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H: Correctness of proof of division in ceiling function
My professor asked me to present a proof to my fellow students tomorrow that
$$\left\lceil\frac{n}{2^k}\right\rceil = \left\lceil\frac{\left\lceil\frac{n}{2^{k-1}}\right\rceil}{2}\right\rceil$$
I tried to prove it as follows, but I am not sure if my proof is correct.
Proof
The above is equivalent to $$\left\lceil\frac{n}{2^{k+1}}\right\rceil = \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
Let us say that $x = \left\lceil\frac{n}{2^{k+1}}\right\rceil$ for ease of notation.
It is then clear that $$x-1 \lt \frac{n}{2^{k+1}} \le x$$
We will now prove what is asked using contradiction. Assume that $$\left\lceil\frac{n}{2^{k+1}}\right\rceil \lt \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
Then there is an integer $a$ such that $$\left\lceil\frac{n}{2^{k+1}}\right\rceil \le a \lt \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
This is equivalent to $$\left\lceil\frac{n}{2^k}\right\rceil \le 2*a \lt \frac{n}{2^k}$$ which can obviously not be true.
QED
Do you see any errors in my proof? Is it clear? I'm a bit doubtful about the last step, where I multiply $a$ by 2.
AI: Your reasoning is a bit shaky in a few places. Let me see if I can help you a bit with the rigor.
To further simplify, let's say $x=\left\lceil\frac{n}{2^{k+1}}\right\rceil$ and $y=\left\lceil\frac{n}{2^k}\right\rceil,$ so we must prove that $x=\left\lceil\frac{y}2\right\rceil$. Since $\frac{n}{2^k}\le y,$ then $\frac{n}{2^{k+1}}\le\frac{y}2,$ and so $x\le\left\lceil\frac{y}2\right\rceil$. Now we can assume that $x\ne\left\lceil\frac{y}2\right\rceil,$ from which it follows that $x<\left\lceil\frac{y}2\right\rceil.$ Then indeed, there is an integer $a$ such that $x\le a<\left\lceil\frac{y}2\right\rceil.$ In fact, $x$ is such an integer! More relevant, though, is the fact that $$\frac{n}{2^{k+1}}\le x<\left\lceil\frac{y}2\right\rceil.$$ Since $x$ is an integer less than $\left\lceil\frac{y}2\right\rceil$, then we have $x<\frac{y}2,$ and so $$2x<y.\tag{$\spadesuit$}$$ On the other hand, $$\frac{n}{2^{k+1}}\le x,$$ so $$\frac{n}{2^k}\le 2x,$$ and since $2x$ is an integer, then $$y=\left\lceil\frac{n}{2^k}\right\rceil\le 2x.\tag{$\clubsuit$}$$ By $(\spadesuit)$ and $(\clubsuit),$ we have our contradiction. |
H: Show that $\lim_{\delta \to 0}(1-\lambda \delta)^{1/\delta} = e^{-\lambda}$
My professor said that
$$\lim_{\delta \to 0}(1-\lambda \delta)^{t/\delta}=e^{-\lambda t}$$
can be shown with L'Hospital's rule. I don't know what he meant. What is the best way to show this (or, more simply, $\lim_{\delta \to 0}(1-\lambda \delta)^{1/\delta} = e^{-\lambda}$)?
If I try as follows
$$\lim_{\delta \to 0}\left(1-\lambda\delta \right)^{1/\delta} = \lim_{\eta \to \infty} \frac{(\eta-\lambda)^\eta}{\eta^\eta},$$
then I'm getting led into confusion trying LHR on the last one.
AI: Another approach: define
$$x:=\frac1\delta\implies \delta\to 0\implies x\to\infty$$
and our limit is
$$\left[\left(1-\frac\lambda x\right)^x\right]^t\xrightarrow[x\to\infty]{}(e^{-\lambda})^t=e^{-\lambda t}$$
We used above the basic
$$\lim_{x\to\infty}\left(1\pm\frac\lambda{f(x)}\right)^{f(x)}=e^{\pm\lambda}$$
for any function $\;f(x)\;$ s.t.
$$\lim_{x\to\infty}f(x)=\infty$$ |
H: find equation of tangent and normal lines (differentiation)
What is the equation of the tangent and normal lines to this function at the point p
$f(x)$ = $x^3$ at the point $p = (2,8)$
AI: First, we take the derivative of $f(x)$
$f'(x) = 3x^2, \tag{1}$
the evaluate it at $x = 2$ to obtain the slope at $(2, 8)$:
$f'(2) = 3(2)^2 = 12. \tag{2}$
Knowing this, we have that the tangent line, that is, the line of slope $12$ passing through the point $(2, 8)$, is given by
$y - 8 = 12(x - 2), \tag{3}$
or
$y = 12x - 16. \tag{4}$
The normal line obeys a similar equation but its slope is $-\frac{1}{12}$; thus
$y - 8 = -\frac{1}{12}(x - 2), \tag{5}$
or
$y = -\frac{1}{12}x + \frac{49}{6} \tag{6}$
are equations for the normal line. QED.
Hope this helps. Cheerio,
and as alwaus,
Fiat Lux!!! |
H: Determine an equation for the tangent to the graph of f(x) at point P.
Determine an equation for the tangent to the graph of f(x) at point P.
Use of CAS tool allowed.
a) f(x)= 3/(1+√x) , P(4, 1)
b) f(x)= √(5-x^2 ) , P(1, 2)
AI: Since you are allowed to use a CAS tool, there is not much to do here.
Here is one such tool: a), and b) |
H: Trouble with caculus problem, parametric equations, I don't know what I'm doing wrong
When a mortar shell is fired with an initial
velocity of v0 ft/sec at an angle α above the
horizontal, then its position after t seconds is
given by the parametric equations
$x = (v0 \cos \alpha)t$ , $y = (v0 \sin \alpha)t − 16t^2$
If the mortar shell hits the ground 4900 feet
from the mortar when α = 75◦, determine v0.
So I've tried various forms of:
\begin{align*}
t = {} & 4900/(v0 \cos 75) \\
0 = {} & (v0 \sin 75)(4900/(v0 \cos 75)) - 16(4900/(v0 \cos 75))^2 \\
4900(v0 \sin 75)/(v0 \cos 75) = {} & 384160000/(v0 \cos 75)^2 \\
v0 \sin 75 = {} & 78400/(v0 \cos 75) \\
v0 = {} & 78400/\sin 75 * v0 * \cos 75 \\
v0^2 = {} & 78400/\sin 75 * \cos 75 \\
v0 = {} & 468.33...i
\end{align*}
which doesn't seem right. And the answer choices are:
v0 = 530 ft/sec
v0 = 560 ft/sec
v0 = 520 ft/sec
v0 = 550 ft/sec
v0 = 540 ft/sec
AI: You seem to have your calculator set to radians, but are entering your angles in degrees.
$$\sqrt{\frac{78400}{\sin75\cos75}} = 468.3359976i $$
However, if we convert $75^\circ$ to radians $75 \frac{\pi}{180} = \frac{5}{12}\pi$
$$\sqrt{\frac{78400}{\sin\left(\frac{5}{12}\pi\right)\cos\left(\frac{5}{12}\pi\right)}} = \dots $$ |
H: Number of steps one needs to infer final height of a walk
Let us have a walk on $\mathbb Z$ of size $2^n$. To compute the final height of the walk, the trivial way is to sum $1$ for an ascending step and $-1$ for a descending step all along the walk. I would like to have some method to infer the final height in a more efficient way.
More formally, let us denote $\bar h = \frac{h}{2^n}$, where $h$ is the final height. Given an error $\epsilon$ and a probability $p$, I'd like to compute $g$ such that $\mathbb P(\lvert g- \bar h \rvert > \epsilon) < 1-p$, in some "efficient" way (i.e in polynomial time in $n$, and the smaller $\epsilon$ and $p$ are, the more I'll give myself time).
What I have is that if I take $i\in \{1,\cdots,2^n\}$, and set $X_i$ to be $1$ if the $i$-th step is ascending and $-1$ if the step is descending, then the expectation of $X_i$ is exactly $\bar h$. I repeat this $k$ times and set $g = \frac{1}{k}\sum_{i=1}^k X_i$. What should be $k$ for $g$ to be close to $\bar h$ with good probability? I guess this should have the flavour of a Chernoff bound, but all I found on this talks about boolean random variables, and I don't know if I can finish we this. How should I conclude?
AI: Hint: Use Chebyshev's inequality, noting that the variance of $h$ is the variance of a binomial distributed variable which has been scaled by a factor of $2$, since you are using the range $\{-1,1\}$ instead of $\{0,1\}$. Also note you should take $g = \mathbb{E}\bar{h} = 0$. |
H: How many different box combinations can you get?
I have 12 books, all different. Four of them are fiction, and eight of them are non-fiction. I want to send you a gift. I'm going to send you five books - two fiction, three non-fiction. You will get them in a big old box, all at once. How many possible gifts can you receive?
AI: Possible gifts you can receive:
$$^4C_2*^8C_3$$ |
H: Why is this defined even when divided by zero
so I've got $$\dfrac{x^3-4x^2+3x}{x^2-1}$$ and want to calculate the asymptotes. There's one a $x=-1$ since the function is not defined there. But the function seems to be defined for $x=1$. How come?
It should be undefined at $x=1$ since $f(1)=1^2-1 = 0$
AI: Factor the numerator: $x^3 - 4x^2 + 3x = x(x-1)(x-3)$. Cancel the common factor in the numerator and denominator: $$\dfrac{x(x-1)(x - 3)}{(x - 1)(x + 1)} = \dfrac{x(x-3)}{x+1}$$
However, please note that the original function is NOT defined at $x = 1$. But no asymptote exists there because it is a removable discontinuity.
Vertical asymptotes occur only when the denominator is zero, but they don't necessarily occur when the denominator is zero.
In the case $x = -1$, the denominator is zero, so we have a potential asymptote at $x = -1$. Indeed, the numerator of the given function is not zero at $x = -1$, and hence an asymptote does in fact exist, and is given by the vertical line $x = -1$.
On the other hand, in the case $x = 1$, while the denominator is zero (and hence we need to determine whether or not an asymptote exists there), we see that the numerator also evaluates to $0$ at $x = 1$. Indeed, we see that, both numerator and denominator share a common factor $x - 1$, which can be canceled, and hence the discontinuity at $x = 1$ can be removed. So, in fact, no asymptote exists at $x = 1$.
Suggestion: graph the original function to get an visual idea of what's happening with the given function. |
H: Is connectedness $\implies $ local connectedness?
Let $X$ be a connected topological space.
Then I wonder that $X $ can be locally connected or not.
I think the title of my question is NOT correct considering a Euclidean space like $\mathbb R^n$.
Do you think so??
Also, conversely, if $X$ is a locally connected, and if some conditions are given additionally, Can $X$ be a connected space?
I think the latter case may be possible (not certainly), but don't know the necessary condition although it is true.
Please somebody tell me answers of them~~
AI: Neither connected nor locally connected implies the other, nor do their negations. Four examples:
$\Bbb R$ is connected and locally connected.
$[1,2] \cup [3,4]$ is locally connected but not connected
The topologist's sine curve is connected but not locally connected.
$\Bbb Q$ is neither |
H: Triangle area division
It is given that in a triangle ABC, a line from A to BC intersects BC at point D. If the ratio in which AD divides BC is given can we say anything about the ratio of areas of triangle ABD and triangle ADC ?
AI: Yes we can find the ratio of the areas of the two triangles.
Area of $\triangle ADC$=$$(AD\times DC\times\sin\theta)\over2$$
Area of $\triangle ABD$=$$(AD\times DB\times \sin(180^\circ-\theta))\over2$$
[$\theta$ is the angle $ADC$]
Divide these two equations and use the ratio you have to get the answer. |
H: If X and θ are both random variables and θ is the parameter of the distribution of X, are X and θ independent?
The answer appears to be no because the distribution of X is defined conditionally by θ which is also assumed to have a distribution as opposed to be a constant. Essentially, the formulation of the probability distribution function of X is explicitly dependent on θ so how could X and θ be independent of each other.
However, I am confused by an equation in the proof of Bayes’ Theorem for random variables:
Theorem:
Suppose that the n random variables X1, . . . , Xn form a random sample from a distribution for which the p.d.f. or the p.f. is f (x | θ). Suppose also that the value of the parameter θ is unknown and the prior p.d.f. or p.f. of θ is ξ(θ). Then the posterior p.d.f. or p.f. of θ is
ξ(θ|x) =[f(x1|θ) . . . f(xn|θ)ξ(θ)]/gn(x)
for θ ∈ Ω,
where gn is the marginal joint p.d.f. or p.f. of X1, . . . , Xn.
In the proof of the theorem it is stated that the joint distribution of x (vector of observed X's) and θ--f(x,θ)--is equal to the product of the conditional joint distribution of x given θ--fn(x|θ)--and the marginal distribution of θ--ξ(θ).
f(x,θ)=fn(x|θ)ξ(θ)
I know that if X and θ are independent random variables, then the joint distribution of X and θ, f(x,θ)=g(x)ξ(θ). The above looks similar except for, of course, the p.d.f. of X is conditional upon θ. So, I wanted to make sure I was not missing the implication that X and θ are actually independent of each other.
Thanks in advance for your help. It seems straightforward, but I couldn't find an explicit answer on the website or in my readings...perhaps because it is so obvious.
AI: Short answer is no - you are not missing anything. Although the condition of indepenence and the formula for the bayesian joint posterior look similar, they are very different, since $\theta$ is present in both equations in the Bayes formulation. If $\theta$ and x are truly indepenent, then fn(x|$\theta$)=fn(x), returning the indpendence condition. But that is not a given.
Qualitatively, you can think about it this way: One uses Bayes theorem to update one's knowledge of one random quantity given knowledge about another random quantity. If x were always independent of $\theta$ then Bayesian inference would be a non-starter. |
H: Show that $x_{n+1}-\dfrac{x_n}{2}$ converges to zero implies that $(x_n)_n$ also converges to zero.
Let $(x_n)_n$ be a sequence of real numbers such that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0.$$ Show that $(x_n)_n$ must also converge to zero.
$\textbf{My Attempt:}$
My original idea was to use the triangle inequality and the fact that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0$$
to show that the sequence $(x_n)_n$ is Cauchy and hence converges to some real number $a$, which would give me $$\lim\limits_{n\rightarrow \infty}x_n=a=\lim\limits_{n\rightarrow \infty}x_{n+1}\implies a-\dfrac a 2= \lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0\implies a=0.$$ However, I'm having trouble showing that $(x_n)_n$ is Cauchy.
Any suggestions/hints would be greatly appreciated. Thanks!
AI: Let $a_n:=x_{n+1}-\frac{x_n}2$; then $a_n\to 0$. We have
$$|x_{n+1}|=\left|a_n+\frac{x_n}2\right|\leqslant |a_n|+\frac{|x_n|}2.$$
Taking the $\limsup$, we get
$$\limsup_{n\to \infty}|x_n|\leqslant \frac 12\limsup_{n\to \infty}|x_n|.$$
Take $M\geqslant 2$ and $n_0$ such that $|a_n|\leqslant 1$ if $n\geqslant n_0$, and assume that $|x_n|\leqslant M$. Then $|x_{n+1}|\leqslant M/2+1\leqslant M$. So the sequence $\{x_n\}$ is bounded and the $\limsup$ is finite. |
H: Proof of $|\int _a ^b \mathbf f | \leq \int _a ^b |\mathbf f|$
Let $\mathbf f:[a,b]\rightarrow \mathbb R ^n$ be continuous. I'm trying to prove the following fact:
$$|\int _a ^b \mathbf f| \leq \int _a ^b |\mathbf f|,$$
where: $$|\mathbf x|^2=\sum _{i=1} ^n x_i ^2.$$
I understand it by intuition: if $\mathbf f$ is interpreted as the velocity of a particle in $\mathbb R ^n$, the first member is the length of the displacement $\mathbf F(b)-\mathbf F(a)$, while the second one is the lenght of the described curve.
