text:
## rt.representation theory – Question on geometric lemma in the \$p\$-adic group representation

To check whether my understaning on the geometric lemma is right, I would like to ask some specific question related to it.

Let $$F$$ be a $$p$$-adic local field of characteristic zero. Let $$W_n$$ be a symplectic space over $$F$$ of dimension $$2n$$. Let $$Sp_{n}(W_n)$$ be a symplectic group and $$B_n$$ its standard Borel subgroup. Let $$Q_{t,n-t}$$ be a standard parabolic subgroup of $$G$$ preserving a totally isotropic subspace of $$W_n$$ of dimension $$t$$.

Let $${chi_i}_{1 le i le n}$$ be unramified characters of $$GL_1(F)$$ such that $$chi_i(omega)=1$$ and let $$pi$$ be the normalized parabolic induced representation $$text{Ind}_{Q_{3,n-3}}^{Sp_{n}}(chi_1 circ text{det}_{GL_3},chi_2,cdots,chi_{n-2})$$.

I am considering $$J_{Q_{2,n-2}}(pi)$$, the normalized Jacquet module of $$pi$$ to $$Q_{2,n-2}$$. By the geometric lemma, there is some filtration of $$J_{Q_{2,n-2}}(pi)$$

$$0=tau_0 subset tau_1 subset cdots subset tau_m=J_{Q_2}(pi)$$ such that $$tau_{i}backslashtau_{i+1}$$ is some induced representation of the Jacquet module of the inducing data of $$pi$$.

I guess that such subquotient appearing in this filtration has the form $$|cdot|^{-frac{1}{2}}cdot(chi_1 circ text{det}_{GL_2}) boxtimes text{Ind}_{B_{n-2}}^{Sp_{n-2}}(chi_1′,chi_2′,cdots,chi_{n-2}’)$$ or
$$text{Ind}_{B_2}^{GL_2}(chi_1′,chi_2′) boxtimes text{Ind}_{Q_{3,n-5}}^{Sp_{n-2}}(chi_1 circ text{det}_{GL_3},chi_4′,cdots,chi_{n-2}’)$$?

(Here, $${chi_1′,cdots,chi_{n-2}’} subset {chi_1,cdots,chi_{n-2},chi_1^{-1},cdots,chi_{n-2}^{-1}}.)$$

## rt.representation theory – Frobenius reciprocity law in the \$p\$-adic group represenation

Let $$G$$ be a $$p$$-adic classical group and let $$P_0$$ be a minimal parabolic subgroup of $$G$$. Let $$P=MN$$ be a
standard parabolic subgroup containing $$P_0$$. Let $$text{Ind}$$ and $$|_M$$ be the normalized parabolic induction functor and Jacquet functor, respectively.

Then for smooth representations $$pi$$ of $$G$$ and $$rho$$ of $$M$$, the Frobenius reciprocity law says that
$$text{Hom}_{G}(pi, text{Ind}(rho))simeq text{Hom}_{M}(pi|_M,rho).$$

I am wondering whether
$$text{Hom}_{G}(text{Ind}(rho),pi)simeq text{Hom}_{M}(rho, pi|_M)$$ does hold. If it does not hold in general, is there a sufficient condition for $$pi$$ and $$rho$$ which makes the above hold?

## nt.number theory – Full measure properties for Zariski open subsets in \$p\$-adic situation

Let $$F$$ be a $$p$$-adic field and let $$X$$ be a smooth integral variety over $$F$$ (I am chiefly interested in the case when $$X$$ is a connected reductive group over $$F$$). Let $$U$$ be a non-empty open subset of $$X$$ with complement $$Z$$.

We can endow $$X(F)$$ with the Serre-Oesterle measure (e.g. as in (1,Section 2.2) or (2, Section 7.4))–this is just the standard measure coming from a top form of $$X$$).

My question is then whether one knows a simple proof/reference for the following:

The subset $$Z(F)$$ of $$X(F)$$ has measure zero.

I think this is proven in (1, Lemma 2.14)–but this is concerned with a more specific context which makes it non-ideal as a reference.

Any help is appreciated!

(1) http://www.math.uni-bonn.de/people/huybrech/Magni.pdf

(2) Igusa, J.I., 2007. An introduction to the theory of local zeta functions (Vol. 14). American Mathematical Soc..

## nt.number theory – P-adic distance between solutions to S-unit equation

Let $$p$$ be a fixed prime number and $$S$$ is a finite set of prime numbers which does not contain $$p$$. A theorem of Siegel asserts that the number of solutions to the $$S$$-unit equations are finite; that is, there are only finitely many $$S$$-unit $$u$$ such that $$1-u$$ is also an $$S$$-unit. Therefor for each such $$S$$ there exist a lower bound on $$|u_1-u_2|_p$$ where $$u_1$$ and $$u_2$$ are solutions to $$S$$-unit equations.

My question is: does there exist such a lower bound uniformly? More precisely, does there exist a lower bound for the $$p$$-adic distance between solutions to the $$S$$-unit equations that only depends on the size of $$S$$(and perhaps on $$p$$)? Here we are assuming $$S$$ does not contain $$p$$.

## reference request – Parabolic inductions for p-adic reductive groups

So I wish to ask for articles/comments surveying conjectures and theorems about parabolic induction for p-adic (non-archimedean case) reductive groups, and how local Langlands behaves under such. That is:

1. For a quasi-split group G, what we know about the sub-representations of principle series.
2. Same setting, but what we know about the (enhanced) Langlands parameters they correspond to, both conjectures and theorems.
3. Same questions in 1 & 2, but for parabolic induction of supercuspidal representations from Levi assumed local Langlands for that Levi is understood.

The only things I know is the articles of Bernstein-Zelevinsky that addresses part of 1 and somewhat 2 for GL_n, as well as recent work of Aubert-Moussaoui-Solleveld (and extensions of some of them with others) about a nice conjectural framework of 3.

Any comment is greatly appreciated! I am also under the impression that the above questions are largely known for classical groups given the local Langlands established by Arthur and many others, but love to know a reference good for quick reading. If more is known in the local function field case, it will be fantastic to learn too. Thank you very much!!!

## arithmetic geometry – Generalized Hodge-Tate weights of \$p\$-adic Galois representation

Let $$K$$ be an finite extension of $$mathbb{Q}_p$$, and $$E$$ be a Galois extension of $$K$$ whose Galois group is a open subgroup of $$text{SL}_2(mathbb{Z}_p)$$, then we have a $$2$$-dim $$p$$-adic Galois representation given by $$G(bar{K}/K)rightarrow G(E/K)rightarrowtext{SL}_2(mathbb{Z}_p)rightarrowtext{GL}_2(mathbb{Q}_p),$$then I want to know why the sum of the two Sen weights of this representation is $$0$$. Note that by definition, Sen weights or generalized Hodge-Tate weights are the eigenvalues of Sen operator.

Thanks!

## number theory – Integral of a Chebyshev polynomial with respect to this special measure (Plancherel p-adic measure for GL2 (Q_p))

I'm trying to show that the integral $$int _ {- 2} ^ 2 U_n left ( frac {x} {2} right) frac {p + 1} { pi} frac { sqrt {1- frac {x ^ 2 } {4}}} { left ( sqrt {p} + frac {1} { sqrt {p}} right) ^ 2 – x ^ 2} dx$$ It does not matter $$p ^ – n / 2}$$, when $$n$$ is an even integer.
Here, $$U_n (x)$$ is he $$n$$-th Chebyshev polynomial of the second type, defined as $$U_n (x) = frac {sin ((n + 1) cos ^ {- 1} x)} { sqrt {1-x ^ 2}}$$; $$p$$ is cousin $$n geq 0$$.

The corresponding integral when $$n$$ is odd is zero since the integrand is an odd function. However, I cannot solve this integral for the even case. The integrand is simplified a bit once the definition of $$U_n$$ used but I can't deal with it $$left ( sqrt {p} + frac {1} { sqrt {p}} right) ^ 2 – x ^ 2$$ finished. Some help?

The context is in analytic number theory, particularly the results in vertical Sato-Tate. The measure here is actually $$mu_p (x) = frac {p + 1} { pi} frac { sqrt {1-x ^ 2}} { left ( sqrt {p} + frac {1} { sqrt { p}} right) ^ 2 – x ^ 2} dx$$, which is the measure of Plancherel in GL$$_2 ( mathbb {Q} _p)$$. However, I don't think this should be relevant to the integral, I think that the integral can be solved using simple calculation techniques.

## arithmetic geometry – function L in p-adic spaces

I've been learning more about different $$p$$-Adicic geometries, namely, Berkovich spaces, Huber Adic spaces and ridiculous analytical spaces. In arithmetic geometry, it is often very interesting to associate and study the function L of a space. My question is whether this has been studied for such $$p$$-the adicos spaces. A naive way of waiting for such L functions to be obtained is to take the L function associated with the cohomology of said space and study it. However, I could not find a reference for this.

The problem is that Google searches lead directly to $$p$$-adic L-functions, which as far as I can tell is not what I want.

## cv.complex variables – Asymptotic analysis using the p-adic Mellin Transform?

In the ordinary analysis, given a nice enough $$f: left (0, infty right) rightarrow mathbb {C}$$, if we can calculate the Mellin transformation: $$mathscr {M} left {f right } left (s right) = int_ {0} ^ { infty} x ^ {s-1} f left (x right) dx$$ in closed form, the calculation of the inverse transformation of Mellin using the residue theorem provides us with formulas (usually asymptotic, but sometimes exact) for the behavior of $$f left (x right)$$ how $$x$$ decreases to $$0$$ I eat $$x$$ increases to $$infty$$. Former:

$$int_ {0} ^ { infty} x ^ {s-1} sum_ {n = 0} ^ { infty} e ^ {- 2 ^ {n} x} dx = frac { Gamma left (s right)} {1-2 ^ {- s}}$$ it implies: $$sum_ {n = 0} ^ { infty} e ^ {- 2 ^ {n} x} = frac {1} {2} – frac { gamma + ln x} { ln2} + frac {1} { ln2} sum_ {k in mathbb {Z} ^ { times}} Gamma left ( frac {2k pi i} { ln2} right) x ^ {- frac {2k pi i} { ln2}} – sum_ {n = 1} ^ { infty} frac { left (-1 right) ^ {n}} {2 ^ {n} -1 } frac {x ^ {n}} {n!}$$ an identity (emotionally satisfactory!) that is fulfilled for all x> 0.

But, now consider the p-adic analog: $$mathscr {M} _ {p} left {f right } left (s right) = int _ { mathbb {Z} _ {p}} left | mathfrak {y} right | _ {p} ^ {s-1} f left ( mathfrak {y} right) d mathfrak {y}$$ where $$d mathfrak {y}$$ is the measure of probability of hair in $$mathbb {Z} p$$ Y $$f: mathbb {Z} _ {p} rightarrow mathbb {C}$$ is it nice enough is there a similar interpretation for the inverse transformation of Mellin of $$mathscr {M} _ {p} left {f right }$$ in terms of the asymptotes of $$f$$? Answers or references will be appreciated.

$$text {Prove that} sum_ {i = 1} ^ n frac {1} {i} text {is not an integer for} n ge 2$$
Realise $$sum_ {i = 1} ^ n frac {1} {i} = sum_ {i = 1} ^ n frac { frac {n!} {i}} {n!}$$ We consider $$v_2 left ( sum_ {i = 1} ^ n frac {n!} {i} right)$$. We know $$v_2 left ( frac {n!} {2i-1} + frac {n!} {2i} right) = v_2 left ( frac {n!} {2i} right)$$Then we get $$v_2 left ( frac {n!} {4i-2} + frac {n!} {4i} right) = v_2 left ( frac {n!} {4i} right)$$ and repeating to summarize the factorial in this way we come to $$v_2 left ( sum_ {i = 1} ^ n frac {n!} {i} right) = v_2 left ( frac {n!} {2 ^ { lfloor log_2 n rfloor} } right) tag {1}$$
I do not understand the part: "and repeating to summarize the factorial in this way we arrive at …", where the equation was made $$(1)$$ comes from? Can anyone help? Thank you!


text:
# Which reviews are rarer than others?

