question stringlengths 200 50k | answer stringclasses 1 value | source stringclasses 2 values |
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# Difference between revisions of "2020 SWMC 6 Problems/Problem 1"
## Problem
Give the only positive value of $x$ in $10x + 5 \leq 14 + x$.
## Solution
To figure out all possible values of $x$, we have to solve the inequality.
First, we have to make sure that only one side has the variable x. To do that, we subtract x from both sides so only the left side has x in it. We have $$9x+5\leq14.$$
Subtracting 5 from both sides gets $9x\leq9$, so $x\leq1$. Therefore the only positive value of $x$ is $\boxed1$. | HuggingFaceTB/finemath | |
SSAT Upper Level Math : Perpendicular Lines
Example Questions
← Previous 1 3 4
Example Question #1 : How To Find Out If Lines Are Perpendicular
Two perpendicular lines intersect at the point . One line passes through point ; the other passes through point . Evaluate .
Explanation:
The line that passes through and has slope
.
The line that passes through and , being perpendicular to the first, has as its slope the opposite reciprocal of , or
Therefore, to find , we use the slope formula and solve for :
Example Question #2 : How To Find Out If Lines Are Perpendicular
A line has the following equation:
Which of the following could be a line that is perpendicular to this given line?
Explanation:
First, put the equation of the given line in the form to find its slope.
Since the slope of the given line is , the slope of the line that is perpendicular must be its negative reciprocal, .
Now, put each answer choice in form to see which one has a slope of .
Example Question #1 : How To Find Out If Lines Are Perpendicular
Which of the following lines is perpendicular to the line ?
Explanation:
Perpendicular lines will have slopes that are negative reciprocals of one another. Our first step will be to find the slope of the given line by putting the equation into slope-intercept form.
The slope of this line is . The negative reciprocal will be , which will be the slope of the perpendicular line.
Now we need to find the answer choice with this slope by converting to slope-intercept form.
This equation has a slope of , and must be our answer.
Example Question #2 : How To Find Out If Lines Are Perpendicular
Which of the following lines is perpindicular to
Explanation:
When determining if a two lines are perpindicular, we are only concerned about their slopes. Consider the basic equation of a line, , where m is the slope of the line. Two lines are perpindicular to each other if one slope is the negative and reciprocal of the other.
The first step of this problem is to get it into the form, , which is . Now we know that the slope, m, is . The reciprocal of that is , and the negative of that is . Therefore, any line that has a slope of will be perpindicular to the original line.
Example Question #1 : How To Find Out If Lines Are Perpendicular
Which of the following equations represents a line that is perpendicular to the line with points and ?
Explanation:
If lines are perpendicular, then their slopes will be negative reciprocals.
First, we need to find the slope of the given line.
Because we know that our given line's slope is , the slope of the line perpendicular to it must be .
Example Question #3 : How To Find Out If Lines Are Perpendicular
Which of the following lines is perpendicular to a line with a slope ?
Not enough information provided
Explanation:
For a given line with a slope , any perpendicular line would have a slope , or the negative reciprocal of .
Given that in this instance, we can conclude that the slope of a perpendicular line would be . Therefore, the equation that contains this slope is
Example Question #4 : How To Find Out If Lines Are Perpendicular
Which of the following lines would be perpendicular to ?
Not enough information provided to solve
Explanation:
For a given line with a slope , any perpendicular line would have a slope , or the negative reciprocal of .
Given that in this instance, we can conclude that the slope of a perpendicular line would be . Given the perpendicular slope, we can now conclude that the perpendicular line is .
Example Question #5 : How To Find Out If Lines Are Perpendicular
A given line has the equation . What is the slope of any line that is perpendicular to this line?
Explanation:
For a given line with a slope , any perpendicular line would have a slope , or the negative reciprocal of .
Given that in this instance, we can conclude that the slope of a perpendicular line would be .
Example Question #2 : Perpendicular Lines
The equation for one line is . What is the slope of the line that is perpendicular to this line?
Explanation:
A line is perpendicular to another if their slopes are negative reciprocals of each other.
Since the slope of the given line is , the negative reciprocal would be .
Example Question #1 : How To Find The Slope Of A Perpendicular Line
What is the slope of any line perpendicular to 2y = 4x +3 ?
2
½
– ½
– 4
– ½
Explanation:
First, we must solve the equation for y to determine the slope: y = 2x + 3/2
By looking at the coefficient in front of x, we know that the slope of this line has a value of 2. To fine the slope of any line perpendicular to this one, we take the negative reciprocal of it:
slope = m , perpendicular slope = – 1/m
slope = 2 , perpendicular slope = – 1/2
← Previous 1 3 4 | HuggingFaceTB/finemath | |
# When scientists use a log graph, do they actually convert x/y w/ log?
1. Aug 28, 2014
### hongiddong
1. The problem statement, all variables and given/known data
For example, a linear plot would with x and y values would just plot normally where x and y increases linearly from 0 to infinity.
When we use a log graph, do we just plug in x and y to a graph in which the x and y go from 0, 10, 100, 1000, or do we convert the numbers such as log(x) = new y and then plug it into a graph that goes from 0, 10, 100, 1000?
2. Relevant equations
3. The attempt at a solution
2. Aug 28, 2014
### HallsofIvy
Staff Emeritus
I'm not sure what your question is. I know it used to be possible, and still may be, to get "semi-log" or "log-log" graph paper. The latter has both axes labeled such that the point labeled "2" is actually at distance "log(2)" from the origin (which is labeled "1"). The "semi-log" paper has one axis like that, the other a regular linear numbering. If you are using such paper, you mark them as they are labeled.
3. Aug 28, 2014
### nrqed
We do not convert the numbers to log(x) = new value.
We plot directly the initial values x without taking a log. The idea is that the scales on the graph paper are distorted in such a way that the log will be taken into account graphically.
The second option is to use *ordinary* graph paper and plot on it the calculated log values of the x.
4. Aug 29, 2014
### hongiddong
Thank you! I understand it now!
5. Aug 29, 2014
### Orodruin
Staff Emeritus
In the "good ol' times" when you would draw graphs by hand, you would typically have log-log paper with preprinted logarithmic scales. This would make it easy to draw your points in the correct places without having to compute the log. Nowadays most people would simply let a computer draw the graphs for them and it no longer matters much. I have seen examples in scientific papers of people plotting in log scale with logarithmic scales, in which case the values on the scales would not be converted, and of people plotting log10(x) in a linear scale (in which case this would be clearly stated). As long as you are clear about what is being plotted, there should be no possibility for misunderstanding.
6. Aug 29, 2014
### Ray Vickson
Probably most office supply stores would not carry log-log or semilog paper anymore, but university stationers might---I haven't checked. If not, there are numerous websites that have such paper downloadable and/or printable.
7. Aug 29, 2014
### Staff: Mentor
All the graphics packages, like excel and kaleidagraph, allow you to change the scale to logarithmic automatically, so you just plot x vs y, and then, from a drop down menu, switch to logarithmic scales with a click of your mouse.
Chet
8. Aug 29, 2014
### Staff: Mentor
Nowadays, there are sites with semilog and log-log paper as a jpeg which you can download and print off. FREE!!
You get to choose how many cycles you need.
EDIT. So now I look back and see Ray Vickson already said as much. | HuggingFaceTB/finemath | |
Study Guide for Physics Test #2
Newton's Laws of Motion
1. State and give an example of each of Newton’s 3 laws of motion.
2. Weight varies with ___________________. Why is this so?
3. Know about balanced and unbalanced forces.
4. What is inertia? What does it mean? Which of Newton’s Laws of Motion deals specifically with “inertia?”
5. What is the formula used in Newton’s 2nd law of motion? What does it mean? What is the unit of measurement?
6. How is weight measured on ANY planet? What is the unit of measurement? What is the formula that is used?
7. The strength of gravity is dependent on two factors. What are they? How are they related? What type of relationship do they have?
8. Essay Question in 3 Parts: Space flight is a direct result of scientists understanding Newton’s laws of motion.
Explain how each law has contributed to space flight.
9. Be able to look at a drawing showing objects in motion or at rest and:
1) Draw vectors of force in the correct places on the illustration;
2) Identify those vectors as to what force it represents; and,
3) Summarize what is happening in the picture by stating how the picture illustrates
one or more of Newton’s Laws of Motion.
10. Be able to solve word problems using this formula: F = m * a
(What is the unit of measurement for FORCE?)
11. Be able to solve word problems using this formula: W = F * D
(What is the unit of measurement for WORK?)
VOCABULARY
1. vector
2. inertia
3. gravity
4. Newton
5. balanced force
6. unbalanced force
7. factors that affect gravitational force
8. how Newton’s Laws of Motion made space travel possible
9. momentum
10. Joules | HuggingFaceTB/finemath | |
math
lim (4sin2x - 3x cos5x)/(3x/2 +(x^2)cscx )
x->0
1. 👍 0
2. 👎 0
3. 👁 195
1. 👍 0
2. 👎 0
2. L'Hoptial's rule applies to this.
I get for the numerator..
8cos2x - 3cos5x - 15x sin5x
and the denominator...
3/2 + 2x csc x + x^2 d cscx/dx
so the limit looks to be
5/ (3/2) = 10/3
check me.
1. 👍 0
2. 👎 0
👨🏫
bobpursley
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value of expression (cos5x + cos3x)/ (sin5x - sin3x) where x = (3.14/8) | HuggingFaceTB/finemath | |
23932
23,932 (twenty-three thousand nine hundred thirty-two) is an even five-digits composite number following 23931 and preceding 23933. In scientific notation, it is written as 2.3932 × 104. The sum of its digits is 19. It has a total of 4 prime factors and 12 positive divisors. There are 11,520 positive integers (up to 23932) that are relatively prime to 23932.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 19
• Digital Root 1
Name
Short name 23 thousand 932 twenty-three thousand nine hundred thirty-two
Notation
Scientific notation 2.3932 × 104 23.932 × 103
Prime Factorization of 23932
Prime Factorization 22 × 31 × 193
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 11966 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 23,932 is 22 × 31 × 193. Since it has a total of 4 prime factors, 23,932 is a composite number.
Divisors of 23932
1, 2, 4, 31, 62, 124, 193, 386, 772, 5983, 11966, 23932
12 divisors
Even divisors 8 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 43456 Sum of all the positive divisors of n s(n) 19524 Sum of the proper positive divisors of n A(n) 3621.33 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 154.7 Returns the nth root of the product of n divisors H(n) 6.60862 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 23,932 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 23,932) is 43,456, the average is 36,21.,333.
Other Arithmetic Functions (n = 23932)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 11520 Total number of positive integers not greater than n that are coprime to n λ(n) 960 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2657 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 11,520 positive integers (less than 23,932) that are coprime with 23,932. And there are approximately 2,657 prime numbers less than or equal to 23,932.
Divisibility of 23932
m n mod m 2 3 4 5 6 7 8 9 0 1 0 2 4 6 4 1
The number 23,932 is divisible by 2 and 4.
• Deficient
• Polite
Base conversion (23932)
Base System Value
2 Binary 101110101111100
3 Ternary 1012211101
4 Quaternary 11311330
5 Quinary 1231212
6 Senary 302444
8 Octal 56574
10 Decimal 23932
12 Duodecimal 11a24
20 Vigesimal 2jgc
36 Base36 igs
Basic calculations (n = 23932)
Multiplication
n×i
n×2 47864 71796 95728 119660
Division
ni
n⁄2 11966 7977.33 5983 4786.4
Exponentiation
ni
n2 572740624 13706828613568 328031822379909376 7850457573195991186432
Nth Root
i√n
2√n 154.7 28.8177 12.4378 7.5127
23932 as geometric shapes
Circle
Diameter 47864 150369 1.79932e+09
Sphere
Volume 5.7415e+13 7.19727e+09 150369
Square
Length = n
Perimeter 95728 5.72741e+08 33845
Cube
Length = n
Surface area 3.43644e+09 1.37068e+13 41451.4
Equilateral Triangle
Length = n
Perimeter 71796 2.48004e+08 20725.7
Triangular Pyramid
Length = n
Surface area 9.92016e+08 1.61537e+12 19540.4
Cryptographic Hash Functions
md5 54366258bb45311e0f0c80f3a3d50ab4 14e3afa78e166772b7227cd7d2b69272941a86f9 9c08b834b9decbcb915cae338793e54b9f5ee70f859030bea4783db2f570e888 848b8e382ea9eaeff725968ed6fa951836c6469023bd35ee90065a33e2011d1e661ed84e33c4e20b650b909d41bd10cd7a1e1e3013a7ee009e2aa4f5a3eb4a3c 8f06d12eeee1b28be3fac561eee0cd05e3a3cde8 | HuggingFaceTB/finemath | |
## September 10, 2014
### Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
#### Posted by Simon Willerton
Guest post by Bruce Bartlett
I recently put an article on the arXiv:
It’s about Chris Schommer-Pries’s recent strictification result from his updated thesis, that every symmetric monoidal bicategory is equivalent to a quasistrict one. Since symmetric monoidal bicategories can be viewed as the syntax for ‘stable 3-dimensional algebra’, one aim of the paper is to write out this stuff out in a diagrammatic notation, like this:
The other aim is to try to strip down the definition of a ‘quasistrict symmetric monoidal bicategory’, emphasizing the central role played by the interchangor isomorphisms. Let me explain a bit more.
## Motivation
Firstly, some motivation. For a long time now I’ve been finishing up a project together with Chris Douglas, Chris Schommer-Pries and Jamie Vicary about 1-2-3 topological quantum field theories. The starting point is a generators-and-relations presentation of the oriented 3-dimensional bordism bicategory (objects are closed 1-manifolds, morphisms are two-dimensional bordisms, and 2-morphisms are diffeomorphism classes of three-dimensional bordisms between those). So, you present a symmetric monoidal bicategory from a bunch of generating objects, 1-morphisms, and 2-morphisms, and a bunch of relations between the 2-morphisms. These relations are written diagrammatically. For instance, the ‘pentagon relation’ looks like this:
To make rigorous sense of these diagrams, we needed a theory of presenting symmetric monoidal bicategories via generators-and-relations in the above sense. So, Chris Schommer-Pries worked such a theory out, using computads, and proved the above strictification result. This implies that we could use the simple pictures above to perform calculations.
## Strictifying symmetric monoidal bicategories
The full algebraic definition of a symmetric monoidal bicategory is quite intimidating, amounting to a large amount of data satisfying a host of diagrams. A self-contained definition can be found in this paper of Mike Stay. So, it’s of interest to see how much of this data can be strictified, at the cost of passing to an equivalent symmetric monoidal bicategory.
Before Schommer-Pries’s result, the best strictification result was that of Gurski and Osorno.
Theorem (GO). Every symmetric monoidal bicategory is equivalent to a semistrict symmetric monoidal 2-category.
Very roughly, a semistrict symmetric monoidal 2-category consists of a strict 2-category equipped with a strict tensor product, plus the following coherence data (see eg. HDA1 for a fuller account) satisfying a bunch of equations:
• tensor naturators, i.e. 2-isomorphisms $\Phi_{f,g} : (f' \otimes g') \circ (f \otimes g) \Rightarrow (f' \circ f) \otimes (g' \circ g)$
• braidings, i.e. 1-morphisms $\beta_{A,B} : A \otimes B \rightarrow B \otimes A$
• braiding naturators, i.e. 2-isomorphisms $\beta_{f,g} : \beta_{A,B} \circ (f \otimes g) \Rightarrow (g \otimes f) \circ \beta_{A,B}$
• braiding bilinearators, i.e. 2-isomorphisms $R_{(A|B, C)} : (id \otimes R_{B,C}) \circ (R_{A,B} \otimes id) \Rightarrow R_{A, B\otimes C}$
• symmetrizors, i.e. 2-isomorphisms $\nu_{A,B} : id_{A \otimes B} \Rightarrow R_{B,A} \circ R_{A,B}$
So — Gurski and Osorno’s result represents a lot of progress. It says that the other coherence data in a symmetric monoidal bicategory (associators for the underlying bicategory, associators for the underlying monoidal bicategory, pentagonator, unitors, adjunction data, …) can be eliminated, or more precisely, strictified.
Schommer-Pries’s result goes further.
Theorem (S-P). Every semistrict monoidal bicategory is equivalent to a quasistrict symmetric monoidal 2-category.
A quasistrict symmetric monoidal 2-category is a semistrict symmetric monoidal 2-category where the braiding bilinearators and symmetrizors are equal to the identity. So - only the tensor naturators, braiding 1-morphisms, and braiding naturators remain!
The method of proof is to show that every symmetric monoidal bicategory admits a certain kind of presentation by generators-and-relations (a ‘quasistrict 3-computad’). And the gismo built out of a quasistrict 3-computad is a quasistrict symmetric monoidal 2-category! Q.E.D.
## Stringent symmetric monoidal 2-categories
In my article, I reformulate the definition of a quasistrict symmetric monoidal 2-category a bit, removing redundant data. Firstly, the tensor naturators $\Phi_{(f',g'),(f,g)}$ are fully determined by their underlying interchangors $\phi_{f,g}$,
(1)$\phi_{f,g} = \Phi_{(f, id), (id, g)} : (f \otimes id) \circ (id \otimes g) \Rightarrow (id \otimes g) \circ (f \otimes id)$
This much is well-known. But also, the braiding naturators are fully determined by the interchangors. So, I define a stringent symmetric monoidal 2-category purely in terms of this coherence data: interchangors, and braiding 1-morphisms. I show that they’re equivalent to quasistrict symmetric monoidal bicategories.
## Wire diagrams
The ‘stringent’ version of the definition is handy, because it admits a nice graphical calculus which I call ‘wire diagrams’. I needed a new name just to distinguish them from vanilla-flavoured string diagrams for 2-categories where the objects of the 2-category correspond to planar regions; now the objects of the 2-category correspond to lines. But it’s really just a rotated version of string diagrams in 3 dimensions. So, the basic setup is as follows:
But to keep things nice and planar, we’ll draw this as follows:
These diagrams are interpreted according to the prescription: tensor first, then compose! So, the interchangor isomorphisms look as follows:
So, what I do is write out the definitions of quasistrict and stringent symmetric monoidal 2-categories in terms of wire diagrams, and use this graphical calculus to prove that they’re the same thing.
That’s good for us, because it turns out these ‘wire diagrams’ are precisely the diagrammatic notation we were using for the generators-and-relations presentation of the oriented 3-dimensional bordism bicategory. For instance, I hope you can see the interchangor $\phi$ being used in the ‘pentagon relation’ I drew near the top of this post. So, that diagrammatic notation has been justified.
Posted at September 10, 2014 8:43 PM UTC
TrackBack URL for this Entry: https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2767
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
The pentagon relation looks like K_4, when the rooted binary trees describe interactions of RP^1’s. These end up being really helpful in studying twistor-string amplitudes.
Posted by: kneemo on September 12, 2014 12:23 AM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
Could you give me a reference on this?
Posted by: Bruce Bartlett on September 18, 2014 11:25 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
All this is great, Bruce. Congratulations to you and your team!
My one tiny complaint is terminological. What you’re calling a ‘symmetrizor’ is basically what Ross Street already called a ‘syllepsis’. (Actually I think his syllepsis is the inverse of your symmetrizor.) I think his term is better, since then there’s a column of the periodic table that goes:
bicategory
monoidal bicategory
braided monoidal bicategory
sylleptic monoidal bicategory
symmetric monoidal bicategory
The syllepsis is the big surprise in this column of the periodic table, just as the braiding is the big surprise in the previous column. I don’t think ‘symmetrizoric monoidal bicategories’ sounds as good as ‘sylleptic monoidal bicategories’.
But this is no big deal…
Posted by: John Baez on September 18, 2014 2:18 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
Agreed, I should have said “syllepsis”. I don’t actually use the term “symmetrizor” in the article. It only appeared in the above post because I suddenly felt at the n-category cafe I should give names to the remaining coherence data in Gurski and Osorno’s strictification theorem. There’s more latitude for that here in an informal setting.
Posted by: Bruce Bartlett on September 18, 2014 11:32 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
I said that the syllepsis is the ‘big surprise’ in the $n = 2$ column of the periodic table of $n$-categories. What I meant is that sylleptic monoidal bicategories provide the ‘new row’ in this column as compared to the $n = 1$ column, where we jump straight from braided monoidal categories to symmetric monoidal categories.
Similarly, the braiding is the ‘big surprise’ in the $n = 1$ column. The syllepsis is a kind of ‘meta-braiding’, in a way that can be made precise. I won’t bother here, except to say that for any $n$, the ‘new row’ in the $n$th column of the periodic table involves a meta-meta-meta-$\cdots$-braiding.
This makes it fun to compare the braiding and the syllepsis.
Chris’s theorem proves something I’d long suspected: every symmetric monoidal bicategory is equivalent to one where the syllepsis is trivial.
This seems a bit unlike the analogous situation in the previous column: not every symmetric monoidal category is equivalent to one in which the braiding is an identity morphism. But I’m not sure this is a fair comparison.
It would be helpful to get a nice supply of sylleptic but not symmetric monoidal bicategories, to understand these issues better. Thanks to the work of Sjoerd Crans, we can obtain these by taking the centers of braided monoidal bicategories. However, I don’t think they’re understood very well!
Sylleptic monoidal bicategories with duals should give invariants of 2d manifolds embedded in 5-space, just as braided monoidal categories with duals give invariants of 1d manifolds embedded in 3-space. This is the ‘almost stable’ situation: add one more dimension of space and the knotting goes away.
Posted by: John Baez on September 18, 2014 2:41 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
Finally: I really like the term ‘stringent’ as a variant of ‘strict’ in this context. As you continue to develop new kinds of semistrict $n$-categories, here are some other terms you can use:
stern, severe, harsh, uncompromising, authoritarian, firm, rigid, tough, austere, inflexible, unyielding, unbending, no-nonsense.
Posted by: John Baez on September 18, 2014 2:45 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
I like ‘stringent’, too. But (I’m sorry) the other terms you suggest, most of them, I really don’t like at all. (Or perhaps you were being tongue-in-cheek?)
‘Firm’ would be good, ‘rigid’ not quite as good (because it’ already used for a different concept). ‘Harsh’ and ‘authoritarian’ and ‘uncompromising’ and ‘no-nonsense’ would be terrible in my opinion!
I wouldn’t mind something along the lines of ‘firm’ such as ‘tight’ or ‘trim’.
Posted by: Todd Trimble on September 18, 2014 10:11 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
I’m sure John is making a joke here. But just to be clear - I was just using the term stringent to refer to a certain form of the definition of a quasistrict symmetric monoidal bicategory. I’m not defining a new mathematical structure, just trying to clearly delineate, for the purpose of the article, two different versions of the same structure. The real choice was Chris Schommer-Pries’s choice (fair enough, in my opinion) of the term quasistrict symmetric monoidal bicategory as something stricter than a semistrict symmetric monoidal bicategory.
Posted by: Bruce Bartlett on September 18, 2014 11:41 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
Yes, I was making a joke. I’m just amused by the idea of papers showing that every tough, no-nonsense $n$-category is equivalent to an authoritarian one, or that firm $n$-categories can be made unyielding.
But if the variety of different strictness conditions on $n$-categories continues to flourish — and I have no reason to think it won’t, or shouldn’t — it seems fine to me if some of the conditions not destined for greatness have goofy names. After all, they can’t all have great names; there just won’t be enough great names. So the choice would be between boring names and funny names.
This applies to names for mathematical concepts in general. The really important concepts get used a lot and deserve great names. The not-so-important concepts get used less, maybe just in a few papers, and deserve either boring or funny names. You can probably guess what I find more enjoyable.
In fact, as we all know, there’s even the danger of using up the great names before the great concepts have been found. Mathematicians are pretty good at eventually saying “oh, I didn’t mean to use such a great word for a concept that, in retrospect, wasn’t quite right”, but it’s a somewhat painful process, so there’s a case to be made for deliberately choosing names that aren’t great when you’re fumbling around in the initial stages of a subject.
For example, in the subject of $k$-tuply monoidal $n$-categories we have a lot of different kinds of ‘flab’, described by the different ‘ators’ Bruce listed—and infinitely many more he didn’t list. So, there’s a game that’s rather interesting to some of us, to see how many of these can safely be assumed to be trivial, in the sense that every $k$-tuply monoidal $n$-category is equivalent to one in which these are trivial. Eventually there will be a few clear ‘winners’, in which maximal amounts of flab of various kinds have been squeezed out… and eventually there may be a magnificent theory of this — unless people decide it’s better to learn to live with flab, because weight-reduction plans of this sort are inherently a waste of time, or ‘evil’. (I can see it either way.) But for the last few decades and maybe the next few we can expect to see some theorems that sound like:
Theorem. Every symmetric monoidal bicategory is equivalent to a semistrict one.
Theorem. Every symmetric monoidal bicategory is equivalent to a quasistrict one.
Theorem. Every symmetric monoidal bicategory is equivalent to a stringent one.
and similarly for other $k$-tuply monoidal $n$-categories.
And in the process of sorting things out, we’ll need some names for things that aren’t so great. So, it seems we get to choose either boring names or funny names.
Of course in academia there’s a great institutional pressure toward boring names, since boringness looks superficially similar to seriousness, and scholars need to convey the impression of doing serious stuff, since we’re getting paid a decent wage to do stuff that doesn’t instantly yield much profit. But mathematicians more than most other academics are willing to countenance funny names for things, probably because our seriousness (and even boringness) is less in question.
Posted by: John Baez on September 21, 2014 3:16 PM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
Nice! Your “wire diagrams” remind me of the string diagrams that Kate Ponto and I used for indexed monoidal categories in this paper. In fact, our notation was adapted from a notation used for symmetric monoidal bicategories by Daniel Schaeppi, by way of the fact that indexed monoidal categories give rise to symmetric monoidal bicategories (and a version of our notation works for general symmetric monoidal bicategories as well). Our notation has two kinds of strings and nodes, one for describing 1-morphisms and one for describing 2-morphisms; while Daniel’s notation omits the strings for 1-morphisms and yours omits the strings for the 2-morphisms.
Posted by: Mike Shulman on September 23, 2014 12:51 AM | Permalink | Reply to this
### Re: Quasistrict Symmetric Monoidal 2-Categories via Wire Diagrams
Yes you’re right it is definitely similar.
Posted by: Bruce Bartlett on September 29, 2014 11:12 AM | Permalink | Reply to this
Post a New Comment | open-web-math/open-web-math | |
1. ## Uniform Variate
Question :
Let X be a Uniform Variate defined on (-k,k). Determine k so that
$P(|X| < 2) = P(|X| > 2)$
Please could please tell me what formula should i use to solve this question?
2. Originally Posted by zorro
Question :
Let X be a Uniform Variate defined on (-k,k). Determine k so that
$P(|X| < 2) = P(|X| > 2)$
Please could please tell me what formula should i use to solve this question?
What have you tried? Where are you stuck?
3. use the fact that those are complementary events.
4. ## I dont know where to start
Originally Posted by mr fantastic
What have you tried? Where are you stuck?
Mr fantastic i dont know where to start ....No clue at all,,,,If possible coldu please guide me
5. Originally Posted by zorro
Mr fantastic i dont know where to start ....No clue at all,,,,If possible coldu please guide me
Draw a picture and use basic geometry.
Alternatively:
$\Pr(|X| < 2) = \Pr(|X| > 2)$
$\Rightarrow \Pr(|X| < 2) = 1 - \Pr(|X| < 2)$ (using matheagle's suggestion)
$\Rightarrow \Pr(|X| < 2) = \frac{1}{2}$.
And you really should be able to use either the pdf or simple geometry to calculate $\Pr(|X| < 2)$.
Hence solve for k.
6. Originally Posted by mr fantastic
Draw a picture and use basic geometry.
Alternatively:
$\Pr(|X| < 2) = \Pr(|X| > 2)$
$\Rightarrow \Pr(|X| < 2) = 1 - \Pr(|X| < 2)$ (using matheagle's suggestion)
$\Rightarrow \Pr(|X| < 2) = \frac{1}{2}$.
And you really should be able to use either the pdf or simple geometry to calculate $\Pr(|X| < 2)$.
Hence solve for k.
Mr fantastic could u please tell me how have u put in the value of................. $1- \Pr(|X| < 2) = \frac{1}{2}$
7. Originally Posted by zorro
Mr fantastic i dont know where to start ....No clue at all,,,,If possible coldu please guide me
I think it is time to ask: Why is it you seem to have no clue where to start with so many of these question? That seems to be the real problem here.
You have at least some of: notes, a text and Google at your disposal.
CB
8. ## I am sorry for these annoying quetions
Originally Posted by CaptainBlack
I think it is time to ask: Why is it you seem to have no clue where to start with so many of these question? That seems to be the real problem here.
You have at least some of: notes, a text and Google at your disposal.
CB
I am sorry for annoying u .....
9. Originally Posted by zorro
I am sorry for annoying u .....
You are not annoying me, but it is best to address the real problem which we are not at present.
Cb
10. Originally Posted by zorro
Mr fantastic could u please tell me how have u put in the value of................. $1- \Pr(|X| < 2) = \frac{1}{2}$
Originally Posted by Mr Fantastic
$\Rightarrow \Pr(|X| < 2) = 1 - \Pr(|X| < 2)$
Add $\Pr(|X| < 2)$ to both sides:
$\Rightarrow 2 \Pr(|X| < 2) = 1$
$\Rightarrow \Pr(|X| < 2) = \frac{1}{2}$
etc. | HuggingFaceTB/finemath | |
# sulfur oxidation number
The sulfur oxyanions with sulfur oxidation numbers between −1 and +6 are unstable in low-temperature aqueous systems with respect to stable sulfide, sulfate and sulfur (Fig. Some mixotrophic bacteria only oxidize thiosulfate to tetrathionate. Remember the sum of the oxidation numbers of all the elements must equal zero because Na 2S 2O 3 is a neutral compound. Structures: NiCl_2(PPh_3)_2 and Ni(NCS)_2(PPh_3)_2... Assigning Oxidation Numbers to Elements in a Chemical Formula, Titration of a Strong Acid or a Strong Base, Hydrogen Peroxide: Preparation, Properties & Structure, D-Block Elements: Properties & Electron Configuration, Ionization Energy: Trends Among Groups and Periods of the Periodic Table, Disproportionation: Definition & Examples, Electrochemical Salt Bridge: Definition & Purpose, Valence Bond Theory of Coordination Compounds, Limiting Reactant: Definition, Formula & Examples, Enthalpy: Energy Transfer in Physical and Chemical Processes, Coordinate Covalent Bond: Definition & Examples, Standard Enthalpy of Formation: Explanation & Calculations, Bond Order: Definition, Formula & Examples, Atomic and Ionic Radii: Trends Among Groups and Periods of the Periodic Table, SAT Subject Test Chemistry: Practice and Study Guide, High School Biology: Homework Help Resource, Holt McDougal Modern Biology: Online Textbook Help, General Studies Earth & Space Science: Help & Review, General Studies Health Science: Help & Review, FTCE Middle Grades General Science 5-9 (004): Test Practice & Study Guide, ILTS Science - Environmental Science (112): Test Practice and Study Guide, ILTS Science - Chemistry (106): Test Practice and Study Guide, SAT Subject Test Biology: Practice and Study Guide, UExcel Anatomy & Physiology: Study Guide & Test Prep, Biological and Biomedical We know Oxygen generally shows a oxidation number of -2. "Thermodynamics and kinetics of sulfide oxidation by oxygen: a look at inorganically controlled reactions and biologically mediated processes in the environment", "Metaproteomics reveals differential modes of metabolic coupling among ubiquitous oxygen minimum zone microbes", "Chemoautotrophic Carbon Fixation Rates and Active Bacterial Communities in Intertidal Marine Sediments", "Ubiquitous Gammaproteobacteria dominate dark carbon fixation in coastal sediments", "Microbial interactions involving sulfur bacteria: implications for the ecology and evolution of bacterial communities", "Quantitative Molecular Analysis of the Microbial Community in Marine Arctic Sediments (Svalbard)", "The life sulfuric: microbial ecology of sulfur cycling in marine sediments", "Single-cell Sequencing of Thiomargarita Reveals Genomic Flexibility for Adaptation to Dynamic Redox Conditions", "Cable bacteria generate a firewall against euxinia in seasonally hypoxic basins", "Natural occurrence of microbial sulphur oxidation by long-range electron transport in the seafloor", "Reclassification of some species of Thiobacillus to the newly designated genera Acidithiobacillus gen. nov., Halothiobacillus gen. nov. and Thermithiobacillus gen. nov", "Oxidation of thiosulfate by Paracoccus denitrificans and other hydrogen bacteria", "Proposal for the reclassification of Thiobacillus novellus as Starkeya novella gen. nov., comb. [59], Most of the chemosynthetic autotrophic bacteria that can oxidize elemental sulfur to sulfate are also able to oxidize thiosulfate to sulfate as a source of reducing power for carbon dioxide assimilation. 21 - Selenous acid, H2SeO3, is reduced by H2S to... Ch. [4] Experimental data from the anaerobic phototroph Chlorobaculum tepidum indicate that microorganisms enhance sulfide oxidation by three or more orders of magnitude. All the Archaea involved in this process are aerobic and belong to the Order Sulfolobales,[19][20] characterized by acidophiles (extremophiles that require low pHs to grow) and thermophiles (extremophiles that require high temperatures to grow). The direct oxidation of sulfite to sulfate by a type of mononuclear molybdenum enzyme known as sulfite oxidoreductase. In oxide: Nonmetal oxides. Some inorganic forms of reduced sulfur, mainly sulfide (H2S/HSâ) and elemental sulfur (S0), can be oxidized by chemolithotrophic sulfur-oxidizing prokaryotes, usually coupled to the reduction of oxygen (O2) or nitrate (NO3â). The formation of sulfate in aerobic conditions entails the incorporation of four oxygen atoms from water, and when coupled with dissimilatory nitrate reduction (DNR) -the preferential reduction pathway under anoxic conditions- can have a contribution of oxygen atoms from nitrate as well. +1 ? To find this oxidation number, it is important to know that the sum of the oxidation numbers of atoms in compounds that are neutral must equal zero. Its simple ion, sulfide carries oxidation status -2. So that hydrogen atom can be reduced to hydrogen's other lower oxidation states such as 0 and -1. [31][32] Some AnSOB, such as the facultative anaerobes Thiobacillus spp., and Thermothrix sp., are chemolithoautotrophs, meaning that they obtain energy from the oxidation of reduced sulfur species, which is then used to fix CO2. Sulfur oxidation involves the oxidation of reduced sulfur compounds such as sulfide (H 2 S), inorganic sulfur (S 0), and thiosulfate (S 2 O 2−3) to form sulfuric acid (H 2 SO 4). © copyright 2003-2021 Study.com. Sulfur oxidizing bacteria (SOB) are aerobic, anaerobic or facultative, and most of them are obligate or facultative autotrophs, that can either use carbon dioxide or organic compounds as a source of carbon (mixotrophs). So, the oxidation number of S 2-will be equal to the charge on the ion. For most microorganisms and oxidation conditions, only small fractionations accompany either the aerobic or anaerobic oxidation of sulfide, elemental sulfur, thiosulfate and sulfite to elemental sulfur or sulfate. Although the biological oxidation of reduced sulfur compounds competes with abiotic chemical reactions (e.g. [12] The large sulfur bacteria (LSB) of the family Beggiatoaceae (Gammaproteobacteria) have been used as model organisms for benthic sulfur oxidation. [53], Among the heterotrophic SOB are included species of Beggiatoa that can grow mixotrophically, using sulfide to obtain energy (autotrophic metabolism) or to eliminate metabolically formed hydrogen peroxide in the absence of catalase (heterotrophic metabolism). Solution for What is the oxidation number of sulfur (S) in sulfuric acid (H2SO4)? [3], The mechanism of bacterial oxidation of tetrathionate is still unclear and may involve sulfur disproportionation, during which both sulfide and sulfate are produced from reduced sulfur species, and hydrolysis reactions.[3]. Given the very small fractionation of 18O that usually accompanies MSO, the relatively higher depletions in 18O of the sulfate produced by MSO coupled to DNR (-1.8 to -8.5 â°) suggest a kinetic isotope effect in the incorporation of oxygen from water to sulfate and the role of nitrate as a potential alternative source of light oxygen. Because there is a lower energetic cost associated with the use of light isotopes, enzymatic processes usually discriminate against the heavy isotopes, and, as a consequence, biological fractionations of isotopes are expected between the reactants and the products. -2 If the sum of the oxidation numbers in a complex ion is equal to the charge on that ion, what is the oxidation number of manganese in MnO 4-? The oxidation number of an element is the "charge"... Iodine displays a wide range of oxidation numbers.... What is the oxidation number of oxygen in O2? The –ide part of the name signifies that it is an anion and hence it must have a negative oxidation number. [3] The AnSOB Cyanobacteria are only able to oxidize sulfide to elemental sulfur and have been identified as Oscillatoria, Lyngbya, Aphanotece, Microcoleus, and Phormidium. S 2-is the sulfide ion, the ion of the element sulfur. Besides, they may have been critical for the evolution of eukaryotic organisms, given that sulfur metabolism could have driven the formation of the symbiotic associations that sustained them[9] (see below). The microbial oxidation of sulfur is an important link in the biogeochemical cycling of sulfur in environments hosting both abundant reduced sulfur species and low concentrations of oxygen, such as marine sediments, oxygen minimum zones (OMZs) and hydrothermal systems.[3]. This page was last edited on 6 January 2021, at 00:36. Related Searches. What is the oxidation number of sulfur in H 2 SO 4? [48] The archaeon Acidianus ambivalens appears to possess both an ADP-dependent and an ADP independent pathway for the oxidation of sulfide. The oxidation number is synonymous with the oxidation state. [58] For the anaerobic oxidation of elemental sulfur, it is thought that the Sox pathway plays an important role, although this is not yet completely understood. The sulfur oxyanions form as intermediates in a number of sedimentary redox processes including the oxic and anoxic oxidation of sulfide and pyrite and the reduction of sulfur compounds. Determine the oxidation number of each element in the following ions or comp… 05:16. It is called electrogenic sulfur oxidation (e-SOx), and involves the formation of multicellular bridges that connect the oxidation of sulfide in anoxic sediment layers with the reduction of oxygen or nitrate in oxic surface sediments, generating electric currents over centimeter distances. [63] The fractionations of oxygen produced by sulfur disproportionation from elemental sulfur have been found to be higher, with reported values from 8 to 18.4â°, which suggests a kinetic isotope effect in the pathways involved in oxidation of elemental sulfur to sulfate, although more studies are necessary to determine what are the specific steps and conditions that favor this fractionation. [54] Other organisms, such as the Bacteria Sphaerotilus natans [55] and the yeast Alternaria [56] are able to oxidize sulfide to elemental sulfur by means of the rDsr pathway. [62] The light isotopes of the elements that are most commonly found in organic molecules, such as 12C, 16O, 1H, 14N and 32S, form bonds that are broken more easily than bonds between the corresponding heavy isotopes, 13C, 18O, 2H, 15N and 34S . Therefore oxidation number of oxygen in SO2Cl2 is -2*2=-4. +7 ? the iron-mediated oxidation of sulfide to iron sulfide (FeS) or pyrite (FeS2)),[10] thermodynamic and kinetic considerations suggest that biological oxidation far exceeds the chemical oxidation of sulfide in most environments. [49] Similarly, both mechanisms operate in the chemoautotroph Thiobacillus denitrificans,[50] which can oxidize sulfide to sulfate anaerobically using nitrate as terminal electron acceptor [51] which is in turn reduced to dinitrogen (N2). The only time this is altered is if … In the chemolithotrophs Thiobacillus denitrificans and Sulfurimonas denitrificans, MSO coupled to DNR has the effect of inducing the SQR and Sox pathways, respectively. Therefore, the oxidation number of sulfur is +4 (it lost four electrons to oxygen) and the oxidation numbers for our compound is as follows: Na +1; S +4; O -2. Oxidation number, also called oxidation state, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.. Each atom that participates in an oxidation-reduction reaction is assigned an oxidation number that reflects its ability to acquire, donate, or share electrons. The oxidation of hydrogen sulfide has been considered one of the most important processes in the environment, given that the oceans have had very low oxygen and high sulfidic conditions over most of the Earth's history. If x is the oxidation number of sulfur in S2Cl2 then: 2x + 2(-1) = 0 ==> x = 1. dissertation, Univ. The table below summarizes the reported fractionations of sulfur isotopes from MSO in different organisms and conditions. The so-called cable bacteria are widespread in shallow marine sediments,[15] and are believed to conduct electrons through structures inside a common periplasm of the multicellular filament,[16] a process that may influence the cycling of elements at aquatic sediment surfaces, for instance, by altering iron speciation. In this work, unamended and elemental S‐amended soils were incubated (30°C) under aerobic conditions for various times. Microbial oxidation of sulfur is the oxidation of sulfur by microorganisms to produce energy. The other groups accumulate elemental sulfur, which they may oxidize to sulfate when sulfide is limited or depleted.[3]. Thus, the atoms in O 2, O 3, P 4, S … 14). Oxidation Number of Sulphur (S) Sulphur (S) also termed as sulfur is a chemical element having oxidation number of -2, +4 and +6. [3] The oxidation of sulfide can proceed aerobically by two different mechanisms: substrate-level phosphorylation, which is dependent on adenosine monophosphate (AMP), and oxidative phosphorylation independent of AMP,[47] which has been detected in several Thiobacilli (T. denitrificans, T. thioparus, T. novellus and T. neapolitanus), as well as in Acidithiobacillus ferrooxidans. [57], Some Bacteria and Archaea can aerobically oxidize elemental sulfur to sulfuric acid. -6 ? The symbiotic SOM provides carbon and, in some cases, bioavailable nitrogen to the host, and gets enhanced access to resources and shelter in return. It should have an ON number of 2- since it is not a peroxide or a superoxide. Similarly, the oxidation number of hydrogen is almost always +1. Then , 2 x (+1) + X + 4 x (-2) = 0 Solving we get, +2 + X - 8 = 0 X = +6 Thus oxidation number of sulfur in H2SO4 is +6. This can be easily determined by the name and position of the ion. -7 The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are always assigned an oxidation state of +2. This lifestyle has evolved independently in sediment-dwelling ciliates, oligochaetes, nematodes, flatworms and bivalves. The fractionation of sulfur and oxygen isotopes during microbial sulfide oxidation (MSO) has been studied to assess its potential as a proxy to differentiate it from the abiotic oxidation of sulfur. The sulfur is the atom whose oxidation number is not covered by one of the rules. [3] Aerobic sulfur oxidizing bacteria are mainly mesophilic, which grow at moderate ranges of temperature and pH, although some of them are thermophilic and/or acidophilic. From all of the SOB, the only group that directly oxidize sulfide to sulfate in abundance of oxygen without accumulating elemental sulfur are the Thiobacilli. I hope this helps! +8 ? Others, such as some filamentous gliding green bacteria (Chloroflexaceae), are mixotrophs. Other zones of the ocean that experience periodic anoxic and sulfidic conditions are the upwelling zones off the coasts of Chile and Namibia, and hydrothermal vents, which are a key source of H2S to the ocean. But, sulfur atom cannot be reduced furthermore because -2 is the lowest oxidation state of sulfur. They are known as 'gradient organisms', that are indicative of hypoxic (low oxygen) and sulfidic (rich in reduced sulfur species) conditions. Atomic sulfur has oxidation number of 0. Become a Study.com member to unlock this In sulfate, $SO_{4}^{2-}$? S-Dealkylation is a minor pathway of metabolism, while direct oxidation of sulfur to a sulfoxide and/or a sulfone is a major pathway. an organism that can obtain energy either by aerobic or anaerobic respiration). The phototrophic oxidation of sulfide to thiosulfate under anoxic conditions also generates negligible fractionations. D. thiozymogenes (chemolithotroph; "cable bacteria"), 0â° (30 °C; compared to the sulfonate functional group); 2 to 4â° (30 °C; compared to the sulfane functional group), Desulfobulbus propionicus (chemoorganotroph), Oxidation of thiosulfate and tetrathionate, CS1 maint: multiple names: authors list (. The oxidation of inorganic compounds is the strategy primarily used by chemolithotrophic microorganisms to obtain energy in order to build their structural components, survive, grow and reproduce. ? [3] The most abundant and studied SOB are in the family Thiobacilliaceae in terrestrial environments, and in the family Beggiatoaceae in aquatic environments . The oxidation number of each atom can be calculated by subtracting the sum of lone pairs and electrons it gains from bonds from the number of valence electrons. [22] On the other hand, the cable bacteria belong to the family Desulfobulbaceae of the Deltaproteobacteria and are currently represented by two candidate Genera, "Candidatus Electronema" and "Candidatus Electrothrix"[27]. (Note that the sum of the oxidation numbers is zero since S2Cl2 is neutral.) Hope it helped! H2SO4 Oxidation number of H = +1 Oxidation number of O = -2 Let oxidation number of sulfur be X. Determining oxidation numbers from the Lewis structure (Figure 1a) is even easier than deducing it from the molecular formula (Figure 1b). It is abundant, multivalent and nonmetallic.Under normal conditions, sulfur atoms form cyclic octatomic molecules with a chemical formula S 8.Elemental sulfur is a bright yellow, crystalline solid at room temperature. Each chloride ion in S2Cl2 has a charge of -1 since chlorine is in Group 7. An example of a sulfur-oxidizing bacterium is Paracoccus. [39] Thiobacillus denitrificans uses oxidized forms on nitrogen as terminal electron acceptor instead of oxygen, and A. ferrooxidans uses ferrous iron. Anaerobic SOB (AnSOB) are mainly neutrophilic/mesophilic photosynthetic autotrophs, which obtain energy from sunlight but use reduced sulfur compounds instead of water as electron donors for photosynthesis. [1][2], Most of the sulfur oxidizers are autotrophs that can use reduced sulfur species as electron donors for carbon dioxide (CO2) fixation. In chemistry, we can say that the total number of electrons gained or lost by an atom to make a chemical bond with the other atom is known as the oxidation number. +6 ? Find the oxidation number of each sulfur in the molecule $\mathrm{H}_{2} \ma… 05:29. The δ18O value of the newly formed sulfate thus depends on the δ18O value of the water, the isotopic fractionation associated with the incorporation of oxygen atoms from water to sulfate and a potential exchange of oxygen atoms between sulfur and nitrogen intermediates and water. [13], Another evolutionary strategy of SOM is to partner up with motile eukaryotic organisms. The most studied have been the genera Sulfolobus, an aerobic Archaea, and Acidianus, a facultative anaerobe (i.e. The modern analog ecosystems are deep marine basins such as those in the Black Sea, near the Cariaco trench and the Santa Barbara basin. Aerobic MSO generates depletions in the 34S of sulfate, that have been found to be as small as â1.5â° and as large as -18â°. AnSOB include some purple sulfur bacteria (Chromatiaceae)[28] such as Allochromatium,[29] and green sulfur bacteria (Chlorobiaceae), as well as the purple non-sulfur bacteria (Rhodospirillaceae)[30] and some Cyanobacteria. 21 - What are the oxidation numbers of sulfur in each... Ch. In the compound sulfur dioxide (SO2), the oxidation number of oxygen is -2. Just as the fractionation of oxygen isotopes reveal, the larger fractionations in sulfate from the disproportionation of elemental sulfur point to a key step or pathway critical for inducing this large kinetic isotope effect. Where N-hydroxylation and N-oxide formation are minor pathways relative to N-dealkylation, the exact opposite is true of sulfur oxidation. ? The branched thiosulfate oxidation pathway, a mechanism in which water-insoluble globules of intermediate sulfur are formed during the oxidation of thiosulfate and sulfide. A. By definition, the oxidation number of an atom is the charge that atom would have if the compound was composed of ions. This limitation has led SOM to develop different morphological adaptations. All rights reserved. 1. They internally store large amounts of nitrate and elemental sulfur to overcome the spatial gap between oxygen and sulfide. [44], Cyanobacteria normally perform oxygenic photosynthesis using water as electron donor. Oxidation number of hydrogen in H 2 is +1. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. Studies of oxidation of elemental S in soils showed that thiosulfate and tetrathionate are produced during S oxidation in soils. {/eq}? Create your account, Let the oxidation number of sulfur is (x). Ch. Three different groups of these enzymes are recognized (the xanthine oxidase, sulfite oxidase (SO) and dimethyl sulfoxide reductase families), and they are present in the three domains of life. read more ", "Enzymology and molecular biology of prokaryotic sulfite oxidation", "Biochemistry and molecular biology of lithotrophic sulfur oxidation by taxonomically and ecologically diverse bacteria and archaea", "Genes involved in hydrogen and sulfur metabolism in phototrophic sulfur bacteria", "Oxidation of sulfide to thiosulfate by Microcoleus chtonoplastes", "Sulphite as Intermediate Sulphur Compound in Anaerobic Sulphide Oxidation to Thiosulphate by Marine Cyanobacteria", "Electron Transfer during Sulphide and Sulphite Oxidation in Thiobacillus denitrificans", "Isolation and characterization of strains CVO and FWKO B, two novel nitrate-reducing, sulfide-oxidizing bacteria isolated from oil field brine", "Effect of Catalase and Cultural Conditions on Growth of Beggiatoa", "Intracellular deposition of sulfur by Sphaerotilus natans", "Ferric iron reduction by sulfur- and iron-oxidizing bacteria", "A periplasmic location for the bisulfiteoxidizing multienzyme system from Thiobacillus versutus", "Oxidation of reduced inorganic sulphur compounds by acidophilic thiobacilli", "Microbiological Fractionation of Sulphur Isotopes", "Isotope effects associated with the anaerobic oxidation of sulfite and thiosulfate by the photosynthetic bacterium,Chromatium vinosum", "Discrimination between 34S and 32S during bacterial metabolism of inorganic sulfur compounds", "Isotope fractionation and sulfur metabolism by pure and enrichment cultures of elemental sulfur-disproportionating bacteria", Microbially induced sedimentary structure, Physical factors affecting microbial life, https://en.wikipedia.org/w/index.php?title=Microbial_oxidation_of_sulfur&oldid=998569880, Creative Commons Attribution-ShareAlike License, Disproportionation, in the presence of iron scavengers, Disproportionation, attenuated isotope effect due to reoxidation by manganese oxides, Anaerobic, coupled to DNR, no intermediates in complete oxidation of sulfide to sulfate (potentially only uses Sox pathway), Anaerobic, disproportionation, in the presence of iron scavengers, Anaerobic, disproportionation, attenuated isotope effect due to reoxidation by manganese oxides, The sulfide:quinone oxidorreductase pathway (SQR), widespread in green sulfur bacteria, that involves the formation of intermediate compounds such as, The rDsr pathway, used by some microorganisms in the. Water as electron donor, are mixotrophs, +2 +4 and +6 of SOM is partner. Are obligate or facultative chemolithoautotrophs water-insoluble globules of intermediate sulfur are formed during the number. And position of the sulfate, lower than -4.3â°, has been found to energy! An organism that can perform a similar process were identified as similar to Thiomicrospira denitrificans and.. Na 2S 2O 3 is a chemical element with sulfur oxidation number oxidation number of,... So_ { 4 } ^ { 2- } [ /math ] comp… 05:16, nematodes, flatworms bivalves... Filamentous gliding green bacteria ( Chloroflexaceae ), the answer is ( x ) anoxic conditions also generates negligible.. N-Dealkylation, the oxidation state of sulfur be x However, the.! In I2 ( S ) in sulfuric acid ( h2so4 ) sulfur has charge. Appears to possess both an ADP-dependent and an ADP independent pathway for the oxidation number incubation were determined sulfur! Sulfate, [ math ] SO_ { 4 } ^ { 2- } [ /math ] +1. Homework and study questions total sulfur oxidation in marine sediments is still unknown SOM is partner. Major pathway Na 2S 2O 3 is a neutral compound this algebraically by setting up equation... Variable x in the molecule$ \mathrm { H } _ { 2 } \ma….. Chemolithotroph ; cable bacteria '' ) Enrichment culture to oxygen causing sulfur to a and/or... Sediment-Dwelling ciliates, oligochaetes, nematodes, flatworms and bivalves has led SOM develop... Atoms in thiosulfate are oxidized to sulfate when sulfide is limited sulfur oxidation number.! Bacteria usually oxidize sulfide to thiosulfate under anoxic conditions also generates negligible fractionations elements must equal zero because Na 2O. Nitrogen as terminal electron acceptor instead of oxygen, and Acidianus, facultative. Sulfate contamination in groundwater near an abandoned mine: Hydrogeochemical modeling, microbiology and isotope geochemistry in is. At 00:36 molecule $\mathrm { H } _ { 2 } \ma… 05:29 of all elements. In SO2 is +4 the following ions or comp… 05:16 of all the elements must equal zero because 2S! An abandoned mine: Hydrogeochemical modeling, microbiology and isotope geochemistry study questions,. Total sulfur oxidation in marine sediments each element in the compound sulfur dioxide ( SO2 ), mixotrophs... Since chlorine is in Group 7 a mechanism in which water-insoluble globules of sulfur... Because Na 2S 2O 3 is a major pathway to water ( ~5â° ) trademarks., Get access to this video and our entire Q & a.! Was last edited on 6 January 2021, at 00:36 oxygen in SO2Cl2 is *! 3 ] synonymous with the symbol S and atomic number 16 has oxidation numbers of all the elements must zero., has been found to produce small fractionations in 18O compared to water ( )., the oxidation number of I in I2 ( S ) in acid! The property of their respective owners is reduced by H2S to... Ch of -2 ( ). Average cell abundances of 108 cells/m3 in organic-rich marine sediments is still unknown a superoxide both atoms. [ 14 ] Recently, a mechanism in which water-insoluble globules of intermediate sulfur are during. Pathway for the oxidation of sulfur to a sulfoxide and/or a sulfone is minor. Recently, a small fractionation in sulfur oxidation number following calculation } \ma… 05:29 most studied have the... Oxygen is -2 * 2=-4 atomic number 16 the most studied have been the genera Sulfolobus, an Archaea! The symbol S and atomic number 16 has oxidation numbers of sulfur be x as 0 and -1 have the! While forming numerous compounds, the ion [ 45 ] [ 46 ] Cyanobacteria... Of SOM is to partner up with motile eukaryotic organisms a oxidation number each... However, the exact opposite is true of sulfur can aerobically oxidize elemental sulfur to sulfuric oxidizes. In the following ions or comp… 05:16 thiosulfate oxidation pathway, a mechanism in which water-insoluble globules intermediate. 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General contribution of microorganisms to produce small fractionations in 18O compared to water ( ~5â° ) h2so4 oxidation number 2-. Thiosulfate and sulfide H 2 so 4 electrons are added to an elemental species, its number..., while direct oxidation of sulfur in each... Ch ADP independent pathway for oxidation... Atoms of only one element was composed of ions Na 2S 2O 3 is a major pathway and.... The SO4−2 sulfur has a charge of ion [ math ] SO_ { 4 ^! Let oxidation number becomes negative an oxidation... What are the oxidation is. Iodide ion to... Ch while direct oxidation of sulfide to sulfate and are obligate facultative. Composed of ions have if the compound was composed of ions anaerobic that. [ 13 ], some bacteria and Archaea can aerobically oxidize elemental sulfur to sulfuric.... Morphological adaptations the Alpha-, Gamma- and Epsilonproteobacterial SOM account for average cell of. Studied have been the genera Sulfolobus, an aerobic Archaea, and so S... Usually oxidize sulfide to sulfate by a type of mononuclear molybdenum enzyme known as sulfite oxidoreductase 2 so 4,. Evolutionary strategy of SOM is to partner up with motile eukaryotic organisms Na! Aerobic Archaea, and A. ferrooxidans uses ferrous iron ) Enrichment culture that microorganisms sulfide... During the oxidation numbers is zero in a neutral compound ions it is equal the that! Edited on 6 January 2021, at 00:36 can vary form -2 to +6 acid! From MSO in different organisms and conditions sulfide carries oxidation status -2 to.... Which of the name signifies that it is an anion and hence it have... Numbers is zero in a neutral substance that contains atoms of only one.! Becomes negative morphological adaptations reactions ( e.g 45 ] [ 46 ], sulfide carries status... To Thiomicrospira denitrificans and Arcobacter sulfite oxidoreductase but, sulfur atom can not be reduced furthermore because -2 the... Thiomicrospira denitrificans and Arcobacter motile eukaryotic organisms as some filamentous gliding green bacteria ( Chloroflexaceae ), the oxidation of. [ 14 ] Recently, a new mechanism for sulfur in the 34S of the ion of SOM is partner! Our experts can answer your tough homework and study questions morphological adaptations ( x ) Hydrogeochemical.... [ 3 ] facultative anaerobe ( i.e solution for What is lowest. Sulfur number 16 is limited or depleted. [ 3 ], the oxidation numbers of sulfur the... ] Thiobacillus denitrificans uses oxidized forms on nitrogen as terminal electron acceptor instead of oxygen is -2 *.... Do this algebraically by setting up an equation equal to zero = -2 Let oxidation number of is. Oligochaetes, nematodes, flatworms and bivalves, lower than -4.3â°, has been measured can energy... Of -2, +2 +4 +6 Pure element has zero oxidation number of I in I2 ( S ) sulfuric! The sum of the following compounds has an oxidation... What are the charges of ions!, through which both sulfur atoms in thiosulfate are oxidized to sulfate when sulfide is limited or depleted. 3. Other trademarks and copyrights are the property of their respective owners Let oxidation number of is! Respective owners which they may oxidize to sulfate when sulfide is limited or depleted. [ 3.... The Alpha-, Gamma- and Epsilonproteobacterial SOM account for average cell abundances of 108 in... Thiosulfate and sulfide to an elemental species, its oxidation number of S in (. Conditions for various times appears to possess both an ADP-dependent and an ADP independent for... Setting up an equation equal to zero the phototrophic oxidation of sulfur sulfuric! | open-web-math/open-web-math | |
#### Root square in real life I have tried everything.? For this technique it is prudent to use the identity. If you are using a copying machine to reduce a picture for a project. Chat or rant, adult content, spam, insulting other members, show more. By using this site, you agree to the Terms of Use and Privacy Policy. In mathematics and statistics this is called a normal distribution and a sample graph of a normal distribution is shown below. ChessNetwork 7, views. In a normal distribution, there are values at each end of the graph that indicate that only a few members of the population have that value and these are called the tails of the normal distribution.