Since :$$|\int _a ^b \mathbf f|^2=\sum _{i=1}^n (\int f_i)^2$$
and $$(\int _a ^b |\mathbf f|)^2=(\int _a ^b \sqrt{\sum _i f_i ^2})^2,$$
I see that this is equivalent to prove the inequality for the second two members. Well... I'm stuck there. Any hints?
AI: This is the proof I've found on Rudin, valid for $\mathbf f\in \mathscr R(\alpha)$:
Let $\mathbf y = \int _a ^b \mathbf f \, \text d \alpha$. Then: $$|\mathbf y|^2 =\sum y_i^2=\sum y_i \int _a ^b f_i \,\text d \alpha=\int _a^b(\sum y_if_i )\,\text d \alpha=\int_a ^b\mathbf y\cdot \mathbf f\,\text d \alpha$$
Using Cauchy-Schwarz's inequality on $\mathbf y\cdot \mathbf f$ proves the result. |
H: Help with two variable equation
I need help with this equation which was given to my son who is in 4rth grade.
Solve in positive integers the equation: $xy=10(x+y)$.
AI: $xy =10x+10y$ $\implies$ $y=\frac {10x}{x-10}$.
Now use hit and trial or you can easily see that y has an integral value when x equals an integral value of 10 except 1. |
H: Localizing the quotient at a prime
I was reading in Dummit and Foote when I came across this statement I don't believe it is true.
Let $R$ be a Dedekind domain (noetherian, integrally closed, domain of dimension 1). They say given a prime ideal $P\neq 0$ then
$$R/P^{m}\cong R_{P}/P^{m}R_{P}.$$
Why is this true? Shouldn't we get
$$ (R/P^{m})_{\overline{P}}\cong R_{P}/P^{m}R_{P}?$$
I don't see why when we localize $R/P^{m}$ at the image of $P$ nothing changes. Thanks for the help.
AI: If $A$ is a local ring with maximal ideal $M$, then $A\simeq A_M$. In your case $A=R/P^m$ is local since $P$ is the only prime containing $P^m$. |
H: Triangle Trig and calculus
The square of the side opposite a MUNDANGLE in a triangle is equal to the sum of the squares of the other two sides added to the product of these two sides multiplied by $\sqrt{3}$. What is MUNDANGLE?
This was an extra credit problem on my test, but my teacher said we had to do it for homework.
If anyone could support me and lead me to through the problem that would be great
AI: We have $\displaystyle a^2=b^2+c^2+bc\sqrt3$
Now using Cosine rule of triangle, $\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{-\sqrt3bc}{2bc}=-\frac{\sqrt3}2=-\cos\frac\pi6=\cos\left(\pi-\frac\pi6\right)$ as $\cos(\pi-x)=-\cos x$
Can you take it from here? |
H: Something to the power of a logarithm
This is probably a very obvious question, but here goes...
An answer in my textbook claims that
$$3^{\log n} = n^{\log 3}$$
and that
$$4n^2 (3/4)^{\log n} = 4n^{\log 3}$$
Why, using more basic laws, is this the case?
(Unfortunately Google confuses this question with changing bases, exponentiation being the inverse of log (which is of course related), and similar matters.)
AI: Hint : $a=b$ implies $\log a=\log b$ and $\log a^b=b\log a$ |
H: Solving for a variable with square roots in equation
I'm working on the following equation:
$$ a= \frac{2}{t^2}[(d+l)^{1/2} - (d^{1/2})]^2 $$
I want to solve for l such that:
$$ l=\frac{1}{2}at^2+(\sqrt{2ad}) t $$
I can't get it in this form. Here's my attempt:
$$ \frac{at^2}{2} = [(d+l)^{1/2} - (d^{1/2})]^2 $$
$$ \sqrt{\frac{at^2}{2}} = [(d+l)^{1/2} - (d^{1/2})] $$
$$ \sqrt{\frac{at^2}{2}} + d^{1/2} = [d+l]^{1/2} $$
$$ (\sqrt{\frac{at^2}{2}} + d^{1/2})^2 = [d+l] $$
$$ (\frac{at^2}{2} + 2\sqrt{\frac{dat^2}{2}} + d = [d+l] $$
$$ \frac{at^2}{2} + 2\sqrt{\frac{dat^2}{2}} = l $$
$$ \frac{at^2}{2} + 2t\sqrt{\frac{da}{2}} = l $$
This of course isn't in the form I'm looking for. The first piece $\frac{at^2}{2} $is good, but the second piece $ 2t\sqrt{\frac{da}{2}}$ does not equal $\sqrt{2ad}t$
Surely I must be making a mistake somewhere?
AI: Umm..actually you answered it yourself. Take the $2$ inside the square root and it becomes $4$ and then cancels out with the $2$ in the denominator. |
H: Property of the limit of function proof
If $c≠0$ and $\lim_{x→c}〖f(x)〗=L$, prove that $\lim_{x→1/c}〖f(1/x)〗=L$
I know that meaning for all $ε>0$, there exist a $δ>0$ such that $0<|x-c|<δ$ implies $|f(x)-L|<ε$. How can I use these to prove the conclusion.
AI: When you say |x-c| < $\delta$ you mean $\delta$ - c < x < $\delta$ + c. Now restate that inequality in terms of 1/x. Can you work from there to construct a new $\delta_{1/x}$ which works for your limit? |
H: Dividing square of 2013x2013
Is the square of 2013×2013 can be divided into rectangles 1×3 in such
a way that the number of rectangles arranged vertically differ by 1
from the number rectangles arranged horizontally? Prove your answer
I totally have no idea how to solve that. I need accurate solutions or good hints.
AI: Since $2013$ is divisble by $3$, show by complete induction that the number of vertical bars that extend upward starting with bottom base at row $k$ must be divisible by $3$. Then you will get that the total number of vertical bars is divisible by $3$, and by symmetry, the number of horizontal bars that extend rightward starting with base column $k$ must be divisible by $3$, so the total number of horizontal bars must also be divisible by $3$, so they cannot differ by $1$. |
H: Inequality and absolute value: $p + |k| \gt |p| + k$
Here is the problem I am confused about. The given relation is:
$p + |k| \gt |p| + k$
It is not mentioned whether $p$ and $k$ are integers.
I need to determine whether $p$ and $k$ are equal or one is greater than other. If the second is true which one is greater.
To solve this inequality problem I was trying to consider the values of $p$ and $k$ with all combinations of signs i.e. $p$ positive and $k$ negative; $p$ negative and $k$ positive; and $p$ and $k$ both negative and positive.
When I assume both $p$ and $k$ are negative, I could not find the solution. The solution is in the book too. But I failed to understand the solution. Can anyone please help me to solve the problem? Thanks.
AI: Here is another approach:
Let $\phi(x) = |x|-x$, and note that the problem is equivalent to determining $p,k$ such that $\phi(k) > \phi(p)$.
Now plot $\phi$ to get
So, from this we can see that $\phi(k)> \phi(p)$ iff $k<p$ and $k <0$. |
H: Is this implication true?
Suppose that a real sequence $u_n$ is such that $$u_{n+1}-u_n \rightarrow0$$
That is not enough to prove that $u_n$ is convergent (take $u_n=ln(n)$)
Now what if $u_n$ is bounded ? I guess it does converge, but how to prove this ? I tried to show that it had only one accumulation point...
AI: Hint: Consider
$$
u_n=\sin(\log(n))
$$
Show that $|u_{n+1}-u_n|\lt\frac1n$ yet $\limsup\limits_{n\to\infty}u_n=1$ and $\liminf\limits_{n\to\infty}u_n=-1$. |
H: why is $a^nb^n$ context-free?
I am writing somthing about Ppumping Lemma. I know that the language L = { $a^nb^n$| n ≥ 0 } is context-free. But I don't understand how this language satisfies the conditions of pumping lemma (for context-free languages) ?
if we pick the string s = $a^pb^p$, |s| > p , |vxy| < p and |vy| > 0.
it seems it will be out of the language when we pump it (pump up or down) or there is something I'm missing.
Any explanation would help.
Edit: I am applying pumping lemma to $a^nb^n$ and it fails to stay in the language for all cases. So, why is it Context free?
AI: The thing is that the lemma only says that for a CFL $L$, a $p$ exists such that any string $s$ of length at least $p$ (i.e., $|s| \ge p$) can be decomposed as $s =uvwxy$ with $|vwx|<p$, $vx \neq \varepsilon$ and $uv^nwx^ny \in L$.
Now, in the example, consider $s=A^nB^n$. Take $p = 3$, $v=A$, $x=B$, $u=A^{n-1}$, $w=\varepsilon$ and $y=B^{n-1}$.
Then clearly $|vwx|=2<3$, $vx = AB \neq \varepsilon$ and $uv^mwx^my = A^{n-1}A^mB^mB^{n-1} = A^{m+n-1}B^{m+n-1} \in L$. |
H: Multivariable Delta Epsilon Proof
I am trying to prove the following limits using the delta-epsilon method. Can you help me out?
$$ \lim_{(x,y)\to(a,b)}(x+y) = a+b$$
AI: Hint:
Since on $\mathbb R^n$ all norms are equivalent, chose the easiest: $\Vert\cdot\Vert_\infty$. This will leave you with two 1-D. $\epsilon$-$\delta$ arguments where
$$\epsilon = (\epsilon_1, \epsilon_2)$$
and you want
$$|x+y-a-b| = |(x-a)+(y-b)| \leq |x-a| + |y-b| < \epsilon_1 + \epsilon_2 = 2\Vert\epsilon\Vert_\infty$$
Get the gist? I can provide more detail if needed.
Now all that remains to prove is
$$\lim_{x\to a} x = a$$
or equivalently (by exchanging symbols)
$$\lim_{y\to b} y = b$$
For the choice of $\delta$, $\delta_\epsilon := \frac{\epsilon}{2}$ suits your needs.
Proof
Using equivalency of norms on $\mathbb R^n$, we have to prove
$$\forall\ \epsilon > 0 \quad \exists\ \delta > 0$$
such that
$$|x+y - (a+b)| < \epsilon \qquad \forall\ (x,y) \in \mathbb R^2 | \Vert (x,y) - (a,b) \Vert_\infty < \delta$$
Now for $\delta := \frac{\epsilon}{2}$ we have
$$|x+y - (a+b)| \leq |x-a| + |y-b| < \delta + \delta = 2\delta = \epsilon$$
q.e.d.
Note
Using the euclidean norm $\Vert\cdot\Vert_2$ we can chose $\delta = \frac{\epsilon}{4}$, since
$$\Vert (x,y) \Vert_2^2 = x^2 + y^2 \leq x^2 + 2|x||y| + y^2 = (|x| + |y|)^2 \leq (2\Vert(x,y)\Vert_\infty)^2$$
(actually we give away a bit here, but that's not important for the task at hand) |
H: Prove that x is rational
Let $x$ be a real number with the properties that $x^3+x$ and $x^5+x$ are rational.
Prove that $x$ is rational. Denote $a=x^3+x$; $b=x^5+x$. We can multiply and add them together until we get the desired result. I also know some non-elementary proofs of this, but have you some nice elementary proofs?
Thank you.
AI: We may take $x>0$. $\frac{x^5+x}{x^3+x}$ is rational, so $bx^4+b = ax^2 +a$ for some integers $a,b$ and $x = \sqrt{k}$, where $k=c+\sqrt{d}$ is the root of a quadratic equation, with $c,d\in\mathbb{Q}$, $d=0$ or $d>0$ not a square.
If $k$ is rational ($d=0$) but not a square, $x^2-1$ is rational and $x^5+x = (x^3+x)(x^2-1)+2x$ is irrational, a contradiction.
If $k$ is irrational, ($d\neq 0$), then expanding $x^3+x$ we get
$$x^3+x = c^3+3cd+c + (3c^2+d+1)\sqrt{d}$$
is irrational, a contradiction. |
H: A question dealing with conditions for which $V_\alpha$ models $ZFC$
I've been reading through models of Set Theory in Kunen's most recent Set Theory text and practicing exercises. He mentions that $V_\alpha$ can be used to satisfy certain axioms of $ZFC$ when $\alpha$ is strongly inaccessible. Here, $V_\alpha$ is the set of all well-founded sets whose rank is less than $\alpha$. From here, he presents an exercise to show why $V_\alpha$ models $ZFC$ under certain conditions.
$(ZFC^-)$ Assume that $0 < \alpha < \beta$ and $V_\alpha \preccurlyeq V_\beta$. Prove that $V_\alpha \models ZFC$ and hence $V_\beta \models ZFC$. You may use the fact, to be proved later, that $V_\alpha \models ZC$ for any limit $\alpha > \omega$.
From here, he gives a hint to show how to do this. I broke up the hint into three parts.
1.) Show $\alpha$ is a limit, since if $\alpha = S(\gamma)$, then $ ``\preccurlyeq" $ would fail with the formula $ ``S(a) \mbox{ exists}" $.
2.) Show that $\alpha > \omega$.
3.) For the Replacement Axiom in $V_\alpha$, if $A \in V_\alpha$ and if $\forall x \in A \exists ! y \varphi(x,y)$ holds, then $\exists B \forall x \in A \exists y \in B \varphi(x,y)$ must hold in $V_\beta$.
I was able to successfully show the first part by using the definition of $V_\alpha$. Thus, $\alpha$ must be a limit ordinal. I'm stuck on second and third parts. For the second part, my guess is to rule out the case when $\alpha = \omega$ by coming up with a formula that is true in $V_\beta$ but not in $V_\omega$, but I can't think of one. For the third part, I'm not sure how to show it satisfies Replacement.
Any help would be greatly appreciated!
AI: Hints:
The axiom of infinity is false in $V_\omega$. Note that since $\alpha<\beta$ we have that $V_\beta\neq V_\omega$. Conclude the same about $\alpha$.
For the axiom of replacement, suppose that $\varphi(u,v)$ is a functional formula (I'm omitting the parameters here) on domain $x\in V_\alpha$. We want to show that $y=\{v\in V_\alpha\mid\exists u\in x.\varphi(u,v)\}$ is in $V_\alpha$.
Next we use the fact that $V_\beta$ is a model of the separation schema (which is a part of $\sf ZC$) to have that $y\in V_\beta$. Conclude that $V_\beta\models\{v\mid\exists u\in x.\varphi(u,v)\}=y$ (by the fact that $\varphi$ is functional on $x$) and use elementarity again. |
H: A proof about complex polar form
I need to proof that $e^{i\theta}e^{i\phi} = e^{i(\theta+\phi)}$ as my homework using Euler's formula. And then using trigonometric sum formulas, I guess.
I would appreciate if you gave me some hints.
AI: Note that
\begin{align}
e^{it} e^{is} &= (\cos{s} + i \sin s) (\cos t + i \sin{t}) \\
&= (\cos{s} \cos{t} - \sin{s} \sin{t}) + i (\sin s \cos t + \sin t \cos s)
\end{align}
Now use the identities
$$\cos{(s + t)} = \cos{s} \cos{t} - \sin{s} \sin{t}$$
and
$$\sin{(s + t)} = \sin s \cos t + \sin t \cos s$$ |
H: Is it possible to get a 'closed form' for $\sum_{k=0}^{n} a_k b_{n-k}$?
