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https://www.physicsforums.com/threads/elastic-potential-energy-of-a-spring-between-2-blocks.276559/ | # Elastic Potential Energy of a Spring Between 2 Blocks
1. Dec 2, 2008
### mac31905
1. The problem statement, all variables and given/known data
Here's the question:
Block A in the figure below has mass 1.00 kg, and block B has mass 3.00 kg. The blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.20 m/s.
Part A: It asked for the final velocity of block A, and I used the mavai + mbvbi = mavaf + mbvbf equation to find A's final velocity of 3.6 m/s.
mavai = 0, mbvbi = 0, ma = 1kg, mb = 3 kg, vbf = -1.20m/s
0 = (1 kg)(Vaf)+(3kg)(-1.20m/s)
0 = 1kg(Vaf) -3.6kgm/s
3.6kgm/s = 1kg(Vaf)
Vaf = 3.6 m/s
Part B: The second part of the question is asking for how much potential energy was stored in the compressed spring... and I have no idea how to find it! I've tried different equations but I have too many unknown variables and don't know where to start.
2. Relevant equations
For Part B:
PE(elastic) = 1/2kx^2 (this seems like the one to use, but how do I find k and x?)
KE = 1/2mv^2 (I don't think I have to find KE to find PE... though I do know m and v.)
GPE = mgh (and nope, this is gravitational potential energy. And plus, I don't have h.)
3. The attempt at a solution
So... if I do indeed use the PE(elastic equation), where in the world do I find k? I know it is the force (spring) constant, but I am unsure as to how to get it from the information given. Also, x is the compressed distance from it's uncompressed state.
*To find the force constant, I'd use this equation: F(spring) = -kx, but I don't have F, nor do I know x.
*And the question says nothing about how long the spring is when it's compressed/not compressed!
It just seems like I am not given enough information, even though that must be false since it's a homework question...
I've been stuck on this for a long time, and I'm so tempted to click "Show answer" on Mastering Physics but that won't help me to understand it, so hopefully one of you guys can help point me in the right direction I just need to figure out k and x and I'm good! I'm not asking for the answer, just an equation or something to get me started. Thanks!!! My test is tomorrow and I'd really appreciate some help!
Last edited: Dec 2, 2008
2. Dec 2, 2008
### naresh
Yes, it is hard to use this equation if you know neither k nor x.
Let's add a new question saying find the KE of the two blocks.
Now (after the spring drops), you have some kinetic energy. What is the source of this kinetic energy?
3. Dec 2, 2008
### mac31905
Hmm... it seems to me that the blocks would be moving after the spring drops. Therefore, the blocks would have some kinetic energy, right? Maybe the source of the KE would be the spring decompressing?
4. Dec 2, 2008
### mac31905
Let me see:
If I use the KE = 1/2mv^2 formula to find the KE of the blocks (if I'm going in the correct direction), then it'd be KE(block a) = 1/2(1kg)(3.6m/s^2); KE(block b) = 1/2(3kg)(-1.20m/s^2)
KEblocka = 6.48
KEblockb = 2.16
So maybe total KE = 6.48+2.16 = 8.64? I have no idea...
5. Dec 2, 2008
### mac31905
HOORAY! I tried my value of 8.64 and it worked! All that stress for a simple equation. Thanks Naresh! I can move on to my other problems now :)
6. Dec 2, 2008
### naresh
I'm glad it worked, but do you understand why it worked? Why should the sum of the kinetic energies give you the potential energy? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8783044815063477, "perplexity": 747.6556433943368}, "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-2017-04/segments/1484560281649.59/warc/CC-MAIN-20170116095121-00525-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://billwolf.space/ | # Welcome
## In a nutshell...
I'm an assistant professor in the Department of Physics and Astronomy at University of Wisconsin - Eau Claire teaching physics and researching stellar astrophysics. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8751561045646667, "perplexity": 1011.4715154048805}, "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-2021-31/segments/1627046154127.53/warc/CC-MAIN-20210731234924-20210801024924-00286.warc.gz"} |
http://mathhelpforum.com/number-theory/186551-prove-equation-solvable-print.html | Prove that this equation is solvable
• August 22nd 2011, 02:23 PM
alexmahone
Prove that this equation is solvable
Given any integers a, b, c and any prime p not a divisor of ab, prove that $ax^2+by^2\equiv c\ (mod\ p)$ is solvable.
• August 24th 2011, 02:04 AM
Opalg
Re: Prove that this equation is solvable
Quote:
Originally Posted by alexmahone
Given any integers a, b, c and any prime p not a divisor of ab, prove that $ax^2+by^2\equiv c\ (mod\ p)$ is solvable.
I can only prove this by quoting a theorem from a recent research paper. However, the theorem is not that hard. Essentially the same result seems to have been proved a long time ago, in this 1893 article by J C Fields (after whom the Fields Medal is named).
If c is a multiple of p then we can take x = y = 0. So we can assume that $c\in\mathbb{Z}_p^*$ (the multiplicative group of nonzero residues mod p). Also, the result is easy to prove in the case p=2. So assume that p is an odd prime.
Let R be the set of quadratic residues mod p, and let N be the set of quadratic non-residues. According to Theorem 2.1 in this paper (pdf file), every element of $\mathbb{Z}_p^*$ belongs to each of the three sets R+R, R+N and N+N. But R is a subgroup of $\mathbb{Z}_p^*$. Its cosets are R and N. The sets $\{ax^2:x\in\mathbb{Z}_p^*\}$ and $\{by^2:y\in\mathbb{Z}_p^*\}$ are also cosets of R. Thus the set $\{ax^2 +by^2:x,y\in\mathbb{Z}_p^*\}$ is equal to one of the sets R+R, R+N or N+N, and therefore contains c. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "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": 0.9702484607696533, "perplexity": 269.88296827409016}, "config": {"markdown_headings": false, "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-2014-10/segments/1393999654758/warc/CC-MAIN-20140305060734-00089-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/newtons-does-a-40-kg-bag-of-stones-weigh.7859/ | # Newtons does a 40 kg bag of stones weigh
1. Oct 27, 2003
### espo
how much in newtons does a 40 kg bag of stones weigh? my answer is 302 newtons is that correct?
Last edited by a moderator: Feb 6, 2013
2. Oct 27, 2003
### chroot
Staff Emeritus
Hi espo,
Just think about Newton's second law,
F = ma
If you have a bag of stones with a mass of 40 kg, and that bag of stones is in Earth's gravitational field (and thus has an acceleration of 9.8 m/s), then it must exert a force of 40 * 9.8 newtons on the ground. This force is what we call weight.
The bag weighs 40 * 9.8 = 392 N. Perhaps you typed 302 accidentally, when you meant to say 392?
- Warren | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9611423015594482, "perplexity": 1712.2096169498511}, "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-2017-04/segments/1484560280319.10/warc/CC-MAIN-20170116095120-00455-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://formulasearchengine.com/wiki/Row_vector | Row vector
In linear algebra, a row vector or row matrix is a 1 × m matrix, i.e. a matrix consisting of a single row of m elements:[1]
${\displaystyle {\mathbf {x} }={\begin{bmatrix}x_{1}&x_{2}&\dots &x_{m}\end{bmatrix}}.}$
The transpose of a row vector is a column vector:
${\displaystyle {\begin{bmatrix}x_{1}\;x_{2}\;\dots \;x_{m}\end{bmatrix}}^{\rm {T}}={\begin{bmatrix}x_{1}\\x_{2}\\\vdots \\x_{m}\end{bmatrix}}.}$
The set of all row vectors forms a vector space (row space) which acts like the dual space to the set of all column vectors (see row and column spaces), in the sense that any linear functional on the space of column vectors (i.e. any element of the dual space) can be represented uniquely as a dot product with a specific row vector.
Notation
Row vectors are sometimes written using the following non-standard notation:
${\displaystyle {\mathbf {x} }={\begin{bmatrix}x_{1},x_{2},\dots ,x_{m}\end{bmatrix}}.}$
Operations
• Matrix multiplication involves the action of multiplying each row vector of one matrix by each column vector of another matrix.
• The dot product of two vectors a and b is equivalent to multiplying the row vector representation of a by the column vector representation of b:
${\displaystyle {\mathbf {a} }\cdot {\mathbf {b} }={\begin{bmatrix}a_{1}&a_{2}&a_{3}\end{bmatrix}}{\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}}.}$
Preferred input vectors for matrix transformations
Frequently a row vector presents itself for an operation within n-space expressed by an n by n matrix M:
v M = p.
Then p is also a row vector and may present to another n by n matrix Q:
p Q = t.
Conveniently, one can write t = p Q = v MQ telling us that the matrix product transformation MQ can take v directly to t. Continuing with row vectors, matrix transformations further reconfiguring n-space can be applied to the right of previous outputs.
In contrast, when a column vector is transformed to become another column under an n by n matrix action, the operation occurs to the left:
p = M v and t = Q p ,
leading to the algebraic expression QM v for the composed output from v input. The matrix transformations mount up to the left in this use of a column vector for input to matrix transformation. The natural bias to read left-to-right, as subsequent transformations are applied in linear algebra, stands against column vector inputs.
Nevertheless, using the transpose operation these differences between inputs of a row or column nature are resolved by an antihomomorphism between the groups arising on the two sides. The technical construction uses the dual space associated with a vector space to develop the transpose of a linear map.
For an instance where this row vector input convention has been used to good effect see Raiz Usmani,[2] where on page 106 the convention allows the statement "The product mapping ST of U into W [is given] by:
${\displaystyle \alpha (ST)=(\alpha S)T=\beta T=\gamma }$."
(The Greek letters represent row vectors).
Ludwik Silberstein used row vectors for spacetime events; he applied Lorentz transformation matrices on the right in his Theory of Relativity in 1914 (see page 143). In 1963 when McGraw-Hill published Differential Geometry by Heinrich Guggenheimer of the University of Minnesota, he uses the row vector convention in chapter 5, "Introduction to transformation groups" (eqs. 7a,9b and 12 to 15). When H. S. M. Coxeter reviewed[3] Linear Geometry by Rafael Artzy, he wrote, "[Artzy] is to be congratulated on his choice of the 'left-to-right' convention, which enables him to regard a point as a row matrix instead of the clumsy column that many authors prefer."
Notes
1. Template:Harvtxt, p. 8
2. Raiz A. Usmani (1987) Applied Linear Algebra Marcel Dekker ISBN 0824776224. See Chapter 4: "Linear Transformations"
References
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• {{#invoke:citation/CS1|citation
|CitationClass=citation }} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8754063844680786, "perplexity": 831.2046689286415}, "config": {"markdown_headings": false, "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-2019-39/segments/1568514575484.57/warc/CC-MAIN-20190922094320-20190922120320-00155.warc.gz"} |
https://newproxylists.com/calculus-and-analysis-plotting-function-and-its-approximation-function/ | calculus and analysis – Plotting function and its approximation function
I have a problem which I have not been able to solve. I want to plot a function and and operator which approximates it when you let w to infinity. I will give all needed information for MWE and my faults.
MathJax
$$operatorname{Fejer}(x):= dfrac{1}{2} operatorname{sinc}^2left(dfrac{x}{2}right) quad (xin mathbb{R})\ operatorname{sinc} x:= begin{cases} dfrac{sin pi x}{pi x}, & xin mathbb{R} backslash {0}\ 1, & x=0 end{cases}\ operatorname{Function}(x):=begin{cases} dfrac{9}{x^2},& x<3\ 2,& -3
Code
sinc(x_) := Piecewise({{1, Equal(x, 0)}, {Sin(Pi x)/(Pi x), True}})
Fejer(x_) := 1/2*sinc(x/2)^2
function(x_) :=
Piecewise(
{{9/(x^2), x < -3},
{2, -3 <= x < -2},
{-1/2, -2 <= x < -1},
{3/2, -1 <= x < 0},
{1, 0 <= x < 1},
{-1, 1 <= x < 2},
{0, 2 <= x < 3},
{-50/(x^4), 3 <= x}})
constant(x_) := 1
With above code I am defining the functions which I gave mathematically above to make it easier for you. Now, I’m trying to define my operator.
operator(w_, kernel_, func_, x_) :=
Sum(
w *
Integrate(func(u), {u, k/w, (k + 1)/w}, Assumptions -> k ∈ Integers) *
kernel(w*x - k),
{k, -Infinity, Infinity})
I’m not sure about the above code. I used Assumption in the integral because I was getting errors like “integral limits may not reals, please add assumption”. I also show it in MathJax so you can understand what I’m trying to do.
Operator
$$(S_wf)(x):=sum_{kin mathbb{Z}} chi(wx-k) wint_{k/w}^{(k+1)/w} f(u)du , quad xin mathbb{R}, , w>0$$
When you take function to br $$1$$ for every $$xin mathbb{R}$$, operator gives $$1$$. Anyway when I running code
operator(w, Fejer, constant, x)
or
operator(5, Fejer, constant, x)
it gives nothing. When I tried plot
Plot(Operator(5, Fejer, cons, x), {x, 0, 5})
it quits the kernel without an error.
When I tried
Plot(Operator(5, Fejer, function, x), {x, 0, 5})
It gives many errors and some of them are:
NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.
NIntegrate::nlim: u = 0.2 k is not a valid limit of integration.
General::stop: Further output of NIntegrate::nlim will be suppressed during this calculation.
Finally, I’m adding a result which I’m trying to reach. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9298198819160461, "perplexity": 4119.344464912201}, "config": {"markdown_headings": false, "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-2021-10/segments/1614178366959.54/warc/CC-MAIN-20210303104028-20210303134028-00466.warc.gz"} |
http://mathhelpforum.com/number-theory/43006-number-solution-x-x-1-mod-n.html | # Math Help - number of solution of x*x =1 (mod n) ??
1. ## number of solution of x*x =1 (mod n) ??
Hi to all,
I am new to this forum. i want to know the number of solution of the equation x*x = 1 (mod n) for given n.
certainly it is 2 if n is prime if n is non prime I cant figure it out .
I would be gratefull if u help me.
2. Originally Posted by hingtingchot
Hi to all,
I am new to this forum. i want to know the number of solution of the equation x*x = 1 (mod n) for given n.
certainly it is 2 if n is prime if n is non prime I cant figure it out .
I would be gratefull if u help me.
$x^2 \equiv 1 (mod \, n) \Longleftrightarrow x^2 -1 \equiv 0(mod \, n) \Longleftrightarrow (x+1)(x-1) \equiv 0(mod \, n)$
3. can u please explain a little bit more ??
4. Originally Posted by kalagota
$x^2 \equiv 1 (mod \, n) \Longleftrightarrow x^2 -1 \equiv 0(mod \, n) \Longleftrightarrow (x+1)(x-1) \equiv 0(mod \, n)$
Originally Posted by hingtingchot
can u please explain a little bit more ??
The equation factors a modulo anything. So there are two solutions: x = 1 and x = -1 = n - 1.
-Dan
5. thnx but tthis is only for those n who are pime the solution is 1 and n-1
but say for 8 the solution may be 1,3,5,7 means number of solution is 4 . how could i figure this out ????
6. Right the faulty assumption in topsquark's answer is there being no zero divisors, which is false in this case. Then what satisfies that equation is a pair of zero divisors different by 2. It is clear -1 and +1 work because then (-1+1)(-1-1)=0 and (1+1)(1-1)=0. When else can this happen though?
It is clear that this is exactly when $n=(x+1)(x-1)$ by the modulo. Suppose there is some other y such that $n=(y+1)(y-1)$. Then $y^2-1=x^2-1$ and so $y=\pm x$.
So if there are two more solutions exactly when n is the product of consecutive even or odd integers. In other words, when n=8,15,24, etc. More generally, when n=x^2-1 for some integer x. The one exception to this is of course x=2 where n=3.
I think this is correct, I haven't thought through it more than a plausibility thing but it seems correct.
EDIT: Ok, thats not correct. I just randomly picked 12 to see that 5^2=25 which is 1 mod 5 because 4*6=24=0 mod 12. So the multiples are also kind of important... whoops! So in this case we're really looking for x-1, x+1 such that the prime divisors of n are contained in the prime divisors of x-1, x+1. I'll think about it some more.
It also seems to be true for multiples of 4, interestingly enough. Divide it in half to get a 0 divisor, then pick $x=n/2\pm1$, and this will clearly work. So two special cases clearly give you 4 divisors. Its been two long since I've thought about the nuances of polynomials in this context. At this point perhaps more experimentation would reveal the pattern?
7. ## Patterns in the solutions
The key is to consider $n = \frac{a^2 - 1}{k}$ for all positive integers a and for all positive integers k that will evenly divide $a^2 - 1$. For every pair of numbers $(a, k)$ with appropriate k, you get another pair of solutions for n: $(a, n - a)$. You only need to consider values of a which are less than half of n after the calculation.
Here are the solutions for select composite numbers:
8 - 1, 3, 5, 7
$3^2 - 1 = 8$ gives $(3, 5)$
12 - 1, 5, 7, 11
$\frac{5^2 - 1}{2} = 12$ gives $(5, 7)$
15 - 1, 4, 11, 14
$4^2 - 1 = 15$ gives $(4, 11)$
20 - 1, 9, 11, 19
$\frac{9^2 - 1}{4} = 20$ gives $(9, 11)$
24 - 1, 5, 7, 11, 13, 17, 19, 23
$5^2 - 1 = 24$ gives $(5, 19)$; $\frac{7^2 - 1}{2} = 24$ gives $(7, 17)$; $\frac{11^2 - 1}{5} = 24$ gives $(11, 13)$
Note also that every solution for a particular number is expressible as $\frac{a^2 - 1}{k}$. For instance, you may obtain the solutions of 11 for 15 by $\frac{11^2 - 1}{8} = 15$ and 14 for 15 by $\frac{14^2 - 1}{13} = 15$, but this is unnecessary since you have obtained them from knowing that 1 and 4 are solutions.
I don't really see a more direct way of determining the solutions explicitly; induction doesn't seem to make much sense in terms of this problem. About all you can say for certain is that there are an even number of solutions for every modulus greater than 2, since the solutions come in pairs. You might also say that every number which is a multiple of 4, with the exception of 4 itself, has at least 4 solutions but some numbers (even those that are not multiples of 4) may have more.
8. ## Re: number of solution of x*x =1 (mod n) ??
To answer this question it is best to look at the structure of the multiplicative group of order n. For a full review on the subject go to this link: Multiplicative group of integers modulo n - Wikipedia, the free encyclopedia
So since we know that satisfying the equation $x^2=1$ is equivalent to finding an element of order at most 2 in this multiplicative group.
Note that in each cyclic group of even order there are exactly 2 such elements. And also note that if you take the direct product of groups the amount of such elements gets multiplied together to make the grand total of such elements.
So all it comes down to is counting the number of 'terms' this multiplicative group is made of.
so let's count: for each odd prime divisor there is exactly 1 term, and for the prime 2 there is either 2 terms if n is a multiple of 8, or 1 if n is even, (or zero if n is odd).
So this gives the formule $2^{\omega(n)} \epsilon$ where $\epsilon$ is 1 if n isn't a multiple of 8, and 2 otherwhise. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "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": 0.9219953417778015, "perplexity": 263.79528982461187}, "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-2015-27/segments/1435375098468.93/warc/CC-MAIN-20150627031818-00005-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://brilliant.org/problems/whats-my-average-velocity/ | # What's my average velocity?
A body covers half the distance with a speed of 20 m/s and the other half with 30 m/s . What is the average speed of the body for the whole journey?
× | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9412307143211365, "perplexity": 234.67353373225637}, "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-2018-17/segments/1524125937113.3/warc/CC-MAIN-20180420022906-20180420042906-00456.warc.gz"} |
http://viveks.wikidot.com/graph-theory | Graphs
A graph is specified by a set $V$ of vertices and a collection $E$ of unordered pair of vertices called edges and denoted by $G = (V, E)$. Several problems can be formulated in terms of graphs, in physics, electrical engineering and computer science. There is an entire branch devoted to the study of graph theoretic algorithms. Sometimes, graphs make problems easier to visualize (although not necessarily easier to solve, since graphs have their own complexities). We begin with some definitions:
The order of a graph is the number of its vertices, and the size is the number of edges. We concern ourselves here with directed and undirected graphs. Suppose $i, j \in V$ and that there is an edge connecting vertex $i$ to vertex $j$. If $G = (V, E)$ is a directed graph or digraph then, $(i, j) \in E$ but $(j, i) \notin E$ (or vice versa), that is, in a digraph there is a sense of direction of an edge. Therefore, we can also define the indegree and outdegree of a vertex in a digraph as the number of edges entering that vertex and the number of edges leaving that vertex, respectively. In an undirected graph all that matters is whether there is an edge connecting two vertices.
Note that if $(i, i) \in E$ then there is a self-loop, that is, an element is connected to itself via an edge (that ends up looking like a loop). A graph with no loops is called a multigraph whereas a pseudograph is a graph having at least one loop. The definition of an undirected graph prohibits undirected graphs from having self loops.
In general, $E \subseteq V \times V$, so $|E| \leq |V|^2$.
A sequence of vertices $\langle u_{1}, u_{2}, \ldots, u_{k} \rangle$ in a given graph is a path if there is no repetition of vertices in the sequence (that is $u_{i} \neq u_{k}$ $\forall i \neq k$) and for every $1\leq i \leq k-1$, $(u_{i}, u_{i+1}) \in E$. Thus, $\langle u_{1}, u_{2}, \ldots, u_{k} \rangle$ is a path from $u_{1}$ to $u_{k}$.
A sequence of vertices $\langle u_{0}, u_{1}, \ldots, u_{k} \rangle$ in a given graph is a cycle if there is no repetition of vertices in the sequence and for every $0\leq i\leq k$, we have $(u_{i \mod k}, u_{(i+1) \mod k}) \in E$. Equivalently, a cycle is sequence of vertices $\langle u_{0}, u_{2}, \ldots, u_{k} \rangle$ such that $u_{0} = u_{k}$ and there is no repetition of vertices in $\langle u_{0}, u_{1}, \ldots, u_{k-1} \rangle$ and $(u_{i}, u_{i+1}) \in E$ $\forall 0\leq i\leq k-1$.
A graph is connected if there exists a path connecting any pair of vertices. A graph is complete if there exists an edge connecting every pair of vertices in the graph. Clearly, a complete graph is connected but the converse is not true in general (e.g. consider a square with no diagonals).
A subgraph of $G = (V, E)$ is a graph $G' = (V', E')$ such that $V' \subseteq V$ and $E' \subseteq E$. By definition, it is also required that $E' \subseteq V' \times V'$. It is obvious that $E' \subseteq V \times V$. A symmetric digraph is a directed graph such that for every edge $(v, w)$ there is also the reverse edge $(w, v)$. Every undirected graph has a corresponding symmetric digraph by interpreting each undirected edge as a pair of directed edges in opposite directions.
A connected component of an undirected graph G is a maximal connected subgraph of G.
Under Construction
page revision: 7, last edited: 16 Oct 2007 08:58 | {"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": 0.8678470849990845, "perplexity": 81.12054492777577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2018-39/segments/1537267156460.64/warc/CC-MAIN-20180920101233-20180920121633-00426.warc.gz"} |
https://www.milania.de/showcase/The_softmax_function_in_the_output_layer_of_neural_networks | Suppose you use a neural network for a classification problem and the neurons in the output layer should return a valid discrete probability distribution. If you set the number of output neurons $$n$$ equal to the number of classes of your classification problem, you have the nice interpretation that the result for each neuron $$y_i$$ gives you the probability that the corresponding input belongs to the class $$\omega_i$$. If the network is confident in its classification, you will see a strong peak in the probability distribution. On the other hand, for a noisy input where the network has not really a clue what it means (or it hasn't learned yet), the resulting distribution will be more broadened.
But how do you transform the weighted input $$u_i = \left\langle \fvec{w}_i, \fvec{x} \right\rangle + b$$ (weight $$\fvec{w}_i$$, input $$\fvec{x}$$ and bias $$b$$) into a valid probability distribution? A common approach is to apply the softmax function on the weighted input
$$\label{eq:Softmax} y_i = \frac{e^{c \cdot u_i}}{\sum_{j=1}^{n} e^{c \cdot u_j}}$$
with the parameter $$c \in \mathbb{R}$$. Using this transformation we ensure that the resulting output vector $$\fvec{y} = (y_1, y_2, \ldots, y_n)$$ satisfies
\begin{equation*} \sum_{i=1}^{n} y_i = 1 \quad \text{and} \quad \forall i : y_i \geq 0 \end{equation*}
and is therefore indeed a valid probability distribution. A common choice is to set $$c=1$$ but it is useful to analyse the result for different values for this parameter. You can do so in the following animation based on an arbitrary example vector
\begin{equation*} \fvec{u} = (-0.2, 1, 2, 0, 0, 2.1, 1.4, 0.8). \end{equation*}
List of attached files:
← Back to the overview page | {"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": 0, "mathjax_display_tex": 1, "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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9435411691665649, "perplexity": 199.28557341212112}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00257.warc.gz"} |
https://icsecbsemath.com/2017/08/26/class-10-sample-problems-quadratic-equations-exercise-7c/ | Solve each of the following equations for $\displaystyle x$
Question 1: $\displaystyle x^2-8x+5=0$
Comparing $\displaystyle x^2-8x+5=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -8 \text{ and } c =5$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-8) \pm \sqrt{(-8)^2-4(1)(5)}}{2(1)}$
$\displaystyle \text{Solving we get } x = 7.32, 0.68$
$\displaystyle \\$
Question 2: $\displaystyle 5x^2+10x-3=0$
Comparing $\displaystyle 5x^2+10x-3=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 5, b = 10 \text{ and } c =-3$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(10) \pm \sqrt{(10)^2-4(5)(-3)}}{2(5)}$
$\displaystyle \text{Solving we get } x = 0.26, -2.26$
$\displaystyle \\$
Question 3: $\displaystyle 2x^2-10x+5=0$
Comparing $\displaystyle 2x^2-10x+5=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = -10 \text{ and } c =5$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-10) \pm \sqrt{(-10)^2-4(2)(5)}}{2(2)}$
$\displaystyle \text{Solving we get } x = 4.44, 0.56$
$\displaystyle \\$
$\displaystyle \text{Question 4: } 4x+ \frac{6}{x} +13=0$
$\displaystyle 4x+ \frac{6}{x} +13=0$ or
$\displaystyle 4x^2+13x+6=0$
Comparing $\displaystyle 2x^2-10x+5=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 4, b = 13 \text{ and } c =6$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(13) \pm \sqrt{(13)^2-4(4)(6)}}{2(4)}$
$\displaystyle \text{Solving we get } x = -0.56, -2.69$
$\displaystyle \\$
Question 5: $\displaystyle x^2-3x-9=0$ [2007]
Comparing $\displaystyle x^2-3x-9=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -3 \text{ and } c =-9$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-9)}}{2(1)}$
$\displaystyle \text{Solving we get } x = 4.85, -1.85$
$\displaystyle \\$
Question 6: $\displaystyle x^2-5x-10=0$ [2013]
Comparing $\displaystyle x^2-5x-10=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -5 \text{ and } c =-10$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}$
$\displaystyle \text{Solving we get } x = 6.53, -1.53$
$\displaystyle \\$
Question 7: $\displaystyle 3x^2-12x-1=0$
Comparing $\displaystyle 3x^2-12x-1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 3, b = -12 \text{ and } c =-1$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-12) \pm \sqrt{(-12)^2-4(3)(-1)}}{2(3)}$
$\displaystyle \text{Solving we get } x = 4.082, -0.082$
$\displaystyle \\$
Question 8: $\displaystyle x^2-16x+6=0$
Comparing $\displaystyle x^2-16x+6=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -16 \text{ and } c =6$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-16) \pm \sqrt{(-16)^2-4(1)(6)}}{2(1)}$
$\displaystyle \text{Solving we get } x = 15.616, 0.384$
$\displaystyle \\$
Question 9: $\displaystyle 2x^2+11x+4=0$
Comparing $\displaystyle 2x^2+11x+4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = 11 \text{ and } c =4$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(2)}$
$\displaystyle \text{Solving we get } x = -0.392, -5.108$
$\displaystyle \\$
Question 10: $\displaystyle x^4-2x^2-3=0$
$\displaystyle x^4-2x^2-3=0$
$\displaystyle \text{Let } x^2 = y$
$\displaystyle \text{Therefore } y^2-2y-3=0$
Comparing $\displaystyle y^2-2y-3=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = -2 \text{ and } c =-3$
$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } 7 = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}$
$\displaystyle \text{Solving we get } y = 3, -1$
$\displaystyle \text{Therefore } x^2 = 3\text{ or } x^2 = -1$ (imaginary number).
$\displaystyle \text{Hence } x = 1.732, -1.732\text{ or } \pm\sqrt{3}$
$\displaystyle \\$
Question 11: $\displaystyle x^4-10x^2+9=0$
$\displaystyle x^4-10x^2+9=0$
$\displaystyle \text{Let } y = x^2$
Therefore we have $\displaystyle y^2-10y+9=0$
Comparing $\displaystyle y^2-10y+9=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = -10 \text{ and } c =9$
$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } 7 = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}$
$\displaystyle \text{Solving we get } y = 9, 1$
$\displaystyle \text{Therefore } x^2 = 9 \Rightarrow x = \pm 3\text{ or } x^2 = 1 \Rightarrow x = \pm 1$ .
$\displaystyle \text{Hence } x = \pm 3, \pm 1$
$\displaystyle \\$
Question 12: $\displaystyle (x^2-x)^2+5(x^2-x)+4=0$
$\displaystyle \text{Let } (x^2-x)=y$
$\displaystyle \text{Therefore } y^2+5y+4=0$
Comparing $\displaystyle y^2+5y+4=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = 5 \text{ and } c =4$
$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } y = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}$
$\displaystyle \text{Solving we get } y = -1, -4$
$\displaystyle \text{When } y = -1 \Rightarrow x^2-x+1=0$
Comparing $\displaystyle x^2-x+1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -1 \text{ and } c =1$
$\displaystyle \text{Therefore } x = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(1)}}{2(1)}$
$\displaystyle \text{Hence } x = Imaginary$
$\displaystyle \text{When } y = -4 \Rightarrow x^2-x+4=0$
Comparing $\displaystyle x^2-x+4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -1 \text{ and } 4 =1$
$\displaystyle \text{Therefore } x = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(4)}}{2(1)}$
$\displaystyle \text{Hence } x = Imaginary$
$\displaystyle \\$
Question 13: $\displaystyle (x^2-3x)^2-16(x^2-3x)-36=0$
$\displaystyle \text{Let } (x^2-3x)=y$
$\displaystyle \text{Therefore } y^2-16y-36=0$
Comparing $\displaystyle y^2-16y-36=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = -16 \text{ and } c =-36$
$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } y = \frac{-(16) \pm \sqrt{(-16)^2-4(1)(-36)}}{2(1)}$
$\displaystyle \text{Solving we get } y = 18, -2$
$\displaystyle \text{When } y = 18 \Rightarrow x^2-3x-18=0$
Comparing $\displaystyle x^2-3x-18=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -3 \text{ and } c =-18$
$\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-18)}}{2(1)}$
$\displaystyle \text{Hence } x = 6, -3$
$\displaystyle \text{When } y = -2 \Rightarrow x^2-x+2=0$
Comparing $\displaystyle x^2-3x+2=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -3 \text{ and } c =2$
$\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(c)}}{2(1)}$
$\displaystyle \text{Hence } x = 2, 1$
$\displaystyle \text{Hence } x = 1, 2, 6, -3$
$\displaystyle \\$
$\displaystyle \text{Question 14: } \sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}} = \frac{5}{2}$
$\displaystyle \sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}} = \frac{5}{2}$
$\displaystyle \text{Let } \sqrt{\frac{x}{x-3}} = y$
Therefore the equation becomes
$\displaystyle y + \frac{1}{y} = \frac{5}{2}$
or $\displaystyle 2y^2-5y+2=0$
Comparing $\displaystyle 2y^2-5y+2=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 2, b = -5 \text{ and } c =2$
$\displaystyle \text{Therefore } y = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(c)}}{2(2)}$
$\displaystyle \text{Hence } y = 2, 0.5$
When y = 2
$\displaystyle \sqrt{\frac{x}{x-3}} = 2$
Squaring both sides
$\displaystyle \frac{x}{x-3} = 4 \Rightarrow 3x=12 \Rightarrow x = 4$
Similarly, when y = 0.5
$\displaystyle \sqrt{\frac{x}{x-3}} = 0.25 \Rightarrow 3x=-3 \Rightarrow x = -1$
$\displaystyle \text{Hence } x = -1, 4$
$\displaystyle \\$
$\displaystyle \text{Question 15: } ( \frac{2x-3}{x-1} )-4( \frac{x-1}{2x-3} )=3$
$\displaystyle ( \frac{2x-3}{x-1} )-4( \frac{x-1}{2x-3} )=3$
$\displaystyle \text{Let } ( \frac{2x-3}{x-1} ) = y$
$\displaystyle \text{Therefore } y - \frac{4}{y} =3$
$\displaystyle y^2-3y-4=0$
$\displaystyle y^2-4y+y-4=0$
$\displaystyle y(y-4)+(y-4)=0$
$\displaystyle (y-4)(y+1)=0 \Rightarrow y = 4, -1$
$\displaystyle \text{When } y = 4$
$\displaystyle \frac{2x-3}{x-1} =4$
$\displaystyle 2x-3=4x-4 \Rightarrow x = 0.5$
$\displaystyle \text{When } y = -1$
$\displaystyle \frac{2x-3}{x-1} =-1$
$\displaystyle 2x-3=-x+1 \Rightarrow x = \frac{4}{3}$
$\displaystyle \text{Hence } x = 0.5, \frac{4}{3}$
$\displaystyle \\$
$\displaystyle \text{Question 16: } ( \frac{3x+1}{x+1} )+( \frac{x+1}{3x+1} )= \frac{5}{2}$
$\displaystyle ( \frac{3x+1}{x+1} )+( \frac{x+1}{3x+1} )= \frac{5}{2}$
$\displaystyle \text{Let } ( \frac{3x+1}{x+1} ) = y$
$\displaystyle y + \frac{1}{y} = \frac{5}{2}$
$\displaystyle 2(y^2+1)=5y$
$\displaystyle 2y^2-5y+2=0$
$\displaystyle 2y^2-4y-y+2=0$
$\displaystyle 2y(y-2)-1(y-2)=0$
$\displaystyle (y-2)(2y-1)=0$
$\displaystyle y = 2, 0.5$
$\displaystyle \text{When } y = 2$
$\displaystyle ( \frac{3x+1}{x+1} ) = 2$
$\displaystyle 3x+1=2x+2 \Rightarrow x =1$
$\displaystyle \text{When } y =0.5$
$\displaystyle ( \frac{3x+1}{x+1} ) = y$
$\displaystyle 6x+2=x+1 \Rightarrow x = - \frac{1}{5}$
$\displaystyle \text{Hence } x = 1, - \frac{1}{5}$
$\displaystyle \\$
$\displaystyle \text{Question 17: } 3\sqrt{\frac{x}{5}} +3\sqrt{\frac{5}{x}} =10$
$\displaystyle 3\sqrt{\frac{x}{5}} +3\sqrt{\frac{5}{x}} =10$
$\displaystyle \text{Let } \sqrt{\frac{x}{5}} = y$
$\displaystyle 3y+ \frac{3}{y} =10$
$\displaystyle 3y^2-10y+3=0$
$\displaystyle 3y^2-9y-y+3=0$
$\displaystyle 3y(y-3)-1(y-3)=0$
$\displaystyle (y-3)(3y-1)=0 \Rightarrow y = 3, \frac{1}{3}$
$\displaystyle \text{When } y = 3$
$\displaystyle \sqrt{\frac{x}{5}} = 3$
$\displaystyle \frac{x}{5} =9 \Rightarrow x = 45$
$\displaystyle \text{When } y = \frac{1}{3}$
$\displaystyle \sqrt{\frac{x}{5}} = \frac{1}{3}$
$\displaystyle \frac{x}{5} = \frac{1}{9} \Rightarrow x = \frac{5}{9}$
$\displaystyle \text{Hence } x = 45, \frac{5}{9}$
$\displaystyle \\$
$\displaystyle \text{Question 18: } 2x- \frac{1}{x} =7$ [2006]
$\displaystyle 2x- \frac{1}{x} =7$
$\displaystyle 2x^2-7x-1=0$
Comparing $\displaystyle 2x^2-7x-1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = -7 \text{ and } c =-1$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-1)}}{2(2)}$
$\displaystyle \text{Solving we get } x = 3.64, -0.14$
$\displaystyle \\$
Question 19: $\displaystyle 5x^2-3x-4=0$ [2012]
Comparing $\displaystyle 5x^2-3x-4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 5, b = -3 \text{ and } c =-4$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(5)(-4)}}{2(5)}$
$\displaystyle \text{Solving we get } x = 1.243, -0.643$
If we want only three significant figures than $\displaystyle x = 1.24, -0.643$
$\displaystyle \\$
Question 20: $\displaystyle (x-1)^2-3x+4=0$ [2014]
$\displaystyle (x-1)^2-3x+4=0$
$\displaystyle x^2+1-2x-3x+4=0$
$\displaystyle x^2-5x+5=0$
Comparing $\displaystyle 5x^2-3x-4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -5 \text{ and } c =5$
$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\displaystyle \text{Therefore } x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(5)}}{2(1)}$
$\displaystyle \text{Solving we get } x = 3.618, 1.382$
If we want only two significant figures than $\displaystyle x = 3.6, 1.3$ | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 482, "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": 0.9958727359771729, "perplexity": 960.7796182584922}, "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-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00560.warc.gz"} |
https://www.physicsforums.com/threads/grav-force-mass-of-an-object.156876/ | # Grav. Force - Mass of an object
• Start date
• #1
ecthelion4
24
0
## Homework Statement
A proton moving at 0.999 of the speed of light orbits a black hole 4362km from the center of the attractor. What is the mass of the black hole?
F=(Gm1*m2)/d^2
## The Attempt at a Solution
Assuming the above is correct, a proton I think has constant mass of 1.6726*[(10)^(-27)], I know the distance, and G is constant. Since what I need is mass of black hole I'm assuming there's a way to find the force using the proton's speed.
• #2
Homework Helper
1,509
2
The gravity of the black hole is acting as the central force. So what is the definition of a central force? Think Newton's second law for circular motion.
• #3
ecthelion4
24
0
super, i figured put the problem, the force would be equal to (m*v^2)/r
now the only thing im missing is v, which is .999 of the speed of light, but what does that mean?
• #4
Homework Helper
1,509
2
Do you know what the speed of light is in a vacuum? It's a constant. The 0.999 is how much of the speed of light the proton is moving at.
• #5
ecthelion4
24
0
ok, so if the speed of light is 299,792,458 m/s, then the proton's speed would be that times .999 right?
• #6
Homework Helper
1,509
2
ok, so if the speed of light is 299,792,458 m/s, then the proton's speed would be that times .999 right?
That's right.
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6K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9771865606307983, "perplexity": 1205.2807124610426}, "config": {"markdown_headings": true, "markdown_code": false, "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-2022-33/segments/1659882573242.55/warc/CC-MAIN-20220818154820-20220818184820-00639.warc.gz"} |
https://projecteuclid.org/euclid.jsl/1294170999 | ## Journal of Symbolic Logic
### Analytic equivalence relations and bi-embeddability
#### Abstract
Louveau and Rosendal [5] have shown that the relation of bi-embeddability for countable graphs as well as for many other natural classes of countable structures is complete under Borel reducibility for analytic equivalence relations. This is in strong contrast to the case of the isomorphism relation, which as an equivalence relation on graphs (or on any class of countable structures consisting of the models of a sentence of ℒω1ω) is far from complete (see [5, 2]).
In this article we strengthen the results of [5] by showing that not only does bi-embeddability give rise to analytic equivalence relations which are complete under Borel reducibility, but in fact any analytic equivalence relation is Borel equivalent to such a relation. This result and the techniques introduced answer questions raised in [5] about the comparison between isomorphism and bi-embeddability. Finally, as in [5] our results apply not only to classes of countable structures defined by sentences of ℒω1ω, but also to discrete metric or ultrametric Polish spaces, compact metrizable topological spaces and separable Banach spaces, with various notions of embeddability appropriate for these classes, as well as to actions of Polish monoids.
#### Article information
Source
J. Symbolic Logic, Volume 76, Issue 1 (2011), 243-266.
Dates
First available in Project Euclid: 4 January 2011
https://projecteuclid.org/euclid.jsl/1294170999
Digital Object Identifier
doi:10.2178/jsl/1294170999
Mathematical Reviews number (MathSciNet)
MR2791347
Zentralblatt MATH identifier
1256.03050
Subjects | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8139217495918274, "perplexity": 589.9488555154953}, "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-2019-35/segments/1566027315551.61/warc/CC-MAIN-20190820154633-20190820180633-00469.warc.gz"} |
https://www.followthesheep.com/?tag=deluxetable | ## Using emulateapj: multi-page tables and deluxetable
`emulateapj` should work fine with `deluxetable`. For tables spanning one column use the normal `deluxetable`, while for tables spanning both columns use `deluxetable*`
To use multi-page tables, follow the instructions given in http://hea-www.harvard.edu/~alexey/emulateapj/emulateapj.cls:
``` %% 3) Multi-page tables cannot be set properly inside the main text; you %% need to move the table to the end of the paper (after the references) and %% issue the command LongTables before it. ```
```LongTable begin{deluxetable*} .... end{deluxetable*} ```
If you need to have landscape tables, you will also need to use the package `lscape` along with forcing `emulateapj` to use `revtex4` in the preamble (using `rotate` inside of `deluxetable` will not work):
``` documentclass[iop,revtex4]{emulateapj} ```
You also must put the table at the end of file with the other tables. For example usage, see the following note from ”emulateapj.cls”:
``` %% 2) The {deluxetable} environment is re-implemented (the problem with the %% the aastex's deluxetable is it does not float). There is also a new %% environment {deluxetable*} (absent in aastex) to set a floating table %% two-column wide. Known problems: %% (a) rotate doesn't work (too difficult to implement). However, %% you can use revtex's turnpage environment %% - load package lscape (usepackage{lscape} in the header) %% - move table at the end of the paper after references %% - clearpage before the table %% - LongTables if the table will span more than 1 page (see next item) %% - put the table inside the landscape environment and clearpage %% at the end: %% clearpage %% LongTables % optionally %% begin{landscape} %% begin{deluxetable} %% .... %% end{deluxetable} %% clearpage %% end{landscape} %%``` | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.889121413230896, "perplexity": 2183.5865740581653}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2019-22/segments/1558232255536.6/warc/CC-MAIN-20190520021654-20190520043654-00367.warc.gz"} |
http://libros.duhnnae.com/2017/jul7/15011230265-The-rate-of-convergence-for-the-renewal-theorem-in-mathbbRd.php | # The rate of convergence for the renewal theorem in $mathbb{R}^d$
The rate of convergence for the renewal theorem in $mathbb{R}^d$ - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
1 MODAL-X - Modélisation aléatoire de Paris X 2 IMB - Institut de Mathématiques de Bordeaux
Abstract : Let $ho$ be a borelian probability measure on $\mathrm{SL} d\mathbb{R}$. Consider the random walk $X n$ on $\mathbb{R}^d\setminus\{0\}$ defined by $ho$ : for any $x\in \mathbb{R}^d\setminus\{0\}$, we set $X 0 =x$ and $X {n+1} = g {n+1} X n$ where $g n$ is an iid sequence of $\mathrm{SL} d\mathbb{R}-$valued random variables of law $ho$. Guivarc-h and Raugi proved that under an assumption on the subgroup generated by the support of $ho$ strong irreducibility and proximality, this walk is transient.In particular, this proves that if $f$ is a compactly supported continuous function on $\mathbb{R}^d$, then the function $Gfx :=\mathbb{E} x \sum {n=0}^{+\infty} fX n$ is well defined for any $x\in \mathbb{R}^d \setminus\{0\}$.Guivarc-h and Le Page proved the renewal theorem in this situation : they study the possible limits of $Gf$ at $0$ and in this article, we study the rate of convergence in their renewal theorem.To do so, we consider the family of operators $Pit {t\in \mathbb{R}}$ defined for any continuous function $f$ on the sphere $\mathbb{S}^{d-1}$ and any $x\in \mathbb{S}^{d-1}$ by\Pit fx = \int {\mathrm{SL} d\mathbb{R}} e^{-it \ln \frac{ \|gx\|}{\|x\|}} f\left\frac{gx}{\|gx\|} ight \mathrm{d} hog\And we prove that, for some $L\in \mathbb{R}$ and any $t 0 \in \mathbb{R} +^\ast$,\\sup {\substack{t\in \mathbb{R}\\ |t| \geqslant t 0}} \frac{ 1 }{|t|^L} \left\| I d-Pit^{-1} ight\| \text{ is finite}\where the norm is taken in some space of hölder-continuous functions on the sphere.
Autor: Jean-Baptiste Boyer -
Fuente: https://hal.archives-ouvertes.fr/
DESCARGAR PDF | {"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": 0.9935623407363892, "perplexity": 773.1872489458357}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2019-04/segments/1547583831770.96/warc/CC-MAIN-20190122074945-20190122100945-00331.warc.gz"} |
http://paperity.org/p/34416526/assessment-of-different-metrics-for-physical-climate-feedbacks | # Assessment of different metrics for physical climate feedbacks
Climate Dynamics, Aug 2013
We quantify the feedbacks from the physical climate system on the radiative forcing for idealized climate simulations using four different methods. The results differ between the methods and differences are largest for the cloud feedback. The spatial and temporal variability of each feedback is used to estimate the averaging scale necessary to satisfy the feedback concept of one constant global mean value. We find that the year-to-year variability, combined with the methodological differences, in estimates of the feedback strength from a single model is comparable to the model-to-model spread in feedback strength of the CMIP3 ensemble. The strongest spatial and temporal variability is in the short-wave component of the cloud feedback. In our simulations, where many sources of natural variability are neglected, long-term averages are necessary to get reliable feedback estimates. Considering the large natural variability and relatively small forcing present in the real world, as compared to the forcing imposed by doubling CO2 concentrations in the simulations, implies that using observations to constrain feedbacks is a challenging task and requires reliable long-term measurements.
This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1007%2Fs00382-013-1757-1.pdf
Daniel Klocke, Johannes Quaas, Bjorn Stevens. Assessment of different metrics for physical climate feedbacks, Climate Dynamics, 2013, 1173-1185, DOI: 10.1007/s00382-013-1757-1 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9163867831230164, "perplexity": 1708.9907750635207}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2017-47/segments/1510934804965.9/warc/CC-MAIN-20171118132741-20171118152741-00358.warc.gz"} |
https://www.arxiv-vanity.com/papers/1409.6167/ | arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
# Quantum multiparameter metrology with generalized entangled coherent state
Jing Liu Department of Physics, Zhejiang University, Hangzhou 310027, China Department of Mechanical and Automation Engineering, The Chinese University of Hong Kong, Shatin, Hong Kong Xiao-Ming Lu Department of Electrical and Computer Engineering, National University of Singapore, 4 Engineering Drive 3, Singapore 117583 Zhe Sun Department of Physics, Hangzhou Normal University, Hangzhou 310036, China Xiaoguang Wang State Key Laboratory of Modern Optical Instrumentation, Department of Physics, Zhejiang University, Hangzhou 310027, China Synergetic Innovation Center of Quantum Information and Quantum Physics, University of Science and Technology of China, Hefei, Anhui 230026, China
###### Abstract
We propose a generalized form of entangled coherent states (ECS) and apply them in a multi-arm optical interferometer to estimate multiple phase shifts. We obtain the quantum Cramér-Rao bounds for both the linear and nonlinear parameterization protocols. Through the analysis, we find that, utilizing the simultaneous estimation, this generalized form of ECS gives a better precision than the generalized NOON states [Phys. Rev. Lett. 111, 070403 (2013)]. Moreover, comparing with the independent estimation, both the linear and nonlinear protocols have the same advantage in the relation to the number of the parameters.
###### pacs:
03.67.-a, 06.90.+v, 42.50.Dv, 42.50.St
## 1 Introduction
Since the pioneer work of Caves in 1981 [1], quantum metrology has made a great progress as a successful application of quantum mechanics to enhance the measurement precision [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]. However, unlike the single-parameter quantum metrology, multiparameter quantum metrology for a long time was not adequately studied. One major reason is that quantum multiparameter Cramér-Rao bound in general cannot be saturated. A decade ago, the condition of this bound to be tight for pure states has been given [16, 17]. Since then, several protocols on multiparameter estimation were proposed in different scenarios [18, 19, 20, 21, 22, 23]. One of these works was given by Humphreys et al. in Ref. [18], where the phase imaging problem was mapped into a multiparameter metrology process and a generalized form of NOON states was used as the input resource. They found that the simultaneous estimation with the generalized NOON states is better than the independent estimation with the NOON state.
On the other hand, the NOON state is not the only state that is available to reach the Heisenberg limit. Another useful state is the so-called entangled coherent state (ECS), which has been widely applied and studied in quantum metrology recently [24, 25, 26, 27, 28, 29]. In a Mach-Zehnder interferometer, ECS has been proved to be more powerful than NOON state in giving a Heisenberg scaling precision [27]. Even in a lossy interferometer, ECS can still beat the shot-noise limit for a not very large loss rate [25, 26]. Thus, it is reasonable for one to wonder that if a generalized form of ECS could give a better theoretical precision than the counterpart of NOON state. This is the major motivation of this work.
In this paper, we apply a generalized form of ECS in linear and nonlinear optical interferometers. By calculating the quantum Fisher information matrix (QFIM), we give the analytical expression of the quantum Cramér-Rao bound (QCRB) for both linear and nonlinear protocols. In the linear protocol, the bound can reach the Heisenberg scaling for most values of the total photon number. Meanwhile, with respect to the parameter number , in both protocols, for most values of photon number, the bounds provide better precisions than that given by independent protocol with ECS or NOON state, which is the same as the generalized NOON state discussed in Ref. [18].
The paper is organized as follows. In Sec. 2, we briefly review the quantum Cramér-Rao bound as well as the quantum Fisher information matrix for multiparameter estimations. In Sec. 3, we introduce a generalized form of entangled coherent state for multiple modes and apply it in linear and nonlinear optical interferometers. Furthermore, the comparison between this state and the generalized NOON state is discussed. In Sec. 4, we discuss the optimal measurement problem to achieve the bound. In Sec. 5, we extend our discussion to random variables and compare the generalized ECS and NOON state with quantum Ziv-Zakai bound. Section 6 is the conclusion.
## 2 Quantum Cramér-Rao bound
In a multiparameter quantum metrology process, the quantum state depends on a set of deterministic parameters . The value of is estimated by processing the observation data obtained by measuring the quantum system. A generalized quantum measurement is characterized by a positive-operator-valued measure with denoting outcomes. According to quantum mechanics, the probability of obtaining the outcome is . Denote the estimator for by , which is a map from the measurement outcome to the estimates. The accuracy of the multiparameter estimation can be measured by the estimation-error covariance matrix: . For (locally) unbiased estimators , the QCRB on the estimation error reads [30, 31]
C≥(νF)−1, (1)
where is the number of the repetition of the experiments, and is the quantum Fisher information matrix (QFIM). Let be the symmetric logarithmic derivative (SLD) for , which is a Hermitian operator satisfying . Then, the QFIM is defined by
Fjk=12Tr[(LjLk+LkLj)ρ]. (2)
Recently, it has been found that similarly with the quantum Fisher information [32], the QFIM can also be expressed in the support of the density matrix [33]. Denote for simplicity henceforth. For a pure state , the elements of QFIM can be expressed as [30, 31]
Fjk=4Re(⟨∂jψ|∂kψ⟩−⟨∂jψ|ψ⟩⟨ψ|∂kψ⟩). (3)
In this work, we use the the total variance as a figure of merit for the multiparameter estimation. Taking the trace on both sides of inequality (1) leads to
|δ^θ|2≥Tr(F−1), (4)
where we have set as we are only interested in the quantum enhancements. For a two-parameter case, Eq. (4) is reduced into , where with denoting the determinant can be treated as an effective quantum Fisher information.
In order to draw conclusion on the best possible estimation error from the QCRB, it is important to know whether the lower bound is achievable. The QCRB for multiparameter estimation is in general not achievable. However, for pure states, the QCRB can be saturated if are satisfied for all , , and [16, 17, 18]. Note that is an SLD operator for . It can be shown that is equivalent to . For a unitary parametrization process, i.e., , this condition can be rewritten as [34]
⟨ψin|[Hj,Hk]|ψin⟩=0,∀j,k, (5)
where is the characteristic operator for the parameterization of the th parameter.
The single-parameter unitary parametrization processes have been detailedly discussed in recent works [34, 35, 36]. For a multiparameter unitary parametrization process, there are two basic methods to sense the multiple parameters: the simultaneous protocols and the sequential protocols, as shown in Fig. 1. Let be a set of Hermitian operators. Then, the simultaneous sensing is described by
UI=exp(d∑j=1iHjθj) (6)
with being the number of the parameters, while the sequential sensing is described by
UII=d∏j=1exp(iHd−j+1θd−j+1). (7)
In this paper, we focus on the phase estimations in the optical multi-arm interferometer, in which is a local operator on the th mode. Thus, all the generating operators are commutative, and these two methods are equivalent. Moreover, in such a case, it is easy to show that , and thus . This implies that the Cramér-Rao bound is theoretically achievable. The element of QFIM can be written as
Fjk=4(⟨ψin|HjHk|ψin⟩−⟨ψin|Hj|ψin⟩⟨ψin|Hk|ψin⟩). (8)
## 3 Generalized entangled coherent state for multiparameter estimation
### 3.1 Generalized entangled coherent state
Entangled coherent state (ECS) has been applied in several single-parameter quantum metrology protocols and proved to be powerful in beating the shot-noise limit [25, 26, 27]. Let be a coherent state and the vacuum state. The ECS is given by [28, 29]
|ECS⟩=N(|α0⟩+|0α⟩), (9)
where is the normalizing factor. Taking the ECS as the input, the QCRB for a parameter sensed by the Hamiltonian with denoting the annihilation operator of the first mode is given by [27]
|δ^θ|2ECS=14N2|α|2[1+|α|2(1−N2)]. (10)
Therefore, for the independent estimations for parameters, the QCRB on the total variance is . In terms of the mean number of total photons involved, i.e., , we obtain
|δ^θ|2ind=d3Ntot[2d+Ntot(N−2−1)]. (11)
When is large, , the total variance
(12)
This bound is lower than , which is given by NOON state in the independent estimation [18].
