 1086 3622 9574 21389 39582 59132 85754 99638 …
The case from your example, $T(5, 2)=2$, can be read from row 5 column 2 of this table. If you omit the diagonal where $N=K$, i.e. drop the trailing 1 from each row, then the resulting triangle is exactly the one you find as the triangular array representation of OEIS sequence A103441. The statement of this sequence sounds a lot like your question, too:
$T(n,k) =$ number of bracelets of $n$ beads (necklaces that can be flipped over) with exactly two colors and $k$ white beads for which the set of distances among the white beads are different.
Unfortunately, there is no closed forula to accompany that sequence. So at least that encyclopedia does not know the formula you are looking for. There is some Mathematica code, but I guess that this does something as brute-force as what I did in Python above. If anyone here does find a closed formula, please make sure that OEIS gets updated with this information as well.
Since your comment below indicates different results on your part, and the first difference is for $N=8,K=4$, here are some details for that case, using the same notation you used in your original question. As you can see, I got $T(8,4)=7$ with the following point sets as wittnesses:
$$\{1,2,3,4\}\to\{1,1,1,2,2,3,5,6,6,7,7,7\}\\
\{1,2,3,5\}\to\{1,1,2,2,3,4,4,5,6,6,7,7\}\\
\{1,2,3,6\}\to\{1,1,2,3,3,4,4,5,5,6,7,7\}\\
\{1,2,4,6\}\to\{1,2,2,3,3,4,4,5,5,6,6,7\}\\
\{1,2,4,7\}\to\{1,2,2,3,3,3,5,5,5,6,6,7\}\\
\{1,2,5,6\}\to\{1,1,3,3,4,4,4,4,5,5,7,7\}\\
\{1,3,5,7\}\to\{2,2,2,2,4,4,4,4,6,6,6,6\}$$
|
H: simple deriving of a function - preparing for hesse-matrix
$ f(x,y) = 2x^2-2xy^2+y^2$
I want to prepare this function for a hesse-matrix. But I'm stuck at deriving.
I get $\frac{\partial f^2}{\partial^2 x} = 4 $
$\frac{\partial f^2}{\partial y^2} = 4x + 4$
$\frac{\partial f^2}{\partial x \partial y} = 4x - 4y + 2y$
which is obviously wrong. but I can't tell why.
AI: When $f=2x^2-2xy^2+y^2$ so $$f_x=4x-2y^2\to f_{xx}=4, ~~f_{xy}=-4y\\
f_y=-4xy+2y\to f_{yy}=-4x+2\\$$ This means that $$D=f_{xx}f_{yy}-f^2_{xy}=4\times(-4x+2)-(-4y)^2$$
|
H: A question about differentiability
So, this is the last subject I have to study for my exam on Friday, and I still can not comprehend how to prove that a function is differentable.
$$f(x,y) = \begin{cases}
(x^2 + y^2) \cdot \sin \left(\frac{1}{\sqrt{x^2 + y^2}}\right) & \text{ if }(x,y) \neq (0,0),\\
0& \text{ if }(x,y) = (0,0).
\end{cases}$$
Is $f(x,y)$ differentiable at $(x,y) = (0,0)$?
I know that I have to calculate the partial derivative according to the definition and then do another limit, but I do not know the formulas for these.
I am very, very sorry for asking such a general question, but I am really in trouble and don't seem to understand it.
I would appreciate very much a thorough answer.
AI: Hint Is $$f(x)=x^2\sin\frac{1}x$$ differentiable at the origin? You're now looking at $f\circ g(x,y)$ where $g(x,y)=\sqrt{x^2+y^2}$. What can you do now? Do you know how to apply the chain rule? Is $g$ differentiable (in the broader sense) at the origin?
