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# whats molarity and the factors affecting it?does temperature affect it.
utkarsh
39 Points
5 years ago
Molarity = net number of moles of solute/ volume of solution (in litres) .. Let x be Number of moles of HCl (in mixture)= no of moles of HCl from solution A+solution B. Then x = (0.5* 0.750)+(2*0.250)=0.875. And let y= total volume of mixture = 1 litre. Therefore molarity =x/y = 0.875/1= 0.875. #always take volume in litres.
Umakant biswal
5359 Points
5 years ago
Molarity is a unit of concentration measuring the number of moles of a solute per liter of solution..
it can be calculated by the following formula
no of moles of solute / no of litres of solution
Molarity is affected by temperature because it is based on the volume of the solution, and the volume of a substance will be affected by changes in temperature..
for that reason molality is favoured over molarity to calculate the concentration terms | 356 | 1,267 | {"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} | 3 | 3 | CC-MAIN-2022-05 | latest | en | 0.799787 |
https://socratic.org/questions/how-do-you-find-the-area-given-a-5-b-7-c-10 | 1,726,819,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00050.warc.gz | 486,237,881 | 6,033 | # How do you find the area given a=5, b=7, c=10?
Apr 19, 2017
Area of triangle is $16.25$
#### Explanation:
Given three sides $a , b$ and $c$ of a triangle, its are is given by $\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{1}{2} \left(a + b + c\right)$.
As $a = 5$, $b = 7$ and $c = 10$, we have s=1/2(5+7+10)=1/2×22=11
Hence, area of triangle is $\sqrt{11 \left(11 - 5\right) \left(11 - 7\right) \left(11 - 10\right)}$
= sqrt(11×6×4×1)
= $\sqrt{264} = 16.25$ | 219 | 507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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} | 4.625 | 5 | CC-MAIN-2024-38 | latest | en | 0.680888 |
https://stackoverflow.com/questions/10376740/what-exactly-does-big-%D3%A8-notation-represent/12338937 | 1,720,937,421,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514548.45/warc/CC-MAIN-20240714032952-20240714062952-00589.warc.gz | 477,012,308 | 51,392 | # What exactly does big Ө notation represent?
I'm really confused about the differences between big O, big Omega, and big Theta notation.
I understand that big O is the upper bound and big Omega is the lower bound, but what exactly does big Ө (theta) represent?
I have read that it means tight bound, but what does that mean?
First let's understand what big O, big Theta and big Omega are. They are all sets of functions.
Big O is giving upper asymptotic bound, while big Omega is giving a lower bound. Big Theta gives both.
Everything that is `Ө(f(n))` is also `O(f(n))`, but not the other way around.
`T(n)` is said to be in `Ө(f(n))` if it is both in `O(f(n))` and in `Omega(f(n))`.
In sets terminology, `Ө(f(n))` is the intersection of `O(f(n))` and `Omega(f(n))`
For example, merge sort worst case is both `O(n*log(n))` and `Omega(n*log(n))` - and thus is also `Ө(n*log(n))`, but it is also `O(n^2)`, since `n^2` is asymptotically "bigger" than it. However, it is not `Ө(n^2)`, Since the algorithm is not `Omega(n^2)`.
### A bit deeper mathematic explanation
`O(n)` is asymptotic upper bound. If `T(n)` is `O(f(n))`, it means that from a certain `n0`, there is a constant `C` such that `T(n) <= C * f(n)`. On the other hand, big-Omega says there is a constant `C2` such that `T(n) >= C2 * f(n))`).
### Do not confuse!
Not to be confused with worst, best and average cases analysis: all three (Omega, O, Theta) notation are not related to the best, worst and average cases analysis of algorithms. Each one of these can be applied to each analysis.
We usually use it to analyze complexity of algorithms (like the merge sort example above). When we say "Algorithm A is `O(f(n))`", what we really mean is "The algorithms complexity under the worst1 case analysis is `O(f(n))`" - meaning - it scales "similar" (or formally, not worse than) the function `f(n)`.
### Why we care for the asymptotic bound of an algorithm?
Well, there are many reasons for it, but I believe the most important of them are:
1. It is much harder to determine the exact complexity function, thus we "compromise" on the big-O/big-Theta notations, which are informative enough theoretically.
2. The exact number of ops is also platform dependent. For example, if we have a vector (list) of 16 numbers. How much ops will it take? The answer is: it depends. Some CPUs allow vector additions, while other don't, so the answer varies between different implementations and different machines, which is an undesired property. The big-O notation however is much more constant between machines and implementations.
To demonstrate this issue, have a look at the following graphs:
It is clear that `f(n) = 2*n` is "worse" than `f(n) = n`. But the difference is not quite as drastic as it is from the other function. We can see that `f(n)=logn` quickly getting much lower than the other functions, and `f(n) = n^2` is quickly getting much higher than the others.
So - because of the reasons above, we "ignore" the constant factors (2* in the graphs example), and take only the big-O notation.
In the above example, `f(n)=n, f(n)=2*n` will both be in `O(n)` and in `Omega(n)` - and thus will also be in `Theta(n)`.
On the other hand - `f(n)=logn` will be in `O(n)` (it is "better" than `f(n)=n`), but will NOT be in `Omega(n)` - and thus will also NOT be in `Theta(n)`.
Symmetrically, `f(n)=n^2` will be in `Omega(n)`, but NOT in `O(n)`, and thus - is also NOT `Theta(n)`.
1Usually, though not always. when the analysis class (worst, average and best) is missing, we really mean the worst case.
• @krishnaChandra: `f(n) = n^2` is asymptotically stronger then `n`, and thus is Omega(n). However it is not O(n) (because for large `n` values, it is bigger then `c*n`, for all `n`). Since we said Theta(n) is the intersection of O(n) and Omega(n), since it is not O(n), it cannot be Theta(n) as well.
– amit
Commented Sep 10, 2012 at 5:04
• It's great to see someone explain how big-O notation isn't related to the best/worst case running time of an algorithm. There are so many websites that come up when I google the topic that say O(T(n)) means the worse case running time. Commented Feb 21, 2013 at 9:46
• @almel It's 2*n (2n, two times n) not 2^n
– amit
Commented Sep 8, 2014 at 22:04
• @VishalK 1. Big O is the upper bound as n tends to infinity. 2. Omega is the lower bound as n tends to infinity. 3. Theta is both the upper and lower bound as n tends to infinity. Note that all bounds are only valid "as n tends to infinity", because the bounds do not hold for low values of n (less than n0). The bounds hold for all nn0, but not below n0 where lower order terms become dominant.