Multiple parameters can also be sensed and estimated by entangled input states, e.g., using a multi-arm interferometer with generalized NOON states as input [18]. Here, we use a generalized form of the ECS instead, as the ECS is more powerful than the NOON states in the two-arm interferometer for a fixed mean number of the total photons. For a multiparameter estimation scenario as shown in Fig. 1, we set the reference beam as mode zero and the parametrized beams as mode to mode . Taking into consideration the symmetry among modes, we generalize the ECS to the multi-mode case as
|ψ⟩=bd∑j=1|α⟩j+c|α⟩0, (13)
where with being the multi-mode vacuum is a state with a coherent state in th mode and vacuums in others. The coefficients and are complex numbers; due to the normalization of , they satisfy
|c|2+(bc∗+cb∗)v+|b|2u=1 (14)
with
u:=d+d(d−1)e−|α|2 and v:=de−|α|2. (15)
In this paper, we will use this generalized form of ECS as the input state to sense the parameters.
### 3.2 Local parameterization
Let us consider that the parameters are sensed via with , where is a positive integer. Taking the generalized ECS Eq. (13) as the input state, it follows that
⟨ψ|Hj|ψ⟩=|b|2f(m,α)and⟨ψ|HjHk|ψ⟩=|b|2f(2m,α)δjk (16)
with . From Eq. (8), the elements of the QFIM are given by
Fjk=4[δjk|b|2f(2m,α)−|b|4f(m,α)2]. (17)
Consequently, the QFIM can be expressed as
F=4|b|2f(2m,α)(\small 1\kern-4.3pt% \normalsize 1−|b|2f(m,α)2f(2m,α)I), (18)
where 11 is the identity matrix, and is the matrix with elements for all and . Noting that , it can be shown that
(19)
with and being real numbers. Thus, we obtain the analytical result for the inverse of the QFIM as
F−1=14|b|2f(2m,α)(\small 1% \kern-4.3pt\normalsize 1+|b|2f(m,α)2f(2m,α)−|b|2f(m,α)2dI). (20)
Tracing both sides of the above equality, we obtain the lower bound on the total variance:
|δ^θ|2≥Tr(F−1)=d4f(2m,α)(1|b|2+1g−|b|2d) (21)
with being a nonnegative number.
Since we are interested in the minimal estimation error, we minimize the QCRB on the total variance over , which is equivalent to minimize the quantity in Eq. (21) over . By noting that takes value in a continuous range and is nonnegative, it follows from Eq. (21) that we only need to investigate those , as will be negative when passes through . In this domain, the minimum of is at the place where the derivative of with respect to is zero, that is
|b|=b⋆:=√g/(√d+d). (22)
Note that the rigor domain of is determined by the normalization condition Eq. (14). Due to an irrelevant global phase in the states in Eq. (13), we can always assume that is real. Equation (14) then becomes
c2+2(Reb)vc+|b|2u−1=0, (23)
which has a solution for only if . Since , we obtain
|b|2≤1u−v2=1d+d(d−1)e−|α|2−d2e−2|α|2≡Γ, (24)
which implies that the domain of is . Therefore, we obtain
min|b|∈[0,√Γ]Tr(F−1)=⎧⎪⎨⎪⎩d(1+√d)24f(m,α)2f(2m,α)2for b⋆≤√Γd4f(2m,α)(1Γ+1g−Γd)for % b⋆>√Γ. (25)
We first consider the linear parametrization protocol, for which . Note that , , and therefore . From Eq. (25), we obtain the lower bound on the total variance:
|δ^θ|2≥Tr(F−1L)=d4|α|2(1+|α|2)(1|b|2+11+|α|−2−d|b|2). (26)
After minimizing over , we obtain
|δ^θ|2≥|δ^θ|2L=d(√d+1)24(1+|α|2)2, (27)
provided that . Figure 2 shows the parameter regime where . The purple areas in both panels represent the regime where is inside the domain of . From Fig. 2(a), it can be found that can be for a large . This is reasonable because when is infinite, and , is always less than . For a large , may be beyond the domain of when is very small. A more experimentally realizable regime is that both of and are not very large and comparable to each other, as shown in Fig. 2(b). In this regime, is inside the domain of for most areas. Especially, when is larger than around , is always reachable, indicating that the corresponding bound can always be reached.
We also consider a nonlinear parametrization protocol for which . It can be shown that and . After minimizing over , we obtain
|δ^θ|2≥|δ^θ|2NL=d(√d+1)24(1+|α|2|α|6+6|α|4+7|α|2+1)2, (28)
provided that . In Fig. 3 we can see that, similarly to Fig. 2, is satisfied in most areas.
### 3.3 Analysis
Here, we give an analysis on the QCRB given by the generalized ECS, and compare them with the ones given by the generalized NOON states, which was applied in multiparameter metrology in Ref. [18]. The generalized NOON states proposed in Ref. [18] reads , where is the state with photons in th mode and vacuum in others. For the linear parametrization protocol , the minimal QCRB on the total variance over all generalized NOON states for a given is [18]
|δ^θ|2sL=d(√d+1)24N2, (29)
where the optimal value of is . For the nonlinear parametrization protocol , through some straightforward calculations, we obtain the minimal QCRB
|δ^θ|2sNL=d(√d+1)24N4, (30)
which is also attained at .
From the expressions of , , and , we find that all these bounds share the same scaling relation with respect to the parameter number ; they are all proportional to . Furthermore, both and provide a advantage compared to the independent estimation with ECS or NOON state, which is the same as [18].
These protocols show different relations to the average total photon number . Obviously, the average total photon number of is . Meanwhile, the average total photon number of is , which is dependent on the values of and . When is sufficiently large such that , we have as a result of implied by the normalization condition Eq. (14). As a matter of fact, when , . For a not very large , choosing for example, at the optimal value of , the difference between and is around . Thus, for most values of and , the average photon number can be approximated as . With this approximation, and .
In Fig. 4, we plot these four QCRBs, , , , and as functions of the average total photon number. Comparing and in linear protocols, in the regime of small average total photon numbers, even may be greater than , the generalized ECS still gives a lower QCRB than the generalized NOON states. However, this advantage reduces when increases. For a very large , the generalize ECS and the generalized NOON states are basically equivalent to each other on the estimation precision. Besides, the nonlinear parametrization process is always better than the linear one for the same input state in this case, as expected. What is more interesting here is that, for a small , the linear protocol with generalized ECS can give a lower bound than the nonlinear counterpart with generalized NOON states. This gives an alternative strategy for small photon number scenario when the nonlinear parametrization is very challenging to perform.
## 4 Measurement
For an entire metrology process, the measurement has to be considered as the QCRB cannot be always saturated for any measurement. As a matter of fact, different measurement strategies would give different classical Cramé-Rao bound and further give different metrology scalings.
For the estimation of a single parameter, the projective measurement with respect to the eigenstates of the SLD operator can be used as the optimal measurement if they are (locally) independent of the parameter under estimation [37, 38, 39]. For the cases where the eigenstates of the SLD operator depend on the true value of the parameter, one has to resort to the adaptive measurement and estimation scheme to asymptotically attain the QCRB [40, 41]. For multiparameter estimations, due to the non-commutativity of the SLDs, the QCRB in general cannot be attained. However, for multiparameter estimation with pure states, the QCRB can be attained if for all , , and [16, 17], which is satisfied for the case consider in this work. In principle, there exists an optimal measurements asymptotically attain the QCRB, although this optimal measurement may be hard to implemented experimentally. General methods to construct such an optimal measurement can be found in Ref. [16, 18].
## 5 Deterministic parameter versus random parameter
During the entire calculation of the paper, we treat the phase shifts as unknown but deterministic signals [42, 43, 44, 45], which means that the true values of phase shifts are always the same during the repetitions of the experiment. In other words, we have independent and identically distributed samples to perform measurements, with the collection of the measurement outcomes the parameters are estimated. For example, during the detection of the gravity, the gravity is commonly treated as a deterministic parameter. Meanwhile, optical interferometry is a major approach for this detection, for example the LIGO (Laser Interferometer Gravitational-Wave Observatory) program. Thus, it is reasonable to treat the phase shifts as deterministic signals.
However, in some different scenarios, for instance, during the measure of the gravitational acceleration in a specific location on earth, its value may be slightly affected by the flow of some underground magma or geology movement. Thus, it is also reasonable to treat the signal as a a random parameter in these scenarios. Recently, Tsang proposed a quantum version of Ziv-Zakai bounds for estimating a random parameter [46]. Using this bound, Giovannetti and Maccone found that for high prior information regime, the accuracy given by sub-Heisenberg strategies is no better than that obtained by guessing according to the prior distribution [47]. Thus, the precision for a random variable and a deterministic parameter may have great differences. For the generalized NOON state, the quantum Ziv-Zakai bound has been given by Zhang and Fan in Ref. [48] as
|δ^θ|2≥max{d(d+√d)280λ2N2,(π2/16−0.5)d(d+√d)2(d+√d−1)N2} (31)
with , where the prior distribution of the random parameters are assumed to be uniform with large width windows. For a large , the previous expression is always larger than the latter on in the braces. Thus, the advantage vanishes if the parameter underestimation is a random variable. However, this bound is still better than that given by independent estimations [48]. Here, we obtain the quantum Ziv-Zakai bound for generalized ECS. Through some straightforward calculations, the bound for linear parametrization process is
|δ^θ|2≥max{d(d+√d)280λ2(|α|2+1)2,(π2/16−0.5)d(d+√d)2(d+√d−1)(|α|2+1)2}. (32)
Similarly with the generalized NOON state, the advantage vanishes in this bound. However, for a not very small value of , , this bound is still lower than Eq. (31), which means even for random variables, the generalized ECS can provide a better precision than generalized NOON state.
## 6 Conclusion
In this paper, we have proposed a generalized form of entangled coherent states and apply them as the input state of a multi-arm interferometer for estimating multiple phase shifts. We have obtained the QCRB on the estimation error for both linear and nonlinear protocols. Similarly with the generalized NOON state, the simultaneous estimation with generalized entangled coherent state can provide a better precision than the independent estimation. Meanwhile, We find that the bound from the generalized entangled coherent state is better than that given by the generalized NOON state.
The authors thank Heng-Na Xiong and Animesh Datta for valuable discussion. This work was supported by the NFRPC through Grant No. 2012CB921602 and the NSFC through Grants No. 11475146. X.-M. L. also acknowledges the support from the Singapore National Research Foundation under NRF Grant No. NRF-NRFF2011-07 and NSFC under Grant No. 11304196. Z. S. also acknowledges the support from NSFC with grants No. 11375003, Natural Science Foundation of Zhejiang Province with grant No. LZ13A040002 and Program for HNUEYT with grants No. 2011-01-011. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9737004637718201, "perplexity": 685.2690309423483}, "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-2020-29/segments/1593655897168.4/warc/CC-MAIN-20200714145953-20200714175953-00264.warc.gz"} |
http://mathhelpforum.com/algebra/149021-elevation-print.html | # Elevation
• Jun 20th 2010, 10:10 PM
winsome
Elevation
He went 4 miles to the west and again went 2 miles to north west. how many miles did he go from the initial point.
• Jun 20th 2010, 10:16 PM
Unknown008
If you made a simple sketch, I'm sure you'll see it better. The two different paths he took make two lines, and the total distance he is from his initial point makes a third line. In all, they make a triangle.
Since you know the directions he took, you'll see that the angle between the first two sides is 90 + 45 = 135 degrees.
Then, use the cosine rule to find the length of the last side of the triangle, which gives $\sqrt{4^2 + 2^2 - 2(4)(2)cos(135)} = 5.6\ miles$
Now, to find the direction from which he is from the initial point is obtained using the sine rule.
• Jun 20th 2010, 10:22 PM
winsome
can you give me its picture ?
• Jun 20th 2010, 10:35 PM
Unknown008
1 Attachment(s)
Here it is:
Attachment 17948
• Jun 20th 2010, 10:41 PM
winsome
Thanks a lot | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "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": 0.9361193776130676, "perplexity": 1071.8017077412178}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2017-17/segments/1492917121453.27/warc/CC-MAIN-20170423031201-00367-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://physics.aps.org/articles/v1/7 | # Viewpoint: New clues in the mystery of persistent currents
, Laboratoire de Physique des Solides Université Paris-Sud, Bldg. 510, Orsay Cedex, France 91405
Published July 28, 2008 | Physics 1, 7 (2008) | DOI: 10.1103/Physics.1.7
#### Effect of Pair Breaking on Mesoscopic Persistent Currents Well above the Superconducting Transition Temperature
H. Bary-Soroker, O. Entin-Wohlman, and Y. Imry
Published July 28, 2008 | PDF (free)
If we apply a voltage pulse to a wire, we know that the current that starts to flow will eventually disappear. This is because electrons scatter inelastically from phonons and other electrons as they flow through most metals, eventually losing all of their energy. However, in metal rings that are smaller than the electron’s phase coherence length—the typical distance the electron travels before it scatters inelastically—it is possible to induce currents that last forever simply by threading the center of the ring with a magnetic flux (Fig. 1) [1]. Until now, the size of these persistent currents, as well as how their direction depends on the orientation of the central flux, has been a mystery. Now, in a paper appearing this week in Physical Review Letters, Hamutal Bary-Soroker, Ora Entin-Wohlman, and Yoseph Imry at the Weizmann Institute in Israel propose a solution to this problem that is so simple, yet compelling, that physicists may wonder why no one has thought of it before [2].
The existence of persistent (i.e., with no energy dissipation) diamagnetic currents on macroscopic scales has been considered a hallmark of superconductivity in metals ever since its discovery in 1911. Persistent currents in superconductors are related to the existence of a state with zero resistance and the fact that a superconductor is a perfect diamagnet (it opposes the applied field). More than seventy years later, in 1983, Büttiker, Landauer, and Imry pointed out that persistent currents should also exist in nonsuperconducting (“normal”) metallic loops threaded by magnetic flux [3]. To observe these dissipationless currents, the temperature must be sufficiently low to reduce the probability of inelastic scattering (from phonons and other electrons) and the circumference of the ring short enough that the phase coherence of the electronic wave functions is preserved around the loop. This typically limits the sample size to a micrometer or so, and the temperature to below 1 K. These persistent currents can be seen as the analogue, albeit on a much greater scale, to the orbital currents in aromatic ring molecules, such as benzene, that give rise to a molecular magnetic moment.
The existence of persistent currents in normal metals is, in fact, a signature of phase coherence in mesoscopic systems and an example of the famous Aharonov-Bohm effect. For a magnetic flux $\mathrm{\Phi }$ threading the loop, the electronic wave function (over the closed loop) acquires a phase proportional to $e/\mathrm{ħ}\phantom{\rule{0}{0ex}}\mathrm{\Phi }$. Since the electronic wave function must be continuous around the loop, the amplitude of the current will adjust so that the acquired phase is an integer multiple of $2\pi$. As a result, the amplitude of the persistent current is a periodic function of the magnetic flux quantum, ${\mathrm{\Phi }}_{0}=h/e\sim 4.14×{10}^{-15}\phantom{\rule{0.333em}{0ex}}\text{T}\phantom{\rule{0.333em}{0ex}}{\text{m}}^{2}$ (in SI units; note that for superconductors, the flux quantum is ${\mathrm{\Phi }}_{0}=h/2e$ because the electrons are paired). The maximum amplitude of the current is of the order $e/{\tau }_{D}$ where ${\tau }_{D}$ is the typical time for an electron to travel around the ring. This yields a current of the order of $1×{10}^{-9}\text{A}$. The magnitude of the current is independent of the cross-sectional area of the ring and is thus much smaller than the persistent current in a superconducting ring of similar dimensions, since the latter results from a coherent motion of all charge carriers condensed in a macroscopic coherent state. Disorder, which gives rise to elastic scattering, reduces, but does not completely destroy, the persistent current in a normal-state metal ring. This means that persistent currents exist in rings that have a finite resistance: if you connect one of these rings to a current generator, the generator will have to do work to push a current through, and, differently from a superconductor, one cannot measure a drop in resistance. Rather, it is only by measuring the very small induced magnetic moment associated with the circulating current that the persistent current can be revealed.
Another fundamental trait of persistent currents in normal rings, which distinguishes them from superconducting persistent currents, is that the direction of the induced current depends on the number of electrons in the ring as well as the details of the disorder potential [3, 4]. In contrast, persistent currents in superconducting rings (at low enough fields) are always diamagnetic.
For all these reasons detection of persistent currents has long been considered an experimental challenge. The moment from an individual ring corresponds to about $100\phantom{\rule{0.333em}{0ex}}{\mu }_{B}$ (Bohr magnetons), which is at the sensitivity limit of the best magnetic detectors (recall that the magnetic moment from a single spin is given by $\mathbf{\text{m}}={\mu }_{B}\phantom{\rule{0}{0ex}}g\mathbf{\text{S}}$). Unfortunately, an ensemble measurement with many rings could not overcome this low signal problem, because each ring was expected to contribute a moment with a random sign. This belief prevailed until it was shown theoretically that if one considers the effects of electron-electron interactions, the ensemble average of persistent currents taken on many rings would not be zero (even though for any individual ring, the sign of the current is random). This ensemble average was also shown to have a ${\mathrm{\Phi }}_{0}/2$ periodicity, half the periodicity of the persistent currents in a single ring, and an average current per loop of $0.05\phantom{\rule{0.333em}{0ex}}e/{\tau }_{D}$ [5]. The average sign of the currents was found to depend on the nature of the interactions: for repulsive interactions, the prediction was that the induced current would generate a moment along the direction of the applied field, while for attractive interactions, a diamagnetic response was expected.
Inspired by these predictions, the first experiment was performed in the early 1990s by measuring the magnetization of an array of about ten million disconnected, micron-sized copper rings [6]. The magnetic signal was, as expected, periodic in ${\mathrm{\Phi }}_{0}/2$, but the amplitude corresponded to an average ring current of $0.3\phantom{\rule{0.333em}{0ex}}e/{\tau }_{D}$, more than five times the theoretical predictions. More surprising, the magnetization was found to be diamagnetic at low magnetic flux. This suggested that attractive electron-electron interactions were important. Attractive interactions between electrons are normally mediated by phonons—the same effect that leads to superconductivity in many simple metals. Yet the mesoscopic rings were made of metals in which the superconducting transition temperature was so low it did not seem possible that phonon interactions were strong enough to lead to such large persistent currents. Copper, for example, has not been found to superconduct, even at temperatures as low as $10\phantom{\rule{0.333em}{0ex}}\mu \text{K}$! How could attractive interactions in copper be so weak that they would not lead to superconductivity at measurable temperatures, but still permit a large persistent current in the normal state?
Bary-Soroker et al.’s solution to this longstanding puzzle of persistent currents is amazingly simple. They consider the role played by a tiny amount of magnetic impurities, of the order of one part per million (1 ppm), which unavoidably exists in any metal. Magnetic impurities give rise to spin-flip scattering, which is one of the most efficient mechanisms that destroy superconductivity. If $1/{\tau }_{S}$ is the rate at which this spin scattering occurs, then as soon as $\mathrm{ħ}/{\tau }_{S}$ rises to the order of ${k}_{B}\phantom{\rule{0}{0ex}}{T}_{c}$ (where ${T}_{c}$ is the superconducting transition temperature), superconductivity is destroyed. This happens when the concentration of magnetic impurities is approximately 1 ppm for a ${T}_{c}\sim 1\text{mK}$. Bary-Soroker et al. suggest that without these pernicious magnetic impurities, superconductivity would occur in pure copper at a transition temperature ${T}_{c0}$ of the order of $1\phantom{\rule{0.333em}{0ex}}\text{mK}$ – about three orders of magnitude higher than the currently estimated upper bound for ${T}_{c}$ in copper. Now, spin-flip scattering also limits the coherence length, ${L}_{s}$, of electrons in the normal state of a metal to ${L}_{s}=\left(D\phantom{\rule{0}{0ex}}{\tau }_{S}{\right)}^{1/2}$, where $D$ is the diffusion coefficient. But ${L}_{s}$ is still of the order of $10\phantom{\rule{0.333em}{0ex}}\mu \text{m}$ for 1 ppm of magnetic impurities. Thus 1 ppm of magnetic impurities is enough to destroy bulk superconductivity in copper, but it has much less of an effect on the phase coherence in micron size rings. This is the case, provided that the metal is at a low enough temperature for which inelastic phase breaking times such as electron-phonon or electron-electron scattering times are longer than ${\tau }_{S}$.
The group’s ideas show that the sign and amplitude of the ensemble average persistent current is determined by the amplitude of the attractive electron-electron interaction, that is, the electron-electron interaction that corresponds to a superconducting transition of the order of $1\phantom{\rule{0.333em}{0ex}}\text{mK}$ (instead of $10\phantom{\rule{0.333em}{0ex}}\mu \text{K}$ as initially believed). The attractive interaction leads to diamagnetic persistent currents, with an amplitude of the order of $e/{\tau }_{D}$, the value measured for persistent currents in copper.
These theoretical findings provide for the first time a complete explanation for the sign and amplitude of persistent currents in mesoscopic rings. But these results will do much more than close the door on a long-standing question. They will likely stimulate new experiments. It is possible, for example, to test these predictions by investigating rings made from known low-${T}_{c}$ materials in which a controlled number of magnetic impurities have been added. Another possibility could be to measure the amplitude of persistent currents in rings made of bilayer materials such as Al/Cu in which the transition temperature can continuously be decreased by decreasing the thickness of the aluminum layer. With such a system, it would be possible to reach a regime where the transition temperature of the bilayer becomes too low to be measurable, but is still known theoretically. Finally, since experiments on mesoscopic rings remain challenging, these results could pave the way to the theoretical investigation of other signatures of pair breaking in mesoscopic rings in the normal state, such as magnetic field dependent transport.
## Acknowledgments
We thank Sophie Guéron and Gilles Montambaux for valuable comments on the manuscript.
### References
1. Y. Imry, Introduction to Mesoscopic Physics[Amazon][WorldCat] 2nd ed. (Oxford University Press, Oxford, 2002)..
2. H. Bary-Soroker, O. Entin-Wohlman and Y. Imry, Phys. Rev. Lett. 101, 057001 (2008).
3. M. Büttiker, Y. Imry, and R. Landauer, Physics Letters A 96, 365 (1983).
4. H-F. Cheung, E. K. Riedel, and Y. Gefen, Phys. Rev. Lett. 62, 587 (1989).
5. A. Schmid, Phys. Rev. Lett. 66, 80 (1991); V. Ambegaokar and U. Eckern, 67, 3192 (1991).
6. L. P. Lévy, G. Dolan, J. Dunsmuir, and H. Bouchiat, Phys. Rev. Lett. 64, 2074 (1990).
### About the Author: Hélène Bouchiat
Hélène studied at Ecole normale supérieure (1977–1981) and completed her Ph.D. (1981–1986) at the Laboratoire de Physique des Solides, Université Paris Sud, in Orsay. From 1987 to 1988 she was a consultant for Bell Laboratories. In 1988 she returned to the Laboratoire de Physique des Solides, where she is now Directeur de Recherche. Her present research interests include the study of quantum transport in mesoscopic systems and carbon nanotubes and low temperature molecular electronics. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 33, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8690210580825806, "perplexity": 643.5878321694308}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2014-41/segments/1410657132883.65/warc/CC-MAIN-20140914011212-00206-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.akustik.rwth-aachen.de/go/id/gkoo/file/abschl_1573/ | # Abschlussarbeit
Schätzung der Einfallsrichtung früher Reflexionen mittels Compressive Sensing
### Steckbrief
Professur:
TA
Status:
abgeschlossen
Art der Arbeit:
Bachelor
#### Betreuer
Bachelorarbeit von Förster, Jonas
Beamforming techniques are nowadays commonly used to detect the direction of arrival (DOA) of sound waves arriving at spherical microphone arrays (SMAs). The conventional plane-wave decomposition beamformer(PWD-BF) can be used to estimate the DOA with a maximal directivity function solving a ℓ2 - minimisation problem. However, since the number of microphones in SMAs is limited by physical constraints, the PWD-BF method suffers from low spatial resolution in the practical use case. The spatial resolution can be improved by applying compressive sensing, the so-called compress- ive beamforming(CB). The main requirement to apply compressive beamforming is sparsity in the solution of the problem. That means that the minimisation problem is underdetermined and can be solved with ℓ1 - minimisation. In reverberant room acoustic scenarios, the condition of sparsity is not fulfilled due to meas- urement noise and too many incoming reflections. Therefore, subspace-based preprocessing methods are used to divide the signals into two parts. The first part is assumed to consist of the direct sound and the early reflections whereas the second part includes late reverberation and measurement noise. The first part is then assumed to be sparse and the directions of arrival of the direct sound and the early reflections can be estimated using CB. In this work, the performance of CB with and without the subspace-based preprocessing methods is compared with the performance of PWD-BF and MUltiple SIgnal Classifica- tion(MUSIC). This is done in two simulation scenarios. In the first scenario, plane-wave sources are generated analytically and measurement noise is added. In the second, the methods are applied on directional room impulse responses. The focus of the analyses is on the influence of measurement noise and late reverberation to the simulations with respect to their effects on sparsity of the problem and the estimation of the DOA of the primary sources and reflections. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8444182276725769, "perplexity": 940.3469022598472}, "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-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00533.warc.gz"} |
http://famouscontractingco.com/dt-meaning-rscc/8927a1-nuclear-fusion-reaction | In nuclear physics, nuclear fusion is a nuclear reaction in which two or more atomic nuclei collide at a very high energy and fuse together into a new nucleus, e.g. It relies on accelerating nucleons to a very high speed by heating them, using 150 million kW•h, which is a difficult process. Afterward there are one proton and one neutron (bound together as the nucleus of deuterium) plus a positron and a neutrino (produced as a consequence of the conversion of one proton to a neutron). It is a nuclear process, where energy is produced by smashing together light atoms. The concept of hybrid fusion can be compared with an accelerator-driven system (ADS), where an accelerator is the source of neutrons for the blanket assembly, rather than nuclear fusion reactions (see page on Accelerator-driven Nuclear Energy). The binding energy B is the energy associated with the mass difference between the Z protons and N neutrons considered separately and the nucleons bound together (Z + N) in a nucleus of mass M. The formula is Click above to view our full fission vs fusion infographic. H + H → D + β + + ν, Deuterium and Tritium are isotopes of hydrogen. Nuclear Fusion is the reaction that makes the Universe shine and lights up all of the stars including the one we are most familiar with, the sun. A Spanish nuclear fusion start-up said its Pulsotron-4 reactor has generated a magnetic field larger than any it had before, while also damaging it. An enormous amount of energy is released in this process and is greater than the nuclear fission reaction. The binding energy of the nucleus is a measure of the efficiency with which its constituent nucleons are bound together. Fusion Reactions The concept of nuclear fusion has been described in Chapter 12. Nuclear fusion. In nuclear and particle physics the energetics of nuclear reactions is determined by the Q-value of that reaction. If light nuclei are forced together, they will fuse with a yield of energy because the mass of the combination will be less than the sum of the masses of the individual nuclei. Meanwhile, the potential peaceful applications of nuclear fusion, especially in view of the essentially limitless supply of fusion fuel on Earth, have encouraged an immense effort to harness this process for the production of power. Nuclear Fusion. A major nuclear fusion reactor powered by lasers has set up a new series of milestones. NIF researchers believe they are close to an important milestone known as 'burning plasma'. In the long road to nuclear fusion, scientists continue to confront one of the more prominent (and literal) bumps: edge localized modes (ELMs). Fusion reactor, also called fusion power plant or thermonuclear reactor, a device to produce electrical power from the energy released in a nuclear fusion reaction. Fusion is the reverse process of nuclear fission. Nuclear fusion is a nuclear reaction through which two light nuclei of atoms, usually hydrogen and its isotopes (deuterium and tritium), are combined forming a heavier nucleus.This binding is usually accompanied by the emission of particles (in case of deuterium nuclei one neutron is emitted). In cases where the interacting nuclei belong to elements with low atomic numbers (e.g., hydrogen [atomic number 1] or its isotopes deuterium and tritium), substantial amounts of energy are released. The UK today embarked on a step toward building the world’s first nuclear fusion power station, by launching a search for a 100-plus hectare site where it can be plugged into the electricity grid. The facility is used for basic science, fusion energy research, and nuclear weapons testing. You can check out the difference between the two in this infographic below. Nuclear fusion, on the other hand, works by forcing atoms together in order to release energy. The National Ignition Facility (NIF), at Lawrence Livermore National Laboratory (LLNL) in … It merges atomic nuclei to create massive amounts of energy—the … The neutrons released by the fusion reactions add to the neutrons released due to fission, allowing for more neutron-induced fission reactions to take place. The vast energy potential of nuclear fusion was first exploited in thermonuclear weapons, or hydrogen bombs, which were developed in the decade immediately following World War II. Generation of fusion energy for practical use also relies on fusion reactions between the lightest elements that burn to form helium. For nuclei lighter than iron-56, the reaction is exothermic, releasing energy. This reaction takes place with elements that have a low atomic number, such as hydrogen. Five years later, Lockheed confirmed that it is still working on the project -- but had made very little progress in nuclear fusion energy. During fusion, atomic nuclei are forced together to form heavier atoms. For a detailed history of this development, see nuclear weapon. It's the same type of reaction that powers hydrogen bombs and the sun. Get exclusive access to content from our 1768 First Edition with your subscription. Within the next three to four years, working in cooperation with a company spun off from MIT in 2017 to commercialize the idea, "Commonwealth Fusion Systems LLC," MIT hopes to have a "SPARC" test reactor built to prove their concept. Our editors will review what you’ve submitted and determine whether to revise the article. It is the opposite reaction of fission, where heavy isotopes are split apart. Fusion reactions occur in stars where two hydrogen nuclei fuse together under high temperatures and … The intensity of this radiation when it reaches the Earth is 1361 W/m 2. The German-born physicist Hans Bethe proposed in the 1930s that the H-H fusion reaction could occur with a net release of energy and provide, along with subsequent reactions, the fundamental energy source sustaining the stars. The vast energy potential of nuclear fusion was first exploited in thermonuclear weapons, or hydrogen … A boosted fission weapon usually refers to a type of nuclear bomb that uses a small amount of fusion fuel to increase the rate, and thus yield, of a fission reaction. The evolution of stars can be viewed as a passage through various stages as thermonuclear reactions and nucleosynthesis cause compositional changes over long time spans. Updates? Nuclear Fusion is the acknowledged world-leading journal specializing in fusion. Nuclear fusion is the great hope for generating clean power. Scientists developing a compact version of a nuclear fusion reactor have shown in a series of research papers that it should work, renewing hopes that the … The use of nuclear fusion reactions for electricity generation remains theoretical. Nuclear fission is the splitting of a large atomic nucleus into smaller nuclei. To obtain a nuclear fusion reaction the atoms involved must overcome an important barrier of electrostatic forces. For nuclei heavier than iron-56, the reaction is endothermic, requiring an external source of energy. Nuclear fusion, process by which nuclear reactions between light elements form heavier elements (up to iron). As one of the leaders in this new field of energy research, Commonwealth Fusion Systems is an IPO prospect I'd want to keep on the lookout for. B = (Zmp + Nmn − M)c2, A better nuclear future? This repulsion is caused by protons that have a positive charge. Answer 1: Nuclear Fusion refers to a reaction which happens when two atoms join together for forming one or more different atomic nuclei and subatomic particles like protons and neutrons. Fusion, the nuclear reaction that powers the Sun and the stars, is a potential source of safe, non-carbon emitting and virtually limitless energy. A major nuclear fusion reactor powered by lasers has set up a new series of milestones. Define nuclear fusion reaction. Tritium (T) has one proton and two neutrons. Nuclear fusion is the source of Sun’s phenomenal energy output. Officials say the ITER nuclear fusion reactor is poised to be the most complicated piece of machinery ever built. Fusion reactions between light elements, like fission reactions that split heavy elements, release energy because of a key feature of nuclear matter called the binding energy, which can be released through fusion or fission. Fusion reactions are of two basic types: (1) those that preserve the number of protons and neutrons and (2) those that involve a conversion between protons and neutrons. The nucleus made by fusion is heavier than either of the starting nuclei. It also produces and consumes tritium within the plant in a closed circuit. The Q-value of the reaction is defined as the difference between the sum of the masses of the initial reactants and the sum of the masses of the final products, in energy units (usually in MeV). Nuclear fusion is a nuclear reaction in which two or more atomic nuclei (e.g. It differs significantly from nuclear fission, which has been our only way of … A viable nuclear fusion reactor — one that spits out more energy than it consumes — could be here as soon as 2025. The energy released is related to Einstein's famous equation, E=mc 2. Sun – The Ultimate Nuclear Fusion Reactor For a nuclear fusion reaction to occur, it is necessary to bring two nuclei so close that nuclear forces become active and glue the nuclei together. where mp and mn are the proton and neutron masses and c is the speed of light. Omissions? Professor of Applied Science, University of California, San Diego. Nuclear fusion is a process where two or more nuclei combine to form an element with a higher atomic number (more protons in the nucleus). In the Sun, with a core temperature close to 15.6 million Kelvin, the predominant pathway, by which more than 99% of solar energy is produced ( through conversion of hydrogen into helium nuclei ), is the Proton-proton (p-p) chain reaction. The entire donut-shaped reactor should be "about the size of a tennis court," says CFS CEO Bob Mumgaard. Nuclear f “ u ” s i o n (combining/f “ u ” sing two atoms together into one new atom) During fusion reactions, nuclei collide and fuse, eventually forming nuclei of heavier elements and producing enormous amounts of energy. If light nuclei are forced together, they will fuse with a yield of energy because the mass of the combination will be less than the sum of the masses of the individual nuclei. On the sun, in a series of nuclear reactions… Given how little progress LockMart seems to have made over the past six years, I'd caution investors not to take even MIT's success for granted. An important fusion reaction for practical energy generation is that between deuterium and tritium (the D-T fusion reaction). Source: chemwiki.ucdavis.edu. MIT and CFS are planning for their prototype ARC reactor to put out about 270 megawatts -- about one-quarter the output of a standard fission reaction nuclear … Fusion reactions occur when two or more atomic nuclei come close enough for long enough that the nuclear force pulling them together exceeds the electrostatic force pushing them apart, fusing them into heavier nuclei. MIT and CFS intend to use "yttrium barium copper oxide (YBCO) high-temperature superconducting magnet technology" to form a magnetic field to contain a reaction in which deuterium and tritium (both isotopes of hydrogen) will be forced to fuse together under high pressure and temperatures of "tens of millions of degrees." CFS might still fail in its endeavor. A fusion reactor produces helium, which is an inert gas. This would be a cleaner, safer, more efficient and more abundant source of power than nuclear fission. Returns as of 12/25/2020. helium. Interior of the U.S. Department of Energy's National Ignition Facility (NIF), located at Lawrence Livermore National Laboratory, Livermore, California. Nuclear fusion is essentially the opposite of nuclear fission. To obtain a nuclear fusion reaction the atoms involved must overcome an important barrier of electrostatic forces. Let us know if you have suggestions to improve this article (requires login). NASA has unlocked nuclear fusion on a tiny scale, with a phenomenon called lattice confinement fusion that takes place in the narrow channels between atoms.In the reaction, the common nuclear … helium. This article focuses on the physics of the fusion reaction and on the principles of achieving sustained energy-producing fusion reactions. 'Burning plasma' entails "a fusion burn sustained by the heat of the reaction … Nuclear Fusion: Nuclear Fusion is a reaction that occurs when two atoms combine together to form one or more different atomic nuclei and subatomic particles like protons and neutrons. Hence, nuclei smaller than iron-56 are more likely to fuse while those heavier than iron-56 are more likely to break apart. On Earth, the most likely fusion reaction is Deuterium–Tritium reaction. They yield millions of times more energy than other sources through nuclear reactions. The HL-2M Tokamak reactor is China's largest and most advanced nuclear fusion experimental research device, and scientists hope that the device can potentially unlock a powerful clean energy source. MIT Validates Science Behind New Nuclear Fusion Reactor Design Research shows Commonwealth Fusion Systems prototype should work, but huge engineering … Nuclear fusion, process by which nuclear reactions between light elements form heavier elements (up to iron). Nuclear fusion in the Iter reactor. That being said, $250 million is far less than the$106 billion market capitalization backing Lockheed Martin's efforts. Image source: Commonwealth Fusion Systems. Nuclear fusion is when two small, light nuclei join together to make one heavy nucleus. Fusion releases energy. In nuclear physics, nuclear fusion is a nuclear reaction in which two or more atomic nuclei collide at a very high energy and fuse together into a new nucleus, e.g. Fission and fusion are two physical processes that produce massive amounts of energy from atoms. As MIT and CFS continue to progress toward construction of their test reactor, an IPO might not be out of the question to accelerate their efforts. Nor will it need so much fuel. Tritium is radioactive (a beta emitter) but its half life is short. By signing up for this email, you are agreeing to news, offers, and information from Encyclopaedia Britannica. The nucleus of deuterium (D) consists of one proton and one neutron, whereas hydrogen only has one proton. Nuclear fusion is the future of nuclear power. In fission, a heavy nucleus is split into smaller nuclei. This produces a tremendous amount of energy in the form of light and heat. Asia. It is summarized in Figure 14-1, which is analogous to Figure 13-2 for nuclear fission. While a 1000 MW coal-fired power plant requires 2.7 million tonnes of coal per year, a fusion plant of the kind envisioned for the second half of this century will only require 250 kilos of fuel per year, half of it deuterium, half of it tritium. In cases where the interacting nuclei belong to elements with low atomic numbers (e.g., hydrogen [atomic number 1] or its isotopes deuterium and tritium ), substantial amounts of energy are released. Nuclear fusion is the reaction in which two or more nuclei combine, forming a new element with a higher atomic number (more protons in the nucleus). where β + represents a positron and ν stands for a neutrino. RSS. Fusion is considered the Holy Grail of energy and is what powers our sun. The promise of nuclear fusion is tantalizing: By utilizing the same atomic process that powers our sun, we may someday be able to generate virtually unlimited amounts of clean energy. The fusion chain believed to be the most practical for use in a nuclear fusion reactor is the following two-step process: $\ce{_1^2H + _1^2H \rightarrow _1^3H + _1^1H},$ $\ce{_1^2H + _1^3H \rightarrow _2^4He + _0^1n}.$ This chain, like the proton-proton chain, produces energy without any radioactive by-product. 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https://math.stackexchange.com/questions/1046406/simplicial-homology-in-hatcher-book | # simplicial homology in Hatcher book.
I was studying simplicial homology in Hatcher's Algebraic topology book.In one paragraph book says following: Some obvious general questions arise: Are the groups $H_n(X)$ independent of the choice of simplicial complex structure on X? In other words, if two complexes are homeomorphic, do they have isomorphic homology groups? More generally, do they have isomorphic homology groups if they are merely homotopy equivalent? To answer such questions and to develop a general theory it is best to leave the rather rigid simplicial realm and introduce the singular homology groups.
My Question is what is the answers of the above mentioned questions?? any help would be appreciated.
The answers are, broadly speaking, "yes": homology is a property that's invariant within a homoeomorphism class, i.e., if $X$ and $Y$ are homeomorphic, they end up having the same homology groups, and the same is true if they're merely homotopy equivalent.
These claims can be proved (with limitations) for simiplicial objects, but the proofs are not very pretty or general, which is why Hatcher says "let's move on to other kinds of homology theory" (which will produce, for "nice" shapes at least, the same groups). Indeed, one of the big theorems is that if two homology theories produces the same values on certain basic shapes, and they both have 5 specific properties, then they produce the same values on a wide category of shapes. (The 5 properties are known as the "Eilenberg Steenrod Axioms.") | {"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": 0.8516292572021484, "perplexity": 370.05232484114055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 5, "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-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00561.warc.gz"} |
http://mathhelpforum.com/trigonometry/123355-trigonometric-equation.html | 1. Trigonometric Equation
Find the general solution:
tanA+tan2A+tan3A=0
2. Using double and triple angle formulas you can say
$\displaystyle \tan(A)+\tan(2A)+\tan(3A) = 0$
$\displaystyle \tan(A)+ \frac{2\tan(A)}{1-\tan(A)} +\frac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}=0$
Make a common denominator, multiply it through both sides and solve.
3. if A = 0
also by observation if $\displaystyle A=0$ than
$\displaystyle tan(0) + tan((2)(0)) + tan((3)(0)) = 0$
$\displaystyle 0+0+0=0$
4. Originally Posted by pickslides
Using double and triple angle formulas you can say
$\displaystyle \tan(A)+\tan(2A)+\tan(3A) = 0$
$\displaystyle \tan(A)+ \frac{2\tan(A)}{1-\tan(A)} +\frac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}=0$
Make a common denominator, multiply it through both sides and solve.
It is better to do so:
$\displaystyle \begin{gathered} \tan A + \tan 2A + \tan 3A = 0; \hfill \\ \frac{{\sin 3A}} {{\cos A\cos 2A}} + \frac{{\sin 3A}} {{\cos 3A}} = 0; \hfill \\ \end{gathered}$
$\displaystyle \begin{gathered} \frac{{\sin 3A\cos 3A + \sin 3A\cos A\cos 2A}} {{\cos A\cos 2A\cos 3A}} = 0; \hfill \\ \sin 3A\left( {\cos 3A + \cos A\cos 2A} \right) = 0. \hfill \\ \end{gathered}$
$\displaystyle 1)~~\sin 3A = 0.$
$\displaystyle \begin{gathered} 2)~~\cos 3A + \cos A\cos 2A = 0; \hfill \\ \cos 3A + \frac{1} {2}\left( {\cos A + \cos 3A} \right) = 0; \hfill \\ 3\cos 3A + \cos A = 0; \hfill \\ \end{gathered}$
$\displaystyle \begin{gathered} 3\cos A\left( {4{{\cos }^2}A - 3} \right) + \cos A = 0; \hfill \\ 3{\cos ^3}A - 2\cos A = 0; \hfill \\ \cos A\left( {\sqrt 3 \cos A - \sqrt 2 } \right)\left( {\sqrt 3 \cos A + \sqrt 2 } \right) = 0. \hfill \\ \hfill \\ \end{gathered}$
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tana tan2a tan4a
Click on a term to search for related topics. | {"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": 0.9916405081748962, "perplexity": 1859.8344938001974}, "config": {"markdown_headings": false, "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-2018-17/segments/1524125938462.12/warc/CC-MAIN-20180420135859-20180420155859-00067.warc.gz"} |
https://homework.cpm.org/category/ACC/textbook/acc7/chapter/7%20Unit%208/lesson/CC3:%207.1.2/problem/7-20 | Home > ACC7 > Chapter 7 Unit 8 > Lesson CC3: 7.1.2 > Problem7-20
7-20.
Use the rectangle at right to answer the following questions. Homework Help ✎
1. Find the area of the entire rectangle. Explain how you found your solution.
The area of a figure is the same as the sum of the area of its parts.
$72+30+48+20$
2. Calculate the perimeter of the figure.
The perimeter of a figure is the sum of its sides.
$54$ units | {"extraction_info": {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.857847273349762, "perplexity": 778.477290428071}, "config": {"markdown_headings": false, "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-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00138.warc.gz"} |
https://homework.cpm.org/category/CC/textbook/CCG/chapter/Ch7/lesson/7.1.4/problem/7-46 | ### Home > CCG > Chapter Ch7 > Lesson 7.1.4 > Problem7-46
7-46.
For each pair of triangles below, determine whether or not the triangles are similar. If they are similar, show your reasoning in a flowchart. If they are not similar, explain how you know.
1. measure of Angle YXZ = 102 degrees, measure of angle BAC = 36 degrees, and Angle ACB is congruent to angle ZYX.
measure of Angle CBA = 102 degrees and measure of Angle XZY = 36 degrees, both by the Triangle Angle Sum Theorem.
All three angles are the same in both triangles.
Triangle A, B, C is similar to triangle X, Y, Z by A, A similarity.
2. The triangles are not similar because corresponding sides do not have the same ratio. One triangle is isosceles and the other is not. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8776477575302124, "perplexity": 853.4430348209652}, "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-2022-33/segments/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00371.warc.gz"} |
https://jp.maplesoft.com/support/help/maplesim/view.aspx?path=LieAlgebrasOfVectorFields%2FLAVF%2FVectorSpaceSum | VectorSpaceSum - Maple Help
VectorSpaceSum
find a LAVF object whose solution space is the sum of the solution spaces of given LAVF objects.
Calling Sequence VectorSpaceSum( L1, L2, ..., depname = vars )
Parameters
L1, L2, ... - a sequence of LAVF objects living on the same space (see AreSameSpace) vars - (optional) a list of new dependent variable names
Description
• Let L1,L2, ... be a sequence of LAVF objects living on the same space. The VectorSpaceSum method finds a LAVF object whose solution space is the vector space sum of solution spaces of L1,L2,....
• By default, the dependent variable names of the returned object are taken from L1. The dependent variable names will be vars if the optional argument depnames = vars is specified.
• This method is front-end to the corresponding method of a LHPDE object. That is, let S1, S2,... be the determining systems of L1,L2,... (i.e. Si = GetDeterminingSystem(Li)), then the call VectorSpaceSum(L1,L2,..) is equivalent to VectorSpaceSum(S1,S2,..). All remaining input arguments will be passed down to its determining system level. See the method VectorSpaceSum of a LHPDE object for more detail.
• This method is associated with the LAVF object. For more detail, see Overview of the LAVF object.
Examples
> $\mathrm{with}\left(\mathrm{LieAlgebrasOfVectorFields}\right):$
> $\mathrm{Typesetting}:-\mathrm{Settings}\left(\mathrm{userep}=\mathrm{true}\right):$
> $\mathrm{Typesetting}:-\mathrm{Suppress}\left(\left[\mathrm{\xi }\left(x,y\right),\mathrm{\eta }\left(x,y\right)\right]\right):$
> $X≔\mathrm{VectorField}\left(\mathrm{\xi }\left(x,y\right)\mathrm{D}\left[x\right]+\mathrm{\eta }\left(x,y\right)\mathrm{D}\left[y\right],\mathrm{space}=\left[x,y\right]\right)$
${X}{≔}{\mathrm{\xi }}{}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\eta }}{}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}$ (1)
The determining system for 2-dim Euclidean
> $\mathrm{E2}≔\mathrm{LHPDE}\left(\left[\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right)+\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x,x\right)=0\right],\mathrm{indep}=\left[x,y\right],\mathrm{dep}=\left[\mathrm{\xi },\mathrm{\eta }\right]\right)$
${\mathrm{E2}}{≔}\left[{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}}{+}{{\mathrm{\eta }}}_{{x}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}{,}{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}{,}{x}}{=}{0}\right]{,}{\mathrm{indep}}{=}\left[{x}{,}{y}\right]{,}{\mathrm{dep}}{=}\left[{\mathrm{\xi }}{,}{\mathrm{\eta }}\right]$ (2)
The determining system for 2-dim translations
> $\mathrm{T2}≔\mathrm{LHPDE}\left(\left[\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0\right],\mathrm{indep}=\left[x,y\right],\mathrm{dep}=\left[\mathrm{\xi },\mathrm{\eta }\right]\right)$
${\mathrm{T2}}{≔}\left[{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]{,}{\mathrm{indep}}{=}\left[{x}{,}{y}\right]{,}{\mathrm{dep}}{=}\left[{\mathrm{\xi }}{,}{\mathrm{\eta }}\right]$ (3)
We first construct LAVFs for E(2) and T(2)
> $\mathrm{LE2}≔\mathrm{LAVF}\left(X,\mathrm{E2}\right)$
${\mathrm{LE2}}{≔}\left[{\mathrm{\xi }}{}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\eta }}{}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[{{\mathrm{\xi }}}_{{y}{,}{y}}{=}{0}{,}{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{-}{{\mathrm{\xi }}}_{{y}}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]\right\}$ (4)
> $\mathrm{LT2}≔\mathrm{LAVF}\left(X,\mathrm{T2}\right)$
${\mathrm{LT2}}{≔}\left[{\mathrm{\xi }}{}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\eta }}{}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]\right\}$ (5)
> $\mathrm{VectorSpaceSum}\left(\mathrm{LE2},\mathrm{LT2}\right)$
$\left[{\mathrm{\xi }}{}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\eta }}{}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}}{=}{-}{{\mathrm{\eta }}}_{{x}}{,}{{\mathrm{\eta }}}_{{x}{,}{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]\right\}$ (6)
> $\mathrm{VectorSpaceSum}\left(\mathrm{LE2},\mathrm{LT2},\mathrm{depname}=\left[\mathrm{\alpha },\mathrm{\beta }\right]\right)$
$\left[{\mathrm{\alpha }}{}\left({x}{,}{y}\right){}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\beta }}{}\left({x}{,}{y}\right){}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\alpha }}{}\left({x}{,}{y}\right){=}{0}{,}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\alpha }}{}\left({x}{,}{y}\right){=}{-}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\beta }}{}\left({x}{,}{y}\right){,}\frac{{{\partial }}^{{2}}}{{\partial }{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\beta }}{}\left({x}{,}{y}\right){=}{0}{,}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\beta }}{}\left({x}{,}{y}\right){=}{0}\right]\right\}$ (7)
Compatibility
• The VectorSpaceSum command was introduced in Maple 2020. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9656106233596802, "perplexity": 2225.888016289981}, "config": {"markdown_headings": true, "markdown_code": false, "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-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00481.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/153367-quotient-groups-matrices.html | # Thread: Quotient Groups - Matrices
1. ## Quotient Groups - Matrices
Let $G=GL_2(Q,*)$ , H={ $\begin{pmatrix} 1&a\\ 0&1 \end{pmatrix} : a\in Q$ } a subgroup of G.
Find [G:H].
How do I do this?! I don't have any idea how to even look at it
Can you please explain to me what is this?
Thanks!
2. Two elements $A,B \in G$ are equivalent modulo $H$ iff
$AB^{-1} \in H$
<=>
there exists $C\in H$ , so that
$A=CB$
If you now write the matrices C and B explicit and calculate CB, you can see that in order for this equation to hold, B and A must have identical second rows.
So you can conclude that if A and B are equivalent, then they must have identical second rows at least or conversely: If the second rows of A and B are different, then A and B are not equivalent.
Since there are infinitely many matrices in G (i.e. invertible matrices with rational entries) which have pairwise different second rows, there are infinitely many elements that are all pairwise not equivalent and so the index [G:H] must be infinite.
3. Originally Posted by Iondor
Two elements $A,B \in G$ are equivalent modulo $H$ iff
$AB^{-1} \in H$ <-- Do you mean that A,B in G are in H iff ...?
<=>
there exists $C\in H$ , so that
$A=CB$
If you now write the matrices C and B explicit and calculate CB, you can see that in order for this equation to hold, B and A must have identical second rows. <-- if A,B are in H, then their second row is 0,1 anyway :O
So you can conclude that if A and B are equivalent, then they must have identical second rows at least or conversely: If the second rows of A and B are different, then A and B are not equivalent.
Since there are infinitely many matrices in G (i.e. invertible matrices with rational entries) which have pairwise different second rows, there are infinitely many elements that are all pairwise not equivalent and so the index [G:H] must be infinite.
I believe there's a misunderstanding here - GL2 is the general linear group of all invertable matrices of size 2x2. What do you mean by "A,B in G are equivalent modulo H"? How can 2 matrices be equivalent?
Thanks
$A$, $B$ are equivalent modulo $H$ means that $AH = BH$ in $G/H$, i.e. they are in the same coset. It can be shown that two elements $A, B$ are in the same coset if and only if $AB^{-1}\in H$. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "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": 0.9271484017372131, "perplexity": 307.2936399672308}, "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-2017-43/segments/1508187823839.40/warc/CC-MAIN-20171020063725-20171020083725-00233.warc.gz"} |
https://documentation.aimms.com/functionreference/algorithmic-capabilities/the-gmp-library/gmp_instance-procedures-and-functions/gmp_instance_createfeasibility.html | Function GMP::Instance::CreateFeasibility(GMP, name, useMinMax)
# GMP::Instance::CreateFeasibility
The function GMP::Instance::CreateFeasibility creates a mathematical program that is the feasibility problem of a generated mathematical program. Its main purpose is to identify infeasibilities in an infeasible problem. The feasibility problem can be used to minimize the sum of infeasibilities, or to minimize the maximum infeasibility.
This function can be used for both linear and nonlinear problems but not for constraint programming problems.
GMP::Instance::CreateFeasibility(
GMP, ! (input) a generated mathematical program
[name], ! (input, optional) a string expression
[useMinMax] ! (input, optional) integer, default 0
)
## Arguments
GMP
An element in the set AllGeneratedMathematicalPrograms.
name
A string that contains the name for the feasibility problem.
useMinMax
If 0, the sum of infeasibilities will be minimized, else the maximum infeasibility will be minimized.