ADD Note that to know the partial derivatives of your function, one really needs to calculate the partial derivatives "by hand". That is $$\eqalign{
& {\left. {\frac{{\partial f}}{{\partial x}}} \right|_{{x_0} = \left( {0,0} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {h,0} \right) - f\left( {0,0} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} h\sin \frac{1}{{\left| h \right|}} = 0 \cr
& {\left. {\frac{{\partial f}}{{\partial y}}} \right|_{{x_0} = \left( {0,0} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0,h} \right) - f\left( {0,0} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} h\sin \frac{1}{{\left| h \right|}} = 0 \cr} $$
So the partial deritvatives exist and are $0$. Thus, we now look at $$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left| {f\left( {x,y} \right) - f\left( {0,0} \right) - \nabla f\left( {0,0} \right) \cdot \left( {x,y} \right)} \right|}}{{\left\| {\left( {x,y} \right)} \right\|}}$$
where ${\left\| {\left( {x,y} \right)} \right\|}=\sqrt{x^2+y^2}$. The above is $$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left| {\left( {{x^2} + {y^2}} \right)\sin \left( {\frac{1}{{\sqrt {{x^2} + {y^2}} }}} \right)} \right|}}{{\sqrt {{x^2} + {y^2}} }} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left| {\sqrt {{x^2} + {y^2}} \sin \left( {\frac{1}{{\sqrt {{x^2} + {y^2}} }}} \right)} \right| = 0$$ so the function is differentiable. We used that $x\sin \frac 1 x\to 0$.
|
H: locally compact Hausdorff space which is not second-countable
I'm trying to find an example of a space that is Hausdorff and locally compact that is not second countable, but I'm stuck. I search an example on the book Counterexamples in Topology, but I can't find anything.
Thank you for any help.
AI: Let $Y$ be an uncountable set, let $p$ be a point not in $Y$, and let $X=\{p\}\cup Y$. Let
$$\mathscr{B}=\{\{x\}:x\in Y\}\cup\{X\setminus F:F\text{ is a finite subset of }Y\}\;;$$
then $\mathscr{B}$ is a base for a Hausdorff, locally compact, compact topology $\tau$ on $X$ that is not second countable. It isn’t even first countable: $p$ has no countable local base. $\langle X,\tau\rangle$ is the one-point compactification of the discrete space $Y$. (Of course $Y$ itself is also an example, but it’s merely locally compact; $X$ is also compact.)
You can make examples with even nicer properties. For instance, $Y\times[0,1]$ is Hausdorff, locally compact, and locally connected but not second countable. The closed long ray has all of those properties and is in addition connected, path connected, and locally path connected. Its one-point compactification loses path connectedness but gains compactness.
|
H: Evaluate $\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx$
Evaluate $$\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx$$ by any method. In short I am interested in any method that overcomes the lack of convergence of this integral and gives an "number" to it.
EDIT
As I'm getting answers regarding convergence test, this should clear the question.
This is integral is divergent. Do other integration theories (Lebesgue, Henstock-Kurweil,..) overcome this problem and actually help assign a non-infinite value to this. (P.S. I'm engineering guy, need incentive to go beyond Riemann and this is a start)
AI: Using the trivial measure, let $\int_{\mathcal{D}} f d\mu = 0$ for any function $f$ and any subset $\mathcal{D}$ of $\mathbb{R}$. Under this definition, your integral is convergent and its value is equal to $0$. But what are the applications of this stupid integral? None. To have a sensible integral, you want, for example, "the area under the line $y=3$ between $0\leq x\leq 2$" to be equal to $2\times 3 = 6$. When we construct an integral that has such natural and useful properties (which is another story), it turns out that your integral does not converge, as shown in the other answers.
|
H: Roots in Banach algebras.
I'm studying positive functionals on normed algebras and I got stuck in the following problem:
Let $A$ be a unital Banach algebra, and $x\in A$ be such that $\Vert x\Vert <1$. Then the series
$$\sum_{k=0}^\infty\dfrac{1}{k!}\left(\dfrac{1}{2}-0\right)\cdots\left(\dfrac{1}{2}-k+1\right)x^k$$
converges (absolutely) to and element $y\in A$ such that $y^2=1_A+x$.