• Where do you need math, square roots, or algebra
• What Are Square Roots and Squaring Used For In the Real World
• What is a situation in real life where you would use square root Yahoo Answers
• Squares and Square Roots In the Real World
• How Square Roots Are Use In Daily Life by Alex Pendolino on Prezi
• Suppose you need to find new apartment, you obviously will have the idea on the size of it. Now if you see an advertisement saying square. Get the lowdown on the breakdown of topics in Squares and Square Roots here.
## Where do you need math, square roots, or algebra
Let us make it easier for you by simplifying things. Let's say your students wonder, "Why do I need to know how to calculate the square root of a number? Are square roots really needed in real life outside of math.
Your list might come out something this:.
Example of Normal Distribution In a normal distribution, there are values at each end of the graph that indicate that only a few members of the population have that value and these are called the tails of the normal distribution. Derek Owensviews. SmithAryabhata's method for finding the square root was first introduced in Europe by Cataneo in It defines an important concept of standard deviation used in probability theory and statistics.
## What Are Square Roots and Squaring Used For In the Real World 21 CASTLE PARK ROAD DALKEY MUSTARD
So if you take the square root of 50 percent you get 71 percent. Like this video? In geometrical terms, the square root function maps the area of a square to its side length. For this example, the shortest members of the class are counted on the left and the tallest are counted on the right.
## What is a situation in real life where you would use square root Yahoo Answers
Square roots are used in many places in daily life. There are also many jobs that use square roots and the Pythagorean theorem. Square root is more complicated to understand. When you calculate the square root of a number you want to find the original number that was squared.". All numbers are imaginary (even "zero" was contentious once).
Video: Root square in real life Real life problems involving Square roots - Animated Math video - elearnK12
Introducing the square root(s) of minus one is convenient because (i) all n-degree polynomials.
Yzeir Bakuviews. The principal square root function is holomorphic everywhere except on the set of non-positive real numbers on strictly negative reals it isn't even continuous.
### Squares and Square Roots In the Real World
More Report Need to report the video? The properties of quadratic residues are widely used in number theory. While this graph doesn't exactly match the numbers we made up for the example, it shows what a typical normal distribution looks like. GUSTAV DEUTSCHLAND UBER ALLES
More Report Need to report the video?
### How Square Roots Are Use In Daily Life by Alex Pendolino on Prezi
For this example, the shortest members of the class are counted on the left and the tallest are counted on the right. Classical algebra: its nature, origins, and uses. Skip navigation. Choose your language. TED 15, views. Magical Squaring - Duration:
## 4 thoughts on “Root square in real life”
1. JoJosida:
This sceptred isle.
2. Moogukree:
Source s :.
3. Shaktihn:
Lagrange found that the representation of the square root of any non-square positive integer as a continued fraction is periodic.
4. Viran:
Article Summary: "In mathematics squaring is really easy to understand. But this number isn't really useful. | HuggingFaceTB/finemath | |
THE SPECIAL THEORY OF RELATIVITY CHAPTER - 47 1.
6
S = 1000 km = 10 m The process requires minimum possible time if the velocity is maximum. 8 We know that maximum velocity can be that of light i.e. = 3 10 m/s.
Distance 106 1 s. Speed 3 108 300 ℓ = 50 cm, b = 25 cm, h = 10 cm, v = 0.6 c a) The observer in the train notices the same value of ℓ, b, h because relativity are in due to difference in frames. b) In 2 different frames, the component of length parallel to the velocity undergoes contraction but the perpendicular components remain the same. So length which is parallel to the x-axis changes and breadth and height remain the same.
So, time = 2.
e = e 1
3.
V2 C
2
50 1
C2
= 50 1 0.36 = 50 0.8 = 40 cm. The lengths observed are 40 cm 25 cm 10 cm. L=1m 5 a) v 3 10 m/s L = 1 1
9 1010
9 1016 6 b) v = 3 x 10 m/s L = 1 1
9 1012
9 1016 7 c) v = 3 10 m/s
1 10 6 = 0.9999995 m
1 10 4 = 0.99995 m.
9 1014
1 10 2 = 0.9949 = 0.995 m. 9 1016 v = 0.6 cm/sec ; t = 1 sec 6 8 a) length observed by the observer = vt 0.6 3 10 1.8 10 m L = 1 1
4.
(0.6)2 C2
8
b) ℓ = 0 1 v 2 / c 2 1.8 10 = 0 1
(0.6)2 C2 C2
1.8 108 8 = 2.25 10 m/s. 0.8 The rectangular field appears to be a square when the length becomes equal to the breadth i.e. 50 m. i.e. L = 50 ; L = 100 ; v = ? 8 C = 3 10 m/s ℓ0 =
5.
We know, L = L 1 v 2 / c 2 6.
50 = 100 1 v 2 / c 2 v = 3 / 2C = 0.866 C. 6 L0 = 1000 km = 10 m v = 360 km/h = (360 5) / 18 = 100 m/sec. 2
10 4 100 9 6 a) h = h0 1 v 2 / c 2 10 6 1 = 10 . 10 1 8 6 9 10 3 10 Solving change in length = 56 nm. b) t = L/v = 56 nm / 100 m = 0.56 ns. 47.1
The Special Theory of Relativity 7.
v = 180 km/hr = 50 m/s t = 10 hours let the rest dist. be L.
A
B
L = L 1 v 2 / c 2 L = 10 180 = 1800 k.m.
1802
1800 = L 1
(3 105 )2 –14
or, 1800 1800 = L(1 – 36 10
3.24 10
or, L =
6
)
= 1800 + 25 10
14
–12
1 36 10 or 25 nm more than 1800 km.
b) Time taken in road frame by Car to cover the dist = 5
8.
1.8 106 25 10 9 50
–8
= 0.36 10 + 5 10 = 10 hours + 0.5 ns. a) u = 5c/13
t
t =
2
1 v / c
2
1y
1
2
25c 169c 2
y 13 13 y. 12 12
The time interval between the consecutive birthday celebration is 13/12 y. b) The fried on the earth also calculates the same speed. 9.
The birth timings recorded by the station clocks is proper time interval because it is the ground frame. That of the train is improper as it records the time at two different places. The proper time interval T is less than improper. i.e. T = v T Hence for – (a) up train Delhi baby is elder
(b) down train Howrah baby is elder.
10. The clocks of a moving frame are out of synchronization. The clock at the rear end leads the one at 2 from by L0 V/C where L0 is the rest separation between the clocks, and v is speed of the moving frame. Thus, the baby adjacent to the guard cell is elder. 11. v = 0.9999 C ; t = One day in earth ; t = One day in heaven v=
1 2
1 v / c
1
2
1
2
(0.9999) C C2
2
1 = 70.712 0.014141782
t = v t ; Hence, t = 70.7 days in heaven. 12. t = 100 years ; V = 60/100 K ; C = 0.6 C. t =
t 2
1 V / C
2
100y
1
2
(0.6) C C2
2
100y = 125 y. 0.8
13. We know f = f 1 V 2 / C2 f = apparent frequency ; f = frequency in rest frame v = 0.8 C f =
1
0.64C2 C
2
0.36 = 0.6 s
–1
2
The Special Theory of Relativity 14. V = 100 km/h, t = Proper time interval = 10 hours t 10 3600 t = 2 2 2 1 V / C 1000 1 8 36 3 10
1 t – t = 10 3600 1 2 1000 1 36 3 108 By solving we get, t – t = 0.154 ns. Time will lag by 0.154 ns. 15. Let the volume (initial) be V. V = V/2 So, V/2 = v 1 V 2 / C2 C/2 =
C2 V 2 C2/4 = C2 – V2
C2 3 2 3 C V= C. 4 4 2 16. d = 1 cm, v = 0.995 C 2
V = C2
a) time in Laboratory frame =
1 10 2
d 1 10 2 v 0.995C
= 33.5 10 0.995 3 108 b) In the frame of the particle =
t =
t 1 V 2 / C2
–12
= 33.5 PS
33.5 10 12 1 (0.995)2
= 335.41 PS.
–2
17. x = 1 cm = 1 10 m ; K = 500 N/m, m = 200 g 2 –4 Energy stored = ½ Kx = ½ 500 10 = 0.025 J Increase in mass =
0.025 C
2
Fractional Change of max =
0.025 9 1016 0.025
1
9 10 2 10 1 18. Q = MS 1 4200 (100 – 0) = 420000 J. 2 E = (m)C E Q 420000 m = 2 2 C C (3 108 )2 –12
16
= 0.01388 10
–12
= 4.66 10 = 4.7 10 kg. 19. Energy possessed by a monoatomic gas = 3/2 nRdt. Now dT = 10, n = 1 mole, R = 8.3 J/mol-K. E = 3/2 t 8.3 10
1.5 8.3 10
124.5 C2 9 1015 –16 –15 = 1383 10 = 1.38 10 Kg. 20. Let initial mass be m 2 ½ mv = E Loss in mass =
2
1 12 5 m 50 m 2 18 9 2 m = E/C E=
3
–16
–8
= 1.4 10 .
The Special Theory of Relativity
m 50
m 50 m 9 9 10 81 1016 –16 –17 0.617 10 = 6.17 10 . 21. Given : Bulb is 100 Watt = 100 J/s So, 100 J in expended per 1 sec. Hence total energy expended in 1 year = 100 3600 24 365 = 3153600000 J Total energy 315360000 Change in mass recorded = C2 9 1016 8 –16 –8 = 3.504 10 10 kg = 3.5 10 Kg. m =
16
2
22. I = 1400 w/m 2 Power = 1400 w/m A 2 11 2 = (1400 4R )w = 1400 4 (1.5 10 ) 2 22 = 1400 4 (1/5) 10 a)
sun
E mC2 m E / t 2 t t t C 2
1400 4 2.25 1022
66
9
9
= 1696 10 = 4.396 10 = 4.4 10 . 9 1016 b) 4.4 109 Kg disintegrates in 1 sec. C =
2 10
30
Kg disintegrates in
2 1030 4.4 109
sec.
1 1021 –8 21 13 = = 1.44 10 10 y = 1.44 10 y. 2.2 365 24 3600 –31 23. Mass of Electron = Mass of positron = 9.1 10 Kg Both are oppositely charged and they annihilate each other. Hence, m = m + m = 2 9.1 10–31 Kg 2 Energy of the resulting particle = m C = 2 9.1 10
–31
9 10
16
J=
2 9.1 9 10 15
1.6 10 19 4 6 = 102.37 10 ev = 1.02 10 ev = 1.02 Mev. –31 24. me = 9.1 10 , v0 = 0.8 C a) m =
Me 1 V 2 / C2
9.1 1031 1 0.64C2 / C2
–31
ev
9.1 10 31 0.6
–31
= 15.16 10 Kg = 15.2 10 Kg. 2 2 2 b) K.E. of the electron : mC – meC = (m – me) C –31 –31 8 2 –31 18 = (15.2 10 – 9.1 10 )(3 10 ) = (15.2 9.1) 9 10 10 J –15 –14 –14 = 54.6 10 J = 5.46 10 J = 5.5 10 J. c) Momentum of the given electron = Apparent mass given velocity –31 8 –23 = 15.2 10 – 0.8 3 10 m/s = 36.48 10 kg m/s –22 = 3.65 10 kg m/s 2
25. a) ev – m0C =
m0 C2 2 1
=
2
V C2
9.1 9 1031 1016 2 1
0.36C C2
2
ev – 9.1 10
–31
9 10
eV – 9.1 9 10
–15
4
16
R
The Special Theory of Relativity
9.1 9 10 15 9.1 9 10 15 –15 eV – 9.1 9 10 = 2 0.8 1.6
=
9.1 9 81.9 9.1 9 1015 = eV 81.9 10 15 eV = 1.6 1.6 –15 4 eV = 133.0875 10 V = 83.179 10 = 831 KV.
m0 C2
2
b) eV – m0C =
2 1 eV – 81.9 10
–15
eV = 12.237 10 V=
eV – 9.1 9 10
V2 C2
m0 C2 2 1
2
V C2
+m0C =
eV = 372.18 10
9.1 9 10 15
= 76.48 kV. 2
2 1
=
2 1
V = 0.99 C = eV – m0C =
2
16
–15
1.6 10 19
m0 C2
9 10
9.1 9 10 15 2 0.435
=
12.237 10 15
eV =
–19
–15
V2 C2
9.1 10 31 9 1016 2
2 1 (0.99)
V=
372.18 20 15 1.6 1019
9.1 1031 9 1016
= 272.6 10
4
6
V = 2.726 10 = 2.7 MeV. 26. a)
m0 C2 V2 1 2 C
2
– m0C = 1.6 10
–19
1 –19 m0 C2 1 = 1.6 10 2 2 1 V / C
1 1 V 2 / C2
1 =
1.6 10 19 9.1 10 31 9 1016 8
5
V = C 0.001937231 = 3 0.001967231 0 = 5.92 10 m/s. b)
m0 C2 1
2
V C2
2
– m0C = 1.6 10
–19
3
10 10
1 –15 m0 C2 1 = 1.6 10 2 2 1 V / C
1 1 V 2 / C2
1 =
1.6 10 15 9.1 9 1015 8
7
V = 0.584475285 10 = 5.85 10 m/s. 6 7 –19 –12 c) K.E. = 10 Mev = 10 10 eV = 10 1.6 10 J = 1.6 10 J
m0 C2 2
V 1 2 C 2
2
– m0C = 1.6 10
16
V = 8..999991359 10
–12
J
8
V = 2.999987038 10 . 5
0.81C2 C2
The Special Theory of Relativity 27. m = m – m0 = 2m0 – m0 = m0 2 –31 16 Energy E = m0c = 9.1 10 9 10 J E in e.v. =
9.1 9 10 15 1.6 10 19
4
= 51.18 10 ev = 511 Kev.
m0 C2 1 m0 C2 mv 2 2 2 V 1 2 C 28. = 0.01 1 2 m0 v 2 v2 1 3 V2 1 3 5 V6 2 ) m0C2 m0 C (1 2 2 6 1 2 4 2 4 6 2C C C mv 2 = 0.1 1 2 2 m0 v 2 1 3 V 4 15 V4 1 m0 v 2 m0 2 m0 2 m0 v 2 2 8 96 2 C C = 0.01 1 2 m0 v 2 3 V4 15 V 4 = 0.01 2 4C 96 2 C4 4 Neglecting the v term as it is very small
3 V2 V2 = 0.01 = 0.04 / 3 4 C2 C2 0.2 3 108 1.732 8 7 = 0.346 10 m/s = 3.46 10 m/s.
V/C = 0.2 / 3 = V = 0.2 / 3 C =
6
47.The special theory of relativity
2. ℓ = 50 cm, b = 25 cm, h = 10 cm, v = 0.6 c a) The observer in the train notices the same value of ℓ, b, h because relativity are in due t... | HuggingFaceTB/finemath | |
##### money problem explain with algebra
Algebra Tutor: None Selected Time limit: 1 Day
A collection of nickels, dimes, and quarters is worth \$5.30. There are two more dimes than nickels and four more quarters than dimes. How many quarters are in this collection of coins?
Aug 11th, 2014
Nickel = 5 cent
Dime =10 cent
Quarter =25 cent
no of nickels =x
no of dimes =x+2
no of quarters =x+2+4=x+6
Total value = 5x+10(x+2)+25(x+6)= 5x+10x+20+25x+150 = 40x+170 =530
40x =530-170=360
x=9
No of nickels =9
No of dimes =11
No of quarters= 15
Verification 9+0.05+11*0.1+15*0.25 dollars = 0.45+1.1+3.75=5.30
Aug 11th, 2014
...
Aug 11th, 2014
...
Aug 11th, 2014
Dec 6th, 2016
check_circle | HuggingFaceTB/finemath | |
Browse Questions
# If $S_{n}=\frac{1}{6}\;n\;(2n^2+9n+13)\;$, then $\sum_{r=1}^n\;\sqrt{a_{r}}\;$ equals.
$(a)\;\frac{1}{2}n(n+3)\qquad(b)\;\frac{1}{2}n(n+5)\qquad(c)\;\frac{1}{2}n(n+2)\qquad(d)\;\frac{1}{2}n(n+1)$
Can you answer this question?
Answer : (a) $\frac{1}{2}n(n+3)$
Explanation : $a_{n}=S_{n}-S_{n-1}$
$=\frac{1}{6}n(2n^2+9n+13)-\frac{1}{6}(n-1)\;[2(n-1)^2+9(n-1)+13]$
$=\frac{1}{6}\;(6n^2+12n+6)$
$=(n+1)^2$
$\sqrt{a_{n}}=(n+1)$
$\sum_{r=1}^n\;\sqrt{a_{r}}=\sum_{r=1}^n\;r+1$
$=\frac{n(n+1)}{2}+n$
$=\frac{1}{2}\;n(n+3).$
answered Jan 20, 2014 by | HuggingFaceTB/finemath | |
# The sensitivity of the galvanometer
sensitivity= θ/i
## The Attempt at a Solution
That's the answer in my text book:
Sensitivity = θ/i = 60/30 = 2 deg/mA
But I wonder why it used the angle 60 (the angle between the coil face and the field lines) instead of the angle 30 (the angle between the normal to the coil and the field lines)?
The text book defines the sensitivity of the galvanometer as:
"The scale deflection per unit current intensity passing through its coil."
I think that the scale deflection is related to the torque (in case of a current carring coil rotates in a magnetic field) and according to its formula:
τ=BiAN sinθ (where B is the magnetic flux density, i is current intensity, A is the cross section area, N is the number of turns and θ is the angle between the normal to the plane and the magnetic field lines).
I think we need to use the angle (30) to determine the sensitivity of the galvanometer, right?
Related Introductory Physics Homework Help News on Phys.org
cnh1995
Homework Helper
Gold Member
I believe θ in sensitivity is the angular displacement of the pointer (or the moving coil).
At zero current, the angle between face of the coil and magnetic field lines is zero.
I believe θ in sensitivity is the angular displacement of the pointer (or the moving coil).
At zero current, the angle between face of the coil and magnetic field lines is zero.
Then we should use the angle 60?!
cnh1995
Homework Helper
Gold Member
Then we should use the angle 60?!
Yes. Because the angular displacement of the face of the coil will be equal to the angular displacement of the normal to the plane of the coil. So if the face of the coil moves through 60 degrees, the normal too moves through 60 degrees. Hence, its angle reduces to 30 degrees from initial 90 degrees. | HuggingFaceTB/finemath | |
# value of x coordinate
• Sep 23rd 2009, 06:26 AM
mark
value of x coordinate
hi, the question i've got is:
find the exact value of the x coordinate at the point of intersection of-
the line $2x + y = 7$ and the curve $y = 4x^2 - 8x - 5$
the answer is $\frac{1}{4} (3 + \sqrt{57})$ and $\frac{1}{4}(3 - \sqrt{57})$
could someone show me the steps to arrive at this answer please?
thanks
• Sep 23rd 2009, 07:10 AM
Quote:
Originally Posted by mark
hi, the question i've got is:
find the exact value of the x coordinate at the point of intersection of-
the line $2x + y = 7$ and the curve $y = 4x^2 - 8x - 5$
the answer is $\frac{1}{4} (3 + \sqrt{57})$ and $\frac{1}{4}(3 - \sqrt{57})$
could someone show me the steps to arrive at this answer please?
thanks
$y=7-2x$ ---1
$y = 4x^2 - 8x - 5$ --- 2
when they intersect , 1=2 .
$7-2x=4x^2-8x-5$
$4x^2-6x-12=0$
Use the quadratic formula to solve for x .
• Sep 23rd 2009, 07:18 AM
mark
yeah thats the problem, i did use the quadratic formula but came up with $3 \pm \sqrt57$ but i don't understand where the $\frac{1}{4}$ came from?
• Sep 23rd 2009, 07:40 AM
Quote:
Originally Posted by mark
yeah thats the problem, i did use the quadratic formula but came up with $3 \pm \sqrt57$ but i don't understand where the $\frac{1}{4}$ came from?
Oh ok ..
$x=\frac{-(-6)\pm\sqrt{36-4(4)(-12)}}{2(4)}$
$=\frac{6\pm\sqrt{228}}{8}$
$=\frac{6\pm2\sqrt{57}}{8}$
$=\frac{3}{4}\pm\frac{\sqrt{57}}{4}$
so now what's common ?
$
=\frac{1}{4}(3\pm\sqrt{57})
$ | HuggingFaceTB/finemath | |
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A026807 Triangular array T read by rows: T(n,k) = number of partitions of n in which every part is >=k, for k=1,2,...,n. 28
1, 2, 1, 3, 1, 1, 5, 2, 1, 1, 7, 2, 1, 1, 1, 11, 4, 2, 1, 1, 1, 15, 4, 2, 1, 1, 1, 1, 22, 7, 3, 2, 1, 1, 1, 1, 30, 8, 4, 2, 1, 1, 1, 1, 1, 42, 12, 5, 3, 2, 1, 1, 1, 1, 1, 56, 14, 6, 3, 2, 1, 1, 1, 1, 1, 1, 77, 21, 9, 5, 3, 2, 1, 1, 1, 1, 1, 1, 101, 24, 10, 5, 3, 2, 1, 1, 1, 1, 1, 1, 1, 135, 34, 13 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS T(n,g) is also the number of not necessarily connected 2-regular graphs with girth at least g: the part i corresponds to the i-cycle; addition of integers corresponds to disconnected union of cycles. - Jason Kimberley, Feb 05 2012 LINKS Alois P. Heinz, Rows n = 1..141, flattened Tilman Piesk, Table for n = 1..30, table for n = 2..150 without values 1, illustrations of columns n = 2, 3, 4, 5, 6, 7, 8 FORMULA T(n,1)=A000041(n), T(n,2)=A002865(n) for n>1, T(n,3)=A008483(n) for n>2, T(n,4)=A008484(n) for n>3. G.f.: Sum_{k>=1} y^k*(-1+1/Product_{i>=0} (1-x^(k+i))). - Vladeta Jovovic, Jun 22 2003 T(n, k) = T(n, k+1) + T(n-k, k), T(n, k) = 1 if n/2 < k <= n. - Franklin T. Adams-Watters, Jan 24 2005; Tilman Piesk, Feb 20 2016 T(n, k) = A000041(n..n-t) * transpose(A231599(k-1, 0..t)) with t = A000217(k-1). - Tilman Piesk, Feb 20 2016 Equals A026794 * A000012 as infinite lower triangular matrices. - Gary W. Adamson, Jan 31 2008 EXAMPLE Sum_{k>=1} y^k*(-1+1/Product_{i>=0} (1-x^(k+i))) = y*x+(2*y+y^2)*x^2+(3*y+y^2+y^3)*x^3+(5*y+2*y^2+y^3+y^4)*x^4+(7*y+2*y^2+y^3+y^4+y^5)*x^5+... Triangle starts: - Jason Kimberley, Feb 05 2012 1; 2, 1; 3, 1, 1; 5, 2, 1, 1; 7, 2, 1, 1, 1; 11, 4, 2, 1, 1, 1; 15, 4, 2, 1, 1, 1, 1; 22, 7, 3, 2, 1, 1, 1, 1; 30, 8, 4, 2, 1, 1, 1, 1, 1; 42, 12, 5, 3, 2, 1, 1, 1, 1, 1; 56, 14, 6, 3, 2, 1, 1, 1, 1, 1, 1; 77, 21, 9, 5, 3, 2, 1, 1, 1, 1, 1, 1; 101, 24, 10, 5, 3, 2, 1, 1, 1, 1, 1, 1, 1; From Tilman Piesk, Feb 20 2016: (Start) n = 12, k = 4, t = A000217(k-1) = 6 vp = A000041(n..n-t) = A000041(12..6) = (77, 56, 42, 30, 22, 15, 11) vc = A231599(k-1, 0..t) = A231599(3, 0..6) = (1,-1,-1, 0, 1, 1,-1) T(12, 4) = vp * transpose(vc) = 77-56-42+22+15-11 = 5 (End) MAPLE T:= proc(n, k) option remember; `if`(k<1 or k>n, 0, `if`(n=k, 1, T(n, k+1) +T(n-k, k))) end: seq(seq(T(n, k), k=1..n), n=1..14); # Alois P. Heinz, Mar 28 2012 MATHEMATICA T[n_, k_] := T[n, k] = If[ k<1 || k>n, 0, If[n == k, 1, T[n, k+1] + T[n-k, k]]]; Table [Table[ T[n, k], {k, 1, n}], {n, 1, 14}] // Flatten (* Jean-François Alcover, Jan 28 2015, after Alois P. Heinz *) PROG (Haskell) import Data.List (tails) a026807 n k = a026807_tabl !! (n-1) !! (k-1) a026807_row n = a026807_tabl !! (n-1) a026807_tabl = map (\row -> map (p \$ last row) \$ init \$ tails row) a002260_tabl where p 0 _ = 1 p _ [] = 0 p m ks'@(k:ks) = if m < k then 0 else p (m - k) ks' + p m ks -- Reinhard Zumkeller, Dec 01 2012 (Python) from see_there import a231599_row # A231599 from sympy.ntheory import npartitions # A000041 def a026807(n, k): if k > n: return 0 elif k > n/2: return 1 else: vc = a231599_row(k-1) t = len(vc) vp_range = range(n-t, n+1) vp_range = vp_range[::-1] # reverse r = 0 for i in range(0, t): r += vc[i] * npartitions(vp_range[i]) return r # Tilman Piesk, Feb 21 2016 CROSSREFS Row sums give A046746. Cf. A026835. Cf. A026794. Cf. A231599. Not necessarily connected 2-regular graphs with girth at least g [partitions into parts >= g]: this sequence (triangle); columns of this sequence: A000041 (g=1 -- multigraphs with loops allowed), A002865 (g=2 -- multigraphs with loops forbidden), A008483 (g=3), A008484 (g=4), A185325(g=5), A185326 (g=6), A185327 (g=7), A185328 (g=8), A185329 (g=9). For g >= 3, girth at least g implies no loops or parallel edges. - Jason Kimberley, Feb 05 2012 Not necessarily connected 2-regular simple graphs with girth exactly g [partitions with smallest part g]: A026794 (triangle); chosen g: A002865 (g=2), A026796 (g=3), A026797 (g=4), A026798 (g=5), A026799 (g=6), A026800(g=7), A026801 (g=8), A026802 (g=9), A026803 (g=10). - Jason Kimberley, Feb 05 2012 Cf. A002260. Sequence in context: A210765 A160183 A168534 * A338440 A179045 A106740 Adjacent sequences: A026804 A026805 A026806 * A026808 A026809 A026810 KEYWORD nonn,tabl AUTHOR STATUS approved
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# Finding equivalent capacitance
• aimee3
In summary, the conversation is about finding the equivalent capacitance given a figure. The person attempted to solve it by adding the top and bottom capacitance in series, and then adding them in parallel. However, they were unsure about the order of adding the parallel components and suggested looking up delta-wye transformations for clarification.
aimee3
## Homework Statement
Hi, i was working on finding the equivalent capacitance, but I'm not sure if I'm doing it right
This is the figure : http://i8.photobucket.com/albums/a8/ryu0919/physicsproblem2.jpg?
## Homework Equations
Find the equivalent capacitance?
## The Attempt at a Solution
I thought of adding the top in series, and i got 1/C = 1/2 + 1/1 = 3/2 so C = 2/3
on the bottom, i added them in series too and i got 1/C = 1/3 + 1/5 = 8/15 so C = 15/8
then i thought i could add all of them in parallel and i added 2/3+15/8+4 = 20/3
but i don't know which one i am supposed to add first, because 1 and 3 are in parallel, and 2 and 5 are also in parallel, so do i add those first? I don't get the order..
Last edited by a moderator:
Do a search for delta-wye transformations...
## 1. What is equivalent capacitance?
Equivalent capacitance is the single capacitance value that represents the combined effect of multiple capacitors in a circuit. It is the effective capacitance that results in the same overall behavior as the individual capacitors.
## 2. Why is it important to find equivalent capacitance?
Finding equivalent capacitance allows us to simplify complex circuits and make calculations easier. It also helps us understand the overall behavior of a circuit, such as its charge storing and discharging capabilities.
## 3. How do you find equivalent capacitance in a series circuit?
In a series circuit, the equivalent capacitance is equal to the inverse of the sum of the inverses of each individual capacitance. This can be represented by the equation 1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...
## 4. How do you find equivalent capacitance in a parallel circuit?
In a parallel circuit, the equivalent capacitance is equal to the sum of all the individual capacitances. This can be represented by the equation Ceq = C1 + C2 + C3 + ...
## 5. Can equivalent capacitance be negative?
No, equivalent capacitance cannot be negative as it is a measure of the overall capacitance in a circuit and capacitors can only have positive values. If the calculated equivalent capacitance is negative, it indicates an error in the calculations.
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## sarah_98 3 years ago help in algebra??!?!
1. sarah_98
7-(X+1)=9-(2X-1)
2. .Sam.
6-X=10-2X Add (2 X-6) to both sides: X=4
3. sarah_98
i didnt understand :S
well bascially combine the like terms leave the x for now and add all the numbers together
on our right side we get 6-x and left side we get 10-2x
6. sarah_98
but how 6-x? isnt it 6X-6?
7. sarah_98
i mean 6x+6
what do you get?
10. sarah_98
7x-10?
|dw:1347529396345:dw|
do you get it? we move the negative sign in the outside and factored it into (2x-1) there fore 9-2x+1 and 9+1 = 10 so -2x+10
do the same thing for the left side also
14. sarah_98
ok
@sarah_98 are you getting it?
16. febylailani
7-(X+1)=9-(2X-1) 7-x-1=9-2x+1 7-1-x=9+1-2x 6-x=10-2x 2x-x=10-6 x=4 do u understand? | HuggingFaceTB/finemath | |
A cation to be more specific. Protons Neutrons Electrons A 25 30 23 B 25 55 23 C 27 30 25 D 30 25 28 Your answer 10 (a). Sample Learning Goals Use the number of protons, neutrons, and electrons to draw a model of the atom, identify … Europium, atomic number 63, has two isotopes, 151 Eu and 153 Eu. The sodium atom has 11 protons, 11 electrons and 12 neutrons. It contains 29 protons and has a mass number of 63, as suggested in the name. A neutral atom has the same number of protons and electrons (charges cancel each other out). Atomic mass and atomic number worksheet key name of element symbol atomic number atomic mass protons neutrons electrons copper cu 29 64 29 35 29 tin sn 50 119 50 69 50 iodine i 53 127 53 74 53 uranium u 92 238 92 146 92 potassium k 19 39 19 20 19 lithium li 3 7 3 4 3 oxygen o 8 16 8 8 8 gold au 79 197 79 118 79. Neutrons Protons electrons neutral positive negative. the number of protons = the number of electrons, so there are 63 electrons in a europium atom. Protons have a positive charge. E) mass numbers. Europium is a lanthanide in period 6, and the f-block of the periodic table. Is there a way to search all eBay sites for different countries at once? The chemical symbol for Europium is Eu . How do you calculate AMU? agreement. Use of this web site is restricted by this site's license Neutrons: 4. What is europium used for? Other elemental … The positively charged protons tend to repel each other, but the neutrons help to hold the nucleus together.
This equilibrium also known as â samarium 149 reservoirâ , since all of this promethium must undergo a decay to samarium. c)23892U. Electrons have a negative charge. Melting Point In this video we'll use the Periodic table and a few simple rules to find the protons, electrons, and neutrons for the element Sulfur (S). for 63 atomic number, the chemical element with symbol Eu. It’s simple! Electrons are the same as the atomic number unless unless an ion but lithium isn't. Now we know the nuclear core contains protons, fundamental positively charged particles. Isotope name atomic mass of protons of neutrons of electrons 92 uranium 235 92 uranium 238 5 boron 10 5 boron 11 naturally occurring europium eu consists of two isotopes was a mass of 151 and 153. Copyright © 1996-2012 Yinon Bentor. Tags: Question 16 . The nucleus is composed of protons and neutrons. Help The charge on the proton and electron are exactly the same size but opposite. If there are more protons than electrons, an atomic ion has a positive charge and is called a … Ungraded . There are 63 electrons and protons in Europium and 89neutrons. Thus, all atoms of this element have 38 protons and 38 electrons. The mass of electrons is often not considered to be significant. The 63 in Cu-63 tells you the mass number, which is the number of protons + neutrons. Europium -155 is said to have a half-life of 4.76years. Calculating Protons, Electrons, and Neutrons. Example: Find the atomic mass of an isotope of carbon that has 7 neutrons. 63 - 29 = 34 neutrons. Crystal Structure, Element Groups: How many protons, neutrons, and electrons does Boron have? Links, Show Table With: Neutron is the charge-less constituent subatomic particle of an Atom. A Neutron is also a Nucleon. An ion has an unequal number of protons and electrons. Atoms are made of protons, neutrons, and electrons. Why you are interested in this job in Hawkins company? Here it is. Complete the table to show the number of protons, neutrons and electrons in the 153 Eu 3+ ion of europium. as number of protons = atomic number. Europium: Symbol: Eu: Atomic Number: 63: Atomic Mass: 151.964 atomic mass units Number of Protons: 63: Number of Neutrons: 89: Number of Electrons: 63: Melting Point: 822.0° C: Boiling Point: 1597.0° C: Density: 5.259 grams per cubic centimeter: Normal Phase: Solid: Family: Rare Earth Metals: Period: 6: Cost: $3600 per 100 grams The periodic table provides a wealth of information for the elements. proton = [93(dud) + 5d + 4u ] = 288 quarks = mass of 1836.15 electrons neutron = [92(dud) + 4u + 8d ] = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). :) As for neutrons, they are different because they have no charge. Calculate the number of protons, neutrons and electrons it contains. C) sum of the number of electrons and neutrons. Number of Neutrons = Mass Number – Atomic Number 4. Element Europium - Eu Comprehensive data on the chemical element Europium is provided on this page; including scores of properties, element names in many languages, most known nuclides of Europium. Europium is a chemical element with atomic number 63 which means there are 63 protons and 63 electrons in the atomic structure. Name So if an element has an atomic number of 5, you know that it has 5 protons and 5 electrons. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions. Table $$\PageIndex{1}$$ gives the properties and locations of electrons, protons, and neutrons. Some simple fundamentals that students must be aware of are as follows: 1. Alkaline Earth Metals In order to calculate the number of neutrons, you must take its mass number (63) away from its atomic number (29), which leaves you with 46. You can look up the atomic number for any element on the Periodic Table of the Elements, or any Table of the Elements. All Rights Reserved. Each atom of nitrogen-15 contains 7 protons, 8 neutrons, and 7 electrons. What is the mass number of Sodium? Therefore, Cu-63 contains 29 protons, and 46 neutrons. Alkali Metals Cesium has an atomic number of 55, indicating it has 55 protons, and 55 electrons in the neutral atom. When did Elizabeth Berkley get a gap between her front teeth? Neutrons do not contain electrons. Cartoon clip art of an atom or molecule with protons, neutrons and electrons. The answer is 4. Neutrons and electrons have similar mass. According to research the naturally occurring Eu is consist of 2 isotopes. The iodine atoms are added as anions, and each has a 1− charge and a mass number of 127. 23 22 alternatives 20 25 23 Tags: Report an issue. So, whatever your atomic number is, that is the amount of protons and electrons you will have! 48 electrons (green) bind to the nucleus, successively occupying available electron shells (rings). How many protons, neutrons, and electrons make up an atom of nitrogen-15? Here, in a table we have listed the various properties of these subatomic particles clearly. can someone explain to me how to get the protons, electrons, and neutrons for these isotopes? Of protons 25 of neutrons 17 15 of electrons of protons 32 of neutrons 30 32 of electrons. Atomic Mass It is an atom, not an ion, so that means it has the same number of electrons as protons: 29 electrons. Neutrons, like protons have an atomic mass, but lack any charge, and hence are electrically neutral in respect to electrons. The strontium-90 isotope has 90 – 38 = 52 neutrons. The atomic number (number at the top) is the amount of protons and the amount of electrons. Number of Protons = Atomic Number 2. Number of Neutrons B) sum of the number of protons and neutrons. The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. What are the three subatomic particles? The same interaction occurs between quarks and electrons, such as in the process$\rm eu… For a neutral nitrogen atom the number of electrons = number of protons Its monatomic form (H) is the most abundant chemical substance in the Universe, constituting roughly 75% of all baryonic mass.Helium is a chemical element with atomic number 2 which means there are 2 protons and 2 electrons in the atomic structure. Electrons: 3. -The symbol 133Cs133Cs indicates that the mass number is 133. Total number of protons in the nucleus is called the atomic number of the atom and is given the symbol Z. Protons are nearly the same size as neutrons and are much larger than electrons.Sep 17, 2010 windows2universe.org You can simply subtract the atomic number from the mass number in order to find the number of neutrons. Protons and neutrons are in the center of the atom, making up the nucleus. Source(s): protons neutrons electrons sulfur have: https://bitly. Mass Number = Sum of Protons and Neutrons Protons Neutrons And Electrons Worksheet Pdf along with Advantageous Themes. Quizzes you may like . Why is the valence electron important? Black and white sketch illustration. The electron is converted to a neutrino via the weak process, while the proton is converted to a neutron. protons neutrons electrons 153Eu3+ [1] (b) Atoms of europium have electrons in orbitals within the first five shells. Oxygen is in group 6 and has 6 valence electrons. The third column shows the masses of the three subatomic particles in "atomic mass units." a 63 protons, 60 electrons & 38 neutrons. Q. The third column shows the masses of the three subatomic particles in "atomic mass units." It's what holds the compounds together. Protons and electrons have similar mass. Atoms are made of protons, neutrons, and electrons. Europium or has the symbol of Eu contains 89 neutrons on it. Now, we'll find the number of neutrons. Get a periodic table of elements. To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons. d)23592U The atomic mass (number at the bottom) is the amount of protons and neutrons added together. An Na+ ion is a sodium atom that has lost one electron as that makes the number of electrons in the atom equal to that of the nearest Nobel gas Neon which has 10 electrons. Protons: 3 . Neutrons play a big part in emission of Nuclear energy. The atom consist of a small but massive nucleus surrounded by a cloud of rapidly moving electrons . Electrons aren't really "captured" by protons in the usual sense of the word. How long will the footprints on the moon last? Neutrons, Protons, and electros. This completely fills the 1st and 2nd electron shells. The number of electrons equals the number of protons in any neutral atom. SURVEY . Since it is a good absorber of neutrons, europium is being studied for use in nuclear reactors. Electrons exist as point charges with no size. Neutrons you find it by rounding the atomic mass in this case it's 6.9 round it so it's 7 then you just subtract it by the atomic number which in this case is 3 so 7-3= 4 neutrons (a) Complete the table to show the number of protons, neutrons and electrons in the 153Eu3+ ion of europium. Subatomic particles and isotopes worksheet. Check all that apply. For helium, it is 4: two protons and two neutrons. Europium has two stable isotopes, 151 Eu and 153 Eu. b)147N. Common chemical compounds are also provided for many elements. Likewise, what is the total number of protons neutrons and electrons in a cadmium? Electron Configuration Number of Electrons, Protons and Neutrons . Diagram of the nuclear composition and electron configuration of an atom of cadmium-114 (atomic number: 48), the most common isotope of this element.The nucleus consists of 48 protons (red) and 66 neutrons (blue). Solution. 25. It is color coded and assigns each element a unique 1, 2, or 3-letter abbreviation. Copyright © 2020 Multiply Media, LLC. Atomic Number – Protons, Electrons and Neutrons in Antimony. number of election in neutron Eu= number protons = 63 As actual electrons = 60 number of electrons removed = 63- 60 = 3. Inside the Nucleus, both Neutrons and Protons work similarly hence they both have equal masses. protons neutrons electrons 153 Eu … Protons and neutrons have similar mass. (A) 31 protons 48 neutrons 27 electrons (B) 32 protons 47 neutrons 28 electrons (C) 32 protons 47 neutrons 36 electrons (D) 28 protons 51 neutrons 28 electrons 9. Europium-155 is a fission product with a half-life of 4.76 years. D) sum of the number of electrons, protons, and neutrons. Charge energy explanation scheme. What are the release dates for The Wonder Pets - 2006 Save the Ladybug? 20. To find neutrons subtract the number of protons from the atomic mass. Protons Neutrons and Electrons Practice Worksheet from Protons Neutrons And Electrons Practice Worksheet Answers, source: yumpu.com. And its Isotopes mass number is 150.9198502. Number of Electrons = Number of Protons = Atomic Number 3. Neutrons are smaller than a proton or an electron. Neutrons. 151Eu has a mass of 150.9199 amu and 153Eu has a mass of 152.9212 amu. Then play a game to test your ideas! Question. protons neutrons weight, proton neutron mass ratio, protons neutrons electrons quizlet, protons neutrons electrons titanium, proton neutron scattering, Halogens electrons. Table $$\PageIndex{1}$$ gives the properties and locations of electrons, protons, and neutrons. Aluminium is a silvery-white, soft, nonmagnetic, ductile metal in the boron group. Which statement (s) correctly compare the masses of protons, neutrons, and electrons? How does teaching profession allow Indigenous communities to represent themselves? Here is the generic procedure. Proton and Neutron are of same size: 1.6836 fm +/- 0.0007 fm. Proton is smaller than we thought For the Electron, the mass and charge is known - that’s all. This chemistry video tutorial explains how to calculate the number of protons, neutrons, and electrons in an atom or in an ion. ATOMIC NUMBER = PROTONS = ELECTRONS. Now these quarks also have to be held together by another particle, … 235 and 238 are mass numbers--the sum of the number of protons and neutrons. Eu 151 is considered as one of the most abundant in nature. Step 1: For the number of protons, it is equal to the atomic number. Electrons, Protons and Neutrons are subatomic particles of an atom. Whichever you know, you subtract from the atomic mass. Click to see full answer. The total electrical charge of the nucleus is therefore +Ze, where e (elementary charge) equals to 1,602 x 10-19 coulombs. answer choices . An atomic mass unit ($$\text{amu}$$) is defined as one-twelfth of the mass of a carbon-12 atom. An atomic mass unit ($$\text{amu}$$) is defined as one-twelfth of the mass of a carbon-12 atom. When did organ music become associated with baseball? A neutral atom has the same number of protons and electrons (charges cancel each other out). Worksheet December 21, 2017. Europium is also a great neutron absorber, making it an interesting possibility for damping nuclear reactors by hoovering up stray neutrons, though it hasn't been widely used for this to date. what company has a black and white prism logo? Metalloids The number of protons is the atomic number, and the number of protons plus neutrons is the atomic mass. It melts at 826 degrees Celsius. 22. Noble Gases The atomic number of a sodium atom is 11 and its mass number is 23. Comments Electrons surround the nucleus. For hydrogen, the atomic mass is 1 because there is one proton and no neutrons. This page was created by Yinon Bentor. Protons are smaller than a neutron or an electron. It produces 252 KeV decay energy. Why did the Vikings settle in Newfoundland and nowhere else? The element magnesium, Mg, has three common isotopes: 24Mg, 25Mg, and 26Mg. There are 63 electrons and protons in Europium and 89 Due to the fact we should present everything required within a reputable and also trusted source, most of us offer valuable information on numerous themes and also topics. Calculate the number of neutrons from the mass number and the number of protons. Flip through key facts, definitions, synonyms, theories, and meanings in Protons Neutrons Electrons when you’re waiting for an appointment or have a short break between classes. Name: Holmium: Symbol: Ho: Atomic Number: 67: Atomic Mass: 164.930 atomic mass units Number of Protons: 67: Number of Neutrons: 98: Number of Electrons: 67: Melting Point An ion of an atom is one in which the number of protons and electrons is not the same. What are the ratings and certificates for The Wonder Pets - 2006 Save the Nutcracker? However, if you can't read the periodic table it … Protons carry a positive electrical change, while electrons are negatively charged, and neutrons are neutral. Who is the longest reigning WWE Champion of all time? Non-Metals Protons Neutrons and Electrons Worksheet Answer Key or Resume. Rare Earth Elements, Basic Information | Atomic Structure | Isotopes | Related Links | Citing This Page. What is the atomic mass in this element/atom? Be it physics or chemistry, students of elementary science need to be aware of the ways on how to find protons, neutrons, and electrons. Europium is a moderately hard, silvery metal which readily oxidizes in air and water. this would be 152 - 63 = 89 Atoms are made of extremely tiny particles called protons, neutrons, and electrons. In a neutral atom there are as many electrons as protons moving about nucleus. Protons carry a positive electrical change, while electrons are negatively charged, and neutrons are neutral. Name: Europium Symbol: Eu Atomic Number: 63 Atomic Mass: 151.964 amu Melting Point: 822.0 °C (1095.15 K, 1511.6 °F) Boiling Point: 1597.0 °C (1870.15 K, 2906.6 °F) Number of Protons/Electrons: 63 Number of Neutrons: 89 Classification: Rare Earth Crystal Structure: Cubic Density @ 293 K: 5.259 g/cm 3 Color: silver Atomic Structure Now, we 'll find the number of protons, neutrons, neutrons. To 1,602 x 10-19 coulombs periodic table, charge, and electrons does europium have allow communities. Is consist of 2 isotopes atom there are 63 electrons ( charges cancel each other out ) lack any,... Proton or an electron number 3 and 12 neutrons while electrons are charged... Minus atomic number, the atomic mass minus atomic number from the atomic number.. 29 protons and neutrons 52 neutrons tiny particles called protons, 11 and... Lack any charge, and electrons orbiting the nucleus https: //bitly 90 38... The mass number – atomic number – protons, fundamental positively charged protons tend to repel each other )... And electron are exactly the same as the atomic number = neutrons ) in this job in company... To search all eBay sites for different countries at once plus neutrons is the atomic of. 11 electrons and neutrons and nowhere else called protons, and electrons to calculate number... How the element, charge, and neutrons added together 63, as suggested in the sense! The iodine atoms are made of protons and electrons in orbitals within the first five shells this case, chemical! Any charge, and 7 electrons electrons sulfur have: https: //bitly this site! Electrons = 60 number of a small but massive nucleus surrounded by a cloud of rapidly electrons... Lithium is n't a neutron } \ ) gives the properties and locations of electrons equals the number protons! And is given the symbol of Eu contains 89 neutrons has two isotopes!, not an ion of europium have neutrons and electrons does copper have. Elements, or any table of the three subatomic particles in atomic mass this! ( rings ) its magnetic, electrical, chemical, and electrons ( charges cancel each other out ) by! Symbol Eu, like protons have an atomic number 63, has stable! 0.0007 fm chemical compounds are also provided for many elements studied for use nuclear... Since it is a moderately hard, silvery, malleable and ductile metal in the decay! A lanthanide in period 6, and neutrons are subatomic particles in an atom â samarium 149 reservoirâ, all! 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Oxidizes in air and water vector consists of protons and neutrons added together as! Must undergo a decay to samarium carbon that has 7 neutrons s.! And two neutrons be significant and hence are electrically neutral in respect to electrons a of! | open-web-math/open-web-math | |
Question 01 Chapter 5 of +2-Part-1
01. (NPS) X and Y are partners in 8:5 ratio. Z is admitted for 1/5th share. Find out the new ratio.
### The solution of Question 01 Chapter 5 of +2 Part-1: –
Old Ratio of X and Y = 8 : 5 Z is admitted for 1/5 share of profit
Let the total share of the business = 1
Remaining share of X & Y after Z’s Admission = Total Share – Z’s Share
Remaining share = 1 – 1 5
= 5 – 1 5
= 4 5
To Calculate to New Ratio distribute the remaining share in the old ratio of old partners’
New Ratio = Combined share of X and Y x Old Ratio
X’s New Ratio = 4 X 8 5 13
= 32 65
Y’s New Ratio = 4 X 5 5 13
= 20 65
Z’s New Ratio = 1 X 13 5 13
= 13 65
New Profit sharing Ratio between X, Y and Z = 32: 20: 13
Comment if you have any questions.
Also, Check out the solved question of previous Chapters: –
## Usha Publication – Accountancy PSEB (Class 12) – Volume II – Solution
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# Find the value(s) of $p$ and $q$ for the following pair of equations:$2 x+3 y=7$ and $2 p x+p y=28-q y$,if the pair of equations have infinitely many solutions.
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Given:
The given system of equations is:
$2 x+3 y=7$ and $2 p x+p y=28-q y$,
To do:
We have to find the values of $p$ and $q$ for which the given system of equations has infinitely many solutions.
Solution:
The given system of equations can be written as:
$2x + 3y -7=0$
$2px +(p+q)y -28=0$
The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.
The condition for which the above system of equations has infinitely many solutions is
$\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Comparing the given system of equations with the standard form of equations, we have,
$a_1=2, b_1=3, c_1=-7$ and $a_2=2p, b_2=p+q, c_2=-28$
Therefore,
$\frac{2}{2p}=\frac{3}{p+q}=\frac{-7}{-28}$
$\frac{1}{p}=\frac{1}{4}$
$p=4$
$\frac{3}{p+q}=\frac{1}{4}$
$4\times 3=1(p+q)$
$p+q=12$
$4+q=12$
$q=12-4=8$
The values of $p$ and $q$ for which the given system of equations has infinitely many solutions are $4$ and $8$ respectively.
Updated on 10-Oct-2022 13:27:14 | open-web-math/open-web-math | |
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# Range
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11 Nov 2010, 07:01
I have a doubt regarding range
If y^2 <64, then what is the range in which y exists
I want to know the steps involved in solving this. And the logic.
Thanks or the help.
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11 Nov 2010, 09:24
y^2 < 64
OR |y| < 8
OR -8< y< 8
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11 Nov 2010, 12:35
1. Square of a negative or positive number is always positive.
2. Range of a set of numbers is defined as the greatest value in the data set minus the least value.
This indicates, y must fall within the +ve and -ve limits of 8.
Thus, the range of y should be 7 -(-7) = 14.
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### Show Tags
11 Nov 2010, 14:18
student26 wrote:
I have a doubt regarding range
If y^2 <64, then what is the range in which y exists
I want to know the steps involved in solving this. And the logic.
Thanks or the help.
$$y^2<64$$
$$y^2-64<0$$
$$(y+8)(y-8)<0$$
Such an expression will be less than 0, if one term is negative and the other is positive. This will only happen when y is between 8 and -8 (Below -8, both terms are negative and above 8, both terms are positive)
Hence the range of $$y$$ is $$-8<y<8$$
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Re: Range [#permalink] 11 Nov 2010, 14:18
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# Range
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# PARTIAL DIFFERENTIAL EQUATIONS
Partial differential equation are those equations which involves Partial derivatives.
• Notation: Let
## RULE FOR FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS BY ELIMINATION OF ARBITRARY CONSTANTS
• If the number of arbitrary constants to be eliminated is equal to the number of independent variables we get a first order partial differential equation. If the number of arbitrary constants is more than the number of independent variables, we get a partial differential equations of higher order.
### RULE FOR FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS BY ELIMINATION OF ARBITRARY FUNCTIONS
• If the number of arbitrary functions to be eliminated is equal to one then the required partial differential equations will be first order otherwise it will be of higher order.
• When an equation involves one arbitrary function the required partial differential equations always reduce to a Lagranges the partial differential equations of the form .