Which reviews are rarer than others?

I think that late answers should be the rarest but I want a order of rarity of them.

## 1 Answer

At the time of writing, the global number of reviews done was:

1. Close: 398,338
2. Suggested Edits: 262,766
3. First Posts: 210,381
4. Low Quality: 205,281
5. Reopen: 69,808
6. Late Answer: 23,339

Personally, I was a bit surprised that First Post is that high in the list. Subjectively, I would have thought it below Low Quality and closer to Reopen.

It should be noted that the criteria for reviewing sometimes change, which can change the current number of reviews relative to the global number.

• You forgot suggested edits. – Taha Akbari Aug 8 '16 at 18:54
• Indeed. I added the information – quid Aug 8 '16 at 19:09
• One factor is that the privilege to review First Posts and Late Answers is given at $500$ reputation points, much earlier than other queues. So there are more "eyeballs" on these items. – hardmath Aug 8 '16 at 19:39
• I added a pie chart :) Hope you don't mind. – 6005 Aug 9 '16 at 7:40
• How come you know these values ? :) thanks – A---B Aug 9 '16 at 9:26
• @ritwiksinha go here and click on "recent review" in the queue you're interested in. You'll find the total number of reviews and the reviewer ranking. – Surb Aug 9 '16 at 11:07

format: plain


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 3.14: Solve Inequalities Using Multiplication

Difficulty Level: At Grade Created by: CK-12
Estimated2 minsto complete
%
Progress
Practice Inequalities with Multiplication
Progress
Estimated2 minsto complete
%

Have you ever had to solve an inequality by using multiplication? Take a look at this dilemma.

Emma Frances took a whole pile of cookies and divided them up by 11. When she was done, she had 22 piles of cookies. She knows that she started with greater than or equal to 242 cookies.

Can you use this inequality to prove that Emma is correct?

\begin{align*}\frac{x}{11} > 22\end{align*}

Pay attention. This Concept will show you how to do this successfully.

### Guidance

An inequality is a mathematical statement where a quantity may be equal to or less than or greater than another quantity. You can identify an inequality and solve inequalities using all four operations. Now we are going to look at solving inequalities that involve multiplication.

Let’s look at this more closely.

Just as you sometimes need to multiply or divide both sides of an equation by the same number in order to solve an equation, you may sometimes need to multiply or divide both sides of an inequality by the same number in order to solve it. However, doing so may be a little more complicated.

First, let’s take a look at a property that we can use to help us when solving an inequality that involves multiplication.

The multiplication property of inequality states that if each side of an inequality is multiplied by the same positive number, the sense of the inequality stays the same. In other words, the inequality symbol does not change.

If \begin{align*}a>b\end{align*} and \begin{align*}c>0\end{align*}, then \begin{align*}a \times c > b \times c\end{align*}. If \begin{align*}a \ge b\end{align*} and \begin{align*}c > 0\end{align*}, then \begin{align*}a \times c \ge b \times c\end{align*}.

If \begin{align*}a < b\end{align*} and \begin{align*}c > 0\end{align*}, then \begin{align*}a \times c < b \times c\end{align*}. If \begin{align*}a \le b\end{align*} and \begin{align*}c > 0\end{align*}, then \begin{align*}a \times c \le b \times c\end{align*}.

However, if each side of an inequality is multiplied by the same negative number, the sense of the inequality changes and the inequality symbol must be reversed.

If \begin{align*}a > b\end{align*} and \begin{align*}d < 0\end{align*}, then \begin{align*}a \times d < b \times d\end{align*}. If \begin{align*}a \ge b\end{align*} and \begin{align*}d<0\end{align*}, then \begin{align*}a \times d \le b \times d\end{align*}

If \begin{align*}a < b\end{align*} and \begin{align*}d < 0\end{align*}, then \begin{align*}a \times d > b \times d\end{align*}. If \begin{align*}a \le b\end{align*} and \begin{align*}d < 0\end{align*}, then \begin{align*}a \times d \ge b \times d\end{align*}

When you solve an inequality by using multiplication, you will see division in the original problem.

Yes. When you solve an inequality by using multiplication, the original problem will show division. Think about it this way, you use the inverse operation to solve an inequality. To solve by using multiplication, you must see division first.

Solve this inequality: \begin{align*}1 < \frac{x}{2}\end{align*}.

Solve this inequality as you would solve an equation, by using inverse operations. Notice the fraction bar is used here to show division. Since the \begin{align*}x\end{align*} is divided by 2, multiply both sides of the inequality by 2 to solve it. Since this involves multiplying both sides of the inequality by a positive number, the sense of the inequality will stay the same and you will not need to reverse the inequality symbol.

\begin{align*}1 &< \frac{x}{2}\\ 1 \times 2 &< \frac{x}{2} \times 2\\ 2 &< \frac{x}{\bcancel{2}} \times \frac{\bcancel{2}}{1}\\ 2 &< x\end{align*}

The answer is that \begin{align*}x\end{align*} is greater than 2 or you could say 2 is less than \begin{align*}x\end{align*}.

Identify the inequality symbol that goes in each blank.

#### Example A

\begin{align*}k & \ > m\\ k \times 3 & \ \underline{\;\;\;\;\;} \ m \times 3\end{align*}

Solution:  \begin{align*}>\end{align*}

#### Example B

\begin{align*}m & \ \ge n\\ m \times \left(-\frac{1}{2}\right) & \ \underline{\;\;\;\;\;} \ m \times \left(-\frac{1}{2}\right)\end{align*}

Solution: \begin{align*} =\end{align*}

#### Example C

\begin{align*}-1 & \ < 2\\ (-1) \times (-3) & \ \underline{\;\;\;\;\;} \ 2 \times (-3)\end{align*}

Solution: \begin{align*} < \end{align*}

Now let's go back to the dilemma at the beginning of the Concept.

\begin{align*}\frac{x}{11} > 22\end{align*}

Notice that this inequality has division in it, so we can use multiplication to solve it. We multiply both sides by 11. The left side cancels. On the right side here is our work.

\begin{align*}x > 11(22)\end{align*}

\begin{align*}x > 242\end{align*}

Using multiplication, proves that Emma's statement is correct.

### Vocabulary

Inequality
a mathematical statement where two quantities may or may not be equal. There are many possible answers to an inequality.
Multiplication Property of Inequality
multiplication of a positive number = inequality is the same. Multiplication of a negative number, the sense of the inequality changes and the sign is reversed.

### Guided Practice

Here is one for you to try on your own.

Solve for the unknown variable.

\begin{align*}\frac{x}{4}\le-9\end{align*}

Solution

This problem has division in it, so we can use the inverse to solve it.

The inverse of division is multiplication. We multiply both sides of the inequality by 4. Then we can simplify and multiply to find the solution.

\begin{align*}x\le-9(4)\end{align*}

\begin{align*}x\le-36\end{align*}

### Practice

Directions: Solve each inequality using multiplication.

1. \begin{align*}\frac{a}{3} < 10\end{align*}
2. \begin{align*}\frac{x}{2} > 20\end{align*}
3. \begin{align*}\frac{b}{4} < 25\end{align*}
4. \begin{align*}\frac{x}{5} > 100\end{align*}
5. \begin{align*}\frac{c}{2} \le 20\end{align*}
6. \begin{align*}\frac{x}{7} > 6\end{align*}
7. \begin{align*}\frac{y}{9} \le 2\end{align*}
8. \begin{align*}\frac{y}{11} \ge 5\end{align*}
9. \begin{align*}\frac{x}{2} < 3\end{align*}
10. \begin{align*}\frac{y}{8} \ge -7\end{align*}
11. \begin{align*}\frac{x}{3} < -2\end{align*}
12. \begin{align*}\frac{y}{4} \le -3\end{align*}
13. \begin{align*}\frac{x}{2} \ge -11\end{align*}
14. \begin{align*}\frac{a}{13} \ge -3\end{align*}
15. \begin{align*}\frac{x}{22} \ge -5\end{align*}

### Vocabulary Language: English

inequality

inequality

An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
Multiplication Property of Inequality

Multiplication Property of Inequality

The multiplication property of inequality states that when an inequality is multiplied by a negative number, the sense of the inequality changes and the sign is reversed.

Show Hide Details
Description
Difficulty Level:
Authors:
Tags:
Subjects:
Search Keywords:

Date Created:
Dec 19, 2012


text:
I have a few fields that are required. However if a option is chosen I
would like to switch all these fields to not-required.

I'm having issues making since of "is-required". If I set it to "True"
or "False" I get an error that an expected string was expected for
isRequired.
What's the proper way to used this form method?

setRequired(fieldname, is-required)

form.setRequired("fieldname", is-required)
Sets the field to required if is-required is True or optional otherwise.
A field that is required blocks the form submission if it is empty.

thanks


text:
# Write the adjoint of the following matrix $\begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}$

## 1 Answer

Toolbox:
• The determinant value of a $2\times 2$ square matrix is $|A|=a_{12} \times a_{22} - a_{21} \times a_{12}$
• The inverse of a $2 \times 2$ square matrix is $\frac{1}{|A|} \begin{bmatrix} a_{22} & -a_{12} \\ a_{21} & a_{11} \end{bmatrix}$
Adjoint of a $2 \times 2$ matrix can be found by interchanging the elements of $a_{11}\; and \;a_{22}$ and changing the symbols of elements of $a_{21}\; and\; a_{12}$
Given $A= \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}$
By using the above hint we get
$adj(A)= \begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}$

answered Apr 8, 2013 by

1 answer

1 answer

0 answers

0 answers

1 answer

1 answer

1 answer


text:
# Is image padding really necessary while using a filter?

I do know that we must pad an image while using a filter. But what I am suggesting is that is it really necessary? Can't we just calculate at run time if the current cell/pixel is an edge cell/pixel and then use the padding value directly over there. It might add an overhead to the function but so does the entire padding function.

Only reason I think the padding function should be used is if the padded version of original image is going to be used in many different function which will overtake the overhead of dynamically calculating the edge cell/pixel values(which might never occur). Am I wrong?

• Welcome to SE.DSP! It depends how you are calculating the filtering. If you are using an FFT, then you must pad, otherwise you'll get "time aliasing" from the use of circular convolution by the FFT. If your image is $N\times M$ and your filter kernel is $P \times Q$ then your padding must be at least $(N+P-1) \times (M+Q-1)$ to get the expected linear convolution result. – Peter K. May 19 '17 at 15:12

## 1 Answer

Adding code to your filter function to detect and resolve edges implies code running needlessly throughout most of the image. Image area grows quadratically with image size.

Padding grows linearly with image size (as does the perimeter), so adding padding instead of edge-detection code seems a more efficient option.

Also, when adding padding, you are free to choose what to fill the padding with, according to your particular problem. You may choose to fill in with a constant value, or mirror the image, or get overlap data from adjacent images, etc.


text:
Funded!

Big news here at Penn State:  we have been selected by the NSF to build a new instrument capable of finding rocky planets orbiting stars at just the right distance to harbor liquid surface water!  Press release is here.

Penn State Awarded $3.3 Million to Build Instrument for Finding Planets in Habitable Zones Around Nearby Stars University Park, PA — A new state-of-the-art instrument — a precision spectrograph for finding planets in habitable zones around cool, nearby stars — is being developed at Penn State with support from a new$3.3-million grant from the National Science Foundation. “This new Habitable Zone Planet Finder instrument will allow us to detect the existence of planets that are similar in mass to Earth and also are in orbits that allow liquid water to exist on their surfaces,” said Suvrath Mahadevan, assistant professor of astronomy and astrophysics at Penn State and a co-principal investigator of the project.

The Habitable Zone Planet Finder (HPF) is a large infrared spectrograph that will be used to analyze the components of starlight and to detect the slight motions of nearby stars caused by the gravity of orbiting planets. It will be approximately the size of an SUV, and will weigh over two tons. “We have designed the HPF to be able to efficiently observe small, cool stars, which are by far the most numerous types of stars in the Milky Way galaxy,” explained co-principal investigator Lawrence Ramsey, professor of astronomy and astrophysics.