This came up when trying to divide series, or rather, express $\frac1{f(x)}$ as a series, knowing that $f(x)$ has a zero of order one at $x=0$, and knowing the Taylor series for $f(x)$ (that is knowing the $b_i$ 's).
I write $$1=\frac1{f(x)}f(x) = (\frac{r}{x}+a_0+a_1x+a_2x^2+...)(b_1x+b_2x^2+b_3x^3+...)$$
And comparing coefficients I get $r= \frac{1}{b_1}$ and more importantly, the equations:
$$0=b_{n+1}r + \sum_{k=0}^{n} a_k b_{n-k}.$$
Is there a closed form for this recursion, i.e. is it possible to extract $a_n = ...$ from this? In the particular case I'm looking at, $b_n = \frac{1}{n!}$.
AI: If you know $f$ has a zero of order $1$ you can write it as $$f(x)=x\left(a_1+a_2x+a_3x^2+\ldots\right),$$
with $a_1\neq0$.
Then $\frac{1}{f}=\frac{1}{x}\frac{1}{a_1+a_2x+a_3x^2+\ldots}$.
To compute the series of $\frac{1}{a_1+a_2x+a_3x^2+\ldots}$ just apply long division of $1$ divided by $a_1+a_2x+a_3x^2+...$. This is an algorithm that allows you to compute, term by term, the series if the quotient.
If we have more information on $f$ other methods could be applied. For $f$ a rational function there is a much better method. |
H: Multiplication by $x^2$ linear maps
$T:P(\mathbb{R}) \mapsto P(\mathbb{R})$ defined by
$(Tp)(x) = x^2p(x)$
Verify that multiplication by $x^2$ is a linear map.
Additivity: $x^2(p+q) = x^2p+x^2q$
Homogeneity: $x^2(ap) = a(x^2p)$
Is this a correct verification?
AI: It seems that you've just written down the definition of linearity, but I expect that the problem is looking for a more thorough proof using the definition of $P(\mathbb{R})$ - so it's really asking why the two equalities you wrote are true.
Let's start with polynomials $f = a_0 + a_1 x + a_2 x^2 + \cdots+ a_n x^n$, and $g = b_0 + b_1 x + \cdots + b_n x^n$ (we don't require $b_n \ne 0$ or $a_n \ne 0$). Then
\begin{align*}
x^2 (f + g) &= x^2 \Big((a_0 + b_0) + (a_1 + b_1) x + \cdots + (a_n + b_n)x^2\Big) \\
&= (a_0 + b_0) x^2 + (a_1 + b_1) x^3 + \cdots + (a_n + b_n) x^{n + 2} \\
&= \Big(a_0 x^2 + a_1x^3 + \cdots + a_n x^{n + 2}\Big) + \Big(b_0 x^2 + b_1 x^3 + \cdots+ b_n x^{n + 2}\Big) \\
&= x^2 (a_0 + \cdots + a_n x^n) + x^2 (b_0 + \cdots + b_n x^n) \\
&= x^2 f + x^2 g
\end{align*}
as desired. Now do something similar for multiplication by a scalar. |
H: Find the sale price given the 40% markup over wholesale and later 20% discount
A retailer pays $\$86$ to a wholesaler for an article. The retail price is set using a $40\%$ rate of markup on selling price. To increase traffic to his store, the retailer marks the article down $20\%$ during a sale. What is the sale price?
How can I calculate the markup on selling price when S is not given?
Also S = C + M but we can't get M because M = E + P. in this case we have no overhead expenses (E) so can someone please help me out. I've been stuck on this for so long!
AI: Hint: Typically a $40\%$ markup means that the cost is $60\%$ of the final selling price. The original selling price is then $\frac {86}{0.6}=143\frac13$. Then if you mark this down by $20\%$, you multiply by $0.80$ and get ??? |
H: Complex power series expansion for $f(z) = \frac{e^z}{a-z}$
I have the following homework problem in my complex analysis class:
Find the complex power series expansion for the function
$$ f(z) = \frac{e^z}{a-z}$$
where $a \in \mathbb{C}$, and $z \ne 0$.
I know the power series expansions for the functions:
$$ g(z) = e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} $$
valid for each $z \in \mathbb{C}$, and
$$ h(z) = \frac{1}{a-z} = \sum_{n=0}^{\infty} \frac{z^n}{a^{n+1}}$$
valid for $\mid \frac{z}{a} \mid < 1$.
However, I do not see how to combine these two expressions to yield the desired expression. More specifically, even though $f(z) = g(z)h(z)$, we do not obtain the correct power series expansion for $f$ by multiplying the power series expansions for $g(z)$ and $h(z)$ term-wise. It doesn't seem like the Cauchy Product of the two series will be of much help either, although I could be mistaken.
Any help would me most appreciated!
AI: The Cauchy product says that the coefficient of $z^n$ will be $\sum_{k=0}^n 1/(a^{k+1}(n-k)!)$ in your series, and it will be valid if $|z| < |a|$. |
H: Finding Revenue Function and Max Revenue
Studying for a midterm.
The demand function for a manufacture's product is $p=1000-\frac1{80} q$
Where $p$ is the
price (in dollars) per unit when $q$ units are demanded (per week) by consumers. Answer
the following questions.
1) Write the Revenue function $R(q)$ in terms of $q$.
2) Find the level of production that will maximize revenue.
3)Suppose there is a fixed cost of $174500, to set up the manufacture and a producing cost of 125 dollars per unit. Find the break even quantities.
First: To find the revenue function.
I know that Revenue=$p*q$ so:
$$R(q)=p*q$$
$$p=1000-\frac1{80}q$$
$$R(q)=(1000-\frac1{80}q)*q$$
$$=1000q-\frac1{80}q^2$$
I believe this is right.
Now to find the level of production to maxime revenue we must find the first derivative of the revenue function.
$$R'(q)=1000-2(\frac1{80}q)$$
$$2(\frac1{80}q)=1000$$
$$\frac1{80}q=500$$
$$q=40000$$
Input this into our demand function:
$$p=1000-\frac1{80}40000$$
$$p=500$$
Now I don't know if this is right, please correct me if I'm wrong.
Now I'm not sure how to find the break even quantities, I would appreciate help, at least to get me started.
Cheers.
AI: Your work is correct. To find the break even quantities, you need to find where the Revenue function is equal to the cost function. Your cost function is $C(q)=174500+125q$. |
H: mathematical induction proof 2 things that are equal
$\begin{align} \forall n \in \mathbb{Z}^+, \sum^n_{p=1} \frac{p^2}{4p^2-1}= \frac{n(n+1)}{2(2n+1)} \end{align}$
I have done mathematical induction proofs with only one phrase, but this one has two. Can someone assist me. The first meaningful case would be 2 right?
AI: No, the first meaningful case is $n=1$: $\dfrac1{4\cdot 1-1} = \dfrac{1\cdot 2}{2\cdot 3}$. Then
$$\frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{4(n+1)^2-1} = \frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{(2n+1)(2n+3)} = \frac{(n+1)(2n^2+5n+2)}{2(2n+1)(2n+3)} = \frac{(n+1)(2n+1)(n+2)}{2(2n+1)(2n+3)}=\frac{(n+1)(n+2)}{2(2n+3)}.$$ |
H: The convergence interval of the series
I want to prove whether $x=4/27$ is convergent or not for the series $$\sum_{n=0}^{\infty}\binom{3n}{2n}x^n$$
I used Raabe's test. But I got limit is 1. So the test is not valid. Please help me which test do I need to use ? Thank you
In fact, if you help me to find its convergence interval, I Will be so happy:)
AI: Call $a_n = \binom{3n}{2n}$. Note
$$ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)}{(2n+2)(2n+1)(n+1)}. $$
For the ratio test, then, we want
$$ \lim_{n \to \infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| = \frac{27}{4} |x| < 1 $$
for absolute convergence. So the series converges absolutely on the interval $\left( -\frac{4}{27}, \frac{4}{27} \right)$. At $x = \frac{4}{27}$, the series becomes
$$ \sum_{n=1}^{\infty} \frac{(3n)!}{(2n)!n!} \frac{2^{2n}}{3^{3n}}. $$
(HINT: use the comparison test) |
H: Natural Logarithm Taylor Series Expansion
f(x)=x$^3$ln(1+2x)
Write the first four non-zero terms of the Taylor Series for the above function with x centered at a=0.
Using this model:
ln(1+x) = Σ$\frac{(-1)^{k}(x)^{k}}{k}$
I get the following...
x$^{3}$ln(1+2x) => x$^3$Σ$\frac{(-1)^{k}(2x)^{k}}{k}$
k=0 -----> Cannot divide by zero.
k=1 -----> x$^3$$\frac{(-1)^{1}(2x)^{1}}{1}$$\frac{(x-a)}{1!}$
-----> (0)$^3$$\frac{(-1)^{1}(2(0))^{1}}{1}$$\frac{(x-0)}{1!}$ = 0
k=2 -----> x$^3$$\frac{(-1)^{2}(2x)^{2}}{2}$$\frac{(x-a)}{1!}$
-----> (0)$^3$$\frac{(-1)^{2}(2(0))^{2}}{1}$$\frac{(x-0)}{2!}$ = 0
and so on...
I keep getting zero for every value of k in the series so I never reach a non-zero term.
I don't even know if I am doing this wrong but I am assuming that I am supposed to get an answer that isn't, "there are no non-zero terms" for this expansion.
AI: As njguliyev said in the comments, you should be substituting $a=0$ rather than $x=0$. Your use of the power series for $\ln$ is correct, however, so the first term is $-2x^4$, as your work suggests. |
H: Proving |sv| = |s||v| and is it true for all dimensions
I've been sitting here quite a while trying to figure out how to prove
$$ |s\vec{\mathbf{v}}| = |s||\vec{\mathbf{v}}| $$
I was not told in what dimension to prove this is, so I am assuming a 3D plane. I think I have a solution but I'm not sure if my method is write. Could someone confirm I am doing it right?
So I started by setting $$\vec{\mathbf{v}} = (x, y, z)$$
$$ |\vec{\mathbf{v}}| = \sqrt{x^2 + y^2 + z^2} $$
(x, y, z) being points in a 3D graph.
Then
$$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2x^2 + |s|^2y^2 + |s|^2z^2} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2(x^2 + y^2 + z^2)} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2} \sqrt{x^2 + y^2 + z^2} $$
$$ |s\vec{\mathbf{v}}| = |s| \sqrt{x^2 + y^2 + z^2} $$
Since
$$ |\vec{\mathbf{v}}| = \sqrt{x^2 + y^2 + z^2} $$
Therefore
$$ |s||\vec{\mathbf{v}}| = |s|\sqrt{x^2 + y^2 + z^2} $$
Hence
$$ |s\vec{\mathbf{v}}| = |s||\vec{\mathbf{v}}| $$
My result does get me what I was looking for but it has happened in the past that for some reason a mistake that I made got me the right answer. Also if this is true for a 3D plane, does that mean that it holds through for all dimensions?
Thanks for any and all help
AI: Your proof looks okay except that I would make the following change
$$ |s\vec{\mathbf{v}}| = \sqrt{(sx)^2 + (sy)^2 + (sz)^2} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{s^2x^2 + s^2y^2 + s^2z^2} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{s^2(x^2 + y^2 + z^2)} $$
$$ |s\vec{\mathbf{v}}| = \sqrt{s^2} \sqrt{x^2 + y^2 + z^2} $$
$$ |s\vec{\mathbf{v}}| = |s| \sqrt{x^2 + y^2 + z^2} $$
Remembering that $\sqrt{s^2} = |s|$. By placing the $|s|$ within the square root in your solution, you have already to some degree presupposed that $|s\vec{\mathbf{v}}| = |s| |\vec{\mathbf{v}}|$.
The proof extends without a problem to more dimensions as long as you use a more generic formulation of the Euclidean norm:
$$|\vec{\mathbf{v}}| := \sqrt{v_1^2 + \cdots + v_n^2}$$ |
H: Conditions necessary for commutators [A,B]=[B,A]?
I know that normally for commutators that [A,B]=-[B,A] where A and B are operators. But under what conditions does [A,B]=[B,A]?
AI: Since it is always true that $[A, B] = -[B, A]$, if we assume that $[A, B] = [B, A]$, we see that
$$[A, B] = -[A, B]$$
so that $[A, B] = 0$. Hence, it is necessary and sufficient that $A$ and $B$ commute. |
H: Show $e^{D}(f(x)) = f(x+1)$ where $D$ is the derivative operator
I would appreciate help showing $e^{D}(f(x)) = f(x+1)$
Where $D$ is the linear operator $D: \mathbb{C}[x] \rightarrow \mathbb{C}[x]$ where (in the context where this statement arose) $x \in \mathbb{N}$;
$f(x) \mapsto \frac{d}{dx} f(x)$
By the Taylor series expansion $e^{D} = \sum_{n=0}^{\infty} \frac{D^n}{n!}$
$(1)$ Then $e^{D} (f(x)) = f(x) + f'(x) + \frac{f''(x)}{2!} +\dots$
I would appreciate help showing that the above line $(1)$ is equal to $f(x +1)$
This is what I have tried, but it feels forced:
for a Taylor series representation of $f(x +1)$ near $x$ I could write
$f(x+1) = f(x) + f'(x)(x+1 - x) + \frac{f''(x)}{2!}(x+1 -x)^2 \dots$
This then is equal to the RHS of line $(1)$.
Could this line of thinking be correct? Thanks very much.
AI: The Taylor series is $$ f(x+h) = \displaystyle \sum_{n \geq 0} \frac{f^{(n)}(x)}{n!}h^n $$ and here $e^A$ is nothing more than a notation for the power series $\sum A^n/n!$. The series is a well-defined operator for any operator $A$ that is "locally nilpotent" (only a finite number of terms are nonzero when applied to any $f$). Using $(hD)^n = h^n D^n$ it is a matter of comparing definitions to see that the operator that takes input any polynomial $g(x)$ and outputs $g(x+h)$ is $e^{hD}$.
That is all for polynomials, in characteristic $0$. |
H: Combinatorics? HELP!
Find the generating function for the number of ways to distribute blank scratch
paper to Alice, Bob, Carlos, and Dave so that Alice gets at least two sheets, Bob
gets at most three sheets, the number of sheets Carlos receives is a multiple of
three, and Dave gets at least one sheet but no more than six sheets of scratch
paper. Without finding the power series expansion for this generating function, determine the coefficients on x2 and x3 in this generating function.
I have no idea how to approach this...
AI: Take each person as a separate variable and multiply together the combinations:
Alice = $x^2+x^3+\dots $
Bob = $1+x+x^2+x^3$
Carlos = $1+x^3+x^6+\dots$
Dave = $x+x^2+x^3+x^4+x^5+x^6$
The product of these could be a generating function that may be one idea to explore though I'd wonder if you need separate variables to know who got how many sheets or not.
Note: The minimum term in the product will be $x^3$ as Alice gets 2 sheets and Dave gets 1 sheet is the minimal solution, so I'd wonder what if the x2 is $x_2$ or $x^2$ as there are a couple of ways to see where a 2 could go here. |
H: Prove that lim sup $x_n$ for infinitely many n.
Let $\{x_n\}$ be a real sequence and $r$ a real number. Prove that $\lim \sup_{n\to\infty}x_n<r$ implies $x_n<r$ for $n$ large enough. Prove that $\lim \sup_{n\to\infty}x_n>r$ implies $x_n>r$ for infinitely many $n$.