### Mathematical formulation
In this section we show how the feasibility problem is constructed. To simplify the explanation we use a linear problem but the same construction applies to a nonlinear problem.
Consider the following problem where $$J$$ denotes the set of variables, $$I_1$$ the set of $$\geq$$ inequalities, $$I_2$$ the set of $$\leq$$ inequalities, and $$I_3$$ the set of equalities.
\begin{split}\begin{aligned} \max \quad & \sum_{j\in J} a_{j} x_j \\ \text{s.t.} \quad & \sum_{j\in J} a_{ij} x_j && \geq b_i \quad i \in I_1 \\ & \sum_{j\in J} a_{ij} x_j && \leq b_i \quad i \in I_2 \\ & \sum_{j\in J} a_{ij} x_j && = b_i \quad i \in I_3 \\ & x \geq 0 \end{aligned}\end{split}
Then if we minimize the sum of infeasibilities the feasibility problem becomes:
\begin{split}\begin{aligned} \min \quad & \sum_{i \in I_1} z^p_i + \sum_{i \in I_2} z^n_i + \\ & \sum_{i \in I_3} (z^p_i + z^n_i) \\ \text{s.t.} \quad & \sum_{j\in J} a_{ij} x_j + z^p_i && \geq b_i \quad && i \in I_1 \\ & \sum_{j\in J} a_{ij} x_j - z^n_i && \leq b_i \quad && i \in I_2 \\ & \sum_{j\in J} a_{ij} x_j + z^p_i - z^n_i && = b_i \quad && i \in I_3 \\ & x, z^p, z^n \geq 0 \end{aligned}\end{split}
If we minimize the maximum infeasibility the feasibility problem becomes:
\begin{split}\begin{aligned} \min \quad & z^m \\ \text{s.t.} \quad & \sum_{j\in J} a_{ij} x_j + z^m && \geq b_i \quad && i \in I_1 \\ & \sum_{j\in J} a_{ij} x_j - z^m && \leq b_i \quad && i \in I_2 \\ & \sum_{j\in J} a_{ij} x_j + z^p_i - z^n_i && = b_i \quad && i \in I_3 \\ & z^m - z^p_i - z^n_i && \geq 0 \quad && i \in I_3 \\ & x, z^p, z^n \geq 0 \end{aligned}\end{split}
## Return Value
A new element in the set AllGeneratedMathematicalPrograms with the name as specified by the name argument.
Note
• The name argument should be different from the name of the original generated mathematical program.
• If the name argument is not specified then AIMMS will name the generated math program as “Feasibility problem of” followed by the name of the GMP.
• If an element with name specified by the name argument is already present in the set AllGeneratedMathematicalPrograms the corresponding generated mathematical program will be replaced (or updated in case the same symbolic mathematical program is involved).
• By using the suffices .ExtendedVariable and .ExtendedConstraint it is possible to refer to the columns and rows that are added to create the feasibility problem. In case the sum of infeasibilities is minimized only variables are added:
• The variable c.ExtendedVariable('PositiveViolation',i) is added for a constraint c(i) with type $$\geq$$.
• The variable c.ExtendedVariable('NegativeViolation',i) is added for a constraint c(i) with type $$\leq$$.
• The variables c.ExtendedVariable('PositiveViolation',i) and c.ExtendedVariable('NegativeViolation',i) are added for an equality constraint c(i).
In case the maximum infeasibility is minimized the following variables and constraints are added:
• The variable mp.ExtendedVariable('MaximumViolation') is added for math program mp.
• The variables c.ExtendedVariable('PositiveViolation',i) and c.ExtendedVariable('NegativeViolation',i) are added for an equality constraint c(i).
• The constraint c.ExtendedConstraint('MaximumViolation',i) is added for an equality constraint c(i).
In the above mathematical formulation,
• c.ExtendedVariable('PositiveViolation',i) corresponds to $$z^p_i$$.
• c.ExtendedVariable('NegativeViolation',i) corresponds to $$z^n_i$$.
• mp.ExtendedVariable('MaximumViolation') corresponds to $$z^m$$. | {"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": 1, "mathjax_asciimath": 1, "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": 1.0000042915344238, "perplexity": 1251.2976242480324}, "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-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00366.warc.gz"} |
http://www.computer.org/csdl/trans/tp/preprint/06587455-abs.html | Subscribe
pp: 1
Yi-Lei Chen , National Tsing Hua University, Hsinchu
Chiou-Ting Candy Hsu , National Tsing Hua University, Hsinchu
Hong-Yuan Mark Liao , Academia Sinica, Taipei and National Chiao Tung University, Hsinchu
ABSTRACT
Tensor completion, which is a high-order extension of matrix completion, has generated a great deal of research interest in recent years. Given a tensor with incomplete entries, existing methods use either factorization or completion schemes to recover the missing parts. However, as the number of missing entries increases, factorization schemes may overfit the model because of incorrectly predefined ranks, while completion schemes may fail to interpret the model factors. In this paper, we introduce a novel concept: complete the missing entries and simultaneously capture the underlying model structure. To this end, we propose a method called Simultaneous Tensor Decomposition and Completion (STDC) that combines a rank minimization technique with Tucker model decomposition. Moreover, as the model structure is implicitly included in the Tucker model, we use factor priors, which are usually known a priori in real-world tensor objects, to characterize the underlying joint-manifold drawn from the model factors. We conducted experiments to empirically verify the convergence of our algorithm on synthetic data, and evaluate its effectiveness on various kinds of real-world data. The results demonstrate the efficacy of the proposed method and its potential usage in tensor-based applications. It also outperforms state-of-the-art methods on multilinear model analysis and visual data completion tasks.
INDEX TERMS
Tensile stress, Equations, Matrix decomposition, Mathematical model, Approximation methods, Visualization, Brain modeling, Machine learning, Multidimensional
CITATION
Yi-Lei Chen, Chiou-Ting Candy Hsu, Hong-Yuan Mark Liao, "Simultaneous Tensor Decomposition and Completion Using Factor Priors", IEEE Transactions on Pattern Analysis & Machine Intelligence, , no. 1, pp. 1, PrePrints PrePrints, doi:10.1109/TPAMI.2013.164 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9235439300537109, "perplexity": 1960.6056185320622}, "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-2015-32/segments/1438042988061.16/warc/CC-MAIN-20150728002308-00167-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/207147-times-f-x-0-a.html | 1. ## #Times f(x)=0
f(x) = x4-4x3-8x2+4
How many times does f(x) = 0? (excuse the bad formulation)
Hint: Gather as much information from the derivative of f(x) as possible. We are only interested in the number of times f(x) equals zero.
Okay so differentiating the function is simple: f ' (x) = 4x3-12x2-16x, but what kind of information do I need to extract in order to determine how many times f(x)=0?
2. ## Re: #Times f(x)=0
Hi Cinnaman,
a few tipps:
f'(x)=0 => x1=-1, x2= 0; x3=4 the event. max- and min points of f(x)
If f''(x1) < 0 => x1 is max. point =>MAX1( x1|f(x1))
If f''(x1) > 0 => x1 is min. point =>MIN1( x1|f(x1))
analog with x2 und x3.
EXTx= Extrem x-point;
xp<xn;
If sign of f(EXTxp) is not = f(EXTxn) => there is a x, beetwen xp und xn with f(x)=0
3. ## Re: #Times f(x)=0
Ah, that makes sense, thank you.
I figured out that there is no x where f(x)=0 between x=-1 and x=0, but there is one between x=0 and x=4.
Is the last step now to take the limit as x approaches inf / -inf and use them as extreme points aswell?
4. ## Re: #Times f(x)=0
Hello, Cinnaman!
$f(x) \:= \:x^4 - 4x^3 - 8x^2 + 4$
How many times does $f(x) = 0$ ?
Hint: Gather as much information from the derivative of $f(x)$ as possible.
We are only interested in the number of times f(x) equals zero.
Set $f'(x) = 0$ and solve.
. . $4x^3 - 12x^2 - 16x \:=\:0 \quad\Rightarrow\quad 4x(x+1)(x-4) \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}1,0,4$
There are horizontal tangents at: $(\text{-}1,1),\:(0,4),\:(4,\text{-}124)$
The Second Derivative Test give us: . $\begin{Bmatrix}(\text{-}1,1) & \text{min.} \\ (0,4) & \text{max.} \\ (4,\text{-}124) & \text{min.} \end{Bmatrix}$
The graph looks like this.
Code:
|
* |
-*-
* * | * *
* * | *
-*- |
-------------+---♥-------♥---
|
| * *
| * *
| -*-
|
Therefore, the graph crosses the x-axis twice. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 7, "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": 0.8736015558242798, "perplexity": 3553.649712024823}, "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-2013-20/segments/1368705069221/warc/CC-MAIN-20130516115109-00026-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/motion-and-friction.9326/ | Homework Help: Motion and friction
1. Nov 20, 2003
latte
i've spent an hour reading this problem over and over again, trying to plug into the formulas but only came out with 2 or more unknown variables. I've not a clue to where to start, please help me.
Here goes..
during the investigation of a traffic accident, police find skid marks 90m long. They determine the coefficient of friction between the car's tires and the roadway to be .5 for the prevailing condition. Estimate the speed of the car when the brakes were applied.
choices are:
a. 9.49 m/s
b. 21
c. 29.7
d. 42
Here's how far i got .
givens:
d=90m
coeff=.5
vf=0
vi?
2. Nov 20, 2003
jamesrc
The only force on the car that you are supposed to consider after the brakes have been applied is kinetic friction between the tires and the road. Set ma = -μN, where m is the mass of the car, a is the acceleration, μ is the coefficient of friction, and N is the normal reaction force of the ground to the car (equal in magnitude to the weight of the car). With a known constant acceleration and stopping distance, there is a kinematic formula you can use to find the initial velocity.
3. Nov 20, 2003
gnome
Here's a clue.
Are you stuck because they didn't give you the mass?
Look: the force exerted on the car by friction is μ times the normal force. In this case, on level ground the normal force is equal to mg. Now, you want to know the rate of deceleration (-acceleration), right? Well, assuming the only force affecting the car's speed was the braking force at the tires, F=ma so:
a = F/m
but here,
F = μmg
OK?
4. Nov 20, 2003
gnome
Good timing, James.
5. Nov 21, 2003
jamesrc
Heh. We should start keeping score on these. I think we're tied now.
6. Nov 21, 2003
latte
C! i believe i got it. thanks! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.927712619304657, "perplexity": 688.6003294804159}, "config": {"markdown_headings": false, "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-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00634.warc.gz"} |
https://math.stackexchange.com/questions/2823838/notation-meaning-braces-in-the-variables-subscript | # notation meaning - braces in the variable's subscript
I am trying to understand the meaning of the braces in the subscript of a variable. I've encountered it in a paper Distributional Reinforcement Learning with Quantile Regression, page 5
Does it mean "as long as the condition is met"? But what is then the alternative expression, when the condition is not met? Have a look at equation (12):
Do I read the equation 12 correctly:
• the $\theta_i$ becomes equal to its current value plus ( $\tau_i$ minus $\delta$ in cases where $r + \gamma z'$ is less than the current $\theta_i$ ), scaled by $\alpha$
Or do these { } braces have another interpretation? Are they used to denote a piece-wise function?
• They introduce the notation at the end of page 3 in the paper. They say "$\delta_{z}$ denotes a Dirac at $z\in\mathbb{R}$". Did you see that? The same notation with braces seems to indicate the same thing for a set of points instead of a single point. – Jackozee Hakkiuz Jun 25 '18 at 3:52
• Thank you, I am pretty sure that's was the right answer. I had an additional discussion with peers and expression seems to mean "1 when {condition is true}, 0 otherwise" – Kari Jun 26 '18 at 13:58 | {"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": 0.9833635091781616, "perplexity": 427.63446702000533}, "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-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00453.warc.gz"} |
http://meetings.aps.org/Meeting/MAR12/Event/165361 | ### Session T41: Swimming, Motility and Locomotion
2:30 PM–5:42 PM, Wednesday, February 29, 2012
Room: 156B
Chair: Paolo Arratia, University of Pennsylvania
Abstract ID: BAPS.2012.MAR.T41.14
### Abstract: T41.00014 : Swimming of bio-inspired micro robots in circular channels
5:06 PM–5:18 PM
Preview Abstract MathJax On | Off Abstract
#### Authors:
Serhat Yesilyurt
(Sabanci University)
Fatma Zeynep Temel
(Sabanci University)
In recent years, bio-inspired micro swimming robots have been attracting attention for use in biomedical tasks such as opening clogged arteries, carrying out minimally invasive surgical operations, and carrying out diagnostic tasks. There have been a number of experimental and modeling studies that address swimming characteristics of micro swimmers with helical tails attached to magnetic heads that rotate and move forward in rotating external magnetic fields. We carried out experimental studies with millimeter long helical swimmers in glass tubes placed in between Helmholtz coils, and demonstrated that swimming speed increases linearly with the frequency of the external field up to the step-out frequency. In order to study interaction of the swimmer with the circular boundary we used a computational fluid dynamics model. In simulations we compared swimming speeds of robots with respect to the frequency of the external magnetic field, wavelength and amplitude of the helical tail, and distance to the channel wall. According to simulation results, as the swimmer gets closer to the boundary swimming speed and efficiency improve. However step-out frequency decreases near the wall due to increased torque to rotate the swimmer.
To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2012.MAR.T41.14 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8048925399780273, "perplexity": 3010.6037290331756}, "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-2014-35/segments/1409535921550.2/warc/CC-MAIN-20140901014521-00051-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-use-the-distributive-property-to-simplify-3-y-6 | Algebra
Topics
# How do you use the distributive property to simplify (3+y)6?
Multiply each term within the parenthesis by $\textcolor{red}{6}$.
$\left(3 + y\right) \textcolor{red}{6}$ becomes:
$\left(\textcolor{red}{6} \times 3\right) + \left(\textcolor{red}{6} \times y\right) = 18 + 6 y$ or $6 y + 18$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "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": 0.8061511516571045, "perplexity": 785.5968443683167}, "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-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00149.warc.gz"} |
http://mathhelpforum.com/algebra/110625-transposing-formulae.html | # Math Help - Transposing of formulae
1. ## Transposing of formulae
Hi there, thanks in advance for any help
Make A the subject of the formula:
Q = C.A square root (2g.h / (1-A^2/x^2))
I have got this far but unsure what to do with the fraction of a fraction... the (x^2) bit.
Q^2 / C^2.A^2 = (2g.h / (1-A^2/x^2))
Thank you again for any assitance
Mike
2. Originally Posted by mikewhant
Hi there, thanks in advance for any help
Make A the subject of the formula:
Q = C.A square root (2g.h / (1-A^2/x^2))
I have got this far but unsure what to do with the fraction of a fraction... the (x^2) bit.
Q^2 / C^2.A^2 = (2g.h / (1-A^2/x^2))
Thank you again for any assitance
Mike
I would first isolate that square root:
$\frac{Q}}{CA}= \sqrt{\frac{2gh}{1- \frac{A^2}{x^2}}$
and then square both sides.
[tex]\frac{Q^2}{C^2A^2}= \frac{2gh}{1- \frac{A^2}{x^2}}
Multiply both numerator and denominator, on the right, by $x^2$
$\frac{Q^2}{C^2A^2}= \frac{2ghx^2}{x^2- A^2}$
Get rid of the fractions by multiplying both sides by the denominators, $C^2A^2$ and $x^2- A^2$
$Q^2(x^2- A^2)= 2ghx^2C^2A^2$
$Q^2x^2- Q^2A^2= 2ghx^2C^2A^2$
Add $Q^2A^2$ to both sides
$Q^2x^2= 2ghx^2C^2A^2+ Q^2A^2$
Factor [tex]A^2[/quote] out of the right side
$Q^2x^2= (2gx^2C^2+ Q^2)A^2$
Divide both sides by $2gx^2C^2+ Q^2$
$\frac{Q^2x^2}{2gx^2C^2+ Q^2}= A^2$
Finally, take the square root of both sides
$A= \sqrt{\frac{Q^2x^2}{2gx^2C^2+ Q^2}}$
3. Thanks alot, great help, i just needed a worked example, now I can do them!
Cheers | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "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": 0.9612497091293335, "perplexity": 2643.5254321460798}, "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-2015-48/segments/1448398446500.34/warc/CC-MAIN-20151124205406-00094-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://nrich.maths.org/387/solution | ### Route to Root
A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?
### Squaresearch
Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?
### Loopy
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
# Sixational
##### Stage: 4 and 5 Challenge Level:
The $n^{th}$ term of a sequence is given by the formula $n^3 + 11n$. Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by $6$.
Congratulations to Julia Collins, age 17, Langley Park School for Girls, Bromley; Kookhyun Lee; Yatir Halevi age 17; Sim S K age 14; Ang Zhi Ping age 16 for your splendid solutions.
This is Kookhyun Lee's solution: First term: $12$, second term: $30$, third term: $60$, fourth term: $108$. The $100$th term is $100^3 +1100 = 1001100$. The $99$th term is $970299 + 1089 = 971388$ so the first term bigger than a million is $1001100$ when $n=100.$
Proof that all the terms are divisible by $6$.
$$n^3 + 11n = n^3 + 12n - n = n(n^2-1) + 12n$$
This must be a multiple of 6 because $n(n^2-1)$ can be written as $(n-1)\times n \times (n+1)$. Any multiple of three consecutive integers is a multiple of $6$ because it contains a multiple of two (an even number) and a multiple of three.
The following solution, uses a different method. It arrived early in the morning on the first day that the question was published from Yatir Halevi, age 17.5, Maccabim and Reut High-School, Israel.
We have a sequence given by the formula $n^3+11n.$ We have to find the first value of $n$ for which $n^3+11n> 10^6$ and we could use
http://www.sosmath.com/algebra/factor/fac11/fac11.html
The second way is a much nicer one. We notice that $100^3$ is $10^6$, so we know for $n=100$ that $n^3+11n$ is bigger than $10^6$, so we check $n=99$ and we get: $971388$ which is smaller than $10^6$. So we have the answer: $n=100$ is the first $n$ for which $n^3+11n$ is bigger than $10^6$.
The next thing we have to prove is that $n^3+11n$ is always divisible by $6$. This we will prove by using modular arithmetic. We will use modulus $6$. For each $n$, we can have a residue of either: $0$, $1$, $2$, $3$, $4$ or $5$. For $n^3$ we get the following residues:
$0$, $1$, $2$, $3$, $4$, $5$ respectively (to $n$). For $11n$ we get the following residues: $0$, $5$, $4$, $3$, $2$, $1$ respectively (to $n$).
Combining $n^3$ and $11n$ (respectively) we get a $0$ residue, because: $0+0=0$ (mod $6$), $1+5=6=0$ (mod $6$), $2+4=6=0$ (mod $6$), $3+3=6=0$ (mod $6$), $4+2=6=0$ (mod $6$), $5+1=6=0$ (mod $6$). This means that we get a zero residue when dividing by $6$, or in other words, $(n^3+11n)$ is a multiple of $6$ or $6$ divides $n^3+11n$. | {"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": 1, "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": 0.949820876121521, "perplexity": 181.54350537520574}, "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-2016-07/segments/1454701160822.87/warc/CC-MAIN-20160205193920-00315-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=21629 | # Simple harmonic motion in a rocket?
by PhysicsPhun
Tags: harmonic, motion, rocket, simple
P: 55 A simple pendulum suspended in a rocket ship has a period T0. Assume that the rocket ship is near the earth in a uniform gravitational field. (If A and E are true, and the others are not, enter TFFFT). A) If the ship accelerates upward, the period decreases. B) If the ship goes into freefall, accelerating downward at 9.81 m/s2, the pendulum will no longer oscillate. C) If the ship moves upward with a constant velocity, the period decreases. D) If the mass of the pendulum doubles, the period increases. E) If the length of the pendulum is doubled, the new period will be the square root of two times T0. I thought I had the answer on this one. But I didn't.. My instinct was A)F B)F C)T D)F E)T I also tried, TTTFT, TFTFT and FTTFT (ABCDE) As you can see I was very confident that C=T D=F E=T.. And I was wrong. Anyone have a better idea?
P: 55 So what i think now is XXFFT Really not sure on the first two, but i think that C=F D=F E=T
P: 88 I'm most probably wrong but I think the answer is given to you at the beginning there. So far I've come up with TFFFT. When the ship accelerates upwards won't that increase the value of the downward acceleration due to gravity? So according to the equation T=sqrt(l/g) (basic equation) wouldn't it cause the period to decrease. And with the last one I'm think i'm sure (haha) that it is right because if you do the maths it works out as square root of 2 times T0. I'm pretty sure that the second one is false. The third one is false as well I think, because if velocity is constant then acceleration is 0. So nothing will happen to the period. As for D, increasing the mass would decrease the period most likely. Hope that helps ya ;)
P: 65
## Simple harmonic motion in a rocket?
K...you guys....remember this equation for the period of a pendulum when solving....
T = 2(pi) * (length/g) ^ (1/2)....that means that when.....
g increases.......the period decreases........
...accelerating upwards means that g will be increased on the ship.....moving upwards at a constant velocity has NO EFFECT whatsoever on g....I believe that the correct answers are....
A. T (Yes....accelerating upwards will increase g...thus decreases T)
B. T (yes....SHM requires a restoring forces...if the pedulum is in free fall...there is NO restoring force...)
C. F (no...when it is moving at a constant velocity....the value of g is not affected)
D. F (no..the mass of a pendulum as no effect on its period...as you can see from the equation)
E. T (yes....the equation proves this to be the case)
P: 88 Cool thanks for the info on the restoring force man ;)
P: 55 Thanks alot for the explanations, you were right on :)
P: 65 np man...anytime :-)
Related Discussions Introductory Physics Homework 3 General Physics 2 Introductory Physics Homework 2 Introductory Physics Homework 2 Introductory Physics Homework 2 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.949013888835907, "perplexity": 1065.4488376967104}, "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-2013-48/segments/1386163052462/warc/CC-MAIN-20131204131732-00038-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1675518/what-does-terence-tao-mean-by-the-statement-primes-behave-randomly | What does Terence Tao mean by the statement “primes behave randomly”?
http://164.67.141.39:8080/ramgen/specialevents/math/tao/tao-20070117.smil
The Riemann hypothesis is, according to Tao, equivalent to the idea that the primes do behave randomly -- they are distributed according to the prime number theorem, with an error term that is exactly what you'd expect from the law of large numbers.
What does this mean?
Edit: You need the latest RealPlayer for the link.
• I think he should have said: "the distribution of primes is still a mystery, so we can generalize their behavior as if it was random" – jameselmore Mar 2 '16 at 23:19
• it is obviously wrong that the primes are randomly distributed, but what T.Tao ment is that if we look only at the growth of $| \pi(x) - \frac{x}{\ln x}| \approx |\sum_{n \le x} (\delta_n(p) - 1)|$ (with $\delta_p(n) = 1$ if $n$ is prime) or at $|\sum_{n \le x} \mu_n|$ or at $|\sum_{n \le x} \lambda(n)|$ it will grow as the cumulated sum of a i.i.d random sequence of $\pm 1$ : this is exactly the Riemann hypothesis. note that $$\ln \zeta(s) = \sum_k \frac{1}{k} \sum_p \delta(p) p^{-sk},\quad\frac{1}{\zeta(s)} = \sum_n \mu(n) n^{-s},\quad\frac{\zeta(2s)}{\zeta(s)} = \sum_n \lambda(n) n^{-s}$$ – reuns Mar 9 '16 at 20:57
Suppose that for each positive integer $n$ you flip a coin that has probability $1/\log n$ of coming up "prime". Then the expected number of primes up to $n$ would be $n/\log n$. But there would be some variation around this expected value – some "error term". The size of the error term would be predicted by The Law of Large Numbers. And it would be the same as what's predicted for actual primes by the Riemann Hypothesis.
• Nothing is outside the realm of Mathematics, least of all randomness. Think: Probability Theory. – Gerry Myerson Feb 28 '16 at 11:42
• Depends on what you mean by "random number generator" and what you mean by "mathematical formalisms". – Gerry Myerson Feb 28 '16 at 11:45
• You seem to be equating "create" with "define". How do you propose to define randomness, if not in mathematical terms? – Gerry Myerson Feb 28 '16 at 22:09
• See Knuth, Seminumerical Algorithms (Volume 2 of The Art Of Computer Programming) for an entertaining discussion of defining "random". But I think I don't agree that there's no formal definition of randomness. And even if it's true, well, there's no formal definition of "point", "line", or "plane", but that doesn't stop us from doing geometry. They may be undefined terms, but they do have precise relations amongst themselves, and that's good enough for Mathematics. – Gerry Myerson Mar 1 '16 at 5:05
• @A.S., at least one of us doesn't know what you're talking about. – Gerry Myerson Mar 3 '16 at 9:50
The Riemann hypothesis is equivalent to the statement that, for all $\epsilon \gt 0$:
$\sum_{1 \le k \le n}{\lambda(k)} = O(n^{\frac{1}{2} + \epsilon})$
where $\lambda$ is the Liouville function, which takes values $\pm 1$ depending on whether its argument has an even or odd number of prime factors. If instead of evaluating this function we were flipping a coin and counting heads as $+1$ and tails as $-1$, this is true almost surely by the law of the iterated logarithm.
• Interesting. Is there a more detailed asymptotics for LHS? Sth like LIL or convergence to a normal? – A.S. Mar 2 '16 at 22:57
• There are lots of results about the related Mertens function. – Dan Brumleve Mar 2 '16 at 23:08
I took it as a general statement, another professor said they were like weeds. If you fill a wall with numbers and highlighted the prime numbers it would look random. There are several pictures and charts even circles on the internet showing prime number relationships. | {"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": 1, "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": 0.8167355060577393, "perplexity": 365.0161251187036}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00071.warc.gz"} |
https://www.esaral.com/q/express-the-given-complex-number-in-the-form-a-ib-65422/ | Express the given complex number in the form a + ib:
Question:
Express the given complex number in the form $a+i b:(5 i)\left(-\frac{3}{5} i\right)$
Solution:
$(5 i)\left(\frac{-3}{5} i\right)=-5 \times \frac{3}{5} \times i \times i$
$=-3 i^{2}$
$=-3(-1) \quad\left[i^{2}=-1\right]$
$=3$ | {"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": 0.9927429556846619, "perplexity": 473.60189656450814}, "config": {"markdown_headings": false, "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-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00184.warc.gz"} |
https://physics.stackexchange.com/questions/397190/what-does-it-mean-by-universe-is-expanding-i-know-that-the-universe-is-expandin | What does it mean by universe is expanding? I know that the universe is expanding because spacetime is expanding but again what does it mean?
I am aware about the balloon expanding analogue, where the galaxies are stuck on the balloon surface and here the balloon is our space time and as the balloon expand our galaxies also go apart.
But here is my doubt.
1. when we say universe is expanding, we say spacetime is expanding, at the same time we say that due to this the galaxies can go apart even greater than the speed of light, since its not the galaxies are moving but the spacetime itself is moving.
2. We know that the universe is expanding by looking at the doppler shift. but again the velocity of source we use in this formula is the actual relative velocity of source so if the source speed is higher than speed of light then how is the use of doppler formula possible?
• "when we say universe is expanding, we say spacetime is expanding" - not spacetime, we say space is expanding. For details, take a look at Metric expansion of space – Hal Hollis Apr 2 '18 at 13:22
• I guess in many place its considered as spacetime (may be a misinterpret) BTW does the term spacetime expansion makes any sense? also if I am not wrong space and time are interlinked (just read it in some book i don't remember) – shinigami Apr 2 '18 at 14:10
• Possible duplicate of Is the universe actually expanding? – Stéphane Rollandin Apr 2 '18 at 15:17
Say some cosmic object spews forth a certain number of photons of a certain frequency, which make their way towards an observer.
The photon flux is a measure of distance: Assuming isotropy, it will decrease with $1/r^2$, constant across a growing spherical surface. The photon frequency is a measure of relative velocity, as evaluated by parallel transport through a curved spacetime.
In Friedmann universes specifically, spacetime can be sliced into spatial hypersurfaces of constant cosmological time, where the distance between 'stationary' points evolves according to a global scale factor. This distance can increase arbitrarily fast, and such 'recession velocities' can be greater than $c$. We can look beyond the Hubble sphere (the place where recession velocities hit the speed of light) without a problem. In contrast, relative velocities will always be smaller than $c$: When relative velocities approach the speed of light, redshift goes to infinity, and we cannot look beyond this cosmological event horizon.
Regarding your second point, note that you cannot just plug-in recession velocities into Doppler's formula and expect the correct answer: There's no distance parallelism in arbitrary spacetimes, and you'd have to integrate infinitesimal Doppler shifts along the photon trajectory. In Friedmann universes, this will yield a simple answer: Cosmological redshift between comoving emitters and observers ends up being anti-proportional to the scale factor at time of emission.
Some historic remarks, prompted by the comments:
Einstein was looking for a relativistic theory of gravity, inventing general relativity. Friedmann then derived a certain class of maximally symmetric solutions to the Einstein equations, described by the FLRW metric and evolving accordig to the Friedmann equations. Einstein wasn't a fan, though, because he believed in a static univere (which necessitated the introduction of the cosmological constant).
Lemaître re-discovered these solutions, and, among other things, derived cosmological redshift. However, this calculation was omitted from the English translation of his paper, which meant few people where aware of it when Hubble derived the law that now bears his name heuristically: He (incorrectly) interpreted cosmological redshift special-relativistically, when the general-relaivistic interpretation had in fact already been proposed...
• I have two more doubt now – shinigami Apr 2 '18 at 13:58
• (a) lets assume a hypothetical case if a galaxy is having a relative velocity and another galaxy is in rest but is moving due to space expansion -- how will we distinguish between that? (b) Did the idea of space expansion was invented for covering up for the fact that few galaxies might have velocities greater than 'c' - which is forbidden by theory of relativity? – shinigami Apr 2 '18 at 14:04
• re (a), you've got a couple of quantities you can use to derive distances (eg photon flux, angular diameter), and redshift as a measure of relative motion (including all types ot 'motion' such as spatial expansion or 'standing still' in a gravity well); they all have to be consistent, and must fit related experimental quantities (eg intensity) – Christoph Apr 2 '18 at 15:05
• re (b), see edit: the evolution of maximally symmetric universes had been calculated from general relativity before observational evidence was found; once the observations were in, we just went with the model that fit the data (though we 'recently' had to re-introduce the cosmological constant to account for accelerated expansion) – Christoph Apr 2 '18 at 15:05
• so if the readings of distances and velocities don't match then its a Space expantion? right – shinigami Apr 2 '18 at 15:14 | {"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": 0.9157475233078003, "perplexity": 554.7324364650907}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00197.warc.gz"} |
https://www.physicsforums.com/threads/photon-energy.781342/ | # Photon energy
Tags:
1. Nov 11, 2014
### Gabriel Hoshino
I was reading a section of a chemistry textbook describing electron energy shells. It compares the electrons to light saying that electrons energies are quantized and so are light energies. Electrons can only jump from one specific energy level to another with no intermediary energy levels. I understand that the same is true for the intensity of light, but I still don't understand how the energy of a photon can have only specific amounts of energy like an electron. Aren't there an infinite number of intermediary frequencies in between two frequencies of light? If that is true than doesn't that mean that the energy of photons doesn't skip around like the energy of electrons? Thanks.
2. Nov 11, 2014
### Staff: Mentor
When light of a particular frequency interacts with anything, it transfers its energy in discrete amounts that are proportional to that frequency. We call this discrete chunks "photons".
3. Nov 11, 2014
### Gabriel Hoshino
But can photons have any energy, or are they like electrons where they can only skip between energy states?
4. Nov 11, 2014
### Staff: Mentor
In general, photons can have any energy. The light emitted by an incandescent light bulb contains photons with a continuous range of energies. Photons that are emitted by transitions between atomic energy levels in a specific type of atom (e.g. in a gas-discharge tube) can have only energies that equal the difference between two of the energy levels.
5. Nov 11, 2014
### Staff: Mentor
Any given photon has only one energy (setting aside for now the phenomenon of gravitational redshift). You can absorb it and then emit another photon with a different energy, but you can't just change the energy of the same photon.
(I should add that we're on very shaky ground even talking about "the same photon" - these aren't like little teeny grains of sand with a distinct existence of their own)
6. Nov 11, 2014
### anorlunda
That is what I thought too, but the Wikipedia article on CMB says, "The photons that existed at the time of photon decoupling have been propagating ever since, though growing fainter and less energetic, since the expansion of space causes their wavelength to increase over time". I have been having trouble understanding the CMB on a per photon basis.
7. Nov 11, 2014
### Staff: Mentor
That's one of the gravitational red shift cases that I didn't want to mess with. :)
In fact, that wikipedia article might be improved if it didn't use the word "photon" in that context, just spoke of radiation propagating outwards and being redshifted.
8. Nov 11, 2014
### Gabriel Hoshino
Awesome, thanks for all of your responses!
Similar Discussions: Photon energy | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8874467015266418, "perplexity": 808.6870883887527}, "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-2017-47/segments/1510934803944.17/warc/CC-MAIN-20171117204606-20171117224606-00393.warc.gz"} |
https://www.physicsforums.com/threads/angular-momentum-impulse-problem.492310/#post-3259576 | # Angular Momentum/Impulse Problem
• Start date
• #1
14
0
1. A 1.8 kg, 20 cm diameter turntable rotates at 160 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?
2. Angular momentum(L)=angular velocity(w)*moment of inertia(I)
3. I'm confused as to whether or not to appraoch this as a conservation of momentum problem or some other way. I think I would need to know the velocities of the blocks before they hit to to this.
• #2
olivermsun
1,248
118
Do the blocks carry with them any angular momentum of their own? (It seems the problem is set up to hint that they do not).
• #3
14
0
Thanks! thats all I needed!
• #4
tiny-tim
Homework Helper
25,836
251
hi cdbowman42!
I think I would need to know the velocities of the blocks before they hit to to this.
hint: what was their angular momentum about the axis, before they hit?
(assuming they fell vertically at speed v)
• #5
283
0
I would think about it in the following way
$$\int \tau dt = \Delta ( I \omega)$$
where $$\int \tau dt$$ is the angular impulse and $$\Delta (I \omega)$$ is the change in angular momentum.
We can see that no vertical torque is exerted on the turntable, so the angular momentum must remain constant.
Then we must have
$$(I \omega)_{initial} = (I \omega)_{final}$$
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3K | {"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": 0, "mathjax_display_tex": 1, "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": 0.9522100687026978, "perplexity": 1803.4576859239503}, "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-2021-43/segments/1634323585696.21/warc/CC-MAIN-20211023130922-20211023160922-00014.warc.gz"} |
https://www.esaral.com/q/the-exact-volumes-of-88550/ | The exact volumes of
Question:
The exact volumes of $1 \mathrm{MNaOH}$ solution required to neutralise $50 \mathrm{~mL}$ of $1 \mathrm{MH}_{3} \mathrm{PO}_{3}$ solution and $100 \mathrm{~mL}$ of $2 \mathrm{MH}_{3} \mathrm{PO}_{2}$ solution, respectively, are:
1. $100 \mathrm{~mL}$ and $100 \mathrm{~mL}$
2. $100 \mathrm{~mL}$ and $50 \mathrm{~mL}$
3. $100 \mathrm{~mL}$ and $200 \mathrm{~mL}$
4. $50 \mathrm{~mL}$ and $50 \mathrm{~mL}$
Correct Option: , 3
Solution:
$\mathrm{H}_{3} \mathrm{PO}_{3}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{HPO}_{3}+2 \mathrm{H}_{2} \mathrm{O}$
$50 \mathrm{ml}^{3} \quad 1 \mathrm{M}$
$1 \mathrm{M} \quad \mathrm{V}=$ ?
$\Rightarrow \frac{\mathrm{n}_{\mathrm{Nao} \mathrm{H}}}{\mathrm{n}_{\mathrm{H}_{1} \mathrm{PO}_{3}}}=\frac{2}{1}$
$\Rightarrow \frac{1 \times \mathrm{V}}{50 \times 1}=\frac{2}{1} \Rightarrow \mathrm{V}_{\mathrm{NaOH}}=100 \mathrm{ml}$
$\mathrm{H}_{3} \mathrm{PO}_{2}+2 \mathrm{NaOH} \rightarrow \mathrm{NaH}_{2} \mathrm{PO}_{3}+\mathrm{H}_{2} \mathrm{O}$
$100 \mathrm{ml}$
2M
$\mathrm{V}=?$ | {"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": 0.9723479151725769, "perplexity": 3221.422160037423}, "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-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00430.warc.gz"} |
http://essayswriting.info/what-is-square-root-of-324/ | # What Is Square Root Of 324
What Is Square Root Of 324. Free online scientific notation calculator. The steps to obtain the square root of 324 using the long division method is provided as follows:
The square root of 648 square feet is about 25.46 feet. What is the length of the sides of a square with and area of 324 centimeters squared? The square root of a number is a number which, when multiplied by itself, results in the original number.
### What Is The Square Number Of Eighteen?
Sqrt(32) factor 32 into its prime factors 32 = 25 to simplify a square root, we extract factors which are squares, i.e.,. 4 • sqrt(2) simplify : Free online scientific notation calculator.
### First Have A Grip On The Square Table.
You can find it using square root division. 0 in a quick and easy way with step by step explanation. The symbol √ is called radix, or radical.
### 1, 4, 9, 36, 81, 324 Since 324 Is On The List Of Perfect Squares Above, It Means That 324 Is A Perfect Square.
Rewrite 324 324 as 182 18 2. A square root of a number 'a' is a number x such that x 2 = a, in other words, a number x whose square is a. Utilize the square root calculator to find the square root of number 324 i.e.
### Following Is The Process To Find The Square Root Of 324 By Prime Factorization Method.
324 18 ⋅ 18 18. Pull terms out from under the radical, assuming positive real. $\rightarrow \sqrt {324} = 2 \times 3 \times 3$ again the right side of the equation on solving it can be written as $\rightarrow \sqrt {324} = 18$ therefore, the square root of $324$ will be.
### Square Root Of 324 = Square Root Of (81 X 4) = Square Root Of 81 X Square Root Of 4.
(this link will show the same work that you can see on this page) you can calculate the square root of any number , just change 324. The square root of 648 square feet is about 25.46 feet. To find the square root of 169, you have to find a number that can be multiplied by itself in order to get the number 169 as the product. | {"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": 0.8360146284103394, "perplexity": 246.59304110146513}, "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-2022-40/segments/1664030337415.12/warc/CC-MAIN-20221003101805-20221003131805-00147.warc.gz"} |
http://healthy-food-life.com/optimazion-4/ | Optimazion
Optimazion is a branch of mathematics that is concerned with minimizing a function’s cost. In this branch, we look for the smallest value that the function can take and find the greatest value for the constraint. The objective function is an equation that must hold no matter what the solution is. Hence, we look for a clear and constant quantity. Once we know what that quantity is, we can easily solve optimization problems.
Typically, a problem involving optimization involves identifying certain extrema. To do this, we need to look at the first derivative of the parent function. The zeros in the parent function correspond to critical points. Using the description of the problem, we can determine the parent function. Once we know the parent function, we can find the maxima. We can also identify extrema by looking at the values of the simplex vertices.
One method of determining optimum solutions is the Lagrange multiplier method. This technique allows us to identify optimal solutions to problems that are constrained by either inequality or equality. We can also use the Karush-Kuhn-Tucker condition to determine the best possible solution. A problem involving optimisation is always a complex problem, which has many possible solutions. This method is based on a mathematical algorithm that is able to determine the best solution in a given problem.
Another approach to finding optimum solutions is to identify certain extrema. The first derivative is a critical point of a function, and the blue dot on it corresponds to the maxima of the parent function. Once you know this, you can use the Karush-Kuhn-Tucker conditions to find a best solution. This technique is more efficient than the other two methods, but it is not as flexible as the first one.
However, a typical optimization problem involves identifying certain extrema. This is done by examining the first derivative of the parent function. The first derivative contains a zero, which corresponds to the maximum and minima of the problem. Then, you can calculate the first extrema of the parent function by considering the problem description. Then, you can calculate the optimal solutions using the Karush-Kuhn-Tucker conditions.
Another way to find an optimal solution is to look for a function with an extreme and a minimum. In other words, we want to find a function that satisfies the criteria of the objective. Then, we need to find the minimum and maximum of this function. Usually, these converge to the same value. For example, a candidate can win if he or she gets the most votes.
Students who are not familiar with the definition of optimazion should read this material before learning it. Usually, students get locked into one solution and try to make every optimization problem conform to the same solution. This is wrong. They should try to find solutions in the context of the problem. So, they need to be clear about what the optimal goal is before solving an optimization problem. They should also understand how constraints can be applied. This way, they can make the best decisions. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8179314732551575, "perplexity": 124.38366639617281}, "config": {"markdown_headings": false, "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-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00334.warc.gz"} |
https://www.physicsforums.com/threads/computing-arc-length-and-cannot-solve-the-integral.573719/ | # Computing Arc Length and Cannot Solve the Integral
1. Feb 3, 2012
### Voodoo583
1. The problem statement, all variables and given/known data
Find the length of the polar curve.
Complete the cardioid r=1+cosθ
2. Relevant equations
L=∫αβ√[f(θ)2+f'(θ)2] dθ
3. The attempt at a solution
Given f(θ)2 is equal to cos2θ+2cosθ+1
and f'(θ)2=sin2θ
I arrive at the integral ∫αβ√[2cosθ+2] dθ which I cannot for the life of me solve for. It's been a while since my calc II class so I flipped through my textbook to see if I could find a solution. I cannot do u substitution as far as I can tell nor integration by parts and I couldn't find anything similar to the form above in the table of integrals.
2. Feb 3, 2012
### tiny-tim
Similar Discussions: Computing Arc Length and Cannot Solve the Integral | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9864848256111145, "perplexity": 527.1350456723649}, "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-2018-09/segments/1518891815843.84/warc/CC-MAIN-20180224152306-20180224172306-00762.warc.gz"} |
https://export.arxiv.org/abs/2009.00469?context=cs.IT | cs.IT
(what is this?)
# Title: Precise Expression for the Algorithmic Information Distance
Authors: Bruno Bauwens
Abstract: We consider the notion of information distance between two objects $x$ and $y$ introduced by Bennett, G\'acs, Li, Vit\'anyi, and Zurek in 1998 as the minimal length of a program that computes $x$ from $y$ as well as computing $y$ from $x$. In this paper, it was proven that the distance is equal to $\max (K(x|y),K(y|x))$ up to additive logarithmic terms, and it was conjectured that this could not be improved to $O(1)$ precision. We revisit subtle issues in the definition and prove this conjecture. We show that if the distance is at least logarithmic in the length, then this equality does hold with $O(1)$ precision for strings of equal length. Thus for such strings, both the triangle inequality and the characterization hold with optimal precision. Finally, we extend the result to sets $S$ of bounded size. We show that for each constant~$s$, the shortest program that prints an $s$-element set $S \subseteq \{0,1\}^n$ given any of its elements, has length at most $\max_{w \in S} K(S|w) + O(1)$, provided this maximum is at least logarithmic in~$n$.
Comments: arXiv admin note: substantial text overlap with arXiv:1807.11087 Subjects: Information Theory (cs.IT) Cite as: arXiv:2009.00469 [cs.IT] (or arXiv:2009.00469v1 [cs.IT] for this version)
## Submission history
From: Bruno Bauwens [view email]
[v1] Sun, 30 Aug 2020 22:26:18 GMT (80kb,D)
Link back to: arXiv, form interface, contact. | {"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": 0.9014956951141357, "perplexity": 798.7673718100712}, "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-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00392.warc.gz"} |
https://www.physicsforums.com/threads/derivative-of-sin-3x.78051/ | # Derivative of sin(3x)
1. Jun 5, 2005
### mattmns
Is the derivative of sin(3x) just 3sin(3x) because of the chain rule? IE, let u=3x, then 3sin(u) => 3sin(3x)
Thanks.
Ok, I am pretty sure that is true. How about, $$\lim_{t\rightarrow 0} tln(t)$$
What is the common approach for this problem?
Last edited: Jun 5, 2005
2. Jun 5, 2005
### OlderDan
It's not true because the derivative of sine is not itself
For the limit, you might look for a minimum of the function. Does it have one? More than one? That should get you started.
3. Jun 5, 2005
### p53ud0 dr34m5
i agree with OlderDan.
$$h(x)=f(g(x))$$
if you have that, then:
$$h'(x)=f'(g(x))*g'(x)$$
so, sine is not the derivative of itself.
4. Jun 5, 2005
### mrjeffy321
I think the derivative of sine is cosine,
so the derivative of sin(3x) would be 3cos(3x)
5. Jun 5, 2005
### mattmns
Bah, that is what I meant, sorry. Thanks, been a while since I did a derivative or a limit | {"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": 0, "mathjax_display_tex": 1, "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": 0.9677594304084778, "perplexity": 3578.1491432427906}, "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-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00306-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathoverflow.net/users/930/js-milne?tab=activity&sort=all&page=2 | # JS Milne
less info
reputation
2239
bio website jmilne.org/math location Ann Arbor, MI, USA, and New Zealand. age member for 5 years seen Oct 18 '10 at 13:35 profile views 6,492
Arithmetic geometry (especially Shimura varieties and abelian varieties).
# 193 Actions
Jul7 awarded Nice Answer Apr29 awarded Nice Answer Nov10 awarded Pundit Oct30 awarded Nice Answer Oct28 awarded Good Answer Oct22 awarded Yearling Oct18 comment Are there any good computer programs for drawing (algebraic) curves? If you have access to Mathematica and it produces the diagram you want, fine, but for more control use gnuplot+tikz. Oct17 comment Awfully sophisticated proof for simple facts Brauer groups and cohomology are certainly overkill for Wedderburn's theorem: if $D$ is a finite division algebra and $L$ is a maximal subfield, then the Noether-Skolem theorem shows that the multiplicative group of $D$ is a union of conjugates of that of $L$; hence $D$=$L$. Oct14 comment Chapters 1--4 of the Artin-Tate notes on Class Field Theory As far as I know, the notes for the first part of the seminar were never written up. Lang missed this part of the seminar because he started as a philosophy student. Oct7 comment who fixed the topology on ideles? Yes, I think the answer is Weil. In his short 1936 paper "Remarques sur des resultats recent de C. Chevalley" he complains about Chevalley's topology, and writes down other constructions and topologies (and mentions Grossencharacters) but doesn't write down anything immediately recognizable (to me) as the ideles. In his 1951 paper (J. Math. Soc. Japan) he defines without comment the ideles with the natural (modern) topology. That is also where he wrote "La recherche d'une interpretation pour C_k ... me semble constituer l'un des problemes fondamentaux..." See also his Commentaries in his CW. Oct3 comment Is a torsion free abelian group finitely generated, if all of its localizations at primes p are finitely generated over Zp? Have you seen the erratum jmilne.org/math/CourseNotes/errata.html#AV ? Sep24 answered Math keyboard: does it exist ? Sep21 comment Are there motives which do not, or should not, show up in the cohomology of any Shimura variety? Blasius has pointed out that the naive generalization of the modularity conjecture fails --- there exist elliptic curves over number fields that are not quotients of the albanese of any Shimura variety --- but I don't know of any reason why the more general version (4) can't be true. (Blasius 2004 MR2058605). Sep11 comment What heuristic evidence is there concerning the unboundedness or boundedness of Mordell-Weil ranks of elliptic curves over $\Bbb Q$? Cassels said it best (JLMS 1966, p.257):... it has been widely conjectured [on the basis of calculations] that there is an upper bound for the rank depending only on the groundfield. This seems to me implausible because the theory makes it clear that an abelian variety can only have high rank if it is defined by equations with very large coefficients. . (For there must be a lot of alternative factorizations to be possible in the arguments of §24.) Sep10 comment Is the category of commutative group schemes abelian? Thanks - fixed. Sep10 revised Is the category of commutative group schemes abelian? added 1041 characters in body Sep10 awarded Nice Answer Sep9 answered Is the category of commutative group schemes abelian? Sep1 comment Is there a classification of embeddings of SL_2 into SP_6 as algebraic groups over Q and R respectively? Assume the base field has characteristic zero. The category of representations of a semisimple Lie algebra is Tannakian, and the algebraic group attached to this category is the simply connected algebraic group with the given Lie algebra. A homomorphism of Lie algebras defines a tensor functor of the Tannakian categories, and hence a homomorphism of the corresponding simply connected algebraic groups. Aug31 comment Why aren't there more classifying spaces in number theory? Continued: It was then found that $H^{1}$ coincided with the group of crossed homomorphisms modulo principal crossed homomorphisms, and $H^{2}$ with the group of equivalence classes of "factor sets", which had been introduced much earlier (e.g., I. Schur, \"{U}ber die Darstellung der endlichen$\ldots$ , 1904; O. Schreier, \"{U}ber die Erweiterungen von Gruppen, 1926; R. Brauer, \"{U}ber Zusammenh\"{a}nge$\ldots$ , 1926). For more on the history, see MacLane 1978 (Origins of the cohomology of groups. Enseign. Math. (2) 24 (1978), no. 1-2, 1--29. MR0497280). | {"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": 0.8206623196601868, "perplexity": 998.1679855425668}, "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-2014-42/segments/1414637898842.15/warc/CC-MAIN-20141030025818-00168-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/32815/is-it-feasible-for-an-unmanned-vehicle-to-travel-from-outside-the-atmosphere-of/32819 | # Is it feasible for an unmanned vehicle to travel from outside the atmosphere of one planet to another without additional propulsion?
Within the Solar System (or any other system for that matter) - Is it feasible for an unmanned vehicle to travel from outside the atmosphere of one planet to another without additional propulsion? That is to say
Given a vehicle in orbit around a planet, say Mercury, suppose a body were launched with the destination Earth with the caveat that it have sufficient velocity to travel the distance so it reaches Earth at velocity necessary to be at Geosynchronous orbit.
Could this be achieved merely by using launch velocity so that the launched body does not need carry it's own propulsion?
-
Without a third body to interact with at the arrival point, then "no", it is not possible to take up orbit when you get there unless you have some thrust (but I'll say a little more about the weasel words in a minute).
The basic problem is that you arrive (by definition from far away) on a hyperbolic orbit and you will, therefore leave on a hyperbolic orbit.
Exceptions:
• You hit the target (that's Richard's "lithobraking") or at least it's atmosphere (areobraking) and lose a lot of energy that way.
• You are going "uphill" and can lose the hyperbolic energy to the Sun's gravity. (I've not actually seen a complete calculation for this, but I think that the mean value theorem requires it to be possible).
• There is a third (or more) body nearby and you can trade some energy with it (them). Again, I haven't seen any calculation and they will be tedious and exacting.
• You have some kind of thrust that is discounted by the question. Perhaps a solar sail.
-
The uphill (mercury-uranus) was merely for illustration... – Everyone Jul 25 '12 at 16:26
Is it possible to define the hyperbolic orbit and velocity at origin such that it experiences lithobraking at target? – Everyone Jul 25 '12 at 16:33
In a word, yes. The concept of the mass driver has this kind of launch in mind. All you need to do is accelerate the vehicle to the escape velocity at the point of launch (in fact, you don't even need to do that - you can simply put the vehicle into a sufficiently elliptical orbit that intersects with the target planet at apoapsis - for infinitely distant targets the velocity required to attain this orbit converges to the escape velocity). What you do when you reach the target planet (get gravitationally captured, crash lithobrake, adjust orbit with thrust, etc) is up to you.
Bear in mind that most space probes would spend almost all of their time in freefall (that is, not under thrust). Ion thrusters are an important exception as far as I know.
-
Suppose the source planet were mercury, and the target Uranus - Would such a vehicle be subject to gravitic perturbation by unavoidable bodies along it's trajectory? – Everyone Jul 25 '12 at 10:18
@Everyone - Yes, but space missions are planned with interactions with other bodies in mind. Also bear in mind that the vehicle is possibly going to be in some kind of orbit around the sun. – Richard Terrett Jul 25 '12 at 10:33
It's also achieved by Martian meteorites on Earth for example. – Martin Beckett Jul 25 '12 at 14:25 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8439755439758301, "perplexity": 771.3217155958666}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2014-41/segments/1410657130272.73/warc/CC-MAIN-20140914011210-00323-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.authorea.com/doi/full/10.22541/au.164924641.18668733 | In this paper, we considered the global regularity for the 2D incompressible anisotropic magnetic B\’{e}nard system with fractional partial dissipation. More precisely, we established the global existence and regularity for the 2D incompressible anisotropic magnetic B\’{e}nard system with only vertical hyperdiffusion $\Lambda_{2}^{2\beta}b_1$ and horizontal hyperdiffusion $\Lambda_{1}^{2\beta}b_2$ and $(-\Delta)^{\alpha}\theta$, where $\Lambda_{1}$ and $\Lambda_{2}$ are directional Fourier multiplier operators with the symbols being $|\xi_1|$ and $|\xi_2|$, respectively. We prove that, for $\beta>1$ and $0<\alpha<1$, this system always possesses a unique global-in-time classical solution when the initial data is sufficiently smooth. | {"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": 0.9641759395599365, "perplexity": 215.99155117323232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2023-14/segments/1679296948867.32/warc/CC-MAIN-20230328135732-20230328165732-00156.warc.gz"} |
http://terrytao.wordpress.com/tag/adjoint-operator/ | You are currently browsing the tag archive for the ‘adjoint operator’ tag.