The series above is just the Taylor series of $\sqrt{1+x}$ centered at $0$. That the series converges absolutely is simple (one can use, for example, the root test). The problem is to show that $y^2=1+x$.
Let $a_k=\dfrac{1}{k!}\left(\dfrac{1}{2}-0\right)\cdots\left(\dfrac{1}{2}-k+1\right)$. It can be shown by induction on $k$ that
$a_k=\dfrac{(-1)^k(2k)!}{(1-2k)(k!)^2(4^k)}$ (Wikipedia)
Since the series $y=\sum_k a_kx^k$ converges absolutely, then
$$y^2=\sum_{n=0}^\infty\sum_{k=0}^n a_k a_{n-k}x^n$$
If we calculate the initial terms, we get $y=1+x+0x^2+0x^3+0x^4+0x^5+\cdots$, but I'm having trouble showing that $\sum_{k=0}^na_ka_{n-k}=0$ for every $n\geq 2$. I've tried induction, but it didn't work for me.
AI: Since $\Vert x\Vert<1$ then its spectrum $\sigma(x)\subset r\mathbb{D}$ for some $r<1$, so we have well defined holomorphic calculus of element $x$
$$
\gamma_x:\mathcal{O}(r\mathbb{D})\to A, f\mapsto \gamma_x(f):=f(x)
$$
which in fact is unital involutive $*$-homomorphism. Note that $h(z)=\sqrt{1+z}\in\mathcal{O}(r\mathbb{D})$ and we get some $y=h(x)\in A$. Finally
$$
y^2=h(x)h(x)=\gamma_x(h)\gamma_x(h)=\gamma_x(h\cdot h)=\gamma_x(1+z)=1+x
$$
|
H: Topology and commutative algebra.
I don't know both of these subjects, but I was wondering if there was any topology in commutative algebra. I don't need any detailed answer (since I don't know any of them yet)...So would it be helpful to know topology when studying commutative algebra, or is there no connection between them?
Thank you in advance
AI: Yes, it is useful to know some general topology to make any headway in commutative algebra. A very powerful technique in commutative algebra is that of completion of a local ring. Such complete rings are often easier to deal with and geometrically they contain important local information.
The process of completion actually borrows amply from general topology. Also topology is the first instance where you start worrying about important properties like compactness, separatedness, properness etc. It will be very difficult to understand the application of some concepts of commutative algebra (in algebraic geometry I guess) without topology.
|
H: Krull dimension of a ring which is finitely generated module over a finitely generated $k$-algebra
Let $k$ be a field and $R=k[y_1,\cdots,y_d]$ where $y_i$ are algebraically independent over $k$. Suppose that $S=k[x_1,\cdots,x_d]$ is a subring of $R$ such that $R$ is a finitely generated $S$-module. It is well known that the Krull dimension of $S$ is equal to the transcendence degree of $k(x_1,\cdots,x_d)$ over $k$.
Question: Is it true that $S$ and $R$ have the same Krull dimension? If yes, how can we see that?
AI: Your question is coming from a more general fact in commutative algebra: Let $S,R$ be commutative rings with $S \subseteq R$ and $R$ integral over $S$. Then $\dim R = \dim S$. The proof if I remember just comes from going up.
Exercise: Use the result I mention above to prove that for affine varieties $X,Y$, the dimension of the product variety $X \times Y$ is equal to the sum of the dimensions of $X$ and $Y$. This requires knowing the isomorphism $A(X\times Y) \cong A(X) \otimes_k A(Y)$ which can be deduced from the Yoneda Lemma.
|
H: finding a basis for $W^\perp$ and understanding it.
Given
$$
w_1 = \begin{bmatrix}
1 \\ -1 \\ 1 \\ 1
\end{bmatrix},w_2= \begin{bmatrix}
0 \\ 1\\2\\3
\end{bmatrix}
$$
let $W$ be the subspace spanned by the given vectors. Find a basis for $W^\perp$ Now my problem is, how do envision this? They do the following:
They use the vectors as rows. Then they say that W is the row space of A, and so it holds that $W^\perp = null(A)$ . and we thus solve for $Ax=0$
Now my problem is: how do i envision this? Why is $W^\perp = null(A)$ I dont like learning these kinds fo things, is there a way to understand this? WHY is this the case, why do they specifically let A use $w_1$ and $w_2$ as the rows?