I am not sure how to prove this. Could anyone give me any hints or tips?
AI: let $l = \limsup_n x_n$. This means $l = \lim_n \sup_{k \ge n} x_k$.
Suppose $l <r$. Choose $N$ large enough so that for $n \ge N$, we have $\sup_{k \ge n} x_k \le l+\frac{l+r}{2}$. It follows that $x_n \le l+\frac{l+r}{2} < r$ for $n \ge N$.
Now notice that $n \mapsto \sup_{k \ge n} x_k$ is non-increasing, hence $\sup_{k \ge n} x_k \ge l$ for all $n$. It follows that for all $\epsilon>0$, the set $\{n | x_n \ge l-\epsilon \}$ is infinite (otherwise $\sup_{k \ge n} x_k \le l-\epsilon$ for some $n$, which contradicts the definition of $l$).
Then choose $\epsilon = l-r >0$ for the second part. |
H: Examples of unit vectors that's in the same direction as vector, let's say $v=(1,2,-3)^T$
Can someone give me some examples of unit vectors that's in the same direction as vector, let's say $v=(1,2,-3)^T$ for:
(i) Euclidean norm
(ii) Weighted norm $||V||^2=2V_1^2+v_2^2+\frac13v_3^2$
(iii) The 1 norm
(iv) The infinite norm
(v) The norm based on their inner product $2v_1w_1-v_1w_1-v_2w_1+2v_2w_2-v_2w_3-v_3w_2+2v_3w_3$.
Ty.
AI: Examples? No. The idea? Yes.
If you have a norm $\|\cdot\|$, then the vector you're looking for is $\|v\|^{-1}v$.
For example (but just this one :-P), the Euclidean norm of your vector is
$$\|v\| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14},$$
so the unit vector in the direction of $v$ is $\sqrt{1/14} v = (1/\sqrt{14}, 2/\sqrt{14}, -3/\sqrt{14})^T$.
Knowing this, others shouldn't be too hard. Good luck and ask if you get stuck.
P.S. "The norm based on their inner product $\langle \cdot, \cdot \rangle$ is given by $\|v\| = \sqrt{\langle v, v \rangle}$. |
H: multiple function recursion
I'm working on a problem that ended up with a recursion in two functions,
$$f_n= f_{n-1}+ f_{n-2}+ 2g_{n-2}$$
$g$ also has a recurrence dependent on $f$,
$$g_n= f_{n-1}+ g_{n-1}$$
Initial conditions $f_0=0=g_0$ and $f_1=1=g_1$, so it appears there is a fibonacci relationship. I'm trying to construct a generating function for $f$, but I'm confused at this point.
Am I going down the wrong path. Does anyone have any ideas?
AI: If you right shift the first and subtract it from the first, you get $$f_n-f_{n-1}=f_{n-1}-f_{n-3}+2g_{n-2}-2g_{n-3}$$ Now right shift your second by $2$ to get $g_{n-2}-g_{n-3}=f_{n-3}$ and substitute in, getting $$f_n=2f_{n-1}+f_{n-3}$$ |
H: Application of Stirling's theorem for the given series
I want to prove whether $x=-4/27$ is convergent or not for the series $$\sum_{n=0}^{\infty}\binom{3n}{2n}x^n$$
I applied alternating series test. But, while using this, I need to apply Stirling's theorem so as to show that $a_n(4/27)\to 0$ as $n \to \infty$.
Please can someone do this? I tried but I cannot. Thanks a lot.
AI: Stirling's approximation says
$$k! \sim \sqrt{2\pi k} \left(\frac{k}{e}\right)^k.$$
Inserting that into $\binom{3n}{2n}$ yields
$$\binom{3n}{2n} \sim \frac{\sqrt{6\pi n} 3^{3n} n^{3n} e^{-3n}}{\sqrt{4\pi n}2^{2n}n^{2n}e^{-2n}\sqrt{2\pi n}n^ne^{-n}} = \frac{\sqrt{6\pi n}}{\sqrt{4\pi n}\sqrt{2\pi n}}\left(\frac{27}{4}\right)^n = \frac{\sqrt{3}}{2\sqrt{\pi n}}\left(\frac{27}{4}\right)^n,$$
so
$$\binom{3n}{2n}\left(\frac{4}{27}\right)^n \sim \frac{\sqrt{3}}{2\sqrt{\pi n}} \to 0.$$ |
H: Let $f: W \rightarrow X$ & $g: W \rightarrow X$ be cont. functions. Prove that if $X$ is Hausdorff then $c(f,g)$ is closed in $W$.
Let $W, V$ be topological spaces, and let $f: W \rightarrow X$ and $g: W \rightarrow X$ be continuous functions. Define the coincidence set of $f$ & $g$ to be the subset $$c(f,g) = \{w \in W | f(w) = g(w)\}$$ of $W$. Prove that if $X$ is Hausdorff then $c(f,g)$ is closed in $W$.
Define $\bar{c}(f,g) = \{(w_1, w_2) \in W \times W | w_1, w_2 \in c(f,g)\}$.
Suppose that $X$ is Hausdorff. Consider the function $\phi := f \times g: W \times W \rightarrow X \times X$. Then $\phi(\bar{c(f,g)}) = \{(f(w),g(w')) | w, w' \in c(f,g) \}$. Since $X$ is Hausdorff, it is also T1. So each $(f(w), g(w)) \in \phi(\bar{c(f,g)})$ is closed. Since the union of closed sets is closed, $\phi(\bar{c(f,g)})$ must be closed.
Since $f$ and $g$ are continuous, $f \times g$ is also continuous. So $\phi(\bar{c(f,g)})$ closed in $X \times X$ $\implies$ $\phi^{-1}\phi(\bar{c(f,g)})$ is closed in $W \times W$.
Then I showed that $\bar{c}(f,g) = \phi^{-1}\phi(c(f,g))$. So $c(f,g)$ must be closed in $W$.
Do you think my approach is correct?
Thanks in advance
AI: Others have pointed out in comments the main problem with your argument: arbitrary unions of closed sets need not be closed. There is a proof using your function $\varphi$ and the fact that since $X$ is Hausdorff, the diagonal in $X\times X$ is closed, but you can also simply show directly that $W\setminus c(f,g)$ is open.
HINT: Suppose that $w\in W\setminus c(f,g)$; then $f(w)\ne g(w)$. $X$ is Hausdorff, so there are open sets $U$ and $V$ in $X$ such that $f(w)\in U$, $g(w)\in V$, and $U\cap V=\varnothing$; now consider the set $f^{-1}[U]\cap g^{-1}[V]$. |
H: Prove that this limsup inequality is strict.
so they are non-negative and bounded.
how do i show this?
AI: I think that $x_n=1+\sin(n)$, $y_n=1+\cos(n)$ work. The limsup of the product is less than 3. The limsup of each separately is 2, so the product is 4. |
H: One-point-functors and the Yoneda Lemma
Let $\mathfrak{Sets}$ be the category of sets, and $\mathfrak{Sch}$ the category of schemes. For any scheme $X$, consider the functor $h_{X}(-)=\mathsf{Hom}_{\mathfrak{Sch}}(-,X):\mathfrak{Sch}\longrightarrow \mathfrak{Sets}$. Consider also the following lemma:
Lemma (Yoneda):
Let $C$ be a category, $A$ an object of $C$ and $F:C^{opp} \longrightarrow \mathfrak{Sets}$ a functor. Let $\mathsf{Nat}(h_{A},F)$ be the set of all natural transformations between $h_{A}$ and $F$. Then there exists a one-to-one correspondance $\mathsf{Nat}(h_{A},F)\cong F(A)$.
I would like to show that any scheme $X$ can be recovered from $h_{X}$ up to unique isomorphism. Can this fact be proven using the Yoneda lemma?
AI: This should come immediately from the Yoneda Lemma. Suppose you have objects $X$ and $Y$. First from Yoneda we know that given any functor $F: C^{opp} \to \textbf{Set}$ that
$$\mathsf{Nat}(h_X,F) = F(X).$$
If you now put $F= h_Y$ then Yoneda in particular says
$$\mathsf{Nat}(h_X,h_Y) = h_Y(X) = \mathsf{Hom}(Y,X).$$
So if you had an isomorphism between the functors $h_X$ and $h_Y$ can you use this to construct an isomorphism between $X$ and $Y$?
By the way, this can be seen as a corollary of Yoneda or just simply as the statement that the representing object is unique upto isomorphism. Yoneda actually says a lot more than the latter statement. |
H: Can the intersection of two non subspace subsets be a subspace?
I feel as though it cannot, but I'm having difficulty explaining why.
AI: Within $\mathbb{R}^1$, let $S=\{0,1\}$ while $T=\{0,2\}$. Neither is a subspace, but their intersection is $\{0\}$, which is a subspace.
For a less silly example, in $\mathbb{R}^2$, let $S=\{(a,b):a\ge b\}$, and $T=\{(a,b):a\le b\}$. Neither is a subspace, but their intersection is $\{(a,a):a\in \mathbb{R}\}$, which is a subspace. |
H: A discontinuous function such that $f(x + y) = f(x) + f(y)$
Is it possible to construct a function $f \colon \mathbb{R} \to \mathbb{R}$ such that $$f(x + y) = f(x) + f(y)$$ and $f$ is not continuous?
AI: "Fix a basis for $\mathbb R$ as a $\mathbb Q$-vector space" (which exists under the axiom of choice, but under weaker axioms I have no idea what happens). The condition $f(x+y) = f(x) + f(y)$ is equivalent to $f$ being $\mathbb Q$-linear, so you're asking if there exists a non-trivial discontinuous map of $\mathbb Q$-vector spaces between $\mathbb R$ and itself. If you map the basis elements to other basis elements in a discontinuous way, you will obtain such a map.
Added : A quick way to see that "a discontinuous way of doing it" exists, is that the set of $\mathbb Q$-linear maps that are also $\mathbb R$-linear has cardinality of the reals, where as the set of all $\mathbb Q$-linear maps (or in other words, the number of maps between the basis elements) has cardinality $|\mathbb R|^{|\mathbb R|}$. To understand why, well of course the basis for $\mathbb R$ as a $\mathbb Q$-vector space has cardinality $\le |\mathbb R|$, but if it was countable, then $\mathbb R$ would be countable because all of its elements would be a linear combination of elements in a countable set with countably many possible coefficients. Since the basis for $\mathbb R$ as a $\mathbb Q$-vector space is uncountable, the set of all maps from a basis to another also is. Therefore there must be at least one map between the two bases which generates a discontinuous linear map.
Hope that helps, |
H: About quadratic residue modular p(p is prime)
The question is this.
Suppose that p is a prime, $p\ge7$. Show that $(\frac np)=(\frac {n+1}p)=1$ for at least one number n in the set {1,2,...,9}.
I think seperating into two cases when n=1 and when n=4 will help me prove it. But I can't think further...
Plz, HELP ME!!!!
AI: Assume that the statement is not true. We clearly have $\left(\frac{1}{p}\right) = \left(\frac{4}{p}\right) = \left(\frac{9}{p}\right) = 1$. By assumption we must then have $\left(\frac{2}{p}\right) = \left(\frac{5}{p}\right) = \left(\frac{10}{p}\right) = -1$. Can you derive a contradiction from this? |
H: Prove that lim is 0.
Prove that $$\lim_{x\to0}\sqrt{|x|}\sin\left(\frac1x+x^{10}\right)=0.$$
How do I show in a rigorous way that this limit as $x\to 0$ equals $0$ ?
Any tips or suggestions would be great!
AI: Since
$|\sin(anything)| \le 1$,
$|\sqrt{|x|}\sin(anything)|
\le |\sqrt{|x|}|$,
and since
$\lim_{x \to 0} |\sqrt{|x|}|
= 0$
(assuming you can prove this),
the limit is $0$. |
H: Understanding the proof that if a set is closed its complement has to be open.
The statement proven here is that: $A \text{ is closed} \implies A^{c} \text{ is open}$.
The proof given in class is this:
Suppose $A^{c}$ is not open, then it must be the case that for some $a \in A^c \implies \nexists N_{\delta}(a) \in A^c $. That is, for such $a \in A^c$ we cannot find a neighborhood of $a$, for all $\delta > 0$, where the neighborhood belongs to the set $A^c$.
What I don't understand from this proof is that why is it the case we can conclude, that $N_{\frac{1}{n}}(a)$ contains points $x_n \in A$, and $x_n$ approaches the point $a$ because $|x_n - a| < \frac{1}{n} \rightarrow 0$, thus concluding that $a$ is a limit point of $A$.
I know that once you prove $a$ is a limit point of $A$, because $A$ is closed, it forces $a \in A$ and hence it creates a contradiction. I guess I'm still confused on this neighborhood concept, what is a good way to explain this proof?
AI: This proof relies on choosing an element from each $N_{\frac{1}{n+1}}(a)$, this sequence must converge to $a$; but as you assumed that each element of the sequence is on $\mathbb{R}\setminus A$, you get an absurd as the set $A$ is closed and a convergent sequence in $\mathbb{R}$ must have an unique limit. Proceed as follows.
If $\mathbb{R}\setminus A$ is not open, then one can find an element $a\in \mathbb{R}\setminus A$ such that for every $\varepsilon > 0$, $N_\varepsilon (a) \not\subset A$. As $\varepsilon$ is arbitrary, for every $n\in\omega$ the set $N_{\frac{1}{n+1}}(a)$ has some element $x_n$ such that $x_n\not\in \mathbb{R}\setminus A$, and, thus is an element of $A$; as $A$ is a closed set, the sequence $\langle x_n \rangle _{n\in\omega}$ converges to some point $x\in A$ which is an absurd, as $x_n\to a.$ |
H: Boolean Algebra - Demorgan Laws
I am given the problem:
!((!A * B) * !((!B + C) * (!C * !D)))
where ! = NOT, * = AND, and + = OR and I tried simplifying it using only Demorgan Laws (no absorption) and I got:
(A + !B) + ((!B + C) + (C + D))
and I was just wondering if it was correct. Any help would be greatly appreciated.
AI: If you just want to do the DeMorgan's law portion of the simplification, (using your notation)
!((!A * B) * !((!B + C) * (!C * !D)))
<==> !(!A*B) + !!((!B + C) * (!C * !D)) (first round of DeMorgan's law usage)
<==> (A + !B) + !!((!B + C) * (!C * !D)) (second round of DeMorgan's law usage)
Now, if we want to finish simplifying,
(A + !B) + !!((!B + C) * (!C * !D))
<==> A + !B + (!B + C) * (!C * !D) (simplifying the !!)
<==> A + !B + (!B * !C * !D) + (!C * C * !D) (distribution)
<==> A + !B + (!B * !C * !D) + 0 (!C * C * !D is equivalent to 0)
<==> A + !B + (!B * !C * !D)
<==> A + !B (using the unavailable absorption law you mentioned) |
H: Complete and elementary proof that $(a^x - 1)/x $ converges as x goes to 0
Anybody who has taken a calculus course knows that
$$\lim_{x \to 0} \frac{a^x - 1}{x}$$
exists for any positive real number $a$, simply because the limit is by definition the derivative of the function $a^x$ at $x = 0$. However, for this argument to be non-circular one must have an independent technique for proving that $a^x$ is differentiable. The standard approach involves the following two steps:
1) Calculate the derivative of $\log_a x$ by reducing it to the calculation of
$$\lim_{h \to 0} (1 + h)^{\frac{1}{h}}$$
2) Apply the inverse function theorem.