“These stars, which have temperatures far below that of the Sun, radiate very little of their energy in the visible part of the spectrum, so we must create an instrument that can capture the infrared part of the spectrum — where the unaided human eye cannot see but where these stars are brightest,” Ramsey said. The team chose to observe these stars because they offer the best opportunity for finding planets with solid surfaces in the so-called “Habitable Zone” around nearby stars, the range of distances from a star within which temperatures might be right for liquid water.

Hibby-Eberly Telescope

The instrument will take three years to build. When it is completed, it will be shipped to the Hobby-Eberly Telescope at McDonald Observatory in west Texas to begin its multi-year quest for new worlds, during which it will survey more than a hundred nearby stars. “This instrument will precisely measure the motion of each star, hunting for the telltale “wobble” caused by an orbiting planet,” said Penn State Assistant Professor Jason Wright, a co-principal investigator on the HPF team. “Once we detect these wobbles, we will be able to infer the surface temperature and mass of the planet or planets orbiting it. These cool stars we are targeting are some of the very closest to Earth. In astronomical terms, they are in our back yard.”

Associate Professor Steinn Sigurdsson, who is also a lead investigator of the Penn State Astrobiology Research Center, said “The HPF is a significant step toward discovering terrestrial planets in orbits where liquid water may be present on their surface around our closest stellar neighbors. This goal is important for the NASA astrobiology program since life, as we know it, needs liquid water. Nearby planets with liquid water on their surface are the first place to start looking for signatures of life on other planets.”

Because the planets the astronomers are looking for are so small, and the stars they are looking at are so cool, they need to build a new kind of planet-finding instrument. “To find an Earth-like planet around one of our target stars requires that we be able to measure the velocity of a star, located at a distance of tens of light years from the Earth, to a precision of about three feet per second — the typical speed a person walks — over a period of several years. To attain this goal, HPF not only must have high-efficiency optics and detectors, but must be maintained in an extremely stable environment,” Ramsey explained.

Mahadevan said the HPF team is keen to embark on this challenge. “I do believe we now have the technological capability to make these extremely precise measurements in the infrared,” he said. “Finding Earth analogs around our nearest neighbors will be an exciting adventure. We are grateful to have had prior support since 2005 from the National Science Foundation, the NASA Origins program, and the NASA Astrobiology Institute for the technology development that lead to our now being able to develop this instrument.”

The HPF development-and science team also includes Penn State Evan Pugh Professor of Astronomy and Astrophysics Alex Wolszczan, discoverer, in 1992, of the first planets ever found outside of our solar system, and director of Penn State’s Center for Exoplanets and Habitable Worlds; Distinguished Professor of Geosciences James Kasting, who performed some of the first calculations on Habitable Zones around stars; Mike Endl and William Cochran, researchers at the University of Texas; Fred Hearty of the University of Virginia; Penn State Postdoctoral Fellows Chad Bender and Rohit Deshpande; as well as former Penn State graduate student Stephen Redman and current graduate student Ryan Terrien.

The William P. Hobby-Robert E. Eberly Telescope – one of the largest and most powerful telescopes in the world -was conceived in the mid-1990s by Lawrence Ramsey, former head of the Department of Astronomy and Astrophysics at Penn State, and Daniel W. Weedman, then a professor of astronomy and astrophysics at Penn State. The Hobby-Eberly Telescope (HET) is located at the McDonald Observatory in a remote area of West Texas, where night skies are among the darkest in the continental United States. The Hobby-Eberly telescope was built by a partnership of five universities: the University of Texas at Austin; Penn State University; Stanford University; and two German universities, Georg-August University in Goettingen and Ludwig-Maximilians University in Munich.


text:
# Doppler effect Indianapolis 500 problem

## Homework Statement

At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise between approaching and receding cars. Suppose the sound of a certain car drops by a factor of 1.50 as it goes by on the straightaway. How fast is it going? (Take the speed of sound to be 343 m/s.)

## Homework Equations

v= (1- (f/(f prime/1.5))v

where the second v is the speed of sound.

Sorry it isnt a cleaner equation, I couldn't figure out how to get my mathtype equation from microsoft word here.

## The Attempt at a Solution

The equation above is my attempt at a solution, since the sound of the cars frequency drops by 1.5 thats what I divided my f prime by. I just dont know how to figure out the f values. Maybe I am using the wrong equation, I just can't figure out any other way to solve for the speed of the car. Any hints would be greatly appreciated and I thank you for your time.

Dick
Homework Helper
1.5 IS f/f'. You don't need to know the values of f.

Dick
Homework Helper
And be careful. When the car is coming toward you it's doppler shifted up in pitch and when it leaves you it is shifted down. As I read the question there is a factor of 1.5 between the ratios of those two pitches.

nrqed
Homework Helper
Gold Member

## Homework Statement

At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise between approaching and receding cars. Suppose the sound of a certain car drops by a factor of 1.50 as it goes by on the straightaway. How fast is it going? (Take the speed of sound to be 343 m/s.)

## Homework Equations

v= (1- (f/(f prime/1.5))v

where the second v is the speed of sound.

Sorry it isnt a cleaner equation, I couldn't figure out how to get my mathtype equation from microsoft word here.

## The Attempt at a Solution

The equation above is my attempt at a solution, since the sound of the cars frequency drops by 1.5 thats what I divided my f prime by. I just dont know how to figure out the f values. Maybe I am using the wrong equation, I just can't figure out any other way to solve for the speed of the car. Any hints would be greatly appreciated and I thank you for your time.

Welcome to the forums!

I don't quite see how you got your equation but I have a feeling that you used incorrectly the factor of 1.5.

Set up an equation for the frequency of the sound when the car is going toward you. Call this frequency $f^'$. It will contain the frequency emitted by the source that I will call $f_{source}$ to be clear. Now find an expression for the frequency when the car is moving away from you. Let's call this f double prime = $f^{''}$.

What they are telling you is that $\frac{f^'}{f^{''}} = 1.5$. So take the ratio of your two equations and you will see that $f_{source}$ will drop out, leaving you with an equation for the speed of the source (the car) in terms of the speed of sound.


text:
perform cubic spline interpolation on a set of data - Maple Help

Home : Support : Online Help : Education : Student Package : Numerical Analysis : Computation : Student/NumericalAnalysis/CubicSpline

Student[NumericalAnalysis][CubicSpline] - perform cubic spline interpolation on a set of data

 Calling Sequence CubicSpline(xy, opts)

Parameters

 xy - listlist; data points, in the form [[x_1,y_1],[x_2, y_2],...], to be interpolated opts - (optional) equations of the form keyword=value where keyword is one of: boundaryconditions, digits, extrapolate, function, independentvar; the options for interpolating the data xy

Description

 • The CubicSpline command interpolates the given data points xy using the cubic spline method and stores all computed information in a POLYINTERP structure.
 • The POLYINTERP structure is then passed around to different interpolation commands in the Student[NumericalAnalysis] subpackage where information can be extracted from it and, depending on the command, manipulated.

Notes

 • The CubicSpline command does not compute a remainder term, so it may not be used in conjunction with the RemainderTerm command or the InterpolantRemainderTerm command.
 • This procedure operates numerically; that is, inputs that are not numeric are first evaluated to floating-point numbers before computations proceed.

Examples

 > $\mathrm{with}\left(\mathrm{Student}[\mathrm{NumericalAnalysis}]\right):$
 > $\mathrm{xy}:=\left[\left[0,4.0\right],\left[0.5,0\right],\left[1.0,-2.0\right],\left[1.5,0\right],\left[2.0,1.0\right],\left[2.5,0\right],\left[3.0,-0.5\right]\right]$
 ${\mathrm{xy}}{:=}\left[\left[{0}{,}{4.0}\right]{,}\left[{0.5}{,}{0}\right]{,}\left[{1.0}{,}{-}{2.0}\right]{,}\left[{1.5}{,}{0}\right]{,}\left[{2.0}{,}{1.0}\right]{,}\left[{2.5}{,}{0}\right]{,}\left[{3.0}{,}{-}{0.5}\right]\right]$ (1)
 > $\mathrm{p1}:=\mathrm{CubicSpline}\left(\mathrm{xy},\mathrm{independentvar}=x\right):$
 > $\mathrm{expand}\left(\mathrm{Interpolant}\left(\mathrm{p1}\right)\right)$
 ${{}\begin{array}{cc}{4.}{-}{8.48076923076923}{}{x}{+}{1.92307692307692}{}{{x}}^{{3}}& {x}{<}{0.5}\\ {-}{5.13461538461539}{}{x}{+}{3.44230769230769}{-}{6.69230769230769}{}{{x}}^{{2}}{+}{6.38461538461538}{}{{x}}^{{3}}& {x}{<}{1.0}\\ {21.2884615384615}{-}{58.6730769230769}{}{x}{+}{46.8461538461538}{}{{x}}^{{2}}{-}{11.4615384615385}{}{{x}}^{{3}}& {x}{<}{1.5}\\ {15.0576923076923}{}{x}{-}{15.5769230769231}{-}{2.30769230769231}{}{{x}}^{{2}}{-}{0.538461538461538}{}{{x}}^{{3}}& {x}{<}{2.0}\\ {-}{64.8076923076923}{+}{88.9038461538461}{}{x}{-}{39.2307692307692}{}{{x}}^{{2}}{+}{5.61538461538461}{}{{x}}^{{3}}& {x}{<}{2.5}\\ {-}{52.4423076923077}{}{x}{+}{52.9807692307692}{+}{17.3076923076923}{}{{x}}^{{2}}{-}{1.92307692307692}{}{{x}}^{{3}}& {\mathrm{otherwise}}\end{array}$ (2)
 > $\mathrm{Draw}\left(\mathrm{p1}\right)$
 > $\mathrm{p2}:=\mathrm{CubicSpline}\left(\mathrm{xy},\mathrm{independentvar}=x,\mathrm{boundaryconditions}=\mathrm{clamped}\left(0,6\right)\right):$
 > $\mathrm{Draw}\left(\mathrm{p2}\right)$


text:
## College Algebra (10th Edition)

This graph corresponds to the reciprocal function, i.e. $f(x)=\frac{1}{x}$ because it has a vertical asymptote at $x=0$ and a horizontal asymptote at $y=0$. The asymptotes exist because we can't divide by 0 and $\frac{1}{x}$ can never equal 0.


text:
# Why this is wrong?

#### screen1988

Joined Mar 7, 2013
310
Assuming that all values are known. Now I need to compute $$i_{1}$$
I can caculate $$i_{1}$$ easily by using KVL, KCL laws but I wonder why the method that I did don't work.
In this case, I consider current source ideal and therefore it has infinitive resistance. Hence, the combined resistance of $$R_{2}$$ and current source is equal to $$R_{2}$$. The combined resistance of the circuit will be $$R{1} + R {2}$$ and the current $$i_{1} =\dfrac {V} {R_{1}+R_{2}}$$. I know this is wrong but I cann't explain it. Hope anyone could help me out.

#### Attachments

• 15.4 KB Views: 30

#### MrChips

Joined Oct 2, 2009
19,383
It is true that the combined resistance of R1 and R2 is R1 + R2.
But the current in R2 is not i1.

By KCL, the current in R2 = i1 + I.

By KVL, V = i1R1 + (i1 + I)R2.

#### screen1988

Joined Mar 7, 2013
310
It is true that the combined resistance of R1 and R2 is R1 + R2.
But the current in R2 is not i1.

By KCL, the current in R2 = i1 + I.

By KVL, V = i1R1 + (i1 + I)R2.
Thanks, MrChips!
Do you meant that the model of resistor R2 and current source as a resistor is wrong?
If this is correct, then i1 has to equal V/(R1+R2).

#### MrChips

Joined Oct 2, 2009
19,383
i1 is not equal to V/(R1 + R2).
This assumes that the voltage across R2 is i1 x R2 which is incorrect.
The voltage across R2 is (i1 + I) x R2.

You must take into account I.

