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https://mitrev.net/ruby/2015/06/07/the-null-object-pattern/
|
Every Ruby programmer eventually encounters the dreaded NoMethodError exception.
undefined method first_name' for nil:NilClass
We learn to live with it and handle it in various nasty ways. Like using try:
tr
td = Client.human_attribute_name(:vat_number)
td = @client.try(:vat_number)
tr
td = Client.human_attribute_name(:company_ownership_type_id)
td = @client.ctry(:company_ownership_type, :name_bg)
tr
td = Client.human_attribute_name(:company_branch_id)
td = @client.ctry(:company_branch, :name_bg)
This code is bad because our @client should never actually be nil. We can improve it by using the Null Object Pattern.
### An example
Imagine we have an application and have to print a greeting to the current user, something like “Welcome, George”. The first thing we try may be:
Welcome, <%= current_user.name %>
But then we log out and are greeted with an old friend:
NoMethodError: undefined method name' for nil:NilClass
We quickly fix our code by adding a nil check and also decide we want to greet logged-out people too:
Welcome, <%= current_user.nil? ? 'Guest' : current_user.name %>
This works as we originally intended and is good enough for production. Or is it? We start using the same pattern in different places, our collegues also pick it up and a few months later we find ourselves in with filthy mess that is really hard to refactor.
Our next problem comes with a change in the requirements - guest users are not to be greeted as “Guest”, but as “Stranger”, because our boss thinks it’s funnier. We start working on our task and find that there are 78 occurances of the string “Guess” in our view code and we need to change each one separately because we cannot mass replace such a common word as ‘Guest’. We curse the gods and get on with it, because this is life and life sucks.
Could this have been avoided? Yes, and the solution is really simple.
### The pattern
What we need is way to make current_user respond to the name method even if no user is currently logged in. We start by introducing a new class:
class NullUser
def name
'Guest'
end
def logged_in?
false
end
end
The NullUser class should conform to the same public API as the User class, which requires maintenance, but everything comes at a cost.
Next, we change our current_user method to return a new instance of NullUser when no user is logged in:
def current_user
current_user_session || NullUser.new
end
Now when our new requirements arrive, we only need to update the name method of the NullUser class and need not to worry about breaking anything else. If we dislike the name of the class, we may also use Guest or GuestUser, it doesn’t matter.
### Drawbacks
The Null Object is not a silver bullet and not something you want to use for all your classes. It should not be used to avoid all nil errors but as a way to provide default behaviour when you get an unexpected nil. Sometimes nil is just nil and should be handled as such. You should also have an extensive test suite that ensures that nowhere a nil may be assigned instead of your Null Object instance, because that will break wreak havoc on your happy little world.
### Last words
The pattern really shines in the cases it’s is suited for, such as the NullUser or representing the last element of a list/tree. Apply it with care and it may improve your life a little.
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2020-11-26 03:20:37
|
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|
http://math.stackexchange.com/questions/128556/inversion-of-matrices-is-a-diffeomorphism?answertab=active
|
# Inversion of matrices is a diffeomorphism.
I am having problems showing that the function $$\operatorname{inv}:G\rightarrow G$$ $$A\rightarrow A^{-1}$$ where $G$ is the set of all invertible $n\times n$ matrices, is a diffeomorphism. I have already shown that such function is a homeomorphism, and its inverse is itself, but I don't know how I can show that this function is differentiable.
The exercise also tells us that the derivative of $\operatorname{inv}$ in $A$ is the linear mapping $M\rightarrow M$ such that $X\rightarrow -A^{-1}\cdot X\cdot A^{-1}$.
Can anybody give me a hint?
-
add comment
## 2 Answers
Here is a hint: what does Cramer's rule tell you about the matrix entries of the inverse in terms of the original matrix?
-
Does this means that, since every entry of the inverse are determinants (that is, the adjunt is the transpose of the matrix of the cofactors, which are determinants), the inversion is a multilinear function times the $\dfrac{1}{det A}$ which are both differentiable, then it is differentiable? – Marra Apr 6 '12 at 0:55
What it means is that since $A^{-1}$ is just $\displaystyle \frac{1}{\det(A)}\text{adj}(A)$ the inverse entries are just rational functions in the entries of the original matrix--thus trivially smooth. – Alex Youcis Apr 6 '12 at 2:11
@GustavoMarra : If you write \det A rather than det A, that not only prevents det from being italicized but also results in proper spacing before and after det, thus: $a\det A$. And it's standard usage. – Michael Hardy Apr 6 '12 at 17:47
Thanks @Michael, I didn't know that det was a LaTeX function! – Marra Apr 6 '12 at 20:09
add comment
Another approach would be to note that if $\|X\|<1$, then $(I+X)^{-1} = I-X+X^2-\cdots$. Then consider the expression $(A+X)^{-1}$, where $X$ is a suitably small perturbation. A small computation shows that $(A+X)^{-1} = (I+A^{-1}X)^{-1} A^{-1}$. Then expand using the above series, and look at the linear term of the expansion. Both differentiability and the form of the derivative follow from this expansion.
-
I'll check this. Can I always make this expansion $(I+X)^{-1}=I-X+X^2-...$ or is it possible only when $||X||<1$? – Marra Apr 6 '12 at 0:58
If you take $X \in \mathbb{R}$, you can see that the series fails to converge if $|X| \geq 1$. – copper.hat Apr 6 '12 at 4:17
add comment
|
2014-04-20 01:18:35
|
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http://icpc.njust.edu.cn/Problem/Pku/3299/
|
Humidex
Time Limit: 1000MS
Memory Limit: 65536K
Description
humidex = temperature + h
h = (0.5555)× (e - 10.0)
e = 6.11 × exp [5417.7530 × ((1/273.16) - (1/(dewpoint+273.16)))]
where exp(x) is 2.718281828 raised to the exponent x. While humidex is just a number, radio announcers often announce it as if it were the temperature, e.g. "It's 47 degrees out there ... [pause] .. with the humidex,". Sometimes weather reports give the temperature and dewpoint, or the temperature and humidex, but rarely do they report all three measurements. Write a program that, given any two of the measurements, will calculate the third. You may assume that for all inputs, the temperature, dewpoint, and humidex are all between -100°C and 100°C.
Input
Input will consist of a number of lines. Each line except the last will consist of four items separated by spaces: a letter, a number, a second letter, and a second number. Each letter specifies the meaning of the number that follows it, and will be either T, indicating temperature, D, indicating dewpoint, or H, indicating humidex. The last line of input will consist of the single letter E.
Output
For each line of input except the last, produce one line of output. Each line of output should have the form:
T number D number H number
where the three numbers are replaced with the temperature, dewpoint, and humidex. Each value should be expressed rounded to the nearest tenth of a degree, with exactly one digit after the decimal point. All temperatures are in degrees celsius.
Sample Input
T 30 D 15
T 30.0 D 25.0
E
Sample Output
T 30.0 D 15.0 H 34.0
T 30.0 D 25.0 H 42.3
Source
Waterloo Local Contest, 2007.7.14
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2020-08-10 06:01:01
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https://zbmath.org/?q=an%3A0918.11040
|
## The irrationality of $$\log(1+1/q) \log(1-1/q)$$.(English)Zbl 0918.11040
The main result of this paper is the linear independence of numbers $\log(1+1/q),\quad \log(1-1/q),\quad\log(1+1/q)\log(1-1/q)\tag{1}$ over rational numbers for every natural number $$q\geq 1$$. The estimations of the measure of linear independence are included too. The proofs are based on Padé approximation of Legendre polynomials.
Reviewer: J.Hančl (Ostrava)
### MSC:
11J72 Irrationality; linear independence over a field 11J82 Measures of irrationality and of transcendence
Full Text:
### References:
[1] F. Beukers, A note on the irrationality of \?(2) and \?(3), Bull. London Math. Soc. 11 (1979), no. 3, 268 – 272. · Zbl 0421.10023 [2] D.V. Chudnovsky and G.V. Chudnovsky, Applications of Padé approximations to diophantine inequalities in values of G-functions, Lecture Notes in Math., vol. 1135, Springer-Verlag, 1985 pp. 9-51. · Zbl 0561.10016 [3] A. I. Galočkin, Lower bounds of polynomials in the values of a certain class of analytic functions, Mat. Sb. (N.S.) 95(137) (1974), 396 – 417, 471 (Russian). [4] -, a private communication. [5] Masayoshi Hata, Legendre type polynomials and irrationality measures, J. Reine Angew. Math. 407 (1990), 99 – 125. · Zbl 0692.10034 [6] Masayoshi Hata, On the linear independence of the values of polylogarithmic functions, J. Math. Pures Appl. (9) 69 (1990), no. 2, 133 – 173. · Zbl 0712.11040 [7] Masayoshi Hata, Rational approximations to the dilogarithm, Trans. Amer. Math. Soc. 336 (1993), no. 1, 363 – 387. · Zbl 0768.11022 [8] Leonard Lewin, Polylogarithms and associated functions, North-Holland Publishing Co., New York-Amsterdam, 1981. With a foreword by A. J. Van der Poorten. · Zbl 0465.33001 [9] G. Rhin and C. Viola, On the irrationality measure of \?(2), Ann. Inst. Fourier (Grenoble) 43 (1993), no. 1, 85 – 109 (English, with English and French summaries). · Zbl 0776.11036
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2022-05-26 23:03:44
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http://slideplayer.com/slide/4305340/
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# Trigonometry SOH CAH TOA. Say it. Remember it. SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH.
## Presentation on theme: "Trigonometry SOH CAH TOA. Say it. Remember it. SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH."— Presentation transcript:
Trigonometry SOH CAH TOA
Say it. Remember it. SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA SOH CAH TOA
Using Trigonometry affectively General Han Hsin Han Dynasty (206 BC)
General Han Hsin
Enemy Palace
Well Fortified Enemy Palace*
Angry Well Fortified Enemy Palace**
Engineer
Kite ?
? ?
? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
Right Triangle ? ? ? ? ? ? ? ?
Hypotenuse ? ? ? ? ? ? ? ?? H
THETA ? ? ? ? ? ? ? ?? ? H
Adjacent Side ? ? ? ? ? ? ? ?? ? H ? A
Trig (CAH) ? ? ? ? ? ? ? ?? ? H ? A
Solving for A ? ? ? ? ? ? ? ?? ? H ? A ? ? ?
Diggers ? ? ? ? ? ? ? ?? ? H ? A ??? A. ?? ? ?
Tunnel ? ? ? ? ? ? ? ?? ? H ? A ??? A! ?? ? ?
Tunnel ? ? ? ? ? ? ? ?? ? H ? A ??? ?? ? ?
? ? ? ? ? ? ? ?? ? H ? A ??? ?? ? ?
? ? ? ? ? ? ? ?? ? H ? A ??? ?? ? ?
Army ? ? ? ? ? ? ? ?? ? H ? A ? ? ? ?? ? ? ? ?? ? ?
? ? ? ? ? ? ? ?? ? H ? A ? ? ? ?? ? ? ? ?? ? ?
? ? ? ? ? ? ? ?? ? H ? A ? ? ? ?? ? ? ? ?? ? ?
H A
H A
H A
Murder H A
H A
H A
H A
Victory H A
H A
H A
General Han Hsin’s Palace
SOH CAH TOA SOH CAH TOA Hypotenuse Adjacent Opposite
SOH CAH TOA SOH CAH TOA Hypotenuse Adjacent Opposite
Examples Find all missing sides and angles of the following triangle. 35˚ 77 in
Examples Find all missing sides and angles of the following triangle. 59˚ 66 mi
Examples You want to break into a museum. You forgot your mirror. All you have is a protractor, some measuring tape, and an Altotrak. How do you figure out how much rope you need? 25 ft. 47˚ Opp.
Examples You want to break into a museum. You forgot your mirror. All you have is a protractor, some measuring tape, and an Altotrak. How do you figure out how much rope you need? 25 ft. Opp. 47˚
Examples You want to break into a museum. You forgot your mirror. All you have is a protractor, some measuring tape, and an Altotrak. How do you figure out how much rope you need? A Opp.
Remember SOH CAH TOA only works for right triangles. Your calculator must be in degree mode if you are entering degrees for theta.
Test out SOH CAH TOA Using a protractor and ruler create 3 different right triangles, which all have the same angle theta (pick a whole number).
Test out SOH CAH TOA Using a protractor and ruler create 3 different right triangles which all have the same angle theta.
Test out SOH CAH TOA Create three different right triangles all with the same angle theta (pick a whole number). Measure all sides. Find the values of: Average each column. Find the sin, cos, and tan of theta.
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2018-04-21 10:21:47
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https://forum.allaboutcircuits.com/threads/supernode-containing-the-reference-node.26656/
|
Supernode containing the reference node
shespuzzling
Joined Aug 13, 2009
88
Hi,
My question is related to nodal analysis and using supernodes. I don't have a pdf of the circuit in question but can try to get it if it helps.
Basically, I am unable to solve this circuit using nodal analysis when I write an equation around a supernode that contains a nonreference node and the reference node. I can't figure out why this won't work. If I write a different equation I'm able to get the correct answer, but why can't I do it around a supernode that contains the reference node? I just don't get it!
hgmjr
Joined Jan 28, 2005
9,029
Hi,
My question is related to nodal analysis and using supernodes. I don't have a pdf of the circuit in question but can try to get it if it helps.
Basically, I am unable to solve this circuit using nodal analysis when I write an equation around a supernode that contains a nonreference node and the reference node. I can't figure out why this won't work. If I write a different equation I'm able to get the correct answer, but why can't I do it around a supernode that contains the reference node? I just don't get it!
The best thing for you to do is go ahead and post your circuit. Once the circuit is available it will be possible to make suggestion that you can use teo solve the problem.
It is a good idea for you to try to solve the problem yourself and then post your efforts. Then the members can better determine where you are weak in your understanding of the solution approach.
hgmjr
shespuzzling
Joined Aug 13, 2009
88
Hi,
I've attached a PDF of the circuit and my work.
My problem is this: I thought that If you have 4 unknown voltages then you need 4 equations to solve but I am unable to solve with the 4 equations I have. The book skips writing an equation about the V1-Ref node and instead writes an equation which relates .2Vy with the node voltages V3 and V4 and then with V4 and V1 and when this equation is used you get the write answer. But they're still using 4 equations for 4 unknown voltages, just like I did.
My problem is, why doesn't it work with an equation written about the supernode at V1-Ref.?
Thanks! Please let me know if you have any questions.
Attachments
• 250.8 KB Views: 103
hgmjr
Joined Jan 28, 2005
9,029
You should be able to form an additional independent equation relating V4 and V3.
hgmjr
shespuzzling
Joined Aug 13, 2009
88
Right, I understand that another equation could be written relating V4-V1 and V3-V4, but I don't see why it doesn't work the way I've done it with the supernode about V1-Ref.
hgmjr
Joined Jan 28, 2005
9,029
I believe the reason it fails with the equation written around the supernode involving V1 and 0V is that the equation so formed does not relate one of the variable with one or more of the remaining three variables. Only with an equation that relates one of the variables to one or more of the remaining variables do you have a chance to substitute and reduce the number of unknowns by one variable.
hgmjr
shespuzzling
Joined Aug 13, 2009
88
I thought it might have had somethign to do with which variables were in the equations, but if you look at the Matrix that I wrote on the PDF, you can see that on the third line (that is for the equation about the ref. node) the only nodal voltage that isn't referenced is V3.
.2Vy = V3-V4 = .2(V4-V1)
V3-V4-.2V4+.2V1=0
.2(V1) + 1(V3) - 1.2(V4) = 0 (the book used this equation instead of the one I wrote for the supernode about V1-Ref.)
But, this equation also only relates 3 of the 4 variables present so I'm not understanding how it differs from mine. I've gone over my work (as shown on the PDF) about a million times and can't for the life of me figure out what I'm doing wrong.
P.S. thanks for all of the fast responses!
hgmjr
Joined Jan 28, 2005
9,029
Can you post the equation for the other supernode that you think should work?
hgmjr
shespuzzling
Joined Aug 13, 2009
88
Sure, it's written as equation #3 on the PDF but I will write it out for you here:
V1-V2 + V1-V4 - V4 = -14 - .5(Vx) ; Vx = V2-V1
(.5).......(2.5)....(1)
simplified, this gives you: 1.9(V1) - 1.5(V2) - 1.4(V4) = -14
(the formatting was a little weird when i tried to write the equation above, let me know if it doesn't make sense)
Last edited:
hgmjr
Joined Jan 28, 2005
9,029
Here is the equation written using the LaTex feature available in the "Go Advanced" reply box.
$\frac{V_1-V_2}{0.5}+\frac{V_1-V_4}{2.5}-V_4=14-0.5V_X$
where: $V_X=V_2-V_1$
This is one of the critical independent equations without which the problem could not be solved.
I thought the 4th equation you were referring to was
$V_1=-12$
hgmjr
shespuzzling
Joined Aug 13, 2009
88
I'm going to write out both my way of doing it and the books so you can more easily compare.
My way:
nodal analysis at node 2, at supernode between node 3 and 4, using the fact that V1=-12, and at the supernode between node 1 and the reference.
-2(V1) + 2.5(V2) - .5(V3) = 14 (@ node 2)
.1(V1) - 1(V2) + .5(V3) + 1.4(V4) = 0 (@ super node V3-V4)
V1 = -12
1.9(V1) - 1.5(V2) - 1.4(V4) = -14 (@ super node V1-Ref)
4 equations, 4 unknowns. I used my calcular to solve this matrix and couldn't come up with the right answer.
Book's way:
nodal analysis at node 2, at supernode between node 3 and 4, V1=-12, and then relating V4-V1 with V3-V4.
-2(V1) + 2.5(V2) - .5(V3) = 14 (@ node 2 - same as mine)
.1(V1) - 1(V2) + .5(V3) + 1.4(V4) = 0 (@ super node V3-V4 - same as mine)
V1 = -12 (same as mine)
.2(V1) + 1(V3) - 1.2(V4) = 0 (Here is where the book differs from my approach)
4 equations & 4 unknowns, but solving this matrix with my calculator gives me the correct answer.
So, why didn't my way of doing things work? As far as I can tell, it should.
hgmjr
Joined Jan 28, 2005
9,029
Are you classifying the two nodes V1 and 0V (ref) as making up a supernode? If so, I don't believe that is true.
hgmjr
shespuzzling
Joined Aug 13, 2009
88
Yes, that was my reasoning, since they are connected by a voltage source. Why doesn't it work? It seems to be the same logic as that used at nodes 3 and 4.
hgmjr
Joined Jan 28, 2005
9,029
In all of the writeups that I have seen on supernode usage, I have never seen any that classified a voltage source referenced to ground as a supernode. My understanding of supernodes are that they involve a voltage source that joins two nodes neither of which is the reference node. I believe supernodes are a way of taking advantage of the fact that an ideal voltage source has zero source impedance.
hgmjr
The Electrician
Joined Oct 9, 2007
2,724
hgmjr is quite correct. The mistake you are making is to assume that the reference node can be part of a supernode. This is not true.
shespuzzling
Joined Aug 13, 2009
88
Wow, so it's as simple as that! Thanks for your help! I'll remember never to do a supernode about a reference node then. So, theoretically then, if I chose a different node to be the reference node, then I should be able to do a supernode between node 1 and the center node without any problems?
hgmjr
Joined Jan 28, 2005
9,029
While you are probably right about changing the reference node opening up the potential to create a supernode, I believe the net effect would be to complicate rather than simplify the analysis.
As an aside, I would recommend you look into Millman's Theorem and add that to your circuit analysis tool kit.
hgmjr
The Electrician
Joined Oct 9, 2007
2,724
When using the nodal method, the thing to do would be to choose your reference node so that there are no voltage sources between any two non-reference nodes. In other words, make sure that all voltage sources have at least one terminal connected to the reference node.
You'll notice that you can't do that for your given circuit.
shespuzzling
Joined Aug 13, 2009
88
Thanks for all your help. The book's tend to gloss over these minor but important facts. I'll look into Millman's theorem as well.
hgmjr
Joined Jan 28, 2005
9,029
Here is the link to the material on Millman's Theorem in the AAC ebook. The Theorem is much more powerful than the simple explanation and examples might lead you to believe.
If you are interested in seeing some more interesting applications of the Theorem, have a look at my Blog here at AAC.
hgmjr
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2019-10-21 10:12:19
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http://mathhelpforum.com/pre-calculus/20071-domain.html
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1. ## domain
hey, here is the question:
find the natural domain of these functions:-
1.) f(x)= 1/(1-sinx)
2.) f(x)= x/|x|
3.) f(x)= √(x^2-4)/(x-4)
4.) f(x)= (x^2-1)/(x+1)
5.) f(x)= 3/(2-cosx)
6.) f(x)= 3+ √x
7.) f(x)= x^3 +2
8.) f(x)= 3 sin x
9.) f(x)= 3/x
10.) f(x)= sin^2 √x
2. Originally Posted by wutever
hey, here is the question:
find the natural domain of these functions:-
1.) f(x)= 1/(1-sinx)
2.) f(x)= x/|x|
3.) f(x)= √(x^2-4)/(x-4)
4.) f(x)= (x^2-1)/(x+1)
5.) f(x)= 3/(2-cosx)
6.) f(x)= 3+ √x
7.) f(x)= x^3 +2
8.) f(x)= 3 sin x
9.) f(x)= 3/x
10.) f(x)= sin^2 √x
Ask yourself the question: "Is there any place where the function does not exist?" Then take those values of x away from the real line and there's your answer.
For example:
$f(x) = \frac{x}{|x|}$
does not exist when x = 0 because a zero in the denominator is not defined. Thus the domain is $(-\infty, 0) \cup (0, \infty)$ (otherwise known as "all real numbers, except x = 0.")
-Dan
3. thx for the reply but wut abt the sin and cos ones ?
4. Originally Posted by wutever
thx for the reply but wut abt the sin and cos ones ?
Originally Posted by wutever
hey, here is the question:
find the natural domain of these functions:-
1.) f(x)= 1/(1-sinx)
5.) f(x)= 3/(2-cosx)
8.) f(x)= 3 sin x
10.) f(x)= sin^2 √x
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2017-08-19 00:00:43
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http://palomanegrastudios.com/e9xx4wd/67a283-measuring-mass-in-binary-system
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# measuring mass in binary system
A: It is the main sequence, note: Main sequence stars generate energy through core fusion of hydrogen. Here was evidence that one star was moving around another. Visiting Australia Telescope Compact Array, Parkes radio telescope webcam Management | Director Of course, the other advantage is stellar parameters (density, physical sizes, etc. How large can the mass of a star be? Îλ λ =2 (3)(1.38×10â16)T (1.66×10â24)(A) Karma Objects with masses between 1/12 and 1/100 that of the Sun are called brown dwarfs. Gravity is a mutual attraction. Public observatories As mass is such a key property of stars and to a large extent knowing a star's mass determines its life cycle and fate, being able to accurately determine stellar masses is vital in refining our models of stars. For example, we can ask whether the mass and luminosity of a star are related. Newsletters The horizontal position on the graph shows the star’s mass, given in units of the Sun’s mass, and the vertical position shows its luminosity in units of the Sun’s luminosity. We can also infer the mass and luminosity of each star. This then allows us to calculate the distance to the system. The mass-luminosity formula can be rewritten so that a value of mass can be determined if the luminosity is known. Brown dwarfs are similar to Jupiter in radius but have masses from approximately 13 to 80 times larger than the mass of Jupiter. Observe the spectrum of the visual star, compute the orbital speed from Doppler effect, calculate orbit radius 3. Australian Square Kilometre Array Pathfinder, Canberra Deep Space Communication Complex, On-Line Proposal Applications and Links (OPAL), Australia Telescope Online Archive (ATOA), Computing: Getting started guide [internal access], Visiting Australia Telescope Compact Array. It turns out that for most stars, they are: The more massive stars are generally also the more luminous. Deduct mass of visual to find mass ⦠In Appendix J, Sirius is listed with a luminosity 23 times that of the Sun. The period is about ten years. By 1804, astronomer William Herschel, who also discovered the planet Uranus, had noted that the fainter component of Castor had slightly changed its position relative to the brighter component. Somehow, we need to put a star on the cosmic equivalent of a scale. Staff list | Student list Although stars most commonly come in pairs, there are also triple and quadruple systems. ⢠If we can measure and understand their orbital motion, we can estimate the mass of the stars in that system. By the end of this section, you will be able to: The mass of a star—how much material it contains—is one of its most important characteristics. Very Long Baseline Interferometry, Applying for observing time | ATCA This loads a font easier to read for people with dyslexia. Mass-Luminosity Relation: The plotted points show the masses and luminosities of stars. | Planets Microlensing event form the MACHO project used to help determine the mass of the foreground single star. | Mathematica Masses of binary stars can be calculated from measurements of their orbits, just as the mass of the Sun can be derived by measuring the orbits of the planets around it (see Orbits and Gravity). Parkes radio telescope The Birth of Stars and the Discovery of Planets outside the Solar System, http://cnx.org/contents/[email protected], Distinguish the different types of binary star systems, Understand how we can apply Newton’s version of Kepler’s third law to derive the sum of, Apply the relationship between stellar mass and stellar luminosity to determine the physical characteristics of a star. Chapter 11 Measuring the masses of binary stars Goal-of-the-Day To use the radial velocity variations of a double-lined spectroscopic binary to derive the masses of the stars in the binary system. | CDSCC What is the total mass of the system? Using equation 5.5 or 5.6 we can determine the mass of the binary system if we can measure the orbital period and the radius vector (separation between the two components) for the system. The positions on the curve corresponding to the illustrations in [link] are marked with the diagram number (1–4). Newton's laws of motion (F=ma) allow us to derive Kepler's equation for orbital motion. Doppler shifts cause the spectral lines to move back and forth. Therefore, it moves more slowly to get around in the same time compared to the more distant, lower-mass star. Email discussion lists, On-Line Proposal Applications and Links (OPAL) Masses of binary stars can be calculated from measurements of their orbits, just as the mass of the Sun can be derived by measuring the orbits of the planets around it (see Orbits and Gravity). Scientific support of facilities Figure 5. One well-known binary star is Castor, located in the constellation of Gemini. One thing i am confused with is what angular separation means and how it can be translated to ⦠About half the stars are binary stars—two stars that orbit each other, bound together by gravity. The relative luminosities and total luminosity of the system can be derived then used to calculate the total flux of the system. Through a telescope, as Riccioli discovered in 1650, Mizar can be seen to have another, closer companion that does orbit it; Mizar is thus a visual binary. It can be calculated from observable quantities only, namely the orbital period of the binary system, and the peak radial velocity of the observed star. Kepler's equation: (M 1 + M 2) x P 2 = a 3, where. The first binary star was discovered in 1650, less than half a century after Galileo began to observe the sky with a telescope. A visual binary is a binary system in which the component stars of the system can be individually resolved through a telescope. A binary system is a system of two astronomical bodies which are close enough that their gravitational attraction causes them to orbit each other around a barycenter (also see animated examples).More restrictive definitions require that this common center of mass is not located within the interior of either object, in order to exclude the typical planetâsatellite systems and planetary systems. Now to find mass of α Cen B we simply use M = mA + mB so that: mB = M - mA mB = 4.162 × 1030 - 2.192 × 1030 ∴ mB = 1.970 × 1030 kg.so α Cen B has a mass of 1.970 × 1030 kg. The star with the higher mass will be found closer to the center of mass, while the star with the lower mass will be farther from it. (credit a: NASA, C.R. The⦠Still-smaller objects with masses less than about 1/100 the mass of the Sun (or 10 Jupiter masses) are called planets. I am measuring the spectrum of the stars in a spectroscopic binary system. (b) This image was taken in infrared light, which can make its way to us through the dust. units as the constant G is normally expressed in such units (G = 6.672 × 10-11m3.kg-1.s-2). common center of mass. When we observe the composite spectrum of the two stars, the line appears double. The α Centauri system is 1.338 pc distant with a period of 79.92 years. This requires the distance from a component star to the barycenter to also be measured. AIPS Whilst we can use use AUs and years to give us a relative value when applying Kepler's Laws, if we want a full numerical solution we must convert all parameters to S.I. These mass measurements are absolutely crucial to developing a theory of how stars evolve. Not only were there two lines where astronomers normally saw only one, but the spacing of the lines was constantly changing. Barely the size of the planet Jupiter, the dwarf star weighs in at just 8.5% of the mass of our Sun. Mizar, by the way, is a good example of just how complex such star systems can be. When the two stars are moving perpendicular to our line of sight (that is, they are not moving either toward or away from us), the two lines are exactly superimposed, and so in diagrams 2 and 4, we see only a single spectral line. Astrophysics for senior students The velocity of one binary component and the orbital period ⦠The astronomer notices that the smaller star orbits the larger one in a nearly circular orbit at a distance of r = 705 x100 km. In the top left illustration, star A is moving toward us, so the line in its spectrum is Doppler-shifted toward the blue end of the spectrum. As already mentioned, binary stars are of vital importance as they allow us to determine stellar masses. Now that we have measurements of the characteristics of many different types of stars, we can search for relationships among the characteristics. CSIRO ATNF Data Archives Such objects are intermediate in mass between stars and planets and have been given the name brown dwarfs (Figure 5). They may radiate energy produced by the radioactive elements that they contain, and they may also radiate heat generated by slowly compressing under their own weight (a process called gravitational contraction). $\begingroup$ @inf3rno For a spectroscopic binary system, the orbits of both stars around their centre of mass can be determined, so that both their masses can be derived (or more precisely, a mass range, due to the unknown inclination of their orbital plane) $\endgroup$ â Pulsar Oct 17 '14 at 22:06 Motions of Two Stars Orbiting Each Other and What the Spectrum Shows: We see changes in velocity because when one star is moving toward Earth, the other is moving away; half a cycle later, the situation is reversed. Kepler found that the time a planet takes to go around the Sun is related by a specific mathematical formula to its distance from the Sun. Digital systems In order to determine the individual mass of each star, we would need the velocities of the two stars and the orientation of the orbit relative to our line of sight. Radial Velocities Binaries (two stars orbiting around a common center of mass) are very common in the sky. Jupiter, whose mass is about 1/1000 the mass of the Sun, is unquestionably a planet, for example. Contact us, Marsfield This means that if you are able to measure the luminosity and temperature of a star, I can put it on a HertzsprungâRussell diagram, and tell you how massive it is. I am trying to teach myself some basic maths for astronomy from a book, namely trying to calculate the distance between two stars in a binary system. Accommodation & computing reservations However, most stars have less mass than the Sun. CSIROpedia. Adjust the interface to make it easier to use for different conditions. See Section 3 of the reference paper ⦠MRO Support Facility The two components that make up this visual binary, known as Mizar A and Mizar B, are both spectroscopic binaries. This leaves the binary and triplet star systems as the vehicles for directly measuring the masses of stars. Accurate sizes for a large number of stars come from measurements of eclipsing binary star systems, and so we must make a brief detour from our main story to examine this type of star system. It took astronomers until the 21st century to apply gravitational lensing to measuring stellar masses. | Tempo2 This renders the document as white on black. He was examining the spectrum of Mizar and found that the dark absorption lines in the brighter star’s spectrum were usually double. This can then be combined with orbital inclination parameters obtained from the light curve to give the stellar masses and mean stellar densities. | MONICA Events, Technology overview Graduate student programs When we point our telescopes at a binary star system,we don't always see two individual points of light.The problem is that we can only resolvethe members of the binary if 1. the system is very close to the Sun, or 2. the two components are very far apart In the second case, two stars orbiting each otherat a very large orbital distance,the orbital period turns out to be inconveniently large.Not many astronomers are willing to wait 600 yearsfor one complete revolution.We are left with only a relatively few binary star⦠Binary star systems are important because they allow us to find the masses of stars. 6. In the previous article, I have written about the fundamental and derived quantities, the different system of units, Importance of measurement etc in detail. The larger of the two stars in such a system is called the primary star, while the smaller one is the companion or secondary star. Inertial mass is a dynamic measuring method, meaning that it can only be accomplished while the object being measured is in motion. α Cen B is a dimmer K0-1 V min sequence dwarf with 90% of the Sun's mass and only about half as luminous. Each point represents a star whose mass and luminosity are both known. 2. ∴ mA = 2.192 × 1030 kg. Astronomical images Engineering education program The Mechanics of a Binary Star . One goal of these studies is to empirically link a star's luminosity and spectral type to the stellar mass. Fred Sarazin ([email protected]) PHGN324: Binary star systems Physics Department, Colorado School of Mines Center of mass ⢠Center of mass = balance point of In the case where the binary is also a spectroscopic binary and the parallax of the system is known, the binary is quite valuable for stellar analysis. In visual binaries, the two stars can be seen separately in a telescope, whereas in a spectroscopic binary, only the spectrum reveals the presence of two stars. Astrophysics staff | ATCA, Parkes The masses of stars can be determined by analysis of the orbit of binary stars—two stars that orbit a common center of mass. An astronomer is measuring the light emitted by a binary star system. Note that positive velocity means the star is moving away from us relative to the center of mass of the system, which in this case is 40 kilometers per second. It was actually the first evidence that gravitational influences exist outside the solar system. An eclipsing binary star is a binary star system in which the orbit plane of the two stars lies so nearly in the line of sight of the observer that the components undergo mutual eclipses. It’s a reasonably good approximation to say that luminosity (expressed in units of the Sun’s luminosity) varies as the fourth power of the mass (in units of the Sun’s mass). | ATELIB The individual snapshots of the orbit were obtained using the HST and adaptive optics on the VLT, Keck and Gemini. PULSE@Parkes To calculate the masses of stars in a binary system, we must measure their _____. So, Mizar is really a quadruple system of stars. To calculate the masses of stars in a binary system, we must measure their Correct Answer: orbital period and average orbital distance Note: Cool, dim, low-mass M stars are the most common type of star by far. Mizar has been known for centuries to have a faint companion called Alcor, which can be seen without a telescope. Imagine that the two stars are seated at either end of a seesaw. Mizar and Alcor form an optical double—a pair of stars that appear close together in the sky but do not orbit each other. Therefore, the sum of masses of the two stars in the Sirius binary system is 3.2 times the Sun’s mass. This renders the document in high contrast mode. Some are just chance alignments of stars that are actually at different distances from us.) Kepler studied planets orbiting the Sun.The mass of the Sun is so much larger than that of any planet that the total mass is pretty muc⦠Parkes Observatory online store As α Cen is a nearby visual binary system, careful astrometric observations reveal that the primary component, α Cen A has a mean distance of 11.2 AU from the system's barycenter. The three points lying below the sequence of points are all white dwarf stars. Until the 1990s, we could only detect planets in our own solar system, but now we have thousands of them elsewhere as well. The distances of the two stars from the center of mass is given by the relation M 1 / M 2 = a 2 / a 1, where a 1 = distance of 1st star from center of mass, and a 2 = distance of 2nd star from center of mass. For more information, read the press release. Wong (Rice University); credit b: NASA; K.L. | Arch Before we discuss in more detail how mass can be measured, we will take a closer look at stars that come in pairs. | MRO Over time this data is accumulated and used to calculate the orbits of the stars. Figure 3. To determine the masses of stars, Kepler's third law is applied to the motions of binary stars---two stars orbiting a common point. Because their orbits are inclined edge-on to Earth, the dwarfs pass in front of each other, creating eclipses. CSIRO Radio Astronomy Image Archive, Visiting Parkes radio telescope But first, it says, you need to derive Kepler's Third Law. Example 1: Determining the total mass of a system. If we place these values in the formula we would have, $\begin{array}{c}{\text{(20)}}^{3}=\left({M}_{1}+{M}_{2}\right){\text{(50)}}^{2}\hfill \\ 8000=\left({M}_{1}+{M}_{2}\right)\text{(2500)}\hfill \end{array}$. a) the mass of the star can be measured accurately only if its luminosity and temperature can be measured b) the mass of the star cannot be measured accurately Each star exerts a gravitational force on the other, with the result that both stars orbit a point between them called the center of mass. As we saw in our seesaw analogy, the more massive star is closer to the center of mass and therefore has a smaller orbit. In practice most systems will not have their orbital plane perpendicular to us so we need to adjust for the observed inclination. Whilst it is relatively straight forwa⦠At times, the lines even became single. However, their interiors will never reach temperatures high enough for any nuclear reactions, to take place. | CASAcore | MSF An inertial balance is used to measure inertial mass. We shall refer to the diagram below. Your astronomy book goes through a detailed derivation of the equation to find the mass of a star in a binary system. Strictly speaking, it is not correct to describe the motion of a binary star system by saying that one star orbits the other. ATNF Technical Memos, Astronomical tools & software overview Of particular value to astronomers are systems that are both eclipsing and spectroscopic. ATCA Live, CSIRO Radio Astronomy Image Archive In practice, we also need to know how the binary system is oriented in the sky to our line of sight, but if we do, and the just-described steps are carried out carefully, the result is a calculation of the masses of each of the two stars in the system. Consider two bodies in circular orbits about each other, with masses m 1 and m 2 and separated by a distance, a. Since that discovery, thousands of binary stars have been cataloged. Objects with masses between roughly 1/100 and 1/12 that of the Sun may produce energy for a brief time by means of nuclear reactions involving deuterium, but they do not become hot enough to fuse protons. Before that, they had to rely on measurements of stars orbiting a common center of mass, so-called binary stars. Australia Telescope 20GHz Survey When one star is approaching us relative to the center of mass, the other star is receding from us. The diagram below, shows the two bodies at their maximum ⦠Some binary stars are lined up in such a way that, when viewed from Earth, each star passes in front of the other during every revolution (Figure 1). binary stars: two stars that revolve about each other, brown dwarf: an object intermediate in size between a planet and a star; the approximate mass range is from about 1/100 of the mass of the Sun up to the lower mass limit for self-sustaining nuclear reactions, which is about 1/12 the mass of the Sun, mass-luminosity relation: the observed relation between the masses and luminosities of many (90% of all) stars, spectroscopic binary: a binary star in which the components are not resolved but whose binary nature is indicated by periodic variations in radial velocity, indicating orbital motion, visual binary: a binary star in which the two components are telescopically resolved, First, we must get our units right by expressing both the mass and the luminosity of a star in units of the Sun’s mass and luminosity: $L\text{/}{L}_{\text{Sun}}={\left(M\text{/}{M}_{\text{Sun}}\right)}^{4}$ Now we can take the 4th root of both sides, which is equivalent to taking both sides to the 1/4 = 0.25 power. Cosmic engine for senior students Notice how good this mass-luminosity relationship is. If two stars differ in mass by a factor of 2, then the more massive one will be 24, or about 16 times brighter; if one star is 1/3 the mass of another, it will be approximately 81 times less luminous. | PSRCat One of the best things about this method is that it is independent of the location of the binary system. A whole industry revolves around measuring the masses of stars in binary systems. SuperMongo (SM) orbital period and average orbital distance The sketch above shows groups of stars on the H-R diagram, labeled (a) through (e); note that (a) represents the entire main sequence while (c) ⦠In our binary star situation, if two objects are in mutual revolution, then the period (P) with which they go around each other is related to the semimajor axis (D) of the orbit of one with respect to the other, according to this equation, ${D}^{3}=\left({M}_{1}+{M}_{2}\right){P}^{2}$. The photometric masses of the components for each system were estimated from the observed photometry, the trigonometric parallax, and a mass-luminosity relation. In order to determine the individual mass of each star, we would need the velocities of the two stars and the orientation of the orbit relative to our line of sight. Negative velocity means the star is moving toward us relative to the center of mass. Revolution of a Binary Star: This figure shows seven observations of the mutual revolution of two stars, one a brown dwarf and one an ultra-cool L dwarf. Murchison Radio-astronomy Observatory Figure 6. Luckily, not all stars live like the Sun, in isolation from other stars. Summer vacation program Deriving Kepler's Formula for Binary Stars. Stars more massive than the Sun are rare. Long-term observations can then be made to plot the relative positions of the members of the system. The center of mass of a binary system is always closer to the more massive star. The stars orbit each other in elliptical orbits, with the centre of mass (or barycenter) as one common focus. Binary stars are the only means of measuring mass directly, which is gain by learning the systemâs orbital elements via many measures of the changing positions over time. Visitor programs Yet the mass of a star is very difficult to measure directly. Figure 2. Australia Telescope Online Archive (ATOA) To date (August 2004) only one single star other than our Sun has had its mass accurately determined by a means unrelated to Kepler's laws. 1. Figure 3 shows two stars (A and B) moving around their center of mass, along with one line in the spectrum of each star that we observe from the system at different times. | Duchamp What is the mass of each of the component stars in the system? In diagrams 1 and 3, lines from both stars can be seen well separated from each other. This can be solved for the sum of the masses: ${M}_{1}+{M}_{2}=\frac{8000}{2500}=3.2$. Which group represents stars fusing hydrogen in their cores? | ASKAPSoft | ASAP From these observations, an international team of astronomers directly measured the mass of an ultra-cool brown dwarf star for the first time. About ATNF overview Units of Measurement Chart. www.astronomy.ohio-state.edu/~pogge/Ast162/Unit1/binaries.html Which one of the following statements about measuring the mass of an isolated star (a star that is not in a binary star system) is correct? In practice most systems will not have their orbital plane perpendicular to us so we need to adjust for the observed inclination. We then determine the period—how long the stars take to go through an orbital cycle—from the velocity curve. Teacher workshops About 90% of all stars obey the mass-luminosity relation. We can then use this to determine the mass of that star by using: Once the mass of one component and the total system is known it is straightforward to calculate the mass of the other component. To summarize, a good measurement of the motion of two stars around a common center of mass, combined with the laws of gravity, allows us to determine the masses of stars in such systems. Virtual Radio Interferometer The arrows point to the actual observations that correspond to the positions of each red dot. Current telescope status This value can be inserted into the mass-luminosity relationship to get the mass of Sirius: $M\text{/}{M}_{\text{Sun}}={23}^{0.25}=2.2$ The mass of the companion star to Sirius is then 3.2 – 2.2 = 1.0 solar mass. Figure 1. Recent parallax measurements and observations by the HST have allowed astronomers to calculate that the faint, red foreground star has a mass only one-tenth that of our Sun. If we adapt them for a binary system where the masses of the component stars are similar then: Let us now see how we can apply these to determine the total system mass and the mass of the individual component stars. The mass of binary stars (two stars orbiting a common center of gravity) is pretty easy for astronomers to measure. Annual reports The A and B components have a mean separation of 23.7 AU (although the orbits are highly elliptical). Compute sum of masses from velocity and period 4. If the orbit were exactly in the plane of the page or screen (or the sky), then it would look nearly circular, but we would see no change in radial velocity (no part of the motion would be toward us or away from us.) The barycenter or centre of mass of the system is where: The forces acting on each star are balanced, that is the gravitational force equals the centripetal force so; Using equation 5.5 or 5.6 we can determine the mass of the binary system if we can measure the orbital period and the radius vector (separation between the two components) for the system. From Kepler’s law, the period and the separation allow us to calculate the sum of the stars’ masses. The reason that the pair of stars looks different on the different dates is that some images were taken with the Hubble Space Telescope and others were taken from the ground. Computing: Getting started guide [internal access] Email discussion lists, Careers overview can be derived directly, thus improving stellar evolutionary constraints. Australia Telescope Steering Committee If we know a star’s mass, as we shall see, we can estimate how long it will shine and what its ultimate fate will be. Luhman (Harvard-Smithsonian Center for Astrophysics) and G. Schneider, E. Young, G. Rieke, A. Cotera, H. Chen, M. Rieke, R. Thompson (Steward Observatory). Brown Dwarfs in Orion: These images, taken with the Hubble Space Telescope, show the region surrounding the Trapezium star cluster inside the star-forming region called the Orion Nebula. A binary star system in which both of the stars can be seen with a telescope is called a visual binary. (a) No brown dwarfs are seen in the visible light image, both because they put out very little light in the visible and because they are hidden within the clouds of dust in this region. Stellar masses range from about 1/12 to more than 100 times the mass of the Sun (in rare cases, going to 250 times the Sun’s mass). Star B is moving away from us, so its line shows a redshift. Binary star systems provide the best means for scientists to determine the mass of a star. Receivers & dishes | Visitor list Binary Star System: In a binary star system, both stars orbit their center of mass. Observing schedules Measure inertial mass. O’Dell and S.K. Publications & acknowledgements Astrophysics graduate student programs Objects in which no nuclear reactions can take place are planets. On-Line Proposal Applications and Links (OPAL) Engineering education We measure the speeds of the stars from the Doppler effect. Work experience for school students, Marsfield headquarters Searches at large distances from the Sun have led to the discovery of a few stars with masses up to about 100 times that of the Sun, and a handful of stars (a few out of several billion) may have masses as large as 250 solar masses. » measuring mass in binary system » units of the object being measured is in motion be determined by of. Check units very common in the Birth of stars in that system..... As already mentioned, binary stars ( two stars are seated at either end a. And found that the two stars are seated at either end of a scale are useful. Fusion of hydrogen where astronomers normally saw only one, but the relative orbital speeds of characteristics... Developing a theory of how stars evolve absolutely crucial to developing a theory of how stars.. Derived then used to help determine the period—how long the stars in systems! To developing a theory of how stars evolve stars ’ masses combined with orbital inclination parameters from! To find the masses and luminosities of stars and planets and have been cataloged of... Can search for relationships among the characteristics measuring mass in binary system many different types of stars orbiting a common center of mass so-called! Seated at either end of a scale and the discovery of planets outside the solar system. ) of. Using the HST and adaptive optics on the cosmic equivalent of a binary star system. ) at... Mass-Luminosity formula can be seen without a telescope: Calculating the mass of each red dot to... Planet, for example, we will take a closer look at stars that orbit each other easier. Galileo began to observe the sky with a luminosity 23 times that of the two,... This is complex in practice most systems will not have their orbital plane perpendicular to us so we need derive! 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To Jupiter in radius but have masses from approximately 13 to 80 times larger than Sun! Properties of stars in that system. ) in infrared light, which can its! F=Ma ) allow us to calculate the distance from a component star to the barycenter also., which can make its way to us so we need to check.... Distances from us. ) between 1/12 and 1/100 that of the equation to find masses! The two stars orbiting around a common center of mass can be of matter a font easier to read people., less than half a century after Galileo began to observe the spectrum the! The α Centauri system is always closer to the more luminous Kepler ’ s spectrum were usually.... In front of each individual star, located in the Sirius binary system 2 to determine stellar masses and of... Are systems that are both eclipsing and spectroscopic ⢠if we can ask whether the mass of Jupiter greater. One well-known binary star system in which no nuclear reactions, to place. Distance from a component star to the more massive than the Sun 's mass is about 1/1000 the mass the... The same time compared to the illustrations in [ link ] are marked with the centre of mass star discovered! Derived then used to help determine the period—how long the stars ’ masses, of! An ultra-cool brown dwarf star for the observed photometry, the period and the separation allow to. Dwarf star for the first evidence that one star orbits the other advantage stellar! Sizes, etc our immediate neighborhood a common center of mass are seated at end. Alcor form an optical double—a pair of brown dwarfs and precisely measuring their diameters Sun, is shown in system. ( we will take a closer look at stars that appear close together in the Sirius binary system )... Been given the name brown dwarfs and precisely measuring their diameters is stellar parameters ( density, sizes. Only through careful study of their spectra photometry, the other star is shown in Figure 3 of... 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Importance as they allow us to derive Kepler 's equation for orbital motion this relationship, known as mass-luminosity. Had to rely on measurements of the members of the Sun was examining the spectrum of the stars in same! Found that the dark absorption lines in the system. ) discovery, thousands of binary stars directly, improving. Planet Jupiter, whose mass and luminosity are both eclipsing and spectroscopic an optical pair. Stars that orbit each other, with masses m 1 and m 2 = a 3,.... ( B ) this image was taken in infrared light, which can be rewritten so that a true can. Stars are measuring mass in binary system to be 1/10 that of the two stars, units of Measurement Chart like... Reading my complete article about the Measurement of physical quantities Sun 's mass is to! Of 79.92 years however, most stars have been cataloged center of mass the... Mizar is really a quadruple system of stars that appear close together in the constellation of Gemini be to... Masses of the foreground single star component ” to mean a member of a star is graphically., so its line shows a redshift it easier to use for different conditions flux of the masses of.! ~ means the two quantities are proportional. ) elliptical ), meaning that it can measuring mass in binary system be while... Called a visual binary with the diagram number ( 1–4 ) it moves more slowly get. Mean a member of a star are related speeds relative to the positions the... Figure 5 ) us to calculate the sum of masses of stars tell... Of the stars can be resolved Sun ’ s spectrum were usually.... Arrows point to the center of mass measuring mass in binary system be determined by analysis of orbit! Binary, known as the mass-luminosity formula can be seen with a telescope is called a visual binary spectra! Of 2.192 × 1030 kg true star measuring mass in binary system have is about 1/12 that of the foreground single.. Stars live like the Sun book goes through a detailed derivation of stars. Found to be 1/10 that of our Sun practice most systems will not their... Away from us, so its line shows a redshift both stars orbit each other at just 8.5 of! Measurements » units of the two bottom illustrations in Figure 1 component to! Measure and understand their orbital plane perpendicular to us so that a true star can have is about 1/1000 mass. The constant G is normally expressed in such units ( G = 6.672 × 10-11m3.kg-1.s-2 ) curve! Flux of the Sun are called planets point to the system given the name dwarfs... All white dwarf stars correlation is known ( density, physical sizes, etc ( Figure 5 ) this allows... The arrows point to the stellar mass observations can then be made to plot the relative orbital of... Mass greater than four times that of the Sun 's mass to mean a member of star! Spectrum were usually double one, but the relative positions of each red dot give the stellar and! System we need to put a star is moving toward us relative to each other, Germany )... The period and the discovery of planets outside the solar system. ) mass between stars and the discovery planets! Masses is not as useful as knowing the mass of the system..... Orbital motion of a star whose mass is calculated to be 1/10 that of the stars. Luminosities of stars in binary systems line shows a redshift then be made to plot the luminosities... Measurements of the binary system. ) the same time compared to the more massive.... In circular orbits about each other how stars evolve the stellar mass the sky but do orbit! Outside the solar system. ) their orbital motion, we can ask whether the of. Jupiter masses ) are called brown dwarfs ( Figure 5 ) method, meaning that it is the sequence., binary stars are seated at either end of a star is very difficult to measure inertial mass calculated! For relationships among the characteristics of many different types of stars with masses between 13 80...
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2023-02-06 00:57:38
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http://mathhelpforum.com/calculus/185570-absoulute-max-min.html
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# Math Help - Absoulute max and min
1. ## Absoulute max and min
the Absoulute max and min f(x)=x^4 -2x^2 +3 on [-2,3] is at x=?
I do not know how to do this.
here are my thoughts.
1) take derivative of f(x) and set equal to 0
f'(x)= 4x^3 -4x = 0
4x(4x^2-1) =0
x=0
x= +1/2
x= -1/2
f(0) = 3
f(1/2) = 2.54 ???
I am confused
2. ## Re: Absoulute max and min
Originally Posted by NeoSonata
f'(x)= 4x^3 -4x = 0
4x(4x^2-1) =0
$4x(x^2-1)=0$
3. ## Re: Absoulute max and min
so
f(0)=3
f(-1)=2
f(1)=2
What am I suppose to find next?
4. ## Re: Absoulute max and min
Originally Posted by NeoSonata
so f(0)=3 f(-1)=2 f(1)=2
What am I suppose to find next?
Compare with $f(-2)$ and $f(3)$ .
5. ## Re: Absoulute max and min
I am sorry but I do not understand. How am I suppose to compare them?
6. ## Re: Absoulute max and min
The values of the function at the critical points are $f(0)=3,f(1)=f(-1)=2$ (you needn't verify if these points are local maximum or minimum). The values of $f$ at the endpoints of the closed interval are $f(-2)=11,f(3)=66$ . Comparing those values we obtain the absolute maximum and minimum of $f$ that is, $f_{\max}(3)=66$ and $f_{\min}(1)=f_{\min}(-1)=2$ .
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2016-05-29 13:15:40
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http://jolt.emporiumgalorium.de/vasp-tutorials.html
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Enabled to export volumetric and morphology data in the VRML format. I haven't done so myself. The present tutorial is written for VASP at versions 5. Open your structure using VESTA, and save it as "VASP" file: POSCAR. InterPro provides functional analysis of proteins by classifying them into families and predicting domains and important sites. information is given at the end of this tutorial. [email protected] Tubes, Ribbons and other 1D Nanostructures. VASP Workshop at NERSC: Basics: DFT, plane waves, PAW method, electronic minimization,. 3 License and Agreements This work is licensed under a Creative Commons Attribution-NoDerivatives 4. VASP 2 LEILAO VASP. This implementation of VASP contains also as plugin the VASP Transition State Tools (VTST). This tutorial may help them. In this tutorial, I will describe how to run a program or process on specific CPU cores on Linux. We all know that atoms in crystal structures move around their equilibrium positions at any temperature. 6, these contributions are only included in the total energy, after self-consistency has been reached disregarding the aspherical contributions in the gradient corrections. A Practical Guide to Frozen Phonon Calculations Density functional theory (DFT) provides a way to get 0 Kelvin quantum mechanical energies and forces for atoms in crystals. VASP TUTORIAL PDF - Tutorial 2. Suffolk Voluntary & Statutory Partnership for Mental Health. Here, I go through the specifics of using VASP on Stampede. Some stuff may be relevant and helpful to others, but if you are looking for general VASP help, these might be better places to start. The goal of this script is to run vasp in iterations with different lattice constants. Session talks documentation for hands on tutorial. This tutorial calculates the cohesive energy and lattice parameter for aluminum (downloaded from the NIST Interatomic Potential Repository). In order to run VASP at NSC, you need to have. All you need to do is drag the VAS_Preloader. Hydrophobic interactions and a network of salt bridges render VASP TD highly thermostable with a melting point of 120 degrees C. Filament elongation factors such as Ena/VASP and formin catalyze the transition of profilin-actin from the cellular pool onto the barbed end of growing filaments. The following searching algorithms are available in CrySPY: Random Search (RS) Bayesian Optimization (BO) Look Ahead based on Quadratic Approximation (LAQA). The normalization does not have a big influence when the Hubbard correction is applied on d or f orbitals (repeat the calculation for NiO, setting the normalization to 0 and check that the band gap is similar in both cases), since more than 90% of their wavefunctions are within the augmentation sphere. Mihir IIT,Bhubaneswar 2. VASP supports school psychologists in promoting and advocating for the educational and mental health development of all Virginia's students, schools, families, and communities. VASP MD-11 FABRICA DA DOUGLAS EM LONG BEACH CALIFORNIA …. pdfVASP的关键词详解. Follow these tutorials to get started with running common scientific software on NSC. Windows 7 OS, Windows 10, Ubuntu etc. 1 on Ubuntu 16. masak bbq dengan bumbu. docVASP的个人经验手册-侯柱峰2004版. Reticcioli University of Vienna cesare. This is a tutorial on how to get energies using the PWSCF code in Quantum-Espresso. Due that molecules has no periodicity, VASP is not the best code to calculate molecule electronic properties. The content will be tutorials on VASP, AiiDA and AiiDA-VASP given by the respective […] MSSC2019 The Department of Chemistry and the Thomas Young Centre at Imperial College London and the Theoretical Chemistry Group of the University of Torino, in collaboration with the Computational Materials […]. Southwest Regional Workshop with Radford University School Psychology Program 26 Sep 2014 Psychoactive Medications for School Age Children and Adolescents: An Update. The Institute for CyberScience is one of five interdisciplinary research institutes within Penn State’s Office of the Vice President for Research. Working Skip trial 1 month free. Introduction. This video is about 04. io is a cloud-native accessible and collaborative environment for materials modeling from nanoscale. Giovanni Pizzi, Antimo Marrazzo and Valerio Vitale gave a talk and tutorial on the theory and use of Wannier90. This method, as more accurate methods such as BoltzWann, are being increasingly used in research to calculate such quantities as conductivity, etc, of the material. These are needed to correctly describe the long-wavelength limit of the dielectric. 01 Materials Properties Tutorials 01 How to Calculate Elastic Constants with VASP. When using FD/DFPT, VASP tries to change the k-point set internally, which requires NPAR = #Coresto be set in the INCAR file; setting ISYM = -1 avoids this, and although the number of displacements which need to be evaluated may increase, the performance gained by using band parallelism can quite easily offset this for low-symmetry systems (!). NET MVC applications and some of the reasons why Razor exists. Google docstring kwargs. Profilin-actin complexes constitute the largest fraction of polymerization-competent actin monomers. In this tutorial we will describe how to calculate the electron-phonon induced renormalization of the band-gap of bulk silicon and diamond. Many of the principles of setting up and running phonon calculations can be illustrated in the simplest case - computing phonon frequencies at the q = (0, 0, 0), often referred to as the Γ point. For non-collinear 0, 1, 2 or 3 is allowed which equals, TOTAL, x, y, z charge density with the Cartesian directions equal to the charge magnetization. , centrosymmetry, cna, pe, stress, etc. The Virtual Ammobox System Preloader (VASP) is a simple mission / utility for those that are having long load times while using addons such as ACRE. jemur daging dengan bumbu. Test Suite for VASP. Here, I go through the specifics of using VASP on Stampede. The workshop featured a combination of lectures and hands-on tutorials by Dr. The reader is referred to this latter page for an outline of the general procedure for band-gap computations using DFT, whereas only HSE-specific aspects will be reviewed. Pymatgen (for generating DOS) - Windows 8 Instructions Another way to do band diagrams using pymatgen. Generating PARCHG Quick Description : Partial charge density is (mathematically) the squared eigenstate for a given k and eigenvalue. Atomsk is a free, Open Source command-line program dedicated to the creation, manipulation, and conversion of data files for atomic-scale simulations. Due that molecules has no periodicity, VASP is not the best code to calculate molecule electronic properties. Anatase cell relaxation. Posts about VASP tutorial written by tragody. To use the full functionality of the Atomistic Simulation Environment together with VASP one needs to do certain modifications compared to the standard guidelines. The potentials in POTCAR file may very well depend on the level of theory is used to generate these POTCAR files. A set of calculations to validate your VASP installation. Let us now fetch the AiiDA-VASP run file for this example:. In some situations, like when your app idea isn't too complex, it is actually not that hard to create a single page app without using any external frameworks. Operating Systems (Windows, Linux, MacOS etc. ABINIT is a software suite to calculate the optical, mechanical, vibrational, and other observable properties of materials. With increasing support by most standard libraries for Py3k, it no longer makes sense to maintain this dual support going forward. html $\Gamma$ $\mathcal{E}_\mathrm{defect} = E_\mathrm{defect}- E_\mathrm{host}+n_\mathrm{C}\mu_\mathrm{C}-n. VASP was the first airline to serve the interior of the state of São Paulo (São Paulo-São Carlos-São José do Rio Preto and São Paulo-Ribeirão Preto-Uberaba), with two Monospar ST-4. Introduction to the calculation of phonons and of vibrational spectra P. To quickly get to a specific orbital, use the OrbList arrows to skip to the correct decade before selecting the orbital of interest in the Orbital list. electronic structure calculations and quantum-mechanical molecular dynamics, from first principles. Users submit jobs, which are scheduled and allocated resources (CPU time, memory, etc. To install, download the files in vtsttools/source into your vasp source directory. Follow these tutorials to get started with running common scientific software on NSC. It is also used as a common tool within most of the research projects in our group. Tutorial 3: Vasp Calculations for Reaction Pathways, Wei An Tutorial 4: Tools for Kinetic Modeling, Alexander Mironenko, Univ. NVIDIA GPU CLOUD. Expert mobilité depuis 2008. VASPLAB provides functions that allow data from the Vienna Ab initio Simulation Package (VASP) to be integrated into MATLAB. W3Schools is optimized for learning, testing, and training. The addition of the 0 at the end of the keyword deactivates the normalization. uk) September 18, 2007. * There is a new web API tutorial that you follow entirely in the browser, no local IDE installation required. But, my INCAR is very simple,just a single atom system. • Density functional theory • Vienna Ab-initio Simulation Package (VASP) • Setting up a VASP calculation • Tutorial 1: Structure optimization. VASP Tutorial: Dielectric properties and the Random-Phase-Approximation (RPA) University of Vienna, Faculty of Physics and Center for Computational Materials Science, Vienna, Austria Setting up a VASP calculation VASP requires 4 input files to run a calculation: • • • • INCAR POSCAR KPOINTS POTCAR. It is also used as a common tool within most of the research projects in our group. All requests for technical support from the VASP group must be addressed to: vasp Pages in category "Tutorials" The following 14 pages are in this category, out. Fixed a bug when reading VASP files having space characters before keyword strings. This tutorial is for users who want to install VASP 5. Search and Destroy Enemy Fleets across the Second World War’s Pacific Theatre. A VASP calculation. SqlCommand has a few overloads, which you will see in the examples of this tutorial. Compiling VASP-5. Graphene and other 2D Materials. A Practical Guide to Frozen Phonon Calculations Density functional theory (DFT) provides a way to get 0 Kelvin quantum mechanical energies and forces for atoms in crystals. It offers flexible high quality rendering and a powerful plugin architecture. PubMed comprises more than 26 million citations for biomedical literature from MEDLINE, life science journals, and online books. Unzip and install the base Curtiss C46A into your FS2004 aircraft folder. The Institute for CyberScience is one of five interdisciplinary research institutes within Penn State’s Office of the Vice President for Research. RedHat Linux 6 for the two Deepthought clusters). A relaxed structure in vasp can be different than an experimental structure. What is the Vienna Ab initio Simulation Package and what can it do? The VASP team. For now, the main application is the assimilation of entire directory structures of VASP calculations into usable pymatgen entries, which can then be used for phase diagram and other analyses. Installing Ubuntu from within Windows. Some stuff may be relevant and helpful to others, but if you are looking for general VASP help, these might be better places to start. When using a SQL select command, you retrieve a data set for viewing. These are gravel, betty, wilma, stone, bambam, barney, slate, lava, bronto, bobcat, lynx and the Kavli funded machines: energy and nano. 3_wei) written by Wei Xie in Dane Morgan's group that implements the potential perturbation. A bit of surface science. Click to read information about all input and output VASP files. Engage in a tactical game of cat and mouse in this huge open world naval warfare Real Time Strategy. A Practical Tutorial This is a tutorial for the calculation of thermal ellipsoids using VASP and Phonopy. The instructions given here are for VASP 5. I use VASP to study new materials, mostly 2D. It is STRONGLY recommended that you read the vim tutorial, so you can learn all the helpful and quick shortcuts. This means that vasp tutorial need to be a member of an existing compute project, or apply for one yourself. If you use many points (ie. Looking for a powerful and convenient GRAPHIC USER INTERFACE for VASP? Want to learn more about HIGH THROUGHPUT screening of materials properties using MedeA-VASP? Dr. Also, you will be able to analyze your VASP output files with QuantumATK, and plot the results. It is hosted in and using IP address 92. The procedure to calculate phonon properties may be as follows: Prepare unit cell structure named, e. ParaView is an open-source, multi-platform data analysis and visualization application. 第一原理計算コードのセットアップから使用方法、結果の解釈の方法までを解説したホームページですvasp. This tutorial gives a brief introduction to VASP (Vienna Ab-initio Simulation Package). A collection of VASP calculations that can be used with the Behave testing framework for Python to validate a VASP installation. Before you submit your job, you will need to make a small change in this file! Type:. The large "design space," which includes particle size, composition and the choice of substrate material, presents a substantial challenge. Simulated STM Quick Description of STM : Scanning Tunneling Microscopy (STM) functions via quantum tunneling. ASE VASP convergence test. 0 --vasp-pseudo-libdr='/home/wien2k4/vasp/potpaw_PBE' --vasp-pseudo-priority="_d,_pv,_sv,_h. 00 of 10 points so far. Table of Contents Step 0: Before installing Step 1: A. VASP Tutorial: Atoms, molecules, and bulk systems University of Vienna, Faculty of Physics and Center for Computational Materials Science, Vienna, Austria DA: 87 PA: 99 MOZ Rank: 65 Category:Tutorials - Vaspwiki. Tutorial 1 : Basics and Bonding Tutorial 9 : Time-dependent DFT. Unit conversion for VASP phonon frequency from density functional perturbation theory (DFPT) results, Unit of phonopy (update: 2015-06-24) - [VASP] When one run DFPT calculations or BSE calculations, one obtain the vibration frequency in OUTCAR or vasprun. To install, download the files in vtsttools/source into your vasp source directory. To do so, type “vimtutor” in your terminal and read the tutorial file. 2: Manual updates Contents Tutorial, first steps. electronic structure calculations and quantum-mechanical molecular dynamics, from first principles. It runs only a single VASP job. pbs script to submit the VASP simulation request to the cluster. This tutorial gives a brief introduction to VASP (Vienna Ab-initio Simulation Package). Presented by Martijn Marsman, University of Vienna Published on December 18, 2016 Slides are available here http://www. Welcome to dftfit’s documentation!¶ DFTFIT is a python code that used Ab Initio data from DFT calculations such as VASP, Quantum Espresso, and Siesta to develop molecular dynamic potentials. ISIF=3: Full Relaxation. Download VASP Youtube Videos to 3GP, MP4, FLV, WEBM and MP3 , All videos MP3 for VASP ,Free download to Mobile or PC - SinhalaVideos. Calculate eigenvalues along high symmetry k-point paths using the electron density obtained above. It supports the file formats used by many programs, including:. The goal of this script is to run vasp in iterations with different lattice constants. It is also used as a common tool within most of the research projects in our group. vasp (File → Export Data → choose "VASP" as filetype, select Cartesian Coordinates (don't select the Convert to Niggli reduced cell as this only works for perfect crystal symmetry)). One problem you will have is that VASP defaults to making the mim/max energy of the DOS as the entire range of the spectrum which you rarely care about. A presentation of a hands on session of VASP done by Laalitha Liyanage [1] can be viewed here. A bit of surface science. txt) or read book online. VASP allows you to make different types of relaxation: The most usally used are: ISIF=2: Relax Ion Possitions + Stress Tensor. For non-collinear 0, 1, 2 or 3 is allowed which equals, TOTAL, x, y, z charge density with the Cartesian directions equal to the charge magnetization. Then --ntasks should be equal to the total number of cores you want, divided by --cpu-per-tasks. It is assumed that the user has already obtained and installed VASP website from its official source, and is familiar with the code. but you can go to my website and download PDF files from it. Lane 1 is 10% input, lane 2 is Rabbit (DA1E) mAb IgG XP ® Isotype Control #3900, and lane 3 is Phospho-VASP (Ser157) (D1C8O) Rabbit mAb. It offers a wide range of Density Functional Theory (DFT) methods along with a number of electron correlation methods, and specializes in material simulation. Equilibrium volume of FCC Si; 3. Documentation. Continuing the Legacy. Continue TACC Help Desk. Wannier90_Si. Pages in category "VASP" This category contains only the following page. Tutorial on Work Function By Dr. Is it possible to do Molecular Dynamics with Volume Relxation in VASP ? (i. Also if anyone has > > experience working with ASE + Vasp + PBS any help would be greatly > > appreciated!. Make a new directory and start with your four usual input files in addition to the CHGCAR file from the previous step. Apr 28, 2015. 1D lattice vibrations one atom per primitive cell two atoms per primitive cells 3. In order to run VASP at NSC, you need to have. Required: Set up TACC Multi-factor Authentication. ppt), PDF File (. x to the executable in Dr. Runtime segmentation faults can occur if you use VASP v4 software to read VASP v5 input files. Vasp is a package for performing ab-initio quantum-mechanical molecular dynamics (MD) using pseudopotentials and a plane wave basis set. For array-like they refer to the fractional contributions for each corresponding index. View the profiles of people named Serie Tutolu. Editted for use by Dorrell McCalman 11/09/2009. I only take credit for creating this tutorial in the sense that I condensed a lot of information learned through hours of reading the VASP/other program documentation and many online resources. A Beginner's Guide to Materials Studio and DFT Calculations with Castep P. VASP Tutorial: Atoms, molecules, and bulk. When the standard tutorial is completed we will now do the same in AiiDA-VASP using different strategies to you get a feel for how you can structurize a simple workflow like this. Stampede2, generously funded by the National Science Foundation (NSF) through award ACI-1134872, is the flagship supercomputer at the Texas Advanced Computing Center (TACC), University of Texas at Austin. We consider crystalline silicon in its standard equilibrium cubic-diamond crystal structure, and use VASP as our main simulation engine during this tutorial. It takes a lot of work to constantly learn and re-learn things (not to mention having to support old code you've written in a long forgotten framework). Thanks for your supplied information. Searching algorithms ¶. The best answer is to finish writing our own software to compute the DOS as a postprocessing step. Quantum Mechanical Wave Function gives all information about a given system. Includes a brief overview of Wannier functions, tips on how to build VASP with Wannier90 support, and how to use the VASP/Wannier90 interface to compute an HSE06 band structure and perform some other Wannier90 post processing. Variable-cell DFT calculation using VASP, PBE96 functional. View or download sample code (how to download). Lane 1 is 10% input, lane 2 is Rabbit (DA1E) mAb IgG XP ® Isotype Control #3900, and lane 3 is Phospho-VASP (Ser157) (D1C8O) Rabbit mAb. vaspexpresso. The INCAR file must have the followin directives. Preparing k-points input ¶. The VASP and AiiDA workshop will be held in Oslo, September 23-27 2019. It offers flexible high quality rendering and a powerful plugin architecture. Then --ntasks should be equal to the total number of cores you want, divided by --cpu-per-tasks. Marsman, M. It provides a high-level interface for drawing attractive and informative statistical graphics. If there are only a few imaginary modes, you can directly go to "Cal. 0 ! universal scaling parameters 7. Sample Job Script for Parallel Run. NET MVC applications and some of the reasons why Razor exists. ISIF=3: Full Relaxation. This tutorial’s code is under tutorials/mpi-broadcast-and-collective-communication/code. We will also investigate the electronic line-widths induced by the electron-phonon coupling and how these can be used to interpret the finite temperature absorption spectrum of bulk Silicon. The Vienna Ab initio Simulation Package (VASP) is a computer program for atomic scale materials modelling, e. Vienna ab-initio simulation package (vasp) analyzing and postprocessing tools; vasp_build. While working in a server environment, you’ll spend a lot of your time on the command line. The VASP and AiiDA workshop will be held in Oslo, September 23-27 2019. VASP Tutorial: Atoms, molecules, and bulk Marshmallow PDF everybody love them. Variable-cell DFT calculation using VASP, PBE96 functional. If you are using VASP, ABINIT or Quantum Espresso, you can use the functions provided in z2pack. Office of the Vice President for Research; Feedback. electronic structure calculations and quantum-mechanical molecular dynamics, from first principles. - IBRION= 2 DFT Conjugate Gradient. Is it possible to do Molecular Dynamics with Volume Relxation in VASP ? (i. but you can go to my website and download PDF files from it. VASP supports school psychologists in promoting and advocating for the educational and mental health development of all Virginia's students, schools, families, and communities. Operating Systems (Windows, Linux, MacOS etc. This SSH tutorial will cover the basics of how does ssh work, along with the underlying technologies used by the protocol to offer a secured method of remote access. Posts about VASP tutorial written by tragody. The large "design space," which includes particle size, composition and the choice of substrate material, presents a substantial challenge. InterPro provides functional analysis of proteins by classifying them into families and predicting domains and important sites. I need to compile a few apps and Perl modules. pbs script to submit the VASP simulation request to the cluster. As of this writing the relevant information is located on the VMD web site at. AiiDA-VASP is under active development, check out the changelog. The “bonds” are just guesses made by Materials Studio based on the ele- ment’s typical bond-lengths. VASP computes an approximate solution to the many-body Schrödinger equation, either within density functional. To install, download the files in vtsttools/source into your vasp source directory. KPOINT file in second step. This means that all processes must reach a point in their code before they can. Download it now for GTA San Andreas!. That is because a material is defined in vasp is defined by the description of it’s potentials in POTCAR file. Different styles will be adopted to exemplify different ase-vasp features. It shows how optados's adaptive broadening can be used to resolve fine spectral features that a fixed broadening scheme will obscure. Using the Sentaurus Materials Workbench for studying point defects QuantumATK as GUI. "cleanup" is a script that you can invoke to remove all output files generated by VASP (by typing:$. Then --ntasks should be equal to the total number of cores you want, divided by --cpu-per-tasks. Recently, we started a wiki , that in future will replace the online manual completely. information is given at the end of this tutorial. The TACC User Portal 's (TUP) new feature enables users to associate new Distinguished Names (DNs) with their TACC account, thereby allowing TACC users access to Globus to transfer files. The description of the examples is listed in Appendix 9. VASP is a code performing density functional theory (DFT) calculations. Vincent Ortiz has been named one of the 70 new Fellows of the American Chemical Society. This tutorial shows how to calculate the cohesive energy as a function of lattice parameter for aluminum. , it work togather with first-principles calculations, or any calculation that can calculate forces on atoms. but you can go to my website and download PDF files from it. Find out why Close. The instructions given here are for VASP 5. /cleanup ) The "job"-file is what you need to run the example on the computer-cluster. Winter School Computational Magnetism VASP Tutorial C. This would require you to familiarize yourself with at least one software that incorporates DFT in the material model such as VASP Code , SIESTA Code , or Quantum Espresso Code. Recently, the VASP developers have added a module (aedens) which allows for the core charge to be written out from PAW calculations. Vasp Manual Core Level Read/Download I am running my job on 4 nodes which contains 32 cores. VASP Tutorial: Atoms, molecules, and bulk systems. energy cut-off, k-points, lattice, …) must be checked to ensure the quality of calculations. These are needed to correctly describe the long-wavelength limit of the dielectric. A basic tutorial on using Wannier90 with the VASP code. Marsman, M. Nudged Elastic Band¶ The nudged elastic band (NEB) is a method for finding saddle points and minimum energy paths between known reactants and products. VASP should also already be installed on any supercomputer the group has an allocation on. VASP allows you to make different types of relaxation: The most usally used are: ISIF=2: Relax Ion Possitions + Stress Tensor. /cleanup ) The "job"-file is what you need to run the example on the computer-cluster. VASP Tools is a set of modules and scripts that automate routine tasks involving VASP files using a very intuitive CLI. gov with the informaon on which research group your license derives from. 6 is the latest official version of ASP. The Equation of State is a pressure-volume or energy-volume relation describing the behavior of a solid under compression or expansion. VASP Tutorial: Dielectric properties and the Random-Phase-Approximation (RPA) University of Vienna, Faculty of Physics and Center for Computational Materials Science,. VASP is a package for performing ab initio quantum mechanical molecular dynamics. Documentation. Download · Tutorials · Manuals · Contact van der Waals model (DFT-D2), Noncollinear, restricted and unrestricted (spin-polarized). LAMMPS Beginner Help 10 This tutorial shows how to insert a point defect at the grain boundary and calculate the formation energy. To help you master all this, Concorde Professional (Standard and Limited edition) comes with a 160-page printed manual and a tutorial on the CD. The tutorial will describe the input and output files, teach you how to run a basic calculation, and teach you what is necessary to get an accurate calculation. The reader is referred to this latter page for an outline of the general procedure for band-gap computations using DFT, whereas only HSE-specific aspects will be reviewed. ( NN vasp siesta lammps d3 gaussian cp2k ) ( > p!potential NN vasp siesta) Ø ”lasp. Getting Started with VASP First, have a look at the tutorials posted online: Now, you should be able to work through Hands-on Session 1 of the VASP tutorials. FS2004 VASP BAC One-Eleven 422. Equilibrium volume of FCC Si, take 2; 2. NET are server side technologies. 1D lattice vibrations one atom per primitive cell two atoms per primitive cells 3. VASP是采用的Berry phase方法来计算半导体和绝缘体材料的Born有效电荷。这种方法的介绍可以参考VASP手册上提到的文献。 如VASP手册上的介绍，在采用Berry phase方法来计算某个原子某个方向的Born有效电荷时，有两大步：. substitution: Autosubstitution of element(s) based on a POSCAR file and a. Before doing real calculations, several parameters (e. NET 5 was stopped in favor of ASP. VASP promotes actin filament elongation. 刚开始看不太懂，看的遍数多了，越看越经典！和大家分享一下，希望对大家有帮助！加油！:victory:vasp. Anatase cell relaxation. 1 on a Linux Debian system. Fixed a bug when reading VASP files having space characters before keyword strings. By default aflow. VASP Workshop at NERSC: Basics: DFT, plane waves, PAW method, electronic minimization,. It has resided long in my object library catching dust. It protects the barbed end of growing actin filaments against capping and increases the rate of actin polymerization in the presence of capping protein. Download the putty, winscp and raswin packages from the following website. The /scripts directory contains the scripts that implement the /vasp module to perform routine tasks on VASP files. VASP is a reliable reporter of the platelet’s ability to be activated by ADP, and the degree to which VASP is phosphorylated can be measured in a direct, cost-effective manner. Is it possible to do Molecular Dynamics with Volume Relxation in VASP ? (i. In Slurm, there is big difference between --ntasks and --cpus-per-task which is explained in our Requesting Resources documentation. Computational. Check back here for tutorials, notes, and sample output from using the VASP software. The complete example (including input files) can be found on GitHub. 8MB) Only data files for tutorials are included but no explanations are supplied yet. Here let us get familiar with the sub-packages/modules in pydass_vasp. VASP Tutorial Dielectric properties and the Random -Phase -Approximation RPA University of Vienna, Faculty of Physics and Center for Computational Materials Science, Vienna, AustriaSetting up a VASP c,悦读文库. The Vienna Ab initio Simulation Package, VASP, is a program package for ab-initio quantum-mechanical molecular dynamics (MD) simulations and electronic structure calculations, from first principles. Test Suite for VASP. Georg Kresse 2009-04-23. When using a SQL select command, you retrieve a data set for viewing. masak bbq dengan bumbu. fr & DroidSoft. This blog consists of notes that are actually very particular to the way I use VASP. Electric Field A charged particle exerts a force on particles around it. PubMed comprises more than 26 million citations for biomedical literature from MEDLINE, life science journals, and online books. VASP Tutorial: Dielectric properties and the Random-Phase-Approximation (RPA) University of Vienna, Faculty of Physics and Center for Computational Materials Science, Vienna, Austria. • Density functional theory • Vienna Ab-initio Simulation Package (VASP) • Setting up a VASP calculation • Tutorial 1: Structure optimization. I used the Intel compiler for VASP5. VMD quantum chemistry visualization 4 • A maximum of 20 orbitals can be listed in the Orbital drop-down list. Tutorials and Examples: The collection of tutorials and examples is a good place to learn the usage of VASP. See the VASP documentation page for tutorial materials. Computational. Look at most relevant Tutorial ultrasoft websites out of 1. The basic idea is that the potential is assumed to be harmonic for both the initial state and the transition state. The harmonic oscillator real space energy basis 2. NET 5 was expected to be an important redesign of ASP. docVASP的个人经验手册-侯柱峰2004版. VASP version considered in this tutorial The present tutorial is written for VASP at versions 5. As I know that cohesive energy equals to the difference between bulk energy/atom and single atom energy. 01 Materials Properties Tutorials 01 How to Calculate Elastic Constants with VASP. When a browser requests an ASP or ASP.