AI: Take a vector ${\bf x}=(x_1,x_2,x_3,x_4)$. You want to find the solutions to the system $$\begin{cases}{\bf x}\cdot {\bf w}_1=0\\{\bf x}\cdot {\bf w}_2=0\end{cases}$$ See? Then find a basis for this.
|
H: Quadratic expression that generate primes
I recently learned that there exist quadratic expression that generate some primes and some of these equations generate more primes than others. In the following video, the person shows the following expression
$$4x^2 -2x+1$$
that generates some primes. So I was wondering, are there infinitely many expression that exist that can produce primes and what makes one quadratic produce more primes than another? In other words, do some quadratic expressions have a special property that causes them to generate more primes than others?
Thanks!
AI: There are infinitely many linear polynomials $f(n)=an+b$ such that $f(n)$ yields prime values at infinitely many integer arguments - I assume this is what "producing primes" means. There are also infinitely many quadratic polynomials that obtain at least one prime value, e.g. $x^2+ax+p$ for various integers $a$ and primes $p$ at $x=0$. Whether or not we know of any quadratics that provably obtain an infinite number of prime values at integer arguments, I'm not sure.
If increase the degree and number of variables, though, we do know of such polynomials that take infinitely many prime values, in fact whose positive values are only prime numbers! See this section of Wikipedia for more information.
Beyond linear polynomials (the subject of the quantitative version of Dirichlet's theorems on primes in arithmetic progressions, which generalizes in a very different direction in algebraic number theory on Cheboratev density), capturing the asymptotic frequency with which systems of polynomials simultaneously yield primes is a very hard and at the moment speculative business.
Bateman-Horn conjecture. Let $f_i(x)$ be a finite family of irreducible polynomials (say with positive leading coefficients). Let $P(n)$ count $k\le n$ for which $f_i(k)$ are prime for all $i$. Then $$P(n)\sim \frac{\displaystyle \prod_p \frac{1-N(p)/p}{(1-1/p)^m}}{(\deg f_1)\cdots(\deg f_m)}\int_2^n\frac{dt}{(\log t)^m}$$ where $N(p)$ counts the solutions to $\{ f_i(x)=0$ in ${\bf F}_p$ (the finite field of integers mod $p$).
A heuristic asymptotic formula concerning the distribution of prime numbers
"Heuristic" here - ubiquitous in analytic number theory - effectively means techniques based on experience, empirical data, intuition, educated guessing, and informal statistical reasoning.
Note that a very particular case of Bateman-Horn is the quantitative version of the twin prime conjecture, the $1$st Littlewood-Hardy conjecture. It also implies the Bunyakovsky_conjecture, the one-polynomial case of BH, and is a quantitative refinement of Shinzel's hypothesis, itself an extension of Bunyakovsky to multiple polynomials. Thus BH is a massively general statement.
Even so, I do not think any particular case of BH has been proven. That includes any quadratic polynomials, which are the specific subject of your question. As for intuition as to why some expressions may generate them more frequently then others (detectable in the difference in the multiplicative constants in the giant formula above), the form of the conjecture suggests that the answer may lie in local-global thinking (another thing that turns up a lot in number theory).
That is to say that there may be good reason to expect that long-term "global" nature of $P(n)$ can be reliably statistically forecasted by the "local" nature of it (the counting function $N(p)$). Given this expectation, "polynomials yield primes at different rates" reduces to "polynomials don't always have the same number of zeros mod every prime $p$," which seems (to me at least) to be not quite as surprising (since it is more finitistic, I guess).
|
H: How many different expressions can you get by inserting parentheses into: $x_{1}-x_{2}-\cdots-x_{n}$?