When studying a mathematical space X (e.g. a vector space, a topological space, a manifold, a group, an algebraic variety etc.), there are two fundamentally basic ways to try to understand the space:
1. By looking at subobjects in X, or more generally maps $f: Y \to X$ from some other space Y into X. For instance, a point in a space X can be viewed as a map from $pt$ to X; a curve in a space X could be thought of as a map from ${}[0,1]$ to X; a group G can be studied via its subgroups K, and so forth.
2. By looking at objects on X, or more precisely maps $f: X \to Y$ from X into some other space Y. For instance, one can study a topological space X via the real- or complex-valued continuous functions $f \in C(X)$ on X; one can study a group G via its quotient groups $\pi: G \to G/H$; one can study an algebraic variety V by studying the polynomials on V (and in particular, the ideal of polynomials that vanish identically on V); and so forth.
(There are also more sophisticated ways to study an object via its maps, e.g. by studying extensions, joinings, splittings, universal lifts, etc. The general study of objects via the maps between them is formalised abstractly in modern mathematics as category theory, and is also closely related to homological algebra.)
A remarkable phenomenon in many areas of mathematics is that of (contravariant) duality: that the maps into and out of one type of mathematical object X can be naturally associated to the maps out of and into a dual object $X^*$ (note the reversal of arrows here!). In some cases, the dual object $X^*$ looks quite different from the original object X. (For instance, in Stone duality, discussed in Notes 4, X would be a Boolean algebra (or some other partially ordered set) and $X^*$ would be a compact totally disconnected Hausdorff space (or some other topological space).) In other cases, most notably with Hilbert spaces as discussed in Notes 5, the dual object $X^*$ is essentially identical to X itself.
In these notes we discuss a third important case of duality, namely duality of normed vector spaces, which is of an intermediate nature to the previous two examples: the dual $X^*$ of a normed vector space turns out to be another normed vector space, but generally one which is not equivalent to X itself (except in the important special case when X is a Hilbert space, as mentioned above). On the other hand, the double dual $(X^*)^*$ turns out to be closely related to X, and in several (but not all) important cases, is essentially identical to X. One of the most important uses of dual spaces in functional analysis is that it allows one to define the transpose $T^*: Y^* \to X^*$ of a continuous linear operator $T: X \to Y$.
A fundamental tool in understanding duality of normed vector spaces will be the Hahn-Banach theorem, which is an indispensable tool for exploring the dual of a vector space. (Indeed, without this theorem, it is not clear at all that the dual of a non-trivial normed vector space is non-trivial!) Thus, we shall study this theorem in detail in these notes concurrently with our discussion of duality. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 14, "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": 0.8838979005813599, "perplexity": 255.04306458580876}, "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-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00106-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://yak14.ru/exponential-decay-radiocarbon-dating-17453.html | Carbon dating has given archeologists a more accurate method by which they can determine the age of ancient artifacts.
Natasha Glydon Exponential decay is a particular form of a very rapid decrease in some quantity.
One specific example of exponential decay is purified kerosene, used for jet fuel.
A fossil found in an archaeological dig was found to contain 20% of the original amount of 14C. I do not get the $-0.693$ value, but perhaps my answer will help anyway.
If we assume Carbon-14 decays continuously, then $$C(t) = C_0e^,$$ where $C_0$ is the initial size of the sample. Since it takes 5,700 years for a sample to decay to half its size, we know $$\frac C_0 = C_0e^,$$ which means $$\frac = e^,$$ so the value of $C_0$ is irrelevant.
If feet of pipe can be represented by the following equation: Suppose that the pollutants must be reduced to 10% in order for the kerosene to be used for jet fuel.
How long does the pipe have to be to ensure that there is only 10% of the pollutants left in the kerosene?
The amount of Carbon 14 contained in a preserved plant is modeled by the equation $$f(t) = 10e^.$$ Time in this equation is measured in years from the moment when the plant dies ($t = 0$) and the amount of Carbon 14 remaining in the preserved plant is measured in micrograms (a microgram is one millionth of a gram).
The stable form of carbon is carbon 12 and the radioactive isotope carbon 14 decays over time into nitrogen 14 and other particles.
Carbon is naturally in all living organisms and is replenished in the tissues by eating other organisms or by breathing air that contains carbon.
Students should be guided to recognize the use of the logarithm when the exponential function has the given base of $e$, as in this problem.
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https://socratic.org/questions/how-do-you-graph-y-x-2-5x-36-3x-using-asymptotes-intercepts-end-behavior | Precalculus
Topics
# How do you graph y=(x^2-5x-36)/(3x) using asymptotes, intercepts, end behavior?
Dec 1, 2016
See the explanation below
#### Explanation:
The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{0\right\}$
As $x \ne 0$
So, $x = 0$ is a vertical asymptote.
The numerator is ${x}^{2} - 5 x - 36$
$= \left(x + 4\right) \left(x - 9\right)$
So, the intercepts when $y = 0$
are $\left(- 4 , 0\right)$ and $\left(9 , 0\right)$
For the limits $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator
${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} {x}^{2} / \left(3 x\right) = {\lim}_{x \to \pm \infty} \frac{x}{3} = \pm \infty$
${\lim}_{x \to {0}^{-}} y = \frac{{0}^{+} + 5 - 36}{3 \cdot {0}^{-}} = + \infty$
${\lim}_{x \to {0}^{+}} y = \frac{{0}^{+} - 5 - 36}{3 \cdot {0}^{+}} = - \infty$
Now you can draw your graph
graph{(x^2-5x-36)/(3x) [-32.47, 32.47, -16.24, 16.25]}
Dec 1, 2016
Asymptotes: $x = 0 \mathmr{and} x - 3 y - \frac{5}{3} = 0.$ x-intercepts by the two branches: -4 and 9. As $x \to 0 , y \to \pm \infty$ and the other asymptote tends to reach the hyperbola, at infinite distance in ${Q}_{1} \mathmr{and} {Q}_{3}$
#### Explanation:
graph{x^2-3xy-5x-36=0 [-20, 20, -10, 10]} The second degree equation $a {x}^{2} + 2 h x y + b {y}^{2} + \ldots = 0$
represents a hyperbola, when $a b - {h}^{2} < 0$ and the second degree
terms reveal the first degree terms in the equations of its
asymptotes.
Here, after cross multiplication,
$x \left(x - 3 y\right) - 5 x - 36 = 0$. Reconstructing as
((x+a)(x-3y+b)+c=0, we find that
$a = 0 , b = - \frac{5}{3} \mathmr{and} c = - 36$.
So, the asymptotes are given by
x=0 and x-3y-5/3=0, meeting at the center C(0, -5/9)#.
x-intercepts by the two branches: -4 and 9.
As $x \to 0 , y \to \pm \infty$ and the other asymptote tends to reach the
hyperbola, at infinite distance in ${Q}_{1} \mathmr{and} {Q}_{3}$
##### Impact of this question
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http://physics.stackexchange.com/questions/92832/why-is-photon-annihilation-associated-with-the-positive-frequency-component-of-t | Why is photon annihilation associated with the POSITIVE frequency component of the electric field?
I'm reading Glauber's paper "The quantum theory of optical coherence". In his work he does not introduce the annihilation and creation operators, but he refers instead to the positive and negative frequency components of the electric field (which probably are proportional to the former, I guess).
He says that "...the positive frequency part $E^{(+)}(\mathbb{r}t)$, may be shown to be a photon annihilation operator..." and then cites Dirac book "The principles of quantum mechanics", 3rd ed. pp 239-242, but even from the reference I couldn't understand why the positive frequency components is associated with the photon annihilation.
PS with the capitalization I don't mean to yell but to stress on the term "positive".
-
If you look at a classical field, you have $2$ possibilities for a field "corresponding" to a positive energy $\omega_k$ : $\Phi_{\pm(x,t)} = e^{\pm i(\omega_kt-\vec k.\vec x)}$. Now, you have to choose a standard, and you may think to the Schrodinger equation where the energy operator is $i \frac{\partial}{\partial t}$ (in units $\hbar=1$). So the standard solution, for a positive energy $\omega_k$ is $e^{- i(\omega_kt-\vec k.\vec x)}$
Now, we want to retrieve some taste of these classical fields by looking at the action of the field operator $A_\mu(x)$ sandwiched between a one-particle state $|k, \lambda\rangle = a^\dagger(k,\lambda) |0\rangle$ and the vaccuum, so we want :
$\epsilon^\lambda_\mu(k) e^{- i(\omega_kt-\vec k.\vec x)} = \langle 0|A_\mu(x)|k, \lambda\rangle$
One sees, that, the decomposition :
$A_\mu(x) = \int d \tilde k \sum\limits_\lambda(\epsilon^\lambda_\mu(k) a(k,\lambda) e^{- i(\omega_kt-\vec k.\vec x)} + (\epsilon^\lambda_\mu(k))^* a^\dagger(k,\lambda) e^{+ i(\omega_kt-\vec k.\vec x) })$
does the job (using the properties of the creation/annihilation operators).
So, annihilation operators are associated to positive energies.
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https://www.neetprep.com/ncert/1485-Mechanical-Properties-Fluids-Mechanical-Properties-Fluids--NCERT-Chapter-PDF | # Chapter Ten
Knowledge of mathematical tools, mechanics is a prerequisite for this chapter
## Mechanical Properties of Fluids
10.1 Introduction
10.2 Pressure
10.3 Streamline flow
10.4 Bernoulli’s principle
10.5 Viscosity
10.6 Reynolds number
10.7 Surface tension
Summary
Points to ponder
Exercises
Appendix
#### 10.1 Introduction
In this chapter, we shall study some common physical properties of liquids and gases. Liquids and gases can flow and are therefore, called fluids. It is this property that distinguishes liquids and gases from solids in a basic way.
Fluids are everywhere around us. Earth has an envelop of air and two-thirds of its surface is covered with water. Water is not only necessary for our existence; every mammalian body constitute mostly of water. All the processes occurring in living beings including plants are mediated by fluids. Thus understanding the behaviour and properties of fluids is important.
How are fluids different from solids? What is common in liquids and gases? Unlike a solid, a fluid has no definite shape of its own. Solids and liquids have a fixed volume, whereas a gas fills the entire volume of its container. We have learnt in the previous chapter that the volume of solids can be changed by stress. The volume of solid, liquid or gas depends on the stress or pressure acting on it. When we talk about fixed volume of solid or liquid, we mean its volume under atmospheric pressure. The difference between gases and solids or liquids is that for solids or liquids the change in volume due to change of external pressure is rather small. In other words solids and liquids have much lower compressibility as compared to gases.
Shear stress can change the shape of a solid keeping its volume fixed. The key property of fluids is that they offer very little resistance to shear stress; their shape changes by application of very small shear stress. The shearing stress of fluids is about million times smaller than that of solids.
#### 10.2 Pressure
A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force. If an elephant were to step on a man’s chest, his ribs would crack. A circus performer across whose chest a large, light but strong wooden plank is placed first, is saved from this accident. Such everyday experiences convince us that both the force and its coverage area are important. Smaller the area on which the force acts, greater is the impact. This impact is known as pressure.
When an object is submerged in a fluid at rest, the fluid exerts a force on its surface. This force is always normal to the object’s surface. This is so because if there were a component of force parallel to the surface, the object will also exert a force on the fluid parallel to it; as a consequence of Newton’s third law. This force will cause the fluid to flow parallel to the surface. Since the fluid is at rest, this cannot happen. Hence, the force exerted by the fluid at rest has to be perpendicular to the surface in contact with it. This is shown in Fig.10.1(a).
The normal force exerted by the fluid at a point may be measured. An idealised form of one such pressure-measuring device is shown in Fig. 10.1(b). It consists of an evacuated chamber with a spring that is calibrated to measure the force acting on the piston. This device is placed at a point inside the fluid. The inward force exerted by the fluid on the piston is balanced by the outward spring force and is thereby measured.
(a) (b)
Fig. 10.1 (a) The force exerted by the liquid in the beaker on the submerged object or on the
walls is normal (perpendicular) to the surface at all points.
(b) An idealised device for measuring pressure.
If F is the magnitude of this normal force on the piston of area A then the average pressure Pav is defined as the normal force acting per unit area.
(10.1)
In principle, the piston area can be made arbitrarily small. The pressure is then defined in a limiting sense as
P = (10.2)
Pressure is a scalar quantity. We remind the reader that it is the component of the force normal to the area under consideration and not the (vector) force that appears in the numerator in Eqs. (10.1) and (10.2). Its dimensions are
[ML
–1T–2]. The SI unit of pressure is N m–2. It has been named as pascal (Pa) in honour of the French scientist Blaise Pascal (1623-1662) who carried out pioneering studies on fluid pressure. A common unit of pressure is the atmosphere (atm), i.e. the pressure exerted by the atmosphere at sea level (1 atm = 1.013 × 105 Pa).
Another quantity, that is indispensable in describing fluids, is the density ρ. For a fluid of mass m occupying volume V,
(10.3)
The dimensions of density are [ML–3]. Its SI unit is kg m–3. It is a positive scalar quantity. A liquid is largely incompressible and its density is therefore, nearly constant at all pressures. Gases, on the other hand exhibit a large variation in densities with pressure.
The density of water at 4oC (277 K) is
1.0
× 103 kg m–3. The relative density of a substance is the ratio of its density to the density of water at 4oC. It is a dimensionless positive scalar quantity. For example the relative density of aluminium is 2.7. Its density is
2.7
× 103 kg m–3. The densities of some common fluids are displayed in Table 10.1.
Table 10.1 Densities of some common fluids
at STP*
Example 10.1 The two thigh bones (femurs), each of cross-sectional area10 cm2 support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.
Answer Total cross-sectional area of the femurs is A = 2 × 10 cm2 = 20 × 10–4 m2. The force acting on them is F = 40 kg wt = 400 N (taking g = 10 m s–2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
#### 10.2.1 Pascal’s Law
The French scientist Blaise Pascal observed that the pressure in a fluid at rest is the same at all points if they are at the same height. This fact may be demonstrated in a simple way.
Fig. 10.2 Proof of Pascal’s law. ABC-DEF is an element of the interior of a fluid at rest. This element is in the form of a right-angled prism. The element is small so that the effect of gravity can be ignored, but it has been enlarged for the sake of clarity.
Fig. 10.2 shows an element in the interior of a fluid at rest. This element ABC-DEF is in the form of a right-angled prism. In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of the gravity is the same at all these points. But for clarity we have enlarged this element. The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element as discussed above. Thus, the fluid exerts pressures Pa, Pb and Pc on this element of area corresponding to the normal forces Fa, Fb and Fc as shown in Fig. 10.2 on the faces BEFC, ADFC and ADEB denoted by Aa, Ab and Ac respectively. Then
Fb sinθ = Fc, Fb cosθ = Fa (by equilibrium)
Ab sinθ = Ac, Ab cosθ = Aa (by geometry)
Thus,
(10.4)
Hence, pressure exerted is same in all directions in a fluid at rest. It again reminds us that like other types of stress, pressure is not a vector quantity. No direction can be assigned to it. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area.
Now consider a fluid element in the form of a horizontal bar of uniform cross-section. The bar is in equilibrium. The horizontal forces exerted at its two ends must be balanced or the pressure at the two ends should be equal. This proves that for a liquid in equilibrium the pressure is same at all points in a horizontal plane. Suppose the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have some net force acting on it. Hence in the absence of flow the pressure in the fluid must be same everywhere in a horizontal plane.
#### 10.2.2 Variation of Pressure with Depth
Consider a fluid at rest in a container. In Fig. 10.3 point 1 is at height h above a point 2. The pressures at points 1 and 2 are P1 and P2 respectively. Consider a cylindrical element of fluid having area of base A and height h. As the fluid is at rest the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top (P1A) acting downward, at the bottom (P2A) acting upward. If mg is weight of the fluid in the cylinder we have
(P2 P1) A = mg (10.5)
Now, if ρ is the mass density of the fluid, we have the mass of fluid to be m = ρV= ρhA so that
P2 P1= ρgh (10.6)
Fig.10.3 Fluid under gravity. The effect of gravity is illustrated through pressure on a vertical cylindrical column.
Pressure difference depends on the vertical distance h between the points (1 and 2), mass density of the fluid ρ and acceleration due to gravity g. If the point 1 under discussion is shifted to the top of the fluid (say, water), which is open to the atmosphere, P1 may be replaced by atmospheric pressure (Pa) and we replace P2 by P. Then Eq. (10.6) gives
P = Pa + ρgh (10.7)
Thus, the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh. The excess of pressure, P Pa, at depth h is called a gauge pressure at that point.
The area of the cylinder is not appearing in the expression of absolute pressure in Eq. (10.7). Thus, the height of the fluid column is important and not cross-sectional or base area or the shape of the container. The liquid pressure is the same at all points at the same horizontal level (same depth). The result is appreciated through the example of hydrostatic paradox. Consider three vessels A, B and C [Fig.10.4] of different shapes. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel.
Fig 10.4 Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all upto the same height.
Example 10.2 What is the pressure on a swimmer 10 m below the surface of a lake?
h = 10 m and ρ = 1000 kg m-3. Take g = 10 m s–2
From Eq. (10.7)
P = Pa + rgh
= 1.01 × 105 Pa + 1000 kg m–3 × 10 m s–2 × 10 m
= 2.01 × 105 Pa
= 2 atm
This is a 100% increase in pressure from surface level. At a depth of 1 km, the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.
#### 10.2.3 Atmospheric Pressure and Gauge Pressure
The pressure of the atmosphere at any point is equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere. At sea level, it is 1.013 × 105 Pa (1 atm). Italian scientist Evangelista Torricelli (16081647) devised for the first time a method for measuring atmospheric pressure. A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury as shown in Fig.10.5 (a). This device is known as ‘mercury barometer’. The space above the mercury column in the tube contains only mercury vapour whose pressure P is so small that it may be neglected. Thus, the pressure at Point A=0. The pressure inside the coloumn at Point B must be the same as the pressure at Point C, which is atmospheric pressure, Pa.
Pa = ρgh (10.8)
where ρ is the density of mercury and h is the height of the mercury column in the tube.
In the experiment it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of ρ in Eq. (10.8). A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent of 1 mm is called a torr (after Torricelli).
1 torr = 133 Pa.
The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar.
1 bar = 105 Pa
An open tube manometer is a useful instrument for measuring pressure differences. It consists of a U-tube containing a suitable liquid i.e., a low density liquid (such as oil) for measuring small pressure differences and a high density liquid (such as mercury) for large pressure differences. One end of the tube is open to the atmosphere and the other end is connected to the system whose pressure we want to measure [see Fig. 10.5 (b)]. The pressure P at A is equal to pressure at point B. What we normally measure is the gauge pressure, which is P Pa, given by Eq. (10.8) and is proportional to manometer height h.
Fig 10.5 (a) The mercury barometer.
(b) The open tube manometer
Fig 10.5 Two pressure measuring devices.
Pressure is same at the same level on both sides of the U-tube containing a fluid. For liquids, the density varies very little over wide ranges in pressure and temperature and we can treat it safely as a constant for our present purposes. Gases on the other hand, exhibits large variations of densities with changes in pressure and temperature. Unlike gases, liquids are, therefore, largely treated as incompressible.
Example 10.3 The density of the atmosphere at sea level is 1.29 kg/m3. Assume that it does not change with altitude. Then how high would the atmosphere extend?
ρgh = 1.29 kg m–3 × 9.8 m s2 × h m = 1.01 × 105 Pa
h = 7989 m 8 km
In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing pressure over
100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.
Example 10.4 At a depth of 1000 m in an ocean (a) what is the absolute pressure?
(b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm
× 20 cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is 1.03 × 103 kg m-3, g = 10 m s–2.)
Answer Here h = 1000 m and ρ = 1.03 × 103 kg m-3.
(a) From Eq. (10.6), absolute pressure
P = Pa + ρgh
= 1.01 × 105 Pa
+ 1.03 × 103 kg m–3 × 10 m s–2 × 1000 m
= 104.01 × 105 Pa
104 atm
(b) Gauge pressure is P Pa = ρgh = Pg
Pg = 1.03 × 103 kg m–3 × 10 ms2 × 1000 m
= 103 × 105 Pa
103 atm
(c) The pressure outside the submarine is P = Pa + ρgh and the pressure inside it is Pa. Hence, the net pressure acting on the window is gauge pressure, Pg = ρgh. Since the area of the window is A = 0.04 m2, the force acting on it is
F = Pg A = 103 × 105 Pa × 0.04 m2 = 4.12 × 105 N
#### 10.2.4 Hydraulic Machines
Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points [Fig. 10.6 (a)]. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.
Archemedes’ Principle
Fluid appears to provide partial support to the objects placed in it. When a body is wholly or partially immersed in a fluid at rest, the fluid exerts pressure on the surface of the body in contact with the fluid. The pressure is greater on lower surfaces of the body than on the upper surfaces as pressure in a fluid increases with depth. The resultant of all the forces is an upward force called buoyant force. Suppose that a cylindrical body is immersed in the fluid. The upward force on the bottom of the body is more than the downward force on its top. The fluid exerts a resultant upward force or buoyant force on the body equal to (P2 P1) × A (Fig. 10.3). We have seen in equation 10.4 that (P2-P1)A = ρghA. Now, hA is the volume of the solid and ρhA is the weight of an equivaliant volume of the fluid. (P2-P1)A = mg. Thus, the upward force exerted is equal to the weight of the displaced fluid.
The result holds true irrespective of the shape of the object and here cylindrical object is considered only for convenience. This is Archimedes’ principle. For totally immersed objects the volume of the fluid displaced by the object is equal to its own volume. If the density of the immersed object is more than that of the fluid, the object will sink as the weight of the body is more than the upward thrust. If the density of the object is less than that of the fluid, it floats in the fluid partially submerged. To calculate the volume submerged, suppose the total volume of the object is Vs and a part Vp of it is submerged in the fluid. Then, the upward force which is the weight of the displaced fluid is ρfgVp, which must equal the weight of the body; ρsgVs = ρfgVpor ρs/ρf = Vp/Vs The apparent weight of the floating body is zero.
This principle can be summarised as; ‘the loss of weight of a body submerged (partially or fully) in a fluid is equal to the weight of the fluid displaced’.
This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. We can say whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is another form of the Pascal’s law and it has many applications in daily life.
Fig 10.6 (a) Whenever external pressure is applied on any part of a fluid in a vessel, it is equally transmitted in all directions.
A number of devices, such as hydraulic lift and hydraulic brakes, are based on the Pascal’s
law. In these devices, fluids are used for transmitting pressure. In a hydraulic lift, as shown in Fig. 10.6 (b), two pistons are separated by the space filled with a liquid. A piston of small
cross-section A1 is used to exert a force F1 directly on the liquid. The pressure P = is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A2, which results in an upward force of P × A2. Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform) F2 = PA2 = . By changing the force at A1, the platform can be moved up or down. Thus, the applied force has been increased by a factor of and this factor is the mechanical advantage of the device. The example below clarifies it.
Fig 10.6 (b) Schematic diagram illustrating the principle behind the hydraulic lift, a device used to lift heavy loads.
Example 10.5 Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?
Answer (a) Since pressure is transmitted undiminished throughout the fluid,
= 90 N
(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.
= 0.67 × 10-2 m = 0.67 cm
Note, atmospheric pressure is common to both pistons and has been ignored.
Example 10.6 In a car lift compressed air exerts a force F1 on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 10.7). If the mass of the car to be lifted is 1350 kg, calculate F1. What is the pressure necessary to accomplish this task? (g = 9.8 ms-2).
Answer Since pressure is transmitted undiminished throughout the fluid,
= 1470 N 1.5 × 103 N
The air pressure that will produce this force is
This is almost double the atmospheric pressure.
Archimedes (287–212 B.C.)
Archimedes was a Greek philosopher, mathematician, scientist and engineer. He invented the catapult and devised a system of pulleys and levers to handle heavy loads. The king of his native city Syracuse, Hiero II, asked him to determine if his gold crown was alloyed with some cheaper metal, such as silver without damaging the crown. The partial loss of weight he experienced while lying in his bathtub suggested a solution to him. According to legend, he ran naked through the streets of Syracuse, exclaiming “Eureka, eureka!”, which means “I have found it, I have found it!”
Hydraulic brakes in automobiles also work on the same principle. When we apply a little force on the pedal with our foot the master piston moves inside the master cylinder, and the pressure caused is transmitted through the brake oil to act on a piston of larger area. A large force acts on the piston and is pushed down expanding the brake shoes against brake lining. In this way, a small force on the pedal produces a large retarding force on the wheel. An important advantage of the system is that the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on all wheels.
#### 10.3 STREAMLINE FLOW
So far we have studied fluids at rest. The study of the fluids in motion is known as fluid dynamics. When a water tap is turned on slowly, the water flow is smooth initially, but loses its smoothness when the speed of the outflow is increased. In studying the motion of fluids, we focus our attention on what is happening to various fluid particles at a particular point in space at a particular time. The flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant in time. This does not mean that the velocity at different points in space is same. The velocity of a particular particle may change as it moves from one point to another. That is, at some other point the particle may have a different velocity, but every other particle which passes the second point behaves exactly as the previous particle that has just passed that point. Each particle follows a smooth path, and the paths of the particles do not cross each other.
Fig. 10.7 The meaning of streamlines. (a) A typical trajectory of a fluid particle.
(b) A region of streamline flow.
The path taken by a fluid particle under a steady flow is a streamline. It is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point. Consider the path of a particle as shown in Fig.10.7 (a), the curve describes how a fluid particle moves with time. The curve PQ is like a permanent map of fluid flow, indicating how the fluid streams. No two streamlines can cross, for if they do, an oncoming fluid particle can go either one way or the other and the flow would not be steady. Hence, in steady flow, the map of flow is stationary in time. How do we draw closely spaced streamlines ? If we intend to show streamline of every flowing particle, we would end up with a continuum of lines. Consider planes perpendicular to the direction of fluid flow e.g., at three points P, R and Q in Fig.10.7 (b). The plane pieces are so chosen that their boundaries be determined by the same set of streamlines. This means that number of fluid particles crossing the surfaces as indicated at P, R and Q is the same. If area of cross-sections at these points are AP,AR and AQ and speeds of fluid particles are vP, vR and vQ, then mass of fluid mP crossing at AP in a small interval of time t is ρPAPvP t. Similarly mass of fluid mR flowing or crossing at AR in a small interval of time t is ρRARvR t and mass of fluid mQ is ρQAQvQ t crossing at AQ. The mass of liquid flowing out equals the mass flowing in, holds in all cases. Therefore,
ρPAPvPt = ρRARvRt = ρQAQvQt (10.9)
For flow of incompressible fluids
ρP = ρR = ρQ
Equation (10.9) reduces to
APvP = ARvR = AQvQ (10.10)
which is called the equation of continuity and it is a statement of conservation of mass in flow of incompressible fluids. In general
Av = constant (10.11)
Av gives the volume flux or flow rate and remains constant throughout the pipe of flow. Thus, at narrower portions where the streamlines are closely spaced, velocity increases and its vice versa. From (Fig 10.7b) it is clear that AR > AQ or vR < vQ, the fluid is accelerated while passing from R to Q. This is associated with a change in pressure in fluid flow in horizontal pipes.
Steady flow is achieved at low flow speeds. Beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent. One sees this when a fast flowing stream encounters rocks, small foamy whirlpool-like regions called ‘white water rapids are formed.
Figure 10.8 displays streamlines for some typical flows. For example, Fig. 10.8(a) describes a laminar flow where the velocities at different points in the fluid may have different magnitudes but their directions are parallel.
Figure 10.8 (b) gives a sketch of turbulent flow.
Fig. 10.8 (a) Some streamlines for fluid flow.
(b) A jet of air striking a flat plate placed perpendicular to it. This is an example
of turbulent flow.
#### 10.4 BERNOULLI’S PRINCIPLE
Fluid flow is a complex phenomenon. But we can obtain some useful properties for steady or streamline flows using the conservation of energy.
Consider a fluid moving in a pipe of varying cross-sectional area. Let the pipe be at varying heights as shown in Fig. 10.9. We now suppose that an incompressible fluid is flowing through the pipe in a steady flow. Its velocity must change as a consequence of equation of continuity. A force is required to produce this acceleration, which is caused by the fluid surrounding it, the pressure must be different in different regions. Bernoulli’s equation is a general expression that relates the pressure difference between two points in a pipe to both velocity changes (kinetic energy change) and elevation (height) changes (potential energy change). The Swiss Physicist Daniel Bernoulli developed this relationship in 1738.
Consider the flow at two regions 1 (i.e., BC) and 2 (i.e., DE). Consider the fluid initially lying between B and D. In an infinitesimal time interval t, this fluid would have moved. Suppose v1 is the speed at B and v2 at D, then fluid initially at B has moved a distance v1t to C (v1t is small enough to assume constant cross-section along BC). In the same interval t the fluid initially at D moves to E, a distance equal to v2t. Pressures P1 and P2 act as shown on the plane faces of areas A1 and A2 binding the two regions. The work done on the fluid at left end (BC) is W1 = P1A1(v1t) = P1V. Since the same volume V passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE) is W2 = P2A2(v2t) = P2V or, the work done on the fluid is –P2V. So the total work done on the fluid is
W1 – W2 = (P1− P2) ∆V
Daniel Bernoulli (1700–1782)
Daniel Bernoulli was a Swiss scientist and mathematician, who along with Leonard Euler had the distinction of winning the French Academy prize for mathematics 10 times. He also studied medicine and served as a professor of anatomy and botany for a while at Basle, Switzerland. His most well-known work was in hydrodynamics, a subject he developed from a single principle: the conservation of energy. His work included calculus, probability, the theory of vibrating strings, and applied mathematics. He has been called the founder of mathematical physics.
Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing the gravitational potential energy. If the density of the fluid is ρ and m = ρA1v1t = ρ∆V is the mass passing through the pipe in time t, then change in gravitational potential energy is
U = ρgV (h2 h1
The change in its kinetic energy is
K = ρ V (v22 v12)
We can employ the work – energy theorem (Chapter 6) to this volume of the fluid and
this yields
(P1 P2) V = ρ V (v22 v12) + ρgV (h2 h1)
We now divide each term by V to obtain
(P1 P2) = ρ (v22 v12) + ρg (h2 h1)
We can rearrange the above terms to obtain
P1 + ρv12 + ρgh1 = P2+ ρv22 + ρgh2 (10.12)
This is Bernoulli’s equation. Since 1 and 2 refer to any two locations along the pipeline, we may write the expression in general as
P + ρv2 + ρgh = constant (10.13)
Fig. 10.9 The flow of an ideal fluid in a pipe of varying cross section. The fluid in a section of length v1t moves to the section of length v2t in time t.
In words, the Bernoulli’s relation may be stated as follows: As we move along a streamline the sum of the pressure (P), the kinetic energy per unit volume and the potential energy per unit volume (ρgh) remains a constant.
Note that in applying the energy conservation principle, there is an assumption that no energy is lost due to friction. But in fact, when fluids flow, some energy does get lost due to internal friction. This arises due to the fact that in a fluid flow, the different layers of the fluid flow with different velocities. These layers exert frictional forces on each other resulting in a loss of energy. This property of the fluid is called viscosity and is discussed in more detail in a later section. The lost kinetic energy of the fluid gets converted into heat energy. Thus, Bernoulli’s equation ideally applies to fluids with zero viscosity or non-viscous fluids. Another restriction on application of Bernoulli theorem is that the fluids must be incompressible, as the elastic energy of the fluid is also not taken into consideration. In practice, it has a large number of useful applications and can help explain a wide variety of phenomena for low viscosity incompressible fluids. Bernoulli’s equation also does not hold for non-steady or turbulent flows, because in that situation velocity and pressure are constantly fluctuating in time.
When a fluid is at rest i.e., its velocity is zero everywhere, Bernoulli’s equation becomes
P1 + ρgh1 = P2 + ρgh2
(P1 P2) = ρg (h2 h1)
which is same as Eq. (10.6).
#### 10.4.1 Speed of Efflux: Torricelli’s Law
The word efflux means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. Consider a tank containing a liquid of density ρ with a small hole in its side at a height y1 from the bottom (see Fig. 10.10). The air above the liquid, whose surface is at height y2, is at pressure P. From the equation of continuity [Eq. (10.10)] we have
v1 A1 = v2 A2
.
Fig. 10.10 Torricelli’s law. The speed of efflux, v1, from the side of the container is given by the application of Bernoulli’s equation. If the container is open at the top to the atmosphere then .
If the cross-sectional area of the tank A2 is much larger than that of the hole (A2 >>A1), then we may take the fluid to be approximately at rest at the top, i.e., v2 = 0. Now, applying the Bernoulli equation at points 1 and 2 and noting that at the hole P1 = Pa, the atmospheric pressure, we have from Eq. (10.12)
Taking y2 – y1 = h we have
(10.14)
When P >>Pa and 2 g h may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand, if the tank is open to the atmosphere, then P = Pa and
(10.15)
This is also the speed of a freely falling body. Equation (10.15) represents Torricelli’s law.
#### 10.4.2 Venturi-meter
The Venturi-meter is a device to measure the flow speed of incompressible fluid. It consists of a tube with a broad diameter and a small constriction at the middle as shown in Fig. (10.11). A manometer in the form of a U-tube is also attached to it, with one arm at the broad neck point of the tube and the other at constriction as shown in Fig. (10.11). The manometer contains a liquid of density ρm. The speed v1 of the liquid flowing through the tube at the broad neck area A is to be measured
from equation of continuity Eq. (10.10) the
speed at the constriction becomes . Then using Bernoulli’s equation (Eq.10.12) for (h1=h2), we get
P1+ ρv12 = P2+ ρv12 (A/a)2
So that
P1- P2 = ρv12 (10.16)
This pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm. The difference in height h measure the pressure difference.
Fig. 10.11 A schematic diagram of Venturi-meter.
P1P2 = ρmgh = ρv12
So that the speed of fluid at wide neck is
v1= (10.17)
The principle behind this meter has many applications. The carburetor of automobile has a Venturi channel (nozzle) through which air flows with a high speed. The pressure is then lowered at the narrow neck and the petrol (gasoline) is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion. Filter pumps or aspirators, Bunsen burner, atomisers and sprayers [See Fig. 10.12] used for perfumes or to spray insecticides work on the same principle.
Fig. 10.12 The spray gun. Piston forces air at high speeds causing a lowering of pressure at the neck of the container.
Example 10.7 Blood velocity: The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery. A = 8 mm2. The narrower part has an area a = 4 mm2. The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery?
Answer We take the density of blood from Table 10.1 to be 1.06 × 103 kg m-3. The ratio of the areas is = 2. Using Eq. (10.17) we obtain
#### 10.4.3 Blood Flow and Heart Attack
Bernoulli’s principle helps in explaining blood flow in artery. The artery may get constricted due to the accumulation of plaque on its inner walls. In order to drive the blood through this constriction a greater demand is placed on the activity of the heart. The speed of the flow of the blood in this region is raised which lowers the pressure inside and the artery may collapse due to the external pressure. The heart exerts further pressure to open this artery and forces the blood through. As the blood rushes through the opening, the internal pressure once again drops due to same reasons leading to a repeat collapse. This may result in heart attack.
#### 10.4.4 Dynamic Lift
Dynamic lift is the force that acts on a body, such as airplane wing, a hydrofoil or a spinning ball, by virtue of its motion through a fluid. In many games such as cricket, tennis, baseball, or golf, we notice that a spinning ball deviates from its parabolic trajectory as it moves through air. This deviation can be partly explained on the basis of Bernoulli’s principle.
(i) Ball moving without spin: Fig. 10.13(a) shows the streamlines around a
non-spinning ball moving relative to a fluid. From the symmetry of streamlines it is clear that the velocity of fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air therefore, exerts no upward or downward force on the ball.
(a)
(b)
(c)
Fig 10.13 (a) Fluid streaming past a static sphere. (b) Streamlines for a fluid around a sphere spinning clockwise. (c) Air flowing past an aerofoil.
(ii) Ball moving with spin: A ball which is spinning drags air along with it. If the surface is rough more air will be dragged. Fig 10.13(b) shows the streamlines of air for a ball which is moving and spinning at the same time. The ball is moving forward and relative to it the air is moving backwards. Therefore, the velocity of air above the ball relative to the ball is larger and below it is smaller (see Section 10.3). The stream lines, thus, get crowded above and rarified below.
This difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spining is called Magnus effect
Aerofoil or lift on aircraft wing: Figure 10.13 (c) shows an aerofoil, which is a solid piece shaped to provide an upward dynamic lift when it moves horizontally through air. The cross-section of the wings of an aeroplane looks somewhat like the aerofoil shown in Fig. 10.13 (c) with streamlines around it. When the aerofoil moves against the wind, the orientation of the wing relative to flow direction causes the streamlines to crowd together above the wing more than those below it. The flow speed on top is higher than that below it. There is an upward force resulting in a dynamic lift of the wings and this balances the weight of the plane. The following example illustrates this.
Example 10.8 A fully loaded Boeing aircraft has a mass of 3.3 × 105 kg. Its total wing area is 500 m2. It is in level flight with a speed of 960 km/h. (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kg m-3]
Answer (a) The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference
P × A = 3.3 × 105 kg × 9.8
= (3.3 × 105 kg × 9.8 m s–2) / 500 m2
= 6.5 ×103 Nm-2
(b) We ignore the small height difference between the top and bottom sides in Eq. (10.12). The pressure difference between them is
then
where v2 is the speed of air over the upper surface and v1 is the speed under the bottom surface.
Taking the average speed
vav = (v2 + v1)/2 = 960 km/h = 267 m s-1,
we have
0.08
The speed above the wing needs to be only 8 % higher than that below.
#### 10.5 VISCOSITY
Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluid motion is like an internal friction analogous to friction when a solid moves on a surface. It is called viscosity. This force exists when there is relative motion between layers of the liquid. Suppose we consider a fluid like oil enclosed between two glass plates as shown in Fig. 10.14 (a). The bottom plate is fixed while the top plate is moved with a constant velocity v relative to the fixed plate. If oil is replaced by honey, a greater force is required to move the plate with the same velocity. Hence we say that honey is more viscous than oil. The fluid in contact with a surface has the same velocity as that of the surfaces. Hence, the layer of the liquid in contact with top surface moves with a velocity v and the layer of the liquid in contact with the fixed surface is stationary. The velocities of layers increase uniformly from bottom (zero velocity) to the top layer (velocity v). For any layer of liquid, its upper layer pulls it forward while lower layer pulls it backward. This results in force between the layers. This type of flow is known as laminar. The layers of liquid slide over one another as the pages of a book do when it is placed flat on a table and a horizontal force is applied to the top cover. When a fluid is flowing in a pipe or a tube, then velocity of the liquid layer along the axis of the tube is maximum and decreases gradually as we move towards the walls where it becomes zero, Fig. 10.14 (b). The velocity on a cylindrical surface in a tube is constant.
On account of this motion, a portion of liquid, which at some instant has the shape ABCD, take the shape of AEFD after short interval of time (t). During this time interval the liquid has undergone a shear strain of x/l. Since, the strain in a flowing fluid increases with time continuously. Unlike a solid, here the stress is found experimentally to depend on ‘rate of change of strain’ or ‘strain rate’ i.e. x/(l t) or v/l instead of strain itself. The coefficient of viscosity (pronounced ‘eta’) for a fluid is defined as the ratio of shearing stress to the strain rate.
(a)
(b)
Fig 10.14 (a) A layer of liquid sandwiched between two parallel glass plates, in which the lower plate is fixed and the upper one is moving to the right with velocity v
(b) velocity distribution for viscous flow in a pipe.
(10.18)
The SI unit of viscosity is poiseiulle (Pl). Its other units are N s m-2 or Pa s. The dimensions of viscosity are [ML-1T-1]. Generally, thin liquids, like water, alcohol, etc., are less viscous than thick liquids, like coal tar, blood, glycerine, etc. The coefficients of viscosity for some common fluids are listed in Table 10.2. We point out two facts about blood and water that you may find interesting. As Table 10.2 indicates, blood is ‘thicker’ (more viscous) than water. Further, the relative viscosity (η/ηwater) of blood remains constant between 0oC and 37oC.
Fig. 10.15 Measurement of the coefficient of viscosity of a liquid.
The viscosity of liquids decreases with temperature, while it increases in the case of gases.
Example 10.9 A metal block of area 0.10 m2 is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 10.15. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m s-1. Find the coefficient of viscosity of the liquid.
Answer The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus, the shear force F is
F = T = mg = 0.010 kg × 9.8 m s–2 = 9.8 × 10-2 N Shear stress on the fluid = F/A = N/m2
Strain rate =
=
= 3.46 ×10-3 Pa s t
Table10.2 The viscosities of some fluids
#### 10.5.1 Stokes’ Law
When a body falls through a fluid it drags the layer of the fluid in contact with it. A relative motion between the different layers of the fluid is set and, as a result, the body experiences a retarding force. Falling of a raindrop and swinging of a pendulum bob are some common examples of such motion. It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion. The other quantities on which the force F depends are viscosity η of the fluid and radius a of the sphere. Sir George G. Stokes (1819–1903), an English scientist enunciated clearly the viscous drag force F as
(10.19)
This is known as Stokes’ law. We shall not derive Stokes’ law.
This law is an interesting example of retarding force, which is proportional to velocity. We can study its consequences on an object falling through a viscous medium. We consider a raindrop in air. It accelerates initially due to gravity. As the velocity increases, the retarding force also increases. Finally, when viscous force plus buoyant force becomes equal to the force due to gravity, the net force becomes zero and so does the acceleration. The sphere (raindrop) then descends with a constant velocity. Thus, in equilibrium, this terminal velocity vt is given by
6πηavt = (4π/3) a3 (ρ-σ)g
where ρ and σ are mass densities of sphere and the fluid, respectively. We obtain
vt = 2a2 (ρ-σ)g / (9η) (10.20)
So the terminal velocity vt depends on the square of the radius of the sphere and inversely on the viscosity of the medium.
You may like to refer back to Example 6.2 in this context.
Example 10.10 The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20oC is 6.5 cm s-1. Compute the viscosity of the oil at 20oC. Density of oil is 1.5 ×103 kg m-3, density of copper is 8.9 × 103 kg m-3
Answer We have vt = 6.5 × 10-2 ms-1, a = 2 × 10-3 m, g = 9.8 ms-2, ρ = 8.9 × 103 kg m-3,
σ =1.5 ×103 kg m-3. From Eq. (10.20)
= 9.9 × 10-1 kg m–1 s–1 t
#### 10.6REYNOLDS NUMBER
When the rate of flow of a fluid is large, the flow no longer remains laminar, but becomes turbulent. In a turbulent flow the velocity of the fluids at any point in space varies rapidly and randomly with time. Some circular motions called eddies are also generated. An obstacle placed in the path of a fast moving fluid causes turbulence [Fig. 10.8 (b)]. The smoke rising from a burning stack of wood, oceanic currents are turbulent. Twinkling of stars is the result of atmospheric turbulence. The waves in the water and in the air left by cars, aeroplanes and boats are also turbulent.
Osborne Reynolds (18421912) observed that turbulent flow is less likely for viscous fluid flowing at low rates. He defined a dimensionless number, whose value gives one an approximate idea whether the flow would be turbulent. This number is called the Reynolds Re.
Re = ρvd/η (10.21)
where ρ is the density of the fluid flowing with a speed v, d stands for the dimension of the pipe, and η is the viscosity of the fluid. Re is a dimensionless number, and therefore, it remains the same in any system of units. It is found that flow is streamline or laminar for Re less than 1000. The flow is turbulent for Re > 2000. The flow becomes unsteady for Re between 1000 and 2000. The critical value of Re (known as critical Reynolds number), at which turbulence sets, is found to be the same for the geometrically similar flows. For example, when oil and water with their different densities and viscosities, flow in pipes of same shapes and sizes, turbulence sets in at almost the same value of Re. Using this fact, a small-scale laboratory model can be set up to study the character of fluid flow. They are useful in designing of ships, submarines, racing cars and aeroplanes.
Re can also be written as
Re = ρv2 / (ηv/d) = ρAv2 / (ηAv/d) (10.22)
= inertial force/force of viscosity.
Thus Re represents the ratio of inertial force (force due to inertia i.e., mass of moving fluid or due to inertia of obstacle in its path) to viscous force.
#### Critical Velocity
The maximum velocity of a fluid in a tube for which the flow remains streamlined is called its critical velocity. From Eq. 10.21, it is
vc = Re × η /(ρ × d).
Turbulence dissipates kinetic energy usually in the form of heat. Racing cars and planes are engineered to precision in order to minimise turbulence. The design of such vehicles involves experimentation and trial and error. On the other hand turbulence (like friction) is sometimes desirable. Turbulence promotes mixing and increases the rates of transfer of mass, momentum and energy. The blades of a kitchen mixer induce turbulent flow and provide thick milk shakes as well as beat eggs into a uniform texture.
###### Example 10.11The flow rate of water from a tap of diameter 1.25 cm is 0.48 L/min. The coefficient of viscosity of water is 10-3 Pa s. After sometime the flow rate is increased to 3 L/min. Characterise the flow for both the flow rates.
Answer Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d2 / 4
v = 4 Q / d2π
Re = 4 ρ Q / π d η
= 4 ×103 kg m3 × Q/(3.14 ×1.25 ×10-2 m ×10-3 Pa s)
= 1.019 × 108 m–3 s Q
Since initially
Q = 0.48 L / min = 8 cm3 / s = 8 × 10-6 m3 s-1,
we obtain,
Re = 815
Since this is below 1000, the flow is steady.
After some time when
Q = 3 L / min = 50 cm3 / s = 5 × 10-5 m3 s-1, we obtain,
Re = 5095
The flow will be turbulent. You may carry out an experiment in your washbasin to determine the transition from laminar to turbulent flow.
#### 10.6 SURFACE TENSION
You must have noticed that, oil and water do not mix; water wets you and me but not ducks; mercury does not wet glass but water sticks to it, oil rises up a cotton wick, inspite of gravity, Sap and water rise up to the top of the leaves of the tree, hair of a paint brush do not cling together when dry and even when dipped in water but form a fine tip when taken out of it. All these and many more such experiences are related with the free surfaces of liquids. As liquids have no definite shape but have a definite volume, they acquire a free surface when poured in a container. These surfaces possess some additional energy. This phenomenon is known as surface tension and it is concerned with only liquid as gases do not have free surfaces. Let us now understand this phenomena.
#### 10.6.1 Surface Energy
A liquid stays together because of attraction between molecules. Consider a molecule well inside a liquid. The intermolecular distances are such that it is attracted to all the surrounding molecules [Fig. 10.16(a)]. This attraction results in a negative potential energy for the molecule, which depends on the number and distribution of molecules around the chosen one. But the average potential energy of all the molecules is the same. This is supported by the fact that to take a collection of such molecules (the liquid) and to disperse them far away from each other in order to evaporate or vaporise, the heat of evaporation required is quite large. For water it is of the order of 40 kJ/mol.
Let us consider a molecule near the surface Fig. 10.16(b). Only lower half side of it is surrounded by liquid molecules. There is some negative potential energy due to these, but obviously it is less than that of a molecule in bulk, i.e., the one fully inside. Approximately it is half of the latter. Thus, molecules on a liquid surface have some extra energy in comparison to molecules in the interior. A liquid, thus, tends to have the least surface area which external conditions permit. Increasing surface area requires energy. Most surface phenomenon can be understood in terms of this fact. What is the energy required for having a molecule at the surface? As mentioned above, roughly it is half the energy required to remove it entirely from the liquid i.e., half the heat of evaporation.
Fig. 10.16 Schematic picture of molecules in a liquid, at the surface and balance of forces. (a) Molecule inside a liquid. Forces on a molecule due to others are shown. Direction of arrows indicates attraction of repulsion. (b) Same, for a molecule at a surface. (c) Balance of attractive (AI and repulsive (R) forces.
Finally, what is a surface? Since a liquid consists of molecules moving about, there cannot be a perfectly sharp surface. The density of the liquid molecules drops rapidly to zero around z = 0 as we move along the direction indicated Fig 10.16 (c) in a distance of the order of a few molecular sizes.
#### 10.6.2 Surface Energy and Surface Tension
As we have discussed that an extra energy is associated with surface of liquids, the creation of more surface (spreading of surface) keeping other things like volume fixed requires additional energy. To appreciate this, consider a horizontal liquid film ending in bar free to slide over parallel guides Fig (10.17).
Fig. 10.17 Stretching a film. (a) A film in equilibrium; (b) The film stretched an extra distance.
Suppose that we move the bar by a small distance d as shown. Since the area of the surface increases, the system now has more energy, this means that some work has been done against an internal force. Let this internal force be F, the work done by the applied force is F.d = Fd. From conservation of energy, this is stored as additional energy in the film. If the surface energy of the film is S per unit area, the extra area is 2dl. A film has two sides and the liquid in between, so there are two surfaces and the extra energy is
S (2dl) = Fd (10.21)
Or, S=Fd/2dl = F/2l (10.22)
This quantity S is the magnitude of surface tension. It is equal to the surface energy per unit area of the liquid interface and is also equal to the force per unit length exerted by the fluid on the movable bar.
So far we have talked about the surface of one liquid. More generally, we need to consider fluid surface in contact with other fluids or solid surfaces. The surface energy in that case depends on the materials on both sides of the surface. For example, if the molecules of the materials attract each other, surface energy is reduced while if they repel each other the surface energy is increased. Thus, more appropriately, the surface energy is the energy of the interface between two materials and depends on both of them.
We make the following observations from above:
(i) Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of the interface between the plane of the liquid and any other substance; it also is the extra energy that the molecules at the interface have as compared to molecules in the interior.
(ii) At any point on the interface besides the boundary, we can draw a line and imagine equal and opposite surface tension forces S per unit length of the line acting perpendicular to the line, in the plane of the interface. The line is in equilibrium. To be more specific, imagine a line of atoms or molecules at the surface. The atoms to the left pull the line towards them; those to the right pull it towards them! This line of atoms is in equilibrium under tension. If the line really marks the end of the interface, as in Figure 10.16 (a) and (b) there is only the force S per unit length acting inwards.
Table 10.3 gives the surface tension of various liquids. The value of surface tension depends on temperature. Like viscosity, the surface tension of a liquid usually falls with temperature.
Table 10.3 Surface tension of some liquids at the temperatures indicated with the heats of the vaporisation
A fluid will stick to a solid surface if the surface energy between fluid and the solid is smaller than the sum of surface energies between solid-air, and fluid-air. Now there is attraction between the solid surface and the liquid. It can be directly measured experimentaly as schematically shown in Fig. 10.18. A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the other side, with its horizontal edge just over water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights are added till the plate just clears water.
Fig. 10.18 Measuring Surface Tension.
Suppose the additional weight required is W. Then from Eq. 10.22 and the discussion given there, the surface tension of the liquid-air interface is
Sla = (W/2l) = (mg/2l ) (10.23)
where m is the extra mass and l is the length of the plate edge. The subscript (la) emphasises the fact that the liquid-air interface tension is involved.
#### 10.6.3 Angle of Contact
The surface of liquid near the plane of contact, with another medium is in general curved. The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as angle of contact. It is denoted by θ. It is different at interfaces of different pairs of liquids and solids. The value of θ determines whether a liquid will spread on the surface of a solid or it will form droplets on it. For example, water forms droplets on lotus leaf as shown in Fig. 10.19 (a) while spreads over a clean plastic plate as shown in Fig. 10.19(b).
(a)
(b)
Fig. 10.19 Different shapes of water drops with interfacial tensions (a) on a lotus leaf (b) on a clean plastic plate.
We consider the three interfacial tensions at all the three interfaces, liquid-air, solid-air and solid-liquid denoted by Sla, Ssa and Ssl , respectively as given in Fig. 10.19 (a) and (b). At the line of contact, the surface forces between the three media must be in equilibrium. From the Fig. 10.19(b) the following relation is easily derived.
Sla cos θ + Ssl = Ssa (10.24)
The angle of contact is an obtuse angle if Ssl > Sla as in the case of water-leaf interface while it is an acute angle if Ssl < Sla as in the case of water-plastic interface. When θ is an obtuse angle then molecules of liquids are attracted strongly to themselves and weakly to those of solid, it costs a lot of energy to create a liquid-solid surface, and liquid then does not wet the solid. This is what happens with water on a waxy or oily surface, and with mercury on any surface. On the other hand, if the molecules of the liquid are strongly attracted to those of the solid, this will reduce Ssl and therefore, cos θ may increase or θ may decrease. In this case θ is an acute angle. This is what happens for water on glass or on plastic and for kerosene oil on virtually anything (it just spreads). Soaps, detergents and dying substances are wetting agents. When they are added the angle of contact becomes small so that these may penetrate well and become effective. Water proofing agents on the other hand are added to create a large angle of contact between the water and fibres.
#### 10.6.4 Drops and Bubbles
One consequence of surface tension is that free liquid drops and bubbles are spherical if effects of gravity can be neglected. You must have seen this especially clearly in small drops just formed in a high-speed spray or jet, and in soap bubbles blown by most of us in childhood. Why are drops and bubbles spherical? What keeps soap bubbles stable?
As we have been saying repeatedly, a liquid-air interface has energy, so for a given volume the surface with minimum energy is the one with the least area. The sphere has this property. Though it is out of the scope of this book, but you can check that a sphere is better than at least a cube in this respect! So, if gravity and other forces (e.g. air resistance) were ineffective, liquid drops would be spherical.
Another interesting consequence of surface tension is that the pressure inside a spherical drop Fig. 10.20(a) is more than the pressure outside. Suppose a spherical drop of radius r is in equilibrium. If its radius increase by r. The extra surface energy is
[4π(r + r) 2- 4πr2] Sla = 8πr r Sla (10.25)
If the drop is in equilibrium this energy cost is balanced by the energy gain due to expansion under the pressure difference (PiPo) between the inside of the bubble and the outside. The work done is
W = (PiPo) 4πr2r (10.26)
so that
(PiPo) = (2 Sla/ r) (10.27)
In general, for a liquid-gas interface, the convex side has a higher pressure than the concave side. For example, an air bubble in a liquid, would have higher pressure inside it.