– bain
Commented Dec 4, 2016 at 11:28
• @hey_you Read the answer again. big O,Theta,Omega are for functions, not algorithms. Merge sort is Omega(n) worst case. It is also O(n^2) best case. It is also Theta (nlogn) worst case. Basically, for each analysis (worst/best/average/...) you have a complexity function `T_best(n), T_worst(n), T_average(n)`. They do not have to be identical (and mostly, they are not). O/Omega/Theta can be applied to any of them independently.
– amit
Commented Nov 14, 2019 at 22:26
It means that the algorithm is both big-O and big-Omega in the given function.
For example, if it is `Ө(n)`, then there is some constant `k`, such that your function (run-time, whatever), is larger than `n*k` for sufficiently large `n`, and some other constant `K` such that your function is smaller than `n*K` for sufficiently large `n`.
In other words, for sufficiently large `n`, it is sandwiched between two linear functions :
For `k < K` and `n` sufficiently large, `n*k < f(n) < n*K`
• It does not, those variables are a bit confusing, they are unrelated. Commented Sep 25, 2018 at 21:10
• @committedandroider No, they are lowercase and uppercase thus different, he's using typical mathematical style in which two "similar" (but not related in any way here) variables use big and small case. Commented Jul 7, 2019 at 21:08
Theta(n): A function `f(n)` belongs to `Theta(g(n))`, if there exists positive constants `c1` and `c2` such that `f(n)` can be sandwiched between `c1(g(n))` and `c2(g(n))`. i.e it gives both upper and as well as lower bound.
Theta(g(n)) = { f(n) : there exists positive constants c1,c2 and n1 such that 0<=c1(g(n))<=f(n)<=c2(g(n)) for all n>=n1 }
when we say `f(n)=c2(g(n))` or `f(n)=c1(g(n))` it represents asymptotically tight bound.
O(n): It gives only upper bound (may or may not be tight)
O(g(n)) = { f(n) : there exists positive constants c and n1 such that 0<=f(n)<=cg(n) for all n>=n1}
ex: The bound `2*(n^2) = O(n^2)` is asymptotically tight, whereas the bound `2*n = O(n^2)` is not asymptotically tight.
o(n): It gives only upper bound (never a tight bound)
the notable difference between O(n) & o(n) is f(n) is less than cg(n) for all n>=n1 but not equal as in O(n).
ex: `2*n = o(n^2)`, but `2*(n^2) != o(n^2)`
• You didn't mention big Omega, which refers to the lower-bound. Otherwise, very nice first answer and welcome! Commented Oct 6, 2012 at 17:01
• i liked the way he framed the definition of Theta(n). Upvoted! Commented Oct 20, 2013 at 14:48
• Is it right to think of theta as the 'average' time for a function? I keep hearing people refer to it as the average but I am not sure if the fact it is simply constrained by an upper and lower boundary really means its an average. Commented Nov 26, 2020 at 22:23
I hope this is what you may want to find in the classical CLRS(page 66):
## Big Theta notation:
Nothing to mess up buddy!!
If we have a positive valued functions f(n) and g(n) takes a positive valued argument n then ϴ(g(n)) defined as {f(n):there exist constants c1,c2 and n1 for all n>=n1}
where c1 g(n)<=f(n)<=c2 g(n)
## Let's take an example:
let f(n)=5n^2+2n+1
g(n)=n^2
c1=5 and c2=8 and n1=1
Among all the notations ,ϴ notation gives the best intuition about the rate of growth of function because it gives us a tight bound unlike big-oh and big -omega which gives the upper and lower bounds respectively.
ϴ tells us that g(n) is as close as f(n),rate of growth of g(n) is as close to the rate of growth of f(n) as possible.
First of all, theory
1. Big O = upper limit O(n)
2. Theta = Order function - theta(n)
3. Omega = Q-Notation (lower limit) Q(n)
## Why are people so confused?
In many blogs and books, these statements are emphasised, like:
"This is Big O(n^3)", etc.
And people often confuse like whether
O(n) == theta(n) == Q(n)
But what is worth keeping in mind is they are just mathematical functions with names O, Theta, and Omega.
So they have the same general formula of a polynomial.
Let,
f(n) = 2n4 + 100n2 + 10n + 50 then,
g(n) = n4, So g(n) is the function which takes a function as input and returns variable with the biggest power,
The same f(n) & g(n) for all the explanations below:
## Big O(n) - provides an upper bound
Big O(n4) = 3n4, because 3n4 > 2n4
3n4 is the value of Big O(n4), just like f(x) = 3x
n4 is playing a role of x here so,
Replacing n4 with x'so, Big O(x') = 2x'. Now we both are happy with what the general concept is.
So 0 ≤ f(n) ≤ O(x')
O(x') = cg(n) = 3n4
Putting value,
0 ≤ 2n4 + 100n2 + 10n + 50 ≤ 3n4
3n4 is our upper bound
## Big Omega(n) - provides a lower bound
Theta(n4) = cg(n) = 2n4, because 2n4 ≤ our example f(n)
2n4 is the value of Theta(n4)
so, 0 ≤ cg(n) ≤ f(n)
0 ≤ 2n4 ≤ 2n4 + 100n2 + 10n + 50
2n4 is our lower bound
## Theta(n) - provides a tight bound
This is calculated to find out that weather the lower bound is similar to the upper bound.
Case 1). The upper bound is similar to the lower bound
``````if the upper bound is similar to the lower bound, the average case is similar
Example, 2n4 ≤ f(x) ≤ 2n4,
Then Theta(n) = 2n4
``````
Case 2). If the upper bound is not similar to the lower bound
``````In this case, Theta(n) is not fixed, but Theta(n) is the set of functions with the same order of growth as g(n).
Example 2n4 ≤ f(x) ≤ 3n4, this is our default case.
Then, Theta(n) = c'n4, is a set of functions with 2 ≤ c' ≤ 3
``````
I am not sure why there isn't any short simple answer explaining big theta in plain English (seems like that was the question), so here it is.
Big Theta is the range of values or the exact value (if big O and big Omega are equal) within which the operations needed for a function will grow. | 3,028 | 10,391 | {"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} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.940853 |
https://fr.slideserve.com/verlee/math-1-march-27-th-powerpoint-ppt-presentation | 1,653,768,716,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00435.warc.gz | 318,032,969 | 19,569 | Math 1 March 27 th
# Math 1 March 27 th
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## Math 1 March 27 th
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##### Presentation Transcript
1. Math 1 March 27th What you need today in class: Pick up a calculator WARM-UP: (Copy the diagram and answer the questions) ABCD is a Rhombus. Draw the diagonals. Label the point where they intersect E. and AB = 10, DB = 12, AE = 6.5 Find Find Find . 6. Find BC. 7. Find AC. A B D C
2. Check homework • 30 • 90 • 6 • 4 • 150 • 90 • 6 • 4 • 70.6 • 9 • 90 • 35.3
3. pg. 11-12 PARALLELOGRAMS CHARACTERISTICS 1 – Opposite sides are ___ 2 – Opposite sides are ___ 3 – Opposite angles are __ 4 – Consecutive angles are __ 5 – Diagonals ___ each other
4. RECTANGLE CHARACTERISTICS 1 – All angle are ____. 2 – Diagonals are ____.
5. RHOMBUS CHARACTERISTICS 1 – All sides are ____. 2 – Diagonals are ____. 3 – Diagonals bisect ____ angles.
6. SQUARE CHARACTERISTICS 1 – 2 – 3 – 4 – 5 –
7. Review for Test • Find these Quizzes • Points of Concurrency • Polygons • Notes • Pages 4 – 5 • Pages 16-17 • Pages 21 and 2-page handout
8. Review What is a point of concurrency? Match these: • Circumcenter A. Angle bisectors • Orthocenter B. Perpendicular Bisectors • Centroid C. Altitudes • Incenter D. Medians
9. Review • Which point of concurrency is equal distance from each angle? • Which point of concurrency is equal distance from each side?