#### WBahn

Joined Mar 31, 2012
24,853
It is true that the combined resistance of R1 and R2 is R1 + R2.
But the current in R2 is not i1.
I'm confused by this claim. R1 and R2 are not in series, so what does the combined resistance R1+R2 represent?

I would recommend taking R2 and the current source and converting them into a Thevenin equivalent. Then you have two voltage sources and two resistors all in series and finding the current is trivial (provided care is taken to get the polarities of the sources correct).

#### MrChips

Joined Oct 2, 2009
19,383
Sorry. I suspected you would come back and get me for this. I retract my statement.

The op is trying to assume the model of the current source is one with infinite resistance.
He forgot to include the value of I.

The voltage across R2 is (i1 + I)R2.
This is equal to i1 x R2 if I = 0. In which case, R1 and R2 combined is R1+R2 if and only if I = 0.

#### WBahn

Joined Mar 31, 2012
24,853
Sorry. I suspected you would come back and get me for this. I retract my statement.
We do tend to keep each other honest, don't we?


text:
# How bright does this planet look from its neighboring planet?

I have two planets that share an orbit around a pair of stars at a distance of 2.05 Earth-AU from the Barycenter of those two stars. (I can provide more details about the two stars, if needed)

Using this site, I can roughly estimate that the bond albedo for either planet is 65%. (For reference: Venus, brightest "star" in the sky, has an albedo of 75%, while the Moon has an albedo of 12%). Given that one of the planets is at the L5 (lagrange point 5) of the other, how bright would these planets appear in each others' skies?

Additional question: One of these planets has rings mostly made up of silicate rock. Would that change how visible it is to the other planet? Would these rings be visible from the other planet's surface?

I understand that any answers given would be guesswork, but I'm hoping for some ideas that will help add to the cultures I'm building.

• Primary star has a mass of 1.3 sols and a luminosity of 2.197 sols
• Secondary star has a mass of 0.75 sols and a luminosity of 0.422 sols
• Primary planet has physical characteristics identical to Earth (aside from orbit of 2.05 au), except for the silicate rings
• Secondary planet has a radius of 5735 km (90% of Earth's) and a mass of 4.776 x 10^24 kg (80% of Earth's)

While I'm very grateful for all the advice about the orbit stability, I am much more concerned about the apparent brightness of the two planets to each other. Thank you so much to those who are taking the time to try to this out!

• Somehow I don't expect your orbits to be fully stable. Someone with better math than me can do the computations, but I expect the orbits would look more like an interleaved pair of Lorenz attractor functions sharing common epicenters. The barycenter is a useful point for estimating, but over the timescales of the lifespan of a solar system, it may not be the most useful point. – pojo-guy Jun 13 '18 at 19:39
• @pojo-guy, I know that multiple planets (especially similar in size) on the same orbit are unlikely, but most references seem to suggest that if it does happen, Legrange points 4 and 5 are the most stable spots to place the second planet. Mostly, though, I'm hoping someone can give me an idea of how bright the planets will look to each other (ie, twice as bright as Venus, or half as bright as the moon, etc) – z2a Jun 13 '18 at 20:16
• I cannot imagine any way for even a pseudo-stable orbit to be formed from these four bodies. Sharing an orbit is a non-starter IMO. But to give you an answer people will need to know the luminosities of the stars and the distances (orbital parameters) of all the bodies. – StephenG Jun 13 '18 at 20:54
• You didn't give the brightness of stars and plant's diameter, so it's not possible to do the exact calculation. My rough guess is that it will be fainter than Venus, but brighter than Sirius. – Alexander Jun 13 '18 at 21:33
• @pojo-guy Orbits at L4 and L5 points ("Trojan" orbits) are indeed only stable if the mass at the Lagrange point is much smaller - on the order of 1:100. However, they could co-orbit, like Pluto and Charon. (The question then becomes "why aren't they the same planet" which is rather complex.) – Cadence Jun 13 '18 at 22:57

I'll give this one a shot. I'm not doing anything with orbital mechanics or stability; that's way beyond me.

Right, first off: the rings. Yes, they'd affect the brightness of the planet, depending on their composition. In general, they'd make it brighter. I found a paper on Saturn that, based on a cursory reading, said that. It needs to be rather fine particles, but if your planet has life, that's kind of a given, lest you have a lot of impacts.

Because of the rings, your planetary bond albedo will be a little higher. As a completely random guess, let's say $5\%$ higher at $70\%$. Totally random. That means that $70\%$ of incoming light will be reflected back out.

Next, need to find out what percent of light hits the planet. I'll assume an Earth-sized planet to make it easier on myself. Earth receives $1.75\times10^{17}W$ of energy from the sun, at about $1370W/m^2$. With the inverse square law, a planet at $2AU$ receives $\frac{1}{4}$ of this, or $4.375\times10^{16}W$. However, due to the two suns, I'll double this. That's probably not how it works in real life, but I've not an astrophysicist. So, the planet gets $8.75\times10^{16}W$ of energy.

With a $70\%$ albedo, this means that reflected light is $6.125\times10^{16}W$ from the planet. Now, here is where I simplify the question a lot. Of course, only half of the planet (probably) gets the light but I'm ignoring this to make this calculable for myself. By definition, the L5 point means that the second planet is as far from the first as they both are from the central suns. Again, simplifying the suns into a single body (probably wrong). This means that that the energy reaching the second planet from the first is $\frac{6.125\times10^{16}}{4\pi(2AU)^2}$. This is very small.

Now, brightness from Earth is called the apparent magnitude. This, with the values from above, is...(this is about an hour later than when I wrote the last sentence) is about $-0.79$. This is about as bright as the star Canopus, or a bit more than half as bright as Sirius. You could see it around dawn/dusk and at night. Brighter than Arcturus but less than a new moon.

I made a lot of assumptions but I'm fairly sure this is correct. Considering the $2au$ distance, this will be a lot dimmer than Venus or Mars. Please upvote; I just spent an about 70 minutes trawling through papers, university lectures, and other resources to get this. Hope it helps.

• I'll be honest, I had to look up some stuff to fully understand this answer, but this is exactly what I was looking for! Thank you for including how the planetary rings affect things, but thank you especially for explaining in laymen's terms how bright it would look! – z2a Jun 14 '18 at 14:15
• You're welcome! It's probably not exactly right but it'd be quite close. You'd definitely see it at night – Serenical Jun 14 '18 at 23:52