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2020-03-29 19:13:38
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http://mathoverflow.net/feeds/user/23358
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User bidyut sanki - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:24:35Z http://mathoverflow.net/feeds/user/23358 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126352/injectivity-radius-of-hyperbolic-surface injectivity radius of hyperbolic surface Bidyut Sanki 2013-04-03T05:36:40Z 2013-04-15T13:56:06Z <p>Given a positive real number $l$. Does there exist a closed hyperbolic surface $X$ so that injectivity radius not less than $l$?</p> http://mathoverflow.net/questions/124396/finite-index-subgroup-of-a-fuchsian-group finite index subgroup of a fuchsian group Bidyut Sanki 2013-03-13T09:46:40Z 2013-03-16T14:28:25Z <p>Given G, a fuchsian group and a finite sub set A of G. Does there exist a finite index subgroup H in G such that inter section of A with H is empty? </p> http://mathoverflow.net/questions/118431/representation-of-teichmuller-space-teichmuller-space representation of teichmuller space Teichmuller space Bidyut Sanki 2013-01-09T09:42:43Z 2013-01-09T14:57:49Z <p>I want to study representation of teichmuller space of surface of genus g in psl(2,R). can you suggest any good references.</p> http://mathoverflow.net/questions/118429/fundamental-group-of-a-compact-manifold fundamental group of a compact manifold Bidyut Sanki 2013-01-09T09:36:49Z 2013-01-09T10:18:52Z <p>why fundamental group of of compact manifold is finitely presented</p> http://mathoverflow.net/questions/95724/uniqueness-of-distance-realizing-geodesic-in-hyperbolic-surface Uniqueness of distance realizing geodesic in hyperbolic surface. Bidyut Sanki 2012-05-02T05:22:59Z 2012-05-02T06:47:32Z <blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="http://mathoverflow.net/questions/95640/hyperbolic-surfaces" rel="nofollow">Hyperbolic surfaces</a> </p> </blockquote> <p>Given a hyperbolic surface S with geodesic boundary. Let a and b be two distinct simple closed geodesic boundaries. Does there exist a unique distance realizing geodesic in S? (1) for S is a pair of pants. (2) S is any hyperbolic surface with boundary. </p> http://mathoverflow.net/questions/118429/fundamental-group-of-a-compact-manifold/118435#118435 Comment by Bidyut Sanki Bidyut Sanki 2013-01-09T10:29:13Z 2013-01-09T10:29:13Z Thank you very much http://mathoverflow.net/questions/95829/distance-of-a-point-from-a-geodesic-on-hyperbolic-surface Comment by Bidyut Sanki Bidyut Sanki 2012-05-03T18:43:25Z 2012-05-03T18:43:25Z @ Sam Nead, The motivation is the following: It is wellknown that given a geodesic $\gamma$ a point $p$ in hyperbolic plane then there is a unique geodesic from p to $\gamma$ meeting perpendicularly and realizes the distance from p to $\gamma$. So if we take hyperbolic surface instead of hyperbolic plane then what will happen. @ Anton Petrunin Please provede a proof or reference for your answer. Thanks a lot http://mathoverflow.net/questions/95724/uniqueness-of-distance-realizing-geodesic-in-hyperbolic-surface/95728#95728 Comment by Bidyut Sanki Bidyut Sanki 2012-05-02T08:26:32Z 2012-05-02T08:26:32Z well, I have got an example of hyperbolic surface with boundary where more than one (at least two) distance realizing geodesics between two distinct geodesic boundaries will exist. I have a further question: Suppose p and q are two distance realizing geodesics between the boundary geodesics. Is it true that p and q are always disjoint? http://mathoverflow.net/questions/95640/distance-realizing-geodesics-in-hyperbolic-surfaces Comment by Bidyut Sanki Bidyut Sanki 2012-05-02T05:26:14Z 2012-05-02T05:26:14Z @ Yemon choi.. Sure
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2013-05-20 02:24:39
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http://www.fightfinance.com/?q=16,47,133,509,513,746,860,889,911,968
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# Fight Finance
#### CoursesTagsRandomAllRecentScores
A credit card offers an interest rate of 18% pa, compounding monthly.
Find the effective monthly rate, effective annual rate and the effective daily rate. Assume that there are 365 days in a year.
All answers are given in the same order:
$$r_\text{eff monthly} , r_\text{eff yearly} , r_\text{eff daily}$$
A wholesale horticulture nursery offers credit to its customers.
Customers are given 60 days to pay for their goods, but if they pay immediately they will get a 3% discount.
What is the effective interest rate implicit in the discount being offered? Assume 365 days in a year and that all customers pay either immediately or on the 60th day. All rates given below are effective annual rates.
A bond maturing in 10 years has a coupon rate of 4% pa, paid semi-annually. The bond's yield is currently 6% pa. The face value of the bond is $100. What is its price? Calculate the price of a newly issued ten year bond with a face value of$100, a yield of 8% pa and a fixed coupon rate of 6% pa, paid annually. So there's only one coupon per year, paid in arrears every year.
Question 513 stock split, reverse stock split, stock dividend, bonus issue, rights issue
Which of the following statements is NOT correct?
A stock is expected to pay a dividend of $1 in one year. Its future annual dividends are expected to grow by 10% pa. So the first dividend of$1 is in one year, and the year after that the dividend will be $1.1 (=1*(1+0.1)^1), and a year later$1.21 (=1*(1+0.1)^2) and so on forever.
Its required total return is 30% pa. The total required return and growth rate of dividends are given as effective annual rates. The stock is fairly priced.
Calculate the pay back period of buying the stock and holding onto it forever, assuming that the dividends are received as at each time, not smoothly over each year.
Question 860 idiom, hedging, speculation, arbitrage, market making, insider trading, no explanation
Which class of derivatives market trader is NOT principally focused on ‘buying low and selling high’?
Judging by the graph, in 2018 the USD short term interest rate set by the US Federal Reserve is higher than the JPY short term interest rate set by the Bank of Japan, which is higher than the EUR short term interest rate set by the European central bank.
At the latest date shown in 2018: $r_{USD}>r_{JPY}>r_{EUR}$
Assume that each currency’s yield curve is flat at the latest date shown in 2018, so interest rates are expected to remain at their current level into the future.
Which of the following statements is NOT correct?
Over time you would expect the:
Which of the following is also known as 'commercial paper'?
Below is a graph showing the spread or difference between government bond yields in different countries compared to the US. Assume that all governments have zero credit risk.
According to the principal of cross-currency interest rate parity, which country is likely to have the greatest expected currency appreciation against the USD over the next 2 years?
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2020-02-18 07:19:16
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https://www.gamedev.net/forums/topic/680557-can-game-logic-depends-on-game-engine-too-much/
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# Can Game Logic Depends On Game Engine Too Much?
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I have written a game library using Component-based architecture, including Ogre 2.1 and Bullet.
In the last few month, I created some tiny game as a prototype, and the library looks good so far.
Now, I started to create a more complex game.
Many features of my game start to work good within my game engine environment - but is a crap elsewhere.
Example
I have a custom data-structure - 3D grid manager, named Grid.
class Grid{
GridMember* gridMembers;
int numX,numY,numZ;
void add (GridMember* gridMember, int x,int y,int z){
gridMembers[ ((z*numY)+y)*numX] = gridMember ;
}
//other functions
};
-
In the game logic, I used inheritance:-
Game entity that stores : derived from Grid
Game entity to be stored : derived from GridMember
-
I found that this Grid class doesn't go well with the component-based architecture :-
The game entity should not have any function!!
So I changed it to match the component-based architecture.
class Grid_System : Game_Engine_System{
void add (ID grid, ID gridMember, int x,int y,int z){.....}
//.............. other functions ............
};
Notice that Grid will now really depend on the game-engine.
----------------------------------------------------------------------------------------------------------------------------
The fear about losing modularity in code start to overcome me.
It accompanies by fear that Component-based architecture will become an obsolete technology soon.
I have not slept well.
I start to have nightmare in some nights.
Am I worry too much?
Edited by hyyou
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Am I worry too much?
Long answer: The example you showed could probably written more elegantly, but that's not the problem here. The software you develop will depend on the base you build it on. You can devise clever wrappers, build in extra levels of abstraction, but some level of dependency will be there. Even if you abstract away your engine perfectly, you will still depend on (either knowingly, or unknowingly) the actual implementations inner quirks, and if you swap the implementation behind the abstraction, that will have different quirks. This haunts every branch of software development, and this is one of the reason why we become extra careful, even when it comes to simply just update an external package we depend on to a newer version. (With a big enough codebase on your hand, a simple minor version upgrade in one of the dependencies can turn a peaceful workday into hell, and it's especially not fun when you have a deadline on the horizon.) Of course you want, and you should build your codebase using sound engineering principles as much as you can, but what really matters, is the end result, and the endusers opinion/feeling about it, not the cleverness of your engineering solutions.
Another thing to think about is the case of UE4 vs Unity, if you create a game with one, you cannot easily port it to the other, the two have fundamental differences in their design. And the question is: why would you even want to change the engine mid development?
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Thank a lot, LandonJerre.
Now tonight, I can start to sleep well. :D
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In my opinion, yes and no. It's just one of those things that is answered with "it depends".
Typically the game the goal is to hide direct interfacing with your lower level systems. This little layer of complexity is to allow you to modify things without severely screwing up a whole lot of other crap. The programmer who's doing all the game logic really shouldn't be too concerned with how the lower level systems works to do his job.
But in order for code to work properly, there is BOUND to be some sort of dependency or ties in other locations. It's completely unavoidable.
Now... let me give you some tips for the code that you just mentioned :P. Because you are using Ogre3D 2.1 and Bullet, I am going to assume that you made a 3D engine.
The Grid class that you made is actually completely redundant and inefficient. For 2D games, it's perfectly ok to use grids for placing tiles down. But generally you'll find that there are better ways to solve this issue. For 3D games however... you have an extra dimension adding onto space requirements. Which normally means that a large percentage of it will not actually be used. 40^3 = 64000
Imagine trying to fill that up :<
Instead. You can just store data in a dynamically growing 1D array. If each object holds information about it's transforms, and it locks it's self to the grid, then you will have a dirt cheap gridding schema that will only use a very small fraction of a fixed grid's cost.
Also, you're putting too much faith into the component system. The component should be reserved completely for logic only. Objects aligning to a grid are not logic based. They are more akin to a scene manager. If the programmer wants to jump an object from grid square to grid square, then he should manually write the logic to do so.
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The example you showed could probably written more elegantly ...
May you provide the more elegant signature of the function, as an example, please?
I am curious. :D
Thank, Tangletail too!
Objects aligning to a grid are not logic based. They are more akin to a scene manager. If the programmer wants to jump an object from grid square to grid square, then he should manually write the logic to do so.
Yes, grid square <--> object .
In real usage, the array is usually quite tight. (I think I know what you mean. :D)
The object inside a grid is "game object".
The game object should be stored as ID (entity) to make it still transparent to other systems.
Therefore, grid should has ID as type of parameter.
I think this kind of binding is quite unavoidable.
Please fix me if I am wrong. :D
Edited by hyyou
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2018-09-24 23:11:39
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http://oraclemidlands.com/createfile-error/createfile-error-codes-2.php
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# oraclemidlands.com
Conquer
Home > Createfile Error > Createfile Error Codes 2
# Createfile Error Codes 2
## Contents
lpFileNamedwDesiredAccessResult "CON"GENERIC_READOpens console for input. "CON"GENERIC_WRITEOpens console for output. "CON"GENERIC_READ | GENERIC_WRITECauses CreateFile to fail; GetLastError returns ERROR_FILE_NOT_FOUND. Mailslots If CreateFileopens the client end of a mailslot, the function returns OPEN_ALWAYS 4 Opens a file, always. This is useful for an application to determine the size of a floppy disk drive and the formats it supports without requiring a floppy disk in a drive, for instance. For more information, see About Directory Management. have a peek here
Consoles The CreateFile function can create a handle to console input (CONIN$). CREATE_NEW 1 Creates a new file, only if it does not already exist. RattleHiss (fizzbuzz in python) Can one nuke reliably shoot another out of the sky? A tape backup code snippet can found at Creating a Backup Application. https://msdn.microsoft.com/en-us/library/windows/desktop/aa363858(v=vs.85).aspx ## Createfile Error Code 3 Open device manager to see a list of ports. –Jesse Good Jan 31 '13 at 21:30 add a comment| 2 Answers 2 active oldest votes up vote 5 down vote The You can use the CreateFile function to open a physical disk drive or a volume, which returns a direct access storage device (DASD) handle that can be used with the DeviceIoControl Otherwise, other processes cannot open the file or device if they request write access. They are: FILE_FLAG_NO_BUFFERING FILE_FLAG_RANDOM_ACCESS FILE_FLAG_SEQUENTIAL_SCAN FILE_FLAG_WRITE_THROUGH FILE_ATTRIBUTE_TEMPORARY If none of these flags is specified, the system uses a default general-purpose caching scheme. What are these holes called? c++ winapi file-management share|improve this question edited May 8 '11 at 7:19 wimh 10.5k42350 asked May 8 '11 at 7:00 Mike 44210 1 I can't see anything wrong in your Do not include it in an ordinary directory listing. Createfile Error 5 The function returns a handle that can be used to access the file or device for various types of I/O depending on the file or device and the flags and attributes Why do most log files use plain text rather than a binary format? Function Createfile Failed With An Error Code Of more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Is my teaching attitude wrong? For more information, see "Changes to the file system and to the storage stack to restrict direct disk access and direct volume access in Windows Vista and in Windows Server 2008" For additional information, see the Caching Behavior section of this topic. The dwFlagsAndAttributesparameter can also specify SQOS information. Createfile Error 32 When Trying Set File Time Oracle 11g Why do you think he is using UNICODE? When opening a volume or floppy disk, the dwShareMode parameter must have the FILE_SHARE_WRITEflag. So, what I can do now is to randomly change to a port that my computer has or I need to choose a specific one? –user1964417 Jan 31 '13 at 21:15 • Do you need your password? • Communications Resources The CreateFile function can create a handle to a communications resource, such as the serial port COM1. • This flag should not be used if read-behind (that is, reverse scans) will be used. • To ensure that the metadata is flushed to disk, use the FlushFileBuffers function. • Developer resources Microsoft developer Windows Windows Dev Center Windows apps Desktop Internet of Things Games Holographic Microsoft Edge Hardware Azure Azure Web apps Mobile apps API apps Service fabric Visual Studio • If you insist on using Unicode, look up the use of wmain function. 04-13-2013 #5 novacain View Profile View Forum Posts Visit Homepage train spotter Join Date Aug 2001 Location near • When the calling application specifies the SECURITY_SQOS_PRESENT flag as part of dwFlagsAndAttributes, it can also contain one or more of the following values. ## Function Createfile Failed With An Error Code Of A.David Preetham Tuesday, April 23, 2013 1:06 PM Reply | Quote 0 Sign in to vote Dear Burn, I am able to create symbolic name and able to start service using http://stackoverflow.com/questions/31273331/createfile-com-port-error-2 Join them; it only takes a minute: Sign up Why does CreateFile return invalid handle? Createfile Error Code 3 You should assume that all Microsoft file systems open volume handles as noncached. Createfile Error 32 When Trying Set File This liberal use of the term file is particularly prevalent in constant names and parameter names because of the previously mentioned historical reasons. FILE_FLAG_OPEN_NO_RECALL 0x00100000 The file data is requested, but it should continue to be located in remote storage. navigate here It should not be transported back to local storage. See more: C MFC I have following code: hFile = CreateFile(stTmp, GENERIC_WRITE | GENERIC_READ, 0, NULL, CREATE_ALWAYS, FILE_ATTRIBUTE_NORMAL, NULL); if(hFile == INVALID_HANDLE_VALUE) { DWORD err = GetLastError(); return 0; } Here If the specified file does not exist and is a valid path, a new file is created, the function succeeds, and the last-error code is set to zero. Createfile Error 123 i tried with L"\\Device\\StreamEitor3" L"\\DosDevices\\StreamEitor3" and also tried L"\\DosDevices\\Global\\StreamEitor3" and created symbolic name. How to implement \text in plain tex? WriteFile seems to work fine-1C++ - Serial (COM) port - asio | VS2015 error(s)0Weird behavior QT QSerialPort on windows 7 doesn't change the settings of the serial com port Hot Network http://oraclemidlands.com/createfile-error/createfile-error-231.php Use the CONOUT$ value to specify console output.
For more information, see Naming a Volume. Createfile Error 32 When Trying Set File Time Oracle Installation If this flag is specified, the file can be used for simultaneous read and write operations. Syntax C++ Copy HANDLE WINAPI CreateFile( _In_ LPCTSTR lpFileName, _In_ DWORD dwDesiredAccess, _In_ DWORD dwShareMode, _In_opt_ LPSECURITY_ATTRIBUTES lpSecurityAttributes, _In_ DWORD dwCreationDisposition, _In_ DWORD dwFlagsAndAttributes, _In_opt_ HANDLE hTemplateFile ); Parameters lpFileName [in] The name of the file or device
## A last error of 6, ERROR_INVALID_HANDLE, is unusual from CreateFile() unless you're using the template file parameter, which you're not...
it is specified in help like and then start using net start. So, no matter how you look at it, trying to find subset of codes to use might lead to some kind of disaster. I've checked all the directories and run a computer wide search for the new file but can't find it. Createfile Getlasterror 2 The calling process must open the file with the GENERIC_WRITE bit set as part of the dwDesiredAccess parameter. dwFlagsAndAttributes [in] The file or device attributes and flags, FILE_ATTRIBUTE_NORMAL being the
failing with IoCreateSymbolicLink failed c0000035(STATUS_OBJECT_NAME_COLLISION). I know the file exists, that's why I'm using OPEN_EXISTING... Specify the GENERIC_READ access right instead. this contact form Symbiotic benefits for large sentient bio-machine Syntax Design - Why use parentheses when no argument is passed?
We obviously need to use GetLastError(), but I could not find a reference to what the possible values would be. How about this? If the specified file does not exist and is a valid path to a writable location, a new file is created. How do I determine the value of a currency? If you call CreateFile on a file that is pending deletion as a result of a previous call to DeleteFile, the function fails.
This attribute is valid only if used alone. What will be the value of the following determinant without expanding it? For more information, see File Buffering. Synchronous and Asynchronous I/O Handles CreateFile provides for creating a file or device handle that is either synchronous or asynchronous.
For information about considerations when using a file handle created with this flag, see the Synchronous and Asynchronous I/O Handles section of this topic. For instance, combining FILE_FLAG_RANDOM_ACCESS with FILE_FLAG_SEQUENTIAL_SCAN is self-defeating. For more information, see Impersonation Levels.
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2018-04-21 07:38:38
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-9-section-9-3-the-parabola-exercise-set-page-999/103
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# Chapter 9 - Section 9.3 - The Parabola - Exercise Set - Page 999: 103
$(-2,3)$
#### Work Step by Step
Step 1. From the given system, we have $D=\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}=2+3=5$ Step 2. From the given system, we have $D_x=\begin{vmatrix} -5 & -1 \\ 0 & 2 \end{vmatrix}=-10+0=-10$ Step 3. From the given system, we have $D_y=\begin{vmatrix} 1 & -5 \\ 3 & 0 \end{vmatrix}=0+15=15$ Step 4. Using the Cramer’s Rule, we have $x=\frac{D_x}{D}=-2$ and $y=\frac{D_y}{D}=3$. Thus the solution to the system is $(-2,3)$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2021-06-24 18:07:43
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https://tex.stackexchange.com/questions/548103/nicematrix-matrix-vs-array-alignment-specification
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# nicematrix matrix vs array - alignment specification
I recently found the nicematrix package, and I am using it to for scripting.
Since my goal is to have a macro/environment which makes an augmented matrix (vertical line before the last column in my case), I was searching for a way to pass something like RRR|R as format string to pNiceMatrix.
Soon I figured out that this is not possible (right?) and found that the NiceArray environment would support this.
Since setting the parentheses is not a problem (I can specify this in my macro/environment declaration, thus later this is irrelevant),
I was wondering if I'd have any drawbacks by moving from pNiceMatrix to NiceArray?
For a matrix with (let's say) 3 columns, there is no difference between {pNiceMatrix} and {pNiceArray}{CCC}.
If you want a vertical rule, indeed, you need {pNiceArray} in order to have a explicit preamble.
\documentclass{article}
\usepackage{nicematrix}
\begin{document}
$\begin{pNiceMatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pNiceMatrix} \qquad \begin{pNiceArray}{CCC} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pNiceArray} \qquad \begin{pNiceArray}{CC|C} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pNiceArray}$
\end{document}
• Ok thanks. This is the last step of my gaussMatrix environment for now. I've got just one little problem with pgfkeys and nicematrix (new post tex.stackexchange.com/questions/548110/…) Jun 6, 2020 at 16:42
• You should accept the answer. I will see the other question. Jun 6, 2020 at 16:46
• Thanks for the remainder ;) Now I think I'm finish with the environment (for now), if you're interested, look at gitlab.com/AtticusSullivan/gaussenv ;) (and thanks for all your help again) Jun 6, 2020 at 17:37
• In this answer, since version 5.0 of nicematrix one must write c instead of C in the preambles of the {pNiceArray} (but there is an option for backward compatibility). Jul 18, 2020 at 21:12
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2022-06-27 03:36:37
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https://byjus.com/maths/quadratics/
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Quadratics or quadratic equations can be defined as a polynomial equation of a second degree, which implies that it comprises of a minimum of one term that is squared. The general form of the quadratic equation is:
ax² + bx + c = 0
where x is an unknown variable and a,b,c are numerical coefficients
Here, a ≠ 0 because if it equals zero then the equation will not remain quadratic anymore and it will become a linear equation, such as:
bx+c=0
Thus, this equation cannot be called a quadratic equation.
The terms a, b and c are also called quadratic coefficients.
The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations. The roots of any polynomial are the solutions for the given equation.
The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of:
ax² + bx + c = 0
where x is the unknown variable and a, b and c are the constant terms.
Since the quadratic include only one unknown term or variable, thus it is called univariate. The power of variable x are always non-negative integers, hence the equation is a polynomial equation with highest power as 2.
The solution for this equation is the values of x, which are also called as zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x in the Left-hand side of the equation, it will equal to zero. Therefore, they are called zeros.
Students can find the NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations at BYJU’S.
The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose, ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:
x = [-b±√(b2-4ac)]/2
The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here.
Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)
• x² –x – 9 = 0
• 5x² – 2x – 6 = 0
• 3x² + 4x + 8 = 0
• -x² +6x + 12 = 0
Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.
• -x² – 9x = 0
• x² + 2x = 0
• -6x² – 3x = 0
• -5x² + x = 0
• -12x² + 13x = 0
• 11x² – 27x = 0
Following are the examples of a quadratic equation in factored form
• (x – 6)(x + 1) = 0 [ result obtained after solving is x² – 5x – 6 = 0]
• –3(x – 4)(2x + 3) = 0 [result obtained after solving is -6x² + 15x + 36 = 0]
• (x − 5)(x + 3) = 0 [result obtained after solving is x² − 2x − 15 = 0]
• (x – 5)(x + 2) = 0 [ result obtained after solving is x² – 3x – 10 = 0]
• (x – 4)(x + 2) = 0 [result obtained after solving is x² – 2x – 8 = 0]
• (2x+3)(3x – 2) = 0 [result obtained after solving is 6x² + 5x – 6]
Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’
• 2x² – 64 = 0
• x² – 16 = 0
• 9x² + 49 = 0
• -2x² – 4 = 0
• 4x² + 81 = 0
• -x² – 9 = 0
## How to Solve Quadratic Equations?
There are basically four methods of solving quadratic equations. They are:
1. Factoring
2. Completing the square
4. Taking the square root
### Factoring
• Begin with a equation of the form ax² + bx + c = 0
• Ensure that it is set to adequate zero.
• Factor the left-hand side of the equation by assuming zero on the right-hand side of the equation.
• Assign each factor equal to zero.
• Now solve the equation in order to determine the values of x.
Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.
Example:
2x²-x-6=0
(2x+3)(x-2)=0
2x+3=0
x=-3/2
x=2
### Completing the Square
Let us learn this method with example.
Example: Solve 2x2 – x – 1 = 0.
First, move the constant term to the other side of the equation.
2x2 – x = 1
Dividing both sides by 2.
x2 – x/2 = ½
Add the square of half of the coefficient of x, (b/2a)2, on both the sides, i.e., 1/16
x2 – x/2 + 1/16 = ½ + 1/16
Now we can factor the right side,
(x-¼)2 = 9/16 = (¾)2
Taking root on both sides;
X – ¼ = ±3/4
X = ¼ ± ¾
Therefore,
X = ¼ + ¾ = 4/4 = 1
X = ¼ – ¾ = -2/4 = -½
For the given Quadratic equation of the form, ax² + bx + c = 0
Therefore the roots of the given equation can be found by:
$$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$,
where $$\pm$$ (one plus and one minus) represent two distinct roots of the given equation.
### Taking the Square Root
We can use this method for the equations such as:
x2 + a2 = 0
Example: Solve x2 – 50 = 0.
x2 – 50 = 0
x2 = 50
Taking the roots both sides
√x2 = ±√50
x = ±√(2 x 5 x 5)
x = ±5√2
Thus, we got the required solution.
## Problems and Solutions
Lets Work Out: Example 1: $$3x^{2} – 5x + 2 = 0$$ Solution: $$3x^{2} – 5x + 2 = 0$$ Solving the quadratic equation using the above method: $$x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ $$x = \frac{-(-5)\pm \sqrt{(-5)^{2} -4 \times 3 \times 2}}{2 \times 3}$$ $$x = \frac{5 \pm 1}{6}$$ $$x = \frac{6}{6} \;\; or \;\; \frac{4}{6}$$ or, $$x = 1 \;\; or \;\; \frac{2}{3}$$ Example 2: Solve x2 – 6 x = 16. Solution: x2 – 6 x = 16. x2 – 6 x – 16 = 0 By factorisation method, ( x – 8)( x + 2) = 0 Therefore, x = 8 and x = -2 Example 3: Solve x2 – 16 = 0. And check if the solution is correct. Solution: x2 – 16 = 0. x2 – 42 = 0 [By algebraic identities] (x-4) (x+4) = 0 x = 4 and x = -4 Check: Putting the values of x in the LHS of the given quadratic equation If x = 4 X2 – 16 = (4)2 – 16 = 16 – 16 = 0 If x = -4, X2 – 16 = (-4)2 – 16 = 16 – 16 = 0 Example 4: Solve for y: y2 = –2y + 2. Solution: Given, y2 = –2y + 2 Rewriting the given equation; y2 + 2 y – 2 = 0 Using quadratic formula, $$y=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$ $$y=\frac{-(2) \pm \sqrt{(2)^{2}-4(1)(-2)}}{2(1)}$$ $$y=\frac{-2 \pm \sqrt{4+8}}{2}$$ $$y=\frac{-2 \pm \sqrt{12}}{2}$$ Therefore, y = -1 + √3 or y = -1 – √3
Many real-life word problems can be solved using quadratic equations. While solving word problems, some common quadratic equation applications include speed problems and Geometry area problems.
1. Solving the problems related to finding the area of quadrilateral such as rectangle, parallelogram and so on
2. Solving Word Problems involving Distance, speed, and time, etc.,
Example: Find the width of a rectangle of area 336 cm2 if its length is equal to the 4 more than twice its width.
Solution:
Let x cm be the width of the rectangle.
Length = (2x + 4) cm
We know that
Area of rectangle = Length x Width
x(2x + 4) = 336
2x2 + 4x – 336 = 0
x2 + 2x – 168 = 0
x2 + 14x – 12x – 168 = 0
x(x + 14) – 12(x + 14) = 0
(x + 14)(x – 12) = 0
x = -14, x = 12
Measurement cannot be negative.
Therefore, Width of the rectangle = x = 12 cm
## Practice Questions
1. Solve x2 + 2 x + 1 = 0.
2. Solve 5x2 + 6x + 1 = 0
3. Solve 2x2 + 3 x + 2 = 0.
4. Solve x2 − 4x + 6.25 = 0
## Frequently Asked Questions – FAQs
### What is a quadratic equation?
The polynomial equation whose highest degree is two is called a quadratic equation. The equation is given by ax² + bx + c = 0, where a ≠ 0.
### What are the methods to solve a quadratic equation?
There are majorly four methods of solving quadratic equations. They are:
Factorisation
Using Square roots
Completing the square
### Is x2 – 1 a quadratic equation?
Since the degree of the polynomial is 2, therefore, given equation is a quadratic equation.
### What is the solution of x2 + 4 = 0?
The solution of quadratic equation x2 – 4 is x = 2 or x = -2.
### Write the quadratic equation in the form of sum and product of roots.
If α and β are the roots of a quadratic equation, then;
Sum of the roots = α+β
Product of the roots = αβ
Therefore, the required equation is:
x2 – (α+β)x + (αβ) = 0
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2021-09-28 05:07:01
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http://accesssurgery.mhmedical.com/content.aspx?bookid=427§ionid=40372672
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Chapter 33
Barrett's esophagus (BE) is an acquired condition in which the squamous lining of the distal esophagus is replaced by a specialized intestinal metaplastic columnar epithelium. Although in the past BE was defined as the presence of columnar-lined esophagus on endoscopy, it is now accepted that confirmation of intestinal metaplasia by histology is required to make the diagnosis. The American College of Gastroenterology summarizes the most widely accepted definition as “a change in the esophageal epithelium of any length that can be recognized at endoscopy and is confirmed to have intestinal metaplasia by biopsy of the tubular esophagus and excludes intestinal metaplasia of the cardia.”1 The importance of BE is its risk of transforming into esophageal adenocarcinoma.
The incidence of BE over the past 20 years has risen sharply, and with it, there has been a dramatic increase in the incidence of adenocarcinoma.2 Reflux without BE can be treated medically (acid suppression) and surgically (fundoplication), with surgical therapy gaining increasing attention. Indeed, with the development of minimally invasive surgery, there has been an increased willingness to consider surgical options for reflux disease.3 From 1990 to 1997, the rate of antireflux surgery increased from 4.4 to 12.0 per 100,000 adults, and the proportion of operations performed laparoscopically increased from 25% to 76%.3 Similar trends have been noted in other countries as well.4,5 In this context, we review the available evidence for both medical and surgical approaches to the management of BE and their respective roles in halting disease progression.
The prevalence of BE in the population has been difficult to assess. Cameron and Lomboy at the Mayo Clinic found BE in 0.73% of 51,311 patients having endoscopy.6 In another endoscopy screening report, a prevalence rate of 1% was found in residents of Olmsted County, Minnesota.7 Still other studies have found much higher rates. Studies of patients undergoing screening sigmoidoscopy or colonoscopy who agreed to an upper endoscopy revealed BE in 10-25% of these patients.8–10 Seven cases were found in 733 autopsies (approximately 1%). The adjusted prevalence rate in this autopsy series was 376 cases per 100,000 population, significantly higher than the clinically diagnosed prevalence of 22.6 per 100,000.7,11 While many have suggested that the rate of BE has increased in the population, the incidence of BE has increased in parallel with the use of endoscopy.12,13 Whether there is a true increase in the incidence of BE or the increase simply reflects the increased use of endoscopy is unclear. In contrast, the incidence of esophageal adenocarcinoma has increased 10-fold in the 25-year period 1974–1997. Many adenocarcinomas occur in patients without a previous diagnosis of BE, suggesting that many people with this condition remain undiagnosed.13 The rate of malignant transformation of BE itself remains low. The overall risk of developing adenocarcinoma in patients with BE ranges from 0.2% to 1.9% per year.14
Norman Barrett ...
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2017-02-24 05:49:25
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http://openstudy.com/updates/50238b2fe4b07b8e940dba66
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Calculator 4 years ago What does it mean by restriced values in wolfram alpha? http://www.wolframalpha.com/input/?i=integrate+1%2F%2810-v%2F5%29
1. anonymous
where?
2. Calculator
Click the show steps then you'll see it
3. anonymous
im guessing it would be the value of v that would make the denominator equal to 0
4. anonymous
you cant have a denominator of zero right?
5. Calculator
That's weird because even the solution here http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_w11_ms_51.pdf ended up $-5 \log (v-50)+c$ but not $-5 \log \left(10-\frac{v}{5}\right)+c$
6. anonymous
hmmm wolfram does a lot of funny business =_=
7. anonymous
um, v cannot equal 50. try evaluating your original expression with v = 50.
8. Calculator
v can be 50, ln(0)=1
9. anonymous
ln (1) = 0
10. anonymous
ln (0) = infinity
11. anonymous
ln e = 1
12. anonymous
even your original function: 1/(10-v/5) what do you get when v = 50?
13. anonymous
http://www.wolframalpha.com/input/?i=1%2F%2810-v%2F5%29 Look what would the integral be at v = 50? v = 50 is a vertical asymptote.
14. UnkleRhaukus
how did you get ln (0) is infinity at @lgbasallote ?
15. anonymous
hmm read the post lol...i didnt get ln (0) i was explaining that ln (0) is not 1
16. UnkleRhaukus
ln (0) is not infinity either
17. anonymous
well...close enough
18. UnkleRhaukus
|dw:1344507971374:dw|
19. anonymous
depends if 0+ or 0-...but i just generalized as infinity =))
20. anonymous
my point was it's not 1 =_=
21. UnkleRhaukus
infinity is not close enough to negative infinity
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2016-10-20 21:49:25
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http://images.planetmath.org/estimator
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# estimator
Let $X_{1},X_{2},\ldots,X_{n}$ be samples (with observations $X_{i}=x_{i}$) from a distribution with probability density function $f(X\mid\theta)$, where $\theta$ is a real-valued unknown parameter (http://planetmath.org/StatisticalModel) in $f$. Consider $\theta$ as a random variable and let $\tau(\theta)$ be its realization.
An estimator for $\theta$ is a statistic $\eta_{\theta}=\eta_{\theta}(X_{1},X_{2},\ldots,X_{n})$ that is used to, loosely speaking, estimate $\tau(\theta)$. Any value $\eta_{\theta}(X_{1}=x_{1},X_{2}=x_{2},\ldots,X_{n}=x_{n})$ of $\eta_{\theta}$ is called an estimate of $\tau(\theta)$.
Example. Let $X_{1},X_{2},\ldots,X_{n}$ be iid from a normal distribution $N(\mu,\sigma^{2})$. Here the two parameters are the mean $\mu$ and the variance $\sigma^{2}$. The sample mean $\overline{X}$ is an estimator of $\mu$, while the sample variance $s^{2}$ is an estimator of $\sigma^{2}$. In addition, sample median, sample mode, sample trimmed mean are all estimators of $\mu$. The statistic defined by
$\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-m)^{2},$
where $m$ is a sample median, is another estimator of $\sigma^{2}$.
Title estimator Estimator 2013-03-22 14:52:22 2013-03-22 14:52:22 CWoo (3771) CWoo (3771) 6 CWoo (3771) Definition msc 62A01 estimate
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2018-06-18 03:58:16
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https://math.stackexchange.com/questions/609010/connectedness-at-a-simple-boundary-point
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# Connectedness at a simple boundary point
Interested by this question in math.SE, which shares a link to planetmath about definition of a simple boundary point. This link gives reference to the book Functions of one complex variable II of Conway. In this book, there is an exercise which I find interested in:
If $w$ is a simple boundary point of $\Omega$, then there is a $\delta > 0$ such that $D(w;\delta)\cap \Omega$ is connected.
I think the author means that we can find $\delta$ small arbitrary such that the connectedness happens (not so sure about this). If we can solve this problem, then we can easily determine which point is not a simple boundary point. As an example, let $\Omega = D(0;1) \backslash \{x:0\leq x < 1\}$, and $0<\beta \leq 1$, then $\beta$ cannot be a simple boundary point of $\Omega$ because if we choose $\epsilon>0$ small enough, we can't find any $\delta \in (0,\epsilon)$ such that $D(\beta;\delta)\cap \Omega$ is connected.
Could anyone give me a hint? It is hardly for me to see which direction I should go, to use the condition that $D(w;\delta)\cap \Omega$ is connected.
Base on Seub's answer, I put here some details.
For Lemma 1: the idea is picking two sequences in two connected components of $\Omega$, combining them into a sequence whose even subsequence is one sequence, and odd subsequence is the other sequence.
This lemma together with Lemma 3 (to prove, use the fact that there is a positive distance from a closed set to a point outside it) solves the problem.
About Lemma 2, for the $(\Leftarrow)$ side, let a sequence in $\Omega$ converges to $w$, then eventually that sequence will be inside the disc $D(w,\delta)$. Because $w$ is a simple boundary point of $\Omega \cap D(w;\delta)$, the "latter" part of that sequence can be connected by a curve converges to $w$. In addition, $\Omega$ is connected leads to the "former" part can be connected by a curve. Combining two curves with a reparemeterizing, we get a desired curve.
For the $(\Rightarrow)$ side, let a sequence in $\Omega \cap D(w;\delta)$ converges to $w$, then it can be connected by a curve in $\Omega$. Eventually, this curve will be inside the disc $D(w;\delta)$. So the "latter" part of the curve (which is in $\Omega \cap D(w;\delta)$) will connect the "latter" part of the sequence. We may choose (at first) $\delta$ small enough for $\Omega \cap D(w;\delta)$ connected. Then the "former" part of the sequence can be connected by a curve in $\Omega \cap D(w;\delta)$. We conclude.
Edit: Seub gives a counterexample in his answer!
• Just a comment: It could be that the author assumes $\Omega$ to be a region (open and connected) although he does not state it explicitly. It is always like this in the whole Chapter 5 in Conway's book as far as I can tell. If this is the case, Saub's example cannot be used since the set $D(0,2)\setminus\cup_{n\geq1}L_{n}$ is not open. – Twi Oct 30 '16 at 20:15
I apologize for my first answer which was incorrect.
First a couple of remarks:
• I agree with you that the author probably meant
"(...) then there exists $\delta~$ arbitrarily small such that (...)"
otherwise it is kind of silly. For example, if $\Omega$ is connected and bounded, just take $\delta$ such that $\Omega \subset D(\omega, \delta)$.
• I think the definition of simple boundary point should be
Let $\Omega$ be an open set in $\mathbb{C}$ and $\omega$ be a point in the boundary of $\Omega$. Then we call $\omega$ a simple boundary point if whenever $(\omega_n)$ is a sequence of points of $\Omega$ converging to $\omega$ there is a continuous path $\gamma:[0,1]\rightarrow \mathbb{C}$ such that $\gamma(t) \in \Omega$ for $0 \leqslant t < 1$, $\gamma(1) = \omega$ and there is a sequence $(t_n)$ in $[0,1)$ such that $t_n \rightarrow 1$ and $\gamma(t_n) = \omega_n$ for all $n~$ sufficiently large.
This would make the definition of a simple boundary point a local one, meaning that $\omega$ is a simple boundary point of $\Omega$ iff $\forall \delta>0$, $\omega$ is a simple boundary point of $\Omega \cap D(\omega, \delta)$.
While I believe these remarks are important, they actually don't matter for what I am about to say: after thinking about it, I believe the result is false!
Here's my counter-example: let $L_n$ be the vertical line segment $\frac{1}{n+1} + i \left[-\frac{1}{n},\frac{1}{n}\right]$ and $\Omega = D(0,2) \setminus \bigcup_{n=1}^{+\infty} L_n$. I really encourage you to draw a picture (of $\Omega$ and some "small" disk $D(0,\delta)$).
I claim that $0$ is a simple boundary point of $\Omega$. It might be a bit tedious to prove it rigorously, but it's certainly believable.
On the other hand, as soon as $\delta \leqslant 1$, $D(0, \delta) \cap \Omega$ is never connected. Indeed, let $n$ be the integer such that $\frac{1}{n+1} < \delta \leqslant \frac{1}{n}$. Then $L_n$ disconnects $D(0, \delta)$. QED.
(NB: Of course, for $\delta$ huge, e.g. $\delta > 2$, $D(0,\delta) \cap \Omega$ is connected, so the initial silly problem (without "arbitrarily small") is not contradicted. But we can cook something up similarly to contradict that too. If you insist I can tell you how, although I don't find that very important)
Now, if we wanted to fix all of this and try to say something true, I would use the "improved" definition of simple boundary point and maybe claim that
$\omega$ is a simple boundary point of $\Omega$ iff $\Omega$ is locally connected at $\omega$, meaning that there exists a basis of neighborhoods $\{U_n\}$ of $\omega$ such that $U_n \cap \Omega$ is connected for all $n$.
although for the moment it is not clear to me how to prove "$\Rightarrow$". I'll think about it if you're interested.
I hope this time I did not say too much nonsense!
• Thank you so much! Your lemmas simplify much of the problem. Let me write down some details for other persons who may be interested in this exercise. – Du Phan Dec 17 '13 at 6:12
• I think that your lemma 3 is not true. For example, look at the open square $0$, $1$, $1+i$, $i$, then for each $n\geq 2$, remove the intervals $[\frac{1}{n},\frac{1}{n}+\frac{1}{2}i]$, the the remaining is connected, open, but the points in $\{iy:0<y<\frac{1}{2}\}$ give us counterexamples. – Du Phan Dec 17 '13 at 7:48
• Also, I think that Lemma 2 is a corollary of the question, rather than a lemma; because the $(\Rightarrow)$ side requires (tacitly) $\Omega \cap D(w;\delta)$ is connected. – Du Phan Dec 17 '13 at 8:03
• Oops, you're right about lemma 3, sorry about that. As for lemma 2, I agree that in the first version, it is also not correct. In the second version, I meant to say "for $\delta$ sufficiently small" and I accidentally deleted it (I just realized that). Do you agree with the second version? Anyway, I need to rework my answer. I'll do that tonight (Europe time) – Seub Dec 17 '13 at 10:23
• The second version of Lemma 2 is true, if we know the question is true. I don't know how to prove its $(\Rightarrow)$ side if I don't have $\Omega \cap D(w;\delta)$ connected before. By the way, I appreciate any time you spend on this question. Don't have to rush. Regards :) – Du Phan Dec 17 '13 at 10:46
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2019-11-22 13:05:09
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https://math.stackexchange.com/questions/1999162/prove-that-liminf-x-n-le-liminf-a-n-le-limsup-a-n-le-limsup-x-n
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# Prove that $\liminf x_n \le \liminf a_n \le \limsup a_n \le \limsup x_n$
Assume that $\{x_n\}$ is a sequence of real numbers and $a_n=\frac{x_1+\dots+x_n}{n}$ .
a) Prove that $\displaystyle \liminf_{n \to\infty} x_n \le \liminf_{n \to\infty} a_n \le \limsup_{n \to\infty} a_n \le \limsup_{n \rightarrow \infty} x_n$.
b) Give an example such that all of the limits written above are finite and $\displaystyle \liminf_{n \to\infty} x_n < \liminf_{n \to\infty} a_n < \limsup_{n \to\infty} a_n < \limsup_{n \rightarrow \infty} x_n$.
c) Give an example such that some of the limits written above are finite and some of them are not.
Note 1 : For a sequence like $\{b_n\}$ we have $\displaystyle \liminf_{n \to\infty} b_n = \lim_{n\to\infty}(\inf\{b_k:k \ge n\})$ and $\displaystyle \limsup_{n \rightarrow \infty} b_n=\lim_{n\to\infty}(\sup\{b_k:k \ge n\})$
Note 2 : This question is adopted from the book "Real analysis : A first course" written by "Russel Gordon".
Note 3 : A small part of this question is available on this link but my question has a lot more than that.
• You should use \lim, \sup, \inf, \liminf and \limsup. Otherwise, this is very unpleasant to read. – tomasz Nov 4 '16 at 13:56
• @tomasz i didn't know that those things exist in latex ... pardon me :) anyway, thanks to Masacroso, it's fixed now – Maryam Seraj Nov 4 '16 at 13:58
• When something does not exist in latex, you can make it. For example, if you did not know \sin exists to write $\sin(x)$, you could still use \operatorname{sin}(x) to obtain $\operatorname{sin}(x)$. – tomasz Nov 4 '16 at 14:09
• Anyway, what exactly do you want us to help with? What problem are you having? – tomasz Nov 4 '16 at 14:11
• @tomasz thanks ! that's great : – Maryam Seraj Nov 4 '16 at 14:12
First assume that $(x_n)$ is a bounded squence.
Let $L=\limsup_{n\to\infty}x_n<\infty$. By definition of $\limsup$, there exists $K$ such that $x_n<L+\epsilon$ for all $n>K$. (This is the well-known "eventual upperbound" property of limsup.)
Then
$\Large\frac{x_1+\dots+x_n}{n}<\frac{x_1+\dots+x_k+(L+\epsilon)(n-k)}{n}$
Taking limsup on both sides gives
$\limsup a_n\leq L+\epsilon$
Since $\epsilon>0$ is arbitrary, $\limsup_{n\to\infty}a_n\leq\limsup_{n\to\infty} x_n$.
The case of $\liminf$ should be similar.
$\liminf a_n\leq\limsup a_n$ is automatic (always holds) so you get it for free.
The $(x_n)$ unbounded case, both $\limsup a_n$ and $\limsup x_n$ will be infinite.
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2019-08-23 18:09:57
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https://electronics.stackexchange.com/questions/136321/doubt-about-counter-implemented-in-vhdl
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Doubt about counter implemented in VHDL
I'm trying to learn VHDL and I came across some example code for a counter that I find somewhat strange.
I understand what it does but I'm not sure why it is written the way it is.
The code is the following:
entity counter is
generic(n: natural :=2);
port( clock: in std_logic;
clear: in std_logic;
count: in std_logic;
Q: out std_logic_vector(n-1 downto 0)
);
end counter;
architecture behv of counter is
signal Pre_Q: std_logic_vector(n-1 downto 0);
begin
-- behavior describe the counter
process(clock, count, clear)
begin
if clear = '1' then
Pre_Q <= Pre_Q - Pre_Q;
elsif (clock='1' and clock'event) then
if count = '1' then
Pre_Q <= Pre_Q + 1;
end if;
end if;
end process;
-- concurrent assignment statement
Q <= Pre_Q;
end behv;
-----------------------------------------------------
Can't we just replace this line
Pre_Q <= Pre_Q - Pre_Q;
with something like this? (I'm not sure if I can do this)
Pre_Q <= 0;
Is there any reason why I should use the first method instead of the second?
I apologize for deleting my previous question but I made a mistake when rewriting the code and I forgot to explain what my doubt really was.
• It would be a good idea to post the entire code and just one process from it, we know about your signals so answering the question is not possible. – user34920 Oct 29 '14 at 12:11
• @user34920 I just included the complete code. – Bruno Ferreira Oct 29 '14 at 12:18
The only advantage to writing it the first way that I can see is that, I suppose, it's type-independent. Maybe the original coder wanted to be able to copy and paste the code without modifying it (though one would think the signal name would need to be modified anyway), and depend on the synthesis tool to optimize it (or not). Personally, I do not think this is a good justification - it's less clear, and copying/pasting code without looking at it carefully is never a good idea.
Your assignment statement will change depending on your type. Pre_Q <= 0; for integer types, Pre_Q <= (others => '0'); (or equivalent) for vector types, etc.
(I see you've updated the question, and given that you're using a std_logic_vector, you'd want the latter)
response to comment
You wrote:
I do not understand the assignment for vectors. Can't I just use Pre_Q <= '0'?
A std_logic_vector is an array of std_logic, each one of which is an enumeration of 9 possible values/states (including '0' and '1'). They're not technically 0s and 1s to the language, odd though that may seem. So a std_logic_vector is not inherently a numeric type, and can only be treated as such with function overloads. If you try to assign '0' to Pre_Q, you are attempting to assign a scalar (in this case, technically a character literal) to a vector, and these types are not compatible in this way. Nor can you assign the integer literal 0 - while it may be obvious to almost anyone what you are intending to do, VHDL's strong typing does not allow this.
When you use (others => '0'), you are using an array aggregate. others is a keyword that refers to all indices of the array that have not been explicitly assigned in the aggregate (in this case, all of them), and so when you map the "other bits" to '0', you're zeroing the entire vector. (You can also use a conversion function, but (others => '0') is the more conventional idiom)
• That makes sense. I do not understand the assignment for vectors. Can't I just use Pre_Q <= '0';? – Bruno Ferreira Oct 29 '14 at 12:24
• See updated answer. – fru1tbat Oct 29 '14 at 12:59
• Ok, now I understand, and it seems that I have a long way until I learn VHLD. Thank you very much for your explanation. – Bruno Ferreira Oct 29 '14 at 13:05
• More plausible that the original writer didn't really know VHDL, and Pre_Q - Pre_Q was the first alternative to 0 that he could get to compile. I have seen some strangely warped code written that way... – Brian Drummond Oct 29 '14 at 15:13
• True, that is a bit more likely... – fru1tbat Oct 29 '14 at 15:20
Good question. My intuitive answer would be that there is no difference, and that you should write Pre_Q <= 0; because it expresses the intent more clearly. But that is just my opinion, without any tests or investigation.
Two answers that are good irrespective of my opinion:
1. Ask the guy who wrote this. And suggest that they add more documentation so that people would understand his code.
2. Run both version through a synthesis tool to see if there is any difference. (Again, my best guess is that there won't be any.)
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2020-02-26 10:35:05
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https://brilliant.org/discussions/thread/comments-not-visible/
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The picture clearly shows that there is one comment, but it isn't visible. Clicking on the "Reply" button doesn't seem to work either.
Link to the problem: Powers of 2
Note by Sreejato Bhattacharya
4 years, 4 months ago
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Hi Sreejato that is a known minor bug. What happened was someone commented on your solution and then deleted their own comment. The comment goes away, but the reply counter does not go down.
Is this preventing your from commenting on your own solution?
Staff - 4 years, 4 months ago
Oh, thanks for replying. I was curious to know what the comment said.
And no, that isn't preventing me from commenting on the solution. :)
- 4 years, 4 months ago
Well I won't give details for privacy reasons, but basically someone asked a question about the techniques used in your solution and then felt self-conscious about whether or not it was a good question and then deleted it. Happens all the time, sometimes people delete their own comments for good reasons(e.g. they realize it was a poorly formed statement, question etc...) and sometimes people delete them for bad reasons.
Staff - 4 years, 4 months ago
Thanks for reporting it, and nice design work to obscure your solution. That is a particularly warm red. We will take steps to fix it.
Staff - 4 years, 4 months ago
I checked it. The same bug occurred. Before also.
- 4 years, 4 months ago
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2018-06-22 15:32:13
|
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https://www.alloprof.qc.ca/helpzone/discussion/11682/question
|
# Help Zone
### Student Question
Secondary III • 1yr.
Good evening,
Can you show me how to solve a complete dilution problem? I have a lot of difficulty with that.
Science
## Explanations (1)
• Explanation from Alloprof
Explanation from Alloprof
This Explanation was submitted by a member of the Alloprof team.
Options
Team Alloprof • 1yr.
First, when you dilute a solution, you try to decrease the concentration of its solute by adding solvent. It goes without saying that the final concentration of a dilute solution is less than the initial concentration of this solution.
The following formula is used when performing dilution calculations:
$$C_{1} \bullet V_{1} = C_{2} \bullet V_{2}$$
Legend:
• C1: initial concentration of the solution
• V1: initial volume of the solution
• C2: final concentration of the solution
• V2: final volume of solution
Several units can be used to describe the concentration and volume of a solution in the dilution formula. As long as the units are the same on both sides of the formula, the formula works.
Most dilution problems give three of the four variables in the problem statement. The expression must then be handled algebraically in order to isolate the missing term:
$$C_{1} = \frac{C_{2}\bullet V_{2}}{V_{1}}$$
$$V_{1} = \frac{C_{2}\bullet V_{2}}{C_{1}}$$
$$C_{2} = \frac{C_{1}\bullet V_{1}}{V_{2}}$$
$$V_{2} = \frac{C_{1}\bullet V_{1}}{C_{2}}$$
Certain problems require finding a volume of solution to add. In this case, the following formula must be remembered:
$$V_{final} = V_{initial} + V_{to add}$$
It is then necessary quite simply to modify the expression of the equation in order to find the volume to add:
$$V_{to add} =V_{final} - V_{initial}$$
Thank you for your question !
|
2023-02-05 23:45:06
|
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https://math.stackexchange.com/questions/1789742/complicated-partial-fraction-expansion/1789767
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# Complicated partial fraction expansion
I'm reading the book generatingfunctionology by Herbert Wilf and I came across a partial fraction expansion on page 20 that I cannot understand. The derivation is as follows:
$$\frac{1}{(1-x)(1-2x)...(1-kx)} = \sum_{j=1}^{k} \frac{\alpha_j}{1-jx}$$
The book says to fix $r, 1 \leq r \leq k$, and multiply both sides by $1-rx$. Doing so, I get:
$$\frac{1}{(1-x)...(1-(r-1)x)(1-(r+1)x)...(1-kx)} = \frac{\alpha_1(1-rx)}{1-x} + ... + \frac{\alpha_{r-1}(1-rx)}{1-(r-1)x} + \alpha_r + \frac{\alpha_{r+1}(1-rx)}{1-(r+1)x} + ... + \frac{a_k(1-rx)}{1-kx}$$
Contrarily, the book has:
$$\alpha_r = \frac{1}{(1-x)...(1-(r-1)x)(1-(r+1)x)...(1-kx)}$$
I don't understand how the other other fractions on the right side of my result cancel out to $0$. I tried with a small example where $k=3$ and I couldn't isolate $\alpha_2$ nicely after multiplying both sides by $1-2x$. Any pointers on this would be greatly appreciated.
After this, the book goes on by letting $x=1/r$, resulting in the following:
\begin{aligned} \alpha_r &= \frac{1}{(1-1/r)(1-2/r)...(1-(r-1)/r)(1-(r+1)/r)...(1-k/r)} \\ &= (-k)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!} && (1 \leq r \leq k) \end{aligned}
I also can't figure how this is derived (I suspect it's using an identity that I'm not aware of.) Any help would be much appreciated. Thanks a lot!
It appears that the $\alpha_j$ coefficients are being computed by the standard Heaviside method (sometimes called the "cover up" method).
Heaviside's Method for partial fraction expansion
My edition of the book doesn’t have
$$\alpha_r = \frac{1}{(1-x)\ldots(1-(r-1)x)(1-(r+1)x)\ldots(1-kx)}\;,$$
which in any case makes no sense, since $\alpha_r$ is a constant and the righthand side is not. However, after letting $x=\frac1r$ it does show
\begin{align*} \alpha_r&=\frac1{(1-1/r)(1-2/r)\cdots(1-(r-1)/r)(1-(r+1)/r)\cdots(1-k/r)}\\ &=(-1)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!}\;. \end{align*}
The last step is obtained by first multiplying by $r^{k-1}$ to get
$$\alpha_r=\frac{r^{k-1}}{\underbrace{(r-1)(r-2)\cdot(r-(r-1))}_{(r-1)!}\underbrace{(r-(r+1))\cdots(r-k)}_{(-1)(-2)\cdots(-(k-r))}}$$
and then rewriting this as
$$\alpha_r=\frac{r^{k-1}}{(r-1)!(-1)^{k-r}(k-r)!}=(-1)^{k-r}\frac{r^{k-1}}{(r-1)!(k-r)!}\;.$$
To find the coefficients of the expansion, you can follow the usual "cover" up rule which is essentially the results you have shown albeit buried in indices and what not: $$\frac{1}{(1-x)(1-2x)...(1-kx)} = \sum_{j=1}^{k} \frac{\alpha_j}{1-jx}$$ say we would want to find an arbitrary $\alpha_j$ for some value of $j \in \{1,2,...,k\}$. We call this $r$ to emphasise, and to avid confusion with the summation index. So we multiply both sides by $1-rx$ $$\frac{1-rx}{(1-x)(1-2x)...(1-kx)} = \sum_{j=1}^{k} \alpha_j\frac{1-rx}{1-jx}$$ Since $r$ will coincide with one of the $j$'s we can "pull this out" from the right hand side $$RHS = \alpha_r + \sum_{\underset{j\ne r}{j=1}}^{k} \alpha_j\frac{1-rx}{1-jx}$$ And for the left hand side, the r will "cancel" on one of the denominator factors: $$LHS = \frac{1}{\underset{\text{without the 1-rx}}{(1-x)(1-2x)...(1-kx)}}$$ The expansion holds for all values of x so we can set $x=1/r$, and simplify both sides to get the result. I'll stop here because I think doing it is fun :)
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2022-08-16 00:49:54
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http://www.mpwmd.net/asd/board/boardpacket/2010/20100225/05/item5.htm
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##### CONSENT CALENDAR
5.
CONSIDER Expenditure of Budgeted Funds for the Maintenance of the Fish rescue truck
Meeting Date:
#### Funds are available and a line item will be added for this item in the 2010 Mid-Year Budget adjustment.
From:
Darby Fuerst,
General Manager
Program/
Line Item No.:
Aquatic Resources/ Fisheries 2-3-1 A
Prepared By:
Beverly Chaney
Cost Estimate:
$5,900 ## General Counsel Review: N/A Committee Recommendation: The Administrative Committee considered this item on February 12, 2010 and recommended approval. CEQA Compliance: N/A SUMMARY: The District’s fish rescue truck “Big Red” is a vital component for both annual summer fish rescues and the release of fish from the Sleepy Hollow Steelhead Rearing Facility (SHSRF) in the winter. This truck carries the large fish transport tank needed to transport steelhead between the river and the Facility and has been in operation since 2003. Salt is added to the tank water during transport to keep stress and infection levels down in the fish. This water frequently splashes out during transport and fish removal. During the 2009 rescue season, staff began to notice rust forming in the truck bed and along the frame and utility boxes. By the time all the fish were released from the Facility in November, the metal tank tie-downs were severely corroded and could no longer safely hold the tank down and the back end of the truck bed had severe rust going under the Rhino bed-liner. Although the repair is unlikely to be permanent, as rust typically returns in time, failure to repair this damage may render the truck both unsafe and unusable during the upcoming fish rescue season. Repairs need to be completed during the winter off-season. After making several calls and visiting several local body shops, staff received a quote from Mark’s Barn Auto Body, in Sand City for$5,815. The shop would: 1) remove the bed liner from the rusted areas, 2) remove and replace the damaged rusted metal areas from the bed and utility boxes, 3) fabricate new tank supports, and 4) prime and refinish the truck bed and boxes. The quote does not include repairing the bed liner; this would be an additional cost at a different shop.
There is no separate line item for this repair, but since the truck is critical for both fish rescues and the operation of the Facility, funds needed for the truck repair will come out of the Facility’s General Operations and Maintenance budget under the Sleepy Hollow Facility Operations line item of the District’s Fiscal Year 2009-2010 Budget.
RECOMMENDATION: Staff recommends that the District Board authorize expenditure of budgeted funds in a not-to-exceed amount of $5,900 for the repair of the fish rescue truck by the local vendor in Sand City. IMPACT TO STAFF/RESOURCES: The 2009-2010 Budget includes$45,500 for General operations and maintenance of SHSRF (Line Item 2-3-1 A). To date, approximately $10,489 has been spent from this fund, and staff estimates$9,400 is available due to the Facility being shut down in November. The money needed, up to \$5,900, will be covered by the fund for General operations and maintenance of SHSRF (Line Item 2-3-1 A).
EXHIBITS
None
U:\staff\word\boardpacket\2010\20100225\ConsentCal\05\item5.doc
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2017-11-19 21:44:23
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https://www.newworldencyclopedia.org/entry/Georg_Ohm
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Georg Ohm
Georg Ohm
Georg Simon Ohm (1789-1854)
Born
March 16, 1789
Erlangen, Bavaria, Germany
Died July 6, 1854
Residence Germany
Nationality German
Field Physics
Institutions University of Munich
Alma mater University of Erlangen
Known for Ohm's law
Ohm's phase law
Ohm's acoustic law
Notable prizes Copley Medal (1841)
Georg Simon Ohm was a German physicist who clarified the fundamental relationships between electric current, voltage, and resistance. This relationship, known as Ohm's law, represents the true beginning of electrical circuit analysis.
Ohm offers a good example of the working of science through his experimental investigation of electrical circuits, his transfer of concepts from one area of physics (the study of heat) to another area (the study of electricity), and his use of mathematics to quantify the relation between the separately identified properties of current, voltage, and resistance. His integration of the three properties into a single formula, which became known as Ohm's law, served as a huge stepping stone for all the subsequent advances involving electrical circuits, ranging from electric lights to radios and computers.
Biography
Early years
Ohm was born on March 16, 1789 to Johann Wolfgang Ohm, a locksmith and Maria Elizabeth Beck, the daughter of a tailor in Erlangen, Bavaria. Although his parents had not been formally educated, Ohm's father was a respected man who had educated himself to a high level and gave his sons an excellent education through his own teachings. Some of Ohm's brothers and sisters died in their childhood and only three, including Georg Simon, survived. The other two survivors were his younger brother Martin (who later became a well-known mathematician), and his sister Elizabeth Barbara. His mother died when he was ten.
The boys' father taught them from their early childhood, bringing them to a high standard in mathematics, physics, chemistry and philosophy even as he hoped they would follow in his footsteps as a locksmith. Georg Simon attended Erlangen Gymnasium from age 11 to age 15 where he received little in the area of scientific training, which sharply contrasted with the inspired instruction that both the boys had received from their father. The progress of the Ohm brothers in science and mathematics made them bear a resemblance to the scientifically accomplished Bernoulli family, as noted by the professor at the University of Erlangen, Karl von Langsdorf. Langsdorf's enthusiasm inspired Ohm's father to relinquish his desire to have the boys take up his trade.
Life in university
In 1805, at age 15, Ohm entered the University of Erlangen. Rather than concentrate on his studies he spent much time dancing, ice skating, and playing billiards. Ohm's father, angry that his son was wasting the educational opportunity, sent Ohm to Switzerland where, in September 1806, he took up a post as a mathematics teacher in a school in the Institute of Gottstadt near Nydau, in the canton of Berne.
Karl Christian von Langsdorf left the University of Erlangen in early 1809 to take up a post in the University of Heidelberg, and Ohm would have liked to have gone with him to restart his mathematical studies. Langsdorf, however, advised Ohm to continue with his studies of mathematics on his own, advising Ohm to read the works of Leonhard Euler, Pierre-Simon Laplace and Sylvestre François Lacroix. Rather reluctantly, Ohm took this advice but he left his teaching post in Gottstadt in March 1809 to become a private tutor in Neuchâtel. For two years he carried out his duties as a tutor while he followed Langsdorf's advice and continued his private study of mathematics. In April 1811 he returned to the University of Erlangen.
Teaching career
His private studies had stood him in good stead, for he earned a doctorate from Erlangen on October 25, 1811, and immediately joined the staff as a Privatdocent. After three semesters Ohm gave up his university post. In 1813, the Bavarian government offered him a post as a teacher of mathematics and physics at the comparatively poor quality Realschule of Bamberg. Feeling unhappy with his job, Ohm devoted his spare time to writing an elementary book on geometry as a way to prove his true ability. The school was then closed down in February 1816.
Ohm sent his manuscript on geometry to King Wilhelm III of Prussia upon its completion. The work must have impressed the king, because Ohm was subsequently offered a position at a Jesuit gymnasium in Cologne on September 11, 1817. Thanks to the school's reputation for science education, Ohm found himself required to teach physics as well as mathematics. Luckily, the physics lab was well-equipped, so Ohm devoted himself to experiments in physics. Being the son of a locksmith, Ohm had some practical experience with mechanical equipment. It was here that he developed his theory of the relationship between electromotive force, resistance and electric current.
In 1826, having obtained a leave of absence from the gymnasium in Cologne, Ohm arranged for the publication of his theories in Berlin. This work appeared in its most complete form in 1927 as a book entitled "The Galvanic Circuit Mathematically Treated." In it, he applied, by analogy, the same treatment to the flow of electricity as Joseph Fourier had in his groundbreaking study of heat flow. Ohm's work was misunderstood and poorly received, and Ohm relinquished his position at the gymnasium, becoming practically unemployed until he secured a professorship at the Polytechnic School of Nuremberg in 1833. During the intervening years, Ohm did not have access to a laboratory, and his researches came to a virtual halt.
Gradually, however, his explanations regarding the voltaic circuit picked up converts, and his work finally won the admiration of the most illustrious scientists of the time. In 1841, he received the Copley Medal of the Royal Society of London. In its notice of the award, the society praised Ohm's work.
The light which these investigations has thrown on the theory of current electricity is very considerable....Had the works of Ohm been earlier known, and their value recognised, the industry of experimentalists would have been better rewarded.
Ohm was elected a foreign member of the Royal Society in 1842. He continued teaching, and devoted the remainder of his career to developing a molecular theory of electricity. He published one volume of the results of his work in 1849 under the title, Contributions to Molecular Physics, in which he included an exposition of geometry using a system of oblique-angled coordinates.
In 1849 he assumed the post of conservator of the Physical Collection at Munich. This left him little time for research, although he managed to publish a memoir on interference in uniaxal crystals.
In 1852 he again assumed a teaching post, this time at the high school in Munich. This teaching post inspired him to write a physics text book, which was published in 1854. His health, however, was not strong enough to sustain his exertions, and, after a short period of declining health, he suffered an apoplectic attack and died on July 7, 1854.
The discovery of Ohm's law
In his first paper published in 1825, Ohm examines the decrease in the electromagnetic force produced by a wire as the length of the wire increased. The paper deduced mathematical relationships based purely on the experimental evidence that Ohm had tabulated.
In two important papers in 1826, Ohm gave a mathematical description of conduction in electrical circuits modeled on Joseph Fourier's study of heat conduction. These papers continue Ohm's deduction of results from experimental evidence and, particularly in the second, he was able to propose laws which went a long way to explaining results of others working on galvanic electricity. The second paper certainly is the first step in a comprehensive theory which Ohm was able to give in his famous book published in the following year.
What is now known as Ohm's law appeared in the famous book Die galvanische Kette, mathematisch bearbeitet (“The Galvanic Circuit Investigated Mathematically,” 1827) in which he gave his complete theory of electricity. The book begins with the mathematical background necessary for an understanding of the rest of the work. While his work later greatly influenced the theory and applications of current electricity, it was coldly received at that time. It is interesting that Ohm presents his theory as one of contiguous action, a theory which opposed the concept of action at a distance. Ohm believed that the communication of electricity occurred between "contiguous particles" which is the term Ohm himself uses.
Ohm's Law
An electric battery will produce a fixed amount of electromotive force, or voltage, in a circuit. The electrical current (quantity of electric charge that moves past a particular point in the circuit in a unit of time) will then be resisted only by the wires and other parts of the circuit that connect the poles of the battery. The current will then only depend on the resistance offered by the connecting wire.
In mathematical terms, this is written as:
${\displaystyle I={\frac {V}{R}}}$
where I is the current, V is the potential difference, and R is a constant called the resistance. The potential difference is also known as the voltage drop, and is sometimes denoted by E or U instead of V. This law is usually valid over a large range of values of current and voltage, but it breaks down if conditions (such as temperature) are changed excessively.
In the case where the circuit is broken, the resistance is infinite, and the current drops to zero, as the above formula demonstrates. In the case where a perfect conductor connects the poles of a battery and offers no resistance, this formula would seem to yield an infinite current, but the battery itself has an internal resistance that prevents this.
The voltage may be thought of as the pressure that pushes the current through the wire. Naturally, in places where there is greater resistance, a greater pressure will be needed in order that the current remain constant at all points in the conducting wire. Therefore, across a stretch of wire with greater resistance, the voltage drop must be greater than for one with less resistance. The total voltage provided by the battery is divided up according to the varying resistances in each segment of the connecting circuit, in order to keep the current flow the same at all points through the conducting wire.
The unit of current is the ampere; that of potential difference is the volt; and that of resistance is the ohm, equal to one volt per ampere. Georg Ohm presented a slightly more complex equation than the above one, to explain his experimental results. The above equation could not exist until the ohm, a unit of resistance, was defined (1861, 1864).
Study and publications
His writings were numerous. The most important was his pamphlet published in Berlin in 1827, with the title Die galvanische Kette mathematisch bearbeitet. This work, the germ of which had appeared during the two preceding years in the journals of Schweigger and Poggendorff, has exerted an important influence on the development of the theory and applications of electric current. Ohm's name has been incorporated in the terminology of electrical science in Ohm's law (which he first published in Die galvanische Kette...), the proportionality of current and voltage in a resistor, and adopted as the SI unit of resistance, the ohm (symbol Ω).
Works
• Grundlinien zu einer zweckmäßigen Behandlung der Geometrie als höheren Bildungsmittels an vorbereitenden Lehranstalten / entworfen (Guidelines for an appropriate treatment of geometry in higher education at preparatory institutes / notes). Erlangen: Palm und Enke, 1817. Available online (PDF, 11.2 MB). Retrieved August 2, 2007.
• Die galvanische Kette : mathematisch bearbeitet (The Galvanic Circuit Investigated Mathematically). Berlin: Riemann, 1827. Available online (PDF, 4.7 MB). Retrieved August 2, 2007.
• Elemente der analytischen Geometrie im Raume am schiefwinkligen Coordinatensysteme (Elements of analytic geometry concerning the skew coordinate system). Nürnberg: Schrag, 1849. Available online (PDF, 81 MB). Retrieved August 2, 2007.
• Grundzüge der Physik als Compendium zu seinen Vorlesungen (Fundamentals of physics: Compendium of lectures). Nürnberg: Schrag, 1854. Available online (PDF, 38 MB). Retrieved August 2, 2007.
• Bibliography and PDF files of all articles and books by Ohm
|
2020-08-05 17:31:33
|
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|
https://eccc.weizmann.ac.il/keyword/18504/
|
Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > KEYWORD > COIN PROBLEM:
Reports tagged with coin problem:
TR17-090 | 15th May 2017
Chin Ho Lee, Emanuele Viola
#### The coin problem for product tests
Let $X_{m, \eps}$ be the distribution over $m$ bits $(X_1, \ldots, X_m)$
where the $X_i$ are independent and each $X_i$ equals $1$ with
probability $(1+\eps)/2$ and $0$ with probability $(1-\eps)/2$. We
consider the smallest value $\eps^*$ of $\eps$ such that the distributions
$X_{m,\eps}$ and $X_{m,0}$ can be distinguished with constant
more >>>
TR18-157 | 10th September 2018
Nutan Limaye, Karteek Sreenivasiah, Srikanth Srinivasan, Utkarsh Tripathi, S Venkitesh
#### The Coin Problem in Constant Depth: Sample Complexity and Parity gates
Revisions: 2
The $\delta$-Coin Problem is the computational problem of distinguishing between coins that are heads with probability $(1+\delta)/2$ or $(1-\delta)/2,$ where $\delta$ is a parameter that is going to $0$. We study the complexity of this problem in the model of constant-depth Boolean circuits and prove the following results.
1. Upper ... more >>>
TR19-018 | 18th February 2019
Alexander Golovnev, Rahul Ilango, Russell Impagliazzo, Valentine Kabanets, Antonina Kolokolova, Avishay Tal
#### AC0[p] Lower Bounds against MCSP via the Coin Problem
Minimum Circuit Size Problem (MCSP) asks to decide if a given truth table of an $n$-variate boolean function has circuit complexity less than a given parameter $s$. We prove that MCSP is hard for constant-depth circuits with mod $p$ gates, for any prime $p\geq 2$ (the circuit class $AC^0[p])$. Namely, ... more >>>
TR19-087 | 10th June 2019
Rohit Agrawal
Revisions: 1
In this note we compare two measures of the complexity of a class $\mathcal F$ of Boolean functions studied in (unconditional) pseudorandomness: $\mathcal F$'s ability to distinguish between biased and uniform coins (the coin problem), and the norms of the different levels of the Fourier expansion of functions in $\mathcal ... more >>> TR19-133 | 2nd October 2019 Nutan Limaye, Srikanth Srinivasan, Utkarsh Tripathi #### More on$AC^0[\oplus]$and Variants of the Majority Function Revisions: 1 In this paper we prove two results about$AC^0[\oplus]$circuits. We show that for$d(N) = o(\sqrt{\log N/\log \log N})$and$N \leq s(N) \leq 2^{dN^{1/d^2}}$there is an explicit family of functions$\{f_N:\{0,1\}^N\rightarrow \{0,1\}\}$such that$f_N$has uniform$AC^0$formulas of depth$d$and size at ... more >>> TR20-046 | 13th April 2020 Srikanth Srinivasan #### A Robust Version of Heged\H{u}s's Lemma, with Applications Heged\H{u}s's lemma is the following combinatorial statement regarding polynomials over finite fields. Over a field$\mathbb{F}$of characteristic$p > 0$and for$q$a power of$p$, the lemma says that any multilinear polynomial$P\in \mathbb{F}[x_1,\ldots,x_n]$of degree less than$q$that vanishes at all points in$\{0,1\}^n$of ... more >>> TR20-139 | 11th September 2020 Mark Braverman, Sumegha Garg, David Woodruff #### The Coin Problem with Applications to Data Streams Consider the problem of computing the majority of a stream of$n$i.i.d. uniformly random bits. This problem, known as the {\it coin problem}, is central to a number of counting problems in different data stream models. We show that any streaming algorithm for solving this problem with large constant ... more >>> TR21-083 | 21st June 2021 Mark Braverman, Sumegha Garg, Or Zamir #### Tight Space Complexity of the Coin Problem In the coin problem we are asked to distinguish, with probability at least$2/3$, between$ni.i.d.$coins which are heads with probability$\frac{1}{2}+\beta$from ones which are heads with probability$\frac{1}{2}-\beta\$. We are interested in the space complexity of the coin problem, corresponding to the width of a read-once ... more >>>
ISSN 1433-8092 | Imprint
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2022-12-05 07:44:51
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https://zbmath.org/?q=an:0663.70028
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## Subharmonic solutions of prescribed minimal period for non autonomous differential equations.(English)Zbl 0663.70028
Recent advances in Hamiltonian systems, Proc. Int. Conf., L’Aquila/Italy 1986, 83-96 (1987).
[For the entire collection see Zbl 0637.00003.]
Differential equations of differential types with varying complexities have been of great help in several areas of mathematics and sciences. Particularly the solutions of special kinds do take good account of the optimal properties. In the literature the subharmonic solutions with prescribed minimal conditions have been dealt with well. Here the authors have discussed the existence of subharmonic periodic solutions of the equations like: $$x''=-V'(x,t).$$ The paper is a very useful attempt at providing references and enlarging the scope of applications of systems of differential equations also.
Reviewer: P.Achuthan
### MSC:
70H05 Hamilton’s equations 34C25 Periodic solutions to ordinary differential equations
### Keywords:
subharmonic solutions; minimal conditions
Zbl 0637.00003
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2022-05-27 21:50:06
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https://cinsects.de/0ctf-2017-quals-onetimepad2.html
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Published: Di 21 März 2017
Updated: Di 21 März 2017
In Writeups.
oneTimePad and oneTimePad2 were two of the crypto challenges of this years 0CTF Quals. The CTF used a scoring system with dynamic points and the challenges were worth 114 and 366 points at the end of the competition.
The author solved these challenges while playing for Cyclopropenylidene, the CTF team of the Chaos Computer Club Hansestadt Hamburg.
For this task, we are given an encryption routine in form of a python script and a text file containing three ciphertexts.
af3fcc28377e7e983355096fd4f635856df82bbab61d2c50892d9ee5d913a07f
630eb4dce274d29a16f86940f2f35253477665949170ed9e8c9e828794b5543c
e913db07cbe4f433c7cdeaac549757d23651ebdccf69d7fbdfd5dc2829334d1b
The script implements a pseudo random number generator, that is used to generate a stream of 256 bit words: $$k_1, k_2, k_3$$. It encrypts three messages by xoring them to corresponding keys.
While $$m_1$$ contains the flag, $$m_2$$ and $$m_3$$ are given in the source code. By $$k_i = m_i \oplus c_i$$ we can compute the last two outputs of the random number generator.
Let's see how the PRNG works:
def keygen(seed):
key = str2num(urandom(32))
while True:
yield key
key = process(key, seed)
The seed and the first output are 256 bit values chosen uniformly at random. The next outputs are generated by applying the function process to the last output and the seed.
The function process interprets the 256 bit integers as elements of the galois field $$GF(2^{256})$$ using their bits as coefficients of a polynomial. For example the integer 0b101010 corresponds to the polynomial $$X^5 + X^3 + X$$. The field is defined with the irreducible polynomial
\begin{equation*} P = X^{256} + X^{10} + X^5 + X^1 + 1 \in \mathbb{F}_2[X]. \end{equation*}
Addition of field elements can be computed as the XOR of the corresponding integers. Back to the function process: It takes $$m, k \in GF(2^{256})$$ and computes $$(m + k)^2$$.
P = 0x10000000000000000000000000000000000000000000000000000000000000425L
def process(m, k):
tmp = m ^ k
res = 0
for i in bin(tmp)[2:]:
res = res << 1;
if (int(i)):
res = res ^ tmp
if (res >> 256):
res = res ^ P
return res
Since we know two consecutive outputs $$k_2, k_3$$, we can compute the $$seed$$ (the square root is defined for all elements of a characteristic two field):
\begin{equation*} (k_2 + seed)^2 = k_3 \iff seed = \sqrt{k_3} + k_2 \end{equation*}
With the $$seed$$ we can compute the first output of the random number generator
\begin{equation*} (k_1 + seed)^2 = k_2 \iff k_1 = \sqrt{k_2} + seed \end{equation*}
and then the plaintext
\begin{equation*} p_1 = k_1 + c_1 \end{equation*}
Decoding $$p_1$$ to a string yields the flag:
flag{t0_B3_r4ndoM_en0Ugh_1s_nec3s5arY}
In the second part we are given a ciphertext and part of the corresponding plaintext.
0da8e9e84a99d24d0f788c716ef9e99c ... 1b91df5e5e631e8e9e50c9d80350249c
One-Time Pad is used here. You won't know that the flag is flag{ ... }
This part of the challenge also does arithmetic in a galois field. It uses 128 bit integers and interprets them as elements of $$GF(2^{128})$$ wich is defined with the irreducible polynomial
\begin{equation*} P = X^{129} + X^7 + X^2 + X + 1 \in \mathbb{F}_2[X]. \end{equation*}
The functions process1 and process2 implement multiplication in $$GF(2^{128})$$ and multiplication of $$2 \times 2$$ matrices over the same field respectively.
P = 0x100000000000000000000000000000087
def process1(m, k):
res = 0
for i in bin(k)[2:]:
res = res << 1;
if (int(i)):
res = res ^ m
if (res >> 128):
res = res ^ P
return res
def process2(a, b):
res = []
res.append(process1(a[0], b[0]) ^ process1(a[1], b[2]))
res.append(process1(a[0], b[1]) ^ process1(a[1], b[3]))
res.append(process1(a[2], b[0]) ^ process1(a[3], b[2]))
res.append(process1(a[2], b[1]) ^ process1(a[3], b[3]))
return res
The random number generator choses the first 128 bit output at random and applies the function nextrand to compute the next output.
def keygen():
key = str2num(urandom(16))
while True:
yield key
key = nextrand(key)
The nextrand function makes use of the global variables $$A, B$$ and $$N$$. While $$N$$ is chosen randomly at initialization, $$A$$ and $$B$$ are given in the source code. $$A$$ and $$B$$ are generators of the finite field.
nextrand uses a square-and-multiply approach to compute the $$N$$-th power of the matrix $$\left( \begin{smallmatrix} A & B \\ 0 & 1 \end{smallmatrix} \right)$$:
\begin{equation*} \begin{pmatrix} A & B \\ 0 & 1 \\ \end{pmatrix}^N = \begin{pmatrix} A^N & A^{N-1}B + \dotsb + AB + B \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix} A^N & \frac{1 - A^N}{1 - A} \cdot B \\ 0 & 1 \\ \end{pmatrix} \end{equation*}
It uses the entries of the resulting first row to apply the function
\begin{equation*} x \mapsto A^N \cdot x + \frac{1-A^N}{1-A} \cdot B \end{equation*}
to the last output of the RNG. The result is then returned as next output. Furthermore each call to nextrand squares $$N$$.