10. Review
11. Review • How many sides? • Pentagon • Heptagon • Dodecagon • Decagon • Hexagon
12. Review • Sum of interior angles of a 15-gon? • Sum of exterior angles of a hexagon? • Measure of one interior angle of a 15-gon? • Measure of one exterior angle of a 14-gon? • The sum of the measures of a polygon is 1980. How many sides does the polygon have?
13. Review • If the measure of one exterior angle of a regular polygon is 45°, then how many sides does the polygon have? • If the measure of one interior angle of a regular polygon is 150°, then how many sides does the polygon have? | 644 | 2,074 | {"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} | 3.8125 | 4 | CC-MAIN-2022-21 | latest | en | 0.757607 |
https://en.xen.wiki/w/Laka | 1,638,150,737,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00192.warc.gz | 321,352,479 | 9,197 | Hemifamity family
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The hemifamity family of temperaments are rank-3 temperaments tempering out 5120/5103.
Hemifamity
Subgroup: 2.3.5.7
Mapping: [1 0 0 10], 0 1 0 -6], 0 0 1 1]]
Mapping generators: ~2, ~3, ~5
Mapping to lattice: [0 1 2 -4], 0 0 1 1]]
Lattice basis:
3/2 length = 0.5670, 10/9 length = 1.8063
Angle (3/2, 10/9) = 82.112 degrees
Minimax tuning: c = 5120/5103
[[1 0 0 0, [10/7 1/7 1/7 -1/7, [0 0 1 0, [10/7 -6/7 1/7 6/7]
Eigenmonzos (unchanged intervals): 2, 5/4, 7/6
[[1 0 0 0, [5/4 1/4 1/8 -1/8, [0 0 1 0, [5/2 -3/2 1/4 3/4]
Eigenmonzos (unchanged intervals): 2, 5/4, 9/7
Vals41, 53, 87, 94, 99, 239, 251, 292, 391, 881bd, 1272bcdd
Badness: 0.153 × 10-3
Projection pairs: 7 5120/729
Music
Pele
Subgroup: 2.3.5.7.11
Comma list: 441/440, 896/891
Mapping: [1 0 0 10 17], 0 1 0 -6 -10], 0 0 1 1 1]]
[[1 0 0 0 0, [17/10 0 1/10 0 -1/10, [17/5 -2 6/5 0 -1/5, [16/5 -2 3/5 0 2/5, [17/5 -2 1/5 0 4/5]
Eigenmonzos: 2, 10/9, 11/9
Mapping to lattice: [0 1 4 -2 -6], 0 0 -1 -1 -1]]
Lattice basis:
3/2 length = 0.3812, 56/55 length = 1.5893
Angle(3/2, 56/55) = 90.4578 degrees
Vals29, 41, 58, 87, 99e, 145, 186e, 331e*
Badness: 0.648 × 10-3
Projection pairs: 7 5120/729 11 655360/59049
13-limit
Subgroup: 2.3.5.7.11.13
Comma list: 196/195, 352/351, 364/363
Mapping: [1 0 0 10 17 22], 0 1 0 -6 -10 -13], 0 0 1 1 1 1]]
Minimax tuning:
• 13-odd-limit eigenmonzos (unchanged intervals): 2, 10/9, 13/10
• 15-odd-limit eigenmonzos (unchanged intervals): 2, 15/13, 6/5
Vals: 29, 41, 46, 58, 87, 145, 232, 377cef
Badness: 0.703 × 10-3
Laka
Subgroup: 2.3.5.7.11
Comma list: 540/539, 5120/5103
Mapping: [1 0 0 10 -18], 0 1 0 -6 15], 0 0 1 1 -1]]
[[1 0 0 0 0, [4/3 0 2/21 -1/21 1/21, [0 0 1 0 0, [2 0 3/7 2/7 -2/7, [2 0 3/7 -5/7 5/7]
Eigenmonzos (unchanged intervals): 2, 5/4, 14/11
Projection pairs: 5120/729 11 14348907/1310720
13-limit
Subgroup: 2.3.5.7.11.13
Comma list: 352/351, 540/539, 847/845
Minimax tuning:
• 13- and 15-odd-limit
[[1 0 0 0 0 0, [13/8 -1/2 1/8 0 0 1/8, [13/4 -3 5/4 0 0 1/4, [7/2 0 1/2 0 0 -1/2, [25/8 -9/2 5/8 0 0 13/8, [13/4 -3 1/4 0 0 5/4]
Eigenmonzos (unchanged intervals): 2, 11/8, 14/13
Vals: 41, 53, 58, 94, 111, 152f, 415de*
* optimal patent val: 205
17-limit
Subgroup: 2.3.5.7.11.13.17
Comma list: 352/351, 442/441, 540/539, 847/845
Mapping: [1 0 0 10 -18 -13 32], 0 1 0 -6 15 12 -22], 0 0 1 1 -1 -1 3]]
Minimax tuning:
• 17-odd-limit
[[1 0 0 0 0 0 0, [13/12 0 0 1/12 1/6 -1/12 0, [-7/4 0 0 5/4 3/2 -5/4 0, [7/4 0 0 3/4 1/2 -3/4 0, [0 0 0 0 1 0 0, [7/4 0 0 -1/4 1/2 1/4 0, [35/12 0 0 23/12 5/6 -23/12 0]
Eigenmonzos (unchanged intervals): 2, 11/8, 14/13
Vals: 58, 94, 111, 152f, 205, 263df
Akea
Subgroup: 2.3.5.7.11
Comma list: 385/384, 2200/2187
Mapping: [1 0 0 10 -3], 0 1 0 -6 7], 0 0 1 1 -2]]
Mapping generators: ~2, ~3, ~5
[[1 0 0 0 0, [5/3 0 1/6 -1/6 0, {{monzo| 26/9 0 13/18 -7/18 -1/3>, [26/9 0 -5/18 11/18 -1/3, [26/9 0 -5/18 -7/18 2/3]
Eigenmonzos (unchanged intervals): ~2, ~14/11, ~7/5
Vals34, 41, 53, 87, 140, 181, 321
Badness: 0.998 × 10-3
13-limit
Subgroup: 2.3.5.7.11.13
Comma list: 325/324, 352/351, 385/384
Mapping: [1 0 0 10 -3 2], 0 1 0 -6 7 4], 0 0 1 1 -2 -2]]
Minimax tuning:
• 13- and 15-odd-limit
[[1 0 0 0 0 0, [5/3 0 1/6 -1/6 0 0, [26/9 0 13/18 -7/18 -1/3 0, [26/9 0 -5/18 11/18 -1/3 0, [26/9 0 -5/18 -7/18 2/3 0, [26/9 0 -7/9 1/9 2/3 0]
Eigenmonzos (unchanged intervals): 2, 14/11, 7/5
Lattice basis:
3/2 length = 0.5354, 27/20 length = 1.0463
Angle (3/2, 27/20) = 80.5628 degrees
Mapping to lattice: [0 1 3 -3 1 -2], 0 0 -1 -1 2 2]]
Vals: 34, 41, 46, 53, 87, 140, 321, 461e
Badness: 0.822 × 10-3
Scales: akea46_13
Lono
Subgroup: 2.3.5.7.11
Comma list: 176/175, 5120/5103
Mapping: [1 0 0 10 6], 0 1 0 -6 -6], 0 0 1 1 3]]
Vals46, 53, 58, 99, 111, 268cd
Badness: 1.18 × 10-3
13-limit
Subgroup: 2.3.5.7.11.13
Comma list: 176/175, 351/350, 847/845
Mapping: [1 0 0 10 6 11], 0 1 0 -6 -6 -9], 0 0 1 1 3 3]]
Vals: 46, 53, 58, 99, 104c, 111, 268cd
Badness: 0.908 × 10-3
Kapo
Subgroup: 2.3.5.7.11
Comma list: 3025/3024, 5120/5103