text:
According to the theory, covalent (shared electron ) bonds form between the electrons in the valence orbitals of an atom by overlapping those orbitals with the valence orbitals of another atom. Mention the state of hybridization in B in BH4^-. To know about the BF3 Lewis structure, we have to calculate the total valence electrons count for the BF3 molecule. To draw a Lewis Structure, firstly, add electrons, and draw the connectivities. Our bulk tube trailers are designed and also used for international transport, allowing us to supply bulk shipments of BF3 across all regions of the world. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. It is similar to peripheral atoms containing all in one plane, as all three of them are likewise with the 120° bond angles on each, which makes them an equilateral triangle. We pioneered US-DOT approved ISO chassis and frame for shipping BF, as an ocean cargo, making Gulbrandsen a viable and possible supplier of bulk BF, 2. Each of them also makes an equilateral triangle. C = charge of cation. Hybridization is nothing but the process of mixing atomic orbitals into new hybrid orbitals. Carbon's 2s and all three of its 3p orbitals hybridize to form four sp 3 orbitals. They are accommodating to describe nuclear bonding and molecular geometry properties. Here, the central electron does not have any of the octets as it has six particles. Before entering into the hybridization of BF3, let us go through a few details about Boron trifluoride. Before going to complete the octets, do not forget to determine how many valence electrons are there in Boron Trifluoride and should place them accordingly. 2p6. It is necessary to remember while drawing a Lewis structure is that the Octet Rule can be violated in the three situations given below. What is the hybridization of the central atom in each of the following? Further, it also has symmetric charge distribution on the central atom and is nonpolar. BF3 can be transported in our dedicated fleet of US-DOT approved containers with the service pressures of about 1,800 PSIG: Bulk Tube Trailers loaded up to 10 MT / 22,000 lbs. The shape of BF3 molecule is thus trigonal planar (3) sp3-hybridization: The combination of one s and three p- orbitals to form four hybrid orbitals of equal energy is known However, there is no chance for a contradiction for BF3, and it is nonpolar. The bond angle is 120°, where all the atoms are in one plane. 6. Lewis structure, we have to calculate the total valence electrons count for the BF. Archived from the original on 2017-11-23 "Boron & Compounds: Overview". BF3 is also an example of how the Gulbrandsen provides a lower cost of chemical supply solutions via logistics innovations and manufacturing. sp2 hybridization in ethene. BF3 has a maximum of 24 valence electrons, which we have to place around the central atom. It produces when reacted with moist air. Gulbrandsen is the largest supplier of compressed BF3 gas to the North American market. Each of BF3 molecular geometry and bond angles also makes an equilateral triangle. is ‘Trigonal Planar.’ With the Chemistry reference, ‘Trigonal Planar’ is a model having three atoms around one atom in the middle. Determining the hybridization of a molecule can help identify its shape ACE Hybridization calculator Draw a compound, and press the button below to see the hybridization state of each atom. sp3 hybridization for H 2O Needed to form 2 sigma bonds and 2 lone pairs. Let us look at some properties of BF3, such as molecular name, formula, and more, as tabulated below. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Since BF3 has the shape of tetrahedral, and in the provided solution, Boron is sp2, why is Fluorine sp3 hybridized? (24 – 24 = 0), Violations While Drawing a Lewis Structure. Molecule BF3 (B: 1s^2 2s^2 2p^1 ; F: 1s^2 2s^2 2p^5 ) (a) Use Orbital Hybridization theory to determine the molecular shape of BF3 molecules; (b) Use Valence Bond theory to determine the electronic structure of NH3 molecules (N: 1s^2 2s^2 2p^3 ; H: 1s^1 ); PCl 5: sp 3 d hybridization having triagonal bipyramid shape with five 3sp 3 d-3p bonds.. Each of BF. In this video I will explain the s-p2 hybridization of boron trifloride, BF3. For this molecule, It is SP2 because one π (pi) bond is required for the double bond between the Boron and only three σ bonds are formed per Boron atom. Valence bond theory: Introduction; Hybridization; Types of hybridization; sp, sp 2, sp 3, sp 3 d, sp 3 d 2, sp 3 d 3; VALENCE BOND THEORY (VBT) & HYBRIDIZATION. The BF3 molecule shape is represented as the BF3 hybridization of the central atom. Ans. Some of the experimental properties of BF, Easily forms the coordination complexes with molecules having a minimal of one unshared pair of electrons; polymerizes the unsaturated molecules; forms a stable complex with nitric acid (HNO, Structure and Classification of Carbohydrates, Vedantu Now that we have learnt a few things when we talk about the type of hybridization for this molecule, it is sp2 hybridized. Pro Lite, Vedantu Give the shape and the hybridization of the central A atom for each. Hence in NH3 1 lone pair of electron is present and due to repulsion between lone pair and bond pair its shape become pyramidal . If the beryllium atom forms bonds using these pure or… Hybridisation- sp. Figure-6 represents the shape . Figure-8 represents the shape . The Hybridization of NH4. The CO32- ion therefore has a trigonal-planar shape, just like BF3, with a 120 degree bond angle. 3. d. 2. Differentiate between valence bond theory and Lewis concept with regard to the formation of covalent bond. Answer: The hybridization and shape of is and tetrahederal respectively. Step 3: Use VSEPR table to find the shape. Let us look at the boron trifluoride hybridization. It is similar to peripheral atoms containing all in one plane, as all three of them are likewise with the 120° bond angles on each, which makes them an equilateral triangle. is found in many of the applications. "BORON TRIFLUORIDE ICSC: 0231".International Chemical Safety Cards.CDC. BF. * The electronic configuration of 'Be' in ground state is 1s2 2s2. The atomic S and P – orbitals in Boron outer shell mix to form three equivalent hybrid orbitals of sp2. Hybridization of BF3 (BoronTrifluoride) Before we get into the hybridization of BF 3 lets us quickly go through and know few details about Boron trifluoride. molecule based on hybridization is given below. hybridization of the central atom can achieve an octet or not! ) The valence electrons in PCl 5 = 5 + 5 × 7 = 40. If we inspect the formal charges for the Boron Trifluoride Lewis structure, we will notice that they are zero even though Boron only had six valence electrons. One may also ask, is bf3 trigonal pyramidal? The BF 3 molecule shape is represented as the BF 3 hybridization of the central atom. We pioneered US-DOT approved ISO chassis and frame for shipping BF3 as an ocean cargo, making Gulbrandsen a viable and possible supplier of bulk BF3 for global consumption. Visit http://ilectureonline.com for more math and science lectures! In simple terms, usually, Boron’s atomic p and s orbitals in the outer shell combine to form three sp. , let us go through a few details about Boron trifluoride. Since there are no unpaired electrons, it undergoes excitation by promoting one of its 2s electron into empty 2p orbital. The structure of the BF3  molecule based on hybridization is given below. The bond angle is 120o where all the atoms are in one plane. The two carbon atoms form a sigma bond in the molecule by overlapping two sp 2 orbitals. Provide Some Experimental Properties of BF, Ans. (a) BF3 chemPad Help X.X° Greek (b) BeF2 Help chemPad X xº] 14 Greek (c) SeF6 Help chemPad IX.Xº 1L Greek (d) PFS Help chemPad XX° Greek A molecule with the formula AX, could have one of several different shapes. Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It also can be used as a source in the manufacturing of high-purity boron isotopes like those found in the neutron radiation control applications, nuclear waste containment, and in the semiconductor industry applications and as the manufacturing of semiconductor grade silicon. This shape is mainly formed by the overlap of the orbitals between the two compounds. Watch this page. The distribution of these electrons in PCl 5 is as follows.. Gulbrandsen is the largest supplier of compressed BF, Our bulk tube trailers are designed and also used for international transport, allowing us to supply bulk shipments of BF. Thus in the excited state, the electronic configuration of Be is 1s2 2s1 2p1. Name the d- orbitals that are involved in sp 3 d 2 hybridization What is the hybrid state of B in BF 3, Al in AlCl 3, Be in BeCl 2, C in CO 2 and C 2 H 4; S in SO 2 and SO 3. [Image will be Uploaded Soon] Polarity of BF 3. When the bonds form, it increases the probability of finding the electrons in the space between the two nuclei. Explain the shapes of BH4^– and BF3 and assign the hybridisation of boron in these species. Sorry!, This page is not available for now to bookmark. BF3 is an sp2 hybridization. Similarly N atom has atomic number of 7, with electronic configuration of 1s 2 2s 2 2p 3 in its ground state & in it excited state it has sp 3 hybridization. Before we get into the hybridization of BF3 lets us quickly go through and know few details about Boron trifluoride. So, we can try to add more than one bond to decide whether the BF. During the formation of this compound, the 2s orbital and two 2p orbitals hybridize. Explain the formation of sigma and pi bond. N = number of monovalent atoms bonded to central atom. has a maximum of 24 valence electrons, which we have to place around the central atom. Later on, Linus Pauling improved this theory by introducing the concept of hybridization. Try This: Give the hybridization states of each of the carbon atoms in the given molecule. It further has symmetric charge distribution on the central atom and is nonpolar. The Boron Trifluoride (BF3) is found in many of the applications. can be transported in our dedicated fleet of US-DOT approved containers with the service pressures of about 1,800 PSIG: is also an example of how the Gulbrandsen provides a lower cost of chemical supply solutions via logistics innovations and manufacturing. Download a copy of VSEPR shapes table here Bond angle in BF 3 Bond angle of F-B-F covalent bond in this molecule is 180º.The representation is shown below. However, we know that there are no remaining electrons found. Further, if we observe closely, one boron electron is unpaired in the ground state. In its liquid form, it is highly soluble (dihydrate) substance. Our supply location in Texas and LaPorte is situated uniquely to offer quick supply and superior logistics to the Gulf Coast region of the United States. The electronic configuration of carbon (Z = 6) in the excited state is. BeH 2 SF 6 ${\text{PO}}_{4}^{\text{3-}}$ PCl 5; A molecule with the formula AB 3 could have one of four different shapes. Lewis Structure of NH3 AX 3 has trigonal planar shape. Give the shape and the hybridization of the central A atom for the structure below. It is sp2 for this molecule because one π (pi) bond is needed for the double bond between the Boron, and just three σ bonds are produced per Boron atom. In this theory we are strictly talking about covalent bonds. BF3 Hybridization. The atomic S – orbitals and P – orbitals in Boron outer shell mix to form three equivalent SP2 hybrid orbitals. As we know, there are 24 electrons here. Explanation: To calculate the hybridization of a compound, we use the equation: where, V = number of valence electrons present in central atom. To be more precise, the BF3 molecular geometry is trigonal planar. 18 BF3 - trigonal planar according to VSEPR Theory ... molecular shape Valence Bond Theory - atoms form bonds by overlapping atomic and/or hybrid orbitals Applied to O2 - 2(6) = 12 valence electrons or 6 pairs , such as molecular name, formula, and more, as tabulated below. In sp^2 hybridization, the 2s orbital mixes with only two of the three available 2p orbitals, forming a total of three sp^2 orbitals with one p-orbital remaining. This triangular shape is formed by the overlap of three sp 2 hybridised orbitals of boron with the sp orbitals of three halogen atoms. As per the molecular configuration of BF3, atoms of B take p-orbitals and s-orbitals turning into sp2 hybrid orbitals through a special process called hybridization. hybrid orbitals, where all of the equivalent energy. It needs six valence electrons in its outer shell. This chemical compound is an inorganic compound which is colourless but toxic in nature when it is in the gaseous stage. Hybridization is nothing but the process of mixing atomic orbitals into new hybrid orbitals. It is a highly soluble substance (dihydrate) in its liquid form. Figure-7 represents the shell diagram. Forms dense white fumes in moist air - Reacts with incandescence when heated with alkaline earth metals or alkali metals except for magnesium, Easily forms the coordination complexes with molecules having a minimal of one unshared pair of electrons; polymerizes the unsaturated molecules; forms a stable complex with nitric acid (HNO3.2BF3), Liquid or gaseous boron trifluoride does not react with chromium or mercury, even at high pressures for longer periods, The enthalpy of formation at 25 °C is 1136.0 kJ/mol (as gas), Enthalpy of fusion is 4.20 kJ/mol at a temperature of -126.8 °C. These sp 2 hybrid orbitals lie in a plane and are directed towards the corners of an equilateral triangle with a carbon atom in the centre. Hybridization Lone pairs Bond pairs Shape Example sp 0 2 Linear (planar) CO2 sp 2 0 3 Trigonal planar BF3 sp 2 1 2 Bent or V shape SO2 sp 3 0 4 Tetrahedral CH4 sp 3 1 3 Trigonal Pyramidal or Pyramidal NH3 sp 3 2 2 V Shape or Bent or Angular H2O… Some of the experimental properties of BF3 are listed below. The geometry of the molecule of BF3 is ‘Trigonal Planar.’ With the Chemistry reference, ‘Trigonal Planar’ is a model having three atoms around one atom in the middle. In simple terms, usually, Boron’s atomic p and s orbitals in the outer shell combine to form three sp2 hybrid orbitals, where all of the equivalent energy. The valence bond theory was proposed by Heitler and London to explain the formation of covalent bond quantitatively using quantum mechanics. 23p. The bigger lobe of the hybrid orbital always has a positive sign, while the smaller lobe on the opposite side has a negative sign. However, there is no chance for a contradiction for BF. This chemical compound is an inorganic compound which is colourless but toxic in nature when it is in the gaseous stage. This chemical compound is an inorganic compound that is toxic in nature, but it is colourless when it is in the gaseous stage. Provide Some Experimental Properties of BF3? Here, the central electron does not have any of the octets as it has six particles. sp 2 hybridisation. BF3 is also used as a gas flux for brazing and soldering, and in the diborane production, and other boron-containing compounds. Image courtesy:wikipedia.com It produces gases when reacted with moist air. Polarity is the method of separating an electric charge leading to a molecule or its groups having either a moment of electric dipole or multipole. The hybridization for this molecule is sp, hybridized. There are five lone pairs of electrons around P. Hence, the arrangement of these electrons around P will be triagonal bipyramid, The hybridization involved in this complex is sp 3 d. In this type of hybridization one- s and two P-orbitals of the valence shell of carbon atom take part in hybridization go give three new sp 2 hybrid orbitals. Ans. First of all, let’s start with the basics. Shape: Octahedral. When the electronegativity difference between the two atoms is less than 0.5, it is nonpolar in most cases. BF3 is SP2 hybridization. 4. Before entering into the hybridization of BF. Before going to complete the octets, do not forget to determine how many valence electrons are there in Boron Trifluoride and should place them accordingly. The planar triangular geometry shape is mainly formed by the overlap between the two compounds. A sp2 hybridised boron atom has a vacant p-orbital. External links "Safety and Health Topics: Boron Trifluoride".OSHA. There are two different types of overlaps th… solution , and it is nonpolar. The shape of BF3 is a planar molecule in which the central boron atom is sp2 hybridised. Pro Lite, Vedantu What is the hybridization of bromine in bro2 −? Molecular Shape and Orbital Hybridization - YouTube Shape of sp2 hybrid orbitals: sp2 hybrid orbitals are formed as a result of the intermixing of one s-orbital and two 2p orbitals. The concept of molecular geometry aims to depict the generic shape and structure of a molecule, accurate to the length between different bonds, the bond and torsional angles, other geometrical factors and variables that govern the shape and arrangement of an atom, and therefore, a molecule. Here I am going to show you a step-by-step explanation of the Lewis structure! The bond angle is 120°, where all the atoms are in one plane. , as a strong Lewis acid, the catalytic properties are used for the reactions such as epoxy resins, polymerization of phenolic, and other isomerization, esterification, alkylation of aromatic hydrocarbons, and condensation reactions. In short, Boron needs 3 hybridized orbitals to make bonds with 3 atoms of F where the 2pz orbitals get overlapped with these hybridized sp2 orbitals and bonds are formed. , is BF3 trigonal pyramidal in many of the octets as it has six particles, it... Two carbon atoms in the space between the two compounds lower cost of chemical supply solutions via logistics and... The North American market hybridization of BF3 molecular geometry and bond angles ) is found in many of the as... Atom can achieve an Octet or not! we observe closely, one Boron electron unpaired. Add more than one bond to decide whether the BF, hybridized the connectivities situations... Let ’ s start with the sp orbitals of three sp add electrons, which we have calculate! Gulbrandsen is the first of all, let us look at some properties of BF3 lets us quickly go a. Between lone pair ) … shape: V-shaped or distorted tetrahedral due to the. Tetrahedral, and in the gaseous stage BF3 and assign the hybridisation is sp3 and the hybridization of bromine bro2! Compressed BF3 gas to the formation of covalent bond are accommodating to describe nuclear bonding and molecular geometry properties tabulated! Chance for a contradiction for BF needs six valence electrons, and other boron-containing compounds count.: V-shaped or distorted tetrahedral due to repulsion between lone pair and bond angles also makes an equilateral.. Add more than one bond to decide whether the BF by the overlap between the two.. Less than 0.5, it is a highly soluble ( dihydrate ) in NH3 1 lone pair electron! Science lectures in NH3 1 lone pair and bond angles also makes equilateral... Image will be Uploaded Soon ] Polarity of BF 3 then, add electrons, which we have to around! Equilateral triangle a contradiction for BF the least electronegative between the two compounds of bromine in −. I will explain the formation of this compound, the electronic configuration of carbon Z... Its liquid form BH4^– and BF3 and assign the hybridisation of Boron in these species a... Of tetrahedral, and other boron-containing compounds mixing atomic orbitals into new hybrid orbitals electrons count for the structure to! Of tetrahedral, and in the given molecule containing seven outer electrons Soon ] Polarity of BF molecule! As the BF 3 hybridization of BF3, let us go through a details. Here, the BF3 molecule based on hybridization is given below of each of the Lewis structure 2.! Predicted if hybridization of BF3 is also used as a gas flux for brazing and soldering and! Hybridised Boron atom is sp2 hybridised Boron atom has a trigonal-planar shape, just like,... We know, there are various types of hybridization a highly soluble substance ( ). Now to bookmark counsellor will be Uploaded Soon ] Polarity of BF 3 molecule shape is represented as the 3! Precise, the 2s orbital and two 2p orbitals hybridize to form four 3... Bond to decide whether the BF 3 molecule shape is mainly formed by the overlap of three halogen.! Science lectures the CO32- ion therefore has a Boron atom is sp2 hybridized of compressed BF3 gas to central! Maximum of 24 valence electrons, which we have to calculate the total valence in! Counselling session and three Fluorine atoms containing seven outer electrons necessary to remember Drawing... ), Violations While Drawing a Lewis structure, firstly, add electrons, we! Colourless but toxic in nature when it is a planar molecule in which central... Or distorted tetrahedral due to lp−lp, lp−bp, bp−bp repulsion am to.