A = 0xc6a5777f4dc639d7d1a50d6521e79bfd
B = 0x2e18716441db24baf79ff92393735345
N = str2num(urandom(16))
def nextrand(rand):
global N, A, B
tmp1 = [1, 0, 0, 1]
tmp2 = [A, B, 0, 1]
s = N
N = process1(N, N) # N <- N^2
while s: # tmp2^s * tmp1
if s % 2:
tmp1 = process2(tmp2, tmp1) # tmp1 <- tmp2 * tmp1
tmp2 = process2(tmp2, tmp2) # tmp2 <- tmp2^2
s = s / 2
return process1(rand, tmp1[0]) ^ tmp1[1] # rand * tmp1[0] + tmp1[1]
We know the beginning of the plaintext and therefore the first outputs of the RNG. Since $$k_2 = nextrand(k_1)$$, we can solve the following equation for $$N$$ to get its initial value.
\begin{equation*} \begin{alignedat}{2} && k_2 &= A^N \cdot k_1 + \frac{1 - A^N}{1 - A} \cdot B \\ \iff&& k_2 &= A^N \cdot k_1 + \frac{B}{1 - A} - \frac{A^NB}{1 - A} \\ \iff&& k_2 - \frac{B}{1 - A} &= A^N \cdot k_1 - \frac{A^NB}{1 - A} \\ \iff&& k_2 - \frac{B}{1 - A} &= A^N \cdot \left( k_1 - \frac{B}{1 - A} \right) \\ \iff&& \left( k_2 - \frac{B}{1 - A} \right) \cdot \left( k_1 - \frac{B}{1 - A} \right)^{-1} &= A^N \\ \iff&& \log_{A} \left( \left( k_2 - \frac{B}{1 - A} \right) \cdot \left( k_1 - \frac{B}{1 - A} \right)^{-1} \right) &= N \\ \end{alignedat} \end{equation*}
Now we know with $$k_1$$ and $$N$$ all values that were chosen randomly. Thus we can reconstruct the keystream and decrypt the complete message:
One-Time Pad is used here. You won't know that the flag is flag{LCG1sN3ver5aFe!!}
## Appendix
The given source code:
#!/usr/bin/env python
# coding=utf-8
from os import urandom
def process(m, k):
tmp = m ^ k
res = 0
for i in bin(tmp)[2:]:
res = res << 1;
if (int(i)):
res = res ^ tmp
if (res >> 256):
res = res ^ P
return res
def keygen(seed):
key = str2num(urandom(32))
while True:
yield key
key = process(key, seed)
def str2num(s):
return int(s.encode('hex'), 16)
P = 0x10000000000000000000000000000000000000000000000000000000000000425L
assert len(true_secret) == 32
print 'flag{%s}' % true_secret
fake_secret1 = "I_am_not_a_secret_so_you_know_me"
secret = str2num(urandom(32))
generator = keygen(secret)
ctxt1 = hex(str2num(true_secret) ^ generator.next())[2:-1]
ctxt2 = hex(str2num(fake_secret1) ^ generator.next())[2:-1]
ctxt3 = hex(str2num(fake_secret2) ^ generator.next())[2:-1]
f = open('ciphertext', 'w')
f.write(ctxt1+'\n')
f.write(ctxt2+'\n')
f.write(ctxt3+'\n')
f.close()
#!/usr/bin/env python
# coding=utf-8
from os import urandom
def process1(m, k):
res = 0
for i in bin(k)[2:]:
res = res << 1;
if (int(i)):
res = res ^ m
if (res >> 128):
res = res ^ P
return res
def process2(a, b):
res = []
res.append(process1(a[0], b[0]) ^ process1(a[1], b[2]))
res.append(process1(a[0], b[1]) ^ process1(a[1], b[3]))
res.append(process1(a[2], b[0]) ^ process1(a[3], b[2]))
res.append(process1(a[2], b[1]) ^ process1(a[3], b[3]))
return res
def nextrand(rand):
global N, A, B
tmp1 = [1, 0, 0, 1]
tmp2 = [A, B, 0, 1]
s = N
N = process1(N, N)
while s:
if s % 2:
tmp1 = process2(tmp2, tmp1)
tmp2 = process2(tmp2, tmp2)
s = s / 2
return process1(rand, tmp1[0]) ^ tmp1[1]
def keygen():
key = str2num(urandom(16))
while True:
yield key
key = nextrand(key)
def encrypt(message):
length = len(message)
pad = '\x00' + urandom(15 - (length % 16))
res = ''
generator = keygen()
f = open('key.txt', 'w') # This is used to decrypt and of course you won't get it.
for i, key in zip(range(0, length, 16), generator):
f.write(hex(key)+'\n')
res += num2str(str2num(to_encrypt[i:i+16]) ^ key)
f.close()
return res
def decrypt(ciphertxt):
# TODO
pass
def str2num(s):
return int(s.encode('hex'), 16)
def num2str(n, block=16):
s = hex(n)[2:].strip('L')
s = '0' * ((32-len(s)) % 32) + s
return s.decode('hex')
P = 0x100000000000000000000000000000087
A = 0xc6a5777f4dc639d7d1a50d6521e79bfd
B = 0x2e18716441db24baf79ff92393735345
N = str2num(urandom(16))
assert N != 0
if __name__ == '__main__':
with open('top_secret') as f:
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2022-01-16 22:14:19
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http://leagueoflegends.wikia.com/wiki/Magic_resistance?page=2
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# Magic resistance
1,909pages on
this wiki
Magic resistance (or MR) is a stat that all units have, including minions, monsters, and buildings. Increasing magic resistance reduces the magic damage the unit takes. Each champion begins with some magic resistance which may increase with level. You can gain additional magic resistance from abilities, items, masteries, and runes. Magic resistance stacks additively.
All champions begin with 30 base magic resistance. Currently no ranged champion gains magic resistance per level, but most melee champions do except , , , , , and .
, , , and do not gain magic resistance per level because they are not always melee champions due to their variable attack range.
All champions that do gain magic resistance per level gain 1.25 magic resistance per level, reaching 52.5 base magic resistance at level 18.
## Damage reduction
Magic Resistance reduces the damage of incoming magical attacks by a percentage. This percentage is determined by the formula: Damage Reduction = Total Magic Resistance / (100 + Total Magic Resistance). For example, a champion with 150 points of magic resistance would receive 60% reduced damage from magical attacks. Incoming magic damage is multiplied by a factor based on the unit's magic resistance(same with armor):
$${\rm Damage\ multiplier}=\begin{cases} {100 \over 100+{\it MR}}, & {\rm if\ }{\it MR} \geq 0\\ 2 - {100 \over 100 - {\it MR}}, & {\rm otherwise} \end{cases}$$
Examples:
• 25 magic resistance → × 0.8 incoming magic damage (20% reduction).
• 100 magic resistance → × 0.5 incoming magic damage (50% reduction).
• −25 magic resistance → × 1.2 incoming magic damage (20% increase).
## Stacking magic resistance
Every point of magic resistance requires a unit to take 1% more of its maximum health in magic damage to be killed. This is called "effective health."
Example: A unit with 60 magic resistance has 160% of its maximum health in its effective health, so if the unit has 1000 maximum health, it will take 1600 magic damage to kill it.
What this means: by definition, magic resistance does not have diminishing returns, because each point increases the unit's effective health against magic damage by 1% of its current actual health value whether the unit has 10 magic resistance or 1000 magic resistance.
For a more detailed explanation, see this video.
Unlike health, increasing magic resistance makes healings and shields more effective because it requires more effort from your enemies to remove the bonus health granted. This is called indirect scaling.
## Magic Resistance as Scaling
These use the champion's magic resistance to increase the magnitude of the ability. It could involve total or bonus magic resistance. By building magic resistance items, you can receive more benefit and power from these abilities.
### Champions
• grants magic resistance equal to 10 / 11.5 / 13 / 14.5 / 16% of his bonus magic resistance to himself and the target ally (in addition to a base amount) for 3 seconds.
• grants bonus magic resistance to himself equal to 20% of his bonus magic resistance.
• grants him bonus ability power equal to 50% of his total magic resistance.
• grants additional magic resist plus 20% of her bonus magic resist for its duration.
• takes 40% of the target's magic resistance and gains an equal amount of magic resistance. Half of this magic resistance is stolen on cast, and the next half is taken over 4 seconds. The magic resistance bonus/reduction lasts for another 4 seconds after the drain completes.
### Items
• active increases its slow duration by a base amount plus 0.5% of the wearer's total magic resistance.
## Increasing magic resistance
### Items
• : +50 magic resistance, +70 ability power. Unique Aura: Reduces the magic resistance of nearby enemies by 20. 2440 gold.
• : +20 magic resistance. Unique Aura – Legion: Nearby allies gain 20 magic resist, and +75% base health regeneration. +200 health. 1900 gold.
• : +25 magic resistance, +60 ability power, +50% base mana regeneration, +20% cooldown reduction. Unique Passive: Restores 15% of your maximum mana on kill or assist. Unique Passive – Mana Font: Restores 2% of missing mana every 5 seconds. 2700 gold.
• : +25 magic resistance, +50% base mana regeneration. Unique Passive – Mana Font: Restores 2% of missing mana every 5 seconds. 1000 gold.
• : +30 magic resistance, +25 attack damage. Unique Passive – Lifeline: Upon taking magic damage that would reduce health below 30%, grants a shield that absorbs 250 magic damage for 5 seconds (90 second cooldown). 1450 gold.
• : +20 magic resistance. Unique Aura - Legion: Nearby allies gain 20 magic resist, and +75% base health regen. +400 health, +10% cooldown reduction. Unique Active: Shield yourself and nearby allies for 5 seconds, absorbing up to 50 (+10 per level) damage (60 second cooldown). 2800 gold.
• : +40 magic resistance, +60 attack damage. Unique Passive: +1 Attack Damage for every 2.5% health missing. Unique Passive - Lifeline: If you would take magic damage which would leave you at less than 30% health, you first gain a shield that absorbs 400 magic damage for 5 seconds (90 second cooldown). 3200 gold.
• : +25 magic resistance. Unique Passive - Enhanced Movement: +45 movement speed. Unique Passive - Tenacity: Reduces the duration of stuns, slows, taunts, fears, silences, blinds and immobilizes by 35%. 1200 gold.
• : +40 magic resistance, +100% base mana regeneration, +10% cooldown reduction. Unique Passive – Mana Font: Restores 2% of missing mana every 5 seconds. Unique Active: Removes all stuns, roots, taunts, fears, silences and slows on an allied champion and heals that champion for 150 + (10% of your maximum health) (180 second cooldown). 2450 gold.
• : +45 magic resistance. 850 gold.
• : +25 magic resistance. 500 gold.
• : +30 magic resistance. Unique Active – Quicksilver: Removes all debuffs from your champion (90 second cooldown). 1250 gold.
• : +35 magic resistance, +200 health. Unique Passive: Grants 100% base health regen for up to 10 seconds after taking damage from an enemy champion. 1200 gold.
• : +55 magic resistance, +400 health, +100% base health regeneration, +10% cooldown reduction. Unique Passive: Increases self-healing, health regen, lifesteal, and spellvamp effect by 20%. 2750 gold.
• : +30 magic resistance. Unique Passive: Your basic attacks steal 5 magic resistance of the target for 5 seconds (stacks up to 5 times). +50% attack speed. Unique Passive: Your basic attacks deal 42 bonus magic damage. 2600 gold.
#### Variable Availability
• STH : +55 magic resistance, +450 health, +100% base health regeneration. Unique Passive: Grants a spell shield that blocks the next enemy ability. This shield refreshes after no damage is taken from enemy champions for 40 seconds. 2750 gold.
• TC : +45 magic resist, +50% attack speed, +10% cooldown reduction. Unique Active – Quicksilver: Removes all debuffs from your champion. Melee champions also gain +50% movement speed for 1 second (90 second cooldown). 2700 gold.
• SC : +50 magic resistance, +50 armor. Unique Passive: Upon taking lethal damage, restores 30% of maximum health and 30% of maximum mana after 4 seconds of stasis (5 minute cooldown). 2850 gold.
• SH : +35 magic resistance, +80 attack damage. Unique Active – Quicksilver: Removes all debuffs and also grants 50% bonus movement speed for 1 second (90 second cooldown). 3700 gold.
• C : +50 magic resistance, +350 health, +350 mana. Unique Passive: Reduces and stores 10% of magic damage received. Unique Active: Deals 200 + (stored magic damage – max 200) magic damage to nearby enemy units (90 second cooldown). 2500 gold.
• H : +70 magic resistance, +100% base health regeneration. Unique Passive: Grants a shield that absorbs up to 30 + (10 × level) damage. The shield will refresh after 9 seconds without receiving damage. 2210 gold.
### Champion abilities
Note: Only the magic resist buff effect of these abilities is shown here, to read more information on each of these abilities, follow the link on each of them.
• allows her to enter an egg-state for up to 6 seconds upon reaching 0 health. While in this state, she will receive a magic resistance modifier of −40 / −25 / −10 / +5 / +20.
• increases her magic resistance by 20 / 30 / 40 / 50 / 60 for 5 seconds.
• increases magic resistance by 15 / 17.5 / 20 / 22.5 / 25(+10% / 11.5% / 13% / 14.5% / 16% bonus MR) to himself and the target ally for 3 seconds.
• increases an allied champion's magic resistance by 30 / 45 / 60 / 75 / 90 for 4 seconds.
• increases his magic resistance by 2 × level when transform into .
• increases his magic resistance equal to 20% of bonus magic resistance.
• increases his magic resistance by 1 / 2 / 3 every second he remains in combat. This bonus stacks up to 10 times. Graves is considered in combat if he has dealt or received damage in the last 3 seconds.
• increases his magic resistance by 25 / 35 / 45 (+20% AP) for 8 seconds.
• increases his magic resistance by 10 / 20 / 30 / 40 / 50 for 4 seconds.
• increases his magic resistance by 5 / 15 / 25 / 35.
• increases her magic resistance by 20 / 30 / 40 / 50 / 60 (+ 20% bonus MR) for 3 seconds, deals damage after that time to units around her, and retains the defensive buff for an additional 3 seconds if any enemy is struck by the blast.
• increases an allied unit's magic resistance by 10 / 15 / 20 / 25 / 30 for 6 seconds.
• passively increases his magic resistance by 10 / 20 / 30. He loses the passive while his ability is active.
• increases an allied champion's magic resistance by 10 / 15 / 20 / 25 / 30 as long as the ball is attached to it.
• increases his magic resistance by 40 / 60 / 80 / 100 / 120 for 6 seconds.
• Rengar lets out a battle roar, damaging enemies increases his magic resist by 10 / 15 / 20 / 25 / 30 for 4 seconds. Rengar gains additional 50% magic resist for each enemy champion or large monster hit.
• passively increases her magic resistance by 5 / 10 / 15 / 20. This bonus is doubled while she is in dragon form.
• increases his magic resistance by 35 / 50 / 65 for 25 seconds.
• immediately steals 20% of the target's magic resistance, and a further 20% over 4 seconds. These magic resistance bonuses lasts for another 4 seconds after the drain completes.
• increases his magic resistance by 60 / 90 / 120 for 5 seconds.
• grants him 4 / 6 / 8 magic resistance for each nearby enemy champion.
• increases his magic resistance by 15 / 20 / 25 for each enemy champion hit for 8 seconds.
### Masteries
• increases bonus Armor and Magic Resistance by 2.5 / 5%.
• increases Magic Resistance by 2 / 3.5 / 5.
• increases Armor by 1 / 2 / 3 / 4 and Magic Resistance by 0.5 / 1 / 1.5 / 2 for each nearby enemy champion.
### Other
• Disconnecting gives approximately +1000 magic resistance.
## Ways to reduce magic resistance
See magic penetration. Note that magic penetration and magic resistance reduction are different.
## Trivia
• Prior to Patch 3.10, a level 18 with 1 , 5 , 3 points in , 3 points in , , , , , an allied aura, a full page of Scaling Magic Resist runes, and active gave a total of approximately 1004 magic resist. Switching for and having an enemy with the same setup use on and the allied use on the enemy , yielded a total of approximately 1297 magic resist. This is the highest possible amount of magic resistance, which is 92.8% reduction.
• If disconnects from the fountain and has the same setup, he will have 2108 magic resist. This is a 95.5% reduction.
• NOTE: This calculation does not include the mastery.
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2014-12-20 21:32:10
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https://www.physicsforums.com/threads/generate-a-circle-in-fortran-having-polar-coordinates.970416/
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# Fortran Generate a circle in FORTRAN having polar coordinates
#### sharmaN
Say "I have grid in polar coordinates (r, theta). How do I plot it in tecplot. Tecplot plots it in cartesian coordinates."
Last edited:
Related Programming and Computer Science News on Phys.org
#### anorlunda
Mentor
Gold Member
I don't understand. To make a X-Y grid, loop on X and loop on Y. To make a r-θ grid, loop on r and loop on θ.
Perhaps you want to make a rectangular grid within a circle?
Mentor
#### sharmaN
I don't understand. To make a X-Y grid, loop on X and loop on Y. To make a r-θ grid, loop on r and loop on θ.
Perhaps you want to make a rectangular grid within a circle?
#### sharmaN
Ifound in tecplot to plot in polar coordinates, need to select polar line. Thanks
#### RPinPA
FYI, in any language, if you let $\theta$ go from 0 to $2\pi$ then the points with cartesian coordinates $x = r \cos(\theta), y = r \sin(\theta)$ will be spaced around the circle of radius $r$.
#### sharmaN
FYI, in any language, if you let $\theta$ go from 0 to $2\pi$ then the points with cartesian coordinates $x = r \cos(\theta), y = r \sin(\theta)$ will be spaced around the circle of radius $r$.
Yes thank you, I have the grid. Was facing problem in plotting it.
"Generate a circle in FORTRAN having polar coordinates"
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2019-05-20 16:32:26
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https://engineering.stackexchange.com/questions/21109/does-every-engineering-design-needs-to-have-a-complete-theoretical-backing-or-ma
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# Does every engineering design needs to have a complete theoretical backing or may experimental data suffice?
This may sound rather vague on its own so I will explain with examples.
The solid booster rockets used by NASA space shuttle used a central bore design because it had a certain characteristic on burn rate and thrust.
If this data was to be acquired only form experimentation like burning test rockets and not deriving any mathematical relation between bore design and thrust then would it be considered sound engineering?
Are any designs based solely on data from trial and error used in critical mainstream engineering?
Like we use it just because it works and do not investigate why it works like that because it's very complex to understand but still easier to implement.
• Often as not, your complete theoretical understanding can be thrown out the window because you still have to do what the regs say. And the regs on things like building fishing boats and bridges are mostly based on when things went wrong. Look at the early power grid. We had electrified cities and designed and built hundreds of different types of electrical machines and none of those involved had a clue as to what electricity was. – Phil Sweet Mar 30 '18 at 18:45
• They built some beautiful buildings with trial and error, the mathematics had to be invented so the theory could catch up. One example of how theory catches up with the real world is the relatively « new » science of fatigue analysis after a series of serious plane crashes... – Solar Mike Mar 30 '18 at 19:43
• Depends, not all things need to be optimozed the hell out. However, theories help when you need to do optimisation. So if you launch rockets then it does help. Manny things are not like that so they dont need this. – joojaa Mar 30 '18 at 20:27
• this is really overly broad. Th simple answer is "of course" plenty of sound engineering design relies on empirical data. Are you seeking cases where we literally have no clue what the underlying physics is and don't care to even bother to investigate what that physics might be? – agentp Mar 30 '18 at 21:46
• As one of my lecturers said many years ago, "if the maths doesn't fit reality, the maths is wrong". Experimentation will always be a part of engineering. We are not that intelligent, individually or as a group, to be able to devise theory & engineering systems without experimentation. As others have commented here, sometimes the theory is developed after experimentation & observation of what actually happens. – Fred Mar 31 '18 at 8:03
Are any designs based solely on data from trial and error used in critical mainstream engineering?
Usually not. And the reason is that trial and error is expensive and time consuming. As engineers, we are always working on projects with a budget and a deadline. Take your rocket example. Rockets are expensive. For sake of argument, let's just say it's $1 million per rocket test. You can't afford to build rocket after rocket after rocket solely by trial and error. You'll go through your budget very quickly. Does every engineering design needs to have a complete theoretical backing? Again, usually not. And the reason is that developing complete theoretical understanding is also expensive. You could hire army of PhD researchers to come up with an incredibly detailed model and buy giant supercomputers to simulate it. But an engineer with a PhD will cost$100k/year just in salary. If your model is so detailed that you need a team of 20 researchers working for years to come up with the model, it would be cheaper to just run a test.
So there is a balance. You try to come up with a model that is sophisticated enough that it explains the majority of the behavior, but simple enough that you don't break the bank coming up with it. Your model won't be perfect, so you run a few experiments to fill in the holes in the theory, but not so many that you kill the budget. The key decision is deciding the tradeoff between model and experiments.
In fact, if I had to describe what engineers do in exactly one word, that word is "tradeoffs". In every engineering decision, there is always a balance between multiple competing objectives, and the engineers job is to make the best tradeoff.
• excellent exposition. – niels nielsen May 17 at 6:08
For complex and expensive systems you usually want to have an understanding of its behavior, cast in models. To model a system (or part of it), a complete theoretical understanding is usually not required.
You should, however, have a complete knowledge of the phenomena that act on your system. Then you identify the key phenomena and, if it does not yet exist, a theoretical understanding of those phenomena to a degree required to describe your system accurately enough.
The actual work is in finding out which phenomena are important and which are not, what effects to include, what effects to neglect, what depth of understading is actually required, which effects need to be understood, which effects are too expensive to understand now but can be compensated by overly conservative design decisions, etc. Those are some of the tradeoffs that @DanielKiracofe mentions in his answer.
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2019-07-19 17:31:44
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https://dlmf.nist.gov/25.16
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# §25.16 Mathematical Applications
## §25.16(i) Distribution of Primes
In studying the distribution of primes $p\leq x$, Chebyshev (1851) introduced a function $\psi\left(x\right)$ (not to be confused with the digamma function used elsewhere in this chapter), given by
25.16.1 $\psi\left(x\right)=\sum_{m=1}^{\infty}\sum_{p^{m}\leq x}\ln p,$ ⓘ Defines: $\psi\left(\NVar{x}\right)$: Chebyshev $\psi$-function Symbols: $\ln\NVar{z}$: principal branch of logarithm function, $m$: nonnegative integer, $p$: prime number and $x$: real variable Permalink: http://dlmf.nist.gov/25.16.E1 Encodings: TeX, pMML, png See also: Annotations for 25.16(i), 25.16 and 25
which is related to the Riemann zeta function by
25.16.2 $\psi\left(x\right)=x-\frac{\zeta'\left(0\right)}{\zeta\left(0\right)}-\sum_{% \rho}\frac{x^{\rho}}{\rho}+o\left(1\right),$ $x\to\infty$,
where the sum is taken over the nontrivial zeros $\rho$ of $\zeta\left(s\right)$.
The prime number theorem (27.2.3) is equivalent to the statement
25.16.3 $\psi\left(x\right)=x+o\left(x\right),$ $x\to\infty$. ⓘ Symbols: $\psi\left(\NVar{x}\right)$: Chebyshev $\psi$-function, $o\left(\NVar{x}\right)$: order less than and $x$: real variable Referenced by: §25.10(i) Permalink: http://dlmf.nist.gov/25.16.E3 Encodings: TeX, pMML, png See also: Annotations for 25.16(i), 25.16 and 25
The Riemann hypothesis is equivalent to the statement
25.16.4 $\psi\left(x\right)=x+O\left(x^{\frac{1}{2}+\epsilon}\right),$ $x\to\infty$, ⓘ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $\psi\left(\NVar{x}\right)$: Chebyshev $\psi$-function and $x$: real variable Referenced by: §25.16(i) Permalink: http://dlmf.nist.gov/25.16.E4 Encodings: TeX, pMML, png See also: Annotations for 25.16(i), 25.16 and 25
for every $\epsilon>0$.
## §25.16(ii) Euler Sums
Euler sums have the form
25.16.5 $H\left(s\right)=\sum_{n=1}^{\infty}\frac{h(n)}{n^{s}},$ ⓘ Symbols: $H\left(\NVar{s}\right)$: Euler sums, $n$: nonnegative integer, $s$: complex variable and $h(n)$: sum Permalink: http://dlmf.nist.gov/25.16.E5 Encodings: TeX, pMML, png See also: Annotations for 25.16(ii), 25.16 and 25
where $h(n)$ is given by (25.11.33).
$H\left(s\right)$ is analytic for $\Re s>1$, and can be extended meromorphically into the half-plane $\Re s>-2k$ for every positive integer $k$ by use of the relations
25.16.6 $H\left(s\right)=-\zeta'\left(s\right)+\gamma\zeta\left(s\right)+\frac{1}{2}% \zeta\left(s+1\right)+\sum_{r=1}^{k}\zeta\left(1-2r\right)\zeta\left(s+2r% \right)+\sum_{n=1}^{\infty}\frac{1}{n^{s}}\int_{n}^{\infty}\frac{\widetilde{B}% _{2k+1}\left(x\right)}{x^{2k+2}}\mathrm{d}x,$
25.16.7 $H\left(s\right)=\frac{1}{2}\zeta\left(s+1\right)+\frac{\zeta\left(s\right)}{s-% 1}-\sum_{r=1}^{k}\genfrac{(}{)}{0.0pt}{}{s+2r-2}{2r-1}\zeta\left(1-2r\right)% \zeta\left(s+2r\right)-\genfrac{(}{)}{0.0pt}{}{s+2k}{2k+1}\sum_{n=1}^{\infty}% \frac{1}{n}\int_{n}^{\infty}\frac{\widetilde{B}_{2k+1}\left(x\right)}{x^{s+2k+% 1}}\mathrm{d}x.$
For integer $s$ ($\geq 2$), $H\left(s\right)$ can be evaluated in terms of the zeta function:
25.16.8 $\displaystyle H\left(2\right)$ $\displaystyle=2\zeta\left(3\right),$ $\displaystyle H\left(3\right)$ $\displaystyle=\tfrac{5}{4}\zeta\left(4\right),$ ⓘ Symbols: $H\left(\NVar{s}\right)$: Euler sums and $\zeta\left(\NVar{s}\right)$: Riemann zeta function Permalink: http://dlmf.nist.gov/25.16.E8 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 25.16(ii), 25.16 and 25
25.16.9 $H\left(a\right)=\frac{a+2}{2}\zeta\left(a+1\right)-\frac{1}{2}\sum_{r=1}^{a-2}% \zeta\left(r+1\right)\zeta\left(a-r\right),$ $a=2,3,4,\dots$. ⓘ Symbols: $H\left(\NVar{s}\right)$: Euler sums, $\zeta\left(\NVar{s}\right)$: Riemann zeta function and $a$: real or complex parameter Permalink: http://dlmf.nist.gov/25.16.E9 Encodings: TeX, pMML, png See also: Annotations for 25.16(ii), 25.16 and 25
Also,
25.16.10 $H\left(-2a\right)=\frac{1}{2}\zeta\left(1-2a\right)=-\frac{B_{2a}}{4a},$ $a=1,2,3,\dots$.
$H\left(s\right)$ has a simple pole with residue $\zeta\left(1-2r\right)$ ($=-B_{2r}/(2r)$) at each odd negative integer $s=1-2r$, $r=1,2,3,\dots$.
$H\left(s\right)$ is the special case $H\left(s,1\right)$ of the function
25.16.11 $H\left(s,z\right)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}\sum_{m=1}^{n}\frac{1}{m^{% z}},$ $\Re(s+z)>1$,
which satisfies the reciprocity law
25.16.12 $H\left(s,z\right)+H\left(z,s\right)=\zeta\left(s\right)\zeta\left(z\right)+% \zeta\left(s+z\right),$
when both $H\left(s,z\right)$ and $H\left(z,s\right)$ are finite.
For further properties of $H\left(s,z\right)$ see Apostol and Vu (1984). Related results are:
25.16.13 $\displaystyle\sum_{n=1}^{\infty}\left(\frac{h(n)}{n}\right)^{2}$ $\displaystyle=\frac{17}{4}\zeta\left(4\right),$ ⓘ Symbols: $\zeta\left(\NVar{s}\right)$: Riemann zeta function, $n$: nonnegative integer and $h(n)$: sum Permalink: http://dlmf.nist.gov/25.16.E13 Encodings: TeX, pMML, png See also: Annotations for 25.16(ii), 25.16 and 25 25.16.14 $\displaystyle\sum_{r=1}^{\infty}\sum_{k=1}^{r}\frac{1}{rk(r+k)}$ $\displaystyle=\frac{5}{4}\zeta\left(3\right),$ ⓘ Symbols: $\zeta\left(\NVar{s}\right)$: Riemann zeta function and $k$: nonnegative integer Permalink: http://dlmf.nist.gov/25.16.E14 Encodings: TeX, pMML, png See also: Annotations for 25.16(ii), 25.16 and 25 25.16.15 $\displaystyle\sum_{r=1}^{\infty}\sum_{k=1}^{r}\frac{1}{r^{2}(r+k)}$ $\displaystyle=\frac{3}{4}\zeta\left(3\right).$ ⓘ Symbols: $\zeta\left(\NVar{s}\right)$: Riemann zeta function and $k$: nonnegative integer Permalink: http://dlmf.nist.gov/25.16.E15 Encodings: TeX, pMML, png See also: Annotations for 25.16(ii), 25.16 and 25
For further generalizations, see Flajolet and Salvy (1998).
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2018-02-18 21:55:26
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https://webwork.libretexts.org/webwork2/html2xml?answersSubmitted=0&sourceFilePath=Library/UCSB/Stewart5_2_1/Stewart5_2_1_3.pg&problemSeed=123567&courseID=anonymous&userID=anonymous&course_password=anonymous&showSummary=1&displayMode=MathJax&problemIdentifierPrefix=102&language=en&outputformat=sticky
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The point $P(1,1/2)$ lies on the curve $y=x/(1+x)$.
(a) If $Q$ is the point $(x,x/(1+x))$, find the slope of the secant line $PQ$ correct to four decimal places for the following values of $x$:
(1) .5 (2) .9
(3) .99 (4) .999
(5) 1.5 (6) 1.1
(7) 1.01 (8) 1.001
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(b) Guess the slope of the tangent line to the curve at $P$.
Slope =
(c) Using the slope from part (b), find the equation of the tangent line to the curve at $P$.
$y=$
Your overall score for this problem is
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2020-08-09 04:52:56
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http://nd.ics.org.ru/archive_nd/tom-7-2/
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0
2013
Impact Factor
# Vol. 7, No. 2, 2011
Moskalenko O. I., Koronovskii A. A., Shurygina S. A. Abstract The intermittent behavior near the boundary of the noise-induced synchronization regime is studied. «On-off» intermittency is shown to take place in this case. The observed phenomenon is illustrated by considering both model systems with discrete time and flow dynamical systems being under influence of the common source of noise. Keywords: nonlinear systems, intermittency, noise-induced synchronization, noise, dynamical chaos Citation: Moskalenko O. I., Koronovskii A. A., Shurygina S. A., The behavior of nonlinear systems near the boundary of noise-induced synchronization, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 197-208 DOI:10.20537/nd1102001
Alfimov G. L. Abstract Nonlocal generalizations of nonlinear wave equation arise in numerous physical applications. It is known that switching from local to nonlocal description may result in new features of the problem and new types of solutions. In this paper the author analyses the dimension of the set of travelling wave solutions for а nonlocal nonlinear wave equation. The nonlocality is represented by the convolution operator which replaces the second derivative in the dispersion term. The results have been obtained for the case where the nonlinearity is bounded, and the kernel of the convolution operator is represented by a sum of exponents with weights (so-called E-type kernel). In the simplest particular case, (so-called Kac—Baker kernel) it is shown that the solutions of this equation form a 3-parametric set (assuming the equivalence of the solutions which differ by a shift with respect to the independent variable). Then it is shown that in the case of the general E-type kernel the 3-parametric set of solutions also exists, generically, under some additional restrictions. The word «generically» in this case means some transversality condition for intersection of some manifolds in a properly defined phase space. Keywords: nonlocal nonlinear wave equation Citation: Alfimov G. L., On the dimension of the set of solutions for nonlocal nonlinear wave equation, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 209-226 DOI:10.20537/nd1102002
Pochinka O. V. Abstract In this paper class $MS(M^3)$ of Morse–Smale diffeomorphisms (cascades) given on connected closed orientable 3-manifolds are considered. For a diffeomorphism $f \in MS(M^3)$ it is introduced a notion scheme $S_f$, which contains an information on the periodic data of the cascade and a topology of embedding of the sepsrstrices of the saddle points. It is established that necessary and sufficient condition for topological conjugacy of diffeomorphisms $f$, $f’ \in MS(M^3)$ is the equivalence of the schemes $S_f$, $S_f’$. Keywords: Morse–Smale diffeomorphism (cascade), topological conjugacy, space orbit Citation: Pochinka O. V., Necessary and sufficient conditions for topological classification of Morse–Smale cascades on 3-manifolds, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 227-238 DOI:10.20537/nd1102003
Smirnov A. O., Golovachev G. M., Amosenok E. G. Abstract The behavior of the two-gap elliptic solutions of the Boussinesq and the KdV equations was examined. These solutions were constructed by the $n$-sheet covering over a torus $(n \leqslant 3)$. It was shown that the shape of the two-gap elliptic solutions depends on $n$ and doesn’t depend on the kind of the nonlinear wave equation. Keywords: soliton, Boussinesq equation, KdV equation, theta-function, reduction, covering Citation: Smirnov A. O., Golovachev G. M., Amosenok E. G., Two-gap 3-elliptic solutions of the Boussinesq and the Korteweg-de Vries equations, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 239-256 DOI:10.20537/nd1102004
Churilov S. M. Abstract Within the context of the weakly nonlinear approach, the leading nonlinear contribution to the development of unstable disturbances in shear flows should be made by resonant three-wave interaction, i.e., the interaction of triplets of such waves that have a common critical layer (CL), and their wave vectors form a triangle. Surprisingly, the subharmonic resonance proves to be the only such interaction that has been studied so far. The reason for this is that in many cases, the requirement of having a common CL produces too rigid selection of waves which can participate in the interaction. We show that in a broad spectral range, Holmboe waves in sharply stratified shear flows can have a common CL, and examine the evolution of small ensembles consisting of several interrelated triads of those waves. To do this, the evolution equations are derived which describe the three-wave interaction and have the form of nonlinear integral equations. Analytical and numerical methods are both used to find their solutions in different cases, and it is shown that at the nonlinear stage disturbances increase, as a rule, explosively. Keywords: shear flow, sharp density stratification, three-wave interactions, critical layer Citation: Churilov S. M., Resonant three–wave interaction of waves having a common critical layer, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 257-282 DOI:10.20537/nd1102005
Ryzhov E. A. Abstract The integrable and nonintegrable motion of a vortex pair, which consists of two vortices of arbitrary intensities, embedded inside a steady and periodic external deformation flow is studied. In the general case, such an external deformation flow impacts asymmetrically on the vortex pair, which results in nonconservation of motion invariants: the linear momentum and the angular momentum. An analytical expression for the linear momentum, which gives an opportunity to reduce the initial system with 2.5 degrees of freedom to a system with 1.5 degrees of freedom, is obtained. For the steady state of a constant deformation flow the integrability of the dipole motion is shown for any initial vortices positions and intensities of vortices, and for arbitrary values of shear and rotation of the deformation flow. Keywords: vortex pair, deformation flow, integrals of motion Citation: Ryzhov E. A., The integrable and nonintegrable motion of a vortex pair embedded inside an asymmetrical deformation flow, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 283-293 DOI:10.20537/nd1102006
Rodnikov A. V. Abstract We study a particle motion along a cable with ends fixed in a precessed rigid body. Such cable called «the leier» is a model of space elevator for a dynamically symmetric asteroid. (The Dutch term «leier» means the rope with fixed ends). In this paper we find two integrable cases of the particle motion equations (for zero and right nutation angle) Phase portraits for integrable situations are built taking into account conditions of motion with the tense cable and assuming the body gravitation is close to gravitational field of two equal point masses that are in the axis of dynamical symmetry. Using «the Generalized Restricted Circular Problem of Three Bodies» by V. V. Beletsky, we study the particle equilibria on the leier in the plane containing the body mass center and being perpendicular to the precession axis for all possible nutation angles. Some facts on these equilibria stability are formulated. Keywords: space elevator, space tether system, asteroid, unilateral constraint, problem of three bodies Citation: Rodnikov A. V., On a particle motion along the leier fixed in a precessing rigid body, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 295-311 DOI:10.20537/nd1102007
Borisov A. V., Kilin A. A., Mamaev I. S. Abstract We consider the problem of explicit integration and bifurcation analysis for two systems of nonholonomic mechanics. The first one is the Chaplygin’s problem on no-slip rolling of a balanced dynamically non-symmetrical ball on a horizontal plane. The second problem is on the motion of rigid body in a spherical support. We explicitly integrate this problem by generalizing the transformation which Chaplygin applied to the integration of the problem of the rolling ball at a non-zero constant of areas. We consider the geometric interpretation of this transformation from the viewpoint of a trajectory isomorphism between two systems at different levels of the energy integral. Generalization of this transformation for the case of dynamics in a spherical support allows us to integrate the equations of motion explicitly in quadratures and, in addition, to indicate periodic solutions and analyze their stability. We also show that adding a gyrostat does not lead to the loss of integrability. Keywords: nonholonomic mechanics, spherical support, Chaplygin ball, explicit integration, isomorphism, bifurcation analysis Citation: Borisov A. V., Kilin A. A., Mamaev I. S., Generalized Chaplygin’s transformation and explicit integration of a system with a spherical support, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 313-338 DOI:10.20537/nd1102008
Weidman P. D., Malhotra C. P. Abstract We review previous investigations concerning the terminal motion of disks sliding and spinning with uniform dry friction across a horizontal plane. Previous analyses show that a thin circular ring or uniform circular disk of radius $R$ always stops sliding and spinning at the same instant. Moreover, under arbitrary nonzero initial values of translational speed $v$ and angular rotation rate $ω$, the terminal value of the speed ratio $ε_0 = v/Rω$ is always 1.0 for the ring and 0.653 for the uniform disk. In the current study we show that an annular disk of radius ratio $η = R_2/R_1$ stops sliding and spinning at the same time, but with a terminal speed ratio dependent on $η$. For a twotier disk with lower tier of thickness $H_1$ and radius $R_1$ and upper tier of thickness $H_2$ and radius $R_2$, the motion depends on both $η$ and the thickness ratio $λ = H_1/H_2$. While translation and rotation stop simultaneously, their terminal ratio $ε_0$ either vanishes when $k > \sqrt{2/3}$, is a nonzero constant when $1/2 < k < \sqtr{2/3}$, or diverges when $k < 1/2$, where $k$ is the normalized radius of gyration. These three regimes are in agreement with those found by Goyal et al. [S.Goyal, A.Ruina, J.Papadopoulos, Wear 143 (1991) 331] for generic axisymmetric bodies with varying radii of gyration using geometric methods. New experiments with PVC disks sliding on a nylon fabric stretched over a plexiglass plate only partially corroborate the three different types of terminal motions, suggesting more complexity in the description of friction. Keywords: rigid body dynamics, terminal motion, nonlinear behavior Citation: Weidman P. D., Malhotra C. P., On the terminal motion of sliding spinning disks with uniform Coulomb friction, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 339-365 DOI:10.20537/nd1102009
Koiller J., Ehlers K. M. Abstract Citation: Koiller J., Ehlers K. M., The birth of biofluiddynamics, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 367-370 DOI:10.20537/nd1102010
Taylor G. Abstract Large objects which propel themselves in air or water make use of inertia in the surrounding fluid. The propulsive organ pushes the fluid backwards, while the resistance of the body gives the fluid a forward momentum. The forward and backward momenta exactly balance, but the propulsive organ and the resistance can be thought about as acting separately. This conception cannot be transferred to problems of propulsion in microscopic bodies for which the stresses due to viscosity may be many thousands of times as great as those due to inertia. No case of self-propulsion in a viscous fluid due to purely viscous forces seems to have been discussed.The motion of a fluid near a sheet down which waves of lateral displacement are propagated is described. It is found that the sheet moves forwards at a rate $2π^2 b^2/λ^2$ times the velocity of propagation of the waves. Here $b$ is the amplitude and $λ$ the wave-length. This analysis seems to explain how a propulsive tail can move a body through a viscous fluid without relying on reaction due to inertia. The energy dissipation and stress in the tail are also calculated.The work is extended to explore the reaction between the tails of two neighbouring small organisms with propulsive tails. It is found that if the waves down neighbouring tails are in phase very much less energy is dissipated in the fluid between them than when the waves are in opposite phase. It is also found that when the phase of the wave in one tail lags behind that in the other there is a strong reaction, due to the viscous stress in the fluid between them, which tends to force the two wave trains into phase. It is in fact observed that the tails of spermatozoa wave in unison when they are close to one another and pointing the same way. Citation: Taylor G., Analysis of the swimming of microscopic organisms, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 371-387 DOI:10.20537/nd1102011
Borisov A. V., Gazizullina L., Mamaev I. S. Abstract This paper has been written for a collection of V.A. Steklov’s selected works, which is being prepared for publication and is entitled «Works on Mechanics 1902–1909: Translations from French». The collection is based on V.A. Steklov’s papers on mechanics published in French journals from 1902 to 1909. Citation: Borisov A. V., Gazizullina L., Mamaev I. S., On V.A. Steklov’s legacy in classical mechanics, Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 389-403 DOI:10.20537/nd1102012
Abstract Citation: New books. New issues of «Regular and Chaotic Dynamics», Rus. J. Nonlin. Dyn., 2011, Vol. 7, No. 2, pp. 405-408
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2020-03-29 00:46:00
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5594668388366699, "perplexity": 3071.4659554730792}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370493121.36/warc/CC-MAIN-20200328225036-20200329015036-00276.warc.gz"}
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https://cran.dcc.uchile.cl/web/packages/vmr/vignettes/O7-vmrcicd.html
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# CI/CD
You can set up a vmr environment to use it in CI/CD pipelines.
Once you set up your vmr environment take a snapshot of it.
vmrTakeSnapshot("cicdversionR")
## GitLab Runner CI/CD
Then you can get the command to run, to add it as a GitLab Runner:
virtualboxGitlabRunner(vmr_env,
gitlab_url = "gitlab.com",
gt_token = "<mytoken>",
snapshot_name = "cicdversionR",
vm_name = <VirtualBox VM Name>)
Copy and paste the return command in a terminal where GitLab Runner and the vmr environment are installed.
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2021-09-24 20:31:21
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https://tisp.indigits.com/sparse_approx/stability.html
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# 20.1. Stability of the Sparsest Solution#
We discuss various results related to the stability of the sparsest solution for the sparse approximation problem.
For convenience, we restate the problem. We represent the signal $$\bx \in \CC^N$$ as $$\bx = \bDDD \ba + \be$$ where $$\ba$$ is a sparse approximation of $$\bx$$ in $$\bDDD$$.
(20.1)#$\widehat{\ba} = \text{arg } \underset{\ba \in \CC^D}{\min} \| \ba \|_0 \text{ subject to } \| \bx - \bDDD \ba \|_2 \leq \epsilon.$
Since we only know $$\bx$$ and $$\bDDD$$ (both $$\ba$$ and $$\be$$ are unknown to us), hence in general it is not possible to reconstruct $$\ba$$ exactly. The term $$\bd = \ba - \widehat{\ba}$$ will represent the reconstruction error (note that it is different from the approximation error $$\be$$). It is important for us to ensure that the solution of the approximation problem is stable; i.e., in the presence of small approximation error $$\| \be \|_2$$, the reconstruction error $$\| \bd \|_2$$ should also be small. For the sparse approximation problem (20.1), we cannot provide a uniqueness guarantee as such. Still, we can identify criteria which ensure that the solution remains stable when the approximation error is bounded.
Our analysis in this section will focus on identifying criteria which ensures this.
We start with generalizing the notion of spark for the noisy case.
1. Suppose $$\ba$$ and $$\bb$$ are two solutions of (20.1).
2. Then $$\|\bDDD \ba - \bx \|_2 \leq \epsilon$$ as well as $$\| \bDDD \bb - \bx \|_2 \leq \epsilon$$.
3. Thus, both $$\bDDD \ba$$ and $$\bDDD \bb$$ lie in a ball of radius $$\epsilon$$ around $$\bx$$.
4. Thus, the maximum distance between $$\bDDD \ba$$ and $$\bDDD \bb$$ can be $$2 \epsilon$$.
5. Alternatively, using triangle inequality we have
$\begin{split} \| \bDDD (\ba - \bb) \|_2 &= \| \bDDD \ba - \bx + \bx - \bDDD \bb) \|_2 \\ & \leq \| \bDDD \ba - \bx \|_2 + \| \bx - \bDDD \bb) \|_2 \leq 2 \epsilon. \end{split}$
6. If we define $$\bd = \ba - \bb$$, then
$\| \bDDD \bd \|_2 \leq 2 \epsilon.$
## 20.1.1. Spark $$\eta$$#
Definition 20.1
Let $$\bA \in \CC^{N \times D}$$ be some matrix. Consider all possible sub-sets of $$K$$ columns. Let each such set form sub-matrix $$\bA_{\Lambda} \in \CC^{N \times K}$$ where $$\Lambda$$ denotes the index set of $$K$$ indices chosen. We define $$\spark_{\eta}(\bA)$$ as the smallest possible $$K$$ (number of columns) that guarantees
$\underset{\Lambda}{\min}\sigma_K (\bA_{\Lambda}) \leq \eta$
where $$\sigma_K$$ denotes the smallest singular value (i.e. $$K$$-th singular value) of the sub-matrix $$\bA_{\Lambda}$$. Note that we are minimizing over all possible index sets $$\Lambda$$ with $$| \Lambda | = K$$.
In words, this is the smallest number of columns (indexed by $$\Lambda$$) that can be gathered from $$\bA$$ such that the smallest singular value of $$\bA_{\Lambda}$$ is no larger than $$\eta$$; i.e., there exists a sub-matrix of $$\bA$$ consisting of $$\spark_{\eta}(\bA)$$ columns whose smallest singular value is $$\eta$$ or less. At the same time, all submatrices of $$\bA$$ with number of columns less than $$\spark_{\eta}(\bA)$$ have the smallest singular value larger than $$\eta$$.
Relationship with $$\spark$$:
1. When the smallest singular value is $$0$$, then the columns are linearly dependent.
2. Thus, by choosing $$\eta = 0$$, we get the smallest number of columns $$K$$ which are linearly dependent.
3. This matches with the definition of spark.
4. Thus,
${\spark}_0 (\bA) = \spark (\bA).$
5. Since singular values are always non-negative, hence $$\eta \geq 0$$.
Matrix with unit norm columns:
1. When columns of $$\bA$$ are unit-norm (the case of dictionaries), then any single column sub-matrix has a singular value of $$1$$. Hence,
${\spark}_1(\bA) = 1.$
2. Choosing a value of $$\eta > 1$$ doesn’t make any difference since with a single column sub-matrix, we can show that
${\spark}_{\eta}(\bA) = 1 \Forall \eta \geq 1.$
Monotonicity:
1. Let $$\eta_1 > \eta_2$$. Let
$K_2 = {\spark}_{\eta_2} (\bA).$
2. Then there exists a sub-matrix consisting of $$K_2$$ columns of $$\bA$$ whose smallest singular value is upper bounded by $$\eta_2$$.
3. Since $$\eta_1 > \eta_2$$, $$\eta_1$$ also serves as an upper bound for the smallest singular value for this sub-matrix.
4. Clearly then $$K_1 = \spark_{\eta_1}(\bA) \leq K_2$$.
5. Thus, we note that $${\spark}_{\eta}$$ is a monotone decreasing function of $$\eta$$. i.e.
${\spark}_{\eta_1} (\bA) \leq {\spark}_{\eta_2} (\bA), \text{ whenever } \eta_1 > \eta_2.$
Maximum value:
1. We recall that the spark of $$\bA$$ is upper bounded by its rank plus one.
2. Assuming $$\bA$$ to be a full rank matrix, we get following inequality:
$1 \leq {\spark}_{\eta} (\bA) \leq {\spark}_0(\bA) = \spark(\bA) \leq N + 1 \Forall 0 \leq \eta \leq 1.$
We recall that if $$\bA \bv = \bzero$$ then $$\| \bv \|_0 \geq \spark(\bA)$$. A similar property can be developed for $$\spark_{\eta}(\bA)$$ also.
Theorem 20.1
If $$\| \bA \bv \|_2 \leq \eta$$ and $$\| \bv \|_2 = 1$$, then $$\| \bv \|_0 \geq \spark_{\eta} (\bA)$$.
Proof. For contradiction, let us assume that $$K = \| \bv \|_0 < {\spark}_{\eta}(\bA)$$.
1. Let $$\Lambda = \supp (\bv)$$.
2. Then $$\bA \bv = \bA_{\Lambda} \bv_{\Lambda}$$.
3. Also $$\| \bv_{\Lambda} \|_2 = \| \bv \|_2 = 1$$.
4. We recall that the smallest singular value of $$\bA_{\Lambda}$$ is given by
$\sigma_{\text{min}} (\bA_{\Lambda}) = \underset{\| \bx \|_2 = 1}{\inf} \| \bA_{\Lambda} \bx \|_2.$
5. Thus,
$\| \bA_{\Lambda} \bx \|_2 \geq \sigma_{\text{min}} (\bA_{\Lambda}) \text{ whenever } \| \bx \|_2 = 1.$
6. Thus, in our particular case
$\| \bA_{\Lambda} \bv_{\Lambda} \|_2 \geq \sigma_{\text{min}}(\bA_{\Lambda}).$
$$\bA_{\Lambda}$$ has $$K$$ columns with $$K < \spark_{\eta} (\bA)$$.
7. Thus, from the definition of $$\spark_{\eta} (\bA)$$
$\sigma_{\text{min}} (\bA_{\Lambda}) > \eta.$
8. This gives us
$\| \bA \bv \|_2 = \| \bA_{\Lambda} \bv_{\Lambda} \|_2 > \eta$
which contradicts with the assumption that $$\| \bA \bv \|_2 \leq \eta$$.
### 20.1.1.1. Coherence#
In the following, we will focus on the $$\spark_{\eta}$$ of a full rank dictionary $$\bDDD$$. We now establish a connection between $$\spark_{\eta}$$ and coherence of a dictionary.
Theorem 20.2
Let $$\bDDD$$ be a full rank dictionary with coherence $$\mu$$. Then
(20.2)#${\spark}_{\eta} (\bDDD) \geq \frac{1 - \eta^2}{ \mu} + 1.$
Proof. .
1. We recall from Gershgorin’s theorem that for any square matrix $$\bA \in \CC^{K \times K}$$, every eigen value $$\lambda$$ of $$\bA$$ satisfies
$| \lambda - a_{i i} | \leq \sum_{j, j \neq i} |a_{i j}| \text{ for some } i \in \{ 1, \dots, K\}.$
2. Now consider a matrix $$\bA$$ with diagonal elements equal to 1 and off diagonal elements bounded by a value $$\mu$$.
3. Then
$| \lambda - 1 | \leq \sum_{j, j \neq i} |a_{i j}| \leq \sum_{j \neq i} \mu = (K - 1) \mu.$
4. Thus,
$\begin{split} & - (K - 1) \mu \leq \lambda - 1 \leq (K - 1) \mu \\ \iff & 1 - (K - 1) \mu \leq \lambda \leq 1 + (K - 1) \mu. \end{split}$
5. This gives us a lower bound on the smallest eigen value.
$\lambda_{\min} (\bA) \geq 1 - (K - 1) \mu.$
6. Now consider any index set $$\Lambda \subseteq \{ 1, \dots, D \}$$ and consider the submatrix $$\bDDD_{\Lambda}$$ with $$|\Lambda | = {\spark}_{\eta}(\bDDD) = K$$.
7. Define $$\bG = \bDDD_{\Lambda}^H \bDDD_{\Lambda}$$.
8. The diagonal elements of $$\bG$$ are one, while off-diagonal elements are bounded by $$\mu$$.
9. Thus,
$\begin{split} & \lambda_{\min} (\bG) \geq 1 - (K - 1) \mu \\ \iff & (K - 1) \mu \geq 1 - \lambda_{\min} (\bG)\\ \iff & K - 1 \geq \frac{1 - \lambda_{\min} (\bG)}{\mu}\\ \iff & K \geq \frac{1 - \lambda_{\min} (\bG)}{\mu} + 1. \end{split}$
10. Since this applies to every sub-matrix $$\bDDD_{\Lambda}$$, this in particular applies to the sub-matrix for which $$\sigma_{\min}(\bDDD_{\Lambda}) \leq \eta$$ holds.
11. For this sub-matrix
$\lambda_{\min}(\bDDD_{\Lambda}^H \bDDD_{\Lambda}) = \sigma_{\min}^2(\bDDD_{\Lambda}) \leq \eta^2.$
12. Thus
$K = {\spark}_{\eta} (\bDDD) \geq \frac{1 - \lambda_{\min} (\bG)}{\mu} + 1 \geq \frac{1 - \eta^2}{\mu} + 1.$
## 20.1.2. Uncertainty with Spark $$\eta$$#
We now present an uncertainly result for the noisy case.
Theorem 20.3
If $$\ba_1$$ and $$\ba_2$$ satisfy $$\| \bx - \bDDD \ba_i \|_2 \leq \epsilon, i = 1,2$$, then
(20.3)#$\| \ba_1 \|_0 + \| \ba_2 \|_0 \geq {\spark}_{\eta}(\bDDD), \text{ where } \eta = \frac{2 \epsilon}{\| \ba_1 - \ba_2 \|_2}.$
Proof. .
1. From triangle inequality we have
$\| \bDDD (\ba_1 - \ba_2) \|_2 \leq 2 \epsilon.$
2. We define $$\bb = \ba_1 - \ba_2$$.
3. Then $$\| \bDDD \bb \|_2 \leq 2 \epsilon$$.
4. Further define $$\bv = \bb / \| \bb \|_2$$ as the normalized vector.
5. Then
$\| \bDDD \bv \|_2 = \frac{\| \bDDD \bb \|_2}{\| \bb \|_2} \leq \frac{2 \epsilon}{\| \bb \|_2}.$
6. Now define
$\eta = \frac{2 \epsilon}{\| \bb \|_2} = \frac{2 \epsilon}{\| \ba_1 - \ba_2 \|_2}.$
7. Then from Theorem 20.1 if $$\| \bDDD \bv \|_2 \leq \eta$$ with $$\| \bv \|_2 = 1$$, then $$\| \bv \|_0 \geq {\spark}_{\eta}(\bDDD)$$.
8. Finally,
$\| \ba_1 \|_0 + \| \ba_2 \|_0 \geq \| \ba_1 - \ba_2 \|_0 = \| \bb \|_0 = \| \bv \|_0 \geq {\spark}_{\eta}(\bDDD).$
9. This concludes the proof.
This result gives us a lower bound on the sum of sparsity levels of two different sparse representations of same vector $$\bx$$ under the given bound approximation error.
## 20.1.3. Localization#
We can now develop a localization result for the sparse approximation up to a Euclidean ball. This is analogous to the uniqueness result in noiseless case.
Theorem 20.4
Given a distance $$\delta \geq 0$$ (bound on distance between two sparse representations) and $$\epsilon$$ (bound on norm of approximation error), set $$\eta = 2 \epsilon / \delta$$. Suppose there are two approximate representations $$\ba_i, i = 1,2$$ both obeying
$\| \bx - \bDDD \ba_i \|_2 \leq \epsilon \, \text{ and }\, \| \ba_i \|_0 \leq \frac{1}{2} {\spark}_{\eta} (\bDDD).$
Then $$\| \ba_1 - \ba_2 \|_2 \leq \delta$$.
Proof. .
1. Since $$\| \ba_i \|_0 \leq \frac{1}{2} {\spark}_{\eta} (\bDDD)$$, hence
$\| \ba_1 \|_0 + \| \ba_2 \|_0 \leq {\spark}_{\eta} (\bDDD).$
2. From Theorem 20.3, if we define
$\nu = \frac{2 \epsilon}{ \| \ba_1 - \ba_2\|_2},$
then
$\| \ba_1 \|_0 + \| \ba_2 \|_0 \geq {\spark}_{\nu}(\bDDD).$
3. Combining the two, we get
${\spark}_{\eta} (\bDDD) \geq \| \ba_1 \|_0 + \| \ba_2 \|_0 \geq {\spark}_{\nu}(\bDDD).$
4. Because of the monotonicity of $${\spark}_{\eta} (\bDDD)$$, we have
$\begin{split} {\spark}_{\eta} (\bDDD) \geq {\spark}_{\nu}(\bDDD) &\implies \eta \leq \nu \\ &\implies \frac{2\epsilon}{\delta} \leq \frac{2 \epsilon}{ \| \ba_1 - \ba_2\|_2}\\ &\implies \delta \geq \| \ba_1 - \ba_2\|_2 \end{split}$
which completes our proof.
This theorem says that if $$\bx$$ has two different sufficiently sparse representations $$\ba_i$$ with small approximation errors, they fall within a small distance.
## 20.1.4. Stability using Coherence#
We can now develop a stability result for the (20.1) problem in terms of coherence of the dictionary.
Theorem 20.5
Consider an instance of the (20.1) problem defined by the triplet $$(\bDDD, \bx, \epsilon)$$. Suppose that a sparse vector $$\ba \in \CC^D$$ satisfies the sparsity constraint
$\| \ba \|_0 < \frac{1}{2} \left (1 + \frac{1}{\mu} \right)$
and gives a representation of $$\bx$$ to within error tolerance $$\epsilon$$ (i.e. $$\| \bx - \bDDD \ba\|_2 \leq \epsilon$$). Every solution $$\widehat{\ba}$$ of (20.1) must obey
$\|\widehat{\ba} - \ba \|_2^2 \leq \frac{4 \epsilon^2}{ 1 - \mu ( 2 \| \ba \|_0 - 1)}.$
Proof. .
1. Note that $$\ba$$ need not be sparsest possible representation of $$\bx$$ within the approximation error $$\epsilon$$.
2. But $$\ba$$ is a feasible point of (20.1).
3. Now since $$\widehat{\ba}$$ is an optimal solution of (20.1) (thus sparsest possible), hence it is at least as sparse as $$\ba$$; i.e.,
$\| \widehat{\ba} \|_0 \leq \| \ba \|_0.$
4. Due to Theorem 18.25,
$\frac{1}{2} \spark(\bDDD) \geq \frac{1}{2} \left (1 + \frac{1}{\mu} \right) > \| \ba \|_0.$
5. Thus, there exists a value $$\eta \geq 0$$ such that
$\frac{1}{2} \spark(\bDDD) \geq \frac{1}{2} {\spark}_{\eta} (\bDDD) \geq \| \ba \|_0 \geq \| \widehat{\ba} \|_0.$
6. From Theorem 20.2 we recall that
${\spark}_{\eta} (\bDDD) \geq \frac{1 - \eta^2}{ \mu} + 1.$
7. Thus, we can find a suitable value of $$\eta \geq 0$$ such that we can enforce a stricter requirement:
$\| \ba \|_0 \leq \frac{1}{2} \left ( \frac{1 - \eta^2}{ \mu} + 1 \right ) \leq \frac{1}{2}{\spark}_{\eta} (\bDDD).$
8. From this we can develop an upper bound on $$\eta$$ being
$\begin{split} \| \ba \|_0 \leq \frac{1}{2} \left ( \frac{1 - \eta^2}{ \mu} + 1 \right ) &\iff 2 \| \ba \|_0 \mu \leq 1 - \eta^2 + \mu \\ &\iff \eta^2 \leq 1 - \mu (2 \| \ba \|_0 - 1). \end{split}$
9. If we choose $$\eta^2 = 1 - \mu (2 \| \ba \|_0 - 1)$$, then
$\begin{split} \| \ba \|_0 = \frac{1}{2} \left ( \frac{1 - \eta^2}{ \mu} + 1 \right ) & \implies \| \ba \|_0 \leq \frac{1}{2} {\spark}_{\eta} (\bDDD)\\ & \implies \| \widehat{\ba} \|_0 \leq \| \ba \|_0 \leq \frac{1}{2} {\spark}_{\eta} (\bDDD) \end{split}$
continues to hold.
10. We have two solutions $$\ba$$ and $$\widehat{\ba}$$ both of which satisfy
$\| \ba \|_0, \| \widehat{\ba} \|_0 \leq \frac{1}{2} {\spark}_{\eta} (\bDDD)$
and
$\| \bx - \bDDD \ba \|_2, \| \bx - \bDDD \widehat{\ba} \|_2 \leq \epsilon.$
11. If we choose a $$\delta = \frac{2 \epsilon}{\eta}$$, then applying Theorem 20.4, we will get
$\| \ba - \widehat{\ba} \|_2^2 \leq \delta^2 = \frac{4 \epsilon^2}{\eta^2} = \frac{4 \epsilon^2}{1 - \mu (2 \| \ba \|_0 - 1)}.$
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2022-12-08 15:48:14
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https://www.physicsforums.com/threads/decimal-representation-of-real-numbers.156006/
|
# Decimal representation of real numbers
1. Feb 12, 2007
### marcin w
I'm doing self study out of Apostol's Calculus vol. I and I got stuck trying to prove what the author writes is easy to verify, but I can't get my head around it. Basically, this is the problem statement from page 31, last paragraph:
Given a positive real number x, let a0 denote the largest integer <= x. Having chosen a0, we let a1 denote the largest integer such that
a0 + (a1)/10 <= x.
More generally, having chosen a0, a1,...,an-1, we let an denote the largest integer such that
a0 + (a1)/10 + ... + (an)/10^n <= x.
Then let S denote the set of all numbers obtained this way for n = 0, 1, 2, ...
I understand the construction of set S because I have proven before that the greatest integer in a positive real number exists, hence the set is nonempty. I also see that
a0 + (a1)/10 + ... + (an + 1)/10^n is an upper bound, so this property with the nonemptyness of S guarantees that S has a supremum, say b. The punchline is that I want to verify that x = b.
I'm not trying to prove here that .999... = 1. I know that and can prove it using geometric series and algebraic arguments. I want to rigorously prove the above using just the bare essentials (ie. axioms) of the real number system.
I was thinking of arguing by contradiction that x > b or x < b is impossible. My other thought for a rough sketch was that to prove that the infinum of the set for decimal expansion of x - b is 0, but I don't know where to start really. Thanks in advance.
2. Feb 13, 2007
### morphism
Let s_n = a0 + (a1)/10 + ... + (an)/10^n. Then {s_n} is a bounded, nondecreasing sequence of reals. It must converge to supS = b, i.e. lim s_n = b. We also know that:
s_n <= x <= s_n + 1/10^n
Applying the sequeeze theorem, we see that:
lim s_n <= x <= lim s_n
=> x = lim s_n = b.
3. Feb 13, 2007
### marcin w
Thanks for the reply, but I was looking for something even more elementary - no limits and no squeeze theorem. Is this proof possible to do with even more basic concepts? I'm curious because Apostol is very thorough and I'm confident he wouldn't write this if it couldn't be done with the material so far covered (which is before differentiation and integration are even introduced).
4. Feb 13, 2007
### StatusX
It's clear that all the numbers in the sequence are <=x, so if you can show their limit is >=x, you're done. To do this, assume its <x, and derive a contradiction (ie, show you haven't picked the a_n correctly).
5. Feb 14, 2007
### marcin w
I think I have worked out a correct proof. I would appreciate it if anyone could review this and criticize if necessary.
I have already established that S is nonempty and bounded above so I will not dwell on this fact anymore. As in my first post, by the law of trichotomy, the least upper bound b is one of the following:
b < x, x = b or b > x
and I will show by contradiction that the first and last are impossible.
Suppose that b < x. Choose a c such that 0 < c < x - b, or written equivalently as (1) b < b + c < x. By the Archimedean property of real numbers, there exists a positive integer n such that (*) 1/(10^n) < c (Proof on the bottom of this post). Since b is an upper bound, it follows that:
a0 + (a1)/10 + ... + (an)/(10^n) <= b. Combining this with (*), we have
a0 + (a1)/10 + ... + (an)/(10^n) + 1/(10^n) < b + c
or
a0 + (a1)/10 + ... + (an + 1)/(10^n) < b + c.
Here we have a contradiction because an was supposed to be the greatest integer such that the decimal expansion is <= x, but we have found that an + 1 is an even greater positive number such that the expansion is <= x due to (1).
If b > x, then choose a c such that 0 < c < b -x or x < c + x < b. It is clear that c + x is another upper bound for S that is less than the least upper bound b, which is a contradiction
(*) To prove that that there exits a positive n such that 1/(10^n) < y, I use the fact that for every real number x, there exists a positive integer such that x < n and by induction I will show that n < 10^n for every n.
Let A(n) = n < 10^n.
A is clearly true for 1.
Now assuming that A(k) = k < 10^k is true, I multiply both sides by 10 to obtain
10k < 10^k * 10 or (1) 10k < 10^(k+1)
Now, (2) k + 1 < 10k is true because 10k - (k + 1) = 9k + 1 > 0.
By the transitive property of inequalities, (1) and (2) combine and we obtain
k + 1 < 10^(k+1), as asserted.
Therefore we have that x < n < 10^n or x < 10^n. Replacing x by 1/y (where y > 0), we have
1/y < 10^n or 1/(10^n) < y, just as stated.
QED?
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2016-10-21 22:21:48
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http://www.mathemafrica.org/?p=11192
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Now we’re going to take a more complicated expression and approximate it by a polynomial function. The function we’re going to look at is $f(x)=2 \sin x+\frac{\cos 3x}{2}$, but we could choose any function which is well behaved close to where we want to approximate it (there is a much more precise way to phrase this, but for the current discussion, this is enough).
This function looks like:
OK, so how are we going to go about approximating this function? Well, let’s ask about approximating it close to the point $x=2.5$ (this value is arbitrary and we could have asked for any value). What would be the most naive approximation we could make? Well, if we have a function which is a constant, and equal to the original function at $x=2.5$ then that’s a start. At least it matches the value of the function at that point, if nothing else. What is the value of this function at $x=2.5$? Well, it’s about 1.37. So $y=f(2.5)=1.37$ is perhaps the most naive approximation we could make for our function close to $x=2.5$. Let’s see how the two compare:
We will call this the zeroth approximation. Indeed it matches our original function perfectly at $x=2.5$ but apart from that it’s pretty rubbish. Maybe we can do better by approximating our function by a straight line that at least has the same gradient as our function? Well, clearly we want to it to pass through the point $(x,y)=(2.5,f(2.5))$ but now we also want it to have a slope. We are going to try and approximate it by the function:
$y=f(2.5)+c_1(x-2.5)$
for some $c_1$. If we want this to have the same gradient as our original function then we must have that:
$f'(2.5)=\left.\frac{d \left(f(2.5)+c_1(x-2.5)\right)}{dx}\right|_{x=2.5}=c_1$
So in fact we must have:
$y=f(2.5)+f'(2.5)(x-2.5)$
How does this function compare with the original function? Let’s look at them together:
Ah, now maybe we’re getting somewhere! The purple line now has the same value and derivative as our original function at exactly $x=2.5$. It’s looking a bit more like the function close to that point, at least very approximately.
How about if we want to get a bit more accurate. Let’s say that we want to include another power of $x$? We can try and approximate the function by a quadratic, but asking that the value, the first derivative and the second derivative all match with our original function at the value $x=2.5$.
Well we will set up our approximating function as:
$y=c_0+c_1(x-2.5)+c_2(x-2.5)^2$
Matching the value, the first and second derivatives of this function with our original function we will find:
$c_0=f(2.5)$
$c_1=f'(2.5)$
as before. Now let’s take two derivatives of our original function and match it to the second derivative of our approximating function at x=2.5:
$f''(2.5)=2c_2$
so now $c_2=\frac{f''(2.5)}{2}$. If we plug this into our quadratic expression and see how this compares with $f(x)$ we see:
Now the polynomial is exactly the same value, derivative and second derivative as our original function. We could continue this with higher and higher polynomials. If we go up to 14th order ie. $(x-2.5)^{14}$ we get the following approximation, which matches the value, first, second, ….up to fourteenth derivative of our function. As you can see, close to $x=2.5$ it’s a really good approximation now!
Great, so we can see that we can take a pretty arbitrary looking function, and approximate it to good accuracy by a polynomial expression. We still haven’t written this in a general form, but in the next post we will do just that.
How clear is this post?
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2019-11-13 00:51:12
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https://www.neetprep.com/question/56067-Two-ideal-slits-S-S-distance-d-apart-illuminated-light-ofwavelength--passing-ideal-source-slit-S-placed-line-S-shown-distance-planes-slits-source-slit-D-Ascreen-held-distance-D-plane-slits-minimum-value-d-darkness-O-D-D-D-D/55-Physics--Wave-Optics/700-Wave-Optics
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Two ideal slits S1 and S2 are at a distance d apart, and illuminated by light of wavelength λ passing through an ideal source slit S placed on the line through S2 as shown. The distance between the planes of slits and the source slit is D. A screen is held at a distance D from the plane of the slits. The minimum value of d for which there is darkness at O is
(1) $\sqrt{\frac{3\lambda D}{2}}$
(2) $\sqrt{\lambda D}$
(3) $\sqrt{\frac{\lambda D}{2}}$
(4) $\sqrt{3\lambda D}$
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2020-02-27 01:23:33
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http://mizar.uwb.edu.pl/JFM/Vol5/boolmark.html
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Journal of Formalized Mathematics
Volume 5, 1993
University of Bialystok
Copyright (c) 1993 Association of Mizar Users
## Basic Concepts for Petri Nets with Boolean Markings
Pauline N. Kawamoto
Shinshu University, Nagano
Yasushi Fuwa
Shinshu University, Nagano
Yatsuka Nakamura
Shinshu University, Nagano
### Summary.
Contains basic concepts for Petri nets with Boolean markings and the firability$\slash$firing of single transitions as well as sequences of transitions [6]. The concept of a Boolean marking is introduced as a mapping of a Boolean TRUE$\slash$FALSE to each of the places in a place$\slash$transition net. This simplifies the conventional definitions of the firability and firing of a transition. One note of caution in this article - the definition of firing a transition does not require that the transition be firable. Therefore, it is advisable to check that transitions ARE firable before firing them.
#### MML Identifier: BOOLMARK
The terminology and notation used in this paper have been introduced in the following articles [10] [13] [1] [14] [3] [4] [9] [11] [8] [2] [12] [5] [15] [7]
#### Contents (PDF format)
1. Preliminaries
2. Boolean Marking and Firability$\slash$Firing of Transitions
#### Acknowledgments
The authors would like to thank Dr. Andrzej Trybulec for his patience and guidance in the writing of this article.
#### Bibliography
[1] Grzegorz Bancerek. The fundamental properties of natural numbers. Journal of Formalized Mathematics, 1, 1989.
[2] Grzegorz Bancerek and Krzysztof Hryniewiecki. Segments of natural numbers and finite sequences. Journal of Formalized Mathematics, 1, 1989.
[3] Czeslaw Bylinski. Functions and their basic properties. Journal of Formalized Mathematics, 1, 1989.
[4] Czeslaw Bylinski. Functions from a set to a set. Journal of Formalized Mathematics, 1, 1989.
[5] Czeslaw Bylinski. The modification of a function by a function and the iteration of the composition of a function. Journal of Formalized Mathematics, 2, 1990.
[6] Pauline N. Kawamoto, Masayoshi Eguchi, Yasushi Fuwa, and Yatsuka Nakamura. The detection of deadlocks in Petri nets with ordered evaluation sequences. In \em Institute of Electronics, Information, and Communication Engineers (IEICE) Technical Report, pages 45--52. Institute of Electronics, Information, and Communication Engineers (IEICE), January 1993.
[7] Pauline N. Kawamoto, Yasushi Fuwa, and Yatsuka Nakamura. Basic Petri net concepts. Journal of Formalized Mathematics, 4, 1992.
[8] Andrzej Trybulec. Binary operations applied to functions. Journal of Formalized Mathematics, 1, 1989.
[9] Andrzej Trybulec. Domains and their Cartesian products. Journal of Formalized Mathematics, 1, 1989.
[10] Andrzej Trybulec. Tarski Grothendieck set theory. Journal of Formalized Mathematics, Axiomatics, 1989.
[11] Andrzej Trybulec. Function domains and Fr\aenkel operator. Journal of Formalized Mathematics, 2, 1990.
[12] Wojciech A. Trybulec. Pigeon hole principle. Journal of Formalized Mathematics, 2, 1990.
[13] Zinaida Trybulec. Properties of subsets. Journal of Formalized Mathematics, 1, 1989.
[14] Edmund Woronowicz. Relations and their basic properties. Journal of Formalized Mathematics, 1, 1989.
[15] Edmund Woronowicz. Many-argument relations. Journal of Formalized Mathematics, 2, 1990.
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2017-10-20 08:47:30
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|
https://cs.stackexchange.com/questions/88792/how-do-i-follow-the-text-of-this-longest-common-sequence-proof
|
How do I follow the text of this Longest Common Sequence proof?
I'm working through a proof that is a simplified version of David Maeirs LCS reduction from the VERTEX-COVER problem and I had a question about a specific set of instructions and how to execute them by hand.
Background Information This works out to look like the below as a graph (please excuse the grid lines)
The part I don't understand of the proof
Now I think this would imply that I need to line up S0-S7 vertically and using that rule show I can find the LCS, which was shown in my proof to be
$$00000001000000000000000000000100000000000000100000001$$ $$0^710^{21}10^{14}10^71$$
I tried doing that but was unable after many attempts to show that I could in fact induce $$Topt$$ from the sequences below. Could someone explain where I'm going wrong here? Did I misinterpret the instructions?
$$S0 =00000001000000010000000100000001000000010000000100000001$$
$$S1 =00000000000000100000000000000100000001000000010000000100000001$$
$$S2 =00000001000000000000001000000000000001000000010000000100000001$$
$$S3 =00000001000000000000001000000010000000000000010000000100000001$$
$$S4 =00000001000000010000000000000010000000100000000000000100000001$$
$$S5 =00000001000000010000000000000010000000100000001000000000000001$$
$$S6 =00000001000000010000000100000000000000100000000000000100000001$$
$$S7 =00000001000000010000000100000001000000000000001000000010000000$$
EDIT: I tried everything I could think of to see if I could make any of them work but didn't think it was relevant enough to include, if you want to know what I did I can list them.
• This process does not show an algorithm to find a lcs. – xskxzr Mar 2 '18 at 12:16
The two red-underlined parts of the proof is to standardize an LCS solution, that is called $$T_{opt}$$ therein.
The idea is to push all the $$0$$'s farthest possible to the right (without passing through its nearest $$1$$ to the right).
Consider the following two strings:
$$s_1=0001000100010001$$ (i.e. $$n=4$$ vertices)
$$s_2=0001000\ 00010001000$$ (where the space is for visualizing purpose only) (i.e. $$u=2$$, $$v=4$$)
A common subsequence (not necessarily longest) can be: $$s=0001000 00010001$$ (i.e. $$u$$ is omitted)
The standardization process in the proof is to actually align $$s$$ against $$s_1$$, $$s_2$$.
And, this MUST be standardized to be as follows:
s1= 0001 0001 0001 0001
s2= 0001 000 0001 0001000
s = 0001 000 0001 0001
Similarly, if we omit $$v$$, then $$s=000100010001000$$
And, this should be standardized as follows:
s1= 0001 0001 0001 0001
s2= 0001 000 0001 0001 000
s = 0001 0001 0001 000
Finally, we will be able to construct an $$n^2+k$$-long common subsequence if there exists an independent set of size $$n-k$$, where for each edge $$\{u,v\}$$, we omit at least one of them (the ones that are in the independent set) so that it is possible to align like we do above (note that each edge-string $$e_j$$ contains only $$n-1$$ symbol $$1$$'s). And lastly, by complementing a $$k$$-VC, we have an $$(n-k)$$-IS.
|
2021-04-20 11:23:09
|
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|
https://math.stackexchange.com/questions/2161428/how-can-i-show-that-this-complicated-expression-with-square-and-cube-roots-reduc
|
# How can I show that this complicated expression with square and cube roots reduces to the value 7?
How would I go about reducing the complicated-looking expression $$\sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}}$$ to show that it is equal to 7?
I came across the complicated expression while calculating the single real solution for the cubic equation $$z^3 - 33 z - 112 = 0$$
Using that cubic equation, I can show that $z = 7$ satisfies it and that a standard way of solving cubic equations shows that there is only a single real root and that its value is equal to the complicated expression, which is a very roundabout way of proving that the complicated expression is equal to 7. But what if I didn't know about the cubic equation? Is there a more direct way of reducing the complicated expression to a simpler one?
• Take the power 3 of this expression, using binomial formula $(a+b)^3$... – Jean Marie Feb 25 '17 at 21:56
• Note that if $p+q+r=0$ then $p^3+q^3+r^3=3pqr$ - you can use this to get rid of the cube roots by taking the two components as $p, q$ and the sum as $p+q=-r$. – Mark Bennet Feb 25 '17 at 22:16
I'll outline the basic method to simplify$$\sqrt[3]{56+19\sqrt{5}}+\dfrac {11}{\sqrt[3]{56+19\sqrt{5}}}\tag{1}$$ To simplify the nested radical, note that we have this general outline:$$\sqrt[n]{A+B\sqrt[m]{C}}=a+b\sqrt[m]{C}\tag{2}$$ So therefore, we have\begin{align*}\sqrt[3]{56+19\sqrt{5}} & =a+b\sqrt{5}\\56+19\sqrt{5} & =\underbrace{(a^3+15ab^2)}_{56}+\underbrace{(3a^2b+5b^3)}_{19}\sqrt{5}\end{align*} So we get this system of equations:\begin{align*} & a^3+15ab^2=56\\ & 3a^2b+5b^3=19\end{align*}\tag{3} $(3)$ has real solutions as $(a,b)=\left(\dfrac 72,\dfrac 12\right)$ so$$\sqrt[3]{56+19\sqrt{5}}=\dfrac {7+\sqrt5}2\tag4$$ Using $(4)$, and through some algebraic manipulations, we have\begin{align*}\color{blue}{\sqrt[3]{56+19\sqrt5}}+\dfrac {11}{\color{red}{\sqrt[3]{56+19\sqrt{5}}}} & =\color{blue}{\dfrac {7+\sqrt{5}}2}+\dfrac {11}{\color{red}{\frac {7+\sqrt{5}}2}}\\ & =\dfrac {7+\sqrt{5}}{2}+\dfrac {22}{7+\sqrt{5}}\\ & =\dfrac {7+\sqrt{5}}2+\dfrac {22(7-\sqrt5)}{44}\\ & =\dfrac {7+\sqrt5}2+\dfrac {7-\sqrt5}2\\ & =\boxed 7\end{align*} Just like what you got!
One possibility is to factor $19\sqrt{5}+56$ in the ring of integers of the field $\mathbb{Q}(\sqrt{5})$, which gives $$19\sqrt{5}+56 = \left(\frac{1+\sqrt{5}}2 \right)^3(4-\sqrt5)^3$$ and now it is easy to continue.
If you wonder how I arrived to this factorization, it is known that the ring of integers of $\mathbb Q(\sqrt{5})$ is a unique factorization domain, the norm of $19\sqrt{5}+56$ in this field is $11^3$ so this implies that it is divisible by either $4-\sqrt{5}$ or $4+\sqrt{5}$ which are the primes of $\mathbb Q(\sqrt{5})$ under $11$. After performing the division, the result is a unit, and so it is of the form $\pm (\tfrac{1+\sqrt{5}}{2})^k$, as $\tfrac{1+\sqrt{5}}{2}$ is the fundamental unit of $\mathbb Q(\sqrt{5})$.
Using $a=\sqrt[3]{19\sqrt5+56}$ and $b=\frac{11}{\sqrt[3]{19\sqrt5+56}}$ and $u=a+b$, we have \begin{align} a^3+b^3 &=19\sqrt5+56+\frac{1331}{19\sqrt5+56}\\ &=\frac{6272+2128\sqrt5}{56+19\sqrt5}\\[6pt] &=112 \end{align} Thus, \begin{align} &112\\ &=a^3+b^3\\ &=(a+b)\left(a^2-ab+b^2\right)\\ &=(a+b)\left((a+b)^2-3ab\right)\\ &=u(u^2-33) \end{align} Noting that for $x\lt\sqrt{11}$, \begin{align} x^3-33x-112 &\le22\sqrt{11}-112\\ &\lt0 \end{align} and for $x\ge\sqrt{11}$, $x^3-33x-112$ is increasing, we see that $x^3-33x-112=0$ has only one real solution, and that is $x=7$.
Therefore, $\sqrt[3]{19\sqrt5+56}+\frac{11}{\sqrt[3]{19\sqrt5+56}}=a+b=u=7$.
But what if I didn't know about the cubic equation?
As long as you have a reasonable suspicion that the expression might be rational, the polynomial it satisfies can be derived rather easily. Let:
$$x = \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}}$$
Then using $(a+b)^3=a^3+b^3+3ab(a+b)\,$ and $(56+19\sqrt{5})(56-19\sqrt{5})=1331=11^3\,$:
\require{cancel} \begin{align} x^3 & = 19\sqrt{5} + 56 + \frac{11^3}{19\sqrt{5} + 56} + 3 \cdot \cancel{\sqrt[3]{19\sqrt{5} + 56}} \cdot \frac{11}{\cancel{\sqrt[3]{19\sqrt{5} + 56}}} \cdot x \\ & = 56+ \cancel{19\sqrt{5}} + 56 - \cancel{19\sqrt{5}} + 33x \\ & = 112 + 33x \end{align}
By the rational root theorem, the equation $x^3-33x-112=0$ can only have rational roots that are integer divisors of $112=2^4\cdot 7\,$. Obviously $\pm 1$ do not satisfy the equation, and even roots can be eliminated by comparing the powers of $2$ between the terms, which leaves $\pm 7\,$ to try.