I find this unsatisfying for two reasons. First, the inverse function theorem is not entirely trivial, even in one variable. Second, the limit in step 1 is quite difficult; in books it is often calculated along the sequence $h = \frac{1}{n}$ where $n$ runs over the positive integers, but doing the full calculation seems to be quite a bit more difficult (if one hopes to avoid circular reasoning).
So I would like a different argument which uses only the elementary theory of limits and whatever algebra is needed. For instance, I would like to avoid logarithms if their use involves an appeal to the inverse function theorem. Is this possible?
AI: Prove that
$$\lim_n \left( 1+\frac{1}{n} \right)^n = \lim_n\left( 1+\frac{1}{n} \right)^{n+1}$$
exists, and call this limit $e$. The reason why these two limit exists is because it can be proven with the Bernoulli inequality that $\left( 1+\frac{1}{n} \right)^n$ is increasing and $\left( 1+\frac{1}{n} \right)^{n+1}$ is decreasing.
It follows that they are both convergent, and their ratio converges to $1$, thus the limit exists.
From here, we also get immediately that
Now, for each $x \in (0, \infty)$ we have $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$. For simplicity I will denote $n:=\lfloor x \rfloor$. Then, we get
$$\left( 1+\frac{1}{n+1} \right)^n \leq \left( 1+\frac{1}{x} \right)^x \leq \left( 1+\frac{1}{n} \right)^{n+1}$$
As $n \to \infty$ when $x \to \infty$, by a squeeze type argument we get
$$\lim_{x \to \infty} \left( 1+\frac{1}{x} \right)^x=e$$
Using $1-\frac{1}{x}=\frac{x-1}{x}=\frac{1}{1+\frac{1}{x-1}}$ we also get
$$\lim_{x \to \infty} \left( 1-\frac{1}{x} \right)^x=e^{-1}$$
and then
$$\lim_{y \to 0} \left( 1+y \right)^\frac{1}{y} =e$$
Let $y=a^x-1$ [Note: no logarithms are used here, we just use the fact that if the limit exists for all $y \to 0$, it also exists for this articular choice of $y$.]
Then, we get
$$\lim_{x \to 0} \left( a^x \right)^\frac{1}{a^x-1} =e$$
Thus, we proved that
$$\lim_{x \to 0} a^\frac{x}{a^x-1} =e \,,$$
exists. At this point logarithms would solve the problem, but you can probably finish the argument without using logarithms. For example, you can prove that $a^y$ is strictly increasing/decreasing (which reduces to $x >y \Rightarrow a^{x-y} >1$) and prove the following lemma:
Lemma If $f$ is strictly monotonic and continuous, and for some $c$ the limit
$$\lim_{x \to c} f(g(x))$$
exists, then $\lim_{x \to c} g(x)$ exists. |
H: Show if $\phi$ is a ring isomorphism of $\mathbb{Z}\to\mathbb{Z}$, then $\phi$ is the identity mapping.
Show if $\phi$ is a RING isomorphism of $\mathbb{Z}\to\mathbb{Z}$, then $\phi$ is the identity mapping.
I don't really know where to start with this one. I know that since $\phi$ is an isomorphism, it is a homomorphism that is both one-to-one and onto. Since it is a homomorphism, $\phi$ satisfies the following.
(i) $\phi(0)=0$
(ii) $\phi(-a)=-\phi(a)$ $\forall a\in\mathbb{Z}$
(iii) $\phi(\mathbb{Z})$ is a subring of $\mathbb{Z}$.
But I don't see how to relate these properties to the conclusion that $\phi$ is the identity mapping.
Any help/hints would be greatly appreciated. ^_^
AI: Hint
$$\phi(1)=\phi(1 \cdot 1) =\phi(1) \phi(1)$$
Also $\phi(1) \neq 0$ (WHY?) |
H: Calculus Trigonometric Integral
I have been trying to solve this particular problem and can not figure out where I have gone wrong in my solution. This is a picture of my solution, if anyone could help point out where I have gone wrong it would be appreciated.
http://i.imgur.com/j0xGdM3.jpg
AI: First of all, your expression in your answer is not necessarily different from the answer. Use the double-angle formula twice to try to recover an expression in terms of powers of $\cos{x}$. But an easy way to see how the given answer is generated is by substitution:
$$\int dx \cos^2{x} \, \sin{2 x} = 2 \int dx \, \sin{x} \, \cos^3{x} = -2 \int d(\cos{x}) \, \cos^3{x} = -\frac12 \cos^4{x} + C$$ |
H: Unique of eigenbasis of self-adjoint operator.
Today I am reading an article: Eigenvalues and sums of Hermitian matrices there is an exercise copy from that article:
Exercise 1 Suppose that the eigenvalues $\lambda_1(A)>\cdots>\lambda_n(A)$ of an $n\times n$ Hermitian matrix are distinct. Show that the associated eigenbasis $u_1(A),\cdots,u_n(A)$ is unique up to rotating each individual eigenvector $u_j(A)$ by a complex phase $e^{i\theta_j}$. In particular, the spectral projections $P_j(A):=u_j(A)^*u_j(A)$ are unique. what happens when there is eigenvalue multiplicity?
Assume $A$ is the matrix of self-adjoint operator $\mathcal{A}$ under some property basis of $V_{\Bbb{C}}$.
I don't know why eigenbasis $u_1(A),\cdots,u_n(A)$ is unique up to rotating. If we multiply a complex number $c_k \not=0$ respectively, then $c_1u_1(A),\cdots,c_nu_n(A)$ is also a basis of $V_{\Bbb{C}}$ consist of $\mathcal{A}$'s eigenvectors,under this basis, $\mathcal{A}$ correspond to a diagonal matrix consist of eigenvalues: $\text{diag}\{\lambda_1,\cdots,\lambda_n\}$. I think it is unique up to multiply a complex number $c_k$ respectively. what's wrong?
AI: Since this is a self-adjoint operator, there is an orthonormal basis of eigenvectors, and that's what they're talking about. If you multiply basis elements by arbitrary nonzero complex numbers you still have a basis of eigenvectors, but it's no longer orthonormal. |
H: Question on types of continuity.
Determine if on the given interval, the function $f(x)=\frac{1}{\sqrt{x}}$, is a) continuous b) uniformly continuous, c) Lipschitz continuous, d) differentiable, and e) $C^1$
The interval $(0, 1]$ and $[1, \infty)$
My work: Let $\epsilon > 0$ given. We must show there exists $\delta>0$ such that for every $x,x_0 \in (0,1], \left|f(x)-f(x_0\right|<\epsilon$ whenever $|x-x_0|<\delta$. Thus $\left|f(x)-f(x_0)\right|=\left|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x_0}}\right|=\left|\frac{\sqrt{x}-\sqrt{x_0}}{\sqrt{xx_0}}\right|=\frac{1}{\sqrt{xx_0}}\left|\sqrt{x}-\sqrt{x_0}\right|$
$\frac{1}{\sqrt{xx_0}}\left|\sqrt{x}-\sqrt{x_0}\right|<\epsilon$
$|\sqrt{x}-\sqrt{x_0}||\sqrt{x}+\sqrt{x_0}|<\sqrt{xx_0}|\sqrt{x}+\sqrt{x_0}|\epsilon$
Note on the interval $\sqrt{xx_0}\leq1$ and $|\sqrt{x}+\sqrt{x_0}|\leq2$,thus $|x-x_0|<2\epsilon$ and we set $\delta = 2\epsilon$ Therefore continuous.
b) We also know that the function is uniformly continuous because our $\delta$ does not depend on x or $x_0$.
c) We can also conclude that the function is Lipschitz continuous because $|x-x_0|<2\delta \Rightarrow |x-x_0|\frac{1}{2}<\delta$
d) ?
e) ?
Is this a sound argument? Also, my book offers no insight into parts d and e so, any help would be appreciated. Is any of this work transferable to $[1, \infty)$?
AI: At the stage where you assume that
$$
\frac{1}{\sqrt{xx_0}}\left|\sqrt{x}-\sqrt{x_0}\right|<\epsilon
$$
you are implicitly assuming that the function is uniformly continuous. As @Arthur's example shows, this is a bad assumption.
To fix, notice that
\begin{align*}
\left|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{a}}\right| &= \frac{1}{\sqrt{a}\sqrt{x}(\sqrt{x}+\sqrt{a})}\left|x-a\right| \\
& \leq \frac{1}{2\sqrt{a}}|x-a|
\end{align*}
Where your assumption messed up was that $\frac{1}{\sqrt{a}}$ can get rather big depending on $a$.
For part $(d)$ you can write down the definition of derivative, simplify things a bit and use the Squeeze Theorem.
For part $(e)$ $C^1$ means that the derivative is continuous. Once you've done $(d)$ you can use your limit definition to show that it is continuous. |
H: Integration trouble on a conservative vector field
I'm trying to integrate $\int_{\mathbf{x}}{\mathbf{F}\cdot{d\mathbf{s}}}$ where
$$\mathbf{F}=x^2\mathbf{i}+\cos{y}\sin{z}\mathbf{j}+\sin{y}\cos{z}\mathbf{k}$$and $$\mathbf{x}:[0,1]\rightarrow\mathbb{R}^3; \mathbf{x}(t)=(t^2+1,e^t,e^{2t})$$
I'm at the point where I have
$$\int_{\mathbf{x}}{\mathbf{F}\cdot{d\mathbf{s}}}=\int_{\mathbf{x}}{x^2dx+\cos{y}\sin{z}dy+\sin{y}\cos{z}dz}$$$$=\int_0^1{(t^2+1)^22tdt+\int_0^1e^t[\cos{e^t}\sin{e^{2t}}+2e^t\cos{e^{2t}}\sin{e^t}]dt}$$
Thought I could use the $\sin{(A+B)}$ rule but the $2e^t$ is in front of the second term. The second thought was substitution, but I can't justify both $e^t$ and $e^{2t}$ terms. What am I missing?
EDIT: corrected typo in vector $\mathbf{x}(t)$. I should also point out that I'm not worried about the integral on the left. That poses no issues....
AI: You are missing the opportunity to use the fundamental theorem of calculus for line integrals. As the title says, the field is conservative: it is the gradient of $x^3/3+\sin y\sin z$. Evaluate the potential function at the endpoints, and you are done. |
H: Good websites/books for geometry exercises?
I'm looking for exercises similar to those seen on putnam exams or olympiad exams, such as finding the area of polygons inscribed other polygons, finding certain angles, etc.
AI: I like Go-geometry, which covers the basics up to IMO level.
I'm not too certain what you would consider Putnam Geometry, in part because the undergraduate competitions don't focus heavily on Euclidean Geometry techniques, but have more to do with calculus ideas. |
H: Generating functions word problem(balloons)
Find the generating function for the number of ways to create a bunch of n balloons selected from white, gold, and blue balloons so that the bunch contains
at least one white balloon, at least one gold balloon, and at most two blue balloons. How many ways are there to create a bunch of 10 balloons subject to these
requirements?
So I have $$\frac{x^2}{(1-x)^2} \cdot \frac{(1-x^3)}{1-x}$$ and I have no idea where to go from here. Have a test on this material in the morning so any help would be greatly appreciated.
AI: You have $$(x^2-x^5)\frac{1}{(1-x)^3}$$
We have the generating function $$\frac{1}{(1-x)^3}=\sum_{n\ge 0} {n+2\choose 2}x^n$$
Multiply by $x^k$ gives $$x^k\frac{1}{(1-x)^3}=\sum_{n\ge 0}{n+2\choose 2}x^{n+k}$$
We reindex using $m=n+k$ to get $$x^k\frac{1}{(1-x)^3}=\sum_{m\ge k}{m-k+2\choose 2}x^m$$
We do this twice, for $k=2$ and $k=5$, then subtract. The result is
$$(x^2-x^5)\frac{1}{(1-x)^3}=\sum_{m\ge 5}{m-2+2\choose 2}-{m-5+2\choose 2}x^m + \sum_{m=2}^4{m-2+2\choose 2}x^m$$
Hence your answer (for $m\ge 5$) is ${m\choose 2}-{m-3\choose 2}$. You want $m=10$, so ${10\choose 2}-{7\choose 2}=24$. |
H: Matrix of transformation
I have just finished understanding the topic of matrix of the transformation, and we just started T-invariant subspaces.
Can anyone please help me with these questions, because I don't know how to approach them. This is not homework by the way, it is just for my understanding.
1) Suppose $B=(w_1, w_2, \dots, w_k, v_1, \dots, v_{n-k})$ is a basis for $V$ and $W$ = Span$(w_1, \ldots, w_k)$ is T-invariant. What does the matrix of $T$ with respect to $B$ look like?
2) Suppose $V$ is the direct sum of $W_1$ and $W_2$, with $W_1,W_2$ both T-invariant. If $B=(u_1,u_2, \ldots, u_n, z_1,z_2, \dots, z_n)$ is a basis for V with the $u$'s being a basis for $W_1$ and the $z$'s being a basis for $W_2$, what does the matrix of T with respect to $B$ look like?
Thanks!!
AI: The columns of your matrix tell you what vectors your basis get sent to under $T$. So, in your first problem, if $W$ is $T$ invariant, then for each $1\leq i\leq k$
$$
T(w_i)=a_1w_1+\cdots +a_kw_k+0v_1+0v_2+\cdots +0v_{n-k}.
$$
Thus the $i$th column of your matrix will be
$$
\begin{pmatrix}
a_1 \\ a_2 \\ \vdots \\ a_k \\ 0 \\ 0 \\ \vdots \\ 0
\end{pmatrix}
$$
for $1\leq i\leq k$. The columns corresponding to the $v_i$ have no restrictions.
For your second question the columns for the $u_i$ will have this form. The columns for the $z_i$ will have $0$'s on top. |
H: Is this absolute value notation or something else?
In this document, in Figure 1 (second to last page) there are several uses of $\| \;\;\|$:
Is this another notation for absolute value, or is this a notation for something to do with vectors/matrices?
AI: As the document says immediately following the introduction:
This Wikipedia page has the definition of the $2$-norm and the $\infty$-norm of a vector. The definition is as follows: for a vector $x=(x_1,\ldots,x_m)\in\mathbb{R}^m$, the $p$-norm of $x$ is
$$\|x\|_p:=(x_1^p+\cdots+x_m^p)^{1/p}$$
if $p$ is finite, and if $p=\infty$, we define
$$\|x\|_\infty:=\max\{x_1,\ldots,x_m\}.$$ |
H: Proving it doesn't exist a homeomorphism between $\mathbb R$ and $\mathbb R^n$, $n>1$.
I have to prove that for $n \ge 2$, there doesn't exist a homeomorphism between $\mathbb R$ and $\mathbb R^n$. Could anyone give me a hint on how could I prove this?
AI: The notion of connectdness is a topological property, meaning that it is invariant under homeomorphisms (this is trivial since the definition involves only the open sets and does not refer to any other property of the space)
Now, assume for the sake of contradiction that there exists a homeomorphism between $\mathbb{R^n}$ and $\mathbb{R}$, say $h \colon \mathbb{R}^n \to \mathbb{R}$.
It is very easy to check that $h \colon \mathbb{R}^n \setminus \{h^{-1}(0)\} \to \mathbb{R}\setminus \{0\}$ is still a homeomorphism (it is bijective and the restriction of continuous functions is continuous hence both this and its inverse are continuous).