I ran into this question and I am finding it very difficult to solve:
How many different expressions can you get by inserting parentheses into:
$$x_{1}-x_{2}-\cdots-x_{n}\quad ?$$
For example:
$$\begin{align*}
x_{1}-(x_{2}-x_{3}) &= x_{1}-x_{2}+x_{3}\\
(x_{1}-x_{2})-x_{3}&=x_{1}-x_{2}-x_{3}\\
x_{1}-(x_{2}-x_{3})-x_{4})&=x_{1}-x_{2}+x_{3}+x_{4}\\
\end{align*}$$
I'm really desperate for a full answer. I've been working on this for 3 hours. Thanks in advance.
AI: The answer is $2^{n-2}$. $x_1$ must always be positive and $x_2$ must always be negative. Then you can pick the signs on all the rest any way you want, starting with $x_3$. For a string of length $n$, start with a string of length $n-1$ that has the signs the way you want up to there. If you want the sign before $x_n$ to be negative, leave it outside the parentheses. If you want it positive, add it to the last set of parentheses if $x_{n-1}$ is inside one, or group it with $x_{n-1}$ if not. Steven Stadnicki's example of $x_1-x_2+x_3+x_4=x_1-(x_2-x_3-x_4)$
|
H: Lagrange Multipliers restriction equation problem
There's a straight line going from B to C in the first quadrant of x,y coordinate system.
B is $(0,s)$, C is $(t,0)$
Let A $(3,3)$ be a point on the line going from A to B.
Find the equation of the line so the distance between B and C will be minimal.
Well, I've started by finding points that their squared distance is minimal \ maximal (Helps me getting rid of un-needed roots) so the distance between B and C is $f(x)=x^2+y^2$.
Now i wanna find a restrict equation and use Lagrange multipliers. But i can't find a proper one.
Guidelines please?
AI: The distance between $B$ and $C$ is $\sqrt{s^2 + t^2}$ so you need to minimize $f(s,t) = s^2 + t^2$. The fact that $A$ lies on the segment connecting $B$ and $C$ gives you a constraint: compare slopes to get $\dfrac{3-s}{3} = \dfrac{3}{3-t}$, which is equivalent to $st - 3s - 3t = 0$. Let $g(s,t) = st - 3s - 3t$ and minimize $f(s,t)$ on the level set $g(s,t) = 0$.
|
H: Do we need measure theory to prove length of diameter and circumference is constant is a circle?
In high school I was taught that $\pi$ is the ratio of the length of circumference and diagonal of a circle. But is it necessary to use some measure theory machinery to define the length of circumference of a circle? Or length of a diameter? And that in a given circle, do I need measure theory to prove that all diameters have equal lengths or can the invariance of length in rotation proved some elementary way? I tried to prove that $\pi$ is constant so I wondered can I avoid using measure theory if I prove everything starting from the Hilbert axioms.
AI: I presume where you wrote "diagonal", you meant "diameter".
Lengths of line segments are distances between the endpoints, and how or whether those are defined depends on what approach you take to foundations of geometry.
The length of a curve, such as a circle, can be defined as the smallest upper bound of all lengths of polygonal paths along the curve.
Say a curve starts at a point $p_1$ (on the circle, pick any point to be the starting point) and you move along an arc to another point $p_2$ on the curve, and then keep moving in the same direction along the curve to $p_3$, and so on until you reach the other end of the curve, which you label $p_n$ (on the circle, you'd have returned to the starting point). Then
$$
\text{distance from $p_1$ to $p_2$} + \text{distance from $p_2$ to $p_3$} + \text{distance from $p_3$ to $p_4$} + \cdots\cdots \\
\cdots\cdots+ \text{distance from $p_{n-1}$ to $p_n$}
$$
is less than or equal to the length of the curve. But now suppose you add some point between $p_1$ and $p_2$, and another between $p_2$ and $p_3$, and so on, and look at the sum you get then. Generally it will be a bigger number. But it is still no bigger than the length of the curve.
The length of the curve is greater than or equal to the lengths of all such "polygonal paths". But the length of the curve is the smallest number that is greater than or equal to the lengths of all such polygonal paths.