See Fig 10.20 (b).
Fig. 10.20 Drop, cavity and bubble of radius r.
A bubble Fig 10.20 (c) differs from a drop and a cavity; in this it has two interfaces. Applying the above argument we have for a bubble
(PiPo) = (4 Sla/ r) (10.28)
This is probably why you have to blow hard, but not too hard, to form a soap bubble. A little extra air pressure is needed inside!
#### 10.6.6 Detergents and Surface Tension
We clean dirty clothes containing grease and oil stains sticking to cotton or other fabrics by adding detergents or soap to water, soaking clothes in it and shaking. Let us understand this process better.
Washing with water does not remove grease stains. This is because water does not wet greasy dirt; i.e., there is very little area of contact between them. If water could wet grease, the flow of water could carry some grease away. Something of this sort is achieved through detergents. The molecules of detergents are hairpin shaped, with one end attracted to water and the other to molecules of grease, oil or wax, thus tending to form water-oil interfaces. The result is shown in Fig. 10.22 as a sequence of figures.
In our language, we would say that addition of detergents, whose molecules attract at one end and say, oil on the other, reduces drastically the surface tension S (water-oil). It may even become energetically favourable to form such interfaces, i.e., globs of dirt surrounded by detergents and then by water. This kind of process using surface active detergents or surfactants is important not only for cleaning, but also in recovering oil, mineral ores etc.
Fig. 10.22 Detergent action in terms of what detergent molecules do.
Example 10.11 The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is 7.30×10-2 Nm-1. 1 atmospheric pressure = 1.01 × 105 Pa, density of water = 1000 kg/m3, g = 9.80 m s-2. Also calculate the excess pressure.
Answer The excess pressure in a bubble of gas in a liquid is given by 2S/r, where S is the surface tension of the liquid-gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4S/r.) The radius of the bubble is r. Now the pressure outside the bubble Po equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
Po = (1.01 × 105 Pa + 0.08 m × 1000 kg m–3
× 9.80 m s
–2)
= 1.01784 × 105 Pa
Therefore, the pressure inside the bubble is
Pi = Po + 2S/r
= 1.01784 × 105 Pa + (2 × 7.3 × 10-2 Pa m/10-3 m)
= (1.01784 + 0.00146) × 105 Pa
= 1.02 × 105 Pa where the radius of the bubble is taken
to be equal to the radius of the capillary tube, since the bubble is hemispherical ! (The answer has been rounded off to three significant figures.) The excess pressure in the bubble is 146 Pa.
#### 10.6.5 Capillary Rise
One consequence of the pressure difference across a curved liquid-air interface is the well-known effect that water rises up in a narrow tube in spite of gravity. The word capilla means hair in Latin; if the tube were hair thin, the rise would be very large. To see this, consider a vertical capillary tube of circular cross section (radius a) inserted into an open vessel of water (Fig. 10.21). The contact angle between water and glass is acute. Thus the surface of water in the capillary is concave. This means that there is a pressure difference between the two sides of the top surface. This is given by
(PiPo) =(2S/r) = 2S/(a sec θ )
= (2S/a) cos θ (10.29)
Fig. 10.21 Capillary rise, (a) Schematic picture of a narrow tube immersed water.
(b) Enlarged picture near interface.
Thus the pressure of the water inside the tube, just at the meniscus (air-water interface) is less than the atmospheric pressure. Consider the two points A and B in Fig. 10.21(a). They must be at the same pressure, namely
P0 + h ρ g = Pi = PA (10.30)
where ρ is the density of water and h is called the capillary rise [Fig. 10.21(a)]. Using Eq. (10.29) and (10.30) we have
h ρ g = (PiP0) = (2S cos θ )/a (10.31)
The discussion here, and the Eqs. (10.26) and (10.27) make it clear that the capillary rise is due to surface tension. It is larger, for a smaller a. Typically it is of the order of a few cm for fine capillaries. For example, if a = 0.05 cm, using the value of surface tension for water (Table 10.3), we find that
h = 2S/(ρ g a)
= 2.98 × 10–2 m = 2.98 cm
Notice that if the liquid meniscus is convex, as for mercury, i.e., if cos θ is negative then from Eq. (10.30) for example, it is clear that the liquid will be lower in the capillary !
#### SUMMARY
1. The basic property of a fluid is that it can flow. The fluid does not have any resistance to change of its shape. Thus, the shape of a fluid is governed by the shape of its container.
2. A liquid is incompressible and has a free surface of its own. A gas is compressible and it expands to occupy all the space available to it.
3. If F is the normal force exerted by a fluid on an area A then the average pressure Pav is defined as the ratio of the force to area
4. The unit of the pressure is the pascal (Pa). It is the same as N m-2. Other common units of pressure are
1 atm = 1.01×105 Pa
1 bar = 105 Pa
1 torr = 133 Pa = 0.133 kPa
1 mm of Hg = 1 torr = 133 Pa
5. Pascal’s law states that: Pressure in a fluid at rest is same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.
6. The pressure in a fluid varies with depth h according to the expression
P = Pa + ρgh
where ρ is the density of the fluid, assumed uniform.
7. The volume of an incompressible fluid passing any point every second in a pipe of non uniform crossection is the same in the steady flow.
v A = constant ( v is the velocity and A is the area of crossection)
The equation is due to mass conservation in incompressible fluid flow.
8. Bernoulli’s principle states that as we move along a streamline, the sum of the pressure (P), the kinetic energy per unit volume (ρv2/2) and the potential energy per unit volume (ρgy) remains a constant.
P + ρv2/2 + ρgy = constant
The equation is basically the conservation of energy applied to non viscuss fluid motion in steady state. There is no fluid which have zero viscosity, so the above statement is true only approximately. The viscosity is like friction and converts the kinetic energy to heat energy.
9. Though shear strain in a fluid does not require shear stress, when a shear stress is applied to a fluid, the motion is generated which causes a shear strain growing with time. The ratio of the shear stress to the time rate of shearing strain is known as coefficient of viscosity, η.
where symbols have their usual meaning and are defined in the text.
10. Stokes’ law states that the viscous drag force F on a sphere of radius a moving with velocity v through a fluid of viscosity is, F = – 6πηav.
11. Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of interface between the liquid and the bounding surface. It is the extra energy that the molecules at the interface have as compared to the interior.
#### Points to ponder
1. Pressure is a scalar quantity. The definition of the pressure as “force per unit area” may give one false impression that pressure is a vector. The “force” in the numerator of the definition is the component of the force normal to the area upon which it is impressed. While describing fluids as a concept, shift from particle and rigid body mechanics is required. We are concerned with properties that vary from point to point in the fluid.
2. One should not think of pressure of a fluid as being exerted only on a solid like the walls of a container or a piece of solid matter immersed in the fluid. Pressure exists at all points in a fluid. An element of a fluid (such as the one shown in Fig. 10.2) is in equilibrium because the pressures exerted on the various faces are equal.
3. The expression for pressure
P = Pa + ρgh
holds true if fluid is incompressible. Practically speaking it holds for liquids, which are largely incompressible and hence is a constant with height.
4. The gauge pressure is the difference of the actual pressure and the atmospheric pressure.
PPa = Pg
Many pressure-measuring devices measure the gauge pressure. These include the tyre pressure gauge and the blood pressure gauge (sphygmomanometer).
5. A streamline is a map of fluid flow. In a steady flow two streamlines do not intersect as it means that the fluid particle will have two possible velocities at the point.
6. Bernoulli’s principle does not hold in presence of viscous drag on the fluid. The work done by this dissipative viscous force must be taken into account in this case, and P2 [Fig. 10.9] will be lower than the value given by Eq. (10.12).
7. As the temperature rises the atoms of the liquid become more mobile and the coefficient of viscosity, η falls. In a gas the temperature rise increases the random motion of atoms and η increases.
8. Surface tension arises due to excess potential energy of the molecules on the surface in comparison to their potential energy in the interior. Such a surface energy is present at the interface separating two substances at least one of which is a fluid. It is not the property of a single fluid alone.
Physical Quantity - Symbol - Dimensions - Unit - Remarks
#### Exercises
10.1 Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
10.2 Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent disolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
10.3 Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally ... with temperatures (increases/decreases)
(b) Viscosity of gases ... with temperature, whereas the viscosity of liquids ... with temperature (increases/decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to ..., while for fluids it is proportional to ... (shear strain/rate of shear strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller)
10.4 Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
(e) A spinning cricket ball in the air does not follow a parabolic trajectory
10.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
10.6 Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density Determine the height of the wine column for normal atmospheric pressure.
10.7 A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
10.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?
10.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?
10.10 In the previous problem, if 15.0 cm of water and spirit each is further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)
10.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
10.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
10.13 Glycerine flows steadily through a horizontal tube of the length of 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = $1.3×{10}^{3}$ kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
10.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m–3.
10.15 Figures 10.23(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?
Figure 10.23
10.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
10.17 A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
10.18 Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.
Fig. 10.24
10.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? The surface tension of mercury at that temperature (20 °C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also, give the excess pressure inside the drop.
10.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10–2 N m–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × ${10}^{5}$ Pa).
10.21 A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. Compute the force necessary to keep the door close.
10.22 A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a). When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) is poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
Fig. 10.25
10.23 Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
10.24 During a blood transfusion, the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Use the density of whole blood from Table 10.1].
10.25 In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
10.26 (a) What is the largest average velocity of blood flow in an artery of radius 2×10–3m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s).
10.27 A plane is in level flight at a constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).
10.28 In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
10.29 Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? The surface tension of mercury at the temperature of the experiment is 0.465 N m–1. Density of mercury = 13.6 × 103 kg m–3.
10.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? The surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 ${\mathrm{ms}}^{-2}$).
Calculator/Computer – Based Problem
10.31 (a) It is known that density ρ of air decreases with height y as:
where ρ0 = 1.25 kg m–3 is the density at sea level and ${\mathrm{y}}_{0}$ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains a constant (isothermal conditions). Also, assume that the value of g remains constant.
(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y0 = 8000 m and ${\mathrm{\rho }}_{\mathrm{He}}$= 0.18 kg m–3].
Appendix 10.1: What is blood pressure?
In evolutionary history, there occurred a time when animals started spending a significant amount of time in the upright position. This placed a number of demands on the circulatory system. The venous system that returns blood from the lower extremities to the heart underwent changes. You will recall that veins are blood vessels through which blood returns to the heart. Humans and animals such as the giraffe have adapted to the problem of moving blood upward against gravity. But animals such as snakes, rats and rabbits will die if held upwards since the blood remains in the lower extremities and the venous system is unable to move it towards the heart.
Fig. 10.26 Schematic view of the gauge pressures in the arteries in various parts of the human body while standing or lying down. The pressures shown are averaged over a heart cycle.
Figure 10.26 shows the average pressures observed in the arteries at various points in the human body. Since viscous effects are small, we can use Bernoulli’s equation, Eq. (10.13),
to understand these pressure values. The kinetic energy term (ρ v2/2) can be ignored since the velocities in the three arteries are small ( 0.1 m s–1) and almost constant. Hence the gauge pressures at the brain PB, the heart PH, and the foot PF are related by
PF = PH + ρghH = PB + ρghB (10.34)
where ρ is the density of blood.
Typical values of the heights to the heart and the brain are hH = 1.3 m and hB = 1.7 m. Taking ρ = 1.06 × 103 kg m–3 we obtain that PF = 26.8 kPa (kilopascals) and PB = 9.3 kPa given that PH = 13.3 kPa. Thus the pressures in the lower and upper parts of the body are so different when a person is standing, but are almost equal when he is lying down. As mentioned in the text the units for pressure more commonly employed in medicine and physiology are torr and mm of Hg. 1 mm of Hg = 1 torr = 0.133 kPa. Thus the average pressure at the heart is PH = 13.3 kPa = 100 mm of Hg.
The human body is a marvel of nature. The veins in the lower extremities are equipped with valves, which open when blood flows towards the heart and close if it tends to drain down. Also, blood is returned at least partially by the pumping action associated with breathing and by the flexing of the skeletal muscles during walking. This explains why a soldier who is required to stand at attention may faint because of insufficient return of the blood to the heart. Once he is made to lie down, the pressures become equalized and he regains consciousness.
An instrument called the sphygmomanometer usually measures the blood pressure of humans. It is a fast, painless and non-invasive technique and gives the doctor a reliable idea about the patient’s health. The measurement process is shown in Fig. 10.27. There are two reasons why the upper arm is used. First, it is at the same level as the heart and measurements here give values close to that at the heart. Secondly, the upper arm contains a single bone and makes the artery there (called the brachial artery) easy to compress. We have all measured pulse rates by placing our fingers over the wrist. Each pulse takes a little less than a second. During each pulse the pressure in the heart and the circulatory system goes through a maximum as the blood is pumped by the heart (systolic pressure) and a minimum as the heart relaxes (diastolic pressure). The sphygmomanometer is a device, which measures these extreme pressures. It works on the principle that blood flow in the brachial (upper arm) artery can be made to go from laminar to turbulent by suitable compression. Turbulent flow is dissipative, and its sound can be picked up on the stethoscope.
The gauge pressure in an air sack wrapped around the upper arm is measured using a manometer or a dial pressure gauge (Fig. 10.27). The pressure in the sack is first increased till the brachial artery is closed. The pressure in the sack is then slowly reduced while a stethoscope placed just below the sack is used to listen to noises arising in the brachial artery. When the pressure is just below the systolic (peak) pressure, the artery opens briefly. During this brief period, the blood velocity in the highly constricted artery is high and turbulent and hence noisy. The resulting noise is heard as a tapping sound on the stethoscope. When the pressure in the sack is lowered further, the artery remains open for a longer portion of the heart cycle. Nevertheless, it remains closed during the diastolic (minimum pressure) phase of the heartbeat. Thus the duration of the tapping sound is longer. When the pressure in the sack reaches the diastolic pressure the artery is open during the entire heart cycle. The flow is however, still turbulent and noisy. But instead of a tapping sound we hear a steady, continuous roar on the stethoscope.
Fig. 10.27 Blood pressure measurement using the sphygmomanometer and stethoscope.
The blood pressure of a patient is presented as the ratio of systolic/diastolic pressures. For a resting healthy adult it is typically 120/80 mm of Hg (120/80 torr). Pressures above 140/90 require medical attention and advice. High blood pressures may seriously damage the heart, kidney and other organs and must be controlled.
Q. 1 A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicate the one that represents the velocity. From the plot shown in figure, indicate the one that represents the velocity(v) of the pebble as a function of time (t).
Q. 2 Which of the following diagrams does not represent a streamline flow?
Q. 3 Along a streamline,
(a) the velocity of a fluid particle remains constant
(b) the velocity of all fluid particles crossing a give position is constant
(c) the velocity of all fluid particles at a given instant is constant
(d) the speed of a fluid article remains constant
Q. 4 An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is
(a) 9:4
(b) 3:2
(c) $\sqrt{3}:\sqrt{2}$
(d) $\sqrt{2}:\sqrt{3}$
Q. 5 The angle of contact at the interface of water-glass is 0°, eth alcohol-glass is 0°, mercury-glass is 140° and methyliodide-glass is 30 A glass capillary is put in a trough containing one of these four liquids is observed, that the meniscus is convex. The liquid in the trough is
(a) water
(b) ethylalcohol
(c) mercury
(d) methyliodide
Q. 6 For a surface molecule,
(a) the net force on it is zero
(b) there is a net downward force
(c) the potential energy is less than that of a molecule inside
(d) the potential energy is more than that of a molecule inside
Q. 7 Pressureis a scalar quantity, because
(a) it is the ratio of force to area and both force and area are vectors
(b) it is the ratio of the magnitude of the force to area
(c) it is the ratio of the component of the force normal to the area
(d) it does not depend on the size of the area chosen
Q. 8 A wooden block with a coin placed on its top, floats in water as shown in
figure.
The distance l and h are shown in the figure. After sometime, the coin falls
into the water. Then,
(a) l decreases
(b) h decreases
(c) l increases
(d) h increases
Q. 9 With increase in temperature, the viscosity of
(a) gases decreases
(b) liquids increases
(c) gases increases
(d) liquids decreases
Q. 10 Streamline flow is more likely for liquids with
(a) high density
(b) high viscosity
(c) low density
(d) low viscosity
Q. 11 Is viscosity a vector?
Q. 12 Is surface tension a vector?
Q. 13 Iceberg floats in water with part of it submerged. Whatis the fraction of the volume of iceberg submerged, if the density of ice is
Q. 14 vessel filled with water is kept on a weighing pan and the scale adjusted to zero. A block of mass M and density $\rho$ is suspended by a massless spring of spring constant k. This block is submerged inside into the water in the vessel. What is the reading of the scale?
Q. 15 A cubical block of density p is floating on the surface of water. Out of its
height L, fraction x is submerged in water. The vessel is in an elevator
accelerating upward with acceleration a. What is the fraction immersed?
Q. 16 The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius r = 2.5$×{10}^{-5}m.$ The surface tension of sap is T=7.28 and the angle of contact is $0°$. Does surface tension alone account for the supply of water to the top of all trees?
Q. 17 The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle $\theta$. If the acceleration is a $m{s}^{-2}$, what will-be the slope of the free surface?
Q. 18 Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury
Q. 19 If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius R, break into N small droplets each of radius r. Estimate the drop in temperature.
Q. 20 The surface tension and vapour pressure of water at $20°C$ is Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20°C$?
Q. 21 (a) Pressure decreases as one ascends the atmosphere.If the density of air is p, what is the change in pressure dp over a differential height dh?
(b) Considering the pressure p to be proportional to the density, find the p at a height h if the pressure on the surface of the earth is ${p}_{0}$
(c) If at what height will be pressure drop to (1/10) the value at the surface of the earth?
(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.
Q. 22 Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases wit! increase in temperature and vanishes at boiling point. Given that thi latent heat of vaporisation for water , the mechanical equivalent of heat , density of water Avagardro’s number and the molecular weigh: of water ${M}_{A}$ = 10 kg for 1 k mole.
(a) Estimate the energy required for one molecule of water to evaporate.
(b) Show that the inter-molecular distance for water is d=${\left[\frac{{M}_{A}}{{N}_{A}}×\frac{1}{{\rho }_{w}}\right]}^{1/3}$and find its value.
(c) 1g of water in the vapour state at 1 atm occupies 1601cm?. Estimate the inter-molecular distance at boiling point, in the vapour state.
(d) During vaporisation a molecule overcomes a force F assumed constant, to go from an inter-molecular distance to d’. Estimate the value of F.
(e) Calculate F / d, which is a measure of the surface tension.
Q. 23 A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can the balloon lift when the outside temperature is 20°C? Assume air in an ideal gas, R=8.314 J mol${e}^{-1}{K}^{-1},$1 atm=1.013$×{10}^{5}$ Pa, the membrane tension is 5 $N{m}^{-1}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8443761467933655, "perplexity": 708.3446355760906}, "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-2022-40/segments/1664030334915.59/warc/CC-MAIN-20220926175816-20220926205816-00312.warc.gz"} |
http://blekko.com/wiki/Mach_number?source=672620ff | # Mach number
"Mach 2" redirects here. For the film, see Mach 2 (film).
An F/A-18 Hornet creating a vapor cone at transonic speed just before reaching the speed of sound
In fluid mechanics, Mach number (M or Ma) is a dimensionless quantity representing the ratio of speed of an object moving through a fluid and the local speed of sound.[1][2]
$\mathrm{M} = \frac {{v}}{{v_\text{sound}}}$
where
M is the Mach number,
v is the velocity of the source relative to the medium, and
vsound is the speed of sound in the medium.
Mach number varies by the composition of the surrounding medium and also by local conditions, especially temperature and pressure. The Mach number can be used to determine if a flow can be treated as an incompressible flow. If M < 0.2–0.3 and the flow is (quasi) steady and isothermal, compressibility effects will be small and a simplified incompressible flow model can be used.[1][2]
The Mach number is named after Austrian physicist and philosopher Ernst Mach, a designation proposed by aeronautical engineer Jakob Ackeret. Because the Mach number is often viewed as a dimensionless quantity rather than a unit of measure, with Mach, the number comes after the unit; the second Mach number is "Mach 2" instead of "2 Mach" (or Machs). This is somewhat reminiscent of the early modern ocean sounding unit "mark" (a synonym for fathom), which was also unit-first, and may have influenced the use of the term Mach. In the decade preceding faster-than-sound human flight, aeronautical engineers referred to the speed of sound as Mach's number, never "Mach 1."[3]
In French, the Mach number is sometimes called the "nombre de Sarrau" ("Sarrau number") after Émile Sarrau who researched explosions in the 1870s and 1880s.[4]
## Overview
The Mach number is commonly used both with objects traveling at high speed in a fluid, and with high-speed fluid flows inside channels such as nozzles, diffusers or wind tunnels. As it is defined as a ratio of two speeds, it is a dimensionless number. At Standard Sea Level conditions (corresponding to a temperature of 15 degrees Celsius), the speed of sound is 340.3 m/s[5] (1225 km/h, or 761.2 mph, or 661.5 knots, or 1116 ft/s) in the Earth's atmosphere. The speed represented by Mach 1 is not a constant; for example, it is mostly dependent on temperature and atmospheric composition and largely independent of pressure. Since the speed of sound increases as the temperature increases, the actual speed of an object traveling at Mach 1 will depend on the fluid temperature around it. Mach number is useful because the fluid behaves in a similar way at the same Mach number. So, an aircraft traveling at Mach 1 at 20°C or 68°F, at sea level, will experience shock waves in much the same manner as when it is traveling at Mach 1 at 11,000 m (36,000 ft) at −50°C or −58F, even though it is traveling at only 86% of its speed at higher temperature like 20°C or 68°F.[6]
## Classification of Mach regimes
While the terms "subsonic" and "supersonic" in the purest verbal sense refer to speeds below and above the local speed of sound respectively, aerodynamicists often use the same terms to talk about particular ranges of Mach values. This occurs because of the presence of a "transonic regime" around M = 1 where approximations of the Navier-Stokes equations used for subsonic design actually no longer apply, the simplest of many reasons being that the flow locally begins to exceed M = 1 even when the freestream Mach number is below this value.
Meanwhile, the "supersonic regime" is usually used to talk about the set of Mach numbers for which linearised theory may be used, where for example the (air) flow is not chemically reacting, and where heat-transfer between air and vehicle may be reasonably neglected in calculations.
In the following table, the "regimes" or "ranges of Mach values" are referred to, and not the "pure" meanings of the words "subsonic" and "supersonic".
Generally, NASA defines "high" hypersonic as any Mach number from 10 to 25, and re-entry speeds as anything greater than Mach 25. Aircraft operating in this regime include the Space Shuttle and various space planes in development.
RegimeMachmphkm/hm/sGeneral plane characteristics
Subsonic<0.8<610<980<270Most often propeller-driven and commercial turbofan aircraft with high aspect-ratio (slender) wings, and rounded features like the nose and leading edges.
Transonic0.8-1.2610-915980-1,470270-410Transonic aircraft nearly always have swept wings, delaying drag-divergence, and often feature design adhering to the principles of the Whitcomb Area rule.
Supersonic1.2–5.0915-3,8401,470–6,150410–1,710Aircraft designed to fly at supersonic speeds show large differences in their aerodynamic design because of the radical differences in the behaviour of flows above Mach 1. Sharp edges, thin aerofoil-sections, and all-moving tailplane/canards are common. Modern combat aircraft must compromise in order to maintain low-speed handling; "true" supersonic designs include the F-104 Starfighter, SR-71 Blackbird and BAC/Aérospatiale Concorde.
Hypersonic5.0–10.03,840–7,6806,150–12,3001,710–3,415Cooled nickel-titanium skin; highly integrated (due to domination of interference effects: non-linear behaviour means that superposition of results for separate components is invalid), small wings, such as those on the X-51A Waverider
High-hypersonic10.0–25.07,680–16,25012,300–30,7403,415–8,465Thermal control becomes a dominant design consideration. Structure must either be designed to operate hot, or be protected by special silicate tiles or similar. Chemically reacting flow can also cause corrosion of the vehicle's skin, with free-atomic oxygen featuring in very high-speed flows. Hypersonic designs are often forced into blunt configurations because of the aerodynamic heating rising with a reduced radius of curvature.
Re-entry speeds>25.0>16,250>30,740>8,465Ablative heat shield; small or no wings; blunt shape
## High-speed flow around objects
Flight can be roughly classified in six categories:
RegimeSubsonicTransonicSonicSupersonicHypersonicHigh-hypersonic
Mach<0.80.8–1.21.01.2–5.05.0–10.0>10.0
For comparison: the required speed for low Earth orbit is approximately 7.5 km/s = Mach 25.4 in air at high altitudes. The speed of light in a vacuum corresponds to a Mach number of approximately 881,000 (relative to air at sea level).
At transonic speeds, the flow field around the object includes both sub- and supersonic parts. The transonic period begins when first zones of M > 1 flow appear around the object. In case of an airfoil (such as an aircraft's wing), this typically happens above the wing. Supersonic flow can decelerate back to subsonic only in a normal shock; this typically happens before the trailing edge. (Fig.1a)
As the speed increases, the zone of M > 1 flow increases towards both leading and trailing edges. As M = 1 is reached and passed, the normal shock reaches the trailing edge and becomes a weak oblique shock: the flow decelerates over the shock, but remains supersonic. A normal shock is created ahead of the object, and the only subsonic zone in the flow field is a small area around the object's leading edge. (Fig.1b)
(a) (b)
Fig. 1. Mach number in transonic airflow around an airfoil; M < 1 (a) and M > 1 (b).
When an aircraft exceeds Mach 1 (i.e. the sound barrier) a large pressure difference is created just in front of the aircraft. This abrupt pressure difference, called a shock wave, spreads backward and outward from the aircraft in a cone shape (a so-called Mach cone). It is this shock wave that causes the sonic boom heard as a fast moving aircraft travels overhead. A person inside the aircraft will not hear this. The higher the speed, the more narrow the cone; at just over M = 1 it is hardly a cone at all, but closer to a slightly concave plane.
At fully supersonic speed, the shock wave starts to take its cone shape and flow is either completely supersonic, or (in case of a blunt object), only a very small subsonic flow area remains between the object's nose and the shock wave it creates ahead of itself. (In the case of a sharp object, there is no air between the nose and the shock wave: the shock wave starts from the nose.)
As the Mach number increases, so does the strength of the shock wave and the Mach cone becomes increasingly narrow. As the fluid flow crosses the shock wave, its speed is reduced and temperature, pressure, and density increase. The stronger the shock, the greater the changes. At high enough Mach numbers the temperature increases so much over the shock that ionization and dissociation of gas molecules behind the shock wave begin. Such flows are called hypersonic.
It is clear that any object traveling at hypersonic speeds will likewise be exposed to the same extreme temperatures as the gas behind the nose shock wave, and hence choice of heat-resistant materials becomes important.
## High-speed flow in a channel
As a flow in a channel becomes supersonic, one significant change takes place. The conservation of mass flow rate leads one to expect that contracting the flow channel would increase the flow speed (i.e. making the channel narrower results in faster air flow) and at subsonic speeds this holds true. However, once the flow becomes supersonic, the relationship of flow area and speed is reversed: expanding the channel actually increases the speed.
The obvious result is that in order to accelerate a flow to supersonic, one needs a convergent-divergent nozzle, where the converging section accelerates the flow to sonic speeds, and the diverging section continues the acceleration. Such nozzles are called de Laval nozzles and in extreme cases they are able to reach hypersonic speeds (Mach 13 (9,896 mph; 15,926 km/h) at 20°C).
An aircraft Machmeter or electronic flight information system (EFIS) can display Mach number derived from stagnation pressure (pitot tube) and static pressure.
## Calculation
The Mach number at which an aircraft is flying can be calculated by
$\mathrm{M} = \frac{v}{v_\text{sound}}$
where:
M is the Mach number
v is velocity of the moving aircraft and
vsound is the speed of sound at the given altitude
Note that the dynamic pressure can be found as:
$q = \frac{\gamma}{2} p\, \mathrm{M}^2$
Assuming air to be an ideal gas, the formula to compute Mach number in a subsonic compressible flow is derived from Bernoulli's equation for M < 1:[7]
$\mathrm{M}=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{q_c}{p}+1\right)^\frac{\gamma-1}{\gamma}-1\right]}\,$
where:
qc is impact pressure (dynamic pressure) and
p is static pressure
$\ \gamma\,$ is the ratio of specific heat of a gas at a constant pressure to heat at a constant volume (1.4 for air).
The formula to compute Mach number in a supersonic compressible flow is derived from the Rayleigh Supersonic Pitot equation:
$\frac{q_c}{p} = \left[\frac{\gamma+1}{2}\mathrm{M}^2\right]^\left(\frac{\gamma}{\gamma-1}\right)\cdot \left[ \frac{\gamma+1}{\left(1-\gamma+2 \gamma\, \mathrm{M}^2\right)} \right]^\left(\frac{1}{ \gamma-1 }\right)$
### Calculating Mach Number from Pitot Tube Pressure
At altitude, for reasons explained, Mach number is a function of temperature. Aircraft flight instruments, however, operate using pressure differential to compute Mach number, not temperature. The assumption is that a particular pressure represents a particular altitude and, therefore, a standard temperature. Aircraft flight instruments need to operate this way because the stagnation pressure sensed by a Pitot tube is dependent on altitude as well as speed.
Assuming air to be an ideal gas, the formula to compute Mach number in a subsonic compressible flow is found from Bernoulli's equation for M < 1 (above):[7]
$\mathrm{M} = \sqrt{5\left[\left(\frac{q_c}{p}+1\right)^\frac{2}{7}-1\right]}\,$
The formula to compute Mach number in a supersonic compressible flow can be found from the Rayleigh Supersonic Pitot equation (above) using parameters for air:
$\mathrm{M} = 0.88128485 \sqrt{\left(\frac{q_c}{p} + 1\right)\left(1 - \frac{1}{7\,\mathrm{M}^2}\right)^{2.5}}$
where:
qc is dynamic pressure measured behind a normal shock
As can be seen, M appears on both sides of the equation. The easiest method to solve the supersonic M calculation is to enter both the subsonic and supersonic equations into a computer spreadsheet such as Microsoft Excel, OpenOffice.org Calc, or some equivalent program to solve it numerically. It is first determined whether M is indeed greater than 1.0 by calculating M from the subsonic equation. If M is greater than 1.0 at that point, then the value of M from the subsonic equation is used as the initial condition in the supersonic equation. Then a simple iteration of the supersonic equation is performed, each time using the last computed value of M, until M converges to a value—usually in just a few iterations.[7] Alternatively, Newton's method can also be used. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "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": 0.8486563563346863, "perplexity": 1198.649819070128}, "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-2014-23/segments/1406510276250.57/warc/CC-MAIN-20140728011756-00185-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/monochromatic-light-frequency-60-10-14-hz-produced-laser-power-emitted-20-10-3-w-estimate-number-photons-emitted-second-average-source-refraction-monochromatic-light_4482 | # Monochromatic Light of Frequency 6.0 × 10^14 Hz is Produced by a Laser. the Power Emitted is 2.0 × 10^−3 W. Estimate the Number of Photons Emitted per Second on an Average by the Source - Physics
Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10−3 W. Estimate the number of photons emitted per second on an average by the source
#### SolutionShow Solution
Energy = hν
power is the energy per unit time therfore the energy given per second is = 2xx10^-3
On putting the values, we get:
number of photons * energy of one photon = total energy
number of photons=(2xx10^-3)/(6.63xx10^-34xx6xx10^14)
=5 xx 10^15 photons
Concept: Refraction of Monochromatic Light
Is there an error in this question or solution? | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "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": 0.8814373016357422, "perplexity": 1141.2329110771775}, "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-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00144.warc.gz"} |
http://mathhelpforum.com/trigonometry/172708-trigonometry-2-a-print.html | # Trigonometry 2
• Feb 26th 2011, 11:09 AM
arm
Trigonometry 2
Find: $S= \frac{\sqrt{3}}{cos 650}+\frac{1}{sin 250}$
• Feb 26th 2011, 11:32 AM
Ackbeet
What ideas have you had so far?
• Feb 26th 2011, 11:52 AM
arm
I only remember general knownledge
• Feb 26th 2011, 11:55 AM
Ackbeet
This is not a homework service. You're expected to put forth some effort, at which point we can help you get unstuck. See Rule # 11 here.
I note that you've posted a lot of problems. That's perfectly fine. However, you need to post some effort in each one, and not expect us to simply solve them for you. That's not what this forum is about.
• Feb 26th 2011, 01:55 PM
mr fantastic
Quote:
Originally Posted by arm
I only remember general knownledge
Then go back and review your classnotes or textbook.
To get help you need to show some effort. Show your work and say where you're stuck. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "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": 0.8003238439559937, "perplexity": 2346.4952021551608}, "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-2017-04/segments/1484560280835.22/warc/CC-MAIN-20170116095120-00250-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.arxiv-vanity.com/papers/1607.01575/ | # On the steady-state behavior of a nonlinear power system model\thanksreffootnoteinfo
[ [ [ Automatic Control Laboratory at the Swiss Federal Institute of Technology (ETH) Zürich, Switzerland.
###### Abstract
In this article, we consider a dynamic model of a three-phase power system including nonlinear generator dynamics, transmission line dynamics, and static nonlinear loads. We define a synchronous steady-state behavior which corresponds to the desired nominal operating point of a power system and obtain necessary and sufficient conditions on the control inputs, load model, and transmission network, under which the power system admits this steady-state behavior. We arrive at a separation between the steady-state conditions of the transmission network and generators, which allows us to recover the steady-state of the entire power system solely from a prescribed operating point of the transmission network. Moreover, we constructively obtain necessary and sufficient steady-state conditions based on network balance equations typically encountered in power flow analysis. Our analysis results in several necessary conditions that any power system control strategy needs to satisfy.
p
thanks: [
footnoteinfo]This research is supported by ETH funds and the SNF Assistant Professor Energy Grant #160573. A preliminary version of part of the results in in this paper has been presented at the 6th IFAC Work-shop on Distributed Estimation and Control in Networked Systems, September 8-9, 2016, Tokyo, Japan.
First]Dominic GroßFirst]Catalin ArghirFirst]Florian Dörfler
ower system dynamics, steady-state behavior, port-Hamiltonian systems.
## 1 Introduction
The electric power system has been paraphrased as the most complex machine engineered by mankind (Kundur, 1994). Aside from numerous interacting control loops, power systems are large-scale, and contain highly nonlinear dynamics on multiple time scales from mechanical and electrical domains. As a result power system analysis and control is typically based on simplified models of various degrees of fidelity (Sauer and Pai, 1998).
A widely accepted reduced power system model is a structure-preserving multi-machine model, where each generator model is reduced to the swing equation modeling the interaction between the generator rotor and the grid, which is itself modeled at quasi-steady-state via the nonlinear algebraic power balance equations, see e.g. (van der Schaft and Stegink, 2016). Despite being based on time-scale separations, quasi-stationarity assumptions, and multiple other simplifications, this model has proved itself useful for power system analysis and control (Kundur, 1994; Sauer and Pai, 1998). Nevertheless, the validity of the simplified model has always been a subject of debate; see (Caliskan and Tabuada, 2015; Monshizadeh et al., 2016) for recent discussions.
The modeling, analysis, and control of power systems has seen a surging research activity in the last years. One particular question of interest concerns the analysis of first-principle nonlinear multi-machine power system models without simplifying generator modeling assumptions and with dynamic (and not quasi-stationary) transmission network models. Fiaz et al. (2013) consider a highly detailed power system model based on port-Hamiltonian system modeling, and they carry out a stability analysis for a single generator connected to a constant linear load. This model can be reduced to the classic swing equation model by replacing the electromagnetic generator dynamics with a static relationship between the mechanical power and electrical power supplied to a generator (van der Schaft and Stegink, 2016).
Caliskan and Tabuada (2014) consider stability analysis of a power system using incremental passivity methods. Their analysis requires, among others, the assumptions of a constant torque and field current at the generators. Unfortunately, their analysis also requires a power preservation property that is hard to verify and whose inherent difficulty is rooted in the specific coordinates used for the analysis. These coordinates are convenient for a single generator but incompatible for multiple generators (Caliskan and Tabuada, 2016).
Barabanov et al. (2017) study a single generator connected to an infinite bus (i.e. in isolation) and improve upon the previous papers by requiring milder conditions to certify stability, though it is unclear if the analysis can be extended to a multi-machine power systems. Related stability results have also been obtained for detailed models of so called grid-forming power converters that emulate the dynamics of generators (Natarajan and Weiss, 2014; Jouini et al., 2016). Finally, detailed generator models with the transmission networked modeled by quasi-stationary balance equations are studied by Stegink et al. (2016); Dib et al. (2009). In particular, Dib et al. (2009) study existence of equilibria to the nonlinear differential-algebraic model.
This paper is organized as follows: In Section 2, we introduce some basic definitions, the first-principle nonlinear dynamical model of a power system, and the synchronous steady-state dynamics. The results on steady-state operation of the power system are presented in Section 3. In Section 4 we state the main result in discuss its implications. The paper closes with some conclusions in Section 5.
## 2 Notation and Problem Setup
### 2.1 Notation
We use to denote the set of real numbers and to denote the set of positive real numbers. The set denotes the unit circle, an angle is a point .For column vectors , we use to denote a stacked vector, and for vectors or matrices , we use
R(θ)\coloneqq[cos(θ)−sin(θ)sin(θ)cos(θ)]=[r(θ)jr(θ)], (1)
where denotes the rotation matrix. Furthermore, we define the matrix , denotes the identity matrix of dimension , denotes the Kronecker product, and denotes the Euclidean norm.
### 2.2 Dynamical Model of a Power System
In this work, we consider a dynamic model of a three-phase power system including nonlinear generator dynamics, transmission line dynamics, and static nonlinear loads derived in Fiaz et al. (2013). The reader is referred to Fiaz et al. (2013) and the references therein for a detailed derivation. The following assumption is required to prove the main result of the manuscript. {assum} It is assumed that the three-phase electrical components (resistance, inductance, capacitance) of each device have identical values for each phase. In addition, all three-phase state variables (i.e., voltages and currents) evolve on a two dimensional subspace of the three-phase -frame called the -frame (Clarke, 1943) during synchronous and balanced steady-state operation. Thus, without loss of generality, we can restrict the analysis to the -frame. The dynamical model used throughout this manuscript is obtained by transforming the model published in Fiaz et al. (2013) into the -frame by applying the Clarke transformation (Clarke, 1943) to each three-phase variable. Moreover, we will work with the co-energy variables, i.e., voltages and currents, instead of the natural Hamiltonian energy variables, i.e., charges and fluxes. We emphasize that the change of coordinates is used for simplicity of notation and all of the results also apply to the model derived in Fiaz et al. (2013) using port-Hamiltonian energy variables in -coordinates.
Because -coordinates can be interpreted as an embedding of the complex numbers into real-valued Euclidean coordinates, the rotation matrix plays the same role that the imaginary unit plays in traditional power system analysis in complex coordinates. We carry out our analysis in the stationary -frame to avoid any limitations which may arise by restricting the analysis to a specific rotating coordinate frame.
#### 2.2.1 Power System Topology
The power system model used in this work consists of generators with index set and AC voltage buses with index set that are interconnected via transmission lines with index set . The topology of the transmission network is described by the oriented incidence matrix of its associated graph, i.e., the voltage buses are the nodes and the transmission lines are the edges of the graph induced by . The incidence matrix of the AC network accounting for -coordinates is denoted by .
#### 2.2.2 Transmission Network
The transmission lines are modeled using the -model (Sauer and Pai, 1998) depicted in Figure 1. The state variables of the power grid model are the line currents and bus voltages , where is the current flowing through a transmission line and is the voltage of a voltage bus . The inputs to the transmission network are the currents , where is the current flowing out of an AC voltage bus and into a generator or a load. The dynamics of the voltage buses and transmission lines are given by
Cddtv =−EiT−iin, (2a) LTddtiT =−RTiT+E⊤v, (2b)
where denotes the matrix of voltage bus capacitances , for , aggregates the line inductances , for all , and denotes the matrix of line resistances , for .
#### 2.2.3 Synchronous Machines
A synchronous machine with index is modeled by
ddtθk =ωk (3a) mkddtωk =−dkωk−τe,k+τm,k (3b) Lk(θk)ddtik =−Rkik+[vkvf,k]−vind,k, (3c)
where aggregates the stator currents and rotor currents , with excitation current , and the damper winding currents and . Moreover, denotes the voltages at the generators voltage bus, is rotational speed of the rotor, its angular displacement, is the electrical torque acting on the rotor, and is a voltage induced by the rotation of the machine. The machine is actuated by the voltage across the excitation winding of the generator and the mechanical torque applied to the rotor. The mechanical and electrical part of the machine are depicted in Figure 2. The inertia and damping of the rotor are denoted by and , and the windings have resistance with stator resistance , and rotor resistance , with excitation winding resistance , and damper winding resistances and . The inductance matrix is defined using the stator inductance , rotor inductance , and mutual inductance :
where , , , and are the inductances of the excitation winding, damper windings, and the mutual inductance of the excitation and damper winding, respectively. With the stator winding inductance and rotor saliency the stator inductance is given by
Ls,k(θk)=[ls,k00ls,k]+R(2θk)[lsa,k00−lsa,k].
Finally, is defined using the mutual inductances , , :
Lm,k(θk)=R(θk)[lsf,klsd,k000−lsq,k].
Finally, the electrical torque acting on the rotor is given by:
τe,k=12i⊤k(Lk(θk)\mathscrlj+\mathscrlj⊤Lk(θk))ik, (4)
and the voltage induced in the machine windings due to the rotation of the machine is given by
vind,k=ωk(Lk(θk)\mathscrlj⊤+\mathscrljL(θ))ik. (5)
In this work, we consider the static load model used in Fiaz et al. (2013). Specifically, loads are included in the model via a load current (flowing out of a voltage bus ) that is a function of the bus voltage , and satisfies the dissipation inequality . We additionally assume that if .
#### 2.2.5 Dynamic Model of the Power System
To obtain the dynamic model of the entire power system, the the synchronous machine model (3) and load models are combined with the transmission network model (2) by defining , where and . With the aggregated vectors , , , the state vector of the nonlinear power system model is , with . Using the vectors , and the inputs are given by , . Moreover, to simplify the notation, we define , , the matrices , , , , , and , which collect the node matrices (e.g., ), and collecting the time constants. We use indicator matrices , , and to describe the interconnection of the components results in the following model of the entire power system:
ddtx=M(x)−1⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣ω−Dω−τe+τm−Ri+I⊤vv+Ifvf−vind−Ivi−EiT−il−RTiT+E⊤v⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=f(x,u), (6)
We formulate the following steady-state dynamics to describe synchronous and balanced steady-state operation of a power system at a nominal frequency as
ddtθk =w, ∀k∈G, (7a) ddtωk =0, ∀k∈G, (7b) ddtik =w\mathscrljik, ∀k∈G, (7c) ddtvk =wjvk, ∀k∈V, (7d) ddtiT,k =wjiT,k, ∀k∈T. (7e)
The steady-state behavior (7) specifies a balanced, synchronous, and sinusoidal operation of each grid component. This results in the steady-state dynamics with nominal frequency
fd(x,w)=(\mathbbl1ngw,\mathbbl0ng,wJgi,wJvv,wJTiT), (8)
where , , and .
In the next section, we derive necessary and sufficient conditions under which the nonlinear power system dynamics (6) coincide with the steady-state dynamics (8) for all time. In particular, we derive conditions on the control inputs and load models, and obtain a separation between the steady-state conditions of the transmission network and generators. This result allows to explicitly recover a steady state of the entire power system and corresponding steady-state inputs from a prescribed steady state of the transmission network. As we will see, this result also implies that the power system admits a non-trivial synchronous steady-state dynamics (8) if and only if there exists a non-trivial solution to the well-known power flow equations.
## 3 Conditions for the Existence of Synchronous Steady-States
We begin our analysis by defining the set on which the vector fields of the dynamics (6) and (8) coincide point-wise in time and the residual dynamics vanish:
S\coloneqq{(x,u,w)∈Rnx×Rnu×R∣∣ρ(x,u,w)=\mathbbl0nx}.
By multiplying with from the left, the residual dynamics can be rewritten as follows:
ρ(x,u,w)=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣ω−\mathbbl1ngwDω+τe−τm(R+wL(θ)Jg)i−I⊤vv−Ifvf+vindwCJvv+Ivi+EiT+il(RT+wLTJT)iT−E⊤v⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦.
We emphasize that the power system operates at the synchronous steady state if and only if holds for all . In this section, we derive conditions which ensure existence of states and inputs such that and obtain conditions which ensure that that all trajectories of the dynamics (6) starting in remain in for all time. Together, these conditions are necessary and sufficient conditions for the power system to admit the synchronous steady-state behavior (8).
We will first derive necessary and sufficient conditions on , , and the load model under which implies that for all , i.e., that is control-invariant. We thereby characterize the steady-state control inputs, admissible synchronous frequency, and class of static load models, for which (8) is a steady-state behavior of the power system (6). Based on these results we will provide conditions for the existence of states and inputs such that .
### 3.1 Control-Invariance of the Set S
To establish control-invariance of , we consider the dynamics obtained by combining the nonlinear power grid dynamics described by with controller (torque and excitation) dynamics and describing the dynamics of the synchronous frequency possibly both depending on time and the system state:
ddt(x,u,w)=(f(x,u),gu(t,x,u,w),gw(t,x,u,w)). (9)
{thm}
(Control-invariance condition) The set is invariant with respect to the dynamics (9) if and only if the control inputs as well as the synchronous frequency take constant values on , i.e., and for all , and the load current satisfies for all . {pf} The set is invariant with respect to (9) if and only if for all . On it holds that , resulting in the following necessary and sufficient condition for invariance of for all :
dρdt=∂ρ∂xfd+∂ρ∂ugu+∂ρ∂wgw=\mathbbl0nx. (10)
Furthermore, the partials of on can be obtained from the following identities:
∂ρ∂u=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣[l]\mathbbl0ng×ng\mathbbl0ng×ngIng\mathbbl0ng×ng\mathbbl03ng×ngIf\mathbbl02nv×ng\mathbbl02nv×ng\mathbbl02nT×ng\mathbbl02nT×ng⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦,∂ρ∂w=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣\mathbbl1ng\mathbbl0ngL(θ)JgiCJvvLTJTiT⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦. (11)
By using it can be verified that
∂τe,k∂θkw+∂τe,k∂ikw\mathscrljik =0, ∂vind,k∂θkw+∂vind,k∂ikw\mathscrljik =w\mathscrljvind,k,
and one obtains
(∂ρ∂x)fd=w⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣\mathbbl02ng(RJg+wJgL(θ)Jg)i−I⊤vJvv+JgvindwCJ2vv+IvJgi+EJTiT+∂il∂vJvv(RT+wLTJT)JTiT−E⊤Jvv⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦.
Next, it can be verified that , and . Moreover, , and commute with diagonal matrices. This results in
By definition of we have for all that
(∂ρ∂x)fd=w⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣\mathbbl02ngJgIfvf∂il∂vJvv−Jvil\mathbbl02nt⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=w⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣\mathbbl02ng\mathbbl02ng∂il∂vJvv−Jvil\mathbbl02nt⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦, (12)
where the last equality follows from . By substituting for the partials in (10) based on (11) and (12) one obtains the following condition for invariance of with respect to the dynamics (9) for all :
⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣\mathbbl0ng×ng\mathbbl0ng×ng\mathbbl1ngIng\mathbbl0ng×ng\mathbbl0ng\mathbbl03ng×ngIfL(θ)Jgi\mathbbl02nv×ng\mathbbl02nv×ngCJvv\mathbbl02nT×ng\mathbbl02nT×ngLTJTiT⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦[gugw]=w⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣\mathbbl0ng\mathbbl0ng\mathbbl02ngJvil−∂il∂vJvv\mathbbl02nt⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦.
This holds if and only if , , and . Moreover, on it holds that , and it follows that
Based on this result we restrict ourselves from now to a constant (possibly zero) frequency corresponding to the nominal synchronous operating frequency of the power system and constant inputs, i.e., and . Letting for all results in the following set parametrized in :
Sω0:={(x,u)∈Rnx×Rnu∣∣ρ(x,u,ω0)=\mathbbl0nx}. (13)
In the next subsection, we will use Theorem 3.1 to characterize the class of static load models that are consistent with the synchronous steady-state behavior (8).
The following result identifies the class of all load models for which the set is control invariant. {thm} Any static load model, i.e. any that satisfies and , which is consistent with the synchronous steady state, i.e. satisfies the steady-state conditions of Theorem 3.1, can be expressed in the form , where , , and . {pf} Without loss of generality any that satisfies if can be expressed as . According to Theorem 3.1 consistency of the load model with the steady state requires that holds on . Moreover, on it holds that . Therefore, in steady-state, the load current and voltage are required to have constant amplitude and rotate with the synchronous frequency , i.e. the relative angle between and and the ratio of their amplitudes is required to be constant. It follows that, for every voltage magnitude , all load currents consistent with the synchronous steady state can be expressed using a relative angle and a gain as follows:
il,k(vk)=μ(∥vk∥)R(δk(∥vk∥))vk. (14)
Theorem 3.1 and Theorem 3.2 show that besides synchronous steady-state operation also requires that the steady-state control inputs are constant and identifies the class of static load models which are consistent with the synchronous steady state. Next, we derive conditions for the existence of states and inputs such that .
### 3.3 Steady-State Analysis and Network Equations
In the following, we establish a connection between the steady-state conditions (13) and Kirchhoff’s equations for the transmission network (also often formulated as power balance/flow equations). The power system (6) can be partitioned into generators with dynamics (3) and the network consisting of the transmission line and voltage bus dynamics. Each generator is interconnected to the network by its terminal voltage (output of the network) and currents injected into the network. In the following we apply this separation to the steady-state conditions (13) and show that the steady-state conditions for the network can be used to fully characterize the synchronous steady state of the full multi-machine system. For notational convenience we define the vector of stator currents . The following equations describe Kirchhoff’s current law at the generator terminal and load buses, as well as Kirchhoff’s voltage law over the network branches:
ρN(is,v,iT)=[(Yl(v)+ω0JvC)v+(is,\mathbbl02nl)+EiT(RT+ω0JTLT)iT−E⊤v].
Based on the vector we define the solution set of the transmission network equations
Nω0:={(is,v,iT)∈Rnz∣∣ρN(is,v,iT)=\mathbbl0nN}. (15)
The following statement formalizes the separation that allows to recover the full system state from a solution to the steady-state equations (15) of the transmission network. In particular, given a solution to the transmission network equations , the remaining states and inputs such that can be recovered. This results in a simpler condition for steady-state operation which no longer depends on the rotor angles and directly links the currents injected by the generators to the network steady-state conditions.
{thm}
(Network generator separation) Consider the sets and defined in (13) and (15). There exists and such that only if . Conversely, for every there exists and such that .
In particular, given , the rotor polarization , and the voltage , where is the stator impedance, all corresponding satisfy
ω0lsf,kif,k =σk∥νk(θk)∥, (16a) jr(θk)∥νk(θk)∥ =σkνk(θk), (16b) as well as ωk=ω0, id,k=iq,k=0, and vf,k =rf,kif,k, (16c) τm,k =Dkω0+12i⊤k(Lk(θk)\mathscrlj+\mathscrlj⊤Lk(θk))ik. (16d)
Moreover, the particular case implies for all angles , and any and any satisfy the steady-state conditions. {pf} To prove the first statement, we note that the last two equations defining are identical to the equations defining . It trivially follows that no and exist such that if | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9521956443786621, "perplexity": 960.5922701646517}, "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-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00488.warc.gz"} |
https://planetmath.org/dimensiontheoremforsymplecticcomplementproof | # dimension theorem for symplectic complement (proof)
We denote by $V^{\star}$ the dual space of $V$, i.e., linear mappings from $V$ to $\mathbb{R}$. Moreover, we assume known that $\dim V=\dim V^{\ast}$ for any vector space $V$.
We begin by showing that the mapping $S:V\to V^{*}$, $a\mapsto\omega(a,\cdot)$ is an linear isomorphism. First, linearity is clear, and since $\omega$ is non-degenerate, $\ker S=\{0\}$, so $S$ is injective. To show that $S$ is surjective, we apply the http://planetmath.org/node/2238rank-nullity theorem to $S$, which yields $\dim V=\dim\mathop{\mathrm{img}}S$. We now have $\mathop{\mathrm{img}}S\subset V^{*}$ and $\dim\mathop{\mathrm{img}}S=\dim V^{\ast}$. (The first assertion follows directly from the definition of $S$.) Hence $\mathop{\mathrm{img}}S=V^{\ast}$ (see this page (http://planetmath.org/VectorSubspace)), and $S$ is a surjection. We have shown that $S$ is a linear isomorphism.