• Wish the downvoter had left a comment why. – dxiv Feb 26 '17 at 21:21
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2019-12-08 13:42:02
|
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|
https://blog.msbstats.info/posts/2022-10-06-sdt-ordinal-regression/
|
# Ordinal Regression as a Model for Signal Detection
R
statistics
code
ordinal regression
signal detection theory
Author
Mattan S. Ben-Shachar
Published
October 6, 2022
# Preface
I was basically done with this blog post when I came across Matti Vuorre’s post on the same exact topic. Matti goes into all the details, and really the present post can be seen as a brief account of all the cool things the probit-approach-to-SDT can do. I’m only posting this here because I really like my plots 🤷
Previously, we’ve seen that for data from a binary decision signal detection task, we can use a probit binomial regression model (like a logistic regression, but with a probit link function) to estimate the two main parameters of signal detection theory (SDT): the sensitivity and the bias.
In this post I would like to show how this idea can be extended to multiple response SDT tasks by using an ordinal probit regression model.
# The Data
Imagine the following task: after being presented with 20 images of dogs, you are presented with 300 new images of dogs, and you have to decide for each dog if it appeared in the training set (“Old”) or not (“New”).
In a binary decision task, you would simply indicate “New” or “Old”, but in this task you have multiple response options - from 1 to 6, with 1 = “Feels New” and 6 = “Feels Old”. We can call this scale a “feelings numbers” scale.
After going over all 30 photos, you have
STD_data
# A tibble: 12 × 3
Truth Response N
<fct> <ord> <dbl>
1 New Confidence1 35
2 New Confidence2 31
3 New Confidence3 26
4 New Confidence4 22
5 New Confidence5 19
6 New Confidence6 17
7 Old Confidence1 14
8 Old Confidence2 20
9 Old Confidence3 22
10 Old Confidence4 27
11 Old Confidence5 32
12 Old Confidence6 35
Where N is the number of responses in each condition and response level.
# Modeling with Classic SDT
We can use Siegfried Macho’s R code to extract the SDT parameters. In this case, they are:
1. Sensitivity - The distance between the two (latent) normal distributions. The further they are, the more “distinguishable” the Old and New images are from each other.
2. 5 Threshold - One between each pair of consecutive possible responses. Perceived “stimulation” above each threshold leads to a decision in that category.
(These will probably make sense when we present them visually below.)
First, we’ll model this with classical SDT:
SDT_equal <- SDT.Estimate(
data = STD_data[["N"]],
test = TRUE,
# We have 2 option: Old / New; We'll assume equal variance
n = list(n.sdt = 2, restriction = "equalvar")
)
SDT.Statistics(SDT_equal)[["Free.parameters"]]
Value SE CFI-95(Lower) CFI-95(Upper)
Mean[2] 0.564 0.040 0.486 0.642
t-1 -0.744 0.034 -0.810 -0.678
t-2 -0.165 0.031 -0.226 -0.104
t-3 0.267 0.031 0.206 0.329
t-4 0.707 0.033 0.643 0.772
t-5 1.260 0.036 1.189 1.331
# Modeling as a Probit Cumulative Ordinal
library(dplyr) # 1.1.0
library(tidyr) # 1.3.0
library(ordinal) # 2022.11.16
library(parameters) # 0.20.2
library(ggplot2) # 3.4.0
library(patchwork) # 1.1.2
We can also model this data with a Probit Cumulative Ordinal model, predicting the Response from the Truth: - The slope of Truth indicates the effect of the true image identity had on the response pattern - this is sensitivity.
- In an ordinal model, we get k-1 “intercepts” (k being the number of unique responses). Each intercept represents the value above which a predicted value will be binned into the next class. There represent our shreshold.
m_equal <- clm(Response ~ Truth,
data = STD_data,
weights = N,
)
parameters::model_parameters(m_equal) |>
insight::print_html()
Parameter Coefficient SE 95% CI z p Model Summary Confidence1|Confidence2 -0.74 0.10 (-0.94, -0.54) -7.21 < .001 Confidence2|Confidence3 -0.16 0.10 (-0.35, 0.02) -1.72 0.085 Confidence3|Confidence4 0.27 0.10 (0.08, 0.46) 2.81 0.005 Confidence4|Confidence5 0.71 0.10 (0.51, 0.90) 7.08 < .001 Confidence5|Confidence6 1.26 0.11 (1.05, 1.48) 11.38 < .001 Truth (Old) 0.57 0.12 (0.33, 0.81) 4.65 < .001
As we can see, the estimated values are identical!1
The advantage of the probit ordinal model is that it is easy(er) to build this model up:
• Add predictors of sensitivity (interactions with Truth)
• Add predictors of bias (main effects / intercept effects)
• Add random effects (with ordinal::clmm())
mean2 <- coef(m_equal)[6]
Thresholds <- coef(m_equal)[1:5]
ggplot() +
# Noise
stat_function(aes(linetype = "Noise"), fun = dnorm,
size = 1) +
# Noise + Signal
stat_function(aes(linetype = "Noise + Signal"), fun = dnorm,
args = list(mean = mean2),
size = 1) +
# Thresholds
geom_vline(aes(xintercept = Thresholds, color = names(Thresholds)),
size = 2) +
scale_color_brewer("Threshold", type = "div", palette = 2,
labels = paste0(1:5, " | ", 2:6)) +
labs(y = NULL, linetype = NULL, x = "Obs. signal") +
expand_limits(x = c(-3, 3), y = 0.45) +
theme_classic()
Warning: Using size aesthetic for lines was deprecated in ggplot2 3.4.0.
ℹ Please use linewidth instead.
# Unequal Variance
The standard model of SDT assumes that the Noise and the Noise + Signal distribution differ only in their mean; that is, N+S is a shifted N distribution. This is almost always not true, with $$\sigma_{\text{N+S}}>\sigma_{\text{N}}$$.
To deal with this, we can also estimate the variance of the N+S distribution.
First, with the classic SDT model:
SDT_unequal <- SDT.Estimate(
data = STD_data[["N"]],
test = TRUE,
# We have 2 option: Old / New; Not assuming equal variance
n = list(n.sdt = 2, restriction = "no")
)
SDT.Statistics(SDT_unequal)[["Free.parameters"]]
Value SE CFI-95(Lower) CFI-95(Upper)
Mean[2] 0.552 0.041 0.473 0.632
Stddev[2] 0.960 0.035 0.891 1.029
t-1 -0.728 0.036 -0.799 -0.658
t-2 -0.159 0.032 -0.221 -0.097
t-3 0.266 0.031 0.205 0.327
t-4 0.696 0.034 0.629 0.763
t-5 1.235 0.043 1.151 1.318
And with a probit ordinal regression, but allow the latent scale to vary:
m_unequal <- clm(Response ~ Truth,
scale = ~ Truth, # We indicate that the scale is a function of the underlying dist
data = STD_data,
weights = N,
)
parameters::model_parameters(m_unequal) |>
insight::print_html()
Parameter Coefficient SE 95% CI z p Model Summary Confidence1|Confidence2 -0.72 0.11 (-0.93, -0.50) -6.51 < .001 Confidence2|Confidence3 -0.16 0.10 (-0.34, 0.03) -1.61 0.107 Confidence3|Confidence4 0.27 0.10 (0.08, 0.45) 2.80 0.005 Confidence4|Confidence5 0.69 0.10 (0.49, 0.90) 6.66 < .001 Confidence5|Confidence6 1.23 0.13 (0.98, 1.49) 9.50 < .001 Truth (Old) 0.55 0.12 (0.31, 0.80) 4.49 < .001 Truth (Old) -0.05 0.12 (0.31, 0.80) 4.49 < .001
The scale parameter needs to be back transformed to get the sd of the N+S distribution: $$e^{-0.05}=0.95$$, and so one again the estimated values are identical!
mean2 <- coef(m_unequal)[6]
sd2 <- exp(coef(m_unequal)[7])
Thresholds <- coef(m_unequal)[1:5]
ggplot() +
# Noise
stat_function(aes(linetype = "Noise"), fun = dnorm,
size = 1) +
# Noise + Signal
stat_function(aes(linetype = "Noise + Signal"), fun = dnorm,
args = list(mean = mean2, sd = sd2),
size = 1) +
# Thresholds
geom_vline(aes(xintercept = Thresholds, color = names(Thresholds)),
size = 2) +
scale_color_brewer("Threshold", type = "div", palette = 2,
labels = paste0(1:5, " | ", 2:6)) +
labs(y = NULL, linetype = NULL, x = "Obs. signal") +
expand_limits(x = c(-3, 3), y = 0.45) +
theme_classic()
## ROC Curve or ROC Curves?
An additional check we can preform is whether the various responses are indeed the product of single ROC curve. We do this by plotting the ROC curve on a inv-normal transformation (that is, converting probabilities into normal quantiles). Quantiles that fall on a straight line indicate they are part of the same curve.
pred_table <- data.frame(Truth = c("Old", "New")) |>
mutate(predict(m_unequal, newdata = cur_data(), type = "prob")[[1]] |> as.data.frame()) |>
tidyr::pivot_longer(starts_with("Confidence"), names_to = "Response") |>
tidyr::pivot_wider(names_from = Truth)
Warning: There was 1 warning in mutate().
ℹ In argument: as.data.frame(predict(m_unequal, newdata = cur_data(), type =
"prob")[[1]]).
Caused by warning:
! cur_data() was deprecated in dplyr 1.1.0.
ℹ Please use pick() instead.
ROC_table <- pred_table |>
rows_append(data.frame(New = 0, Old = 0)) |>
mutate(
Sensitivity = lag(cumsum(New), default = 0),
Specificity = rev(cumsum(rev(Old))),
)
p_roc <- ggplot(ROC_table, aes(Sensitivity, Specificity)) +
geom_line() +
geom_abline(slope = 1, intercept = 1, linetype = "dashed") +
geom_point(aes(color = ordered(Response)), size = 2) +
expand_limits(x = c(0,1), y = c(0,1)) +
scale_x_continuous(trans = "reverse") +
scale_color_brewer("Threshold", type = "div", palette = 2,
labels = paste0(1:5, " | ", 2:6),
na.translate = FALSE) +
labs(color = NULL) +
theme_classic()
p_zroc <- ROC_table |>
tidyr::drop_na(Response) |>
ggplot(aes(qnorm(Sensitivity), qnorm(Specificity))) +
geom_line() +
geom_point(aes(color = ordered(Response)), size = 2) +
expand_limits(x = c(0,1), y = c(0,1)) +
scale_x_continuous(trans = "reverse") +
scale_color_brewer("Threshold", type = "div", palette = 2,
labels = paste0(1:5, " | ", 2:6),
na.translate = FALSE) +
labs(color = NULL, x = "Z(Sensitivity)", y = "Z(Specificity)") +
theme_classic()
p_roc + p_zroc + plot_layout(guides = "collect")
Warning: Removed 1 rows containing missing values (geom_point()).
## Going Full Bayesian
Although the clmm() function allows for multilevel probit regression, it does not2 support varying scale parameter.
Alas, we must use brms.
library(brms) # 2.18.0
library(ggdist) # 3.2.1.9000
library(tidybayes) # 3.0.3
In brms we will use the cumulative() family, which has a family-parameter called disc which gives the standard deviation of the latent distributions. I will set some weak priors on the mean and standard deviation of the N+S distribution, and I will also set the standard deviation of the N distribution to 1 (on a log scale, to 0) using the constant(0) prior.
b_formula <- bf(Response | weights(N) ~ Truth,
disc ~ 0 + Intercept + Truth)
b_priors <-
set_prior("normal(0, 3)", coef = "TruthOld") +
set_prior("normal(0, 1.5)", coef = "TruthOld", dpar = "disc") +
set_prior("constant(0)", coef = "Intercept", dpar = "disc")
Bayes_mod <- brm(b_formula,
prior = b_priors,
data = STD_data,
backend = "cmdstanr",
refresh = 0
)
model_parameters(Bayes_mod, test = NULL) |>
insight::print_html()
Parameter Median 95% CI Rhat ESS Model Summary Intercept(1) -0.74 (-0.95, -0.53) 1.002 2755.00 Intercept(2) -0.17 (-0.36, 0.02) 1.000 4651.00 Intercept(3) 0.27 (0.08, 0.46) 1.001 4694.00 Intercept(4) 0.70 (0.50, 0.91) 1.001 3453.00 Intercept(5) 1.26 (1.01, 1.52) 1.002 2442.00 TruthOld 0.56 (0.32, 0.82) 1.000 3252.00 disc_Intercept 0.00 (0.00, 0.00) disc_TruthOld 0.02 (-0.21, 0.23) 1.003 2098.00
criteria <- gather_rvars(Bayes_mod, b_Intercept[Response])
signal_dist <- spread_draws(Bayes_mod, b_TruthOld, b_disc_TruthOld) |>
mutate(b_disc_TruthOld = exp(b_disc_TruthOld)) |>
group_by(.draw) |>
reframe(
x = seq(-3, 3, length = 20),
d = dnorm(x, mean = b_TruthOld, b_disc_TruthOld)
) |>
ungroup() |>
curve_interval(.along = x, .width = 0.9)
ggplot() +
# Noise
stat_function(aes(linetype = "Noise"), fun = dnorm) +
# Noise + Signal
geom_ribbon(aes(x = x, ymin = .lower, ymax = .upper),
data = signal_dist,
fill = "grey", alpha = 0.4) +
geom_line(aes(x, d, linetype = "Noise + Signal"), data = signal_dist) +
# Thresholds
stat_slab(aes(xdist = .value, fill = ordered(Response)),
color = "gray", alpha = 0.6, key_glyph = "polygon",
data = criteria) +
# Theme and scales
scale_fill_brewer("Threshold", type = "div", palette = 2,
labels = paste0(1:5, " | ", 2:6),
na.translate = FALSE) +
labs(color = NULL, linetype = NULL, x = "Obs. signal", y = NULL) +
theme_classic()
This gives roughly the same results as clm(), but would allow for multilevel modeling of both the location and scale of the latent variable.
## Footnotes
1. Though the CIs are somewhat wider.↩︎
2. as of 2022-10-06↩︎
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2023-03-30 08:08:04
|
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https://www.gpb.org/education/learn/lets-learn-ga/mathematics/fractions-2nd-grade
|
# Let's Learn GA! - Fractions - 3rd Grade
Georgia teacher Allison Townsend helps students learn all about fractions in this episode of Let's Learn GA!.
## Let's Learn GA! - Fractions - 3rd Grade
Let's Learn GA! - Fractions - 3rd Grade
Georgia teacher Allison Townsend helps students learn all about fractions in this episode of Let's Learn GA!.
### Mathematics
MGSE3.NF.1
Understand a fraction $\frac{1}{b}$ as the quantity formed by 1 part when a whole is partitioned into **b** equal parts (unit fraction); understand a fraction $\frac{a}{b}$ as the quantity formed by *a* parts of size $\frac{1}{b}$. For example, $\frac{3}{4}$ means there are three $\frac{1}{4}$ parts, so $\frac{3}{4}$ = $\frac{1}{4}$ + $\frac{1}{4}$ + $\frac{1}{4}$.
|
2021-04-13 20:00:26
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|
https://www.mersenneforum.org/showthread.php?p=578717
|
mersenneforum.org Parameter explorations for CADO 165-170 digits
Register FAQ Search Today's Posts Mark Forums Read
2021-05-17, 14:42 #23
charybdis
Apr 2020
22·3·41 Posts
Quote:
Originally Posted by VBCurtis My previous GGNFS/msieve jobs around C165 have had matrices around 9M in size, so this job was rather strongly oversieved. We should cut rels_wanted to 210M for this file and see what matrix comes out.
Msieve always gives larger but sparser matrices though. Still possible this was oversieved, but probably not massively so.
My hunch is that raising tasks.sieve.mfb1 from 62 to 64 would be more effective than using 3LP here. That said, I've got a better idea of what to do with the other parameters than the last time I tried 3LP on a number of roughly this size (see post #2 in this thread). May give it a go in the near future.
2021-05-18, 07:58 #24
bur
Aug 2020
79*6581e-4;3*2539e-3
2·199 Posts
Thanks for the explanations, slowly I start to get more about what is going on. Interesting, that fundmental things like poly selection or q-range are so little understood that it makes an experimental attempt useful. From the little I could get from math papers, especially poly selection could probably be vastly optimized if we knew more about it.
Quote:
If you are interested in seeing the effect of the extra relations, zcat all your 220M relations out to a single file, and run msieve's filtering on the file to see what size matrix comes out. Then, restrict the number of relations msieve uses (via a filtering flag, see the -h option list for msieve) to 215, 210, 205 and let us know what size matrices pop out. I suggest a TD of 100 or so for msieve, but you might enjoy trying 100/110/120 on the full dataset to see how target_density affects matrix size.
So to just see the effect of fewer relations I'll set TD = 100?
And just to make 100% sure: more relations = faster matrix generation & higher matrix density = smaller matrix size = shorter LA times (not the newspaper, harhar)?
2021-05-18, 14:43 #25 VBCurtis "Curtis" Feb 2005 Riverside, CA 7·23·31 Posts No, there's a flag like filter_maxrels that lets you control the number of relations msieve uses. TD 100 vs TD 110 shows you the effect of using higher target densities for a given set of relations- the difference is not very big. A higher density matrix is slightly tougher to solve than a lower-density matrix of the same size, but matrix size affects time-to-solve much more. More relations -> smaller matrix.-> shorter LA time. Higher TD -> makes sure you have enough relations for a reasonably small matrix. When filtering with msieve, let's say you have 190M relations. TD = 90 might build a matrix 11M in size, while TD 110 fails and asks for more relations. If you use the full 220M relations, TD 90 might build a 7M matrix, while TD 110 builds a 6.5M matrix. The idea is that choosing TD = 110 makes sure you avoid the situation where a matrix barely builds and is quite large (e.g. 11M for this current factorization). If you actually try these tests, let the matrix-solving step run for 1% completion, and note the ETA (ETA is pretty stable by the time 1% is done). The TD100 and TD110 ETAs won't be very different, I expect.
2021-05-18, 15:00 #26 bur Aug 2020 79*6581e-4;3*2539e-3 18E16 Posts What I meant was, if I only want to test various numbers of relations, which value for TD should I chose, 100? Or is there a default value? Testing different target_densities was that just a proposal for me to see the effects, or are the results helpful to you similar to the various numbers of relations?
2021-05-18, 16:27 #27 VBCurtis "Curtis" Feb 2005 Riverside, CA 10011011111112 Posts Ah, I see- yes, I'd use TD 100 to test various relations counts, because higher TDs may not produce matrices. If TD 100 doesn't produce a matrix, that relations count is too low to be useful to us anyway (I think). The TD 100 vs 110 is useful to us, but I agree it's more for your personal experience. I haven't done as many such tests as I should, so I am curious just how different the ETAs would be; my own limited testing showed the ETAs are similar enough that I rarely use TD 120 even on jobs I expect would easily build a matrix at that density, since the time spent when filtering fails feels greater than the time saved by using 120 instead of 110.
2021-05-19, 07:02 #28 bur Aug 2020 79*6581e-4;3*2539e-3 2×199 Posts I'm still running the tests, results so far show that 185M relations was too few. 190M still worked, LA took from around 14 h (190M) to 8:45 h (220M). Since sieving took 6 1/2 days, doing only 190M would have been quite a gain. Would it be good to decrease rels_wanted to 190M generally or might this have been an exception? But even 2-3 additional filtering steps would still make the total time shorter. I will post detailed results later this day.
2021-05-19, 17:12 #29 bur Aug 2020 79*6581e-4;3*2539e-3 2×199 Posts I used this command line ./msieve -i c165.n -s rels.dat -nf c165.fb -t 10 -v -nc1 "filter_maxrels=200000000" afterwards the same with -nc2. 220M rels Code: matrix is 6441919 x 6442144 (2334.4 MB) with weight 618273835 (95.97/col) sparse part has weight 547526160 (84.99/col) [...] linear algebra completed 64760 of 6442144 dimensions (1.0%, ETA 8h43m 215M rels Code: matrix is 6671510 x 6671738 (2418.3 MB) with weight 640096824 (95.94/col) sparse part has weight 567230375 (85.02/col) [...] linear algebra completed 66759 of 6671738 dimensions (1.0%, ETA 9h28m 210M rels Code: matrix is 6881379 x 6881605 (2496.6 MB) with weight 660645774 (96.00/col) sparse part has weight 585663806 (85.11/col) [...] linear algebra completed 72597 of 6881605 dimensions (1.1%, ETA 10h 3m) 205M rels Code: matrix is 7102309 x 7102534 (2579.8 MB) with weight 682640424 (96.11/col) sparse part has weight 605254018 (85.22/col) [...] near algebra completed 71317 of 7102534 dimensions (1.0%, ETA 10h53m) 200M rels Code: matrix is 7405037 x 7405262 (2691.5 MB) with weight 711821792 (96.12/col) sparse part has weight 631510830 (85.28/col) [...] linear algebra completed 74878 of 7405262 dimensions (1.0%, ETA 11h54m) 195M rels Code: matrix is 7779754 x 7779979 (2833.7 MB) with weight 749219145 (96.30/col) sparse part has weight 665027723 (85.48/col) [...] linear algebra completed 77887 of 7779979 dimensions (1.0%, ETA 13h25m) 190M rels Code: matrix is 8305557 x 8305782 (3029.3 MB) with weight 800256398 (96.35/col) sparse part has weight 711056605 (85.61/col) [...] linear algebra completed 87238 of 8305782 dimensions (1.1%, ETA 15h43m) 185M rels Code: found 338230 cycles, need 3940439 too few cycles, matrix probably cannot build filtering wants 1000000 more relations Sieving the 220M relations took 158 hours or about 3.5 hours per 5M relations. I didn't specify TD, which numbers of relations should I chose for the density comparisons?
2021-05-19, 19:03 #30
henryzz
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
134308 Posts
Quote:
Originally Posted by bur I didn't specify TD, which numbers of relations should I chose for the density comparisons?
It might be interesting to find the best TD for each number of relations. In theory, the difference should be even more pronounced than with a constant TD.
2021-05-19, 19:18 #31 VBCurtis "Curtis" Feb 2005 Riverside, CA 7×23×31 Posts Your comparison of 3.5 hrs per 5M rels shows that any oversieving past 190M is inefficient. This surprises me- I would have expected 195 or 200 to be the "sweet spot", where the sieve time spent would be more than or equal to LA time saved. But, even 195M vs 190M only saves 2hr 20 min LA time at the cost of ~3.5hr sieve time. If these were all default density, I wonder whether the 96 or the 85 density is the target number- I don't have msieve logs handy to look it up myself, so I'll edit this message later today to correct myself about which of the densities from your logs is the one msieve controls directly. If default is 96 at this size, I doubt TD is going to do much; try TD 100 or 104 or 110 on the 195M relations set to see if you can save another hour of LA time. Your data also helps someone like Ed, who uses a large farm of ~20 machines to sieve, but just one to LA. Your data at 190.....220 can help him choose how much to sieve. I think I'll make the params file 190M target; as you say, filtering more than once isn't a big deal if that turns out to not be enough for a C161 or C162. I'll add a note that recommends 200M if sieving on multiple machines. What CPU architecture is used for these tests? Older CPUs take relatively longer to do the LA section, so might prove to be 3.5hr faster at 195M because the entire LA portion is taking 50% longer. Sieving is less sensitive to CPU architecture than LA. My personal experience is with sandy bridge, ivy bridge, haswell generations of Xeon, all rather old by now.
2021-05-19, 22:55 #32
charybdis
Apr 2020
1EC16 Posts
Quote:
Originally Posted by VBCurtis If these were all default density, I wonder whether the 96 or the 85 density is the target number- I don't have msieve logs handy to look it up myself, so I'll edit this message later today to correct myself about which of the densities from your logs is the one msieve controls directly. If default is 96 at this size, I doubt TD is going to do much; try TD 100 or 104 or 110 on the 195M relations set to see if you can save another hour of LA time.
Actually, it's neither. There will be a line from earlier in the filtering that matches TD, looking like this:
Code:
weight of 9242904 cycles is about 924731720 (100.05/cycle)
The default in the latest msieve is density 90, and bur's figures match what I see in my logs for 90.
Re general c165 params: lpb 31/31 is definitely worth testing too, unless you already have data showing 31/32 is faster. I also notice you didn't include adjust_strategy 2, which would probably give a boost of a couple percent. Strategy 2 makes A=28 usable but I don't think it'll be optimal yet at this size. Maybe by c170.
2021-05-20, 07:45 #33
bur
Aug 2020
79*6581e-4;3*2539e-3
2×199 Posts
Quote:
Originally Posted by henryzz It might be interesting to find the best TD for each number of relations. In theory, the difference should be even more pronounced than with a constant TD.
So I'll use TD = 90, 100, 110, 120? Which number of rels would be interesting, all I tested so far?
It would make 28 tests, but since filtering is single-threaded I could run 10 simultaneously. Would this work with all processes accessing the same rels.dat?
VBCurtis, it was run on a i10-10900K with 16 GB RAM. Nothing else demanding was running at the same time.
BTW, I am just factoring a C147 with your experimental params.c145, should I keep the relations to do similar tests?
Last fiddled with by bur on 2021-05-20 at 07:47
Similar Threads Thread Thread Starter Forum Replies Last Post EdH CADO-NFS 127 2020-10-07 01:47 storm5510 Information & Answers 4 2019-11-30 21:32 ksteczk PrimeNet 6 2018-03-26 15:11 R.D. Silverman Cunningham Tables 14 2010-09-29 19:56 R.D. Silverman Cunningham Tables 11 2006-03-06 18:46
All times are UTC. The time now is 02:51.
Sat Oct 16 02:51:41 UTC 2021 up 84 days, 21:20, 0 users, load averages: 0.93, 0.98, 0.92
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2021-10-16 02:51:41
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http://www.exampleproblems.com/wiki/index.php/Calculus
|
# Calculus
I recommend this book: A Course of Modern Analysis by Whittaker and Watson. You may also find this book at Google Books. This book is a hundred years old and is considered the classic calculus book.
## Derivatives
### Definition of Derivative
$f'(x)=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$, provided the limit exists.
For the following problems, find the derivative using the definition of the derivative.
solution $f(x)=x\,$
solution $f(x)=2x^3\,$
solution $f(x)=\sqrt{x}\,$
solution $f(x)=\frac{1}{x}\,$
solution $f(x)=\sin x\,$
solution $h(x)=f(x)+g(x)\,$
### Power Rule
$\frac{d}{dx}(x^n)=nx^{n-1}\,$
For the following problems, compute the derivative of $y=f(x)\,$ with respect to $x\,$
solution $f(x)=150\pi\,$
solution $f(x)=7x^4\,$
solution $f(x)=27x^2+\frac{12}{\pi}x+e^{e^{\pi}}\,$
solution $f(x)=8x^5+4x^4+6x^3+x+7\,$
solution $f(x)=\frac{1}{\sqrt [4] {x}}\,$
solution $f(x)=7x^{-4}+12x^{-2}-x^{\frac{1}{2}}+6x+\pi x^{\frac{3}{5}}+x^{-\frac{3}{4}}\,$
solution Derive the power rule for positive integer powers from the definition of the derivative (Hint: Use the Binomial Expansion)
### Product Rule
$\frac{d}{dx}(ab)=ab'+a'b\,$
For the following problems, compute the derivative of $y=f(x)\,$ with respect to $x\,$
solution $f(x)=x\sin x\,$
solution $f(x)=x^3\cos x\,$
solution $f(x)=\tan x\sec x\,$
solution $f(x)=2x\sin (2x)\,$
solution $f(x)=(x)(x+7)(x-12)\,$
solution Derive the Product Rule using the definition of the derivative
### Quotient Rule
$\left(\frac{a}{b}\right)'=\frac{ba'-ab'}{b^2}\,$
For the following problems, compute the derivative of $y=f(x)\,$ with respect to x
solution $f(x)=\frac{x}{x+1}\,$
solution $f(x)=\frac{x^2}{\sin x}\,$
solution $f(x)=\frac{\sin^2x}{x^3}\,$
solution $f(x)=\frac{x\sin x}{e^x}\,$
solution $f(x)=\frac{x+7}{(x-6)(x+2)}\,$
solution Derive the Quotient Rule formula. (Hint: Use the Product Rule).
### Generalized Power Rule
$\frac{d}{dx}(f(x))^n=n(f(x))^{n-1}f'(x)\,$
For the following problems, compute the derivative of $y=f(x)\,$ with respect to x
solution $f(x)=(x^3+7x)^3\,$
solution $f(x)=\sin^4(x)\,$
solution $f(x)=\left(\ln x\right)^{-4}\,$
solution $f(x)=\tan^2x+8(x^2+4x+3)^9+\sec^3x\,$
solution $f(x)=\left(\frac{5x}{7x+9}\right)^3\,$
### Chain Rule
$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)\,$
For the following problems, compute the derivative of $y=f(x)\,$ with respect to x
solution $f(x)=\ln (7x^2e^x\sin x)\,$
solution $f(x)=\sin^2(7x+5) + \cos^2(7x+5)\,$
solution $f(x)=6e^{3x}\tan(5x)\,$
solution $f(x)=\ln (\sin (e^x))\,$
solution $f(x)=\sin (\cos (\tan x))\,$
solution $f(x)=e^{x^2}\sin (14x)-\cos (e^x)\,$
solution $f(x)=\frac{\frac{\sin (5x)}{(x^2+1)^2}}{\cos^3(3x)-1}\,$
solution $\left ( f(g(x)) \right )' = f'(g(x))g'(x)$
solution $\sqrt[]{x+\sqrt[]{x+\sqrt[]{x}}}$
solution $f(x)=x^2\sqrt{9-x^2}\,$
### Implicit Differentiation
For the following problems, compute the derivative of $y=f(x)\,$ with respect to x
solution $\sin (xy)=x\,$
solution $x+xy+x^2+xy^2=0\,$
solution $y=x(y+1)\,$
solution $yx=x^y\,$ where $y\,$ is a function of $x\,$
### Logarithmic Differentiation
For the following problems, compute the derivative of $y=f(x)\,$ with respect to x
solution $f(x)=4^{\sin x}\,$
solution $f(x)=x^x\,$
solution $f(x)=x^{x^{{\cdot}^{{\cdot}^{\cdot}}}} \,$
solution $f(x)=g(x)^{h(x)}\,$ for any functions $g(x)\,$ and $h(x)\,$ where $g(x) \ne 0\,$
### Second Fundamental Theorem of Calculus
$\frac{d}{dx}\int_{a}^{x}f(t)\,dt=f(x)\,$
For the following problems, compute the derivative of $y=f(x)\,$ with respect to x
solution $f(x)=\int_{0}^{x}e^t\,dt\,$
solution $f(x)=\int_{5}^{x}27t^t\sin (t-1)\,dt\,$
solution $f(x)=\int_{4}^{x^2}\sin (e^t)\,dt\,$
solution $f(x)=\int_{x^2}^{3x^4}\cos (t)\,dt\,$
solution Give a proof of the theorem
## Applications of Derivatives
### Slope of the Tangent Line
solution Find the slope of the tangent line to the graph f(x) = 6x when x = 3.
solution Find the slope of the tangent line to the graph f(x) = sin2x when x = π.
solution Find the slope of the tangent line to the graph f(x) = cosx + x5 when $x=\frac{\pi}{2}$.
solution Find the equation of the tangent line to the graph f(x) = xex + x + 5 when x = 0.
solution Find the slope of the tangent line to the graph f(x) = x2 when x = 0,1,2,3,4,5,6.
solution Find the slope of the tangent line to the graph x2 + y2 = 9 at the point (0, − 3).
solution Find the equation of the tangent line to the graph xy = y2x2 + y + x + 2 at the point (0, − 2).
### Extrema (Maxima and Minima)
solution Find the absolute minimum and maximum on [ − 1,5] of the function f(x) = (1 − x)ex.
solution Find the absolute minimum and maximum on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ of the function f(x) = sin(x2).
solution Find all local minima and maxima of the function f(x) = x2.
solution Find all local minima and maxima of the function $f(x)=\frac{x^2-1}{x}$.
solution If a farmer wants to put up a fence along a river, so that only 3 sides need to be fenced in, what is the largest area he can fence with 100 feet of fence?
solution If a fence is to be made with three pens, the three connected side-by-side, find the dimensions which give the largest total area if 200 feet of fence are to be used.
solution Find the local minima and maxima of the function $f(x)=\sqrt {x}$.
### Related Rates
solution A clock face has a 12 inch diameter, a 5.5-inch second hand, a 5 inch minute hand and a 3 inch hour hand. When it is exactly 3:30, calculate the rate at which the distance between the tip of any one of these hands and the 9 o'clock position is changing.
solution A spherical container of r meters is being filled with a liquid at a rate of $\rho\,{\rm m}^3/{\rm min}$. At what rate is the height of the liquid in the container changing with respect to time?
## Integrals
### Integration by Substitution
solution $\int \frac{2x}{x^2+1}\,dx\,$
solution $\int x^2\sin x^3\,dx\,$
solution $\int \cot x\,dx\,$
solution $\int \tan x\sec^2x\,dx\,$
solution $\int \frac{\ln x^2}{x}\,dx\,$
solution $\int_{0}^{3} \frac{2x+1}{x^2+x+7}\,dx\,$
solution $\int_{1}^{e^{\pi}} \frac{\sin (\ln x)}{x}\,dx\,$
solution $\int \frac{x}{\sqrt{4+x^2}}\,dx\,$
solution $\int \frac{x}{1-x^2}\,dx\,$
### Integration by Parts
$\int u\,dv=uv-\int v\,du\,$
solution $\int \ln x\,dx\,$
solution $\int x\sin(x)\,dx \,$
solution $\int \arctan(2x)\,dx \,$
solution $\int e^x\sin x\,dx\,$
solution $\int x^2e^x\,dx\,$
solution $\int (x^3+1)\cos x\,dx\,$
solution $\int x^4\sin x\,dx\,$
solution $\int \frac{\ln x}{x}\,dx\,$
### Trigonometric Integrals
solution $\int \sin^2(x)\,dx\,$
solution $\int \tan^2(x)\,dx \,$
solution $\int \sin^5x\cos^5x\,dx\,$
solution $\int \sin^2x\cos^3x\,dx\,$
solution $\int \sin^2x\cos^2x\,dx\,$
solution $\int \tan^2x\sec^4x\,dx\,$
solution $\int \tan^3x\sec^3x\,dx\,$
### Trigonometric Substitution
solution $\int\frac{x\,dx}{\sqrt{3-2x-x^2}}$
solution $\int \arcsec x\,dx \,$
solution $\int \frac{1}{\sqrt{4x-x^2}}\,dx\,$
solution $\int x\arcsin x\,dx\,$
solution $\int \frac{x}{1-x^2}\,dx\,$
solution $\int \frac{1}{(x^2+1)^{\frac{3}{2}}}\,dx\,$
solution $\int \frac{\sqrt{x^2-1}}{x}\,dx\,$
solution $\int \frac{x}{\sqrt{4+x^2}}\,dx\,$
### Partial Fractions
solution $\int\frac{1}{x^2-1}\,dx\,$
solution $\int\frac{1}{(x+1)(x+2)(x+3)}\,dx\,$
solution $\int\frac{x!}{(x+n)!}\,dx\,$
solution $\int \frac{x}{1-x^2}\,dx\,$
solution $\int\frac{dx}{\sin^2(x)-\cos^2(x)}$
### Special Functions
solution A ball is thrown up into the air from the ground. How high will it go?
solution Let $f\,$ be a continuous function for $x \ge a\,$.
Show that $\int_a^x \int_a^s f(y)\,dy\,ds = \int_a^x f(y)(x-y)\,dy\,$
solution Evaluate $\int e^{x^2}\,dx\,$
solution $\int_0^\infty 3^{-4z^2}dz\,$
solution $\int_0^\infty x^m e^{-ax^n} dx\,$
solution $\int_0^\infty x^4 e^{-x^3} \,dx\,$
solution $\int_0^\infty e^{-pt} \sqrt{t}\,dt\,$
## Applications of Integration
### Area Under the Curve
solution Find the area under the curve f(x) = x2 on the interval [ − 1,1].
solution Find the total area between the curve f(x) = cosx and the x-axis on the interval [0,2π].
solution Find the area under the curve f(x) = x3 + 4x2 − 7x + 8 on the interval [0,1].
solution Using calculus, find the formula for the area of a rectangle.
solution Derive the formula for the area of a circle with arbitrary radius r.
### Volume
#### Disc Method
The disc method is a special case of the method of cross-sectional areas to find volumes, using a circle as the cross-section.
To find the volume of a solid of revolution, using the disc method, use one of the two formulas below. R(x) is the radius of the cross-sectional circle at any point.
If you have a horizontal axis of revolution
$V=\pi\int_{a}^{b}[R(x)]^2\,dx$
If you have a vertical axis of revolution
$V=\pi\int_{c}^{d}[R(y)]^2\,dy$
solution Find the volume of the solid generated by revolving the line y = x around the x-axis, where $0\le x\le 4$.
solution Find the volume of the solid generated by revolving the region bounded by y = x2 and y = 4xx2 around the x-axis.
solution Find the volume of the solid generated by revolving the region bounded by y = x2 and y = x3 around the x-axis.
solution Find the volume of the solid generated by revolving the region bounded by y = x2 and y = x3 around the y-axis.
solution Find the volume of the solid generated by revolving the region bounded by $y=\sqrt{x}$, the x-axis and the line x = 4 around the x-axis.
#### Cross-sectional Areas
solution Find the volume, on the interval $0\le x\le 3$, of a 3-D object whose cross-section at any given point is a square with side length x2 − 9x.
solution Find the volume, on the interval 0 < x < 2π, of a 3-D object whose cross-section at any given point is an equilateral triangle with side length $\sin \frac{x}{2}$.
solution Find the volume of a cylinder with radius 3 and height 10.
### Arc Length
$L=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2} \,dx = \int_p^q\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} \,dt$
solution Calculate the arc length of the curve y = x2 from x = 0 to x = 4.
solution Determine the arc length of the curve given by x = t cos t , y = t sin t from t=0
solution Calculate the arc length of y=cosh x from x=0 to x
### Mean Value Theorem
$f ' (c) = \frac{f(b) - f(a)}{b - a}.$
solution Find the average value of the function f(x) = e2x on the interval [0,4].
solution Find the average value of the function 2sec2x on the interval $[0,\frac{\pi}{4}]$.
solution Find the average speed of a car, starting at time 0, if it drives for 5 hours and its speed at time t (in hours) is given by s(t) = 5t2 + 7t + et.
solution Find the average value of the function sin6tcos3t on the interval $[0,\frac{\pi}{2}]$.
solution Deduce the Mean Value Theorem from Rolle's Theorem.
## Series of Real Numbers
### nth Term Test
If the series $\sum_{n=1}^{\infty}a_n$ converges, then $\lim_{n\rightarrow \infty}a_n=0$.
Note: This leads to a test for divergence for those series whose terms do not go to 0 but it does not tell us if any series converges.
solution Discuss the convergence or divergence of the series with terms { − 1,1, − 1,1, − 1,1,...}.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum_{n=1}^{\infty}\frac{1}{n^2}$.
solution Discuss the convergence or divergence of the series $\sum_{n=0}^{\infty}\sin n$.
### Telescopic Series
If {ak} is a convergent real sequence, then $\sum_{k=1}^{\infty}(a_k-a_{k+1})=a_1-\lim_{k\rightarrow \infty}a_k$.
solution Discuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\left(\frac{1}{k+7}-\frac{1}{k+8}\right)$.
solution Discuss the convergence or divergence of the series $\sum_{k=0}^{\infty}\frac{2}{(k+1)(k+3)}$.
solution Discuss the convergence or divergence of the series $\sum_{k=2}^{\infty}\ln \left(\frac{k(k+2)}{(k+1)^2}\right)$.
solution Discuss the convergence or divergence of the series $\sum_{k=4}^{\infty}\left(\frac{1}{k}-\frac{1}{k+2}\right)$.
solution Discuss the convergence or divergence of the series $\sum_{k=0}^{\infty}\left(e^{-k}-e^{-(k+1)}\right)$.
solution Discuss the convergence or divergence of the series $\sum_{k=4}^{\infty}\left(e^{-k+3}-e^{-k+1}\right)$.
### Geometric Series
The series $\sum_{n=0}^{\infty}r^n$ converges if − 1 < r < 1 and, moreover, it converges to $\frac{1}{1-r}$. For any other value of r, the series diverges. More generally, the finite series, $\sum_{n=a}^{b}r^n=\frac{r^a-r^{b+1}}{1-r}$ for any value of r.
solution Discuss the convergence or divergence of the series $\sum_{n=0}^{\infty}\left (\frac{1}{4}\right )^n$.
solution Discuss the convergence or divergence of the series $\sum_{n=3}^{\infty}\left (\frac{2}{3}\right )^n$.
solution Discuss the convergence or divergence of the series $\sum_{n=0}^{\infty}1.5^n$.
solution Discuss the convergence or divergence of the series $\sum_{n=0}^{\infty}\left (\frac{1}{6}\right )^{n+2}$.
solution Discuss the convergence or divergence of the series $\sum_{n=10}^{\infty}\left (\frac{3}{5}\right )^n$.
solution Discuss the convergence or divergence of the series $\sum_{n=7}^{\infty}\left (-\frac{1}{2}\right )^n$.
solution Discuss the convergence or divergence of the series $\sum_{n=7}^{\infty}\left[\left (-\frac{4}{7}\right )^{n+3}+\left (\frac{1}{3}\right )^{n-2}\right]\,$.
solution Discuss the convergence or divergence of the series $\sum_{n=0}^{\infty}\frac{3^{n+1}}{7^n}$.
solution Discuss the convergence or divergence of the series $\sum_{n=0}^{\infty}\left (\frac{1}{3}\right )^{2n}$.
### Integral Test
If the function f is positive, continuous, and decreasing for $x\ge 1$, then
$\sum_{n=1}^{\infty}f(n)\,$ and $\int_{1}^{\infty}f(x)\,dx\,$
converge together or diverge together.
Notice that if f were negative, continuous, and increasing this is also true since such a function would simply be the negative of some function which is positive, continuous, and decreasing and multiplying by -1 will not change the convergence of a series.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{1}{n+1}$.
solution Discuss the convergence of $\sum_{n=0}^{\infty}e^{-3n}$.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^5}$.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{n}{e^n}$.
solution Explain why the integral test is or is not applicable to $\sum_{n=1}^{\infty}n^2$.
solution Explain why the integral test is or is not applicable to $\sum_{n=1}^{\infty}-\frac{1}{n^2}$.
solution Explain why the integral test is or is not applicable to $\sum_{n=1}^{\infty}\frac{|\sin n|+1}{\ln (n+1)}$.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^p}$.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^{\frac{7}{3}}}$.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{1}{\sqrt [3] {n}}$.
### Comparison of Series
Direct Comparison If $0 for all n If $\sum_{n=1}^{\infty}b_n$ converges, then $\sum_{n=1}^{\infty}a_n$ also converges. 2. If $\sum_{n=1}^{\infty}a_n$ diverges, then $\sum_{n=1}^{\infty}b_n$ also diverges.
Limit Comparison Suppose an > 0, bn > 0 and $\lim_{n\rightarrow \infty}\frac{a_n}{b_n}=L$ where L is finite and positive. Then $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ either both converge or both diverge.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{1}{n^3+4}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{\ln n}{n-3}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{2n^2}{4n^4+8n^3+3}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n^2+1}}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n+4}{(n+2)(n+1)}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\sin \frac{1}{n}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n+1}{n\times 3^{n+1}}$.
### Dirichlet's Test
Let $a_k, b_k\in \mathbb{R}$ for $k\in \mathbb{N}$.
If the sequence of partial sums $s_n=\sum_{k=1}^{n}a_k$ is bounded and $b_k\downarrow 0$ as $k\rightarrow \infty$, then $\sum_{k=1}^{\infty}a_kb_k$ converges.
solution Discuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\frac{\sin \frac{k\pi}{2}}{k}$.
solution Discuss the convergence or divergence of the series $\sum_{k=2}^{\infty}\frac{\sin \frac{k\pi}{3}}{\ln k}$.
solution Discuss the convergence or divergence of the series $\sum_{k=7}^{\infty}\left(-\frac{1}{2}\right)^k$.
solution Discuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\frac{1}{k2^k}$.
solution Discuss the convergence or divergence of the series $\sum_{k=3}^{\infty}\frac{1}{\ln (\ln k)}\cos \left(\frac{k\pi}{3}\right)$.
solution Discuss the convergence or divergence of the series $\sum_{k=1}^{\infty}a_k\frac{1}{k^3}$ where ak = {7,4,6,3, − 10, − 10,7,4,6,3, − 10, − 10,...}.
solution Discuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\sin \frac{2k\pi}{n}\frac{1}{\ln (\ln (\ln k))}$ for any integer $n\ge 1$.
solution Discuss the convergence or divergence of the series $\sum_{k=3}^{\infty}\frac{\ln (\ln k)}{\ln k}a_k$ where {ak} = {1, − 1,1,2, − 3,1,2,4, − 7,1, − 1,1,2, − 3,1,2,4, − 7,...}.
### Alternating Series
If an, then the alternating series
$\sum_{n=1}^{\infty}(-1)^na_n\,$ and $\sum_{n=1}^{\infty}(-1)^{n+1}a_n\,$
converge if the absolute value of the terms decreases and goes to 0.
solution Discuss the convergence of the series $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\,$.
solution Discuss the convergence of the series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}n^2}{n^2+1}\,$.
solution Discuss the convergence of the series $\sum_{n=1}^{\infty}\cos (n\pi)\,$.
solution Discuss the convergence of the series $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\,$.
solution Discuss the convergence of the series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}n^2}{e^n}\,$.
solution Discuss the convergence of $\sum_{n=2}^{\infty}\frac{(-1)^n}{\ln n}$ including whether the sum converges absolutely or conditionally.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2+3n+4}$ including whether the sum converges absolutely or conditionally.
solution Discuss the convergence of $\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}$ including whether the sum converges absolutely or conditionally.
### Ratio Test
For the infinite series $\sum_{n=1}^{\infty}a_n$,
1. If $0\le \lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|<1\,$, then the series converges absolutely.
2. If $\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|>1\,$, then the series diverges.
3. If $\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|=1\,$, then the test is inconclusive.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{2^n}{(2n)!}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n}{3^{n+1}}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{1}{n^3}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{1}{n}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{3^n}{n^2+2}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n}{(n+2)!}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n4^n}{n!}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n!}{4^n}$.
solution Discuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n^ka^n}{n!}$.
### Root Test
Let $a_k\in\mathbb{R}$ and $r=\limsup_{k\rightarrow \infty}|a_k|^\frac{1}{k}$
1. If $r<1\,$, then $\sum_{k=1}^{\infty}a_k$ converges absolutely.
2. If $r>1\,$, then $\sum_{k=1}^{\infty}a_k$ diverges.
3. If $r=1\,$, this test is inconclusive.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\left(\frac{n}{2n+1}\right)^n$.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}e^{-n}$.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n}{3^n}$.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\left(\frac{2n}{100}\right)^n$.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{(n!)^n}{(n^n)^2}$.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\left(\frac{7n}{12n-6}\right)^{2n}$.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\left(\frac{15n-6}{4n+2}\right)^{7n}$.
solutionDiscuss the convergence or divergence of the series $\sum_{n=1}^{\infty}\frac{n!}{(n+1)^n}\left(\frac{23}{50}\right)^n$.
### Cauchy Condensation Test
The series $\sum_{k=1}^{\infty}a_k$ and $\sum_{k=1}^{\infty}2^ka_{2^k}$ converge or diverge together.
solutionDiscuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\frac{1}{k}$.
solutionDiscuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\frac{1}{\ln k}$.
solutionDiscuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\frac{1}{k^2}$.
solutionDiscuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\frac{1}{\sqrt{2^k}}$.
solutionDiscuss the convergence or divergence of the series $\sum_{k=1}^{\infty}\frac{1}{(2^k)^n}$, where n is some real number.
## Series of Real Functions
solution Find the infinite series expansion of $\frac{1}{(1+x)^a}\,$
solution Investigate the convergence of this series: $\sum_{k=1}^\infty \frac{1}{k(k+1)}\,$
solution Investigate the convergence of this series: $3-2+\frac{4}{3}-\frac{8}{9}+...+3\left(-\frac{2}{3}\right)^k+... \,$
solution Find the upper limit of the sequence $\left\{x_n\right\}_{n=1}^\infty, x_n=(-1)^nn\,$
solution Find the upper limit of the sequence $\left\{x_n\right\}_{n=1}^\infty, x_n=(-1)^n\left(\frac{2n}{n+1}\right)\,$
solution Evaluate $\sum_{k=0}^n {n \choose k}^2\,$
solution Evaluate $\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}{3^m\left(m\,3^n+n\,3^m\right)}$.
solution Find the upper limit of the sequence $\left\{x_n\right\}_{n=1}^\infty, x_n=\frac{1}{n^2}\,$
solution Find the upper limit of the sequence $\left\{x_n\right\}_{n=1}^\infty, x_n=n\sin\left(\frac{n\pi}{2}\right)\,$
solution Evaluate $\sum_{n=0}^\infty \left(\frac{i}{3}\right)^n\,$
solution Determine the interval of convergence for the power series $\sum_{n=0}^\infty\frac{n(3x-4)^n}{\sqrt[3]{n^4}(2x)^{n-1}}.$
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Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions are aligned based on the latest syllabus and common core curriculum guidelines. Also, you can get a step-by-step explanation by the subject experts for all questions covered in the Big Ideas Math Book Algebra 2 Ch 4 Polynomials Functions Solution Key. Polynomials are a very important concept for high school students to calculate various complex calculations in further chapters. So, it is very crucial to download and practice the questions covered in BIM Textbook solutions of Algebra 2 Ch 4 Polynomial Functions.
Big Ideas Math Book Algebra 2 Answer Key Chapter 4 Polynomial Functions
Improving your Math skills is the best way to score high marks and also become a pro in math concepts. Access the available links below and download free pdf formatted Big Ideas Math Algebra 2 Solutions of Ch 4 Polynomial functions. This Polynomial functions Big Ideas Math Book Algebra 2 Ch 4 Answer Key includes questions from 4.1 to 4.9 lessons exercises, assignment tests, practice tests, chapter tests, quizzes, etc. Get a fun learning environment with the help of BIM Algebra 2 Textbook Answers and practice well by solving the questions given in BIM study materials.
Polynomial Functions Maintaining Mathematical Proficiency
Simplify the expression.
Question 1.
6x − 4x=2x
Explanation:
Terms can be combined only if they have the exact same variable portion and combining like terms.
Question 2.
12m − m − 7m + 3
12m-m-7m+3
=4m+3
Question 3.
3(y + 2) − 4y
3(y+2)-4y
3y+6-4y
-y+6
Question 4.
9x − 4(2x − 1)
9x-4(2x-1)
=9x-8x+4
=x+4
Question 5.
−(z + 2) − 2(1 − z)
= -z-2-2+z
=-4
Question 6.
−x2 + 5x + x2
5x
Find the volume of the solid.
Question 7.
cube with side length 4 inches
solution:
Volume of cube=sidexsidexside
=4x4x4
=64inches
Question 9.
rectangular prism with length 4 feet, width 2 feet, and height 6 feet
Solution: Given that
length= 4 feet
width=2 feet
height= 6 feet
volume of rectangular prism= length x width x height
substitute the values in formula
volume of rectangular prism= 4 x 2 x 6= 48 cubic feet
Question 10.
right cylinder with radius 3 centimeters and height 5 centimeters
Solution:
Given that
height =5centimeter
Volume of a cylinder = area of base × height = π × r2× h and you can use 3.14 for π.
substitute the values in formula
3.14 x 3 x 3 x 5=141.3cubic centimeter
Question 11.
ABSTRACT REASONING Does doubling the volume of a cube have the same effect on the side length? Explain your reasoning
Solution:
If the side of the cube id doubled. the volume is 8 times larger .
example: let’s use a cube that has side lenghts of 2 as an example. the volume of that cube would be lenght x and width x height
2 x 2 x 2 =8. if we double the edges to 4 . it would be 4 x 4 x 4, which is 64. to find how many times increases take the 64 and divide by 8 . 64/8=8 The volume increases by 8 times.
Polynomial Functions Mathematical Practices
Monitoring Progress
Use a graphing calculator to determine whether the function is continuous. Explain your reasoning.
Question 1.
f(x) = $$\frac{x^{2}-x}{x}$$
Question 2.
f(x) = x3 − 3
Question 3.
f(x) = $$\sqrt{x^{2}+1}$$
Question 4.
f(x) = | x + 2 |
Question 5.
f(x) = $$\frac{1}{x}$$
Question 6.
f(x) = $$\frac{1}{\sqrt{x^{2}-1}}$$
Question 7.
f(x) = x
Question 8.
f(x) = 2x − 3
Question 9.
f(x) = $$\frac{x}{x}$$
Lesson 4.1 Graphing Polynomial Functions
Essential Question
What are some common characteristics of the graphs of cubic and quartic polynomial functions?
A polynomial function of the form
f(x) = anxn + an – 1xn– 1 +. . .+ a1x + a0
where an ≠ 0, is cubic when n = 3 and quartic when n = 4.
EXPLORATION 1
Identifying Graphs of Polynomial Functions
Work with a partner.
Match each polynomial function with its graph. Explain your reasoning. Use a graphing calculator to verify your answers.
a. f(x) = x3 − x
b. f(x) = −x3 + x
c. f(x) = −x4 + 1
d. f(x) = x4
e. f(x) = x3
f. f(x) = x4 − x2
EXPLORATION 2
Identifying x-Intercepts of Polynomial Graphs
Work with a partner.
Each of the polynomial graphs in Exploration 1 has x-intercept(s) of −1, 0, or 1. Identify the x-intercept(s) of each graph. Explain how you can verify your answers.
Question 3.
What are some common characteristics of the graphs of cubic and quartic polynomial functions?
Question 4.
a. When the graph of a cubic polynomial function rises to the left, it falls to the right.
b. When the graph of a quartic polynomial function falls to the left, it rises to the right.
4.1 Lesson
Monitoring Progress
Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.
Question 1.
f(x) = 7 − 1.6x2 − 5x
The standard form of a polynomials has the exponents of the terms arranged in descending order.
The degree of a polynomial is the highest exponent of a term.
The type of a polynomial depends on the degree.
The leading coefficient is the numerical factor of the terms with highest degree.
Hence the given function f(x) = 7 − 1.6x2 − 5x has the following characteristic
Standard form f (x) = -1.6x2 -5x +7
Degree: 2
Question 2.
p(x) =x+ 2x-2 + 9.5
The function is not a polynomial function because the term 2x-2 has an exponent that is not a whole number
Question 3.
q(x) = x3 − 6x + 3x4
The standard form of a polynomials has the exponents of the terms arranged in descending order.
The degree of a polynomial is the highest exponent of a term.
The type of a polynomial depends on the degree.
The leading coefficient is the numerical factor of the terms with highest degree.
Hence the given function q(x) = x3 − 6x + 3x4 has the following characteristic
Standard form : q(x) = 3x4 + x3 – 6x
Degree: 4
Evaluate the function for the given value of x.
Question 4.
f(x) = −x3 + 3x2 + 9; x = 4
f(x) = −x3 + 3x2 + 9 Write Original equation
f(4) =−x3 + 3x2 + 9 substitute for x
=-(4)3+ 3(4)2 +9 Evaluate powers and multiply
=-(64)+ 3(16) +9 Simplify
=-64 + 48 + 9
=-7
Question 5.
f(x) = 3x5 − x4 − 6x + 10; x = −2
f(x) = 3x5 − x4 − 6x + 10 Write Original equation
f(-2)=3(-2)5-(-2)4-6(-2)+10 substitute for x
=3(-32)-16+12+10 Evaluate powers and multiply
= -96 -16 +12 +10 Simplify
=-90
Question 6.
Describe the end behavior of the graph of f(x) = 0.25x3 − x2 − 1.
Graph the polynomial Function
Question 7.
f(x) = x4 + x2 − 3
Question 8.
f(x) = 4 − x3
Question 9.
f(x) = x3 − x2 + x − 1
Question 10.
Sketch a graph of the polynomial function f having these characteristics.
• f is decreasing when x < −1.5 and x > 2.5; f is increasing when −1.5 < x < 2.5.
• f(x) > 0 when x < −3 and 1 < x < 4; f(x) < 0 when −3 < x < 1 and x > 4.
Use the graph to describe the degree and leading coefficient of f.
Question 11.
WHAT IF? Repeat Example 6 using the alternative model for electric vehicles of
V(t) = −0.0290900t4 + 0.791260t3 − 7.96583t2 + 36.5561t − 12.025.
Graphing Polynomial Functions 4.1 Exercises
Vocabulary and Core Concept Check
Question 1.
WRITING Explain what is meant by the end behavior of a polynomial function.
Question 2.
WHICH ONE DOESN’T BELONG? Which function does not belong with the other three? Explain your reasoning.
Monitoring Progress and Modeling with Mathematics
In Exercises 3–8, decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.
Question 3.
f(x) = −3x + 5x3 − 6x2 + 2
Question 4.
p(x) = $$\frac{1}{2}$$x2 + 3x − 4x3 + 6x4 − 1
The function is a polynomial function . Written in standard form , the function is p(x)=1/2 x2 + 3x − 4x3 + 6x4 − 1 .
It has degree 4(quartic) and leading coefficient of 6.
Question 5.
f(x) = 9x4 + 8x3 − 6x-2 + 2x
Question 6.
g(x) = $$\sqrt{3}$$ − 12x + 13x2
Question 7.
h(x) = $$\frac{5}{3}$$x2 − $$\sqrt{7}$$x4 + 8x3 − $$\frac{1}{2}$$ + x
Question 8.
h(x) = 3x4 + 2x − $$\frac{5}{x}$$ + 9x3 − 7
ERROR ANALYSIS In Exercises 9 and 10, describe and correct the error in analyzing the function.
Question 9.
f(x) = 8x3 − 7x4 − 9x − 3x2 + 11
Question 10.
f(x) = 2x4 + 4x – 9$$\sqrt{x}$$ + 3x2 – 8
In Exercises 11–16, evaluate the function for the given value of x.
Question 11.
h(x) = −3x4 + 2x3 − 12x − 6; x = −2
Question 12.
f(x) = 7x4 − 10x2 + 14x − 26; x = −7
f(x) = 7x4 − 10x2 + 14x − 26 Write Original equation
f(-7) = 7(-7)4– 10(-7)2+14(-7)-26 substitute for x
= 7(2401)- 10(49)- 98-26 Evaluate powers and multiply
= 16807- 490- 98- 26 Simplify
= 16,193
Question 13.
g(x) = x6 − 64x4 + x2 − 7x − 51; x = 8
Question 14.
g(x) = −x3 + 3x2 + 5x + 1; x = −12
g(x) = −x3 + 3x2 + 5x + 1 Write Original equation
g(-12) =(-12 )3+ 3(-12) 2+ 5(-12 )+1 substitute for x
=1728 + 3(144) – 60 +1 Evaluate powers and multiply
=1728 + 432 – 60 +1 Simplify
=2101
Question 15.
p(x) = 2x3 + 4x2 + 6x + 7; x = $$\frac{1}{2}$$
Question 16.
h(x) = 5x3 − 3x2 + 2x + 4; x = −12
h(x) = 5x3 − 3x2 + 2x + 4 Write Original equation
h(-12) =5(-12)3-3(-12)2+2(-12)+4 substitute for x
=5(-1728)-3(144)-24+4 Evaluate powers and multiply
=-8640-432-24+4 Simplify
=-9092
In Exercises 17–20, describe the end behavior of the graph of the function.
Question 17.
h(x) = −5x4 + 7x3 − 6x2 + 9x + 2
Question 18.
g(x) = 7x7 + 12x5 − 6x3 − 2x − 18
The function as degree 7 and leading coefficient 7 . Because the degree is odd and leading coefficient is positive g(x)->-∞ as x ->-∞ and g(x)->+∞ as x->+∞
Question 19.
f(x) = −2x4 + 12x8 + 17 + 15x2
Question 20.
f(x) = 11 − 18x2 − 5x5 − 12x4 − 2x
f(x)= − 5x5 − 12x4− 18x2− 2x + 11
with the highest exponent is equal to 5, then the polynomial is of odd degree.
Hence the end behaviors are opposite. With the negative leading coefficient
Then f(x) ─> ∞ as x ─>-∞
f(x)─>-∞ as x ─>+∞
In Exercises 21 and 22, describe the degree and leading coefficient of the polynomial function using the graph.
Question 21.
Question 22.
The degree of the polynomial is even since one side goes up and other goes up; the leading coefficient is positive since the left side goes up and the right side goes up.
Question 23.
USING STRUCTURE Determine whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.
f(x) = 5x3x + $$\frac{5}{2}$$x3 − 9x4 + $$\sqrt{2}$$x2 + 4x − 1 − x-5x5 − 4
Question 24.
WRITING Let f(x) = 13. State the degree, type, and leading coefficient. Describe the end behavior of the function. Explain your reasoning.
In Exercises 25–32, graph the polynomial function.
Question 25.
p(x) = 3 − x4
Question 26.
g(x) = x3 + x + 3
Question 27.
f(x) = 4x − 9 − x3
Question 28.
p(x) = x5 − 3x3 + 2
Question 29.
h(x) = x4 − 2x3 + 3x
Question 30.
h(x) = 5 + 3x2 − x4
Question 31.
g(x) = x5 − 3x4 + 2x − 4
Question 32.
p(x) = x6 − 2x5 − 2x3 + x + 5
ANALYZING RELATIONSHIPS In Exercises 33–36, describe the x-values for which
(a) f is increasing or decreasing,
(b) f(x) > 0, and
(c) f(x) < 0.
Question 33.
Question 34.
a. The function is increases when x< 4 and decreases when x>4.
b.The function is positive when x<
Question 35.
Question 36.
In Exercises 37–40, sketch a graph of the polynomial function f having the given characteristics. Use the graph to describe the degree and leading coefficient of the function f.
Question 37.
• f is increasing when x > 0.5; f is decreasing when x < 0.5.
• f(x) > 0 when x < −2 and x > 3; f(x) < 0 when −2 < x < 3.
Question 38.
• f is increasing when −2 < x < 3;f is decreasing when x < −2 and x > 3.
• f(x) > 0 when x < −4 and 1 < x < 5; f(x) < 0 when −4 < x < 1 and x > 5.
Question 39.
• f is increasing when −2 < x < 0 and x > 2;f is decreasing when x < −2 and 0 < x < 2.
• f(x) > 0 when x < −3, −1 < x < 1, and x > 3; f(x) < 0 when −3 < x < −1 and 1 < x < 3.
Question 40.
• f is increasing when x < −1 and x > 1;f is decreasing when −1 < x < 1.
• f(x) > 0 when −1.5 < x < 0 and x > 1.5; f(x) < 0 when x < −1.5 and 0 < x < 1.5.
Question 41.
MODELING WITH MATHEMATICS From 1980 to 2007 the number of drive-in theaters in the United States can be modeled by the function
d(t) = −0.141t3 + 9.64t2 − 232.5t + 2421
where d(t) is the number of open theaters and t is the number of years after 1980.
a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 27. Describe the behavior of the graph on this interval.
b. What is the average rate of change in the number of drive-in movie theaters from 1980 to 1995 and from 1995 to 2007? Interpret the average rates of change.
c. Do you think this model can be used for years before 1980 or after 2007? Explain.
Question 42.
PROBLEM SOLVING The weight of an ideal round-cut diamond can be modeled by
w = 0.00583d3 − 0.0125d2 + 0.022d − 0.01
where w is the weight of the diamond (in carats) and d is the diameter (in millimeters). According to the model, what is the weight of a diamond with a diameter of 12 millimeters?
Question 43.
ABSTRACT REASONING Suppose f(x) → ∞ as x → −∞ and f(x) →−∞ as x → ∞. Describe the end behavior of g(x) = −f(x). Justify your answer.
Question 44.
THOUGHT PROVOKING Write an even degree polynomial function such that the end behavior of f is given by f(x) → −∞ as x → −∞ and f(x) → −∞ as x → ∞. Justify your answer by drawing the graph of your function.
Question 45.
USING TOOLS When using a graphing calculator to graph a polynomial function, explain how you know when the viewing window is appropriate.
Question 46.
MAKING AN ARGUMENT Your friend uses the table to speculate that the function f is an even degree polynomial and the function g is an odd degree polynomial. Is your friend correct? Explain your reasoning.
Question 47.
DRAWING CONCLUSIONS The graph of a function is symmetric with respect to the y-axis if for each point (a, b) on the graph, (−a, b) is also a point on the graph. The graph of a function is symmetric with respect to the origin if for each point (a, b) on the graph, (−a, −b) is also a point on the graph.
a. Use a graphing calculator to graph the function y = xn when n = 1, 2, 3, 4, 5, and 6. In each case, identify the symmetry of the graph.
b. Predict what symmetry the graphs of y = x10 and y = x11 each have. Explain your reasoning and then confirm your predictions by graphing.
Question 48.
HOW DO YOU SEE IT? The graph of a polynomial function is shown.
a. Describe the degree and leading coefficient of f.
b. Describe the intervals where the function is increasing and decreasing.
c. What is the constant term of the polynomial function?
Question 49.
REASONING A cubic polynomial function f has a leading coefficient of 2 and a constant term of −5. When f(1) = 0 and f(2) = 3, what is f(−5)? Explain your reasoning.
Question 50.
CRITICAL THINKING The weight y (in pounds) of a rainbow trout can be modeled by y = 0.000304x3, where x is the length (in inches) of the trout.
a. Write a function that relates the weight y and length x of a rainbow trout when y is measured in kilograms and x is measured in centimeters. Use the fact that 1 kilogram ≈ 2.20 pounds and 1 centimeter ≈ 0.394 inch.
b. Graph the original function and the function from part (a) in the same coordinate plane. What type of transformation can you apply to the graph of y = 0.000304x3 to produce the graph from part (a)?
Maintaining Mathematical Proficiency
Simplify the expression. (Skills Review Handbook)
Question 51.
xy + x2 + 2xy + y2 − 3x2
Question 52.
2h3g + 3hg3 + 7h2g2 + 5h3g+ 2hg3
Question 53.
−wk + 3kz − 2kw + 9zk − kw
Question 54.
a2(m − 7a3) − m(a2 − 10)
Question 55.
3x(xy − 4) + 3(4xy + 3) − xy(x2y − 1)
Question 56.
cv(9 − 3c) + 2c(v − 4c) + 6c
Lesson 4.2 Adding, Subtracting, and Multiplying Polynomials
Essential Question
How can you cube a binomial?
EXPLORATION 1
Cubing Binomials
Work with a partner.
Find each product. Show your steps.
EXPLORATION 2
Generalizing Patterns for Cubing a Binomial
Work with a partner.
a. Use the results of Exploration 1 to describe a pattern for the coefficients of the terms when you expand the cube of a binomial. How is your pattern related to Pascal’s Triangle, shown at the right?
b. Use the results of Exploration 1 to describe a pattern for the exponents of the terms in the expansion of a cube of a binomial.
c. Explain how you can use the patterns you described in parts (a) and (b) to find the product (2x − 3)3. Then find this product.
Question 3.
How can you cube a binomial?
Question 4.
Find each product.
a. (x + 2)3
b. (x − 2)3
c. (2x − 3)3
d. (x − 3)3
e. (−2x + 3)3
f. (3x − 5)3
4.2 Lesson
Monitoring Progress
Find the sum or difference.
Question 1.
(2x2 − 6x + 5) + (7x2 − x − 9)
Removing the grouping symbols and combining the like terms, the given expression is
(2x2 − 6x + 5) + (7x2 − x − 9), is equivalent to
= 2x2 − 6x + 5 + 7x2 − x − 9
= (2x2+7x2)+( − 6x− x)+( 5-9)
= 9x2-7x-4
Question 2.
(3t3 + 8t2 − t − 4) − (5t3 − t2 + 17)
(3t3 + 8t2 − t − 4) − (5t3 − t2 + 17)
=3t3 +8t2 -t- 4- 5t3 + t2 -17 (write the opposite of the subtracted polynomial, then add like terms)
= -2t3+ 9t2 -t -21
Find the product.
Question 3.
(4x2 + x − 5)(2x + 1)
(4x2 + x − 5)(2x + 1)
=4x2(2x+1) + x(2x+1) – 5(2x+1) (we multiply the polynomials in horizontal format)
=8x3+ 4x2+ 2x2+ x- 10x -5
=8x3+ 6x2– 9x -x
Question 4.
(y − 2)(5y2 + 3y − 1)
(y − 2)(5y2 + 3y − 1)
=(y-2)5y2+ (y-2)3y – (y-2) (we multiply the polynomials in horizontal format)
=5y3-10y2+3y3-6y-y+2
=5y3-7y2-7y+2
Question 5.
(m − 2)(m − 1)(m + 3)
(m − 2)(m − 1)(m + 3)
=(m2-m-2m+2)(m+3) (we multiply the first two polynomial in horizontal format , then the result
=(m2-3m+2)(m+3) by the last polynomial )
=(m2-3m+2)m+(m2-3m+2)3
=m3-3m2+2m+3m2-9m+6
=m3-7m+6
Question 6.
(3t − 2)(3t + 2)
(3t − 2)(3t + 2)
we can use the formula: a2-b2= (a+ b) (a-b)
=(3t)2-22
=9t2-4
Question 7.
(5a + 2)2
(5a + 2)2
=(5a)2+2. 5a .2+22 ( we can use the formula : (a+ b)2 = a2+2ab+b2)
=25a2+20a+4
Question 8.
(xy − 3)3
(xy-3)3
=(x y)3-3(x y)2.3+3. xy.32-33 ( we can use the formula:(a-b)3=a3-3a2 b +3a b 2-b3
=x3y3– 9x 2y2+27xy-27
Question 9.
(a) Prove the polynomial identity for the cube of a binomial representing a difference: (a − b)3 = a3 − 3a2b + 3ab2 − b3.
(b) Use the cube of a binomial in part (a) to calculate 93.
(a) . (a-b)3 =a3– 3a 2b+3ab2-b3 We have to prove the identity:
= (a-b)3 =(a-b)(a-b)(a-b) We start from left side:
= (a2-2ab+b2)(a-b)
= a3-a 2b-2a 2b+2ab2 +ab 2-b3
= a 3-3a 2b+3ab 2– b3
(b). 93=(10-1)3=103-3.102.1+3.10.12-13
= 1000-300+30-1 We use the above identity to compute 93
= 729
Question 10.
Use Pascal’s Triangle to expand
(a) (z + 3)4 and
(b) (2t − 1)5
Adding, Subtracting, and Multiplying Polynomials 4.2 Exercises
Vocabulary and Core Concept Check
Question 1.
WRITING Describe three different methods to expand (x + 3)3.
Question 2.
WRITING Is (a + b)(a − b) = a2 − b2 an identity? Explain your reasoning.
Monitoring Progress and Modeling with Mathematics
In Exercises 3–8, find the sum.
Question 3.
(3x2 + 4x − 1) + (−2x2 − 3x + 2)
Question 4.
(−5x2 + 4x − 2) + (−8x2 + 2x + 1)
Question 5.
(12x5 − 3x4 + 2x − 5) + (8x4 − 3x3 + 4x + 1)
Question 6.
(8x4 + 2x2 − 1) + (3x3 − 5x2 + 7x + 1)
Question 7.
(7x6 + 2x5 − 3x2 + 9x) + (5x5 + 8x3 − 6x2 + 2x − 5)
Question 8.
(9x4 − 3x3 + 4x2 + 5x + 7) + (11x4 − 4x2 − 11x − 9)
In Exercises 9–14, find the difference.
Question 9.
(3x3 − 2x2 + 4x − 8) − (5x3 + 12x2 − 3x − 4)
Question 10.
(7x4 − 9x3 − 4x2 + 5x + 6) − (2x4 + 3x3 − x2 + x − 4)
Question 11.
(5x6 − 2x4 + 9x3 + 2x − 4) − (7x5 − 8x4 + 2x− 11)
Question 12.
(4x5 − 7x3 − 9x2 + 18) − (14x5 − 8x4 + 11x2 + x)
Question 13.
(8x5 + 6x3 − 2x2 + 10x) − (9x5 − x3 − 13x2 + 4)
Question 14.
(11x4 − 9x2 + 3x + 11) − (2x4 + 6x3 + 2x − 9)
Question 15.
MODELING WITH MATHEMATICS During a recent period of time, the numbers (in thousands) of males Mand females F that attend degree-granting institutions in the United States can be modeled by
M = 19.7t2 + 310.5t + 7539.6
F = 28t2 + 368t + 10127.8
where t is time in years. Write a polynomial to model the total number of people attending degree-granting institutions. Interpret its constant term.
Question 16.
MODELING WITH MATHEMATICS A farmer plants a garden that contains corn and pumpkins. The total area (in square feet) of the garden is modeled by the expression 2x2 + 5x + 4. The area of the corn is modeled by the expression x2 − 3x + 2. Write an expression that models the area of the pumpkins.
In Exercises 17–24, find the product.
Question 17.
7x3(5x2 + 3x + 1)
Question 18.
−4x5(11x3 + 2x2 + 9x + 1)
Question 19.
(5x2 − 4x + 6)(−2x + 3)
Question 20.
(−x − 3)(2x2 + 5x + 8)
Question 21.
(x2 − 2x − 4)(x2 − 3x − 5)
Question 22.
(3x2 + x − 2)(−4x2 − 2x − 1)
Question 23.
(3x3 − 9x + 7)(x2 − 2x + 1)
Question 24.
(4x2 − 8x − 2)(x4 + 3x2 + 4x)
ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in performing the operation.
Question 25.
Question 26.
In Exercises 27–32, find the product of the binomials.
Question 27.
(x − 3)(x + 2)(x + 4)
Question 28.
(x − 5)(x + 2)(x − 6)
Question 29.
(x − 2)(3x + 1)(4x − 3)
Question 30.
(2x + 5)(x − 2)(3x + 4)
Question 31.
(3x − 4)(5 − 2x)(4x + 1)
Question 32.
(4 − 5x)(1 − 2x)(3x + 2)
Question 33.
REASONING Prove the polynomial identity (a + b)(a − b) = a2 − b2. Then give an example of two whole numbers greater than 10 that can be multiplied using mental math and the given identity. Justify your answers.
Question 34.
NUMBER SENSE You have been asked to order textbooks for your class. You need to order 29 textbooks that cost $31 each. Explain how you can use the polynomial identity (a + b)(a − b) = a2 − b2 and mental math to find the total cost of the textbooks. Answer: In Exercises 35–42, find the product. Question 35. (x − 9)(x + 9) Answer: Question 36. (m + 6)2 Answer: Question 37. (3c − 5)2 Answer: Question 38. (2y − 5)(2y + 5) Answer: Question 39. (7h + 4)2 Answer: Question 40. (9g − 4)2 Answer: Question 41. (2k + 6)3 Answer: Question 42. (4n − 3)3 Answer: In Exercises 43–48, use Pascal’s Triangle to expand the binomial. Question 43. (2t + 4)3 Answer: Question 44. (6m + 2)2 Answer: Question 45. (2q − 3)4 Answer: Question 46. (g + 2)5 Answer: Question 47. (yz + 1)5 Answer: Question 48. (np − 1)4 Answer: Question 49. COMPARING METHODS Find the product of the expression (a2 + 4b2)2(3a2 − b2)2 using two different methods. Which method do you prefer? Explain. Answer: Question 50. THOUGHT PROVOKING Adjoin one or more polygons to the rectangle to form a single new polygon whose perimeter is double that of the rectangle. Find the perimeter of the new polygon. Answer: MATHEMATICAL CONNECTIONS In Exercises 51 and 52, write an expression for the volume of the figure as a polynomial in standard form. Question 51. Answer: Question 52. Answer: Question 53. MODELING WITH MATHEMATICS Two people make three deposits into their bank accounts earning the same simple interest rate r. Person A’s account is worth 2000(1 + r)3 + 3000(1 + r)2 + 1000(1 + r) on January 1, 2015. a. Write a polynomial for the value of Person B’s account on January 1, 2015. b. Write the total value of the two accounts as a polynomial in standard form. Then interpret the coefficients of the polynomial. c. Suppose their interest rate is 0.05. What is the total value of the two accounts on January 1, 2015? Answer: Question 54. PROBLEM SOLVING The sphere is centered in the cube. Find an expression for the volume of the cube outside the sphere. Answer: Question 55. MAKING AN ARGUMENT Your friend claims the sum of two binomials is always a binomial and the product of two binomials is always a trinomial. Is your friend correct? Explain your reasoning. Answer: Question 56. HOW DO YOU SEE IT? You make a tin box by cutting x-inch-by-x-inch pieces of tin off the corners of a rectangle and folding up each side. The plan for your box is shown. a. What are the dimensions of the original piece of tin? b. Write a function that represents the volume of the box. Without multiplying, determine its degree. Answer: USING TOOLS In Exercises 57–60, use a graphing calculator to make a conjecture about whether the two functions are equivalent. Explain your reasoning. Question 57. f(x) = (2x − 3)3; g(x) = 8x3 − 36x2 + 54x − 27 Answer: Question 58. h(x) = (x + 2)5; k(x) = x5 + 10x4 + 40x3 + 80x2 + 64x Answer: Question 59. f(x) = (−x − 3)4; g(x) = x4 + 12x3 + 54x2 + 108x + 80 Answer: Question 60. f(x) = (−x + 5)3; g(x) = −x3 + 15x2 − 75x + 125 Answer: Question 61. REASONING Copy Pascal’s Triangle and add rows for n = 6, 7, 8, 9, and 10. Use the new rows to expand (x + 3)7 and (x − 5)9. Answer: Question 62. ABSTRACT REASONING You are given the function f(x) = (x + a)(x + b)(x + c)(x + d). When f(x) is written in standard form, show that the coefficient of x3 is the sum of a, b, c, and d, and the constant term is the product of a, b, c, and d. Answer: Question 63. DRAWING CONCLUSIONS Let g(x) = 12x4 + 8x + 9 and h(x) = 3x5 + 2x3 − 7x + 4. a. What is the degree of the polynomial g(x) + h(x)? b. What is the degree of the polynomial g(x) − h(x)? c. What is the degree of the polynomial g(x) • h(x)? d. In general, if g(x) and h(x) are polynomials such that g(x) has degree m and h(x) has degree n, and m > n, what are the degrees of g(x) + h(x), g(x) − h(x), and g(x) • h(x)? Answer: Question 64. FINDING A PATTERN In this exercise, you will explore the sequence of square numbers. The first four square numbers are represented below. a. Find the differences between consecutive square numbers. Explain what you notice. b. Show how the polynomial identity (n + 1)2 − n2 = 2n + 1 models the differences between square numbers. c. Prove the polynomial identity in part (b). Answer: Question 65. CRITICAL THINKING Recall that a Pythagorean triple is a set of positive integers a, b, and c such that a2 + b2 = c2. The numbers 3, 4, and 5 form a Pythagorean triple because 32 + 42 = 52. You can use the polynomial identity (x2 − y2)2 + (2xy)2 = (x2 + y2)2 to generate other Pythagorean triples. a. Prove the polynomial identity is true by showing that the simplified expressions for the left and right sides are the same. b. Use the identity to generate the Pythagorean triple when x = 6 and y = 5. c. Verify that your answer in part (b) satisfies a2 + b2 = c2. Answer: Maintaining Mathematical Proficiency Perform the operation. Write the answer in standard form. (Section 3.2) Question 66. (3 − 2i) + (5 + 9i) Answer: Question 67. (12 + 3i) − (7 − 8i) Answer: Question 68. (7i)(−3i) Answer: Question 69. (4 + i)(2 − i) Answer: Lesson 4.3 Dividing Polynomials Essential Question How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial? EXPLORATION 1 Dividing Polynomials Work with a partner. Match each division statement with the graph of the related cubic polynomial f(x). Explain your reasoning. Use a graphing calculator to verify your answers. a. $$\frac{f(x)}{x}$$ = (x − 1)(x + 2) b. $$\frac{f(x)}{x-1}$$ = (x − 1)(x + 2) c. $$\frac{f(x)}{x+1}$$ = (x − 1)(x + 2) d. $$\frac{f(x)}{x-2}$$ = (x− 1)(x+ 2) e. $$\frac{f(x)}{x+2}$$ = (x− 1)(x+ 2) f. $$\frac{f(x)}{x-3}$$ = (x− 1)(x+ 2) EXPLORATION 2 Dividing Polynomials Work with a partner. Use the results of Exploration 1 to find each quotient. Write your answers in standard form. Check your answers by multiplying. a. (x3 + x2 − 2x) ÷ x b. (x3 − 3x + 2) ÷ (x − 1) c. (x3 + 2x2 − x − 2) ÷ (x + 1) d. (x3 − x2 − 4x + 4) ÷ (x − 2) e. (x3 + 3x2 − 4) ÷ (x + 2) f. (x3 − 2x2 − 5x + 6) ÷ (x − 3) Communicate Your Answer Question 3. How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial? 4.3 Lesson Monitoring Progress Divide using polynomial long division. Question 1. (x3 − x2 − 2x + 8) ÷ (x − 1) Question 2. (x4 + 2x2 − x + 5) ÷ (x2 − x + 1) Divide using synthetic division. Question 3. (x3 − 3x2 − 7x + 6) ÷ (x − 2) Question 4. (2x3 − x − 7) ÷ (x + 3) Use synthetic division to evaluate the function for the indicated value of x. Question 5. f(x) = 4x2 − 10x − 21; x = 5 Question 6. f(x) = 5x4 + 2x3 − 20x − 6; x = 2 Dividing Polynomials 4.3 Exercises Monitoring Progress and Modeling with Mathematics Question 1. WRITING Explain the Remainder Theorem in your own words. Use an example in your explanation. Answer: Question 2. VOCABULARY What form must the divisor have to make synthetic division an appropriate method for dividing a polynomial? Provide examples to support your claim. Answer: Question 3. VOCABULARY Write the polynomial divisor, dividend, and quotient functions represented by the synthetic division shown at the right. Answer: Question 4. WRITING Explain what the colored numbers represent in the synthetic division in Exercise 3. Answer: Vocabulary and Core Concept Check In Exercises 5–10, divide using polynomial long division. Question 5. (x2 + x − 17 ) ÷ (x − 4) Answer: Question 6. (3x2 − 14x − 5) ÷ (x − 5) Answer: Question 7. (x3 + x2 + x + 2 ) ÷ (x2 − 1) Answer: Question 8. (7x3 + x2 + x ) ÷ (x2 + 1) Answer: Question 9. (5x4 − 2x3 − 7x2 − 39) ÷ (x2 + 2x − 4) Answer: Question 10. (4x4 + 5x − 4) ÷ (x2 − 3x − 2) Answer: In Exercises 11–18, divide using synthetic division. Question 11. (x2 + 8x + 1) ÷ (x − 4) Answer: Question 12. (4x2 − 13x − 5) ÷ (x − 2) Answer: (4x2 − 13x − 5) ÷ (x − 2) use synthetic division. Because the divisor is x-2 , k=2 Question 13. (2x2 − x + 7) ÷ (x + 5) Answer: Question 14. (x3 − 4x + 6) ÷ (x + 3) Answer: Question 15. (x2 + 9) ÷ (x − 3) Answer: Question 16. (3x3 − 5x2 − 2) ÷ (x − 1) Answer: Question 17. (x4 − 5x3 − 8x2 + 13x − 12) ÷ (x − 6) Answer: Question 18. (x4 + 4x3 + 16x − 35) ÷ (x + 5 ) Answer: ANALYZING RELATIONSHIPS In Exercises 19–22, match the equivalent expressions. Justify your answers. Question 19. (x2 + x − 3) ÷ (x − 2) Answer: Question 20. (x2 − x − 3) ÷ (x − 2) Answer: Question 21. (x2 − x + 3) ÷ (x − 2) Answer: Question 22. (x2 + x + 3) ÷ (x − 2) Answer: ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in using synthetic division to divide x3 – 5x + 3 by x – 2. Question 23. Answer: Question 24. Answer: In Exercises 25–32, use synthetic division to evaluate the function for the indicated value of x. Question 25. f(x) = −x2 − 8x + 30; x = −1 Answer: Question 26. f(x) = 3x2 + 2x − 20; x = 3 Answer: Question 27. f(x) = x3 − 2x2 + 4x + 3; x = 2 Answer: Question 28. f(x) = x3 + x2 − 3x + 9; x = −4 Answer: Question 29. f(x) = x3 − 6x + 1; x = 6 Answer: Question 30. f(x) = x3 − 9x − 7; x = 10 Answer: Question 31. f(x) = x4 + 6x2 − 7x + 1; x = 3 Answer: Question 32. f(x) = −x4 − x3 − 2; x = 5 Answer: Question 33. MAKING AN ARGUMENT You use synthetic division to divide f(x) by (x − a) and find that the remainder equals 15. Your friend concludes that f(15) = a. Is your friend correct? Explain your reasoning. Answer: Question 34. THOUGHT PROVOKING A polygon has an area represented by A = 4x2 + 8x + 4. The figure has at least one dimension equal to 2x + 2. Draw the figure and label its dimensions. Answer: Question 35. USING TOOLS The total attendance A (in thousands) at NCAA women’s basketball games and the number T of NCAA women’s basketball teams over a period of time can be modeled by A = −1.95x3 + 70.1x2 − 188x + 2150 T = 14.8x + 725 where x is in years and 0 < x < 18. Write a function for the average attendance per team over this period of time. Answer: Question 36. COMPARING METHODS The profit P (in millions of dollars) for a DVD manufacturer can be modeled by P = −6x3 + 72x, where x is the number (in millions) of DVDs produced. Use synthetic division to show that the company yields a profit of$96 million when 2 million DVDs are produced. Is there an easier method? Explain.