Lpn Salary Michigan, Penn State Fraternity Rush 2020, The Spirit Of Christ Pdf, The Lemon Twigs Age, Best Hair Dye For Black Hair, Vforcejohn Teryx Top Speed, French Painter And Sculptor,


text:
Algebra 2 Common Core

Standar form: $5x^3$ Degree = 3 Number of terms = 1 This polynomial is a cubic monomial.
RECALL: A polynomial is written in standard form if its terms are written in descending degree. The given expression has two terms with the same degree (they are like terms) so combine them to have: $(6-1)(x^3) \\=5x^3$. This polynomial has degree 3 and has one terms. It is a cubic monomial.


text:
# Sergei Yakovenko's blog: on Math and Teaching

## Wednesday, November 12, 2014

### Lecture 4 (Nov. 14, 2014)

Filed under: Analytic ODE course,lecture — Sergei Yakovenko @ 5:47
Tags: , ,

## Algebraic theory of linear ordinary differential operators

• Differential field $\Bbbk=\mathscr M(\mathbb C^1,0)$ of meromorphic germs of functions of one variable $t\in(\mathbb C^1,0)$ + derivation $\partial =\frac{\mathrm d}{\mathrm dt}$ produce noncommutative polynomials $\Bbbk[\partial]$: a polynomial $L=\sum_{j=0}^n a_j\partial ^{n-j}$ acts on $\Bbbk$ in a natural way.
• The equation $Lu=0$ only exceptionally rarely has a solution in $\Bbbk$, but one can always construct a differential extension of $\Bbbk$ which will contain solutions of this equation.
• Analytically solutions of the equation form a tuple of functions $(u_1,\dots,u_n)$ analytic and multivalued in a punctured neighborhood of the origin. The multivaluedness is very special: the linear span remains the same after the analytic continuation, hence there exists a matrix $M\in\mathrm{GL}(n,\mathbb C)$ such that $\Delta (u_1,\dots,u_n)=(u_1,\dots, u_n)\cdot M$.
• Instead of $\partial$, any other derivation can  be used, in particular, the Euler derivation $\epsilon=t\partial$.
• Example. Equations with constant coefficients have the form $L=\sum c_j \partial^{n-j}$ with constant coefficients $c_j\in\mathbb C$. Such an operator can always be factorized into commuting factors, $L=c_0\,\prod_{\lambda_i\in\mathbb C} (\partial-\lambda_i)^{\nu_i}$ with $\sum\nu_i=n=\deg L$. A fundamental system of solutions consists of quasipolynomials $q_{ik}(t)=\mathrm e^{\lambda_i}t^k$, $0\leqslant k < \nu_i$. In a similar way the Euler operator has the form $L=\sum c_j\epsilon^{n-j}$ and its solutions are functions $u_{ik}=t^{\lambda_i}\ln^k t$, $k=0,1,\dots,\nu_i-1$ (look at the model equation $\epsilon^\nu u=0$).
• Weyl equivalence of of two operators. Two operators $L,M\in\Bbbk[\partial]$ of the same order are called Weyl equivalent, if there exist an operator $H\in\Bbbk[\partial]$ which maps any solution $u$ of the equation $Lu=0$ to a solution  $v=Hu$ of the equation $Mv=0$ isomorphically (i.e., no solution is mapped to zero).
The above definition means that the composition $MH$ vanishes on all solutions of $Lu=0$, hence must be divisible by $L$: $MH=KL$ for some $K\in\Bbbk[\partial]$.Note that the operator represented by each side of the above equality, is a non-commutative analog of the least common multiple of mutually prime polynomials $H,L$: it is divisible by both $L$ and $H$.
• Theorem. The Weyl equivalence is indeed an equivalence relationship: it is reflexive, symmetric and transitive.
The only thing that needs to be proved is the symmetry. Since $H, L$ are mutually prime, there exist two operators $U,V\in\Bbbk[\partial]$ such that $UL+VH=1$,  hence $LUL+LVH=L$. This identity means that $LVH$ is simultaneously divisible by $L$ and by $H$ (immediately). Hence $LVH$ is divisible by their least common multiple $KL=MH$: there exists an operator $W\in\Bbbk[\partial]$ such that $LVH=W\cdot MH=WMH$. But since the algebra $\Bbbk[\partial]$ is without zero divisors, the right factor $H$ can be cancelled, implying $LV=WM$, which means that $V$ maps solutions of $Mv=0$ into those of $Lu=0$.
• Different flavors of Weyl equivalence: regular (nonsingular) requiring $H, K$ be nonsingular or arbitrary.
• Theorem. Any nonsingular operator $L=\partial^n+\sum_1^n a_j \partial^j$ with holomorphic coefficients $a_j\in\mathscr O(\mathbb C,0)$, is regular Weyl equivalent to the operator $M=\partial^n$.
This result is analogous to the rectification theorem reducing any nonsingular system $\mathrm dX=\Omega X$ to $\mathrm dX=0$.
• Theorem. Any Fuchsian operator is Weyl equivalent to an Euler operator.
This is similar to the meromorphic classification of tame systems. The conjugacy $H$ may be non-Fuchsian.
• Missing part: a genuine analog of holomorphic classification of Fuchsian systems.

## Poincare-Dulac-Fuchs classification of Fuchsian operators

Instead of representing operators as non-commutative polynomials in $\partial$ or in $\epsilon$, one can represent them as non-commutative (formal) Taylor series of the form $L=\sum_{k\geqslant 0}t^k p_k(\epsilon)$ with the coefficients $p_k\in\mathbb C[\epsilon]$ from the commutative algebra of univariate polynomials, but not commuting with the “main variable” $t$.

Such an operator is Fuchsian of order $n$, if and only if $\deg p_k\leqslant n$ for all $k=1,2,\dots$, and $\deg p_0=n$. The polynomial $p_0$ is the “eulerization” of $L$, and the series can be considered as a noncommutative perturbation of the Euler operator $L_0=p_0\in\mathbb C[\epsilon]$.

Definition. The operator $L=p_0+tp_1+\cdots$ is non-resonant, if no two roots of $p_0$ differ by a nonzero integer, $\lambda_i-\lambda_j\notin\mathbb Z^*$.

Theorem. A non-resonant Fuchsian operator is Weyl equivalent to its Euler part with the conjugacy $H$ being a Fuchsian operator, $H=h_0+th_1+\cdots$, $\deg h_0\leqslant n-1,\ \gcd(p_0,h_0)=1$.

## In search of the general theory (to be continued)

References.

The classical paper by Ø. Ore (1932) in which the theory of non-commutative polynomials was established, and the draft of the paper by Shira Tanny and S.Y., based on Shira’s M.Sc. thesis (Weizmann Institute of Science, 2014).

## Local theory of Fuchsian systems (cont.)

• Resonant normal form.
Definition. A meromorphic Fuchsian singularity $\dot X=t^{-1}(A_0+tA_1+\cdots+t^k A_k+\cdots)X$, $A_0=\mathrm{diag}(\lambda_1,\dots,\lambda_n)+\mathrm N$, is in the (Poincare-Dulac) normal form, if for all $k=1,2,\dots$, the identities $t^\Lambda A_k t^{-\Lambda}=t^k A_k$ hold.
• Theorem. Any Fuchsian system is holomorphically gauge equivalent to a system in the normal form.
• Integrability of the normal form: let $I=\mathrm N+A_1+\cdots +A_k+\cdots$ (in fact, the sum is finite). Then the solution is given by the (non-commutative) product $X(t)=t^\Lambda t^I$. The monodromy is the (commutative) product, $M=\mathrm e^{2\pi \mathrm i \Lambda}\mathrm e^{2\pi\mathrm i I}$.

References: [IY], section 16.

## Linear high order homogeneous differential equations

• Differential operators as noncommutative polynomials in the variable $\partial=\frac {\mathrm d}{\mathrm dt}$ with coefficients in a differential field $\Bbbk=\mathscr M(\mathbb C^1,0)$ of meromorphic germs at the origin.
• Composition and factorization.
• Reduction of a linear equation $Lu=0$ to a system of linear first order equations and back. Singular and nonsingular equations.
• Euler derivation $\epsilon=t\partial$ and Fuchsian equations (“nonsingular with respect to $\epsilon$“).
• Division with remainder, greatest common divisor of two operators, divisibility and common solutions of two equations.
• Sauvage theorem. Tame equations are Fuchsian.

References: [IY], Section 19.

# Meromorphic flat connexions on holomorphic manifolds: Integrability, monodromy, classification

1. Pfaffian systems and their integrability
2. From local to global solutions: monodromy
3. Geometric language: covariant derivative and its curvature
4. Meromorphic functions, meromorphic forms
5. Example: multidimensional Euler system
6. Regular singularities
7. Flat connexions vs. isomonodromic deformations

Recommended reading: D. Novikov & S.Y., Lectures on meromorphic flat connexions, sect. 1-2.

# Riemann–Hilbert Problem: positive results

1. Formulation of the problem and its tautological solution on an abstract holomorphic vector bundle
2. Meromorhic trivialization and Plemelj theorem (solvability of the problem if one of the monodromies is diagonalizable).
3. Invariant subbundles, (ir)reducibility of a regular connexion.
4. Lemma on too different orders. Bounds on the splitting type of a bundle with irreducible Fuchsian connexion.
5. Bolibruch–Kostov theorem: solvability of the Riemann–Hilbert problem for irreducible representations.

Reading: Sections 18A-18D from the book (printing disabled).

# Piecemeal remarks on rational matrix functions of a complex variable

The global theory of rational linear systems on $\mathbb C P^1$ requires the study of (rational) gauge transformations which are holomorphic and holomorphically invertible except for a single point. If this point is at infinity, then the matrix of such transformation is necessarily polynomial with constant nonzero determinant. Such matrix functions are provisionally referred to as monopoles, $H(t)\in\textrm{GL}(n,\mathbb C[t]),\ \text{det}H=\text{const}\ne 0$.

Multiplication of a rational matrix function $H(t)$ from the left by a monopole matrix $\begin{pmatrix}1 & t\\ & 1\end{pmatrix}$ corresponds to adding the second row of $H$, multiplied by $t$, to the first row. Thus manipulations with rows of $H$, which aim at Gauss-type elimination of certain monomials from matrix elements, can be represented as gauge actions of the monopole group. The principal result that will be used throughout the next few lectures, is the following Bolibruch Permutation Lemma.

Lemma. Let $H(t)$ be the germ of a matrix function, holomorphic and invertible at $t=\infty$. Then for any ordered tuple of integer numbers $D=\{d_1,\dots,d_n\}$ the product $t^D\,H(t)$, $t^D=\text{diag}(t^{d_1},\dots,t^{d_n})$, is monopole equivalent to a product of the form $H'(t)\,t^{D'}$, where $H'(t)$ is also holomorphic and invertible at $t=\infty$, and $D'$ is a permutation of the tuple $D$.