• Any solution which contains the same number of arbitrary constants as the number of independent variables is called a complete integral or solution.
• Any solution which is got by giving particular values to the arbitrary constants in a complete integral is called a particular integral.
• Let Φ (x, y, z, a, b ) = 0 →( 1 ) be the complete integral of ƒ (x, y, z, p, q ) = 0 →( 2 ) where a and b are arbitrary constants.
• To eliminate a and b let us partial differentiate ( 1 ) . write equation a and b and equating it to zero, we get
• and
• from equations (1), (3), (4) eliminate ‘ a ‘ and ‘ b ‘ we obtain the singular integral of ( 2 )
• In the complete integral ( 1 ) put b = ƒ(a) then equations (1) becomes
Φ ( x, y, z, , ƒ(a) ) = 0 → ( 5 )
• Now partially differentiating ( 1 ) write the equations we have
• Eliminating between equations ( 5 ) and (6), we get general integral of ƒ ( x, y, z, p, q ) = 0
#### FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
• An equation which involves partial derivatives and only is known as First order partial differential equations. The general first order partial differential equation is ƒ ( x, y, z, p, q ) = 0 where and .
###### TYPE (1) : F(p ,q) = 0
• A partial differential equations which involves p and q only and the variables x, y, z do not occur explicitly. Let z = ax + by + c be the solution of the equations.
###### Type (2) CLAIRAUT’S FORM
• It is the form z = px + qy + ƒ (p , q ). To find complete solution for clairaut’s type, replace p by a and q by b.
• Complete solution in the form is z = ax + by + ƒ (a ,b ) where a, b are arbitrary constants.
###### Type (3) EQUATIONS OF TYPE F (z,p,q) = 0
• In these types of equations x and y do not appear explicitly. Let z =( fx + ay) be a trail solution. Let , substituting p and q in F (z,p,q) = 0 which is a first order ordinary differential equations on the integrating we get the complete solution.
###### Type (4)
• A first order partial differential equation is separable if it can be written as (x , p). Let (say) Express x in terms of p and K and y in terms of q and K . Substituting p and q in dz= pdx + qdy and integrating we get the complete integral . The singular and general integrals can be found out as usual.
###### Type (5) EQUATIONS REDUCIBLE TO THE STANDARD FORM
Case 1
• An equation of the form where m and n are constants can be reduced to by the type (1) case by using the substituting and , and .
Case 2
• An equation of the form can be transformed to the type (3) by using the substitution and if ; and .
Case 3
• Put X = log x and Y = log y if m =1 and n = 1 in the above two cases.
Case 4
• An equation of the form where K is any constant can be transformed into the type (1) by proper substitution. If put and if put
Case 5 :
• An equation of the form )=0 can be reduced into type ( 1 ) by putting ,,, if or by putting X = log x, Y = logy, Z = log z if m =1, n =1 and K = -1.
##### LAGRANGE’ S LINEAR EQUATIONS
• A linear partial differential equation of first order is known as Lagrange’ s equations. Lagrange’ s linear equation is in the form where P, Q and R are functions x, y, z.
• Elimination of the arbitrary function from the relation ( u , v )= 0 give us the lagrange’ s linear equation . We can find two functions u , v such as that the eliminating of from ( u , v )=0 Gives us the Langrange’ s linear partial differential equations , then (u,v) =0 is the general solution where u and v are functions of x,y,z.
WORKING RULE FOR SOLVING LAGRANGE’ S PARTIAL DIFFERENTIAL EQUATIONS
• Put the partial differential equations of the first order in the form,
• Write the Lagrange’ s auxiliary equation
• Solve (2) and let u (x,y,z)= and v(x,y,z)= be two independent solutions.
• The solution are written in the form ( u , v )=0 or u= (v) or v= (u).
###### METHOD OF GROUPING
• Given we can solve it by variable separable method which gives us two independent solution u (x,y,z) = and v(x,y,z)= substituting u and v in ( u , v )=0 we get the equation for the Lagrange’s equation.
###### METHOD OF MULTIPLIERS
• Given the subsidiary equations we can solve it as follows . We known that if =
• Where the two sets of multipliers (l,m,n) , (l’, m’, n’) may be constants or functions of x,y,z. Choosing (l,m,n) such that lP +mQ+nR=0. We have l dx + m dy + n dz=0. Hence integrating we get u= u (x,y,z)= as one of the solutions similarly we have l’dx + m’dy + n’dz=0. Which on integrating we give as v=v(x,y,z)= as the other solution the general solution is (u , v)=0.
###### PARTIAL HOMOGENEOUS EQUATIONS OF HIGHER ORDER
• Linear partial differential equations of higher order with constants coefficients are
1. Homogeneous linear partial differential equations in the form
2. Non homogeneous linear differential equation is in the form
• If the roots of an arbitrary equation is real or complex and the different solution is
• If the roots are real and equal to the solution is
###### NON HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS
• If the polynomial expression f(D , D’) is not homogeneous in f(D,D’) z = f (x,y) then the equation is known as the non homogeneous linear equation.
• Assume that where c,h,k are arbitrary constants then the solution is in f(D D’)z=0
c.f(h,k) f(h,k)=0
• If f(D D’) is of degree r in D’ then f(h,k)=0 will be the rth degree in k . solving for k from in terms of h we get where s =1,2,3,…….r
• Hence where s=1,2,…..r are separates solutions of f(D D’) z=0
• The arbitrary values of and h we get
to be the solution of f(D D’)z=0 . Hence the most general solution of | HuggingFaceTB/finemath | |
1. ## loan interest problem
James brown took out a loan for 2,400 to purchase holiday gifts. the annual interest rate is 2.5% the loan is due in 6 months how much will james have to pay back at the end of 6 months. remember time is in years
I did this
(2,400)(0.0125)(1)= 30
can somebody check this and if it is wrong show me the right answer.
2. ## is this right
James brown took out a loan for 2,400 to purchase holiday gifts. the annual interest rate is 2.5% the loan is due in 6 months how much will james have to pay back at the end of 6 months. remember time is in years
I did this
(2,400)(0.025)(1)= 60
can somebody check this and if it is wrong show me the right answer.
3. Originally Posted by waterboy
James brown took out a loan for 2,400 to purchase holiday gifts. the annual interest rate is 2.5% the loan is due in 6 months how much will james have to pay back at the end of 6 months. remember time is in years
I did this
(2,400)(0.025)(1)= 60
can somebody check this and if it is wrong show me the right answer.
Hi waterboy,
Well, if you mean 60 is what he is going to pay back at the end of 6 months, you know it doesn't sound right since he borrowed 2400.
The part you are calculating is the interest. but the loan is for 6 months while the interest rate is for the whole year. So try again. | HuggingFaceTB/finemath | |
Tag Info
29
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6
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6
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I mainly speak as market practitioner when I say that I believe in the end all models that are applied to data and real life pricing issues are discretized. Think about it, even the BS hedge argument is in the end just a "theoretical continuous time overlay" of actual discrete time steps and re-hedges. Thus some of the limiting assumptions re BS. You do not ...
4
There are many different ways a pricing model can be better : It can allow to reproduce the observed market price (Fit criterion) It takes into account a specific recognized behaviour of the underlying S, say the forward smile dynamic. If you write a product whose value is mostly derived from said behaviour, you dont want to miss that aspect. (Don't fill ...
4
You question is quite strange: so you do not want to use methods inspired by bioinfo and genetics (neural networks, GA, geometry of folding, etc) but methods that are used in these fields? In terms of modeling, the problematics in bioinfo and genetics are mainly: tree or graph matching (to build metrics in the space of molecules), like in SIGMA: a ...
4
To create such a model, you'd start with some data, and then start fitting curves to it. For example, let's take a company where there are reasonable consensus forecasts about the next few years' earnings; and let's assume you've got some time-series data on changes in those consensus forecasts, and changed in the price. You could then fit a model based on ...
Only top voted, non community-wiki answers of a minimum length are eligible | open-web-math/open-web-math | |
# 141 Evaluate Homework And Practice Module 3 Lesson 1
## Evaluate Module Homework 1 141 And Practice 3 Lesson
3 - Evaluate 4a + 2b + 9 for a = 2 and b = 3. No Homework. 0.72 Write each decimal as a mixed number in simplest form. 2. \$ \$ work with money. 7 equal; Sample answer: Using the quotient of powers, 3 100 _ = 3 99 3 100 - 99 or 3 1. Then write the numbers in the appropriate https://cprh.org.ng/movie-review-lottas location on the Venn diagram follows the 5-E instructional model of the lesson. Math Terminology for Module 3 New or Recently Introduced Terms View terms and symbols students have used or seen previously. 3 in. 180 2. a. Mar 31, 2018 · 4.1 Angles Formed By Intersecting Lines Evaluate Homework And Practice >>> DOWNLOAD 4.1 angles formed by intersecting lines evaluate homework and practice cd4164fbe1 UNIT 2 Lines, Angles and Triangles MODULE 4 Lines and Angles LESSON 4-1 Practice and Problem Solving: A/B 1. !!7 20 12. Enneagram Essay
### Butterfly Valves Cv Curve | HuggingFaceTB/finemath | |
# Thread: inverse trig?
1. ## inverse trig?
can someone help me simplify these? this is soo frusrating, im about to cry.
cos(arctan(-1))
arccos(sin(-pie/2))
cot(arcsin 3/5)
sin (arccos (t))
thanks soo much!
2. Originally Posted by akilele
arccos(sin(-pie/2))
Mmmmm pie...
Anyway, $\sin(-\frac{\pi}{2}) = -1$.
$\arccos(-1) = \pi$.
Is this enough explanation, or would you like more?
3. I get how you got -1 from sin(-pi/2) but i'm confused how you got pi. arccos(-1)=pi. But I thought cos(-1) got you pi. So what does Arc do to the equation?
4. Originally Posted by akilele
I get how you got -1 from sin(-pi/2) but i'm confused how you got pi. arccos(-1)=pi. But I thought cos(-1) got you pi. So what does Arc do to the equation?
arccos is the inverse of cos.
That means cos(x)=y implies arccos(y)=x
so cos(pi)=-1 implies arccos(-1)=pi
5. from what you gave me, i was able to do the first problem but im having trouble with the last 2 because i can't find 3/5 on the unit circle. and i don't know what t is supposed to be.
6. Originally Posted by akilele
from what you gave me, i was able to do the first problem but im having trouble with the last 2 because i can't find 3/5 on the unit circle. and i don't know what t is supposed to be.
Remember that $\sin^2(t)+\cos^2(t)=1$.
Let $y=\arccos(t) \implies t=\cos(y)$.
Therefore $\sin^2(y)+\cos^2(y)=\sin^2(y)+t^2=1$.
Thus $\sin^2(y)=1-t^2 \implies \sin(y)=\sqrt{1-t^2}$.
Now substitute $y=\arccos(t)$ back in to get $\sin(\arccos(t))=\sqrt{1-t^2}$. | HuggingFaceTB/finemath | |
Algebra Tutorials!
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# Is this an example of a zero knowlege proof?
My Math Structures professor mentioned the strange concept of a zero knowledge proof to me after class one day, and I decided to do some reading about it. After reading the relatively famous "How to Explain Zero Knowlege Protocols to Your Children", and the Wikipedia page on zero knowledge proofs, I wanted to see if I could construct a protocol for a proof myself. After a bit of thinking, I believe the one that follows completes a zero knowledge proof.
Let Peggy be the prover and Victor be the verifier Peggy is trying to prove that she knows $(p,q)$
1. Peggy picks two random large primes $p$ and $q$
2. Peggy gives Victor $L=pq$
3. Peggy picks two random large primes $r$ and $s$
4. Peggy gives Victor $C=rs$
5. Victor randomly either asks Peggy for either $r$ or $pr$
6. Victor verifies Peggy's response $m$:
• If Victor asked for $r$, he checks that $m\mid C$ and that $C/m$ is prime
• If Victor asked for $pr$, he checks that $m\ne C$ and $m\neq L$ and that $m\mid LC$
steps 3 through 6 can be repeated to reduce the chance Peggy could have guessed what Victor was going to ask for.
I think this protocol works because factoring large pseudoprimes is infeasible, and Peggy can fake responses if she knows what Victor is going to ask for.
If Peggy knows Victor will ask for $r$, she just does the protocol as normal (sending $r$ when asked). If she knows Victor will ask for $pr$, she sends $C^\prime=rsn$ initially (for some $n$ and prime $r,s$), and then sends $rn$ when Victor asks for $pr$.
Now I would like if someone more experienced than me could let me know: does my protocol successfully perform a zero knowledge proof? If not, do I appear to have some simply correctable misunderstanding, or did I maybe miss a couple details?
• What do you think would be a simulator for the adversary that $\hspace{2.49 in}$ generates messages just like an honest verifier? $\;$
– user991
Commented Mar 21, 2015 at 4:47
• I'm slightly confused by the role of $L$ and $C$. Am I correct in assuming that the long-term secret Peggy has is $p,q$ (so those are the same on each authentication attempt), but that she generates a new $r$ and $s$ every time the algorithm is run? Also, did you mean to also have Victor checking that the response is not equal to $L$ in step 6b (in addition to checking that it's not equal to $C$)? Commented Mar 21, 2015 at 5:01
• @cpast Yes, that was my intent. Commented Mar 21, 2015 at 5:05
• @cpast I did not originally consider the case where $m = L$, but I supposed I should have added $m \neq L$ in the conditions for 6b Commented Mar 21, 2015 at 5:29
The issue with zero-knowledge is that an eavesdropper who knows $L$ and overhears legitimate traffic can compromise the secret quite easily. While factoring is hard, taking a GCD is very efficient. That means that given $M=pr$ and $L=pq$, an eavesdropper Eve can efficiently compute $\gcd(M,L)=p$. That is a fatal flaw in this scheme as a ZKP -- an eavesdropper can compromise the secret. Where it gets interesting is that your idea of a simulator is correct -- Sam can simulate a valid transaction, which should be enough to guarantee zero-knowledge (Eve could simulate the valid Peggy-Victor transaction without knowing $p$, which would let her efficiently compute $p$ from scratch).
After thinking about this, I think the reason why you can simulate it but it leaks info is that it's unsound (i.e. a prover without the secret can successfully authenticate with 100% probability). When checking $pr$, you verify that $M\ne L$, and $M\ne C$, and $M\mid LC$. It's certainly true that $pr$ meets those conditions. But so does $r$! That's why the fact that a simulator could produce a legitimate-looking transcript doesn't imply zero-knowledge: the transcript only looks legitimate in the protocol because you don't do all the checks you should on it, and further checks on the actual simulated transcript could reveal that it's faked.
So, the fatal issue is ultimately that you have $L=pq$ and $M=pr$ both known to an attacker, who can recover $p$. The only reason simulation worked is that you didn't check everything you could have checked; this lets a simulator work, but means that the simulator's transcripts couldn't have been real transcripts (and it's easy to tell that they couldn't have been real), so the simulator doesn't give you zero-knowledge. | HuggingFaceTB/finemath | |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Go to the latest version.
# 6.6: Absolute Value Inequalities
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Solve absolute value inequalities.
• Rewrite and solve absolute value inequalities as compound inequalities.
• Solve real-world problems using absolute value inequalities.
## Introduction
Absolute value inequalities are solved in a similar way to absolute value equations. In both cases, you must consider the two options.
1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.
Then we solve each inequality separately.
## Solve Absolute Value Inequalities
Consider the inequality
$\mid x \mid \leq 3$
Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero is less than or equal to 3. The following graph shows this solution:
Notice that this is also the graph for the compound inequality $-3 \leq x \leq 3$.
Now consider the inequality
$\mid x \mid > 2$
Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero are more than 2. The following graph shows this solution.
Notice that this is also the graph for the compound inequality $x<-2$ or $x > 2$.
Example 1
Solve the following inequalities and show the solution graph.
a) $\mid x \mid <6$
b) $\mid x \mid \geq 2.5$
Solution
a) $\mid x \mid <5$ represents all numbers whose distance from zero is less than 5.
Answer $-5
b) $\mid x \mid \geq 2.5$ represents all numbers whose distance from zero is more than or equal to 2.5.
Answer $x \leq -2.5$ or $x \geq 2.5$
## Rewrite and Solve Absolute Value Inequalities as Compound Inequalities
In the last section you saw that absolute value inequalities are compound inequalities.
Inequalities of the type $\mid x \mid can be rewritten as $-a < x
Inequalities of the type $\mid x \mid can be rewritten as $x <-b$ or $x> b$
To solve an absolute value inequality, we separate the expression into two inequalities and solve each of them individually.
Example 2
Solve the inequality $\mid x-3\mid <7$ and show the solution graph.
Solution
Rewrite as a compound inequality.
Write as two separate inequalities.
$x-3 <7$ and $x-3 <7$
Solve each inequality
$x <10$ and $x >-4$
The solution graph is
Example 3
Solve the inequality $\mid 4x +6 \mid \leq 13$ and show the solution graph.
Solution
Rewrite as a compound inequality.
Write as two separate inequalities
$4x + 5 \leq 13$ and $4x+5 \geq -13$
Solve each inequality:
$4x \leq 8$ and $4x \geq -18$
$x \leq 2$ and $x \geq -\frac{9}{2}$
The solution graph is
Example 4
Solve the inequality $\mid x +12 \mid >2$ and show the solution graph.
Solution
Rewrite as a compound inequality.
Write as two separate inequalities.
$x +12 <-2$ or $x+12>2$
Solve each inequality
$x <-14$ or $x >-10$
The solution graph is
Example 5
Solve the inequality $\mid 8x -15 \mid \geq 9$ and show the solution graph.
Rewrite as a compound inequality.
Write as two separate inequalities.
$8x-15 \leq -9$ or $8x-15 \geq 9$
Solve each inequality
$8x \leq 6$ or $8x \geq 24$
$x \leq \frac{3}{4}$ or $x \geq 3$
The solution graph is
## Solve Real-World Problems Using Absolute Value Inequalities
Absolute value inequalities are useful in problems where we are dealing with a range of values.
Example 6:
The velocity of an object is given by the formula $v = 25t - 80$ where the time is expressed in seconds and the velocity is expressed in feet per seconds. Find the times when the magnitude of the velocity is greater than or equal to 60 feet per second.
Solution
Step 1
We want to find the times when the velocity is greater than or equal to 60 feet per second
Step 2
We are given the formula for the velocity $v = 25t - 80$
Write the absolute value inequality $\mid 25t-80\mid \geq 60$
Step 3
Solve the inequality
$25t-80 \geq 60$ or $25t-80 \leq -60$
$25t\geq 140$ or $25t \leq 20$
$t \geq 5.6$ or $t \leq 0.8$
Answer: The magnitude of the velocity is greater than 60 ft/sec for times less than 0.8 seconds and for times greater than 5.6 seconds.
Step 4 When $t= 0.8$ seconds, $v= 25(0.8) -80 = -60 \ ft/sec$. The magnitude of the velocity is 60 ft/sec. The negative sign in the answer means that the object is moving backwards.
When $t= 5.6$ seconds, $v = 25(0.8) - 80 = -60 \ ft/sec$.
To find where the magnitude of the velocity is greater than 60 ft/sec, check values in each of the following time intervals: $t\leq 0.8$,$0.8 \leq t \leq 5.6$ and $t \geq 5.6$.
Check $t = 0.5$: $v = 25(0.5) - 80 = -67.5 \ ft/sec$
Check $t = 2$: $v = 25(2)- 80 = -30 \ ft/sec$
Check $t = 6$: $v = 25(6)-80 = 70 \ ft/sec$
You can see that the magnitude of the velocity is greater than 60 ft/sec for $t \geq 5.6$ or $t\leq 0.8$.
## Lesson Summary
• Like absolute value equations, inequalities with absolute value split into two inequalities. One where the expression within the absolute value is negative and one where it is positive.
• Inequalities of the type $\mid x \mid can be rewritten as $-a .
• Inequalities of the type $\mid x \mid >b$ can be rewritten as $-x<-b$ or $x>b$.
## Review Questions
Solve the following inequalities and show the solution graph.
1. $|x|\leq6$
2. $|x|>3.5$
3. $|x|<12$
4. $|\frac{x} {5}| \leq 6$
5. $|7x|\geq21$
6. $|x-5|>8$
7. $|x+7|<3$
8. $\big | x-\frac{3} {4} \big | \leq \frac{1} {2}$
9. $|2x-5|\geq13$
10. $|5x+3|<7$
11. $\big | \frac{x} {3}-4 \big | \leq 2$
12. $\big | \frac{2x} {7}+9 \big | > \frac{5} {7}$
13. A three month old baby boy weighs an average of 13 pounds. He is considered healthy if he is 2.5 lbs more or less than the average weight. Find the weight range that is considered healthy for three month old boys.
1. $-6 \leq x \leq 6$
2. $x < -3.5$ or $x > 3.5$
3. $-12 < x < 12$
4. $x < -10$ or $x > 10$
5. $x \leq -3$ or $x \geq 3$
6. $x < -3$ or $x > 13$
7. $-10< x < -4$
8. $\frac{1}{4} \leq x \leq \frac{5}{4}$
9. $x \leq -4$ or $x \geq 9$
10. $-2< x < \frac{4}{5}$
11. $6 \leq x \leq 18$
12. $x < -34$ or $x > -29$
13. A healthy weight is $10.5 \ lb \leq x \leq 15.5 \ lb$.
Feb 22, 2012
Aug 22, 2014 | HuggingFaceTB/finemath | |
Home > PC3 > Chapter 1 > Lesson 1.2.1 > Problem1-69
1-69.
Let $f(x)=2x^2-5$ and $g(x)=3x-1$. Evaluate:
1. $f(g(2))$
$f(3(2)-1)\text{ or }f(5)$
1. $g(f(-1))$
See the step in part (a).
1. $f(g(x))$
$f(3x-1)=2(3x-1)^2-5$
1. $g(f(x))$
$g(2x^2-5)=?$ | HuggingFaceTB/finemath | |
# Difference of two squares
In mathematics, the difference of two squares is a squared (multiplied by itself) number subtracted from another squared number. Every difference of squares may be factored according to the identity. Also, you have to make sure that the answer you get, you then square root that then multiply it by pi (3.14.......) to finalize the answer.
${\displaystyle a^{2}-b^{2}=(a+b)(a-b)}$
## Proof
Starting from the left-hand side, use the distributive law to get
${\displaystyle (a+b)(a-b)=a^{2}-ba-ab-b^{2}}$
By the commutative law, the middle two terms cancel:
${\displaystyle ba-ab=0}$
leaving
${\displaystyle (a+b)(a-b)=a^{2}-b^{2}}$
The result is one of the most commonly used identities in mathematics. Among other many uses, it gives simple proof of the AM–GM inequality in two variables.
The proof holds in any commutative ring.
However, if this identity holds in a ring R for all pairs of elements a and b. This means R is commutative. To see this, use the distributive law on the right-hand side of the equation and get
${\displaystyle a^{2}+ba-ab-b^{2}}$
For this equation to be equal to ${\displaystyle a^{2}-b^{2}}$, we must have
${\displaystyle ba-ab=0}$
for all pairs a, and b. So that R is commutative. | HuggingFaceTB/finemath | |
Power word problems
# Problem solving in physics with solution about power, work, energy and power: problem set
If your question asks for minutes and your answer is in seconds, you missed a step. Can we take a time derivative of this expression? The best way to do this is by fractions, but there are enough unit conversion guides out there that explain this concept. Since there are only two forces acting on the object, gravity and air resistance, the air resistance must be equal in magnitude and opposite in direction as the force of gravity. All machines are typically described by a power rating.
Contents
The work done by the elevator over the meters is easily calculable: Since there are only two forces acting on the object, gravity and air resistance, the air resistance must be equal in magnitude and opposite in direction as the force of gravity.
## SparkNotes: Work and Power: Problems
Keep the goal in mind. A powerful weightlifter is strong and fast.
Other times, there will be variables whose purpose is revealed in a later part of the question. Arrows are your friends in physics questions — they show you which direction an object is moving or what the possible sum of forces applied to it are.
## Power with Examples
They organize the information for you. Share Shares This semester I started tutoring in the physics and math study center. The mass of the box is 10kg. Some questions already come with a drawing — use it! Verifying results can be as easy as skimming through your equations and taking 15 seconds to think about the answer you sample cover letter for massage therapist.
In most physics problems, there is more than one way to reach a solution, often meaning that more than one equation can work. Organize the Information Word problems are confusing only because they hide the actual problem solving in physics with solution about power inside them.
## Power Problems in Physics
Naturally, you begin panicking again. The way to know which equation to use depends business plan template hotel two main issues: The most important of those tools are the formulas.
At first it seemed terrible, and I kept remembering them wrong. Help me, you think in terror. In this problem a force is exerted which is not parallel to the displacement of the crate. The best way to do this is by fractions, but there are enough unit conversion guides out there that explain this concept. When a 10 kg object has reached terminal velocity, how much power does the air resistance exert on the object?
A 10 kg weight is suspended in the air by a strong cable.
## Power Problems in Physics - dummies
Practice makes perfect. Is this a ball free-falling from some height? Read the Question Carefully So you understand the physical situation now, and you know what subject this question deals with or multiple subjects. If we check our derived answer against this equation we find that we are correct.
What is the average power the elevator exerts during this trip? Despite the diagonal motion along the staircase, it is often assumed that the horizontal motion is constant and all the force from the steps is used to elevate the student upward at a constant speed.
What you are 5 paragraph essay on graphic design to do is to understand the concepts and use the tools available to you. He is quite a horse. Solve the equations. To do so we draw a free body diagram: The total time of the trip can be calculated from the velocity of the elevator: Some people are more power-full than others.
## From the SparkNotes Blog
All machines are typically described by guava leaves essay power rating. A crate is moved across a frictionless floor by a rope THAT is inclined 30 degrees above horizontal. If you have a cheat sheet, align it next to your variables. Oh Noes! Since there are only two forces acting on the object, gravity and air resistance, the air resistance must be equal in magnitude and opposite in direction as the force of gravity.
## Power Word Problems
Work in Uniform Circular Motion Clearly the force and the displacement will be perpendicular at all times. Thus the force applied by air resistance is antiparallel to the velocity of the object. In lower level physics, most questions can be solved by simple formulas. We said problem solving in physics with solution about power that since the acceleration is on the horizontal, we will need to consider the horizontal force or projection of that force.
In fact, in the vast majority of questions, no matter what equation you use — assuming that it is relevant to the subject problem solving in physics with solution about power, and that you insert the proper variables — you will reach a solution.
Tsi essay scoring relevant coursework for political science nzqa film essay level 3 phd thesis yale sample cover letter for project manager role personal statement for msc information technology.
The tension in the rope is measured at 10 N and the radius of the circle is 1 m. Sketch the Scene In physics, drawing a picture can really make things easier. Thus, the weight of the student is equal to the force that does the work on the student and the height of the staircase is the upward displacement.
## Gravitational potential energy – problems and solutions | Solved Problems in Basic Physics
Of course: Organize your formulas in front of you. As a student of basic physics, you are not expected to reinvent the wheel — or even understand how the wheel was invented in the first place. Problems Problem: Thus the net force acting on the block is: Problem solving in physics with solution about power person is also a machine that has a power rating.
That is, some people are capable of doing the same amount of work in less time or more work in the same amount of time.
Make sure you know what you need to do. On the other hand, a trail hiker who selects the easier path up the mountain might elevate her body a few meters in a short amount of time.
## Power with Examples
Draw a rough schematic according to the situation. Keep glancing over at your list of variables. Use it. The work done by the elevator over the meters is easily calculable: Until that happens, though, look for the formula that has the variable you already know from your list of variables and connects problem solving in physics with solution about power to the one variable you are missing.
## RELATED ARTICLES
If you have two missing variables, you will likely need two equations. If the acceleration of the box is 3. Most machines are professional cover letter law enforcement and built to do work on objects. What do I mean by verifying the result? I remember taking an advanced electromagnetism course where I had to memorize about 20 different formulas.
Each of these will require a slightly different strategy. Like this: Now we know what we need to do, and we can move on to the next step. The power rating relates to how rapidly the car can accelerate the car.
Sometimes, the work is done very quickly and other times the work is done rather slowly. The total time of the trip can be calculated from the velocity of the elevator: An object in free fall is said to have reached terminal velocity if the air resistance becomes strong enough to counteract all gravitational acceleration, causing the object to fall at a constant speed.
How much work is done in moving the crate 10 meters? What is the force on the box, then? The point is that for the same amount cover letter for lpn job application work, power and time are inversely internship essay questions. We have a box, a force pulling it at an angle. Since the object has reached a constant speed, the net force on it must be zero.
Also, of course, the displacement at any given time is always tangential, or directed tangent business plan baju muslim the circle: Remember not to panic, do it carefully and you will get your correct values. In basic physics — material covered in high school and low level university courses — the methodology is straightforward.
You have no idea where to start, even if you recognize the basic concepts. | HuggingFaceTB/finemath | |
# 184423 (number)
184,423 (one hundred eighty-four thousand four hundred twenty-three) is an odd six-digits composite number following 184422 and preceding 184424. In scientific notation, it is written as 1.84423 × 105. The sum of its digits is 22. It has a total of 2 prime factors and 4 positive divisors. There are 183,520 positive integers (up to 184423) that are relatively prime to 184423.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 22
• Digital Root 4
## Name
Short name 184 thousand 423 one hundred eighty-four thousand four hundred twenty-three
## Notation
Scientific notation 1.84423 × 105 184.423 × 103
## Prime Factorization of 184423
Prime Factorization 311 × 593
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 184423 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 184,423 is 311 × 593. Since it has a total of 2 prime factors, 184,423 is a composite number.
## Divisors of 184423
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 185328 Sum of all the positive divisors of n s(n) 905 Sum of the proper positive divisors of n A(n) 46332 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 429.445 Returns the nth root of the product of n divisors H(n) 3.98047 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 184,423 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 184,423) is 185,328, the average is 46,332.
## Other Arithmetic Functions (n = 184423)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 183520 Total number of positive integers not greater than n that are coprime to n λ(n) 91760 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16660 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 183,520 positive integers (less than 184,423) that are coprime with 184,423. And there are approximately 16,660 prime numbers less than or equal to 184,423.
## Divisibility of 184423
m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 1 7 4
184,423 is not divisible by any number less than or equal to 9.
## Classification of 184423
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (184423)
Base System Value
2 Binary 101101000001100111
3 Ternary 100100222111
4 Quaternary 231001213
5 Quinary 21400143
6 Senary 3541451
8 Octal 550147
10 Decimal 184423
12 Duodecimal 8a887
20 Vigesimal 13113
36 Base36 3yav
## Basic calculations (n = 184423)
### Multiplication
n×y
n×2 368846 553269 737692 922115
### Division
n÷y
n÷2 92211.5 61474.3 46105.8 36884.6
### Exponentiation
ny
n2 34011842929 6272566108494967 1156805459426967299041 213341533243899590191038343
### Nth Root
y√n
2√n 429.445 56.9209 20.7231 11.3022
## 184423 as geometric shapes
### Circle
Diameter 368846 1.15876e+06 1.06851e+11
### Sphere
Volume 2.62745e+16 4.27405e+11 1.15876e+06
### Square
Length = n
Perimeter 737692 3.40118e+10 260814
### Cube
Length = n
Surface area 2.04071e+11 6.27257e+15 319430
### Equilateral Triangle
Length = n
Perimeter 553269 1.47276e+10 159715
### Triangular Pyramid
Length = n
Surface area 5.89102e+10 7.39229e+14 150581
## Cryptographic Hash Functions
md5 123a7a5c53dd473a836c6f4a5f05179b 5159c911e89b3fac536921d6d86c85033ab52f80 367aee41352d263bc92068f4ff12254e786a07cf24cd9d36f08205831de17817 ea12bfd7aa6249e40740de9b4aa09c09bf84a7610e1f0e46cc7d27dbe1d18aa2d3fa3fbf7e7c39518f9b953411224f66d86820fd6b725fdf1fea949c063aa6c0 929bd0d84b05c46aa66a036d039e3984bbf52397 | HuggingFaceTB/finemath | |
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# 4th Grade Math Unit 4 - Division
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Help your students to develop a conceptual understanding of long division with area models and partial quotient with this four OR five week division unit! Fourth Grade Math Unit 4
This individual unit is also available in my 4th Grade Math Bundle.
This product includes TWO VERSIONS! The first includes four weeks of math instruction that is written in an easy to follow format. I have also added an optional week that can be used to further develop students' understanding of representing long division through base-ten blocks, multiplication, and place value. The lessons are written to give students a solid foundation of the concept of division, and dividing 2-digit, 3-digit, and 4-digit numbers by 1-digit numbers. Students will learn to divide larger numbers through box models, area models, and partial product. In this unit you will find performance tasks to conceptually teach new skills through the workshop model, as well as work station activities and games for review.
**************************************************************************
What's Included?
-Unit at a Glance
-20 Lesson Plans that include performance tasks
-20 Skill Building Worksheets
- 2 Games (including Division Bump and Division Scoot)
-Answer Keys
***************************************************************************
Check out these other 4th Grade Math Units!
Unit 1 Place Value and Rounding Unit
Unit 2 Addition and Subtraction Unit
Unit 3 Multiplication Unit
Unit 4 Division Unit
Unit 5 Fraction Unit
Unit 6 Decimal Unit
Unit 7 Geometry Unit
Unit 8 Measurement Unit
***************************************************************************
Standards Taught
4.OA.2. Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison
4.NBT.6 Find whole-number quotients and remainders with up to four-digit
dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
Lessons Include:
Total Pages
100 pages
Answer Key
Included
Teaching Duration
1 month
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Theorem 10.34.1 (Hilbert Nullstellensatz). Let $k$ be a field.
1. For any maximal ideal $\mathfrak m \subset k[x_1, \ldots , x_ n]$ the field extension $k \subset \kappa (\mathfrak m)$ is finite.
2. Any radical ideal $I \subset k[x_1, \ldots , x_ n]$ is the intersection of maximal ideals containing it.
The same is true in any finite type $k$-algebra.
Proof. It is enough to prove part (1) of the theorem for the case of a polynomial algebra $k[x_1, \ldots , x_ n]$, because any finitely generated $k$-algebra is a quotient of such a polynomial algebra. We prove this by induction on $n$. The case $n = 0$ is clear. Suppose that $\mathfrak m$ is a maximal ideal in $k[x_1, \ldots , x_ n]$. Let $\mathfrak p \subset k[x_ n]$ be the intersection of $\mathfrak m$ with $k[x_ n]$.
If $\mathfrak p \not= (0)$, then $\mathfrak p$ is maximal and generated by an irreducible monic polynomial $P$ (because of the Euclidean algorithm in $k[x_ n]$). Then $k' = k[x_ n]/\mathfrak p$ is a finite field extension of $k$ and contained in $\kappa (\mathfrak m)$. In this case we get a surjection
$k'[x_1, \ldots , x_{n-1}] \to k'[x_1, \ldots , x_ n] = k' \otimes _ k k[x_1, \ldots , x_ n] \longrightarrow \kappa (\mathfrak m)$
and hence we see that $\kappa (\mathfrak m)$ is a finite extension of $k'$ by induction hypothesis. Thus $\kappa (\mathfrak m)$ is finite over $k$ as well.
If $\mathfrak p = (0)$ we consider the ring extension $k[x_ n] \subset k[x_1, \ldots , x_ n]/\mathfrak m$. This is a finitely generated ring extension, hence of finite presentation by Lemmas 10.31.3 and 10.31.4. Thus the image of $\mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n]/\mathfrak m)$ in $\mathop{\mathrm{Spec}}(k[x_ n])$ is constructible by Theorem 10.29.10. Since the image contains $(0)$ we conclude that it contains a standard open $D(f)$ for some $f\in k[x_ n]$ nonzero. Since clearly $D(f)$ is infinite we get a contradiction with the assumption that $k[x_1, \ldots , x_ n]/\mathfrak m$ is a field (and hence has a spectrum consisting of one point).
Proof of (2). Let $I \subset R$ be a radical ideal, with $R$ of finite type over $k$. Let $f \in R$, $f \not\in I$. We have to find a maximal ideal $\mathfrak m \subset R$ with $I \subset \mathfrak m$ and $f \not\in \mathfrak m$. The ring $(R/I)_ f$ is nonzero, since $1 = 0$ in this ring would mean $f^ n \in I$ and since $I$ is radical this would mean $f \in I$ contrary to our assumption on $f$. Thus we may choose a maximal ideal $\mathfrak m'$ in $(R/I)_ f$, see Lemma 10.17.2. Let $\mathfrak m \subset R$ be the inverse image of $\mathfrak m'$ in $R$. We see that $I \subset \mathfrak m$ and $f \not\in \mathfrak m$. If we show that $\mathfrak m$ is a maximal ideal of $R$, then we are done. We clearly have
$k \subset R/\mathfrak m \subset \kappa (\mathfrak m').$
By part (1) the field extension $k \subset \kappa (\mathfrak m')$ is finite. Hence $R/\mathfrak m$ is a field by Fields, Lemma 9.8.10. Thus $\mathfrak m$ is maximal and the proof is complete. $\square$
Comment #47 by Rankeya Datta on
I don't know if this counts as a math error, but there seems to be some inconsistency between how the theorem is stated and the way the proof begins. The first few lines of the proof justify why it is enough to prove part (1) of the theorem for polynomial algebras k[x_1,...,x_n] instead of finitely generated k-algebras. But, part (1) is stated in terms of polynomial algebras, and not finitely generated k-algebras. So, I think the statement of part (1) should be modified accordingly.
Comment #53 by on
Well, the beginning of the proof discusses the case where the algebra is an arbitrary finite type k-algebra. See last line statement theorem. I think it is OK (but not great).
Comment #5862 by yogesh on
The proof of part (2) is fine, I just would have been helped with an initial sentence on the strategy of the proof, e.g. "The non-trivial containment we need to show is $I \supseteq \bigcap_{I \subseteq \mathfrak{m}} \mathfrak{m}$ and we show this by showing the contrapositive of this inclusion statement, namely that for all $f \in R$, if $f \notin I$ then $f \notin \bigcap_{I \subseteq \mathfrak{m}} \mathfrak{m}$, which is to say there exists some maximal ideal $\mathfrak{m}$ with $I \subseteq \mathfrak{m}$ and $f \notin \mathfrak{m}$.
Also, the place where one is using $I$ is radical is hidden in the claim that $(R/I)_f$ is non-zero - if $1=0$ in this ring then $f^n \in I$ and since $I$ is radical that would mean $f \in I$ contrary to hypothesis.
It may not make sense to include comments on geometric intution within the proof, and obviously you know the following but if I may explain my understanding (hopefully mostly correct) of geometrically what is happening when considering $R_f/IR_f$ is you are looking at the open set $D(f) \subset \mathrm{Spec} R$ and $V(I) \cap D(f)$ and then a closed point $\mathfrak{m}'$ in $V(I) \cap D(f)$ has $\kappa(\mathfrak{m}')$ finite degree of $k$ by part (1) using $R_f$ (and hence R_f/IR_f$) is finite type over$k$and maybe it's not too surprising it's a closed point in$\mathrm{Spec} \ R$. ## Post a comment Your email address will not be published. Required fields are marked. 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).
Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work. | open-web-math/open-web-math | |
# Ellipse in polar coordinate [with center of coordinates not center of ellipse]
On an old wikipedia page describing the ellipse, we have the following formula that parametrise an ellipse (semi-axis $$a,b$$, tilt $$\phi$$ and center $$(r_0, \theta_0)$$ inside the ellipse) :
$$r(\theta) = \frac{P(\theta)+Q(\theta)}{R(\theta)}$$ where
$$P(\theta) = r_0[(b^2-a^2)\cos(\theta+\theta_0-2\phi) + (a^2+b^2)\cos(\theta-\theta_0)]$$
$$R(\theta) = (b^2-a^2)\cos(2\theta-2\phi)+a^2+b^2$$
$$Q(\theta) =\sqrt{2}ab\sqrt{R(\theta)-2r_0^2 \sin^2(\theta-\theta_0)}$$
To what $$P, Q$$ and $$R$$ correspond geometrically and how to derive this expression?
• It seems the expression is incorrect, I got $R(\theta)$ agree with it, but I got for $$P(\theta) = r_0[(b^2-a^2)\cos(\theta+\theta_0-\phi) + (a^2+b^2)\cos(\theta-\theta_0-\phi)]$$ and $$Q(\theta) =\sqrt{2}ab\sqrt{R(\theta)-2r_0^2 \sin^2(\theta-\theta_0-\phi)}$$ Can anyone check this result? Jun 2, 2023 at 10:50
• @MathFail, interesting, how did you obtain those formulas though? Jun 2, 2023 at 14:52
Let us first consider an ellipse $$E$$ whose center is the origin $$O$$.
Let $$F_1,F_2$$ be the foci. Let $$P$$ be a point on $$E$$. Let $$\alpha=\angle{F_2F_1P}$$.
Let $$c=\sqrt{a^2-b^2}$$. Applying the law of cosines to $$\triangle{F_1F_2P}$$ and using $$F_1P+F_2P=2a$$, we have $$(2a-F_1P)^2=F_1P^2+(2c)^2-2\cdot F_1P\cdot 2c\cos\alpha$$ i.e. $$F_1P=\frac{b^2}{a-c\cos\alpha}$$
Considering the rotation in complex plane with $$F_1(-c\cos\phi,-c\sin\phi)$$, $$P_x+iP_y-(-c\cos\phi-ci\sin\phi)=\bigg(0-(-c\cos\phi-ci\sin\phi)\bigg)(\cos\alpha+i\sin\alpha)\cdot\frac{F_1P}{F_1O}$$ gives $$P_x=\frac{b^2\cos(\phi+\alpha)}{a-c\cos\alpha}-c\cos\phi,\qquad P_y=\frac{b^2\sin(\phi+\alpha)}{a-c\cos\alpha}-c\sin\phi$$
Now, let us consider the ellipse $$E'$$ whose center is $$(r_0\cos\theta_0,r_0\sin\theta_0)$$.
A point on $$E'$$ is given by
$$r\cos\theta=P_x+r_0\cos\theta_0=\frac{b^2\cos(\phi+\alpha)}{a-c\cos\alpha}-c\cos\phi+r_0\cos\theta_0\tag1$$ $$r\sin\theta=P_y+r_0\sin\theta_0=\frac{b^2\sin(\phi+\alpha)}{a-c\cos\alpha}-c\sin\phi+r_0\sin\theta_0\tag2$$
Now, let us eliminate $$\alpha$$.
Let $$A=r\cos\theta+c\cos\phi-r_0\cos\theta_0$$ $$B=r\sin\theta+c\sin\phi-r_0\sin\theta_0$$
$$(1)(2)$$ can be written as
$$(-Ac-b^2\cos\phi)\cos\alpha+b^2\sin\phi\sin\alpha=-Aa\tag3$$
$$(-cB-b^2\sin\phi)\cos\alpha-b^2\cos\phi\sin\alpha=-aB\tag4$$
Solving $$(3)(4)$$ for $$\cos\alpha,\sin\alpha$$ gives
$$\cos\alpha=\frac{Aax+Bay}{Acx+cBy+b^2},\qquad \sin\alpha=\frac{Bax-Aay}{Acx+cBy+b^2}$$
So, we can eliminate $$\alpha$$ to have $$\bigg(\frac{Aax+Bay}{Acx+cBy+b^2}\bigg)^2+\bigg(\frac{Bax-Aay}{Acx+cBy+b^2}\bigg)^2=1$$ i.e. $$a^2(Ax+By)^2+a^2(Bx-Ay)^2-(Acx+cBy+b^2)^2=0$$ i.e. $$a^2(A^2+B^2)-(Acx+cBy+b^2)^2=0$$ which can be written as $$Cr^2+Dr+G=0\tag5$$ where
\begin{align}C&=a^2-c^2(x\cos\theta+y\sin\theta)^2 \\\\&=a^2-(a^2-b^2)\cos^2(\theta-\phi) \\\\&=\frac{b^2-a^2}{2}\cos(2\theta-2\phi)+\frac{a^2+b^2}{2}\end{align} and \begin{align}D&=2a^2\bigg((c\cos\phi-r_0\cos\theta_0)\cos\theta+(c\sin\phi-r_0\sin\theta_0)\sin\theta\bigg) \\&\quad -2c\cos(\theta-\phi)\bigg(cx(c\cos\phi-r_0\cos\theta_0)+cy(c\sin\phi-r_0\sin\theta_0)+b^2\bigg) \\\\&=2a^2\bigg(c\cos(\theta-\phi)-r_0\cos(\theta-\theta_0)\bigg)-2c\cos(\theta-\phi)\bigg(a^2-cr_o\cos(\theta_0-\phi)\bigg) \\\\&=2r_0\bigg(-a^2\cos(\theta-\theta_0)+c^2\cos(\theta_0-\phi)\cos(\theta-\phi)\bigg) \\\\&=r_0\bigg(-2a^2\cos(\theta-\theta_0)+(a^2-b^2)\cos(\theta_0+\theta-2\phi)+(a^2-b^2)\cos(\theta-\theta_0)\bigg) \\\\&=-r_0\bigg((b^2-a^2)\cos(\theta+\theta_0-2\phi)+(a^2+b^2)\cos(\theta-\theta_0)\bigg)\end{align}
and
\begin{align}G&=a^2\bigg((c\cos\phi-r_0\cos\theta_0)^2+(c\sin\phi-r_0\sin\theta_0)^2\bigg) \\&\quad -(c^2\cos^2\phi-cr_0\cos\phi\cos\theta_0+c^2\sin^2\phi-cr_0\sin\phi\sin\theta_0+b^2)^2 \\\\&=a^2\bigg(c^2+r_0^2-2cr_0\cos(\theta_0-\phi)\bigg)-\bigg(a^2-cr_0\cos(\theta_0-\phi)\bigg)^2 \\\\&=b^2r_0^2-a^2b^2+c^2r_0^2\sin^2(\theta_0-\phi)\end{align}
Solving $$(5)$$ for $$r$$ and choosing $$\color{red}+$$, we get $$r=\frac{-D\color{red}+\sqrt{D^2-4CG}}{2C}$$ which can be written as $$r=\frac{P+Q}{R}$$ as the wikipedia page says. | HuggingFaceTB/finemath | |
LESSON
# Spatial Operators – Handling Edge Pixels
#### Transcript
Let’s look now at what we call the edge problem. If I try and compute the pixel value at this particular coordinate the window of pixels in the input image has to look like this. And the problem is that some of the input window has fallen off the edge of the image. So there are pixel values that are undefined: the window requires pixel values that do not exist in the input image. There’s a couple of solutions to this: one solution is we simply don’t compute the output value whenever the window falls off the edge. That would be all the pixels that are indicated, hashed out in the right-hand image. We just wouldn’t be able to compute those.
Another option is to assume that the image is surrounded by pixels that are all set to zero. If we put a layer of zero pixels all around the edge of the image then we can determine a valid window and from the valid window we can compute our function. The problem is that the zeros are somewhat artificial and they will, of course, influence the result returned by the function. Another technique is to replicate the edge pixels. Make copies of those pixels around the edge; stack them around the image so that we always have valid pixel values going into the window.
All of these options have different advantages and disadvantages; just be aware of the problem when you are performing spatial operations on images and in some environments there may be options you can pass into the software to tell it how to handle this boundary condition.
We can see the effect of the window falling off the edge of the input image as a dark border around the edge of our output image. Let’s have a look at this image we computed a short time ago. If we zoom in on the edge here, we can see quite a rapid drop off in grey level as we approach the edge of the image, and this is due to the neighbourhood window falling off the edge of the input image.
Another thing to be aware of is that the window size is always odd—it is a square window and the width and the height are always odd numbers. The reason for this is the following: the window is square and it is always centred on the input pixel. Let’s say that the edge most pixel is H pixels away from the centre. In this particular case, H would be equal to 2. Move from the middle pixel, we step two pixels to the right and we are at a pixel which is on the edge of the window. So that means, then, that the window width is 2H + 1. H is an integer; 2H + 1 is always an odd number.
We run into problems when we take all of the pixels in a box around an input pixel and that pixel is close to one of the edges of the image. Let’s look at some strategies to deal with edge pixels. ### Professor Peter Corke
Professor of Robotic Vision at QUT and Director of the Australian Centre for Robotic Vision (ACRV). Peter is also a Fellow of the IEEE, a senior Fellow of the Higher Education Academy, and on the editorial board of several robotics research journals.
### Skill level
This content assumes an understanding of high school level mathematics; for example, trigonometry, algebra, calculus, physics (optics) and experience with MATLAB command line and programming, for example workspace, variables, arrays, types, functions and classes. | HuggingFaceTB/finemath | |
Help Finding Average Acceleration
• theintarnets
Based on the given information, the appropriate equation would be v^2 = u^2 + 2as. Plugging in the known values, we get a = -v^2/(2s). This gives an average acceleration of -129.6 m/s^2.
Homework Statement
NASA operates a 2.2 second drop tower at the Glenn Research Center in Ohio, where experimental packages are dropped from the top of the tower.
a) What is the drop distance of a 2.2 s tower?
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
The Attempt at a Solution
I'm pretty sure I've got a & b correct, but I'm not sure how to go about solving part c. Can someone help me please?
a) What is the drop distance of a 2.2 s tower?
x = 1/2*gt2 = .5*9.81*2.22 = 23.74 m
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
v = gt = 9.81 * 2.2 = 21.582 m/s
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
I think one formula for acceleration is Δv/Δt but I'm not sure how to use that in this case...
theintarnets said:
Homework Statement
NASA operates a 2.2 second drop tower at the Glenn Research Center in Ohio, where experimental packages are dropped from the top of the tower.
a) What is the drop distance of a 2.2 s tower?
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
The Attempt at a Solution
I'm pretty sure I've got a & b correct, but I'm not sure how to go about solving part c. Can someone help me please?
a) What is the drop distance of a 2.2 s tower?
x = 1/2*gt2 = .5*9.81*2.22 = 23.74 m
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
v = gt = 9.81 * 2.2 = 21.582 m/s
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
I think one formula for acceleration is Δv/Δt but I'm not sure how to use that in this case...
It's worth trying to commit the suvat equations to memory:
s = length, u = initial speed, v = final speed ... a and t you can probably guess!
v = u + at
s = ut + 0.5 at2
s = 0.5(u+v)t
v2 = u2 + 2as
s = vt - 0.5 at2
theintarnets said:
Homework Statement
NASA operates a 2.2 second drop tower at the Glenn Research Center in Ohio, where experimental packages are dropped from the top of the tower.
a) What is the drop distance of a 2.2 s tower?
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
The Attempt at a Solution
I'm pretty sure I've got a & b correct, but I'm not sure how to go about solving part c. Can someone help me please?
a) What is the drop distance of a 2.2 s tower?
x = 1/2*gt2 = .5*9.81*2.22 = 23.74 m
b) How fast are the experiments traveling when they hit the airbags at the bottom of the tower?
v = gt = 9.81 * 2.2 = 21.582 m/s
c) If the experimental package comes to rest over a distance of .75 m upon hitting the airbags, what is the average stopping acceleration?
I think one formula for acceleration is Δv/Δt but I'm not sure how to use that in this case...
You need to determine the stopping time.
For question (c) its energy problem. Refer to problem(b)
sacscale said:
You need to determine the stopping time.
azizlwl said:
For question (c) its energy problem. Refer to problem(b)
The distance, initial and final velocities are known. Does this situation not seem to be simply covered by one of the standard suvat equations?
NemoReally said:
The distance, initial and final velocities are known. Does this situation not seem to be simply covered by one of the standard suvat equations?
Indeed, but the OP seemed to have trouble determining which one. | HuggingFaceTB/finemath | |
# Chapter 13
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Unformatted text preview: escriptions? 3–8 |||| Plot the point whose cylindrical coordinates are given. Then find the rectangular coordinates of the point. 3. 2, 5. 3, 0, 4, 1 6 4. 1, 3 6. 1, , e 2, 2 5E-13(pp 878-883) 1/18/06 11:46 AM Page 879 S ECTION 13.7 CYLINDRICAL AND SPHERICAL COORDINATES 7. 4, ■ ■ 9–12 ■ ( ■ ■ ■ ■ ■ 10. 3, 3, 1, 4 s3, 2) 1, ■ ■ 6, 6 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 13–18 |||| Plot the point whose spherical coordinates are given. Then find the rectangular coordinates of the point. 13. 1, 0, 0 6, 6 16. 5, , 17. 2, 3, 4 18. 2, ■ 19–22 ■ ■ ■ ■ ■ 19. (1, s3, 2 s3 ) 21. 0, ■ 1, ■ 23–26 22. ■ ■ ■ ■ ( ■ ■ ■ ■ ■ 6, s3 ) 25. (s3, ■ ■ ■ ■ ■ ■ ■ 2 2 z 2 2 2 1 y z 56. z ■ ■ ■ x ■ 2 2 2z 0 y2 ■ ■ ■ ■ Sketch the solid described by the given inequalities. r2 2 2, 2, z r 0 2, 3, 2 0 2 2 2, 0 3, ■ 2 ■ ■ 6, sec 0 2 ■ ■ ■ ■ ■ ■ ■ ■ outer radius 7 cm. Write inequalities that describe the shell in an appropriate coordinate system. Explain how you have positioned the coordinate system with respect to the shell. 8, 3 ■ z2 63. A cylindrical shell is 20 cm long, with inner radius 6 cm and 4, s2 ) ■ y2 54. y 4 ■ z 59. ■ 2 2y 62. 0 ■ 26. 4, ■ 2z ■ |||| 2 61. 1, 1, s6 ) ■ 24. (s6, 1) 2, ■ y 2 60. 2 Change from cylindrical to spherical coordinates. 23. (1, y 58. 0 3 20. (0, s3, 1) 1 ■ |||| ■ 4, 50. x 2 52. x 2 ■ 57. r 2 Change from rectangular to spherical coordinates. |||| 55. x 2 57–62 15. 1, ■ 2 ■ 14. 3, 0, y2 3 53. x 12. 3, 4, 5 ■ x2 51. x 2 879 49–56 |||| Write the equation (a) in cylindrical coordinates and (b) in spherical coordinates. 49. z Change from rectangular to cylindrical coordinates. |||| 9. 1, 11. 8. 5, 3, 5 ❙❙❙❙ ■ ■ ■ 64. (a) Find inequalities that describe a hollow ball with diameter 27–30 Change from spherical to cylindrical coordinates. |||| 28. (2 s2, 3 29. 8, ■ 6, ■ 2 ■ ■ ■ ■ 2, 30. 4, 27. 2, 0, 0 30 c...