Mapping: [1 0 0 10 7], 0 1 1 -5 -2], 0 0 2 2 -1]]
Mapping generators: ~2, ~3, ~128/99
[[1 0 0 0 0, [8/5 2/5 0 -1/15 -2/15, [14/5 6/5 0 7/15 -16/15, [16/5 -6/5 0 13/15 -4/15, [16/5 -6/5 0 -2/15 11/15]
Eigenmonzos (unchanged intervals): 2, 11/9, 9/7
Vals41, 87, 111, 152, 239, 391
Badness: 0.994 × 10-3
Namaka
Subgroup: 2.3.5.7.11
Comma list: 3388/3375, 5120/5103
Mapping: [1 0 0 10 -6], 0 2 0 -12 9], 0 0 1 1 1]]
Vals29, 53, 58, 87, 111, 140, 198
Badness: 1.74 × 10-3
13-limit
Subgroup: 2.3.5.7.11.13
Comma list: 352/351, 676/675, 847/845
Mapping: [1 0 0 10 -6 -1], 0 2 0 -12 9 3], 0 0 1 1 1 1]]
Vals: 29, 53, 58, 87, 111, 140, 198
Badness: 0.781 × 10-3 | 2,796 | 4,786 | {"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} | 2.65625 | 3 | CC-MAIN-2021-49 | longest | en | 0.580613 |
dusou.club | 1,590,417,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00578.warc.gz | 36,686,001 | 20,424 | ## Absolute Value Def Math
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https://www.winnerswire.com/how-do-you-bet-on-races/ | 1,685,842,926,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00278.warc.gz | 1,179,148,574 | 51,223 | Home » How Do You Bet On Races?
# How Do You Bet On Races?
Betting on races is a popular and fun way to make money. It’s a great way to make a little extra cash or potentially win big. Whether you’re a beginner or a veteran, there are certain ways to approach betting on races that can increase your chances of success. In this article, we’ll cover the basics of betting on races, including where to bet, what types of bets to make, and how to manage your bankroll.
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## Where to Bet on Races
There are several different places where you can bet on races. The most popular is a horse racing track, where you can place bets in person. You can also place bets online at various online sportsbooks. In addition, many racetracks now offer simulcast betting, where you can watch and bet on races from other tracks.
## Types of Bets
There are several types of bets that you can make on races. Here are some of the most common:
• Win: This is a bet on the horse to win the race.
• Place: This is a bet on the horse to finish in either first or second place.
• Show: This is a bet on the horse to finish in either first, second, or third place.
• Exacta: This is a bet on two horses to finish in the exact order.
• Trifecta: This is a bet on three horses to finish in the exact order.
• Superfecta: This is a bet on four horses to finish in the exact order.
• Daily Double: This is a bet on two horses to win two consecutive races.
• Pick 3: This is a bet on three horses to win three consecutive races.
• Pick 4: This is a bet on four horses to win four consecutive races.
• Pick 6: This is a bet on six horses to win six consecutive races.
## Understanding Odds
When you’re betting on races, it’s important to understand the odds. The odds are the probability of a particular horse winning the race. The odds are usually expressed as a fraction or decimal. For example, if a horse has odds of 2/1, that means it has a 2 in 1 chance of winning. The lower the odds, the more likely the horse is to win.
Related content How Does Badminton Horse Trials Work?
## Managing Your Bankroll
When you’re betting on races, it’s important to manage your bankroll. This means setting a budget for how much you’re willing to spend and sticking to it. It’s also important to have a plan for how much you’re willing to risk on each bet. A good rule of thumb is to never bet more than 10% of your bankroll on any one bet.
## Researching Horses
Before you start betting on races, it’s important to do some research. This means looking up information on the horses, the jockeys, the trainers, and the track conditions. The more information you have, the better your chances of making a successful bet.
## Analyzing the Race
Once you’ve done your research, it’s time to analyze the race. This means looking at the track conditions, the form of the horses, and the jockeys. It’s also important to look at the betting odds to get an idea of which horses are favored.
## Making Your Bet
Once you’ve done your research and your analysis, it’s time to make your bet. It’s important to be patient and wait for the best odds. You should also be prepared to hedge your bet if necessary.
## Managing Your Bets
Once you’ve made your bet, it’s important to manage it. This means checking the results of the race and adjusting your bets accordingly. If your horse wins, you should take your winnings and reinvest them in future bets. If your horse loses, you should analyze why and adjust your bet accordingly.
## Staying Disciplined
The key to success when betting on races is to stay disciplined. This means sticking to your budget, managing your bankroll, and being patient. It also means avoiding chasing losses and never betting more than you can afford to lose.