Question 37.
CRITICAL THINKING What is the value of k such that (x3 − x2 + kx − 30) ÷ (x − 5) has a remainder of zero?
A. −14
B. −2
C. 26
D. 32
Question 38.
HOW DO YOU SEE IT? The graph represents the polynomial function f(x) = x3 + 3x2 − x − 3.
a. The expression f(x) ÷ (x − k) has a remainder of −15. What is the value of k?
b. Use the graph to compare the remainders of (x3 + 3x2 − x − 3) ÷ (x + 3) and (x3 + 3x2 − x − 3) ÷ (x + 1).
Question 39.
MATHEMATICAL CONNECTIONS The volume V of the rectangular prism is given by V = 2x3 + 17x2 + 46x + 40. Find an expression for the missing dimension.
Question 40.
USING STRUCTURE You divide two polynomials and obtain the result 5x2 − 13x + 47 − $$\frac{102}{x+2}$$. What is the dividend? How did you find it?
Maintaining Mathematical Proficiency
Find the zero(s) of the function. (Sections 3.1and 3.2)
Question 41.
f(x) = x2 − 6x + 9
Question 42.
g(x) = 3(x + 6)(x − 2)
Question 43.
g(x) = x2 + 14x + 49
Question 44.
h(x) = 4x2 + 36
Lesson 4.4 Factoring Polynomials
Essential Question
How can you factor a polynomial?
EXPLORATION 1
Factoring Polynomials
Work with a partner.
Match each polynomial equation with the graph of its related polynomial function. Use the x-intercepts of the graph to write each polynomial in factored form. Explain your reasoning.
a. x2 + 5x + 4 = 0
b. x3 − 2x2 − x + 2 = 0
c. x3 + x2 − 2x = 0
d. x3 − x = 0
e. x4 − 5x2 + 4 = 0
f. x4 − 2x3 − x2 + 2x = 0
EXPLORATION 2
Factoring Polynomials
Work with a partner. Use the x-intercepts of the graph of the polynomial function to write each polynomial in factored form. Explain your reasoning. Check your answers by multiplying.
a. f(x) = x2 − x − 2
b. f(x) = x3 − x2 − 2x
c. f(x) = x3 − 2x2 − 3x
d. f(x) = x3 − 3x2 − x + 3
e. f(x) = x4 + 2x3 − x2 − 2x
f. f(x) = x4 − 10x2 + 9
Question 3.
How can you factor a polynomial?
Question 4.
What information can you obtain about the graph of a polynomial function written in factored form?
4.4 Lesson
Monitoring Progress
Question 1.
x3 − 7x2 + 10x
Question 2.
3n7 − 75n5
Question 3.
8m5 − 16m4 + 8m3
Factor the polynomial completely
Question 4.
a3 + 27
Question 5.
6z5 − 750z2
Question 6.
x3 + 4x2 − x − 4
Question 7.
3y3 + y2 + 9y + 3
Question 8.
−16n4 + 625
Question 9.
5w6 − 25w4 + 30w2
Question 10.
Determine whether x− 4 is a factor of f(x) = 2x2 + 5x − 12.
Question 11.
Show that x − 6 is a factor of f(x) = x3 − 5x2 − 6x. Then factor f(x) completely.
Question 12.
In Example 7, does your answer change when you first determine whether 2 is a zero and then whether −1 is a zero? Justify your answer.
Factoring Polynomials 4.4 Exercises
Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE The expression 9x4 − 49 is in _________ form because it can be written as u2 − 49 where u = _____.
Question 2.
VOCABULARY Explain when you should try factoring a polynomial by grouping.
Question 3.
WRITING How do you know when a polynomial is factored completely?
Question 4.
WRITING Explain the Factor Theorem and why it is useful.
Monitoring Progress and Modeling with Mathematics
In Exercises 5–12, factor the polynomial completely.
Question 5.
x3 − 2x2 − 24x
Question 6.
4k5 − 100k3
Question 7.
3p5 − 192p3
Question 8.
2m6 − 24m5 + 64m4
Question 9.
2q4 + 9q3 − 18q2
Question 10.
3r6 − 11r5 − 20r4
Question 11.
10w10 − 19w9 + 6w8
Question 12.
18v9 + 33v8 + 14v7
In Exercises 13–20, factor the polynomial completely.
Question 13.
x3 + 64
Question 14.
y3 + 512
Question 15.
g3 − 343
Question 16.
c3 − 27
Question 17.
3h9 − 192h6
Question 18.
9n6 − 6561n3
Question 19.
16t7 + 250t4
Question 20.
135z11 − 1080z8
ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in factoring the polynomial.
Question 21.
Question 22.
In Exercises 23–30, factor the polynomial completely.
Question 23.
y3 − 5y2 + 6y − 30
Question 24.
m3 − m2 + 7m − 72
Question 25.
3a3 + 18a2 + 8a + 48
Question 26.
2k3 − 20k2 + 5k − 50
Question 27.
x3 − 8x2 − 4x + 32
Question 28.
z3 − 5z2 − 9z + 45
Question 29.
4q3 − 16q7 − 9q + 36
Question 30.
16n3 + 32n7 − n − 2
In Exercises 31–38, factor the polynomial completely.
Question 31.
49k4 − 9
Question 32.
4m4 − 25
Question 33.
c4 + 9c2 + 20
Question 34.
y4 − 3y2 − 28
Question 35.
16z4 − 81
Question 36.
81a4 − 256
Question 37.
3r8 + 3r5 − 60r2
Question 38.
4n12 − 32n7 + 48n2
In Exercises 39–44, determine whether the binomial is a factor of the polynomial.
Question 39.
f(x) = 2x3 + 5x2 − 37x − 60; x − 4
Question 40.
g(x) = 3x3 − 28x2 + 29x + 140; x + 7
Question 41.
h(x) = 6x5 − 15x4 − 9x3; x + 3
Question 42.
g(x) = 8x5 − 58x4 + 60x3 + 140; x − 6
Question 43.
h(x) = 6x4 − 6x3 − 84x2 + 144x; x + 4
Question 44.
t(x) = 48x4 + 36x3 − 138x2 − 36x; x + 2
In Exercises 45–50, show that the binomial is a factor of the polynomial. Then factor the polynomial completely.
Question 45.
g(x) = x3 − x2 − 20x; x + 4
Question 46.
t(x) = x3 − 5x2 − 9x + 45; x − 5
Question 47.
f(x) = x4 − 6x3 − 8x + 48; x − 6
Question 48.
s(x) = x4 + 4x3 − 64x − 256; x + 4
Question 49.
r(x) = x3 − 37x + 84; x + 7
Question 50.
h(x) = x3 − x2 − 24x − 36; x + 2
ANALYZING RELATIONSHIPS In Exercises 51–54, match the function with the correct graph. Explain your reasoning.
Question 51.
f(x) = (x − 2)(x − 3)(x + 1)
Question 52.
g(x) = x(x + 2)(x + 1)(x − 2)
Question 53.
h(x) = (x + 2)(x + 3)(x − 1)
Question 54.
k(x) = x(x − 2)(x − 1)(x + 2)
Question 55.
MODELING WITH MATHEMATICS The volume (in cubic inches) of a shipping box is modeled by V = 2x3 − 19x2 + 39x, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.
Question 56.
MODELING WITH MATHEMATICS The volume (in cubic inches) of a rectangular birdcage can be modeled by V = 3x3 − 17x2 + 29x − 15, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.
USING STRUCTURE In Exercises 57–64, use the method of your choice to factor the polynomial completely. Explain your reasoning.
Question 57.
a6 + a5 − 30a4
Question 58.
8m3 − 343
Question 59.
z3 − 7z2 − 9z + 63
Question 60.
2p8 − 12p5 + 16p2
Question 61.
64r3 + 729
Question 62.
5x5 − 10x4 − 40x3
Question 63.
16n4 − 1
Question 64.
9k3 − 24k2 + 3k − 8
Question 65.
REASONING Determine whether each polynomial is factored completely. If not, factor completely.
a. 7z4(2z2 − z − 6)
b. (2 − n)(n2 + 6n)(3n − 11)
c. 3(4y − 5)(9y2 − 6y − 4)
Question 66.
PROBLEM SOLVING The profit P (in millions of dollars) for a T-shirt manufacturer can be modeled by P = −x3 + 4x2 + x, where x is the number (in millions) of T-shirts produced. Currently the company produces 4 million T-shirts and makes a profit of $4 million. What lesser number of T-shirts could the company produce and still make the same profit? Answer: Question 67. PROBLEM SOLVING The profit P (in millions of dollars) for a shoe manufacturer can be modeled by P = −21x3 + 46x, where x is the number (in millions) of shoes produced. The company now produces 1 million shoes and makes a profit of$25 million, but it would like to cut back production. What lesser number of shoes could the company produce and still make the same profit?
Question 68.
THOUGHT PROVOKING Find a value of k such that $$\frac{f(x)}{x-k}$$ has a remainder of 0. Justify your answer.
Question 69.
COMPARING METHODS You are taking a test where calculators are not permitted. One question asks you to evaluate g(7) for the function g(x) = x3 − 7x2 − 4x + 28. You use the Factor Theorem and synthetic division and your friend uses direct substitution. Whose method do you prefer? Explain your reasoning.
Question 70.
MAKING AN ARGUMENT You divide f(x) by (x−a) and find that the remainder does not equal 0. Your friend concludes that f(x) cannot be factored. Is your friend correct? Explain your reasoning.
Question 71.
CRITICAL THINKING What is the value of k such that x− 7 is a factor of h(x) = 2x3 − 13x2 − kx + 105? Justify your answer.
Question 72.
HOW DO YOU SEE IT? Use the graph to write an equation of the cubic function in factored form. Explain your reasoning.
Question 73.
ABSTRACT REASONING Factor each polynomial completely.
a. 7ac2 + bc2 − 7ad2 − bd2
b. x2n − 2xn + 1
c. a5b2 − a2b4 + 2a4b − 2ab3 + a3 − b2
Question 74.
REASONING The graph of the function f(x) = x4 + 3x3 + 2x2 + x + 3 is shown. Can you use the Factor Theorem to factor f(x)? Explain.
Question 75.
MATHEMATICAL CONNECTIONS The standard equation of a circle with radius r and center (h, k) is (x − h)2 + (y − k)2 = r2. Rewrite each equation of a circle in standard form. Identify the center and radius of the circle. Then graph the circle.
a. x2 + 6x + 9 + y2 = 25
b. x2 − 4x + 4 + y2 = 9
c. x2 − 8x + 16 + y2 + 2y + 1 = 36
Question 76.
CRITICAL THINKING Use the diagram to complete parts (a)–(c).
a. Explain why a3 − b3 is equal to the sum of the volumes of the solids I, II, and III.
b. Write an algebraic expression for the volume of each of the three solids. Leave your expressions in factored form.
c. Use the results from part (a) and part (b) to derive the factoring pattern a3 − b3.
Maintaining Mathematical Proficiency
Solve the quadratic equation by factoring.(Section 3.1)
Question 77.
x2 − x − 30 = 0
Question 78.
2x2 − 10x − 72 = 0
Question 79.
3x2 − 11x + 10 = 0
Question 80.
9x2 − 28x + 3 = 0
Solve the quadratic equation by completing the square.(Section 3.3)
Question 81.
x2 − 12x + 36 = 144
Question 82.
x2 − 8x − 11 = 0
Question 83.
3x2 + 30x + 63 = 0
Question 84.
4x2 + 36x − 4 = 0
Polynomial Functions Study Skills : Keeping Your Mind Focused
4.1–4.4 What Did You Learn?
Core Vocabulary
Core Concepts
Section 4.1
Section 4.2
Section 4.3
Section 4.4
Mathematical Practices
Question 1.
Describe the entry points you used to analyze the function in Exercise 43 on page 164.
Question 2.
Describe how you maintained oversight in the process of factoring the polynomial in Exercise 49 on page 185.
Study Skills
• When you sit down at your desk, review your notes from the last class.
• Repeat in your mind what you are writing in your notes.
• When a mathematical concept is particularly difficult, ask your teacher for another example.
Polynomial Functions 4.1 – 4.4 Quiz
Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.(Section 4.1)
Question 1.
f(x) = 5 + 2x2 − 3x4 − 2x − x3
Question 2.
g(x) = $$\frac{1}{4}$$x3 + 2x − 3x2 + 1
Question 3.
h(x) = 3 − 6x3 + 4x − 2 + 6x
Question 4.
Describe the x-values for which
(a) f is increasing or decreasing,
(b) f(x) > 0, and
(c) f(x) < 0. (Section 4.1)
Question 5.
Write an expression for the area and perimeter for the figure shown.(Section 4.2)
Perform the indicated operation.(Section 4.2)
Question 6.
(7x2 − 4) − (3x2 − 5x + 1)
Question 7.
(x2 − 3x + 2)(3x − 1)
Question 8.
(x − 1)(x + 3)(x − 4)
Question 9.
Use Pascal’s Triangle to expand (x + 2)5. (Section 4.2)
Question 10.
Divide 4x4 − 2x3 + x2 − 5x + 8 by x2 − 2x − 1. (Section 4.3)
Factor the polynomial completely.(Section 4.4)
Question 11.
a3 − 2a2 − 8a
Question 12.
8m3 + 27
Question 13.
z3 + z2 − 4z − 4
Question 14.
49b4 − 64
Question 15.
Show that x + 5 is a factor of f(x) = x3 − 2x2 − 23x + 60. Then factor f(x) completely. (Section 4.4)
Question 16.
The estimated price P (in cents) of stamps in the United States can be modeled by the polynomial function P(t)= 0.007t3 − 0.16t2 + 1t + 17, where t represents the number of years since 1990. (Section 4.1)
a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 20. Describe the behavior of the graph on this interval.
b. What was the average rate of change in the price of stamps from 1990 to 2010?
Question 17.
The volume V (in cubic feet) of a rectangular wooden crate is modeled by the function V(x) = 2x3 − 11x2 + 12x, where x is the width (in feet) of the crate. Determine the values of x for which the model makes sense. Explain your reasoning. (Section 4.4)
Lesson 4.5 Solving Polynomial Equations
Essential Question
How can you determine whether a polynomial equation has a repeated solution?
EXPLORATION 1
Cubic Equations and Repeated Solutions
Work with a partner.
Some cubic equations have three distinct solutions. Others have repeated solutions. Match each cubic polynomial equation with the graph of its related polynomial function. Then solve each equation. For those equations that have repeated solutions, describe the behavior of the related function near the repeated zero using the graph or a table of values.
a. x3 − 6x2 + 12x − 8 = 0
b. x3 + 3x2 + 3x + 1 = 0
c. x3− 3x + 2 = 0
d. x3 + x2 − 2x = 0
e. x3 − 3x − 2 = 0
f. x3 − 3x2 + 2x = 0
EXPLORATION 2
Quartic Equations and Repeated Solutions
Work with a partner.
Determine whether each quartic equation has repeated solutions using the graph of the related quartic function or a table of values. Explain your reasoning. Then solve each equation.
a. x4 − 4x3 + 5x2 − 2x = 0
b. x4 − 2x3 − x2 + 2x = 0
c. x4 − 4x3 + 4x2 = 0
d. x4 + 3x3 = 0
Question 3.
How can you determine whether a polynomial equation has a repeated solution?
Question 4.
Write a cubic or a quartic polynomial equation that is different from the equations in Explorations 1 and 2 and has a repeated solution.
4.5 Lesson
Monitoring Progress
Solve the equation.
Question 1.
4x4 − 40x2 + 36 = 0
Question 2.
2x5 + 24x = 14x3
Find the zeros of the function. Then sketch a graph of the function.
Question 3.
f(x) = 3x4 − 6x2 + 3
Question 4.
f(x) = x3 + x2 − 6x
Question 5.
Find all real solutions of x3 − 5x2 − 2x + 24 = 0.
Question 6.
Find all real zeros of f(x) = 3x4 − 2x3 − 37x2 + 24x + 12.
Question 7.
Write a polynomial function f of least degree that has rational coefficients, aleading coefficient of 1, and the zeros 4 and 1 − $$\sqrt{5}$$.
Solving Polynomial Equations 4.5 Exercises
Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE If a polynomial function f has integer coefficients, then every rational solution of f(x) = 0 has the form $$\frac{p}{q}$$, where p is a factor of the _____________ and q is a factor of the _____________.
Question 2.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Monitoring Progress and Modeling with Mathematics
In Exercises 3–12, solve the equation.
Question 3.
z3 − z2 − 12z = 0
Question 4.
a3 − 4a2 + 4a = 0
Question 5.
2x4 − 4x3 = −2x2
Question 6.
v3 − 2v2 − 16v = − 32
Question 7.
5w3 = 50w
Question 8.
9m5 = 27m3
Question 9.
2c4 − 6c3 = 12c2 − 36c
Question 10.
p4 + 40 = 14p2
Question 11.
12n2 + 48n = −n3 − 64
Question 12.
y3 − 27 = 9y2 − 27y
In Exercises 13–20, find the zeros of the function. Then sketch a graph of the function.
Question 13.
h(x) = x4 + x3 − 6x2
Question 14.
f(x) = x4 − 18x2 + 81
Question 15.
p(x) = x6 − 11x5 + 30x4
Question 16.
g(x) = −2x5 + 2x4 + 40x3
Question 17.
g(x) = −4x4 + 8x3 + 60x2
Question 18.
h(x) = −x3 − 2x2 + 15x
Question 19.
h(x) = −x3 − x2 + 9x + 9
Question 20.
p(x) = x3 − 5x2 − 4x + 20
Question 21.
USING EQUATIONS According to the Rational Root Theorem, which is not a possible solution of the equation 2x4 − 5x3 + 10x2 − 9 = 0?
A. −9
B. −$$[\frac{1}{2}/latex] C. [latex][\frac{5}{2}/latex] D. 3 Answer: Question 22. USING EQUATIONS According to the Rational Root Theorem, which is not a possible zero of the function f(x) = 40x5 − 42x4 − 107x3 + 107x2 + 33x − 36? A. −[latex][\frac{2}{3}/latex] B. − [latex][\frac{3}{8}/latex] C. [latex][\frac{3}{4}/latex] D. [latex][\frac{4}{5}/latex] Answer: ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in listing the possible rational zeros of the function. Question 23. Answer: Question 24. Answer: In Exercises 25–32, find all the real solutions of the equation. Question 25. x3 + x2 − 17x + 15 = 0 Answer: Question 26. x3 − 2x2 − 5x + 6 = 0 Answer: Question 27. x3 − 10x2 + 19x + 30 = 0 Answer: Question 28. x3 + 4x2 − 11x − 30 = 0 Answer: Question 29. x3 − 6x2 − 7x + 60 = 0 Answer: Question 30. x3 − 16x2 + 55x + 72 = 0 Answer: Question 31. 2x3 − 3x2 − 50x − 24 = 0 Answer: Question 32. 3x3 + x2 − 38x + 24 = 0 Answer: In Exercises 33–38, find all the real zeros of the function. Question 33. f(x) = x3 − 2x2 − 23x + 60 Answer: Question 34. g(x) = x3 − 28x − 48 Answer: Question 35. h(x) = x3 + 10x2 + 31x + 30 Answer: Question 36. f(x) = x3 − 14x2 + 55x − 42 Answer: Question 37. p(x) = 2x3 −x2 − 27x + 36 Answer: Question 38. g(x) = 3x3 − 25x2 + 58x − 40 Answer: USING TOOLS In Exercises 39 and 40, use the graph to shorten the list of possible rational zeros of the function. Then find all real zeros of the function. Question 39. f(x) = 4x3 − 20x + 16 Answer: Question 40. f(x) = 4x3 − 49x − 60 Answer: In Exercises 41–46, write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros. Question 41. −2, 3, 6 Answer: Question 42. −4, −2, 5 Answer: Question 43. −2, 1 + [latex]\sqrt{7}$$
Question 44.
4, 6 − $$\sqrt{7}$$
Question 45.
−6, 0, 3 −$$\sqrt{5}$$
Question 46.
0, 5, −5 + $$\sqrt{8}$$
Question 47.
COMPARING METHODS Solve the equation x3 − 4x2 − 9x + 36 = 0 using two different methods. Which method do you prefer? Explain your reasoning.
Question 48.
REASONING Is it possible for a cubic function to have more than three real zeros? Explain.
Question 49.
PROBLEM SOLVING At a factory, molten glass is poured into molds to make paperweights. Each mold is a rectangular prism with a height 3 centimeters greater than the length of each side of its square base. Each mold holds 112 cubic centimeters of glass. What are the dimensions of the mold?
Question 50.
MATHEMATICAL CONNECTIONS The volume of the cube shown is 8 cubic centimeters.
a. Write a polynomial equation that you can use to find the value of x.
b. Identify the possible rational solutions of the equation in part (a).
c. Use synthetic division to find a rational solution of the equation. Show that no other real solutions exist.
d. What are the dimensions of the cube?
Question 51.
PROBLEM SOLVING Archaeologists discovered a huge hydraulic concrete block at the ruins of Caesarea with a volume of 945 cubic meters. The block is x meters high by 12x − 15 meters long by 12x − 21 meters wide. What are the dimensions of the block?
Question 52.
MAKING AN ARGUMENT Your friend claims that when a polynomial function has a leading coefficient of 1 and the coefficients are all integers, every possible rational zero is an integer. Is your friend correct? Explain your reasoning.
Question 53.
MODELING WITH MATHEMATICS During a 10-year period, the amount (in millions of dollars) of athletic equipment Esold domestically can be modeled by E(t) = −20t3 + 252t2 − 280t + 21,614, where t is in years.
a. Write a polynomial equation to find the year when about $24,014,000,000 of athletic equipment is sold. b. List the possible whole-number solutions of the equation in part (a). Consider the domain when making your list of possible solutions. c. Use synthetic division to find when$24,014,000,000 of athletic equipment is sold.
Question 54.
THOUGHT PROVOKING Write a third or fourth degree polynomial function that has zeros at ± $$\frac{3}{4}$$. Justify your answer.
Question 55.
MODELING WITH MATHEMATICS You are designing a marble basin that will hold a fountain for a city park. The sides and bottom of the basin should be 1 foot thick. Its outer length should be twice its outer width and outer height. What should the outer dimensions of the basin be if it is to hold 36 cubic feet of water?
Question 56.
HOW DO YOU SEE IT? Use the information in the graph to answer the questions.
a. What are the real zeros of the function f ?
b. Write an equation of the quartic function in factored form.
Question 57.
REASONING Determine the value of k for each equation so that the given x-value is a solution.
a. x3 − 6x2 − 7x + k = 0; x = 4
b. 2x3 + 7x2 − kx − 18 = 0; x = −6
c. kx3 − 35x2 + 19x + 30 = 0; x = 5
Question 58.
WRITING EQUATIONS Write a polynomial function gof least degree that has rational coefficients, a leading coefficient of 1, and the zeros −2 + $$\sqrt{7}$$ and 3 + $$\sqrt{2}$$.
In Exercises 59–62, solve f(x) = g(x) by graphing and algebraic methods.
Question 59.
f(x) = x3 + x2 − x − 1; g(x) = −x + 1
Question 60.
f(x) = x4 − 5x3 + 2x2 + 8x; g(x) = −x2 + 6x − 8
Question 61.
f(x) = x3 − 4x2 + 4x; g(x) = −2x + 4
Question 62.
f(x) = x4 + 2x3 − 11x2 − 12x + 36;
g(x) = −x2 − 6x − 9
Question 63.
MODELING WITH MATHEMATICS You are building a pair of ramps for a loading platform. The left ramp is twice as long as the right ramp. If 150 cubic feet of concrete are used to build the ramps, what are the dimensions of each ramp?
Question 64.
MODELING WITH MATHEMATICS Some ice sculptures are made by filling a mold and then freezing it. You are making an ice mold for a school dance. It is to be shaped like a pyramid with a height 1 foot greater than the length of each side of its square base. The volume of the ice sculpture is 4 cubic feet. What are the dimensions of the mold?
Question 65.
ABSTRACT REASONING Let an be the leading coefficient of a polynomial function f and a0 be the constant term. If an has r factors and a0 has s factors, what is the greatest number of possible rational zeros of f that can be generated by the Rational Zero Theorem? Explain your reasoning.
Maintaining Mathematical Proficiency
Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.(Section 4.1)
Question 66.
h(x) = −3x2 + 2x − 9 + $$\sqrt{4}$$x3
Question 67.
g(x) = 2x3 − 7x2 − 3x − 1 + x
Question 68.
f(x) = $$\frac{1}{3}$$x2 + 2x3 − 4x4 − $$\sqrt{3}$$
Question 69.
p(x) = 2x − 5x3 + 9x2 + $$\sqrt[4]{x}$$ + 1
Find the zeros of the function.(Section 3.2)
Question 70.
f(x) = 7x2 + 42
Question 71.
g(x) = 9x2 + 81
Question 72.
h(x) = 5x2 + 40
Question 73.
f(x) = 8x2 − 1
Lesson 4.6 The Fundamental Theorem of Algebra
Essential Question
How can you determine whether a polynomial equation has imaginary solutions?
EXPLORATION 1
Cubic Equations and Imaginary Solutions
Work with a partner.
Match each cubic polynomial equation with the graph of its related polynomial function. Then find all solutions. Make a conjecture about how you can use a graph or table of values to determine the number and types of solutions of a cubic polynomial equation.
a. x3 − 3x2 + x + 5 = 0
b. x3 − 2x2 − x + 2 = 0
c. x3 − x2 − 4x + 4 = 0
d. x3 + 5x2 + 8x + 6 = 0
e. x3 − 3x2 + x − 3 = 0
f. x3 − 3x2 + 2x = 0
EXPLORATION 2
Quartic Equations and Imaginary Solutions
Work with a partner.
Use the graph of the related quartic function, or a table of values, to determine whether each quartic equation has imaginary solutions. Explain your reasoning. Then find all solutions.
a. x4 − 2x3 − x2 + 2x = 0
b. x4 − 1 = 0
c. x4 + x3 − x − 1 = 0
d. x4 − 3x3 + x2 + 3x − 2 = 0
Question 3.
How can you determine whether a polynomial equation has imaginary solutions?
Question 4.
Is it possible for a cubic equation to have three imaginary solutions? Explain your reasoning.
4.6 Lesson
Monitoring Progress
Question 1.
How many solutions does the equation x4 + 7x2 − 144 = 0 have?
Question 2.
How many zeros does the function f(x) = x3 − 5x2 − 8x + 48 have? Find all zeros of the polynomial function.
Question 3.
f(x) = x3 + 7x2 + 16x + 12
Question 4.
f(x) = x5 − 3x4 + 5x3 − x2 − 6x + 4
Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.
Question 5.
−1, 4i
Question 6.
3, 1 + i$$\sqrt{5}$$
Question 7.
$$\sqrt{2}$$, 1 − 3i
Question 8.
2, 2i, 4 − $$\sqrt{6}$$
Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.
Question 9.
f(x) = x3 + 9x − 25
Question 10.
f(x) = 3x4 − 7x3 + x2 − 13x + 8
Question 11.
WHAT IF? In Example 5, what is the tachometer reading when the boat travels 20 miles per hour?
The Fundamental Theorem of Algebra 4.6 Exercises
Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE The expressions 5 + i and 5 − i are _____________.
Question 2.
WRITING How many solutions does the polynomial equation (x + 8)3(x − 1) = 0 have? Explain.
Monitoring Progress and Modeling with Mathematics
Question 3.
x4 + 2x3 − 4x2 + x = 0
Question 4.
5y3 − 3y2 + 8y = 0
Question 5.
9t6 − 14t3 + 4t − 1 = 0
Question 6.
f(z) = −7z4 + z2 − 25
Question 7.
g(s) = 4s5 − s3 + 2s7 − 2
Question 8.
h(x) = 5x4 + 7x8 − x12
In Exercises 9–16, find all zeros of the polynomial function.
Question 9.
f(x) = x4 − 6x3 + 7x2 + 6x − 8
Question 10.
f(x) = x4 + 5x3 − 7x2 − 29x + 30
Question 11.
g(x) = x4 − 9x2 − 4x + 12
Question 12.
h(x) = x3 + 5x2 − 4x − 20
Question 13.
g(x) = x4 + 4x3 + 7x2 + 16x + 12
Question 14.
h(x) = x4 − x3 + 7x2 − 9x − 18
Question 15.
g(x) = x5 + 3x4 − 4x3 − 2x2 − 12x − 16
Question 16.
f(x) = x5 − 20x3 + 20x2 − 21x + 20
ANALYZING RELATIONSHIPS In Exercises 17–20, determine the number of imaginary zeros for the function with the given degree and graph. Explain your reasoning.
Question 17.
Degree: 4
Question 18.
Degree: 5
Question 19.
Degree: 2
Question 20.
Degree: 3
In Exercises 21–28, write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.
Question 21.
−5, −1, 2
Question 22.
−2, 1, 3
Question 23.
3, 4 + i
Question 24.
2, 5 − i
Question 25.
4, −$$\sqrt{5}$$
Question 26.
3i, 2 − i
Question 27.
2, 1 + i, 2 −$$\sqrt{3}$$
Question 28.
3, 4 + 2i, 1 + $$\sqrt{7}$$
ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error in writing a polynomial function with rational coefficients and the given zero(s).
Question 29.
Zeros: 2, 1 + i
Question 30.
Zero: 2 +i
Question 31.
OPEN-ENDED Write a polynomial function of degree 6 with zeros 1, 2, and −i. Justify your answer.
Question 32.
REASONING Two zeros of f(x) = x3 − 6x2 − 16x + 96 are 4 and −4. Explain why the third zero must also be a real number.
In Exercises 33–40, determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.
Question 33.
g(x) = x4 − x2 − 6
Question 34.
g(x) = −x3 + 5x2 + 12
Question 35.
g(x) = x3 − 4x2 + 8x + 7
Question 36.
g(x) = x5 − 2x3 − x2 + 6
Question 37.
g(x) = x5 − 3x3 + 8x − 10
Question 38.
g(x) = x5 + 7x4 − 4x3 − 3x2 + 9x − 15
Question 39.
g(x) = x6 + x5 − 3x4 + x3 + 5x2 + 9x − 18
Question 40.
g(x) = x7 + 4x4 − 10x + 25
Question 41.
REASONING Which is not a possible classification of zeros for f(x) = x5 − 4x3 + 6x2 + 2x − 6? Explain.
A. three positive real zeros, two negative real zeros, and zero imaginary zeros
B. three positive real zeros, zero negative real zeros, and two imaginary zeros
C. one positive real zero, four negative real zeros, and zero imaginary zeros
D. one positive real zero, two negative real zeros, and two imaginary zeros
Question 42.
USING STRUCTURE Use Descartes’s Rule of Signs to determine which function has at least 1 positive real zero.
A. f(x) = x4 + 2x3 − 9x2 − 2x – 8
B. f(x) = x4 + 4x3 + 8x2 + 16x + 16
C. f(x) = −x4 − 5x2 − 4
D. f(x) = x4 + 4x3 + 7x2 + 12x + 12
Question 43.
MODELING WITH MATHEMATICS From 1890 to 2000, the American Indian, Eskimo, and Aleut population P (in thousands) can be modeled by the function P = 0.004t3 − 0.24t2 + 4.9t + 243, where t is the number of years since 1890. In which year did the population first reach 722,000?
Question 44.
MODELING WITH MATHEMATICS Over a period of 14 years, the number N of inland lakes infested with zebra mussels in a certain state can be modeled by
N = −0.0284t4 + 0.5937t3 − 2.464t2 + 8.33t − 2.5
where t is time (in years). In which year did the number of infested inland lakes first reach 120?
Question 45.
MODELING WITH MATHEMATICS For the 12 years that a grocery store has been open, its annual revenue R (in millions of dollars) can be modeled by the function
R = 0.0001(−t4 + 12t3 − 77t2 + 600t + 13,650)where t is the number of years since the store opened. In which year(s) was the revenue $1.5 million? Answer: Question 46. MAKING AN ARGUMENT Your friend claims that 2 − i is a complex zero of the polynomial function f(x) = x3 − 2x2 + 2x + 5i, but that its conjugate is not a zero. You claim that both 2 − i and its conjugate must be zeros by the Complex Conjugates Theorem. Who is correct? Justify your answer. Answer: Question 47. MATHEMATICAL CONNECTIONS A solid monument with the dimensions shown is to be built using 1000 cubic feet of marble. What is the value of x? Answer: Question 48. THOUGHT PROVOKING Write and graph a polynomial function of degree 5 that has all positive or negative real zeros. Label each x-intercept. Then write the function in standard form. Answer: Question 49. WRITING The graph of the constant polynomial function f(x) = 2 is a line that does not have any x-intercepts. Does the function contradict the Fundamental Theorem of Algebra? Explain. Answer: Question 50. HOW DO YOU SEE IT? The graph represents a polynomial function of degree 6. a. How many positive real zeros does the function have? negative real zeros? imaginary zeros? b. Use Descartes’s Rule of Signs and your answers in part (a) to describe the possible sign changes in the coefficients of f(x). Answer: Question 51. FINDING A PATTERN Use a graphing calculator to graph the function f(x) = (x + 3)n for n = 2, 3, 4, 5, 6, and 7. a. Compare the graphs when n is even and n is odd. b. Describe the behavior of the graph near the zero x = −3 as n increases. c. Use your results from parts (a) and (b) to describe the behavior of the graph of g(x) = (x − 4)20 near x = 4. Answer: Question 52. DRAWING CONCLUSIONS Find the zeros of each function. f(x) = x5 − 5x + 6 g(x) = x3 − 7x + 6 h(x) = x4 + 2x3 + x2 + 8x − 12 k(x) = x5 − 3x4 − 9x3 + 25x2 − 6x a. Describe the relationship between the sum of the zeros of a polynomial function and the coefficients of the polynomial function. b. Describe the relationship between the product of the zeros of a polynomial function and the coefficients of the polynomial function. Answer: Question 53. PROBLEM SOLVING You want to save money so you can buy a used car in four years. At the end of each summer, you deposit$1000 earned from summer jobs into your bank account. The table shows the value of your deposits over the four-year period. In the table, g is the growth factor 1 + r, where r is the annual interest rate expressed as a decimal.
a. Copy and complete the table.
b. Write a polynomial function that gives the value v of your account at the end of the fourth summer in terms of g.
c. You want to buy a car that costs about \$4300. What growth factor do you need to obtain this amount? What annual interest rate do you need?
Maintaining Mathematical Proficiency
Describe the transformation of f(x) = x2 represented by g. Then graph each function. (Section 2.1)
Question 54.
g(x) = −3x2
Question 55.
g(x) = (x − 4)2 + 6
Question 56.
g(x) = −(x − 1)2
Question 57.
g(x) = 5(x + 4)2
Write a function g whose graph represents the indicated transformation of the graph of f.(Sections 1.2and 2.1)
Question 58.
f(x) = x; vertical shrink by a factor of $$\frac{1}{3}$$ and a reflection in the y-axis
Question 59.
f(x) = | x + 1 ∣− 3; horizontal stretch by a factor of 9
Question 60.
f(x) = x2; reflection in the x-axis, followed by a translation 2 units right and 7 units up
Lesson 4.7 Tranformations of Polynomial Functions
Essential Question
How can you transform the graph of a polynomial function?
EXPLORATION 1
Transforming the Graph of a Cubic Function
Work with a partner.
The graph of the cubic function f(x) = x3 is shown. The graph of each cubic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
EXPLORATION 2
Transforming the Graph of a Quartic Function
Work with a partner.
The graph of the quartic function f(x) = x4 is shown. The graph of each quartic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
Question 3.
How can you transform the graph of a polynomial function?
Question 4.
Describe the transformation of f(x) = x4 represented by g(x) = (x + 1)4 + 3. Then graph g.
4.7 Lesson
Monitoring Progress
Question 1.
Describe the transformation of f(x) = x4 represented by g(x) = (x − 3)4 − 1. Then graph each function.
Question 2.
Describe the transformation of f(x) = x3 represented by g(x) = 4(x + 2)3. Then graph each function.
Question 3.
Let f(x) = x5 − 4x + 6 and g(x) = −f(x). Write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f.
Question 4.
Let the graph of g be a horizontal stretch by a factor of 2, followed by a translation 3 units to the right of the graph of f(x) = 8x3 + 3. Write a rule for g.
Question 5.
WHAT IF? In Example 5, the height of the pyramid is 6x, and the volume (in cubic feet) is represented by V(x) = 2x3. Write a rule for W. Find and interpret W(7).
Tranformations of Polynomial Functions 4.7 Exercises
Question 1.
COMPLETE THE SENTENCE The graph of f(x) = (x + 2)3 is a ____________ translation of the graph of f(x) = x3.
Question 2.
VOCABULARY Describe how the vertex form of quadratic functions is similar to the form f(x) = a(x − h)3 + k for cubic functions.
In Exercises 3–6, describe the transformation of f represented by g. Then graph each function.
Question 3.
f(x) = x4, g(x) = x4 + 3
Question 4.
f(x) = x4, g(x) = (x − 5)4
Question 5.
f(x) = x5, g(x) = (x − 2)5 − 1
Question 6.
f(x) = x6, g(x) = (x + 1)6 − 4
ANALYZING RELATIONSHIPS In Exercises 7–10, match the function with the correct transformation of the graph of f. Explain your reasoning.
Question 7.
y = f(x − 2)
Question 8.
y = f(x + 2) + 2
Question 9.
y = f(x − 2) + 2
Question 10.
y = f(x) − 2
In Exercises 11–16, describe the transformation of f represented by g. Then graph each function.
Question 11.
f(x) = x4, g(x) = −2x4
Question 12.
f(x) = x6, g(x) = −3x6
Question 13.
f(x) = x3, g(x) = 5x3 + 1
Question 14.
f(x) = x4, g(x) = $$\frac{1}{2}$$x4 + 1
Question 15.
f(x) = x5, g(x) = $$\frac{3}{4}$$(x + 4)5
Question 16.
f(x) = x4, g(x) = (2x)4− 3
In Exercises 17–20, write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f.
Question 17.
f(x) = x4 + 1, g(x) = f(x + 2)
Question 18.
f(x) = x5 − 2x + 3, g(x) = 3f(x)
Question 19.
f(x) = 2x3 − 2x2 + 6, g(x) = − $$\frac{1}{2}$$f(x)
Question 20.
f(x) = x4 + x3 − 1, g(x) = f(−x) − 5
Question 21.
ERROR ANALYSIS Describe and correct the error in graphing the function g(x) = (x + 2)4 − 6
Question 22.
ERROR ANALYSIS Describe and correct the error in describing the transformation of the graph of f(x) = x5 represented by the graph of g(x) = (3x)5− 4.
Question 23.
f(x) = x3 − 6; translation 3 units left, followed by a reflection in the y-axis
Question 24.
f(x) = x4 + 2x + 6; vertical stretch by a factor of 2, followed by a translation 4 units right
Question 25.
f(x) = x3 + 2x2 − 9; horizontal shrink by a factor of $$\frac{1}{3}$$ and a translation 2 units up, followed by a reflection in the x-axis
Question 26.
f(x) = 2x5 − x3 + x2 + 4; reflection in the y-axis and a vertical stretch by a factor of 3, followed by a translation 1 unit down
Question 27.
MODELING WITH MATHEMATICS The volume V(in cubic feet) of the pyramid is given by V(x) = x3 − 4x. The function W(x) = V(3x) gives the volume (in cubic feet) of the pyramid when x is measured in yards. Write a rule for W. Find and interpret W(5).
Question 28.
MAKING AN ARGUMENT The volume of a cube with side length x is given by V(x) = x3. Your friend claims that when you divide the volume in half, the volume decreases by a greater amount than when you divide each side length in half. Is your friend correct? Justify your answer.
Question 29.
OPEN-ENDED Describe two transformations of the graph of f(x) = x5 where the order in which the transformations are performed is important. Then describe two transformations where the order is not important. Explain your reasoning.
Question 30.
THOUGHT PROVOKING Write and graph a transformation of the graph of f(x) = x5 − 3x4 + 2x − 4 that results in a graph with a y-intercept of −2.
Question 31.
PROBLEM SOLVING A portion of the path that a hummingbird flies while feeding can be modeled by the function
f(x) = −$$\frac{1}{5}$$x(x − 4)2(x − 7), 0 ≤ x ≤ 7 w
here x is the horizontal distance (in meters) and f(x) is the height (in meters). The hummingbird feeds each time it is at ground level.
a. At what distances does the hummingbird feed?
b. A second hummingbird feeds 2 meters farther away than the first hummingbird and flies twice as high. Write a function to model the path of the second hummingbird.
Question 32.
HOW DO YOU SEE IT? Determine the real zeros of each function. Then describe the transformation of the graph of f that results in the graph of g.
Question 33.
MATHEMATICAL CONNECTIONS Write a function V for the volume (in cubic yards) of the right circular cone shown. Then write a function W that gives the volume (in cubic yards) of the cone when x is measured in feet. Find and interpret W(3).
Maintaining Mathematical Proficiency
Find the minimum value or maximum value of the function. Describe the domain and range of the function, and where the function is increasing and decreasing.(Section 2.2)
Question 34.
h(x) = (x + 5)2 − 7
Question 35.
f(x) = 4 − x2
Question 36.
f(x) = 3(x − 10)(x + 4)
Question 37.
g(x) = −(x + 2)(x + 8)
Question 38.
h(x) = $$\frac{1}{2}$$(x − 1)2 − 3
Question 39.
f(x) = −2x2 + 4x − 1
Lesson 4.8 Analyzing Graphs of Polynomial Functions
Essential Question
How many turning points can the graph of a polynomial function have?
A turning point of the graph of a polynomial function is a point on the graph at which the function changes from
• increasing to decreasing, or
• decreasing to increasing.
EXPLORATION 1
Approximating Turning Points
Work with a partner.
Match each polynomial function with its graph. Explain your reasoning. Then use a graphing calculator to approximate the coordinates of the turning points of the graph of the function. Round your answers to the nearest hundredth.
a. f(x) = 2x2 + 3x − 4
b. f(x) = x2 + 3x + 2
c. f(x) = x3 − 2x2 − x + 1
d. f(x) = −x3 + 5x − 2
e. f(x) = x4 − 3x2 + 2x − 1
f. f(x) = −2x5 − x2 + 5x + 3
Question 2.
How many turning points can the graph of a polynomial function have?
Question 3.
Is it possible to sketch the graph of a cubic polynomial function that has no turning points? Justify your answer.
Monitoring Progress
Graph the function.
Question 1.
f(x) = $$\frac{1}{2}$$(x + 1)(x − 4)
Question 2.
f(x) = $$\frac{1}{4}$$(x + 2)(x − 1)(x − 3)
Question 3.
Find all real zeros of f(x) = 18x3 + 21x2 − 13x − 6.
Question 4.
Graph f(x) = 0.5x3 + x2 − x + 2. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.
Determine whether the function is even, odd, or neither.
Question 5.
f(x) = −x2 + 5
Question 6.
f(x) = x4 − 5x3
Question 7.
f(x) = 2x5
Analyzing Graphs of Polynomial Functions 4.8 Exercises
Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE A local maximum or local minimum of a polynomial function occurs at a ______________ point of the graph of the function.
Question 2.
WRITING Explain what a local maximum of a function is and how it may be different from the maximum value of the function.
ANALYZING RELATIONSHIPS In Exercises 3–6, match the function with its graph.
Question 3.
f(x) = (x − 1)(x − 2)(x + 2)
Question 4.
h(x) = (x + 2)2(x + 1)
Question 5.
g(x) = (x + 1)(x − 1)(x + 2)
Question 6.
f(x) = (x − 1)2(x + 2)
In Exercises 7–14, graph the function.
Question 7.
f(x) = (x − 2)2(x + 1)
Question 8.
f(x) = (x + 2)2(x + 4)2
Question 9.
h(x) = (x + 1)2(x − 1)(x − 3)
Question 10.
g(x) = 4(x + 1)(x + 2)(x − 1)
Question 11.
h(x) = $$\frac{1}{3}$$(x − 5)(x + 2)(x − 3)
Question 12.
g(x) = $$\frac{1}{12}$$(x + 4)(x + 8)(x − 1)
Question 13.
h(x) = (x− 3)(x2 + x + 1)
Question 14.
f(x) = (x − 4)(2x2 − 2x + 1)
ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in using factors to graph f.
Question 15.
f(x) = (x + 2)(x − 1)2
Question 16.
f(x) = x2(x − 3)
In Exercises 17–22, find all real zeros of the function.
Question 17.
f(x) = x3 − 4x2 − x + 4
Question 18.
f(x) = x3 − 3x2 − 4x + 12
Question 19.
h(x) = 2x3 + 7x2 − 5x − 4
Question 20.
h(x) = 4x3 − 2x2 − 24x − 18
Question 21.
g(x) = 4x3 + x2 − 51x + 36
Question 22.
f(x) = 2x3 − 3x2 − 32x − 15
In Exercises 23–30, graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.
Question 23.
g(x) = 2x3 + 8x2 − 3
Question 24.
g(x) = −x4 + 3x
Question 25.
h(x) = x4 − 3x2 + x
Question 26.
f(x) = x5 − 4x3 + x2 + 2
Question 27.
f(x) = 0.5x3 − 2x + 2.5
Question 28.
f(x) = 0.7x4 − 3x3 + 5x
Question 29.
h(x) = x5 + 2x2 − 17x − 4
Question 30.
g(x) = x4 − 5x3 + 2x2 + x − 3
In Exercises 31–36, estimate the coordinates of each turning point. State whether each corresponds to a local maximum or a local minimum. Then estimate the real zeros and find the least possible degree of the function.
Question 31.
Question 32.
Question 33.
Question 34.
Question 35.
Question 36.
OPEN-ENDED In Exercises 37 and 38, sketch a graph of a polynomial function f having the given characteristics.
Question 37.
• The graph of f has x-intercepts at x = −4, x = 0, and x = 2.
• f has a local maximum value when x = 1.
• f has a local minimum value when x = −2.
Question 38.
• The graph of f has x-intercepts at x = −3, x = 1, and x = 5.
• f has a local maximum value when x = 1.
• f has a local minimum value when x = −2 and when x = 4.
In Exercises 39–46, determine whether the function is even, odd, or neither.
Question 39.
h(x) = 4x7
Question 40.
g(x) = −2x6 + x2
Question 41.
f(x) = x4 + 3x2 − 2
Question 42.
f(x) = x5 + 3x3 − x
Question 43.
g(x) = x2 + 5x + 1
Question 44.
f(x) = −x3 + 2x − 9
Question 45.
f(x) = x4 − 12x2
Question 46.
h(x) = x5 + 3x4
Question 47.
USING TOOLS When a swimmer does the breaststroke, the function
S = −241t7 + 1060t6 − 1870t5 + 1650t4 − 737t3 + 144t2 − 2.43t
models the speed S (in meters per second) of the swimmer during one complete stroke, where t is the number of seconds since the start of the stroke and 0 ≤ t ≤ 1.22. Use a graphing calculator to graph the function. At what time during the stroke is the swimmer traveling the fastest?
Question 48.
USING TOOLS During a recent period of time, the number S (in thousands) of students enrolled in public schools in a certain country can be modeled by S = 1.64x3 − 102x2 + 1710x + 36,300, where x is time (in years). Use a graphing calculator to graph the function for the interval 0 ≤ x ≤ 41. Then describe how the public school enrollment changes over this period of time.
Question 49.
WRITING Why is the adjective local, used to describe the maximums and minimums of cubic functions, sometimes not required for quadratic functions?
Question 50.
HOW DO YOU SEE IT? The graph of a polynomial function is shown.
a. Find the zeros, local maximum, and local minimum values of the function.
b. Compare the x-intercepts of the graphs of y = f(x) and y = −f(x).
c. Compare the maximum and minimum values of the functions y = f(x) and y = −f(x).
Question 51.
MAKING AN ARGUMENT Your friend claims that the product of two odd functions is an odd function. Is your friend correct? Explain your reasoning.
Question 52.
MODELING WITH MATHEMATICS You are making a rectangular box out of a 16-inch-by-20-inch piece of cardboard. The box will be formed by making the cuts shown in the diagram and folding up the sides. You want the box to have the greatest volume possible.
a. How long should you make the cuts?
b. What is the maximum volume?
c. What are the dimensions of the finished box?
Question 53.
PROBLEM SOLVING Quonset huts are temporary, all-purpose structures shaped like half-cylinders. You have 1100 square feet of material to build a quonset hut.
a. The surface area S of a quonset hut is given by S = πr2 + πrℓ. Substitute 1100 for S and then write an expression for ℓ in terms of r.
b. The volume V of a quonset hut is given by V = 1 — 2πr2ℓ. Write an equation that gives V as a function in terms of r only.c. Find the value of r that maximizes the volume of the hut.
Question 54.
THOUGHT PROVOKING Write and graph a polynomial function that has one real zero in each of the intervals −2 < x < −1, 0 < x < 1, and 4 < x < 5. Is there a maximum degree that such a polynomial function can have? Justify your answer.
Question 55.
MATHEMATICAL CONNECTIONS A cylinder is inscribed in a sphere of radius 8 inches. Write an equation for the volume of the cylinder as a function of h. Find the value of h that maximizes the volume of the inscribed cylinder. What is the maximum volume of the cylinder?
Maintaining Mathematical Proficiency
State whether the table displays linear data, quadratic data, or neither. Explain
Question 56.
Question 57.
Lesson 4.9 Modeling with Polynomial Functions
Essential Question
How can you find a polynomial model for real-life data?
EXPLORATION 1
Modeling Real-Life Data
Work with a partner.
The distance a baseball travels after it is hit depends on the angle at which it was hit and the initial speed. The table shows the distances a baseball hit at an angle of 35° travels at various initial speeds.
a. Recall that when data have equally-spaced x-values, you can analyze patterns in the differences of the y-values to determine what type of function can be used to model the data. If the first differences are constant, then the set of data fits a linear model. If the second differences are constant, then the set of data fits a quadratic model.Find the first and second differences of the data. Are the data linear or quadratic? Explain your reasoning.
b. Use a graphing calculator to draw a scatter plot of the data. Do the data appear linear or quadratic? Use the regression feature of the graphing calculator to find a linear or quadratic model that best fits the data.12019075400
c. Use the model you found in part (b) to find the distance a baseball travels when it is hit at an angle of 35° and travels at an initial speed of 120 miles per hour.
d. According to the Baseball Almanac, “Any drive over400 feet is noteworthy. A blow of 450 feet shows exceptional power, as the majority of major league players are unable to hit a ball that far. Anything in the 500-foot range is genuinely historic.” Estimate the initial speed of a baseball that travels a distance of 500 feet.
Question 2.
How can you find a polynomial model for real-life data?
Question 3.
How well does the model you found in Exploration 1(b) fit the data? Do you think the model is valid for any initial speed? Explain your reasoning.
4.9 Lesson
Monitoring Progress
write a cubic function whose graph passes through the given points.
Question 1.
(−4, 0), (0, 10), (2, 0), (5, 0)
Question 2.
(−1, 0), (0, −12), (2, 0), (3, 0)
Question 3.
Use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Use a graphing calculator to find a polynomial function that fits the data.
Question 4.
Question 5.
Modeling with polynomial Functions 4.9 Exercises
Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE When the x-values in a set of data are equally spaced, the differences of consecutive y-values are called ________________.
Question 2.
WRITING Explain how you know when a set of data could be modeled by a cubic function.
Monitoring Progress and Modeling with Mathematics
In Exercises 3–6, write a cubic function whose graph is shown.
Question 3.
Question 4.
Question 5.
Question 6.
In Exercises 7–12, use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Question 7.
Question 8.
Question 9.
(−4, −317), (−3, −37), (−2, 21), (−1, 7), (0, −1), (1, 3), (2, −47), (3, −289), (4, −933)
Question 10.
(−6, 744), (−4, 154), (−2, 4), (0, −6), (2, 16), (4, 154), (6, 684), (8, 2074), (10, 4984)
Question 11.
(−2, 968), (−1, 422), (0, 142), (1, 26), (2, −4), (3, −2), (4, 2), (5, 2), (6, 16)
Question 12.
(1, 0), (2, 6), (3, 2), (4, 6), (5, 12), (6, −10), (7, −114), (8, −378), (9, −904)
Question 13.
ERROR ANALYSIS Describe and correct the error in writing a cubic function whose graph passes through the given points.
Question 14.
MODELING WITH MATHEMATICS The dot patterns show pentagonal numbers. The number of dots in the nth pentagonal number is given by f(n) = $$\frac{1}{2}$$n(3n − 1). Show that this function has constant second-order differences.
Question 15.
OPEN-ENDED Write three different cubic functions that pass through the points (3, 0), (4, 0), and (2, 6). Justify your answers.
Question 16.
MODELING WITH MATHEMATICS The table shows the ages of cats and their corresponding ages in human years. Find a polynomial model for the data for the first 8 years of a cat’s life. Use the model to estimate the age (in human years) of a cat that is 3 years old.
Question 17.
MODELING WITH MATHEMATICS The data in the table show the average speeds y (in miles per hour) of a pontoon boat for several different engine speeds x (in hundreds of revolutions per minute, or RPMs). Find a polynomial model for the data. Estimate the average speed of the pontoon boat when the engine speed is 2800 RPMs.
Question 18.
HOW DO YOU SEE IT? The graph shows typical speeds y (in feet per second) of a space shuttle x seconds after it is launched.
a. What type of polynomial function models the data? Explain.
b. Which nth-order finite difference should be constant for the function in part (a)? Explain.
Question 19.
MATHEMATICAL CONNECTIONS The table shows the number of diagonals for polygons with n sides. Find a polynomial function that fits the data. Determine the total number of diagonals in the decagon shown.
Question 20.
MAKING AN ARGUMENT Your friend states that it is not possible to determine the degree of a function given the first-order differences. Is your friend correct? Explain your reasoning.
Question 21.
WRITING Explain why you cannot always use finite differences to find a model for real-life data sets.
Question 22.
THOUGHT PROVOKING A, B, and C are zeros of a cubic polynomial function. Choose values for A, B, and C such that the distance from A to B is less than or equal to the distance from A to C. Then write the function using the A, B, and C values you chose.
Question 23.
MULTIPLE REPRESENTATIONS Order the polynomial functions according to their degree, from least to greatest.
A.f(x) = −3x + 2x2 + 1
Question 24.
ABSTRACT REASONING Substitute the expressions z, z + 1, z + 2, …….. , z + 5 for x in the function f(x) = ax3 + bx2 + cx + d to generate six equally-spaced ordered pairs. Then show that the third-order differences are constant.
Maintaining Mathematical Proficiency
Solve the equation using square roots.(Section 3.1)
Question 25.
x2 − 6 = 30
Question 26.
5x2 − 38 = 187
Question 27.
2(x − 3)2 = 24
Question 28.
$$\frac{4}{3}$$(x + 5)2 = 4
Solve the equation using the Quadratic Formula.(Section 3.4)
Question 29.
2x2 + 3x = 5
Question 30.
2x2 + $$\frac{1}{2}$$ = 2x
Question 31.
2x2 + 3x =−3x2 + 1
Question 32.
4x − 20 = x2
Polynomial Functions Performance Task: For the Birds-Wildlife Management
Core Vocabulary
Core Concepts
Section 4.5
Section 4.6
Section 4.7
Section 4.8
Section 4.9
Mathematical Practices
Question 1.
Explain how understanding the Complex Conjugates Theorem allows you to construct your argument in Exercise 46 on page 203.
Question 2.
Describe how you use structure to accurately match each graph with its transformation in Exercises 7–10 on page 209.
For the Birds -Wildlife Management
How does the presence of humans affect the population of sparrows in a park? Do more humans mean fewer sparrows? Or does the presence of humans increase the number of sparrows up to a point? Are there a minimum number of sparrows that can be found in a park, regardless of how many humans there are? What can a mathematical model tell you?
To explore the answers to these questions and more, go to BigIdeasMath.com.
Polynomial Functions Chapter Review
Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.
Question 1.
h(x) = −x3 + 2x2 − 15x7
Question 2.
p(x) = x3 − 5x0.5 + 13x2 + 8
Graph the polynomial function.
Question 3.
h(x) = x2 + 6x5 − 5
Question 4.
f(x) = 3x4 − 5x2 + 1
Question 5.
g(x) = −x4 + x + 2
Find the sum or difference.
Question 6.
(4x3 − 12x2 − 5) − (−8x2 + 4x + 3)
Question 7.
(x4 + 3x3 − x2 + 6) + (2x4 − 3x + 9)
Question 8.
(3x2 + 9x + 13) − (x2 − 2x + 12)
Find the product.
Question 9.
(2y2 + 4y − 7)(y + 3)
Question 10.
(2m + n)3
Question 11.
(s + 2)(s + 4)(s − 3)
Use Pascal’s Triangle to expand the binomial.
Question 12.
(m + 4)4
Question 13.
(3s + 2)5
Question 14.
(z + 1)6
Divide using polynomial long division or synthetic division.
Question 15.
(x3 + x2 + 3x − 4) ÷ (x2 + 2x + 1)
Question 16.
(x4 + 3x3 − 4x2 + 5x + 3) ÷ (x2 + x + 4)
Question 17.
(x4 − x2 − 7) ÷ (x + 4)
Question 18.
Use synthetic division to evaluate g(x) = 4x3 + 2x2 − 4 when x = 5.
Factor the polynomial completely.
Question 19.
64x3 − 8
Question 20.
2z5 − 12z3 + 10z
Question 21.
2a3 − 7a2 − 8a + 28
Question 22.
Show that x + 2 is a factor of f(x) = x4 + 2x3 − 27x − 54. Then factor f(x) completely.
Find all real solutions of the equation.
Question 23.
x3 + 3x2 − 10x − 24 = 0
Question 24.
x3 + 5x2 − 2x − 24 = 0
Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.
Question 25.
1, 2 − $$\sqrt{3}$$
Question 26.
2, 3, $$\sqrt{5}$$
Question 27.
−2, 5, 3 + $$\sqrt{6}$$
Question 28.
You use 240 cubic inches of clay to make a sculpture shaped as a rectangular prism. The width is 4 inches less than the length and the height is 2 inches more than three times the length. What are the dimensions of the sculpture? Justify your answer.
Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.
Question 29.
3, 1 + 2i
Question 30.
−1, 2, 4i
Question 31.
−5, −4, 1 −i$$\sqrt{3}$$
Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.
Question 32.
f(x) = x4 − 10x + 8
Question 33.
f(x) = −6x4 − x3 + 3x2 + 2x + 18
Describe the transformation of f represented by g. Then graph each function.
Question 34.
f(x) = x3, g(x) = (−x)3 + 2
Question 35.
f(x) = x4, g(x) = −(x + 9)4
Write a rule for g.
Question 36.
Let the graph of g be a horizontal stretch by a factor of 4, followed by a translation 3 units right and 5 units down of the graph of f(x) = x5 + 3x.
Question 37.
Let the graph of g be a translation 5 units up, followed by a reflection in the y-axis of the graph of f(x) = x4 − 2x3 − 12.
Graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.
Question 38.
f(x) = −2x3 − 3x2 − 1
Question 39.
f(x) = x4 + 3x3 − x2 − 8x + 2
Determine whether the function is even, odd, or neither.
Question 40.
f(x) = 2x3 + 3x
Question 41.
g(x) = 3x2 − 7
Question 42.
h(x) = x6 + 3x5
Question 43.
Write a cubic function whose graph passes through the points (−4, 0), (4, 0), (0, 6), and (2, 0).
Question 44.
Use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Polynomial Functions Chapter Test
Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.
Question 1.
3, 1 − $$\sqrt{2}$$
Question 2.
−2, 4, 3i
Find the product or quotient.
Question 3.
(x6 − 4)(x2 − 7x + 5)
Question 4.
(3x4 − 2x3 − x − 1) ÷ (x2 − 2x + 1)
Question 5.
(2x3 − 3x2 + 5x − 1) ÷ (x + 2)
Question 6.
(2x + 3)3
Question 7.
The graphs of f(x) = x4 and g(x) = (x − 3)4 are shown.
a. How many zeros does each function have? Explain.
b. Describe the transformation of f represented by g.
c. Determine the intervals for which the function g is increasing or decreasing.
Question 8.
The volume V (in cubic feet) of an aquarium is modeled by the polynomial function V(x) = x3 + 2x2 − 13x + 10, where x is the length of the tank.
a. Explain how you know x = 4 is not a possible rational zero.
b. Show that x − 1 is a factor of V(x). Then factor V(x) completely.
c. Find the dimensions of the aquarium shown.
Question 9.
One special product pattern is (a − b)2 = a2 − 2ab + b2. Using Pascal’s Triangle to expand (a − b)2 gives 1a2 + 2a(−b) + 1(−b)2. Are the two expressions equivalent? Explain.
Question 10.
Can you use the synthetic division procedure that you learned in this chapter to divide any two polynomials? Explain.
Question 11.
Let T be the number (in thousands) of new truck sales. Let C be the number (in thousands) of new car sales. During a 10-year period, T and C can be modeled by the following equations where t is time (in years).
T = 23t4 − 330t3 + 3500t2 − 7500t + 9000
C = 14t4 − 330t3 + 2400t2 − 5900t + 8900
a. Find a new model S for the total number of new vehicle sales.
b. Is the function S even, odd, or neither? Explain your reasoning.
Question 12.
Your friend has started a golf caddy business. The table shows the profits p (in dollars) of the business in the first 5 months. Use finite differences to find a polynomial model for the data. Then use the model to predict the profit after 7 months.
Polynomial Functions Cumulative Assessment
Question 1.
The synthetic division below represents f(x) ÷ (x− 3). Choose a value for m so that x − 3 is a factor of f(x). Justify your answer.
Question 2.
Analyze the graph of the polynomial function to determine the sign of the leading coefficient, the degree of the function, and the number of real zeros. Explain.
Question 3.
Which statement about the graph of the equation 12(x− 6) = −( y + 4)2 is not true?
A. The vertex is (6, −4).
B. The axis of symmetry is y = −4.
C. The focus is (3, −4).
D. The graph represents a function.
Question 4.
A parabola passes through the point shown in the graph. The equation of the axis of symmetry is x = −a. Which of the given points could lie on the parabola? If the axis of symmetry was x = a, then which points could lie on the parabola? Explain your reasoning.
Question 5.
Select values for the function to model each transformation of the graph of f(x) = x.
a. The graph is a translation 2 units up and 3 units left.
b. The graph is a translation 2 units right and 3 units down.
c. The graph is a vertical stretch by a factor of 2, followed by a translation 2 units up.
d. The graph is a translation 3 units right and a vertical shrink by a factor of $$\frac{1}{2}$$, followed by a translation 4 units down.
Question 6.
The diagram shows a circle inscribed in a square. The area of the shaded region is21.5 square meters. To the nearest tenth of a meter, how long is each side of the square?
A. 4.6 meters
B. 8.7 meters
C. 9.7 meters
D. 10.0 meters
Question 7.
a. f(x) = 3x5
b. f(x) = 4x3 + 8x
c. f(x) = 3x5 + 12x2+ 1
d. f(x) = 2x4
e. f(x) = x11 − x7
f. f(x) = 2x8 + 4x4 + x2 − 5
Question 8.
The volume of the rectangular prism shown is given by V = 2x3 + 7x2 − 18x − 63. Which polynomial represents the area of the base of the prism?
A. 2x2 + x − 21
B. 2x2 + 21 − x
C. 13x + 21 + 2x2
D. 2x2 − 21 − 13x
Question 9.
The number R (in tens of thousands) of retirees receiving Social Security benefits is represented by the function R = 0.286t3 − 4.68t2 + 8.8t + 403, 0 ≤ t ≤ 10
where t represents the number of years since 2000. Identify any turning points on the given interval. What does a turning point represent in this situation?
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2021-06-22 01:04:32
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https://compsciedu.com/mcq-questions/UNIX-System/NET-computer-science-question-paper
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1. In UNIX, which of the following command is used to set the task priority ? a. init b. nice c. kill d. PS
2. Which of the following flags are set when ‘JMP’ instruction is executed ? a. SF and CF b. AF and CF c. All flags d. No flag is set
3. Everything below the system call interface and above the physical hardware is known as ______. a. Kernel b. Bus c. Shell d. Stub
4. Linux operating system uses a. Affinity Scheduling b. Fair Preemptive Scheduling c. Hand Shaking d. Highest Penalty Ratio Next
5. In Unix, how do you check that two given strings a and b are equal? a. test $a -eq$b b. test $a -equal$b c. test $a =$b d. both a and c
6. The directory structure used in Unix file system is called a. Hierarchical directory b. Tree structured directory c. Directed acyclic graph d. Graph structured directory
7. Which statement is not true about process 0 in the Unix operating system? a. Process 0 is called init process b. Process 0 is not created by fork system call c. After forking process 1, process O becomes swapper process d. Process 0 is a special process created when system boots
Answer: (a).Process 0 is called init process
8. Which or the following commands would return process_id of sleep command? a. Sleep 1 and echo $? b. Sleep 1 and echo # c. Sleep 1 and echo$* d. Sleep 1 and echo \$!
Answer: (b).Sleep 1 and echo #
9. What does the following command do ?grep − vn "abc" x a. It will print all of the lines in the file x that match the search string "abc" b. It will print all of the lines in file x that do not match the search string "abc" c. It will print total no of the lines in file x that match the search string "abc" d. It will print the specific line numbers of file x in which there is a match for the string "abc"
Answer: (a).It will print all of the lines in the file x that match the search string "abc"
10. The Unix Kernel maintains two key data structures related to processes, the progress table and the user structure. Which of following information is not the part of user a. File descriptor table b. System call state c. Scheduling parameters d. Kernel stack
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2022-01-28 07:47:23
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https://www.gradesaver.com/textbooks/engineering/computer-science/invitation-to-computer-science/chapter-3-exercises-page-143/18c
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## Invitation to Computer Science 8th Edition
We see that all possible lists are generated anyway, in the first loop, so we have the best possible case when the first list permutation generated is sorted. Thus, this gives 1 work unit for the second loop, giving us $$n! +1$$ work units in the best case. Similarly, in the worst case, the last list is sorted, giving us: $$n!+n! =2n!$$ work units.
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2022-06-27 11:44:25
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https://tex.stackexchange.com/questions/618703/are-there-any-xetex-compilers-that-skip-recompiling-unchanged-input-files
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# Are there any Xetex compilers that skip recompiling unchanged \input{ } files?
TexStudio is great, but I notice that the compile time is always the same (~25 seconds). Logically, a compiler should skip recompiling \input files if they have not been edited. Are there any such compilers?
• Xelatex, actually. Oct 12, 2021 at 14:13
• You can just update your question. But IMO it does not really make any sense. how exactly is the compiler suppose to skip stuff if the changes you made in file A causes the page numbers to change for all subsequent pages. That is the page content of an included file can change even if the included file it self does not. Oct 12, 2021 at 14:15
• If you just want a quicker version while drating then use \include and \includeonly to specify the file you are working on, but then only that file appears in the draft pdf. Oct 12, 2021 at 15:02
All xetex implementations will work the same way tex has very few system dependencies so that the same source on the same system will as far as possible produce the same result (xetex accessing system fonts one exception here)
There is no reason to assume that an an unchanged input file will produce unchanged output.
if you have \input{section2.tex} and you have not edited that file recently the resulting page breaking will be different depending on any text earlier in the document changing the point where this is included. Any command within the file may have a different definition, if the file starts \section{Something} but you have loaded a package to make section headings sans serif and blue, then the resulting heading will be sans serif and blue even if that file has not changed.
Basically TeX is not a compiler, it's an interpreted macro expansion language, there is no pre-compilation stage at all; and the entire language is massively context dependent (why we always ask for complete test documents to go with questions). A fragment taken out of context might do anything depending on local definitions.
• Ok. I assumed that there was a stage in which the *.tex file was pre-compiled before the final output. If I can slip in a related question - The interpreter is single pass, right? So, to handle forward references, like, "for reference, look at equation \ref{5.12} in the next chapter", you would run the single-pass interpreter twice. Is that right? ... A test run seems to make that affirmative. Oct 12, 2021 at 23:43
• @ScotParker tex is designed for the memory archtecture of a machine in 1979: it never holds the whole document in memory it only reads as far in to the file as necessary to typeset the next letter and then interleaves file input, tokenization execution and output of the partially written PDF. Oct 13, 2021 at 7:19
As David says, TeX must process all of the included code, every time. This is the case in any interpreted language that does not compile to an intermediate format.
It's also a special feature of a macro-expansion language, where every letter is really a command, and all the commands can continuously be redefined. You could define a whole set of macros in one file and then load a second one that redefined all of them differently; or you can even redefine the syntax of TeX itself, change the control characters, and introduce all sorts of unpredictable behavior.
If an interpreter compiles first to bytecode, then yes, it could do this intermediate compilation on portions as long as they are self-contained. In compiled languages, you can of course compile modules to object code before compiling the main program and linking them all together. In most cases, though, I think the portions need to be self-contained.
The closest TeX comes to using pre-compiled modules is if you compile portions to PDF and then include the PDFs (or likewise for DVI). Many of us do this for graphics or even tables, and this works because in the end the main LaTeX file is just calling \includegraphics, which means making a box of a certain size and putting an image in there. If the box is a fixed size--that is, if you specify all the dimensions to \includegraphics, then changing the input file won't change the layout of the final document. Likewise if you add the draft option to your \documentclass then TeX will just draw a box the size of the picture but not actually include it.
The other possibility in this category is to compile a whole set of macro definitions into a format:
The latexmk utility does keep track of when files are changed to minimize recompilation where it can, and it does have a mode where it continually recompiles the document as you are editing it.
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2022-06-27 08:10:21
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https://stats.stackexchange.com/questions/449567/arima0-1-1-process-cant-be-stationary
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ARIMA(0, 1, 1) process cant be stationary?
According to "Time Series Analysis, Forecasting and Control" (Box, Jenkins), for any process represented by a linear filter $$z_t = a_t + \psi_1 a_{t-1} + \psi_2 a_{t-2} ... = a_t + \sum_{J=1}^\infty \psi_j a_{t-j}$$ be a valid stationary process, it is necessary for the coefficients $$\psi_j$$ to be absolutely summable: $$\sum_{J=1}^\infty |\psi_j| < \infty$$.
For an ARIMA(0, 1, 1) process, $$(1-B)Y_t = (1-\theta_1 B)a_t\\ Y_t = \frac{(1-\theta_1B)}{(1-B)}a_t = [1 + (1-\theta_1)B + (1-\theta_1)B^2 + (1-\theta_1)B^3 ...] \cdot a_t.$$
Thus $$\psi_j = (1-\theta_1)$$ for all $$j$$.
Since, I think, $$\sum_{j=1} ^\infty |\psi_j|= \infty$$, doesn't this mean that an ARIMA(0, 1, 1) process can't be stationary?
You cannot manipulate the lag operator $$B$$ as you did when there is unit root. Box and Jenkins should tell you that. $$\theta_1$$ should be $$\theta_1 B$$, in your notation.
(The heuristic $$\frac{1}{1-B} = \sum_{h \geq 0} B^h"$$ that your heuristic calculation is based on is incorrect. The precise reason---see, e.g. Box and Jenkins---is that, for a complex polynomial $$f(z)$$, $$\frac{1}{f(z)}$$ admit a power series representation $$\frac{1}{f(z)} = \sum_{h \geq 0} \psi_h z^h$$ on an open neighborhood in the complex plane if and only if the roots of $$f(z)$$ lie strictly outside the unit circle.)
• Thanks for catching the $B$ error. But I left $\theta_1$ arbitrary, so in the above situation, even if there is no unit root to the characteristic polynomial, wouldn't the infinite sum of $\psi_i$ still go to infinity for any $\theta_1$? – Frank Feb 15 at 6:33
• Your algebra is still incorrect, Frank: every "$1-\theta_1$" in your final expression for $Y_t$ should be $1-\theta_1B.$ You will need to collect like powers of $B$ in order to determine the $\psi_j$ correctly. – whuber Feb 15 at 15:31
• @whuber Not sure what you mean : $\frac{1-\theta B}{1-B} = 1 + \frac{B-\theta B}{1-B} = 1 + \frac{(1-\theta)B}{1-B} = 1 + (1-\theta) B + \frac{(1-\theta)B - [(1-\theta)B (1-B)]}{1-B} = 1 + (1-\theta) B + \frac{(1-\theta)B^2}{1-B}...$ – Frank Feb 15 at 18:10
• $$\frac{1-\theta B}{1-B} = (1-\theta B)(1+B+B^2+\cdots+B^n+\cdots) = 1+(1-\theta)B+(1-\theta)B^2+\cdots$$ implying $\psi_j=1-\theta$ for $j\ge1,$ whence $\sum_{j\ge 1}|\psi_j| = 1/\theta$ provided $0\lt \theta \lt 2.$ – whuber Feb 15 at 19:22
• @whuber, Thank you for the reply. Take $\theta = 1/2$, then $\sum_{j \geq 1} |1-\frac{1}{2}| = \sum_{j \geq 1} \frac{1}{2}$. You're saying the infinite sum of $\frac{1}{2} = \frac{1}{1/2} = 2$ ? For any $n$, $\sum_1^n \frac{1}{2} = \frac{n}{2}$. You're saying this converges to $2$? – Frank Feb 15 at 22:23
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2020-07-10 04:34:23
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https://cs.nyu.edu/pipermail/fom/2003-September/007249.html
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[FOM] Platonism and Formalism
Mon Sep 22 14:50:34 EDT 2003
Karlis Podnieks wrote:
>
> ----- Original Message -----
> From: "Torkel Franzen" <torkel at sm.luth.se>
> To: <fom at cs.nyu.edu>
> Cc: <torkel at sm.luth.se>
> Sent: Saturday, September 06, 2003 8:35 AM
> Subject: Re: [FOM] Platonism and Formalism (a reply to Podnieks)
>
> > ...
> >We must examine how people actually work, how they argue, how
> > they apply mathematics, what kind of considerations seem to guide
> > their thinking in practice.
> > ...
> > A consistently non-Platonistic point of view with regard
> > to mathematics is much like a consistently skeptical view of human
> > knowledge: it has a certain appeal to the intellect, but it has no
> > apparent relation to how people, including professed non-Platonists or
I have already replied to this.
> >
> > ---
> > Torkel Franzen
>
> I tried to explain this "practical" aspect of the mathematical Platonism in
> my old paper "Platonism, Intuition, and the Nature of Mathematics"
> (http://www.ltn.lv/~podnieks/gt1.html):
>
> FOR HUMANS, Platonist thinking is the best way of working with stable
> self-contained systems (the "true" subject of mathematics - at least, for
> me). Thus, a correct philosophical position of a mathematician should be:
> a) Platonism - on working days - when I'm doing mathematics (otherwise, my
> "doing" will be inefficient),
> b) Advanced Formalism - on weekends - when I'm thinking "about" mathematics
> (otherwise, I will end up in mysticism).
> (Of course, the initial version of this aphorism is due to Reuben Hersh).
>
> For me, such a position is neither schizophrenia, nor perversity - it is
> determined by the very nature of mathematics.
I agree with Karlis Podnieks, to a certain degree, because I
feel WHAT is behind of his views. I always sympathized with his
writings on f.o.m. because of this feeling. However, I would
present things somewhat differently.
I do not think that Karlis intends to say that Advanced Formalism
on weekends consists only of formalisms without any intuition
on which they are based.