Let us next define the mapping $T:V\to W^{*}$, $a\mapsto\omega(a,\cdot)$. Applying the http://planetmath.org/node/2238rank-nullity theorem to $T$ yields
$\displaystyle\dim V$ $\displaystyle=$ $\displaystyle\dim\ker T+\dim\mathop{\mathrm{img}}T.$ (1)
Now $\ker T=W^{\omega}$ and $\mathop{\mathrm{img}}T=W^{*}$. To see the latter assertion, first note that from the definition of $T$, we have $\mathop{\mathrm{img}}T\subset W^{*}$. Since $S$ is a linear isomorphism, we also have $\mathop{\mathrm{img}}T\supset W^{*}$. Then, since $\dim W=\dim W^{*}$, the result follows from equation 1. $\Box$
Title dimension theorem for symplectic complement (proof) DimensionTheoremForSymplecticComplementproof 2013-03-22 13:32:52 2013-03-22 13:32:52 matte (1858) matte (1858) 5 matte (1858) Proof msc 15A04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 34, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.998893678188324, "perplexity": 170.6894970577407}, "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-2019-13/segments/1552912202506.45/warc/CC-MAIN-20190321072128-20190321094128-00135.warc.gz"} |
https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.blas.zgeap.html | # naginterfaces.library.blas.zgeap¶
naginterfaces.library.blas.zgeap(side, trans, perm, k, b)[source]
zgeap permutes the rows or columns of a complex rectangular matrix using an int array of permutations.
For full information please refer to the NAG Library document for f06vj
https://www.nag.com/numeric/nl/nagdoc_28.7/flhtml/f06/f06vjf.html
Parameters
sidestr, length 1
Specifies the operation to be performed.
and
.
and
.
and
.
and
.
transstr, length 1
Specifies the operation to be performed.
and
.
and
.
and
.
and
.
permint, array-like, shape
The indices which define the interchanges in the representation of . It is usual to have , but this is not necessary.
kint
, the number of columns of if , or the number of rows of if .
bcomplex, array-like, shape
Note: the required extent for this argument in dimension 1 is determined as follows: if : ; if : ; otherwise: .
Note: the required extent for this argument in dimension 2 is determined as follows: if : ; if : ; otherwise: .
The original matrix ; is if , or if .
Returns
bcomplex, ndarray, shape
The permuted matrix .
Raises
NagValueError
(errno )
On entry, error in parameter .
Constraint: or .
(errno )
On entry, error in parameter .
Constraint: or .
(errno )
On entry, error in parameter .
Constraint: .
(errno )
On entry, error in parameter .
Constraint: .
(errno )
On entry, error in parameter .
Constraint: .
Notes
No equivalent traditional C interface for this routine exists in the NAG Library.
zgeap performs one of the permutation operations
where is a complex matrix, and is a permutation matrix.
is represented in the form
where is the permutation matrix that interchanges items and ; that is, is the unit matrix with rows and columns and interchanged. If , .
Let denote the number of rows of if , or the number of columns of if : the function does not require to be passed as an argument, but assumes that , for .
This function requires the indices to be supplied in an int array; zgeapr() performs the same operation with the indices supplied in a float array. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "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": 0.9066489934921265, "perplexity": 2887.119011668}, "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-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00015.warc.gz"} |
https://www.physicsforums.com/threads/kinetic-velocity-independ-of-direction.283070/ | # Kinetic velocity independ. of direction?
1. Jan 5, 2009
### razored
Why is the velocity in the kinetic energy formula independent of the direction? I can't seem to figure out why. This is not a homework problem.
2. Jan 5, 2009
### jgens
Kinetic energy is defined to be the amount - magnitude - of work neccessary to accelerate an object to a given speed; hence, only the magnitude is important - regarding kinetic energy.
Mathematically, it's independant because the quantity is squared.
3. Jan 5, 2009
### Sisplat
The kinetic energy formula depends only of v.v (the length of the vector v squared). Obviously the length of a vector doesn't depend on its direction.
It doesn't matter what direction you run in, you use up the same amount of energy*.
* in a very idealized sports hall.
4. Jan 5, 2009
### razored
Mathematically speaking, no problem because it is squared; however, when we throw a ball 45 degrees above horizontal and then 45 degrees below, it just intuitively for me does not make sense that they contain the same kinetic energy. hmm.
5. Jan 5, 2009
### Sisplat
If you did it in the absence of gravity, they would have the same kinetic energy. You would also have trouble defining the term horizontal.
If you do your experiment on earth the problem is different.
Initially, at the exact moment you throw the two balls, they would have the same kinetic energy.
But after any time has passed, the ball 45 degrees above the horizon will be loosing kinetic energy and gaining gravitational potential energy (calculated with mass*gravity*height in a homogeneous - the same everywhere - gravitational field).
The other ball would gain kinetic energy and loose potential energy.
I hope this helps.
6. Jan 5, 2009
### jgens
Read my definition of kinetic energy.
Definition: Kinetic energy is defined to be the amount - magnitude - of work neccessary to accelerate an object to a given velocity. Therefore, kinetic energy is direction independent.
Sisplat also explained a physical scenario that illustrates why kinetic energy is always greater than or equal to zero.
7. Jan 5, 2009
### razored
I seem to comprehend it a bit more now. Thanks a bunch.
Similar Discussions: Kinetic velocity independ. of direction? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8908640146255493, "perplexity": 840.3351070475841}, "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-2017-17/segments/1492917124371.40/warc/CC-MAIN-20170423031204-00546-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://mathedu.academy/Blog/?tag=notation-exponentielle | # The Exponential Notation for the Floating Real Numbers
Our scientific calculators deal sometimes with numbers with a $$E$$ inside: this is the so called Exponential, or Scientific Notation.
But what does that mean, and how to read such numbers? | {"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": 0, "mathjax_display_tex": 1, "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": 0.9490170478820801, "perplexity": 1082.5724127807298}, "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-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00167.warc.gz"} |
http://thebeardedmathman.com/tag/square-roots/ | ## Multiplying and Dividing Square Roots, Rationalizing the Denominator
1.7.2 square root operations continued
Square Roots
Multiplication and Division
At some point square roots should no longer be considered an operation but rather the most efficient way to express a number. For example, the best way to write one hundred trillion is $1×{10}^{14}$. The best way to express the number times itself that is two is as $\sqrt{2}.$
That provides insight when we consider multiplying a rational number and an irrational number together. It is not confusing for some irrational numbers, like π. Nobody confused 3π because we understand that symbol is the best way to write the number. There’s not a way to rewrite multiples of π other than by writing the multiple in front.
However, $3\sqrt{2}$ is often written as $\sqrt{6}$. There are reasons explained by the order of operations which tell us why this is false, but understanding what the square root of two is perhaps offers the simplest insight.
$\sqrt{2}\approx 1.414$
$3\sqrt{2}=\sqrt{2}+\sqrt{2}+\sqrt{2}$
$3\sqrt{2}\approx 1.414+1.414+1.414$
4.242
The square root of six is approximately 2.449. Not the same thing at all.
The following, however, is true:
$\sqrt{2}×\sqrt{3}=\sqrt{6}$
and
$\sqrt{2×3}=\sqrt{6}$.
The following generalization can be used. Sometimes it is best to write things one way versus another, and it is up to you to decide if rewriting an expression offers insight.
$\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$
If two numbers are both square roots you can multiply their radicands together. But you cannot multiply the radicand of a square root with rational number like we saw above.
Division is a little more nuanced, but only when your denominator is a fraction.
This generalization is true for division:
$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
For example:
$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{2}.$
This can be calculated two ways.
$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{\frac{8}{4}}=\sqrt{2}.$
or
$\frac{\sqrt{8}}{\sqrt{4}}=\frac{2\sqrt{2}}{2}=\sqrt{2}.$
But you cannot divide rational numbers into the radicand, or the radicand of a square root into a rational number. Remember, square roots, when simplified, are the most efficient way of writing irrational numbers. If we used k to represent the square root of two, these types of confusing things would not be happening.
Nobody would confuse what is happening with
$\frac{6}{k}$. We simply cannot evaluate that because 6 and k do not have common factors. When k is written as the square root of two, sometimes people just see a 2 and reduce.
The only issue with division of square roots occurs if you end up with a square root in the denominator.
$\frac{5}{\sqrt{2}}$
Denominators must be rational and the square root of two is irrational. However, there’s an easy fix. Remember that $\sqrt{2}×\sqrt{2}=\sqrt{4},$ and $\sqrt{4}=2.$ It is also true that:
$\frac{\sqrt{2}}{\sqrt{2}}=1$ .
To Rationalize the Denominator, which means make the denominator a rational number, we just multiply as follows:
$\frac{5}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{2}$
Sometimes we end up with something like this:
$\frac{5}{3\sqrt{2}}$
Three is a rational number and is perfectly okay in the denominator. If you multiply by the fraction $\frac{3\sqrt{2}}{3\sqrt{2}},$ you can still get the simplified equivalent, but you’ll have extra reducing to do at the end. Instead, just multiply by the irrational portion.
$\frac{5}{3\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{6}$.
In summary, to divide or multiply with square roots, you can multiply or divide the radicands. However, if you’re multiplying or dividing rational numbers and square roots, you cannot combine the radicands and the rational numbers.
Practice Problems:
Perform the indicated operations:
$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(5\sqrt{7}\right)\left(3\sqrt{14}\right)\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\sqrt{15}\right)\left(\sqrt{3}\right)\\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3\sqrt{2}}{\sqrt{8}}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\sqrt{5}}{\sqrt{3}}\\ \\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{\sqrt{8}}\cdot \frac{\sqrt{2}}{6}\end{array}$
## Addition and Subtraction of Square Roots
Mathematical Operations and Square Roots
Part 1
In this section we will see why we can add things like $5\sqrt{2}+3\sqrt{2}$ but cannot add things like $2\sqrt{5}+2\sqrt{3}$. Later we will see how multiplication and division work when radicals (square roots and such) are involved.
Addition and Subtraction: Addition is just repeated counting. The expression $5\sqrt{2}$ means $\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}$, and the expression So if we add those two expressions, $5\sqrt{2}+3\sqrt{2},$ we get $8\sqrt{2}$ . Subtraction works the same way.
Consider the expression $2\sqrt{5}+2\sqrt{3}$. This means $\sqrt{5}+\sqrt{5}+\sqrt{3}+\sqrt{3}.$ The square root of five and the square root of three are different things, so the simplest we can write that sum is $2\sqrt{5}+2\sqrt{3}$.
A common way to describe when square roots can or cannot be added (or subtracted) is, “If the radicands are the same you add/subtract the number in front.” This is not a bad rule of thumb, but it treats square roots as something other than numbers.
$5×3+4×3=9×3$
The above statement is true. Five groups of three and four groups of three is nine groups of three.
$5\sqrt{3}+4\sqrt{3}=9\sqrt{3}$
The above statement is also true because five groups of the numbers squared that is three, plus four more groups of the same number would be nine groups of that number.
However, the following cannot be combined in such a fashion.
$3×8+5×2$
While this can be calculated, we cannot add the two terms together because the first portion is three $–$ eights and the second is five $–$ twos.
$3\sqrt{8}+5\sqrt{2}$
The same situation is happening here.
Common Mistake: The following is obviously wrong. A student learning this level of math would be highly unlikely to make such a mistake.
$7×2+9×2=16×4$
Seven $–$ twos and nine $–$ twos makes a total of sixteen $–$ twos, not sixteen $–$ fours. You’re adding the number of twos you have together, not the twos themselves. And yet, this is a common thing done with square roots.
$7\sqrt{2}+9\sqrt{2}=16\sqrt{4}$
This is incorrect for the same reason. The thing you are counting does not change by counting it.
Explanation: Why can you add $5\sqrt{2}+3\sqrt{2}$? Is that a violation of the order of operations (PEMDAS)? Clearly, the five and square root of two are multiplying, as are the three and the square root of two. Why does this work?
Multiplication is a short-cut for repeated addition of one particular number. Since both terms are repeatedly adding the same thing, we can combine them.
But if the things we are repeatedly adding are not the same, we cannot add them together before multiplying.
What About Something Like This: $3\sqrt{40}-9\sqrt{90}$?
Before claiming that this expression cannot be simplified you must make sure the square roots are fully simplified. It turns out that both of these can be simplified.
$3\sqrt{40}-9\sqrt{90}$
$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$
The dot symbol for multiplication is written here to remind us that all of these numbers are being multiplied.
$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$
$3\cdot 2\cdot \sqrt{10}-9\cdot 3\cdot \sqrt{10}$
$6\sqrt{10}-27\sqrt{10}$
$-21\sqrt{10}$
What About Something Like This: $\sqrt{7+7}$ versus $\sqrt{7}+\sqrt{7}.$
Notice that in the first expression there is a group, the radical symbol groups the sevens together. Since the operation is adding, this becomes:
$\sqrt{7+7}=\sqrt{14}$.
Since the square root of fourteen cannot be simplified, we are done.
The other expression becomes:
$\sqrt{7}+\sqrt{7}=2\sqrt{7}.$
Summary: If the radicals are the same number, the number in front just describes how many of them there are. You can combine (add/subtract) them if they are the same number. You are finished when you have combined all of the like terms together and all square roots are simplified.
Practice Problems: Perform the indicated operation.
$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{25}-5\sqrt{5}+5\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{48}+3\sqrt{3}\\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\sqrt{75}+8\sqrt{24}+\sqrt{75}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{200}+8\sqrt{8}-2\sqrt{32}\\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2\sqrt{98}+16\sqrt{2}\end{array}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 58, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8662310838699341, "perplexity": 376.84016440241953}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2019-26/segments/1560627999783.49/warc/CC-MAIN-20190625011649-20190625033649-00016.warc.gz"} |
https://www.physicsforums.com/threads/calculating-power-of-a-rotating-shaft.216969/ | # Calculating Power of a rotating shaft
1
Hi, it's been a number of years since I've even thought about physics, so this is a very simple question. I have a rig set up where I have a power source that is rotating a shaft. Attached to this shaft is a wheel with a string, I have this string going through a pulley in the ceiling, and pulling a hanging weight. What I need to do is calculate the power that I'm getting out of this device, I remember how to calculate work and power, but can't for the life of me figure out where the radius of the wheel comes into play, and the torque. Any help would be appreciated
Cheers,
Corey
2. ### FredGarvin
5,087
If you have the power, you can calculate the torque (neglecting friction) from:
$$P = T \dot{\theta}$$
where
$$P$$ = Power
$$T$$ = Torque
$$\dot{\theta}$$ = Angular speed
Just make sure your units are consistent! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.862939178943634, "perplexity": 326.7115934181384}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988718.8/warc/CC-MAIN-20150728002308-00031-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://electronics.stackexchange.com/questions/123370/theory-question-about-j-imaginary-unit-ac-circuit-analysis | # Theory question about “j” imaginary unit (AC circuit analysis)
I have just started to learn about AC network analysis and have some questions about "j" (or "i" on my calculator), the imaginary unit. My book doesn't go into a great deal about this, and jumps right into formulas and substitutions (more practical approach, not theoretical). So, what exactly does J represent?
I see that if I draw a complex-plane (y-axis being imaginary, x-axis being real), and draw a unit circle on it, a 90° angle is $\sqrt{-1}$, which is "j". I see that I can use this substitution in phasor form when, say, solving for the voltage across a capacitor when the current through it is known:
$$V = \frac{I}{j \omega C}$$
Can someone help me understand this?
To be honest, this question is pretty vague because I'm not even sure how to ask about what J is; it's that foreign to me. I would like a common-sense explanation (big-picture) of it's meaning and purpose in AC circuit analysis. I'm not necessarily looking for a rigorous mathematical explanation (although any necessary mathematical explanation is welcome).
-
Algebra is case-sensitive. J and j are different things. – TRiG Jul 30 '14 at 16:49
You might want to look at the questions under the complex-numbers tag on math.SE: math.stackexchange.com/questions/tagged/… – The Photon Jul 30 '14 at 19:07
Of course what you find on math.SE will leave open the really interesting question: Why are complex numbers useful in engineering? – The Photon Jul 30 '14 at 19:09
@The Photon: The answer is on Wikipedia: en.wikipedia.org/wiki/Phasor I can summarize it here, but given the dynamics of voting on SE sites, it would be "wasted bullets". – Respawned Fluff Jan 13 at 18:03
@RespawnedFluff, Are you responding to what you wanted to respond to? – The Photon Jan 13 at 18:04
## 3 Answers
If you put a minus sign in front of the number "5" it becomes "-5".
Try and look at this differently. Try thinking that it rotates the number "5" (tied to the origin by a piece of string of length 5) through 180 degrees to become "-5"
OK so far? Negative signs are the same as rotating thru 180 degrees...
Why not extend this further to produce something you can "stick" in front of a positive number that rotates it thru 90 degrees - in EE this is usually called "j" and it acts to rotate a value (about the origin) thru 90 degrees counter-clock wise i.e. if you did it twice (j*j) you'd get 180 degrees ("-").
From this gem of knowledge you can therefore say j*j = -1 therefore, j = $\sqrt{-1}$
Just as a minus sign can rotate any positive value thru 180 degrees it can rotate any vector or phasor thru 180 degrees. The same applies to the j operator - it rotates any vector or phasor thru 90 degrees counter clockwise.
EDIT - forgot part of question: -
substituting j into the impedance of a capacitor. Remember the basic formula for a capacitor is Q=CV and therefore differentiating the variables we get: -
$I = \dfrac{dQ}{dt} = C\dfrac{dV}{dt}$
This tells us that for a sinewave applied voltage across a capacitor, the current will also be a sinewave but differentiated into a cosine like this: -
If you tried to calculate the impedance (V/I) of a capacitor from the V-I relationship you'd get into trouble because when I passes thru zero, V is NOT zero so you get infinities. If on the other hand you apply a "j" to bring current in phase with voltage the math works out fine - current and voltage are aligned and impedance based on instantaneous values of V/I makes sense.
I'm aware that you are just starting out so I've tried to keep this both accurate and simple (maybe too simple for some?).
If you look at the inductor, the "j" can be applied to the voltage to align it with the current hence "j" is in the numerator for inductive reactance and j is in the denominator for capacitive reactance. There are subtleties lying around here that hopefully will make sense as you learn more - it's actually no coincidence that "j" appears to "follow" omega when it comes to impedances - my explanation doesn't cover that and neither does your question!
-
I found your answer to be very helpful, especially with your mention of using j to bring the waveforms in phase; this helped me understand its use because I remember that voltage leads current by 90* for pure inductance, and vice-versa for pure cap. Thanks! – eestack Jul 31 '14 at 5:08
In pure maths we use $i$ to represent the prime square root of $-1$.
The other square root of $-1$ being $-i$.
If you imagine a number line with real numbers placed horizontally. We can now add a second number line going vertically containing the imaginary numbers.
We have now created a system of complex where every point on the plane is represented by a real and imaginary part e.g. $4 + i \ 3$ represents a point that is 4 units along the real axis and 3 units up the imaginary axis.
Because a point in two dimensional space can now be represented as a single number, calculations involving 2-dimensional vectors are simplified.
In electronics, when considering systems supplied by a single frequency sine wave, we are taught initially to draw phasor diagrams. Then later to use complex numbers to get to deal with these problems.
We also use $j$ instead of $i$ but the meaning is identical. It’s just to avoid confusion because in electronics $i$ is often used for current.
If you would like a little more insight take a look at this question: What are imaginary numbers? from the Mathematics Stack Exchange site.
Or take a look here: A Visual, Intuitive Guide to Imaginary Numbers.
-
Thanks for your help and references to some additional reading! – eestack Jul 31 '14 at 5:09
In mathematics imaginary unit is a very helpful number used to solve equations with higher than 2 order. It was introduced just.... to the test, and it works pretty until today. This provides for obtaining at least one root in every polynomial.
In electronics imaginary unit represents the energy stored in our circuit. So, in capacitor, it is the energy stored in it. It also represents phase shift in circuit, when we are dealing with sinusoidal signals.
I think you should more precise your question, or just write questions which bother you in points.
For example... If your circuit's impedance will be represented only by imaginary unit, not by real, your bill for energy will be... zero :)
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https://settheory.mathtalks.org/moritz-muller-on-the-relative-strength-of-finitary-combinatorial-principles/ | # Moritz Müller: On the relative strength of finitary combinatorial principles
Talk held by Moritz Müller (KGRC) at the KGRC seminar on 2017-12-14 at 4pm .
Abstract: Define a finitary combinatorial principle to be a first-order sentence which is valid in the finite but falsifiable in the infinite. We aim to compare the strength of such principles over a weak arithmetic. We distinguish “weak” and “strong” principles based on their behaviour with respect to finite structures that are only partially defined. The talk sketches a forcing proof of a theorem stating that over relativized $T^1_2$ “weak” principles do not imply “strong” ones. | {"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": 0.8488912582397461, "perplexity": 1442.785138383037}, "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-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00292.warc.gz"} |
https://threesixty360.wordpress.com/2008/11/15/generating-pythagorean-triples/ | ## Generating Pythagorean Triples
by
A recent post at jd2718 noted that for all Pythagorean triples $(a, b, c)$, that is to say, for all positive integer solutions to $a^2 + b^2 = c^2$, it turns out that 60 divides the product $abc$.
In the course of exploring proofs of that result, I suggested a proof using the representation $(2uv, u^2-v^2, u^2+v^2)$, which generates all primitive Pythagorean triples (triples which have no common factors). Proving that 60 divides their product proves the general result, as every Pythagorean triple has the form $(ka, kb, kc)$, where k is a positive integer and $(a, b, c)$ is primitive.
In the ensuing discussion, a question arose as to how we know that all possible Pythagorean triples can be found in this way. Since it is easier for me to post LaTeX code to this blog rather than in the comments on the jd2718 blog, I’ll present the proof here.
Suppose that $a^2 + b^2 = c^2$, with positive integers $a, b, c$ having no common factors. This implies that exactly one of a and b is even (since if both are even, then so is c, leading to a contradiction, while if both a and b are odd, then both $a^2$ and $b^2$ are congruent to 1 mod 4, while $c^2$ would be congruent to 0 mod 4, not 2 mod 4).
So without loss of generality, assume that a is even. Thus $a^2 = (c-b)(c+b)$ is also even, and since $(c-b)$ and $(c+b)$ differ by an even number, both of those factors must be even.
Furthermore, exactly one of $(c-b)$ and $(c+b)$ is 2x(an odd number), since if both factors are divisible by 4, it would follow that both b and c would be even (since b and c are $\frac{ (c+b)\pm (c-b)}{2}$.) Likewise, any odd common factor of $(c-b)$ and $(c+b)$ would also then be a common factor of both b and c, from which it follows that
• $(c+b)$ and $(c-b)$ have no odd common factors
• one of these expressions is of the form $2 u^2$ (for some odd u)
• and hence the other expression is of the form $2 v^2$, where $v^2 = \frac{a^2}{4u^2}$, where u and v are positive integers
It now follows that $c = u^2 + v^2$, and $b = u^2 - v^2$. We can find a using $a^2=(c+b)(c-b) = (2u^2)(2v^2)$, and thus $a = 2uv$.
Conclusion: every primitive Pythagorean triple has the form $(2uv, u^2-v^2, u^2+v^2)$, for positive integers u and v.
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### 24 Responses to “Generating Pythagorean Triples”
1. jd2718 Says:
why must one of $(c+b) and (c-b)$ be of the form $2u^2$ ?
I see twice an odd number, not twice an odd square. What have I missed?
(oh, and for the wordpress installation of LaTeX in wordpress comments, if it’s easy, I just do it. If it’s hard, I write a dummy post and preview it.. I still screw up, but not all the time)
Jonathan
2. TwoPi Says:
All of the prime factors of $a^2$ occur to even powers. Any odd prime factor of $c+b$ (say) cannot also be a factor of $c-b$, and so must occur to an even power in $c+b$. Thus both $c+b$ and $c-b$ must be a power of 2 multiplied by an odd square.
3. Todd Trimble Says:
I much prefer the classic geometric proof, based on stereographic projection. First, one notes that primitive Pythagorean triples $(a, b, c)$ are in bijective correspondence with rational pairs $(a/c, b/c)$ on the unit circle. So it suffices to parametrize rational points $(u, v)$ on the unit circle.
Consider the point where the line through $(0, 1)$ and $(u, v)$ intersects the line $y = -1$. Since these two lines have rational coefficients, the point of intersection has rational coordinates, say $(2t, -1)$ [the factor of 2 makes the arithmetic come out nicer]. You can work out what $(x, y)$ is in terms of $t$: just note that the point $(x, y)$ on the circle is the intersection of two lines, the first one through $(0, 1)$ and $(2t, -1)$, which is
$\displaystyle y - 1 = -\frac1{t}x$
and the second is the line perpendicular to that through $(0, -1)$, which is
$y + 1 = t x$.
[That is, we use here a famous result from Euclidean geometry: that a right triangle is formed when you draw the chords from a point on the circle to the endpoints of a diameter.]
Solving for the intersection of those two lines, one quickly finds
$\displaystyle (u, v) = (\frac{2t}{t^2 + 1}, \frac{t^2 - 1}{t^2+1}$
Putting $t = m/n$, this gives
$\displaystyle (u, v) = (\frac{2mn}{m^2 + n^2}, \frac{m^2 - n^2}{m^2 + n^2})$
and voila! there’s your Pythagorean triple. Neat, isn’t it?
4. Todd Trimble Says:
Edit: the (x, y) in the fourth and fifth lines of the second paragraph should have been (u, v).
5. Carnival of Mathematics #44 « Maxwell’s Demon Says:
[…] can learn how to bound binomial coefficients at the Endeavour, or generate Pythagorean triples at 360. To stretch your mathematical muscles a little more look for Terry Tao, considering polynomials […]
6. Harbey Says:
Here is the original Pythagorean formula: For any pair of positive integers “y > x” odd and coprime,
where the product “xy = i” is the odd leg and “(y2 − x2) / 2 = p” is the even leg,
being “(y2 + x2) / 2 = h” the hypotenuse and the radius of the incenter is “r = x(y − x) / 2”.
Remembering that “h − p = x2” is easy to get “h + p = y2” from a known primitive triple.
By the way “n + m = y”, then “(y − x) / 2 = n” and “y − ((y − x) / 2) = m”.
7. David Gillies Says:
(2 u v, u^2 – v^2, u^2 + v^2) (or (u v, (u^2 – v^2)/2, (u^2 + v^2)/2) for that matter) is not primitive for non-coprime u, v.
Harbey’s form is correct.
It is trivial to show that 8|(u^2 – v^2) for odd u, v
u^2 – v^2 = (u + v)(u – v)
=(2m + 1 + 2n + 1)(2m + 1 – 2n – 1)
=4(m + n + 1)(m – n)
2|(m + n + 1) or (m – n)
8|(u^2 – v^2)
4|((u^2 – v^2)/2)
The divisibility of one of the legs by 3 arises from the fact that 2 is not a quadratic residue mod 3. A similar approach shows that 5 divides one of the sides, whence the divisibility by 60 of the product of the sides.
8. Juxtaposition: Millionaire Triangles « 360 Says:
[…] What would come next? Is there even an ordering for Pythagorean triples? (Well, maybe: TwoPi wrote earlier about how every Pythagorean triple can be written as , for positive integers u and v, so we could […]
9. Keith Raskin Says:
Re:Juxtaposition: Mill’s question above: Is there an ordering for Pythagorean triples?
Please check out my hastily written online paper, “Ordering the Primitive Pythagorean Triples by Leg Difference and Size Using Generalized Pell Sequences”, in the Rational Argumentator!
10. Ξ Says:
Thanks Keith! I found a copy of your paper here.
11. Keith Raskin Says:
This question came up in a blog: Can there be two Pythagorean triples of the form {a,b,c} and {2a,d,c}. After much pondering and failed attempts to prove it impossible (see Natural Blogarithms — comments on Pythagorean triples — for details), my last response was this:
Fascinating question, Chris. It amounts to being able to stack two inscribed rectangles in a circle (of whole number radius) with the same heights and expect whole number lengths. Stacking on a horizontal diameter
We could generalize it to triples of the form {a,b,c}, {na,d,c} and get similarly inscribed rectangles, one with height n -1 times the other.
I suspect that not only would an irrational crop up, but that the big transcendental pi itself would come out to contradict this.
It also amounts to getting that the sinp = a/c, sinq =2a/c, cosp = b/c, cosq = d/c.
Anybody out there know if there’s a proof regarding this?
12. Math Teachers at Play and some other stuff « 360 Says:
[…] speaking of reading things, after reading Keith Raskin’s comment I headed over to Natural Blogarithms to look for Pythagorean Triple stuff, and I was immediately […]
13. Keith Raskin Says:
Answer (I think) to prime number problem referred to above by 360:
Time for a solution yet? Given A and B are primes.
Either A or B must be 2 in order for A + B to be prime (odd > 2)
It must be B in order for A – B to be prime (assuming primes are positive: 2, 3, 5 …)
The only prime bracketed between primes like this: A – 2, A, A + 2
is 5, since A must be 1 or 2 mod 3, and either A – 2 or A + 2 is 0 mod 3, divisible by 3, which is only okay if it is 3 itself. A can’t be 3, since 1 is not prime; that’s how I know it’s got to be 1 or 2 mod 3.
So, A = 5, b = 2
A + B + A – B + A + B = 3A + B = 17 prime.
14. Keith Raskin Says:
My latest thoughts: If there exist Pyth. triples of the form {a, b, c}, {2a, d, c}; and if we can assume (which I think we can) that they are primitive and that the a and 2a can be represented as the even legs in the Euclidean Algorithm, then
{a, b, c} = {n^2 – m^2, 2nm, n^2 + m^2}
{2a, d, c} = {v^2 – u^2, 4nm, n^2 + m^2}.
This implies that (n^2 + m^2)^2 – (4nm)^2 is a square, since it’s root must be v^2 – u^2 (an odd leg).
So, n^4 – 14n^2m^2 + m^4 is a square, while n^4 – 14n^2m^2 + 49m^4 must also be a square, as its a squared binomial.
This corresponds to x^2 – 14x + 1 and x^2 – 14x + 49 being squares, where x = n^2 and m= 1. They differ by 48. The only squares that differ by 48 are 1 and 49, 16 and 64, and 121 and 169, which respectively correspond to n^2 = 0, 15, and 20, none of which are permissible.
Don’t know if this really leads anywhere.
Anyone with knowledge or rumor of a proof, please come forward.
15. Keith Raskin Says:
Pythagorean Triples Summary and Clarification:
Regarding the question about whether or not two Pythagorean triples of the form {a,b,c} and {2a,d,c} exist. I don’t think so — still don’t have a proof — but the question leads to lots of interesting questions about squares and rationals.
Let’s assume such triples exist. We can also assume without loss of generality that they are both primitive triples. If we now assume that a is odd, we get a contradiction. Observe:
{a,b,c} = {u^2 – v^2, 2uv, u^2 + v^2}
{2a,d,c} = {2xy, x^2 – y^2, x^2 + y^2}
for some u,v and x,y such that each pair is relatively prime and of opposite parity.
That means 2xy = 2a ⇒ xy = a, which means a is even, since either x or y is even.
Therefore a must be even and
{a,b,c} = {2uv, u^2 – v^2, 2uv, u^2 + v^2}
{2a,d,c} = {2xy, x^2 – y^2, x^2 + y^2}
where 2xy = 4uv and x^2 + y^2 = u^2 + v^2.
From here we get that c – 2a, c – a , c + a, and c + 2a are squares, because these are really
x^2 – 2xy + y^2, u^2 – 2uv + v^2, u^2 + 2uv + v^2, and x^2 + 2xy + y^2
OR (x – y)^2, (u – v)^2, (u + v)^2, and (x + y)^2.
I suspect that no set of squares are spaced out like that. In other words, I seriously doubt that there exist four squares of the form n^2, n^2 + k, n^2 + 3k, and n^2 + 4k. However, that’s just my mathematical intuition. I just don’t think that a parabola or a square root function will bend through those four linearly spaced points without hitting an irrational.
If such triples do exist, though, we would get the following bonus set of triples:
{4m, 4(u^2 – 4uv + v^2) – 1, 4(u^2 – 4uv + v^2) + 1}
{4(u – v), 4(u – v)^2 – 1, 4(u – v)^2 + 1}
{4(u + v), 4(u + v)^2 – 1, 4(u + v)^2 + 1}
{4w, 4(u^2 + 4uv + v^2) – 1, 4(u^2 + 4uv + v^2) + 1}
for m, w that equal the square roots of c – 2a, c + 2a, respectively.
16. Keith Raskin Says:
LOOKS LIKE WE HAVE OUR PROOF (THANKS TO HENK)!
I buy Henk’s assumptions about the factors of c. They must be primitive hypotenuses as long as they are prime, which we can assume. This has been proven by Euler and the proof can be found in Richard Friedberg’s “An Adventurer’s Guide to Number Theory.”
AND it looks like the identities (below, also provided by Henk) provide our contradiction! If we assume that x and z are odd, then {xw – yz, yw + xz, kl} and {yw – xz, xw + yz, kl} represent
{a, b, c} and {d, 2a, c}. Therefore a = 2yz = xw – yz. So, 3yz = xw.
This is impossible, since x, y and w,z are relatively prime and x cannot be equal to w or z (nor can y be w or z). Yet if 3yz = xw, then x = 3z. But this forces y = w. Impossible!
Maybe I’m missing something, as I’m coming up with this as I’m writing this submission (which I really shouldn’t be doing!) But this looks like the contradiction I’ve been looking for.
THANKS HENK!!!
PROBLEM RESOLVED?
identities:
(x^2 + y^2)(w^2 + z^2) = (xw – yz)^2 + (yw + xz)^2
= (yw – xz)^2 + (xw + yz)^2
17. Keith Raskin Says:
Re: My August 30th post
I think I thought I had good reason to assert that neither x nor y could equal to z or w, but i can’t see that reason at all now.
Don’t know if there’s a solid contradiction there after all.
18. Keith Raskin Says:
HEY! How about a contest???
Who can either unearth or craft a proof that no pair of Pythagorean triples of the form {a, b, c} and {2a, d, c} exists?
Without loss of generality, I believe, we can safely assume that the triples are primitive and the doubled leg is the even one, ie a is even.
We could extend this to any or all n such that no pair of Pythagorean triples of the form {a, b, c} and {na, d, c} exists.
Whaddayou think???
19. Keith Raskin Says:
Full Disclosure: At the time I posted the above challenge a fairly esteemed blogger outlined a proof using the Fibonacci Box method that triples cannot be transformed to get the above result. I argued that I needed more than inability to transform; I needed proof of the impossibility of the existence of the triples above. We agreed to disagree.
But perhaps the blogger is right and there is a proof resting on the use of Fibonacci Boxes and/or Ternary Trees.
20. Keith Raskin Says:
Please post the above comment under Keith Raskin.
Thanks!
21. Keith Raskin Says:
The proof I’m discussing directly above using the Fibonacci Box method is definitely invalid for our purposes. We need to go beyond proving that modified triples will not result in the same hypotenuse, unless the modifications are totally unlimited, which they are not in the proof given.
22. Keith Says:
Corollaries and conjecture:
If a pair of triples of the form {a,b,c} and {2a,d,c} exist, then it is easy to show that
c – 2a, c -a, c + a and c + 2a are squares. Just look at their forms in Euclid’s algorithm,
given that a must be even (must equal 2mn, c – a = m^2 – 2mn + n^2,
c – 2a = r^2 – 2rs + s^2, etc.)
Let these squares be represented as M^2, N^2, O^2 and P^2. The first and last two
differ by a = N^2 – M^2, and the middle two differ by 2a. So, we can represent them
as M^2, N^2, 3N^2 – 2M^2 and 4N^2 – 3M^2, with M and N odd. It appears from
inspection of tables that when 3N^2 – 2M^2 or 4N^2 – 3M^2 is a square, the other
misses being an odd square by a factor of 8. In other words, either O or P must be irrational, it appears. In still other words, if you take an interval with perfect square
endpoints and advance it by double its length, one of the new endpoints is not a
perfect square; the image of the original interval under the square root function has
integral length, while the image of the new one has irrational length. If this conjecture is proven true, it would suffice in proving that the original triples could not
exist.
Also, if we represent O as N +q and P as N + p, then we get that q^2 + 2Nq over
p^2 + 2Np must equal 2/3. There are some inherent constraints: p and q are even
integers, with p bigger, and N must be greater than or equal to 1/2q^2 + Nq + 1.
If that quotient can never be exactly 2/3 for qualifying entries of p, q and N, then no
such original triples exist.
23. Keith Raskin Says:
Corollaries and conjecture:
If a pair of triples of the form {a,b,c} and {2a,d,c} exist, then it is easy to show that c – 2a, c -a, c + a and c + 2a are squares. Just look at their forms in Euclid’s algorithm, given that a must be even (must equal 2mn, c – a = m^2 – 2mn + n^2, c – 2a = r^2 – 2rs + s^2, etc.)
Let these squares be expressed as M^2, N^2, O^2 and P^2. The first and last two differ by a = N^2 – M^2, and the middle two differ by 2a. So, we can express them as M^2, N^2, 3N^2 – 2M^2 and 4N^2 – 3M^2, with M and N odd. It appears from inspection of tables that when 3N^2 – 2M^2 or 4N^2 – 3M^2 is a square, the other misses being an odd square by a factor of 8. In other words, either O or P must be irrational, it appears. In still other words, if you take an interval with perfect square endpoints and advance it by double its length, one of the new endpoints is not a perfect square; the image of the original interval under the square root function has integral length, while the image of the new one has irrational length. If this conjecture is proven true, it would suffice in proving that the original triples could not exist.
Also, if we represent O as N +q and P as N + p, then we get that q^2 + 2Nq over p^2 + 2Np must equal 2/3. There are some inherent constraints: p and q are even integers, with p bigger, and N must be greater than or equal to 1/2q^2 + Nq + 1. If that quotient can never be exactly 2/3 for qualifying entries of p, q and N, then no such original triples exist. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 55, "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": 0.8624800443649292, "perplexity": 844.2218080363972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2017-09/segments/1487501172447.23/warc/CC-MAIN-20170219104612-00234-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=676054 | Slingshot Physics
by FredericChopin
Tags: physics, slingshot
P: 57 I am investigating how the displacement of an elastic band affects the distance an attached object will travel (the relationship between how far back you pull a slingshot and the distance an object will travel). What I am trying to do is find a formula where distance travelled is a function of elastic band displacement (in other words, a formula which says "x = ... d..."), but I ended up very confused. I tried equation manipulation with the following equations: W = F*d W = ΔEKinetic EKinetic = (1/2)*m*v2 and EElastic Potential = (1/2)*k*x2 I got close to the final equation, but I couldn't make it. Here are a few things to keep in mind: * In the final equation, where "x = ... d...", I don't want the force, F, to be in the equation. * We are assuming that initial kinetic energy was 0 (meaning that the initial velocity was 0). * We are assuming that the effects of friction and air resistance are negligible. Thank you.
P: 2,817 I don't think the F of the elastic band is constant with distance. You could try hooke's law on it and measure the F values for various distances of pullback. Then you can use that to compute the work involved.
P: 72 I didn't quite understand the question..
HW Helper
P: 7,055
Slingshot Physics
Quote by FredericChopin In the final equation, where "x = ... d...", I don't want the force, F, to be in the equation.
You won't need F, but you will need the "sprint constant" k for the sling shot. Note this is an approximation, since the graph of force versus stretch of rubber bands is not a straight line, but a curve. Example graph from a wiki article, which is also somewhat idealized. Usually the slope for the initial stretch is steeper, then decreases to a near straight line, then increases again at the limit of stretch (permanent deformation can occur if stretched close to the limit):
Elastic_hysteresis.htm
Archived web page showing a stretch versus tension graph for latex rubber used to launch radio control gliders (tension at 300% is about 175 lbs per square inch cross sectional area):
rubberdata.htm
P: 57 Thank you very much, but I have found a solution: First, due to conservation of energy, the elastic potential energy of the elastic band is equal to the kinetic energy of release. So: EElastic Potential = (1/2)*k*x2 EKinetic = (1/2)*m*v2 (1/2)*k*x2 = (1/2)*m*v2 (1/2)*x2 = (m*v2)/(2k) x2 = (2m*v2)/(2k) x = √((m*v2)/(k)) By finding v2 using the equations of motion, the mass, m, will be the function of spring displacement, x, while v2 and k remain constant (v2 will change depending on the desired displacement for the object to travel in the x or y axis).
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http://mathhelpforum.com/calculus/227850-area-between-two-curves-simple-question.html | # Math Help - Area between two curves (simple question)
1. ## Area between two curves (simple question)
Hi,
I'm trying to find the area between the curves:
y=x^2 + 4x +1 and
y = 3x+3
A sketch shows me that y =3x+3 is above y=x^2 +4x +1 and I have found the intersection points are x=-2 and x=1.
Does this then mean that I just need to evaluate the integral of (3x+3)-(x^2+4x+1) from -2 to 1?
I know that some of this area lies underneath the x-axis and not sure if I need to be accounting for that or not.
Thanks!
2. ## Re: Area between two curves (simple question)
Hi Andy, you don't have to worry if one of the graphs is below the x-axis. In fact, both graphs can be below the x axis and the definite integral ∫ [higher function] - [lower function] dx = area between the curves still holds. So you're right, just evaluate the integral you said before.
3. ## Re: Area between two curves (simple question)
Thanks very much | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9363343119621277, "perplexity": 857.456494669068}, "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-2016-30/segments/1469257826773.17/warc/CC-MAIN-20160723071026-00274-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/the-dirac-delta-function.874444/ | The Dirac Delta Function
Tags:
1. Jun 4, 2016
Summer95
1. The problem statement, all variables and given/known data
Differential equation: $Ay''+By'+Cy=f(t)$ with $y_{0}=y'_{0}=0$
Write the solution as a convolution ($a \neq b$). Let $f(t)= n$ for $t_{0} < t < t_{0}+\frac{1}{n}$. Find y and then let $n \rightarrow \infty$.
Then solve the differential equation with $f(t)=\delta(t-t_{0})$.
2. Relevant equations
Convolution (Boas)
Laplace Transforms (Boas)
3. The attempt at a solution
So when I go through the first part with $f(t)= n$ for $t_{0} < t < t_{0}+\frac{1}{n}$ and do convolution I get $y=\frac{1}{A(b-a)}\int_0^t(e^{-a(t-\tau)}-e^{-b(t-\tau)})f(\tau)d\tau$ which has different cases depending on t:
0 if $t<t_{0}$
$\frac{n}{A(b-a)}(\frac{1}{a}(1-e^{a(t_{0}-t)})-\frac{1}{b}(1-e^{b(t_{0}-t)}))$ if $t_{0}<t<t_{0}+\frac{1}{n}$
$\frac{n}{A(b-a)}(\frac{1}{a}(e^{-a(t-t_{0}-\frac{1}{n})}-e^{a(t_{0}-t)})-\frac{1}{b}(e^{-b(t-t_{0}-\frac{1}{n})}-e^{b(t_{0}-t)}))$ if $t>t_{0}+\frac{1}{n}$
I don't understand what happens as $n\rightarrow \infty$. I know it should become
$\frac{1}{A(b-a)}(e^{a(t_{0}-t)}-e^{b(t_{0}-t)})$ for $t>t_{0}$ because that is what I get when I use $f(t)=\delta(t-t_{0})$ from the beginning. But what happens to all the extra terms? And the n out front that goes to infinity?
2. Jun 4, 2016
Ray Vickson
Is $f(t)$ supposed to be zero for $t < t_0$ and $t > t_0+\frac{1}{n}$?
3. Jun 4, 2016
Summer95
Yes! Sorry I forgot to specify that.
Draft saved Draft deleted
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http://www.ellipsix.net/blog/2012/12/yes-virginia-the-universe-is-probably-real.html | ellipsix informatics
2012
Dec
25
## Yes, Virginia, the universe is (probably) real
It's been far too long since I posted anything — almost three weeks, in fact, since the end of NaBloWriMo. So I took the time to make a solid writeup of the speculation that we may be living in a computer. Consider it my Christmas present to the blogosphere. Happy holidays everyone!
Have you ever wondered if everything you know might only exist inside some hyperintelligent alien's computer?
It's a fun possibility to consider, that we might be living in a computer simulation, and we might even be able to figure out how to manipulate it. It even makes for some great science fiction (as well as some terrible science fiction :-P perhaps). But in reality, figuring out whether the universe is a simulation isn't as easy as popping a red pill and waking up in an alternate reality with one less layer of abstraction. If we are in a simulation, the way it's going to reveal itself is in the tiny details of physical processes that take place on sub-subatomic distance scales.
There is actually a branch of physics, lattice QCD, that is entirely devoted to simulating reality. Right now it only works for very small volumes, barely large enough to fit a proton, but that's already enough to give us a sense of what kinds of tiny details, or artifacts, can prevent a simulation from being completely accurate. These artifacts are the subject of a recent paper by Silas Beane, Zohreh Davoudi, and Martin Savage. It's a nice concise piece of work: they imagine that the universe is a computer simulation on a cubic grid, similar to simple lattice QCD models, and just calculate some of the observable consequences of this grid.
Unfortunately, when this hit the popular press (and the blogosphere), everything went to hell. Nearly every time I've seen this research reported, it completely misrepresents the original paper. People have been saying everything from that the scientists are proposing a test for whether the universe is a simulation, to that they've actually shown that it is. But none of that is true! To clear things up, I wanted to examine the paper itself, and clear up what exactly these physicists did conclude.
# Fields on a grid
I'll start with a primer on how a simulation on a grid works. The grid itself is just a set of points in space (or spacetime) that you're going to include in your simulation. Generically, you can pick the points any way you want, but the simplest way, and the one discussed in the BDS paper, is to just pick them evenly spaced in every dimension. The distance between adjacent points in each dimension is the grid spacing, and the smaller that is, the more accurate the simulation will be.
On the grid, you have some number of fields. Normally a field is a thing that has a value at every point in space. But when you're working on a grid, the points that are part of the grid are effectively the only points in space. So your simulated fields have values at every point on the grid. In reality, if you weren't doing a simulation, the fields would also have values at points between the grid points, but for the simulation, you have to ignore that and hope the field doesn't do anything too weird in the spaces.
Once you have the grid (the points in spacetime) and the fields (the values at those points), the actual process of simulation is simple, in principle. At each time step, you go through each point on the grid and use some set of rules to calculate new values for the next time step. The particular set of rules you use depends on what physical system you're trying to simulate. If it's a weather simulation, you would use some discrete approximation of the laws of fluid dynamics. If it's a cosmological simulation, you would use some discrete approximation of general relativity. And if you're trying to simulate the structure of a proton, as they do in lattice QCD, you would use some discrete approximation of the laws of quantum chromodynamics. The point is, when you're doing a simulation on a grid, you always need a discrete approximation of the physical laws that govern your system — you have to use modified laws that only use the points on your grid, not the nonexistent (in your simulation) points in between. And that's the inherent limitation of any grid simulation: you have to ignore the fine details of the behavior of the fields between grid points. It's those fine details that cause the slight differences that would expose a grid simulation as a simulation.
# Details of the grid model
Now let's turn to the actual paper. Beane, Davoudi, and Savage started with the quantum electrodynamic description of a charged particle in an EM field, which in our ordinary, continuous reality is given by this expression, the Lagrangian density:
The details of what this expression means don't really matter. But since I know some of my readers who don't know quantum field theory won't be happy unless they can make some sense of it:
• is the charged particle field — let's say it's the electron
• is the charged particle's (electron's) mass
• is the particle's charge, which also determines how strongly the charged particle and the EM field interact
• is the electromagnetic vector potential
So for example, the term represents an electron () interacting with the electromagnetic field (). It's that term that determines how electrons react to electromagnetic forces. But like I said, most of that doesn't really matter.
What does matter is how you have to modify that expression to make it work on a grid. The modified expression looks like this:
where is the grid spacing, is a function of the electromagnetic vector potential, and and are basically constant coefficients. The most important change is that we have a new term, , which involves an additional, different interaction between electrons and the electromagnetic field! This is an example of a simulation artifact. It modifies the way in which electrons respond to EM fields, and so it could have observable consequences.
# What could we see?
In the rest of the paper, the scientists identify three ways in which this particular simulation artifact could show itself, if it really exists. Before we get into that, though, it's important to pause and remember what we're dealing with here: this is a toy model. It's only the very simplest way in which one aspect of the universe (i.e. the electromagnetic force) could be simulated, and bear in mind that we don't even have any reason to believe the universe is a simulation! Nobody honestly expects this model to be real, it's only meant to show how these kinds of calculations could work.
OK, now that that's out of the way, what did they find?
## Modified muon magnetic moment
An electron or muon, because it has charge and angular momentum, acts like a magnet. In physics terms, it has a magnetic dipole moment . The magnitude of this dipole moment — the strength of the magnet — is expressed as
where is a fixed physical constant, the Bohr magneton, and is a number which depends on the particle: for the electron, for the muon. You can think of as measuring the electron's or muon's magnetic moment in units of . This number has been measured very precisely and also can be predicted very precisely from QED; in fact, the measurement of is the most precisely confirmed prediction in all of physics!
If the extra "grid term" from the Lagrangian density were real, it would modify the above equation to
In this case, QED still predicts the value of , but when we think we're measuring , we would actually be measuring . That means, if this theory were true, the measured and calculated values of should differ by . It just so happens that a slight difference of this type does exist for : its measured value is , whereas the QED prediction is . That's a difference of , and if you set that equal to , it tells you how big the grid spacing would have to be to account for that difference:
(There are some subtleties I'm glossing over here, but that's the gist.) So if the universe is a simulation on a cubic grid, the spacing can't be any bigger than that, otherwise our measurement of the muon magnetic moment would be even more different from the QED prediction than it is. Effectively, we have an upper bound on the grid spacing — one that we're still (un)comfortably far away from being able to see directly. But nothing prevents the grid spacing from being smaller, or nonexistent; it's entirely possible that the difference between the measured and predicted magnetic moments is due to something else entirely.
## Rydberg constant
The Rydberg constant is the amount of energy it takes to break apart a hydrogen atom into a free proton and a free electron. This amount of energy can be fairly precisely measured, and then we can work backwards from that value to get a value for , the electromagnetic coupling. (It's related to the electron's charge by ; either one of those variables, or , can be used to represent how strongly electrons interact with EM fields.)
We can also get a value for from the electron's magnetic moment. But it turns out that, if we live in a simulation on a cubic grid, what we measure using the Rydberg constant method and what we measure using the magnetic moment method aren't quite the same thing! The difference between those two measurements is related to the grid spacing .
Unfortunately the paper is pretty light on details when it comes to this method, so there's not much I can say, but I can give the punch line: the difference between the two measured values of is . That's actually consistent with zero: it seems entirely likely that there is no difference at all, which is exactly what we'd expect to see for a fully continuous, non-simulated universe.
## Cosmic ray cutoff
Cosmic rays are pretty much what they sound like, except that they're not exactly rays. They're PARTICLES FROM SPACE!!!! And some of them have very high energies, many orders of magnitude more than what we can produce in any particle accelerator on Earth (like the LHC).
It turns out, though, that ultra-high-energy particles, specifically those with energy higher than about , cannot exist on a cubic grid! That's because of an effect called aliasing, which will be familiar to anyone who's worked with sound engineering or digital signal processing. Aliasing basically means that any wave with a wavelength smaller than twice the grid spacing behaves exactly like a different wave with a wavelength larger than twice the grid spacing, like the blue wave and the orange wave in this picture.
And since wavelength is inversely proportional to energy for a massless particle, , along each of the grid axes there is an upper limit of on the energies of waves that can be simulated. As long as you're talking about particles whose masses can be neglected (which is definitely true for these ultra-high-energy cosmic rays), no individual object in the simulation can ever act like it has any more energy than . (This limit is the spatial equivalent of the Nyquist frequency in signal processing.)
If we're going to consider the possibility that the universe is a simulation, this energy limit that comes from aliasing has to compete with a different energy limit, the GZK cutoff, which arises from an entirely different reason. According to special relativity, a cosmic ray proton with more than a certain energy () can collide with a photon from the cosmic microwave background with enough energy to decay into a pion and a slower proton. These CMB photons are everywhere — they quite literally permeate all of space — and so any cosmic ray with enough energy almost certainly will collide with one of them, and will be stopped long before it gets to Earth.
The reason for that value of is clearest if you think about a collision between a proton and a CMB photon from the proton's perspective. It sees a highly blueshifted CMB: instead of a uniform thermal distribution of photons with energies corresponding to the temperature of the universe, , it sees photons coming from one direction with much higher energies. If the proton is moving fast enough, it sees those photons coming with enough energy to excite the proton and turn it into a delta baryon, which then emits a pion to decay into a slower-moving proton. In this way, an energetic proton will repeatedly lose energy until it slows down enough that it doesn't interact with the CMB anymore. The end result is that cosmic rays with energies more than the GZK cutoff should be quite rare. And in fact they are: recent observations by the Auger and HiRes cosmic ray telescopes, among others, do actually see many fewer cosmic rays than expected with energies above .
So in a universe simulated on a cubic grid, we have two different cutoffs that limit the number of cosmic rays above a certain energy: the GZK cutoff, which kicks in at , and the grid cutoff, which kicks in at the unknown value . Depending on the value of , there are three (but really two) possibilities:
• : the grid cutoff is less than the GZK cutoff. In this case there are no cosmic rays with high enough energies to see the GZK cutoff at all. This is clearly not true in our universe because, as I mentioned, various telescopes have actually seen those high-energy cosmic rays.
• : the grid cutoff is greater than the GZK cutoff. In this case there are few to no cosmic rays with high enough energies to see the grid cutoff. This could conceivably be true in our universe, if you accept the idea that the universe might be a simulation on a cubic grid. If you solve the inequality for , you get an upper limit of .
• : the grid cutoff and the GZK cutoff are about equal. This case is actually going to look similar to the last one, and also similar to what we would see if there is no grid cutoff at all, because in each one we just see a dropoff in the number of cosmic rays above . However, there is one key feature that sets this option apart from the others: the grid cutoff is different in different directions! In the lingo of physics, we say the grid cutoff is not rotationally invariant; in the language of astrophysics, we say it's anisotropic. Hopefully it's intuitively clear why this would be true: the spacing between points on the grid is different if you're moving along the direction of an axis as opposed to if you're moving diagonally. And that means that the grid cutoff will behave differently depending on which direction a particle is moving relative to the grid. If we actually live in this kind of simulation, then, we can expect to see the highest energy particles traveling only in certain directions in space.