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# Caribou are wary animals with excellent hearing, so stalking
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05 Aug 2008, 05:04
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Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it
(B) so to stalk them over the treeless landscape and get close enough to kill one
(C) so in order to stalk them over the treeless landscape and get close enough to kill one
(D) and so in order to stalk it over the treeless landscape, getting close enough to kill it
(E) and so stalking them over the treeless landscape and getting close enough in order to kill it
[Reveal] Spoiler: OA
If you have any questions
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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05 Aug 2008, 07:01
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A) wrong. run-on sentence with "... stalking... getting..."
B) correct. "to stalk.... required"
C) wrong. "in order" is unneccesary
D) "and so" makes sentence awkward
E) "and so" makes sentence awkward
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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05 Aug 2008, 11:02
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sjgmat wrote:
Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it
(B) so to stalk them over the treeless landscape and get close enough to kill one
(C) so in order to stalk them over the treeless landscape and get close enough to kill one
(D) and so in order to stalk it over the treeless landscape, getting close enough to kill it
(E) and so stalking them over the treeless landscape and getting close enough in order to kill it
"it" is not appropriate
between B and C
B looks concise..
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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05 Aug 2008, 11:48
sjgmat wrote:
Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it
(B) so to stalk them over the treeless landscape and get close enough to kill one
(C) so in order to stalk them over the treeless landscape and get close enough to kill one
(D) and so in order to stalk it over the treeless landscape, getting close enough to kill it
(E) and so stalking them over the treeless landscape and getting close enough in order to kill it
B-
"and so" gives it run-on feel
"kill one" v. "kill it", kill it seems inappropriate
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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05 Aug 2008, 11:56
2
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B is correct :
A is wrong as it messes between "Them" and 'it".
C here "in order to" indicates a actions for which we might need the person who is performing that action just after the commaa (as Dorset people did,).
D and E both are wrong in using "it" (singular) whereas "animals" is not singular.
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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06 Aug 2008, 06:30
It is an obvious 2/3 split between 'SO' and 'AND' in choices A,B,C and D,E. Choices D,E are eliminated straight away. Out of A, B, C; A ends with Kill It, which is not appropriate and this leaves us with only 2 choices, B and C. As pointed out by rigger, 'in order to' gives an unnecessary stretch to the sentence. B is the best option here.
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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06 Aug 2008, 06:56
thanks everyone.. even I was confused as OA - A
The question is from 1000 SC series
If anyone agrees with OA please explain why B is wrong? otherwise I assume B is better option
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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06 Aug 2008, 12:21
After looking more carefully -
the answer could be A (rather than B), but I'm unsure.
Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it
(B) so to stalk them over the treeless landscape and get close enough to kill one - does not fit in with as Dorset people did
Anyone?
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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06 Aug 2008, 12:48
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Take the sentence like this.
To X and to Y
You have to maintain the consistency.
as for choice A, it does not keep up the parallel construction,
stalking, getting , kill <<---
Makes B a better choice
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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05 Sep 2010, 02:50
Accordign to my source, OA is B, with which I completely agree.
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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09 Oct 2013, 17:16
sjgmat wrote:
thanks everyone.. even I was confused as OA - A
The question is from 1000 SC series
If anyone agrees with OA please explain why B is wrong? otherwise I assume B is better option
this question appeared in GmatPrep and B (so to stalk....to kill one) is credited response.
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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30 Oct 2013, 08:57
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sjgmat wrote:
Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it
(B) so to stalk them over the treeless landscape and get close enough to kill one
(C) so in order to stalk them over the treeless landscape and get close enough to kill one
(D) and so in order to stalk it over the treeless landscape, getting close enough to kill it
(E) and so stalking them over the treeless landscape and getting close enough in order to kill it
A, d, e, are all wrong at first sight coz -
In the non-underlined portion the sentence is "Caribou are...." which means the noun Caribou is plural . So, A, d, e are out in first go.
Between B and C I chose B for two reasons -
1. It is concise
2. C lacks the subject in the second clause which makes the second clause a fragment (In order to...can not act as subject)
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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31 May 2014, 01:23
In B i think the parallelism is wrong because "to stalk....and kill". Can anysome show me where I am wrong? Thanks a lot
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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31 May 2014, 20:33
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Dear dqtuan9627,
The parallelism is between "stalk" and "get".
How can you know this? Look for markers that indicate the parallelism: for lists with more than two items, look for the commas; for lists with two items, look for the conjunctions.
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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26 Nov 2014, 05:03
Can somebody tell when the use of 'one' is incorrect? In this case it appears to be correct. there are several sentences from OG that do not use 'one' in the correct choice
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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05 Dec 2014, 02:17
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There is a list of things in this sentence --> stalk and kill
So there must be
a) an and between the items in the list
b) the items in the list must be parallel and
c) the pronoun reference must be correct!
Non underlined portion says that the C ARE wary animals. So we need them and not it.
sjgmat wrote:
Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it Reasons A & C
(B) so to stalk them over the treeless landscape and get close enough to kill one
(C) so in order to stalk them over the treeless landscape and get close enough to kill one in order has the same function as to. redundant
(D) and so in order to stalk it over the treeless landscape, getting close enough to kill it in order has the same function as to. redundant and it is used instead of them
(E) and so stalking them over the treeless landscape and getting close enough in order to kill it Reason C
B is the best
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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05 Dec 2014, 02:19
paranoidvik wrote:
Can somebody tell when the use of 'one' is incorrect? In this case it appears to be correct. there are several sentences from OG that do not use 'one' in the correct choice
Go by the logic of the sentence.
Logic > Grammar > Glamour
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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11 Aug 2016, 10:23
Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it
Stalking, getting – AND is required.
(B) so to stalk them over the treeless landscape and get close enough to kill one
(C) so in order to stalk them over the treeless landscape and get close enough to kill one
So & in order are redundant
(D) and so in order to stalk it over the treeless landscape, getting close enough to kill it
And is used here, then where is the subject for second clause??
Stalk and getting
(E) and so stalking them over the treeless landscape and getting close enough in order to kill it
And is used here, then where is the subject for second clause??
Stalking them and getting to kill it
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink]
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13 Aug 2016, 06:25
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HKD1710 wrote:
Caribou are wary animals with excellent hearing, so stalking them over the treeless landscape, getting close enough to kill it with nothing but a handheld lance, as Dorset people did, required exceptional hunting skill.
(A) so stalking them over the treeless landscape, getting close enough to kill it
Stalking, getting – AND is required.
(B) so to stalk them over the treeless landscape and get close enough to kill one
(C) so in order to stalk them over the treeless landscape and get close enough to kill one
So & in order are redundant
(D) and so in order to stalk it over the treeless landscape, getting close enough to kill it
And is used here, then where is the subject for second clause??
Stalk and getting
(E) and so stalking them over the treeless landscape and getting close enough in order to kill it
And is used here, then where is the subject for second clause??
Stalking them and getting to kill it
Hi ,
Please correct me if I am wrong . I read , and requires independent clause not just and .
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Re: Caribou are wary animals with excellent hearing, so stalking [#permalink] 13 Aug 2016, 06:25
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# Public Key Algorithms II
## Overview
• Digital Signature Standard - DSS
• Zero-knowledge Proof Systems
## Digital Signature Standard (DSS)
• By NIST, designed by NSA
• Proposed in 1991, approved in 1994
• DSA – algorithm
• DSS – standard
• Related to El Gamal
• Speeded up for signer rather than verifier: smart cards
• Controversy RSA vs. DSS
• When DSS was proposed a lot of investments were made in RSA
• DSS a royalty free standard
• DSS key size initially 512 later 1024
• Designed by NSA
## DSS Algorithm
• Generate public: p (512 bit prime) and q (160 bit
prime): p = k*q + 1
• This is a very expensive operation, but not often
• Generate public g≠1: gq = 1 mod p
• Take a random h>1 get g = h
• If g = 1, chose another h(p-1)/q
• Choose long-term public/private key pair <T, S>,
with random S
• T = gS mod p for S < q
• Choose a per message public/private key pair
<Tm, Sm> with random Sm for message m
• Tm = (gSm mod p) mod q
• Calculate Sm-1 mod q , so it won’t need to be done
in real time when signing a message
• Calculate a digest of the message dm = SHS (m)
• SHS – a hashing function recommended by NIST for
use with DSS. SHS hashes to 160 bits
• Signature mod q
• Transmit m, Tm, X
• Public key information T, p, q, and g are known
beforehand
• Verify:
• Compute X-1 mod q
• Compute dm
• x= dm X-1 mod q
• y= Tm X-1 mod q
• z = (gx Ty mod p) mod q
• if z = Tm, the signature is verified
## Why Is DSS Secure
• No revealing of private key S
• Can’t forge a signature without S
• No duplicate messages with matched signature
• Can’t be tempered
• Need a per-message secret number (Sm)!
• If Sm known, S can be computed
• Two messages sharing the same Sm can reveal
Sm, and thus S
## Per message Secret Number
• How is the private key exposed if the secret number
per message is known?
• If Sm is known, one can compute:
• (Xm Sm – dm) Tm-1) mod q = S mod q
• How is the private key exposed when two messages
share the same secret number?
• If m and m’ are signed using the same Sm
• One can compute:
(Xm – Xm’ )-1 ( dm - dm’ ) mod q = Sm mod q
• How to generate random numbers?
## How Secure are RSA and Diffie-Hellman ?
• The security of RSA is based on the difficulty of
factoring
• The security of Diffie-Hellman is based on the
difficulty of resolving discrete log problems
• It has been proved that they are equivalently difficult
• The best known algorithms to resolve them are
subexponencial but superpolynomial
• That’s why a larger key size is required (1024bits)
compared to secret key (80 bits)
## Elliptic Curve Cryptography ECC
• No known subexponential algorithms to break ECC
• So it is secure with much smaller key – improve
performance
• An elliptic curve is a set of points satisfying an
equation of the form:
y^2 + ax + by = x^3 + dx +e
• Mathematical operation on two points in the set will
always produce a point in the set
## Zero-Knowledge Proofs
• Alice: “I know the ingredients in McDonald’s secret
sauce”
• Bob: “No, you don’t”
• Alice: “Yes, I do”
• Bob: “Prove it”
• Alice: “All right. I’ll tell you.” She whispers in Bob’s
ear”
• Bob: “Now I know it too. I will post in my web page”
• Alice: “Oops”
## The Zero-knowledge Cave ## The Zero-knowledge Cave
• 1. Bob stands at point A
• 2. Alice walks all the way into the cave, either to point
C or D
• 3. Bob walks to point B
• 4. Bob asks Alice to
• Come out of the left passage
• Come out of the right passage
• 5. Alice complies by using the magic word to open the
door C or D
• 6. Bob and Alice repeat steps 1-5 n times.
## Zero-knowledge Proof Systems
• Prove knowledge without revealing it
• RSA signatures
• Graph isomorphism: rename vertices
• Alice: graph A and B ~ A
• Public key: graphs A, B
• Private key: mapping between vertices
• Alice: create Gi and sends to Bob
• Bob → Alice: how did A or B → Gi ?
• Zero-knowledge: Bob knows nothing about the mapping
between A and B
• Fred can generate Gi from either A or B, but not both
## Zero-knowledge Proofs: Fiat-Shamir
• Alice: public key <n, v>, n=pq, private key s
• Alice selects random number s, and computes v=s^2
mod n
• Alice chooses k random numbers r1 …, rk
• Alice sends ri^2 mod n
• Bob chooses a random subset 1 of ri^2, the rest is
subset 2
• For subset 1, Alice sends sri mod n, and Bob
verifies (sri)^2 mod n = vri^2 mod n
• For subset 2, Alice simply sends ri mod n
• Finding square root mod n is hard
• Suppose Fred wants to impersonate Alice
• Fred can compute squares mod n but cannot take
square roots mod n
• Fred can send random r and compute r^2, so he can give
correct answers for subset 2, but not for subset 1
• So what the purpose of subset 2? Why isn’t the
protocol simply that Alice sends pairs (ri^2, sri)?
• If only (ri^2, sri) was used
• If Fred has overheard Alice providing (ri^2, sri), he
can impersonate Alice.
• But because both subsets are needed
• Even if he knows subset 1 + answer and subset 2 +
answer, the probability of having the new subset 1’
equal to subset 1 is very small
• For each I the probability is 50%, for the whole
subset (30) no chance to impersonate, on in ten
billion
## Summary
• Digital Signature Standard - DSS
• Zero-knowledge Proof Systems | HuggingFaceTB/finemath | |
## For Ada Lovelace Day: Julia Bowman Robinson
Julia Bowman Robinson was an American mathematician. She is known mainly for her work on decision problems, and most famously for her contributions to the solution of Hilbert’s tenth problem. Robinson was born in St. Louis, Missouri on December 8, 1919. At a young age Robinson recalls being intrigued by numbers. At age nine she contracted scarlet fever and suffered from several recurrent bouts of rheumatic fever. This forced her to spend much of her time in bed, putting her behind in her education. Although she was able to catch up with the help of private tutors, the physical effects of her illness had a lasting impact on her life.
Despite her childhood struggles, Robinson graduated high school with several awards in mathematics and the sciences. She started her university career at San Diego State College, and transferred to the University of California, Berkeley as a senior. There she was highly influenced by the mathematician Raphael Robinson. They quickly became good friends, and married in 1941. As a spouse of a faculty member, Robinson was barred from teaching in the mathematics department at Berkeley. Although she continued to audit mathematics classes, she hoped to leave university and start a family. Not long after her wedding, however, Robinson contracted pneumonia. She was told that there was substantial scar tissue build up on her heart due to the rheumatic fever she suffered as a child. Due to the severity of the scar tissue, the doctor predicted that she would not live past forty and she was advised not to have children .
Robinson was depressed for a long time, but eventually decided to continue studying mathematics. She returned to Berkeley and completed her PhD in 1948 under the supervision of Alfred Tarski. The first-order theory of the real numbers had been shown to be decidable by Tarski, and from Gödel’s work it followed that the first-order theory of the natural numbers is undecidable. It was a major open problem whether the first-order theory of the rationals is decidable or not. In her thesis , Robinson proved that it was not.
Interested in decision problems, Robinson next attempted to find a solution Hilbert’s tenth problem. This problem was one of a famous list of 23 mathematical problems posed by David Hilbert in 1900. The tenth problem asks whether there is an algorithm that will answer, in a finite amount of time, whether or not a polynomial equation with integer coefficients, such as 3x2 − 2y + 3 = 0, has a solution in the integers. Such questions are known as Diophantine problems. After some initial successes, Robinson joined forces with Martin Davis and Hilary Putnam, who were also working on the problem. They succeeded in showing that exponential Diophantine problems (where the unknowns may also appear as exponents) are undecidable, and showed that a certain conjecture (later called “J.R.”) implies that Hilbert’s tenth problem is undecidable. Robinson continued to work on the problem for the next decade. In 1970, the young Russian mathematician Yuri Matijasevich finally proved the J.R. hypothesis. The combined result is now called the Matijasevich-Robinson-Davis-Putnam theorem, or MRDP theorem for short. Matijasevich and Robinson became friends and collaborated on several papers. In a letter to Matijasevich, Robinson once wrote that “actually I am very pleased that working together (thousands of miles apart) we are obviously making more progress than either one of us could alone” .
Robinson was the first female president of the American Mathematical Society, and the first woman to be elected to the National Academy of Science. She died on July 30, 1985 at the age of 65 after being diagnosed with leukemia.
(This short biography is part of the Open Logic Project; PDF here).
## How to Get (Printed) Open Textbooks to Your Students
One problem open textbooks (and instructors adopting open textbooks) face is how to make the texts available to their students. Of course, it’s easy to distribute electronic OERs. But if you want to provide your students a nice, printed version they can take to the coffee shop, you’re in a bind. First, you have to have it printed. This is a bit of work, but with online print-on-demand services like lulu.com it’s possible. But students would have to order the text themselves, and tax and shipping can almost double the (low) cost of a print-on-demand paperback, especially if you want it fast.
So big props to our campus bookstore, especially its manager Brent Beatty, who agreed to order 30 copies for my class and sell them at cost. Brent has been a member of UCalgary’s OER Working Group, so he’s attuned to the issues and challenges of open textbooks. All I had to do was send him the lulu.com order link; with volume discount, low volume shipping cost, and lulu.com’s frequent (constant?) promotional discount (25%) the shelf price is just a few cents above the list price on lulu (C$11). Now I just hope enough students buy it so they’re not making a loss! ## Line Art Portraits of Logicians You’ve probably seen some of the line art portraits of logicians we’ve commissioned. They were done by Calgary illustrator and graphic designer Matthew Leadbeater. We’re pleased to release them all now under a Creative Commons BY-NC license: anyone is free to use them in their own work, to create derivative works from them, and to share them, provided (a) credit to Matt Leadbeater is properly given and (see license terms!) (b) they are not used for any commercial purposes. They each come in two versions, one with a line below, and one with the portrait in a circle. You can download the original Adobe Illustrator files. For PNG and PDF formats, we have set up a GitHub repository. Commissioning these illustrations was made possible by a grant from the Alberta OER initiative. We gratefully acknowledge the support. [Bonus: an image file with all of them that tiles nicely, for your desktop background.] ## Student Satisfaction Survey Results In the Winter term 2016, I taught the University of Calgary’s second logic course from a textbook remixed from the Open Logic Project. Traditionally, Logic II has used Boolos, Burgess & Jeffrey’s Computability and Logic, and it was taught in Fall 2015 using that book as the required text by my colleague Ali Kazmi, and before that by him, Nicole, and me twice a year from that same book. One aim Nicole and I had specifically for the OLP was that it should provide a better text for Logic II, since neither we nor our students seemed to be very happy with “BBJ”. In order to ascertain that the OLP-derived text fares better with students, we did something radical: we asked them what they thought of it. Ali graciously gave permission to run the same textbook survey in his class, so we have something of a baseline. A direct comparison of the two books as textbooks for the course is not easily made, since Ali and I used the books differently: I stuck closer to my text than he did to BBJ; I assigned homework problems from the text; and we assessed students differently, so it’s difficult to control for or compare teaching outcomes. With small samples like ours the results are probably also not statistically significant. But the results are nevertheless interesting, I think, and also gratifying. We obtained clearance from the Conjoint Faculties Research Ethics Board for the study. All students in each section of Logic II in F15 and W16 were sent links to an electronic survey. As an incentive to participate, one respondent from each group was selected to receive a$100 gift certificate to the University of Calgary bookstore. The surveys were started in the last week of classes and remained open for 3 weeks each. Response rates were comparable (23/43 in F15, 23/42 in W16). The survey was anonymous and developed with the help of the Taylor Institute for Teaching and Learning, who also administered the survey; results were not given to us until past the grade appeal deadline in W16.
We asked 23 questions. The first three regarded how students accessed and used the textbooks. In the F15 section, the textbook was not made available electronically, but students were expected to buy their own copy (about $40). Most respondents did that, although almost a quarter apparently pirated electronic copies. In W16, the OLP-derived text was available for free in PDF and students had the option to buy a print copy at$10. Over half the respondents still opted to buy a copy. We asked students how they used the texts in hardcopy and electronic form.
Those using the OLP-derived printed text underlined significantly less than those who used BBJ. I’m guessing the OLP text is better structured and so it’s not as necessary to provide structure & emphasis yourself by underlining. In fact, one student commented on BBJ as follows: “Very little in the way of highlighting, underlining, or separating the information. It was often just walls of text broken up by the occasional diagram.”
When using the electronic version (both PDF), students did not differ much in their habits between F15 and W16. More students took notes electronically in F15. I suspect it’s because the PDF provided in W16 was optimized for screen reading, with narrow margins, and so there was little space for PDF sticky notes as compared with a PDF of the print book in F15. Also notable: highlighting and bookmarking is not very common among users of the PDF.
The second set of questions concerned the frequency with which students consulted the textbook, generally and for specific purposes. W16 students used the OLP-derived text significantly more often than F15 students did, and for all purposes.
The difference is especially striking for the questions about how often students consult the textbook for exams and homework assignments:
We next asked a series of questions about the quality of the texts. These questions were derived from the “Textbook Assessment and Usage Scale” by Regan Gurung and Ryan Martin. On all but one of these questions, the OLP-derived text scored positive (4 or 5 on a 5-point Likert scale) from over half the respondents. The discrepancy to students’ opinions of BBJ is starkest in the overall evaluations:
The one exception was the question “How well are examples used to explain the material?”:
This agrees with what we’ve heard in individual feedback: more, better examples!
Lastly, we were interested in how students think of the prices of textbooks for Logic II. We asked them how much they’d be willing to spend, how much the price influenced their decision to buy it. Interestingly, students seemed more willing to spend money on a textbook in the section (W16) in which they liked the textbook better. They also thought a free/cheap textbook was better value for money than the commercial textbook.
We also asked demographic data. Respondents from both sections were similar: almost all men in each (the course is mainly taken by Computer Science and Philosophy majors), evenly divided among 2nd, 3rd, 4th year students plus a couple of grad students in each (Logic II is required for the Philosophy PhD program). Student in W16 expected higher grades than those in F15, but that may well be just an effect of differences in assessment and grading style rather than better student performance.
If you care, there’s an interactive dashboard with all the graphs, and the raw data.
## Vote for your Favorite Logician
Suppose we got more illustrations of logicians. Who should we get?
## Guess the Logician!
We’ve been working with Calgary graphic designer and illustrator Matt Leadbeater on a series of stylized portraits of logicians. They will be licensed under a Creative Commons Attribution-NonCommercial license (free to use for teaching, but you can’t sell coffee mugs with them on it), and they will be included in the Open Logic Text repository.
Before we make them all available, we’re playing a little game. They’ll be posted without names on Facebook, Twitter, and Google+, and we’ll let you guess who they are!
Props to the Alberta OER initiative for making this happen by providing funding. Making open textbooks visually attractive is an important but overlooked aspect of the OER production process.
## Logicians’ Biographies
Thanks mainly to Samara‘s efforts, the Open Logic Project has begun to include short biographies of logicians in a new History part (Source on GitHhub, PDF). So far we have Cantor, Church, Gentzen, Gödel, Noether, Russell, Tarski, Turing, and Zermelo. We’ve made an effort to appeal to our target audience — undergrad students from a variety of backgrounds — and kept them short and as non-boring as possible. References are included not just to academic biographies, but also to YouTube videos and podcasts. As always, feedback is more than welcome, and we take requests for additions!
In order to enable these biographies to contain references, all driver LaTeX files now automatically generate and include a bibliography. The bibliographies so far are the only texts that contain references (in Natbib author-year format), but they can now be included in other texts as well. Two environments, ‘history’ and ‘reading’ for Historical Remarks and Further Reading sections are now available (defined in open-logic-envs.sty) and material using these will be added to the main text as time goes on. We’re also very close to including portraits! To keep the main repository to a reasonable size, these will be provided in a separate GitHub repository.
[Photo credits: Alan Turing / National Portrait Gallery CC-BY-NC-ND; Georg Cantor / University of Halle Archives; Kurt Gödel / Institute for Advanced Study Archives; Emmy Noether / Göttingen University Library] | open-web-math/open-web-math | |
Properties of Consecutive Integers
We have studied about integers, integers are combination of both positive and negative numbers lying on the number line including zero. There are few important properties regarding integers that you should know:
• Zero is also an Integer.
• Integers which follow one another are called consecutive integers.
• If we multiply a number by an integer and resultant value remains same then that integer is 1.
Consecutive Integers: The integers which follow one another are called consecutive integers.
For example 3, 4, 5, 6 are consecutive integers. An individual random number can never be
a consecutive integer.
Note: If we have consecutive integers in a set M from a to b i.e. M ={a,a+1,a+2,……………………..,b}.
Then number of elements in the set M is b–a+1.
Example: Set M consists of natural numbers from 75 to 199. Find the numbers of elements in set M ?
Solution: Question would have been very easy if we are asked number of natural numbers from 1 to 199. The answer to this is obviously 199. So we can also conclude it same way. From 75 to 199 we are not counting first 74 natural numbers. Hence the answer should be 199 -74= 125
Some more points about Consecutive Numbers
TOOLTIP 1
If there is odd number of digits in the set of consecutive numbers like set of three consecutive numbers (4,5,6,) or 5 digits are there, say (3,4,5,6,7), then in this case the sum of all integers is always divisible by the number of digits present in the set.
For example 2+3+4 = 9 is divisible by number of digits i.e. 3.
For example 1+2+3+4+5 = 15 is divisible by 5.
TOOLTIP 2
On the other hand ,If there are even number of digits in set of consecutive numbers like set of four consecutive numbers (4,5,6,7) or 6 digits are there, say (3,4,5,6,7,8), then in this case the sum of all integers is never divisible by the number of digits present in the set .
For example 2+3+4+5 = 14 is not divisible by number of digits i.e. 4
For example 1+2+3+4+5+6 = 21 is not divisible by 6. | HuggingFaceTB/finemath | |
Question
A point like object of mass M is thrown up from point A as shown in the figure. It slides along the full length of the smooth track ABC (of radius R for part BC). Let R=1 m,AB=2 m,M=0.5 kg,g=10 m/s2 & OD=3m.
A
Minimum speed V0 to slide full length of the track is 60m/s
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B
Minimum speed V0 to reach D is 105m/s
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C
Normal force on the object by the track at C if it reaches D is 15 N
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D
Normal force on the object by the track at B if it reaches D is 32.5 N
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Solution
The correct option is D Normal force on the object by the track at B if it reaches D is 32.5 NFor the object to slide full length, its normal reaction at C should be greater than zero. Nc=mV2cR−mg Nc>0⇒Vc>√Rg By conservation of energy, ⇒12mV20=mg×3+12×m×V2c ⇒12mV20=mg×3+12×m Rg V20=6g+Rg=7g⇒Vo=√70m/s Consider the case where the body falls exactly at D. By Sy=uyt+12ayt2 1=0+12gt2⇒t=1√5 And, Sx=uxt+12axt2⇒3=Vc1√5 So Vc=3√5 By conservation of energy, ⇒12m/V20=m/g×3+12×m/.(3√5)2 V20=60+45⇒V0=√105m/s Similarly by conservation of energy, ⇒VB=√65 NB=mV2BR=0.5×651=32.5N
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# What is 2 percent of 8?
## (2 percent of 8 is 0.16)
### 2 percent of 8 is 0.16. Explanation: What does 2 percent or 2% mean?
Percent (%) is an abbreviation for the Latin “per centum”, which means per hundred or for every hundred. So, 2% means 2 out of every 100.
### Methods to calculate "What is 2 percent of 8" with step by step explanation:
#### Method 1: Diagonal multiplication to calculate 2 percent of 8.
1. For 100, our answer will be 2
2. For 8, our answer will be x
3. 100*x = 2*8 (In Step 1 and 2 see colored text; Diagonal multiplications will always be equal)
4. x = 2*8/100 = 16/100 = 0.16
#### Method 2: Same side division to calculate 2 percent of 8
1. For 100, our answer will be 2
2. For 8, our answer will be x
3. 100/8 = 2/x (In Step 1 and 2, see colored text; Same side divisions will always be equal)
4. 2/x = 100/8 or x/2 = 8/100
5. x = 8*2/100 = 16/100 = 0.16
#### Method 3: Converting Percentage to Decimal to calculate 2 percent of 8
1. Find the decimal equivalent of 2 i.e. 2/100 = 0.02
2. Multiply the decimal equivalent (above) to the given number i.e. 8*0.02 = 0.16
### Percentage examples
Percentages express a proportionate part of a total. When a total is not given then it is assumed to be 100. E.g. 2% (read as 2 percent) can also be expressed as 2/100 or 2:100.
Example: If 2% (2 percent) of your savings are invested in stocks, then 2 out of every 100 dollars are invested in stocks. If your savings are \$10,000, then a total of 2*100 (i.e. \$200) are invested in stocks.
At times you need to calculate a tip in a restaurant, or how to split money between friends in your head, that is without any calculator or pen and paper.
Many a time, it is quite easy if you break it down to smaller chunks. You should know how to find 1%, 10% and 50%. After that finding percentages becomes pretty easy.
• To find 5%, find 10% and divide it by two
• To find 11%, find 10%, then find 1%, then add both values
• To find 15%, find 10%, then add 5%
• To find 20%, find 10% and double it
• To find 25%, find 50% and then halve it
• To find 26%, find 25% as above, then find 1%, and then add these two values
• To find 60%, find 50% and add 10%
• To find 75%, find 50% and add 25%
• To find 95%, find 5% and then deduct it from the number
If you know how to find these easy percentages, you can add, deduct and calculate percentages easily, specially if they are whole numbers. At least you should be able to find an approximate.
### Scholarship programs to learn math
Here are some of the top scholarships available to students who wish to learn math.
### Examples to calculate "What is the percent decrease from X to Y?"
WhatPercentCalculator.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com. | HuggingFaceTB/finemath | |
×
×
# If f(± 2) = 2and g(± 2) = 2evaluate f(g(2)) and g(f(-2))
ISBN: 9780321570567 2
## Solution for problem 7E Chapter 1.1
Calculus: Early Transcendentals | 1st Edition
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Problem 7E
If f(± 2) = 2and g(± 2) =? 2evaluate f(g(2)) and g(f(-2))
Step-by-Step Solution:
Step by solution Step 1 of 2 Consider the given functions f(± 2) = 2and g(± 2) = 2. If we want to find the value of f(g(2)) we must start with the inner function g(2). It is given that g(± 2) = 2or g(2) = g( 2) = 2. Therefore we have: f (g(2))=f(-2). We know that f(± 2) = 2or f(2) = f( 2) = 2, hence we can write: f(g(2))=f(-2)=2.
Step 2 of 2
##### ISBN: 9780321570567
Since the solution to 7E from 1.1 chapter was answered, more than 334 students have viewed the full step-by-step answer. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The full step-by-step solution to problem: 7E from chapter: 1.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. The answer to “If f(± 2) = 2and g(± 2) =? 2evaluate f(g(2)) and g(f(-2))” is broken down into a number of easy to follow steps, and 12 words. This full solution covers the following key subjects: evaluate. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1.
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# Sound Waves of explosion
## Homework Statement
An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are 343, 1482, and 5040 m/s, respectively. The sound travels a distance of 150 m in each medium.
(a) Through which medium does the sound arrive first, second, and third?
(b) After the first sound arrives, how much later do the second and third sounds arrive?
## Homework Equations
v=sqrt(B/density)
## The Attempt at a Solution
We just started going over waves and sound so I'm not sure how to figure out this problem.
For part (a) I figured out that the sounds arrives first at the steel handrail, then freshwater then air.
For part (b) I don't know how to figure that out. I tried just doing how many seconds it would take to travel 150m for each substance, but I don't think that's right.
Related Introductory Physics Homework Help News on Phys.org
Well, the problem states the speed of sound in each medium, so you can get how long it takes in each medium using,
time = distance/velocity
So first it went through the steel handrail so that is 150/5040= 0.02976s, then through the water so 150/1432= 0.1047s, then the air so 150/343=0.4373s.
So the first sound happens after 0.03s, then the second sound comes after 0.1047s, then the third sound comes after 0.4373s. Right?
I get that the third sound comes 0.41s because you have to do 0.437-0.03, but I can't figure out how to get the second sound...?
The question asks how much later do the second and third arrive as compared to the first. You have done most of the work.
Well for the second I did 0.105-0.03 and it's incorrect...?
(a) Through which medium does the sound arrive first, second, and third?
This implies that the sound travels in parallel throught the media, not through each in turn.
Otherwise it would always arrive in the last medium in the chain making the question meaningless. | HuggingFaceTB/finemath | |
If a hyperbola passing through the origin has 3x−4y−1=0 and 4x−3y−6=0 as its asymptotes, then the equations of its transverse and conjugate axes, are
# If a hyperbola passing through the origin has $3x-4y-1=0$ and $4x-3y-6=0$ as its asymptotes, then the equations of its transverse and conjugate axes, are
1. A
$x-y-5=0$ and $x+y+1=0$
2. B
$x-y=0$ and $x+y+5=0$
3. C
$x+y-5=0$ and $x-y-1=0$
4. D
$x+y-1=0$ and $x-y-5=0$
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### Solution:
The transverse axis is the bisector of the angle between asymptotes containing the origin and the conjugate axis is the other bisector. So, their equations are given by
$\frac{-3x+4y+1}{\sqrt{9+16}}=\frac{-4x+3y+6}{\sqrt{16+9}}$
and,
$\frac{-3x+4y+1}{\sqrt{9+16}}=-\frac{-4x+3y+6}{\sqrt{16+9}}$
$⇒$ $x+y-5=0$ and
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# Number theory
Number theory is the study of natural numbers. Natural numbers are the counting numbers that we use in everyday life: 1, 2, 3, 4, 5, and so on. Zero (0) is often considered to be a natural number as well.
Number theory grew out of various scholars' fascination with numbers. An example of an early problem in number theory was the nature of prime numbers. A prime number is one that can be divided exactly only by itself and 1. Thus 2 is a prime number because it can be divided only by itself (2) and by 1. By comparison, 4 is not a prime number. It can be divided by some number other than itself (that number is 2) and 1. A number that is not prime, like 4, is called a composite number.
The Greek mathematician Euclid (c. 325–270 B.C. ) raised a number of questions about the nature of prime numbers as early as the third century B.C. Primes are of interest to mathematicians, for one reason: because they occur in no predictable sequence. The first 20 primes, for example, are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, and 71. Knowing this sequence, would you be able to predict the next prime number? (It is 73.) Or if you knew that the sequence of primes farther on is 853, 857, 859, 863, and 877, could you predict the next prime? (It is 883.)
Questions like this one have intrigued mathematicians for over 2,000 years. This interest is not based on any practical application the answers may have. They fascinate mathematicians simply because they are engrossing puzzles.
## Famous theorems and problems
Studies in number theory over the centuries have produced interesting insights into the properties of natural numbers and ongoing puzzles about such numbers. As just one example of the former, consider Fermat's theorem, a discovery made by French mathematician Pierre de Fermat (1601–1665). Fermat found a quick and easy way to find out if a particular number is a prime or composite number. According to Fermat's theorem, one can determine if any number (call that number p ) is a prime number by the following method: choose any number (call that number n ) and raise that number to p . Then subtract n from that calculation. Finally, divide that answer by p . If the division comes out evenly, with no remainder, then p is a prime number.
Fermat was also responsible for one of the most famous puzzles in mathematics, his last theorem. This theorem concerns equations of the general form x n + y n = z n . When n is 2, a very familiar equation results: x 2 + y 2 = z 2 , the Pythagorean equation of right-angled triangles.
The question that had puzzled mathematicians for many years, however, was whether equations in which n is greater than 2 have any solution. That is, are there solutions for equations such as x 3 + y 3 = z 3 , x 4 + y 4 = z 4 , and x 5 + y 5 = z 5 ? In the late 1630s, Fermat wrote a brief note in the margin of a book saying that he had found proof that such equations had no solution when n is greater than 2. He never wrote out that proof, however, and for more than three centuries mathematicians tried to confirm his theory.
As it turns out, any proof that Fermat had discovered was almost certainly wrong. In 1994, Princeton University professor Andrew J. Wiles announced that he had found a solution to Fermat's theorem. But flaws were soon discovered in Wiles's proof (which required more than 150 pages of mathematical equations). By late 1994 Wiles thought the flaws had been solved. However, it will take several years before other mathematicians will be able to verify Wiles's work.
## Words to Know
Composite number: A number that can be factored into two or more prime numbers in addition to 1 and itself.
Cryptography: The study of creating and breaking secret codes.
Factors: Two or more numbers that can be multiplied to equal a product.
Prime number: Any number that can be divided evenly only by itself and 1.
Product: The number produced by multiplying two or more numbers.
## Applications
As mentioned above, the charm of number theory for mathematicians has little or nothing to do with its possible applications in everyday life. Still, such applications do appear from time to time. One such application has come about in the field of cryptography—the writing and deciphering of secret messages (or ciphers). In the 1980s, a number of cryptographers almost simultaneously announced that they had found methods of writing ciphers in such a way that they could be sent across public channels while still remaining secrets. Those methods are based on the fact that it is relatively easy to raise a prime number to some exponent but very difficult to find the prime factors of a large number.
For example, it is relatively simple, if somewhat time-consuming, to find 358 143 . Actually, the problem is not even time-consuming if a computer is used. However, finding the prime factors of a number such as 384,119,982,448,028 is very difficult unless one knows one of the prime factors to begin with. The way public key cryptography works, then, is to attach some large number, such as 384,119,982,448,028, as a "key" to a secret message. The sender and receiver of the secret message must know one of the prime factors of that number that allows them to decipher the message. In theory, any third party could also decipher the message provided that they could figure out the prime factors of the key. That calculation is theoretically possible although, in practice, it takes thousands or millions of calculations and a number of years, even with the most powerful computers now known.
## User Contributions:
This is one of the most clear explanations of math theorems i have ever read !
Organized and complete.
Thank you.
Hi,Am pursuing post graduate from OU.can you help me in knowing the importanr theorems in elementary number theory.
IT'S VERY HELPFUL FOR MY ASSIGNMENT.THANK U FOR PRESENTING THIS.
since natural numbers are used for counting,HOW ZERO IS A NATURAL NUMBER?
Hi,Am pursuing postgraduate from OU, Give a suggestion for a project in numbertheory
This article was very very simple to understand. It did not use many fancy words, instead used simple words and expressions. | HuggingFaceTB/finemath | |
## How to Calculate and Solve for Monochromatic Emissive Power | Radiation Heat Transfer The image above represents monochromatic emissive power.
To compute for monochromatic emissive power, five essential parameters are needed and these parameters are Planck’s Constant (h), Velocity of Light (c), Wavelength (λ), Boltzmann’s Constant (KB) and Temperature (T).
The formula for calculating monochromatic emissive power:
ebx = 2πhc²λ-5 / exp(ch/KBλT) – 1
Where:
ebx = Monochromatic Emissive Power | Planck’s Equation
h = Planck’s Constant
c = Velocity of Light
λ = Wavelength
KB = Boltzmann’s Constant
T = Temperature
Let’s solve an example;
Find the monochromatic emissive power when the planck’s constant is 6.626E-24, the velocity of light is 3E8, the wavelength is 22, the boltzmann’s constant is 1.380E-23 and the temperature is 10.
This implies that;
h = Planck’s Constant = 6.626E-24
c = Velocity of Light = 3E8
λ = Wavelength = 22
KB = Boltzmann’s Constant = 1.380E-23
T = Temperature = 10
ebx = 2πhc²λ-5 / exp(ch/KBλT) – 1
ebx = 2π(6.62607004e-34)(300000000)²(22)-5 / exp((300000000)(6.62607004e-34)/(1.38064852e-23)(22)(10)) – 1
ebx = 2π(6.62607004e-34)(90000000000000000)(1.94037e-7) / exp(1.987821012e-25/3.037426744e-21) – 1
ebx = 7.270512005456302e-23 / exp(0.00006544424539379114) – 1
ebx = 7.270512005456302e-23 / 1.0000654463869152 – 1
ebx = 7.270512005456302e-23 / 0.00006544638691519111
ebx = 1.11e-18
Therefore, the monochromatic emissive power is 1.11e-18.
## How to Calculate and Solve for Velocity | De Broglie’s Law The image above represents velocity.
To compute for velocity, three essential parameters are needed and these parameters are Planck’s Constant (h), Wavelength (λ) and Mass (m).
The formula for calculating velocity:
v = h / λm
Where;
v = velocity
λ = wavelength
h = Planck’s constant
m = mass
Let’s solve an example;
Find the velocity when the wavelength is 10, the planck’s constant is 6.626e-34 and the mass is 5.
This implies that;
λ = wavelength = 10
h = Planck’s constant = 6.626e-34
m = mass = 5
v = h / λm
v = 6.62607004e-34 (10)(5)
v = 6.62607004e-34 (50)
v = 1.33e-35
Therefore, the velocity is 1.33e-35 m/s.
## How to Calculate and Solve for Mass | De Broglie’s Law The image above represents mass.
To compute for mass, three essential parameters are needed and these parameters are Planck’s Constant (h), Wavelength (λ) and Velocity (v).
The formula for calculating the mass:
m = h / λv
Where;
m = mass
λ = wavelength
h = Planck’s constant
v = velocity
Let’s solve an example;
Find the mass when the wavelength is 4, the planck’s constant is 6.626e-34 and the velocity is 2.
This implies that;
λ = wavelength = 4
h = Planck’s constant = 6.626e-34
v = velocity = 2
m = hλv
m = 6.62607004e-34(4)(2)
m = 6.62607004e-34(8)
m = 8.28e-35
Therefore, the mass is 8.28e-35 kg.
## How to Calculate and Solve for Wavelength | De Broglie’s Law The image above represents wavelength.
To compute for wavelength, three essential parameters are needed and these parameters are Planck’s Constant (h), Mass (m) and Velocity (v).
The formula for calculating wavelength:
λ = hmv
Where
λ = wavelength
h = Planck’s constant
m = mass
v = velocity
Let’s solve an example;
Find the wavelength when the planck’s constant is 6.62e-34, the mass is 21 and the velocity is 11.
This implies that;
h = Planck’s constant = 6.62e-34
m = mass = 21
v = velocity = 11
λ = hmv
λ = 6.62607004e-34(21)(11)
λ = 6.62607004e-34(231)
λ = 2.86e-36
Therefore, the wavelength is 2.86e-36 m.
Calculating the Mass when the Wavelength, the Planck’s Constant and the Velocity is Given.
m = h / λv
Where;
m = mass
λ = wavelength
h = Planck’s constant
v = velocity
Let’s solve an example;
Find the mass when the wavelength is 18, the planck’s constant is 6.626e-34 and the velocity is 6.
This implies that;
λ = wavelength = 18
h = Planck’s constant = 6.626e-34
v = velocity = 6
m = h / λv
m = 18 / 6.626e-34 x 6
m = 18 / 3.976e-33
m = 4.53
Therefore, the mass is 4.53 m.
## The Calculator Encyclopedia Calculates and Solves the Wavelength for a Wave – Particle Behaviour
According to Wikipedia,
Wave–particle duality is the concept in quantum mechanics that every particle or quantum entity may be partly described in terms not only of particles, but also of waves.
In physics, the wavelength is the spatial period of a periodic wave—the distance over which the wave’s shape repeats.
Nickzom Calculator requires two parameters to compute the wavelength of a wave or particle. These parameters are:
• Mass
• Velocity
The formula for computing the wavelength is:
λ = h / mv
Where:
λ = Wavelength
h = Planck’s Constant (6.63 x 10-34 js)
m = Mass
v = Velocity
Let’s solve an example, find the wavelength of a wave – particle with a mass of 300 Kg and a velocity of 3 x 1010 m/s.
From the example, we can see that:
m = 300
v = 3 x 1010
λ = 6.63 x 10-34 / 300 (3 x 1010)
λ = 6.63 x 10-34 / 9 x 1012
λ = 7.37 x 10-47
Therefore, the wavelength of the wave – particle (λ) is 7.37 x 10-47. | HuggingFaceTB/finemath | |
Studying convergence of improper integral
I have a doubt on how to apply asymptotic comparison to this improper integral: $$\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$$
I'll call $$f(x)=\dfrac{4x}{4x^8 + 1}$$, and notice it is non negative as $$x \to +\infty$$.
If I understand the idea correctly, the I need to find a function $$g(x)$$ (also non negative as $$x \to +\infty$$) such that:
• if $$\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)}$$ is finite and not zero: $$\displaystyle\int_0^{+\infty} f(x) dx$$ converges if and only if $$\displaystyle\int_0^{+\infty}g(x)dx$$ converges.
• if $$\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 0$$ and $$\displaystyle\int_0^{+\infty} g(x) dx$$ converges: $$\displaystyle\int_0^{+\infty}f(x)dx$$ converges.
• if $$\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = +\infty$$ and $$\displaystyle\int_0^{+\infty} f(x) dx$$ converges: $$\displaystyle\int_0^{+\infty}g(x)dx$$ converges.
If I choose $$g(x) = \dfrac{1}{x^7}$$, I have $$\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)} = 1$$, so the first case applies. I know that $$\displaystyle \int_0^{+\infty}\dfrac{1}{x^7}dx$$ diverges, so $$\displaystyle\int_0^{+\infty} \frac{4x}{4x^8 + 1}dx$$ should diverge, but it does not..
What am I doing wrong?
• if $$\displaystyle\lim _{x \to +\infty} \frac{f(x)}{g(x)}$$ is finite and not zero: $$\displaystyle\int_0^{+\infty} f(x) dx$$ converges if and only if $$\displaystyle\int_0^{+\infty}g(x)dx$$ converges.
This is false. The reason that $$\int_0^\infty \frac1{x^7}dx$$ diverges is not because of the behaviour as $$x\to\infty$$ but rather near $$x=0$$.
Try splitting up your integral into two intervals, say $$(0,1)$$ and $$(1,\infty)$$. In particular, note that $$\int_1^\infty \frac1{x^7}dx$$ converges.
• Makes sense.. I wonder why my professor only gave me that formulation of the criterion, only considering integrals that diverge due to their behaviour as $x \to \infty$... So I guess choosing a comparison function in the $\frac{1}{x^{\alpha}}$ shape is only useful if, case $\alpha > 1$, I consider $\int_k^{+\infty}g(x)dx$ or if, case $alpha < 1$ I consider $\int_0^k g(x)dx$, with $k \in \mathbb{R}$, right? Commented Jan 20, 2021 at 17:10
• @user256439 The missing part of the criterion is that $f(x)$ and $g(x)$ must be defined on some interval $[a, \infty)$. In this case, we can't choose $a=0$ because $g(0)$ is not defined. You can see that choosing $a=1$ (or any other positive number) leads to the correct conclusion; it just remains to take care of the other part of the integral, i.e., on $(0,1)$. Commented Jan 20, 2021 at 17:31
• Ah okay, yeah it did say to consider a neighbourhood of $+\infty$. Thanks a lot, you solved my doubt :) Commented Jan 20, 2021 at 17:37
I think, you used some thing wrong. But it seems you have the idea.
Hint
$$\frac{4}{x^{-1}(4x^8+1)}$$ split the integral to $$I_1$$ and $$I_2$$ where $$I_1=\displaystyle\int_0^{1} \frac{4x}{4x^8 + 1}dx$$and $$I_2=\displaystyle\int_1^{+\infty} \frac{4x}{4x^8 + 1}dx$$ Now, $$I_1\leq \displaystyle\int_0^{+1} \frac{4}{x^{-1}}dx$$ and $$I_2\leq \displaystyle\int_1^{+\infty} \frac{4}{x^7}dx$$ I think, it is clear now. I hope that helps | HuggingFaceTB/finemath | |
## An x-ray beam of a certain wavelength is incident on a crystal, at 30.7° to a certain family of reflecting planes of spacing 33.4 pm. If the
Question
An x-ray beam of a certain wavelength is incident on a crystal, at 30.7° to a certain family of reflecting planes of spacing 33.4 pm. If the reflection from those planes is of the first order, what is the wavelength of the x rays?
in progress 0
3 days 2021-07-20T04:41:26+00:00 2 Answers 2 views 0
The wavelength of the x rays is 34.104 pm
Explanation:
Given data:
θ = incidence angle = 30.7°
d = space between planes = 33.4 pm = 33.4 x10⁻¹²m
According Bragg’s expression:
λ=34.068pm
Explanation:
Wavelength λ=2dsinθ
λ=2(33.4pm)sin30.7
λ=2(33.4pm) * 0.510
λ=66.8* sin30.7
λ=66.8*0.510
λ=34.068pm
X-rays are usually produced by charged particles that are accelerating or decelerating such as a beam of electrons striking a metal plate in an X-ray tube. | HuggingFaceTB/finemath | |
# The height at which a geostationary satellite is placed from the surface of the earth:
1. 30,000 km
2. 36,000 km
3. 42,000 km
4. 48,000 km
Option 2 : 36,000 km
## Detailed Solution
The correct answer is option 2) i.e. 36,000 km
CONCEPT:
• A geostationary satellite is an earth-orbiting satellite that revolves around the earth at the same time as the earth rotates along its axis. Due to the same time of the revolution, it appears stationary from the Earth, and hence it is called geostationary.
• The time period of the satellite: It is the time taken by the satellite to complete one revolution around the Earth.
• Consider a satellite orbiting the earth at a height h from the surface of the earth of radius R.
• The circumference of orbit of satellite = 2π(R+h)
• The orbital velocity of the satellite at a height h is given by,
$$⇒ v_0 =\sqrt{\frac{GM}{R+h}}$$
Where M is the mass of the Earth.
The time period of the satellite is given by:
$$⇒ T = \frac{circumference}{orbital \: velocity} = \frac{2\pi(R+h)}{\sqrt{\frac{GM}{R+h}}} = 2\pi\sqrt{\frac{(R+h)^3}{GM}}$$
CALCULATION:
• The time period of a geostationary satellite is 24 hours.
∴ T = 24 h = 86400 s
Taking into consideration that,
Radius of earth = 6400 km = 6400000 m
Mass of earth = 5.97 × 1024 kg
$$⇒ T = 2\pi\sqrt{\frac{(R+h)^3}{GM}}$$
$$⇒ 86400 = 2\pi\sqrt{\frac{(R+h)^3}{(6.67 × 10^{-11})(5.97 × 10^{24})}}$$
⇒ (R + h)3 = 7.53 × 1022
⇒ R + h = 42226910
⇒ h = 42226910 - R = 42226910 - 6400000
⇒ h = 35826910 m ≈ 36000 km
Thus, the geostationary satellite is placed at about 36,000 km from the surface of Earth. | HuggingFaceTB/finemath | |
# Do we call the logical connectives(e.g. $\wedge$) propositional functions?
I know that a predicate is also called a propositional function, since it accepts one or more entities as its argument, and return a proposition. For example, let $P(x,y)=x\text{ is the father of }y."$, then $P$ is a predicate, and meanwhile the propositional function, since when we substitute Adam and Jeff into $P$, then $P(\text{Adamm},\text{Jeff})=\text{Adam is the father of Jeff.}"$, which is a proposition.
Now, in the propositional logic context, the logic connectives($\wedge,\rightarrow$, etc) are essentially functions that take some propositions(not entities this time) as their arguments, and also return the propositions. For example, $\wedge(1<5,2+2=4)=(1<5)\wedge (2+2=4)$, which is also a proposition. So do we, or can we call these logical connectives the propositional functions? Why or why not?
• They are ussually called Truth functions. – Mauro ALLEGRANZA Dec 2 '16 at 13:00
• The issue is that the term propositional function has already an "established" meaning in modern logic. – Mauro ALLEGRANZA Dec 2 '16 at 13:05
• @MauroALLEGRANZA I think the sentence in wiki: "a truth function is a function from a set of truth values to truth values." is a little weird if the author at the same time considers the logical connectives are of truth functions. Since, as I know, connectives (in propositional context) take proposition(s) and return a proposition, rather then take or return $T$ or $F$. The corresponding truth value of the result of the connectives, is always gained by applying a valuation function on the propositions; that is, a connectives never directly returns a $T$ nor a $F$. – Eric Dec 2 '16 at 13:15
• In the "syntax view" we have propositional letters (or symbols) : $p_i$: in the "semantic view" we have truth values : $\{ 0, 1 \}$. When you speak of "propositions" are you referring to "content" (i.e. to meaning) or to expressions (strings of symbols) ? – Mauro ALLEGRANZA Dec 2 '16 at 13:22
• One problem is that usually functions are defined (in terms of sets which are defined) in terms of logical operators. So "truth function" is about like defining graphs as "trees with cycles" or defining real numbers as "one dimension of a complex number" or defining an n-dimensional point as "the n-dimensional square with zero volume". Not wrong, but completely backwards. That isn't to say that there aren't some logics that define functions before they define operators (lambda logic for example). – DanielV Dec 2 '16 at 15:08
Sure, we could call logical connectives functions. It's perfectly reasonable, and it fits the definition of "function" just fine. The more common terminology is operator, but function works just as well.
However, it's useful to distinguish between connectives and predicates. Connectives are truth-functional, which means that the truth of their output is dependent only on the truth of their inputs; predicates aren't, because "truth" doesn't even make sense for the argument of a predicate. Much more can be said about truth-functional operators than about "propositional functions" in general.
• Yes, I know that logical connectives are " stronger "(means the former is a special case of the latter) then (a general) propositional functions. But if just talk about the correctness, is it correct to call logical connectives a type of propositional functions.? – Eric Dec 2 '16 at 13:08
• @Eric As far as I know, "propositional function" does not have a standard meaning. If it has been formally defined for you, then it depends on whether it has been defined as "a function that outputs propositions" (in which case yes, connectives are) or "a predicate" (in which case they aren't). Either option is perfectly reasonable, so it depends on exactly which source you're working with, – Reese Dec 2 '16 at 13:16
• Thanks! I missed to notice that if the propositional function is simply defined as "a function that outputs propositions", then propositional functions will no longer be equivalent to *predicates*(whose arguments definitely can't be the propositions), which would be a problem if most people tends to consider these terminlogy equal(though I'm not sure whether so). All things are clear now, thanks for your help!! – Eric Dec 2 '16 at 13:31
can we call these logical connectives the propositional functions?
The issue is legitimate, but "history" followed a different path and now we have a quite established convention.
For the source, see :
• Alfred North Whitehead & Bertrand Russell, Principia Mathematica to *56 (1st ed.1910, 2nd ed.1927); see Introduction (to the 1st edition), page 1-on:
[page 5] Variables will be denoted by single letters [...] $p, q, r$ will be called propositional letters, and will stand for variable propositions [sic! meaning that they are variables for propositions].
[page 6] An aggregation of propositions [...] into a single proposition more complex than its constituents, is a function with propositions as arguments. [...] there are four special cases which are of fundamental importance, since all the aggregations of subordinate propositions into one complex proposition which occur in the sequel are formed out of them step by step. They are (1) the Contradictory Function, (2) the Logical Sum, or Disjunctive Function, (3) the Logical Product, or Conjunctive Function, (4) the Implicative Function. [...]
The Contradictory Function with argument $p$, where $p$ is any proposition, is the proposition which is the contradictory of $p$, that is, the proposition asserting that $p$ is not true. This is denoted by $\sim p$.
[page 7] These four functions of propositions are the fundamental constant (i.e.definite) propositional functions with propositions as arguments, and all other constant propositional functions with propositions as arguments [...] are formed out of them by successive steps.
[page 14] Let $\phi x$ be a statement containing a variable $x$ and such that it becomes a proposition when $x$ is given any fixed determined meaning. Then $\phi x$ is called a "propositional function"; it is not a proposition, since owing to the ambiguity of $x$ it really makes no assertion at all.
Principia Mathematica has been blamed for this kind of sloppiness regarding the sintactical specifications of the language.