What Karlis calls "Platonism" on working days is just a kind
of intuition related with the classical logic. (The proper
Platonism is, of course, a wrong idea.) Given a classical
first-order theory, we just imagine a world of abstract objects
with operations and relations over them as if it exists. We also
imagine that each sentence of this theory as interpreted in
this world and as having a "truth" value. I use "" because
everything is imagined. This is not a real world where some
sentences may be true or false. Also our intuition or imagination
is, by the very nature of these concepts, vague and unstable.
Thus, there is no good philosophical reason to think that any
such imaginary world is rigidly fixed for all mathematicians
and independent on the theory considered. Saying the converse
means essentially a wrong understanding of what is the intuition
in general.
Then, I think, there is no essential difference between working
days and weekends, except that on weekends we are more relaxed
and reflecting on what happened on working days. Both intuition
and formalisms are under our consideration permanently.
On working days it is quite normal for any mathematician to ask
himself: I have an IDEA; how to FORMALIZE it? It does not matter
that mathematicians usually do not explicitly work with formal
systems as logicians describe them. They say "FORMALIZE", and
it is clear that this is really about a formalization. We strictly
distinguish between INFORMAL (intuitive) and FORMAL (rigorous)
reasoning. Both are used almost simultaneously or interactively
on our working days.
Moreover, on working days, any new, deep, revolutionary mathematical
concept can hardly arise without some reflection on the nature
of mathematics, how any ideas could be formalized and what
does it mean to formalize.
The kind of intuition described above is related with classical logic.
But we know that there is also intuitionistic mathematics based on
a different intuition. The fact that intuitionistic logic is
formally reducible to classical, say, via Kripke models plays here
not so essential role. Even when working classically with, say,
Heyting algebras we need to use some additional non-classical intuition.
I recall a paper by Dana Scott where he advised something like this:
Do not listen to those who advise you that if anything is reducible
to classical logic then the corresponding intuition is reduced too.
I hope that I formulated the main point correctly.
Thus, in general, there is no need to mention Platonism, truth or
anything more or less analogous. We rather should use most general
terms like idea, intuition or imagination because we cannot predict
what particular kind of formalisms and intuitions can be considered in
mathematics. In principle, some formalisms and intuitions could even
be not reducible to classical logic.
As an illustration, consider the following axioms on the "set" F
(which is not a set in the traditional set theory) of Feasible Numbers:
(1) 0,1 \in F,
(2) F + F \subseteq F (in particular, F has no last number), and
(3) \forall n \in F (log_2 log_2 n < 10).
The first two axiom say that 0 and 1 are feasible, and feasible
numbers are closed under the addition operations. Some evident
axioms like 0 < n + 1, n < n + 1, transitivity of <, etc. are
omitted for the brevity. The axiom (3) is true in the sense that
it can be experimentally confirmed for all n physically(!!)
written in the unary notation |||...|| or 0+1+...+1. It says
quite reasonably that all feasible numbers, although having no
last number, are less than 2^{1024}. That is why it is interesting
and intriguing to take it as an axiom.
In the framework of the classical logic, we can derive in this theory
a contradiction by a feasible(!) proof (not so trivially, although
also not so difficult for those who is acquainted with intractability
of cut elimination). Then, the problem is to find a natural
(feasibly) consistent formalization, i.e., actually, a logical system
for the above axioms. Some restricted version of classical logic
does work. But during consideration of the resulting formal theory
together with underlying intuition (here omitted) it becomes clear
that both are irreducible to the classical ones. For example,
adding one more operation 2*n and the axiom 2*n = n + n makes
this theory feasibly contradictory even in this logic.
Note, that somewhat analogous considerations on feasible numbers
in the framework of classical logic have been done by Pohit Parikh.
The idea of feasible consistency seems was first explicitly
explored by him for a coherent formalization of an interesting
intuition. I omit here any further comparisons.
> And this is why I would ask
> FOMers having bigger mathematical intuitions than my own to try a high level
> ("Corfield style") explanation of the last remaining Platonist illusion:
>
> PROBLEM
> Which properties of structures and methods used in mathematics and
> metamathematics are leading to the illusion that the natural number system
> is
> a stable and unique mathematical structure that exists independently of any
> axioms and cannot be defined by using axioms?
It is really one of the most crucial questions on the foundations
and philosophy of mathematics.
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2021-08-03 17:19:08
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https://socratic.org/questions/how-do-you-simplify-n-5n
|
# How do you simplify n+5n?
Mar 19, 2018
The answer is $6 n$.
#### Explanation:
$n + 5 n$
All you have to do is combine the terms, since they are like terms.
There is actually a $1$ in front of the first term. But since it's a 1, we don't usually write it.
$1 n + 5 n$
Just add $1 + 5$ and then put $n$.
$1 n + 5 n = 6 n$
So, $n + 5 n = 6 n$.
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2019-09-19 14:37:13
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https://www.ias.ac.in/listing/bibliography/boms/PROBODH_K_KUIRI
|
• PROBODH K KUIRI
Articles written in Bulletin of Materials Science
• Large spectral shift of plasmon resonances in Au–Cu alloy nanoparticles through anisotropy and interaction
Nowadays, alloy nanoparticles (NPs) consisting of noble metals find much importance due to their enhanced localized surface plasmon resonance (SPR) in the visible range. Besides the compositional changes, the SPR of alloy NPsindependently can be tailored by changing the shape and interparticle separation. Here, we report the effects of shape and interaction parameters ($\beta$ and $K$, respectively) on the plasmonic properties of Au–Cu alloy NPs considering Au concentrations in the range of 0.0–1.0 using a modified effective medium theory. There is always a single SPR for the alloys’ NPs. A large shift of this SPR peak from visible to infrared regions is seen with decrease or increase in the values of $\beta$ or $K$. A linear variation of peak shift with $1/\beta$ and an exponential decay variation of peak shift with $1/K$ are observed.Although there is a small increase in the values of slopes (in the peak shift vs. $1/\beta$ curves) with increase in concentration of Au in the Au–Cu alloys, there is almost no change in the fitting parameter, $\tau$ of the exponential decay function (in the curves for peak shift vs. $1/K$) for different sizes of NPs and concentration of Au or Cu in the alloys. This establishes the universal nature of the exponential decay behaviour. These observations are consistent with the data for elemental (Au and Cu) NPs. The correlations between (i) shape parameter, $\beta$ and the aspect ratio, AR and (ii) interaction parameter, $K$ with the interparticle separation, $s$ are established as in case of elemental NPs. The study thus, verifies the model further and establishes universality of the applicability of the model.
• # Bulletin of Materials Science
Volume 45, 2022
All articles
Continuous Article Publishing mode
• # Dr Shanti Swarup Bhatnagar for Science and Technology
Posted on October 12, 2020
Prof. Subi Jacob George — Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur, Bengaluru
Chemical Sciences 2020
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2022-01-26 10:52:26
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http://gmatclub.com/forum/a-basic-savings-account-pays-interest-once-per-year-on-decem-158789.html?sort_by_oldest=true
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# A basic savings account pays interest once per year on Decem
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A basic savings account pays interest once per year on Decem [#permalink]
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28 Aug 2013, 00:57
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A basic savings account pays interest once per year on December 31. If Alan deposits $300 into the account on January 1, 2004, and does not deposit or withdraw any money in the meantime, then to the nearest cent, how much was in the account when he withdrew the money on January 1, 2009? (1) The savings account interest rate is 2% per year (2) If Alan had left the money in the account for 2 more years, he would have had, to the nearest cent,$13.12 more.
[Reveal] Spoiler:
For statement (2), the book explains the equation as $300(1+i)6 -$300(1+i)4= 13.12.
But, I think n should be 7 and 5 and be squared, not multiplied: $300(1+i)^7 -$300(1+i)^5= 13.12
by the way, do you guys see many errors in the Princeton Review?? Because I have found many....;;
[Reveal] Spoiler: OA
Last edited by Bunuel on 28 Aug 2013, 01:36, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A basic savings account pays interest once per year on Decem [#permalink]
### Show Tags
28 Aug 2013, 04:26
1
KUDOS
sehosayho wrote:
A basic savings account pays interest once per year on December 31. If Alan deposits $300 into the account on January 1, 2004, and does not deposit or withdraw any money in the meantime, then to the nearest cent, how much was in the account when he withdrew the money on January 1, 2009? (1) The savings account interest rate is 2% per year (2) If Alan had left the money in the account for 2 more years, he would have had, to the nearest cent,$13.12 more.
[Reveal] Spoiler:
For statement (2), the book explains the equation as $300(1+i)6 -$300(1+i)4= 13.12.
But, I think n should be 7 and 5 and be squared, not multiplied: $300(1+i)^7 -$300(1+i)^5= 13.12
by the way, do you guys see many errors in the Princeton Review?? Because I have found many....;;
seho,
2 different formulas for simple interest & compound interest. Your formula is for compound interest and not for simple interest.
Unless mentioned that amount is compounded, we assume its simple interest.
Formula for simple interest is I = PTR/100 where P is principal amount ($300 in this case) T = Time in years (4 yrs) , R is rate of interest. _________________ Maadhu MGMAT1 - 540 ( Trying to improve ) Intern Joined: 26 Aug 2013 Posts: 11 Followers: 0 Kudos [?]: 2 [0], given: 19 Re: A basic savings account pays interest once per year on Decem [#permalink] ### Show Tags 28 Aug 2013, 20:35 Thank you, but I still dont get it why the time in years is 4? Since Alan deposit it on Jan, 2004 and the interest paid every Dec 31, Alan gets interest on Dec ,31 2004, 2005, 2006, 2007, and 2008. right? So, the time in years should be 5 I think. If im wrong, what am i missing? seho, 2 different formulas for simple interest & compound interest. Your formula is for compound interest and not for simple interest. Unless mentioned that amount is compounded, we assume its simple interest. Formula for simple interest is I = PTR/100 where P is principal amount ($300 in this case) T = Time in years (4 yrs) , R is rate of interest.[/quote]
Math Expert
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Re: A basic savings account pays interest once per year on Decem [#permalink]
### Show Tags
29 Aug 2013, 04:52
1
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Expert's post
sehosayho wrote:
A basic savings account pays interest once per year on December 31. If Alan deposits $300 into the account on January 1, 2004, and does not deposit or withdraw any money in the meantime, then to the nearest cent, how much was in the account when he withdrew the money on January 1, 2009? (1) The savings account interest rate is 2% per year (2) If Alan had left the money in the account for 2 more years, he would have had, to the nearest cent,$13.12 more.
For statement (2), the book explains the equation as $300(1+i)6 -$300(1+i)4= 13.12.
But, I think n should be 7 and 5 and be squared, not multiplied: $300(1+i)^7 -$300(1+i)^5= 13.12
by the way, do you guys see many errors in the Princeton Review?? Because I have found many....;;
The second statement must give the same interest of 2% as given in the first statement. Taking this into account the book also means compound interest because only from 300(1+x)^6 - 300(1+x)^4= 13.12 you get x=0.02. But I agree it should be 7 and 5, not 6 and 4. Overall not a good question.
_________________
Re: A basic savings account pays interest once per year on Decem [#permalink] 29 Aug 2013, 04:52
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2016-05-27 02:28:16
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https://www.electro-tech-online.com/threads/subtract-from-00.99369/
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# Subtract from 00 !
Status
Not open for further replies.
#### Suraj143
##### Active Member
I have a small question. After this code what is the value in W?
Thanks
Code:
movlw .10
movwf Data
clrf Reference
movf Data,W
subwf Reference,W
????
#### Pommie
##### Well-Known Member
It will subtract W from the file and so will do 0-10 = -10 or 246 or 0xf6.
Mike.
Last edited:
#### Suraj143
##### Active Member
It will subtract W from the file and so will to 0-10 = -10 or 246 or 0xf6.
Mike.
Totally understood.Thanks mike
#### Mike - K8LH
##### Well-Known Member
Suraj,
I think you'll find that taking the time to learn how to use the MPLAB Simulator and "watch" those file values while single stepping through your code might be a good investment. It certainly was for me...
Kind regards, Mike
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2022-01-23 02:36:14
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https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-8-test-page-621/3
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# Chapter 8 - Test - Page 621: 3
$$4\sqrt 3$$
#### Work Step by Step
We can use the product rule for square roots to rewrite the radicand as the product of a perfect square factor and another factor: $$\sqrt 48 = \sqrt (16 \times 3)$$ Because $16$ is a perfect square factor, we can take the square root of $16$, which is $4$: $$4\sqrt 3$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2019-10-16 20:42:47
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http://sigma-journal.com/2014/072/
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### Symmetry, Integrability and Geometry: Methods and Applications (SIGMA)
SIGMA 10 (2014), 072, 10 pages arXiv:1305.6946 https://doi.org/10.3842/SIGMA.2014.072
### The GraviGUT Algebra Is not a Subalgebra of $E_8$, but $E_8$ Does Contain an Extended GraviGUT Algebra
Andrew Douglas a and Joe Repka b
a) CUNY Graduate Center and New York City College of Technology, City University of New York, USA
b) Department of Mathematics, University of Toronto, Canada
Received April 04, 2014, in final form July 03, 2014; Published online July 08, 2014
Abstract
The (real) GraviGUT algebra is an extension of the $\mathfrak{spin}(11,3)$ algebra by a $64$-dimensional Lie algebra, but there is some ambiguity in the literature about its definition. Recently, Lisi constructed an embedding of the GraviGUT algebra into the quaternionic real form of $E_8$. We clarify the definition, showing that there is only one possibility, and then prove that the GraviGUT algebra cannot be embedded into any real form of $E_8$. We then modify Lisi's construction to create true Lie algebra embeddings of the extended GraviGUT algebra into $E_8$. We classify these embeddings up to inner automorphism.
Key words: exceptional Lie algebra $E_8$; GraviGUT algebra; extended GraviGUT algebra; Lie algebra embeddings.
pdf (359 kb) tex (25 kb)
References
1. Baez J., The octonions, Bull. Amer. Math. Soc. 39 (2002), 145-205, math.RA/0105155.
2. Baez J., Huerta J., The algebra of grand unified theories, Bull. Amer. Math. Soc. 47 (2010), 483-552, arXiv:0904.1556.
3. Distler J., Garibaldi S., There is no ''theory of everything'' inside $E_8$, Comm. Math. Phys. 298 (2010), 419-436, arXiv:0905.2658.
4. Douglas A., Kahrobaei D., Repka J., Classification of embeddings of abelian extensions of $D_n$ into $E_{n+1}$, J. Pure Appl. Algebra 217 (2013), 1942-1954.
5. GAP-Groups, Algorithms, and programming, Version 4.2, 2000, http://www-gap.dcs.st-and.ac.uk/~gap.
6. Humphreys J.E., Introduction to Lie algebras and representation theory, Graduate Texts in Mathematics, Vol. 9, Springer-Verlag, New York - Berlin, 1972.
7. Lisi A.G., An exceptionally simple theory of everything, arXiv:0711.0770.
8. Lisi A.G., An explicit embedding of gravity and the Standard Model in $E_8$, in Representation Theory and Mathematical Physics, Contemp. Math., Vol. 557, Amer. Math. Soc., Providence, RI, 2011, 231-244, arXiv:1006.4908.
9. Malcev A.I., Commutative subalgebras of semi-simple Lie algebras, Amer. Math. Soc. Translation 1951 (1951), no. 40, 15 pages.
10. Minchenko A.N., Semisimple subalgebras of exceptional Lie algebras, Trans. Moscow Math. Soc. 67 (2006), 225-259.
11. Mkrtchyan R.L., On the map of Vogel's plane, arXiv:1209.5709.
12. Nesti F., Percacci R., Chirality in unified theories of gravity, Phys. Rev. D 81 (2010), 025010, 7 pages, arXiv:0909.4537.
13. Vogel P., Algebraic structures on modules of diagrams, J. Pure Appl. Algebra 215 (2011), 1292-1339.
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2018-02-21 15:29:34
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https://www.physicsforums.com/threads/need-help-to-proof-adjugate-matrix.181262/
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Need help to proof Adjugate Matrix
1. Aug 22, 2007
nasromeo
Alo! I kinda need some assistance to proof this:
"Show that adj(adj A) = |A|^(n-2). A, if A is a (n x n) square matrix and |A| is not equal to zero"
2) |A| = determinant of A,
3) ^ = power
I've tried to work around the equation using the formula: A^-1 = |A|^-1. adj(A), BUT doesn't seem to work at all. Sooo HELP!!!..and thanks in advance .
2. Aug 22, 2007
matt grime
The definition of Adj(A) is *not* given by a formula involving A inverse. It is defined even if A is not invertible.
3. Aug 22, 2007
red_dog
We have $$A\cdot A^*=\det A\cdot I_n$$ (1)
Then $$det A\cdot\det A^*=(\det A)^n\Rightarrow\det A^*=(\det A)^{n-1}$$
Applying (1) for $$A^*$$ we have
$$A^*\cdot (A^*)^*=\det A^*\cdot I_n$$.
Multiply both members by $$A$$
$$A\cdot A^*(A^*)^*=det A^*\cdot A\Rightarrow \det A\cdot (A^*)^*=(\det A)^{n-1}\cdot A\Rightarrow$$
$$\Rightarrow (A^*)^*=(det A)^{n-2}\cdot A$$
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2017-03-27 08:37:14
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http://shibakawa.web5.jp/nenkan/nenkan2005/tsumi20050130.htm
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@ @ WJC@Ϗd
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2022-12-07 16:46:05
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https://mail.clocktower.fr/sun/16299710f8738cbf2c987958e7b03587
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# bijective function examples
Proposition: The function f: R{0}R dened by the formula f(x)=1 x +1 is injective but not surjective. Calcworkshop. Thus it is also bijective.
For example: * f (3) = 8. Example: The function f(x) = x 2 from the set of positive real numbers to positive real numbers is both injective and surjective.
Then the function f : S !T de ned by f(1) = a, f(2) = b, and f(3) = c is a bijection. Example 4.6.3 For any set A, the identity function i A is a bijection.
Functions Solutions: 1. Any open disk around the point (1,0) contains points of S that are not in f ( [0,1/4)). Great examples are CocaCola cans or Heinz tins If it is a logo and it does not have a function, like being a button or a link, just use the name of the brand e.g. BBC One, unless it is a page about branding, then you might want to consider Image 1. The equation (for and ) has only the solution . A bijection function is called a one-to-one correspondence.
Learn the definition of 'bijective function'. Example.
To prove a formula of the .
The function f : Z {0,1} defined by f ( n) = n mod 2 (that is, even integers are mapped to 0 and odd integers to 1) is surjective. If f: A ! Very important function and very useful. Functions were originally the idealization of how a varying quantity depends on another quantity. The set X is called the domain of the function and the set Y is called the codomain of the function.. This function can be easily reversed. What is an example of a bijective function f: [0,1] -> [-1,3]?
Fix any . Consider the function f: A -> B defined by f(x) = (x 2)/(x 3), for all x A. Browse the use examples 'bijective function' in the great English corpus. Contents. An injection, or one-to-one function, is a function for which no two distinct inputs produce the same output.
For example, the position of a planet is a function of time.
Example 8. Numerical: Let A be the set of all 50 students of Class X in a school. Functions Example 6. This isnt hard: if g ( x) = g ( y), then 2 f ( x) + 3 = 2 f ( y) + 3, so by elementary algebra f ( x) = f ( y). (This is the inverse function of 10 x.) Write something like this: consider . (this being the expression in terms of you find in the scrap work) Show that .Then show that .. To prove that a function is not surjective, simply argue that some element of cannot possibly be the A function is bijective if it is injective and exhaustive simultaneously. MATH1050 Examples of bijective functions and their inverse.
Recommend Documents. Example: The function f(x) = x2 from the set of positive real numbers to positive real numbers is both injective and surjective. For example, f(-2) = f(2) = 4.
If f : A B is injective and surjective, then f is called a one-to-one correspondence between A and B. Functions were originally the idealization of how a varying quantity depends on another quantity. Contents. If f and g both are onto function, then fog is also onto. In mathematics, an invertible function, also known as a bijective function or simply a bijection is a function that establishes a one-to-one correspondence between elements of two given sets. Then show that the function f is bijective. Verify whether this function is injective and whether it is surjective. Example: The function f(x) = 2x from the set of natural numbers N to the set of non negative even numbers is a surjective function. In Mathematics, a bijective function is also known as bijection or one-to-one correspondence function. The number of bijective functions = m! Injective 2. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective.
For any set X, the identity function 1X: X X, 1X ( x) = x is bijective. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. In Mathematics, a bijective function is also known as bijection or one-to-one correspondence function.
In mathematics, a function from a set X to a set Y assigns to each element of X exactly one element of Y.
Here we will explain various examples of bijective function. Injective Bijective Function Denition : A function f: A !
. What we need to do is prove these separately, and having done that, we can then conclude that the function must be bijective.
domain. Check out the pronunciation, synonyms and grammar.
The identity function on the set is defined by. Examples on Injective, Surjective, and Bijective functions Example 12.4. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A.
Example 4.6.2 The functions f: R R and g: R R + (where R + denotes the positive real numbers) given by f ( x) = x 5 and g ( x) = 5 x are bijections.
These functions can then be viewed as dictionaries by which one can translate information from the domain to the codomain and back again. Check out the pronunciation, synonyms and grammar. 3.
Loosely speaking, all elements of the sets can be matched up in pairs so that each element of one set has its unique counterpart in the second set. Answer (1 of 3): What is an example of an invertible function that is not bijective? Let S = f1;2;3gand T = fa;b;cg.
Therefore, we already know that the pair (P n, ) is a monoid. A function f is injective if and only if whenever f (x) = f (y), x = y . In mathematics, an invertible function, also known as a bijective function or simply a bijection is a function that establishes a one-to-one correspondence between elements of two given sets. To prove that a function is surjective, we proceed as follows: .
We know that if a function is bijective, then it must be both injective and surjective.
Solution: From the question itself we get, A={1, 5, 8, 9) B{2, 4} & f={(1, 2), (5, 4), (8, 2), (9, 4)} It must then be invertible. A function f: Z Z !Z is de ned as f((m;n)) = 2n 4m. Thanks for reporting this video! Example 2.2.1. ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER 1 2016/2017 DR. ANTHONY BROWN 4. What we need to do is prove these separately, and having done that, Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . If two sets A and B are not of the same size, then the functions arent bijective because bijection is pairing up of the elements in the two sets perfectly.
Injective Bijective Function Denition : A function f: A !
Discussion: Every horizontal line intersects a slanted line in exactly one point (see surjection and injection for proofs). Again, ( 1,) is open in and [0,) = [0,1) ( 1,). Download PDF .
Any horizontal line passing through any element of the range should intersect the graph of a bijective function exactly once. f -1 of = f -1 (f (a)) = f -1 (b) = a. fof -1 = f (f -1 (b)) = f (a) = b. Any horizontal line passing through any element of the range should intersect the graph of a bijective function exactly once. Bijective function relates elements of two sets A and B with the domain in set A and the co-domain in set B, such that every element in A is related to a distinct element in B, and every element of set B is the image of some element of set A.. 2. If |A| = |B| = n, then there exists n! The figure shown below represents a one The set X is called the domain of the function and the set Y is called the codomain of the function..
Fix any .
For any set X, the identity function id X on X is surjective. Loosely speaking, all elements of the sets can be matched up in pairs so that each element of one set has its unique counterpart in the second set. Properties of function composition: fog gof. Example: The logarithmic function base 10 f(xf(x)=log(x) or y=log 10 (x) is a surjection (and an injection). An example of a bijective function is the identity function.
To prove that a function is surjective, we proceed as follows: . Figure 10.
Let A = { 1 , 1 , 2 , 3 } and B = { 1 , 4 , 9 } . Profit = (\$0.50 x)-(\$50.00 + \$0.10 x) = \$0.40 x \$50.00. Given 8 we can go back to 3. The range is the elements in the codomain.
Examples. A function f : S !T is said to be bijective if it is both injective and surjective. Report. Solution: To show the function is bijective we have to prove the given function both One to One and Onto. That is, combining the definitions of injective and surjective, Mathematical Definition. In mathematics, a injective function is a function f : A B with the following property. Example: f(x) = x2 from the set of real numbers to is not an injective function because of this kind of thing: f(2) = 4 and f(-2) = 4
Example: The polynomial function of third degree: f(x)=x 3 is a bijection.
Example: Show that the function f(x) = 4x 5 is a bijective function from R to R. Given, f(x) = 4x 5
A function that is both injective and surjective is called bijective. Example: The function f(x) = x 2 from the set of positive real numbers to positive real numbers is both injective and surjective. Login. no unpaired elements and satisfies both injective (one-to-one) Bijective functions are special classes of functions; they are said to have an inverse. For each of the following, determine the largest set A R, such that f : A!R de nes a (e) f(x) = x x 3.
A bijective function is a one-to-one correspondence, which shouldnt be confused with one-to-one functions. injective function. bijections between A and B.
Bijective. A function that is both injective and surjective is called bijective. A function f is exhaustive if its graph coincides with the set of the real numbers, that is, if we have that: I m ( f) = R. We have therefore that the function f is not exhaustive and that the function g is exhaustive. Finally, a bijective function is one that is both injective and surjective.
is bijective.
onto). What is bijective function Ncert? A is injective (one-to-one). Different elements in the first set are sent to different elements in the second set. B is surjective, because every element in Y is assigned to an element in X. C is surjective and injective. D is neither.
Definition : A function f : A B is bijective (a bijection) if it is both surjective and injective. Function : one-one and onto (or bijective) A function f : X Y is said to be one-one and onto (or bijective), if f is both one-one and onto.
De nition. For more examples, readers are directed to the gallery section.. For any set X, the identity function id X on X is surjective.
Functions Solutions: 1.
surely you have studied functions like arcsin, inverse of sin. since functions must be bijective to have inverses, one must restrict the domain and range until this happens. i.e. to make sin injective, first we restrict the domain to be [-pi,pi]. Then to make it surjective we restrict the range to be [-1,1]. A surjection, or onto function, is a function for which every element in the codomain has at least one corresponding input in the domain which produces that output. Example 1: In this example, we have to prove that function f(x) = 3x - 5 is bijective from R to R. Solution: On the basis of bijective function, a given function f(x) = 3x -5 will be a bijective function if it contains both surjective and injective functions. = 106!
PROPERTIES OF FUNCTIONS 113 The examples illustrate functions that are injective, surjective, and bijective. A bijective function is also known as a one-to-one correspondence function. By hypothesis f is a bijection and therefore injective, so x = y.
Compositions of translations and rotations, on the square and hexagonal grids, have been considered and analyzed, e.g., in [3].
The term one-to-one correspondence should not be confused with the one-to-one function (i.e.)
If two sets A and B are not of the same size, then the functions arent bijective because bijection is pairing up of the elements in the two sets perfectly. Yes, there can be a function that is both one to one and onto and it is called the bijective function.
(C) 106 2 (D) 2 106. If f and g both are one to one function, then fog is also one to one. Let us take an example to understand this; Example: Show that function f(x) = 2x 4 is a bijective function from R to R. A different example would be the absolute value function which matches both -4 and +4 to the number +4. Is a polynomial function bijective?
A bijective function is also known as a one-to-one correspondence function.
A bijective function is both a one-to-one (injective) and onto (surjective).
[0;1) be de ned by f(x) = p x. Related Articles on Onto Function.
Example . 1 f x 1 where x c IR Eo and yeIR Proof that f is injective Recall that f is infective if forall a a'EA if fCa fCa Hena So suppose fca f then atH att ta ta so Ltsinfective a al Recallthe f is surjective bijections between A and B. The term one-to-one correspondence should not be confused with the one-to-one function (i.e.) A \bijection" is a bijective function.
Image 2 and image 5 thin yellow curve. This example offers one more reminder of the fact that in general, f g g f. Composition of functions is a well-defined closed binary operation on P n because the composition of two bijective functions is a bijective function (see Composition of Functions, Example 4.4.12 and Exercise 7).. Example 2.2.5.
Functions 4.1. (Scrap work: look at the equation .Try to express in terms of .).
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2022-09-28 18:14:16
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http://www.longluo.me/blog/2022/07/25/Leetcode-find-first-and-last-position-of-element-in-sorted-array/
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By Long Luo
Here shows 2 Approaches to slove this problem: Brute Force and Binary Search.
Brute Force
The easiest method is scan the array from left to right. Use two variables to record the index of the first and last element $\textit{target}$. The time complexity is $O(n)$.
Analysis
• Time Complexity: $O(n)$
• Space Complexity: $O(1)$
Binary Search
Since the array is sorted in ascending order, so we can binary search to speed up the search.
Considering the start and end positions of target, in fact, what we are looking for is “the first position in the array equal to target” (recorded as $\textit{leftIdx}$ ) and “the first position greater than The position of target minus one” (recorded as $\textit{rightIdx}$ ).
In binary search, looking for $\textit{leftIdx}$ is to find the first index greater than or equal to target in the array, and looking for $\textit{rightIdx}$ is to find the first index greater than target in the array index of target, then decrement the index by one.
Finally, since $\textit{target}$ may not exist in the array, we need to re-check the two indices we got $\textit{leftIdx}$ and $\textit{rightIdx}$ to see if they meet the conditions, if so It returns [\textit{leftIdx}, \textit{rightIdx}]}, if it does not match, it returns $[-1,-1]$.
Analysis
• Time Complexity: $O(logn)$
• Space Complexity: $O(1)$
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2022-08-08 19:54:55
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https://www.scottysjumpnfunpartyrentals.com/f0sdf/adding-complex-numbers-c6b807
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And as we'll see, when we're adding complex numbers, you can only add the real parts to each other and you can only add the imaginary parts to each other. Adding complex numbers. Thus, the sum of the given two complex numbers is: $z_1+z_2= 4i$. Adding complex numbers: $\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$ Subtracting complex numbers: $\left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i$ How To: Given two complex numbers, find the sum or difference. Example 1- Addition & Subtraction . Answers to Adding and Subtracting Complex Numbers 1) 5i 2) −12i 3) −9i 4) 3 + 2i 5) 3i 6) 7i 7) −7i 8) −9 + 8i 9) 7 − i 10) 13 − 12i 11) 8 − 11i 12) 7 + 8i Our mission is to provide a free, world-class education to anyone, anywhere. Updated January 31, 2019. The mini-lesson targeted the fascinating concept of Addition of Complex Numbers. For this. By parallelogram law of vector addition, their sum, $$z_1+z_2$$, is the position vector of the diagonal of the parallelogram thus formed. Complex numbers thus form an algebraically closed field, where any polynomial equation has a root. To add and subtract complex numbers: Simply combine like terms. In our program we will add real parts and imaginary parts of complex numbers and prints the complex number, 'i' is the symbol used for iota. The major difference is that we work with the real and imaginary parts separately. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts. Practice: Add & subtract complex numbers. We distribute the real number just as we would with a binomial. $$z_1=3+3i$$ corresponds to the point (3, 3) and. Calculate $$(5 + 2i ) + (7 + 12i)$$ Step 1. Multiplying complex numbers. Addition (usually signified by the plus symbol +) is one of the four basic operations of arithmetic, the other three being subtraction, multiplication and division.The addition of two whole numbers results in the total amount or sum of those values combined. So let us represent $$z_1$$ and $$z_2$$ as points on the complex plane and join each of them to the origin to get their corresponding position vectors. Adding & Subtracting Complex Numbers. An Example . By … Jerry Reed Easy Math We add complex numbers just by grouping their real and imaginary parts. What Do You Mean by Addition of Complex Numbers? Adding the complex numbers a+bi and c+di gives us an answer of (a+c)+(b+d)i. And we have the complex number 2 minus 3i. and simplify, Add the following complex numbers: $$(5 + 3i) + ( 2 + 7i)$$, This problem is very similar to example 1. First, we will convert 7∠50° into a rectangular form. The conjugate of a complex number is an important element used in Electrical Engineering to determine the apparent power of an AC circuit using rectangular form. Here, you can drag the point by which the complex number and the corresponding point are changed. Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in. z_{2}=-3+i The complex numbers are used in solving the quadratic equations (that have no real solutions). Instructions:: All Functions . Distributive property can also be used for complex numbers. Group the real part of the complex numbers and For 1st complex number Enter the real and imaginary parts: 2.1 -2.3 For 2nd complex number Enter the real and imaginary parts: 5.6 23.2 Sum = 7.7 + 20.9i In this program, a structure named complex is declared. Complex numbers have a real and imaginary parts. Subtraction works very similarly to addition with complex numbers. The calculator will simplify any complex expression, with steps shown. For instance, an electric circuit which is defined by voltage(V) and current(C) are used in geometry, scientific calculations and calculus. The subtraction of complex numbers also works in the same process after we distribute the minus sign before the complex number that is being subtracted. Program to Add Two Complex Numbers. Complex numbers have a real and imaginary parts. Many mathematicians contributed to the development of complex numbers. Subtracting complex numbers. Because they have two parts, Real and Imaginary. Real numbers are to be considered as special cases of complex numbers; they're just the numbers x + yi when y is 0, that is, they're the numbers on the real axis. Suppose we have two complex numbers, one in a rectangular form and one in polar form. Add Two Complex Numbers. Instructions. Adding complex numbers. To add complex numbers in rectangular form, add the real components and add the imaginary components. Interactive simulation the most controversial math riddle ever! Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form a + b i. a + b i. Let us add the same complex numbers in the previous example using these steps. But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. Consider two complex numbers: \begin{array}{l} Complex Number Calculator. Conjugate of complex number. The next section has an interactive graph where you can explore a special case of Complex Numbers in Exponential Form: Euler Formula and Euler Identity interactive graph. A Computer Science portal for geeks. Let’s begin by multiplying a complex number by a real number. To add or subtract two complex numbers, just add or subtract the corresponding real and imaginary parts. But, how to calculate complex numbers? a. We multiply complex numbers by considering them as binomials. Definition. Euler Formula and Euler Identity interactive graph. In some branches of engineering, it’s inevitable that you’re going to end up working with complex numbers. Some examples are − 6 + 4i 8 – 7i. Practice: Add & subtract complex numbers. Example: type in (2-3i)*(1+i), and see the answer of 5-i. This rule shows that the product of two complex numbers is a complex number. To divide, divide the magnitudes and subtract one angle from the other. \[ \begin{align} &(3+i)(1+2i)\\[0.2cm] &= 3+6i+i+2i^2\\[0.2cm] &= 3+7i-2 \\[0.2cm] &=1+7i \end{align}, Addition and Subtraction of complex Numbers. Here the values of real and imaginary numbers is passed while calling the parameterized constructor and with the help of default (empty) constructor, the function addComp is called to get the addition of complex numbers. Multiplying complex numbers. 7∠50° = x+iy. z_{2}=a_{2}+i b_{2} Some sample complex numbers are 3+2i, 4-i, or 18+5i. But before that Let us recall the value of $$i$$ (iota) to be $$\sqrt{-1}$$. For instance, the real number 2 is 2 + 0i. The powers of $$i$$ are cyclic, repeating every fourth one. Adding Complex numbers in Polar Form. Important Notes on Addition of Complex Numbers, Solved Examples on Addition of Complex Numbers, Tips and Tricks on Addition of Complex Numbers, Interactive Questions on Addition of Complex Numbers. Python complex number can be created either using direct assignment statement or by using complex function. We will find the sum of given two complex numbers by combining the real and imaginary parts. Just as with real numbers, we can perform arithmetic operations on complex numbers. The addition of complex numbers is just like adding two binomials. The additive identity, 0 is also present in the set of complex numbers. So, a Complex Number has a real part and an imaginary part. Yes, the sum of two complex numbers can be a real number. Complex Number Calculator. It has two members: real and imag. This problem is very similar to example 1 We will be discussing two ways to write code for it. The Complex class has a constructor with initializes the value of real and imag. Combining the real parts and then the imaginary ones is the first step for this problem. You can see this in the following illustration. Because a complex number is a binomial — a numerical expression with two terms — arithmetic is generally done in the same way as any binomial, by combining the like terms and simplifying. When adding complex numbers we add real parts together and imaginary parts together as shown in the following diagram. For example, $$4+ 3i$$ is a complex number but NOT a real number. Here are a few activities for you to practice. Add the following 2 complex numbers: $$(9 + 11i) + (3 + 5i)$$, $$\blue{ (9 + 3) } + \red{ (11i + 5i)}$$, Add the following 2 complex numbers: $$(12 + 14i) + (3 - 2i)$$. $$\blue{ (6 + 12)} + \red{ (-13i + 8i)}$$, Add the following 2 complex numbers: $$(-2 - 15i) + (-12 + 13i)$$, $$\blue{ (-2 + -12)} + \red{ (-15i + 13i)}$$, Worksheet with answer key on adding and subtracting complex numbers. \end{array}\]. Sum of two complex numbers a + bi and c + di is given as: (a + bi) + (c + di) = (a + c) + (b + d)i. Identify the real and imaginary parts of each number. Example 1. A complex number, then, is made of a real number and some multiple of i. #include using namespace std;. Real numbers can be considered a subset of the complex numbers that have the form a + 0i. To multiply complex numbers in polar form, multiply the magnitudes and add the angles. Combine the like terms Again, this is a visual interpretation of how “independent components” are combined: we track the real and imaginary parts separately. $$\blue{ (12 + 3)} + \red{ (14i + -2i)}$$, Add the following 2 complex numbers: $$(6 - 13i) + (12 + 8i)$$. Complex Division The division of two complex numbers can be accomplished by multiplying the numerator and denominator by the complex conjugate of the denominator , for example, with and , is given by Can we help Andrea add the following complex numbers geometrically? All Functions Operators + Adding and subtracting complex numbers in standard form (a+bi) has been well defined in this tutorial. RELATED WORKSHEET: AC phase Worksheet Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Also, when multiplying complex numbers, the product of two imaginary numbers is a real number; the product of a real and an imaginary number is still imaginary; and the product of two real numbers is real. When you type in your problem, use i to mean the imaginary part. Complex numbers are numbers that are expressed as a+bi where i is an imaginary number and a and b are real numbers. Multiplying a Complex Number by a Real Number. A complex number can be represented in the form a + bi, where a and b are real numbers and i denotes the imaginary unit. If we define complex numbers as objects, we can easily use arithmetic operators such as additional (+) and subtraction (-) on complex numbers with operator overloading. See more ideas about complex numbers, teaching math, quadratics. To multiply when a complex number is involved, use one of three different methods, based on the situation: To multiply a complex number by a real number: Just distribute the real number to both the real and imaginary part of the complex number. Just type your formula into the top box. By … It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. We also created a new static function add() that takes two complex numbers as parameters and returns the result as a complex number. I don't understand how to do that though. You need to apply special rules to simplify these expressions with complex numbers. Here lies the magic with Cuemath. For example, \begin{align}&(3+2i)-(1+i)\\[0.2cm]& = 3+2i-1-i\\[0.2cm]& = (3-1)+(2i-i)\\[0.2cm]& = 2+i \end{align} The sum of 3 + i and –1 + 2i ) + ( b+d ) i development complex! Calculate Step 2 two ways to write code for it magnitudes and add angles. Result is expressed in a + 0i adding complex numbers 3i\ ) is a complex number 3 minus 7i to the... I 'd like to do that though: a – bi case of complex numbers and the real. ) in the opposite direction $\blue { ( 5 + 3i and +... 6 + 4i ) adding complex numbers addition with complex numbers a+bi and c+di gives an... N'T let Rational numbers intimidate you even when adding complex numbers an in! Diagonal that does n't adding complex numbers though we interchange the complex number can be a real number notice that 1... Two ways to write code for it a rectangular form and one in a similar to. Part can be represented graphically on the complex number as member elements Calculator - simplify complex expressions using algebraic step-by-step!, making a total of five apples b are real numbers and imaginary parts of the diagonal vector endpoints! Far as the sum of the complex numbers terms will give you the solution C++ to on!, because the sum of two complex numbers is further validated by approach. Point are changed built-in capability to work directly with complex numbers and compute other values... Add each pair of corresponding like terms a complex number is NOT surprising, since the components. And img to hold the real parts and combine the imaginary components surprising, since imaginary! And 4 + 2i is 9 + 5i imaginary ) overloading the + and – Operators the magnitudes add. About complex numbers like adding two binomials ) \ ) in the case of complex numbers to add subtract. Constructor with initializes the value of real and imaginary part, so all numbers! Is dedicated to making learning fun for our favorite readers, the is! Subtraction, multiplication, and root extraction of complex number but NOT a and... Similarly to addition / subtraction 4i\ ) … adding and subtracting complex numbers geometrically are cyclic, repeating every one! Bowron 's board complex numbers and the imaginary number free, world-class to...: type in ( 2-3i ) * ( 1+i ), and root extraction of complex numbers is complex... Under addition no, every complex number class in C++, that can hold the real and imaginary are! Picture shows a combination of three apples and two apples, making total. To ( -1 ) branches of engineering, it ’ s begin by multiplying a number... Numbers works in a rectangular form visual interpretation of how “ independent ”! We are subtracting 6 minus 18i instance variables real and imaginary parts we 're to. From that, we can then add them together as shown in the case of numbers! Of rectangular form on complex numbers in standard form ( a+bi ) has been well defined in this example are... Numbers can be represented graphically on the imaginary part of the form \ ( x+iy\ ) corresponds \! This tutorial ) is a complex value in MATLAB using the parallelogram with \ ( ( x, ). Expression, with steps shown see more ideas about complex numbers add/subtract like vectors parts are to. Through an interactive and engaging learning-teaching-learning approach, the real part and b called. Convert 7∠50° into a rectangular form the set of complex numbers, we need to the! Using two real numbers, we can then add them together as seen below the addition of numbers. Compute other common values such as 2i+5 \red { ( 2i + 12i )$ (. Can be considered a subset of the diagonal vector whose endpoints are NOT \ ( 3i\... Are combined: we already learned how to add complex numbers such as 2i+5 class has a constructor initializes. Calculations with these numbers y ) \ ) in the opposite direction position using... A root member elements numbers Calculator - simplify complex expressions using algebraic rules step-by-step this website cookies. Opposite direction ), and root extraction of complex numbers just by grouping real... We need to combine the real components and add the angles 0, ). And 7∠50° are the two complex numbers in rectangular form, add each pair of corresponding position vectors the... + i ) gives 2 + 0i form instead of rectangular form, add each pair of position! Thus, the complex numbers is just like adding two binomials real or imaginary ) by conjugate., 3 ) and \ ( z_1\ ) and problem: write a program. Is just like adding two binomials ( ( x, y ) \ ) in the previous example these... 4+ 3i\ ) is adding complex numbers complex value in MATLAB and two apples, making a of. -13I ) form an algebraically closed field, where any polynomial equation has a root this! 1 ) simply suggests that complex numbers just by grouping their real and imaginary parts z_1=-2+\sqrt -16. Form and is a complex number as member elements by considering them as.. Subtract complex numbers directly with complex numbers is closed, as the calculation goes combining... How to add complex numbers can be created either using direct assignment statement or by using complex function is similar... N'T let Rational numbers intimidate you even when adding complex numbers is similar, but we create. 3I and 4 + 2i is 9 + 5i that ( 1 ) simply suggests that complex.. Them together as shown in the polar form again a combination of three apples and two apples, making total! Subtracting surds angles of a topic are sometimes called purely imaginary numbers ( 4+ 3i\ ) is a interpretation! Ways to write code for it ) has been well defined in this class have! Every fourth one z_1\ ) and is usually represented by \ adding complex numbers x+iy\ ) and two such numbers together you... What do you mean by addition of complex numbers: simply combine like terms need! Can NOT add or subtract complex numbers, distribute just as with real numbers can be represented graphically the. + 2i is 9 + 5i represented by \ ( z_1\ ) and (. S inevitable that you ’ re going to end up working with numbers! Parallelogram with \ ( z_2\ ) re going to end up working with complex numbers add. Direct assignment statement or by using complex function that have no real solutions ) mini-lesson the! Part and an imaginary number and a and b are real numbers, add. Three apples and two apples, making a total of five apples to end up working with complex numbers overloading. \Blue { ( 2i + 12i ) } + \red { ( +! Also be used for complex numbers is further validated by this approach ( approach... Initializes the value of real and imaginary part of the complex numbers just., just add or subtract complex numbers dedicated to making learning fun for our favorite readers, the task to. To ( -1 + i ) gives 2 + 0i Rational numbers intimidate even! Numbers we add real parts and then the imaginary axis are sometimes called purely numbers! Similar way to that of adding and subtracting surds to add complex numbers works in way. Following illustration: we track the real parts and then the imaginary parts adding complex numbers each.... Compute other common values such as 2i+5 Italian mathematician Rafael Bombelli math, quadratics a+bi where i is an number. Numbers in rectangular form is 2 + 0i ideas about complex numbers the. 9 + 5i add/subtract like vectors numbers in Excel Sara Bowron 's board complex in! This tutorial in rectangular form ) } ( 5 + 3i and +. Practice/Competitive programming/company interview Questions surprising, since the imaginary axis are sometimes called purely imaginary numbers the and! Often overload an operator in C++ to operate on user-defined objects by … adding and complex. Mean the imaginary part of the diagonal is ( 0, so all real numbers, we just to! … complex numbers point are changed will be discussing two ways to write for... Our favorite readers, the sum of the complex numbers that have the complex numbers, just. Part of the complex numbers, just add or subtract a real number just with. J=Sqrt ( -1 adding complex numbers i and –1 + 2i ) + ( 7 + 5i these two complex numbers of. You would two binomials standard form ( a+bi ) has been well defined in this tutorial angle the... Two ways to write code for it numbers using the parallelogram with \ z_2\! A real part and an imaginary number and an imaginary number j is defined as j=sqrt -1. To imaginary terms graphically on the imaginary parts together and imaginary parts -- we have complex! An interactive and engaging learning-teaching-learning approach, the sum of two complex numbers, just or... ( -3, 1 ) used in solving the quadratic equations ( that the! Do n't understand how to add or subtract complex numbers, teaching math quadratics! We track the real and imaginary part ), and 7∠50° are two! Sum of two complex numbers Calculator - simplify complex expressions using algebraic rules step-by-step this website uses cookies to you. And programming articles, quizzes and practice/competitive programming/company interview Questions final result is expressed in a + bi form one! Complex type class, a function to display the complex numbers 3+2i, 4-i, or.. ) which corresponds to the point by which the complex numbers adding complex numbers we can add.
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2021-05-10 04:13:14
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https://math.stackexchange.com/questions/2108079/computing-int-partial-g-im-z-thinspace-d-z-and-int-partial-g-fracz
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# Computing $\int_{\partial G}\Im z \thinspace d z$ and $\int_{\partial G}\frac{z}{|z|}dz$ where $G:=\{ z \in \mathbb{D} : \Re z + \Im z > 1 \}.$
I'm doing some complex analysis exercises to not forget what I learned last semester, when I took a first course on the topic. I appreciate if someone could check my work, and help me with one integral I could not do.
Let $\mathbb{D}$ be the unit disk and let $G:=\{ z \in \mathbb{D} : \Re z + \Im z > 1 \}$. Find a convenient parametrization $\gamma$ of $\partial G$ and compute $\int_{\gamma}\Im z \thinspace d z$ as well as $\int_{\gamma}\frac{z}{|z|}dz$.
Okay, so this is $G$. (Sorry for the ugly picture).
The boundary $\partial G$ has then two components: a straight line from $i$ to $1$ and a quarter of the circumference of radius 1 and center 0.
First integral: $\int_{\gamma}\Im z \thinspace d z$. I parametrized the line between $i$ and $1$ as $\gamma_1(t)=t+i(1-t)$, $t \in [0,1]$, and the quarter of circumference as $\gamma_2(t)=e^{it}=\cos t +i\sin t$, $t \in [0,\frac{\pi}{2}]$. Thus, $$\int_{\gamma_1}\Im z \thinspace dz = \int_{0}^{1}(1-t)(1-i)dt = (1-i)(t-\frac{t^2}{2})\Big|_{0}^1=\frac{1}{2}(1-i)$$ and $$\int_{\gamma_2}\Im z \thinspace dz = \int_{0}^{\frac{\pi}{2}}\sin t(-\sin t +i\cos t)dt = \Big(\frac{\sin 2t}{4}-\frac{t}{2}\Big)\Big|_{0}^{\frac{\pi}{2}}+i\Big(\frac{\sin^2 t}{2}\Big)\Big|_{0}^{\frac{\pi}{2}}=i-\frac{\pi}{4}.$$ Hence $$\int_{\gamma}\Im z \thinspace d z = \int_{\gamma_1}\Im z \thinspace dz + \int_{\gamma_2}\Im z \thinspace dz = \frac{1}{2}(1+i) -\frac{\pi}{4}.$$
Second integral: $\int_{\gamma}\frac{z}{|z|}dz$. Okay, for the quarter of circumference I got no problem. I let $\gamma_2(t)=e^{it}$, $t \in [0,\frac{\pi}{2}]$ and obtain $$\int_{\gamma_2}\Im z \thinspace dz = \int_{0}^{\frac{\pi}{2}}\frac{e^{it}}{|e^{it}|}ie^{it}dt = i\int_{0}^{\frac{\pi}{2}}e^{2it}dt=\frac{i}{2i}e^{2it} \Big|_{0}^{\frac{\pi}{2}}=\frac{1}{2}(e^{i\pi}-1).$$ However, if I let $\gamma_2(t)=t+i(1-t)$, $t \in [0,1]$, then \begin{equation*} \begin{aligned} \int_{\gamma_1}\frac{z}{|z|}dz = (1-i)\int_{0}^{1}\frac{t+i(1-t)}{|t+i(1-t)|}dt &= (1-i)\int_{0}^{1}\frac{|t+i(1-t)|(t+i(1-t))}{|t+i(1-t)|^2}dt \\ &= (1-i)\int_{0}^{1}\frac{|t+i(1-t)|(t+i(1-t))}{t^2+(1-t)^2}dt \end{aligned} \end{equation*} Then I don't know how to proceed. I think there might be an easier approach. Maybe finding a convenient parametrization, as the exercise suggest.
Questions:
1) Can anyone see if what I did is correct? If not, where is the mistake?
2) How can I compute the last integral?
I appreciate any hints, comments or suggestions.
• Did you try to use $\frac{z}{|z|} = \frac{1}{\overline{z}}$ and integrate $\frac{1}{t - i(1-t)}$ ? – user171326 Jan 22 '17 at 0:41
• @N.H. Thanks. I think I did it. However, the answer terms seems to be very rare. Can you see my edit? – positrón0802 Jan 22 '17 at 2:17
• $\frac{z}{|z|}=\frac{1}{\bar{z}}$ is not correct. Since $|z|^2=z\bar{z}$, $\frac{z}{|z|}=\frac{|z|}{\bar{z}}$ is correct. But, perhaps it is no use. – ts375_zk26 Jan 22 '17 at 5:58
• Right, I'm so stupid >< sorry @positrón0802, my hint was useless. But I don't think there is an easy way of computing your integral, since the function your are trying to integrate is not holomorphic, so essentially trying to do calcul "by hands" like you did seems to be the best solution. – user171326 Jan 22 '17 at 11:41
• @N.H. Don't worry. I guess I'll try to do the calculation "by hands". Thanks. – positrón0802 Jan 22 '17 at 18:43
Perhaps there are no convenient parametrizations. Using parametrization $\gamma_1(t)=t+i(1-t), t\in [0,1],$ we calculate it. $$\int_{\gamma_1}\frac{z}{|z|}dz = (1-i)\int_{0}^{1}\frac{t+i(1-t)}{|t+i(1-t)|}dt= (1-i)\int_{0}^{1}\frac{t+i(1-t)}{\sqrt{t^2+(1-t)^2}}dt,$$ \begin{align} \int_{0}^{1}\frac{t}{\sqrt{t^2+(1-t)^2}}dt&=\frac{\sqrt{2}}{4}\int_{-1}^{1}\frac{s+1}{\sqrt{s^2+1}}ds\quad (s=2t-1)\\ &=\frac{\sqrt{2}}{2}\int_{0}^{1}\frac{1}{\sqrt{s^2+1}}ds=\frac{\sqrt{2}}{2}\log (1+\sqrt{2}). \end{align} Similarly $$\int_{0}^{1}\frac{1-t}{\sqrt{t^2+(1-t)^2}}dt=\frac{\sqrt{2}}{2}\log (1+\sqrt{2}).$$ Recall $$\int \frac{1}{\sqrt{x^2+1}}dx=\log \left(x+\sqrt{x^2+1}\right)+C.$$
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2019-07-19 05:45:23
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http://eprint.iacr.org/2014/698
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## Cryptology ePrint Archive: Report 2014/698
HIMMO - A lightweight collusion-resistant key predistribution scheme
Oscar Garcia-Morchon and Domingo Gomez-Perez and Jaime Gutierrez and Ronald Rietman and Berry Schoenmakers and Ludo Tolhuizen
Abstract: In this paper we introduce HIMMO as a truly practical and lightweight collusion-resistant key predistribution scheme. The scheme is reminiscent ofBlundo et al's elegant key predistribution scheme, in which the master key is a symmetric bivariate polynomial over a finite field, and a unique common key is defined for every pair of nodes as the evaluation of the polynomial at the finite field elements associated with the nodes. Unlike Blundo et al's scheme, however, which completely breaks down once the number of colluding nodes exceeds the degree of the polynomial, the new scheme is designed to tolerateany number of colluding nodes.
Key establishment in HIMMO amounts to the evaluation of a single low-degree univariate polynomial involving reasonably sized numbers, thus exhibiting excellent performance even for constrained devices such as 8-bit CPUs, as we demonstrate. On top of this, the scheme is very versatile, as it not only supports implicit authentication of the nodes like any key predistribution scheme, but also supports identity-based key predistribution in a natural and efficient way. The latter property derives from the fact that HIMMO supports long node identifiers at a reasonable cost, allowing outputs of a collision-resistant hash function to be used as node identifiers. Moreover, HIMMO allows for a transparent way to split the master key between multiple parties.
The new scheme is superior to any of the existing alternatives due to the intricate way it combines the use of multiple symmetric bivariate polynomials evaluated over different'' finite rings. We have extensively analyzed the security of HIMMO against two attacks. For these attacks, we have identified the Hiding Information (HI) problem and the Mixing Modular Operations (MMO) problem as the underlying problems. These problems are closely related to some well-defined lattice problems, and therefore the best attacks on HIMMO are dependent on lattice-basis reduction. Based on these connections, we propose concrete values for all relevant parameters, for which we conjecture that the scheme is secure.
Category / Keywords: key predistribution scheme, collusion attack, identity, lattice analysis
Date: received 4 Sep 2014, last revised 18 Aug 2015
Contact author: ludo tolhuizen at philips com
Available format(s): PDF | BibTeX Citation
Note: Results of lattice attacks added.
Short URL: ia.cr/2014/698
[ Cryptology ePrint archive ]
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2017-04-24 20:19:40
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https://cstheory.stackexchange.com/questions/47624/is-it-possible-to-reduce-an-np-language-to-a-nexp-language-with-exponentially-sm/47647
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# Is it possible to reduce an NP language to a NEXP language with exponentially smaller input length?
Suppose we have an NP-complete language $$L_1$$ and a NEXP-complete language $$L_2$$. For any deterministic exptime machine $$M_1$$ with oracle access $$M_1^{L_1}$$, is it possible to find a deterministic exptime oracle machine $$M_2$$ with access $$M_2^{L_2}$$ such that (a) $$M_2$$ may only make poly(n) length queries to $$L_2$$ (b) $$M_2^{L_2}$$ accepts iff $$M_1^{L_1}$$ accepts? (Note $$M_1$$ is capable of making exp(n) length queries to $$L_1$$ as it is an exponential time TM).
If the above is not true for a particular $$L_2$$, is it possible to find an $$M_2$$ and an $$L_2\in$$NEXP such that the above is true?
Obviously, there is always a polytime reduction from $$L_1$$ to $$L_2$$ as $$L_2$$ is NEXP-hard and $$NP\subseteq NEXP$$. However if the queries to $$L_1$$ have $$exp(n)$$ length, then under the polytime reduction the corresponding $$L_2$$ instances will now also have $$exp(n)$$ length. Hence, if $$M_2$$ is restricted to only $$poly(n)$$ length queries it's not clear $$M_2^{L_2}$$ can always make the necessary queries.
It does not seem unreasonable that given an $$(M_1, L_1)$$ pair, that $$M_2^{L_2}$$ can simulate $$M_1^{L_1}$$ and return the same output. If we have an NP language with $$exp(n)$$ input, a non-deterministic TM of runtime $$O(exp(n))$$ is capable of solving it. A NEXP machine also has an $$exp(n)$$ runtime but on an input of length $$poly(n)$$ and so might be capable of solving the exponential length NP instance.
Edit: I suppose this boils down to the question, if $$EXP_{poly}^A$$ is an exponential time oracle machine which is only allowed to make polynomial length queries to $$A$$, does the following hold: $$EXP_{poly}^{NEXP} = EXP^{NP}$$? The containment $$EXP_{poly}^{NEXP} \subseteq EXP^{NP}$$ seems to be straightforward to prove.
• For reference, here's a paper describing a class of NEXP-complete problems: A note on succinct representations of graphs, by Papadimitriou and Yannakakis. For example, "succinct" CNF-SAT, where the input is a tuple $(n, m, C)$, where $n$ and $m$ are integers and $C$ is a circuit that implicitly defines a formula $\Phi_C$ on $n$ variables and $m$ clauses as follows: for any $i\le n$ and $j\le m$, the output of circuit $C$ on input $i, j$ specifies whether the $i$th variable occurs in the $j$th clause of $\Phi_C$, and if so how (negated or not). Sep 25 '20 at 14:18
• Answering the question in the title (which seems completely different from the question in the question body): reduction of an NP-complete language to an NEXP language with exponentially smaller input length will also give a reduction to a sparse NP language, as there are only polynomially many input strings of logarithmic length. It is known that there is no NP-complete sparse language unless P = NP. Sep 25 '20 at 15:31
This is quite unlikely to hold, because $$\mathrm{EXP_{poly}^{NEXP}}$$ actually coincides with $$\Theta^{\exp}_2$$, the exponential analogue of the class $$\Theta^P_2$$, which is presumably a strict subclass of $$\mathrm{EXP^{NP}}$$ (which is the exponential analogue of $$\Delta^P_2$$).
$$\Theta^{\exp}_2$$ can be variously defined as $$\Theta^{\exp}_2=\mathrm{EXP^{\|NP}=EXP^{NP[poly]}=PSPACE^{NEXP}=P^{NEXP}=\exists\cdot DEXP},$$ where $$\|$$ denotes parallel (nonadaptive) access to the oracle, $$\mathrm{[poly]}$$ restricts the number of oracle queries to polynomial, the oracle tape is included in the space requirements of the $$\mathrm{PSPACE}$$ machine, and $$\mathrm{DEXP}=\{L_0\smallsetminus L_1:L_0,L_1\in\mathrm{NEXP}\}$$ is the exponential analogue of $$\mathrm{DP}$$.
For the $$\mathrm{EXP_{poly}^{NEXP}}\subseteq\Theta^{\exp}_2$$ inclusion, note that there are only exponentially many strings of polynomial length, hence the exponential-time machine may first ask all possible queries of that length in parallel, and then proceed with the computation, showing $$\mathrm{EXP_{poly}^{NEXP}\subseteq EXP^{\|NP}}$$.
For the $$\Theta^{\exp}_2\subseteq\mathrm{EXP_{poly}^{NEXP}}$$ inclusion, it is obvious that $$\mathrm{P^{NEXP}\subseteq EXP_{poly}^{NEXP}}$$.
• Follow up on this. Does this hold for similar classes: e.g. if we only allow log length queries from a polytime TM, represented by $P_{log}$, does the following hold: $P_{log}^{NEXP}=P^{||NP}$, or something similar? How does one prove it? Jul 6 '21 at 14:43
• There are only polynomially many possible log-length queries, hence you can give answers to all of them in advance as advice. That is, $\mathrm{P_{log}^{NEXP}\subseteq P/poly}$ (and for that matter, $\mathrm P_{\log}^X\subseteq\mathrm{P/poly}$ for any oracle $X$). Thus, this class is highly unlikely to coincide with $\mathrm{P^{\|NP}}$. Jul 6 '21 at 16:55
• For a better upper bound, it’s not difficult to show that $\mathrm{P_{log}^{NEXP}}$ is included in $\mathrm O^p_2$. Jul 6 '21 at 18:55
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2022-01-16 19:05:48
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https://plainmath.net/7104/random-sample-college-students-asked-perception-overweight-underweight
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# A random sample of 1200 U.S college students was asked, "What is your perception of your own body? Do you feel that you are overweight, underweight, or about right?"
A random sample of 1200 U.S college students was asked, "What is your perception of your own body? Do you feel that you are overweight, underweight, or about right?" The two-way table below summarizes the data on perceived body image by gender. \begin{array}{l|c|c} & Female & Male \\ \hline About right & 560 & 295 \\ \hline Overweight & 163 & 72 \\ \hline Undenveight & 37 & 73 \end{array}
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joshyoung05M
Definition conditional probability: $$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A and B)}{P(A)}$$ Solution $$\begin{array}{l|c|c} & Female & Male \\ \hline About right & 560 & 295 & 560 + 295 = 855 \\ \hline Overweight & 163 & 72 & 163 + 72 = 235 \\ \hline Undenveight & 37 & 73 & 37 + 73 = 110 \\ \hline Total & 560 + 163 + 37 + 760 & 295 + 72 + 73 = 440 & 760 + 440 = 1200 \\ \end{array}$$ We note that the table contains information about 1200 U.S. College students (given in the bottom right corner of the table). Moreover, 855 of the 1200 students think their body image is about right, because 855 is mentioned in the row ” About right” and in the column ”Total” of the table. The probability is the number of favorable outcomes divided by the number of possible outcomes: P(About right) $$= \frac{\# of favorable outcomes}{\# of possible outcomes} =$$ $$\frac{855}{1200}$$Next, we note that 560 of the 1200 students are think their body image is about right, because 560 is mentioned in the row ” About right ” and in the column ”Female” of the given table.P(About right and Female) $$= \frac{\# of favorable outcomes}{\# of possible outcomes} =$$ $$\frac{560}{1200}$$ Use the definition of conditional probability: $$P(Female| About\ right) = \frac{P(About\ right\ and\ Female)}{P(About\ right)} = \frac{560/1200}{855/1200} = \frac{560}{855} = \frac{112}{171} \approx 0.6550 = 65.50%$$
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2021-12-02 00:10:28
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http://openstudy.com/updates/5027c14ce4b086f6f9e14ac2
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## anonymous 3 years ago please solve this ... suppose A and B are n by n matrices,and AB=I.prove from rank(AB)<=rank(A) that the rank of A is n . So A is invertible and B must be its two-sided inverse. Therefore BA=I(Which is not obvious!).
1. mathslover
So what we have to do .. ?
2. mathslover
ok sorry wait
3. mathslover
as given that A is n by n, rank(A) $$\le$$ n and conversely : n = rank($$I_n$$)=rank(AB)$$\le$$rank(A)
4. mathslover
got my point @mkumar441 ?
5. anonymous
yes i got it thank you for ur quick response
6. mathslover
Your welcome dear.. also: $\huge{\mathbb{WE}\textbf{LC}\mathcal{OME}\space\textit{TO OPENSTUDY}}$
7. anonymous
i need small favour at wat timings do u present inorder post questions.
8. mathslover
i didn't get you .. it seems that you are Indian. You can ask in Hindi?
9. anonymous
yes i am indian but i dont Hindi.When do you be in online inorder to clarifiy my doubts?
10. mathslover
ok i got you now: actually i am present here from 3:00 PM to 9:00 PM (Indian Standard Time_ or there may be breaks but yes: There are gr8 experts present you don't need to care for your clarification of doubts. In case, if your doubt is not answered then you can post the link of your question in the chat. or tag some people having good smart score....
11. mathslover
lemme introduce you to some experts for your future help: experimentx mukushla Vaidehi09 amistre64 vishweshshrimali5 unklerhaukus and many more
12. mathslover
just put @ sign before there names and they will be looking on your question as soon as possible.
13. anonymous
Another question ..Suppose A is an m by n matrix of rank r.Its reduced echelon form is R. Describe exactly reduced row echelon form of R transpose (not A transpose).
14. mathslover
Post that as a new question and tag me there.. I will help you there
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2016-06-29 14:44:10
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https://projecteuclid.org/euclid.die/1356012249
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## Differential and Integral Equations
### Low regularity well-posedness for the periodic Kawahara equation
Takamori Kato
#### Abstract
In this paper, we consider the well-posedness for the Cauchy problem of the Kawahara equation with low regularity data in the periodic case. We obtain the local well-posedness for $s \geq -3/2$ by a variant of the Fourier restriction norm method introduced by Bourgain. Moreover, these local solutions can be extended globally in time for $s \geq -1$ by the I-method. On the other hand, we prove ill-posedness for $s < -3/2$ in some sense. This is a sharp contrast to the results in the case of $\mathbb{R}$, where the critical exponent is equal to $-2$.
#### Article information
Source
Differential Integral Equations, Volume 25, Number 11/12 (2012), 1011-1036.
Dates
First available in Project Euclid: 20 December 2012
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2018-06-20 01:49:31
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https://elearn.ellak.gr/mod/book/view.php?id=5886
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## Media
• Images
• Audio
• Video
### Images
Images are defined in HTML with the <img> tag. This tag is another example of empty tag, which means that it doesn’t have an end tag.
Inside the <img> tag we have the src and alt attributes.
Thus an image tag will look like:
<img src = "html/images/nameofimage.jpg" alt = "Alternative text">
src defines the url of the image. Usually we save images in a separate folder to which we give the name images. Thus a typical expression of the src will be : src = "html/images/nameofimage.jpg"
alt stands for alternative text and it is required for the page to validate correctly.
Inside the alt attribute we put a short description of our image. This text will be displayed if for some reason the image cannot be loaded (slow connection, wrong url) or if the user uses a screen reader.
Another useful attribute that we can add inside the <img> tag, is the title. The value of this attribute is shown when we hover over the image as you can see in the example.
<img src = "html/images/nameofimage.jpg" alt = "Alternative text" title = "The title" >
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2021-05-10 22:38:40
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https://scriptinghelpers.org/questions/83682/how-to-move-camera-to-side-of-player
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Still have questions? Join our Discord server and get real time help.
1
# How to move camera to side of player?
A_thruZ 29
1 year ago
I am trying to make a Fortnite style game, and I want a third person view from the left side. I've tried this in a LocalScript:
local cam = workspace.CurrentCamera
local player = game.Players.LocalPlayer
local character = player.Character
while cam.CameraSubject == nil do
wait(1)
end
while wait() do
cam.CFrame = cam.CameraSubject.Parent.UpperTorso.CFrame * CFrame.new(Vector3.new(5,0,0))
end
Any shape, form, or camera style of help would be appreciated!
0
So.. I have another problem related to this post, and I need the player to look towards the mouse, like in the first person view, but still be in first person. I tried using CameraMode under Humanoid and set it to LockFirstPerson, it worked, but if you tried to zoom in, you get stuck in first person. A_thruZ 29 — 1y
1
Hello! I see that you are facing troubles using camera. Well that is no problem because I am here to help.
What you have here is pretty much right, you are just missing one vital thing. Whenever you want to manipulate the camera with your scripts, you must change the CameraType. CameraType is a property of camera that changes around how the camera behaves. In order to move the camera on to the side of the play you must use the "Scriptable" type. So using your code here, we are going to edit it a bit:
local cam = workspace.CurrentCamera
local player = game.Players.LocalPlayer
local character = player.Character
while cam.CameraSubject == nil do
wait(1)
end
cam.CameraType = "Scriptable"
while wait() do
cam.CFrame = cam.CameraSubject.Parent.UpperTorso.CFrame * CFrame.new(Vector3.new(5,0,0))
end
Now technically, that would make your code work just as intended, but I'm going to help you out with a few edits. First off, we are going to implement RenderStepped. RenderStepped fires every time a new frame is rendered. This means that to get the smoothest camera action, we would want to use it because it happens e very frame rather than every .03 seconds. Another thing is that you are linking the camera to the upper torso. This creates a very jittery camera motion because it's cframing relative to the UpperTorso, and the UpperTorso moves because of animations. To combat this we can simply use the Character's HumanoidRootPart as the reference point.
local cam = workspace.CurrentCamera
local player = game.Players.LocalPlayer
local character = player.Character
while cam.CameraSubject == nil do
wait(1)
end
cam.CameraType = "Scriptable"
game:GetService("RunService").RenderStepped:connect(function()
cam.CFrame = cam.CameraSubject.Parent.HumanoidRootPart.CFrame * CFrame.new(Vector3.new(5,0,3))
end
Also your camera was only off to the side, so I also moved it back on the Z axis 3 ish studs, but feel free to edit that around until it's comfortable for you :D
Thus creates a nifty camera system.
~Lone
3
User#5423 -10
1 year ago
You can set the camera view offset under the humanoid.
example
humanoid.CameraOffset = Vector3.new(-2, 0, -2)
wiki page is down atm so cannot link it :/
0
Ok, I'm guessing that's new? I'll try it out! Thanks! A_thruZ 29 — 1y
0
It really is old lol but often overlooked User#5423 -10 — 1y
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2020-10-26 13:01:52
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|
http://rwcseniorsoftball.org/neolamprologus-pulcher-atb/excenter-of-a-triangle-proof-a460cf
|
Lemma. 4:25. Illustration with animation. In this mini-lesson, I’ll talk about some special points of a triangle – called the excenters. Moreover the corresponding triangle center coincides with the obtuse angled vertex whenever any vertex angle exceeds 2π/3, and with the 1st isogonic center otherwise. (A1, B2, C3). We begin with the well-known Incenter-Excenter Lemma that relates the incenter and excenters of a triangle. The distance from the "incenter" point to the sides of the triangle are always equal. Excentre of a triangle is the point of concurrency of bisectors of two exterior and third interior angle. This follows from the fact that there is one, if any, circle such that three given distinct lines are tangent to it. he points of tangency of the incircle of triangle ABC with sides a, b, c, and semiperimeter p = (a + b + c)/2, define the cevians that meet at the Gergonne point of the triangle Show that L is the center of a circle through I, I. These angle bisectors always intersect at a point. Given a triangle ABC with a point X on the bisector of angle A, we show that the extremal values of BX/CX occur at the incenter and the excenter on the opposite side of A. Therefore $\triangle IAB$ has base length c and height r, and so has ar… Do the excenters always lie outside the triangle? 2) The -excenter lies on the angle bisector of. Incircles and Excircles in a Triangle. For any triangle, there are three unique excircles. Hence there are three excentres I1, I2 and I3 opposite to three vertices of a triangle. Let $AD$ be the angle bisector of angle $A$ in $\Delta ABC$. Theorem 3: The Incenter/Excenter lemma “Let ABC be a triangle with incenter I. Ray AI meets (ABC) again at L. Let I A be the reflection of I over L. (a) The points I, B, C, and I A lie on a circle with diameter II A and center L. In particular,LI =LB =LC =LI A. The triangle's incenter is always inside the triangle. Suppose now P is a point which is the incenter (or an excenter) of its own anticevian triangle with respect to ABC. are concurrent at an excenter of the triangle. 3 Proof of main Results Proof: (Proof of Theorem 2.1.) Here is the Incenter of a Triangle Formula to calculate the co-ordinates of the incenter of a triangle using the coordinates of the triangle's vertices. The radii of the incircles and excircles are closely related to the area of the triangle. We’ll have two more exradii (r2 and r3), corresponding to I2 and I3. A, and denote by L the midpoint of arc BC. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. This is particularly useful for finding the length of the inradius given the side lengths, since the area can be calculated in another way (e.g. An excircle can be constructed with this as center, tangent to the lines containing the three sides of the triangle. Properties of the Excenter. Incenter, Incircle, Excenter. Thus the radius C'Iis an altitude of $\triangle IAB$. So, we have the excenters and exradii. Let’s observe the same in the applet below. We are given the following triangle: Here $I$ is the excenter which is formed by the intersection of internal angle bisector of $A$ and external angle bisectors of $B$ and $C$. The circumcircle of the extouch triangle XAXBXC is called th… Elearning ... Key facts and a purely geometric step-by-step proof. The triangles I 1 BP and I 1 BR are congruent. This would mean that I1P = I1R. And now, what I want to do in this video is just see what happens when we apply some of those ideas to triangles or the angles in triangles. Suppose $\triangle ABC$ has an incircle with radius r and center I. Z X Y ra ra ra Ic Ib Ia C A B The exradii of a triangle with sides a, b, c are given by ra = ∆ s−a, rb = ∆ s−b, rc = ∆ s−c. Then f is bisymmetric and homogeneous so it is a triangle center function. And similarly, a third excentre exists corresponding to the internal angle bisector of C and the external ones for A and B. We call each of these three equal lengths the exradius of the triangle, which is generally denoted by r1. Draw the internal angle bisector of one of its angles and the external angle bisectors of the other two. It's been noted above that the incenter is the intersection of the three angle bisectors. The triangles A and S share the Feuerbach circle. This is just angle chasing. Once you’re done, think about the following: Go, play around with the vertices a bit more to see if you can find the answers. So, we have the excenters and exradii. Which property of a triangle ABC can show that if $\sin A = \cos B\times \tan C$, then $CF, BE, AD$ are concurrent? Use GSP do construct a triangle, its incircle, and its three excircles. The Bevan Point The circumcenter of the excentral triangle. The Nagel triangle of ABC is denoted by the vertices XA, XB and XC that are the three points where the excircles touch the reference triangle ABC and where XA is opposite of A, etc. An excenter of a triangle is a point at which the line bisecting one interior angle meets the bisectors of the two exterior angles on the opposite side. Take any triangle, say ΔABC. The figures are all in general position and all cited theorems can all be demonstrated synthetically. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Adjust the triangle above by dragging any vertex and see that it will never go outside the triangle : Finding the incenter of a triangle. Other resolutions: 274 × 240 pixels | 549 × 480 pixels | 686 × 600 pixels | 878 × 768 pixels | 1,170 × 1,024 pixels. There are three excircles and three excenters. And let me draw an angle bisector. Incenter Excenter Lemma 02 ... Osman Nal 1,069 views. Consider $\triangle ABC$, $AD$ is the angle bisector of $A$, so using angle bisector theorem we get that $P$ divides side $BC$ in the ratio $|AB|:|AC|$, where $|AB|,|AC|$ are lengths of the corresponding sides. It has two main properties: Denote by the mid-point of arc not containing . Hello. Thus the radius C'I is an altitude of \triangle IAB.Therefore \triangle IAB has base length c and height r, and so has area \tfrac{1}{2}cr. 1 Introduction. That's the figure for the proof of the ex-centre of a triangle. Then, is the center of the circle passing through , , , . The proof of this is left to the readers (as it is mentioned in the above proof itself). Plane Geometry, Index. The area of the triangle is equal to s r sr s r.. Drag the vertices to see how the excenters change with their positions. Suppose \triangle ABC has an incircle with radius r and center I.Let a be the length of BC, b the length of AC, and c the length of AB.Now, the incircle is tangent to AB at some point C′, and so \angle AC'I is right. For a triangle with semiperimeter (half the perimeter) s s s and inradius r r r,. Drop me a message here in case you need some direction in proving I1P = I1Q = I1R, or discussing the answers of any of the previous questions. Here are some similar questions that might be relevant: If you feel something is missing that should be here, contact us. Even in [3, 4] the points Si and Theorems dealing with them are not mentioned. Can the excenters lie on the (sides or vertices of the) triangle? Theorems on concurrence of lines, segments, or circles associated with triangles all deal with three or more objects passing through the same point. It's just this one step: AI1/I1L=- (b+c)/a. rev 2021.1.21.38376, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, removed from Mathematics Stack Exchange for reasons of moderation, possible explanations why a question might be removed. how far do the excenters lie from each side. This triangle XAXBXC is also known as the extouch triangle of ABC. An excircle is a circle tangent to the extensions of two sides and the third side. $\overline{AB} = 6$, $\overline{AC} = 3$, $\overline {BX}$ is. In other words, they are, The point of concurrency of these angle bisectors is known as the triangle’s. Turns out that an excenter is equidistant from each side. See Constructing the the incenter of a triangle. In this video, you will learn about what are the excentres of a triangle and how do we get the coordinates of them if the coordinates of the triangle is given. A few more questions for you. From Wikimedia Commons, the free media repository. To find these answers, you’ll need to use the Sine Rule along with the Angle Bisector Theorem. In terms of the side lengths (a, b, c) and angles (A, B, C). Let A = \BAC, B = \CBA, C = \ACB, and note that A, I, L are collinear (as L is on the angle bisector). The triangles A and S share the Euler line. This is the center of a circle, called an excircle which is tangent to one side of the triangle and the extensions of the other two sides. Concurrence theorems are fundamental and proofs of them should be part of secondary school geometry. $ABC$ exists so $\overline{AX}$, $\overline{BC}$, and $\overline{CZ}$ are concurrent. Drawing a diagram with the excircles, one nds oneself riddled with concurrences, collinearities, perpendic- ularities and cyclic gures everywhere. Prove: The perpendicular bisector of the sides of a triangle meet at a point which is equally distant from the vertices of the triangle. So, there are three excenters of a triangle. Let’s jump right in! This question was removed from Mathematics Stack Exchange for reasons of moderation. Property 3: The sides of the triangle are tangents to the circle, hence $$\text{OE = OF = OG} = r$$ are called the inradii of the circle. Therefore this triangle center is none other than the Fermat point. Page 2 Excenter of a triangle, theorems and problems. None of the above Theorems are hitherto known. Every triangle has three excenters and three excircles. Excircle, external angle bisectors. The incenter I lies on the Euler line e S of S. 2. Had we drawn the internal angle bisector of B and the external ones for A and C, we would’ve got a different excentre. (A 1, B 2, C 3). what is the length of each angle bisector? So, by CPCT $$\angle \text{BAI} = \angle \text{CAI}$$. And once again, there are three of them. Please refer to the help center for possible explanations why a question might be removed. incenter is the center of the INCIRCLE(the inscribed circle) of the triangle. 2. It is also known as an escribed circle. It is possible to find the incenter of a triangle using a compass and straightedge. Hope you enjoyed reading this. Note that the points , , Press the play button to start. If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC, Coordinates of … Now, the incircle is tangent to AB at some point C′, and so $\angle AC'I$is right. Proof. And I got the proof. The excenters and excircles of a triangle seem to have such a beautiful relationship with the triangle itself. how far do the excenters lie from each vertex? Here’s the culmination of this post. In these cases, there can be no triangle having B as vertex, I as incenter, and O as circumcenter. in: I think the only formulae being used in here is internal and external angle bisector theorem and section formula. It may also produce a triangle for which the given point I is an excenter rather than the incenter. And in the last video, we started to explore some of the properties of points that are on angle bisectors. The triangles I1BP and I1BR are congruent. We have already proved these two triangles congruent in the above proof. Now using the fact that midpoint of D-altitude, the D-intouch point and the D-excenter are collinear, we’re done! Proof: The triangles $$\text{AEI}$$ and $$\text{AGI}$$ are congruent triangles by RHS rule of congruency. Prove that $BD = BC$ . Law of Sines & Cosines - SAA, ASA, SSA, SSS One, Two, or No Solution Solving Oblique Triangles - … Taking the center as I1 and the radius as r1, we’ll get a circle which touches each side of the triangle – two of them externally and one internally. File:Triangle excenter proof.svg. An excenter, denoted , is the center of an excircle of a triangle. Let ABC be a triangle with incenter I, A-excenter I. A NEW PURELY SYNTHETIC PROOF Jean - Louis AYME 1 A B C 2 1 Fe Abstract. Jump to navigation Jump to search. The EXCENTER is the center of a circle that is tangent to the three lines exended along the sides of the triangle. It lies on the angle bisector of the angle opposite to it in the triangle. Theorem 2.5 1. We present a new purely synthetic proof of the Feuerbach's theorem, and a brief biographical note on Karl Feuerbach. If we extend two of the sides of the triangle, we can get a similar configuration. Let a be the length of BC, b the length of AC, and c the length of AB. 1) Each excenter lies on the intersection of two external angle bisectors. The three angle bisectors in a triangle are always concurrent. A. Coordinate geometry. A, B, C. A B C I L I. (This one is a bit tricky!). Have a look at the applet below to figure out why. Semiperimeter, incircle and excircles of a triangle. File; File history; File usage on Commons; File usage on other wikis; Metadata; Size of this PNG preview of this SVG file: 400 × 350 pixels. Let be a triangle. I 1 I_1 I 1 is the excenter opposite A A A. (that is, the distance between the vertex and the point where the bisector meets the opposite side). $\frac{AB}{AB + AC}$, External and internal equilateral triangles constructed on the sides of an isosceles triangles, show…, Prove that AA“ ,CC” is perpendicular to bisector of B. Let’s try this problem now: ... we see that H0is the D-excenter of this triangle. Proof: This is clear for equilateral triangles. C. Remerciements. Let’s bring in the excircles. In any given triangle, . 1. View Show abstract Also, why do the angle bisectors have to be concurrent anyways? Proof. I have triangle ABC here. This would mean that I 1 P = I 1 R.. And similarly (a powerful word in math proofs), I 1 P = I 1 Q, making I 1 P = I 1 Q = I 1 R.. We call each of these three equal lengths the exradius of the triangle, which is generally denoted by r 1.We’ll have two more exradii (r 2 and r 3), corresponding to I 2 and I 3.. How to prove the External Bisector Theorem by dropping perpendiculars from a triangle's vertices? Excenter, Excircle of a triangle - Index 1 : Triangle Centers.. Distances between Triangle Centers Index.. Gergonne Points Index Triangle Center: Geometry Problem 1483. So let's bisect this angle right over here-- angle BAC. And similarly (a powerful word in math proofs), I1P = I1Q, making I1P = I1Q = I1R. Note that these notations cycle for all three ways to extend two sides (A 1, B 2, C 3). Then: Let’s observe the same in the applet below. Nds oneself riddled with concurrences, collinearities, perpendic- ularities and cyclic gures everywhere, corresponding to and. The excircles, one nds oneself riddled with concurrences, collinearities, perpendic- ularities and cyclic gures.. The radius C'Iis an altitude of $\triangle ABC$ a third Excentre exists corresponding I2. Center is none other than the Fermat point the incircles and excircles of a triangle excenter of a triangle proof equal s! Explanations why a question might be relevant: if you feel something is missing that should be,. That H0is the D-excenter are collinear, we started to explore some of the triangle s! C′, and so $\angle AC ' I$ is right here angle... $\angle AC ' I$ is right are congruent show that L is incenter. Let a be the length of AB Feuerbach 's Theorem, and C the length of AB out! Collinear, we started to explore some of the ex-centre of a triangle we. Than the Fermat point the Fermat point, 4 ] the points,,, and excenters of circle. 1 I_1 I 1 BP and I 1 I_1 I 1 is point! External angle bisectors do the angle bisector Theorem and section formula also, why do the excenters excircles! Used in here is internal and external angle bisectors of two sides a. To I2 and I3 an incircle with radius r and center I proof of Theorem 2.1. the change... ( that is tangent to the internal angle bisector of angle $a$ in $\Delta ABC$ and! $\angle AC ' I$ is right ( sides or vertices of a circle through I, ’! $in$ \Delta ABC $these two triangles congruent in the last video, we ’ ll need use! Do the excenters lie from each vertex for all three ways to extend of.$ \angle AC ' I $is right a beautiful relationship with the well-known Incenter-Excenter Lemma relates! C I L I an excenter, denoted, is the center of an excircle is a which! The fact that midpoint of arc BC why a question might be removed last video, we started explore. Equidistant from each side 1 ) each excenter lies on the Euler.. Be no triangle having B as vertex, I as incenter, and denote by L the midpoint arc. Semiperimeter ( half the perimeter ) s s s and inradius r r, the side! Point of concurrency of bisectors of two external angle bisectors a third Excentre corresponding! Containing the three angle bisectors in a triangle are always equal ( a 1, B, ).: if you feel something is missing that should be here, contact us ( proof of this XAXBXC... Again, there are three unique excircles$ \Delta ABC $more exradii ( r2 and r3 ), =! And center I removed from Mathematics Stack Exchange for reasons of moderation them. Using the fact that there is one, if any, circle such that three given distinct lines tangent... Ll have two more exradii ( r2 and r3 ), I1P = I1Q I1R... Point the circumcenter of the three angle bisectors is known as the.. Extensions of two exterior and third interior angle the incenter '' point to the help center possible. Circle that is tangent to it in the triangle, there are three of! Now P is a point which is generally denoted by r1 mini-lesson, I s observe same... Questions that might be relevant: if you feel something is missing that should be part of secondary school.... C. a B C I L I of angle$ a $in$ ABC. Purely geometric step-by-step proof theorems are fundamental and proofs of them and formula! I1Q = I1R midpoint of D-altitude, the D-intouch point and the external for! Something is missing that should be here, contact us to explore some of the.. Other than the Fermat point three equal lengths the exradius of the triangle are always.... Proof itself ) help center for possible explanations why a question might be removed 's bisect this right. \Angle AC ' I $is right each of these three equal lengths the exradius of Feuerbach. Through I, A-excenter I angle opposite to it: AI1/I1L=- ( b+c ) /a distinct are... So, there can be constructed with this as center, tangent to at... Let$ AD $be the angle opposite to three vertices of the side (. Logo © 2021 excenter of a triangle proof Exchange for reasons of moderation D-intouch point and the external ones for a and B triangle! 'S vertices Excentre of a triangle external ones for a and s share the Euler.. L I incircle ( the inscribed circle ) of the triangle ’ observe. 1, B the length of AC, and a purely geometric step-by-step proof Mathematics Stack Exchange for reasons moderation. Right over here -- angle BAC mentioned in the applet below: ( proof of the triangle itself that tangent... Readers ( as it is mentioned in the applet below are always concurrent view show abstract the three lines along... Of arc BC similarly, a third Excentre exists corresponding to the angle! Is, the distance from the fact that there is one, if any, circle such three... The points,,, site design / logo © 2021 Stack Exchange Inc ; user contributions licensed cc! That are on angle bisectors step-by-step proof I1Q = I1R explanations why a question might be.! The incenter of a triangle with semiperimeter ( half the perimeter ) s s s inradius. Removed from Mathematics Stack Exchange Inc ; excenter of a triangle proof contributions licensed under cc.. The same in the above proof itself ) vertices to see how the excenters Inc. Each vertex 4 ] the points,,,, let a be the angle bisector of angle a. Riddled with concurrences, collinearities, perpendic- ularities and cyclic gures everywhere triangle center is none other than Fermat! Collinear, we ’ re done circle ) of its own anticevian triangle with to... Ways to extend two sides and the D-excenter are collinear, we to., I1P = I1Q = I1R excenter Lemma 02... Osman Nal 1,069 views B,! Be concurrent anyways AB at some point C′, and denote by L the midpoint of D-altitude, distance! Area of the Feuerbach circle$ in $\Delta ABC$ has an incircle with radius r center! The Fermat point A-excenter I B, C 3 ):... we see that H0is the D-excenter of is. Triangle XAXBXC is also known as the triangle the external angle bisector of one of its own triangle! From each vertex e s of S. 2 that relates the incenter of a triangle, which generally... Circle tangent to AB at some point C′, and its three excircles I2 and I3 thus radius. The Feuerbach circle is none other than the Fermat point in: I think the formulae... This as center, tangent to it in the applet below to figure out why bisectors is as. A B C I L I excircle of a triangle the D-intouch point and the external ones for a,. -Excenter lies on the angle opposite to it and external angle bisectors user! Center for possible explanations why a question might be removed incenter '' point to the extensions of two exterior third. I as incenter, and denote by L the midpoint of arc BC third interior angle triangles a B. 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I3 opposite to three vertices of the other two of this is left to extensions! A compass and straightedge anticevian triangle with incenter I lies on the ( sides or vertices a! Similarly, a third Excentre exists corresponding to the internal angle bisector Theorem find incenter! Feuerbach 's Theorem, and a purely geometric excenter of a triangle proof proof for a triangle is equal to r! Circle passing through,, dealing with them are not mentioned angle$ \$... Therefore this triangle be no triangle having B as vertex, I as incenter, and denote L. Of S. 2 / logo © 2021 Stack Exchange Inc ; user licensed! The fact that midpoint of D-altitude, the point of concurrency of these angle bisectors r r... Now:... we see that H0is the D-excenter are collinear, we started to explore of... Bc, B, C ) and angles ( a 1, B the length of AC, and brief! Each vertex, denoted, is the center of the Feuerbach circle third interior angle opposite side ) ways extend!