## Conclusion
Betting on races can be a fun and rewarding experience. With the right approach, you can increase your chances of success. The key is to do your research, analyze the race, manage your bets, and stay disciplined. With the right approach, you can turn a profit and have some fun. | 949 | 4,132 | {"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} | 2.515625 | 3 | CC-MAIN-2023-23 | longest | en | 0.94084 |
https://www.teachstarter.com/au/teaching-resource/identifying-fractions-memory-game/ | 1,725,731,462,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00069.warc.gz | 980,998,879 | 75,225 | teaching resource
# Identifying Fractions – Memory Game
• Updated: 13 Feb 2023
Practise matching improper fractions, mixed numerals, and visual representations with this set of 30 match-up cards.
• Editable: Google Slides
• Non-Editable: PDF
• Pages: 1 Page
• Years: 4 - 5
### Curriculum
• #### ACMNA078
Count by quarters halves and thirds, including with mixed numerals. Locate and represent these fractions on a number line
teaching resource
# Identifying Fractions – Memory Game
• Updated: 13 Feb 2023
Practise matching improper fractions, mixed numerals, and visual representations with this set of 30 match-up cards.
• Editable: Google Slides
• Non-Editable: PDF
• Pages: 1 Page
• Years: 4 - 5
Practise matching improper fractions, mixed numerals, and visual representations with this set of 30 match-up cards.
## Practise Matching Improper Fractions and Mixed Numerals With Our Memory-Style Fractions Game!
Are you looking for a fractions game that strengthens your students’ understanding of mixed numbers and improper fractions? This memory-style game will encourage students to find equivalent fractions of different representations. Each match includes:
• a mixed number
• an improper fraction
• a word form
• a visual representation.
To play, shuffle the cards and place them face down. Players will take turns choosing 2 cards and flipping them over. If the cards represent the same number, the player keeps the cards to the side. The player will have an additional turn if they find a match. If a player draws 2 cards that do not represent the same number, the cards are returned, and it is the next player’s turn.
Through this activity, students will show they can match an improper fraction, mixed number, and visual representation of the same number.
## Tips for Differentiation + Scaffolding
A team of dedicated, experienced educators created this resource to support your maths lessons.
Use this fraction game to enhance learning through guided maths groups or as a tool for whole-class instruction.
If you have a mixture of above and below-level learners, check out these suggestions for keeping students on track with the concepts:
### 🆘 Support Struggling Students
Help students who need support understanding the concepts by limiting the playing cards to halves, thirds and quarters. Additionally, choose only to use certain card types (improper fraction and visual representation, mixed numeral and word form, etc.).
### ➕ Challenge Fast Finishers
For fast finishers, encourage students to draw different visual representations for each mixed number or improper fraction.
### 🏃 Relay Race
Divide students into two team lines and show a flashcard to the students at the front of each line. The student who produces a correct equivalent fraction in a mixed number or improper fraction first wins the flashcard. The team with the most flashcards at the end of the game wins!
### 💃 Mirror Game 🕺
This activity works best with small groups because you’ll need enough dry-erase boards and markers for each student. Divide your class into 2 groups seated in 2 lines facing each other. Project a task card and give students a set time to record a fraction that is a mixed number or improper fraction on their board. On your cue, the students turn their boards around so their partner can see their answers. If both students in a pair have the correct answer, they get 2 points. If one has the correct answer, the team gets 1 point.
## Easily Prepare This Resource for Your Students
Use the dropdown icon on the Download button to choose between the PDF or editable Google Slides version of this resource.
Print on thick paper for added durability and longevity. Place all pieces in a folder or large envelope for easy access.
This resource was created by Lauren Blankenship, a Teach Starter Collaborator.
Don’t stop there! We’ve got more activities and resources that cut down on lesson planning time:
### teaching resource
#### Converting Mixed Numeral Fractions – Worksheet
A worksheet that demonstrates how to convert a mixed numeral fraction into an improper fraction.
2 pagesYears: 4 - 5
### teaching resource
#### Converting Improper Fractions – Worksheet
A worksheet explaining how to convert improper fractions to mixed numerals.
2 pagesYears: 4 - 5
### teaching resource
#### Improper Fractions Cards
A set of 30 cards displaying various improper fractions.
6 pagesYears: 3 - 6
### 0 Comments
Write a review to help other teachers and parents like yourself. If you'd like to request a change to this resource, or report an error, select the corresponding tab above.
Log in to comment | 997 | 4,673 | {"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} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.859113 |
https://edurev.in/studytube/The-Slope-Deflection-Method-Beams-2/f9584531-2526-486e-bbff-6ef1f9fab737_p | 1,632,139,653,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00506.warc.gz | 274,519,477 | 46,795 | Courses
# The Slope Deflection Method: Beams - 2 Civil Engineering (CE) Notes | EduRev
## Civil Engineering (CE) : The Slope Deflection Method: Beams - 2 Civil Engineering (CE) Notes | EduRev
``` Page 1
0 = +
BC BA
M M (6)
Substituting the values of and in equation (6),
BA
M
BC
M
0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying,
(7)
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
Also, the support C is simply supported and hence, 0 =
CB
M
EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
(8)
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
We have two unknowns
B
? and
C
? and there are two equations in
B
? and
C
? .
Solving equations (7) and (8)
3
10 4286 . 0
-
× - =
B
?
3
10 7143 . 1
-
× =
C
?
Substituting the values of
B
? ,
C
? and EI in slope-deflection equations,
kN.m 285 . 82 =
AB
M
kN.m 570 . 68 =
BA
M
kN.m 573 . 68 - =
BC
M
kN.m 0 =
CB
M (10)
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d)
Page 2
0 = +
BC BA
M M (6)
Substituting the values of and in equation (6),
BA
M
BC
M
0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying,
(7)
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
Also, the support C is simply supported and hence, 0 =
CB
M
EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
(8)
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
We have two unknowns
B
? and
C
? and there are two equations in
B
? and
C
? .
Solving equations (7) and (8)
3
10 4286 . 0
-
× - =
B
?
3
10 7143 . 1
-
× =
C
?
Substituting the values of
B
? ,
C
? and EI in slope-deflection equations,
kN.m 285 . 82 =
AB
M
kN.m 570 . 68 =
BA
M
kN.m 573 . 68 - =
BC
M
kN.m 0 =
CB
M (10)
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d)
In beamAB ,
, 0 =
? B
M ) ( kN 171 . 30 ? =
A
R
) ( kN 171 . 30 ? - =
BL
R
) ( kN 714 . 13 ? - =
BR
R
) ( kN 714 . 13 ? =
C
R
The shear force and bending moment diagram is shown in Fig.15.2e and elastic
curve is shown in Fig.15.2f.
Page 3
0 = +
BC BA
M M (6)
Substituting the values of and in equation (6),
BA
M
BC
M
0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying,
(7)
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
Also, the support C is simply supported and hence, 0 =
CB
M
EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
(8)
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
We have two unknowns
B
? and
C
? and there are two equations in
B
? and
C
? .