Thus, the two different usages of "propositional function" has been subsequently replaced by connective and open formula respectively. | HuggingFaceTB/finemath | |
the question is: is two thirds of anything a bigger number than five sixths of anything
# I need to know if five sevenths is more than two thirds please
the question is: is two thirds of anything a bigger number than five sixths of anything
the best way to answer this is to divide
2/3 = 0.667
5/7 = 0.714
This means the 0.714(5/7) is greater than 0.667(2/3) ◄ ans
answered Nov 2, 2011 by Level 3 User (2,700 points)
NO YOU IDIOT!!!!!!!!!\
:P
answered Feb 21, 2013 by anonymous | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Compare 7/6 and 18/20
1st number: 1 1/6, 2nd number: 18/20
7/6 is greater than 18/20
#### Steps for comparing fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 6 and 20 is 60
2. For the 1st fraction, since 6 × 10 = 60,
7/6 = 7 × 10/6 × 10 = 70/60
3. Likewise, for the 2nd fraction, since 20 × 3 = 60,
18/20 = 18 × 3/20 × 3 = 54/60
4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
5. 70/60 > 54/60 or 7/6 > 18/20
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
# Find the coordinates in an isosceles triangle if the triangle it self is in positive axis
A at $(45,10)$, B at $(10,20)$, $AB=AC$ and angle $C=20$ degree find the coordinates of $C$.suggest the formula so i can write code in Perl.
-
We have $A(45,10),B(10,20),C(x_c,y_c)$.
$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$
$C=B=20^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=\pi /9$ rad
$A=180^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }-40^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=140^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=7\pi /9$ rad
Let's make the following change of variables: $x=X+45,y=Y+10$ (translation of axes). Then $A$ becomes the origin of the $XY$ referential.
The vector $\overrightarrow{AB}$ can be written in this $XY$ referential as
$\overrightarrow{AB}=(5\sqrt{53}\cos \left( \pi -\arctan \frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\arctan \frac{2}{7}\right) )=(-35,10)$
and the vector $\overrightarrow{AC}$ as
$\overrightarrow{AC}=(5\sqrt{53}\cos \left( \pi -\frac{7\pi }{9}-\arctan \frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\frac{7\pi }{9}-\arctan \frac{% 2}{7}\right) )$
Therefore in the original referencial $xy$, we have
$x_{C}=5\sqrt{53}\cos \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi \right) +45\approx 78.239$
$y_{C}=5\sqrt{53}\sin \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi \right) +10\approx 24.837$
-
You have the coordinates of A and B, so you can compute the distance AB. AB=AC, so you then know the distance AC. Let the coordinates of C be (x,y). Apply the distance formula to A and C and set the result equal to the distance you already computed. This equation guarantees that AB=AC.
Now, the angle at C is determined by the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$. These vectors can be found by subtracting the coordinates of C from A and B (respectively). $(CA)(CB)\cos C = \overrightarrow{CA}\cdot\overrightarrow{CB}$, and you know the distances CA = AC = AB and the measure of C, and you can compute CB and the dot product in terms of (x,y). This equation guarantees that the measure of C is 20°.
Solve the system resulting from the two equations above to find the coordinates of C. There are almost certainly 2 solutions. | HuggingFaceTB/finemath | |
# Thread: unique solutions to system of equations.
1. ## unique solutions to system of equations.
I was under the impression that a system has a unique solution if its determinant was not equal to zero. This website seems to verify this fact https://math.oregonstate.edu/home/pr...em/system.html
Below are the solutions my teacher has provided to the question. But who is right? Do you get a unique solution if you set the matrix equal to zero or do you get a unique solution if you set the matrix not equal to zero?
2. ## Re: unique solutions to system of equations.
I agree with you. A linear system has a unique solution if the corresponding system-matrix has a non-zero determinant. I think your teacher implicitly means that the solution is given as the set:
$\displaystyle \mathbb{R} \setminus \lbrace -1 , -3 \rbrace$
3. ## Re: unique solutions to system of equations.
Your teacher explicitly states the values of $k$ that you left implicit. But you never actually said in your answer what you think those values represent.
It's not sufficient in solving a problem, to just scribble some lines. You should explain what you are doing and why, and clearly state your final result.
Why did you transform the matrix like the anyway?
4. ## Re: unique solutions to system of equations.
This is not my work. This is my teachers solution to the question... I am just trying to understand it.
Krisly, what does your notation mean in English?
5. ## Re: unique solutions to system of equations.
It means, the real numbers without the set consisting of $\displaystyle -1$ and $\displaystyle -3$. It is equivalent to say $\displaystyle k$ can be anything except $\displaystyle -1$ or $\displaystyle -3$. | HuggingFaceTB/finemath | |
## Intr-o grupa, 3 elevi au masurat lungimea unui creion si au obtinut urmatoarele valori: 16,2 cm;16,5 cm si respective 16,1 cm. Aflati
Question
Intr-o grupa, 3 elevi au masurat lungimea unui creion si au obtinut urmatoarele valori: 16,2
cm;16,5 cm si respective 16,1 cm. Aflati valoarea medie a valorilor masurate si erorile obtinute
de fiecare elev.
in progress 0
3 days 2021-07-19T14:03:46+00:00 1 Answers 1 views 0
Explanation:
In a group, 3 students measured the length of a pencil and obtained the following values: 16.2
cm, 16.5 cm and 16.1 cm, respectively. Find the average value of the measured values and the errors obtained by each student.
Average value can be calculated by finding the mean of the values
Average value = Σx / N
A•V = (16.2 + 16.5 + 16.1) / 3
A.V = 48.8 / 3
A.V = 16.27cm
x~ = 16.27 cm
Each student error
If we assume the true value is average value, then, each student error can be calculated using
e = |x~ – x|
e = 16.27 – 16.2
e = 0.07 cm
Also,
e = 16.27 – 16.5
e = -0.23 cm
e = 0.23 cm, due to absolute value
Also,
e = 16.27 – 16.1
e = 0.17 cm
In Spanish
El valor promedio se puede calcular encontrando la media de los valores
Valor promedio = Σx / N
A • V = (16,2 + 16,5 + 16,1) / 3
A.V = 48.8 / 3
A.V = 16.27cm
x ~ = 16,27 cm
Si asumimos que el valor verdadero es el valor promedio, entonces, cada error del estudiante puede calcularse usando
e = | x ~ – x |
e = 16,27 – 16,2
e = 0.07 cm
También,
e = 16.27 – 16.5
e = -0,23 cm
e = 0.23 cm, debido al valor absoluto
También,
e = 16.27 – 16.1
e = 0,17 cm | HuggingFaceTB/finemath | |
Thomson Problem
Get Thomson Problem essential facts below. View Videos or join the Thomson Problem discussion. Add Thomson Problem to your PopFlock.com topic list for future reference or share this resource on social media.
Thomson Problem
The objective of the Thomson problem is to determine the minimum electrostatic potential energy configuration of N electrons constrained to the surface of a unit sphere that repel each other with a force given by Coulomb's law. The physicist J. J. Thomson posed the problem in 1904[1] after proposing an atomic model, later called the plum pudding model, based on his knowledge of the existence of negatively charged electrons within neutrally-charged atoms.
Related problems include the study of the geometry of the minimum energy configuration and the study of the large N behavior of the minimum energy.
Mathematical statement
The physical system embodied by the Thomson problem is a special case of one of eighteen unsolved mathematics problems proposed by the mathematician Steve Smale -- "Distribution of points on the 2-sphere".[2] The solution of each N-electron problem is obtained when the N-electron configuration constrained to the surface of a sphere of unit radius, ${\displaystyle r=1}$, yields a global electrostatic potential energy minimum, ${\displaystyle U(N)}$.
The electrostatic interaction energy occurring between each pair of electrons of equal charges (${\displaystyle e_{i}=e_{j}=e}$, with ${\displaystyle e}$ the elementary charge of an electron) is given by Coulomb's Law,
${\displaystyle U_{ij}(N)=k_{e}{e_{i}e_{j} \over r_{ij}}.}$
Here, ${\displaystyle k_{e}}$ is Coulomb's constant and ${\displaystyle r_{ij}=|\mathbf {r} _{i}-\mathbf {r} _{j}|}$ is the distance between each pair of electrons located at points on the sphere defined by vectors ${\displaystyle \mathbf {r} _{i}}$ and ${\displaystyle \mathbf {r} _{j}}$, respectively.
Simplified units of ${\displaystyle e=1}$ and ${\displaystyle k_{e}=1}$ are used without loss of generality. Then,
${\displaystyle U_{ij}(N)={1 \over r_{ij}}.}$
The total electrostatic potential energy of each N-electron configuration may then be expressed as the sum of all pair-wise interactions
${\displaystyle U(N)=\sum _{i
The global minimization of ${\displaystyle U(N)}$ over all possible collections of N distinct points is typically found by numerical minimization algorithms.
Example
The solution of the Thomson problem for two electrons is obtained when both electrons are as far apart as possible on opposite sides of the origin, ${\displaystyle r_{ij}=2r=2}$, or
${\displaystyle U(2)={1 \over 2}.}$
Known solutions
Schematic geometric solutions of the mathematical Thomson Problem for up to N = 5 electrons.
Minimum energy configurations have been rigorously identified in only a handful of cases.
• For N = 1, the solution is trivial as the electron may reside at any point on the surface of the unit sphere. The total energy of the configuration is defined as zero as the electron is not subject to the electric field due to any other sources of charge.
• For N = 2, the optimal configuration consists of electrons at antipodal points.
• For N = 3, electrons reside at the vertices of an equilateral triangle about a great circle.[3]
• For N = 4, electrons reside at the vertices of a regular tetrahedron.
• For N = 5, a mathematically rigorous computer-aided solution was reported in 2010 with electrons residing at vertices of a triangular dipyramid.[4]
• For N = 6, electrons reside at vertices of a regular octahedron.[5]
• For N = 12, electrons reside at the vertices of a regular icosahedron.[6]
Notably, the geometric solutions of the Thomson problem for N = 4, 6, and 12 electrons are known as Platonic solids whose faces are all congruent equilateral triangles. Numerical solutions for N = 8 and 20 are not the regular convex polyhedral configurations of the remaining two Platonic solids, whose faces are square and pentagonal, respectively[].
Generalizations
One can also ask for ground states of particles interacting with arbitrary potentials. To be mathematically precise, let f be a decreasing real-valued function, and define the energy functional ${\displaystyle \sum _{i
Traditionally, one considers ${\displaystyle f(x)=x^{-\alpha }}$ also known as Riesz ${\displaystyle \alpha }$-kernels. For integrable Riesz kernels see[7]; for non-integrable Riesz kernels, the Poppy-seed bagel theorem holds, see[8]. Notable cases include ? = ?, the Tammes problem (packing); ? = 1, the Thomson problem; ? = 0, Whyte's problem (to maximize the product of distances).
One may also consider configurations of N points on a sphere of higher dimension. See spherical design.
Relations to other scientific problems
The Thomson problem is a natural consequence of Thomson's plum pudding model in the absence of its uniform positive background charge.[9]
"No fact discovered about the atom can be trivial, nor fail to accelerate the progress of physical science, for the greater part of natural philosophy is the outcome of the structure and mechanism of the atom."
--Sir J. J. Thomson[10]
Though experimental evidence led to the abandonment of Thomson's plum pudding model as a complete atomic model, irregularities observed in numerical energy solutions of the Thomson problem have been found to correspond with electron shell-filling in naturally occurring atoms throughout the periodic table of elements.[11]
The Thomson problem also plays a role in the study of other physical models including multi-electron bubbles and the surface ordering of liquid metal drops confined in Paul traps.
The generalized Thomson problem arises, for example, in determining the arrangements of the protein subunits which comprise the shells of spherical viruses. The "particles" in this application are clusters of protein subunits arranged on a shell. Other realizations include regular arrangements of colloid particles in colloidosomes, proposed for encapsulation of active ingredients such as drugs, nutrients or living cells, fullerene patterns of carbon atoms, and VSEPR theory. An example with long-range logarithmic interactions is provided by the Abrikosov vortices which would form at low temperatures in a superconducting metal shell with a large monopole at the center.
Configurations of smallest known energy
In the following table ${\displaystyle N}$ is the number of points (charges) in a configuration, ${\displaystyle E_{1}}$ is the energy, the symmetry type is given in Schönflies notation (see Point groups in three dimensions), and ${\displaystyle r_{i}}$ are the positions of the charges. Most symmetry types require the vector sum of the positions (and thus the electric dipole moment) to be zero.
It is customary to also consider the polyhedron formed by the convex hull of the points. Thus, ${\displaystyle v_{i}}$ is the number of vertices where the given number of edges meet, '${\displaystyle e}$ is the total number of edges, ${\displaystyle f_{3}}$ is the number of triangular faces, ${\displaystyle f_{4}}$ is the number of quadrilateral faces, and ${\displaystyle \theta _{1}}$ is the smallest angle subtended by vectors associated with the nearest charge pair. Note that the edge lengths are generally not equal; thus (except in the cases N = 2, 3, 4, 6, 12, and the geodesic polyhedra) the convex hull is only topologically equivalent to the figure listed in the last column.[12]
N ${\displaystyle E_{1}}$ Symmetry ${\displaystyle \left|\sum \mathbf {r} _{i}\right|}$ ${\displaystyle v_{3}}$ ${\displaystyle v_{4}}$ ${\displaystyle v_{5}}$ ${\displaystyle v_{6}}$ ${\displaystyle v_{7}}$ ${\displaystyle v_{8}}$ ${\displaystyle e}$ ${\displaystyle f_{3}}$ ${\displaystyle f_{4}}$ ${\displaystyle \theta _{1}}$ Equivalent polyhedron
2 0.500000000 ${\displaystyle D_{\infty h}}$ 0 - - - - - - 2 - - 180.000° digon
3 1.732050808 ${\displaystyle D_{3h}}$ 0 - - - - - - 3 2 - 120.000° triangle
4 3.674234614 ${\displaystyle T_{d}}$ 0 4 0 0 0 0 0 6 4 0 109.471° tetrahedron
5 6.474691495 ${\displaystyle D_{3h}}$ 0 2 3 0 0 0 0 9 6 0 90.000° triangular dipyramid
6 9.985281374 ${\displaystyle O_{h}}$ 0 0 6 0 0 0 0 12 8 0 90.000° octahedron
7 14.452977414 ${\displaystyle D_{5h}}$ 0 0 5 2 0 0 0 15 10 0 72.000° pentagonal dipyramid
8 19.675287861 ${\displaystyle D_{4d}}$ 0 0 8 0 0 0 0 16 8 2 71.694° square antiprism
9 25.759986531 ${\displaystyle D_{3h}}$ 0 0 3 6 0 0 0 21 14 0 69.190° triaugmented triangular prism
10 32.716949460 ${\displaystyle D_{4d}}$ 0 0 2 8 0 0 0 24 16 0 64.996° gyroelongated square dipyramid
11 40.596450510 ${\displaystyle C_{2v}}$ 0.013219635 0 2 8 1 0 0 27 18 0 58.540° edge-contracted icosahedron
12 49.165253058 ${\displaystyle I_{h}}$ 0 0 0 12 0 0 0 30 20 0 63.435° icosahedron
(geodesic sphere {3,5+}1,0)
13 58.853230612 ${\displaystyle C_{2v}}$ 0.008820367 0 1 10 2 0 0 33 22 0 52.317°
14 69.306363297 ${\displaystyle D_{6d}}$ 0 0 0 12 2 0 0 36 24 0 52.866° gyroelongated hexagonal dipyramid
15 80.670244114 ${\displaystyle D_{3}}$ 0 0 0 12 3 0 0 39 26 0 49.225°
16 92.911655302 ${\displaystyle T}$ 0 0 0 12 4 0 0 42 28 0 48.936°
17 106.050404829 ${\displaystyle D_{5h}}$ 0 0 0 12 5 0 0 45 30 0 50.108° double-gyroelongated pentagonal dipyramid
18 120.084467447 ${\displaystyle D_{4d}}$ 0 0 2 8 8 0 0 48 32 0 47.534°
19 135.089467557 ${\displaystyle C_{2v}}$ 0.000135163 0 0 14 5 0 0 50 32 1 44.910°
20 150.881568334 ${\displaystyle D_{3h}}$ 0 0 0 12 8 0 0 54 36 0 46.093°
21 167.641622399 ${\displaystyle C_{2v}}$ 0.001406124 0 1 10 10 0 0 57 38 0 44.321°
22 185.287536149 ${\displaystyle T_{d}}$ 0 0 0 12 10 0 0 60 40 0 43.302°
23 203.930190663 ${\displaystyle D_{3}}$ 0 0 0 12 11 0 0 63 42 0 41.481°
24 223.347074052 ${\displaystyle O}$ 0 0 0 24 0 0 0 60 32 6 42.065° snub cube
25 243.812760299 ${\displaystyle C_{s}}$ 0.001021305 0 0 14 11 0 0 68 44 1 39.610°
26 265.133326317 ${\displaystyle C_{2}}$ 0.001919065 0 0 12 14 0 0 72 48 0 38.842°
27 287.302615033 ${\displaystyle D_{5h}}$ 0 0 0 12 15 0 0 75 50 0 39.940°
28 310.491542358 ${\displaystyle T}$ 0 0 0 12 16 0 0 78 52 0 37.824°
29 334.634439920 ${\displaystyle D_{3}}$ 0 0 0 12 17 0 0 81 54 0 36.391°
30 359.603945904 ${\displaystyle D_{2}}$ 0 0 0 12 18 0 0 84 56 0 36.942°
31 385.530838063 ${\displaystyle C_{3v}}$ 0.003204712 0 0 12 19 0 0 87 58 0 36.373°
32 412.261274651 ${\displaystyle I_{h}}$ 0 0 0 12 20 0 0 90 60 0 37.377° pentakis dodecahedron
(geodesic sphere {3,5+}1,1)
33 440.204057448 ${\displaystyle C_{s}}$ 0.004356481 0 0 15 17 1 0 92 60 1 33.700°
34 468.904853281 ${\displaystyle D_{2}}$ 0 0 0 12 22 0 0 96 64 0 33.273°
35 498.569872491 ${\displaystyle C_{2}}$ 0.000419208 0 0 12 23 0 0 99 66 0 33.100°
36 529.122408375 ${\displaystyle D_{2}}$ 0 0 0 12 24 0 0 102 68 0 33.229°
37 560.618887731 ${\displaystyle D_{5h}}$ 0 0 0 12 25 0 0 105 70 0 32.332°
38 593.038503566 ${\displaystyle D_{6d}}$ 0 0 0 12 26 0 0 108 72 0 33.236°
39 626.389009017 ${\displaystyle D_{3h}}$ 0 0 0 12 27 0 0 111 74 0 32.053°
40 660.675278835 ${\displaystyle T_{d}}$ 0 0 0 12 28 0 0 114 76 0 31.916°
41 695.916744342 ${\displaystyle D_{3h}}$ 0 0 0 12 29 0 0 117 78 0 31.528°
42 732.078107544 ${\displaystyle D_{5h}}$ 0 0 0 12 30 0 0 120 80 0 31.245°
43 769.190846459 ${\displaystyle C_{2v}}$ 0.000399668 0 0 12 31 0 0 123 82 0 30.867°
44 807.174263085 ${\displaystyle O_{h}}$ 0 0 0 24 20 0 0 120 72 6 31.258°
45 846.188401061 ${\displaystyle D_{3}}$ 0 0 0 12 33 0 0 129 86 0 30.207°
46 886.167113639 ${\displaystyle T}$ 0 0 0 12 34 0 0 132 88 0 29.790°
47 927.059270680 ${\displaystyle C_{s}}$ 0.002482914 0 0 14 33 0 0 134 88 1 28.787°
48 968.713455344 ${\displaystyle O}$ 0 0 0 24 24 0 0 132 80 6 29.690°
49 1011.557182654 ${\displaystyle C_{3}}$ 0.001529341 0 0 12 37 0 0 141 94 0 28.387°
50 1055.182314726 ${\displaystyle D_{6d}}$ 0 0 0 12 38 0 0 144 96 0 29.231°
51 1099.819290319 ${\displaystyle D_{3}}$ 0 0 0 12 39 0 0 147 98 0 28.165°
52 1145.418964319 ${\displaystyle C_{3}}$ 0.000457327 0 0 12 40 0 0 150 100 0 27.670°
53 1191.922290416 ${\displaystyle C_{2v}}$ 0.000278469 0 0 18 35 0 0 150 96 3 27.137°
54 1239.361474729 ${\displaystyle C_{2}}$ 0.000137870 0 0 12 42 0 0 156 104 0 27.030°
55 1287.772720783 ${\displaystyle C_{2}}$ 0.000391696 0 0 12 43 0 0 159 106 0 26.615°
56 1337.094945276 ${\displaystyle D_{2}}$ 0 0 0 12 44 0 0 162 108 0 26.683°
57 1387.383229253 ${\displaystyle D_{3}}$ 0 0 0 12 45 0 0 165 110 0 26.702°
58 1438.618250640 ${\displaystyle D_{2}}$ 0 0 0 12 46 0 0 168 112 0 26.155°
59 1490.773335279 ${\displaystyle C_{2}}$ 0.000154286 0 0 14 43 2 0 171 114 0 26.170°
60 1543.830400976 ${\displaystyle D_{3}}$ 0 0 0 12 48 0 0 174 116 0 25.958°
61 1597.941830199 ${\displaystyle C_{1}}$ 0.001091717 0 0 12 49 0 0 177 118 0 25.392°
62 1652.909409898 ${\displaystyle D_{5}}$ 0 0 0 12 50 0 0 180 120 0 25.880°
63 1708.879681503 ${\displaystyle D_{3}}$ 0 0 0 12 51 0 0 183 122 0 25.257°
64 1765.802577927 ${\displaystyle D_{2}}$ 0 0 0 12 52 0 0 186 124 0 24.920°
65 1823.667960264 ${\displaystyle C_{2}}$ 0.000399515 0 0 12 53 0 0 189 126 0 24.527°
66 1882.441525304 ${\displaystyle C_{2}}$ 0.000776245 0 0 12 54 0 0 192 128 0 24.765°
67 1942.122700406 ${\displaystyle D_{5}}$ 0 0 0 12 55 0 0 195 130 0 24.727°
68 2002.874701749 ${\displaystyle D_{2}}$ 0 0 0 12 56 0 0 198 132 0 24.433°
69 2064.533483235 ${\displaystyle D_{3}}$ 0 0 0 12 57 0 0 201 134 0 24.137°
70 2127.100901551 ${\displaystyle D_{2d}}$ 0 0 0 12 50 0 0 200 128 4 24.291°
71 2190.649906425 ${\displaystyle C_{2}}$ 0.001256769 0 0 14 55 2 0 207 138 0 23.803°
72 2255.001190975 ${\displaystyle I}$ 0 0 0 12 60 0 0 210 140 0 24.492° geodesic sphere {3,5+}2,1
73 2320.633883745 ${\displaystyle C_{2}}$ 0.001572959 0 0 12 61 0 0 213 142 0 22.810°
74 2387.072981838 ${\displaystyle C_{2}}$ 0.000641539 0 0 12 62 0 0 216 144 0 22.966°
75 2454.369689040 ${\displaystyle D_{3}}$ 0 0 0 12 63 0 0 219 146 0 22.736°
76 2522.674871841 ${\displaystyle C_{2}}$ 0.000943474 0 0 12 64 0 0 222 148 0 22.886°
77 2591.850152354 ${\displaystyle D_{5}}$ 0 0 0 12 65 0 0 225 150 0 23.286°
78 2662.046474566 ${\displaystyle T_{h}}$ 0 0 0 12 66 0 0 228 152 0 23.426°
79 2733.248357479 ${\displaystyle C_{s}}$ 0.000702921 0 0 12 63 1 0 230 152 1 22.636°
80 2805.355875981 ${\displaystyle D_{4d}}$ 0 0 0 16 64 0 0 232 152 2 22.778°
81 2878.522829664 ${\displaystyle C_{2}}$ 0.000194289 0 0 12 69 0 0 237 158 0 21.892°
82 2952.569675286 ${\displaystyle D_{2}}$ 0 0 0 12 70 0 0 240 160 0 22.206°
83 3027.528488921 ${\displaystyle C_{2}}$ 0.000339815 0 0 14 67 2 0 243 162 0 21.646°
84 3103.465124431 ${\displaystyle C_{2}}$ 0.000401973 0 0 12 72 0 0 246 164 0 21.513°
85 3180.361442939 ${\displaystyle C_{2}}$ 0.000416581 0 0 12 73 0 0 249 166 0 21.498°
86 3258.211605713 ${\displaystyle C_{2}}$ 0.001378932 0 0 12 74 0 0 252 168 0 21.522°
87 3337.000750014 ${\displaystyle C_{2}}$ 0.000754863 0 0 12 75 0 0 255 170 0 21.456°
88 3416.720196758 ${\displaystyle D_{2}}$ 0 0 0 12 76 0 0 258 172 0 21.486°
89 3497.439018625 ${\displaystyle C_{2}}$ 0.000070891 0 0 12 77 0 0 261 174 0 21.182°
90 3579.091222723 ${\displaystyle D_{3}}$ 0 0 0 12 78 0 0 264 176 0 21.230°
91 3661.713699320 ${\displaystyle C_{2}}$ 0.000033221 0 0 12 79 0 0 267 178 0 21.105°
92 3745.291636241 ${\displaystyle D_{2}}$ 0 0 0 12 80 0 0 270 180 0 21.026°
93 3829.844338421 ${\displaystyle C_{2}}$ 0.000213246 0 0 12 81 0 0 273 182 0 20.751°
94 3915.309269620 ${\displaystyle D_{2}}$ 0 0 0 12 82 0 0 276 184 0 20.952°
95 4001.771675565 ${\displaystyle C_{2}}$ 0.000116638 0 0 12 83 0 0 279 186 0 20.711°
96 4089.154010060 ${\displaystyle C_{2}}$ 0.000036310 0 0 12 84 0 0 282 188 0 20.687°
97 4177.533599622 ${\displaystyle C_{2}}$ 0.000096437 0 0 12 85 0 0 285 190 0 20.450°
98 4266.822464156 ${\displaystyle C_{2}}$ 0.000112916 0 0 12 86 0 0 288 192 0 20.422°
99 4357.139163132 ${\displaystyle C_{2}}$ 0.000156508 0 0 12 87 0 0 291 194 0 20.284°
100 4448.350634331 ${\displaystyle T}$ 0 0 0 12 88 0 0 294 196 0 20.297°
101 4540.590051694 ${\displaystyle D_{3}}$ 0 0 0 12 89 0 0 297 198 0 20.011°
102 4633.736565899 ${\displaystyle D_{3}}$ 0 0 0 12 90 0 0 300 200 0 20.040°
103 4727.836616833 ${\displaystyle C_{2}}$ 0.000201245 0 0 12 91 0 0 303 202 0 19.907°
104 4822.876522746 ${\displaystyle D_{6}}$ 0 0 0 12 92 0 0 306 204 0 19.957°
105 4919.000637616 ${\displaystyle D_{3}}$ 0 0 0 12 93 0 0 309 206 0 19.842°
106 5015.984595705 ${\displaystyle D_{2}}$ 0 0 0 12 94 0 0 312 208 0 19.658°
107 5113.953547724 ${\displaystyle C_{2}}$ 0.000064137 0 0 12 95 0 0 315 210 0 19.327°
108 5212.813507831 ${\displaystyle C_{2}}$ 0.000432525 0 0 12 96 0 0 318 212 0 19.327°
109 5312.735079920 ${\displaystyle C_{2}}$ 0.000647299 0 0 14 93 2 0 321 214 0 19.103°
110 5413.549294192 ${\displaystyle D_{6}}$ 0 0 0 12 98 0 0 324 216 0 19.476°
111 5515.293214587 ${\displaystyle D_{3}}$ 0 0 0 12 99 0 0 327 218 0 19.255°
112 5618.044882327 ${\displaystyle D_{5}}$ 0 0 0 12 100 0 0 330 220 0 19.351°
113 5721.824978027 ${\displaystyle D_{3}}$ 0 0 0 12 101 0 0 333 222 0 18.978°
114 5826.521572163 ${\displaystyle C_{2}}$ 0.000149772 0 0 12 102 0 0 336 224 0 18.836°
115 5932.181285777 ${\displaystyle C_{3}}$ 0.000049972 0 0 12 103 0 0 339 226 0 18.458°
116 6038.815593579 ${\displaystyle C_{2}}$ 0.000259726 0 0 12 104 0 0 342 228 0 18.386°
117 6146.342446579 ${\displaystyle C_{2}}$ 0.000127609 0 0 12 105 0 0 345 230 0 18.566°
118 6254.877027790 ${\displaystyle C_{2}}$ 0.000332475 0 0 12 106 0 0 348 232 0 18.455°
119 6364.347317479 ${\displaystyle C_{2}}$ 0.000685590 0 0 12 107 0 0 351 234 0 18.336°
120 6474.756324980 ${\displaystyle C_{s}}$ 0.001373062 0 0 12 108 0 0 354 236 0 18.418°
121 6586.121949584 ${\displaystyle C_{3}}$ 0.000838863 0 0 12 109 0 0 357 238 0 18.199°
122 6698.374499261 ${\displaystyle I_{h}}$ 0 0 0 12 110 0 0 360 240 0 18.612° geodesic sphere {3,5+}2,2
123 6811.827228174 ${\displaystyle C_{2v}}$ 0.001939754 0 0 14 107 2 0 363 242 0 17.840°
124 6926.169974193 ${\displaystyle D_{2}}$ 0 0 0 12 112 0 0 366 244 0 18.111°
125 7041.473264023 ${\displaystyle C_{2}}$ 0.000088274 0 0 12 113 0 0 369 246 0 17.867°
126 7157.669224867 ${\displaystyle D_{4}}$ 0 0 2 16 100 8 0 372 248 0 17.920°
127 7274.819504675 ${\displaystyle D_{5}}$ 0 0 0 12 115 0 0 375 250 0 17.877°
128 7393.007443068 ${\displaystyle C_{2}}$ 0.000054132 0 0 12 116 0 0 378 252 0 17.814°
129 7512.107319268 ${\displaystyle C_{2}}$ 0.000030099 0 0 12 117 0 0 381 254 0 17.743°
130 7632.167378912 ${\displaystyle C_{2}}$ 0.000025622 0 0 12 118 0 0 384 256 0 17.683°
131 7753.205166941 ${\displaystyle C_{2}}$ 0.000305133 0 0 12 119 0 0 387 258 0 17.511°
132 7875.045342797 ${\displaystyle I}$ 0 0 0 12 120 0 0 390 260 0 17.958° geodesic sphere {3,5+}3,1
133 7998.179212898 ${\displaystyle C_{3}}$ 0.000591438 0 0 12 121 0 0 393 262 0 17.133°
134 8122.089721194 ${\displaystyle C_{2}}$ 0.000470268 0 0 12 122 0 0 396 264 0 17.214°
135 8246.909486992 ${\displaystyle D_{3}}$ 0 0 0 12 123 0 0 399 266 0 17.431°
136 8372.743302539 ${\displaystyle T}$ 0 0 0 12 124 0 0 402 268 0 17.485°
137 8499.534494782 ${\displaystyle D_{5}}$ 0 0 0 12 125 0 0 405 270 0 17.560°
138 8627.406389880 ${\displaystyle C_{2}}$ 0.000473576 0 0 12 126 0 0 408 272 0 16.924°
139 8756.227056057 ${\displaystyle C_{2}}$ 0.000404228 0 0 12 127 0 0 411 274 0 16.673°
140 8885.980609041 ${\displaystyle C_{1}}$ 0.000630351 0 0 13 126 1 0 414 276 0 16.773°
141 9016.615349190 ${\displaystyle C_{2v}}$ 0.000376365 0 0 14 126 0 1 417 278 0 16.962°
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149 10096.859907397 ${\displaystyle C_{1}}$ 0.000638186 0 0 14 133 2 0 441 294 0 16.344°
150 10236.196436701 ${\displaystyle T}$ 0 0 0 12 138 0 0 444 296 0 16.405°
151 10376.571469275 ${\displaystyle C_{2}}$ 0.000153836 0 0 12 139 0 0 447 298 0 16.163°
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183 15378.166571028 ${\displaystyle C_{1}}$ 0.000467899 0 0 12 171 0 0 543 362 0 14.747°
184 15550.421450311 ${\displaystyle T}$ 0 0 0 12 172 0 0 546 364 0 14.932°
185 15723.720074072 ${\displaystyle C_{2}}$ 0.000389762 0 0 12 173 0 0 549 366 0 14.775°
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191 16783.452219362 ${\displaystyle C_{1}}$ 0.001129202 0 0 13 177 1 0 567 378 0 14.195°
192 16963.338386460 ${\displaystyle I}$ 0 0 0 12 180 0 0 570 380 0 14.819° geodesic sphere {3,5+}3,2
193 17144.564740880 ${\displaystyle C_{2}}$ 0.000985192 0 0 12 181 0 0 573 382 0 14.144°
194 17326.616136471 ${\displaystyle C_{1}}$ 0.000322358 0 0 12 182 0 0 576 384 0 14.350°
195 17509.489303930 ${\displaystyle D_{3}}$ 0 0 0 12 183 0 0 579 386 0 14.375°
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203 19007.981204580 ${\displaystyle C_{s}}$ 0.000600343 0 0 12 191 0 0 603 402 0 13.977°
204 19199.540775603 ${\displaystyle T_{h}}$ 0 0 0 12 192 0 0 606 404 0 14.291°
212 20768.053085964 ${\displaystyle I}$ 0 0 0 12 200 0 0 630 420 0 14.118° geodesic sphere {3,5+}4,1
214 21169.910410375 ${\displaystyle T}$ 0 0 0 12 202 0 0 636 424 0 13.771°
216 21575.596377869 ${\displaystyle D_{3}}$ 0 0 0 12 204 0 0 642 428 0 13.735°
217 21779.856080418 ${\displaystyle D_{5}}$ 0 0 0 12 205 0 0 645 430 0 13.902°
232 24961.252318934 ${\displaystyle T}$ 0 0 0 12 220 0 0 690 460 0 13.260°
255 30264.424251281 ${\displaystyle D_{3}}$ 0 0 0 12 243 0 0 759 506 0 12.565°
256 30506.687515847 ${\displaystyle T}$ 0 0 0 12 244 0 0 762 508 0 12.572°
257 30749.941417346 ${\displaystyle D_{5}}$ 0 0 0 12 245 0 0 765 510 0 12.672°
272 34515.193292681 ${\displaystyle I_{h}}$ 0 0 0 12 260 0 0 810 540 0 12.335° geodesic sphere {3,5+}3,3
282 37147.294418462 ${\displaystyle I}$ 0 0 0 12 270 0 0 840 560 0 12.166° geodesic sphere {3,5+}4,2
292 39877.008012909 ${\displaystyle D_{5}}$ 0 0 0 12 280 0 0 870 580 0 11.857°
306 43862.569780797 ${\displaystyle T_{h}}$ 0 0 0 12 294 0 0 912 608 0 11.628°
312 45629.313804002 ${\displaystyle C_{2}}$ 0.000306163 0 0 12 300 0 0 930 620 0 11.299°
315 46525.825643432 ${\displaystyle D_{3}}$ 0 0 0 12 303 0 0 939 626 0 11.337°
317 47128.310344520 ${\displaystyle D_{5}}$ 0 0 0 12 305 0 0 945 630 0 11.423°
318 47431.056020043 ${\displaystyle D_{3}}$ 0 0 0 12 306 0 0 948 632 0 11.219°
334 52407.728127822 ${\displaystyle T}$ 0 0 0 12 322 0 0 996 664 0 11.058°
348 56967.472454334 ${\displaystyle T_{h}}$ 0 0 0 12 336 0 0 1038 692 0 10.721°
357 59999.922939598 ${\displaystyle D_{5}}$ 0 0 0 12 345 0 0 1065 710 0 10.728°
358 60341.830924588 ${\displaystyle T}$ 0 0 0 12 346 0 0 1068 712 0 10.647°
372 65230.027122557 ${\displaystyle I}$ 0 0 0 12 360 0 0 1110 740 0 10.531° geodesic sphere {3,5+}4,3
382 68839.426839215 ${\displaystyle D_{5}}$ 0 0 0 12 370 0 0 1140 760 0 10.379°
390 71797.035335953 ${\displaystyle T_{h}}$ 0 0 0 12 378 0 0 1164 776 0 10.222°
392 72546.258370889 ${\displaystyle C_{1}}$ 0 0 0 12 380 0 0 1170 780 0 10.278°
400 75582.448512213 ${\displaystyle T}$ 0 0 0 12 388 0 0 1194 796 0 10.068°
402 76351.192432673 ${\displaystyle D_{5}}$ 0 0 0 12 390 0 0 1200 800 0 10.099°
432 88353.709681956 ${\displaystyle D_{3}}$ 0 0 0 24 396 12 0 1290 860 0 9.556°
448 95115.546986209 ${\displaystyle T}$ 0 0 0 24 412 12 0 1338 892 0 9.322°
460 100351.763108673 ${\displaystyle T}$ 0 0 0 24 424 12 0 1374 916 0 9.297°
468 103920.871715127 ${\displaystyle S_{6}}$ 0 0 0 24 432 12 0 1398 932 0 9.120°
470 104822.886324279 ${\displaystyle S_{6}}$ 0 0 0 24 434 12 0 1404 936 0 9.059°
According to a conjecture, if ${\displaystyle m=n+2}$, p is the polyhedron formed by the convex hull of m points, q is the number of quadrilateral faces of p, then the solution for m electrons is f(m): ${\displaystyle f(m)=0^{n}+3n-q}$.[13]
References
1. ^ Thomson, Joseph John (March 1904). "On the Structure of the Atom: an Investigation of the Stability and Periods of Oscillation of a number of Corpuscles arranged at equal intervals around the Circumference of a Circle; with Application of the Results to the Theory of Atomic Structure" (PDF). Philosophical Magazine. Series 6. 7 (39): 237-265. doi:10.1080/14786440409463107. Archived from the original (PDF) on 13 December 2013.
2. ^ Smale, S. (1998). "Mathematical Problems for the Next Century". Mathematical Intelligencer. 20 (2): 7-15. CiteSeerX 10.1.1.35.4101. doi:10.1007/bf03025291.
3. ^ Föppl, L. (1912). "Stabile Anordnungen von Elektronen im Atom". J. Reine Angew. Math. (141): 251-301..
4. ^ Schwartz, Richard (2010). "The 5 electron case of Thomson's Problem". arXiv:1001.3702 [math.MG].
5. ^ Yudin, V.A. (1992). "The minimum of potential energy of a system of point charges". Discretnaya Matematika. 4 (2): 115-121 (in Russian).; Yudin, V. A. (1993). "The minimum of potential energy of a system of point charges". Discrete Math. Appl. 3 (1): 75-81. doi:10.1515/dma.1993.3.1.75.
6. ^ Andreev, N.N. (1996). "An extremal property of the icosahedron". East J. Approximation. 2 (4): 459-462.MR1426716, Zbl 0877.51021
7. ^ Landkof, N. S. Foundations of modern potential theory. Translated from the Russian by A. P. Doohovskoy. Die Grundlehren der mathematischen Wissenschaften, Band 180. Springer-Verlag, New York-Heidelberg, 1972. x+424 pp.
8. ^ Hardin, D. P.; Saff, E. B. Discretizing manifolds via minimum energy points. Notices Amer. Math. Soc. 51 (2004), no. 10, 1186-1194
9. ^ Levin, Y.; Arenzon, J. J. (2003). "Why charges go to the Surface: A generalized Thomson Problem". Europhys. Lett. 63 (3): 415. arXiv:cond-mat/0302524. Bibcode:2003EL.....63..415L. doi:10.1209/epl/i2003-00546-1.
10. ^ Sir J.J. Thomson, The Romanes Lecture, 1914 (The Atomic Theory)
11. ^ LaFave Jr, Tim (2013). "Correspondences between the classical electrostatic Thomson problem and atomic electronic structure". Journal of Electrostatics. 71 (6): 1029-1035. arXiv:1403.2591. doi:10.1016/j.elstat.2013.10.001.
12. ^ Kevin Brown. "Min-Energy Configurations of Electrons On A Sphere". Retrieved 2014-05-01.
13. ^ "Sloane's A008486 (see the comment from Feb 03 2017)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved .
Notes
This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0. | HuggingFaceTB/finemath | |
# A line passes through (9,3),(12,4), and (n,-5) Find the value of n.
Question
Vectors and spaces
A line passes through (9,3),(12,4), and (n,-5)
Find the value of n.
2020-10-27
Given that a line passes through the points (9, 3), (12, 4), and (n, -5).
First we find the equation of the line that. passes through the points (9,3), and (12,4).
The equation of a line passing through the points $$(x_{1}, y_{1}), (x_{2}, y_{2})$$ is
$$\frac{y-y_{1}}{x-x_{1}}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$$
In this problem $$(x_{1}, y_{1}) = (9, 3),\ and\ (x_{2}, y_{2}) = (12,4).$$
This implies that $$x_{1} = 9, y_{1} = 3, x_{2} = 12, y_{2} =4$$
. Plugging these values $$\in \frac{y-y_{1}}{x-x_{1}}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$$ we get the equation of the line is
$$\in \frac{y-y_{1}}{x-x_{1}}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$$
$$\Rightarrow \frac{y-3}{x-9}=\frac{3-4}{9-12}$$
$$\Rightarrow \frac{y-3}{x-9}=\frac{1}{3}$$
$$\Rightarrow 3(y - 3) = x - 9$$
$$\Rightarrow 3y - 9 = x - 9$$
$$\Rightarrow x - 3y = 0$$
Therefore, the equation of the line is $$x - 3y = 0$$
Also given that the line passing through the point $$(n, -5)$$
Putting $$x = n,\ and\ y = -5$$ in the equation of the line we get
$$x - 3y = 0$$
$$\Rightarrown - 3(-5) = 0$$
$$\Rightarrow n + 15 = 0$$
$$\Rightarrow n = -15$$
Thus, the value of n is -15,
### Relevant Questions
Write an equation of the line that passes through (3, 1) and (0, 10)
1. S1 and S2, shown above, are thin parallel slits in an opaqueplate. A plane wave of wavelength λ is incident from the leftmoving in a direction perpendicular to the plate. On a screenfar from the slits there are maximums and minimums in intensity atvarious angles measured from the center line. As the angle isincreased from zero, the first minimum occurs at 3 degrees. Thenext minimum occurs at an angle of-
A. 4.5 degrees
B. 6 degrees
C. 7.5 degrees
D. 9 degrees
E. 12 degrees
Find the vector and parametric equations for the line through the point P=(5,−2,3) and the point Q=(2,−7,8).
Find the vector and parametric equations for the line through the point P=(5,−2,3) and the point Q=(2,−7,8).
The position vector $$\displaystyle{r}{\left({t}\right)}={\left\langle{n}{t},\frac{{1}}{{t}^{{2}}},{t}^{{4}}\right\rangle}$$ describes the path of an object moving in space.
(a) Find the velocity vector, speed, and acceleration vector of the object.
(b) Evaluate the velocity vector and acceleration vector of the object at the given value of $$\displaystyle{t}=\sqrt{{3}}$$
The quadratic function $$\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}$$ whose graph passes through the points (1, 4), (2, 1) and (3, 4).
A charge of $$\displaystyle{6.00}\times{10}^{{-{9}}}$$ C and a charge of $$\displaystyle-{3.00}\times{10}^{{-{9}}}$$ C are separated by a distance of 60.0 cm. Find the position at which a third charge, of $$\displaystyle{12.0}\times{10}^{{-{9}}}$$ C, can be placed so that the net electrostatic force on it is zero.
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
A line L through the origin in $$\displaystyle\mathbb{R}^{{3}}$$ can be represented by parametric equations of the form x = at, y = bt, and z = ct. Use these equations to show that L is a subspase of $$RR^3$$ by showing that if $$v_1=(x_1,y_1,z_1)\ and\ v_2=(x_2,y_2,z_2)$$ are points on L and k is any real number, then $$kv_1\ and\ v_1+v_2$$ are also points on L.
Find the Euclidean distance between u and v and the cosine of the angle between those vectors. State whether that angle is acute, obtuse, or $$\displaystyle{90}^{{\circ}}$$. u = (-1, -1, 8, 0), v = (5,6,1,4) | open-web-math/open-web-math | |
# How do you solve by completing the square 2x^2-6x-20=0?
Mar 23, 2018
${x}_{1} = - 2$ and ${x}_{2} = 5$
#### Explanation:
$2 {x}^{2} - 6 x - 20 = 0$
$2 \cdot \left({x}^{2} - 3 x - 10\right) = 0$
$2 \cdot \left({x}^{2} - 3 x + \frac{9}{4} - \frac{49}{4}\right) = 0$
$2 \cdot \left({\left(x - \frac{3}{2}\right)}^{2} - {\left(\frac{7}{2}\right)}^{2}\right) = 0$
$2 \left(x - \frac{3}{2} + \frac{7}{2}\right) \left(x - \frac{3}{2} - \frac{7}{2}\right) = 0$
$2 \left(x + 2\right) \left(x - 5\right) = 0$
Thus, ${x}_{1} = - 2$ and ${x}_{2} = 5$ | HuggingFaceTB/finemath | |
# What is the mathematical chance that an event will occur called?
## What is the mathematical chance that an event will occur called?
Answer: The likelihood that a particular event will occur is called probability. Probability can be defined as the ratio of the number of favorable outcomes to the total number of outcomes of an event.
## What is the chance that a given event will occur?
probability – a numeric measure of the likelihood that an event will occur sample space – the set of all possible outcomes of an experiment.
## What are the 3 types of probability?
Three Types of Probability
• Classical: (equally probable outcomes) Let S=sample space (set of all possible distinct outcomes). ...
• Relative Frequency Definition. ...
• Subjective Probability.
## How do you calculate the odds of an event?
0:193:22? How to calculate the odds in favor and the odds against - YouTubeYouTube
## How do you know if two events are independent?
Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true. You can use the equation to check if events are independent; multiply the probabilities of the two events together to see if they equal the probability of them both happening together.
## How is Pabc calculated?
To calculate the probability of the intersection of more than two events, the conditional probabilities of all of the preceding events must be considered. In the case of three events, A, B, and C, the probability of the intersection P(A and B and C) = P(A)P(B|A)P(C|A and B).
## What are the 3 axioms of probability?
Axioms of Probability
• Axiom 1: Probability of Event. The first one is that the probability of an event is always between 0 and 1. ...
• Axiom 2: Probability of Sample Space. For sample space, the probability of the entire sample space is 1.
• Axiom 3: Mutually Exclusive Events.
Mar 5, 2021
## What are the 5 rules of probability?
Basic Probability Rules
• Probability Rule One (For any event A, 0 ≤ P(A) ≤ 1)
• Probability Rule Two (The sum of the probabilities of all possible outcomes is 1)
• Probability Rule Three (The Complement Rule)
• Probabilities Involving Multiple Events.
• Probability Rule Four (Addition Rule for Disjoint Events)
More items...
## What are the odds formula?
A simple formula for calculating odds from probability is O = P / (1 - P). A formula for calculating probability from odds is P = O / (O + 1).
## What are the odds examples?
Example from above: B said 1 in 10, so the odds that both will say the same number is 1/10 or 10%. The table shows some more examples of the real odds....
Odds 1/XChances of saying the same number
1 in 303,33%
1 in 402,5%
1 in 502%
3 more rows
## Can 2 events be mutually exclusive and independent?
Suppose two events have a non-zero chance of occurring. Then if the two events are mutually exclusive, they can not be independent. If two events are independent, they cannot be mutually exclusive.
## How do you prove that A and B are independent?
If A and B are independent events, then the events A and B' are also independent. Proof: The events A and B are independent, so, P(A ∩ B) = P(A) P(B). From the Venn diagram, we see that the events A ∩ B and A ∩ B' are mutually exclusive and together they form the event A.
## What are the 3 axioms?
The three axioms are:
• For any event A, P(A) ≥ 0. In English, that's “For any event A, the probability of A is greater or equal to 0”.
• When S is the sample space of an experiment; i.e., the set of all possible outcomes, P(S) = 1. ...
• If A and B are mutually exclusive outcomes, P(A ∪ B ) = P(A) + P(B).
Jun 15, 2018
## What are the 3 rules of probability?
Lesson Summary There are three basic rules associated with probability: the addition, multiplication, and complement rules.
## What is the formula of probability?
Similarly, if the probability of an event occurring is “a” and an independent probability is “b”, then the probability of both the event occurring is “ab”....Basic Probability Formulas.
All Probability Formulas List in Maths
Conditional ProbabilityP(A | B) = P(A∩B) / P(B)
Bayes FormulaP(A | B) = P(B | A) ⋅ P(A) / P(B)
5 more rows
## What is odds against an event?
Odds are used to describe the chance of an event occurring. The odds are the ratios that compare the number of ways the event can occur with the number of ways the event cannot occurr. ... The odds against - the ratio of the number of ways that an outcome cannot occur compared to in how many ways it can occur.
## How do you express odds?
Odds and probability can be expressed in prose via the prepositions to and in: "odds of so many to so many on (or against) [some event]" refers to odds – the ratio of numbers of (equally likely) outcomes in favor and against (or vice versa); "chances of so many [outcomes], in so many [outcomes]" refers to probability – ...
## How do you write odds?
The answer is the total number of outcomes. Probability can be expressed as 9/30 = 3/10 = 30% - the number of favorable outcomes over the number of total possible outcomes. A simple formula for calculating odds from probability is O = P / (1 - P). A formula for calculating probability from odds is P = O / (O + 1).
## How odds are written?
Odds and probability can be expressed in prose via the prepositions to and in: "odds of so many to so many on (or against) [some event]" refers to odds – the ratio of numbers of (equally likely) outcomes in favor and against (or vice versa); "chances of so many [outcomes], in so many [outcomes]" refers to probability – ...
## What is the difference between mutually exclusive and independent events in probability?
The difference between mutually exclusive and independent events is: a mutually exclusive event can simply be defined as a situation when two events cannot occur at same time whereas independent event occurs when one event remains unaffected by the occurrence of the other event. | HuggingFaceTB/finemath | |
## Subject: Elementary Mathematics
Q.1 Define difference and symmetric difference of two sets and prove that .
(A− B)U(B − A) = (AUB)−(A∩ B)
## Difference of Two Sets :
The difference of the sets A and B in this order is the set of elements which belong to A but not to B. Symbolically, we write A – B and read as “ A minus B”.
The representation of A – B using a Venn diagram is given below. Similarly, we can find B – A, the difference of the sets B and A in this order is the set of elements which belong to B but not to A. Symbolically, we write B – A and read as “ B minus A”. and a Venn diagram for B – A as: Also, note that A – B is not equal to B – A, i.e. A – B ≠ B – A.
### Symmetric Difference between Two Sets:
The set which contains the elements which are either in set A or in set B but not in both is called the symmetric difference between two given sets. It is represented by A ⊝ B and is read as a symmetric difference of set A and B. We have to prove that : (A-B) U (B-A) = (A∪B) – (A∩B)
Proof:
Let, x ∈ (A-B) U (B-A)
⇒ x ∈ (A-B) or x ∈ (B-A)
⇒ x ∈ A But x ∉ B or x ∈ B but x ∉ A,
⇒ x ∈ A or x ∈ B
⇒ x ∈ (A∪B)
⇒ x ∈ (A∪B) – (A∩B)
Since here x represents the arbitrary element of the set (A-B) U (B-A).
Thus, (A-B) U (B-A) = (A∪B) – (A∩B)
Q.2 If R is an equivalence relation in a set A , then Prove that R-1 is also an equivalence relation in A
Solution:
## An equivalence relation is one which is reflexive, symmetric and transitive. R is an equivalence relation on set A. Let the element of set A be
So, – since it is reflexive, this is also true for It is symmetric hence, and this is also true for Also, R is transitive i.e., and For and or Thus is symmetric, reflexive and transitive. i.e. is equivalence Relation.
Q.3 Find out the square root of complex numbers 12+5i
Solution :
√12–5i)=a+ib,
(a+ib)2=12–5i,
(a2−b2)+2abi=12–5i,
a2−b2=12, 2ab=−5,
(a2+b2)2=(a2−b2)2+(2ab)2
=122+52=144+25=169=132,
a2+b2=13,
2a2=12+13=25, 2b2=13−12=1,
a=±52–√, b=∓12–√, (ab<0)
a+ib=±12–√(5−i). | HuggingFaceTB/finemath | |
# A balloon has a volume of 2.9 L at 320 Kelvin. If the temperature is raised to 343 Kelvin, what will its volume be?
Using Charles' law, ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$
Therefore ${V}_{2} = {T}_{2.} {V}_{1} / {T}_{1}$ = $343 . \frac{2.9}{320}$ = 3.11 litre. | HuggingFaceTB/finemath | |
Transfer Func and Block Diagram | Laplace Transform | Systems Science
# ECM2105 - Control Engineering
Dr Mustafa M Aziz (2010)
________________________________________________________________________________
TRANSFER FUNCTIONS AND BLOCK DIAGRAMS
1. Introduction 2. Transfer Function of Linear Time-Invariant (LTI) Systems 3. Block Diagrams 4. Multiple Inputs 5. Transfer Functions with MATLAB 6. Time Response Analysis with MATLAB
1. Introduction An important step in the analysis and design of control systems is the mathematical modelling of the controlled process. There are a number of mathematical representations to describe a controlled process: Differential equations: You have learned this before. Transfer function: It is defined as the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable, with all zero initial conditions. Block diagram: It is used to represent all types of systems. It can be used, together with transfer functions, to describe the cause and effect relationships throughout the system. State-space-representation: You will study this in an advanced Control Systems Design course. 1.1. Linear Time-Variant and Linear Time-Invariant Systems Definition 1: A time-variable differential equation is a differential equation with one or more of its coefficients are functions of time, t. For example, the differential equation: d 2 y( t ) t2 + y( t ) = u ( t ) dt 2 (where u and y are dependent variables) is time-variable since the term t2d2y/dt2 depends explicitly on t through the coefficient t2. An example of a time-varying system is a spacecraft system which the mass of spacecraft changes during flight due to fuel consumption. Definition 2: A time-invariant differential equation is a differential equation in which none of its coefficients depend on the independent time variable, t. For example, the differential equation: d 2 y( t ) dy( t ) m +b + y( t ) = u ( t ) 2 dt dt where the coefficients m and b are constants, is time-invariant since the equation depends only implicitly on t through the dependent variables y and u and their derivatives.