It is also known that continuous functions map connected spaces into connected spaces. Since $\mathbb{R}\setminus \{0\}$ is clearly not connected (I am sure that you can find a separation at first glance :D ) then all we need to show is that $\mathbb{R}^n \setminus \{h^{-1}(0)\}$ is connected. It is very intuitive but to make it a little bit more rigorous we can note that it is path connected: take two points, say $x$ and $y$: if the line between them does not pass through $0$ then we are done; otherwise pick any point $z$ that is not on that line and consider the path that goes from $x$ to $z$ with a line and from $z$ to $y$ with another line. This concludes :D |
H: Selecting marbles from two jars
Jar $A$ contains $3$ red and $3$ black marbles, and Jar $B$ contains $4$ red and $6$ black marbles. If a marble is randomly selected from each Jar, what is the probability that the marbles will be the same color?
I'm having difficulty approaching this problem. It seems like I should use Baye's Theorem to solve this problem, that is, let $A = \{\text{select a marble randomly from Jar A and Jar B}\}$ and $B = \{\text{the color of the marbles is the same}\}$. I think I'm suppose to find $P(A|B)$ by Baye's Theorem. However, I get stuck here as I need $P(A)$ and $P(B|A)$. Perhaps my approach is incorrect.
AI: \begin{align*}
&\,\mathbb{P}(\text{marbles are the same color})=\mathbb{P}(\text{both marbles are black or both marbles are red})\\
=&\,\mathbb{P}(\text{both marbles are black})+\mathbb{P}(\text{both marbles are red})\\
=&\,\mathbb{P}(\text{marble from A is black and marble from B is black})\\+&\,\mathbb{P}(\text{marble from A is red and marble from B is red})\\
=&\,\mathbb{P}(\text{marble from A is black})\times\mathbb{P}(\text{marble from B is black})\\+&\,\mathbb{P}(\text{marble from A is red})\times\mathbb{P}(\text{marble from B is red})\\
=&\,\frac{3}{6}\times\frac{6}{10}+\frac{3}{6}\times\frac{4}{10}=\frac{1}{2}.
\end{align*} |
H: find a basis for the subspace $S$ = {$p \in \mathbb{P}_3$ |$ p$(5) = 0}
This is as far as I've gotten:
$p$($x$) = $a$ + $bx$ + $c$$x^2$ + $dx^3$ then $p$($5$) = $a$ + $5b$ + $25c$ + $125d$ = $0$
which gets the vector $(1, 5, 25, 125)$ but then I'm not sure how to a basis from this vector. I tried setting some other $p$($n$)$= 0$ and solving a system of linear equations but that didn't get me anywhere.
What do?
AI: Assuming that $P_3$ is the set of degree $\le 3$ polynomials over $\mathbb{R}$, note that the space has a basis of four elements. But $S$ is a proper subspace, and so has a basis with at most three elements; hence it is sufficient to find three linearly independent polynomials of degree at most $3$.
Such a set could be $\{x - 5, x^2 - 5x, x^3 - 5x^2\}$. |
H: Creating a System Impulse Response in Matlab
Preface: I'm extremely new to Matlab.
Ok, so I have a sound file that I loaded in Matlab.
Two variables are loaded:
xx - the speech waveform - 16001 samples
fs - the sampling frequency of the speech waveform = 22050 Hz
Since fs = 1/T, T = 0.00004535147 seconds
Since there are 16001 samples, the total time for the file is 16001 * T = 0.72566893424 seconds
I plotted the waveform with the x-axis representing time, by using the following command:
plot(0:1/fs:0.72566893424,xx);
At this point, I'm trying to create a discrete-time system impulse response that creates an "echo" of the original waveform. (The echo should be delayed by 1-second and 1/4 the amplitude of the original waveform)
My question is, how do I generate an impulse response in Matlab?
At this time, I've tried to create another array, h, by adding 22050 zeros to the beginning (a 1 second delay), multiplying the whole array by 1/4, then convolving it with the original waveform... all I've managed to achieve is a sound that is roughly equivalent to the Dark Lord himself speaking.
Am I on the right track? This is homework, so I don't want an "answer", but rather steps to understand what I'm doing a little better, I need to learn the material. Sorry if this rambles a bit, Matlab is fairly foreign to me.
AI: An impulse response should be independent of the stimulus. For example, given input signal $f(x)$ and impulse response $h(x)$, the output signal $y(x)$ is generated as follows:
$$y(x) = f(x) {\Large *} h(x)$$
This relationship holds true for any input signal we choose to apply. Let's say you want an impulse response that makes $y(x) = f(x)$. The the solution is $h(x) = \delta(x)$ where $\delta$ signifies the Dirac Delta Function.
$$y(x) = f(x) {\Large *} \delta(x) = f(x)$$
Now let's say that I want my output to be a scaled version of my output.
$$y(x) = f(x) {\Large *} \left(\tfrac{1}{2}\delta(x)\right) = \tfrac{1}{2}f(x)$$
What if I want my output to be a delayed version of my output?
$$y(x) = f(x) {\Large *} \delta(x-\tau) = f(x-\tau)$$
For linear time-invariant systems, you are free to make any linear combination of impulse responses that you would like. That means that
$$y(x) = f(x) {\Large *} \left(\tfrac{1}{2}\delta(x) + \delta(x-\tau)\right) = \tfrac{1}{2}f(x) + f(x-\tau) \qquad (1)$$
So the output in this situation is A scaled copy with no delay added to a full-scale copy of the signal with a delay of $\tau$ seconds.
Finally, how do we use dirac delta function in Matlab? Let's say that I want to implement equation (1) in Matlab. This corresponds the following impulse response, which I break into two separate impulse responses:
$$h(x) = h_1(x) + h_2(x) = \tfrac{1}{2}\delta(x) + \delta(x-\tau)$$
Since we are working in discrete time, I need my $\tau$ to be in terms of samples. Let's say that I am looking for a delay of 1000 samples.
tau = 1000;
h1 = [0.5, zeros(1, tau)];
h2 = [zeros(1, tau) 1];
h = h1 + h2;
y = conv(f,h);
Notice that in the impulse response h2 I placed the 1 at the index 1001 rather than 1000. This is because putting a 1 at the first index corresponds to no delay, so I needed to add 1000 to that in order to delay my signal. Also, I made h1 longer than it needs to be. In fact, h1 could have just been h1 = 0.5, since trailing zeros will have no effect in the convolution. However, I wanted to make it the same length as h2 so that I could add the two of them together.
Hopefully this is enough to get you started. |
H: There are 2013 random points on the plain, prove exactly 8 lies within a circle
This was posed as a homework problem to me and I have made progress, but I don't think I have a solution.
So there are a lot of random points on the plane and we need to find a circle that contains exactly eight of them. The way I would attack this is to list the distances of each point pairs. Now take a pair (among perhaps several) that has a minimal distance.
I think that even if the minimal distances are equal, the most you can do with equal minimal distances without having a circle covering them is seven points (suspiciously one short of that eight!) by picking the centre of a circle and then six equidistant points on the circle (belonging to a 60 degree central degree).
AI: You can find such a circle with almost any center in the plane you want. Imagine picking some point and slowly expanding the radius of a circle until you have eight points inside the circle. The only time you fail is when there are at least two points equidistant from the center you pick, and as you expand the circle they take you from less than eight points within the circle to nine or more. Each pair of points has a line of points they are equidistant from. There are a little more than 2 million pairs of points. Pick any point not on any of these two million lines as a center and as the radius expands you will cross the points one at a time. As the lines have zero area, there are still plenty of centers to choose from. When you have crossed eight and not nine, you have your circle. |
H: Show that if a and b are positive integers, and $a^3 | b^2$, then a | b
Show that if a and b are positive integers, and $a^3 | b^2$, then a | b.
If p is a prime divisor of a, and $p^r$ is the highest power of p dividing a. Then $p^{3r} | a^3$, and so $p^{3r} | b^2$. If $p^s$ is the highest power of p dividing b, then 3r ≤ 2s. And so we have, r ≤ (2/3)s < s.
But how come this implies that $p^r | b$, and eventually a | b.
AI: Since $a^2\mid a^3$, we have $a^2\mid b^2$. Then let $c=b/a$. We want to show $c\in \mathbb{Z}$. Note that $c$ satisfies $x^2-\frac{b^2}{a^2}$. This is monic in $\mathbb{Z}[x]$, and since any rational solution to a monic in $\mathbb{Z}[x]$ is actually an integer, we get that $c\in\mathbb{Z}$.
Thm: If $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ is such that $a_i\in \mathbb{Z}$, and we have that for some $c\in \mathbb{Q}$ $f(c)=0$, then $c\in \mathbb{Z}$
Proof: Write $c=b/a$ with $c$ a reduced fraction (meaning $a$ and $b$ are coprime). Then we have:
$$
(b/a)^n+a_{n-1}(b/a)^{n-1}+...+a_0=0
$$Multiply everything by $a^n$. You get that $b^n=aD$ for $D\in \mathbb{Z}$. Hence, $a\mid b^n$. If $a$ is not $\pm 1$, then there is a prime $p$ such that $p\mid a$. This means that $p\mid b^n$ which then implies that $p\mid b$. This contradicts the fact that $a$ and $b$ are coprime. Hence we get that $a=\pm 1$, so $c\in \mathbb{Z}$ just as we wanted. |
H: Additional Law Word Problem
Forgive me if this is really basic:
Tammy is a general contractor and has submitted two bids for two projects (A and B). The probability of getting project A is 0.65. The probability of getting project B is 0.77. The probability of getting at least one of the projects is 0.90. What is the probability that she will get both projects?
Is this a simple question using the addition law or am I missing something? I calculated that her probability of getting both would be 0.52.
(0.65 + 0.77 - 0.90) = .52
AI: \begin{align*}
\mathbb{P}(A)=&\,0.65,\\
\mathbb{P}(B)=&\,0.77,\\
\mathbb{P}(A\cup B)=&\,0.90,\\
\mathbb{P}(A\cap B)=&\,\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cup B)=0.52.
\end{align*}
Your reasoning is correct. |
H: What functions model this relationship?
I'm currently working a bit on an AI, and in order for it to function, it must be able to quickly predict where a point will be in space, given any distance. The movement of this point may be modeled as a linear graph 1 up until it reaches a maximum y value, at which point the movement begins to decrease with the slope -1 until the minimum y value.
The equation for the upward cycle of the function is $y=min+x-?$, but when the function is on a downward cycle, it becomes $y=max-x+?$. The $?$ in the equation represents a value which basically cancels out all growth in $x$ beyond the $min$ and $max$ respectively. For some reason, I strongly suspect that differential calculus is the means by which this can be solved, but I am currently only on a pre-calculus level of mathematics.
What I need is a function that is periodic and bounces between the min and max like a sine wave (or a periodic absolute value graph), but which has no curve. Does there exist any such function? If not, how may I go about finding an equation for this relationship?
AI: You either want http://en.wikipedia.org/wiki/Square_wave http://en.wikipedia.org/wiki/Sawtooth_wave or http://en.wikipedia.org/wiki/Triangle_wave. I don't think that they are expressible as "equations" in the way you're thinking, though they are certainly well defined functions. I would also guess that most programming languages would allow you to make such a function quite easily. |
H: Trying to simplify $A'B'C'D + A'B'CD + A'BC'D + AB'CD + ABCD$
My solution so far:
$A'(B'C'D + B'CD + BC'D) + A(B'CD + BCD)$
$= A'(C'D(B' + B) + B'CD) + A(CD(B'+B))$
$= A'(C'D(1) + B'CD) + A(CD(1))$
$= A'C'D + A'B'CD + ACD$
$= D(A'C' + AC) + A'B'CD$
$= D(A\text{ xnor }C) + A'B'CD$
$= D[(A\text{ xnor }C) + A'B'C]\text{ <<< Can't seem to get any further than this.}$
Did I perhaps screw up somewhere? Or not do it as well as I could have? Maybe missed using a rule?
Thanks!
K
AI: I'd start by plotting the Karnaugh map.
$A′B′C′D+A′B′CD+A′BC′D+AB′CD+ABCD$ is an sum of five minterms, thus the Karnaugh map will show five ones (I've left out zeroes for greater visual contrast):
$$
\begin{matrix}
& A'B' & A'B & AB & AB' \\
C'D' & \\
C'D & 1 & 1 \\
CD & 1 & & 1 & 1 \\
CD' &
\end{matrix}
$$
This can be covered as a sum of three products, of which two are essential, in two ways.
$$
F = A'C'D + ACD + B'CD = ((A\ xnor\ C) + B'C)D \\
F = A'C'D + ACD + A'B'D = ((A\ xnor\ C) + A'B')D
$$
Which are both one term shorter than your original: $((A\ xnor\ C) + A'B'C)D$. Note how you used disjoint terms as a covering for the function. This often happens when you simplify as expression without looking at the Karnaugh map.
Starting from the same Karnaugh map, you can also choose to start with a product of sums. As a product of three sums, all essential:
$$
F = D(A'+C)(A+B'+C')
$$
Or if you prefer, an inverted sum of products:
$$ F = not(D'+AC'+A'BC) $$ |
H: In the field $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$, find the inverse of $f(x)=x^3+x+3$.
In the field $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$, find the inverse of $f(x)=x^3+x+3$.
I know how to find the inverses of elements within sets, rings, and fields. I know what to do if the field was just $\mathbb{Z}_7$, but the fact that the field is $\mathbb{Z}_7[x]/\langle x^4+x+1\rangle$ confuses me. I don't know where to start.
AI: The computation here is the same in ${\mathbb Q}$ as in
${\mathbb Z}_7$.
You look for a solution of the form
$$
z=a+bx+cx^2+dx^3 \tag{1}
$$
You then have
$$
z(x^3+x+1)=dx^6+cx^5+(b+d)x^4+(a+c+3d)x^3+(b+3c)x^2+(a+3b)x+3a=Q(x) \tag{2}
$$
Next, divide the result by $x^4+x+1$ :
$$
Q(x)=(x^4+x+1)(dx^2+cx+b+d)+R(x) \tag{3}
$$
where the remainder $R(x)$ equals
$$
R(x)=(a+c+2d)x^3+(b+2c-d)x^2+(a+2b-c-d)x+(3a-b-d) \tag{4}
$$
Then, solve the system
$$
a+c+2d=b+2c-d=a+2b-c-d=0, \ \ 3a-b-d=1 \tag{5}
$$
This will lead you to the solution
$$
a=\frac{11}{47}, b=\frac{-8}{47}, c=\frac{1}{47}, d=\frac{-6}{47},
z=\frac{11-8x+x^2-6x^3}{47} \tag{6}
$$ |
H: A set that is transitive but not well-ordered by $\in$?
I am trying to find a transitive set which is not well-ordered by $\in$. This question raises when I read Jech's Set Theory, in which an ordinal number is defined as a transitive and $\in$-well-ordered set.
Thanks!
AI: $V_\omega$ is a very easy example; more generally, $V_\alpha$ for $\alpha\ge 2$. In particular, $\{0,1,2,\{1\}\}$ is a small example. |
H: Prove that $\sup \{-x \mid x \in A\} = -\inf\{x\mid x \in A\}$
I need to prove that $\sup \{-x \mid x \in A\} = -\inf\{x \mid x \in A\}$ and am having trouble moving the $-x$ out of the $\sup$ to $\inf$. Another thing is that I don't quite know how to prove $b = \inf$ (see below). Any help would be appreciated! Here my attempt:
Let $a = \sup\{-x \mid x \in A\}$
Then by definition $a \geq -x$ for $x \in A \implies -a\leq x$ for $x \in A$
Thus $-a$ is a lower bound of $A$.