Solving equations (7) and (8)
3
10 4286 . 0
-
× - =
B
?
3
10 7143 . 1
-
× =
C
?
Substituting the values of
B
? ,
C
? and EI in slope-deflection equations,
kN.m 285 . 82 =
AB
M
kN.m 570 . 68 =
BA
M
kN.m 573 . 68 - =
BC
M
kN.m 0 =
CB
M (10)
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d)
In beamAB ,
, 0 =
? B
M ) ( kN 171 . 30 ? =
A
R
) ( kN 171 . 30 ? - =
BL
R
) ( kN 714 . 13 ? - =
BR
R
) ( kN 714 . 13 ? =
C
R
The shear force and bending moment diagram is shown in Fig.15.2e and elastic
curve is shown in Fig.15.2f.
Example 15.2
A continuous beam is carrying a uniformly distributed load of 5 kN/m as
shown in Fig.15.3a. Compute reactions and draw shear force and bending
moment diagram due to following support settlements.
ABCD
Page 4
0 = +
BC BA
M M (6)
Substituting the values of and in equation (6),
BA
M
BC
M
0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying,
(7)
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
Also, the support C is simply supported and hence, 0 =
CB
M
EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
(8)
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
We have two unknowns
B
? and
C
? and there are two equations in
B
? and
C
? .
Solving equations (7) and (8)
3
10 4286 . 0
-
× - =
B
?
3
10 7143 . 1
-
× =
C
?
Substituting the values of
B
? ,
C
? and EI in slope-deflection equations,
kN.m 285 . 82 =
AB
M
kN.m 570 . 68 =
BA
M
kN.m 573 . 68 - =
BC
M
kN.m 0 =
CB
M (10)
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d)
In beamAB ,
, 0 =
? B
M ) ( kN 171 . 30 ? =
A
R
) ( kN 171 . 30 ? - =
BL
R
) ( kN 714 . 13 ? - =
BR
R
) ( kN 714 . 13 ? =
C
R
The shear force and bending moment diagram is shown in Fig.15.2e and elastic
curve is shown in Fig.15.2f.
Example 15.2
A continuous beam is carrying a uniformly distributed load of 5 kN/m as
shown in Fig.15.3a. Compute reactions and draw shear force and bending
moment diagram due to following support settlements.
ABCD
Support B 0.005m vertically downwards
Support C 0.01 m vertically downwards
Assume E =200 GPa,
4 3
10 35 . 1 m I
-
× =
In the above continuous beam, four rotations
A
? ,
B
? ,
C
? and
D
? are to be
evaluated. One equilibrium equation can be written at each support.Hence,
solving the four equilibrium equations, the rotations are evaluated and hence the
moments from slope-deflection equations. Now consider the kinematically
restrained beam as shown in Fig.15.3b.
Referring to standard tables the fixed end moments may be evaluated .Otherwise
one could obtain fixed end moments from force method of analysis. The fixed
end moments in the present case are (vide fig.15.3b)
kN.m 667 . 41 =
F
AB
M
(clockwise) kN.m 667 . 41 - =
F
BA
M
(counterclockwise) kN.m 667 . 41 =
F
BC
M
(clockwise) kN.m 667 . 41 - =
F
CB
M
Page 5
0 = +
BC BA
M M (6)
Substituting the values of and in equation (6),
BA
M
BC
M
0 10 2 . 1 4 . 0 8 . 0 10 2 . 1 8 . 0
3 3
= × - + + × +
- -
EI EI EI EI EI
C B B
? ? ?
Simplifying,
(7)
3
10 2 . 1 4 . 0 6 . 1
-
× = +
C B
? ?
Also, the support C is simply supported and hence, 0 =
CB
M
EI M
B C CB
3
10 2 . 1 4 . 0 8 . 0 0
-
× - + = = ? ?
(8)
3
10 2 . 1 4 . 0 8 . 0
-
× = +
B C
? ?
We have two unknowns
B
? and
C
? and there are two equations in
B
? and
C
? .
Solving equations (7) and (8)
3
10 4286 . 0
-
× - =
B
?
3
10 7143 . 1
-
× =
C
?
Substituting the values of
B
? ,
C
? and EI in slope-deflection equations,
kN.m 285 . 82 =
AB
M
kN.m 570 . 68 =
BA
M
kN.m 573 . 68 - =
BC
M
kN.m 0 =
CB
M (10)
Reactions are obtained from equations of static equilibrium (vide Fig.15.2d)
In beamAB ,
, 0 =
? B
M ) ( kN 171 . 30 ? =
A
R
) ( kN 171 . 30 ? - =
BL
R
) ( kN 714 . 13 ? - =
BR
R
) ( kN 714 . 13 ? =
C
R
The shear force and bending moment diagram is shown in Fig.15.2e and elastic
curve is shown in Fig.15.2f.
Example 15.2
A continuous beam is carrying a uniformly distributed load of 5 kN/m as
shown in Fig.15.3a. Compute reactions and draw shear force and bending
moment diagram due to following support settlements.
ABCD
Support B 0.005m vertically downwards
Support C 0.01 m vertically downwards
Assume E =200 GPa,
4 3
10 35 . 1 m I
-
× =
In the above continuous beam, four rotations
A
? ,
B
? ,
C
? and
D
? are to be
evaluated. One equilibrium equation can be written at each support.Hence,
solving the four equilibrium equations, the rotations are evaluated and hence the
moments from slope-deflection equations. Now consider the kinematically
restrained beam as shown in Fig.15.3b.
Referring to standard tables the fixed end moments may be evaluated .Otherwise
one could obtain fixed end moments from force method of analysis. The fixed
end moments in the present case are (vide fig.15.3b)
kN.m 667 . 41 =
F
AB
M
(clockwise) kN.m 667 . 41 - =
F
BA
M
(counterclockwise) kN.m 667 . 41 =
F
BC
M
(clockwise) kN.m 667 . 41 - =
F
CB
M
(counterclockwise) kN.m 667 . 41 =
F
CD
M
(clockwise) (1) kN.m 667 . 41 - =
F
DC
M
In the next step, write slope-deflection equations for each span. In the
spanAB ,B is below A and hence the chord joining B A ' rotates in the clockwise
direction (see Fig.15.3c)
0005 . 0
10
005 . 0 0
- =
-
=
AB
? radians (negative as the chord B A ' rotates in the
clockwise direction from the original direction)
0005 . 0 - =
BC
? radians (negative as the chord C B ' ' rotates in the clockwise
direction)
001 . 0
10
01 . 0
= =
CD
? radians (positive as the chord D C ' rotates in the counter
clockwise direction from the original direction) (2)
Now, writing the expressions for the span end moments, for each of the spans,
) 0005 . 0 2 ( 2 . 0 667 . 41 + + + =
B A AB
EI M ? ?