That is how the circumference is defined.
|
H: Finite dimensional subspace of Hilbert space and basis
Let $H$ be infinite-dimensional Hilbert space with basis functions $b_i$.
Let $B_n = \text{span}\{b_1, ...,b_n\}$.
So $\text{dim}(N) = n$.
Let $c_i$ be another basis for $H$. Is it true that
$$B_n = \text{span}\{c_{j_1}, ..., c_{j_n}\}$$
for some indices ${j_i}$?
I think so since $B_n$ has dimension $n$ so anything in the set can be written as sum of $n$ basis functions?
AI: It is not even true for finite-dimensional spaces. Let $B = \mathrm{span}(e_1) \subseteq \mathbb{R}^2$ and look at the basis $c_1 = (1,1)$, $c_2 = (1,-1)$ of $\mathbb{R}^2$.
|
H: Numerical Analysis, build a contractive function
I have a question regarding Numerical Analysis. I've never been asked these sorts of questions before and don't even know where to begin.
The goal of this exercise is to find a value alpha such that:
$$f(\alpha)=0$$
When $f$ is:
$$f(x) = x-0.2\sin(x)-0.5$$
Do the following steps to achieve this goal:
1) Build a contractive function $\Phi$ that satisifies:
$$\Phi(\alpha)=\alpha$$
2) Calculate $\alpha$ using the fixed-point method.
I don't know how to do step 1 and would like some help.
AI: The $\Phi$ almost suggests itself:
$$f(\alpha)=0\iff \alpha-0.2\sin\alpha-0.5=0\iff \alpha=0.2\sin\alpha+0.5 $$
so try $\Phi(x)=0.2\sin x+0.5$. Fortunately, $|\Phi'(x)|=|0.2\cos x|\le 0.2<1$, so this is a contraction.
|
H: How to show that this limit is identical to..
Suppose we have expressions
$$f_1=\frac{x(y-1)}{x(2y-1)-y}$$
and
$$f_2=\frac{xw/y+(1-x)w/(1-y)+4}{8}-\sqrt{\left(\frac{xw/y+(1-x)w/(1-y)+4}{8}\right)^2-1/4(1+xw/y)}.$$
Moreover, we can assume $w>0$, $x,y\in[0,1]$ and $x\ge y$. As you can probably guess, $f_2$ is a solution to a quadratic equation.
In numerical calculations I almost accidentally discovered that
$$\lim_{w\to\infty} f_2\to f_1,$$
which is not at all obvious from the expressions. Is that true - what's the limit of $f_2$ and how to calculate it?
AI: Your expression is of the form
$$a-\sqrt{a^2-b}$$
which, when $a$ is "much" larger than $b$, is approximately $b/(2 a)$. In your case,
$$a=\frac18 \left (\left [ \frac{x}{y} + \frac{1-x}{1-y}\right] w + 4\right )$$
$$b = \frac14 \left ( 1+ \frac{x}{y} w\right )$$
Now, as $w \to \infty$:
$$\frac{b}{2 a} = \frac{ 1+ \frac{x}{y} w}{\left [ \frac{x}{y} + \frac{1-x}{1-y}\right] w + 4} \sim \frac{\frac{x}{y}}{\frac{1-x}{1-y}+\frac{x}{y}}$$
which, when simplified, produces the sought-after expression.
What I didn't answer is, when is this true, and it should be valid for all $x,y \in [0,1]$ because $a$ remains much larger than $b$ as $w \to \infty$.
ADDENDUM
I made the assertion
$$a-\sqrt{a^2-b} \sim \frac{b}{2 a}$$
This comes from the expression for the Taylor series for the function
$$\sqrt{1+z} = 1+\frac12 z + O\left(z^2\right)$$
where the $O$ term is an error term for small $z$. Then
$$a-\sqrt{a^2-b} = a-a \sqrt{1-\frac{b}{a^2}} \sim a - a \left (1 - \frac{b}{2 a^2}\right ) = \frac{b}{2 a}$$
|
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