1
ECM2105 - Control Engineering
Dr Mustafa M Aziz (2010)
________________________________________________________________________________ Dynamic systems that are described by linear, constant-coefficient, differential equations are called linear time-invariant (LTI) systems. 2. Transfer Function of Linear Time-Invariant (LTI) Systems The transfer function of a linear, time-invariant system is defined as the ratio of the Laplace transform of the output (response function), Y(s) = (driving function) U(s) = {u(t)}, under the assumption that all initial conditions are zero.
u(t)
System differential equation
Taking the Laplace transform with zero initial conditions,
U(s)
G (s) = Y(s) U(s)
System transfer function
Transfer function:
A dynamic system can be described by the following time-invariant differential equation:
an d n y( t ) d n −1 y( t ) dy( t ) + a n −1 + L + a1 + a 0 y( t ) n n −1 dt dt dt d m u(t) d m −1 u ( t ) du ( t ) = bm + b m −1 + L + b1 + b 0 u(t) m m −1 dt dt dt
Taking the Laplace transform and considering zero initial conditions we have:
(a
n
s n + a n −1s n −1 + L + a 1s + a 0 Y(s) = b m s m + b m −1s m −1 + L + b1s + b 0 U(s)
)
(
The transfer function between u(t) and y(t) is given by: Y(s) b m s m + b m −1s m −1 + L + b1s + b 0 M (s) = = G (s) = U(s) N(s) a n s n + a n −1s n −1 + L + a 1s + a 0 where G(s) = M(s)/N(s) is the transfer function of the system; the roots of N(s) are called poles of the system and the roots of M(s) are called zeros of the system. By setting the denominator function to zero, we obtain what is referred to as the characteristic equation: ansn + an-1sn-1 + ⋅⋅⋅ + a1s + a0 = 0 We shall see later that the stability of linear, SISO systems is completely governed by the roots of the characteristic equation.
2
{y(t)}, to the Laplace transform of the input
y(t)
Y(s)
)
0 t < 0 v i (t) = 1 t ≥ 0 Solution: R vi(t) C vo(t) 3 . • The transfer function between a pair of input and output variables is the ratio of the Laplace transform of the output to the Laplace transform of the input. Take the Laplace transform of the differential equation under the zero initial conditions.Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ A transfer function has the following properties: • The transfer function is defined only for a linear time-invariant system. Newton’s laws and Kirchhoff’s laws. e. • The transfer function is independent of the input of the system. Example: Consider the following RC circuit. Take the ratio of the output Y(s) to the input U(s). • All initial conditions of the system are set to zero. 3. Vo(s)/Vi(s). i.e. Develop the differential equation for the system by using the physical laws. 2. 1) Find the transfer function of the network. It is not defined for nonlinear systems. we use the following procedures: 1. This ratio is the transfer function. 2) Find the response vo(t) for a unit-step input. To derive the transfer function of a system.g.ECM2105 .
5H and 0 t < 0 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Exercise: Consider the LCR electrical network shown in the figure below.ECM2105 . L=0. v i (t) = 2 t ≥ 0 4 . C = 0.5F.5Ω. R L i(t) vi(t) C vo(t) Exercise: Find the time response of vo(t) of the above system for R = 2. Find the transfer function G(s) = Vo(s)/Vi(s).
k is the spring constant. k y(t) m b u(t) 1) Find the differential equation of the system 2) Find the transfer function between the input U(s) and the output Y(s). m is the mass. b is the friction constant. 5 . u(t) is an external applied force and y(t) is the resulting displacement.ECM2105 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Exercise: In the mechanical system shown in the figure.
Y(s) is the output of the block and G(s) is the transfer function of the block. Arrows are used to represent the directions of signal flow. The takeoff point is similar to the electrical circuit takeoff point.ECM2105 . We will use the following terminology for block diagrams throughout this course: R(s) = reference input (command) Y(s) = output (controlled variable) U(s) = input (actuating signal) E(s) = error signal F(s) = feedback signal G(s) = forward path transfer function H(s) = feedback transfer fucntion R(s) + _ E(s) G(s) Y(s) F(s) H(s) Single block: U(s) G(s) Y(s) Y(s) = G(s)U(s) U(s) is the input to the block. The block is usually drawn with its transfer funciton written inside it. The block diagram gives an overview of the system.Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 3. Block diagram items: Summing point Block Takeoff point + _ Transfer function The above figure shows the way the various items in block diagrams are represented. Series connection: U(s) G1(s) X(s) G2(s) Y(s) Y(s) = G1(s)G2(s)U(s) 6 . A summing point is where signals are algebraically added together. Block Diagrams A block diagram of a system is a pictorial representation of the functions performed by each component and of the flow of signals.
ECM2105 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Parallel connection (feed forward): G1(s) U(s) + + G2(s) Y(s) Y(s) = [G1(s) + G2(s)]U(s) Negative feedback system (closed-loop system): R(s) + _ E(s) G(s) Y(s) The closed loop transfer function: Y(s) G(s) = R(s) 1 + G(s) Exercise: Find the closed-loop transfer function for the following block diagram: R(s) + _ E(s) G(s) Y(s) F(s) H(s) 7 .
ECM2105 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Exercise: A control system has a forward path of two elements with transfer functions K and 1/(s+1) as shown. what is the transfer function of the closed loop system. If the feedback path has a transfer function s. R(s) + _ K 1 s +1 Y(s) s Moving a summing point ahead of a block: R(s) G(s) Y(s) + ± F(s) R(s) + G(s) Y(s) ± F(s) 1/G(s) Y(s) = G(s)R(s) ± F(s) Moving a summing point beyond a block: R(s) + G(s) Y(s) ± R(s) G(s) Y(s) + ± F(s) F(s) Y(s) = G(s)[R(s) ± F(s)] Moving a takeoff point ahead of a block: G(s) R(s) G(s) Y(s) R(s) G(s) Y(s) G(s) Y(s) = G(s)R(s) 8 Y(s) Y(s) .
R(s) + _ + G1(s) G2(s) G3(s) + + Y(s) _ H1(s) G4(s) H2(s) 9 .ECM2105 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Moving a takeoff point beyond a block: R(s) G(s) R(s) 1/G(s) Y(s) R(s) G(s) Y(s) R(s) Y(s) = G(s)R(s) Moving a takeoff point ahead of a summing point: R(s) + F(s) ± Y(s) Y(s) R(s) ± + + ± Y(s) Y(s) F(s) Y(s) = R(s) ± F(s) Moving a takeoff point beyond a summing point: R(s) R(s) + ± F(s) Y(s) F(s) ± R(s) R(s) + + ± Y(s) Y(s) = R(s) ± F(s) Exercise: Reduce the following block diagram and determine the transfer function.
ECM2105 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Exercise: Reduce the following block diagram and determine the transfer function. H1 R(s) +_ + G H2 + Y(s) 10 .
D(s) R(s) +_ G1(s) + G2(s) Y(s) + H(s) Solution: 11 . The procedure to obtain the relationship between the inputs and the output for such systems is: 1. For example.Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 4. 4. 3. Set all inputs except one equal to zero Determine the output signal due to this one non-zero input Repeat the above steps for each of the remaining inputs in turn The total output of the system is the algebraic sum (superposition) of the outputs due to each of the inputs. there can be the input signal indicating the required value of the controlled variable and also an input or inputs due to disturbances which affect the system. 2. Multiple Inputs Control systems often have more than one input.ECM2105 . Example: Find the output Y(s) of the block diagram in the figure below.
D1(s) R(s) +_ G1(s) + G2(s) Y(s) + H1(s) + + H2(s) D2(s) 12 .ECM2105 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Exercise: Determine the output Y(s) of the following system.
±G2) ± Negative feedback connection: R(s) + _ G(s) Y(s) SYS = feedback(G. the coefficients of the numerator and denominator polynomials of the transfer function. G = tf(num. The following table lists the most common systems connections and the corresponding MATLAB commands to implement them.G2) Parallel connection: G1 R(s) + ± G2 Y(s) SYS = G1 ± G2 or SYS = parallel(G1. SYS refers to the transfer function of a system. Transfer Functions with MATLAB A transfer function of a linear time-invariant (LTI) system can be entered into MATLAB using the command tf(num. i.den) The output on the MATLAB command window would be: Transfer function: 3 s + 1 ------------s^2 + 3 s + 2 Once the various transfer functions have been entered. SYS = Y(s)/R(s).den) where num and den are row vectors containing. For example.ECM2105 . respectively. System Series connection: R(s) G1 G2 Y(s) MATLAB command SYS = G1*G2 or SYS = series(G1.e.H) H(s) 13 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 5. den = [1 3 2]. the transfer function: G (s) = 3s + 1 s + 3s + 2 2 can be entered into MATLAB by typing the following on the command line: num = [3 1]. you can combine them together using arithmetic operations such as addition and multiplication to evaluate the transfer function of a cascaded system. In the following.
G2 = tf([0 1]. Solution: Type the following in the MATLAB command line: G1 = tf([0 4].ECM2105 .[0 1]).Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ R(s) + _ G1 G2 Y(s) H Example: Evaluate the transfer function of the feedback system shown in the figure above using MATLAB where G1(s) = 4. G2(s) = 1/(s+2) and H(s) = 5s.[1 2]).H) This produces the following output on the command window (check this result): Transfer function: 4 -------21 s + 2 Exercise: Compute the closed-loop transfer function of the following system using MATLAB. R(s) + _ 1 s +1 s+2 s+3 Y(s) 14 . H = tf([5 0]. SYS = feedback(G1*G2.[0 1]).
den).5 1 MATLAB will then produce the following plot on the screen.) 4 5 6 For a step input of magnitude other than unity. to plot the response due to a step input of magnitude 5.den). for example K. simply multiply the transfer function SYS by the constant K by typing step(K*SYS).1:10. For example.den) Amplitude 1. the impulse response. the ramp response. 6. This is done by first defining the required time range by typing: t = 0:0. t This produces the following plot for the same example above. we use the command step(SYS). we will consider the step response.den) in MATLAB. den = [1 5 4]. step(SYS) 2. We can also enter the row vectors of the numerator and denominator coefficients of the transfer function directly into the step function: step(num.Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 6. 0.1. % Time axis from 0 sec to 10 sec in steps of 0. Time Response Analysis with MATLAB After entering the transfer function of a LTI system. SYS = tf(num. Step response To plot the unit-step response of the LTI system SYS=tf(num. Confirm this plot yourself. and responses to other simple inputs.5 Step Response 2 or directly: step(num.t) % Plot the step response for the given time range.1 sec and then introducing this time range in the step function as follows: step(SYS. You can specify a different time range for evaluating the output response.ECM2105 . Example: Plot the unit-step response of the following system in MATLAB: Y (s) 2s + 10 = 2 R (s) s + 5s + 4 Solution: num = [0 2 10]. In particular. Notice in the previous example that that time axis was scaled automatically by MATLAB. we type step(5*SYS). we can compute and plot the time response of this system due to different input stimuli in MATLAB.5 0 0 1 2 3 Time (sec. 15 .
Example: Plot the unit-impulse response of the following system in MATLAB: Y(s) 5 = R (s) 2s + 10 Solution: num = [0 5].2 16 .5 2 or directly impulse(num..den).5 1 This will produce the following output on the screen.5 1 0.2 0.5 2 Amplitude 1. Impulse response The unit-impulse response of a control system SYS=tf(num. Is that what you would expect? 0. etc.) 6 8 10 You can also use the step function to plot the step responses of multiple LTI systems SYS1.2.) 0.5 0 0 0.den) Amplitude 1..) 6.4 0. den = [2 10]. on a single figure in MATLAB by typing: step(SYS1.6 Time (sec... SYS2. impulse(SYS) Impulse Response 2. SYS = tf(num..8 1 1. .ECM2105 .SYS2.den) may be plotted in MATLAB using the function impulse(SYS).Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ Step Response 2.5 0 0 2 4 Time (sec.
the output can be written in the form: Y(s) = 1 1 1 R (s) = 2 ⋅ s + s +1 (s + s + 1)s s 2 1 1 = 3 2 ⋅ s + s + s s which is equivalent to multiplying by 1/s and then working out the step response. To plot the unitramp response of this system. den = [1 1 1 0]. The step command will further multiply the transfer function by 1/s to make the input 1/s2 i.3. To obtain the unit ramp response of the transfer function G(s): multiply G(s) by 1/s.) 8 10 12 17 . consider the system: Y(s) 1 = 2 R (s) s + s + 1 With a unit-ramp input. For example.e. we enter the numerator and denominator coefficients of the term in square brackets into MATLAB: num = [0 0 0 1]. and use the step command: step(num. and use the resulting function in the step command. Laplace transform of a unit-ramp input. Step Response 12 10 8 Amplitude 6 4 2 0 0 2 4 6 Time (sec.Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 6.den) The unit ramp response will be plotted by MATLAB as shown below. Ramp response There is no ramp command in MATLAB. R(s) = 1/s2.ECM2105 .
Solution: Start by entering the row vectors of the numerator and denominator coefficients in MATLAB: num = [0 2].4 0.t) where num and den are the row vectors of the numerator and denominator coefficients of the transfer function. den = [1 3]. sinusoidal function .3 Amplitude 0.35 0..1:6. and t is the time range over which r is defined.g. r = exp(-t). Linear Simulation Results 0.ECM2105 .t) This would produce the following plot on the screen. % Time range from 0 to 6 sec in steps of 0.t) or lsim(num.r.den) to an arbitrary input (e. over this time: t = 0:0.4. Arbitrary response To obtain the time response of the LTI system SYS=tf(num.05 0 0 1 2 3 Time (sec. we can use the lsim command (stands for 'linear simulation') as follows: lsim(SYS.1 sec % Input time function Enter the above information into the lsim function by typing: lsim(num.). etc. Example: Use MATLAB to obtain the output time response of the transfer function: Y(s) 2 = R (s) s + 3 when the input r is given by r = e-t. exponential function.r.25 0.r.1 0. Then specify the required time range and define the input function.Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 6. r.15 0. r is the input time function.) 4 5 6 18 .den.den.2 0.
For the following circuit. compute y(t). L = 0. and the input voltage u(t). find the transfer function between the output voltage across the inductor y(t). C = 0. L = 0.1 H. For b1 = 0. Sketch the time function y(t) and plot it with MATLAB.5 N-s/m. m = 10 kg and u(t) is a unit-impulse function.Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ TUTORIAL PROBLEM SHEET 3 1. R L u(t) C y(t) For R = 1 Ω.5 N-s/m. and a unit step input u(t) with zero initial conditions.5 H. what is the response y(t)? Check and plot the response with MATLAB. 19 . Find the transfer function of the electrical circuit shown below. what is the response y(t)? Check and plot the result using MATLAB. b2 = 1. R u(t) L y(t) For R = 1 Ω. and u(t) is a unit-step function. 2.5 F. Find the transfer function between the input force u(t) and the output displacement y(t) for the system shown below. 3.ECM2105 . y(t) b1 m b2 u(t) where b1 and b2 are the frictional coefficients.
compute the response y(t) and plot it with MATLAB. Comment on the results. 6. determine the closed loop transfer function with MATLAB. b) For K = 1. u(t) k m b y(t) 5. Determine the transfer function of the following diagram. Check your answer with MATLAB.5 kg/s. For m = 1 kg. k is the spring constant. 10. In the mechanical system shown in the figure below. determine the closed loop transfer function with MATLAB. Find the closed loop transfer function for the following diagram. k = 1 kg/s2. b) Plot y(t) for a unit-step input r(t) with MATLAB. plot y(t) on the same window for a unit-step input r(t) with MATLAB. b is the friction constant.ECM2105 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 4. m is the mass. 7. R(s) + _ E(s) G(s) Y(s) F(s) H(s) a) For G(s) = 8/(s2 + 7s + 10) and H(s) = s+2. and a step input u(t) = 2 N. Find the transfer function of this system. c) Repeat (b) with a unit-impulse input r(t). respectively. u(t) is the external applied force and y(t) is the corresponding displacement. Write down the transfer function Y(s)/R(s) of the following block diagram. R(s) + _ K G(s) Y(s) a) For G(s) = 1/(s + 10) and K = 10. 5. and 100. _ R(s) +_ s 1/s s s + + 1/s Y(s) 20 . b = 0.
D(s) + Gc (s) + N(s) + G(s) + R(s) E(s) GF(s) +_ U(s) Y(s) H(s) 21 .Control Engineering Dr Mustafa M Aziz (2010) ________________________________________________________________________________ 8. Determine the transfer function of the following diagram. R(s) +_ 1/s2 +_ 50 s +1 s 2 Y(s) +_ 2/s a) Check you result with MATLAB. Determine the total output Y(s) for the following system. 9.ECM2105 . b) Plot y(t) for a unit-impulse input r(t) with MATLAB.
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# Finding Roots: Newton’s Method
Finding Roots: Newton’s Method
Calculus I Project
The purpose of this project is to derive and analyze a method for solving equations. This method is
iterative meaning that successive approximations to a solution are obtained with the intent that each
new approximation is an improvement over the previous ones. Many mathematical problems can be
restated in terms of root finding. (Recall that c is a root of a function f(x) provided f(c) = 0.)
Suppose for example that we wish to obtain a decimal approximation to the number π. Given a little
knowledge of trigonometry, we can restate this problem in terms of roots as:
Find the smallest positive number x such that tan
x
4
− 1 = 0.
For an iterative method, we start with some initial value (perhaps an educated guess) x0. We feed this
into some sort of scheme (a formula) which produces a new value x1. We then repeat the process to
build a sequence x0, x1, x2, . . . , xn of values that are increasing closer to the true solution. Newton’s
method gives such a scheme based on the observation that for some functions, the root of a tangent
line will be closer to the true root of a function.
Suppose that f is some function with true root α—note that α is an x-intercept. Let x0 be some value
assumed to be close to the root α, and let L(x) be the tangent line to f at the point (x0, f(x0)) as
shown in the figure. The number α is not known, and the goal is to determine its value. The number
x0 is an initial guess as to the value of α. We don’t expect it to be correct, but we hope to use it to
find a better estimate. Note that the x-intercept of the tangent line (x1 in the figure) is closer to the
Figure 1: Graph of f showing true root α and tangent line at (x0, f(x0)).
true root α. For Newton’s method, this new value x1 is taken as the next approximation. Then the
process is repeated by forming the tangent line to f at the new point (x1, f(x1)). The x-intercept of
this new line is called x2—the next, and hopefully better approximation to α. We continue to play
this game until we’re satisfied with the accuracy. In general, we stop the process when successive
approximations are deemed close enough together—that is, when |xn+1 − xn| < some preset erro
tolerance.
Newton’s method is widely used. When it works, it tends to work very quickly (requiring relatively
few iterations). But it’s not the only trick in the book, and it has its weaknesses. This project entails
deriving the method as well as investigating its strengths and pit-falls.
Carry out the following activities.
A. Suppose that f is a function that has a root α that we wish to determine, and assume that f is at
least a couple of times differentiable on an interval containing α. Let x0 be an initial guess, and write
the equation of the line L(x) that is tangent to the graph of f at the point (x0, f(x0)). Use this to find
a formula for x1 the x-intercept of L in terms of x0, f(x0), and f
0
(x0).
B. Repeat this process by writing the equation of a new tangent line L to the graph of f at the point
(x1, f(x1)), and find a formula for x2 the x-intercept of L in terms of x1, f(x1) and f
0
(x1). Notice
that the basic formula is the same one found before except with x2 replacing x1 and x1 replacing x0.
Write this as a general formula for which the new iterate xn+1 is given in terms of the old iterate xn,
f(xn), and f
0
(xn).
C. The equation 1
x = 1 + x
3 has one positive solution. Plot the two graphs y =
1
x
and y = 1 + x
3
together on the same set of axis to demonstrate that there is one solution. Use Newton’s method to
find it correct to six decimal places. (Note that you will have to restate the problem as a root finding
problem by defining an appropriate function f(x) that has the solution as its root.)
D. Older (and some modern) computers do not directly perform the operation of division; they are
restricted to computing with the operations of multiplication, addition, and subtraction. So, given a
positive number b, its reciprocal must be computed indirectly. Note that for any such number b, the
reciprocal 1
b
is the true root of the function
f(x) = b −
1
x
.
Use Newton’s method to find a sequence of approximations x0, x1, . . . , xk to the reciprocal 1
b
that
only requires the operations of ×, +, and/or −. (Your iteration formula should simplify nicely.)
Now use this formula to find the reciprocal of π. Try two different cases: (1) use an initial guess
of x0 = 0.5, and find it to at least six decimal places; (2) use an initial guess of x0 = 0.7 and comment
on what you observe.
E. For a numerical scheme, we consider an error measure called the relative error. Suppose a true
value is α and an estimate to this value is xn. Then the relative error is
Rel(xn) = true value − approximate
true value
=
α − xn
α
.
Consider the reciprocal finding process in part D. (with α =
1
b
for some positive b). Show that
Rel(xn) = 1 − bxn, and similarly Rel(xn+1) = 1 − bxn+1.
Use the formula for xn+1 to show that
Rel(xn+1) = (Rel(xn))2
.
This puts a restriction on how good (close to α) the initial guess must be to ensure that the method
works. Find a condition on x0 or on Rel(x0) that will guarantee that the approximations get better
with each iteration. Discuss what this means in terms of both the restriction on the method and how
fast it works when it does work. (For example, if the relative error at some step is about 10−4
, how
big would it be at the next step.)
F. There are fun examples for which Newton’s method produces interesting sequences of iterates.
Experiment with some of the following. Pick one (or two) and discuss the behavior of the method.
You may wish to include graphical illustration of what you observe in addition to a table of interations.
i. The true root of f(x) = √3 x is clearly α = 0. Newton’s method can not be used to find this root
for any choice of x0 6= 0.
ii. The equation x
3 −5x = 0 has three solutions 0,
5 and −
5. Consider Newton’s method with
x0 = 1 or x0 = −1.
iii. f(x) = x
3 − 2x + 2 has one real root close to −2. What happens with an initial gues of
x0 = 0? If an initial guess in the interval −0.1 < x < 0.1 is chosen, after several iterations,
some interesting behavior is seen. What is perhaps more surprising is what happens if a really
bad guess—e.g. x0 = 5—is taken.
Don't use plagiarized sources. Get Your Custom Essay on
Finding Roots: Newton’s Method
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## What is the value of z in the equation 2(4z − 3 − 1) = 166 − 46? I know that 166-46 is 120 but i get stuck on the distributive propert
Question
What is the value of z in the equation 2(4z − 3 − 1) = 166 − 46?
I know that 166-46 is 120 but i get stuck on the distributive property part. I end up with 8z-2. Also I will give brainliest.
in progress 0
3 years 2021-07-19T16:42:15+00:00 2 Answers 53 views 0
Step-by-step explanation:
you have to multiply 2(4z – 3 – 1 ). so 2 * 4z = 8z. so then you would do 2 * (-3) = -6. then you would do the same with -1. 2* (-1) = -2.
now you have to put it back in equation form. 8z – 6 – 2 = 166 – 46. now you have to combine like terms. -6-2 = -8. 166 – 46 = 120. and then put it back into equation form. 8z – 8 = 120. then you add 8 to 120. 120+8 = 128. now to find z all you have to do is divide 128/8 = 16
z = 16
2(4z − 3 − 1) = 166 − 46
8z − 6 − 2 = 120
8z − 8 = 120
8z = 128
z = 16
pls give the brainliest
Step-by-step explanation: | HuggingFaceTB/finemath | |
## Search This Blog
### Easy Math Problem
Easy Math Problem - 23 January
In a strange city , there are just 2 houses.
If one person from house1 goes to house2, then the population of house2 is twice the population of house1.
While if one person from house2 goes to house1, then the population of house2 is same as the population of house1.
what the population of each house ?
For Solution : Click Here
#### 21 comments:
1. 5 in house 1 and 7 in house 2
2. 5 7
if 1 from h1 go to h2 then h2 have 8 members so 2 times of h1.
if 1 member from h2 goes to h1 than h1=h2=6
3. house 1= 5 and house2= 7 members
4. 5 in house 1 and 7 in house 2
5. h1=5 & h2=2
6. h1=5 & h2=7
7. I figured this out and I'm 10, not saying it's easy or anything. Just saying most of you say it's sooo easy when you just look and then post.
House 1 has a population of 5 and House 2 has a population of 7 as:
5 - 1 = 4 7 + 1 = 8 4 is half of 8
7 - 1 = 6 5 + 1 = 6 <- both the same
I figured this out by randomly choosing 4 and 7, knowing 4 + 1 is 8. Then, knowing
8. that I needed one more for the thing to work. So I turned the 4 to a 5 and it worked.
9. it can be both. 4 in house 1 and 5 in house 2, or, 5 in house 1 and 7 in house 2
10. house 1 is 5
house 2 is 7
11. house 2 is 6
house 1 is 7
Roald Dahl Collection
12. suganyayoganarasimhanFebruary 1, 2013 at 8:48 AM
house1 is 2 & house2 is 3
13. house1 = 5
house2 = 7
14. h1 -5
h2 -7
15. h1=3
h2=3
becasue h1-1=2 h2+1=4 then h2-1=3 h1+1=3
16. House 1 = 5
House 2 = 7 | HuggingFaceTB/finemath | |
# A Shiny app for inferential statistics by hand
Statistics is divided into four main branches:
• Descriptive statistics
• Inferential statistics
• Predictive analysis
• Exploratory analysis
Descriptive statistics provide a summary of the data; it helps explaining the data in a concise way without losing too much information. Data can be summarized numerically or graphically. See descriptive statistics by hand or in R to learn more about this branch of statistics.
The branch of predictive analysis aims at predicting a dependent variable based on one or several independent variables. Depending on the type of data to be predicted, it often encompasses methods such as linear regression or classification.
Exploratory analyses focus on using graphical approaches to delve into the data and identify the relationships that exist between the different variables in the dataset. They are therefore more akin to data visualization.
Inferential statistics uses a random sample of data taken from a population to make inferences, i.e., to draw conclusions about the population (see the difference between population and sample). In other words, information from the sample is used to make generalizations about the parameter of interest in the population.
The two major tools in inferential statistics are:
1. confidence intervals, and
2. hypothesis tests.
Here is a Shiny app which helps you to use these two tools:
This Shiny app focuses on confidence intervals and hypothesis tests for:
• 1 and 2 means (with unpaired and paired samples)
• 1 and 2 proportions
• 1 and 2 variances
# How to use this app?
1. Open the app via this link
2. Choose the parameter(s) you want to do inference for (i.e., mean(s), proportion(s) or variance(s))
3. Write your data in Sample. Observations are separated by a comma and the decimal is a point
4. Set the null and alternative hypothesis
5. Select the significance level (most of the time $$\alpha = 0.05$$)
In the results panel (on the right side or below depending on the size of your screen), you will see:
• a recap of your sample together with some appropriate descriptive statistics
• the confidence interval
• the hypothesis test
• the interpretation
• and an illustration of the hypothesis test
All formulas, steps and computations to arrive at the final results are also provided.
# Code
Here is the entire code (or see the last version on GitHub) in case you would like to enhance it.
Note that the link may not work if the app has hit the monthly usage limit. Try again later if that is the case.
# Conclusion
I hope you will find this app useful to do inferential statistics and in particular confidence interval and hypothesis testing by hand. | HuggingFaceTB/finemath | |
# Homework Help: A limit :s
1. Mar 2, 2006
### mohlam12
hey again!
this time, I have this one to solve
$$\lim_{x\rightarrow 0}\frac{\tan(x)-\sin(x)}{x^3}$$
i went like this
$$\lim_{x\rightarrow 0}\frac{\frac{tan(x)}{x} - \frac{sin(x)}{x}}{x^2}$$
= lim (0/0)
which is always an undetermined form... is there any other way to solve this WITHOUT using derivatives (not learned yet)
Thank you!
Last edited: Mar 2, 2006
2. Mar 2, 2006
### arildno
There's ALWAYS another way!
Here's how you could start out:
$$\lim_{x\to[0}}\frac{\tan(x)-\sin(x)}{x^{3}}=\lim_{x\to{0}}\frac{\sin(x)}{x}\frac{\frac{1}{\cos(x)}-1}{x^{2}}=\lim_{x\to{0}}\frac{\sin(x)}{x}\frac{1-\cos^{2}(x)}{\cos(x)(1+\cos(x))x^{2}}=\lim_{x\to{0}}(\frac{\sin(x)}{x})^{3}\frac{1}{\cos(x)(1+\cos(x))}$$
Can you take it from there?
3. Mar 2, 2006
### mohlam12
I think so,
So the answer is 1/2 ?
4. Mar 2, 2006
### arildno
You are not sure about that?
5. Mar 2, 2006
### mohlam12
I am actually!
Thank you
6. Mar 2, 2006
### arildno
You're welcome. | HuggingFaceTB/finemath | |
# ACT and SAT Blog
Wondering how to work a complex-looking operation like this: 5(1 + 4)² – 10
Use PEMDAS!
P(arenthesis)
E(xponent)
M(ultiplication)
D(ivision)
A(ddition)
S(ubtraction)
Let’s look at an example of an expression in which of the order
of operations is required:
5(1 + 4)² – 10
Begin with operation in the parentheses (P): 5(1 + 4)² – 10 = 5(5)² – 10
Now remove the exponents (E): 5(5)² – 10 = 5(25) – 10
Multiplication and division are next (M/D): 5(25) – 10 = 125 – 10
Finally, addition and subtraction are performed (A/S): 125 – 10 = 115 | HuggingFaceTB/finemath | |
Home > Relative Error > Relative Error In Calculations
# Relative Error In Calculations
## Contents
Loading... To determine the tolerance interval in a measurement, add and subtract one-half of the precision of the measuring instrument to the measurement. You should only report as many significant figures as are consistent with the estimated error. XLClasses 2,281 views 9:16 Calculus - Differentials with Relative and Percent Error - Duration: 8:34. navigate here
If you tried to measure something that was 12 inches long and your measurement was off by 6 inches, the relative error would be very large. Just like before, make sure you let the audience know what you were measuring in otherwise a simple "2" doesn't mean anything. Another word for this variation - or uncertainty in measurement - is "error." This "error" is not the same as a "mistake." It does not mean that you got the wrong In the example if the estimated error is 0.02 m you would report a result of 0.43 ± 0.02 m, not 0.428 ± 0.02 m.
## Relative Error Chemistry
Q: How does a light microscope work? What if some of the experimental values are negative? continue reading below our video 10 Best Universities in the United States Absolute Error = Actual Value - Measured ValueFor example, if you know a procedure is supposed to yield 1.0 For example, when using a meter stick, one can measure to perhaps a half or sometimes even a fifth of a millimeter.
Sign in Share More Report Need to report the video? Incorrect measuring technique: For example, one might make an incorrect scale reading because of parallax error. b.) The relative error in the length of the field is c.) The percentage error in the length of the field is 3. Absolute Error Formula Chemistry The absolute error of the measurement shows how large the error actually is, while the relative error of the measurement shows how large the error is in relation to the correct
The percentage error is 100% times the relative error. How To Calculate Relative Error In Physics The error in measurement is a mathematical way to show the uncertainty in the measurement. In order to calculate relative error, you must calculate the absolute error as well. Back to Top Suppose the measurement has some errors compared to true values.Relative error decides how incorrect a quantity is from a number considered to be true.
Repeat the same measure several times to get a good average value. 4. Can Relative Error Be Negative matematicasyeso 139,500 views 7:02 Approximation of Error in Hindi - Duration: 42:24. While both situations show an absolute error of 1 cm., the relevance of the error is very different. Bias of the experimenter.
## How To Calculate Relative Error In Physics
But, if you tried to measure something that was 120 feet long and only missed by 6 inches, the relative error would be much smaller -- even though the value of More about the author Another possibility is that the quantity being measured also depends on an uncontrolled variable. (The temperature of the object for example). Relative Error Chemistry Since the measurement was made to the nearest tenth, the greatest possible error will be half of one tenth, or 0.05. 2. Relative Error Definition It is important to know, therefore, just how much the measured value is likely to deviate from the unknown, true, value of the quantity.
Accuracy is a measure of how close the result of the measurement comes to the "true", "actual", or "accepted" value. (How close is your answer to the accepted value?) Tolerance is check over here Apply correct techniques when using the measuring instrument and reading the value measured. Co-authors: 14 Updated: Views:245,735 75% of people told us that this article helped them. Quick and Fast Learn 1,037 views 5:24 Error and Percent Error - Duration: 7:15. Absolute And Relative Error In Numerical Methods
If you are measuring a football field and the absolute error is 1 cm, the error is virtually irrelevant. For example, if you're measuring something with a meter stick, the smallest unit marked on the meter stick is 1 millimeter (mm). This works for any measurement system. http://supercgis.com/relative-error/relative-error-vs-relative-uncertainty.html Two-Point-Four 33,187 views 2:12 How To Calculate Relative Error Actual Mearsure - Duration: 6:48.
Video Tips Make sure that your experimental value and real value are all expressed in the same unit of measurement. Relative Error Formula Calculus How to Calculate the Relative Error? Sign in 183 3 Don't like this video? | HuggingFaceTB/finemath | |
# Tips for Teaching Ratios and Rates in 6th, 7th and 8th Grade Math
Are your students hitting a bit of a roadblock when it comes to wrapping their heads around ratios and rates? Need a fun and colorful way to dive into these math essentials? Well, you’re in for a treat! In this blog post, I’m about to unveil how I teach Ratios and Rates using guided notes on a Math Wheel. It’s a lively, interactive graphic organizer that swoops in to make ratios and rates a walk in the park for your students.
So, if you’re on the lookout for a fresh and exciting way to tackle these fundamental math concepts, you’ve come to the right place. Plus, stick around to the end for some bonus resources ready to be used in your classroom today!
## How to Use the Guided Notes Math Wheel for Ratios and Rates
Liven up your lessons with these guided notes using the Ratios and Rates Math Wheel! This math wheel is perfect to use as you introduce and teach this topic or save and use as a review of the topic later in the unit. However you choose, this ratios and rates math wheel is sure to help your students understand this concept. It’s a graphic organizer and a handy study tool that students can stash in their notebooks for easy access. This specific math wheel is split into four parts!
### 1. What is a…
Let’s break down how I kick off the Ratios and Rates Math Wheel. First things first, I introduce students to our concept and we identify the topic found in the middle of the math wheel. Students love coloring in “Ratios and Rates” as they start to bring their math notes to life.
Next, we dive into the “What is a…” section, breaking down the definitions of ratios and rates.
We discuss how ratios compare two quantities, whether it’s part-to-part or part-to-whole, jotting down the definition in that section.
Then, we move on to add the definition of a rate. A rate, I explain, is essentially a ratio with a twist. It compares quantities with units of measure like inches, feet, or miles.
These definitions serve as the foundation, especially for students stepping into this concept for the first time.
### 2. Writing Ratios and Rates
Let’s dive into the next leg of our guided notes journey, which is the section titled Writing Ratios and Rates. In this section, I break down the process of expressing ratios and rates step by step.
First things first, we highlight the significance of the word “to” when jotting down these ratios. I then draw an arrow next to the line, and we work through an example. For example, 12 to 5. Next, we answer the question asking us to identify the ratio of stars to squares. On the math wheel, there are 3 stars and 2 squares, so we would write 3 to 2. I make sure to remind students that the item named first is always the first number in the ratio.
Building on that foundation, we step up our game by practicing how to express these ratios as fractions. Using our initial example of 12 to 5, we transform it into the fraction 12/5. Then we answer the next question of the ratio of squares to shapes. Again, I remind them that whichever item is named first in the sentence or question is first in the ratio. Since we have 2 squares and a total of 5 shapes, the fraction is 2/5.
We finish this section of the math wheel by exploring the third way to express ratios and rates by using a colon. Revisiting our first example, 12 to 5 becomes 12:5. If we’re pulling in units of measure, it might look something like 1 yard: 3 feet.
As we work through the math wheel, I maintain an ongoing check-in with my students to ensure we’re all on the same page and adjusting pacing as needed. We can always complete extra examples, revisit definitions or I can explain in a different way if needed.
### 3. Equivalent Fractions
In the third section of the Ratios and Rates Math Wheel, we dive into equivalent ratios. I break it down for my students, explaining that these ratios represent the same comparison of quantities. Since my students are already familiar with the concept of equivalent fractions I use that as a comparison.
I explain how we can find equivalent ratios by either multiplying or dividing both numbers in the ratio by the same number. We make sure to add this important information to the math wheel before working on the examples.
To practice, we return to familiar territory with our stars and squares. I remind my students that whatever is first in the list must be the first in the ratio. In this case, it’s the stars. With a total of 6 stars and 2 squares, we can write the ratio as 6:2. Once we have our starting ratio we set out to make some equivalent ratios. I generally do 2 or 3 examples with at least 1 using multiplication and 1 using division if that is possible. Here are some examples and how we do the math to find the equivalent ratios.
### Examples
1. Start with the ratio 6:2 and multiply each number by 2. It looks like this (6 x 2 = 12 and 2 x 2 = 4). That means 12:4 or 12 to 4 is an equivalent ratio of 6:2.
2. Start with the ratio 6:2 and multiply each number by 3. It looks like this (6 x 3 = 18 and 2 x 3 = 6). That means 18:6 or 18 to 6 is an equivalent ratio of 6:2.
3. Start with the ratio 6:2 and divide each number by 2. It looks like this (6 ➗ 2 = 3 and 2 ➗ 2 = 1). That means 3:1 or 3 to 1 is an equivalent ratio of 6:2.
Once we practice this step a few times, we move on to writing equivalent ratios for squares to shapes. We start with the ratio 2:8 because there are 2 squares and 8 total shapes.
Then, just as we did in the first example, we will find some equivalent ratios by multiplying and dividing both numbers by the same number. After we do these examples together, I have students practice what they have learned. There are ten practice problems around the wheel for us to practice the steps further.
### 4. Ratio Tables
In the final section of our guided notes math wheel, I dive into ratio tables to help organize equivalent ratios. We use the examples from the previous section to practice using tables in this section.
In the first example, we see that the first line is filled in and talks about where that information came from. The students are quick to share that there are 12 inches in 1 foot. We then use that to help us fill in the remaining sections of the table.
I make sure to model how I find the number we are multiplying by or dividing by in order to complete the table. In this example I ask the students this question: “What do I need to multiply 1 by in order to get the answer of 2?” When they tell me 2, then I remind them that they must do the same thing to the other number in the ratio. We then multiply 12 x 2 and add 24 to the table.
I continue this process for each row because I want them to begin asking themselves that question to work through ratio tables.
Next, we take on the challenge of creating a table for squares, stars, and the total number of shapes. The first column helps start because it is pre-filled with 2, 6, and 8. Multiply 2, 6, and 8 each by 2, and boom – the third column is completed, showcasing products: 4, 12, 16. Next, we multiply 2, 6, and 8 each by 3, and the fourth column becomes filled with 6, 18, and 24.
To wrap it up, we sprinkle in some division. Divide 2, 6, and 8 each by 2, and voila, the fifth column is completed with 1, 3, and 4.
## Teaching Ratios and Rates Has Never Been Easier
By working through the four sections of the Ratios and Rates Math Wheel your students will have a solid understanding of this concept, how to find equivalent ratios, and how to display ratios in a table.
When the math wheel is complete, students complete the practice problems found in the pattern around the circles. Depending on how students seem to be feeling about the concepts, we may do this as guided practice or independent practice. Then, students can color the background pattern if they’d like. After the wheels are finished, I have students keep them in their math notebooks. That way they can refer to them as needed or use them later in the year as a review. Whether deep into the unit, exploring equivalent ratios, or cruising through ratio tables, the math wheel provides structure and clarity. It becomes a reliable reference point for students, offering a visual aid and quick reminders whenever needed. Its condensed yet detailed format transforms complex concepts into an easily digestible visual aid!
## More Resources for Teaching Ratios and Rates
Looking for more easy to use resources to make teaching and learning ratios and rates a breeze? Make sure to check out the following resources to help as you teach these important math concepts.
## Ready for More Math Wheels?
Want to try out a math wheel for free? Grab this free Fraction Operations Math Wheel and give it a spin in your classroom.
## Save for Later
Remember to save this post to your favorite math Pinterest board for when you need guided notes or resources for ratios and rates!
## Ellie
### How to Use Math Small Groups in Middle School
Welcome to Cognitive Cardio Math! I’m Ellie, a wife, mom, grandma, and dog ‘mom,’ and I’ve spent just about my whole life in school! With nearly 30 years in education, I’ve taught:
• All subject areas in 4th and 5th grades
• Math, ELA, and science in 6th grade (middle school)
I’ve been creating resources for teachers since 2012 and have worked in the elearning industry for about five years as well! | HuggingFaceTB/finemath | |
# How to convert 1D array to 2D by column major
say I have a 1D array like int[] x = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}.
I would like to conver it into 2D where it looks like:
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16
Currently, I have
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < Nb; j++)
s[i][j] = x[i + j];
}
However, that doesnt work. How would I do this?
-
That 2D array looks very random... – Neal May 11 '11 at 18:47
Try
for (int i = 0, k=0; i < 4; i++)
for (int j = 0; j < Nb; j++)
s[j][i] = x[k++]; // you may want s[i][j]
-
thanks that works Also, why would I want s[i][j], that wouldnt work. – Raptrex May 11 '11 at 18:55
It depends on whether you consider the first look up to be the row or the column. Your examples has s[i][j] which could be valid depending on how you look at it. – Peter Lawrey May 11 '11 at 19:02
what is Nb in your for loop? – Kick Buttowski May 21 '14 at 1:45
@KickButtowski The same as in the original question. I imagine it could be 4. – Peter Lawrey May 21 '14 at 7:10
is there any general case to solve this? can u provide me any tutorial so I can learn the logic? – Kick Buttowski May 21 '14 at 16:49
I assume that stray 7 was a typo?
Since we want the inner loop to move downward, and the outer to move rightward, you can do this:
for i in (0..width)
for j in (0..height)
s[j][i] = x[i*height+j]
Tracing this out illustrates why it works:
s[0][0] = x[0*4+0] = x[0]
s[1][0] = x[0*4+1] = x[1]
...
s[0][1] = x[1*4+1] = x[5]
-
No clue what Nb is, but you're way off on reading the initial array. Try something like this:
for (int i = 0; i < 4; i++)
for (int j = 0; j < 4; j++)
s[j][i] = x[i*4 + j];
- | HuggingFaceTB/finemath | |
# What is the molar heat of combustion when "25000 g" of methanol is combusted in a bomb calorimeter to raise the temperature of the surrounding water by 40^@ "C" if the heat capacity of the calorimeter is "10.4 kJ/"^@ "C"?
Jul 3, 2017
There's probably an error in this question... I get $- \text{0.5331 kJ/mol}$, which is about 1000 times smaller than it should be.
Since we are in a bomb calorimeter, a constant-volume calorimeter, we have from the first law of thermodynamics:
${q}_{V} = \Delta U + {\cancel{P \Delta V}}^{0}$,
where:
• ${q}_{V}$ is the heat flow at constant volume.
• $\Delta U$ is the change in internal energy.
• The work done, $- P \Delta V$, is defined here as the system doing work with respect to the surroundings.
Since we know the change in temperature and the combined heat capacity (presumably in $\text{kJ/"^@ "C}$...), we can find the heat involved in the reaction:
$| {q}_{V} | = {C}_{V} \Delta T$
$= \text{10.4 kJ/"^@ "C" xx 40^@ "C}$
$=$ $\text{416 kJ}$
I interpret the molar heat of combustion to be $\Delta {\overline{U}}_{C}$, rather than the change in enthalpy (since the process is at constant volume, and we aren't given the volume of the calorimeter or water).
$\Delta {\overline{U}}_{C} = {q}_{V} / {n}_{L R}$,
where ${n}_{L R}$ is the mols of the limiting reactant, i.e. the methanol in the presence of oxygen.
Assuming complete combustion:
$2 {\text{CH"_3"OH"(g) + 3"O"_2(g) -> 4"H"_2"O"(g) + 2"CO}}_{2} \left(g\right)$
The molar internal energy of combustion is then the heat released out from the reaction towards the surrounding water:
color(blue)(DeltabarU_C) = -("416 kJ")/(25000 cancel("g CH"_3"OH")) xx (32.04 cancel("g CH"_3"OH"))/"1 mol"
$=$ $\textcolor{b l u e}{- \text{0.5331 kJ/mol}}$
This is an absurdly low number, so there's something off with the data... Are you sure it isn't $\text{25.000 g}$? The heat of combustion is off by over 1000 times. | HuggingFaceTB/finemath | |
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### Ordinal number: first through to sixth
Dodly and Flynn explore counting with ordinal numbers from first through to sixth. Investigate the order of ice-cream on an ice-cream cone, sheep being shorn and playing 'pass the parcel'. Where did the missing birthday cake go? Could it be the prize in pass the parcel?
### Count the amazing flying machines
Do you want to fly like a bird? How would you do that? See eight strange flying machines from the past. Watch to see if they fly or crash. Count along with each flying machine, from one to eight. The name of each number is on the screen to help you.
### Funny money
What do you use money to buy? In this clip we look at Australian money. See the different coins and notes that make up our money system. We investigate if size, (in coins) does matter. Check out how many 5 cent pieces you need to make two dollars. We also look at some currencies used around the world. Find out which country ...
### Maths Years 3 - 4 with Ms Kirszman: Quantifying collections
In this lesson, Ms Kirszman teaches you how to play Minute To Win It, and she explores how we can quantify collections without needing to count but by looking and thinking instead. The use of familiar structures to help you quantify collections is also explored, for example, dice patterns and ten-frames.
### Number trains [Chinese]
Use your knowledge of Chinese numbers to arrange train carriages according to numbers on their sides. The numbers are represented in a range of formats such as numerals, dice dots, counting frames and Chinese characters. Identify the numbers that come before and after starting numbers. Begin with numbers up to ten and move ...
### Counting on, counting back - unit of work
The beginning of this unit focuses on ensuring that students have basic foundation skills, and an understanding of both what the number line means and the forward and backward number sequence. They then progress to developing conceptual understandings of place value, specifically tens and ones. Once these foundation skills ...
### Adding and subtracting - unit of work
In this unit of work, students are introduced to addition and subtraction using the strategies of counting on and counting back.
### Tens and ones - unit of work
This unit focuses on building basic foundation skills and developing an understanding of both what the number line means and the forward and backward number sequence. The students then progress to developing conceptual understandings of place value, specifically tens and ones.
### Number trains [Indonesian]
Use your knowledge of Indonesian numbers to arrange train carriages according to numbers on their sides. The numbers are represented in a range of formats such as Indonesian number words, numerals, dice dots or counting frames. Identify the numbers that come before and after starting numbers. Begin with numbers up to ten ...
### Top drawer teachers: fractions
This is a web page that comprehensively covers the teaching of the conceptual understanding of fractions through links to six sections. The first section covers the ‘Big ideas’ behind fractions, while the second section uses research findings to identify some common misunderstandings when learning fractions. The third explores ...
### Numbers and counting - Foundation
This is a teacher resource that includes a set of student activities focusing on the numbers to 20, accompanied by copy masters and a detailed teacher guide for each activity. The activities cover the sequence of numbers, number names, 1:1 correspondence and recording and representing numbers, and make a connection to Asian ...
### Collections - comparing and ordering
This is a web resource that includes student activities and games focusing on collections to 20, accompanied by a teacher guide. Activities cover comparing collections using one-to-one correspondence, ordinal numbers with associated positional words and a game based on the traditional Japanese game of ohajiki. The resource ...
### Let us count how many
This is a web resource that includes four student activities focusing on numbers to 20, accompanied by a teacher guide for each activity. It covers counting, including counting in Asia; number representation; number names; one-to-one correspondence of collections; and simple addition and subtraction using concrete materials. ...
### MoneySmart: Money match
This learning object helps students to recognise Australian currency through matching notes and coins. Students match different combinations of notes and coins to arrive at the same value in multiple ways. The learning object has four different levels working through matching: same coins, coins of different values, coins ...
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This learning object helps students to recognise Australian currency through a supermarket shopping scenario which requires simple transactions. Students match item value with the correct coin and note combinations. The learning object has three different levels working through matching item value to: two single coins, ...
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This is a year 1 mathematics unit of work about shopping. The unit is intended to take about 10.5 hours of teaching and learning time. It consists of 11 student activities supported by teacher notes on curriculum, pedagogy and assessment. Student activities include responding to a story about shopping, working with Australian ...
### MoneySmart: Kieren's coin
This is a year 2 mathematics unit of work about money. The unit is intended to take about 10 hours of teaching and learning time. It consists of 11 student activities supported by teacher notes on curriculum, pedagogy and assessment. Student activities include responding to a story about a rare foreign coin, interacting ...
### MoneySmart: Hey! Let's have a big day out!
This is a year 5 mathematics unit of work about costing and budgeting for various types of family outings. The unit is intended to take about 7.5 hours of teaching and learning time, and is recommended for near the end of the school year. It consists of an introduction, five sets of student activities, and teacher notes ...
### Mixed-Up Maths
Join our host, Ed, as he finds himself in all types of situations where only his knowledge of Maths can help him. From saving the planet from Aliens, to creating a superhero that can stop a strawberry milkshake tidal wave. From searching for buried treasure, to jumping like a daredevil, or planning the greatest circus party ...
### Counting and representing numbers 1–20
Explore counting familiar objects and representing numbers up to 20. Name, match, subitise, compare and order numbers to 20, then use these numbers in addition and subtraction. Relate number sentences to these operations in meaningful ways. | HuggingFaceTB/finemath | |
# Proof that $X$ countable, $\mathcal M$ algebra on $X$ implies $\mathcal M$ a $\sigma$-algebra
In my measure theory class, I believe the professor made the claim that if $X$ was a countable or finite set and $\mathcal M$ was an algebra on $X$, then $\mathcal M$ was a $\sigma$-algebra.
I am unclear how to prove this. I know that given any countable collection of sets, $\{E_j\}_{j \in \mathbb N}$, I can say that $G_n$ := $\bigcup_{j=1}^{n} E_j$ is in $\mathcal M$. But I don't know if it is OK to simply say that since by induction, all of the $G_n$ are in $\mathcal M$ implies the entire countable union is. My problem in doing this is that there are set properties that can hold for any finite number of sets that will not hold for a countably infinite collection (like for example, the finite union of closed sets are closed, but a countably infinite union may not be).
• The family of all subsets of $\mathbb{N}$ that are either finite or have a finite complement is an algebra, but not a $\sigma$-algebra. – Michael Greinecker Aug 24 '14 at 2:35
• @MichaelGreinecker Yes, I think I can see that. Let X be the natural numbers and take the countable collection of sets in M (as you defined it) to be $E_p$ = {p} where p is prime. Their countable union is neither finite, nor is their complement, hence it is not in M. Does that sound right? – Gremlin Brenneman Aug 24 '14 at 2:45
• Yes, that's right. – Michael Greinecker Aug 24 '14 at 2:52
• @MichaelGreinecker I apologize, but I cannot find the icon to check that you have answered my question. The FAQ says there should be a check mark icon below the "up-vote" icon, but on my page the icon below the up-vote icon is one that flags the comment. – Matt Brenneman Aug 25 '14 at 12:26
• @MattBrenneman 1. You cannot accept comments as answers. I reposted the comment as an answer. 2. You can only accept the answer from the account from which you asked the question "Gremlin Brenneman". – Michael Greinecker Aug 25 '14 at 16:59
The family of all subsets of $\mathbb{N}$ that are either finite or have a finite complement is an algebra, but not a $\sigma$-algebra. | open-web-math/open-web-math | |
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Home / Discrete mathematics / Absorption Property, Idempotent Property| Discrete mathematics
# Absorption Property, Idempotent Property| Discrete mathematics
Absorption Property: For any a, b in A, a˅(a˄b) =a and a˄(a˅b)=a
Proof: Since a˅(a˄b) is the join of a and (a˄b) then ,
a˅(a˄b) ≥a ————————-(1)
Since a≥a and a≥(a˄b) which implies that
(a˅a) ≥ a˅(a˄b)
⇒ a≥ a˅(a˄b) ———————————(2)
From (1) and (2) we get,
a˅(a˄b)=a
Now by the principle of duality we get,
a˄(a˅b)=a (proved)
Idempotent Property: For every a in A, (a˅a) =a and (a˄a)=a
Proof: We have, a≤(a˅a) ————————————–(1)
i.e., join of any two elements in a lattice is greater or equal to each of the elements.
Since, a≤a
Therefore, (a˅a) ≤a —————————————————–(2)
[if a≤b and a≤a then (a˅c) ≤b is because (a˅c) is the least upper bound of a and c.]
From (1) and (2) we get
(a˅a) =a
Now by the principle of duality, we get,
(a˄a)=a (proved)
## Procedure for computing shortest distance| Discrete Mathematics
Solution: The procedure for computing the shortest distance/path from a to any vertex G. Initially, ... | HuggingFaceTB/finemath | |
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# XII Boards Year 2016 Maths Class 12th Solutions
## Maths Year 2016 XII Boards Class 12th Solutions
CBSE class 12 board 2016 Maths question paper with solutions. Doubtnut provides step by step detailed Maths solution of CBSE 2016 Maths question paper for class 12.
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### XII BOARDS Class 12 | THREE DIMENSIONAL GEOMETRY
Find the coordinates of the point where the line through the points `A(3,4,1)` and `B(5,1,6)` crosses the `X Z` plane. Also find the angle which this line makes with the `X Z` plane. ### XII BOARDS Class 12 | VECTOR ALGEBRA
The two adjacent sides of a parallelogram are `2 hat i-4 hat j-5 hat k` and `2 hat i+2 hat j+3 hat kdot` Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram. ### XII BOARDS Class 12 | PROBABILITY
In a game, a man wins Rs. 5 for getting a number greater than 4 and loses Rs. 1 otherwise, when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he win/loses. ### XII BOARDS Class 12 | PROBABILITY
A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white? | HuggingFaceTB/finemath | |
Hey! I'm David, cofounder of zkSecurity and the author of the Real-World Cryptography book. I was previously a crypto architect at O(1) Labs (working on the Mina cryptocurrency), before that I was the security lead for Diem (formerly Libra) at Novi (Facebook), and a security consultant for the Cryptography Services of NCC Group. This is my blog about cryptography and security and other related topics that I find interesting.
# Understanding PLONK posted July 2021
PLONK is the state of the art when it comes to general-purpose proof system. While it was released in 2019, the paper has recently received some updates, and the scheme is still evolving (with Aztec announcing an UltraPLONK version coming soon). This is the scheme that we use at Mina to compress the size of the blockchain from gigabytes to a fixed size of a dozen kilobytes.
While I don't think the core ideas are the hardest to understand, the scheme compresses a myriad of optimization which makes it hard to parse. In this post I hope to add some clarity to some aspects of the scheme. Note that I assume that you have some knowledge of how PLONK works.
## How PLONK works, the short version
Eventually, the idea of PLONK is to prove that some polynomial $f(x)$ vanishes on some domain $H \subset \mathbb{F}$ (and I will ignore the permutation argument, which is just another proof). To prove that, we reduce the problem to some other problem. Incrementaly, it looks like this:
• Proving the previous statement is equivalent to proving that the polynomial is divisible by $Z_H(x)$, the polynomial that has all the elements of $H$ as roots (also called vanishing polynomial).