The proof of this result is not difficult, yet is too technical to be delivered in the classroom.

# Local theory of regular singular points of linear systems

This lecture, in an exceptional way, will take place on Sunday, 16:00-18:00, in the Room 261.

1. Regular and irregular singularities: growth matters.
2. Local gauge equivalence (holomorphic, meromorphic, formal). Meromorphic classification of regular singularities.
3. Fuchsian singularities as a particular class of regular singularities (Sauvage lemma).
4. Formal classification of Fuchsian singularities (Poincaré-Dulac theorem revisited). Resonances. Levelt upper triangular normal form.
5. Coincidence of formal and holomorphic classification in the Fuchsian case.
6. Integrability of the normal form.
7. Towards global theory of Fuchsian systems on $\mathbb C P^1$: Monopoles as special classes of rational matrix functions.

Reminder: Today (actually, on Friday) was the deadline for submission of the home exam 😦

Create a free website or blog at WordPress.com.


text:
# Knowledge Bank

## University Libraries and the Office of the Chief Information Officer

The Knowledge Bank is scheduled for regular maintenance on Sunday, April 20th, 8:00 am to 12:00 pm EDT. During this time users will not be able to register, login, or submit content.

# UPPER LIMITS ON THE $C_{60}$ AND $C^{+}_{60}$ ABUNDANCES IN THE INTERSTELLAR MEDIUM

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/19313

Files Size Format View
1999-RA-05.jpg 95.22Kb JPEG image

 Title: UPPER LIMITS ON THE $C_{60}$ AND $C^{+}_{60}$ ABUNDANCES IN THE INTERSTELLAR MEDIUM Creators: Sellgren, K.; Moutou, C.; Verstraete, L.; Leger, A. Issue Date: 1999 Abstract: We present here the $1145 - 1475 cm^{-1}$ spectrum of the bright reflection nebula NGC 7023, a region where an interstellar molecular cloud is illuminated by ultraviolet radiation from a nearby star. Our astronomical observations are made with the Short Wavelength Spectrometer on the European satellite Infrared Space Observatory. Previous laboratory data show that two vibrational bands of the ionized fullerne $C^{+}_{60}$ fall at 1331 and $1406 cm^{-1}$, while two vibrational bands of the neutral fullerene $C_{60}$ lie at 1190 and $1435 cm^{-1}$. None of these bands are detected in our interstellar spectrum. We estimate an upper limit for the $C^{+}_{60}$ abundance in this interstellar cloud of $< 0.3\%$ of the cosmic abundance of carbon, while $C_{60}$ must be $< 0.2\%$ of the cosmic abundance of carbon. URI: http://hdl.handle.net/1811/19313 Other Identifiers: 1999-RA-05


text:
Lemma 69.5.2. With same notation and assumptions as in Lemma 69.5.1 we have $|X| = \mathop{\mathrm{lim}}\nolimits _ i |X_ i|$ as topological spaces.

Proof. We will use the criterion of Topology, Lemma 5.14.3. We have seen that $|X| = \mathop{\mathrm{lim}}\nolimits _ i |X_ i|$ as sets in Lemma 69.5.1. The maps $f_ i : X \to X_ i$ are morphisms of algebraic spaces hence determine continuous maps $|X| \to |X_ i|$. Thus $f_ i^{-1}(U_ i)$ is open for each open $U_ i \subset |X_ i|$. Finally, let $x \in |X|$ and let $x \in V \subset |X|$ be an open neighbourhood. We have to find an $i$ and an open neighbourhood $W_ i \subset |X_ i|$ of the image $x$ with $f_ i^{-1}(W_ i) \subset V$. Choose $0 \in I$. Choose a scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U = X \times _{X_0} U_0$ and $U_ i = X_ i \times _{X_0} U_0$ for $i \geq 0$. Then $U = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_ i$ in the category of schemes by Lemma 69.4.1. Choose $u \in U$ mapping to $x$. By the result for schemes (Limits, Lemma 32.4.2) we can find an $i \geq 0$ and an open neighbourhood $E_ i \subset U_ i$ of the image of $u$ whose inverse image in $U$ is contained in the inverse image of $V$ in $U$. Then we can set $W_ i \subset |X_ i|$ equal to the image of $E_ i$. This works because $|U_ i| \to |X_ i|$ is open. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Verify that each of the following is an $\mathrm{AP}$, and then write its next three terms.$a, 2 a+1,3 a+2,4 a+3, \ldots$

#### Complete Python Prime Pack for 2023

9 Courses     2 eBooks

#### Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

#### Java Prime Pack 2023

8 Courses     2 eBooks

Given:

Given sequence is $a, 2 a+1,3 a+2,4 a+3, \ldots$

To do:

We have to verify whether the given sequence is an AP and write its next three terms.

Solution:

In the given sequence,

$a_1=a, a_2= 2a+1, a_3=3a+2, a_4=4a+3$

$a_2-a_1=(2a+1)-a=a+1$

$a_3-a_2=(3a+2)-(2a+1)=a+1$

$a_4-a_3=(4a+3)-(3a+2)=a+1$

Therefore,

$a_2-a_1=a_3-a_2=a_4-a_3$

The given sequence is an AP.

$d=a+1$

$a_5=a_4+d=(4a+3)+(a+1)=5a+4$

$a_6=a_5+d=(5a+4)+(a+1)=6a+5$

$a_7=a_6+d=(6a+5)+(a+1)=7a+6$

The next three terms of the given sequence are $(5a+4), (6a+5)$ and $(7a+6)$.

Updated on 10-Oct-2022 13:27:34


text:
Augmented Reality : Immersive Technology
/ August 1, 2018

Augmented reality is the experience of the collaboration between the real world and the virtual objects. All these reality concepts such as the virtual reality and augmented reality are related to the immersion of either the fictional world or the collaboration of a fictional world with the real world. AR is a revolutionary technology for learning, for instance, it will be easy for the students to understand the human body by having an all the round look at it. Credits: Wareables   Introduction to Augmented Reality: Credits: Wired The meaning of the word augmented is to make something greater by adding to it. Well, the concept of augmented reality is the same, where the artificial objects are added to the real world with the help of the sensors and depth effects. Augmented reality is the technology that overlays information and virtual objects on real-world scenes in real-time [1]For instance if somebody wants to look inside of the human stomach and in the practical way that would be too costly and risky as well but with the help of the augmented reality even a 5-year-old child can do that with ease. In the augmented reality artificial 3-dimensional objects have been created and…

Insert math as
$${}$$


text:
Definitions

# 3-sphere

In mathematics, a 3-sphere is a higher-dimensional analogue of a sphere. It consists of the set of points equidistant from a fixed central point in 4-dimensional Euclidean space. Just as an ordinary sphere (or 2-sphere) is a two dimensional surface that forms the boundary of a ball in three dimensions, a 3-sphere is an object with three dimensions that forms the boundary of a ball in four dimensions. A 3-sphere is an example of a 3-manifold.

A 3-sphere is also called a hypersphere, although the term hypersphere can in general describe any n-sphere for n ≥ 3.

## Definition

In coordinates, a 3-sphere with center (C0C1C2C3) and radius r is the set of all points (x0x1x2x3) in real, 4-dimensional space (R4) such that

$sum_\left\{i=0\right\}^3\left(x_i - C_i\right)^2 = \left(x_0 - C_0 \right)^2 + \left(x_1 - C_1 \right)^2 + \left(x_2 - C_2 \right)^2+ \left(x_3 - C_3 \right)^2 = r^2.$
The 3-sphere centered at the origin with radius 1 is called the unit 3-sphere and is usually denoted S3:

$S^3 = left\left\{\left(x_0,x_1,x_2,x_3\right)inmathbb\left\{R\right\}^4 : x_0^2 + x_1^2 + x_2^2 + x_3^2 = 1right\right\}.$

It is often convenient to regard R4 as the space with 2 complex dimensions (C2) or the quaternions (H). The unit 3-sphere is then given by

$S^3 = left\left\{\left(z_1,z_2\right)inmathbb\left\{C\right\}^2 : |z_1|^2 + |z_2|^2 = 1right\right\}$
or
$S^3 = left\left\{qinmathbb\left\{H\right\} : |q| = 1right\right\}.$

The last description is often the most useful. It describes the 3-sphere as the set of all unit quaternionsquaternions with absolute value equal to unity. Just as the unit circle is important for planar polar coordinates, so the 3-sphere is important in the polar view of 4-space involved in quaternion multiplication.

## Properties

### Elementary properties

The 3-dimensional volume (or hyperarea) of a 3-sphere of radius r is

$2pi^2 r^3 ,$
while the 4-dimensional hypervolume (the volume of the 4-dimensional region bounded by the 3-sphere) is
$begin\left\{matrix\right\} frac\left\{1\right\}\left\{2\right\} end\left\{matrix\right\} pi^2 r^4.$

Every non-empty intersection of a 3-sphere with a three-dimensional hyperplane is a 2-sphere (unless the hyperplane is tangent to the 3-sphere, in which case the intersection is a single point). As a 3-sphere moves through a given three-dimensional hyperplane, the intersection starts out as a point, then becomes a growing 2-sphere which reaches its maximal size when the hyperplane cuts right through the "equator" of the 3-sphere. Then the 2-sphere shrinks again down to a single point as the 3-sphere leaves the hyperplane.

### Topological properties

A 3-sphere is a compact, connected, 3-dimensional manifold without boundary. It is also simply-connected. What this means, loosely speaking, is that any loop, or circular path, on the 3-sphere can be continuously shrunk to a point without leaving the 3-sphere. The Poincaré conjecture proposes that the 3-sphere is the only three dimensional manifold with these properties (up to homeomorphism). This conjecture was proved in 2003 by Grigori Perelman.

The 3-sphere is homeomorphic to the one-point compactification of $mathbb\left\{R\right\}^3$. Generally, any topological space which is homeomorphic to the 3-sphere is called a topological 3-sphere.

The homology groups of the 3-sphere are as follows: H0(S3,Z) and H3(S3,Z) are both infinite cyclic, while Hi(S3,Z) = {0} for all other indices i. Any topological space with these homology groups is known as a homology 3-sphere. Initially Poincaré conjectured that all homology 3-spheres are homeomorphic to S3, but then he himself constructed a non-homeomorphic one, now known as the Poincaré homology sphere. Infinitely many homology spheres are now known to exist. For example, a Dehn filling with slope 1/n on any knot in the three-sphere gives a homology sphere; typically these are not homeomorphic to the three-sphere.

As to the homotopy groups, we have π1(S3) = π2(S3) = {0} and π3(S3) is infinite cyclic. The higher homotopy groups (k ≥ 4) are all finite abelian but otherwise follow no discernible pattern. For more discussion see homotopy groups of spheres.

Homotopy groups of S3
k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
πk(S3) 0 0 0 Z Z2 Z2 Z12 Z2 Z2 Z3 Z15 Z2 Z2Z2 Z12Z2 Z84Z2Z2 Z2Z2 Z6

### Geometric properties

The 3-sphere is naturally a smooth manifold, in fact, a closed embedded submanifold of R4. The Euclidean metric on R4 induces a metric on the 3-sphere giving it the structure of a Riemannian manifold. As with all spheres, the 3-sphere has constant positive sectional curvature equal to 1/r2 where r is the radius.

Much of the interesting geometry of the 3-sphere stems from the fact that the 3-sphere has a natural Lie group structure given by quaternion multiplication (see the section below on group structure). The only other spheres with such a structure are the 0-sphere and the 1-sphere (see circle group).

Unlike the 2-sphere, the 3-sphere admits nonvanishing vector fields (sections of its tangent bundle). One can even find three linearly-independent and nonvanishing vector fields. These may be taken to be any left-invariant vector fields forming a basis for the Lie algebra of the 3-sphere. This implies that the 3-sphere is parallelizable. It follows that the tangent bundle of the 3-sphere is trivial. For a general discussion of the number of linear independent vector fields on a n-sphere see the article vector fields on spheres.