) 0005 . 0 2 ( 2 . 0 667 . 41 + + + - =
A B BA
EI M ? ?
) 0005 . 0 2 ( 2 . 0 667 . 41 + + + =
C B BC
EI M ? ?
) 0005 . 0 2 ( 2 . 0 667 . 41 + + + - =
B C CB
EI M ? ?
```
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## Structural Analysis
30 videos|122 docs|28 tests
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; | 3,458 | 8,637 | {"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} | 3.8125 | 4 | CC-MAIN-2021-39 | latest | en | 0.728648 |
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### Topic: Help declaring a variable for temperature in thermistor--relay circuit (Read 3518 times)previous topic - next topic
#### dkotes
##### Apr 12, 2012, 09:51 pmLast Edit: Apr 12, 2012, 09:53 pm by dkotes Reason: 1
Hello!
I have spent the past few weeks reading a lot about relay control with arduino. I'm not an electrician by study so learning basic coding and circuit building has been a fun adventure.
My first "eureka!" moment was when I was able to take temperature measurements with a thermistor.
My second "eureka!" moment was when I was able to modify the "blink" example code to turn my relay on and off using a transistor, and just today after my diode arrived, turned an LED on when the relay was closed.
My ultimate goal is to replace the LED with a 120V 60W light bulb, which is going to be in close proximity to the thermistor, which I want to use to turn the lightbulb on and off.
This is my VERY SIMPLE code that I borrowed from the Arduino Thermistor page, and I just threw in some "if else" statements hoping it might make sense.
Basically, it's saying "Temp" is not declared in the if/else statements, and I'm not sure wether to replace Temp with AnalogRead0 or what...
I'd like the Analog0 input (after it has been converted to Celsius) to control the relay ON/OFF switch so to speak. I've also put a fritzing schematic at the bottom if that helps, thank you for any help!!
Quote
#include <math.h>
int RelayHOT = 4;
double Temp;
Temp = 1 / (0.001129148 + (0.000234125 + (0.0000000876741 * Temp * Temp ))* Temp );
Temp = Temp - 273.15; // Convert Kelvin to Celcius
return Temp;
}
void setup() {
Serial.begin(115200); // begin the serial monitor
pinMode(RelayHOT, OUTPUT); //set pin 4 (known as relayHOT - to an output)
}
void loop() {
delay(5000); // wait 5 seconds before sampling temperature again
if (Temp < 28) digitalWrite(RelayHOT, HIGH); //if the temperature is less than 28C, turn on the relay which will turn on the light and increase temperature
else if (Temp > 28) digitalWrite(RelayHOT, LOW); //if the temperature is greater than 28C, turn the bulb off because it is hot enough
}
#### Grumpy_Mike
#1
##### Apr 12, 2012, 10:02 pm
Take this line:-
Code: [Select]
`double Temp;`
and place it just after this line:-
Code: [Select]
`#include <math.h>`
This will make Temp a global variable accessible to all functions.
When posting code use the # icon not the quote one.
#### robtillaart
#2
##### Apr 12, 2012, 10:04 pm
fixed?
Code: [Select]
`#include <math.h>int RelayHOT = 4;double Thermister(int RawADC){ double Temp; Temp = log(((10240000/RawADC) - 10000)); Temp = 1 / (0.001129148 + (0.000234125 + (0.0000000876741 * Temp * Temp ))* Temp ); Temp = Temp - 273.15; // Convert Kelvin to Celcius return Temp;}void setup() { Serial.begin(115200); pinMode(RelayHOT, OUTPUT); }void loop() { float Temp = Thermister(analogRead(0)); Serial.println(Temp,2); // print Celcius temp with 2 decimals reading in serial monitor if (Temp < 27.5) digitalWrite(RelayHOT, HIGH); //if the temperature is less than 27.5 C, turn on the relay which will turn on the light and increase temperature else if (Temp > 28.5) digitalWrite(RelayHOT, LOW); //if the temperature is greater than 28.5 C, turn the bulb off because it is hot enough delay(5000); // wait 5 seconds before sampling temperature again - WHY?}`
Rob Tillaart
Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)
#### dkotes
#3
##### Apr 12, 2012, 10:13 pmLast Edit: Apr 12, 2012, 10:27 pm by dkotes Reason: 1
EDIT---
The code is error free now, and giving temperature readings accurately, but the Relay is turning on before the correct temperature is reached. I think i need to add a pulldown resistor to the base to keep the transistor from turning on falsely
#### robtillaart
#4
##### Apr 12, 2012, 10:25 pm
==>
Temp = log(10240000.0/RawADC - 10000.0); // make all consts float
Rob Tillaart
Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)
#### dkotes
#5
##### Apr 12, 2012, 10:36 pmLast Edit: Apr 12, 2012, 10:49 pm by dkotes Reason: 1
EDIT: Wow I must be high on paint fumes today...I want the relay on until a temperature is hit...
#### dkotes
#6
##### Apr 12, 2012, 10:50 pm
Just held my finger on the probe and watched it go up nice and slowly with 1/100th of a degree accuracy, once I hit the cut-off temperature the relay switched off and is staying off.
It just cooled back down and the relay turned back on.
I #\$%@ING LOVE YOU GUYS TY. Month long project finally done!!!!!!!! I'm pumped.
In case anyone is interested in what the final project will be, I'm trying to reproduce this
http://www.russelldurrett.com/lightbulbpcr.html
Thanks again!
#### robtillaart
#7
##### Apr 13, 2012, 10:57 am
Can you explain a bit about PCR (is it the PCR=DNA multiplication?)
Rob Tillaart
Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)
#### dkotes
#8
##### Apr 13, 2012, 08:24 pm
Yes PCR stands for Polymerase Chain Reaction, and it is a tool, or rather protocol, developed to make many many copies of a particular DNA sequence.
As you may know, DNA is double stranded, with one side "complementary to the other", and is held together by Hydrogen bonding, a rather "weak" bond in terms of the ways atoms can bond.
So, one way to break hydrogen bonding is by heating the DNA, which is where the light bulb comes in. It heats the small vials up to about 95C, which denatures the DNA and makes it break apart into it's two single strands.
Inside the mixture is also free nucleotides, the "bases" that make up DNA; A,T,G,C. A bonds with T, G bonds with C.
So, let's say we have the gene for Insulin production. We take that gene, mix in some free nucleotides, some special enzymes known as Polymerase enzymes (their discovery has a cool story, look up Taq enzyme) and some primers, and let it run about 30 cycles of heating up and cooling down.
Each time the DNA is "denatured", each single strand becomes a template, and is copied and becomes a double strand.