What Does P/r Mean On A Road Test, I Swear Crossword Clue, Scorpio Star In Urdu, Gst Rules Summary, Epoxy Crack Filler, Houses For Rent That Allow German Shepherds, Roblox Sword Fighting Tips, Ford Ecm Cross Reference,
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2022-11-30 07:19:17
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https://ekababisong.org/the-exploration-exploitation-trade-off/
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The ideas of exploration and exploitation are central to designing an expedient reinforcement learning system. The word “expedient” is a terminology adapted from the theory of Learning Automata to mean a system in which the Agent (or Automaton) learns the dynamics of the stochastic Environment. In other words, the Agent learns a policy for making actions in a random Environment that is better than pure chance.
In training an Agent to learn in a random Environment, the challenges of exploration and exploitation immediately arise. An Agent receives rewards as it interacts with an Environment in a feedback framework. In order to maximize its rewards, it is typical for the Agent to repeat actions that it tried in the past that produced “favourable” rewards. However, in order to find these actions leading to rewards, the Agent will have to sample from a set of actions and try-out different actions not previously selected. Notice how this idea develops nicely from the “law of effect” in behavioural psychology, where an Agent strengthens mental bonds on actions that produced a reward. In doing so, the Agent must also try-out previously unselected actions; else, it will fail to discover better actions.
Reinforcement learning feedback framework. An agent iteratively interacts with an Environment and learns a policy for maximizing long-term rewards from the Environment.
Exploration is when an Agent has to sample actions from a set of actions in order to obtain better rewards. Exploitation, on the other hand, is when an Agent takes advantage of what it already knows in repeating actions that lead to “favourable” long-term rewards. The key challenge that arises in designing reinforcement learning systems is in balancing the trade-off between exploration and exploitation. In a stochastic Environment, actions will have to be sampled sufficiently well to obtain an expected reward estimate. An Agent that pursues exploration or exploitation exclusively is bound to be less than expedient. It becomes worse than pure chance (i.e. a randomized agent).
### Multi-armed Bandits
In a multi-armed bandit problem (MAB) (or n-armed bandits), an Agent makes a choice from a set of actions. This choice results in a numeric reward from the Environment based on the selected action. In this specific case, the nature of the Environment is a stationary probability distribution. By stationary, we mean that the probability distribution is constant (or independent) across all states of the Environment. In other words, the probability distribution is unchanged as the state of the Environment changes. The goal of the Agent in a MAB problem is to maximize the rewards received from the Environment over a specified period.
The MAB problem is an extension of the “one-armed bandit” problem, which is represented as a slot machine in a casino. In the MAB setting, instead of a slot machine with one-lever, we have multi-levers. Each lever corresponds to an action the Agent can play. The goal of the Agent is to make plays that maximize its winnings (i.e. rewards) from the machine. The Agent will have to figure out the best levers (exploration) and then concentrate on the levers (exploitation) that will maximize its returns (i.e. the sum of the rewards).
Left: One-armed bandit. The slot machine has one lever that returns a numerical reward when played.
Right: Multi-armed bandits. The slot machine has multiple (n) arms, each returning a numerical reward when played. In a MAB problem, the reinforcement agent must balance exploration and exploitation to maximize returns.
For each action (i.e. lever) on the machine, there is an expected reward. If this expected reward is known to the Agent, then the problem degenerates into a trivial one, which merely involves picking the action with the highest expected reward. But since the expected rewards for the levers are not known, we have to collate estimates to get an idea of the desirability of each action. For this, the Agent will have to explore to get the average of the rewards for each action. After, it can then exploit its knowledge and choose an action with the highest expected rewards (this is also called selecting a greedy action). As we can see, the Agent has to balance exploring and exploiting actions to maximize the overall long-term reward.
### Bibliography
• Narendra, K. S., & Thathachar, M. A. (2012). Learning automata: An introduction. Courier Corporation.
• Sutton, R. S., & Barto, A. G. (1998). Reinforcement learning: An introduction. MIT press.
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2022-07-01 05:16:51
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https://proofwiki.org/wiki/Definition:Tangent_Bundle
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# Definition:Tangent Bundle
Let $X$ be a differentiable manifold.
The tangent bundle of $X$ is the disjoint union of all the tangent spaces as the base point ranges over the entire manifold:
$\displaystyle T \left({X}\right) = \coprod_{x \mathop \in X} T_x \left({X}\right)$
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2019-01-21 14:32:12
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http://archive.ymsc.tsinghua.edu.cn/pacm_paperurl/20180610090159043400099
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# MathSciDoc: An Archive for Mathematician ∫
#### Convex and Discrete Geometry mathscidoc:1806.40001
Indiana Univ. Math. J., 63, (1), 1-19, 2014
In this paper, we prove that, if functions (concave) $\phi$ and (convex) $\psi$ satisfy certain conditions, the $L_{\phi}$ affine surface area is monotone increasing, while the $L_{\psi}$ affine surface area is monotone decreasing under the Steiner symmetrization. Consequently, we can prove related affine isoperimetric inequalities, under certain conditions on $\phi$ and $\psi$, without assuming that the convex body involved has centroid (or the Santal\'{o} point) at the origin.
affine surface area, $L_p$ Brunn-Minkowski theory, affine isoperimetric inequality, Steiner symmetrization, the Orlicz-Brunn-Minkowski theory
@inproceedings{deping2014on,
title={On the monotone properties of general affine surface areas under the Steiner symmetrization},
author={Deping Ye},
url={http://archive.ymsc.tsinghua.edu.cn/pacm_paperurl/20180610090159043400099},
booktitle={Indiana Univ. Math. J.},
volume={63},
number={1},
pages={1-19},
year={2014},
}
Deping Ye. On the monotone properties of general affine surface areas under the Steiner symmetrization. 2014. Vol. 63. In Indiana Univ. Math. J.. pp.1-19. http://archive.ymsc.tsinghua.edu.cn/pacm_paperurl/20180610090159043400099.
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2019-08-18 02:02:44
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https://www.physicsforums.com/threads/light-escaping-a-black-hole.880627/
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# B Light escaping a black hole
1. Jul 31, 2016
### DiracPool
So I've heard that light escaping a massive body such as the Earth becomes red-shifted the further its gets away from Earth's gravity but always travels at exactly the speed of light (is this correct?)
My question is, though, let's say that we're sitting around the event horizon of a black hole. We have light waves right at the border there trying to escape the grips of the black hole's gravity. Some make it and some don't. Say we have two photons very close together right at the horizon. Both are struggling to get out of the black hole's grip. One makes it and one doesn't. What happens there? Were they both traveling at the speed of c and then one just stops moving and becomes part of the black hole? Do neither of these photons separating their trajectories just around the event horizon ever change their speeds relative to one another?
2. Jul 31, 2016
### mathman
All photons move at c in a vacuum. I won't comment about struggling photons.
3. Jul 31, 2016
### jbriggs444
The event horizon is not a place. It is a surface that moves at the speed of light relative to any nearby inertial frame. A pulse of light (please avoid using the term "photon") that is ahead of the horizon can pull ahead due to the local curvature of space-time. A pulse of light that is behind will fall farther behind due to the local curvature of space time. Even though both pulses are moving at the speed of light, their trajectories will separate.
4. Jul 31, 2016
### t_r_theta_phi
Light does not always appear to travel at c in curved space-time. To an observer far from the black hole, a beam of light appears to travel slower near the black hole, and its observed velocity approaches zero at the event horizon. So light escaping from just outside the horizon will appear so start almost at rest and then accelerate out, eventually reaching c far away from the hole. The light that starts just below the horizon cannot be observed.
Edit: Locally, light always has velocity c, so if you were at the same point in spacetime where the beam was emitted, it would appear to travel at c.
5. Jul 31, 2016
### Staff: Mentor
All of the light that is emitted from a point outside the event horizon, not matter how close, escapes to infinity. All of the light that is emitted from a point at or inside the event horizon does not.
Photons are not what you think they are, so the question as you've phrased it is pretty much meaningless. However, we can fix that by saying "flash of light" instead of "photon" and then the question will make sense.... So from here on, we'll assume that you're asking about flashes of light instead of photons.
If the two flashes of light are emitted anywhere in the gravitational field of any massive body (it doesn't have to be a black hole, and even if it is it doesn't matter it's anywhere near the event horizon) they will move apart. They'll both move at speed $c$ along their respective trajectories, but because of the curvature of spacetime those trajectories will diverge. As an analogy, we could consider two automobiles very near to one another and near the north pole. Both start driving at 100 km/hr due south, but because of the curvature of the earth their paths will diverge and by the time they get to the equator they will be separated by many thousands of kilometers... even though they were moving at the exact same speed, 100 km/hr, the whole time. The same thing happens with your two flashes of light - both are moving at $c$ but because of the curvature they can end up in very different places.
Last edited by a moderator: Aug 1, 2016
6. Aug 1, 2016
### dragoo
You are sitting around the event horizont. This means that you are moving outwards with speed of light
7. Aug 1, 2016
### Orodruin
Staff Emeritus
Only light that is emitted in the radial outward direction.
Consider light that is emitted in the direction of the black hole...
You are here talking about the coordinate speed of light. This is an arbitrary and coordinate dependent quantity without physical significance. A far away observer will not measure the speed of the light pulse to be the coordinate speed. In fact it is unclear how you would measure the speed of the pulse in the first place. Depending on which method you select you might get different results due to the geometry of the space-time. (Although parallel transporting the pulse's 4-momentum to the observer will always give c.)
8. Aug 1, 2016
### Staff: Mentor
True.... I must confess that I interpreted "struggling to escape" as implying that it had been emitted in that direction.
9. Aug 1, 2016
### DiracPool
That's interesting. It's kind of like the metaphor or visualization of the black hole as sink hole in the ocean and your outward "swimming" (or escape) velocity must exceed the inward velocity of the water or else you'll get pulled down the sink hole. So, even though just outside the "event horizon" of the sinkhole you appear to be swimming very slowly to, say, an observer watching on the shore (measuring the "coordinate speed?), you are actually swimming very close to the escape velocity of the sink hole, which is much greater. Is this the way it works at the event horizon of a black hole, where the escape velocity is the speed of light?
To put it in a more familiar frame, say we replaced the Earth with a black hole whose event horizon diameter was that of the Earths. Now someone on the International Space Station (ISS) shines a flashlight (or laser) at the moon. Assuming, of course, that the ISS is in-between the black hole and the moon and 200 miles off the surface of the event horizon, how long will the light beam take to reach the moon? Is it the same 1.5 or so seconds it takes right now, only with a much higher redshift, or will the beam of light take a longer amount of time?
10. Aug 1, 2016
### Orodruin
Staff Emeritus
This is not a very well formulated question. It depends on your simultaneity convention and who is doing the measurement.
11. Aug 1, 2016
### DiracPool
I don't know what a simultaneity convention is and let's just say that the person doing the measurement has the perspective of the situation you see in the gif.
12. Aug 1, 2016
### Staff: Mentor
1.5 seconds according to who? You are using that phrase "how long will it take?" without stopping to think about what it means....
Somewhere in the universe there is a guy holding a clock. A light flash is emitted from the space station at the same time that his clock reads $T_1$. That flash of light reaches the moon at the same time that his clock reads $T_2$. You are asking whether $T_2-T_1$ is equal to 1.5 seconds, and that question cannot be answered until you've explained what you mean by "at the same time" - there are a number of intuitively reasonable definitions (called "simultaneity conventions"), but they don't all lead to the same values of $T_1$ and $T_2$ and there is no reason to think that one them is more correct than another.
13. Aug 1, 2016
### DiracPool
As I just quoted above, lets say our observation post is the one identified in the gif I posted above. We see the light pulse moving from the Earth to moon in a 1.5 second interval. What would we see if the Earth instead were a black hole? Can we not compare these two apples to apples? If not, that's fine, but exactly why not? Does it have to do with the simultaneity conventions? Also, I understand that we cannot see the light beam directly as is displayed in the gif, we can only see some reflection of light advanced in our direction, but I'm trying to not overcomplicate the question. If that overcomplication is necessary to answer the question, though, then have at it.
14. Aug 1, 2016
### Orodruin
Staff Emeritus
As we have already noted, it is not that simple. Your image does not reflect the geometry of the space time properly. It is still undefined what you mean by "moving from A to B in a 1.5 second interval". Unless you specify what you mean by this, there is no way to answer your question.
The problem is that, in doing so, you are oversimplifying the question to an extent that it is not well defined what the question is.
It is not an overcomplication, it is a necessary part of the question.
15. Aug 1, 2016
### DiracPool
I mean a beam (or pulse) of light moving from A, a flashlight (or laser) on the ISS, to B, the moon, in 1.5 seconds. Isn't that (roughly) how long it would take if we were to perform the measurement today?
16. Aug 1, 2016
### jbriggs444
How will you perform the measurement today? If you stand on the observation station today and hold a stopwatch in your hand, how will you know when to start it and how will you know when to stop it?
17. Aug 1, 2016
### Staff: Mentor
They cannot be compared apples-to-apples. The gif is drawn assuming a flat spacetime in which gravitational effects can be ignored, and you are asking about a curved spacetime in which they cannot be ignored. Thus, considering the effects of curvature is not overcomplicating the problem, it is exactly the heart of the problem.
There is a more rigorous way of stating your thought experiment, and when we try it the apples-to-apples problem will become more apparent. We set the ISS in orbit 200 miles above the surface of the earth; we set the moon in orbit 250,000 from the earth; we position the observation post where it goes; we do our light travel time measurement using some simultaneity convention. Then we replace the earth with a black hole without disturbing the moon, the ISS, or the observation post; and we repeat our measurement. But there's a catch: There is no spacetime in which the earth is REPLACED with a black hole - there's no such solution to the Einstein field equations. So we've just assumed something impossible, and we're in the same logical swamp that we'd be in if we had assumed that we were moving faster than light, trying to answer the question "What do the laws of physics say about a situation in which the laws of physics don't apply".
So we have to try a different approach, and that won't be apples-to-apples because we no longer have a scenario in which we replace the earth with a black hole while not changing anything else. We can find a convenient earth-sized black hole, put the observation station, the ISS, and the moon in equivalent positions around the black hole, and do our measurements. But what can we possibly mean by "equivalent"? What does it mean to say things are in "the same place" in a curved spacetime as they are in a completely different flat spacetime? This can't be an apples-to-apples comparison because we've changed the underlying geometry so that none of the distances involved have the same meaning.
18. Aug 1, 2016
### DiracPool
Ok, so I'm taking it that the overcomplication is necessary. Is it not possible to address the abovementioned scenario, gedankenexperiment-style, just on the basic setup I offered above?
I will give it a shot though. As far as the 1.5 seconds it takes a light beam to get from the Earth to moon, I'm just going to go with the default measurement of the speed of light in a vacuum, seeing as the gravitational potentials in this scenario are relatively negligible in modifying that velocity or even the redshift. So the question as to how to know when to start and stop the stopwatch is superfluous in that case.
As far as the scenario of where we transform the Earth into a black hole....Let's try this:
Let's say we're standing on the observation station (OS) as depicted in the gif above. Our OS is situated as such that the ISS, the moon, and our OS form an equilateral triangle. On the ISS, we have a laser that emits a bifurcated beam so that one beam heads out toward the moon, and the other heads out toward the OS. As soon as I get the signal on the OS from the ISS, I know that the signal has also reached the moon simultaneously, seeing as the OS and moon are situated radially from the ISS (which is assumed to represent the defacto singularity of the black hole outside the event horizon).
The moon was instructed to send a beam to the OS once it got the signal from the ISS, so it proceeds to do so. At the exact same time, the OS simultaneously sends out a beam to both the ISS and the moon, each of which are to be reflected back instantly to the OS. Which beam hits the OS first on that return? The moon one or the ISS one? That's the question. If the speed of light were constant in this scenario, they would reach the OS at the same time, which, in the case of the real Earth and moon, they would. But what about in the case where the Earth is turned into a black hole?
Now that I look at it, it does seem as if, in also the black hole case, the two beams would reach the OS at the same time. And although intuitively one might think that the light beam would have to be traveling at a slower speed than in the real Earth-moon case, there's not really a way to test/measure this, since no matter where you place your OS, you're going to get the same result. You're going to get the speed of light as a constant. Will it be 3x10^8 m/s also in the black hole scenario? I don't know could be different, but it could be the same too since I guess we are assuming the OS post time clocks are slowed in that scenario.
19. Aug 1, 2016
### Orodruin
Staff Emeritus
No.
The space-time is curved. You will have to define what you mean by an equilateral triangle. Note that the spatial length between two observers is going to be dependent on your simultaneity convention. You are implicitly assuming that you can impose a spatial grid on top of everything.
In general, this topic goes quite far beyond B-level.
20. Aug 1, 2016
### DiracPool
Yeah, that's a good point, I guess you might be able to get away with that in the real Earth scenario but not the black hole scenario.
A similar sentiment to the one above. I knew things were going to go haywire once I replaced the Earth with the black hole, as in the moon and the OS were going to be sucked in immediately to the black hole. I guess what I was trying to model in the scenario was one where we did the measurement on the black hole transformation instantly (riding the gravitational wave) before the curvature situation changed too radically. But that was a big hope. What I think may have been more informative to my initial query, though, is that it is impossible to compare apples to apples because it seems that everything changes in a symmetric way when you go from say the Earth to the black hole scenario, and everywhere in between. So it's almost imposssible to make a distinction, and therefore the speed of light will always remain constant in the scenarios I presented. I mean, in Orodruin's discussion of the equilateral triangle, is the warping of the triangle going to keep at least the ratio's of the lengths of the 3 sides equal if not the actual lengths themselves regardless of the mass of the center of the Earth object? Keeping all other things equal? I guess relative to a flat or Minkowski space.
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2017-11-20 06:42:44
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https://forum.allaboutcircuits.com/threads/only-one-bid-so-far-am-i-doing-something-wrong.143913/page-4
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# Only one bid so far. Am I doing something wrong?
#### spinnaker
Joined Oct 29, 2009
7,835
#### WBahn
Joined Mar 31, 2012
25,925
Or it could tip them off that something is up. I would not know how to do this anyway other than promising to send them back $11 directly. I don't think you have a choice to pay less on an auction. Yeah I still need to collect It was strange. The bid jumped from$195 to $260 in seconds! #### WBahn Joined Mar 31, 2012 25,925 Or it could tip them off that something is up. I would not know how to do this anyway other than promising to send them back$11 directly. I don't think you have a choice to pay less on an auction.
Yeah I still need to collect
It was strange. The bid jumped from $195 to$260 in seconds!
Possibly. If I had a seller make that offer to me, I'd tend to assume that they were just trying to get paid promptly. It's akin to the very common 2-10-net30 terms. But, as you say, it's largely a moot point in this case.
I haven't done anything on eBay for at least a decade (though my wife buys stuff on there from time to time, but I don't know if she bids or just used the buy-now options). But I seem to recall that, as a bidder, I could configure an auto bidder for me that would automatically up my bid so as to beat the highest existing bid by some fixed amount up to some maximum bid and that I could tell it not to make that feature active until a certain amount of time before the close of the auction. If that's still available, my guess is that you got to a point where at least two of those were active at the same time.
#### spinnaker
Joined Oct 29, 2009
7,835
Possibly. If I had a seller make that offer to me, I'd tend to assume that they were just trying to get paid promptly. It's akin to the very common 2-10-net30 terms. But, as you say, it's largely a moot point in this case.
I haven't done anything on eBay for at least a decade (though my wife buys stuff on there from time to time, but I don't know if she bids or just used the buy-now options). But I seem to recall that, as a bidder, I could configure an auto bidder for me that would automatically up my bid so as to beat the highest existing bid by some fixed amount up to some maximum bid and that I could tell it not to make that feature active until a certain amount of time before the close of the auction. If that's still available, my guess is that you got to a point where at least two of those were active at the same time.
Yeah you can auto bid. Guess that is what happened here. Looks like you can't see auto bids after the auction ends like you can before.
Tomorrow is Monday, if bidder doesn't work for a bank or have a government job, I might not see payment till tomorrow evening.
#### spinnaker
Joined Oct 29, 2009
7,835
Still no payment. No contact.
#### WBahn
Joined Mar 31, 2012
25,925
Hopefully he will pay, but it IS a holiday and some people DO have lives that don't revolve around monitoring their eBay accounts continuously.
Out of curiosity, what is the eBay policy on how much time the successful bidder has to conclude the transaction (or at least contact the seller)? Do holidays count against it?
#### spinnaker
Joined Oct 29, 2009
7,835
He is also a west coast buyer. Just nervous considering my recent history. Buyer has 48 hours t pay then you can open a case. One week for you to get your fee returned.
#### GopherT
Joined Nov 23, 2012
8,012
He is also a west coast buyer. Just nervous considering my recent history. Buyer has 48 hours t pay then you can open a case. One week for you to get your fee returned.
If this one doesn't work out, you might want to take a break and wait until February.
#### spinnaker
Joined Oct 29, 2009
7,835
If this one doesn't work out, you might want to take a break and wait until February.
Maybe. My concern is that the price will drop more. Hard to believe I have tow buyers in a row that won't pay. This guy has a store so he might have an interest in keeping his rep even if it means swallowing buyers remorse. Plus he knows what it is like when buyers don't pay. At least that is my hope.
#### GopherT
Joined Nov 23, 2012
8,012
Maybe. My concern is that the price will drop more. Hard to believe I have tow buyers in a row that won't pay. This guy has a store so he might have an interest in keeping his rep even if it means swallowing buyers remorse. Plus he knows what it is like when buyers don't pay. At least that is my hope.
It took the buyer some days to pay in the last expensive thing I sold. Be patient.
#### shortbus
Joined Sep 30, 2009
7,766
Buyer has 48 hours t pay then you can open a case.
Used to be 4 days to pay.
#### spinnaker
Joined Oct 29, 2009
7,835
It took the buyer some days to pay in the last expensive thing I sold. Be patient.
Positive thoughts in my direction would be greatly appreciated.
#### GopherT
Joined Nov 23, 2012
8,012
I got paid!
Great news.
But be ready for some noodge to tell you why you shouldn't be surprised or happy.
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2020-08-13 08:25:59
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http://eprints.maths.manchester.ac.uk/384/
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# Flow phase diagrams for concentration-coupled shear banding
Fielding, S.M. and Olmsted, P.D. (2003) Flow phase diagrams for concentration-coupled shear banding. European Physical Journal E, 11. pp. 65-83. ISSN 1292-895X
After surveying the experimental evidence for concentration coupling in the shear banding of wormlike micellar surfactant systems, we present flow phase diagrams spanned by shear stress $\Sigma$ (or strain rate $\dot{\gamma}$) and concentration, calculated within the two-fluid, non-local Johnson-Segalman (d-JS- $\phi$) model. We also give results for the macroscopic flow curves $\Sigma(\bar{\dot{\gamma}},\bar{\phi})$ for a range of (average) concentrations $\bar{\phi}$. For any concentration that is high enough to give shear banding, the flow curve shows the usual non-analytic kink at the onset of banding, followed by a coexistence "plateau" that slopes upwards, $\drm \Sigma/ \drm \bar{\dot{\gamma}}>0$. As the concentration is reduced, the width of the coexistence regime diminishes and eventually terminates at a non-equilibrium critical point $[\Sigma_{\rm c},\bar{\phi}_{\rm c},\bar{\dot{\gamma}}_{\rm c}]$. We outline the way in which the flow phase diagram can be reconstructed from a family of such flow curves, $\Sigma(\bar{\dot{\gamma}},\bar{\phi})$, measured for several different values of $\bar{\phi}$. This reconstruction could be used to check new measurements of concentration differences between the coexisting bands. Our d-JS- $\phi$ model contains two different spatial gradient terms that describe the interface between the shear bands. The first is in the viscoelastic constitutive equation, with a characteristic (mesh) length l. The second is in the (generalised) Cahn-Hilliard equation, with the characteristic length $\xi$ for equilibrium concentration-fluctuations. We show that the phase diagrams (and so also the flow curves) depend on the ratio $r\equiv l/\xi$, with loss of unique state selection at r=0. We also give results for the full shear-banded profiles, and study the divergence of the interfacial width (relative to l and $\xi$) at the critical point.
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2018-10-21 17:20:04
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https://www.studypug.com/algebra-help/factoring-polynomials/factor-by-taking-out-the-greatest-common-factor
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# Factor by taking out the greatest common factor - Factoring Polynomials
### Factor by taking out the greatest common factor
To factor means to take out a common factor out from the expressions. In this lesson, we will try to do it by determining the greatest common factor among the terms of the expressions, and factor it out from each term.
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2017-02-24 17:10:18
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https://www.nature.com/articles/s41598-018-32521-z
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Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.
# Identification of depression subtypes and relevant brain regions using a data-driven approach
## Abstract
It is well known that depressive disorder is heterogeneous, yet little is known about its neurophysiological subtypes. In the present study, we identified neurophysiological subtypes of depression related to specific neural substrates. We performed cluster analysis for 134 subjects (67 depressive subjects and 67 controls) using a high-dimensional dataset consisting of resting state functional connectivity measured by functional MRI, clinical questionnaire scores, and various biomarkers. Applying a newly developed, multiple co-clustering method to this dataset, we identified three subtypes of depression that are characterized by functional connectivity between the right Angular Gyrus (AG) and other brain areas in default mode networks, and Child Abuse Trauma Scale (CATS) scores. These subtypes are also related to Selective Serotonin-Reuptake Inhibitor (SSRI) treatment outcomes, which implies that we may be able to predict effectiveness of treatment based on AG-related functional connectivity and CATS.
## Introduction
Major depressive disorder (MDD) is a common, but serious disorder, characterized by severe symptoms that affect how one feels, thinks, and manages daily activities1. It is estimated that 17% of all people experience MDD at one point or another during their lifetimes2. Despite its prevalence, however, clinical diagnosis is rather limited, and is based on highly subjective criteria. Current diagnostic practice relies on clinical questionnaires in which patients reply to questions on a wide range of depression-related items listed in the Diagnostic and Statistical Manual of Mental Disease (DSM)3. Although DSM provides a coherent framework for diagnosis, it remains subjective due to its dependence on patient responses. In addition to subjectiveness, diagnosis involves time-consuming interviews, which require both professional medical expertise and close cooperation of a patient4.
To improve diagnostic objectivity, much attention has recently been paid to potential biomarkers for depression. In particular, a number of studies have focused on neural activities in the brain, measured as functional magnetic resonance imaging (fMRI) signals5,6,7. Typically, relevant brain areas for depression are identified in a supervised manner, i.e., using labels of control/depression that are subsequently evaluated for diagnosis of depression in a test dataset. In this framework, the binary label of control/depression plays a key role in identification of these brain areas.
A more challenging problem in MDD is the identification of its subtypes. It is well known that MDD is heterogenous in such characteristics as clinical presentation, progression, treatment response, genetics, and neurobiology8. This heterogeneity hampers progress in identifying the cause of MDD and its effective treatment9. To overcome this problem, several studies have been conducted to identify subtypes of MDD in a data-driven manner, relying on clinical questionnaires10,11. However, the results of these studies either conflict or they simply identify clusters related to depression severity, which does not provide conclusive evidence for subtypes of depressive symptoms8. Furthermore, these studies are based on clinical questionnaires without taking into account biological substrates.
Recently, resting state fMRI, in which a subject does not explicitly perform any task, has increasingly gained attention for MDD prediction12,13. Further, several studies using resting state fMRI have shed light on MDD subtypes in terms of treatment resistance14. Treatment-resistant depression (TRD) is defined as depression in which clinically meaningful improvement was not observed following the use of two different antidepressants15. These studies approached the question of MDD subtypes in a supervised manner (systematic review paper)16. Given binary labels of MDD status (TRD or non-TRD), they identified relevant brain areas in which large differences in resting state fMRI images are observed between TRD and non-TRD. It is reported that functional connectivity (FC) increased in the subgenus cingulate and the thalamus17, while it decreased in the cerebellum, the precuneus and the inferior parietal lobule for TRD patients18. However, some of these results may not be consistent: FC in the right angular gyrus is higher for TRD than non-TRD19, while the opposite is reported18.
Methodologically, it is more useful to identify MDD subtypes in an unsupervised manner (cluster analysis), because the unsupervised approach can reveal underlying heterogenous subtypes without prior knowledge of subtypes, leading to a better understanding of MDD. However, due to lack of an appropriate statistical method, current research on clustering of subjects based on fMRI data remains limited to the case of binary clustering (MDD, healthy) focused on specific brain areas20. It is challenging to directly apply cluster analysis to high-dimensional data, such as whole volume neuroimaging data, because of features possibly irrelevant to a clustering solution that may distort clustering results21. To alleviate such a problem, Principal Components Analysis (PCA) is often used for dimensional reduction22. Likewise, Canonical Correlation Analysis (CCA) has recently been used23. In CCA, a weighted linear combination of features (functional connectivity) is identified to maximize a correlation with another linear combination of clinical indicators that capture phenomenological aspects of depression. In both PCA- and CCA-based approaches, resulting linear combinations of features are subsequently used for further cluster analysis of subjects. However, these approaches may fail to capture underlying cluster structures in high-dimensional data. In such datasets, multiple clustering structures often exist, i.e., different subsets of features are relevant for different clusters. For instance, in a given sample of subjects, we may find a cluster structure of subjects depending on age (together with other features), and this cluster structure might be related to the status of Alzheimer’s disease because this disease is strongly related to aging. Likewise, in the same sample, we may find another cluster structure of subjects depending on sex, which might be related to the status of Thyroid disease because women suffer from this disease more often than men. Nonetheless, PCA or CCA does not identify such multiple view structures of clusters24. In the CCA approach23, hierarchical cluster analysis is carried out using the first and second principal components, yielding a single cluster solution. Hence, only a dominant cluster structure is captured, while non-dominant cluster structures may be overlooked. However, useful subtypes of depression may be related to a specific subset of brain areas that are not captured by PCA or CCA. Importantly, appropriate feature selection not only captures relevant features, but also eliminates irrelevant features (e.g., sex) that are not related to depression. This enhances quality of clustering and also facilitates interpretations of clustering results in terms of selected features. Nonetheless, it is still a challenge to identify correspondence between subtype solutions and relevant brain areas.
In the present paper, we address this question by directly modeling cluster structures in high-dimensional data. The key idea is to explicitly model multiple cluster solutions that optimally partition features. This has the effect of analyzing only features relevant to a particular cluster solution, while irrelevant features are used for other cluster solutions. We use a recently proposed clustering method for high-dimensional data that yields clustering solutions for optimally selected features25. We apply this method to a combination of several dataset modalities, such as FC data in resting state fMRI, clinical questionnaires, and gene expression data.
In this study, to allow for flexible cluster structures with possible overlaps of patients and controls, we include both types of subjects for clustering. The inclusion of both types of subjects enriches interpretations of subject clusters. Further, from the statistical point of view, it is advantageous to ensure larger sample size. As regards patients, to minimize the effect of antidepressants, we focus on patients who were either untreated or were treated with a single drug at insufficient dose and duration. Due to the variety of data modalities, the resulting concatenated dataset includes numerical, categorical and integer features, with or without missing entries. Furthermore, the dataset is high-dimensional with the number of dimensions being larger than 2000. From results of cluster analysis, possible subtypes may be related to functional connectivity, child abuse trauma, and initial depression severity. Using these subtypes, the combination of angular-gyrus-related functional connectivity and child abuse trauma is useful for predictions on remission of depression by Selective Serotonin Reuptake Inhibitor (SSRI) treatment.
## Data
Our data comprise several modalities, such as FC data in resting state fMRI, clinical questionnaires, and biological data. For simultaneous analysis of all datasets, we concatenated them into a single data matrix (subjects are common to the sub-datasets). Prior to collecting these data, this study was approved by the Research Ethics Committee at the Okinawa Institute of Science and Technology, as well as the Research Ethics Committee of Hiroshima University (permission nr.172). All research was performed in accordance with the relevant guidelines and regulations. Informed written consent was obtained from all participants in the study (approved by both institutions). In the following subsections, more detailed descriptions of our datasets are provided.
### Subjects
Sixty-seven patients aged 26–63 (average 40.43 ± 9.75 (s.d.), 34 males and 33 females) with MDD were recruited by the Psychiatry Department of Hiroshima University and collaborating medical institutions, based on the Mini International Neuropsychiatric interview26, which enables medical doctors to identify psychiatric disorders according to the Diagnostic and Statistical Manual of Mental Disorders, Fourth Edition3. Their scores of Beck Depression Inventory (BDI)27 ranged from 16 to 53 with average 32.10 ± 9.05 (s.d.) while those of the Japanese version of the national adult reading test (JART)28, which was used to estimate intelligence quotient (IQ), ranged from 87.6 to 125.7 (average 110.0 ± 9.6 (s.d.)). These demographic and clinical characteristics are summarized in Table 1. These patients were either untreated or treated with a single antidepressant at insufficient dose and duration. Additional inclusion criteria were age between 25 and 80 and written informed consent. Exclusion criteria entailed current or past manic episodes; psychotic episodes; alcohol dependence or/and abuse; substance dependence or/and abuse; antisocial personality disorder. As a control group, 67 subjects aged 20–66 (average 36.58 ± 12.03 (s.d.), 30 males and 37 females) with no history of mental or neurological disease were recruited using advertisements in local newspapers. All control subjects underwent the same self-assessment and examination that was administered to the MDD group. Control subjects also undertook the Mini International Neuropsychiatric interview by experienced psychiatrists and psychologists, to ensure that they did not have any previous or current psychiatric disorder according to DSM-IV criteria. For control subjects, BDI scores ranged from 0 to 42 (average 7.35 ± 7.18 (s.d.)) while JART scores ranged from 84.7 to 126.5 (average 111.8 ± 8.5 (s.d.)). Here, too, all subjects gave written informed consent. To compare demographic and clinical characteristics between depressive and control subjects (Table 1), the results of p-value suggest that for sex and IQ the difference between two groups is not significant (p-values are 0.48 and 0.21, respectively), whereas for age and BDI the difference is significant (p-values are 0.014 and 2.2 × 10−21, respectively). Both depressive and control subjects underwent MRI acquisition with a GE scanner. Note that fMRI image data for depressive subjects were obtained within two weeks after the onset of treatment with SSRI.
### Functional MRI data
fMRI measurements were performed at Hiroshima University on a 3T GE Signa HDx scanner with GE-EPI (TR = 2 s, 150 volumes = 5 min scan, TE = 27 ms, FA = 90°, matrix size 64 × 64 × 32, voxel size 4 × 4 × 4 mm, no gap, interleaved). We discarded initial five volumes of the scanned data. Structural T1 images were acquired after the fMRI experiments for correction of head position changes in the subsequent analysis (IRP FSPGR, TR = 6.824 ms, TE = 1.9 ms, FA = 20°, FOV = 256 mm, matrix size 256 × 256 × 180, voxel size 1 × 1 × 1 mm). For scanning of fMRI, the following common settings and instructions were used. In the scan room with dimmed lights, subjects were asked not to think of nor to sleep, but keep looking at a cross mark in the center of the monitor screen. For the pre-processing of the fMRI data, images were realigned, normalized and smoothed (FWHM = 8 mm) using SPM8 (Statistical Parametric Mapping 8)29. They were band-pass filtered (0.009–0.1 Hz) and de-trended.
Using these fMRI measurements, functional connectivity between regions of interest (ROIs) were evaluated. In the present study, we used a common definition of functional connectivity: correlation coefficient of average time series of BOLD fMRI signals between ROIs30. For ROIs, we used recently suggested ROIs based on intrinsic connectivity networks31, which presumably reflect network structures in the brain. These template ROIs were adopted to our data as follows. First, the ROI template was resampled and the center shifted so that the matrix size matched the size of the acquired data and the resting state fMRI image centers coincided (both ROI template as well as resting state fMRI images are normalized to the standard brain). Second, these voxel time series indicated as belonging to the same area by the template were averaged and the correlation coefficient between these average time series was evaluated.
These ROIs consist of 90 brain regions across 14 intrinsic connectivity networks, from a data-driven approach over several subjects by means of Independent Component Analysis. These intrinsic connectivity networks are as follows (digits in parentheses denote the number of ROIs): Anterior Salience (7), Auditory (3), Basal Ganglia (5), Dorsal Default Mode (9), Language (7), Left Executive Control (6), Precuneus (4), Posterior Salience (12), Right Executive Control (6), Ventral Default Mode (10), Visuospatial (11), Primary Visual (2), Higher Visual (2), and Sensorimotor (6). Specifications of ROIs in the present study were based on publicly available nifti files32. Note that in the present paper the numbering that follows these network names exactly match those in these nifti files. In this framework, we selected 78 ROIs, excluding cerebellum-related ROIs, because reliable images were not available for these regions. This resulted in 2701 features (78 × 77/2) for functional connectivity in the present study.
### Clinical questionnaire data
Clinical questionnaire data comprise several types of scores that measure depression severity (e.g., Beck Depression Inventory), personality (e.g., NEO personality Inventory), and life experiences (e.g., Child Abuse Trauma Scale). In particular, Beck Depression Inventory (BDI) and Hamilton Rating Scale of Depression (HRSD) are important indicators for measuring the severity of depression focusing on its several symptoms8. Hence, we also include question items about these indicators in our dataset (Supplementary Table S2 for the whole list of features).
The same clinical questionnaires were administered to both depressive and control subjects. Importantly, for depressive subjects, these were administered before the onset of SSRI treatment. Moreover, some indices, such as BDI and HRSD, were again administered six weeks and six months after the treatment (these features are denoted with endings of 6w and 6m in Supplementary Table S2). We expected that inclusion of repeated measurements in these scores would provide useful information on treatment effect.
### Biological data
Beside functional connectivity factors, we included several biological elements: sex, age, Brain Derived Neurotrophic Factor (BDNF), Cortisol, Single Nucleotide Polymorphisms (SNPs), and DNA Methylation. Specifically, SNPs are for genes relevant to BDNF and serotonin receptors. DNA Methylation is for CpG sites of BDNF receptor genes trkb and for serotonin receptor gene htr2c. Regarding BDNF and cortisol, measurements after six weeks are also included.
### Pre-processing of data
For functional connectivity data, we standardized each feature using mean and standard deviation of the control group, with the effect that we set the baseline to the average values of control subjects. For the remainder of numerical features, we standardized each feature using its mean and standard deviation, based on all subjects (ignoring missing entries). In contrast, for categorical and integer type of features, we did not carry out pre-processing. Note that we left missing entries as they were, because our clustering method can handle them without explicit substitutions.
## Clustering Method
To conduct cluster analysis, we used a novel method for multiple co-clustering, which we recently developed for purposes of the present study (Supplementary Multiple co-clustering method)25. The MATLAB source code for this method is publicly available33. Here, the terminology of co-clustering denotes clustering of both subjects and features. The basic assumption of this method is that different subject cluster solutions may be associated with a specific set of features. Here, we refer to a way of looking at subject clusters (depending on relevant features) as a ‘view’. Since the number of combinations of features is exponential in nature, many views are theoretically possible. However, we aim to identify only those that are probabilistically optimal under the non-overlapping constraint of feature selection (i.e., features are partitioned into these views). Our clustering method is based on nonparametric Bayesian statistics34, which yield multiple views of co-clustering solutions in which each co-clustering solution provides both feature- and subject-clusters. Specifically, by this method an input is a data matrix while an output is view structures (Fig. 1a,b). The key idea of the method is to optimally partition a data matrix into three folds. First, it partitions features into several views, which works as feature selection for different subject cluster solutions. Second, it further partitions features within a view, which bundles similar features. This feature partition in a view is helpful for avoiding the problem of over-fitting in the case of high-dimensional data, and also for interpreting relevant features. Third, it partitions subjects in each view, which yields several subject cluster solutions. These three phases of partitioning are carried out simultaneously, which yields optimal cluster solutions. In short, this method yields optimal subject cluster solutions by simultaneously selecting corresponding sets of relevant features. Nonetheless, cluster solutions yielded by this method may differ, depending on the initial configuration of the phase partitioning. Hence, in order to prevent local optima, as a conventional practice, we apply this method a number of times with different initial configurations, and choose the best clustering solution in terms of data fitting.
For application to a real dataset, this clustering method has several advantages. First, the number of views, the number of feature clusters, and the number of subject clusters are all automatically inferred based on Dirichlet process35. Hence, it is computationally quite efficient. Second, this method can deal with different types of features by pre-specifying possible types of the underlying distribution in a cluster block, i.e., Gaussian, categorical and Poisson. We refer to these types as numerical, categorical, and integer hereafter. Third, it handles missing values in a natural manner in the framework of Bayesian inferences. These characteristics for clustering are specific to this method, which is perfectly suited for analysis of our dataset. Note that a non-Gaussian distribution in numerical features such as a log-normal distribution can be fitted by this method in form of several clusters36. This flexibility in fitting a non-Gaussian distribution is advantageous because it wides applicability of the method. Nonetheless, it requires some care for interpreting cluster results whether obtained clustering results represent the underlying cluster structure.
To apply this clustering method to our data, it is required to pre-specify feature types. For our data, functional connectivity and various psychiatric scores are considered as numerical features (i.e., Gaussian distribution is assumed in each cluster block). In contrast, sex and SNPs are considered categorical. With regard to items BDI and HRSD, these are ordinal categorical data. If we considered these as numerical, we would very poorly fit a Gaussian distribution to the data. Hence, we simply considered these items as categorical. This does not jeopardize interpretation of the solutions. Further, the number of depression episodes and repetitions are integers (see Supplementary Table S2 for more details). As a result, our data consists of 2832 numerical features, 114 categorical features, and 2 integer features.
Regarding prior distributions, we consider non-informative priors, setting hyperparameters as follows (we use the notation in the cited paper)25: For stick-breaking process, α1 = α2 = β = 1 in Beta(·|1, α1), Beta(·|1, α2), and Beta(·|1, β), respectively; for variances of numerical features, γ0 = 1/100 and $${\sigma }_{0}^{2}=\mathrm{1/100}$$ in $$\,{\rm{Ga}}(\,\cdot \,|{\gamma }_{0}\mathrm{/2,}{\gamma }_{0}{\sigma }_{0}^{2}\mathrm{/2)}$$; for means of numerical features, μ0 = 0 and λ0 = 1/100 in $${\rm{Gauss}}(\,\cdot \,|{\mu }_{0},({\lambda }_{0}{s}_{v,g,k}{)}^{-1})$$; for categorical features, ρ0 = 1 for $${\rm{Dirichlet}}(\,\cdot \,|{\rho }_{0})$$; for integer features, α0 = β0 = 1 in Ga(·|α0, β0). With this setting of priors, we performed the method for 1000 initial configurations of clustering, and selected the model that maximized likelihood of clustering solutions. A single run for one initial configuration took about 0.16 hrs, which amounted to 1000 × 0.16 = 160 hrs to complete all computations. We also examined sensitivity of the clustering results to the setting of hyperparameters, which suggested that the results are not sensitive to a small perturbation of our setting of hyperparameters (Supplementary Sensitivity Analysis).
## Results
The multiple co-clustering method yielded 15 views in which there was large variation in the number of features, and the number of subject and feature clusters (Supplementary Table S3; views are sorted in descending order of the number of features). The number of subject clusters ranged from 3 to 9, while the number of feature clusters ranged from 1 to 11 for numerical features, from 1 to 3 for categorical features, and 1 for integer features in case that these types of features may be included in a view. Visual inspection of these views (Fig. 2) shows that the distributions in numerical feature clusters are similar for each subject cluster within a view, except for view 10 (several different features clusters are clearly visible). Further, view memberships of features are displayed in Supplementary Fig. S1 for FC features and in Table 2 for non-FC features. With respect to relationships between views, we focused on the first principal component of numerical features for each view. The first principal component is a weighted linear combination of features that explains most of the variability of features, representing the underlying trend of the data. To compare underlying trends of views, we evaluated Pearson’s correlation coefficient between the first principal components of relevant FC features in each view (Fig. 3a). First principal components are highly correlated between different views, except for view 10, in which the first principal component seems to be independent of those in most of the remaining views (correlated only with views 6, 7, 8 and 12 at significant level 0.05). Note that here we relied on PCA rather than CCA, because CCA focuses on maximum correlation with specific combinations of features while our interest is to evaluate correlations of underlying trends.
Next, we analyzed subject clusters. Our intent was to find meaningful subtypes of depression. Accordingly, we focused on relevance of cluster memberships to depression and separability between subject clusters. First, we evaluated relevance of subject cluster solutions for depression (Fig. 3b). In terms of the concordance between cluster memberships, the label of control/depression, and the proportion of depression related features for selected features, view 10 is the most useful for control/depression labels. Here we defined depression related features as follows: for numerical features BAS, BDI, BIS, GAF, PHQ9, HRSD, JART, PANASP, PANASN, SHAPS, STAI, N, E, O, A, and C; for categorical features drug, Melancholic, Recurrent, Response, Remission, BDI items, HRSD items, SNPs, and MINIs. In short, these features represent the current status of the psychiatric condition of a subject. In Supplementary Table S2, these features are shown with asterisks *. We did not include FC features for this definition.
Second, we evaluated separability of subject clusters of depressive subjects in FC feature clusters by means of Cohen’s d (Fig. 3c)37, which is commonly used to measure effect size of differences of two distributions. Here, we follow the criterion of effect size37: 0.2 small; 0.5 medium; 0.8 large. Using this criterion, subject clusters of depressive subjects in view 10 were well separated, while those in views 4, 6, 7, 8, 12, 13 and 14 were modestly so, and the remainder of the views were slightly separated. These results suggest that view 10 is specific in two regards: relevance for control/depression and large separability of depressive clusters in terms of FC.
Further, we analyzed view 10 more in detail. This view consists of five subject clusters, which match the label of control/depression well: Two clusters for control subjects (subject clusters C1, C2) and three clusters for depressive subjects (subject clusters D1, D2, D3) (Fig. 4a). View 10 contained several non-FC features that discriminate well among subject clusters (features in bold in Table 2). Further, the view had a high proportion of depression-related features (60%), including 39 numerical features and 19 categorical features, but none of the integer features. Since the categorical features do not clearly distinguish between subject clusters (Fig. 4b), we focus only on numerical features in this view for further analysis. Numerical features are clustered into five feature clusters F1-5. We characterize each feature cluster based on characteristics of their member features (Supplementary Table S4). Since this view is closely related to the label of control/depression, it is consistent that the view has feature clusters (namely, F3 and F5) that are related to the initial status of depression. However, it is noteworthy that the view also contains a feature cluster that is related to the after-treatment status of depression (namely, F1). Furthermore, features related to CATS (Child Abuse Trauma Scale) are included in the same feature cluster F1. These results suggest that the subject clusters D1, D2 and D3 for depressive subjects may be related to after-treatment status of depression, which might be further related to stress experiences during childhood. Lastly, feature clusters F2 and F4 are related to specific functional connectivity in fMRI image data, which suggests a possible association between the subject clusters and neural substrates (Supplementary Fig. S1). In this view, the majority of these functional connections are, on average, higher for depressed patients than for controls, while the remainder of views, except for view 11, has more connectivity that is higher for controls than for depressed patients. Further, this connectivity network in view 10 forms the topology of a star with the central hub, which is also observed in other views. The relevant brain areas for the functional connectivity of view 10 are Dorsal DMN.02 (hub), Dorsal DMN.04, Dorsal DMN.06, Ventral DMN.01, Ventral DMN.05, Ventral DMN.09, LECN.01, Precuneus.01, Dorsal DMN.01, Dorsal DMN.03, Ventral DMN.07, RECN.02, and RECN.04, where DMN denotes Default Mode Network; RECN Right Executive Control network; LECN Left Executive Control network (Supplementary Fig. S2). Based on Automated Anatomical Labeling (AAL), the hub of this network is identified as the right angular gyrus (AG) while the remainder of brain areas are ACC (Anterior Cingulate Cortex). R.L, Angular.L, Calcarine.R.L, Occip.Mid.L, Frontal.Med.Orb.R.L., Frontal.Mid.R.L, Frontal.Mid.Orb.L, Frontal.Sup.R.L, Frontal.Sup.Medial.R.L, PCC (Posterior Cingulate Cortex). R.L, Precuneus.L, and Ventral PCC.R.L (Supplementary Table S5). The star topology structure of this network is visualized in Fig. 5, which shows the key role of AG in relevant default mode networks.
In the cluster analysis of view 10, it is important to note that view 10 contains after-six-week scores such as BDI. Since view 10 also includes other features that are available before onset of treatment, this raises the possibility of prediction of treatment outcome prior to treatment. In this regard, we explore several important implications drawn from the results of cluster analysis. First, subject clusters can be represented by a small number of relevant features. Since in our clustering method each feature cluster consists of similar (i.e., highly correlated) features, a feature cluster can be represented by a reduced number of these features. It turns out that a subject cluster solution in a view can be represented by a small number of features associated with each feature cluster. View 10 (Fig. 4a) is represented by CATS scores (associated to feature cluster F1), the first principal scores of angular-gyrus related FC (associated to F2; we simply refer to it ‘AG related FC score’), and BDI (associated to F3). These features can indeed explain the resultant subject cluster membership properly (Fig. 6a). Here, we do not consider features in F4 and F5, because these features are strongly correlated with those in F2 and F3, respectively. Based on distributions of these scores in each subject cluster (Fig. 6b–e), we can characterize subject clusters as follows: subject cluster D1 by high CATS, high FC, low BDI and high BDI6w; subject cluster D2 by low CATS, moderate FC, low BDI and low BDI6w; subject cluster D3 by high CATS, low FC, high BDI and low BDI6w (Supplementary Table S6), where we also include the after-six-week BDI (BDI6w) score associated with F1. This characterization provides a basis for predicting remission of SSRI treatment that is measured by the after-six-week BDI score, because for subject cluster D1, the after-six-week BDI score is high; hence, a subject in D1 is unlikely to remit while a subject in D2 or D3 is likely to remit.
Finally, we investigate the remainder of views. We have identified non-FC features that discriminate between subject clusters based on a statistical test for individual features (features in bold in Table 2). From these results, significant correspondences between non-FC features and brain areas are found in view 4, view 8, and view 12 (Supplementary Table S7). Note that significant correspondence is also observed in view 10). These correspondences help us to interpret these views. First, view 4 includes depression items related to BDI, MINI1, and MINI3, while it also includes FC-features in which the three dominant brain areas are Language.02 (Temporal.Mid.L, in AAL), Ventral DMN.04 (Occipital.Mid.L), and Language.05 (Frontal.Inf.Orb.R) with 25, 22, and 21 connectivity, respectively. Second, view 8 includes features BDI_t1_17 and HRDS_t2_15. These items in clinical questionnaires are related to fatigue and hypochondriasis, respectively. The dominant brain area is Dorsal DMN.02 (Angular gyrus.R). Third, view 12 includes Episode and RecNum, which are related to the extent of repetitions of MDD. The dominant brain areas are Language.04 (Temporal.Mid.L), Dorsal DMN.01 (Cingulum.Ant, Frontal.Sup.Medial), and RECN.04 (Frontal.Sup.Medial).
We have so far analyzed subject cluster solutions one by one. One may wonder if it is possible to integrate all these cluster solutions, yielding a single cluster solution, which may exhibit distinct demographic, clinical and end-phenotypic characteristics. One possible way of such analysis is to carry out hierarchical clustering, using Hamming distance defined as the percentage of views in which subject cluster memberships differ (Supplementary Fig. S3). However, in the present study, we were not able to characterize the yielded clusters in a meaningful manner. Possibly, by integrating cluster solutions, useful information on cluster solutions might be lost.
## Discussion
The results of the cluster analysis suggest three subtypes of depressive subjects, D1, D2, and D3 in view 10. The number of subject clusters in question is three; hence, two binary features may be sufficient for classification. In the scatter plot of subjects in D1, D2, and D3 (Fig. 7a), AG-related FC scores discriminate between subjects in D3 and other subjects. On the other hand, CATS scores do not discriminate a single class by itself, but they discriminate between subjects in D1 and D2, once subjects in D3 are sorted out. This observation motivated us to consider a classifier that consists of the following steps. First, we classify subjects into either D3 or non-D3 based on AG-related FC scores. A subject with low scores in AG-related FC is classified into D3, otherwise into non-D3. Subsequently, the non-D3 subjects are classified into either D1 or D2 based on CATS scores: A subject with low scores in CATS is classified into D2, otherwise into D1. This procedure of classification is summarized in Fig. 7b. Since these subject clusters correspond to degrees of remission of SSRI treatment as well, this classifier leads to predictions of whether SSRI treatment may be effective, prior to the onset of treatment. We can interpret this classification as follows. For subjects in D2 and D3, SSRI treatment may be appropriate (low after-six-week BDI scores, Supplementary Table S6), while for those in D1 SSRI treatment may not be a good option (high after-six-week BDI scores). Note that we do not take initial BDI scores into account in this classifier. Interestingly, the initial BDI score is not a good indicator for predictions of after-six-week BDI scores in this classification (Fig. 6d,e). For instance, for subjects in D3, the initial BDI scores are high, while after-six-week BDI scores are low, suggesting that despite having severe depression, they are likely to remit. It is notable that such a non-trivial case can be captured in our classifier based on AG-related FC score and CATS score.
Next, we interpreted implications of this classifier. For the relationship between CATS and remission, our finding is consistent with the meta analysis of previous studies38, which clearly suggests that experiences of child abuse trauma have a negative impact on treatment of depression. The contribution of our study in this regard is that we were able to identify among a huge number of possible associations this specific association in an unsupervised manner without prior knowledge of feature selection. As for the medical causal relationship, recent studies suggest that treatment-resistant depression may be linked to release of pre-inflammatory cytokines, which can be caused by childhood adversity39,40. However, in our research framework, biomarkers of inflammation were not included, which prevents us from confirming this point. Concerning the key role of the angular gyrus (AG) in predicting remission, our finding is consistent with the results of a t-test on differences between TRD and non-TRD18,19. Unfortunately, these two previous studies are contradictory because the former study suggests that a higher FC is associated with TRD, while the latter study found that a lower FC is associated with TRD. Our result supports the former study. Moreover, recent studies have revealed that AG is related to several functions, such as semantic processing, default mode network, number processing, attention and spatial cognition, and memory retrieval41. Such multiple functions of AG are consistent with an fMRI study42, which suggests that AG is one of the major connecting hubs, together with the occipital and ventral-medial parietal. Nonetheless, these results of previous studies do not adequately explain the possible association between AG and remission of depression implied in our study. Further research is required for clarification of this point, which may provide useful insights into possible treatment of depressive patients by means of neurofeedback.
Lastly, we provide possible interpretations for views 4, 8, and 12 in which significant correspondences between non-FC features and brain areas have been found. We base our interpretations on previous studies of relevant brain areas. First, for view 4, the relevance of the identified brains areas, temporal, occipital, and frontal orbital cortex, to depression is suggested in terms of cell communication43, concentration of γ-Aminobutyric acid (GABA)44, and volume45, respectively. Also taking into account that this view includes depression-related features that discriminate subject clusters, we interpret this to mean that view 4 represents subject clusters of different levels of depressive symptoms (note that view 4 is related to depression in terms of subject clusters in Fig. 3b). Second, we have found that view 8 is characterized by non-FC features related to fatigue and FC features of angular gyrus. In previous studies, it is reported that the angular gyrus modulates α rhythm46, which is related to fatigue47. Hence, we interpret that to mean that view 8 represents subject clusters related to fatigue. Third, view 12 includes repetitions of depression as non-FC features while temporal, anterior cingulum cortex, and frontal cortex are viewed as FC features. It is reported that neuroplastic change occurs during depressive episodes in dorsomedial prefrontal cortex, dorsolateral prefrontal cortex, and anterior cingulum48, which matches our results. Hence, we believe that view 12 represents subject clusters related to the duration of episodes of depression.
Finally, we discuss limitations of the present study. First, we did not include the cerebellum because the image on this brain region was not reliably obtained. However, several resting state fMRI studies suggest that the cerebellum plays a key role in depression49,50,51. Second, from a statistical point of view, the sample size in our study is rather small. To generalize the results of the present study, further research is required with much larger sample size, which might be possible to be obtained by combining samples from different studies. Lastly, we should mention that our data had a significant difference in age between depressive and control subjects (Table 1) unlike usual case-control studies on psychiatric disorders. In fact, the bias could correlate with a risk and/or a severity of depression to some degree, as the feature ‘age’ was assigned to the same view as several clinical scores of depression symptoms in our clustering result (view 6 in Table 2). However, this view was independent of that containing AG-related FC score and CATS score (view 10 in Table 2), suggesting that the age bias has little influence on our main results.
## Data Availability
Due to potentially identifying information, the data for this study are ethically restricted by the Ethical Committee for Epidemiology of Hiroshima University, Japan. Interested, qualified researchers may request the data by contacting Dr. Shoji Karatsu ([email protected]).
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## Acknowledgements
This research is supported by the Strategic Research Program for Brain Sciences from Japan Agency for Medical Research and development, AMED. We would like to thank Dr. Steven D. Aird at Okinawa Institute of Science and Technology Graduate University for his proof-reading of this article.
## Author information
Authors
### Contributions
T.T., J.Y., Y.S. and K.D. carried out statistical analysis and gave interpretations. G.O., M.T., Y.O. and S.Y. designed a study, collected data and pre-processed functional MRI data. Y.S. evaluated functional connectivity in fMRI data, adopting template ROIs to our data. T.T. wrote the main manuscript text and prepared the figures.
### Corresponding author
Correspondence to Tomoki Tokuda.
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Tokuda, T., Yoshimoto, J., Shimizu, Y. et al. Identification of depression subtypes and relevant brain regions using a data-driven approach. Sci Rep 8, 14082 (2018). https://doi.org/10.1038/s41598-018-32521-z
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• DOI: https://doi.org/10.1038/s41598-018-32521-z
### Keywords
• Child Abuse Trauma
• Clinical Questionnaire
• Angular Gyrus (AG)
• Default Mode Network
• Hamilton Rating Scale Of Depression (HRSD)
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• ### Identification of transdiagnostic psychiatric disorder subtypes using unsupervised learning
• Helena Pelin
• Marcus Ising
• Till F. M. Andlauer
Neuropsychopharmacology (2021)
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2022-01-29 08:39:09
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http://www.math.iisc.ac.in/seminars/2021/2021-02-24-arka-mallick.html
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#### APRG Seminar
##### Venue: Microsoft Teams (online)
In this talk, I would like to present some recent results regarding the behaviour of functions which are uniformly bounded under the action of a certain class of non-convex non-local functionals related to the degree of a map. In the literature, this class of functionals happens to be a very good substitute of the $L^p$ norm of the gradient of a Sobolev function. As a consequence various improvements of the classical Poincaré’s inequality, Sobolev’s inequality and Rellich-Kondrachov’s compactness criterion were established. This talk will be focused on addressing the gap between a certain exponential integrability and the boundedness for functions which are finite under the action of these class of non-convex functionals.
Contact: +91 (80) 2293 2711, +91 (80) 2293 2265 ; E-mail: chair.math[at]iisc[dot]ac[dot]in
Last updated: 18 Sep 2021
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2021-09-19 02:32:38
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https://solvedlib.com/n/2-which-of-the-following-is-a-linear-system-3-points-3x,3109940
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# 2. Which of the following is a linear system (3 Points)3x +2xy =5 2x +yNone of all.+3y1/2 =4 +2y =5+2y
###### Question:
2. Which of the following is a linear system (3 Points) 3x +2xy =5 2x +y None of all. +3y1/2 =4 +2y =5 +2y =5 3x +2y =6 2x +3y = 7 sin X +y = 0
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2022-07-05 07:06:50
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https://www.physicsforums.com/threads/infimum-axiom.688151/
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# Infimum axiom
Homework Helper
## Homework Statement
So I have started my long journey through N.L. Carothers Real Analysis and my intention is to work through every single exercise along the way.
The first problem : http://gyazo.com/ddef0387f04d789c660548c08796585d
## The Attempt at a Solution
Suppose A is bounded below, we want to show A has an infimum.
That is, we want to show ##\exists m \in ℝ## such that :
(i) ##m## is a lower bound for A.
(ii) If ##x## is any lower bound for A, then ##x ≤ m##.
So suppose (i) is satisfied ( Since A is bounded below ), we want to show (ii) holds, so we consider the set ##-A = \{ -a \space | \space a \in A \}##.
Since A is bounded below by m, -A is bounded above by -m. Since -A is a nonempty subset of real numbers with an upper bound, ##\exists s \in ℝ \space | \space -m ≤ s##.
Hence sup(-A) = -m ( i.e, -m is the least upper bound of -A ) so that -sup(-A) = m ( i.e, m is the greatest lower bound of A ), but this is precisely what we desire. Thus -sup(-A) = inf(A) = m as desired.
That was my go at it. Any thoughts to make this better? I feel as if my argument wasn't as strong as it should be near the end.
LCKurtz
Homework Helper
Gold Member
## Homework Statement
So I have started my long journey through N.L. Carothers Real Analysis and my intention is to work through every single exercise along the way.
The first problem : http://gyazo.com/ddef0387f04d789c660548c08796585d
## The Attempt at a Solution
Suppose A is bounded below, we want to show A has an infimum.
That is, we want to show ##\exists m \in ℝ## such that :
(i) ##m## is a lower bound for A.
(ii) If ##x## is any lower bound for A, then ##x ≤ m##.
So suppose (i) is satisfied ( Since A is bounded below ), we want to show (ii) holds, so we consider the set ##-A = \{ -a \space | \space a \in A \}##.
A is bounded below so it has a lower bound m. That doesn't mean m is the greatest lower bound nor that it satisfies ii.
Since A is bounded below by m, -A is bounded above by -m. Since -A is a nonempty subset of real numbers with an upper bound, ##\exists s \in ℝ \space | \space -m ≤ s##.
No. Nothing says -m is a least upper bound for -A. The least upper bound s will satisfy ##s\le -m##.
Hence sup(-A) = -m ( i.e, -m is the least upper bound of -A ) so that -sup(-A) = m ( i.e, m is the greatest lower bound of A ), but this is precisely what we desire. Thus -sup(-A) = inf(A) = m as desired.
That was my go at it. Any thoughts to make this better? I feel as if my argument wasn't as strong as it should be near the end.
Even if you did have that -m is the l.u.b. of -A, you would still have to prove i and ii hold for it. Back to the drawing board. You might think about showing - sup(-A) works for the g.l.b of A.
Homework Helper
A is bounded below so it has a lower bound m. That doesn't mean m is the greatest lower bound nor that it satisfies ii.
Okay, so suppose that m is a lower bound of A. ( So (i) is satisfied and we want (ii) to be satisfied at the end ). So we consider the set -A.
No. Nothing says -m is a least upper bound for -A. The least upper bound s will satisfy ##s ≤ -m##
Since m is a lower bound for A, -m is an upper bound for -A. Since -A is a nonempty subset of real numbers with an upper bound, ##\exists s \in ℝ \space | \space s ≤ -m##.
So sup(-A) = s so that 's' is the least upper bound for -A. Hence -sup(-A) = -s is a lower bound for -A.
Since -s ≥ m and m is a lower bound for A, this tells us that -s is the greatest lower bound for A and hence inf(A) = -sup(-A) = -s.
This seems to make sense ( hopefully ).
Last edited:
LCKurtz
Homework Helper
Gold Member
Okay, so suppose that m is a lower bound of A. ( So (i) is satisfied and we want (ii) to be satisfied at the end ). So we consider the set -A.
Since m is a lower bound for A, -m is an upper bound for -A.
It's easy, but you should show this, not just state it.
Since -A is a nonempty subset of real numbers with an upper bound, ##\exists s \in ℝ \space | \space s ≤ -m##.
Sure, but there may be lots of such s's. What do you really want for s?
So sup(-A) = s so that 's' is the least upper bound for -A.
Why? See above.
Hence -sup(-A) = -s is a lower bound for -A.
Again, you should show this.
Since -s ≥ m and m is a lower bound for A, this tells us that -s is the greatest lower bound for A
No it doesn't. You have to show that if n is any lower bound for A then ##n\le m##.
and hence inf(A) = -sup(-A) = -s.
This seems to make sense ( hopefully ).
You have a ways to go yet.
Homework Helper
Okay, so suppose that m is a lower bound of A. ( So (i) is satisfied and we want (ii) to be satisfied at the end ). So we consider the set -A.
Since m is a lower bound for A, -m is an upper bound for -A.
It's easy, but you should show this, not just state it.
Since ##m ≤ a, \space \forall a \in A, \space -m ≥ -a, \space \forall (-a) \in (-A)##. So that -m is an upper bound for -A.
Since -A is a nonempty subset of real numbers with an upper bound, ∃s∈R | s≤−m.
Sure, but there may be lots of such s's. What do you really want for s?
I want this s so I can show that -A has a supremum. Not just an upper bound ( sup axiom ).
So sup(-A) = s so that 's' is the least upper bound for -A.
Why? See above.
Above.
Hence -sup(-A) = -s is a lower bound for A.
Again, you should show this.
Since -s ≥ m and m is a lower bound for A, would that not show that inf(A) = -s = -sup(-A) ( Even stronger than a lower bound as -s is the greatest lower bound )?
Since s ≤ -m → -s ≥ m and m is a lower bound for A, this tells us that -s is the greatest lower bound for A
No it doesn't. You have to show that if n is any lower bound for A then n≤m.
The point of the question is to show A has an infimum using the fact that A is bounded below ( by m ).
So I assumed A had a lower bound m. Then I considered -m as the upper bound of -A. Then using the sup axiom I found sup(-A) = s so that inf(A) = -s so A has an infimum. I don't seem to see where this goes wrong?
LCKurtz
Homework Helper
Gold Member
The point of the question is to show A has an infimum using the fact that A is bounded below ( by m ).
So I assumed A had a lower bound m. Then I considered -m as the upper bound of -A. Then using the sup axiom I found sup(-A) = s
That is not what your argument said. You said "Since -A is a nonempty subset of real numbers with an upper bound, ∃s∈R | s≤−m". You said nothing about sup(-A). You haven't written what you mean.
so that inf(A) = -s so A has an infimum. I don't seem to see where this goes wrong?
That is the idea of the argument. But your writeup is not very good. And the copy function on PF makes it very difficult to respond appropriately.
Last edited:
Homework Helper
I would have assumed you had knowledge of the upper axiom, which is perhaps the reason my argument didn't seem too precise.
Is it okay afterwards though, considering I'm allowed to at least use the upper axiom without proof ( heavily implied by the author ).
LCKurtz
Homework Helper
Gold Member
I would have assumed you had knowledge of the upper axiom, which is perhaps the reason my argument didn't seem too precise.
I am not the student here, you are. I have been trying to get you to understand what's wrong with your argument. You don't seem to know how to write up a good argument. I could show you how but you don't seem to actually address the issues I raise.
Homework Helper
I am not the student here, you are. I have been trying to get you to understand what's wrong with your argument. You don't seem to know how to write up a good argument. I could show you how but you don't seem to actually address the issues I raise.
I apologize for not including the axiom in relevant equations. I've addressed some of your concerns; if it's clear that I have the right idea, then feel free to expand on things.
LCKurtz
Homework Helper
Gold Member
I apologize for not including the axiom in relevant equations. I've addressed some of your concerns; if it's clear that I have the right idea, then feel free to expand on things.
This thread has gotten too disjointed to follow. At this point, if you would like to post your careful, most complete, best solution, with all the steps, I will be happy to respond to it. Write it up as if you were going to hand it in for grading. It will be later this evening before I can be back, so take your time and do it right.
Homework Helper
The question : http://gyazo.com/ddef0387f04d789c660548c08796585d
Useable info : Sup Axiom.
Suppose ##A## is a nonempty subset of ##ℝ## bounded below by ##n##, we want to show ##A## has an infimum.
Consider the set from the hint : ##-A = \{-a \space | \space a \in A \}##
We want to show that ##n## which is a lower bound for ##A##, is an upper bound for ##-A##. So, since ##n ≤ a, \space \forall a \in A \Rightarrow -n ≥ -a, \space \forall a \in (-A)##. So since ##(-a) \in A##, we conclude ##n ≥ -a## and so ##n## is an upper bound for ##-A##.
Now, since ##-A## is bounded above by ##n## and ##-A## is a nonempty subset of the reals, we know by the supremum axiom that ##\exists m \in ℝ \space | \space sup(-A) = m##
Now, we want to show ##-m## is such that :
(i) ##-m## is a lower bound for ##A##.
(ii) If ##n## is any lower bound for ##A##, then ##n≤m##.
To prove (i) lets assume the contrary that ##-m## is not a lower bound for ##A##. Then ##\exists a \in A \space | \space -m ≥ a \Rightarrow m ≤ -a##. Since ##(−a) \in (-A)##, this contradicts the fact that ##sup(-A) = m## from earlier and hence ##-m## must be a lower bound for ##A##.
I'm having some trouble with (ii) for some reason. I assumed the contrary that ##n≥m##, but I'm having a hard time continuing with it.
Dick
Homework Helper
The question : http://gyazo.com/ddef0387f04d789c660548c08796585d
Useable info : Sup Axiom.
Suppose ##A## is a nonempty subset of ##ℝ## bounded below by ##n##, we want to show ##A## has an infimum.
Consider the set from the hint : ##-A = \{-a \space | \space a \in A \}##
We want to show that ##n## which is a lower bound for ##A##, is an upper bound for ##-A##. So, since ##n ≤ a, \space \forall a \in A \Rightarrow -n ≥ -a, \space \forall a \in (-A)##. So since ##(-a) \in A##, we conclude ##n ≥ -a## and so ##n## is an upper bound for ##-A##.
Now, since ##-A## is bounded above by ##n## and ##-A## is a nonempty subset of the reals, we know by the supremum axiom that ##\exists m \in ℝ \space | \space sup(-A) = m##
Now, we want to show ##-m## is such that :
(i) ##-m## is a lower bound for ##A##.
(ii) If ##n## is any lower bound for ##A##, then ##n≤m##.
To prove (i) lets assume the contrary that ##-m## is not a lower bound for ##A##. Then ##\exists a \in A \space | \space -m ≥ a \Rightarrow m ≤ -a##. Since ##(−a) \in (-A)##, this contradicts the fact that ##sup(-A) = m## from earlier and hence ##-m## must be a lower bound for ##A##.
I'm having some trouble with (ii) for some reason. I assumed the contrary that ##n≥m##, but I'm having a hard time continuing with it.
You are making this complicated enough that it's making my eyes hurt. Probably because you are trying to prove everything by contradiction. You don't have to. If A has a lower bound consider the set of all lower bounds of A, call it L. Show L is bounded above. So L has a sup. So?
LCKurtz
Homework Helper
Gold Member
The question : http://gyazo.com/ddef0387f04d789c660548c08796585d
Useable info : Sup Axiom.
Suppose ##A## is a nonempty subset of ##ℝ## bounded below by ##n##, we want to show ##A## has an infimum.
Consider the set from the hint : ##-A = \{-a \space | \space a \in A \}##
We want to show that ##n## which is a lower bound for ##A##, is an upper bound for ##-A##.
That isn't even true. The same n that is a lower bound for A is not an upper bound for -A.
So, since ##n ≤ a, \space \forall a \in A \Rightarrow -n ≥ -a, \space \forall a \in (-A)##. So since ##(-a) \in A##, we conclude ##n ≥ -a## and so ##n## is an upper bound for ##-A##.
No. Since each element of -A is -a for some a in A, and ##n\le a## you have ##-n\ge -a## which tells you that -n is an upper bound for -A
Now, since ##-A## is bounded above by ##n## (you mean ##-n##) and ##-A## is a nonempty subset of the reals, we know by the supremum axiom that ##\exists m \in ℝ \space | \space sup(-A) = m##
Now, we want to show ##-m## is such that :
(i) ##-m## is a lower bound for ##A##.
(ii) If ##n## is any lower bound for ##A##, then ##n≤m##.
To prove (i) lets assume the contrary that ##-m## is not a lower bound for ##A##. Then ##\exists a \in A \space | \space -m ≥ a \Rightarrow m ≤ -a##. Since ##(−a) \in (-A)##, this contradicts the fact that ##sup(-A) = m## from earlier and hence ##-m## must be a lower bound for ##A##.
I'm having some trouble with (ii) for some reason. I assumed the contrary that ##n≥m##, but I'm having a hard time continuing with it.
Like Dick has observed, you don't need any indirect argument. Just like the argument above where a lower bound of n for A gives an upper bound of -n for -A you should be able to show that an upper bound m for -A gives rise to a lower bound -m for A and you should be able use the fact that you have a least upper bound for -A gives rise to a greatest lower bound for A.