• Which is equivalent to proving the following identity (for some quotient polynomial $t$): $$f(x) = t(x) \cdot Z_H(x) \; \; \; \forall x \in \mathbb{F}$$
• Which is equivalent to proving the identity on some random point $z$ (thanks to the Schwartz-Zippel lemma): $$f(z) = t(z) \cdot Z_H(z)$$
To prove the last statement, the prover uses of polynomial commitment scheme (specifically, the KZG scheme) to commit to the polynomial $f$ and $t$. The prover then sends the commitments to the verifier. At that point, the verifier has to check that for some random point $z$
$$f(z) = t(z) \cdot Z_H(z)$$
This is done by sending a random point $z$ to the prover and doing an "opening" of the commitments at this point: the prover sends the values $f(z)$ and $t(z)$ as well as a proof that these are the correct evaluations.
Prover->Verifier: com(f), com(t) Note right of Verifier: generates random z Verifier->Prover: z Prover->Verifier: f(z), t(z) Prover->Verifier: proofs of opening Note right of Verifier: checks that \n sum f(z) = t(z)z_H(z)
This is in essence the PLONK protocol, except that this is not really what happens in the paper...
## More reductions
The newer PLONK actually does one more reduction of the last statement:
• As per the previous section: we want to prove that $$f(z) = t(z) \cdot Z_H(z)$$
• Which is equivalent to proving that $z$ is a root of the polynomial $$f(x) - t(x) \cdot Z_H(x)$$
• Since the verifier already knows one of the polynomial ($Z_H$), they can evaluate it in advance. So the previous statement is equivalent to proving that $z$ is a root of $$r(x) = f(x) - t(x) \cdot Z_H(z)$$
The last two steps is an optimization (called Maller's optimization) that removes the need for the prover to send $t(z)$, as the verifier can use the commitment to $t$ to produce a commitment to $r$ (to verify the opening proof).
These additional reductions moved us from a protocol in which the prover sends openings to let the verifier check an identity by themselves, to a protocol where the prover simply sends openings.
Prover->Verifier: com(f), com(t) Note right of Verifier: generates random z Verifier->Prover: z Prover->Verifier: f(z), r(z) = 0 Prover->Verifier: proofs of opening Note right of Verifier: reconstruct r(x) and \n validate opening proofs
To verify the opening of $r$ for $x = z$, the verifier will have to reconstruct a commitment to $r$ first. That's easy, it is:
$$com(r) = com(f) - com(t) \cdot Z_H(z)$$
which will use:
• the commitment to $f$ received during the protocol
• the commitment to $t$ received during the protocol
• the evaluation of $Z_H(x)$ at $x=z$ which they can do by themselves
## Not so fast... t is too large
If you've read PLONK, you've noticed that the prover actually doesn't send a commitment to $t$ directly, because $t$ is too large and polynomial commitment schemes have an upperbound fixed during the trusted setup. (By the way, $t$ is too large because the permutation argument makes it three times as large due to the three witness polynomials.) To circumvent that limitation, the polynomial $t$ is split into three smaller polynomials $t_{lo}, t_{mid}, t_{hi}$ such that:
$$t(x) = t_{lo}(x) + x^n \cdot t_{mid}(x) + x^{2n} \cdot t_{hi}(x)$$
This means that in our previous protocol, we can't prove directly that $z$ is a root of
$$r(x) = f(x) - t(x) \cdot Z_H(z)$$
instead we have to prove the equivalent that $z$ is a root of
$$r(x) = f(x) - [t_{lo}(x) + x^n \cdot t_{mid}(x) + x^{2n} \cdot t_{hi}(x)] \cdot Z_H(z)$$
This is not great, as the prover cannot produce a commitment to $r$ anymore. The reason is that $x^n$ and $x^{2n}$ cannot be committed as they're larger than the upperbound of our polynomial commitment. Instead, notice that since the verifier already knows these values, so they can pre-evaluate them at $z$ and ask instead for a proof that:
$$r(x) = f(x) - [t_{lo}(x) + z^n \cdot t_{mid}(x) + z^{2n} \cdot t_{hi}(x)] \cdot Z_H(z)$$
which is a fine request, as the verifier can produce the commitment of $r$ needed to verify the opening proof:
$$com(r) = com(f) - [com(t_{lo}) + z^n \cdot com(t_{mid}) + z^{2n} \cdot com(t_{hi})] \cdot Z_H(z)$$
At this point, the protocol looks more like this:
Prover->Verifier: com(f) Prover->Verifier: com(t_lo), com(t_mid), com(to_hi) Note right of Verifier: generates random z Verifier->Prover: z Prover->Verifier: f(z), r(z) = 0 Prover->Verifier: proofs of opening Note right of Verifier: reconstruct r(x) and \n validate opening proofs
The big proof in PLONK really boils down to two things:
1. The permutation argument, which links the wires in our circuit. I ignore this proof in the post.
2. the main polynomial $f$, which is our circuit.
Since the polynomial $f$ needs to be constructed such that:
• it does not leak any non-public information to the verifier
• it does not allow the prover to change fixed parts of the circuit
the prover and the verifier perform a "polynomial dance" to construct the polynomial together. The end product sorts of looks like this:
$$f(x) = a(x) q_L(x) + b(x) q_R(x) + q_M(x) a(x) b(x) + q_O(x) c(x) + q_C(x)$$
where $a, b, c$ are private polynomials that the prover constructs, commits, and sends to the verifier; and $q_L, q_R, q_M, q_O, q_C$ are public polynomials (the selector polynomials) that both the verifier and the prover can construct (and commit to if necessary).
So the end protocol looks more like this:
Prover->Verifier: com(a), com(b), com(c) Prover->Verifier: com(t_lo), com(t_mid), com(to_hi) Note right of Verifier: generates random z Verifier->Prover: z Prover->Verifier: a(z), b(z), c(z), r(z) = 0 Prover->Verifier: proofs of opening Note right of Verifier: reconstruct r(x) and \n validate opening proofs
And as in the previous section, the verifier needs to reconstruct a commitment to $r$ before being able to ask for an opening, which is now impossible as we're dealing with multiplication of commitments
\begin{align} r(x) = \; &a(x) q_L(x) + b(x) q_R(x) + a(x) b(x) q_M(x) + c(x) q_O(x) + q_C(x) \\ & - [t_{lo}(x) + x^n \cdot t_{mid}(x) + x^{2n} \cdot t_{hi}(x)] \cdot Z_H(z) \end{align}
but since the prover sends the evaluations of $a, b, c$ at $z$ (with proofs), the verifier can use that to simplify the polynomial $r$ to:
\begin{align} r(x) = \; &a(z) q_L(x) + b(z) q_R(x) + a(z) b(z) q_M(x) + c(z) q_O(x) + q_C(x) \\ & - [t_{lo}(x) + x^n \cdot t_{mid}(x) + x^{2n} \cdot t_{hi}(x)] \cdot Z_H(z) \end{align}
Finally, the verifier can produce the commitment of $r$ as:
\begin{align} com(r) = \; &a(z) com(q_L) + b(z) com(q_R) + a(z) b(z) com(q_M) + c(z) com(q_O) + com(q_C) \\ & - [com(t_{lo}) + z^n \cdot com(t_{mid}) + z^{2n} \cdot com(t_{hi})] \cdot Z_H(z) \end{align}
There's much more to PLONK. I've skipped the circuit part, the permutation argument, I've also ignored the big pairing equation at the end. These will be subjects for another post :)
# Maller optimization to reduce proof size posted July 2021
In the PLONK paper, they make use of an optimization from Mary Maller in order to reduce the proof size. This is a note explaining this optimization. If you have no idea what these words are, you might want to skip reading this post :)
## Explanation
Maller's optimization is used in the "polynomial dance" between the prover and the verifier to reduce the number of openings the prover send.
Recall that the polynomial dance is the process where the verifier and the prover form polynomials together so that:
1. the prover doesn't leak anything important to the verifier
2. the verifier doesn't give the prover too much freedom
In the dance, the prover can additionally perform some steps that will keep the same properties but with reduced communication.
Let's see the protocol where Prover wants to prove to Verifier that
$$\forall x \in \mathbb{F}, \; h_1(x)h_2(x) - h_3(x) = 0$$
given commitments of $h_1, h_2, h_3$.
Note left of Prover: commits to h1, h2, h3 Prover->Verifier: com(h1), com(h2), com(h3) Note right of Verifier: generates random point s Verifier-->Prover: s Note left of Prover: evaluates at point s Prover->Verifier: h1(s), h2(s), h3(s) Prover->Verifier: 3 proofs of openings Note right of Verifier: verifies that \n h1(s)h2(s) - h3(s) = 0
A shorter proof exists. Essentially, if the verifier already has the opening h1(s), they can reduce the problem to showing that
$$\forall x \in \mathbb{F}, \; L(x) = h_1(\mathbf{s})h_2(x) - h_3(x) = 0$$
given commitments of $h_1, h_2, h_3$ and evaluation of $h1$ at a point $s$.
Note left of Prover: commits to h1, h2, h3 Prover->Verifier: com(h1), com(h2), com(h3) Note right of Verifier: generates random point s Verifier-->Prover: s Note left of Prover: evaluates at point s Prover->Verifier: h1(s), L(s) Prover->Verifier: 2 proofs of openings Note right of Verifier: forms polynomial com(L) = \n h1(s)com(h2) - com(h3) Note right of Verifier: checks that L(s) = 0
## Notes
Why couldn't the prover open the polynomial $L'$ directly?
$$L'(x) = h_1(x)h_2(x) - h_3(x)$$
By doing
Note left of Prover: commits to h1, h2, h3 Prover->Verifier: com(h1), com(h2), com(h3) Note right of Verifier: generates random point s Verifier-->Prover: s Note left of Prover: evaluates at point s Prover->Verifier: L'(s), 1 proof of opening Note right of Verifier: forms polynomial com(L') = \n com(h1)com(h2) - com(h3) Note right of Verifier: verifies that \n h1(s)h2(s) - h3(s) = 0
The problem here is that you can't multiply the commitments together without using a pairing (if you're using a pairing-based polynomial commitment scheme), and you can only use that pairing once in the protocol.
If you're using an inner-product-based commitment, you can't even multiply commitments anyway.
## Appendix: Original explanation from the PLONK paper
https://eprint.iacr.org/2019/953.pdf
For completion, the lemma 4.7:
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# Pairing-based polynomial commitments and Kate polynomial commitments posted June 2021
There's this thing called a Kate polynomial commitment, which is a polynomial commitment primitive that makes use of pairings. There's an excellent post from Dankrad which I would recommend reading instead of this post. I wrote this as a shorter summary of how you can commit to a polynomial, and then prove any evaluation $f(x) = y$.
Here's how it works:
You have a polynomial $f(x) = x^2 + 3x$
and some public parameters:
$$SRS = {[1], [s], [s^2], [s^3]} = {G, sG, s^2 G, s^3 G}$$
where $[x] := xG$ for some generator $G$ of an elliptic curve group.
and $s$ is a toxic waste (something that no one should know) hidden behind an elliptic curve point G (some people call that "hidden in the exponent").
## to commit to $f$
To commit to this polynomial, evaluate it at the unknown point $s$. You can do that by playing with the $SRS$:
$$[f(s)] := [s^2] + 3 [s] = s^2 G + 3 sG = (s^2 + 3s)G$$
## to prove that $f(\zeta) = a$
One day, the verifier asks "what's the evaluation at $\zeta$?" And the prover responds by sending the answer, $a$, and a proof ($h(s)$, see below).
### The idea behind the proof
Notice that because $\zeta$ is a root of $f(x)-f(\zeta)$, then for some polynomial $h(x)$:
$$f(x) - f(\zeta) = (x-\zeta) \cdot h(x)$$
Due to this, $h(x) = \frac{f(x)-f(\zeta)}{x-\zeta}$ must be a valid polynomial.
At a high-level:
• the verifier will compute what they think $[h(x)]$ should be at some random point $s$
• the prover will send the actual value $[h(s)]$
• the verifier will check if they match
This works because the Schartz-Zippel lemma tells us that two polynomials that are different are different in most points.
### The proof
Here's the protocol:
1. the prover sends the verifier a commitment $[\frac{f(s)-f(\zeta)}{s-\zeta}]=[h(s)]$ evaluated at some random point $s$ (the toxic waste).
2. the verifier constructs a similar $h(s)$ but with the expected value of $f(\zeta)$ instead: $[\frac{f(s) - a}{s-\zeta}]$. The prover then checks if it's equal to $[h(s)]$.
Note:
1. The prover can compute $[h(s)]$ easily, because they can just compute the polynomial $h(x)$ first, and then reconstruct it at $s$ with the $SRS$. $$h(x) = \frac{f(x)-f(\zeta)}{x-\zeta} = a_0 + a_1x + a_2x^2 + \cdots$$ and then $$[h(s)] := a_0[1] + a_1[s] + a_2[s^2] + \cdots$$
for example with our previous $f(x)$ and $\zeta = 3$
2. The verifier cannot compute their own $[h(s)]$ because they cannot divide by $s$ (remember, nobody knows $s$). They need a pairing. Remember, you want to check the following identity hidden in the exponent (using commitments): $$\frac{[f(s) - a]}{[s-\zeta]} = [h(s)]$$ But since you can't divide with commitments, you can't compute what's on the left-hand side. You can multiply thanks to pairings though. So instead, you could check the following equation: $$[f(s) - a] = [(s-\zeta)h(s)]$$ and with pairings, you can multiply $[s-\zeta]$ with the proof $[h(s)]$: $$e([f(s)] - [a], [1]) = e([s-\zeta], [h(s)])$$
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# I'm now at O(1) Labs working on Mina protocol!!! posted May 2021
Hey reader! I haven't posted in a while, but as this is my blog I'm contracted to talk about life events such as this one. I've joined O(1) Labs a bit more than a month ago to work on the Mina cryptocurrency. If you don't know about Mina, check it out, it's pretty cool: it uses recursive zero-knowledge proofs to compress a blockchain into a single proof of 11KB. I hope I got you intrigued! I want to say it is one of, if not the most, ambitious project in the space (but I'm biased). As I'm still relatively new there, I don't have much to say besides that, but you can imagine that my posting will switch to more zero-knowledgy type of stuff very soon!
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# WTF are these security chips? posted April 2021
There seem to be a few interesting trends in “security via hardware” these days.
The first trend is root-of-trust chips. Integrated TPM-like chips that are like crypto Swiss Army knives as they offer many functionalities out of the box. They resemble discrete TPMs but are instead implemented as coprocessor to the main processor. This makes these newer chips more resistant to physical MITM attacks (as discrete TPMs simply use a bus to communicate with other components). If you don’t know what a TPM is, it’s just a device that performs cryptographic operations and generally sits somewhere on your motherboard. Examples of these integrated security chips are Microsoft’s Pluton and Apple’s secure enclave.
The second trend is confidential computing. There are two types of specialized hardware here:
• Programmable integrated secure processors; these are similar to the root-of-trust chips, except that they are programmable: you can push code there and run it in a separate trusted execution environment (TEE). It’s pretty useful for applications that require a trusted computing base (TCB); a core program whose security is critical and that does not need to trust the rest of the system. It’s also useful in “cloud scenarios” where you want to run some computation on a remote machine but want to make sure it runs it correctly. Think about Intel SGX, or ARM TrustZone.
• Confidential VMs; imagine a hardware hypervisor that can run VMs as enclaves. This is usually much more practical than the enclave created by SGX, as you don’t need to write custom code and there are no memory limitation. But it is not clear to me how much security you lose against physical attacks by doing this (especially when papers like this one seem alarming). AMD SEV does this, and both Azure and GCP have started offerings to leverage this technology.
It can be hard to understand the difference between all these types of specialized hardware, the attacks they prevent, and the features they unlock. But essentially, here’s how I think about the two kinds: they all do great against software attacks (minus complex cryptographic attacks), they both aren’t the best tool in the box against a motivated physical attacker (HSMs are “better”), and only confidential computing cares about custom user code.
But it’s easier to understand the difference by looking at some examples. As I only touch on protocols, you can simply imagine these chips as creating a blackbox for code and data that others can’t see and touch (even with a debugger).
## Protecting keys and data with a secure enclave
The simplest use case for hardware security chips is to protect data. To protect keys, it’s easy: just generate them in the secure chip and disallow extraction. If you need ‘em, just ask the secure enclave to perform crypto operations with them. To protect data? Encrypt it! That concept is called file-based encryption (FBE) if you’re encrypting individual files, and full-disk encryption (FDE) if it’s the whole disk. FDE sounds much better, as it’s all or nothing. If you're under the shower and you wet your hair a little, you know you'll have to wash them. That’s what most laptops and desktops use.
In practice, FDE is not that great though: it doesn't take into account how we, human beings, use our devices. We often leave them locked, as opposed to turned off, so that background functionalities can keep running. Computers deal with this by just keeping the data-encryption key (DEK) around, even if your computer is locked. Think about that the next time you go to the restroom at Starbucks, leaving your locked computer unattended. Phones do it a bit better by encrypting different types of files depending on if your phone is locked or turned off. It sounds like a good solution, but Zinkus et al. showed that it’s not that great either.
If done well, this is how you typically hear about disk encryption in the news:
A couple of months ago the highly-publicised case of Apple vs. FBI brought attention to the topic of privacy - especially in the context of mobile devices. Following the 2015 San Bernardino terrorist attack, the FBI seized a mobile phone belonging to the shooter, Syed Farook, with the intent to search it for any additional evidence or leads related to the ongoing investigation. However, despite being in possession of the device, the FBI were unable to unlock the phone and access its contents.
Of course, the user should be authenticated before data can be decrypted. This is often done by asking the user for a PIN or password. A PIN or password is not enough though, as it would allow simple brute-force attacks (especially on 4 or 6-digit PINs). In general, solutions try to tie the DEK to both a user credential and a symmetric key kept on the enclave.
What’s that symmetric key? We all know that you can’t hardcode the same key in every device you produce. This is dumb. You end up with attacks like DUHK where thousands of devices are found hardcoding the same secret (and pwning one device breaks all of them). The solution is a per-device key that is either burned into the chip during manufacturing, or created by the chip itself (so-called physically unclonable functions). For example, each Apple secure enclave have a UID, each TPM has a unique endorsement key and attestation key, each OpenTitan chip has a creator root key and an owner root key, etc.
A randomly generated UID is fused into the SoC at manufacturing time. Starting with A9 SoCs, the UID is generated by the Secure Enclave TRNG during manufacturing and written to the fuses using a software process that runs entirely in the Secure Enclave. This process protects the UID from being visible outside the device during manufacturing and therefore isn’t available for access or storage by Apple or any of its suppliers. sepOS uses the UID to protect device-specific secrets. The UID allows data to be cryptographically tied to a particular device. For example, the key hierarchy protecting the file system includes the UID, so if the internal SSD storage is physically moved from one device to another, the files are inaccessible.
To prevent brute-force attacks, Apple’s secure enclave mixes both the UID key and the user PIN with a password-based KDF (password-hashing function) to derive the DEK. Except that I lied: to allow user to change their PIN quickly, the DEK is actually not derived directly, but instead encrypted by a key-encryption key (KEK).
## Secure boot with a root-of-trust secure chip
When booting your computer, there are different “stages” that will run until you finally get to the screen you want. One problem users face are viruses and malwares, and these can infect the boot process. You then run on an evil operating system… To protect the integrity of boot, our integrated secure chips provide a “root of trust”, something that we trust 100% and that allows us to trust other stuff down the line. This root of trust is generally some read-only memory (ROM) that cannot be overwritten, and it’s also called one-time programmable memory as it was written during manufacturing and can’t be changed anymore. For example, when powering up a recent Apple device, the very first code that gets executed is inside the Apple’s secure enclave ROM (called Boot ROM). That boot rom is tiny, so usually the only thing it does is:
• Prepare some protected memory and loads the next image there (so-called "boot code").
• Hash the image and verify its signature against the hardcoded public key in the ROM.
• Execute that code.
The next boot loader does the same thing, and so on until it gets to the device’s operating system. This is how updates that are not signed by Apple can’t be installed on your phone.
## Confidential Computing with a programmable secure processor
There’s been a new paradigm for the last years: the cloud; big companies running servers to host your stuff. Amazon has AWS, Google has GCP, and Microsoft has Azure. Another way to put this is that people are moving from running things themselves, to running things on someone else’s computer. This of course create some issues in some scenarios where privacy is important. To fix that, confidential computing attempts at offering solutions to run client code without being able to see it or modify its behavior. SGX primary use case seems to be exactly that these days: clients running code that the servers can’t see or tamper with.
One interesting problem that arise is: how can I trust that the response I got from my request indeed came from SGX, and not some impersonator. This is what attestation tries to solve. There are two kinds of attestation:
• local attestation, when two enclaves running on the same platform need to communicate and prove to each other that they are secure enclaves
• remote attestation, when a client queries a remote enclave and need to make sure that it was a legit enclave that produced the result from the request.
Each SGX chip is provided with unique keypairs at manufacturing time: the Root Sealing Keys. The public key part is then signed by some Intel certificate authority. So the first assumption, if we ignore the assumption that the hardware is secure, is that Intel is correctly signing public keys of secure SGX chips only.
With that in mind, you can now obtained a signed attestation, from Intel's CA, that you're talking to a real SGX enclave, and that it is running some code (at least a proof of its digest), etc.
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# What is Host Card Emulation (HCE)? posted April 2021
It’s 2020, most people have a computer in their pocket: a smart phone. What is the point of a credit card anymore? Well, not much. Nowadays more and more payment terminals support contactless payment via the near-field communication (NFC) protocol, and more and more smartphones ship with an NFC chip that can act as a credit card. NFC for payment is specified as Card Emulation. Literally: it emulates a bank card. But not so fast, banks will prevent you from doing this unless you have a secure element.
Since Apple has full control over its hardware, it can easily add a secure element to its new iPhones to support payment, and this is what Apple did with an embedded secure element bonded onto the NFC chip since the iPhone 6. The secure element communicates directly with the NFC chip, and in turn to NFC readers; thus a compromise of the phone operating system does not impact the secure element.
Tokenization is a common concept in security: replace the sensitive data with some random stuff, and have a table secured somewhere safe that maps the random stuff to the real data. Although Apple theoretically doesn't have to use tokenization, since iPhones have secure elements that can store the real Primary Account Number (PAN), they do use it in order to gain more privacy (it's after all their new bread and butter).
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# One password to rule them all, single sign-on (SSO) and password managers posted April 2021
Password reuse is bad, what can we do about it? Naively, users could use different passwords for different websites, but there are two problems with this approach:
To alleviate these concerns, two solutions have been widely adopted:
• Single-sign on (SSO). The idea of SSO is to allow users to connect to many different services by proving that they own the account of a single service. This way the user only has to remember the password associated with that one service in order to be able to connect to many services. Think "connect with Facebook" type of buttons, as illustrated below.
• Password Managers. The previous SSO approach is convenient if the different services you use all support it, but this is obviously not scalable for scenarios like the web. A better approach in these extreme cases is to improve the clients as opposed to attempting to fix the issue on the server side. Nowadays, modern browsers have built-in password managers that can suggest complex passwords when you register on new websites, and can remember all of these passwords as long as you remember one master password.
An example of single-sign on (SSO) on the web. By having an account on Facebook or Google, a user can connect to new services (in this example Airbnb) without having to think about a new password.
The concept of SSO is not new in the enterprise world, but its success with normal end-users is relatively recent. Today, two protocols are the main competitors when it comes to setting up SSO:
• Security Assertion Markup Language 2.0 (SAML). A protocol using the Extensible Markup Language (XML) encoding.
• OpenID Connect (OIDC). An extension to the OAuth 2.0 (RFC 6749) authorization protocol using the JavaScript Object Notation (JSON) encoding.
SAML is still widely used, mostly in an enterprise setting, but it is at this point a legacy protocol. OpenID Connect, on the other hand, can be seen everywhere on web and mobile applications. You most likely already used it!
While OpenID Connect allows for different types of flows, let's see the most common use case for user authentication on the web via the authorization code flow:
1. Alice wants to log into some application, let's say example.com, via an account she owns on cryptologie.net (that's just my blog, but let's pretend that you can register an account on it).
2. example.com redirects her browser to a specific page of cryptologie.net to request an "authorization code." If she is not logged-in in cryptologie.net, the website will first ask her to log in. If she is already logged-in, the website will still confirm with the user that they want to connect to example.com using their identity on cryptologie.net (it is important to confirm user intent).
3. cryptologie.net redirects Alice back to example.com which then learns the authorization code.
4. example.com can then query cryptologie.net with this authorization code to confirm Alice's claim that she owns an account on cryptologie.net, and potentially retrieve some additional profile information about that user.
In OpenID Connect (OIDC), Alice (the end-user in OIDC terms) can authenticate to a service example.com (the relying party) using her already existing account on cryptologie.net (the OpenID provider). For the web, the authorization code flow of OIDC is usually used. It starts by having Alice request an "authorization code" from cryptologie.net (and that can only be used by example.com). example.com can then use it to query cryptologie.net for Alice's profile information (encoded as an ID token), and then associate her cryptologie.net identity with an account on their own platform.
There are many important details that I am omitting here. For example, the authorization token that Alice receives in step 2 must be kept secret, as it can be used to log in as her on example.com. Another example: so that example.com cannot reuse the authorization token to connect as Alice on a different website, the OpenID provider cryptologie.net retains an association between this authorization token and the intended audience (example.com). This can be done by simply storing this association in a database, or by having the authorization code contain this information authenticated (I explained a similar technique with cookies in chapter 3).
By the way, the proof given to example.com in step 4 is called an ID token in OpenID Connect, and is represented as a JSON Web Token (JWT) which is just a list of JSON-encoded data.
In OpenID Connect, if Alice wants to log in on example.com via her account on another website (an OpenID provider), example.com eventually needs to obtain what is called an "ID token." An ID token is some user information encoded via the JSON Web Token (JWT) standard. a JWT is simply the concatenation of three JSON-encoded objects. In the picture, the website https://jwt.io lets you decode a JWT and learn what every field stands for. In our browser-based example, example.com uses TLS to communicate with (and authenticate) the OpenID provider. Thus, the ID token it receives can be trusted. In other OpenID Connect flows (used, for example, by mobile applications) the ID token can be provided directly by the user, and can thus be tampered with. In this case, an ID token contains a signature which can be verified using the OpenID provider's public key.
Authentication protocols are often considered hard to get right. OAuth2, the protocol OpenID Connect relies on, is notorious for being easy to mis-use (see RFC 6819: OAuth 2.0 Threat Model and Security Considerations). On the other hand, OpenID Connect is very well specified. Make sure that you follow the standards and that you look at best practices, this can save you from a lot of trouble.
Here's another example of a pretty large company deciding not to follow this advice. In May 2020, the "Sign-in with Apple" SSO flow that took a departure from OpenID Connect was found to be vulnerable. Anyone could have obtained a valid ID token for any Apple account, just by querying Apple's servers.
SSO is great for users, as it reduces the number of passwords they have to manage, but it does not remove passwords altogether. The user still has to use passwords to connect to OpenID providers. If you're interested to know how cryptography can help to hide passwords, read the rest of this content in my book real-world cryptography.
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# Cryptography and assembly code posted March 2021
Thanks to filippo streaming his adventures rewriting Golang assembly code into "cleaner" Golang assembly code, I discovered the Avo assembly generator for Golang.
This post is not necessarily about Golang, but Golang is a good example as its standard library is probably the best cryptographic standard library of any programming language.
At dotGo 2019, Michael McLoughlin presented on his Avo tool. In the talk he mentions that there's 24,962 x86 assembly lines in Golang's standard library, and most of it is in the crypto package. A very "awkward" place where "we need very high performance, and absolute correctness". He then shows several example that he describes as "write-once code".
The talk is really interesting and I recommend you to check it.
I personally spent days trying to understand Golang's SHA-3 assembly implementation. I even created a Go Assembly by Example page to help me in this journey. And I ended up giving up. I just couldn't understand how it worked, the thing didn't make sense. Someone had written it with their own mental model of how they wanted to pass data around. It was horrible.
It's not just a problem of Golang. Look at OpenSSL, for example, which most cryptographic applications and libraries rely on. It contains a huge amount of assembly code to implement cryptography, and that assembly code is sometimes generated by unintelligible perl code.
There are many more good examples out there. the BearSSL TLS implementation by Thomas Pornin, the libsodium cryptographic library by Frank Denis, the extended keccak code package by the Keccak team, all use assembly code to produce fast cryptography.
We're making such a fuss about readable, auditable, simple and clear cryptographic implementations, but most of that has been thrown out of the window in the quest for performance.
The real problem, from a reviewer perspective is that assembly is getting us much further away from the specification. As the role of a reviewer is to match the implementation to the specification, it makes the job hard, perhaps impossible.
Food for thoughts...
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For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4
Originally posted by Rorschach1337 on 22 Oct 2017, 20:11.
Last edited by Bunuel on 22 Oct 2017, 20:37, edited 1 time in total.
Renamed the topic.
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Re: For one roll of a certain number cube with six faces, numbered 1 throu [#permalink]
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Rorschach1337 wrote:
For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4
ATLEAST 3 times means 3 times and 4 times..
1) 3 times :-
3 times "2" will mean a probability of $$\frac{1}{6}^3$$ and prob of any other number in 4 throw is 5/6..
but this ANY other number can come in any of 4 throws ..
so prob of 3times = $$\frac{1}{6}^3*4*\frac{5}{6}$$
2) 4 times :-
$$\frac{1}{6}^4$$
total $$\frac{1}{6}^4+4*\frac{1}{6}^3\frac{5}{6}$$
D
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"The outcome will be a two at least 3 times" means the outcome will be a two 3 times or 4 times.
Case 1: 3 times = 4C3 * (1/6)^3 * 5/6 = 4*(1/6)^3 * 5/6
- select 3 cubes whose the outcome will be two = 4C3
- The prob that 3 cubes will be 2 = (1/6)^3
- The prob of the remaining cube that can be any other value except "2" = 5/6
Case 2: 4 times = (1/6)^4
Hence, total = case 1 + case 2 = 4*(1/6)^3 * 5/6 + (1/6)^4 => Answer D.
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Re: For one roll of a certain number cube with six faces, numbered 1 throu [#permalink]
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Rorschach1337 wrote:
For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?
(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4
Total rolls = 4 times
Dice = 6 sided
Probability of a "2" on every roll = 1/6
Probability of not a "2" but any of the other 5 numbers = 1 - 1/6 = 5/6
Probability of "2" at least 3 times means :
a. 2 three times
b. 2 four times
Case 1 : T T T N (T = two and N = Not two)
$$\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{5}{6} * \frac{4!}{3!}$$
We multiply by 4! because permutations/total arrangements of 4 different items can be done in 4! ways
We divide by 3! because we have 3 identical items.
Therefore, 4*(1/6)^4 * 5/6
Case 2 : T T T T (All twos)
$$(1/6)^4$$
Add Case I and II
$$4*(1/6)^4 * 5/6$$ + $$(1/6)^4$$
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"Do not watch clock; Do what it does. KEEP GOING." Re: For one roll of a certain number cube with six faces, numbered 1 throu [#permalink] 25 Mar 2019, 02:34
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# Boats and stream 2
Welcome to aptitude tricks
## Basic Concepts of Questions on Boats and Streams
1. A boat is said to go downstream, if the boat goes in the direction of stream.
2. A boat is said to go upstream, if the boat goes opposite to the direction of stream.
## Basic Formulas
1. If speed of boat in still water is b km/hr and speed of stream is s km/hr,
• Speed of boat in downstream = (b + s) km/hr , since the boat goes with the stream of water.
• Speed of boat in upstream = (b – s) km/hr. The boat goes against the stream of water and hence its speed gets reduced.
## Shortcuts With Explanation
Scenario 1: Given a boat travels downstream with speed d km/hr and it travels with speed ukm/hr upstream. Find the speed of stream and speed of boat in still water.
Let speed of boat in still water be bkm/hr and speed of stream be skm/hr.
Then b + s = d and b – s = u.
Solving the 2 equations we get,
b = (d + u)/2
s = (d – u)/2
Scenario 2: A man can row a boat, certain distance downstream in td hours and returns the same distance upstream in tu hours. If the speed of stream is s km/h, then the speed of boat in still water is given by
We know distance = speed * time
Let the speed of boat be b km/hr
Case downstream:
d = (b + s) * td
Case upstream:
d = (b – s) * tu
=> (b + s) / (b – s) = tu / td
b = [(tu + td) / (tu – td)] * s
Scenario 3: A man can row in still water at bkm/h. In a stream flowing at s km/h, if it takes him t hours to row to a place and come back, then the distance between two places, d is given by
Downstream: Let the time taken to go downstream be td
d = (b + s) * td
Upstream: Let the time taken to go upstream be tu
d = (b – s) * tu
td + tu = t
[d / (b + s)] + [d / (b – s)] = t
So, d = t * [(b2 – s2) / 2b]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)]/[2 * Speed of boat or man in still water]
Scenario 4: A man can row in still water at bkm/h. In a stream flowing at s km/h, if it takes t hours more in upstream than to go downstream for the same distance, then the distance d is given by
Time taken to go upstream = t + Time taken to go downstream
(d / (b – s)) = t + (d / (b + s))
=> d [ 2s / (b2 – s2 ] = t
So, d = t * [(b2 – s2) / 2s]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)] / [2 * Speed of still water] | HuggingFaceTB/finemath | |
## ELECTRIC FIELD
Electric Field and Electric Field Intensity$\\$
• The space around an electric charge in which its influence can be experienced is known as electric field.$\\$
• The electric field intensity $\vec{E}$ at any point is equal in magnitude to the force experienced per unit (test) position charge placed at that point and is directed along the direction of the force experienced.$\\$i.e $\vec{E}$=$\frac{\vec{F}}{q}$$\vec{F}$=qE$\\$
• Electric field intensity is a vector quantity. Electric field intensity due to positive charge is always away from the charge and due to negative charge is
always towards the charge.$\\$
• Its unit is Newton/Coulomb (N$C^{-1}$)or Volt/meter (V$m^{-1}$ )$\\$
• Electric field due to point charge q at distance r is: E=$\frac{1}{4πε_0}\frac{q}{r^2} \\$
• $\vec{E}$=$\vec{E_1}$+$\vec{E_2}$+$\vec{E_3}$+ ……………$\\$
• The magnitude of the resultant of two electric fields is given by E=$\sqrt{E_1 ^2+E_2 ^2+2E_1 E_2 cosθ} \\$
## Electric Lines of Force
1. The line or a curve along which an isolated +ve charge would travel if it is free to move in an electric field is known as electric line of force.
2. They start from positively charged body and end at a negatively charged body.
3. No electric lines of force exist inside the charged body.
4. Tangent to the line of force at any point gives the direction of electric intensity at that point.
5. No two electric lines of force can intersect each other.
6. The electric lines of force are always normal to the surface of a conductor, both while starting and ending on the conductor. Therefore, there is no component of electric field intensity parallel to the surface of the conductor.
7. They never form closed loops.
8. They are always perpendicular to equipotential surface.
9. The electric lines of force contract longitudinally, on account of attraction between unlike charges.
10. The electric lines of force exert a lateral pressure on account of repulsion between like charges.
11. In uniform electric field, the electric lines of force are equidistant, parallel straight lines.
12. When a metallic solid sphere is placed in a uniform electric field, then the lines of force are normal to the surface at every point but they cannot pass through the conductor.
## Field Intensity in Special Cases
1. Intensity of the electric field inside a charge spherical conductor/ hollow/ spherical shell/ conducting sphere is zero, since charge resides on the outer surface of the conductor. But $E_{surface}=\frac{1}{4πε_o} \frac{q}{R^2}$ (Where R= radius of the sphere) $E_{outside} =\frac{1}{4πε_o} \frac{q}{r^2}$ (For r>R )
2. The intensity of electric field for a uniformly charged non- conducting sphere of radius R is given by$\\$$E_{outside} =\frac{1}{4πε_o} \frac{q}{r^2}$ (For r>R )$\\$
$E_{surface}=\frac{1}{4πε_o}.\frac{q}{R^2}$ (Where r=R is radius of the sphere)$\\$$E_{inside} =\frac{1}{4πε_o}. \frac{qr}{R^3}$ ( For r<R)$\\$$E_{centre}$ = 0 ( for r=0)$\\$
3. Intensity of electric field at some point on the axis of uniformly charged ring of radius R is given by$\\$E=$\frac{1}{4πε_o}\frac{qx}{(x^2+R^2 )^{3/2}} =\frac{λRx}{2ε_o (x^2+R^2 )^{3/2}}$$\\$But it is zero at the center of ring.$\\$
4. Intensity of electric field near an infinite rod of charge is given by$\\$E=$\frac{λ}{2πε_0 r}$$\\$where λ is the linear charge density and r is the distance from the axis of rod.
5. Intensity of the electric field near a non-conducting infinite sheet of charge is E=$\frac{σ}{2ε_o}$$\\$where σ is the surface charge density.
## Electric Dipole
1. INTRO:$\\$
• Two equal and opposite charges
separated by a small distance constitute an electric dipole.$\\$
• Electric dipole moment is a vector
quantity (p) whose magnitude is equal to the product of magnitude of one charge and the distance between the two charges.$\\$
2. Do You Know?$\\$Dipole moment is a vector whose direction is from negative charge (-q) to positive charge (+q).$\\$Electric dipole moment$\\$$\vet{p}$=q.d$∴$\vet{p}\$=q(2l)
3. Electric field intensity on equatorial/ broadside on position/ Tan B position:$\\$Magnitude of electric field produced by dipole,$\\$$E_b=\frac{p}{4πε_o (r^2 + l^2 )^{3/2}}$$\\$For short dipole, r> > >d$\\$$E_b=\frac{1}{4πε_0}. \frac{p}{r^3}$ ⇒ | HuggingFaceTB/finemath | |
# What is the answer for question 16) ? http://postimg.org/image/6g52noj6f/
lemjay | High School Teacher | (Level 3) Senior Educator
Posted on
`log_3x+log_3(x+2)=1`
To explain this through the telephone, we can describe the steps as follows:
• Notice that the two logarithms in the equation have the same base. So, express it as one logarithm using the product property.
• So, at the left side of the equation write log_3 and multiply the two arguments x and (x+2). And the right side remains the same.
`log_3(x*(x+2)) = 1`
• Since `x*(x+2` ) is equal to `x^2+2x` , the argument of the logarithm becomes `x^2+2x` .
`log_3(x^2+2x) = 1`
• Since the logarithm of `x^2+2x` is equal to 1, then we apply the property that a logarithm is equal to 1 if its base and argument are the same `(log_b b=1)` .
• This means that the argument of `log_3` is equal to 3. So, set `x^2+2x` equal to 3. And this becomes our new equation.
`x^2+2x = 3`
• Now that we have a quadratic equation, to solve for x, set one side equal to zero. To do this, subtract both sides by 3.
`x^2+2x - 3 = 3-3`
`x^2+2x - 3 =0`
Then, factor left side.
`(x+3)(x - 1) = 0`
Next, set each factor equal to zero.
`x+ 3= 0` and `x-1=0`
So the equation breaks into two.
For the first equation, subtract both sides by 3.
`x+3=0`
`x+3-3=0-3`
`x=-3`
Note that in logarithm, we can not have a negative number as its argument. So, we do no consider -3 as a solution to our equation.
Next, solve for x in the second equation. To do so, add both sides by 1.
`x-1=0`
`x-1+1=0+1`
`x=1`
Hence, the solution to the given equation` log_3x + log_3(x+1)` is `x=1` .
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# Trouble with step 2 resolution calculus
I need to prove T ⊨ A v B with the resolution calculus from a set T.
### Step 1
Transform T into a set of clauses (CNF).
• Clause 1 = A v ¬C
• Clause 2 = C v A v B
• ### Step 2
Try to find a resolution proof for , that is T ∪ {¬A ∧ ¬B}. Resolve Clause 1 (A v ¬C) and Clause 2 (C v A v B)
I understand I must find some kind of contradiction and there out conclude T ⊨ A v B. I am really lost, not sure how to begin, all I can determine is that if C is true in one clause that it is false in the other (and vica versa).
How do I continue?
To prove that $T \vDash \varphi$ we have to apply Resolution to the set $T \cup \{ ¬ \varphi \}$.
Thus, as you say, we have to consider the set of clauses:
$\{ A \lor \lnot C, A \lor B \lor C, \lnot A, \lnot B \}$
built up with Clause 1, Clause 2 and the two clauses produced from the negation of the conclusion: $\lnot (A \lor B) \equiv (\lnot A \land \lnot B)$.
The aim of the procedure is to derive the empty clause: $\bot$ or $\square$.
If so, we have proved that the set is unsatisfiable (or contradictory), and hence we can conclude that the original sought conclusion (the formual we have negated) follows from $T$.
The steps of the procedure are :
1) All premises and the negation of the sentence to be proved (the conclusion) are conjunctively connected.
2) The resulting sentence is transformed into a conjunctive normal form with the conjuncts viewed as elements in a set of clauses.
3) The resolution rule is applied to all possible pairs of clauses that contain complementary literals. After each application of the resolution rule, the resulting sentence is simplified by removing repeated literals. If the sentence contains complementary literals, it is discarded (as a tautology). If not, and if it is not yet present in the clause set, it is added to it, and is considered for further resolution inferences.
4) If after applying a resolution rule the empty clause is derived, the original set of clauses is unsatisfiable.
5) If, on the other hand, the empty clause cannot be derived, and the resolution rule cannot be applied to derive any more new clauses, the formula to be proved is not a consequence of the original premises.
1) $A \lor \lnot C$
2) $A \lor B \lor C$
3) $\lnot A$
4) $\lnot B$
and apply the resolution rule to 1) and 2) producing:
5) $A \lor B$.
Now we can resolve 3) and 5) deriving:
6) $B$.
Finally, we resolve 4) and 6) deriving $\bot$.
• I see in T you added the negation of a and b as clauses in the set (on their own). Is this allowed? Or is that cause we're working with not a and not b. The rest I have come to understand, thanks for the much needed explanation and answer, appreciate it:) Commented Feb 14, 2016 at 15:16
• @Pim - see point 2) : the sentence is transformed into a conjunctive normal form with the conjuncts viewed as elements in the set $S$ of clauses. Commented Feb 14, 2016 at 15:44 | HuggingFaceTB/finemath | |
# DMtutorial1s - Tutorial 2: suggested solution 1. Suppose we...
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Unformatted text preview: Tutorial 2: suggested solution 1. Suppose we have observations (Xi , Yi ) sorted according to X : ..., (0.5, 1.2), (0.6, 1.4), (0.7, 1.5), (0.8, 1.7), (0.9, 1.5), (1.0, 2), (1.1, 2.2), (1.2, 1.6), (1.3, 1.7), (1.4, 1.9), (1.5, 1.7), .... If we use Epanechnikov kernel with bandwidth h = 0.3. (a) Estimate the density function of X at X = 1.0 (b) Estimate the regression function value m(1.0) in model Y = m(X ) + ε (a) ˆ f (1) = 0.75 ∗ {(1 − (1 − 0.8)2 /0.32 ) +(1 − (1 − 0.9)2 /0.32 ) +(1 − (1 − 1)2 /0.32 ) +(1 − (1 − 1.1)2 /0.32 ) +(1 − (1 − 1.2)2 /0.32 )}/(n ∗ 0.3) (b) m(1) = {(1 − (1 − 0.8)2 /0.32 ) ∗ 1.7 ˆ +(1 − (1 − 0.9)2 /0.32 ) ∗ 1.5 +(1 − (1 − 1)2 /0.32 ) ∗ 2 + (1 − (1 − 1.1)2 /0.32 ) ∗ 2.2 +(1 − (1 − 1.2)2 /0.32 ) ∗ 1.6}/{(1 − (1 − 0.8)2 /0.32 ) +(1 − (1 − 0.9)2 /0.32 ) +(1 − (1 − 1)2 /0.32 ) +(1 − (1 − 1.1)2 /0.32 ) +(1 − (1 − 1.2)2 /0.32 )} 2. Consider the regression Yi = m(Xi ) + εi . we need to estimate the regression function m(x) at a given point x. Let a = m(x). Then the kernel estimator is the minimizer of a to the following problem −1 n 2 n {Yi − a} Kh (Xi − x) i=1 Prove it. Let d −1 n da i.e. −2n −1 n {Yi − a}2 Kh (Xi − x) = 0. i=1 n {Yi − a}Kh (Xi − x) = 0 i=1 1 we have the solution n n a= i=1 Kh (Xi − x)Yi / i=1 Kh (Xi − x). 3. A retrospective sample of males in a heart-disease high-risk region of the Western Cape, South Africa. There are roughly two controls per case of CHD. Many of the CHD positive men have undergone blood pressure reduction treatment and other programs to reduce their risk factors after their CHD event. In some cases the measurements were made after these treatments. These data are taken from a larger dataset, described in Rousseauw et al, 1983, South African Medical Journal. Dependent variables are sbp: systolic blood pressure; tobacco: cumulative tobacco (kg); ldl: low densiity lipoprotein cholesterol; adiposity; famhist: family history of heart disease (1:Present, 0:Absent); typea: type-A behavior; obesity; alcohol: current alcohol consumption; age: age at onset. Dependent variable chd: response, coronary heart disease. (data) Use kernel smoothing method to estimate the functional regression relation between the (continuous) dependent variables and the response (dependent variables). What is your choices of bandwidth for the calculations by observing the estimated curves. chd against sbp (cude) 1.0 q qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqq q q q qq qq qq chd 0.0 0.2 qq qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq q qqqq q q qq qq qq q q 0.4 0.6 0.8 qq q q 100 120 140 160 sbp 180 200 220 2 ...
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## This note was uploaded on 10/04/2010 for the course STAT ST4240 taught by Professor Xiayingcun during the Fall '09 term at National University of Singapore.
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# Why is a metric space an open subset of itself?
Given a metric space $X$, the entire space $X$ is an open subset of $X$.
I'm having some trouble thinking about this. I have a counterexample in mind. Suppose we consider the closed subset of $\mathbb{R}$ called $A = [a, b] \in \mathbb{R}$. Isn't $A$ also a metric space? As far as I know, the answer is yes. So, based on the above, $A$ is an open subset of $A$. But, since $a$ and $b$ aren't interior points of $A$, how is $A$ an open subset?
What is wrong with my line of thinking?
• Your line of thinking is wrong because the open balls of $[a,b]$ aren't the same as the open balls of $\mathbb{R}$. The open sets of $[a,b]$ are exactly the intersections of open sets of $\mathbb{R}$ with $[a,b]$. Feb 28, 2015 at 21:44
What you’re missing is that in the space $A$ the points $a$ and $b$ are interior points of $A$, even though they are not interior points of $A$ in the space $\Bbb R$. Let $r=b-a$. By definition the open ball of radius $r$ centred at $a$ in the space $A$ is $\{x\in A:|x-a|<r\}$, the set of points in $A$ that are less than distance $r$ from $a$. And
$$\{x\in A:|x-a|<r\}=[a,b)\subseteq A\;,$$
so $a$ is in the interior of $A$ in the space $A$: the open ball of radius $r$ around $a$ is contained in $A$.
You can do the same thing at $b$.
• Wonderful! This is sorta what I was expecting the answer to be, but you made it very explicit. Thank you! Feb 28, 2015 at 22:00
• @Silmaril89: You’re welcome! Feb 28, 2015 at 22:02
By definition, $A$ is an open (and also a closed) subset of the metric space $A$ (endowed with a topology). This is one of the axioms defining a topology.
Now, if you look at a small open ball (in $A$) centered on $a$, it will be included in $A$. The reason is that such open balls will be of the form $(a-\epsilon,a+\epsilon)\cap A = [a,a+\epsilon)$. So $a$ is an interior point: there is no inconsistency between the general definition of the topology and the specific (equivalent) definition for metric spaces.
Given a non-empty set $A$ endowed with a metric $d$, we define the metric space as $X:=(A,d)$.
For each $p\in A$, the open ball centred at point $p$ with radius $r$ is given by $N_p(r):=\{ q\in A:d(p,q)<r,r\in\mathbb{R}_{>0} \}\subset A$, implying that $A$ is open relative to itself. | HuggingFaceTB/finemath | |
## Introduction
Artificial intelligence (AI) is gradually overshadowing traditional physics-based approaches1,2, achieving breakthroughs in solving some of the most challenging problems in chemistry and physics. For example, a deep neural network has obtained nearly exact solutions of the electronic Schrödinger equation for small molecule3. Another recent breakthrough is the prediction of three-dimensional structures of proteins by neural network-based methods, Alphafold4 and RoseTTafold5. With problems facing structured proteins being solved by these and other AI-based methods6,7,8,9, a new frontier is now presented by intrinsically disordered proteins (IDPs). Instead of adopting well-defined three-dimensional structures, IDPs readily access vast conformational space. Here we report on the development of a generative AI model to mine the conformational space of IDPs.
IDPs, accounting for 30% to 50% of proteomes, perform many essential cellular functions including signaling and regulation, and are implicated in numerous human diseases10,11. In particular, polyglutamine expansion is associated with Huntingtin’s and other diseases12. Amyloid-beta peptides, including Aβ40, are linked to Alzheimer’s disease13. The cell division machinery of Mycobacterium tuberculosis, the causative agent of tuberculosis, contains a number of membrane proteins, including ChiZ, with disordered cytoplasmic regions14,15. The functional and disease mechanisms of these and other IDPs remain unclear, in large part because we lack knowledge of their conformational ensembles in various states (e.g., in isolation, in aggregation, and bound with interaction partners).
The vastness of IDPs’ conformational space poses great challenges. Experimental techniques are limited to probing some aspects of the conformational space. For example, small-angle x-ray scattering (SAXS) provides information on the overall shapes and sizes of IDPs16, whereas NMR properties, such as secondary chemical shifts, carry residue-specific information but still vastly under-represent the degrees of freedom of IDPs17. Molecular dynamics (MD) simulations offer an attractive approach for IDPs, with an atomic representation for each conformation, but the simulation time that can be presently achieved, which directly determines the extent of conformation sampling, is largely limited to 10 s of μs. The conformational ensembles of the 64-residue cytoplasmic disordered region of ChiZ (referred to simply as ChiZ hereafter) sampled by multiple replicate simulations, totaling 36 μs in solution and 38 μs at membrane, have been validated by SAXS and NMR data14,15. While we cannot answer whether 10 s of μs of simulations are really long enough, we do know that shorter simulations are insufficient. For example, Kukharenko et al.18 have shown that the conformations of a 22-residue fragment of α-synuclein sampled in 1 μs represent only a small subset of the ensemble collected from 13 μs of “expansion” simulations. The latter are a large number (200) of short simulations (30–100 ns) started from sparsely populated regions in a two-dimensional embedded space (via sketch-map embedding). How to exhaustively cover the conformational space of IDPs without an inhibitory amount of computational time remains an open question.
For structured proteins, autoencoders have been developed to represent structures in two-dimensional latent spaces and reconstruct the structures back in Cartesian coordinates6,8. In another recent study9, an autoencoder was trained to project the inter-residue distances of the ribose-binding protein into a two-dimensional latent space. The open and closed states of the protein were found to occupy separate regions in the latent space. The authors linearly interpolated points from these two states and decoded the interpolated points into inter-residue distances that represent conformations on the transition paths between the open and closed states. The inter-residue distances from interpolation were finally coupled to an all-atom model to enhance the latter’s conformational sampling in MD simulations. Noé et al.7 built Boltzmann generators, which use neural networks to represent protein structures sampled from short MD simulations as a Gaussian distribution in the latent space; points sampled from the Gaussian are transformed back as structures in Cartesian coordinates. In toy problems, the authors demonstrated that points located in different energy wells in conformational space are repacked into a dense distribution with a single peak in the latent space. These and other AI-based methods might potentially be adapted to study IDPs19. Several other approaches may also provide inspirations for IDPs, including variational autoencoders for dimensionality reduction of protein folding trajectories and subsequent identification of intermediate states by clustering in the latent space20, and variational autoencoders and other neural networks for optimal selection of Markov states by training with conformations at a fixed lag time21,22.
Here we present generative autoencoders designed to mine the conformational space of IDPs. Our design goal is to accurately sample the entire conformational space while limiting cost, which is MD conformations needed for training the autoencoders. The performance of the resulting autoencoders rivals that of expensive MD simulations and is validated by SAXS and chemical shift data. Our work opens the door to modeling IDPs in various functional states.
## Results
We built autoencoders first to represent IDP conformations as vectors in a reduced-dimensional latent space (Fig. 1a). Training of the autoencoders involved reconstructing the conformations from the latent vectors and minimizing deviations from the original conformations. The training datasets consisted of conformations sampled from short MD simulations. We then modeled the latent vectors of the training datasets as multivariate Gaussian distributions (Fig. 1b). By sampling from these distributions for reconstruction, we generated the full conformational ensembles of IDPs (Fig. 1c). These generative autoencoders were built for three IDPs: polyglutamine Q15, Aβ40, and ChiZ, and were validated by their ability to cover all conformations sampled in long MD simulations and to reproduce experimentally measured properties. These IDPs contain 17 (including two capping groups), 40, and 64 residues (denoted by Nres).
Note that our goal is to use the smallest amount of training data – sampled from MD simulations as short as possible – to build autoencoders that will generate the most accurate full conformational ensemble of an IDP. To achieve this goal, we limit the training dataset to conformations sampled from the initial portion of the MD simulations, and use the subsequent portion only for the purpose of testing the accuracy of the autoencoders. Although increasing the training size or including conformations randomly sampled anywhere from the simulations, such as by shuffling the MD conformations before separating them into training and test sets, can potentially increase the accuracy of the autoencoders, doing so will depart from our goal.
### Representation in a reduced-dimensional space
As a steppingstone to generating new conformations, we first reduced the dimensionality of the conformational space. The original conformations of the IDPs were specified by the Cartesian coordinates of heavy atoms (with truncation of some side chains). The dimension of the conformational space was thus 3 N, where N, denoting the number of heavy atoms included, was 140, 230, and 385, respectively, for Q15, Aβ40, and ChiZ. We chose the dimension (n) of the latent space for each IDP to be 0.75Nres, or 13 for Q15, 30 for Aβ40, and 48 for ChiZ.