If we were to actually observe a directional distribution of these ultra-high-energy cosmic rays, that would be huge. Nearly all of modern physics is based in part on the principle of rotational invariance, which means that fundamentally, there are no special directions. If we saw high energy cosmic rays traveling only along the axes of a cube, that would blow that out the window, and suddenly we'd have to take a whole new look at the physics of high energy particles! It wouldn't change anything in our everyday lives, but it would be really cool.
But of course, no sign of any of this has ever been observed. And judging by the results presented in this paper, we're a long way away from finding any simulation artifacts, even if they do exist. So for the foreseeable future, the continuum of reality remains just that. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8310292363166809, "perplexity": 325.55532342737155}, "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-2017-13/segments/1490218191986.44/warc/CC-MAIN-20170322212951-00350-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://brilliant.org/discussions/thread/primitive-root/ | ×
# Primitive root
If $$n$$ is a positive integer, the integers between 1 and $$n − 1$$ that are coprime to $$n$$ (or equivalently, the congruence classes coprime to $$n$$) form a group with multiplication modulo n as the operation; it is denoted by Zn× and is called the group of units modulo n or the group of primitive classes modulo n. As explained in the article multiplicative group of integers modulo n, this group is cyclic if and only if n is equal to 2, 4, p^k, or 2p^k where pk is a power of an odd prime number. A generator of this cyclic group is called a primitive root modulo n, or a primitive element of Zn^×.
Note by Sattik Biswas
6 months ago
Sort by:
What is the [3], [4], [5] supposed to mean? · 6 months ago
Sorry it was a mistake, I removed it... by the way root ta ki bhabe likbo goh? LaTex?
$$\sqrt{x}$$ · 6 months ago
You still have [5] in there. On Brilliant, communicating in English is recommended.
I'll edit your comment to make a $$\sqrt{x}$$. Click on edit and check that for reference. · 6 months ago
ahhhh....i am sorry again..i will remove it..since its my first note I am making too many mistakes. · 6 months ago
What matters is that you're posting notes, which is great! · 6 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9297587871551514, "perplexity": 924.5185461520501}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660746.32/warc/CC-MAIN-20160924173740-00289-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://cstheory.stackexchange.com/questions/47114/minimizing-the-gaps-with-incremental-capacity | # Minimizing the gaps with incremental capacity
There are a single job, a machine and a set of $$n$$ slots. The machine has a capacity that increments by $$\zeta(t)$$ every slot $$t=1,2,\ldots,n$$. Initially (before the first slot), the machine has 0 capacity i.e., the available capacity $$C(t)$$ at slot $$t$$ is $$C(t):=\sum_{s\leq t}\zeta(s)$$ (if the job was not scheduled at $$t$$ or before $$t$$). If the job is scheduled at slot $$t$$, then it will consume $$c(t)$$ units of the available capacity $$C(t)$$. If the job is not scheduled for a period of $$x$$ consecutive slots, then a penalty of $$\lfloor x/2\rfloor$$ occurs.
EDIT
• We have to guarantee that $$C(t)=\sum_{s\leq t}\zeta(s)-\sum_{s\in S}c(s)\geq 0$$ for all $$t$$ where $$S\subseteq\{1,2,\ldots,t\}$$ is the set of slots where the job was scheduled.
• For the penalty: there is a penalty for every contiguous unused block.
Here, is an example to illustrate the problem. Say $$n=8$$ and the job is scheduled at time $$1$$, $$4$$, and $$8$$. Here, we have a penalty of $$\lfloor{2/2}\rfloor=1$$ between time $$1$$ and $$4$$ since the job is not scheduled for a period of 2 consecutive slots ($$2$$ and $$3$$). Also, we have a penalty of $$\lfloor{3/2}\rfloor=1$$ between time $$4$$ and $$8$$ since the job is not scheduled for a period of 3 consecutive slots ($$5$$, $$6$$ and $$7$$). Thus, the objective here is $$1+1$$.
Given $$\zeta(t)$$, $$c(t)$$ for all $$t=1,2,\ldots,n$$, the objective is to schedule the job during the $$n$$ slots in order to minimize the sum of penalties while respecting the capacity of machine in all scheduled slots. Is this problem NP-hard?
I tried to reduce the knapsack problem to it but I did not succeed yet. Also, I tried to solve the problem in polynomial-time using dynamic programming but failed also due to the incremental capacity.
• For (i) and (ii), yes, we have to guarante that $C(t)=\sum_{s\leq t}\zeta(s)-\sum_{s\in S}c(s)\geq 0$ for all $t$. I don't get the difference between the last part of (iii) and (iv) but there is a penalty for every contiguous unused block. I will edit the question and give an example to illustrate my meaning. – zdm Jun 26 at 20:44
• yes for every maximal contiguous block. – zdm Jun 26 at 21:36
Here's a poly-time dynamic-programming algorithm.
Lemma 1. The problem in the post has a poly-time dynamic-programming algorithm.
Proof sketch. Fix an input $$(\zeta, c)$$ over time slots $$\{1,2,\ldots, n\}$$.
For each $$t, p\in \{0, 1,\ldots, n\}$$, define subproblem $$M(t, p)$$ as follows. Consider the problem restricted to the first $$t$$ time slots. (That is, the problem over time slots $$\{1,2,\ldots, t\}$$, with $$\zeta$$ and $$c$$ restricted to those time slots.) For this restricted problem, consider just those solutions that have total penalty $$p$$ and (if $$t\ge 1$$) do use slot $$t$$. Define $$M(t, p)$$ to be the maximum, over all such solutions, of the capacity $$C(t)$$ achieved by that solution. (Or $$-\infty$$ if there are no such solutions.)
The desired answer is $$\min \{ p + \lfloor (n-t)/2\rfloor : t,p\in\{0,\ldots,n\},\, M(t, p) \ne -\infty\}$$.
Then $$M(0, 0) = 0$$ and $$M(0, p) = -\infty$$ for $$p>0$$. For $$t>0$$, the following recurrence relation holds.
$$M(t, p)$$ is the maximum, over $$s\in\{0,1,\ldots,t-1\}$$, of
$$\begin{cases} M(s, p-\lfloor(t-s-1)/2\rfloor) -c(t) + \sum_{i=s+1}^t \zeta(i) & \scriptsize\textit{ (if that quantity is well-defined and non-negative)} \\ -\infty & \scriptsize\textit{ (otherwise). } \end{cases}$$ Here's the intuition. Consider the possible solutions for the first $$t$$ time slots that use slot $$t$$ and achieve total penalty $$p$$. Partition these solutions according to their last slot used, say, slot $$s$$, before slot $$t$$. (Or $$s=0$$ if slot $$t$$ is the first slot used.) Given $$s$$, such a solution consists of some solution $$S_s$$ for slots $$1,2,\ldots, s$$ (with slot $$s$$ used if $$s>0$$), followed by unused slots $$s+1,s+2,\ldots, t-1$$, followed by the used slot $$t$$.
The penalty incurred for the size-$$(t-s)$$ block of unused slots $$s+1,\ldots, t-1$$ is $$\lfloor (t-s)/2\rfloor$$. So the cumulative penalty incurred by $$S_s$$ must be $$p$$ minus this. The capacity $$C(t)$$ at time $$t$$ must be the capacity $$C(s)$$ achieved by $$S_s$$ at time $$s$$ plus the additional capacity added for unused slots $$s+1,\ldots, t$$, minus $$\zeta(t)$$ for using slot $$t$$. So $$C(t)$$ will be maximized when $$C(s)$$ is maximized (over all solutions $$S_s$$ with appropriate penalty). This is why the recurrence relation holds.
There are $$O(n^2)$$ subproblems, and for each the right-hand side of the recurrence can be evaluated in time $$O(n)$$ (with appropriate preprocessing), so this yields an $$O(n^3)$$-time dynamic-programming algorithm. $$~~\Box$$.
• In the reccurrence relation, did you mean $-c(t)+\sum_{i=s+1}^{t}\zeta(i)$ instead of $-\zeta(t)+\sum_{i=s+1}^{t}c(i)$ – zdm Jul 6 at 2:05
• Yes, got them backwards. Also had an off-by-one error. Edited to fix them both. – Neal Young Jul 6 at 2:56
• Hi. If we have two jobs instead of one and we have $\zeta_1$ and $\zeta_2$ and $c_1$ and $c_2$ for job 1 and job 2 respectively. If the two jobs cannot be scheduled at the same slot, can we still solve the problem using dynamic programming? – zdm Nov 7 at 23:23 | {"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": 92, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9118162393569946, "perplexity": 335.9012160248318}, "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-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00040.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_The_Basics_of_GOB_Chemistry_(Ball_et_al.)/05%3A_Introduction_to_Chemical_Reactions/5.1%3A_The_Law_of_Conservation_of_Matter | # 5.1: The Law of Conservation of Matter
Skills to Develop
• Correctly define a law as it pertains to science.
• State the law of conservation of matter.
In science, a law is a general statement that explains a large number of observations. Before being accepted, a law must be verified many times under many conditions. Laws are therefore considered the highest form of scientific knowledge and are generally thought to be inviolable. Scientific laws form the core of scientific knowledge. One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved.
With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter.
### Law of Conservation of Mass
According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.
$\overbrace{\underbrace{\ce{HgO (s)}}_{\text{100 g}}}^{\text{Mercuric oxide}} \rightarrow \underbrace{\overbrace{\ce{Hg (l) }}^{\text{Mercury}}}_{\text{92.6 g}} + \underbrace{\overbrace{\ce{O2 (g)}}^{\text{Oxygen}}}_{\text{7.4 g}}$
The law of conservation of mass is also known as the "law of indestructibility of matter."
Example $$\PageIndex{1}$$
If heating 10 grams of $$\ce{CaCO3}$$ produces 4.4 g of $$\ce{CO2}$$ and 5.6 g of $$\ce{CaO}$$, show that these observations are in agreement with the law of conservation of mass.
A sample of calcium carbonate (CaCO3). Image used with permission (Public Domain; Walkerma).
SOLUTION
• Mass of the reactants: $$10 \,g$$
• Mass of the products: $$4.4 \,g+ 5.6\, g = 10\, g$$.
Because the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.
What does this mean for chemistry? In any chemical change, one or more initial substances change into a different substance or substances. Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and type of atoms after the chemical change as were present before the chemical change.
Before looking at explicit examples of the law of conservation of matter, we need to examine the method chemists use to represent chemical changes.
Exercise $$\PageIndex{1}$$
1. What is the law of conservation of matter?
2. How does the law of conservation of matter apply to chemistry?
The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant
The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.
### Summary
The amount of matter in a closed system is conserved.
### Exercises
1. Express the law of conservation of matter in your own words.
2. Explain why the concept of conservation of matter is considered a scientific law. | {"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": 1, "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": 0.9091677069664001, "perplexity": 455.42089007315843}, "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-2019-18/segments/1555578624217.55/warc/CC-MAIN-20190424014705-20190424040705-00504.warc.gz"} |
https://www.physicsforums.com/threads/volume-by-rotating-a-curve.362102/ | # Homework Help: Volume by rotating a curve
1. Dec 9, 2009
### mikhailpavel
1. The problem statement, all variables and given/known data
Hey i have a problem here with volume by cylindrical shells. i wanted to find the given volume of the solid obtained by rotating the region bounded by the curves x=1+(y-2)^2 and x=2 about the x axis.
2. Relevant equations
we tried to integrating 2 phi f(y) dy with upper limit 3 and lower limit 1.
3. The attempt at a solution
we got the answer 33.5 but i dont think it is the correct one.
immediate help will be appreciable.
2. Dec 9, 2009
### Staff: Mentor
You are using cylindrical shells, but your formula is incorrect. A typical shell in this problem has a volume of 2pi*y*(x2 - x1)*dy
x2 is the x-value on the line x = 2, and x1 is the x-value on the parabola. You have graphed the region being revolved, right?
3. Dec 10, 2009
### mikhailpavel
can u tell me if my upper and lower limits are right because still i am not getting the right answer..i think!
4. Dec 10, 2009
### HallsofIvy
Yes, the two curves intersect at (2, 1) and (2, 3) so y ranges between 1 and 3. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9070200324058533, "perplexity": 700.3152580328936}, "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-2018-51/segments/1544376825363.58/warc/CC-MAIN-20181214044833-20181214070333-00318.warc.gz"} |
http://mathhelpforum.com/geometry/137933-quadrilateral.html | Math Help - Quadrilateral
1. Quadrilateral
What's the maximum of the sum of the lengths of the diagonals in a convex quadrilateral whose perimeter is 4?
2. This is trial and error -
1. A square of side 1 has sum of diagonals 2 srqt(2) = 2.8+.
2. A flattened square (think parallelogram of side 1 with 2 parallel sides both nearly on the x-axis) has 1 diagonal of length 2 and the other of length 0, so sum is 2.
3. A rectangle of sides (2-x,x,2-x,x), with x just a little bit larger than 0 has diagonals of about 2 and 2, so the sum is 4.
So I'd guess the maximum length of 2 diagonals of a quadrilateral of perimeter 4 is 4.
3. Originally Posted by qmech
This is trial and error -
1. A square of side 1 has sum of diagonals 2 srqt(2) = 2.8+.
2. A flattened square (think parallelogram of side 1 with 2 parallel sides both nearly on the x-axis) has 1 diagonal of length 2 and the other of length 0, so sum is 2.
3. A rectangle of sides (2-x,x,2-x,x), with x just a little bit larger than 0 has diagonals of about 2 and 2, so the sum is 4.
So I'd guess the maximum length of 2 diagonals of a quadrilateral of perimeter 4 is 4.
OK but it's easy to show that it cannot be more than 4 because of the triangle inequality. Just take the two triangles with one common diagonal side and write the 2 inequalities, add them and you get that $2e. This goes for the other diagonal so $e+f<4$. But I'm asked for the maximum and where it attains. As you said we can make it sufficiently close to 4 but we can't reach it. Therefore I guess there is no maximum. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "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": 0.8126547336578369, "perplexity": 324.7403710179051}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2015-18/segments/1429246647589.15/warc/CC-MAIN-20150417045727-00186-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://glossary.informs.org/ver2/mpgwiki/index.php?title=Minimal_inequality | # Minimal inequality
In integer programming, a valid inequality is minimal if it not dominated by any valid inequality. Originally, this was limited to not being able to decrease any coefficient and remain valid. For example, suppose $LaTeX: 2x_1 + x_2 \ge 1$ is a valid inequality. Then, if we decrease the first coefficient to obtain $LaTeX: x_1 + x_2 = 1,$ either this is not valid or it dominates the former, rendering it non-minimal.
More generally, suppose $LaTeX: a^Tx \ge b$ is a valid inequality, and we consider $LaTeX: (a',b')$ such that $LaTeX: a' \le t a$ and $LaTeX: b' \ge t b$ for some positive $LaTeX: t$ such that $LaTeX: (a',b') \neq t (a,b).$ If $LaTeX: (a')^T x \ge b'$ is a valid inequality, it dominates the original one because
For example, $LaTeX: 4x_1 + 2x_2 \ge 3$ dominates $LaTeX: 2x_1 + x_2 \ge 1$ (use $LaTeX: t = 2$), so if this is valid, $LaTeX: 2x_1 + x_2 \ge 1$ cannot be minimal. Every facet-defining inequality is minimal, but not conversely. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "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": 0.9793038964271545, "perplexity": 222.3733146485847}, "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-2019-47/segments/1573496667177.24/warc/CC-MAIN-20191113090217-20191113114217-00401.warc.gz"} |
https://www.physicsforums.com/threads/is-my-justification-acceptable.927699/ | # Is my justification acceptable?
• I
• Thread starter Tio Barnabe
• Start date
• #1
Tio Barnabe
I wanted to argue that $$\frac{d}{d \cos \theta} \sin \theta, \ \theta \in [0, \pi]$$ should be ignored in the given interval, because integration of it leads to a divergent integral. Is this acceptable as a reason?
## Answers and Replies
• #2
Mark44
Mentor
35,140
6,892
I wanted to argue that $$\frac{d}{d \cos \theta} \sin \theta, \ \theta \in [0, \pi]$$ should be ignored in the given interval, because integration of it leads to a divergent integral. Is this acceptable as a reason?
First off, ##\frac{d}{d \cos \theta} \sin \theta## is pretty unwieldy, as ##\sin(\theta)## isn't a function of ##\cos(\theta)## at first glance. However, you could write the derivative as ##\frac d {d(\cos(\theta))} (\pm \sqrt{1 - \cos^2{\theta}})##, and then use the chain rule. So, no, I don't see that it's reasonable to ignore it.
Second, integration and differentiation are different operations, so the fact that the integral of some function on some interval is divergent doesn't have any bearing here.
Likes Tio Barnabe, NFuller, FactChecker and 1 other person
• #3
FactChecker
Gold Member
6,384
2,523
In addition to what @Mark44 said, the idea that something can be ignored because it is divergent is wrong. If it was very small compared to other terms, that would be different, but being too large makes it impossible to ignore.
Likes Tio Barnabe
• #4
660
239
Just to further elaborate @Mark44's point, it isn't very hard to actually calculate the derivative and show that it does exist. So to drive home the point, here is the result.
Using the transformation ##\xi(\theta)=\text{cos}(\theta)##
$$\text{sin}(\theta)=\pm\sqrt{1-\xi^{2}}$$
so
$$\frac{d}{d\text{cos}(\theta)}\text{sin}(\theta)=\pm\frac{d}{d\xi}\sqrt{1-\xi^{2}}=\pm\frac{\xi}{\sqrt{1-\xi^{2}}}=\pm\frac{\text{cos}(\theta)}{\text{sin}(\theta)}=\pm\text{cot}(\theta)$$
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• #5
Tio Barnabe
Thank you to everyone
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1K | {"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": 0, "mathjax_display_tex": 1, "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": 0.9563779234886169, "perplexity": 2062.952729164577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00452.warc.gz"} |
https://www.learncram.com/physics/equation-of-continuity/ | # Equation of Continuity | Definition, Derivation – Hydrodynamics
Equation of Continuity Physics:
If a liquid is flowing in streamline flow in a pipe of non-uniform cross-sectional area, then rate of flow of liquid across any cross-section remains constant.
A continuity equation in physics is an equation that describes the transport of some quantity. It is particularly simple and powerful when applied to a conserved quantity, but it can be generalized to apply to any extensive quantity.
We are giving a detailed and clear sheet on all Physics Notes that are very useful to understand the Basic Physics Concepts.
## Equation of Continuity | Definition, Derivation – Hydrodynamics
Equation of Continuity Derivation:
i.e. a1v1 = a2v2 ⇒ av = constant
or
a ∝$$\frac{1}{v}$$
The velocity of liquid is slower where area of cross-section is larger and faster where area of cross-section is smaller.
The falling stream of water becomes narrower, as the velocity of falling stream of water increases and therefore its area of cross-section decreases. Deep water appears still because it has large cross-sectional area.
Hydrodynamics:
In physics, hydrodynamics of fluid dynamics explains the mechanism of fluid such as flow of liquids and gases. It has a wide range of applications such as evaluating forces and momentum on aircraft, prediction of weather, etc. | {"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": 0, "mathjax_display_tex": 1, "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": 0.9683239459991455, "perplexity": 991.248907443473}, "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-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00670.warc.gz"} |
http://rufuspollock.com/2008/12/10/theories-of-contextual-judgement-in-relation-to-well-being-and-other-outcomes/ | # Theories of Contextual Judgement in Relation to Well-Being and Other Outcomes
DECEMBER 10, 2008
I’ve just posted online a new paper on “Theories of Contextual Judgement in Relation to Well-Being and Other Outcomes”. This is a more a review-type effort and summarizes my thoughts (and reading of the existing literature) from the last year or so in relation to “relative” utility, status races and general contextual judgement/utility.
Abstract:
The paper presents an overview of existing theories on contextual judgement/utility situating them within a general framework. We consider the extent to which these theories are empirically distinguishable and apply them to some well-known economic questions such as the relation of happiness and the income distribution, status races and quitting. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8174432516098022, "perplexity": 1882.7457865649199}, "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-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00342.warc.gz"} |
https://www.uea.ac.uk/mathematics/news-and-events/pure-seminars/2014-15 | ## Spring 2015 Spring 2015
Seminars take place on Monday afternoons 14:00 - 15:00. Everyone is welcome.
### Talks
• January 26, (ARTS 01.01), Mark Blyth (UEA), Axisymmetric Solitary Waves on a Ferrofluid Jet.
• February 2 (JSC 0.01), Joseph Grant (UEA), Braid Groups and Quiver Mutation
• February 16, (ARTS 2.02), Jacob Hilton (Leeds), The Topological Pigeonhole Principle for Ordinals.
• February 23, (ITCS 01.26), Hans Jakob Rivertz (Visiting Professor UEA and Sørtrøndelag University College), Self-Similarity of Dihedral Tilings.
• March 2, (ARTS 2.01), Kaisa Kangas (Helsinki), From Strongly Minimal Sets to Zariski-like Structures.
• March 9, (ARTS 2.01), Alexei Vernitski (Essex), Knot Semigroups: Semigroup-Theory Problems and Knot-Theory Problems
• March 16, 15:00, (EFRY 01.02), Joint PhD Research Seminars
• April 20 (LT1), Teresa Conde (Oxford), The Quashihereditary Structure of a Schur-Like Algebra.
• April 27, (SCI 3.05), András Pongrácz (Middlesex University), The Topological Monoid Structure of End(F).
• May 11, (SCI 3.05), Philip Welch (Bristol), Turing Abound: Black Holes, Neural Nets, Bangs, Shrieks, Transfinite Ordinals and the Kitchen Sink.
• May 18, (SCI 3.05), Stefanie Zegowitz (Exeter), Closed Orbits in Quotient Systems.
• June 1, (SCI 3.05), Prof Alessandro Beraducci (universita'di Pisa),
• June 8, (SCI 3.05), Aaron Chan (Uppsala University), Tilting Theory of Brauer Graph Algebras and Curves of Riemann Surfaces.
• June 8, (SCI 3.05), Jeremie Guilhot (Tours), Kazhdan-Lusztig Cells in (Affine) Weyl Group.
### Abstracts
Joseph Grant: The possible ways to braid a given number of strings give a group, known as the braid group. Artin gave generators and relations for these groups, which can be encoded in a graph of Dynkin type A. This presentation can be generalized to other graphs.
A quiver is a directed graph. Mutation of quivers was introduced by Fomin and Zelevinsky as part of their theory of cluster algebras. I will describe joint work with Robert Marsh which links presentations of braid groups with mutation of quivers, and explain how this is related to triangulations of polygons.
The main result of this talk involves no complicated tools. I will assume the audience knows what a group is but, until the last ten minutes or so, I will explain any other concepts I use. However, the results are connected to some of the most important objects in modern mathematics and mathematical physics, as I will outline towards the end of the talk.
Jacob Hilton: The ordinary pigeonhole principle states that if you put more than n items into n containers, then one container must contain 2 or more items. The infinite pigeonhole principle states that if you put infinitely many items into finitely many containers, then one container must contain infinitely many items. What happens if you divide up a countable ordinal into finitely many containers? Must there be some container that contains a set of order type omega+1? We will study this question and see how the answer changes when we topologize the ordinals. We will also take a glance at joint work with Andrés Caicedo on Ramsey-style analogues of this question.
Hans Jakob Rivertz: Aperiodic tilings have been studied since the seventies. In 1972, Roger Penrose discovered a tessellation of the plane with pentagonal symmetry. This tessellation has two kinds of rhombic tiles. One thin tile with angle 36 degree and one fat tile with angle 72 degree. Although, this tessellation has no translational symmetries it has almost translational symmetries, i.e. translations that preserves all tiles but an arbitrary sparse set of tiles. There is also a subdivision of the tiles which give a new tessellation of the same type. A remarkable property is that the vertex set of the tessellation has a subset which is a vertex set of a tessellation with bigger tiles but of the same type as the original.
In a joint paper with J.-H. Eschenburg, we show that for any prime number $n = 2r+1\geq 5$ there exist $r$ planar tilings with self-similar vertex set and the symmetry of a regular n-gon ($D_n$-symmetry). The tiles are the rhombi with angle $\pi k/n$ for $k = 1,...,r.$
Kaisa Kangas: In the context of first order logic, strongly minimal theories provide a setting where mathematical structures can be classified based on dimension-like invariants. Assuming the existence of a Zariski-like topology, Zilber's trichotomy principle holds in this setting: a strongly minimal set is either trivial, interprets a vector space over some division ring or interprets an algebraically closed field. Finally, we look at generalizations of this setting beyond the first order context.
Teresa Conde: Given an arbitrary algebra $A$, we may associate to it a special endomorphism algebra $R_A$, introduced by Auslander, and further studied by Smalø. The algebra $R_A$ contains $A$ as an idempotent subalgebra and is quasihereditary with respect to a heredity chain constructed by Dlab and Ringel. In this talk we will discuss the nice properties of $R_A$ that stem from this heredity chain and mention some questions that arise from this setting.
Alexei Vernitski: Recently I introduced knot semigroups as a semigroup-theory construction inspired by knot groups. Then I started studying a number of questions which can be asked about knot semigroups. Here are some. Is it true that a semigroup of a knot is merely the set of 'positive' elements of its knot group? Can every knot semigroup be embedded in a group? Do all knot semigroups have a solvable word problem? Is a knot semigroup a knot invariant? Can we define semigroups of braids, links and tangles? How are knot semigroups related to Garside monoids (this is another knot-inspired semigroup construction)? How are knot semigroups related to the semigroup of knots with the operation of connected sum? How are knot semigroups related to graph semigroups?
András Pongrácz: The significance of algebraic invariants, such as the automorphism group, endomorphism monoid and polymorphism clone, in the study of countable omega-categorical structures was observed long ago, notably by Ryll-Nardzewski, who showed that omega-categoricity itself only depends on the permutation group structure of Aut(F). Later on, Ahlbrandt and Ziegler have shown that a countable omega-categorical F is uniquely determined up to first-order bi-interpretability by the topological group structure of Aut(F). Here, the topology is the one induced by pointwise convergence: closed sets are subsets of the full symmetric group Sym(F) that are closed under pointwise convergence. It was investigated in the 80's and 90's whether the condition of the Ahlbrandt-Ziegler theorem can be relaxed to the abstract group structure. We say that a group of the form Aut(F) has reconstruction if whenever it is isomorphic to another closed subgroup H of Sym(F) (as abstract group), it is also isomorphic to H as a topological group. Surprisingly, many subgroups of Sym(F) have this property, and it is in fact consistent with ZF that for all countable structures F the group Aut(F) has reconstruction.
With Manuel Bodirsky and Michael Pinsker, we initiated the study of the analogous concept of reconstruction for endomorphism monoids and polymorphism clones of omega-categorical structures. This is in particular motivated by generalizations of the Ahlbrandt-Ziegler theorem for these invariants, but also has potential applications in theoretical computer science. We have shown that the polymorphism clone of the random graph has reconstruction. In my talk, I will present the most important notions and results related to reconstruction, and briefly explain the connection of these results with theoretical computer science.
Philip Welch: There has been much talk (some of it rather loose) in the last few years concerning the possibility of "hypercomputation". This can start out from discussions as to whether Turing's machine model is appropriate for "modern" computation, or from claims that we can - perhaps only theoretically - compute "beyond the Turing limit", that is somehow fix up a device that enables us to 'compute' the Turing halting problem. On closer inspection most of such devices are seen to be formally cheating in some way or other, usually by assuming infinite precision measurement of some kind.
We look at the mathematics of one model that at least is not cheating in that sense: these are computations in Malament-Hogarth spacetimes. We investigate as a purely logical exercise the limits of computation in such manifolds.
Discrete transfinite machine models can be defined that run through ticks of time that are transfinitely ordered. These are purely logico-mathematical models, but are of interest as the associated recursion theory for such models ties in with work of Moschovakis, Harrington, Kechris, Gandy and Normann from the 1970's in generalized recursion theories. If time permits we shall discuss these too.
Stefanie Zegowitz We study the relationship between pairs of topological dynamical systems (X, T) and (X', T') where (X',T') is the quotient of (X, T) under the action of a finite group G. We describe three phenomena concerning the behaviour of closed orbits in the quotient system, and the constraints given by these phenomena. We find upper and lower bounds for the extremal behaviour of closed orbits in the quotient system in terms of properties of G and show that any growth rate in between these bounds can be achieved. Further, we study the asymptotic behaviour of the dynamical analogue of the Prime Number Theorem and Mertens' Theorem in the context of quotient systems.
Aaron Chan A ribbon graph is a combinatorial object one could use to construct bounded Riemann surfaces; it is a tool which is used in several branches of differential geometry. One can also construct finite dimensional algebras - the Brauer graph algebras out of ribbon graphs. Thus, one could ask the following questions. Given a homological problem of a Brauer graph algebra, what is its translation in the combinatorial theory of the defining ribbon graph? Similarly, what is its translation in the geometric theory of the associated Riemann surface? In this talk, I will describe an on-going collaboration with Takahide Adachi and Takuma Aihara, where we found that many tilting complexes of a Brauer graph algebra can be translated into nice collections of curves on the associated Riemann surface.
Jeremie Guilhot: A Coxeter group is a group generated by involutions plus some relations. The two most important kind of Coxeter groups are Weyl groups (which are finite groups generated by reflections) and affine Weyl groups (which are infinite groups generated by affine reflections). The aim of this talk is to introduce the notion of Kazhdan-Lusztig cells in a (affine) Weyl group. In the first part, we will concentrate on a small example (the symmetric group of order 3) and try to explicitly compute some cells. In the second part, we will discuss cells in affine Weyl groups.
## Autumn 2014 Autumn 2014
Seminars take place on Monday afternoons 14:00 - 15:00. Everyone is welcome.
### Talks
• October 6, (LT1), Carlos De la Mora (UEA), L-functions Here and There
• October 13, 16:00, (SCI 0.31), Alex Zalesski (UEA), Restriction of the Steinberg of Orthogonal Groups to the Orthogonal Subgroup of One Less Dimension
• October 20, (LT1), Tom Ward (Durham), Attempting a Polya-Carlson Dichotomy in Algebraic Dynamics
• October 27, 15:00, (SCI 3.05), Pure and Applied Joint PhD Maths Student Seminars, Ruari Walker - KLR Algebras and VV Algebras and Davide Maestrini - Vortex Clustering and Negative Temperature States in a 2D Bose-Einstein Condensate
• November 3, (ARTS 3.07), Keith Brown (UEA), Properly Stratified Quotients of Khovanov-Lauda-Rouquier Algebras
• November 10, (LT3), Sarah Scherotzke (Bonn), Graded Quiver Varieties and Derived Categories
• November 17, (ARTS 0.100), Baiying Liu (Utah), On Fourier Coefficients of Automorphic Forms
• November 24, (EFRY 01.08), Rob Kuriinczuk (Bristol), Rankin-Selberg Local Factors Modulo-1
• December 8, (TPSC 0.1), Faculty joint pure and applied seminar, David Aspero (MTH), Extremely Large Cardinals in the Absence of Choice
• December 15, (SCI 3.05), Nicola Gambino (Leeds), From Type Theory and Homotopy Theory to Univalent Foundations of Mathematics
### Abstracts
Carlos De la Mora: Let F be a number field and let Ḟ be the algebraic closure of F. Given ρ a representation of Gal(Ḟ/F) we can form its Artin L-function. It is known that Artin's L-functions have a meromorphic continuation to the complex plane and that they have an Euler product. Since Artin's L-functions have an Euler product we can study one prime at the time. Indeed, to each prime ideal we can attach two canonical complex functions, these are the local L-factor and the local epsilon factor. We now change gears and consider G to be a connected reductive group defined over F, A be the ring of adeles over F and let π be an automorphic representation of G(A). If V is the set of valuations of F, for v ϵ V we can complete F, we denote by Fv the completion of F with respect to v. It is known that π = ¤' v ϵ Vπv where πv is a smooth representation of G(Fv) for all v. One of the conjectures by Langlands is that to every smooth representation πv of G(Fv) one should be able to attach two canonical complex functions, these should be a local L-factor and an epsilon factor. Moreover if ρ is a representation of Gal(Ḟ/F) there should exist an automorphic cuspidal representation π of G(A) such that the local L-factor and epsilon factor of ρ and the local L-factor and epsilon factor of π agree for each valuation. The aim of this talk is to explain all the above in more detail.
Alex Zalesski: I consider the Stenberg character of the special orthogonal group over a finite field and the Steinberg character of it. The problem under discussion is to find the irreducible constituents of the restriction of this character to orthogonal group of one less dimension. The main result is a rather satisfactory solution to this problem.
Tom Ward: A subtle and beautiful classical result in complex variables is the Polya-Carlson theorem, which gives a strong rigidity to the possible behaviour of complex power series with integer coefficients. Many examples - and a few theorems - suggest a similar dichotomy may exist for a collection of functions arising in dynamical systems. I will give an overview of what is known here, and explain a wider context into which this dichotomy should fit - if it is true. Many of the arguments amount to quite concrete calculations with power series.
Ruari Walker: A new family of graded algebras have been introduced by Khovanov, Lauda and independently by Rouquier, the representation theory of which is closely related to that of the affine Hecke algebra of type A. They are often called KLR algebras. More recently, Varagnolo and Vasserot have defined a new family of graded algebras whose representation theory is related to the representation theory of the affine Hecke algebras of type B. These algebras can be thought of as type B analogues of the KLR algebras. I plan to explain this in a little more detail, show how the KLR algebras relate to the VV algebras and compare their module categories via Morita equivalence.
Davide Maestrini: In this work we investigate the question of clustering of like signed vortices in a two-dimensional Bose-Einstein condensate. Such clustering can be understood in terms of negative temperature states of a vortex gas. Due to the long-range nature of the Coloumb-like interactions in point vortex flows, these negative temperature states strongly depend on the shape of the geometry in which this clustering phenomena is considered. We analyze the problem of clustering of portices in a number of different regions. We present a theory to uncover the regimes for which clustering of like signed vortices can occur and compare our predictions with numerical simulations of a point vortex gas. We also extend our results to the Gross-Pitaevskii model of a Bose gas by performing numerical simulations for a range of vortex configurations using parameters that are relevant to current experiments.
Keith Brown: Introduced in 2008 by Khovanov and Lauda, and independently by Rouquier, the KLR algebras, are a family of infinite dimensional graded algebras which categorify the negative part of the quantum group associated to a graph~$\Gamma$. These algebras are known to have nice homological properties, in particular they are affine quasi-hereditary. In this talk I'll explain what it means to be affine quasi-hereditary and how this relates to properties of finite dimensional algebras. I'll then introduce a finite dimensional quotient of the KLR algebra which preserves some of the homological structure of the original algebra and provide a bound on its finitistic dimension. If time permits I'll explore a particular example in which the homological algebra is particularly nice. This work will form part of my PhD thesis, supervised by Dr. Vanessa Miemietz.
Sarah Scherotzke (Bonn): Nakajima's quiver varieties are important geometric objects in representation theory that can be used to give geometric constructions of quantum groups. Very recently, graded quiver varieties also found application to monoidal categorification of cluster algebras. Nakajima's original construction uses geometric invariant theory. In my talk, I will give an alternative representation theoretical definition of graded quiver varieties. I will show that the geometry of graded quiver varieties is governed by the derived category of the quiver $Q$. This approach brings about some new results on geometric properties of quiver varieties.
Baiying Liu (Utah): Fourier coefficients play an important role in the study of modular forms. In this talk, mainly focusing on general linear groups, I briefly introduce how to obtain Fourier coefficients and what kinds of Fourier coefficients people usually consider in the setting of automorphic forms. More explicitly, consider the standard maximal unipotent subgroup Un of GLn, consisting of upper triangular matrices. The question is that how to obtain Fourier coefficients of an automorphic form \phi along Un. When n = 2, Un is abelian, we know that taking abelian Fourier expansion is enough. But, when n >= 3, Un is not abelian. First, I will introduce the idea of Piatetski-Shapiro and Shalika, to obtain non-degenerate Whittaker Fourier coefficients of cuspidal automorphic forms along Un, using the property that Un is solvable. Then, I will show how to apply the same idea to obtain Fourier coefficients of automorphic forms in the residual spectrum of GLn. If time permitting, I will also briefly introduce a general setting of attaching Fourier coefficients to nilpotent orbits.
Rob Kuriinczuk (Bristol): I will briefly explain what an L-function is, how Rankin-Selberg local factors arise in the theory of automorphic L-functions, and discuss some recent work defining l-modular Rankin-Selberg local factors. This is joint work with Nadir Matringe.
David Aspero (UEA): We will stroll through the upper reaches of the large cardinal hierarchy, especially in a non-AC context.
Nicola Gambino (Leeds): Voevodsky's Univalent Foundations of Mathematics programme is an ambitious, long-term, project that seeks to develop a new approach to the foundations of mathematics on the basis of recently-discovered connections between type theory and homotopy theory. This programme aims also at facilitating the use of computer systems for the verification of mathematical proofs. The talk will consist of two parts. In the first part, I will give an introduction to the Univalent Foundations of Mathematics programme, without assuming prior familiarity with type theory or homotopy theory. In the second part, I will explain how the type-theoretic counterpart of the topological notion of contractibility can be used to characterise certain free constructions by means of a universal property that circumvents the use of infinite sets of coherence conditions. | {"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": 0.8610121607780457, "perplexity": 810.0396583286063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00168.warc.gz"} |
http://ptsymmetry.net/?p=1496 | ## Causality and phase transitions in PT-symmetrical optical systems
A. A. Zyablovsky, A. P. Vinogradov, A. V. Dorofeenko, A. A. Pukhov, A. A. Lisyansky
We discuss phase transitions in PT-symmetric optical systems. We show that due to frequency dispersion of the dielectric permittivity, an optical system can have PT-symmetry at isolated frequency points only. An assumption of the existence of a PT-symmetric system in a continuous frequency interval violates the causality principle. Therefore, the ideal symmetry-breaking transition cannot be observed by simply varying the frequency.
http://arxiv.org/abs/1401.4043
Optics (physics.optics); Mesoscale and Nanoscale Physics (cond-mat.mes-hall); Quantum Physics (quant-ph) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8947980999946594, "perplexity": 4474.68848914222}, "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-2018-47/segments/1542039742338.13/warc/CC-MAIN-20181115013218-20181115035218-00477.warc.gz"} |
https://www.centiserver.org/centrality/Holevo_Quantity-Edge_Centrality/ | # Holevo Quantity - Edge Centrality
#### Definition
This method is a novel edge centrality measure based on the quantum information theoretical concept of Holevo quantity. This is a measure of the difference in Von Neumann entropy between the original graph and the graph where $e$ has been removed. In other words, it can be seen as a measure of the contribution of $e$ to the Von Neumann entropy of $G$
For a graph $G = (V,E)$, the Holevo edge centrality of $e \in E$ is:
$$HC(c)=x (\{ ( {m-1\over m} ,H_{\bar e}),({1\over m} , H_e)\})$$
were $m=|E|$, $H_{\bar e}$ and $H_e$ are the subgraphs over edge sets $\{e\}$ and $E \{e\}$, respectively. $(m− 1)/m$ is constant for all the edges and thus can be safely ignored.
#### References
• Lockhart J., Minello G., Rossi L., Severini S., Torsello A., 2016. Edge centrality via the Holevo quantity. Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics), 10029 LNCS, pp.143-152. DOI: 10.1007/978-3-319-49055-7_13 | {"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": 1, "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": 0.8249157071113586, "perplexity": 731.4454397324361}, "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-2021-43/segments/1634323587655.10/warc/CC-MAIN-20211025061300-20211025091300-00458.warc.gz"} |
http://mathhelpforum.com/advanced-math-topics/183226-motorcyclist-s-problem-reformed-feasibility-general-case-hard.html | # Thread: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
1. ## The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
n walkers ${A}_{i}$ start at the same time from X to Y with speeds ${a}_{i}, i=1,2,...,n.\ {a}_{i}\neq{a}_{j}$, if $i\neq j$. At the same time, a motorcyclist M starts at Y to carry them. The motorcyclist's speed is m and it can go forwards or backwards. However, the motorcycle can only carry only one walker at a time, although it can choose to drop the walker on it at any time. The motorcylist's goal is to find a plan that allows all the n walkers to arrive at Y at the same time (i.e., a feasible plan). The problem and some definitions are illustrated in the attachment.
Now in case of n walkers, define $B^{n+2}\in {\mathbb{R}}_{++}^{n+2}$ to be the set of feasible configurations (i.e., configurations for which a feasible plan exists). Then after a little thought we know $B^{3}={\mathbb{R}}_{++}^{3}$, $B^{4}={\mathbb{R}}_{++}^{4}$.
Question 1: What is $B^{5}$ ?
Question 2: What is ${B}^{n}$ ? ( $n>5$) (But I really doubt there exists a closed-form expression for ${B}^{n}$ in terms of ${a}_{i}$, m, L and n! What do you think?)
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for the original problem (which asks you to find a feasible plan that takes least time) and some progresses made there please refer to http://www.mathhelpforum.com/math-he...es-182882.html
Attached Thumbnails
2. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
The following 3 properties of $B^{n+2}$ should be easy to prove:
1) If (a1,a2,...,an,m,L) $\in$ $B^{n+2}$, then(ra1,ra2,...,ran,rm,rL) $\in$ $B^{n+2}$, for any $r>0$. (This means that $B^{n+2}$ is homogenous of degree one. Hence we may fix one component of the configuration to 1)
2) If (a1,a2,...,an,m,L) $\in$ $B^{n+2}$, then(ra1,ra2,...,ran,rm,L) $\in$ $B^{n+2}$, for any $r>0$. (So we can further fix one more component to 1)
3) If (a1,a2,...,an,m,L) $\in$ $B^{n+2}$, then (a1,a2,...,an,bm,L) $\in$ $B^{n+2}$, for any b>1.
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and I suspect the following 2 are also corret (cannot prove them though):
4) If (a1,a2,...,an,m,L) $\in$ $B^{n+2}$, then (da1,a2,...,an,m,L) $\in$ $B^{n+2}$, for some d<1, where we assume a1>ai, for i=2,3,...,n
5) Any (a1,a2,...,an,m,L) $\in$ ${\mathbb{R}}_{++}^{n+2}$, there exists k large enough such that (a1,a2,...,an,km,L) $\in$ $B^{n+2}$.
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A $B^{n+2}$ that satisfies 1)~5) may indeed be strange...
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Edit: 1) and 5) are added. Corrected (ra1,ra2,...,ran,m,L) in 2) for (ra1,ra2,...,ran,rm,L). "any d<1" in property 4) is changed to "some d<1"
3. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Hi godelproof,
What do you say is the least time in hours to get the walkers over the finish line at the same time of day.
4. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by bjhopper
Hi godelproof,
What do you say is the least time in hours to get the walkers over the finish line at the same time of day.
Hi, I believe I've discussed that in http://www.mathhelpforum.com/math-he...es-182882.html, at #10 there. This least time is given there as $T_{min}$. But that formular is correct only if m is large enough. And the general case? We don't know yet...
5. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by godelproof
Question 1: What is $B^{5}$ ?
Question 2: What is ${B}^{n}$ ? ( $n>5$) (But I really doubt there exists a closed-form expression for ${B}^{n}$ in terms of ${a}_{i}$, m, L and n! What do you think?)
I'll post my opinions for Q1. Check to see if there's any problem...
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Consider the configuration $({a}_{1},\ {a}_{2},\ {a}_{3},m,L)$. Without loss of generality, let ${a}_{3}>{a}_{2}>{a}_{1}$. Let ${a}_{3} (otherwise no feasible plan exists).
Conjecture: Let the motorcycle go straight to ${A}_{1}$ and carry him forward to catch ${A}_{2}$. A feasible plan exists if and only if the motorcycle is able to catch ${A}_{2}$ before or just when she reaches Y.
Is the above conjecture true?
If it is, then ${\mathbb{B}}^{5}$ looks like this:
${\mathbb{B}}^{5}$={ $({a}_{1},{a}_{2},{a}_{3},m,L)|m>{a}_{3}>{a}_{2}>{a }_{1},\ 2{a}_{2}\leq{a}_{1}+m$}.
An illustration is given below for m=10. For any m, just shift the the blue lines up or down.
Attached Thumbnails
6. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by bjhopper
Hi godelproof,
What do you say is the least time in hours to get the walkers over the finish line at the same time of day.
least time takes about 4.5 hours
7. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
and a question for you two too, Wilmer and BJH!
True or false:
If ${A}_{1}, {A}_{2}, {A}_{3}$ and M all start out at X at the same time, then
${a}_{1}<{a}_{2}<{a}_{3} $\Longrightarrow$ a plan can be found to get the three walkers to Y simultaneously. (M can't carry a walker to a point before X)
8. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
TRUE; 179 fully integer cases (including pick-ups and drop-offs) keeping distance XY < 1000.
Smallest case: XY = 132; m = 6, a = 1, b = 2, c = 3 ; total time = 42
9. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by Wilmer
TRUE; 179 fully integer cases (including pick-ups and drop-offs) keeping distance XY < 1000.
Smallest case: XY = 132; m = 6, a = 1, b = 2, c = 3 ; total time = 42
No, that begs the question! For ANY a, b, c and m starting at X, as long as a<b<c<m, a plan exists:
XY=7 m=10.5 a=1 b=2 c=10 plan exists
XY=1 m=100.5 a=1 b=99 c=100 plan exists
XY=3 m=3.14 a=3 b=3.1 c=3.11 plan exists
ect...
True or false?
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Interestingly, although you've only considered integer solutions, for any noninteger configuration and solutions, we can always multiply them by some large integer and transform it to a problem with integer solutions! (See property 1) in #2). What does this mean? It means the problem has a solution iff when you multiply its configuration by some large integer, it has an integer solution! Amazing!!!
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And here's a challenge for everybody:
m=50, a=10, b=45, c=48, XY=whatever you like; a b c at X, m at Y
Can you find a plan? (noninteger plans are allowable)
(If you can find a plan for this, my conjecture in #5 would be incorrect)
10. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by godelproof
I'll post my opinions for Q1. Check to see if there's any problem...
-------------------------------------------------------------------
Consider the configuration $({a}_{1},\ {a}_{2},\ {a}_{3},m,L)$. Without loss of generality, let ${a}_{3}>{a}_{2}>{a}_{1}$. Let ${a}_{3} (otherwise no feasible plan exists).
Conjecture: Let the motorcycle go straight to ${A}_{1}$ and carry him forward to catch ${A}_{2}$. A feasible plan exists if and only if the motorcycle is able to catch ${A}_{2}$ before or just when she reaches Y.
If the motorcycle just catches $A_2$; by that time A3 has already finished the journey since, ${a}_{2}
Is the above conjecture true?
If it is, then ${\mathbb{B}}^{5}$ looks like this:
${\mathbb{B}}^{5}$={ $({a}_{1},{a}_{2},{a}_{3},m,L)|m>{a}_{3}>{a}_{2}>{a }_{1},\ 2{a}_{2}\leq{a}_{1}+m$}.
An illustration is given below for m=10. For any m, just shift the the blue lines up or down.
.
11. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by Sudharaka
If the motorcycle just catches ; by that time A3 has already finished the journey since ${a}_{2}<{a}_{3}$
Not true, my friend! On its way to meet ${A}_{1}$, M will meet ${A}_{3}$ first. M then carry him backward for some distance, drop him somewhere before M meets ${A}_{1}$. And M then carries ${A}_{1}$ to catch ${A}_{2}$ at destination Y, while ${A}_{3}$ just arrive there at the same time too! See the attachment!
Attached Thumbnails
12. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by godelproof
No, that begs the question! For ANY a, b, c and m starting at X, as long as a<b<c<m, a plan exists:
Didn't realise you meant "ANY"...
So a case like a=1, b=2, c=9, m=10 (everybody at X) could be one of the "ANY"!
Looks like possible this way (but not 100% sure):
1: M carries A for u hours; A continues, M returns
2: M meets C after v hours, then carries C back for w hours ; (u+v+w hours so far)
3: M drops off C, picks up B and carries B for x hours; B continues, M returns
4: M meets C after y hours, then carries C to Y taking z hours ; (u+v+w+x+y+z hours so far)
5: A and B and C arrive at Y at same time
I get a headache just thinking of it!!
By the way, you posted your "challenge" as an "edit": so will only be seen if someone happens
to be looking at the "edited" post.
13. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by Wilmer
Didn't realise you meant "ANY"...
So a case like a=1, b=2, c=9, m=10 (everybody at X) could be one of the "ANY"!
Looks like possible this way (but not 100% sure):
1: M carries A for u hours; A continues, M returns
2: M meets C after v hours, then carries C back for w hours ; (u+v+w hours so far)
3: M drops off C, picks up B and carries B for x hours; B continues, M returns
4: M meets C after y hours, then carries C to Y taking z hours ; (u+v+w+x+y+z hours so far)
5: A and B and C arrive at Y at same time
I get a headache just thinking of it!!
By the way, you posted your "challenge" as an "edit": so will only be seen if someone happens
to be looking at the "edited" post.
Haha, I appologize! I didn't mean to make you headache~ It's true for ANY 0<a<b<c<m, I proved it. But I will not post the proof (So to save you from another headache).
Why I ask you this? Because I can use this knowledge to prove my conjecture in #5. Now I've proved it to be true. But unless someone challenges me, I will not post the proof, either.
As for my challenge question, the answer is you can't find a solution for any XY
---------------------------------------------------------------------------
So we now know ${B}^{5}$ is as stated in #5. What is ${B}^{n}$ (n>5)? Well, perhaps that question just goes beyond my ability to answer... but would love to see someone come and work it out!
-----------------------------------------------
Edit: red letters
14. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by godelproof
Not true, my friend! On its way to meet ${A}_{1}$, M will meet ${A}_{3}$ first. M then carry him backward for some distance, drop him somewhere before M meets ${A}_{1}$. And M then carries ${A}_{1}$ to catch ${A}_{2}$ at destination Y, while ${A}_{3}$ just arrive there at the same time too! See the attachment!
Yeah I see. Didn't thought about the case where M carrying the walker backwards.
15. ## Re: The motorcyclist's problem reformed: feasibility in the general case (Hard!!!)
Originally Posted by godelproof
Not true, my friend! On its way to meet ${A}_{1}$, M will meet ${A}_{3}$ first. M then carry him backward for some distance, drop him somewhere before M meets ${A}_{1}$. And M then carries ${A}_{1}$ to catch ${A}_{2}$ at destination Y, while ${A}_{3}$ just arrive there at the same time too! See the attachment!
Hi godelproof,
So the motorist will meet A3 and carry him backwards on his way to meet A1. This is a must; if not your conjecture is useless. Agreed? After that he will carry A1 forwards. Suppose he cannot meet A2, then obviously A2 will reach the destination Y before the motorist and A1. Is this the proof of your conjecture? By the way, how did you write, $2{a}_{2}\leq{a}_{1}+m$ ?
Page 1 of 2 12 Last | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 87, "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": 0.8739041090011597, "perplexity": 1996.8355743724037}, "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-2013-48/segments/1387345767540/warc/CC-MAIN-20131218054927-00069-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://stats.stackexchange.com/questions/47890/can-you-use-proportions-as-a-covariate-in-a-cox-proportional-hazards-model | # Can you use proportions as a covariate in a Cox proportional hazards model?
In R's survival::coxph function, can I mix a covariate representing proportions (in the range 0.0-0.5) with an integer covariate (in the range 1-15), or should I transform the first one also to integers (0-50)?
-
You can mix them. As in other forms of regression, changing the scale will change the parameter estimates, but only in the same way that changing height from meters to feet would change it.
The meaning of the resulting model will be the same.
But you have to keep track of what you've done to the variables.
-
@PeterFlom makes good points: this kind of rescaling will not materially affect the fit of your model, but it will change the interpretation of the coefficients.
If you use the proportions in your model (stored as 0 to 0.5) then the hazard ratio (HR) will refer to the relative hazard difference for each one unit increase in that proportion (i.e. as though a difference of 100%!) .
So this variable might be a good candidate for rescaling (so that you can talk about a hazard ratio for each one percentage point difference, for instance, or a ten percentage point difference.) Same model: one is often easier to explain than another because the units are more interpretable.
Example: modelling risk of dying from myocardial infarction based on someone's weight (and let's assume that there is a relationship).
You could use weight in grams or weight in kilograms as the scaling for the predictor: both provide the same "information" in the model, but the HR when using grams will look barely different from one (HR for each one gram increase in weight -- most software probably couldn't print enough digits to make sense of such an HR,) while the HR when using kilograms will be more easily interpretable (one might even rescale weight into five or ten-kilo units in this scenario.)
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https://pythoninchemistry.org/sim_and_scat/classical_methods/bonds | # Bonds
Tha ability to simulate atoms interacting through the van der Waals and charged interactions is useful for the study of non-covalent materials. However, to truely probe the chemistry of a system we must also be able to model the bonded interactions.
The potential energy of the bonded interactions of a system is usually made up of bonds, angles, dihedrals, and other higher order processes,
We will focus just on the first two, more information about the higher order processes can be found in relevent textbooks (see Home). The bond lengths are typically modelled with a harmonic potential energy function,
where, $b_0$ is the equilibrium bond length and $K_b$ is the force constant for the bond. These must be determined (see Parameterisation), and $b$ is the measured bond length. An example of this function can be seen below, using the parameters of a carbon-carbon bond.