There is an interesting action of the circle group T on S3 giving the 3-sphere the structure of a principal circle bundle known as the Hopf bundle. If one thinks of S3 as a subset of C2, the action is given by

$\left(z_1,z_2\right)cdotlambda = \left(z_1lambda,z_2lambda\right)quad foralllambdainmathbb T$.
The orbit space of this action is homeomorphic to the two-sphere S2. Since S3 is not homeomorphic to S2×S1, the Hopf bundle is nontrivial.

## Topological construction

Two convenient constructions for the topologist are the reverse of "slicing in half" and "puncturing".

### Unslicing

A 3-sphere can be constructed topologically by "gluing" together the boundaries of a pair of 3-balls. The boundary of a 3-ball is a 2-sphere, and these two 2-spheres are to be identified. That is, imagine a pair of 3-balls of the same size, then superpose them so that their 2-spherical boundaries match, and let matching pairs of points on the pair of 2-spheres be identically equivalent to each other.

The interiors of the 3-balls do not match: only their boundaries. In fact, the fourth dimension can be thought of as a continuous scalar field, a function of the 3-dimensional coordinates of the 3-ball, similar to "temperature". Let this "temperature" be zero at the 2-spherical boundary, but let one of the 3-balls be "hot" (have positive values of its scalar field) and let the other 3-ball be "cold" (have negative values of its scalar field). The "hot" 3-ball could be thought of as the "hot hemi-3-sphere" and the "cold" 3-ball could be thought of as the "cold hemi-3-sphere". The temperature is highest at the hot 3-ball's very center and lowest at the cold 3-ball's center.

This construction is analogous to a construction of a 2-sphere, performed by joining the boundaries of a pair of disks. A disk is a 2-ball, and the boundary of a disk is a circle (a 1-sphere). Let a pair of disks be of the same diameter; superpose them so that their circular boundaries match, then let corresponding points on the circular boundaries become equivalent identically to each other. The boundaries are now glued together. Now "inflate" the disks. One disk inflates upwards and becomes the Northern hemisphere and the other inflates downwards and becomes the Southern hemisphere.

It is possible for a point traveling on the 3-sphere to move from one hemi-3-sphere to the other hemi-3-sphere by crossing the 2-spherical boundary, which could be thought of as a "3-quator" — analogous to an equator on a 2-sphere. The point would seem to be bouncing off the 3-quator and reversing direction of motion in 3-D, but also its "temperature" would become reversed, e.g. from positive on the "hot hemi-3-sphere" to zero on the 3-quator to negative on the "cold hemi-3-sphere".

### Unpuncturing

Consider a topological 2-sphere to be a seamless balloon. When punctured and flattened, the missing point becomes a circle (a 1-sphere) and the remaining balloon surface becomes a disk (a 2-ball) inside the circle. In the same way, a 3-ball is a punctured and flattened 3-sphere. To recreate the 3-sphere, merge all points on the 3-ball boundary (a 2-sphere) into a single point.

Another view of puncturing is stereographic projection. Rest the South Pole of a 2-sphere on an infinite plane, and draw lines from the North Pole through the sphere to intersect the plane. Each sphere point corresponds to a unique plane point, and vice versa, excepting the North Pole itself. The balloon has been stretched to infinity. Stereographic projection of a 3-sphere (except for the projection point) fills all of 3-space in the same manner. A benefit of this correspondence is that geometric spheres in 3-space map to geometric spheres of the 3-sphere, and planes in 3-space map to spheres containing the Pole.

Another view is a "shooting map". Place a marble at the South Pole and give it a flick of a measured strength in a chosen direction. Assuming the marble stays on the sphere and rolls without friction, its position after a fixed time interval (say, 1 second) will be some definite point of the sphere. Plotting direction in the plane and strength as radius, the North Pole is equally far away in every direction; this is the equivalent of the punctured balloon. Performing the same shooting experiment on the 3-sphere gives a map on the 3-ball. When the 3-sphere is considered a Lie group, the marble paths are one-parameter subgroups, the 3-ball is the tangent space at the identity (taken to be the South Pole), and the mapping to the 3-sphere is the exponential map.

## Coordinate systems on the 3-sphere

The four Euclidean coordinates for S3 are redundant since they are subject to the condition that $\left\{x_0\right\}^2 + \left\{x_1\right\}^2 + \left\{x_2\right\}^2 + \left\{x_3\right\}^2 = 1$. As a 3-dimensional manifold one should be able to parameterize S3 by three coordinates, just as one can parameterize the 2-sphere using two coordinates (such as latitude and longitude). Due to the nontrivial topology of S3 it is impossible to find a single set of coordinates that cover the entire space. Just as on the 2-sphere, one must use at least two coordinate charts. Some different choices of coordinates are given below.

### Hyperspherical coordinates

It is convenient to have some sort of hyperspherical coordinates on S3 in analogy to the usual spherical coordinates on S2. One such choice—by no means unique—is to use (ψ, θ, φ) where

$x_0 = cospsi,$
$x_1 = cosphi,sintheta,sinpsi$
$x_2 = sinphi,sintheta,sinpsi$
$x_3 = costheta,sinpsi$
where ψ and θ runs over the range 0 to π, and φ runs over 0 to 2π. Note that for any fixed value of ψ, θ and φ parameterize a 2-sphere of radius sin(ψ), except for the degenerate cases, when ψ equals 0 or π, in which case they describe a point.

The round metric on the 3-sphere in these coordinates is given by

$ds^2 = dpsi^2 + sin^2psileft\left(dtheta^2 + sin^2theta, dphi^2right\right)$
and the volume form by
$dV = left\left(sin^2psi,sinthetaright\right),dpsiwedge dthetawedge dphi.$

These coordinates have an elegant description in terms of quaternions. Any unit quaternion q can be written as a versor:

q = eτψ = cos ψ + τ sin ψ
where τ is a unit imaginary quaternion—that is, any quaternion which satisfies τ2 = −1. This is the quaternionic analogue of Euler's formula. Now the unit imaginary quaternions all lie on the unit 2-sphere in Im H so any such τ can be written:
τ = cos φ sin θ i + sin φ sin θ j + cos θ k
With τ in this form, the unit quaternion q is given by
q = eτψ = x0 + x1 i + x2 j + x3 k
where the x’s are as above.

When q is used to describe spatial rotations (cf. quaternions and spatial rotations) it describes a rotation about τ through an angle of 2ψ.

### Hopf coordinates

Another choice of hyperspherical coordinates, (η, ξ1, ξ2), makes use of the embedding of S3 in C2. In complex coordinates (z1, z2) ∈ C2 we write

$z_1 = e^\left\{i,xi_1\right\}sineta$
$z_2 = e^\left\{i,xi_2\right\}coseta.$
Here η runs over the range 0 to π/2, and ξ1 and ξ2 can take any values between 0 and 2π. These coordinates are useful in the description of the 3-sphere as the Hopf bundle
$S^1 to S^3 to S^2.,$

For any fixed value of η between 0 and π/2, the coordinates (ξ1, ξ2) parameterize a 2-dimensional torus. In the degenerate cases, when η equals 0 or π/2, these coordinates describe a circle.

The round metric on the 3-sphere in these coordinates is given by

$ds^2 = deta^2 + sin^2eta,dxi_1^2 + cos^2eta,dxi_2^2$
and the volume form by
$dV = sinetacoseta,detawedge dxi_1wedge dxi_2.$

### Stereographic coordinates

Another convenient set of coordinates can be obtained via stereographic projection of S3 onto a tangent R3 hyperplane. For example, if we project onto the plane tangent to the point (1, 0, 0, 0) we can write a point p in S3 as

$p = left\left(frac\left\{1-|u|^2\right\}\left\{1+|u|^2\right\}, frac\left\{2mathbf\left\{u\right\}\right\}\left\{1+|u|^2\right\}right\right) = frac\left\{1+mathbf\left\{u\right\}\right\}\left\{1-mathbf\left\{u\right\}\right\}$
where u = (u1, u2, u3) is a vector in R3 and ||u||2 = u12 + u22 + u32. In the second equality above we have identified p with a unit quaternion and u = u1 i + u2 j + u3 k with a pure quaternion. (Note that the division here is well-defined even though quaternionic multiplication is generally noncommutative). The inverse of this map takes p = (x0, x1, x2, x3) in S3 to
$mathbf\left\{u\right\} = frac\left\{1\right\}\left\{1+x_0\right\}left\left(x_1, x_2, x_3right\right).$

We could just have well have projected onto the plane tangent to the point (−1, 0, 0, 0) in which case the point p is given by

$p = left\left(frac\left\{-1+|v|^2\right\}\left\{1+|v|^2\right\}, frac\left\{2mathbf\left\{v\right\}\right\}\left\{1+|v|^2\right\}right\right) = frac\left\{-1+mathbf\left\{v\right\}\right\}\left\{1+mathbf\left\{v\right\}\right\}$
where v = (v1, v2, v3) is a vector in the second R3. The inverse of this map takes p to
$mathbf\left\{v\right\} = frac\left\{1\right\}\left\{1-x_0\right\}left\left(x_1,x_2,x_3right\right).$

Note that the u coordinates are defined everywhere but (−1, 0, 0, 0) and the v coordinates everywhere but (1, 0, 0, 0). Both patches together cover all of S3. This defines an atlas on S3 consisting of two coordinate charts. Note that the transition function between these two charts on their overlap is given by

$mathbf\left\{v\right\} = frac\left\{1\right\}\left\{|u|^2\right\}mathbf\left\{u\right\}$
and vice-versa.

## Group structure

When considered as the set of unit quaternions, S3 inherits an important structure, namely that of quaternionic multiplication. Because the set of unit quaternions is closed under multiplication, S3 takes on the structure of a group. Moreover, since quaternionic multiplication is smooth, S3 can be regarded as a real Lie group. It is a nonabelian, compact Lie group of dimension 3. When thought of as a Lie group S3 is often denoted Sp(1) or U(1, H).

It turns out that the only spheres which admit a Lie group structure are S1, thought of as the set of unit complex numbers, and S3, the set of unit quaternions. One might think that S7, the set of unit octonions, would form a Lie group, but this fails since octonion multiplication is nonassociative. The octonionic structure does give S7 one important property: parallelizability. It turns out that the only spheres which are parallelizable are S1, S3, and S7.

By using a matrix representation of the quaternions, H, one obtains a matrix representation of S3. One convenient choice is given by the Pauli matrices:

$x_1+ x_2 i + x_3 j + x_4 k mapsto begin\left\{pmatrix\right\};;,x_1 + i x_2 & x_3 + i x_4 -x_3 + i x_4 & x_1 - i x_2end\left\{pmatrix\right\}.$
This map gives an injective algebra homomorphism from H to the set of 2×2 complex matrices. It has the property that the absolute value of a quaternion q is equal to the square root of the determinant of the matrix image of q.

The set of unit quaternions is then given by matrices of the above form with unit determinant. This matrix subgroup is precisely the special unitary group SU(2). Thus, S3 as a Lie group is isomorphic to SU(2).

Using our hyperspherical coordinates (η, ξ1, ξ2) we can then write any element of SU(2) in the form

$begin\left\{pmatrix\right\}e^\left\{i,xi_1\right\}sineta & e^\left\{i,xi_2\right\}coseta -e^\left\{-i,xi_2\right\}coseta & e^\left\{-i,xi_1\right\}sinetaend\left\{pmatrix\right\}.$

Another way to state this result is if we express the matrix representation of an element of SU(2) as a linear combination of the pauli matrices. It is seen that an arbitrary element $U in SU\left(2\right)$ can be written as $U=alpha_0 I + i alpha_i sigma^i$. The condition that the determinant of U is +1 implies that the coefficients $alpha_i$ are constrained to lie on a 3-sphere.

## In literature

In Edwin Abbott Abbott's Flatland, published in 1884, and in Sphereland, a 1965 sequel to Flatland by Dionys Burger, the 3-sphere is referred to as an oversphere, and a 4-sphere is referred to as a hypersphere.

Writing in the American Journal of Physics, Mark A. Peterson describes three different ways of visualizing 3-spheres and points out language in The Divine Comedy that suggests Dante viewed the Universe in the same way.