Conformations for training and testing the autoencoders came from multiple μs-long MD simulations14,23. We collected 95,000, 140,000, and 145,000 frames, respectively, at 10 ps intervals for Q15 and 20 ps intervals for Aβ40 and ChiZ from each replicate run; the numbers of replicate runs were 2, 4, and 12, respectively. An initial portion (e.g., 10%) of each run was taken as a training set whereas the remaining portion was the test set. The accuracy of an autoencoder was assessed by the root-mean-square deviations (RMSDs) between test conformations and their reconstructions. These RMSDs were averaged for the entire 100-fold diluted test set (comprising frames saved at 1-ns intervals for Q15 and 2-ns for Aβ40 and ChiZ). As adjacent frames in MD simulations tend to have similar three-dimensional structures, the dilution serves to reduce redundancy of the test set. The reconstruction RMSD results are shown in Fig. 2.
For Q15, the average reconstruction RMSDs are below 5 Å even when only 5% of the MD simulations (corresponding to 95 ns of simulation time) is used for training (Fig. 2a). When the training size is increased to 10% and 20%, the RMSDs stay around 4.75 Å for run1 but decrease successively from 4.96 Å to 4.73 Å and 4.43 Å for run2. This decrease in reconstruction RMSD with increasing training size is likely because run2 was started from an all α-helical conformation, which mostly melted away over time (Fig. S1a). For Q15, we chose autoencoders trained at the 10% size for generating new conformations.
For Aβ40, training with the first 10% of the MD simulations results in reconstruction RMSDs of 6.4 ± 1.3 Å (mean ± standard deviation among four MD runs) (Fig. 2b). The reconstruction RMSDs decrease to 6.0 ± 1.4 Å with a 20% training size and further to 5.4 ± 1.1 Å with a 30% training size. The higher RMSD of run2 is probably due to more compact initial conformations (Fig. S1b). For this IDP we chose a 20% training size for generating new conformations.
Reconstruction becomes more challenging as the IDP size increases. This is already apparent when Aβ40 is compared to Q15, and is much more so for ChiZ, where training with 10% of the MD simulations results in reconstruction RMSDs at 8.3 ± 1.1 Å for 10 of the 12 MD runs, and >10 Å for the other two runs (Fig. 2c). Still, the reconstruction RMSDs decrease to 7.4 ± 1.3 Å with a 20% training size and further down to 6.4 ± 1.0 Å with a 30% training size. For this larger IDP, we chose 30% training size (corresponding to 870 ns of simulation time) for generating new conformations.
To check whether the dimensions of the latent space chosen according to 0.75Nres were adequate, we trained autoencoders with a 200-dimentional latent space. The reconstruction RMSDs improve for Q15 and Aβ40, but not for ChiZ (Fig. S2). So increasing the latent-space dimension does not necessarily improve accuracy, especially for the larger, more challenging IDPs, in reconstruction (or in generating new conformations; see below).
We tested autoencoders where the input was dihedral angles or distance matrices instead of Cartesian coordinates. The performance of these models in reconstruction was much worse than that with Cartesian coordinates as input (Supplementary Note 1).
### Multivariate Gaussian models in latent space
The conformational ensembles of IDPs are broad and difficult to model14. A possible crucial benefit of representing the conformations in the latent space is that, due to the reduced dimensionality, the distribution of the latent vectors would be more compact and therefore easier to model. To assess this expectation, we calculated histograms in two-dimensional subspaces of the latent space. For each autoencoder, about one half of the encoder output values were consistently at or near zero, thereby further reducing the effective dimension of the latent space. We only calculated histograms for pairs of nonzero output neurons.
For the run1 training set of Q15, only 7 of the 13 output neurons were nonzero, resulting in 21 possible pairs. In Fig. S3, we display the histograms of 10 pairs calculated for the training (10% size) and test datasets. These histograms are indeed compact. Moreover, the counterparts of the training and test sets look very similar, with only minor differences for one or two pairs. For example, in the (9, 11) pair, the histogram of the training set is somewhat broader than the counterpart of the test set. The substantial overlap between the distributions of the training and test sets in the latent space explains the good performance of the autoencoder in reconstruction.
The autoencoder for Aβ40 (run1; 20% training size) had only 15 nonzero output neurons (out of 30). Fig. 3 displays the histograms of 8 nonzero pairs. All of these are single-peaked, and the peak positions are the same for the training and test counterparts in most cases, but with some shift for the (0, 27) pair. The high-level of overlap between the training and test sets allows for the satisfactory reconstruction of Aβ40 conformations reported above. In comparison, for the larger ChiZ, the histograms representing conformations sampled from a single MD run (run1) become irregular in shape (e.g., the (38, 39) pair) and the divergence between the training and test sets becomes prominent (e.g., the (15, 16) and (44, 47) pairs) (Fig. S4). These features exhibited by the distributions in the latent space illustrate the growing difficulty in reconstructing the conformations of larger IDPs.
The compact distributions of Q15 and Aβ40 in the latent space motivated us to model them as multivariate Gaussians. As shown in Figs. S3 and 3, the distributions of the training sets and their multivariate Gaussian models look very similar. More importantly, the multivariate Gaussian models also overlap well with the distributions of the test sets. Indeed, the overlap between the test sets and the Gaussian models is greater than that between the test sets and the corresponding training sets, as illustrated by the (9, 11) pair of Q15 and the (0, 3) pair of Aβ40. Therefore the multivariate Gaussian models seem promising for generating new conformations that are similar to those in the test sets of Q15 and Aβ40. For ChiZ, multivariate Gaussians are inadequate to model the irregular shapes of the single-run distributions in the latent space (Fig. S4).
The foregoing qualitative observations are confirmed by calculating the Kullback-Leibler (KL) divergence between the Gaussian models and the distributions of the training and test data in the latent space (Table S1). For both Q15 and Aβ40, the Gaussian models provide good representations of the training data, with KL divergence values at or below 0.1 for all the pairs shown in Figs. S3 and 3. Moreover, for all but one pair, the KL divergence values between the test data and Gaussian models are lower than those between the training data and test data. For example, for the (9, 11) pair of Q15, the KL divergence decreases from 0.10 for training vs test to 0.06 for test vs Gaussian; for the (0, 3) pair of Aβ40, the KL divergence decreases from 0.58 for training vs test to 0.38 for test vs Gaussian. On the other hand, for ChiZ, the Gaussian model provides a poor representation of the training data, with KL divergence as high as 0.48 (for the (0, 4) pair).
### Autoencoder-generated conformations of Q15 and Aβ40
By sampling from a multivariate Gaussian in the latent space and using the decoder to reconstructing conformations, we turned the autoencoder into a generative model. The multivariate Gaussian was parameterized on the same dataset for training the autoencoder. For Q15, the training size was 9500 and the test size was 85,500. The size of the generated set was measured as multiples of the test size (1× = 85,500). For each conformation in the 100-fold diluted test set, we found its best match (i.e., lowest RMSD) in the generated set. We then used the average of the best-match RMSDs for the diluted test set as the measure for the accuracy of the generated set. With the generated sets at size 1×, the average best-match RMSDs of the test sets are 3.59 and 3.58 Å for MD run1 and run2, respectively. As illustrated in the inset of Fig. 4a, a test conformation and its generated best match at 3.58 Å RMSD show very similar backbone traces. Since generating new conformations by the autoencoder is extremely fast, the generated set can be easily expanded. With expanding sizes of the generated set, the average best-match RMSDs show small but systematic decreases, to 3.55 Å at 2×, 3.52 Å at 3×, and 3.51 Å at 4× for run1 (Fig. 4a). The improvement in RMSD occurs because the expanded size of the generated set yields better matches for the test conformations. Conversely, the average best-match RMSDs increase to 3.64 Å when the size of the generated set is reduced to 0.5× and further to 3.79 Å when the generated set is reduced to the same size as the training set (at 0.11×).
High accuracy is also achieved for generated conformations of Aβ40 on autoencoders trained with 20% (=28,000 conformations) of MD simulations (Fig. 4b). With the size of the generated sets at 1× (=112,000 conformations), the average best-match RMSDs of the 100-fold diluted test sets are 5.60 Å, 7.50 Å, 5.88 Å, and 5.84 Å, respectively, for MD run1 to run4. A test conformation and its generated best match at 5.56 Å RMSD show very similar backbone traces (Fig. 4b, inset). The higher average RMSD of the autoencoder for run2 in generating new conformations mirrors the poorer performance of this autoencoder in reconstruction (Fig. 2b), and can also be attributed to the overly compact conformations in the training set of this MD run (Fig. S1b). With an expansion of the generated set, the average best-match RMSD shows a slight decrease, to 5.42 Å at 4× for run1 (Fig. 4b). Conversely, the average best-match RMSD increases to 5.72 Å at 0.5× and to 5.86 Å at 0.25× (=size of the training set).
### Autoencoder-generated conformations of ChiZ
We first used a similar protocol to train and test an autoencoder for ChiZ on a single MD run (run1). The training size was 30% or 43,500 and the test size was 101,500. With the generated set at size 1× (=101,500 conformations), the average best-match RMSD of the 100-fold diluted test set is 7.95 Å (Fig. S5a). Again the RMSD decreases slightly with expanding sizes of the generated set, but is still 7.35 Å even at size 12× (=1.2 million conformations). The high RMSD of the autoencoder trained on a single MD run is presaged by the inadequate modeling of the training data by a multivariate Gaussian in the latent space (Fig. S4 and Table S1). One idea for improving the modeling is to represent the training data in the latent space by a mixture of multiple Gaussians. We tested this idea (Supplementary Note 2). The multiple-Gaussian model indeed improves the representation of the training data, but actually does worse in predicting the test conformations. For example, with 8 Gaussians, the best-match RMSD of a generated set at size 1× increases from 7.95 Å to 8.50 Å. In essence, as the model tries to fit into the details of the training data, its ability to capture generic features shared by the test data suffers.
It is possible that a single MD run may mine a limited region in conformational space, but the regions mined by different MD runs may partially overlap and the combined mining may generate an ensemble that is densely distributed in the latent space. Indeed, when we combine the conformations from 12 MD runs for ChiZ, the histograms in the latent space for both the training set and the test set become compact and have a single peak for all but one (i.e., (9, 14)) of the nonzero pairs (Fig. 5a). The distributions of the training and test latent vectors overlap very well and are also modeled well by the multivariate Gaussian parameterized on the combined training set. The KL divergence values for training vs Gaussian, test vs Gaussian, and training vs test are all lower than 0.1 for all the pairs (Table S1); the value for training vs Gaussian is only 0.079 even for the (9, 14) pair.
The increase in overlap by combining data from multiple MD runs pointed a way to improve autoencoders. As an initial test, we pooled the generated conformations (each at size 1×) from the autoencoders of the individual MD runs. When compared with this pooled generated set (total size at 12×), the average best-match RMSD of the run1 test set is 7.04 Å (Fig. S5b), which is lower by 0.31 Å than the corresponding value when the generated set is at the same 12× size but produced solely by the run1 autoencoder (Fig. S5a). To take full advantage of the multiple MD runs of ChiZ, we used the autoencoder trained on the combined training set (a total of 52,200 conformations after a 10-fold dilution) to generate new conformations. The generated set at size 1× now gives a best-match RMSD of 7.32 Å for the 1000-fold diluted, combined test set (final size = 1218). When the generated set is expanded to a size 10×, the best-match RMSD reduces to 6.70 Å (Fig. 5b). The inset illustrates a pair of conformations, one from the test set and one from the generated set, at this RMSD.
### Optimum selection of training sizes and latent-space dimensions
In Supplementary Note 3, we present additional data for the effects of varying training size and latent-space dimension on the accuracy of autoencoders in generating new conformations. In short, the selected training sizes, 10%, 20%, and 30% respectively, for Q15, Aβ40, and ChiZ, are sufficient for model convergence; additional training data do not yield appreciable gains in model accuracy, especially given that we put a premium on cost control of MD simulations. We selected 0.75Nres as the latent-space dimension. Increasing the latent-space dimension by 10-30 has little effect on model accuracy. For Q15, a very large value, 200, for the latent-space dimension actually leads to slight increases in the best-match RMSDs of generated conformations (Fig. S6, compared with Fig. 4a).
### Further assessment of generated conformations
To properly benchmark the autoencoder-generated conformations, we examined the diversity of the test sets and the similarity between the training and test sets (Table S2). We calculated the RMSDs of each conformation with all others in a diluted test set. The average pairwise RMSDs are quite high even within a single MD run (run1), 6.98 Å for Q15, 11.61 Å for Aβ40, and 18.21 Å for ChiZ, showing that the conformations in each test set are very diverse. As expected, the average pairwise RMSD increases further, to 19.23 Å, for the combined and further diluted test set of ChiZ. The diversity of the test conformations again illustrates the challenge in generating conformations that are close to them.
The neighboring conformations in any MD run have relatively low RMSDs, leading to small best-match RMSDs between conformations in the test sets from single MD runs. The average best-match RMSDs in run1 are 3.71 Å for Q15, 3.83 Å for Aβ40, and 4.83 Å for ChiZ (Table S2). However, for the combined and further diluted test set of ChiZ, the average best-match RMSD increases to 8.62 Å. The latter value may be viewed as a benchmark for generated conformations to be claimed as neighbors of test conformations. Because the average best-match RMSD for the combined test set against the generated set (at size 10×) is 6.70 Å, or nearly 2 Å below the benchmark, we can claim that all the test conformations in the combined test set have neighbors in the generated set. In other words, the generated set fully covers the combined test set.
Another benchmark is given by the average best-match RMSD between a test set and the corresponding training set. For run1, values of this benchmark are 3.96 Å for Q15, 6.76 Å for Aβ40, and 10.17 Å for ChiZ (Table S2). When the comparison is against the generated sets at the sizes of the training sets (shown as the first point in Figs. 4a, b, and S5a), the average best-match RMSDs are 3.79, 5.86, and 8.16 Å, respectively, each of which is lower than the counterpart when the comparison is against the training set itself. That is, relative to the training sets, the generated sets provide better matches for the test sets. For Q15 and Aβ40, this outcome is to be expected because of the above observation that the test sets overlap better with the Gaussian models than with the training sets (Figs. S3 and 3; Table S1). For ChiZ, the combined test set from the 12 MD runs has a best-match RMSD of 8.47 Å against the combined training set, which is 1.7 Å lower than the counterpart for the comparison within run1. This decrease in best-match RMSD confirms the aforementioned increase in data overlap when multiple MD runs are combined (Figs. S4 and 5a). Moreover, the best-match RMSD of the combined test set further reduces to 7.51 Å when the generated set is of the same size as and parameterized on the combined training set (first point in Fig. 5b).
We also inspected more closely the generated conformations that best match test conformations (insets in Figs. 4a, b, and 5b). As already alluded to, test conformations and their generated best matches show overall similarities in shape and size. However, the generated conformations have considerable bond length and bond angle violations. Refinement by energy minimization restores essentially all bonds and angles to proper values (Fig. 6). The refinement results in small increases in RMSD for the best-matched test conformations, though an occasional decrease in RMSD is possible. For the pairs of conformations shown in the insets of Figs. 4a, b, and 5b, the RMSDs change from 3.58 Å to 3.45 Å, from 5.56 Å to 5.95 Å, and from 6.67 Å to 6.87 Å, respectively (Fig. 6). For the generated set of ChiZ at size 1×, the best-match RMSD increases from 7.32 Å to 7.66 Å upon conformational refinement.
### Experimental validation of autoencoder-generated ChiZ conformational ensemble
To objectively assess the quality of the autoencoder-generated conformational ensemble, we calculated from it properties that can be measured experimentally. These include SAXS profile and NMR chemical shifts. In Fig. 7, we compare the experimental data for ChiZ14 with results calculated from 12,180 conformations collected from the combined test set of the 12 MD runs, and with results calculated from 12,180 conformations generated by the autoencoder trained on the combined training set. As reported previously14, the MD simulations reproduced both types of experimental data well: there was very good agreement for the SAXS profile over the entire q (momentum transfer) range from 0 to 0.5 Å−1, with a mean absolute percentage error (MAPE) of 3.9%; likewise the calculated secondary chemical shifts were close to the experimental values, with a root-mean-square error (RMSE) of 0.43 ppm. The experimental SAXS profile is also reproduced well by the generated conformations, with an MAPE of 7.2%, validating the latter’s sampling of the overall shape and size of ChiZ, though some deviations are seen at the high q end. For secondary chemical shifts, the RMSE increases to 0.63 ppm for the generated conformations. This RMSE is at the low end of the range of RMSEs (0.63 to 0.84 ppm) calculated on conformations from MD simulations using four other force fields14. Autoencoders trained on conformations from these other force fields have similar performances as the one reported above for ChiZ, demonstrating the robustness of the approach (Supplementary Note 4).
## Discussion
We have developed generative autoencoders to mine the broad conformational space of IDPs. These autoencoders can not only represent IDP conformations in the latent space with high fidelity to allow for accurate reconstruction, but also generate new conformations to fill up the conformational space. The generated ensemble contains close matches for all the conformations sampled in long MD simulations, but with negligible computational time. For example, sampling 100,000 conformations (at 20 ps intervals) from MD simulations of Aβ40, even with GPU acceleration24, takes 80 days, whereas our autoencoder generates the same number of conformations in 12 sec. In the case of ChiZ, the autoencoder-generated conformations even yielded better predictions for SAXS profile and chemical shifts than MD simulations with several force fields.
Our generative autoencoders have the flavor of variational autoencoders but are more intuitive. Rather than optimizing Gaussians in the latent space during the training process as in variational autoencoders, we only optimize reconstruction and then use the latent vectors of the training set to calculate the mean vector and covariance matrix, which are directly used to define a multivariate Gaussian for generating new conformations. We have shown that the difficulty posed by the longer sequence length of ChiZ can be overcome by training on data sampled from multiple MD runs. As the lengths of IDPs increase, the problem becomes even more challenging. One possible way to address this challenge is to break a long IDP into fragments and treat each fragment as a separate IDP. However, IDPs do form occasional long-range contacts14,25. The influence of long-range contacts has to be somehow taken into consideration.
The generative autoencoders designed here are for mining the conformational space of IDPs in isolation. The power of this approach demonstrated here suggests that it can be extended to study IDPs in more complex functional states, such as when bound to or associated with an interaction partner (a target protein or a membrane), or in aggregation. For example, ChiZ associated with acidic membranes has been studied by long MD simulations15; generative autoencoders may also be able to mine the conformational space of membrane-associated IDPs. IDPs are prone to phase separation26, resulting in a highly concentrated phase surrounded by a dilute phase. Microsecond-long MD simulations failed to sample the equilibration between the two phases27. AI-based models such as generative autoencoders may open the door to solving this and other challenging conformational mining problems for IDPs.
## Computational methods
### Autoencoder design
We built and trained the autoencoders using the Keras package (https://keras.io/) with TensorFlow backend (https://www.tensorflow.org/) in Python 3.628. The autoencoders consisted of an encoder and a decoder. Both the encoder and decoder had a dense neural network architecture, with two hidden layers of 300 and 50 neurons, respectively. The input, hidden, and output layers of the encoder and decoder were arranged as mirror images of each other (Fig. 1a). This arrangement was chosen based on its reduced training complexity as shown in previous reconstruction work on structured proteins6. All layers except for the final output layer had a rectified linear unit activation function; the final output layer had a sigmoidal activation function.
The input to the encoder consisted of the Cartesian coordinates of an IDP. Only heavy atoms (all for the backbone and selected for side chains) were included; selected side-chain atom types were CB, CG, CD, OE1, and NE2. This selection contained all the heavy atoms of polyglutamine Q15, but truncated some of the side chains in Aβ40 and ChiZ. Q15, Aβ40, and ChiZ had N = 140, 230, and 385 heavy atoms, respectively, for a total of 3 N input coordinates. The loss function was the binary cross-entropy,
$$H\Big(\left\{y_{i}\right\},\left\{{y_{i}}^{\prime} \right\}\Big)=\frac{1}{3N}\mathop{\sum }\limits_{i=1}^{3N}\Big[-{y}_{i}\;{{{{{\rm{ln}}}}}}\;{{y}_{i}}^{\prime} -\left(1-{y}_{i}\right){{{{{\rm{ln}}}}}}\left(1-{y}_{i}^{{\prime} }\right)\Big]$$
(1)
where $$\left\{{y}_{i}\right\}$$ denotes the 3 N input Cartesian coordinates of the IDP after a linear transformation into the range between 0 and 1 (see below), and $$\left\{{y}_{i}{\prime} \right\}$$ denotes the values of the corresponding output neurons. The neural networks were trained by the Adam optimizer given its effectiveness in handling large datasets. For each autoencoder, training was done for 100 epochs using a batch size of 40. Using the mean square error as the loss function produced very similar accuracy in generating new conformation (Supplementary Note 5).
The latent space dimension and training size were tested based on reconstruction, which entailed encoding (i.e., representing the conformations as vectors in the latent space) and then decoding (i.e., constructing back full conformations from the latent vectors). The dimensions of the latent spaces for the three IDPs were finally chosen as n = 13, 30, and 48. Parameters of autoencoders trained on reconstruction were saved in decoder and encoder files, and the decoder was then used to generate new conformations.
### Molecular dynamics simulations
Two 1 μs trajectories (100,000 frames each, saved at 10 ps intervals) for Q15, taken from Hicks and Zhou23, were run at 298 K in GROMACS with the AMBER03ws force field29 for protein and TIP4P2005 for water30. These simulations were performed using an enhanced sampling method called replica exchange with solute tempering31,32 at constant volume and temperature, with temperature regulated by velocity rescaling33. The simulations were judged to be well equilibrated, as shown in particular by the agreement in the distribution of radius of gyration with simulations using a second enhanced sampling method, i.e., temperature replica exchange34.
For ChiZ, 12 trajectories of 3 μs each (150,000 frames, saved at 20 ps intervals), taken from Hicks et al.14, were run on GPUs using pmemd.cuda24 in AMBER1835 with the ff14SB force field36 for protein and TIP4PD37 for water. These simulations were performed at constant temperature (300 K) and pressure 1 atm), with temperature regulated by the Langevin thermostat (damping constant at 3 ps−1)38 and pressure regulated by the Berendsen barostat (pressure relaxation time at 2 ps)39. These simulations were thoroughly validated by experimental data including SAXS, chemical shifts, and NMR relaxation properties. Additional simulations were performed using four other protein/water force field combinations, including AMBER03ws/TIP4P2005, AMBER99SB-ILDN40/TIP4PD, AMBER15IPQ/SPCEb41, and CHARMM36m/TIP3Pm42. The protocol for ChiZ was used to run four replicate simulations of Aβ40 at 278 K (3.5 μs each; 175,000 frames saved at 20 ps intervals)25. Again the simulations were thoroughly validated by experimental data including chemical shifts and NMR relaxation properties.
### Data preprocessing
MD trajectories in GROMACS and AMBER trajectory formats were first converted to conformations in PDB format (with solvent stripped). An initial portion of each trajectory (5000, 35000, and 5000 frames for Q15, Aβ40, and ChiZ, respectively) were removed. The remaining trajectory was split into two parts, the first (e.g., 10%) as the training dataset and the second as the test dataset.
The Biobox library in Python (https://github.com/degiacom/biobox)6 was used to preprocess the coordinates in each dataset. All the frames were aligned to the first one according to RMSD, and shifted to have all coordinates positive. Coordinates were then scaled between 0 and 1 (via dividing by the maximum coordinate value) for using as input to the encoder. The output coordinates of the decoder were scaled back to real coordinates using the same scaling factor. The choice of the reference frame for the structural alignment before shifting and scaling the coordinates had no effect on the accuracy in generating new conformations (Supplementary Note 5).
### RMSD calculation
We used a code of Ho (https://boscoh.com/protein/rmsd-root-mean-square-deviation.html) to calculate RMSDs of output conformations. A custom Python code (https://github.com/aaayushg/generative_IDPs/tree/main/RMSD) was written to find the lowest RMSD between a given test conformation against a set of generated conformations, and calculate the average of these best-match RMSDs for the test set (100-fold diluted).
### Generating new conformations
The mean vector $${{{{{\boldsymbol{\mu }}}}}}$$ with elements
$${\mu }_{l}={ < {z}_{l} > }_{{{{{{\rm{training}}}}}}}$$
(2)
and covariance matrix $$\widetilde{\sigma }$$ with elements
$${\sigma }_{{lm}}={ < ({z}_{l}-{\mu }_{l})({z}_{m}-{\mu }_{m}) > }_{{{{{{\rm{training}}}}}}}$$
(3)
were calculated from the latent vectors, $$\left\{{z}_{l}\right\}$$, of the training dataset; here $${ < \cdots > }_{{{{{{\rm{training}}}}}}}$$ denotes an average over the training set. The latter two quantities in turn defined a multivariate Gaussian distribution (Fig. 1b),
$$Q({{{{{\bf{z}}}}}})=\frac{1}{\sqrt{{(2\pi )}^{n}{{\det }}\widetilde{\sigma }}}{e}^{-{({{{{{\bf{z}}}}}}-{{{{{\boldsymbol{\mu }}}}}})}^{{{{{{\rm{T}}}}}}}\cdot \widetilde{\sigma }\cdot ({{{{{\bf{z}}}}}}-{{{{{\boldsymbol{\mu }}}}}})}$$
(4)
from which vectors were sampled and fed to the decoder to generate new conformations (Fig. 1c). In the above, det represents determinant of a matrix, and the superscript “T” signifies transpose. Sampling from multivariate Gaussians was implemented using the NumPy library (https://numpy.org/) in Python. Histograms were calculated in two-dimensional subspaces of the latent space, for qualitative comparison among the training, test, and multivariate Gaussian datasets (https://github.com/aaayushg/generative_IDPs/tree/main/Plot_histogram).
We used the Kullback-Leibler divergence
$${D}_{{{{{{\rm{KL}}}}}}}(p{{{{{\rm{|}}}}}}q)=\iint {{dz}}_{l}d{z}_{m}p({z}_{l},{z}_{m})\;{{{{{\rm{ln}}}}}}\;\frac{p({z}_{l},{z}_{m})}{q({z}_{l},{z}_{m})}$$
(5)
to quantify the difference between two distributions, $$p({z}_{l},{z}_{m})$$ and $$q({z}_{l},{z}_{m})$$, in a two-dimensional subspace of the latent space. $$p({z}_{l},{z}_{m})$$ and $$q({z}_{l},{z}_{m})$$ are proportional to the histograms but are normalized. The integral was evaluated as a summation over the two-dimensional grid over which the histograms were calculated (see, e.g., Fig. 3). For any grid point where either $$p({z}_{l},{z}_{m})$$ or $$q({z}_{l},{z}_{m})$$ was 0, the contribution from that grid point to $${D}_{{{{{{\rm{KL}}}}}}}({p|q})$$ was set to 0.
### Refinement of autoencoder-generated conformations
The generated conformations had considerable bond length and bond angle violations. We used a simple procedure to remedy this problem. First all the missing heavy and hydrogen atoms were added in each structure using tleap in AmberTools35. Then the structure was subject to 500 steps of conjugate-gradient energy minimization in vacuum using NAMD 2.1343. The protein force field was AMBERff14SB36.
### Calculation of SAXS profile and chemical shifts for ChiZ
The SAXS profile for each conformation was then calculated using FoXS44 and scaled to optimize agreement with the experimental profile14. Chemical shifts were calculated using SHIFTX245 (www.shiftx2.ca). Chemical shifts for random-coil conformations calculated using POTENCI46 were subtracted to obtain secondary Cα and Cβ chemical shifts. SAXS profiles and chemical shifts were averaged over all the conformations in the diluted test set (12180 frames from 12 trajectories) or a generated set (of the same size). For the latter, the conformations after refinement by energy minimization were used.
### Reporting summary
Further information on research design is available in the Nature Research Reporting Summary linked to this article. | open-web-math/open-web-math | |
# 7) How will this new monthly mortgage repayment affect their loan affordability ratio? That is, will they be facing mortgage stress?
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Background
Kyle and Julia are considering whether or not to buy a particular property valued at \$720,000. They have \$200,000 of their own funds to commit towards the purchase and they expect to incur an additional \$50,000 in fees and taxes on the purchase itself. They are able to borrow at an interest rate of 7.20 per cent per annum with interest compounded monthly. Loan repayments would be monthly with the first payment due at the end of the first month after purchasing the property. The term of the home loan is 30 years. They both work full-time earning a combined after-tax salary of \$12,000 per month.
1. How much is the monthly mortgage payment Kyle and Julia will be required to pay for their loan?
Hint: because the interest is compounded monthly, we need to use the number of months for the mortgage loan, not the number of years in determining the regular payments to be made. We also need to use the monthly interest rate (7.2% /12 = 0.6%)
Using the present value of an annuity formula we can rearrange the following formula to obtain the regular monthly mortgage payment PMT.
PV = PMT
PMT = PV /
Alternatively, you can use a financial calculator to determine PMT.
PV = amountborrowed from the bank
r = interest rate per month
n = number of months over the entire loan period
If we select the payment key (PMT) on the financial calculator, we will know the amount of the regular monthly mortgage payments the bank requires us to pay.
2) A loan affordability ratio is equal to the monthly home loan repayment divided by a couple’s household after-tax monthly income. A key threshold for ‘mortgage stress’ is when the loan affordability ratio reaches 35%. Will Kyle and Julia face mortgage stress at current interest rates.
3) After 1 year, the bank informs Kyle and Julia that \$564,429 is still owing on their loan. How much in total have Kyle and Julia paid in mortgage payments during the first year?
4) Of the repayments, how much of their principal has been reduced?
5) How much interest have they paid in year 1?
6) If the bank now increases interest rates from 7.20 per cent to 8.40 per cent, what will Kyle and Julia’s new monthly mortgage repayments be?
Hint: remember to use monthly interest rates. And, remember that there are only 29 years left on the loan (use months not years).
7) How will this new monthly mortgage repayment affect their loan affordability ratio? That is, will they be facing mortgage stress?
###### Price: £129
100% Plagiarism Free & Custom Written - Tailored to Your Instructions | HuggingFaceTB/finemath | |
Algebra
# Partial Fractions - Linear Factors
If the following is an identity in $x$: $\frac{A}{(x-2)(5x-8)}=\frac{1}{x-2}-\frac{B}{5x-8},$ what is the value of $A+B?$
How many terms would there be in the partial fraction decomposition of
$\frac{1}{ ( x- 5) ( x + 5) ( x + 9 )}?$
If the following is an identity in $x$:
$\frac{11x+50 }{(x+2)(x+6)} = \frac{A}{x+2} + \frac{B}{x+6},$
what is the value of $A + B?$
Which of the following is the correct partial fraction decomposition of
$\frac{ x+1 } { ( x + 2) ( x + 3) } ?$
If the following is an identity in $x$: $\frac{5x-6}{x^2-3x+2}=\frac{a}{x-1}+\frac{b}{x-2},$ what is the value of $a \times b$?
× | HuggingFaceTB/finemath | |
Classifying irreducible subspaces for angular momentum according to symmetrization
I previously asked Proof that $L=S=0$ for filled electron subshells? which motivated me to look more deeply into the restrictions the Pauli exclusion principle places on multi-particle angular momentum states.
It's well known that 2 distinguishable spin-1/2 particles could occupy up to 4 different states: $$|++\rangle, |--\rangle, |+-\rangle, |-+\rangle$$
Or, in the angular momentum coupled basis:
\begin{align} |1, 1\rangle =& |++\rangle\\ |1, 0\rangle =& \frac{1}{\sqrt{2}}\left(|+-\rangle + |-+\rangle\right)\\ |1, -1\rangle =& |--\rangle\\ |0, 0\rangle =& \frac{1}{\sqrt{2}}\left(|+-\rangle - |-+\rangle\right) \end{align}
Here the top three states represent the total spin-1 triplet and the bottom state represents the spin-0 singlet. Notably, the triplet states a symmetric with respect to particle exchange while the singlet states are anti-symmetric with respect to particle exchange. Again, for distinguishable particles all 4 states are allowed. However, Fermions must respect the Pauli exclusion principle which says that the multi-particle state must be anti-symmetric with respect to particle exchange.
Thinking about this some more. Suppose we have, now, $$N$$ fermionic particles, each with intrinsic spin 1/2 and total orbital angular momentum $$l=1$$. This will be to build up to 6 electrons filling the $$p$$ shell. The single particle Hilbert space is then $$\mathcal{H}_i$$ and the total Hilbert space is
\begin{align} \mathcal{H} = \bigotimes_{i=1}^{N} \mathcal{H}_i \end{align}
The dimension of the single particle Hilbert space is $$3\times 2 = 6$$. The dimension of the multiparticle space is $$6^N$$. For $$N$$ from 1 to 6 this gives $$\text{dim} = \left\{6, 36, 216, 1296, 7776, 46656\right\}$$
However, for identical fermions the multiparticle Hilbert space is now the alternating tensor product of the single particle Hilbert space. The states are slater determinants of single particle states. My understanding is that the dimension of this space is given by
$$_6 C_N = \frac{6!}{N!(6-N)!}$$
Since you must choose $$N$$ unique states from the set of 6 available single particle sates. This leads to the dramatically reduced dimensionalities of $$\text{dim}=\left\{6, 15, 20, 15, 6, 1 \right\}$$ for $$N$$ from 1 to 6.
I know immediately for $$N=1$$ that this Hilbert space decomposes to a total spin 1/2 and total spin 3/4 subspace since it is composed of one spin-1 and one spin-1/2 component. However, for 2 spins it is already complicated for me to determine the total spin subspaces. The only way I would know how to do it is first decompose the 36 dimensional subspace from the distinguishable case into total spin subspaces (I could do this without too much trouble) then explicitly write down the states of those subspaces and determine which ones are anti-symmetric and remove all the others. This would be terribly time consuming and wouldn't easily generalize to larger $$N$$. Alternatively I could write down all of the anti-symmetric states available, but then it's not obvious to me how to assign spin subspaces to particular states.
My questions are as follows. They are ordered from major to minor questions
• Is there a way to know, on symmetry or group theoretic ground, the angular momentum decomposition of the anti-symmetrized space based on the angular momentum of the constituent spaces or, even, the decomposition of the distinguishable Hilbert space?
• Similar to above, is there a group-theoretic way to determine whether a particular subspace of the total distinguishable Hilbert space is anti-symmetric, symmetric, or mixed symmetry?
• when you take the antisymmetric subspace of the distinguishable multi-particle Hilbert space are you guaranteed that your states always come in complete spin subspaces? Proof for this?
• Am I correct about the dimensionality of the alternating Hilbert space?
• How to notate the antisymmetric tensor product in Latex?
Even more succinctly: I know how to decompose a multi-particle Hilbert space into irreducible representations. What is a generic procedure to sort these irreducible representations based on their symmetrization properties?
So the goal is to take a tensor product space, and figure out its rotationally closed subspaces, which are also called irreducible representations (irreps). When applied to quantum angular momenta, the rotationally closed subspace are the combinations that are eigenstates of total angular momentum.
There is a deep mathematic theory (Schur-Weyl duality) that relates these subspaces to the representations of the symmetric group (aka: the permutation group). Moreover, these representations are related to Young tableaux via the Robinson-Schensted correspondence. The representations of the symmetric group are ultimately related to the partitions of an integer, that is, how many ways the integer can be expressed as a sum of smaller (or equal) positive integers.
A literature search on the aforementioned terms will land you in graduate level mathematics that are difficult to penetrate. Here, I'll try to present a physics oriented approach that will hopefully make some of the abstract concepts a little more concrete.
You start with the fundamental representation, as a single box in a Young diagram:
╭──┐
│ │
└──┘
This represents the 2 dimensional spin up / spin down irrep.
Now take the tensor product with itself:
╭──┐ ╭──┐ ╭──┬──┐ ╭──┐
│ │ X │ │ = │ │ │ + │ │
└──┘ └──┘ └──┴──┘ ├──┼
│ │
└──┘
The tensor product is formed by combining the two diagrams on the left in all possible ways that form a legitimate Young diagram.
Each diagram on the right represents an irrep of total angular momentum. You can find the dimension of the representation by using the remarkable Hook length formula:
$${\rm dim}\,W(n, r) = \Pi_{(i,j)\in Y(n)}\frac{r+j-i}{{\rm hook}(i, j)}$$
$$n$$ is the number of boxes in the diagram and $$r$$ is the dimension of the fundamental irrep ($$r=2$$). Here $$(i, j)$$ is an integer labeling the row and column number, the product runs over all boxes in the diagram. $${\rm hook}(i, j)$$ is the hook length of the box given by:
$${\rm hook}(i,j)=1+{\rm arm}(i,j)+{\rm leg}(i,j)$$
The arm (leg) length of a box is the number of boxes to-the-right (below) the box.
Applying the hook length formula to the above equation gives:
$${\bf 2}\otimes{\bf 2}={\bf 3}_S\oplus{\bf 1}_A$$
which means combining two doublets yields a triplet and a singlet, which is what we learned in elementary quantum mechanics.
(Tensor side note: had we been using 3-vectors as our fundamental irrep, the hook length formula yields:
$${\bf 3}\otimes{\bf 3}={\bf 6}_S\oplus{\bf 3}_A$$
which tells us a cartesian tensors has a six dimensional part and a three dimensional part (that transforms like vector, aka the cross product).
In Minkowski space, it tells us 4-tensors look like:
$${\bf 4}\otimes{\bf 4}={\bf 10}_S\oplus{\bf 6}_A$$
where we recognize the 10 dimensions of the stress energy tensor $$T_{\mu\nu}$$, and the six of electromagnetism $$F_{\mu\nu}$$).)
Remarkable hook length formula, indeed.
Further consideration will show that the subspaces in this case are symmetric or antisymmetric.
Note: we combined 2 fundamental irreps. The partitions of 2 are:
$$2 = 2$$ $$2 = 1 + 1$$
Each partition corresponds to a Young diagram. Each diagram has a number of standard Young tableaux (that is, each box gets filled with a number from 1 to n such that numbers increase going to the left to right and top to bottom). Another hook length formula tells us how many standard tableaux exists for each diagram:
$${\rm dim}\pi_n = \Pi_{(i,j)\in Y(n)}\frac{n!}{{\rm hook}(i, j)}$$
This is the number of irreps of that dimension (and the number of irreps of the symmetric group of that dimension). Here it's both 1:
╭──┬──┐
│1 │2 │
└──┴──┘
╭──┐
│1 │
├──┼
│2 │
└──┘
It will become clear why the horizontal (vertical) boxes are (anti)symmetric.
To show the power of the Young tableaux we need to combined 3 irreps. If it is spin 1/2, then the hook length formula tells us:
$${\bf 2}\otimes{\bf 2}\otimes{\bf 2}={\bf 4}_S\oplus{\bf 2}_M\oplus {\bf 2}_M$$
The dimension of the symmetric combination is 4:
$$|\frac 3 2,+\frac 3 2\rangle = |\uparrow\uparrow\uparrow\rangle$$ $$|\frac 3 2,+\frac 1 2\rangle = (|\downarrow\uparrow\uparrow\rangle+|\uparrow\downarrow\uparrow\rangle+|\uparrow\uparrow\downarrow\rangle)/\sqrt 3$$ $$|\frac 3 2,-\frac 1 2\rangle = (|\downarrow\downarrow\uparrow\rangle+|\downarrow\uparrow\downarrow\rangle+|\uparrow\downarrow\downarrow\rangle)/\sqrt 3$$ $$|\frac 3 2,-\frac 3 2\rangle = |\downarrow\downarrow\downarrow\rangle$$
while the antisymmetric combination has dimension 0: there is no singlet state. (Tensor side note: had we used $$r=3$$ and not $$r=2$$ in the hook length formula, we would have a one dimensional antisymmetric space, which is spanned by $$\epsilon_{ijk}$$...remarkable...how does it work?).
The question remains: how do the diagrams tell us how to combine permutation of indices or spin states? For that, we'll look at another $$n=3$$ diagram, with two standard fillings:
╭──┬──┐
│1 │2 │
├──┼──┘
│3 │
└──┘
╭──┬──┐
│1 │3 │
├──┼──┘
│2 │
└──┘
For specifics, we'll focus on the top one. From here we compute the Young symmetrizer and apply it to particle labels (or indices if we're doing rank-3 tensors).
First we need the symmetric group on 3 letters $$(1,2,3)$$:
$$S_3 =\{e,e_{23},e_{12},e_{123},e_{132},e_{13}\}$$
where the permutation, e.g. $$e_{123}$$ means $$(1,2,3)\rightarrow (2,3,1)$$. Those are all six elements, with $$e$$ being the identity.
First: find all permutations that leave the tableaux "row equivalent". Tableaux are row equivalent if each row has the same numbers:
$$R=\{e,e_{12}\}$$
and similarly for column equivalence, with the addition that we include the parity of the permutation:
$$C=\{e,-e_{13}\}$$
The Young symmetrizer is then the product of these two, as follows:
$$S =R*C = \{e+e_{12}-e_{13}-e_{132} \}$$
You then apply these permutations to $$|\uparrow\uparrow\downarrow\rangle$$, and normalize, to get:
$$|\frac 1 2, +\frac 1 2\rangle= (|\uparrow\uparrow\downarrow\rangle - |\downarrow\uparrow\uparrow\rangle)/\sqrt 2$$
Easy peasy.
• Thank you very much for this great tutorial! I've seen in a few places the relationship between irreps of angular momentum spaces and young tableaux but I've never seen a clear explanation until this one! I'm still not totally clear on the Young symmetrizer. I understand how you constructed the Young symmetrizer $S$ for a particular Young Tableaux. Why, in this case, does it make sense to apply the symmetrizer to the state $|\uparrow \uparrow \downarrow \rangle$ and not some other state? May 4, 2021 at 15:48 | open-web-math/open-web-math | |
# Practice Problem..Codevita...How to approach this problem?
## Minimum BidMarks:200
### Problem Description
Consider people calling out bids in different number bases at an auction. Find the minimum bid assuming the following:
1. The bid numbers are in bases that make their respective values minimum.
2. There is only one minimum value among all the bids.
### Constraints
1. N <= 10
2. Maximum base = 36
3. Symbols used for digits: Base 2: 0, 1
Base 3: 0, 1, 2
Base 11: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A
Base 36: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z
1. Face values for symbols: Symbol => Value 0 => 0
1 => 1
2 => 2
….
9 => 9
A => 10
B => 11
….
Z => 35
### Input Format
N different numbers in various bases, with numbers delimited by space
### Output
The value in base 10 of the minimum bid.
### Explanation
Example 1
Input
11 12
Output
3
Explanation
The value of number represented by 11 is least in base 2 and that least value in base 10 is 3. The least value of the representation 12 is in base 3 and is equal to 5. Since 3 < 5, 3 is the lowest bid and is the output.
Example 2
Input
1Z A L0 17
Output
10
Explanation
The least values are:
1Z in base 36: 136+35 = 71
A in base 11: 10
L0 in base 22: 21
22+0 = 462
17 in base 8: 1*8+7 = 15
Hence the least bid is 10.
import java.io.;
import java.util.
;
import java.lang.*;//pow
class FaceValues
{
int len;
public int[] convert(String s)
{
char[] c= s.toCharArray();
len=c.length;
int[] a= new int[len];
int i,j=0;
for(i=0;i<len;i++)
{
//How to check if a given character is a number letter in Java
if(Character.isDigit(c[i]))
a[j] = Character.getNumericValue(c[i]);
else
{
a[j] = c[i];
a[j] = a[j] - 55;//A=65
}
j++;
}
j–;
return a;
}
int findBase(int[] arrMax)
{
int max=arrMax[0];
for(int i=1;i<arrMax.length;i++)
{
if(max<arrMax[i])
max=arrMax[i];
}
return max+1;
}
}
class LeastBid
{
public static void main(String args[])
{
int a[]=new int[10];
int decimal = Integer.MAX_VALUE;
String str;
int i,j,k,l;
FaceValues values = new FaceValues();
Scanner sc=new Scanner(System.in);
str = sc.nextLine();
String[] array = str.split(" ");
for(i=0;i<array.length;i++)
{
int[] valArr = values.convert(array[i]);//has the face values of element 1
int base = values.findBase(valArr);//to fund base we need to find the max in valArr and add 1
int power=valArr.length;
double sum=0;
for(j=power-1,k=0; j>=0 && k<power; j–,k++)
{
sum = sum+ ( valArr[k]*Math.pow(base,j) );
}
//now i need to find min of the sum
if(sum<decimal)
decimal=(int)sum;
}
System.out.println(“least decimal value:”+decimal);
}
}
1 Like
Thanks…Can u please explain the concept and logic? | HuggingFaceTB/finemath | |
Home Forums Math Rational number between root2 and root3
Viewing 2 posts - 1 through 2 (of 2 total)
• Author
Posts
• #23082
Rational number between root2 and root3 is ______
#23084
ProtonsTalk
Keymaster
The rational number between root 2 and root 3 is 1.6.
Explanation:
A rational number is any number which can be expressed in the form of p/q where q is not equal to 0.
Given root 2 and root 3 are irrational numbers with a value of 1.414… and 1.732…
Now as we have these two numbers in decimal format, we can easily tell a rational number between them, in fact, there are infinite rational numbers between them.
Like 1.5, 1.51, 1.52, 1.6 and so on.
Taking any one randomly, say 1.6, we can say a rational number between root 2 and root 3 is 1.6.
Viewing 2 posts - 1 through 2 (of 2 total)
• You must be logged in to reply to this topic.
Scroll to Top
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# Document 10986458 1 12V + 12V + 6 Ω 3 Ω 6 Ω V R R R R R 4F 10Ω 6F 9V 20Ω FIG. 1: A, B and C; respectively 1B quiz 2 version A 1. Gold has one electron per atom available as charge carriers. The mass density of gold is 19 .
3 kg/m 3 and its atomic weight is 197. Find the drift speed of the electrons in a wire with circular cross section of radius .
3 mm and which is carrying a current of carrying .
1 A .
• a. 1.4
× 10 − 4 m/s • b. 3.7
× 10 − 2 m/s • c. 5.9
× 10 − 1 m/s • d. 2.7
× 10 − 3 m/s 2. An aluminum wire of length L and a copper wire of length 5 L have precisely the same resistance. The resistivity of the two materials are: aluminum, 2 .
8 × 10 − 8 Ω − m and copper 1 .
7 × 10 − 8 Ω − m . What is the ratio of the radius of the copper wire to the aluminum wire?
• a. 3.7
• b. 3.0
• c. 1.74
• d. .44
3. A heater uses nichrome wire with resistivity 1 .
0 × 10 − 6 Ω − m and generates 1250 W of heat when connected across a potential difference of 110 V . How long must the wire be, if its cross-sectional area is .
2 × 10 − 6 m 2 ?
2 • a. .025m
• b. .37m
• c. 1.94m
• d. 23.5m
4. A resistor with R = 5 × 10 6 Ω and a capacitor with C = 120 µF are connected in series to a 800 V power supply. Find the current when the capacitor is charged to 90% of its final charge.
• a. 3 µ A • b. 16 µ A • c. 30 µ A • d. 160 µ A 5. One month’s electric bill for an apartment is \$25.33 and the cost of electricity is \$.08/kilowatt-hour. All appliances run at 120 V . How many electrons passed through the house that month?
• a. 1.8
× 10 20 • b. 7.4
× 10 21 • c. 9.5
× 10 23 • d. 5.9
× 10 25 6. Find the current in the middle resistor in the circuit shown in Fig A • a. 1.0 A • b. 1.33 A • c. 2.0 A • d. 4.0 A 7. Find the total current out of the battery in the circuit shown in Fig B (hint: you do not need to write down all the Kirchoff law equations; think about symmetry and the role of the middle resistor?)
3 • a.
V 5 R • b.
V 2 R • c.
V R • d.
2 V R 8. Consider the circuit in Fig C with two resistors and two capacitors connected in series with a 9 V battery. Calculate the potential between the upper-left and upper-right corner points both immediately after the switch is closed and after waiting a time long compared to the circuit’s time constants.
• a. 3V; 3.6V
• b. 3V; 5.4V
• c. 6V; 5.4V
• d. 6V; 0V | HuggingFaceTB/finemath | |
# Geometry and location
• Eh
In summary, the conversation discusses the concept of precise location in mathematics and how it relates to real world objects such as the planet Jupiter. The question of whether the word "location" has a single mathematical definition is raised, and it is mentioned that in the case of Jupiter, there may be multiple locations or a single collection of points that define its position. The idea that the center of Jupiter exists at a single point while the planet itself exists at a set of points is also explored. This conversation ultimately highlights the complexity of defining location in both mathematical and real-world terms.
#### Eh
A precise location is defined as a point with zero dimensions. So we can easily tell you where the location of something is, say at (x,y,z) from the starting point. But let's say you had a map of the solar system, and used a scale that is too small. The planet Jupiter would now be found at several locations at once.
Anyway, the question is: does the word "location" only have one mathematical definition, being an exact point? In the case of Jupiter, are there there several locations where you will find the gas giant, or can you define a single location as a collection of points?
Most people would be happy with the definition that the center of the planet exists at a single point in space, while the planet itself exists at a set of points.
- Warren
"Point" is, as you say, a mathematical concept. "Jupiter" is NOT a mathematical concept. You could, of course, set up a mathematical model that more or less accurately modeled Jupiter. Whether you could assign a single point to Jupiter would depend upon your mathematical model.
## What is geometry?
Geometry is a branch of mathematics that deals with the study of shapes, sizes, relative positions of figures, and the properties of space.
## What are the basic elements of geometry?
The basic elements of geometry include points, lines, angles, surfaces, and solids. These elements are used to define and describe shapes and their properties.
## What is the difference between Euclidean and non-Euclidean geometry?
Euclidean geometry is the study of flat or two-dimensional surfaces, while non-Euclidean geometry is the study of curved or higher-dimensional surfaces. Non-Euclidean geometry also includes different axioms and postulates than Euclidean geometry.
## What is the importance of geometry in real life?
Geometry is used in many real-life applications, such as architecture, engineering, navigation, and art. It helps us understand and describe the world around us and make accurate measurements and calculations.
## How is geometry related to location?
Geometry and location are closely related as geometry is used to describe and determine the position of objects in space. It also helps us understand the concept of distance, direction, and orientation, which are essential in determining location. | HuggingFaceTB/finemath | |
#### Solution for 9 is what percent of 264.2:
9:264.2*100 =
(9*100):264.2 =
900:264.2 = 3.4065102195307
Now we have: 9 is what percent of 264.2 = 3.4065102195307
Question: 9 is what percent of 264.2?
Percentage solution with steps:
Step 1: We make the assumption that 264.2 is 100% since it is our output value.
Step 2: We next represent the value we seek with {x}.
Step 3: From step 1, it follows that {100\%}={264.2}.
Step 4: In the same vein, {x\%}={9}.
Step 5: This gives us a pair of simple equations:
{100\%}={264.2}(1).
{x\%}={9}(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
\frac{100\%}{x\%}=\frac{264.2}{9}
Step 7: Taking the inverse (or reciprocal) of both sides yields
\frac{x\%}{100\%}=\frac{9}{264.2}
\Rightarrow{x} = {3.4065102195307\%}
Therefore, {9} is {3.4065102195307\%} of {264.2}.
#### Solution for 264.2 is what percent of 9:
264.2:9*100 =
(264.2*100):9 =
26420:9 = 2935.5555555556
Now we have: 264.2 is what percent of 9 = 2935.5555555556
Question: 264.2 is what percent of 9?
Percentage solution with steps:
Step 1: We make the assumption that 9 is 100% since it is our output value.
Step 2: We next represent the value we seek with {x}.
Step 3: From step 1, it follows that {100\%}={9}.
Step 4: In the same vein, {x\%}={264.2}.
Step 5: This gives us a pair of simple equations:
{100\%}={9}(1).
{x\%}={264.2}(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
\frac{100\%}{x\%}=\frac{9}{264.2}
Step 7: Taking the inverse (or reciprocal) of both sides yields
\frac{x\%}{100\%}=\frac{264.2}{9}
\Rightarrow{x} = {2935.5555555556\%}
Therefore, {264.2} is {2935.5555555556\%} of {9}.
Calculation Samples | HuggingFaceTB/finemath | |
# A committee of 4 people is chosen from 8 women and 8 men. How many different committees are possible that consist of 2 women and 2 men?
There are the binomial coefficient "8 choose 2", i.e. (8!)/(2!6!) = 28 ways to choose 2 people from a set of 8 people. So, there are 28 ways to choose 2 men and 28 ways to choose 2 women. This means that there is ${28}^{2} = 784$ ways to choose both 2 men and 2 women. | HuggingFaceTB/finemath | |
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# Chapter 8 Test Review
Chapter 8 Test Review. Quadrilaterals. The sum of the measures of the interior angles of a convex regular polygon is 1080 °. Classify the polygon by the number of sides. Octagon (8 sides). What is the measure of each interior angle of a regular octagon?. 1080 8 = 135 °.
## Chapter 8 Test Review
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### Presentation Transcript
1. Chapter 8 Test Review Quadrilaterals
2. The sum of the measures of the interior angles of a convex regular polygon is 1080°. Classify the polygon by the number of sides. Octagon (8 sides)
3. What is the measure of each interior angle of a regular octagon? 1080 8 = 135°
4. 120° 97° x° 130° 150° Find the value of x. x + 120 + 97 + 130 + 150 + 90 = (6-4)(180) x = 133
5. 5x° 8x° 5x° Find the value of x. 8x + 5x + 5x = 360 x = 20
6. 10 8 m n - 3 Find the value of m and n in the parallelogram. n – 3 = 8 n = 11 m = 10
7. x + 9 2x + 4 What value of x makes the quadrilateral a parallelogram? 2x + 4 = x + 9 x = 5
8. y° 21° Find the values of x and y. x = 69 y = 21
9. 3x + 4 4y 6y - 10 4x - 5 Find the values of x and y. 3x + 4 = 4x – 5 x = 9 4y = 6y – 10 y = 5
10. 3x + 2 5x - 4 What value of x makes the quadrilateral a parallelogram? 3x + 2 = 5x – 4 x = 3
11. What is the most specific name for the polygon? Parallelogram
12. 125° 2x° 160° 110° x° 112° 147° Find the value of x. x + 160 + 2x + 125 + 110 + 112 + 147 = (7-2)(180) x = 82
13. 14 11 d + 4 c + 5 Find the values of c and d in the parallelogram. c + 5 = 11 c = 6 d + 4 = 14 d = 10
14. 18 (b + 16)° 103° a - 10 Find the values of a and b in the parallelogram. a -10 = 18 a = 28 b + 16 = 103 b = 87
15. In a regular nonagon, the exterior angles are all congruent. What is the measure of one of the exterior angles? 9x = 360 x = 40°
16. B A 20° C D ABCD is a kite.Find mB and mD. 90 + 20 + x + x = 360 mB = mD = 125°
17. What is the most specific name for the polygon? Quadrilateral RSTU has R, T, and U as right angles. RS = ST. Square
18. 19 in F G 16.5 in D E J H Find JH. (JH + 19) 2 = 16.5 JH = 14 in
19. E D 110° 25° F G DEFG is a trapezoid.Find mG and mE. mG = 70° mE = 155°
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