Assume $b$ is a lower bound of $A$
Then $b \leq -a
\implies a \geq -b$
and $a = \sup \{−x \mid x\in A\} = -\inf(A)$
AI: After you say that $-a\leq x$ for all $x\in A$, you can then conclude that
$$
-a \leq b:= \inf A
$$
and hence
$$
a \geq -b
$$
Similarly, $b\leq x$ for all $x\in A$, and hence $-b \geq -x$ for all $x\in A$, and so
$$
-b \geq a
$$ |
H: How to solve $9\sin 2x-40\cos 2x=\frac{41}{\sqrt{2}}$
Solve the following question:
\begin{eqnarray}
\\9\sin 2x-40\cos 2x&=&\frac{41}{\sqrt{2}}\\
\end{eqnarray}
I know that there is a formula for solving the above question like :
\begin{eqnarray}
\cos a=\frac{A}{\sqrt{A^2+B^2}} , \space\space
\sin a=\frac{B}{\sqrt{A^2+B^2}} \\
\\A\sin 2x-B\cos 2x=\sqrt{A^2+B^2}\cos \left(x-a\right)\\
\end{eqnarray}
If I do not use this formula, how can I solve the above question?
AI: Put $9=r\cos y,40=r\sin y$ where $y>0$
Squaring & adding we get $\displaystyle r^2=1681=41^2\implies r=41$
$\displaystyle \cos y=\frac9r=\frac9{41},\sin y=\frac{40}{41}$
$\displaystyle\implies 9\cos2x-40\sin2y=\frac{41}{\sqrt2}\iff\cos2x\cos y-\sin2x\sin y=\frac1{\sqrt2}$
$\displaystyle\implies \cos(2x+\arccos\frac9{41})=\frac1{\sqrt2}=\cos\frac\pi4$
$\displaystyle\implies 2x+\arccos\frac9{41}=2n\pi\pm\frac\pi4$ where $n$ is an integer |
H: For a Gaussian Random walk where $x_n$ is the sum of $n$ normal random variables, what is $P(x_1 >0, x_2 >0)$?
I know that the events $x_1 >0$ and $x_2 >0$ are not independent, but I can't think of a way to find a conditional probability so I can solve this.
Thanks!
AI: Presumably (but this ought to be in the question), $x_2=x_1+y$ where $(x_1,y)$ is i.i.d. and standard normal. In particular, one assumes the increments are centered and with the same variance. Then, a picture in the $(x_1,y)$ plane reveals that $(x_1\gt0,y+x_1\gt0)$ is the angular sector going from the North to the South-East through North-East and East.
Thus, this sector is $\frac38$ of the whole. The distribution of $(x_1,y)$ is invariant by every rotation centered at $(0,0)$ hence $P[x_1\gt0,x_2\gt0]=\frac38$. |
H: Why left continuity does not hold in general for cumulative distribution functions?
Definition: The c.d.f. $F$ of a random variable $X$ is a function defined for each real number $x$ as follows:$$F(x)=\Pr(X\leq x) \text{ for } -\infty<x<\infty$$
Let $$F(x^-)=\lim_{y\rightarrow x,\,y<x}F(y)$$ and $$F(x^+)=\lim_{y\rightarrow x,\,y>x}F(y)$$
Property of cumulative distribution function: A c.d.f. is always continuous from the right; that is , $F(x)=F(x^+)$ at every point $x$.
Proof: Let $y_1>y_2>\dots$ be a sequence of numbers that are decreasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the intersection of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty} \Pr(X\leq y_n)=F(x^+).$$
Now I think the left inequality can also be proved in the similar way as:
Let $y_1<y_2<\dots$ be a sequence of numbers that are increasing such that $$\lim_{n\rightarrow \infty}y_n=x.$$Then the event $\{X\leq x\}$ is the union of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$ .Hence, $$F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty}\Pr(X\leq y_n)=F(x^-).$$
Where am I wrong?
AI: You write:
Then the event $\{X\leq x\}$ is the union of all the events $\{X\leq y_n\}$ for $n=1,2,\dots$.
This is the faulty step. To wit:
If $y_n\lt x$ for every $n$ and $y_n\to x$, then $\bigcup\limits_n\{X\leqslant y_n\}$ is equal to $\{X\lt x\}$, not to $\{X\leqslant x\}$.
You might want to check that $x$ is in $(-\infty,y_n]$ for no $n$ whatsoever hence $x$ is not in $\bigcup\limits_n(-\infty,y_n]$, and in fact $\bigcup\limits_n(-\infty,y_n]$ is equal to $(-\infty,x)$, not to $(-\infty,x]$. |
H: If a group $G$ has odd order, then the square function is injective.
Suppose $G$ has odd order, show the function $f:G\rightarrow G$ defined by $f(x)=x^2$ is injective. This proposition is easily provable if we assume $G$ is Abelian, but I don't know how to start this without the assumption of being Abelian.
AI: Let $|G|=2n-1$. Assume $a^2=b^2$. Then $a=a\cdot a^{2n-1}=(a^2)^n=(b^2)^n=b$. |
H: Computing an inverse modulo $25$
Supposed we wish to compute $11^{-1}$ mod $25$. Using the extended Euclid algorithm, we find that $15 \cdot 25 - 34 \cdot 11 =1$. Reducing both sides modulo $25$, we have $-34 \cdot 11 \equiv 1$ mod $25$. So $-34 \equiv 16 $ mod $25$ is the inverse of $11$ mod $25$.
im realy confused about this problem, can someone go over this please and explain how this is done
AI: Note that $25 | 15 \cdot 25$, so $15 \cdot 25 \equiv 0 \pmod{25}$. On the other hand, we see that
$$1 \equiv -15 \cdot 25 - 34 \cdot 11 \equiv 0 + (-34) \cdot 11 \pmod{25}$$
Finally,
$$-34 \equiv -34 + 50 \equiv 16 \pmod{25}$$
so we can rewrite the above as
$$1 \equiv 16 \cdot 11 \pmod{25}$$
and $16$ is the desired inverse.
Note that by direct computation,
$$16 \cdot 11 = 176 = 7 \cdot 25 + 1$$
Alternatively, we can use the Euclidean algorithm and find that
\begin{align*}
25 &= 2 \cdot 11 + 3 \\
11 &= 3 \cdot 3 + 2 \\
3 &= 1 \cdot 2 + 1
\end{align*}
Thus,
\begin{align*}
1 &= 3 - 1 \cdot 2 \\
&= 3 - 1 \cdot (11 - 3 \cdot 3) = 4 \cdot 3 - 1 \cdot 11 \\
&= 4 \cdot (25 - 2 \cdot 11) - 1 \cdot 11 = 4 \cdot 25 - 9 \cdot 11
\end{align*}
So $1 \equiv -9 \cdot 11 \equiv 16 \cdot 11 \pmod{25}$ as before. |
H: Why, precisely, is $\{r \in \mathbb{Q}: - \sqrt{2} < r < \sqrt{2}\}$ clopen in $\mathbb{Q}$?
Just going over some old notes and I realized I always took this for granted without actually fleshing out exactly why it is true.
The set of all r is closed in $\mathbb{Q}$, because the set of all r is just all of the rationals in that interval, and obviously contains all of its limits. If you take its complement, the irrational numbers between $- \sqrt{2}$ and $\sqrt{2}$, what precisely can we say about this to conclude that the original set is clopen? i.e. why precisely is the complement open?
Thanks.
AI: $\Bbb Q$ inherits ist topology from $\Bbb R$. So if a set $U$ is open in $\Bbb R$, then $\Bbb Q \cap U$ is open in $\Bbb Q$, and the same thing goes for closed sets.
Now, the funny thing is that your set is both the intersection of a closed interval in $\Bbb R$ with $\Bbb Q$ (namely $[-\sqrt 2, \sqrt 2] \cap \Bbb Q$), and it's the intersection of an open interval with $\Bbb Q$ (namely $(-\sqrt 2, \sqrt 2)\cap \Bbb Q$). Thus, from the topology it ingerits from $\Bbb R$, your set is both an open set and a closed set. That is why it's a clopen set. |
H: Any counterexample to answer this question on elementary geometry?
Question: See the figure below. If AB=BC and DE=EF, is the line DF parallel to the line AC?
This should be an elementary problem. But I can't construct a counterexample to disprove the above question. If the answer is negative, please give a counterexample. Thanks.
AI: Hint:
Let $D'F'$ be a segment analogous to $DF$ such that $E \in D'F'$ and $D'F' \parallel AC$.
Then $\triangle DED'$ and $\triangle FEF'$ are congruent and so $DD' \parallel FF'$.
I hope this helps $\ddot\smile$ |
H: Writing real invertible matrices as exponential of real matrices
Every invertible square matrix with complex entries can be written as the exponential of a complex matrix. I wish to ask if it is true that
Every invertible real matrix with positive determinant can be written as the exponential of a real matrix. (We need +ve determinant condition because if $A=e^X$ then $\det A=e^{\operatorname{tr}(X)} > 0$.) If not is there a simple characterization of such real matrices (with +ve determinant) which are exponentials of other matrices ?
AI: No, a real matrix has a real logarithm if and only if it is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times. So, it is possible that a matrix with positive determinant is not the matrix exponential of a real matrix. Here are two counterexamples: $\pmatrix{-1&1\\ 0&-1}$ and $\operatorname{diag}(-2,-\frac12,1,\ldots,1)$. For more details, see
Walter J. Culver, On the existence and uniqueness of the real logarithm of a matrix, Proceedings of the American Mathematical Society, 17(5): 1146-1151, 1966. |
H: Does this figure represent a cumulative distribution function?
Is this a c.d.f.?
I have no problem for random variable $X$ at $-\infty<X<x_2$. But if p.d.f. were continuous in interval $x_2\leq X<\infty$ , then c.d.f. should have been continuous. If probability mass function were discrete at $x_3$, then the c.d.f. would have a jump at $x_3$, but would remain constant thereafter upto $x_4$. The given figure does not have these both properties.
Is this figure an example of c.d.f.? If yes, how is it?
(Definition: The c.d.f. $F$ of a random variable $X$ is a function defined for each real number $x$ as follows:$$F(x)=Pr(X\leq x) \text{ for } -\infty<x<\infty\text{ )}$$
AI: This is a cumulative distribution function, because:
$F(x)\leq F(y)$ if $x \leq y$
$\lim\limits_{x \to {-\infty}}F(x)=0$ and $\lim\limits_{x \to {+\infty}}F(x)=1$
$F(x)$ is continuous to the right
You have a discrete variable at $x_3$, however the variable is continuous in $[x_3,+\infty)$ so it dones't have to be a constant.
That just means that $P\{x=x_3\}\ne0$ |
H: A supposed incorrect theorem
Hi i'm new here and here's my question:
Suppose $A\subseteq$$P(A)$. Prove that $P(A)\subseteq$$P(P(A))$.
I am using the book How to prove it by Velleman and i was wondering if anyone had a link to the soluions manual pdf if there ever was one?? I've googled it but it seems that no one has worked out the exercises and posted the solutions online and if they have it's not for all the chapters. It's a great book and i'd really appreciate it.
Also, i understand that a set cannot be a subset of its powerset so i don't understand what the question is tring to prove??
AI: There's nothing wrong with a set being a subset of its powerset. Sets with that property are called transitive and all ordinals satisfy that property.
Assume $A\subseteq\mathcal{P}(A)$ and $z\in\mathcal{P}(A)$. We just have to show $z\in\mathcal{PP}(A)$.
Since $z\in\mathcal{P}(A)$, $z\subseteq A$ and with our hypothesis $z\subseteq A\subseteq\mathcal{P}(A)$.
$z$ is a subset of $\mathcal{P}(A)$ and that just means $z\in\mathcal{PP}(A)$. |
H: Finding the mistake in precalc equation
$$\sqrt{x+3}+\sqrt{x-1} = -1;\tag{1}$$ $$ \mathbb{D}=[1,\infty)\tag{2}$$
$$x+3=(-1-\sqrt{x-1})^2\tag{3}$$
$$x+3=1+2\sqrt{x-1}+x-1\tag{4}$$
$$x_{1/2} = \pm\sqrt{\frac 3 2}+1\tag{5}$$
$$\mathbb{D} -> x=\sqrt{\frac 3 2}+1\tag{6} $$
According to the solution there shouldn't be any value for x.
Where is my mistake?
AI: If you look eqution (1), you see that left hand of equation is always positive, thus there isn't no solution for this equation. |
H: Adding weightage to number.
I have four numbers
x=50
y=30
z=20
sum of x+y+z=100
if i want to add a number n where
n=(x+y)/2
Edit:
n=45
such that x+y+z+n=100
now x+y+z=100.If any number (45 here) is added, then the sum should be always 100. i.e we need to subtract number(say m) from each x,y,z,n such that sum x+y+z+n should be equal to 100.
is it possible to do that.
AI: Let $a = n/3$. Then
\begin{align*}
(x-a) + (y-a) + (z-a) + n &= x - n/3 + y - n/3 + z - n/3 + n \\
&= x + y + z + n - n = x + y + z.
\end{align*} |
H: Conditional expected value of a product of two independent normal variables
I'm trying to work out the following conditional expectation:
$E[\epsilon_t z_t|\epsilon_t + z_t = k, \epsilon_{t-1}, z_{t-1}, \epsilon_{t-2}, z_{t-2},...]$
where $k$ is known and $\epsilon_{t}$ and $z_t$ are independent and normal distributed random variables with zero means and variances equal to $\sigma^2$ and $\tau^2$ respectively.
Any hints?
Regards,
Paul
AI: Assume without loss of generality that the random variables $x$ and $y$ are independent and centered normal with variances $\sigma^2$ and $\tau^2$ respectively, and that one looks for $$
E[xy\mid x+y].
$$
Let $z=x+y$, then $t=\tau^2x-\sigma^2y$ is independent of $z$ since $(z,t)$ is centered normal and $E[zt]=0$, and
$$
(\sigma^2+\tau^2)\cdot(x,y)=(\sigma^2z+t,\tau^2z-t),
$$
hence
$$
(\sigma^2+\tau^2)^2xy=\sigma^2\tau^2z^2+(\tau^2-\sigma^2)zt-t^2.
$$
Note that $E[z^2\mid z]=z^2$, $E[t^2\mid z]=E[t^2]=\tau^2\sigma^2(\sigma^2+\tau^2)$, $E[zt\mid z]=zE[t\mid z]$ and $E[t\mid z]=E[t]=0$. Hence,
$$
(\sigma^2+\tau^2)^2E[xy\mid z]=\sigma^2\tau^2z^2-\tau^2\sigma^2(\sigma^2+\tau^2),
$$
that is,
$$
E[xy\mid x+y]=\frac{\sigma^2\tau^2}{(\sigma^2+\tau^2)^2}\left((x+y)^2-(\sigma^2+\tau^2)\right).
$$ |
H: What does the variable $v$ in characteristic function of a random variable signify?
The characteristic function of a random variable $X$ is given as:
$$E(e^{jvX}) = \int _{-\infty} ^{\infty} e^{jvx} p(x) dx $$
This is interpreted as either mean of function $e^{jvx}$ or fourier transform of pdf $p(x)$. I know that $x$ represents random variable, but what does $v$ represents? What is its physical significance?
AI: For a heuristic (or "physical") interpretation consider the following "real version" of the characteristic function instead:
$$E_X(\omega):=\int_{-\infty}^\infty \cos(\omega x) p(x)\ dx\ .$$
Without the factor $\cos(\omega x)$ all values $x$ that the random variable $X$ can take are "equally important". Introducing this factor modulates this importance with frequency $\omega$. If it so happens that $X$ tends to prefer values which are spaced with period ${2\pi\over\omega_0}$ for some $\omega_0$ then this will show up in $|E_X(\omega_0)|$ becoming large.