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
def bonded(kb, b0, b):
"""
Calculation of the potential energy of a bond.
Parameters
----------
kb: float
Bond force constant (units: eV/Å^2)
b0: float
Equilibrium bond length (units: Å)
b: float
Bond length (units: Å)
Returns
float
Energy of the bonded interaction
"""
return kb / 2 * (r - b0) ** 2
r = np.linspace(1, 2, 100)
plt.plot(r, bonded(274965.16, 1.522, r))
plt.xlabel(r'$r$/Å')
plt.ylabel(r'$E$/eV')
plt.show()
Using this harmonic function the bond length is modelled such that minimum if at the equilibrium bond length and both increasing and decreasing the length will increase the energy equally.
The potential energy of an angle between two atoms that are separated by a third is modeled with a similar harmonic function,
where, $\theta$ is the measured angle, $K_\theta$ is the force constant for the angle, and $\theta_0$ is the equilibrium angle size. This has the same shape as the bond potential energy curve shown above.
Activate the interactive MyBinder environment (by clicking the “Interact” button at the top). This will allow the code cell below to be run. This simulation shows a single diatomic molecule interacting through a single bond modelled with a harmonic potential model. Again, you can vary the temperature of the simulation and observe how the bond vibration changes.
import bond
%matplotlib notebook
bond.simulation(100)
For the sake of simplicity, the remainder of this resource will only focus on non-bonded interactions. However, it is important to be aware that all aspects discussed are equally applicable to bonded interactions. | {"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": 1, "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": 0.8950635194778442, "perplexity": 828.4244606112732}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2019-09/segments/1550247481612.36/warc/CC-MAIN-20190217031053-20190217053053-00411.warc.gz"} |
https://www.radicalsys.com/pl/pomoc-techniczna/baza-wiedzy/french-vista-program-files-directory-program-files-or-programmes/ | # Wyszukaj w bazie
## French Vista Program Files Directory - Program Files or Programmes?
Hint Ref: 020803110008
Hint Date: 11/03/2008
Hint Details:
With a French installation of Vista, there can be some confusion over the fact that the Space Manager installation installs into the c:\Program Files directory, but when exploring the directory structure, only the directory c:\Programmes can be seen.
Note that both 'Program Files' and 'Programmes' are the same location and which one is used will depend on how you are doing the browsing.
Also note that in some Vista installations, the location C:\Program Files\Spaceman32 can sometimes be redirected to c:\users\<userName>\AppData\VirtualStore\Program Files\Spaceman32.
This is not a good idea if Space Manager is to be run by multiple users. To run Space Manager from the usual directory location, Right Click on the Icon for the program or it's shortcut and go to Properties. On the Compatibility tab at the bottom is a check box to select to Run as Administrator. There is also an option at the top to run it in a mode from a prior operating system. If issues continue to be experienced, try setting this to XP service pack 2 mode. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8020000457763672, "perplexity": 2945.643492760783}, "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-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00383.warc.gz"} |
https://math.stackexchange.com/questions/1594340/center-of-arc-with-two-points-radius-and-normal-in-3d | # Center of Arc with Two Points, Radius, and Normal in 3D
I'm struggling to get the math to work out on this. I need to derive an alorithm for a program where I'm representing geometric entities. In this case, it's an arc. I would like to create the arc in 3D space using two points, the radius, and a normal. Any help would be greatly appreciated.
Edit:
To clarify....
• The two points, radius, and normal are given.
• It is a circular arc
• The two points are the end points of the arc.
• The radius is the radius of the circle defining the arc. Thus as would be expected, the segments from the two end points of the arc to the center would have a length equal to the radius.
• The normal is the normal unit vector to the plane of the circle defining the arc, thus yes perpendicular to the plane of the arc.
I'm just trying to derive an algorithm to find the center point of the circle defining the arc with the aforementioned given in 3D space. Any help would be greatly appreciated.
• Please be more clear. Are you given the two points, radius and normal and you want to find a corresponding circular arc? Are the two points the ends of the arc? Is the "radius" just a line segment that has the same length as a radius? Is the "normal" in the plane of the arc, perpendicular to that plane, or other? Is finding the center point of the arc enough do say we "found" the arc? – Rory Daulton Dec 30 '15 at 20:48
• I've updated the question per your comments. – bjhuffine Dec 30 '15 at 21:14
• Thank you, I have retracted my close vote due to your original unclear question. One last point: the "radius" you are given is a positive real number, the size of any radius of the circular arc? – Rory Daulton Dec 31 '15 at 0:15
• I assume you're asking if the radius is considered a variable parameter to the algorithm? Yes it is and yes it should be assumed to be any positive realy number. – bjhuffine Dec 31 '15 at 13:03
• The point of my question is that a "radius" can be a line segment from the center to the rim of a circle. It also can be a positive real number, the length of such a line segment. It is now clear that you do not mean the line segment. – Rory Daulton Dec 31 '15 at 13:33
The plane of the arc is found from one of the points with $d=\hat{n} \cdot \vec{p}_1$ such that any point $\vec{r}=(x,y,z)$ belongs to the plane if $$\hat{n} \cdot \vec{r} = d$$ where $\hat{n}$ is the unit normal vector.
The midpoint of the two points is $$\vec{p}_m = \frac{\vec{p}_1 + \vec{p}_2}{2}$$
Next we need two unit vectors on the plane to define planar coordinates. Use $$\hat{u} = {\rm unitvector}(\vec{p}_2-\vec{p}_1)$$ $$\hat{v} = \hat{n} \times \hat{u}$$
We need the center of the arc point on the plane, and we can find it by moving from the mid point $\vec{p}_m$ along the direction $\vec{v}$ by $$h = \frac{ \ell^2}{2 r}$$ where $\ell = \| \vec{p}_2 - \vec{p}_1 \|$.
The figure below shows the arc and the height calculation from the chord length $\ell$ and radius $r$.
So the center is at $$\vec{p}_c = \vec{p}_m + (h-r) \hat{v}$$
The included angle of the arc is $$\theta = 2 \sin^{-1} \left( \frac{\ell}{2 r} \right)$$
Finally the arc is parametrically defined as $$\vec{r}(t) = \vec{p}_c + {\rm Rot}(\hat{n}, -\frac{\theta}{2} + t \theta) \hat{v} r$$
• I tried working through what you have here and you seem to be very, very close. The equation for the sagitta, h, is off though. It should be $h=r-\sqrt{r^2-\ell^2}$ where $\ell$ is actually the half chord length. I had to research this part after finding the issue when debugging my code. Also, I haven't verified the effect on your angle and parametric equation formulas. – bjhuffine Dec 31 '15 at 15:08
• You are probably right. What I calculated and what I drew was different and thus in the inconsistency. Anyway I think you get the idea enough to work it out on your own. – ja72 Dec 31 '15 at 16:56 | {"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": 1, "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": 0.8444898724555969, "perplexity": 233.64536514454417}, "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-2019-39/segments/1568514575674.3/warc/CC-MAIN-20190922201055-20190922223055-00369.warc.gz"} |
https://istopdeath.com/find-the-area-rectangle-11in8in/ | # Find the Area rectangle (11in)(8in)
l=11inw=8in
The area of a rectangle is equal to the length times the width.
(length)⋅(width)
Substitute the values of the length l=11 and width w=8 into the formula for the area of a rectangle.
11⋅8
Multiply 11 by 8.
88(in)2
Find the Area rectangle (11in)(8in)
Scroll to top | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8412742018699646, "perplexity": 1294.3793153112001}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00536.warc.gz"} |
https://cstheory.stackexchange.com/questions/18652/how-does-this-computational-method-work/18665 | # How does this computational method work?
The last computational method example Knuth gives in 1.1 of Vol. 1 of 'The Art of Computer Programming' is defined by the following:
Let $A$ be a finite set of letters
Let $A^*$ be the set of all strings on A
Let $N$ be a nonnegative integer
Let $Q$ be the set of all $(\sigma,j)$ where $\sigma$ is in $A^*$ and $j$ is an integer $0\leq j\leq N$. ($Q$ denotes the states of computation in a computational method)
Let $I$ be the subset of $Q$ with $j=0$. ($I$ denotes the inputs in a computational method)
Let $\Omega$ be the subset of $Q$ with $j=N$
Let $f$ be a function, defined by the strings $\theta_j, \phi_j$ and the integers $a_j, b_j$ for $0\leq j < N$
$f(\sigma,j) = (\sigma, a_j)$ if $\theta_j$ does not occur in $\sigma$
$f(\sigma,j) = (\alpha\phi_j\omega, b_j)$ if $\alpha$ is the shortest possible string for which $\sigma=\alpha\theta_j\omega$
$f(\sigma,N)$ = $(\sigma,N)$.
What I don't understand is how the function works other than the last part where every output points to itself. Can someone please shed some light on what the first two function definitions are doing.
Before we look at what f does, let's look at some of Knuth's definitions earlier in the chapter.
A computational method is a quadruple ($Q, I, \Omega, f$) where $I,\Omega \subseteq Q$ and $f : Q \mapsto Q$; and $f$ has the property of leaving $\Omega$ pointwise fixed, i.e.: $\forall q \in \Omega, f(q) = q$.
The intended, informal meanings of these terms are the following:
$\fbox{Q}$ the states of the computation
$\fbox{I}$ the input
$\fbox{$\Omega$}$ the output
$\fbox{f}$ the computational rule
Although irrelevant to our discussion, Knuth then defines the notion of algorithm as follows:
An algorithm is a computational method that terminates in finitely many steps for all $x \in I$.
On page 8 he gives an explication of a gcd algorithm in terms of this formal notion. Now, getting back to f, it would be helpful if we keep the following auxiliary definitions in sight:
Occurs $(\theta, \sigma) =_{df} \exists \alpha,\omega \in A^*(\sigma = \alpha\theta\omega)$.
Shortest $(\sigma, \Sigma, \phi) =_{df} \phi(\sigma) \land \forall\tau\in\Sigma(\phi(\tau) \rightarrow [\tau] > [\sigma])$, where [σ] is the length of σ.
With these, we're ready to make sense of f. It maps strings σ and natural numbers j to:
$(\sigma, \alpha_j)$ $~~~~~~~~$if$~~$ $\lnot Occurs(\theta_j,\sigma)$
$(\alpha\phi_j\omega, b_j)$ $~~$if$~~$ $\exists\alpha\in A^*(\sigma = \alpha\theta_j\omega \land Shortest(\alpha, A^*, [\lambda x.\sigma = x\theta_j\omega])$
$(\sigma, N)$ $~~~~~~~~$ otherwise
Informally, we can describe f's behavior as follows. If no two strings $\alpha$ and $\omega$ exist in $A^*$ that can be wrapped around $\theta_j$ to make $\sigma$, then f returns $\sigma$ with its index $a_j$ (i'm not sure why Knuth differentiates between $a_j$ and $j$ s.t. $0 \leq j \leq N$). Else, if there exist strings $\alpha$ and $\omega$ that can be wrapped around $\theta_j$ in the following way: $\alpha\theta_j\omega$, to make $\sigma$, then f returns: the sequence of the shortest $\alpha$ that makes that equality true, $\phi_j$, $\omega$, along with the index $b_j$. In all the other cases, f returns $\sigma$ with the index N.
Hope this helps. If you find errors, feel free to edit this post or add a comment below.
• Can you explain what \tau is? I'm having trouble understanding that line. Also, how do I interpret that part with a lambda in it? Oct 3 '13 at 8:21
• (1/2) $\tau$, $\sigma$ are strings; $\Sigma$ is a set of strings (in the definition of f it gets called with $A^*$). The definition of Shortest says: given a string $\sigma$, a set of strings $\Sigma$, and a certain property $\phi$, we say that $\sigma$ is the shortest string in $\Sigma$ that satisfies the property $\phi$ iff (i) $\phi(\sigma)$, and (ii) every string $\tau \in \Sigma$ that also satisfies $\phi$ has a length ($[\tau]$) greater than the length of $\sigma$ ($[\sigma]$). Oct 3 '13 at 18:06
• (2/2) Then in the definition of f, Shortest gets called with $\alpha$, $A^*$, and $[\lambda x. \sigma = x \theta_j \omega]$. The last argument is an implicit way of defining the property/predicate "$x\theta_j \omega$ holds for x". I assumed most were familiar with the lambda notation, so I used it to define the property. Another way would be to define the property explicitly somewhere before the definition of f by saying: P(x) is true iff $\sigma = x\theta_j \omega$. Then in the definition of f, we could just say: Shortest($\alpha$, $A^*$, P). Oct 3 '13 at 18:20 | {"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": 1, "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": 0.9390782117843628, "perplexity": 250.1652190217218}, "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-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00682.warc.gz"} |
http://mathhelpforum.com/latex-help/222824-test-print.html | # Test
• Oct 9th 2013, 04:04 PM
Mathlv
Test
The Problem: Let S = (0,1)
Prove for each $\varepsilon$ > 0 there exist an x $\in$ S such that x < $\varepsilon$.
Solution:
For every $\varepsilon$ > 0, 0 < $\frac{\varepsilon }{2}$ < $\varepsilon$
Consider $\frac{\varepsilon }{2}$ and x = $\frac{1}{2}$
then
x < $\varepsilon$
$\Leftrightarrow$ $\frac{1}{2}$ < $\frac{\varepsilon }{2}$
$\Leftrightarrow$ 1 < $\varepsilon$
Thus
$\varepsilon$ > 0 and $\varepsilon$ = 1 > x = $\frac{1}{2}$
Therefore x <
$\varepsilon$
Q.E.D
• Oct 9th 2013, 05:47 PM
SlipEternal
Re: Test
See my post in the other forum. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "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": 0.9939796924591064, "perplexity": 1696.4838015880578}, "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-2016-44/segments/1476988725475.41/warc/CC-MAIN-20161020183845-00142-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://arxiv.org/abs/1507.08349 | cs.IT
(what is this?)
# Title: Converse Bounds for Entropy-Constrained Quantization Via a Variational Entropy Inequality
Abstract: We derive a lower bound on the smallest output entropy that can be achieved via vector quantization of a $d$-dimensional source with given expected $r$th-power distortion. Specialized to the one-dimensional case, and in the limit of vanishing distortion, this lower bound converges to the output entropy achieved by a uniform quantizer, thereby recovering the result by Gish and Pierce that uniform quantizers are asymptotically optimal as the allowed distortion tends to zero. Our lower bound holds for all $d$-dimensional memoryless sources having finite differential entropy and whose integer part has finite entropy. In contrast to Gish and Pierce, we do not require any additional constraints on the continuity or decay of the source probability density function. For one-dimensional sources, the derivation of the lower bound reveals a necessary condition for a sequence of quantizers to be asymptotically optimal as the allowed distortion tends to zero. This condition implies that any sequence of asymptotically-optimal almost-regular quantizers must converge to a uniform quantizer as the allowed distortion tends to zero.
Comments: 26 pages, 1 figure. Submitted to IEEE Transactions on Information Theory. Most important changes with respect to previous version: i) changed title (the old title was "Rate-distortion bounds for high-resolution vector quantization via Gibbs's inequality"); ii) added necessary conditions for a sequence of quantizers to be asymptotically optimal (Theorem 7 and Corollary 8) Subjects: Information Theory (cs.IT) Cite as: arXiv:1507.08349 [cs.IT] (or arXiv:1507.08349v3 [cs.IT] for this version)
## Submission history
From: Tobias Koch [view email]
[v1] Thu, 30 Jul 2015 00:03:28 GMT (160kb,D)
[v2] Fri, 1 Jul 2016 10:35:51 GMT (110kb,D)
[v3] Fri, 24 Mar 2017 14:58:23 GMT (122kb,D) | {"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": 0.8788238763809204, "perplexity": 1477.5643688327643}, "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-2017-30/segments/1500549424945.18/warc/CC-MAIN-20170725002242-20170725022242-00271.warc.gz"} |
https://www.asmedigitalcollection.asme.org/HT/proceedings-abstract/HT2007/42754/531/323995 | The turbulent flow and the heat transfer in a unitary cell of a cross corrugated plate pattern heat exchanger has been studied using Chen’s high-Re k-ε model, Suga’s low-Re k-ε model, the RSM and the V2F model at a Reynolds number of 4930. The ability of these models in predicting the mean Nusselt number and Fanning friction factor has been investigated. The V2F model predicted higher heat transfer and friction factors than the other models. It was observed that the upper and lower flow in the unitary cell interact throw a shear process. This in turn initiates a complex secondary flow pattern which promotes the heat transfer. The V2F model predicted the strongest shear process. This may explain the fact that it also predicted the highest values of heat transfer and friction factor compared to the other models. The shear flow also caused high levels of turbulent kinetic energy in the centre of the unitary cell. The observed secondary motion is believed to be an efficient means of increasing the heat transfer coefficient with limited pressure drop penalty. It is also demonstrated that despite the geometrical complexity, high quality computational grids can be created and thereby details of the flow and heat transfer phenomena can be studied. The RSM appeared to be instable and gave results similar to Chen’s k-ε model. Therefore, its use is not motivated for such applications.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.8812710046768188, "perplexity": 821.2453727325719}, "config": {"markdown_headings": false, "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-2023-06/segments/1674764500303.56/warc/CC-MAIN-20230206015710-20230206045710-00640.warc.gz"} |
http://mathonline.wikidot.com/a-sufficient-condition-for-the-differentiability-of-function | A Suff. Condition for the Differentiability of Functions from Rn to Rm
# A Sufficient Condition for the Differentiability of Functions from Rn to Rm
So far we have stated many properties of functions $\mathbf{f} : S \to \mathbb{R}^m$ (where $S \subseteq \mathbb{R}^n$) that are differentiable at a point $\mathbb{c} \in S$. For example, on the Differentiable Functions from Rn to Rm are Continuous page we saw that $\mathbf{f}$ being differentiable at $\mathbf{c}$ implies $\mathbf{f}$ is continuous at $\mathbf{c}$.
We have not yet developed any conditions under which we can determine whether a function $\mathbf{f}$ is differentiable or not (apart from using the definition outright). The following theorem gives us a sufficient (though not necessary) condition for differentiability of a function $\mathbf{f}$ at a point $\mathbf{c}$.
Theorem 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and let $\mathbf{f} : S \to \mathbb{R}^m$ where $\mathbf{f} = (f_1, f_2, ..., f_m)$. If at least one of the partial derivatives $D_1 \mathbf{f}$, $D_2 \mathbf{f}$, …, $D_n \mathbf{f}$ exist at $\mathbf{c}$ and if the remaining $n - 1$ partial derivatives are continuous at $\mathbf{c}$ and exist on an open ball centered at $\mathbf{c}$ then $\mathbf{f}$ is differentiable at $\mathbf{c}$.
Let's look at an example of applying this theorem. Consider the function $f : \mathbb{R}^3 \to \mathbb{R}$ defined for all $(x, y, z) \in \mathbb{R}^3$ by:
(1)
\begin{align} \quad f(x, y, z) = x^2 + y^2 + z^3 \end{align}
We claim that $f$ is differentiable at every point $(x, y, z) \in \mathbb{R}^3$.
First $S = \mathbb{R}^3$ is an open subset of $\mathbb{R}^3$ and $\mathbf{c} = (x, y, z) \in \mathbb{R}^3 = S$.
The three partial derivatives of $f$ are:
(2)
So at least one of the partial derivatives of $D_1 f$, $D_2 f$, $D_3 f$ exist at $\mathbf{c} = (x, y, z)$ (they all exist), and the remaining partial derivatives are certainly continuous on an open ball centered at $\mathbf{c} = (x, y, z)$ (in fact all of the partial derivatives above are continuous on every open ball centered at $\mathbf{c} = (x, y, z)$).
So by Theorem 1 we have that $f$ is differentiable at $\mathbf{c}$ for every $\mathbf{c} = (x, y, z) \in \mathbb{R}^3$. Since $\mathbf{c}$ was arbitrary, this means that $f$ is differentiable on all of $\mathbb{R}^3$. | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9938732385635376, "perplexity": 63.90653639099038}, "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-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00555.warc.gz"} |
http://math.stackexchange.com/users/43741/mack?tab=activity&sort=comments | # Mack
less info
reputation
837
bio website location Valdosta, ga age 21 member for 1 year, 9 months seen Jul 3 at 19:46 profile views 308
I am just a man who has an insatiable desire for knowledge.
"For the rest, brethren, whatever is true, whatever is worthy of reverence and is honorable and seemly, whatever is just, whatever is pure, whatever is lovely and lovable, whatever is kind and winsome and gracious, if there is any virtue and excellence, if there is anything worthy of praise, think on and weigh and take account of these things [fix your minds on them]."~ Philippians 4:8
Jun25 comment Considering Vectors Geometrically You seem to be saying that the geometric picture follows from the way in which vector addition is defined, and the fact that $\mathbb{R}^n$ itself has a geometric interpretation? Jun13 comment The role of 'arbitrary' in proofs I particularly like when you said "This just means we are not assuming anything about x..." That's an interesting idea. May23 comment Fourier's Heat Law In Integral Form Why wouldn't we use Fourier's law to find the total heat flow? I used the Heat Equation (the PDE) to find the temperature distribution $T(x,t)$; then I was going to use Fourier's law to find the total heat flow. Isn't your final equation just the Heat Equation integrated over $x$? I still don't see how that would give the heat flow. May23 comment Fourier's Heat Law In Integral Form I'm sorry, but I do not quite see how I can determine the amount of heat that has flown from (or into) the rod during a certain time interval from the final equation.. Could you help me with that? Feb8 comment Integrating With Respect To $x$ So, it would not be correct to use the differential of $y$, $dy = \frac{dy}{dx} dx$? Dec28 comment Satisfiability Problem: Determining Which People To Invite I don't believe that that is so. According to the answer key, which you seem to agree with, if Kanti attends the party, then Samir MUST attend--and there is no exception. But still, the answer key claims that p only if q is the same as q implies p, yet the people who answered my question given in the link state that p only if q is the same as p implies q. Essentially my question is, how can the answer key be correct, it would appear that it is incorrect. Dec25 comment Conditional Statements: “only if” This is a very interesting answer, something of which I was seeking. Dec24 comment Conditional Statements: “only if” Okay, I believe I am beginning to understand now. Thank you. Dec24 comment Conditional Statements: “only if” Why introduce the negations? Isn't that what you are doing, "arbitrarily" introducing negations? Why can't the statement be translated as is? Dec24 comment Conditional Statements: “only if” So, the word "only" introduces a negation? Why is that so? Dec24 comment Conditional Statements: “only if” I have read them carefully, and probably have done so for over a year. I understand what sufficient conditions and necessary conditions are. I understand the conditional relationship in almost all of its forms, except the form "q only if p" What I do not understand is, why is p the necessary condition and q the sufficient condition. I am not asking, what are the sufficient and necessary conditions, rather, I am asking why. Dec24 comment Conditional Statements: “only if” This is the problem I am having. When you have the statement "q if p," it translates to "p implies q;" and this makes sense: q can only be true if p is true. Now, when I see the statement "p only if q," I simply see this as a stronger version of "q if p," and should thus be translated in the same way. Dec9 comment Calculus.Integration of definite integral How come my LateX does not look as appealing as others? Dec7 comment Calculus.Integration of definite integral @Nabla Why would you have to apply it twice? Wouldn't once suffice? Dec7 comment Calculus.Integration of definite integral Is this the integral $\int_{-b}^{b} (\frac{\pi}{a^4})(y^2 - b^2)^4 dy$? I ask, because I am having a little difficulty interpreting what you wrote. Apr25 comment Solve for $x$: $4x = 6~(\mod 5)$ @AyushKhaitan What precisely is an integral value? Apr25 comment Solve for $x$: $4x = 6~(\mod 5)$ It is? How so?. Apr25 comment Solve for $x$: $4x = 6~(\mod 5)$ I don't see that written anywhere in your answer. Apr25 comment Solve for $x$: $4x = 6~(\mod 5)$ I'm not quite certain of what you are doing. Are you solving for a specific x-value? Apr23 comment Using The Pigeon-Hole Principle So there difference being less than n--that is, $j - k < n$--proves that when taking the modulus the result is distinct? I'm sorry, I don't quite see it. | {"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": 0.8963505029678345, "perplexity": 672.4256319121823}, "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-2014-23/segments/1404776417380.9/warc/CC-MAIN-20140707234017-00084-ip-10-180-212-248.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/197563-proof-p-1-mod-4-if-p-x-2-1-a-print.html | # Proof p=1 mod 4 if p|x^2+1
• April 19th 2012, 11:25 AM
ILikeSerena
Proof p=1 mod 4 if p|x^2+1
Problem statement
Let n be a whole number of the form $n=x^2+1$ with $x \in Z$, and p an odd prime that divides n.
Proof: $p \equiv 1 \pmod 4$.
Attempt at a solution
The only relevant case is if p=3 (mod 4).
If I try to calculate mod 3, or mod 4, or mod p, I'm not getting anywhere.
Help?
• April 20th 2012, 03:50 PM
Sylvia104
Re: Proof of p=1 mod 4 if p|(x^2+1)
Use Euler's criterion. $p\mid x^2+1$ $\implies$ $-1$ is a quadratic residue $\mod p$ $\implies$ $1=\left(\frac{-1}p\right)=(-1)^{\frac{p-1}2}$ $\implies$ $\frac{p-1}2$ is even $\implies$ $p\equiv1\mod4.$
• April 20th 2012, 04:08 PM
ILikeSerena
Re: Proof of p=1 mod 4 if p|(x^2+1)
Thanks!
I was not aware of Euler's criterion yet.
Now I am! (Bow)
• April 21st 2012, 05:41 AM
TDA120
Re: Proof p=1 mod 4 if p|x^2+1
Sylvia, sorry, I don’t get your advice. What’s Euler’s criterion?
• April 21st 2012, 05:47 AM
Sylvia104
Re: Proof of p=1 mod 4 if p|(x^2+1)
• April 21st 2012, 05:55 AM
ILikeSerena
Re: Proof p=1 mod 4 if p|x^2+1
Quote:
Originally Posted by TDA120
Sylvia, sorry, I don’t get your advice. What’s Euler’s criterion?
Welcome to MHF, TDA120! :)
@Sylvia: Not bad, a response time of 6 minutes! :D
• April 21st 2012, 01:05 PM
ILikeSerena
Re: Proof p=1 mod 4 if p|x^2+1
Let me make an attempt to explain in a little more detail.
(Burn me down if I'm wrong.)
From Fermat's little theorem we know that $a^{p-1} \equiv 1 \pmod p$, if p is prime and is not a multiple of p.
The question becomes, what about $a^{p-1 \over 2}$, assuming p is odd?
That would be either -1 or +1.
And that is basically what Euler's criterion says.
The notation $\left({a \over p}\right)$ is just a fancy way of writing $a^{p-1 \over 2}$.
• April 23rd 2012, 01:52 PM
Deveno
Re: Proof p=1 mod 4 if p|x^2+1
here is the way i look at it:
odd primes come in two flavors, 4n+3 and 4n+1 (since all odd numbers only come in these two flavors).
saying p|(x2+1) is another way of saying that we have a square root of -1 in Zp.
so if we can show that if p = 4n + 3 there is no such root, we are done.
now, since p is prime U(Zp) is a multiplicative group of order p - 1. if p = 4n + 3, then p - 1 = 4n + 2.
this is not divisible by 4, so in this case, U(Zp) can have no element a of order 4, or else |<a>| = 4 would divide |U(Zp)| = 4n + 2, by Lagrange's theorem.
note that in Zp[x], x4 - 1 = (x2 + 1)(x2 - 1). since 1 and -1 account for the roots of x2 - 1, we are left with the roots of x2 + 1. but if a2 = -1, then a is of order 4, and we have no such elements.
• April 23rd 2012, 02:05 PM
ILikeSerena
Re: Proof p=1 mod 4 if p|x^2+1
Thanks!
Actually, I recently got the hint: what is the order of x in Z/pZx?
When I thought about it (for a long while!), I realized that if p|x2+1, that means that x2=-1 mod p.
In turn that means that |x|=4 in Z/pZx.
Since |x| must divide the order of Z/pZx (Lagrange), it follows that 4|p-1.
Therefore p=1 mod 4. | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "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": 0.9360288977622986, "perplexity": 1184.316168936681}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2016-22/segments/1464049276567.28/warc/CC-MAIN-20160524002116-00061-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://www.onemathematicalcat.org/algebra_book/online_problems/is_num_perfect_square.htm | IDENTIFYING PERFECT SQUARES
• PRACTICE (online exercises and printable worksheets)
Take the whole numbers and square them:
$0^2 = 0$
$1^2 = 1$
$2^2 = 4$
$3^2 = 9$
and so on.
The resulting numbers $\,0, 1, 4, 9, 16, 25, 36, \ldots\,$ are called perfect squares.
DEFINITION perfect square
A number $\,p\,$ is called a perfect square if and only if
there exists a whole number $\,n\,$ for which $\,p = n^2\,$.
In other words:
How do you get to be a perfect square?
Answer: By being equal to the square of some whole number.
(Recall that the whole numbers are $\,0, 1, 2, 3, \ldots\,$)
In this exercise, you will decide if a given number is a perfect square.
The key is to rename the number (if possible) as a whole number, squared!
You may want to review this section first: Equal or Opposites?
EXAMPLES:
Question: Is $\,9\,$ a perfect square?
Solution: Yes. $\,9 = 3^2$
Question: Is $\,7\,$ a perfect square?
Solution: No. The number $\,7\,$ can't be written as a whole number, squared.
Question: Is $\,17^2\,$ a perfect square?
Solution: Yes. The number $\,17\,$ is a whole number, so $\,17^2\,$ is a whole number, squared.
Question: Is $\,17^4\,$ a perfect square?
Solution: Yes. Rename as $\,(17^2)^2\,$. The number $\,17^2\,$ is a whole number, so $\,(17^2)^2\,$ is a whole number, squared.
Question: Is $\,(-6)^2\,$ a perfect square?
Solution: Yes. Rename as $\,6^2\,$. The number $\,6\,$ is a whole number, so $\,6^2\,$ is a whole number, squared.
Question: Is $\,-6^2\,$ a perfect square?
Solution: No. Recall that $\,-6^2 = (-1)(6^2) = (-1)(36) = -36\,$. A perfect square can't be negative.
Be careful!
The numbers $\,-6^2\,$ and $\,(-6)^2\,$ represent different orders of operation, and are different numbers!
Question: Is $\,(-7)^{12}\,$ a perfect square?
Solution: Yes. Rename: $\,(-7)^{12} = 7^{12} = (7^6)^2\,$.
The number $\,7^6\,$ is a whole number, so $\,(7^6)^2\,$ is a whole number, squared.
Question: Is $\,-4\,$ a perfect square?
Solution: No. A perfect square can't be negative.
Master the ideas from this section
Writing Expressions in the form $\,A^2$ | {"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": 0.8464263677597046, "perplexity": 837.1459711264516}, "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-2017-22/segments/1495463608067.23/warc/CC-MAIN-20170525121448-20170525141448-00125.warc.gz"} |
http://www.winsteps.com/winman/table23_0.htm | # Table 23.0 Variance components for items
The dimensionality analysis of Table 23 stratifies the items into three clusters for each Contrast (Principal Component). The items are listed in Table 23.2. For each Contrast, each person is measured on each cluster of items. These measures are then correlated for each pair of clusters. If the correlations approach 1.0, then empirically the clusters of items are measuring approximately the same thing. Since measurement error (low test reliability) lessens the correlations, the disattenuated correlations are also reported.
Table 23.0 Variance components scree plot for items
Table 23.2, 23.12 Item Principal components analysis/contrast of residuals
Table 23.3, 23.13 Item contrast by persons
Table 23.6, 23.16 Person measures for item clusters in contrast. Cluster Measure Plot for Table 23.6.
Table 23.99 Largest residual correlations for items
Extreme scores produce estimated measures with high standard errors. Correlations are reported excluding and also including (+Extreme) person estimates for extreme scores.
Table of STANDARDIZED RESIDUAL variance in Eigenvalue units = ACT information units
Eigenvalue Observed Expected
Total raw variance in observations = 50.9521 100.0% 100.0%
Raw variance explained by measures = 25.9521 50.9% 50.7%
Raw variance explained by persons = 10.3167 20.2% 20.2%
Raw Variance explained by items = 15.6354 30.7% 30.6%
Raw unexplained variance (total) = 25.0000 49.1% 100.0% 49.3%
Unexplned variance in 1st contrast = 4.6287 9.1% 18.5%
Unexplned variance in 2nd contrast = 2.9434 5.8% 11.8%
Unexplned variance in 3rd contrast = 2.2957 4.5% 9.2%
Unexplned variance in 4th contrast = 1.7322 3.4% 6.9%
Unexplned variance in 5th contrast = 1.6327 3.2% 6.5%
STANDARDIZED RESIDUAL VARIANCE SCREE PLOT
VARIANCE COMPONENT SCREE PLOT
+--+--+--+--+--+--+--+--+--+--+--+
100%+ T +
| |
V 63%+ +
A | M |
R 40%+ U +
I | |
A 25%+ I +
N | P |
C 16%+ +
E | |
10%+ +
L | 1 |
O 6%+ +
G | 2 |
| 4%+ 3 +
S | 4 5 |
C 3%+ +
A | |
L 2%+ +
E | |
D 1%+ +
| |
0.5%+ +
+--+--+--+--+--+--+--+--+--+--+--+
TV MV PV IV UV U1 U2 U3 U4 U5
VARIANCE COMPONENTS
Table 23.0 shows a variance decomposition of the observations for the items. This is not produced for PRCOMP=O.
If your Table says "Total variance in observations", instead of "Total raw variance in observations", then please update to the current version of Winsteps, or produce this Table with PRCOMP=R.
Extreme items and persons (minimum possible and maximum possible raw scores) are omitted from this computation because their correlations are 0.
Simulation studies, and the empirical results of Winsteps users, indicated that the previous computation of "variance explained" was over-optimistic in explaining variance. So a more conservative algorithm was implemented. Technically, the previous computation of "variance explained" used standardized residuals (by default). These are generally considered to have better statistical properties than the raw residuals. But the raw residuals (PRCOMP=R) were found to provide more realistic explanations of variance, so the current Winsteps computation uses raw residuals for "variance explained" in the top half of the variance table.
The "Unexplained variance" is controlled by PRCOMP=, which defaults to standardized residuals (PRCOMP=S). Set PRCOMP=R to express the entire table in terms of raw residuals.
Table of STANDARDIZED RESIDUAL variance in Eigenvalue units = ITEM information units Eigenvalue Observed percentage of total variance Observed percentage of unexplained variance Expected percentage of total variance Total raw variance in observations = 50.9 100.0% ← Expected values if these data fit the Rasch model perfectly → ← If these match reasonably, then the measures explain the expected amount of variance in the data → 100.0% Raw variance explained by measures = 25.9 50.9% 46.5% Raw variance explained by persons = 10.3 20.2% 18.5% Raw Variance explained by items = 15.6 30.7% 28.0% Raw unexplained variance (total) = 25.0 = count of items (or persons) 49.1% 100.0% 53.5% Unexplned variance in 1st contrast = 4.6 9.1% 18.5% ← Use simulations to estimate the Rasch-model-expected values SIFILE= Unexplned variance in 2nd contrast = 2.9 5.8% 11.8% Unexplned variance in 3rd contrast = 2.3 4.5% 9.2% Unexplned variance in 4th contrast = 1.7 3.4% 6.9% Unexplned variance in 5th contrast = 1.6 3.2% 6.5%
Table of STANDARDIZED RESIDUAL variance: the standardized residuals form the basis of the "unexplained variance" computation, set by PRCOMP=
in Eigenvalue units: variance components are rescaled so that the total unexplained variance has its expected summed eigenvalue.
ITEM information units: the eigenvalue units are rescaled to match the number of items, so these values are equivalent to "strength in item units".
Observed: variance components for the observed data
Expected: variance components expected for these data if they exactly fit the Rasch model, i.e., the variance that would be explained if the data accorded with the Rasch definition of unidimensionality.
Total raw variance in observations: total raw-score variance in the observations
Raw variance explained by measures: raw-score variance in the observations explained by the Rasch item difficulties, person abilities and polytomous scale structures.
Raw variance explained by persons: raw-score variance in the observations explained by the Rasch person abilities (and apportioned polytomous scale structures)
Raw variance explained by items: raw-score variance in the observations explained by the Rasch item difficulties (and apportioned polytomous scale structures)
Raw unexplained variance (total): raw-score variance in the observations not explained by the Rasch measures
Unexplned variance in 1st, 2nd, ... contrast: variance that is not explained by the Rasch measures is decomposed into Principal Component Analysis, PCA, components = Contrasts. The size of the first, second, ... contrast (component) in the PCA decomposition of standardized residuals (or as set by PRCOMP=), i.e., variance that is not explained by the Rasch measures, but that is explained by the contrast.
The important lines in this Table are "contrasts". If the first contrast is much larger than the size of an Eigenvalue expected by chance, usually less than 2 - www.rasch.org/rmt/rmt191h.htm - please inspect your Table 23.3 to see the contrasting content of the items which is producing this large off-dimensional component in your data, or Table 24.3 to see the contrasting persons. The threat to Rasch measurement is not the ratio of unexplained (by the model) to explained (by the model), or the amount of explained or unexplained. The threat is that there is another non-Rasch explanation for the "unexplained". This is what the "contrasts" are reporting.
How Variance Decomposition is done ...
1.A central person ability and a central item difficulty are estimated. When the central ability is substituted for the estimated person abilities for each observation, the expected total score on the instrument across all persons equals the observed total score. Similarly, when the central ability is substitute for the estimated item difficulties for each observation, the expected total score on the instrument across all items equals the observed total score.
2.For each observation, a central value is predicted from the central person ability and the central item difficulty and the estimated rating scale (if any).
In the "Observed" columns:
3."Total raw variance in observations =" the sum-of-squares of the observations around their central values.
4."Raw unexplained variance (total)=" is the sum-of-squares of the difference between the observations and their Rasch predictions, the raw residuals.
5."Raw variance explained by measures=" is the difference between the "Total raw variance" and the "Raw unexplained variance".
6."Raw variance explained by persons=" is the fraction of the "Raw variance explained by measures=" attributable to the person measure variance (and apportioned rating scale structures).
7."Raw variance explained by items=" is the fraction of the "Raw variance explained by measures=" attributable to the item measure variance (and apportioned rating scale structures).
8.The reported variance explained by the items and the persons is normalized to equal the variance explained by all the measures. This apportions the variance explained by the rating scale structures.
9.The observation residuals, as transformed by PRCOMP=, are summarized as an inter-person correlation matrix, with as many columns as there are non-extreme persons. This correlation matrix is subjected to Principle Components Analysis, PCA.
10.In PCA, each diagonal element (correlation of the person with itself) is set at 1.0. Thus the eigenvalue of each person is 1.0, and the total of the eigenvalues of the matrix is the number of persons. This is the sum of the variance modeled to exist in the correlation matrix, i.e., the total of the unexplained variance in the observations.
11.For convenience the size of the "Raw unexplained variance (total)" is rescaled to equal the total of the eigenvalues. This permits direct comparison of all the variance terms.
12.The correlation matrix reflects the Rasch-predicted randomness in the data and also any departures in the data from Rasch criteria, such as those due to multidimensionality in the persons.
13.PCA reports components. If the data accord with the Rasch model, then each person is locally independent and the inter-person correlations are statistically zero. The PCA analysis would report each person as its own component. Simulation studies indicate that even Rasch-conforming data produce eigenvalues with values up to 2.0, i.e., with the strength of two persons.
14.Multidimensionality affects the pattern of the residuals. The residual pattern should be random, so the "contrast" eigenvalue pattern should approximately match the eigenvalue pattern from simulated data. When there is multidimensionality the residuals align along the dimensions, causing the early contrast eigenvalues to be higher than those from random (simulated) data. So multidimensionality inflates the early PCA contrasts above the values expected from random data, and correspondingly must lower the later ones, because the eigenvalue total is fixed.
15."Unexplned variance in 1st contrast =" reports the size of the first PCA component. This is termed a "contrast" because the substantive differences between persons that load positively and negatively on the first component are crucial. It may reflect a systematic second dimension in the persons.
16."Unexplned variance in 2nd contrast =". Consecutively smaller contrasts are reported (up to 5 contrasts). These may also contain systematic multi-dimensionality in the persons.
In the "Expected" columns:
17."Raw variance explained by measures=" is the sum-of-squares of the Rasch-predicted observations (based on the item difficulties, person abilities, and rating scale structures) around their central values.
18."Raw variance explained by persons=" is the fraction of the "Raw variance explained by measures=" attributable to the person measure variance (and apportioned rating scale structures).
19."Raw variance explained by items=" is the fraction of the "Raw variance explained by measures=" attributable to the item measure variance (and apportioned rating scale structures).
20.The reported variance explained by the items and the persons is normalized to equal the variance explained by all the measures. This apportions the variance explained by the rating scale structures.
21."Raw unexplained variance (total)=" is the summed Rasch-model variances of the observations around their expectations, the unexplained residual variance predicted by the Rasch model.
22."Total raw variance in observations =" is the sum of the Rasch-model "Raw variance explained by measures=" and the "Raw unexplained variance (total)="
23.The "Model" and the "Empirical" values for the "Total raw variance in observations =" are both rescaled to be 100%.
24.Use the SIFILE= option in order to simulate data. From these data predicted model values for the contrast sizes can be obtained.
STANDARDIZED RESIDUAL VARIANCE SCREE PLOT
VARIANCE COMPONENT SCREE PLOT
+--+--+--+--+--+--+--+--+--+--+--+
100%+ T +
| |
V 63%+ +
A | M |
R 40%+ U +
I | |
A 25%+ I +
N | P |
C 16%+ +
E | |
10%+ +
L | 1 |
O 6%+ +
G | 2 |
| 4%+ 3 +
S | 4 5 |
C 3%+ +
A | |
L 2%+ +
E | |
D 1%+ +
| |
0.5%+ +
+--+--+--+--+--+--+--+--+--+--+--+
TV MV PV IV UV U1 U2 U3 U4 U5
VARIANCE COMPONENTS
Scree plot of the variance component percentage sizes, logarithmically scaled:
On plot On x-axis Meaning T TV total variance in the observations, always 100% M MV variance explained by the Rasch measures P PV variance explained by the person abilities I IV variance explained by the item difficulties U UV unexplained variance 1 U1 first contrast (component) in the residuals 2 U2 second contrast (component) in the residuals, etc.
For the observations (PRCOMP=Obs), a standard Principal Components Analysis (without rotation, and with orthogonal axes) is performed based on the scored observations.
Table of OBSERVATION variance (in Eigenvalue units)
-- Observed --
Raw unexplained variance (total) = 13.0 100.0%
Unexplned variance in 1st contrast = 9.9 76.1%
Unexplned variance in 2nd contrast = .9 7.2%
Unexplned variance in 3rd contrast = .6 4.3%
Unexplned variance in 4th contrast = .3 2.6%
Unexplned variance in 5th contrast = .3 2.0%
Here "contrast" means "component" or "factor".
Approximate relationships between the KID measures
PCA ACT Pearson Disattenuated Pearson+Extr Disattenuated+Extr
Contrast Clusters Correlation Correlation Correlation Correlation
1 1 - 3 0.1404 0.2175 0.1951 0.2923
1 1 - 2 0.2950 0.4675 0.3589 0.5485
1 2 - 3 0.8065 1.0000 0.8123 1.0000
2 1 - 3 0.5319 0.7042 0.5640 0.7464
2 1 - 2 0.6546 0.9232 0.6782 0.9434
2 2 - 3 0.6271 0.8799 0.6672 0.9217
3 1 - 3 0.2320 0.4686 0.2765 0.5197
3 1 - 2 0.8083 1.0000 0.8080 1.0000
3 2 - 3 0.4327 0.7342 0.5060 0.7971
4 1 - 3 0.6955 1.0000 0.7088 1.0000
4 1 - 2 0.7467 1.0000 0.7496 1.0000
4 2 - 3 0.7731 1.0000 0.7601 1.0000
5 1 - 3 0.5854 0.9622 0.6242 1.0000
5 1 - 2 0.6667 1.0000 0.6986 1.0000
5 2 - 3 0.7665 1.0000 0.7913 1.0000
For explanation, see Table 23.1
Example 1:
We are trying to explain the data by the estimated Rasch measures: the person abilities and the item difficulties. The Rasch model also predicts random statistically-unexplained variance in the data. This unexplained variance should not be explained by any systematic effects.
Table of RAW RESIDUAL variance (in Eigenvalue units)
Empirical Modeled
Total raw variance in observations = 19.8 100.0% 100.0%
is composed of
Raw variance explained by measures = 7.8 39.3% 39.1%
Raw unexplained variance (total) = 12.0 60.7% 100.0% 60.9%
Nothing is wrong so far. The measures are central, so that most of the variance in the data is unexplained. The Rasch model predicts this unexplained variance will be random.
Raw variance explained by measures = 7.8 39.3% 39.1%
is composed of
Raw variance explained by persons = 5.7 28.9% 28.8%
Raw Variance explained by items = 2.1 10.4% 10.3%
Nothing is wrong so far. The person measures explain much more variance in the data than the item difficulties. This is probably because the person measure S.D. is bigger than the item difficulty S.D. in Table 3.1.
Raw unexplained variance (total) = 12.0 60.7% 100.0% 60.9%
is composed of
Unexplned variance in 1st contrast = 2.6 13.1% 21.5%
Unexplned variance in 2nd contrast = 1.4 6.9% 11.4%
Unexplned variance in 3rd contrast = 1.3 6.4% 10.6%
Unexplned variance in 4th contrast = 1.1 5.6% 9.3%
Unexplned variance in 5th contrast = 1.1 5.4% 8.9%
Now we have multidimensionality problems. According to Rasch model simulations, it is unlikely that the 1st contrast in the "unexplained variance" (residual variance) will have a size larger than 2.0. Here it is 2.6. Also the variance explained by the 1st contrast is 13.1%. this is larger than the variance explained by the item difficulties 10.4%. A secondary dimension in the data appears to explain more variance than is explained by the Rasch item difficulties. Below is the scree plot showing the relative sizes of the variance components (logarithmically scaled).
VARIANCE COMPONENT SCREE PLOT
+--+--+--+--+--+--+--+--+--+--+--+
100%+ T +
| |
V 63%+ +
A | U |
R 40%+ +
I | M |
A 25%+ P +
N | |
C 16%+ +
E | 1 |
10%+ I +
L | |
O 6%+ 2 3 +
G | 4 5 |
| 4%+ +
S | |
C 3%+ +
A | |
L 2%+ +
E | |
D 1%+ +
| |
0.5%+ +
+--+--+--+--+--+--+--+--+--+--+--+
TV MV PV IV UV U1 U2 U3 U4 U5
VARIANCE COMPONENTS
Example 2:
Question: My Rasch dimension only explains 45.5% of the variance in the data and there is no clear secondary dimension. How can I increase the "variance explained"?
Reply: If there is "no clear secondary dimension" and no excessive amount of misfitting items or persons, then your data are under statistical control and your "variance explained" is as good as you can get without changing the sample or the instrument.
Predicted Explained-Variance
A Rasch model predicts that there will be a random aspect to the data. This is well understood. But what does sometimes surprise us is how large the random fraction is. The Figure shows the proportion of "variance explained" predicted to exist in dichotomous data under various conditions.
The x-axis is the absolute difference between the mean of the person and item distributions, from 0 logits to 5 logits. The y-axis is the percent of variance in the data explained by the Rasch measures. Each plotted line corresponds to one combination of standard deviations. The lesser of the person S.D. and the item S.D. is first, 0 to 5 logits, followed by "~". Then the greater of the person S.D. and the item S.D. Thus, the arrows indicate the line labeled "0-3". This corresponds to a person S.D. of 0 logits and an item S.D. of 3 logits, or a person S.D. of 0 logits and an item S.D. of 3 logits. The Figure indicates that, with these measure distributions about 50% of the variance in the data is explained by the Rasch measures. When the person and item S.D.s, are around 1 logit, then only 25% of the variance in the data is explained by the Rasch measures, but when the S.D.s are around 4 logits, then 75% of the variance is explained. Even with very wide person and item distributions with S.D.s of 5 logits only 80% of the variance in the data is explained.
In general, to increase the variance explained, there must be a wide range of person measures and/or of item difficulties. We can obtain this in three ways:
1. Increase the person S.D.: Include in the sample more persons with measures less central than those we currently have (or omit from the sample persons with measures in the center of the person distribution)
2. Increase the item S.D.: Include in the test more items with measures less central than those we currently have (or omit from the test items with measures in the center of the item distribution)
3. Make the data more deterministic (Guttman-like) so that the estimated Rasch measures have a wider logit range:
a.) Remove "special causes" (to use quality-control terminology) by trimming observations with extreme standardized residuals.
b.) Reduce "common causes" by making the items more discriminating, e.g., by giving more precise definitions to rating scale categories, increasing the number of well-defined categories, making the items more similar, etc.
However, for a well-constructed instrument administered in a careful way to an appropriate sample, you may already be doing as well as is practical.
For comparison, here are some percents for other instruments: exam12.txt 78.7% (FIM sample chosen to exhibit a wide range of measures) exam1.txt 71.1% (Knox Cube Test) example0.txt 50.8% (Liking for Science) interest.txt 37.5% (NSF survey data - 3 category rating scale) agree.txt 30.0% (NSF survey data - 4 category rating scale) exam5.txt 29.5% (CAT test) - as CAT tests improve, this % will decrease! coin-toss 0.0%
In Winsteps Table 23.0, the "Model" column gives the "Variance Explained" value that you could expect to see if your data had perfect fit to the Rasch model with the current degree of randomness. The "Model" value is usually very close to the empirical value. This is because some parts of your data underfit the model (too little variance explained) and some parts overfit (too much variance explained).
Relationship to Bigsteps and earlier versions of Winsteps:
My apologies for the difficulties caused by the change in computation to "Variance Explained".
The earlier computation was based on the best statistical theory available to us at the time. Further developments in statistical theory, combined with practical experience, indicated that the previous computation was too generous in assigning "variance explained" to the Rasch measures. The current computation is more accurate.
Set PRCOMP=R (raw residuals) in Bigsteps and earlier versions of Winsteps, and you will obtain approximately the same explained/unexplained variance proportions as the current version of Winsteps.
Research in the last couple of years has demonstrated that PRCOMP=R gives a more realistic estimate of the variance explained than PRCOMP=S (standardized residuals). PRCOMP=S overestimates the explained variance as a proportion of the total variance.
But for decomposing the unexplained variance into "contrasts", PRCOMP=S is better. So this mixed setting is now the default for Winsteps.
The Eigenvalue reported for the 1st contrast has not changed. If this is much larger than the size of an Eigenvalue expected by chance, usually less than 2 - www.rasch.org/rmt/rmt191h.htm - Please inspect your Table 23.3 to see the contrasting content of the items which is producing this large off-dimensional component in your data.
Help for Winsteps Rasch Measurement Software: www.winsteps.com. Author: John Michael Linacre
The Languages of Love: draw a map of yours!
Forum Rasch Measurement Forum to discuss any Rasch-related topic
Rasch Publications
Rasch Measurement Transactions (free, online) Rasch Measurement research papers (free, online) Probabilistic Models for Some Intelligence and Attainment Tests, Georg Rasch Applying the Rasch Model 3rd. Ed., Bond & Fox Best Test Design, Wright & Stone
Rating Scale Analysis, Wright & Masters Introduction to Rasch Measurement, E. Smith & R. Smith Introduction to Many-Facet Rasch Measurement, Thomas Eckes Invariant Measurement with Raters and Rating Scales: Rasch Models for Rater-Mediated Assessments, George Engelhard, Jr. & Stefanie Wind Statistical Analyses for Language Testers, Rita Green
Rasch Models: Foundations, Recent Developments, and Applications, Fischer & Molenaar Journal of Applied Measurement Rasch models for measurement, David Andrich Constructing Measures, Mark Wilson Rasch Analysis in the Human Sciences, Boone, Stave, Yale
in Spanish: Análisis de Rasch para todos, Agustín Tristán Mediciones, Posicionamientos y Diagnósticos Competitivos, Juan Ramón Oreja Rodríguez
Winsteps Tutorials Facets Tutorials Rasch Discussion Groups
Coming Winsteps & Facets Events
May 22 - 24, 2018, Tues.-Thur. EALTA 2018 pre-conference workshop (Introduction to Rasch measurement using WINSTEPS and FACETS, Thomas Eckes & Frank Weiss-Motz), https://ealta2018.testdaf.de
May 25 - June 22, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
June 27 - 29, 2018, Wed.-Fri. Measurement at the Crossroads: History, philosophy and sociology of measurement, Paris, France., https://measurement2018.sciencesconf.org
June 29 - July 27, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Further Topics (E. Smith, Winsteps), www.statistics.com
July 25 - July 27, 2018, Wed.-Fri. Pacific-Rim Objective Measurement Symposium (PROMS), (Preconference workshops July 23-24, 2018) Fudan University, Shanghai, China "Applying Rasch Measurement in Language Assessment and across the Human Sciences" www.promsociety.org
Aug. 10 - Sept. 7, 2018, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Oct. 12 - Nov. 9, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
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John "Mike" L.'s Wellness Report: I'm 72, take no medications and, March 2018, my doctor is annoyed with me - I'm too healthy! According to Wikipedia, the human body requires about 30 minerals, maybe more. There are 60 naturally-occurring minerals in the liquid Mineral Supplement which I take daily. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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": 0.9193592071533203, "perplexity": 1166.3449832756535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2018-17/segments/1524125947822.98/warc/CC-MAIN-20180425135246-20180425155246-00203.warc.gz"} |
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