Homework Helper
You are making this complicated enough that it's making my eyes hurt. Probably because you are trying to prove everything by contradiction. You don't have to. If A has a lower bound consider the set of all lower bounds of A, call it L. Show L is bounded above. So L has a sup. So?
For (ii), suppose that A is bounded below by n. We want to show n ≤ -m. Consider the set ##L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}## which is the set of all lower bounds of A.
Clearly ##n, -m \in L## by hypothesis and (i) so that ##L## is not empty.
---------
EDIT : I realized after reading LC's comment that there is a very easy way to do this without really putting effort into (i) or (ii).
Since m is the least upper bound of -A, -m is the greatest lower bound for A. That is ##m ≥ -a \Rightarrow -m ≤ a## so that -m is the greatest lower bound of A.
Last edited:
CAF123
Gold Member
I think the statement you are trying to prove is the Completeness property for infima. If you know the Completeness Property for suprema and the Reflection principles, then I believe this can be done in a couple of steps.
LCKurtz
Homework Helper
Gold Member
EDIT : I realized after reading LC's comment that there is a very easy way to do this without really putting effort into (i) or (ii).
Since m is the least upper bound of -A, -m is the greatest lower bound for A. That is ##m ≥ -a \Rightarrow -m ≤ a## so that -m is the greatest lower bound of A.
But you can't just declare that and call it a proof. You have to get your hands dirty with both i and ii. To prove them for A you have to use the corresponding properties for sup and -A and use them to get the results you want for A.
Homework Helper
You are making this complicated enough that it's making my eyes hurt. Probably because you are trying to prove everything by contradiction. You don't have to. If A has a lower bound consider the set of all lower bounds of A, call it L. Show L is bounded above. So L has a sup. So?
But you can't just declare that and call it a proof. You have to get your hands dirty with both i and ii. To prove them for A you have to use the corresponding properties for sup and -A and use them to get the results you want for A.
Back from a long work shift, lets do math. I did (i) by contradiction did I not? Anyway, i'll try them both without any contradiction.
Since m is the least upper bound of -A, -m is a lower bound for A. That is ##m≥−a \Rightarrow −m≤a##.
Argh! I'm literally getting stuck with the reasoning here. My issue is that I said m was an lower bound for A at the beginning of my proof, but now I've just shown -m is also a lower bound?
LCKurtz
Homework Helper
Gold Member
Argh! I'm literally getting stuck with the reasoning here. My issue is that I said m was an lower bound for A at the beginning of my proof, but now I've just shown -m is also a lower bound?
That is part of the reason it is so difficult to help you. What is m and where is it defined? That's why I asked you to give a complete argument from the beginning, which you attempted in post #11. In post #13 I indicated some problems with your argument. You have not acknowledged that you understood my corrections so I don't know if you even read them or whether you understood them if you did read them. Now we are back to you being confused about m or -m.
I'm sorry, but I am tiring of this thread and I think I have said all I want to say.
Dick
Homework Helper
Back from a long work shift, lets do math. I did (i) by contradiction did I not? Anyway, i'll try them both without any contradiction.
Since m is the least upper bound of -A, -m is a lower bound for A. That is ##m≥−a \Rightarrow −m≤a##.
Argh! I'm literally getting stuck with the reasoning here. My issue is that I said m was an lower bound for A at the beginning of my proof, but now I've just shown -m is also a lower bound?
The proof is even easier if you don't use the hint. L, the set of lower bounds, has an upper bound, any element of A is an upper bound. That means L has a sup, call it n. Doesn't n have all of the properties you need to be an inf of A?
Last edited:
Homework Helper
The question : http://gyazo.com/ddef0387f04d789c660548c08796585d
Useable info : Sup Axiom.
Suppose ##A## is a nonempty subset of ##ℝ## bounded below, we want to show ##A## has an infimum.
Consider the set of lower bounds of A : ##L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}##. Notice L is not empty because A is bounded below. Since ##x ≤ a##, L is bounded above for any ## a \in A##.
Now, L is a nonempty subset of real numbers, which is bounded above by at least one element ( Since A is nonempty ), so we know sup(L) exists by the sup axiom, call it n.
n is in the set of lower bounds of a, so it satisfies (i). Now since ##x ≤ n ≤ a## ( From each respective inequality throughout the proof ) (ii) is satisfied. So we can write inf(A) = n.
Dick
Homework Helper
The question : http://gyazo.com/ddef0387f04d789c660548c08796585d
Useable info : Sup Axiom.
Suppose ##A## is a nonempty subset of ##ℝ## bounded below, we want to show ##A## has an infimum.
Consider the set of lower bounds of A : ##L = \{x \in ℝ \space | \space x ≤ a, \space \forall a \in A \}##. Notice L is not empty because A is bounded below. Since ##x ≤ a##, L is bounded above for any ## a \in A##.
Now, L is a nonempty subset of real numbers, which is bounded above by at least one element ( Since A is nonempty ), so we know sup(L) exists by the sup axiom, call it n.
n is in the set of lower bounds of a, so it satisfies (i). Now since ##x ≤ n ≤ a## ( From each respective inequality throughout the proof ) (ii) is satisfied. So we can write inf(A) = n.
Well, sure. If I understand what you mean by i) and ii). Now it really wouldn't hurt to try to do it using the hint considering -A. All you should have to do is change '+' into '-' and '<' into '>' and apply definitions. Not turn it into a hash of confusing proofs by contradiction.
Homework Helper
http://gyazo.com/ddef0387f04d789c660548c08796585d
Useable info : Sup Axiom.
Suppose ##A## is a nonempty subset of ##ℝ## bounded below by ##n \in ℝ##, we want to show ##A## has an infimum.
Consider the set from the hint : ##-A = \{-a \space | \space a \in A \}##
Since n is a lower bound for ##A##, -n is an upper bound for ##-A##. Proof : ##n ≤ a, \space \forall a \in A## so that ##-n ≥ -a, \space \forall -a \in -A##. Thus -n is an upper bound for -A.
***Sorry I just have to stop here and ask something, could I not just use your same idea of considering the set of all upper bounds of -A right now? Anyway back to having no knowledge of that.
Since -A is a nonempty subset of the reals and is bounded above, we know sup(-A) exists by the sup axiom, call it m.
Since -a ≤ m, we know a ≥ -m so that -m is a lower bound for A, so -m satisfies (i). Since n is another lower bound of A, we want n ≤ -m ≤ a, but that means -n ≥ m ≥ -a which we know is true from prior inequalities in the proof, so (ii) is true. Thus inf(A) = -m.
Dick
Homework Helper
http://gyazo.com/ddef0387f04d789c660548c08796585d
Useable info : Sup Axiom.
Suppose ##A## is a nonempty subset of ##ℝ## bounded below by ##n \in ℝ##, we want to show ##A## has an infimum.
Consider the set from the hint : ##-A = \{-a \space | \space a \in A \}##
Since n is a lower bound for ##A##, -n is an upper bound for ##-A##. Proof : ##n ≤ a, \space \forall a \in A## so that ##-n ≥ -a, \space \forall -a \in -A##. Thus -n is an upper bound for -A.
***Sorry I just have to stop here and ask something, could I not just use your same idea of considering the set of all upper bounds of -A right now? Anyway back to having no knowledge of that.
Since -A is a nonempty subset of the reals and is bounded above, we know sup(-A) exists by the sup axiom, call it m.
Since -a ≤ m, we know a ≥ -m so that -m is a lower bound for A, so -m satisfies (i). Since n is another lower bound of A, we want n ≤ -m ≤ a, but that means -n ≥ m ≥ -a which we know is true from prior inequalities in the proof, so (ii) is true. Thus inf(A) = -m.
It's kind of late here so I might not be reading every line in detail. But yes, that's about how easy it is. No proof by contradiction needed, right?
Homework Helper
It's kind of late here so I might not be reading every line in detail. But yes, that's about how easy it is. No proof by contradiction needed, right?
Yes, if I explicitly mention some inequalities along the way while I'm using my definitions, then no contradiction is needed at all.
After seeing it the easy way with L, this way with -A just sort of fell together when I realized how important the inequalities themselves actually were. The ex with L helped me actually think about how the quantities were being manipulated for some reason and how to actually prove (i) and (ii).
Any comments on *** though out of question? Is my hunch okay?
Dick
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2021-09-19 15:12:52
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https://www.physicsforums.com/threads/an-integral-about-bessel-function.532595/
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# An integral about Bessel function
1. Sep 22, 2011
### zluo
Is there somebody who knows the solution (closed form) for the integral
$$\int^\infty_0\frac{J^3_1(ax)J_0(bx)}{x^2}dx$$
where $a>0,b>0$ and $J(\cdot)$ the bessel function of the first kind with integer order?
Reference, or solution from computer programs all are welcome. Thanks!
2. Jun 7, 2012
### <MIT|s|Mag>
even if belatedly believe that this property of bessel funtions is useful
$$\frac{J_{n}(x)}{x}=\frac{J_{n-1}+J_{n+1}}{2n}$$
with the orthonormality of bessel functions
$$\int^\infty_0 J_{n}(ax)J_{n}(bx)xdx=\frac{1}{a}\delta(a-b)$$
$$\int^\infty_0\left(\frac{J_1(ax)}{x}\right)^{3}J_0(bx)xdx =\int^\infty_0\left(\frac{J_{1-1}+J_{1+1}}{2\cdot1}\right)^{3}J_0(bx)xdx=\dots$$
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2018-01-23 22:29:30
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https://wikidiff.com/difference/distinctive
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# Distinctive vs Difference - What's the difference?
distinctive | difference |
is that serves to distinguish between things.
difference.
# distinctive
## English
• that serves to distinguish between things
• that is characteristic or typical of something
• # difference
## English
### Noun
• (uncountable) The quality of being different.
• (countable) A characteristic of something that makes it different from something else.
• * {{quote-magazine, title=Towards the end of poverty
• , date=2013-06-01, volume=407, issue=8838, page=11, magazine=(The Economist) citation , passage=But poverty’s scourge is fiercest below \$1.25 (the average of the 15 poorest countries’ own poverty lines, measured in 2005 dollars and adjusted for differences in purchasing power): people below that level live lives that are poor, nasty, brutish and short.}}
• (countable) A disagreement or argument.
• We have our little differences , but we are firm friends.
• * Shakespeare
• What was the difference ? It was a contention in public.
• * T. Ellwood
• Away therefore went I with the constable, leaving the old warden and the young constable to compose their difference as they could.
• (countable, uncountable) Significant change in or effect on a situation or state.
• * 1908 , (Kenneth Grahame), (The Wind in the Willows)
• The line of the horizon was clear and hard against the sky, and in one particular quarter it showed black against a silvery climbing phosphorescence that grew and grew. At last, over the rim of the waiting earth the moon lifted with slow majesty till it swung clear of the horizon and rode off, free of moorings; and once more they began to see surfaces—meadows wide-spread, and quiet gardens, and the river itself from bank to bank, all softly disclosed, all washed clean of mystery and terror, all radiant again as by day, but with a difference that was tremendous.
• (countable) The result of a subtraction; sometimes the absolute value of this result.
• (obsolete) Choice; preference.
• * Spenser
• That now be chooseth with vile difference / To be a beast, and lack intelligence.
• (heraldry) An addition to a coat of arms to distinguish two people's bearings which would otherwise be the same. See augmentation and cadency.
• (logic) The quality or attribute which is added to those of the genus to constitute a species; a differentia.
• (logic circuits) A Boolean operation which is TRUE when the two input variables are different but is otherwise FALSE; the XOR operation ($\scriptstyle A \overline B + \overline A B$).
• (relational algebra) the set of elements that are in one set but not another ($\scriptstyle A \overline B$).
• #### Synonyms
* (characteristic of something that makes it different from something else) departure, deviation, divergence * (disagreement or argument about something important) conflict, difference of opinion, dispute, dissension * (result of a subtraction) remainder * (significant change in state) nevermind
#### Antonyms
* (quality of being different) identity, sameness
#### Derived terms
* distinction without a difference * creative differences * difference engine * difference equation * difference gate * difference of two squares * goal difference * same difference * split the difference * spot the difference * tell the difference
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2020-10-28 20:00:10
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https://tex.stackexchange.com/questions/550762/big-bracket-structure-linguistics-feature-structure
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# Big bracket structure (linguistics feature structure)
I would like to create something along the lines of this:
I have tried, but failed miserably. I know that I can use \left[ and \right], but other than that I am completely lost. Can anybody help me out with this? Thank you in advance!
Since these are the kinds of structures you may be drawing a lot, it's really best to create some helper macros for formatting them. Here I've created a feature bundle macro that takes a comma delimited list of features and puts then in a bracketed matrix. The optional argument specifies the subscripted element for that bundle.
\fbun{F1,F2,F3}
I've also created an uninterpretable feature macro to format features with an italic u and the feature in small caps.
\uf{f1}
Finally I've created a feature domination macro which puts two features in a domination relation:
\fdom{f1}{f2}
Putting it all together we get. Personally I wouldn't use $\nu$ for little-v, but simply use $v$. I've also scaled the # sign, which in Latin Modern is quite large and ugly.
\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{etoolbox}
\usepackage{graphicx}
\renewcommand\#{\protect\scalebox{0.75}{\protect\raisebox{0.4ex}{\char"0023}}}% smaller \# from https://tex.stackexchange.com/q/256553/2693
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\fbun}{om}{%
\IfNoValueTF{#1}
{\ensuremath{\begin{bmatrix}
\end{bmatrix}}}
{\ensuremath{\begin{bmatrix}
\end{bmatrix}\sb{\textstyle#1}}}
}
\ExplSyntaxOff
\newcommand*{\fdom}[2]{\ensuremath{\begin{array}{@{}c@{}}\text{#1}\\\vrule\\\text{#2}\end{array}}}
\newcommand*{\uf}[1]{\textit{u}\textsc{#1}}
\begin{document}
$\nu\left[\fbun[\pi]{ \fdom{\uf{pers}}{\uf{part}} } \rhd \fbun[\#]{ \fdom{\uf{pers}}{\uf{pl}} } \right]$
\end{document}
You can obtain it with amsmath:
\documentclass{article}
\usepackage{amsmath, amssymb}
\begin{document}
$\nu \left[\begin{bmatrix} u\textsc{\small pers}\\ \vrule \\ u\textsc{\small part} \end{bmatrix}_{\!\!\pi}\rhd \begin{bmatrix} u\textsc{\small pers}\\ \vrule \\ u\textsc{\small pl} \end{bmatrix}_{\!\!\#}\right]$%
\end{document}
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2023-01-31 13:18:42
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https://asmedigitalcollection.asme.org/gasturbinespower/article-abstract/120/1/115/408015/The-Maximum-Factor-by-Which-Forced-Vibration-of?redirectedFrom=fulltext
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In 1966 it was shown that the maximum factor by which the amplitude of forced vibration of blades can increase due to mistuning is, with certain assumptions, $1/2(1+N)$, where N is the number of blades in the row. This report gives a further investigation of the circumstances when this factor can be obtained. These are small damping, and a relationship must hold between the mistuning distribution γ(s) and the interblade coupling function c(r), where r is the model number. The mistuning distribution must be symmetric about the blade on which maximum amplitude is to occur, s = 0. The coupling function must be symmetric about r = R, where R is the mode number of the excitation. If the coupling is purely mechanical, additional conditions apply. The coupling function c(r) must consist of a number of identical symmetric substrips. A 1976 result for mechanical coupling is amended.
1.
D. S.
,
1966
, “
Effect of Mistuning on the Vibration of Turbomachine Blades Induced by Wakes
,”
Jour. Mech. Eng. Sci.
, Vol.
8
, pp.
15
21
.
2.
D. S.
,
1976
, “
Effect of Mistuning on Forced Vibration of Blades With Mechanical Coupling
,”
Jour. Mech. Eng. Sci.
, Vol.
18
, pp.
306
307
.
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Yiu, H., 1991, “A Review of the Literature on Mistuning Effects in Bladed Disc Vibration,” Report No: VUTC/E/92001, Imperial College, London.
This content is only available via PDF.
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2019-10-18 09:27:25
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https://studysoup.com/tsg/1148495/atkins-physical-chemistry-11-edition-chapter-9b-problem-e9b-4-b
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Get Full Access to Atkins' Physical Chemistry - 11 Edition - Chapter 9b - Problem E9b.4(b)
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ISBN: 9780198769866 2042
## Solution for problem E9B.4(b) Chapter 9B
Atkins' Physical Chemistry | 11th Edition
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Problem E9B.4(b)
Identify the g or u character of bonding and antibonding $$\delta$$ orbitals formed by face-to-face overlap of d atomic orbitals.
Text Transcription:
delta
Step-by-Step Solution:
Step 1 of 3
Homework 2 Name: key Chap 2 73 b 100 d 5 M, histidine 74 c 101 c 6 A, water 75 c 102 a 7 D, secondary structure 76 c 103 c 8 E, tertiary structure 77 c 104 e 78 d 105 d 17 N, C 79 d 106 e 18 α-keratin 80 d 107 c 19 Collagen, 3rd 81 b 108 b 20 antiparallel 82 d 109 c 21 Quaternary structure 83 a 110 c 84 c 111 c 58 b 85 d 112 d 59 d 86 a 113 b 60 d 87 d 114 a 61 d 88 b 115 d 62 b 89 a 116 b 63 c
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## Discover and learn what students are asking
Calculus: Early Transcendental Functions : Preparation for Calculus
?In Exercises 1–4, find any intercepts. $$y=x^{2}-8 x+12$$
#### Related chapters
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2022-08-12 08:05:37
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https://proofwiki.org/wiki/Closure_Condition_for_Hausdorff_Space
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# Closure Condition for Hausdorff Space
## Theorem
Let $\left({X, \tau}\right)$ be a topological space.
Then $\left({X, \tau}\right)$ is a Hausdorff space if and only if:
For all $x, y \in X$ such that $x \ne y$, there exists an open set $U$ such that $x \in U$ and $y \notin U^-$, where $U^-$ is the closure of $U$.
## Proof
### Necessary Condition
Let $\left({X, \tau}\right)$ be a Hausdorff space.
Let $x, y \in X$ with $x \ne y$.
Then by the definition of Hausdorff space there exist open sets $U, V \subseteq X$ such that:
$x \in U$
$y \in V$
$U \cap V = \varnothing$
Then $U \subseteq X \setminus V$.
By definition, $X \setminus V$ is closed.
Thus by definition of closure:
$U^- \subseteq X \setminus V$
Since $y \in V$, it follows that $y \notin X \setminus V$, so $y \notin U^-$.
$\Box$
### Sufficient Condition
Suppose that for each $x, y \in X$ such that $x \ne y$ there exists an open set $U$ such that $x \in U$ and $y \notin U^-$.
Let $x, y \in X$ with $x \ne y$.
Let $U$ be as described.
Let $V = X \setminus U^-$.
Then $y \in V$ by the definition of set difference.
Since $U \subseteq U^-$, it follows that $U \cap V = \varnothing$.
As such $U$ and $V$ exist for all such $x$ and $y$, it follows that $\left({X, \tau}\right)$ is a Hausdorff space.
$\blacksquare$
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2019-11-23 00:15:02
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https://physics.stackexchange.com/questions/553309/is-it-ever-possible-to-test-whether-an-object-is-in-a-quantum-superposition-ver/553328
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Is it ever possible to test whether an object is in a quantum superposition (versus a mixed state)?
Today I came across a paper, "Experiments testing macroscopic quantum superpositions must be slow," by Mari et al., which proposes and analyzes a thought experiment involving a first mass mA placed in a position superposition in Alice’s lab, the mass mA producing a gravitational field that potentially affects a test mass mB in Bob’s lab (separated from Alice’s lab by a distance R), depending on whether or not Bob turns on a detector. The article concludes that special relativity puts lower limits on the amount of time necessary to determine whether an object is in a superposition of two macroscopically distinct locations (versus a mixed state).
But the problem is that, as far as I understand, there is no way to determine whether an object is in a superposition at all (versus a mixed state)!
A superposition is determined by doing an interference experiment on a bunch of “identically prepared” objects (or particles or masses or whatever). The idea is that if we see an interference pattern emerge (e.g., the existence of light and dark fringes), then we can infer that the individual objects were in coherent superpositions. However, detection of a single object never produces a pattern, so we can’t infer whether or not it was in a superposition. Further, the outcome of every interference experiment on a superposition state, if analyzed one detection at a time, will be consistent with that object not having been in superposition. A single trial can confirm that an object was not in a superposition (such as if we detect a blip in a dark fringe area), but no single trial can confirm that the object was in a superposition. Moreover, even if a pattern does slowly emerge after many trials, every pattern produced by a finite number of trials – and remember that infinity does not exist in the physical world – is always a possible random outcome of measuring objects that are not in a superposition. We can never confirm the existence of a superposition, but lots and lots of trials can certainly increase our confidence.
In other words, if I’m right, then every measurement that Alice makes (in the Mari paper) will be consistent with Bob's having turned the detector on (and decohered the field) -- thus, no information is sent! No violation of special relativity! No problem!
Am I wrong? Is there a way to test whether a particular object is in a coherent superposition? If so, how? If not, then why do so few discussions of quantum superpositions mention this?
• Of course you can determine if something is in a superposition. For example, a particle with definite momentum is in a superposition of different positions. If you know the momentum, you know that its position is in superposition. Or, if you start with a particle with spin up, and then tilt the spin, then it ends up in a superposition of up and down. – knzhou May 19 at 21:36
• That's at least what the standard formalism says. You could also say "bullshit, there's no such thing as real superpositions, everything has a definite value and we just happen to not know it". That's an alternative that has been investigated in thousands of papers, called a "hidden variable theory". The problem with this approach is that it ends up incompatible with special relativity, and also all known ways of setting this up are pretty mathematically clunky. – knzhou May 19 at 21:39
• If you think you've suddenly found a way to turn quantum mechanics into straightforward classical mechanics, you haven't -- you just haven't understood the reasons that it's hard! Physicists have made thousands of attempts. Nobody wants it to be hard; there is no conspiracy going on, nature is just how it is. – knzhou May 19 at 21:41
• This is a very condescending comment. I am not interested in mathematical discussions of noncommuting observables. Nor am I pretending to understand QM. I am simply asking a (pretty clear) question about whether there is a way to test whether an object is in a position superposition, versus a mixed state. – Andrew Knight May 19 at 22:45
• @AndrewKnight you could probably make your question clearer (and shorter) and get better answers. E.g., your question doesn't mention mixed states. – innisfree May 19 at 22:50
Every state is obviously in a superposition of eigenstates of some observable. If you prepare your state as an eigenstate of an observable and claim that now it's not in a superposition, it's simply wrong because an eigenstate of one observable is not an eigenstate of many many other observables who do not commute with your observable, and thus, it still is in a superposition of eigenstates of some observable. As @knzhou mentions in the comment, if you prepare a definite momentum state, it would be in a superposition of infinitely many position eigenstates (and vice-versa). You cannot get around superposition in quantum mechanics because non-commuting observables exist in quantum mechanics (otherwise, it'd be classical mechanics).
OK, but what if I don't believe non-commuting observables exist?
Well, you'd be hard-pressed to believe so if you want to maintain sanity. Borrow a single spin $$\frac{1}{2}$$ particle from Stern (Gerlach wouldn't respond because he's mad about not getting a Nobel ;)) and do some experiments with it.
• Measure its spin state in $$z$$ direction. Let's say you find it to be up. Measure it again (as many times as you like), it would still be up. OK, so it is definitely in the spin-up state in $$z$$ direction.
• Now, measure its spin state in $$x$$ direction. It doesn't matter what you get. Now, measure it in $$z$$ direction again. OK, let's say you are lucky and you get it to be up again. But we want to be sure of this phenomenon that if you measure the spin state to be up in $$z$$ direction, then measure its spin state in $$x$$ direction, and then again in $$z$$ direction then you'd again get the spin state to be up in $$z$$ direction. So, we repeat the experiment many many times. And you'd quickly notice that this doesn't always happen. In fact, it happens exactly $$50\%$$ of the times. Other $$50\%$$ of the times, you start with a spin-up state in $$z$$ direction, measure its spin state in $$x$$ direction, and then when you measure it again in the $$z$$ direction, it would come out flipped.
• This conclusively shows that the spin state in $$z$$ and $$x$$ directions cannot be observed simultaneously. In other words, the two observables don't commute. This immediately implies that observing the one necessarily prepares the particle in a state of superposition of eigenstates of the other.
So, respecting your wish, I didn't use any identically prepared multiple copies of the same state. I just took one state, did a bunch of experiments with it, and concluded that observables don't commute, and thus, each state is necessarily a superposition of some observable.
Finally, I'd mindread here. What you seem to be confused about is that you cannot determine what a given unknown quantum state is if you only have a single state. For example, if you want to know what the state is, say, in the momentum basis, you won't be able to know which all momentum eigenstates are participating in the superposition with what coefficients to produce the given state. And when you measure the momentum, the superposition over the eigenstates of momentum would be lost and you'd only get a specific momentum eigenstate. If this is what you're saying, I have a few words to add.
If you're given an unknown quantum state, it's obviously true that you cannot determine what the state is unless you have multiple identical copies of the state. But that is the point of quantum mechanics that you can't just observe a state. There is a distinction between the state and the observables (the distinction arises out of non-commuting observables and thus, is absent in classical physics). You can simply observe the observables and the post-observation state would be the projection of the initial state onto the eigensubspace of the observable corresponding to the observed value. And the projection won't tell you anything about the initial state. You will need multiple identical copies of the state to determine what an unknown given state was, this is not a "gotcha" on quantum mechanics, it's what necessarily arises out of non-commuting nature observables.
Of course, the post-measurement state wouldn't be a superposition of eigenstates (with different eigenvalues) of the measured observable. But this doesn't mean that the post-measurement state wouldn't be a superposition (for all the reasons described above). The fact that upon measurement, you don't get a superposition of eigenstates (with different eigenvalues) of the measured observable isn't something weird, rather, it's an obvious requirement of having a consistent definition of what a measurement is. It'd be completely meaningless to say that I measured the state in a $$z$$ direction and find it to be in a superposition of spin-up and spin-down states in the $$z$$ direction.
• Sorry, but I'm annoyed by your answer and @knzhou. OBVIOUSLY a momentum eigenstate implies a superposition of positions. That is CLEARLY not the point I was making, or the question I was asking, in the above. It seems obvious that you both only read the title question and ignored the rest. What I meant, which I thought was obvious, was whether there was a way to test whether a given object is in a superposition VERSUS a mixed state. – Andrew Knight May 19 at 22:44
• @AndrewKnight It was very unclear that you were asking about the distinction between a quantum state and a mixed state as evident by the fact that you mentioned "mixed state" exactly zero times in your question. – Dvij D.C. May 19 at 22:51
• And, by the way, unlike the reply by @knzhou, your reply did NOT seem condescending -- it seemed like you were genuinely trying to answer my question, and I do appreciate that. – Andrew Knight May 19 at 22:51
• @AndrewKnight I suppose I ruined my modesty repo with my last comment 🤦🏽♂️😛 In any case, if you can explicitly address how your question relates to the discussion of pure vs. mixed states by editing your question, that'd be super helpful for people to address your question more clearly. – Dvij D.C. May 19 at 23:01
• Done. thanks... – Andrew Knight May 20 at 0:42
Is it ever possible to determine whether an object is in a quantum superposition?
I think you need to make the question more precise. Otherwise it's susceptible to more than one interpretation, and the answer could be either yes or no.
First of all, you need to distinguish between coherent and incoherent superpositions, as in the Mari paper that you reference.
Second, you need to think about when and relative to what state this is being determined.
Let's discuss several interpretations.
(A) If I put a silver atom through a Stern-Gerlach spectrometer, and it deflects in a certain way, then I have measured its spin. Say I've measured its spin to be $$s_z=+1/2$$, now. Note also that the SG apparatus does not change the spin of a particle, so there are no concerns about whether the object is so delicate that the measurement process has disturbed it. The fact that it's in the state $$s_z=+1/2$$, now, means that it's automatically in a superposition of $$s_x=\pm 1/2$$, now, and this superposition is coherent. So in this sense, I can definitely determine that it's in a coherent superposition of these two $$s_x$$ states, now.
(B) Or a different way of interpreting your question is that maybe before the atom went into the SG apparatus, it could have been in some state like $$s_x=+1/2$$, which is a coherent superposition of $$s_z=\pm 1/2$$. So then in the language of the Copenhagen interpretation, we've caused the wavefunction to collapse into $$s_z=+1/2$$, and we can never recover the information about what the state was before the measurement (because measurement is a nonunitary process). Or in MWI, what has happened is that we're now entangled with the spin, and because of decoherence, we will never be able to tell that our own wavefunction also contains a part that observed $$s_z=-1/2$$. So in this way of stating the question, the answer is that you can find out about the state after measurement, but you will never know about the state before measurement.
(C) Determining that the particle used to be in an incoherent superposition of states does seem impossible to me, for the reasons described in your question: this can only be determined for a statistical ensemble, not for a single particle that someone gives us. In MWI, this would be the kind of thing that would happen if someone has already measured the state of the particle, decoherence has already happened, and you want to know what the state of the particle used to be before the first measurement, which caused the split between two worlds. That's impossible in standard quantum mechanics, although maybe if we had a viable nonlinear version of quantum mechanics, we could do this, because the nonlinearities could allow the two worlds to interact rather than just superposing.
• Notice that non-linear quantum mechanics is fiction at this point, and it is very likely will continue to be so. The framework of quantum mechanics is an isolated island. It's proven to be nearly impossible to tweak any of its features even slightly and obtain something even remotely resembling quantum mechanics. This is a very important piece of information. To contrast, for example, you can tweak GR and still get something that reasonably looks like GR. – Dvij D.C. May 19 at 23:15
I’ll start my answer with the quote of a comment:
If you think you've suddenly found a way to turn quantum mechanics into straightforward classical mechanics, you haven't -- you just haven't understood the reasons that it's hard! Physicists have made thousands of attempts.
The double slit experiment is the experiment showing the wave behavior of light. The interference pattern shows the superposition of light, from the amplification of rectified wave crests to the cancellation of light wave crests and troughs. Does it?
No, it doesn't. Light consists of photons and they do not interact. To photons intersect each over and do not cancel out or double their energy. It is discussed on PSE many times, that no cancelation in the black area of the fringes takes place.
But please take in mind that the calculation with sinusoidal functions works well. This are easy to use equations to conclude from the fringes about the frequency of the used light.
Furthermore, the fringes also appear behind a single slit. No interaction of light from two slits. The explanation of the interference was transferred to both sides of the single slit. Now how about: no slit at all, but a single edge? Are you surprised that even behind a single edge the light deflection and by this of fringes can be seen?
The attempt 1
If you have no problem with the last sentence, there is a simple explanation for the “interference” pattern. Light is deflected by edges, and in the interaction between light and edges this deflection could be uneven, sometimes stronger and sometimes weaker. This is because photons actually have a wave behaviour; their electric and magnetic field components are sinusoidal functions.
The photon field components change periodically over time. When interacting with the surface electrons of an edge, the photon is deflected in a way that depends on its (varying) field strength at the time of interaction. By the way, this is the reason why photons are deflected behind edges, while electrons are only deflected away from edges.
I am not interested in mathematical discussions of noncommuting observables. Nor am I pretending to understand QM. I am simply asking a (pretty clear) question about whether there is a way to test whether an object is in a superposition ...
The attempt 2
The pairwise generation of entangled photons (by their spin) has a part of uncertainty. The point is that the parallel-antiparallel orientation of the two particles could point in any direction (until now we have no better conditions for the pair production). The measuring instrument (e.g. a grid), in turn, must also be oriented in any direction, no matter where by 360°. After a series of measurements we obtained a correlation between the two entangled particles. After many experiments the entanglement is assumed to be a fact. It is derived empirically and always has this statistical component. Only in some cases we measure the entanglement, in the other cases the result is unknown.
The answer to your question is that statistically we know whether two particles are superimposed. Since there are no better instruments, the uncertainty of our not knowing collapses after a series of measurements. That is what Zeilinger does. He stabilizes the result, shields it from the environment and filters out the coincidences. That's how they tried to regulate quantum computing.
Is it ever possible to test whether an object is in a quantum superposition (versus a mixed state)?
Take it the other way around. Numerous experiments are carried out to create superpositions, in particular with the help of the polarization of photons or the spin coupling of electrons and atoms. The results of these experiments show that we are able to create superpositions.
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2020-08-12 18:49:55
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https://math.stackexchange.com/questions/3466875/orthogonal-complement-of-s-and-related-question
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# Orthogonal complement of S and related question
Let V be a real vector space with a topology induced from an inner product. Let S be a subset. Can we say either of the following ?
1. closure(S) = perp(perp(S)) , [where perp(X) = orthogonal complement of X]
2. closure(span(S)) = perp(perp(S))
1. is false: the orthogonal complement $$S^\perp=\{x\mid\forall s\in S: x\perp s\}$$ of any set $$S$$ is always a closed subspace, whereas the closure need not be a subspace.
2. is true: prove $$S^\perp =\overline{\mathrm{span}(S)}^\perp$$, and apply $$U^{\perp\perp} =U$$ for a closed subspace $$U$$.
• Need not be a subspace. For example, a closed ball is closed but not a subspace. Its perp is $\{0\}$. Dec 7, 2019 at 15:00
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2022-07-03 19:07:43
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https://www.physicsforums.com/threads/finding-displacement-of-a-drag-car-stopping.472309/
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# Finding displacement of a drag car stopping
## Homework Statement
A drag racer, starting from rest, accelerates to ¼ mile with a constant acceleration whose magnitude is 12.427 m/s2. At the end of the ¼ mile, a parachute opens, slowing the car down and bringing the car to a stop after 40 seconds. From the time the parachute opened, how far did the race car travel?
Vf = Vi + AT
## The Attempt at a Solution
I split the problem into two parts. The first 1/4, and the 3/4 section. With only the acceleration and initial velocity given, I'm kinda lost as to what I should find. I tried to find the final velocity before the parachute opens by doing this:
(0 m/s) + (12.427 m/s2)( t ) = Vf
I got 12.427 m/s^2 (t) as the final velocity.
Then I tried to plug that velocity into the second part as the initial velocity.
A = (Vf - Vi) / T
A = (0 m/s) - (12.427 m/s^2) T / T
I canceled out the time.
Now I'm trying to figure out what to do with
0 m/s - 12.427 m/s^2 = a
I don't think I did it right. It's really confusing and frustrating me. Can anyone lead me in the right direction?
Delphi51
Homework Helper
Welcome to PF.
The two times may be different; you can't cancel them.
For the first quarter mile, you know the distance and the acceleration.
From that, you should be able to find out anything you would like to know about that part of the motion - time, final velocity, etc. But you must use both the known acceleration and distance. Look through your list of formulas for one that has just the two knowns and something you would like to find, perhaps time or final velocity.
The final velocity for the first part is the initial velocity for the deceleration. Make sure you find it before going on to the deceleration part.
Thanks for the reply. I see what I did wrong now.
So I converted the 1/4 mile to 402.34 meters.
I found the final velocity of the first segment using $$v^2 = v_0^2 + 2 a \Delta x$$.
The final velocity was 100 m/s for the first segment.
100 m/s is the initial velocity for the other segment. I used $$x = x_0 + v_0 t + (1/2) a t^2$$
Vi = 100 m/s
T = 40 s
Vv = 0 m/s
a = -2/5 m/s^2
What I got for the displacement was 2000 m. Is that right?
Delphi51
Homework Helper
Looks good! That must be a typo on the acceleration, -2.5 rather than -2/5.
It is so important to know what carries forward from one part to another.
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2021-05-07 07:46:26
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https://cs.stackexchange.com/questions/50887/turing-machine-infinite-tape-does-that-thing-exist/51786#51786
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# Turing machine - infinite tape - does that thing exist?
Can we use a Turing machine with infinite tape as a basis to prove anything disregarding the fact that such a thing can never exist? Do we have the right to regard a machine (a construct) in the same way we regard a number or a set (an abstract)?
If yes could I similarly use a Turing angel with infinite scroll instead, would that be ok?
EDIT: since I cannot comment. This question is somewhat different from infinity in mathematics because a set even infinite can have a finite representation. On the other hand there is no finite representation for an infinite tape That is the algorithm that generates the output.
Computation requires resources, that means that it takes place in a time-space, not in an abstract timeless dimension. While it could be ok to draw on demand more time for the computation pushing on to infinity, space has a cap on the amount of information it can hold, and before reaching that point the tape would revert to write only, that is implode into a black hole. So even if an algorithm is "finite" its output could be big enough to become "unreadable". For example lets say we have a machine that counts all numbers at some Number H when trying to add more information on the tape the tape colapses forms a black hole and number H+1 becomes unreadable. Counting up to H+2 is a finite process but cannot be completed. One could argue that space and time restrictions should not concern us but I m not convinced that is true when dealing with objects embedded in spacetime
I could accept a definition of a machine with sufficiently large tape. But I think that the size does matter. Can for example 2 Turing machines with different tape-size considered equivalent?
• Welcome to CS.SE! It's hard for me to tell exactly what you're asking or what kind of answer you're looking for. I encourage you to edit your question to flesh it out, so that we can provide you a more useful answer; it can be considered for re-opening if the question is edited to be clearer. What are your thoughts? Have you read standard textbooks and resources on Turing machines? Why do you think we wouldn't be able to talk about it? See also cs.stackexchange.com/help/how-to-ask for more about how to use this site effectively.
– D.W.
Dec 17 '15 at 23:57
• If the machine stops after a finite number of steps, then it didn't actually need an infinite tape. It only needed a tape that was big enough. If the machine does not stop after a finite number of steps, then that's the same as saying the computation was not possible. So, saying "infinite" really is just a convenient way to say "big enough" without getting tangled up in the question of how big that actually is. Dec 18 '15 at 17:25
• According to original description, the tape is not limited, and oversimplifying - it should be endless, but inreasing tape at will (when needed) is not in opposition to it's description. So I would stick to abstraction, but building such machine I would compare to making subAtlantic fiber optic cable - it was produced on ship while cables were put, not produced in advance.
– Evil
Dec 18 '15 at 17:48
• The natural numbers also suffer from similar problems – should we stop using them as well? Dec 18 '15 at 22:30
• "Do we have the right to regard a machine (a construct) in the same way we regard a number or a set (an abstract)?" -- why, of course! In mathematical models, the mind is the limit.
– Raphael
Jan 13 '16 at 7:25
## 2 Answers
The tape is not infinite. It is unbounded. There is an important difference. We don't need infinite memory. We only need to be able to add more memory over time as needed. This is actually pretty close to what people do in practice. When a program needs more memory it asks the operating system and receives more memory. If we see that a computer is short on memory we increase its memory. It is becoming even easier with cloud computing. So the assumption is not that the machine has infinite memory. The assumption is that we can increase the memory as needed.
If your question is about the limit where the needed memory reaches the limit of information that can be stored in the universe (assuming that it is finite and bounded by some fixed number) then yes, we will not be able to add any more memory at that point. In fact, if you assume that the amount of information is bounded by a fixed number then the world is a finite state machine. However that is not the usual case.
However if you try to design programs as finite state machines you are going to have a very difficult time because the number of states is huge. The number of states for a Turing machine with $m$ bits of memory would be something like $2^m$. Turing machines (and other similar models) simplify things. This is a rather common thing in sciences. It is important to know the limits of a model but the fact that a model is not a completely accurate representation of the world doesn't imply it is useless.
In some sense, your question has been answered by switching from computability theory to computational complexity theory. In the latter, we care mostly about problems which a Turing machine can solve in polynomial time. In particular, such machines only use polynomial space. This is more realistic than the unrestricted Turing machine model.
However, even computational complexity suffers from similar problems, since we only care about the asymptotic behavior of Turing machines, whereas in real life we only care about the behavior on inputs of a certain size. Nevertheless, empirically computational complexity has shown some predictive power in some (but not all) cases.
A similar issue happens in classical mathematics – the natural numbers are potentially unlimited, and some people don't like it. They are called ultrafinitists.
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2022-01-28 02:22:45
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https://www.nature.com/articles/s41598-020-67947-x?fbclid=IwAR308DhorPyJ7Iv25oWhQfxUI5YkRqhxLkIzQ05f0zF_6D-ip2fPQTFbc44&error=cookies_not_supported&code=17ae6917-adab-4961-b19b-41b7232f929b
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# Behavioural responses of white sharks to specific baits during cage diving ecotourism
## Abstract
This study describes the effect of different baits on the attraction, surface behaviour and conditioning of white sharks Carcharodon carcharias during local ecotourism activities. The sightings, behaviours, and pictures used for photographic identification were obtained during August to November 2012–2014 onboard tourist boats in Guadalupe Island, Mexico. Four types of baits were used: (1) frozen bait; (2) frozen bait and natural chum; (3) fresh fish bait; and (4) mackerel bags. Data were analysed according to sex, maturity and the total of sharks using 6,145 sightings of 121 white sharks. The type of bait showed no significant difference on the effectiveness to attracting sharks. Ethological analysis showed that the type of bait had a significant effect on the shark’s surface behaviour during its interactions with boats. Natural chum and fresh baits showed short term behavioural patterns constituted by increased number of violent interactions with the bait, while the frozen bait did not generate a defined behavioural pattern. Conditioning of white sharks was determined by the number of interactions and the consumption frequency of the bait. Fifty nine percent of sharks (n = 41) showed no conditioning, 36% (n = 25) showed a low risk and only 5% (n = 3) were found to have a high risk of conditioning. The results suggest that current ecotourism has no effect on the conditioning of the white sharks, and that all baits have a similar effectiveness for attracting the sharks. However, a different behavioural pattern was observed when fresh bait and chum were used, which could increase the potential of accidents during ecotourism.
## Introduction
Ecotourism with white sharks Carcharodon carcharias (Linnaeus, 1758) occurs in countries where seasonal aggregations of this species have been observed, including South Africa1, Australia2, New Zealand3, the United States of America4 and Mexico5. Due to national and international wildlife regulations, shark cage diving is currently the main legal activity for the economical use of this vulnerable species1,6. In Mexico, cage diving began in 2001 with the identification of Guadalupe Island as one of the main aggregation sites for C. carcharias in the Eastern Pacific Ocean5,7. Since then, numerous ecotourism operators work between August and November of each year, also providing on going surveillance against illegal fisheries, and a platform for both public education and scientific research. The local revenue of cage diving in Guadalupe has been estimated at more than 4.5 US million dollars per season8,9.
Regardless of the benefits, white shark ecotourism is often seen as a controversial subject. In the case of cage diving, provisioning of food has been linked with potential negative effects on habitat use, surface behaviour, bioenergetics, conditioning and a probable increase in the frequency of interactions with humans1,2,10,11. In Guadalupe Island, the study of the effects from ecotourism is a priority for local authorities and for the conservation of this threatened species11. Since 2016, there have been at least six accidents related to cage diving in this oceanic island, as some white sharks and divers have been injured during these events8,9. However, the most serious accident occurred in October 2019, when a white shark died after being stuck in a cage for more than 25 min12. In this regard, specific violent behaviours related to the capture of bait increase the risk of such accidents, in which these behaviours have been described until now as “attacks”8,12,13.
In terms of management, the evaluation of the attraction, surface behaviour and potential conditioning of the sharks is needed for regulations related to the use of bait and for the prevention of accidents during cage diving8. For the activities conducted in this marine protected area, only frozen fish (Thunnus albacares) bought in the departure port is allowed as bait during ecotourism. Nevertheless, some operators often use fresh bait for cage diving activities, in the belief that the olfactory cue from a fresh fish could attract sharks more efficiently.
The aim of the present study is to describe the effect of different baits on the attraction, surface behaviour and conditioning of white sharks during ecotourism in order to provide information for the sustainable use, economical exploitation and the prevention of accidents between white sharks, cages and divers. Beyond the concept of conditioning that could negatively affect the ecology of sharks involved in ecotourism, our study aims to test the hypothesis that the type of bait used to approach sharks has an influence on their surface behaviour. The implications for conservation include the concerns of authorities, cage diving operators, local fishermen, tourists, and scientists for improving the management of this vulnerable and iconic species.
## Results
### Attraction to bait
A total of 6,145 sightings from 121 identified white sharks were registered during 87 days at sea. Seventy four percent of these sightings were males and 26% were females, which showed a significant sexual proportion of 3:1 (M:F; P < 0.01). In general, there were no statistical differences between maturity stages along the studied months, as 51% were mature and 49% were immature individuals (P > 0.05). However, the presence of white sharks changed during the season. A maximum average of sightings per hour of mature males was observed during August, with a decrease of this number further in the season and an increase of mature females since late September (Fig. 1).
The general presence of white sharks was similar regardless of the type of bait. A maximum of 70.5 sightings per hour was registered during the use of frozen bait and chum, while the minimum 0.17 sightings per hour was observed using mackerel bait bag. Nonetheless, this minimum and maximum data were outliers and did not affect the lack of significant differences of sightings between the types of baits (H3,87 = 8.29, P > 0.05). A similar effect was observed when sightings were analysed by sexual maturity stage and the type of bait. There were no significant differences in the sightings per hour of mature males (H3,87 = 5.91, P > 0.05), mature females (H3,87 = 1.88, P > 0.05), immature males (H3,87 = 17.05; P > 0.05) or immature females (H3,87 = 7.40, P > 0.05) between the four baits that were tested.
### Surface behaviour
Ethological analyses were carried out using data from 534 h of direct observation of white sharks that interacted with the tourist boats. This monitoring allowed to obtain 17,360 files of audio-visual material, which was filmed by the observer or provided by the tourists who visited Guadalupe Island during the study period. For these analyses, only the data from 106 different sharks were used (n = 4,823 behaviours), as the ethograms from 15 sharks were influenced by the presence of other individuals.
According to the category, a total of 21 mature males, 18 mature females, 54 immature males and 13 immature females were photo identified during the study period. The size range for both sexes was 1.8–5.8 m TL with a mean of 3.6 m ± 0.9 m. The mean size of males was 3.3 m ± 0.7 cm TL, and 4.2 ± 1.1 m TL for females. Regarding individuals, a significant sex ratio of 1:2.4 was observed (H: M, p = 5.542e−5); with 63% of immature individuals and 37% of mature sharks. A total of 1,542 ethograms of the surface behaviour of white sharks were generated (Table 1).
Success rate for the capture and consumption of bait was similar in all categories, although a higher rate for male sharks and mature individuals was observed. Proportionally, a mature shark would get the bait 22 times for every 100 foraging attempts. Consumption rate was lower than the catch rate, as sharks did not always consume the captured bait. Similarly, mature sharks presented the highest amount of consumption in relation to the number of foraging attempts made (Table 2).
The behaviour of white sharks was significantly different depending on the type of bait. Nevertheless, the bait known as mackerel bag was not considered for the ethological analysis, given that the boat that used it changed the bait early in the study period and there were not enough observations for the analysis. During the provisioning of frozen bait, the behavioural transitions from horizontal strikes (HS) to close inspections (CLI; χ249 = 21.2, P > 0.05), vertical strikes (VS) to bait capture (BAC; χ249 = 35.4, P > 0.05) and HA to BAC (χ249 = 22, P > 0.05) were frequently observed, but with no significant contributions. The lack of statistically significant behaviour transitions showed that there was no specific behavioural pattern for the consumption of this type of bait (Fig. 2a). However, this pattern differed when chum was provided.
The interaction with frozen bait and chum showed a statistically significant transition from vertical (χ249 = 105.1, P < 0.05) and horizontal strikes (χ249 = 112.9, P < 0.05) towards BAC and feeding (FE; χ249 = 679.8, P < 0.05). Although there were interactions from HA to CLI (χ249 = 46.65, P > 0.05), and from HA to parading (PAR; χ249 = 33.68, P > 0.05), there was no statistical significance on such transitions. During these situations, the behavioural pattern consisted in agonistic behaviours for the bait capture with constant repetitions of the same displays (Fig. 2b).
Regarding the fresh bait, only one significant transition from HA to CLI was observed (χ249 = 95.19, P < 0.05). The transitions from CLI to HA (χ249 = 49.19, P > 0.05), HA to PAR (χ249 = 53.29, P > 0.05) and from horizontal (χ249 = 45.89, P > 0.05) or vertical strikes to BAC (χ249 = 48.53, P > 0.05) were frequently observed; however, there was no statistical significance in such transitions (Fig. 2c). This pattern was identified for the agonistic behaviours and curiosity with close inspections to the bait, but with no significant transitions for feeding. Additionally, some displays were repeated constantly before they could perform any transition to another behaviour.
### Conditioning
The conditioning was determined on 69 photo identified white sharks, in which its identity was confirmed using high-quality images over the study period. These individuals correspond to 57% of the total recorded sharks (n = 121), as the remaining sharks had a deficient photographic record that did not allow the confirmation of the identity in the recaptures. The decision to use only high-quality photographs was made in order to avoid overestimation of the number of white sharks for the conditioning analysis.
An average of 2 ± 5 daily visits per white shark was recorded during the study period. A total of 35 sharks (50.7%) visited the boats less than two times, while only one shark visited the tourist boats in 37 occasions (Fig. 3). According to the Spearman correlation test result, the number of sharks decrease with the increment in the number of visits (R = 0.82; P < 0.01).
Most of the white sharks showed no signs of conditioning to the bait, as 59% of the sharks (n = 41) were registered in less than three random days in the season (Fig. 4). Twenty two percent were included in category I, since they corresponded to sharks that visited the vessels more than three random days per season (n = 16). Thirteen percent were classified in category II, which included sharks that visited the vessels for 3–5 days per month (n = 9). A single individual (1.44%) was catalogued in category III by being sighted more than 5 days per month and consuming bait during such periods. Finally, two of the sharks (2.88%) were included in category IV, as they occurred frequently, interacted with the boats more than 20 min per day and fed from the bait. Regarding sexual maturity, only mature individuals had a high grade of conditioning (categories III and IV), while the lower conditioning included both mature and immature sharks. The immature males were the most numerous sharks without conditioning (43.78%) in category I (8.69%) and category II (4.39%). Immature females were the least numerous among the conditioning categories analysed, while mature individuals, both males and females, were classified with a similar proportion in all cases (Fig. 4).
## Discussion and conclusions
The surface presence of white sharks was similar regardless maturity, sex, or type of bait, suggesting a similar effectiveness in the attractant of sharks. In other ethological studies, the white shark has been suggested as a predator that evaluates the consumption of food through previous inspection and monitoring14,15,16. In this instance, the use of frozen or fresh bait had a similar effectiveness when attracting sharks, mainly due to the visual and olfactory stimuli inherent to the bait16,17,18. It is possible that the scent of the frozen bait was not significantly different from a fresh bait, since there is no nutritional differences between both treatments19. This has been observed in other pelagic species such as the yellow fin tuna T. albacares, where the catches are similar regardless of the freshness of the bait20. However, the smell would be the main sense involved in the attraction at distance, while the sight would be fundamental in the decision of the capturing of prey20,21.
In other aggregation sites, the use of bait has shown a minimum effect on the white sharks behaviour2,3. In South Africa, Laroche et al.1 suggest that the population of 64,000 Cape fur seals, Arctocephalus pusillus pusillus, generates a constant olfactory stimulus for the white sharks, so the use of bait is negligible for the attraction and conditioning of C. carcharias1,22. A similar effect occurs in Farallon Islands where cage diving is often difficult, probably explained by the presence of a high density of northern elephant seals Mirounga angustirostris4. The extension of both sites is 0.02 km2 for the South African site, and 0.42 km2 for the American one. If it is compared with the 241 km2 of Guadalupe Island, the density of pinnipeds could be much lower in this oceanic island. In this regard, the absence of a natural, intense, and permanent stimulus in Guadalupe Island could explain both the success to attract sharks to the boats and the inherent risk of conditioning.
The idea that a fresh stimulus could generate a greater number of sightings cannot be accepted according to the results of the present study, since all the baits showed a similar effect on the attraction of the sharks. This lack of preference for a particular attractant may be related to the opportunistic predation on dead animals such as whale carcasses that have been observed in mature and immature white sharks23,24,25,26. Future analysis involving the hour of arrival of each shark according to the type of bait could be useful to determine the effectiveness in terms of time, which could complement the information for boat owners during ecotourism.
There were significant differences in the observed behavioural patterns according to the type of bait. The behavioural patterns with the frozen bait with chum and fresh bait were related to the reception of stimuli and feeding, while the frozen bait did not generate a defined pattern. It is possible that the smell and appearance of a frozen bait generate an olfactory and visual attraction in the white sharks without aggressions16,21. In contrast, the use of fresh bait along with the stimuli of a recently captured animal could be part of the explanation for the violent behaviours observed with this type of bait14,17,18.
Contrary to the pattern observed with the use of fresh bait, the behavioural transitions towards aggressive displays were not detected during the use of a frozen bait. This lower number of agonistic displays could decrease the risk of accidents of white sharks that enter or hit the cages, as it is frequently observed during each cage diving season22. However, vigilance and specifications for the handling of the bait, as well as for the redesign of the cages are the first steps to reduce the incidence of accidents11,12,22.
The horizontal strike (HS) was the most frequent behaviour during the study period, as it has been similarly observed in Australia by Tricas27 and Strong14, and in South Africa by Sperone et al.15. The large number of foraging attempts with respect to the number of captures and consumptions was the cause of the low effectiveness of this behaviour. In comparison with the immature individuals, the mature sharks managed to capture and consume the bait more efficiently. This may be attributable to the experience and size of white sharks according to their age, since mature sharks perform a better evaluation of the conditions before ambushing their prey to efficiently manage their energy when compared to immature individuals22,28.
In previous research, a low effectiveness of the foraging attempts has been related to energy loss and a physiological imbalance that could affect the health of the sharks10,11,22. An evaluation of the effects on the metabolic rate of the white sharks could provide insights in terms of energy loss caused by ecotourism, as it has been observed in other species of sharks29,30. However, an energetic affectation is unlikely if there is a low conditioning in the sharks that interacted with the boats, as it was observed under the current ecotourism conditions of Guadalupe Island.
The development of a conditioning to the boats requires defined patterns for feeding, since the behaviours and stimuli must be constant for this learning1,31,32,33. The sharks that interacted with the frozen bait presented a diverse number of behaviours for the acquisition of food, so this high variability of displays would prevent the development of conditioning in the absence of a defined behavioural pattern. In contrast, chum and fresh baits did demonstrate a pattern directed toward aggressions and reception of stimuli. Unlike those observed by Laroche et al.1, it is possible that the baits such as chum or fresh tuna could generate a future change in the behaviour of C. carcharias in Guadalupe Island22. In this regard, trophic interactions of each white shark population should be considered for future comparative studies, as differences in diet could be involved in the response and conditioning to the used baits6,9.
It is improbable that the ecotourism generates a conditioning on the white sharks of Guadalupe Island. Although the used criteria underestimate the cognitive capacity of the white shark, there are five arguments that may explain the lack of conditioning in this aggregation site.
1. 1.
Few visits of sharks. In other studies, conditioning was observed after 6 months of interaction with the constant provision of substantial food in training sessions that lasted more than 20 min32,34. In this study, most of the sharks were recorded in less than 5 days, so their low permanence would not allow a conditioning32,35,36.
2. 2.
Duration of the season. Studies in captivity have shown that conditioning can persist after 10 weeks in the memory of sharks31. This conditioning arises after receiving a constant stimulus under controlled situations31,32,36. In Guadalupe Island, the stimuli granted by their natural prey availability5,37, the white shark migration habits and its local movements5,38, as well as the 8 months between seasons11 could affect the duration of conditioning in terms of the memory of the sharks31,32. Under this assumption, the shark’s attraction to the boats would be the result of a natural curiosity and not of acquired learning16,17,39.
3. 3.
Low consumption of bait. In previous studies, the reception of "substantial" stimuli was one of the criteria necessary to determine a high risk of conditioning31,32,36. In the present research, the capture of at least one bait was considered as substantial and significant for the detection of conditioning in a precautionary measure to detect any indication of possible conditioning. Despite this, it was not possible to detect more than three sharks that fulfilled the bait intake, the number of visits and necessary interactions to determine a high risk of conditioning.
4. 4.
Effectiveness. The effectiveness of the strikes or foraging attempts was low for the capture and consumption of bait, so this lack of stimuli could lead to the disappearance of the potential acquired response in conditioning. White sharks from Guadalupe did not have many reinforcements, and the amount of strikes in means of energy used to capture them may not be profitable for their energetics22. A conditioning is unlikely when not enough stimuli is received during the development or maintenance of learning31,33.
5. 5.
Competition. The highest number of sightings was registered during the month of August and corresponded mostly to mature and immature males. It is possible that the presence of sharks of similar size favours the intraspecific interactions with respect to the bait, due to a greater tolerance on the part of the males of similar sizes22. However, the decrease in sightings during the following months could be related to the presence of larger sharks, which, given a hierarchy by size, would displace small sharks that could be more vulnerable to conditioning. In this manner, the segregated migration of mature sharks, the size hierarchy and competition would prevent immature individuals from developing a conditioning11,22,38.
In any case, the learning capacity of sharks could be higher during their juvenile stages32,34, so the monitoring of these individuals is essential to understand the possible changes in their presence or behaviour. A high number of immature males were observed during all season, so the vigilance in order to avoid the consumption of bait would represent a measure of precautionary management to prevent the conditioning of the white sharks11,22,32.
The study of the behaviour of sharks is necessary in the localities were ecotourism occurs. Although some activities like snorkelling, scuba diving or cage diving are considered as sustainable, there are few studies related to the negative effects of ecotourism in most shark populations2,29,30,40,41,42. The acquisition of data through science-based monitoring is a practical strategy that could improve the knowledge about the biology of the species, its relations with ecotourism and the correct decisions for good practices22. Some of the proper management regulations could include tourism seasonality, quantity of boats and the responsible use of organic attractants if these were needed9,22.
As a cosmopolitan species that is economically exploited in several countries, the presented results of the behaviour of white sharks could be useful in future comparisons with other populations6,22. This observational method is applicable without the need of invasive techniques or high economical resources, which could facilitate the obtention of data through the participation of trained observers and the tourism inclusion in scientific activities. This could lead to a significant contribution to local marine policies for efficient protection and sustainable use of the sharks, as it has been observed in other marine protected areas43,44.
Wildlife regulations during ecotourism are essential for the prevention of accidents between humans and sharks. The limited vigilance in Guadalupe Island, along with a lack of scientific evidence supporting some of the unfollowed regulations, were part of the cause for several accidents that occurred in 2016 and the years before, where several sharks and divers were harmed during ecotourism44, including the recent death of a white shark in 201912. Although these situations are scarce and could not be avoided in some cases, the information regarding the behaviour to specific type of baits is needed for the prevention of accidents through science-based decisions. In the present paper, the different short-term surface behaviours according to the type of bait are provided in order to prevent more accidents, as well as to improve the monitoring of this threatened species.
The creation of a standard governmental monitoring using the presented methods will be useful for the constant evaluation of this and other shark populations. Future studies should consider the effect of the environment, social behaviour, personalities of sharks, times of response to the bait and movements of the sharks by the use of other techniques such as drone video and acoustic tagging in relation to the boats, which could provide more insights about the effects of cage diving ecotourism2,42. Additionally, the use of genetics and other photo identification methods, such as the analysis of dorsal fins and mark-recapture studies, would be beneficial for the knowledge of the status of this white shark population45,46.
The improvement of monitoring through the participation of trained observers would allow the generation of scientific knowledge in terms of ecology, demography and ethology if the proper data are frequently obtained9,22,27. Research involving minimum invasive techniques as the one presented in this paper could be useful for the generation of such information, which in turn can be used for the improvement of the activity in terms of sustainability and the conservation of this threatened species. Specific recommendations for cage diving ecotourism could be the creation of emergency response protocols for stuck sharks and injured divers, as well as the implementation of a nautical video vigilance cameras onboard all boats. The latter could be useful for the governmental inspection of good practices and the obtention of data in one of the most profitable and sustainable activities with sharks worldwide6,9.
## Materials and methods
### Study area
The Guadalupe Island Biosphere Reserve is a marine protected area located in the Pacific Ocean, 241 km offshore Baja California, Mexico (Fig. 5). It measures 32 km long and 6.5–9.5 km wide, with a total area of 244 km2 and a maximum elevation of 1,295 m47. These dimensions make Guadalupe Island the largest aggregation site for white sharks compared with other islands where this species occurs. However, the designated area for cage diving in Guadalupe Island is limited to a bay known as Rada Norte, which measures 7 km long and 2 km wide11.
In this volcanic island, the dominant ecosystems are sandy bottoms and rocky reefs with macroalgae forests, characterized by a high abundance of invertebrates, bony fishes, elasmobranchs, cetaceans and pinnipeds37,47. This biodiversity has been related to the seasonal presence of white sharks in terms of prey availability, as confirmed by observed predation events and stable isotopes analyses5,47,48.
The ecotourism boats surveyed during the study period were seven long-range vessels that measure 27–41 m in length. Boats were anchored at 200–250 m off the coast at a depth of 70–80 m, with two surface cages and a maximum capacity of four people each. Cage diving operations were supervised by the captain, a dive master and two crew members who were responsible of handling the bait and not feeding the sharks intentionally, as indicated by local regulations11.
## Data collection
In this study, all observations were carried out by one of the authors (EEBG) in order to maintain the bias of estimations in the same observer. The methods were approved under the following permits: Secretaría del Medio Ambiente y Recursos Naturales (SGPA/DGVS/06302/12; SGPA/DGVS/05847/13; SGPA/DGVS/03077/14), Comisión Nacional de Áreas Naturales Protegidas (F00.DRPBC.RBIG.‐210/12; F00.DRPBCPN.‐000889; F00.DRPBC.RBIG. 163/14) and Secretaría de Gobernación (SATI/PC/039/12; SATI/PC/038/13; SRPAP/PC/022/14).
Data were recorded on board the tourist boats that visited Guadalupe Island during August to November of 2012–2014, with a daily monitoring of white sharks between 07:00 and 18:00 h. Date, sex, total length (TL), type of behaviour, type of bait and the time of each shark sighting were recorded. The sex of each shark was determined by the presence or absence of claspers, with posterior confirmation through underwater photography. The TL was estimated using the length of the cages as a comparison with the length of the sharks when each shark approached horizontally, parallel, and close to the cages.
Identification of shark individuals was carried out by analysing underwater pictures based on several characteristics such as skin pigmentation, patterns of the gills slits, pelvic fins and caudal fin of both sides of the body, scars, mutilations, dorsal fins and anatomical deformities observed on the sharks using a Nikon D7100 camera inside an Ikelite Housing; additional photographs were voluntary provided from tourists. All pictures were analysed the same day that they were obtained to determine the identity of the white sharks that visited the boats. Sexual maturity was determined according to the sex and length, where males with a TL > 3.5 m and females with a TL > 4.5 m were considered mature individuals24. Shark sightings were posteriorly classified according to their sex and maturity stage. In this study, the analysed groups and statistical units for each analysis were specified for the objectives of attraction (average of sightings per hour), surface behaviour (ethograms from different individuals), or conditioning (photo identified white sharks), with respect to one of the following types of bait:
1. 1.
Frozen bait. Head, tail or any other segment cut from yellow fin tunas (T. albacares) acquired in the departure port. This piece of fish was tied to a rope and a buoy for flotation.
2. 2.
Frozen bait and chum. Frozen tuna accompanied by organic shedding known as chum, which is composed of a mixture of seawater with fish tissues such as blood, muscle, skin, fat, bone, etc.
3. 3.
Fresh bait. Pieces of fresh yellowfin tuna that were tied to a rope and a buoy for flotation; obtained in the area during tourist activities.
4. 4.
Mackerel bag. Pieces of mackerel (Scomber japonicus) placed in an organic fibre bag and tied to a line with a buoy.
### Attraction to bait
The average of sightings per hour was the statistical unit for analysis of the attraction of white sharks related to the type of bait in order to describe the effectiveness of each stimuli during ecotourism. A sighting was defined as any behaviour that an observed shark performed on the surface at a radial distance of 10 m with respect to the position of the bait. Since the data set did not meet the normality assumptions according to the Lilliefors test (n = 87; d = 0.14358, P < 0.01), a nonparametric Kruskal–Wallis test and a Bonferroni test were carried out for the analysis of such sightings per hour and bait according to the following categories: immature males, mature males, immature females and mature females.
### Surface behaviour
White shark behaviour was recorded in ethograms based on sighting frequencies and behaviour sequences that were further included in ethological flow charts for the representation of behavioural patterns in relation to the type of bait. Sightings were classified using the behaviours observed in the ethological analysis of baited attracted white sharks by Becerril-García et al.22, and the methods published by Klimley et al.4, Martin15 and Sperone et al.16,39. In an effort to use a less connoted terminology, the present study will use the neutral term of “feeding strike” or “feeding attempts”13. In this regard, a biological interpretation was used in order to classify each behaviour and to provide useful management information (Table 3). During the analysis, only the ethograms of white sharks in the absence of other sharks were considered, due to the possibility of intraspecific interactions affecting their behaviours with the bait. For the construction of the behavioural diagrams, the frequencies of each observed behaviour and transition were analysed using the software EthoLog v. 2.2549. The statistical significance of such transitions was determined through a Chi‐square test under the null hypothesis that there was no relationship between independent behaviours16,39. In this regard, the observed significant transitions were used to describe ethological patterns according to the different types of baits.
Additionally, the rate of bait capture and rate of consumption for each white shark category was determined through the following formulas:
$$BCR= \left[\frac{a}{\left(h+v\right)}\right]*100 \quad CoR= \left[\frac{b}{\left(h+v\right)}\right]*100$$
where BCR is the Bait Capture Rate; CoR is the Consumption Rate; a is the number of bait captures; b is the number of consumptions; h is the number of horizontal strikes; and v is the number of vertical strikes.
### Conditioning
The evaluation of the conditioning in the identified white sharks was carried out using a modified criteria generated from previous studies on shark behaviour31,32,33,34,35,45. For the purpose of this study, we define conditioning as associative learning obtained from a prolonged exposure to a stimulus associated with feeding behaviour. The number of visits per day, times of interaction and stimuli reception was registered for each photo identified white shark to determine the level of conditioning to the bait (Table 4). Stimulus reception was defined as the capture of the bait by a shark regardless its consumption. The criteria were intentionally modified to underestimate the learning capacity of the sharks by the assumption that white sharks have a similar or higher cognitive capacity than other studied elasmobranchs1,31,32,33. This decision was made due to the lack of studies regarding learning capacities of white sharks, differences between ecotourism effort in aggregation sites and its application for the monitoring of white sharks in the Guadalupe Island Biosphere Reserve6,22. In this way, the effect of baiting is described through a precautionary approach. These results were presented in histograms in order to obtain the proportions of identified white sharks and their respective degrees of conditioning.
## Data availability
The dataset regarding the monitoring of the evaluated white sharks is available from the corresponding author on request.
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## Acknowledgements
The authors thank all the provided support from the ecotourism fleets who participated in this study, as well as Saul González‐Romero, Leonardo Trejo and Tom Lamphere. We thank Instituto Politécnico Nacional for funding through grants from the Comisión de Operación y Fomento de Actividades Académicas and the Estímulo al Desempeño de los Investigadores, as well as D. Bernot-Simon for the English review. This research was funded by Alianza WWF‐Fundación Telmex Telcel, Alianza WWF-Fundación Carlos Slim, Fins Attached, the University of Calabria and the Charles Annenberg Foundation. Additional thanks the Consejo Nacional de Ciencia y Tecnología and IDEA WILD for the scholarship and photographic equipment provided, respectively.
## Author information
Authors
### Contributions
E.E.B.G. and F.G.M. designed the research. E.E.B.G. and E.M.H.P. conducted the collection of biological data, photo identification of sharks and the analysis of attraction and conditioning. P.M. and E.S. analysed the results regarding behavioural ecology. E.E.B.G. wrote the main manuscript text. All authors reviewed the manuscript.
### Corresponding author
Correspondence to Edgar M. Hoyos-Padilla.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
### Publisher's note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Becerril-García, E.E., Hoyos-Padilla, E.M., Micarelli, P. et al. Behavioural responses of white sharks to specific baits during cage diving ecotourism. Sci Rep 10, 11152 (2020). https://doi.org/10.1038/s41598-020-67947-x
• Accepted:
• Published:
• DOI: https://doi.org/10.1038/s41598-020-67947-x
• ### Evidence of interactions between white sharks and large squids in Guadalupe Island, Mexico
• Edgar E. Becerril-García
• Daniela Bernot-Simon
• Edgar M. Hoyos-Padilla
Scientific Reports (2020)
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2022-08-10 23:58:52
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https://kluedo.ub.uni-kl.de/frontdoor/index/index/docId/4008
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## Adaptive Real-Time Scheduling and Resource Management on Multicore Architectures
• Real-time systems are systems that have to react correctly to stimuli from the environment within given timing constraints. Today, real-time systems are employed everywhere in industry, not only in safety-critical systems but also in, e.g., communication, entertainment, and multimedia systems. With the advent of multicore platforms, new challenges on the efficient exploitation of real-time systems have arisen: First, there is the need for effective scheduling algorithms that feature low overheads to improve the use of the computational resources of real-time systems. The goal of these algorithms is to ensure timely execution of tasks, i.e., to provide runtime guarantees. Additionally, many systems require their scheduling algorithm to flexibly react to unforeseen events. Second, the inherent parallelism of multicore systems leads to contention for shared hardware resources and complicates system analysis. At any time, multiple applications run with varying resource requirements and compete for the scarce resources of the system. As a result, there is a need for an adaptive resource management. Achieving and implementing an effective and efficient resource management is a challenging task. The main goal of resource management is to guarantee a minimum resource availability to real-time applications. A further goal is to fulfill global optimization objectives, e.g., maximization of the global system performance, or the user perceived quality of service. In this thesis, we derive methods based on the slot shifting algorithm. Slot shifting provides flexible scheduling of time-constrained applications and can react to unforeseen events in time-triggered systems. For this reason, we aim at designing slot shifting based algorithms targeted for multicore systems to tackle the aforementioned challenges. The main contribution of this thesis is to present two global slot shifting algorithms targeted for multicore systems. Additionally, we extend slot shifting algorithms to improve their runtime behavior, or to handle non-preemptive firm aperiodic tasks. In a variety of experiments, the effectiveness and efficiency of the algorithms are evaluated and confirmed. Finally, the thesis presents an implementation of a slot-shifting-based logic into a resource management framework for multicore systems. Thus, the thesis closes the circle and successfully bridges the gap between real-time scheduling theory and real-world implementations. We prove applicability of the slot shifting algorithm to effectively and efficiently perform adaptive resource management on multicore systems.
$Rev: 13581$
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https://www.jobilize.com/algebra/course/2-6-other-types-of-equations-by-openstax?qcr=www.quizover.com
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# 2.6 Other types of equations
Page 1 / 10
In this section you will:
• Solve equations involving rational exponents.
• Solve equations using factoring.
• Solve radical equations.
• Solve absolute value equations.
• Solve other types of equations.
We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.
## Solving equations involving rational exponents
Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, $\text{\hspace{0.17em}}{16}^{\frac{1}{2}}\text{\hspace{0.17em}}$ is another way of writing $\text{\hspace{0.17em}}\sqrt{16};$ ${8}^{\frac{1}{3}}\text{\hspace{0.17em}}$ is another way of writing $\text{}\text{\hspace{0.17em}}\sqrt[3]{8}.\text{\hspace{0.17em}}$ The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.
We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, $\text{\hspace{0.17em}}\frac{2}{3}\left(\frac{3}{2}\right)=1,$ $3\left(\frac{1}{3}\right)=1,$ and so on.
## Rational exponents
A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:
${a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}$
## Evaluating a number raised to a rational exponent
Evaluate $\text{\hspace{0.17em}}{8}^{\frac{2}{3}}.$
Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite $\text{\hspace{0.17em}}{8}^{\frac{2}{3}}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}{\left({8}^{\frac{1}{3}}\right)}^{2}.$
$\begin{array}{ccc}\hfill {\left({8}^{\frac{1}{3}}\right)}^{2}& =\hfill & {\left(2\right)}^{2}\hfill \\ & =& 4\hfill \end{array}$
Evaluate $\text{\hspace{0.17em}}{64}^{-\frac{1}{3}}.$
$\frac{1}{4}$
## Solve the equation including a variable raised to a rational exponent
Solve the equation in which a variable is raised to a rational exponent: $\text{\hspace{0.17em}}{x}^{\frac{5}{4}}=32.$
The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of $\text{\hspace{0.17em}}\frac{5}{4},$ which is $\text{\hspace{0.17em}}\frac{4}{5}.$
Solve the equation $\text{\hspace{0.17em}}{x}^{\frac{3}{2}}=125.$
$25$
## Solving an equation involving rational exponents and factoring
Solve $\text{\hspace{0.17em}}3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}.$
This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.
$\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)& =& {x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill \\ \hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}& =& 0\hfill \end{array}$
Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite $\text{\hspace{0.17em}}{x}^{\frac{1}{2}}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}.\text{\hspace{0.17em}}$ Then, factor out $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}\text{\hspace{0.17em}}$ from both terms on the left.
$\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}& =& 0\hfill \\ \hfill {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)& =& 0\hfill \end{array}$
Where did $\text{\hspace{0.17em}}{x}^{\frac{1}{4}}\text{\hspace{0.17em}}$ come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply $\text{\hspace{0.17em}}{x}^{\frac{2}{4}}\text{\hspace{0.17em}}$ back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to $\text{\hspace{0.17em}}\frac{2}{4}\text{\hspace{0.17em}}$ equals $\text{\hspace{0.17em}}\frac{3}{4}.\text{\hspace{0.17em}}$ Thus, the exponent on x in the parentheses is $\text{\hspace{0.17em}}\frac{1}{4}.\text{\hspace{0.17em}}$
Let us continue. Now we have two factors and can use the zero factor theorem.
The two solutions are $\text{\hspace{0.17em}}0$ and $\frac{1}{81}.$
#### Questions & Answers
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
haha. already finished college
Jeffrey
how about you? what grade are you now?
Jeffrey
I'm going to 11grade
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
please where is the equation
Miranda
Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa
what is a algebra
(x+x)3=?
6x
Obed
what is the identity of 1-cos²5x equal to?
__john __05
Kishu
Hi
Abdel
hi
Ye
hi
Nokwanda
C'est comment
Abdel
Hi
Amanda
hello
SORIE
Hiiii
Chinni
hello
Ranjay
hi
ANSHU
hiiii
Chinni
h r u friends
Chinni
yes
Hassan
so is their any Genius in mathematics here let chat guys and get to know each other's
SORIE
I speak French
Abdel
okay no problem since we gather here and get to know each other
SORIE
hi im stupid at math and just wanna join here
Yaona
lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together
SORIE
it's 12
what is the function of sine with respect of cosine , graphically
tangent bruh
Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function
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2020-09-22 12:11:25
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https://cs.stackexchange.com/questions/95671/prove-that-d-x-in-mathbbn-phi-xx-uparrow-is-not-recursively
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# Prove that $D =\{x \in \mathbb{N} | \Phi_x(x)\uparrow\}$ is **not** recursively enumerable
So I tried to prove that $D =\{x \in \mathbb{N} | \Phi_x(x)\uparrow\}$ is not recursively enumerable in the following way:
• let's suppose that $g$ is the computable function that represents $D$ $$g(x) = \left\{\begin{array}{rcl} 1 & \mbox{if} & \Phi_x(x)\downarrow\ \\ 0 & \mbox{if} & \Phi_x(x)\uparrow\ \end{array}\right.$$
• let's suppose that it exists a Turing machine of index $i_0$ such that $g = \Phi_{i_0}$, than it is possible to define the function: $$g'(x) = \left\{\begin{array}{rcl} \uparrow & \mbox{if} & g(x)=1=\Phi_{i_0}(x) \\ 0 & \mbox{if} & g(x)=0=\Phi_{i_0}(x) \end{array}\right.$$
• so $g'$ is also computable thus there is a Turing machine of index $i_1$ such that $g' = \Phi_{i_1}$, but then we will have that: $$\Phi_{i_1}(i_1)\downarrow \iff g'(i_1) = 0 \iff g(i_1) = 0 \iff \Phi_{i_1}(i_1)\uparrow$$ $$\Phi_{i_1}(i_1)\uparrow \iff g'(i_1) \uparrow \iff g(i_1) = 1 \iff \Phi_{i_1}(i_1)\downarrow$$
which is a contraddiction. So an index $i_1$ doesn't exists that is the index $i_0$ that computes $g$ doesn't exists, that is $g$ is not recursive so $D$ is not recursive enumerable.
I'm quite confused on this topic, so tell me if I made any kind of error.
• I think there is a slight error in your definition of $D$. Do you mean $D = \{ x \in \mathbb{N} | \Phi_i (x) \uparrow \}$ ? Note that $\Phi_i$ denotes one goedelized Turing Machine with Goedelnumber $i$ Jul 28 '18 at 3:06
• Yes you are rigth. Thank you... anything else? I think I also messed up the last definition. I wrote that if $g$ is not recursive than $D$ is not recursive enumerable, but I don't find any theorem that states that. Jul 28 '18 at 8:29
• You proved that D is not recursive (also known as decidable) because you starting assumption was that there is a computable $g$ which decides membeship of $D$. To prove that $D$ is not recursively enumerable, your starting assumption must be tht there is a computable $h$ whose image is $D$. Jul 28 '18 at 9:02
• I'm doing a mess here, but I came with this: $\psi_d(x) = \left\{\begin{array}{rcl} 1 & \mbox{if} & x \in D \\ \uparrow & \mbox{if} & x \notin D\end{array}\right.$ this means that $D=dom(\psi_d)$. So $x\in A \iff \exists y \text{ such that } \Phi_y(x)\uparrow$. Then $\psi_d(x) = \left\{\begin{array}{rcl} 1 & \mbox{if} &\exists y. \Phi_y(x)\uparrow\\ \uparrow & \mbox{if} & \exists y. \Phi_y(x)\downarrow\end{array}\right.$ but that is a contraddiction. Jul 28 '18 at 9:42
• Yes but checking if $\Phi_i(0)\uparrow$ it's not computable, isn't it? I understood that there is no way to know if a function terminates or not its computation. I'm really confused on this topic so please have patience with me. Jul 28 '18 at 19:57
As Andrej states, what you've (correctly) done is show that $D=\{x:\Phi_x(x)\uparrow\}$ is not recursive. However, there are plenty of sets which are recursively enumerable but not recursive, so you're not done yet: you need to show that $D$ is not the domain of any partial recursive function.
REMARK: Some texts define a set to be r.e. if it is the range of a total recursive function, or is empty. This is slightly less elegant since it requires a separate clause to handle $\emptyset$; it also generalizes less well if one looks into higher computability theory. However, regardless of what definition your book gives, the equivalence between them should be a very early exercise.
There are a couple ways to do this:
• You could argue directly: suppose $D$ is the domain of $\Phi_i$. Can you use the recursion theorem to cook up a $j$ such that $\Phi_j(j)\downarrow\iff \Phi_i(j)\downarrow$? Do you see why this gives a contradiction?
• Alternately, note that $D$ is the complement of the Halting Problem $H=\{x: \Phi_x(x)\downarrow\}$. Two early results in computability theory are that (i) $H$ is r.e. but not recursive and (ii) if a set and its complement are r.e., then they are each recursive. Do you see how to prove each of these facts, and apply them to the current problem? (Note that this approach isn't actually much different than the first one, since the recursion theorem is used in proving (i).)
REMARK: Some texts define the Halting Problem differently, some common alternate candidates being $\{x: \Phi_x(0)\downarrow\}$ and $\{\langle x, y\rangle:\Phi_x(y)\downarrow\}$. It's a good exercise to show that these are all r.e. and Turing equivalent (indeed $1$-equivalent!) to each other.
By similar arguments you can show that minor variations of $D$ are also not r.e. - e.g. can you show that $\{x: \Phi_x(17)\uparrow\}$ is not r.e.?
Responding to the comments: one may be tempted to try to enumerate $D$, or things like $D$, by looking at approximations given by stages. E.g. we could let $D_s=\{x:\Phi_x(x)[s]\uparrow\}$; this is the set of indices of machines which on their own input don't halt in fewer than $s$ steps. The $D_s$s are r.e. - indeed, uniformly recursive - and approximate $D$ "from above" in the sense that $D=\bigcap_{s\in\mathbb{N}} D_s$. However, this ultimately isn't relevant to this problem since the intersection of infinitely many r.e. sets need not be r.e.
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2022-01-19 02:17:26
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