In this way $\omega$ (the $v$ in your question) can be considered as a "sensor setting" chosen to investigate whether the given PDF $p$ has some inbuilt periodicity at frequency $\omega$. The Fourier inversion theorem guarantees that the collection of all sensor readings $E_X(\omega)$ $(\omega\in{\mathbb R})$ is sufficient to reproduce $p$ as a function of $x$. |
H: Verify proof that $\varnothing \in F$ implies $\cap F = \varnothing$
Hi so I want to know if my proof attempt is correct, so here's the question:
Suppose that $F$ is a family of sets. Prove that if $\varnothing\in F$ then $\cap F=\varnothing$.
Proof.
Suppose $\varnothing\in F$ and suppose $\cap F\ne\varnothing$. But this means that $x\in\varnothing$, since $\varnothing\in F$. Since $A$ was an arbitrary element of $F$, $x\in\varnothing$. But this is a contradiction since $\varnothing$ is the empty set. Thus $\cap F=\varnothing$.
Therefore if $\varnothing\in F$ then $\cap F=\varnothing$.
AI: Your proof is valid. However, it seems that the sentence starting with "Since $A$" is superfluous -- also, you haven't specified what $A$ is.
One thing that you could improve would be to expand on the second sentence by letting the reader know what $x$ is. E.g.:
But this means that for some $x$, $x \in \varnothing$, since $\varnothing \in F$.
If the course this is homework for is introductory, it is also good to recall the definition of any concept you're using. So you could add a sentence like:
By definition of $\cap F$, we have $x \in \cap F$ for some $x$ if and only if $x \in A$ for every $A \in F$.
to make it explicit where you draw the conclusion $x \in \varnothing$ from. |
H: Question about $ m = \prod_i (p_i-1)^{b_i} $ and the unique factorization theorem.
The fundamental theorem of arithmetic says that every integer $n>1$ is of the form
$$ n = \prod_i {p_i}^{a_i} $$
where $p_i$ is the $i$ th prime and $a_i$ is a nonnegative integer.
My question is how many $m$ satisfy $ 1<m<n $ and
$$ m = \prod_i {(p_i-1)}^{b_i} $$
where $p_i$ is the $i$ th ODD prime and $b_i$ is a nonnegative integer.
Lets call $f(n)$ the amount of $m$ that satisfy the above.
What is a good approximation to $f(n)$ ?
AI: You are asking for the counting function of the semigroup generated by the "shifted primes", the numbers $p-1$. Kevin Ford, 1998, showed that this counting function is on the order of $${x\over\log x}\exp(C_1(L_3(x)-L_4(x))^2+C_2L_3 (x)+C_3L_4(x))$$ where I write $L_3(x)$ and $L_4(x)$ for $\log\log\log x$ and $\log\log\log\log x$, respectively. See page 9 of this set of slides. I don't have bibliographic details for the 1998 paper.
EDIT: There is a link to the paper on Kevin Ford's website, so anyone who wants to can now have a look at the proof. The result is given as Corollary 15. The bibliographic details are Kevin Ford, The distribution of totients, The Ramanujan Journal 2, no. 1-2 (1998) 67-151, but the link goes to "the updated 2012 version, with various corrections and simplifications to the original paper." |
H: Derivative of f(x)=arth(lnx)
I'm struggling with finding derivative of f(x)=arth(lnx).
I've done following:
x=th(y)
f'(x)=(arth(lnx))'=1/(thlnx)'=1/(1/(xch^2(lnx)))=(x(ch^2(lnx)))/(ch^2x-sh^2x)=(ch^2(x(lnx)))/(ch^2(1-th^2x))=(x(lnx))/(1-x^2)
so in conclusion I get f'(arth(lnx))=(x(lnx))/(1-x^2)
but I think it's not correct. Any ideas?
AI: I assume you mean $\text{arctanh}{(\ln{x})}$. In this case, use the fact that
$$\text{arctanh}(z) = \frac12 \ln{\left ( \frac{1+z}{1-z} \right )}$$
$$\frac{d}{dx} \text{arctanh}{(\ln{x})} = \frac12 \left (\frac{1}{1+\ln{x}} + \frac{1}{1-\ln{x}} \right ) \frac{d}{dx} \ln{x} = \frac1{2 x} \left (\frac{1}{1+\ln{x}} + \frac{1}{1-\ln{x}} \right )$$
Simplifying this a bit, I get
$$\frac{1}{x (1-\ln^2{x})}$$ |
H: dominating function for $(1-\frac{x^2}n)^n(1+\sqrt{nx})$ in the dominated convergence theorem
Compute the following limit:
$\displaystyle\lim_{n\to\infty}\int_{-\sqrt{n}}^{\sqrt{n}}\Bigl(1-\frac{x^2}{n}\Bigr)^{n}(1+\sqrt{n}x)dx$
Hint: $\displaystyle\int_{-\infty}^{\infty}e^{-x^2}dx=\pi$
so I have a sequence $f_{n}(x)=\Bigl(1-\frac{x^2}{n}\Bigr)^{n}(1+\sqrt{n}x)$ and I have to find $g\in L^{1}$ s.t $|f_{n}(x)|\le g$
it is known that $\displaystyle\lim_{n\to\infty}\Bigl(1-\frac{x^2}{n}\Bigr)^{n}=e^{-x^{2}}$ but $\displaystyle\lim_{n\to\infty}(1+\sqrt{n}x)=\infty$
how to bound this sequence above?
AI: You cannot apply the dominated convergence theorem, since the integrands aren't dominated by an integrable function.
The trick is to split the integral,
$$\int_{-\sqrt{n}}^{\sqrt{n}} \left(1 - \frac{x^2}{n}\right)^n(1 + \sqrt{n} x)\,dx = \int_{-\sqrt{n}}^{\sqrt{n}} \left(1 - \frac{x^2}{n}\right)^n\,dx + \sqrt{n}\int_{-\sqrt{n}}^{\sqrt{n}} \left(1 - \frac{x^2}{n}\right)^nx\,dx.$$
You can apply the dominated convergence theorem to the first, and the second integral is $0$ since the integrand is odd, and the interval symmetric about $0$. |
H: Sturm-Liouville Problem --Suppose that the functions $p(x), p′(x), q(x)$, and $r(x)$ in $[p(x)y′]′ − q(x)y + \lambda r(x)y = 0, \quad (a < x < b)$
Suppose that the functions $p(x), p′(x), q(x)$, and $r(x)$ in $[p(x)y′]′ − q(x)y + \lambda r(x)y = 0, \quad (a < x < b)$. Question pasted as image given below:
AI: Here you have 3 cases:
1-) $\lambda=0$
The solution is $y=c_1+c_2x$, applying boundary conditions, $c_1=c_2=0$ hence $y=0$
2-) $\lambda<0$
The solution is $y=c_1e^{\sqrt{-\lambda} x}+c_2e^{-\sqrt{-\lambda} x}$, applying boundary conditions, $c_1=c_2=0$ hence $y=0$
3-) $\lambda>0$
The solution is $y=A\sin{\sqrt{\lambda} x}+B\cos{\sqrt{\lambda} x}$. Applying $y'(0)=0$ we obtain $A=0$, applying $y(L)=0$ we obtain $B\cos{\sqrt{\lambda} L}=0$. If $B=0$ we obtain trivial solution. So $\cos{\sqrt{\lambda} L}=0$, which implies $\sqrt{\lambda}=\frac{n\pi}{2L},n=1,3,5,...$
Therefore eigenvalues are $\lambda=(\frac{n\pi}{2L})^2$ and corresponding eigenfunctions are $y_n=B_n\cos\frac{n\pi}{2L}x$ |
H: Evaluating the convolution using the convolution integral
I am having trouble evaluating the convolution of two signals using the convolution integral.I want to find the convolution of two signals x and h where,
$$
x(t) = \begin{cases}
e^{-at} & \text{$t > 0$} \\
0 & \text{$t < 0$ } \\
\end{cases}
$$
$$
h(t) = \begin{cases}
e^{-bt} & \text{$t > 0$} \\
0 & \text{$t < 0$ } \\
\end{cases}
$$
using the convolution integral
$$
y(t) = x(t)*h(t) = \int_{-\infty}^{\infty} h(\tau)x(t - \tau) d\tau
$$
Which will mean that:
$$
h(\tau) = \begin{cases}
e^{-b\tau} & \text{$\tau > 0$} \\
0 & \text{$\tau < 0$ } \\
\end{cases}
$$
$$
x(t - \tau) = \begin{cases}
e^{-at}e^{a \tau} & \text{$t > \tau$} \\
0 & \text{$t < \tau$ } \\
\end{cases}
$$
But how do I proceed from here? I don't know how to handle the $x(t - \tau)$ function which is non-zero only when $t > \tau$.
AI: Write each of the signals as
$$e^{-k \tau} \theta(\tau)$$
where $k$ is either of $a$ or $b$, and $\theta(\tau)$ is the Heaviside step function, zero when $\tau < 0$ and $1$ when $\tau > 0$. The convolution integral may then be written as
$$\int_{-\infty}^{\infty} d\tau \, e^{-a \tau} \theta(\tau) \, e^{-b (t-\tau)} \theta(t-\tau)$$
Now, the product of the two Heavisides in the integral is zero outside the interval $[0,t]$. Therefore, we may write the convolution integral as
$$\int_0^t d\tau \, e^{-a \tau} \, e^{-b (t-\tau)} = e^{-b t} \int_0^t d\tau \, e^{-(a-b) \tau} $$
which is
$$\frac{1}{a-b} e^{-b t} \left (1-e^{-(a-b) t} \right ) = \frac{e^{-b t}-e^{-a t}}{a-b}$$ |
H: Differentiation of function with chain rule
the following expression is part of a function I have to differentiate:
$y = \tan^3(5x^4-7)$
I tried using the chain-rule, so:
$ y' = 3\tan^2(5x^4-7)\cdot(20x^3)$
is this correct?
AI: Not quite correct; we need to also differentiate the "tangent" component.
In general, the chain rule for a composition of functions $y = f(g(x))$ is given by: $$y'= f'(g(x)) g'(x).\,$$
In your case, we actually have $y = \Big(f(g(x)\Big)^3$, which as you correctly noticed, involves the power rule, as well. So we have $$g(x) = 5x^4 - 7 \implies g'(x) = 20x^3$$ $$\;f(g(x)) = \tan(g(x))\implies f'(g(x)) = \sec^2(g(x)) = \sec^2(5x^4 - 7)$$
Putting it all together gives us:
$$\begin{align} y = \Big(f(g(x)\Big)^3 \implies y' & = 3\Big(f(g(x)\Big)^2\cdot f'(g(x))\cdot g'(x) \\ \\ & = 3\Big(\tan(5x^4 - 7)\Big)^2\cdot \sec^2(5x^4 - 7)\cdot (20x^3) \\ \\ & = 60x^3 \tan^2(5x^4 - 7)\sec^2(5x^4 - 7)\end{align}$$ |
H: Functions satisfying $(b-a)f'(\tfrac{a+b}{2}) = f(b)- f(a)$
Let $f$ be a differientable real function such that $$(b-a)f'(\tfrac{a+b}{2}) = f(b)- f(a)$$
for all reals $a,b$.
Is $f$ polynomial of degree $\leq 2$ ?
AI: The answer is Yes.
The condition on $f$ is equivalent to
$$f(x+h)-f(x-h)=2f'(x)h\tag{1}$$
for all reals $x$ and $h$. Letting $h=1$ in $(1)$, we can conclude that $f$ is twice differentiable. Therefore, we can differentiate both sides of $(1)$ with respect to $h$ twice, which gives us
$$f''(x+h)-f''(x-h)=0. \tag{2}$$
Since $x$ and $h$ are arbitrary in $(2)$, $f''$ must be a constant function, i.e. $f$ is a polynomial of degree no more than $2$. |
H: Trying to simplify the expression $A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$
So far I've got:
$A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$
$= A'C'D'(B' + B) + C'D(A'B + AB')$
$= A'C'D'(1) + C'D(A \;\text{ XOR }\; B)$
$= C'[A'D' + D(A \;\text{ XOR }\; B)]$
Did I do this correctly? Is there a simpler solution?
Thanks
K
AI: Yes, your work is correct.
Simplification of a Boolean expression depends on context, and what form you are seeking in your "simplification": for example, conjunctive normal form (product of sums) or disjunctive normal form (sum of products), etc. Typically, one does not introduce "xor" $\oplus$ unless expressing the entire function in terms of $\oplus$, $\land$, $'$.
See, for example, the following, without the use of $\oplus$ (xor):
$$\begin{align} A'C'D'(B' + B) + A'BC'D + AB'C'D &= A'C'D' + A'BC'D + AB'C'D \\ \\
&= C'[A'D' + D(A'B + AB')]\end{align}$$
But again, all your manipulations are indeed correct. |
H: when $p \mid ab \Rightarrow p \mid a$ or $p \mid b$?
I want to know when this statement is true ($p$ is a prime):
$p \mid ab \Rightarrow p \mid a$ or $p \mid b$ ?
AI: Given $p$ is a prime, then your statement$$p \mid ab \implies p \mid a\;\lor p\mid b$$ is always true.
But the proposition is not necessarily true when $p$ is not prime.
For example, suppose $p = 6$ and $a = 2, b = 3$.
Then $6\mid (2\cdot 3)$, but it is not true that $6\mid 2$ or that $6\mid 3$. |
H: How to estimate the value of this supremum?
Could you please guide me through how to estimate or calculate this supremum
$$\sup_{x\in\mathbb R}\left|x \cdot \arctan(nx)-x \cdot \frac{\pi}{2}\right|=?$$
AI: Use that
$$
\arctan x=\int_0^x\frac{dt}{1+t^2}\quad\text{and}\quad \frac{\pi}{2}=\int_\infty^x\frac{dt}{1+t^2}
$$
to write
$$
\Bigl|x\,\arctan(nx)-\frac{\pi}{2}\,x\Bigr|=|x|\int_{nx}^\infty\frac{dt}{1+t^2}.
$$
It is clear, as hinted to you by Harald, that as $x\to-\infty$, the above expression is not bounded.
Now I will make a wild guess: you are trying to prove that $x\,\arctan(n\,x)$ converges uniformly to $\pi\,x/2$ on $[0,\infty)$. In that case you do not need the exact value of the $\sup$; an upper bound might be enough. If $x>0$, then
$$
x\int_{nx}^\infty\frac{dt}{1+t^2}\le x\int_{nx}^\infty\frac{dt}{t^2}=\frac{1}{n}.
$$ |
H: Probability that coin will fall into a square
So the exercise is this:
We have and infinite chessboard and we have a coin. Every grid is of length and width $a$, whereas the coin has diameter $2 \cdot r<a$. We throw a coin into a chessboard and we want to know with what probability the coin will falll into the grid.
So let $S_{1}$=area of green rectangle, $S_{2}=S_{1}+$ area of the red border.
So the probability in my opinion is
$ P(a)= \frac{S_{1}}{S_{2}} $.
Is this in any shape or form correct?
AI: The rounded corners of your red area are a mistake.
The best way to look at the problem is to reformulate
The coin falls completely inside a single square.
to
The center of the coin is at least one coin radius from any grid line.
You can then shade the area of a grid square where the coin center cannot fall -- this shaded area will look the same in every square. If the coin radius is $r$ and the grid square sides have length $d$, there's a small square of side length $d-2r$ in each square where the center can fall without the coin extending beyond the grid square.
So the probablity of staying within the square becomes
$$ \frac{(d-2r)^2}{d^2} $$
assuming that the position of the coin's center relative to the square it happens to land on is uniformly distributed. |
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