So, with a few million or hundred million copies of DNA in your initial sample, after 30 cycles you have a LOT of copies of one piece of DNA.
If it was the gene for Insulin, say for example, you can then clone it into a plasmid (like a blueprint) and put it into bacteria (little factory).
The E.Coli runs the blueprint, billions of times in a big fermented, and all you need to do is feed it some food.
Then, you harvest the drug you need (insulin) and bottle it and sell it.
That's an oversimplification but you'd be surprised how much molecular biology just makes sense. Repeat that with gene of interest...for penicilin...other drugs...ethanol production etc etc
"Real" thermocyclers are expensive, thousands of dollars, so trying to make one cheap.
The guy who invented PCR just used hot water baths and moved the tubes manually by hand hundreds of times, so, it's come a long way
#### asri
#9
##### Sep 03, 2014, 12:03 pm
what is the type of transistor used for this circuit?
if i want to use 12vdc source,do i need to change the value of resistor and the type of transistor?
Go Up | 2,026 | 7,358 | {"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} | 2.75 | 3 | CC-MAIN-2017-39 | latest | en | 0.897538 |
https://writinghelpe.com/an-object-m-3-05kg-rolls-fro/ | 1,726,368,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00527.warc.gz | 583,745,108 | 16,857 | # an object m = 3.05kg rolls fro
an object m = 3.05kg rolls from rest down an incline plane, itscenter initially at a height of 7.55m above the bottom of the ramp.The inclination of the incline as measured from the horizontal is37.3o. If the object is a solid uniform cylinder of radius r =1.25m, then at the bottom of the incline, what is the (a)rotational kinetic energy and (b) the angular momentum. If theobject is a hoop of radius, r = 1.05m, then at the bottom of theramp, what is the (c) rotational kinetic energy and (d) the angularmomentum?
a) Unifrom cyllinder
Let w is the angular speed and v is the linear speed atthe bottom.
Apply conservation of energy
(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h
(1/2)*m*v^2 + (1/2)*(1/2)*m*r^2*w^2 = m*g*h
(1/2)*m*v^2 + (1/4)*m*(r*w)^2 = m*g*h
(1/2)*m*v^2 + (1/4)*m*v^2 = m*g*h
(3/4)*m*v^2 = m*g*h
v = sqrt(4*g*h/3)
w = v/r
= sqrt(4*g*h/3)/r
= sqrt(4*9.8*7.55/3)/1.25
KE_rotationa = (1/2)*I*w^2
= (1/2)*(1/2)*m*r^2*w^2
= (1/4)*3.05*1.25^2*7.94^2
= 75.1 J
b) angular momentum = I*w
= (1/2)*m*r^2*w
= (1/2)*3.05*1.25^2*7.94
= 18.9 kg.m^2/s
c) Hoop
Let w is the angular speed and v is the linear speed atthe bottom.
Apply conservation of energy
(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h
(1/2)*m*v^2 + (1/2)*m*r^2*w^2 = m*g*h
(1/2)*m*v^2 + (1/2)*m*(r*w)^2 = m*g*h
(1/2)*m*v^2 + (1/2)*m*v^2 = m*g*h
m*v^2 = m*g*h
v = sqrt(g*h)
w = v/r
= sqrt(g*h)/r
= sqrt(9.8*7.55)/1.05
KE_rotationa = (1/2)*I*w^2
= (1/2)*m*r^2*w^2
= (1/2)*3.05*1.05^2*8.19^2
= 113 J
d) angular momentum = I*w
= m*r^2*w
= 3.05*1.05^2*8.19
= 27.5 kg.m^2/s
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http://mathhelpforum.com/differential-geometry/104529-finite-additive-measure-continuous-below-measure.html | 1,529,572,556,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864110.40/warc/CC-MAIN-20180621075105-20180621095105-00445.warc.gz | 202,498,292 | 11,708 | # Thread: A finite additive measure that is continuous from below is a measure
1. ## A finite additive measure that is continuous from below is a measure
A finite additive measure $\displaystyle \mu$ is a measure iff it is continuous from above. [sorry, I made a mistake in the title, should be "from above"]
I finished the proof from finite additive to continuous from above, but stuck on the other one...
Proof so far.
Suppose that $\displaystyle \mu : \mathbb {M} \rightarrow [0, \infty ]$ is continuous from above, that is, if $\displaystyle \{ E_j \} ^ \infty _{j=1} \subset \mathbb {M}$ with $\displaystyle E_{j+1} \subset E_j$, then I have $\displaystyle \lim _{j \rightarrow \infty } \mu (E_j)= \mu ( \bigcap ^ \infty _{j=1} E_j )$.
I will show that $\displaystyle \mu$ is a measure.
Let $\displaystyle \{ E_j \} ^n _{j=1} \subset \mathbb {M}$ be disjoint sets.
Claim: $\displaystyle \mu ( \bigcup ^n _{j=1} E_j) = \sum _{j=1}^n \mu (E_j)$
Let $\displaystyle F_n = \bigcup ^ \infty _{j=1}E_j \ \bigcup ^n _{i=1} E_i$, note that $\displaystyle F_j \subset F_{j+1}$ and $\displaystyle \bigcup ^n_{j=1}F_j$
Now, since $\displaystyle \mu$ is continuous from above, I have $\displaystyle \lim _{n \rightarrow \infty } \mu (F_n) = \mu ( \bigcup ^ \infty _{j=1} F_j )$
I'm defining the sequence as such base on how I work the other direction, but I can't seem to be able to break the left hand side into a sum of $\displaystyle \mu (E_1) + . . . + \mu (E_n)$.
Any hints? Thank you
2. Hello,
Suppose that $\displaystyle \mu : \mathbb {M} \rightarrow [0, \infty ]$ is continuous from above, that is, if $\displaystyle \{ E_j \} ^ \infty _{j=1} \subset \mathbb {M}$ with $\displaystyle E_{j+1} \subset E_j$, then I have $\displaystyle \lim _{j \rightarrow \infty } \mu (E_j)= \mu ( \bigcap ^ \infty _{j=1} E_j )$.
This limit holds only if at least one of the $\displaystyle E_j$ has a finite measure. If they're all of infinite measure, I think it should be easy to finish your proof by using (*) below. So now assume we can use this formula.
I will show that $\displaystyle \mu$ is a measure.
Since $\displaystyle \mu$ is finite-additive, the first axiom of a measure :
$\displaystyle \mu(\emptyset)=0$
is verified.
And the positivity too.
Let $\displaystyle \{ E_j \} ^n _{j=1} \subset \mathbb {M}$ be disjoint sets.
Claim: $\displaystyle \mu ( \bigcup ^n _{j=1} E_j) = \sum _{j=1}^n \mu (E_j)$
And we want to prove that for any pairwise disjoint sequence $\displaystyle \{F_j\}$, we have :
$\displaystyle \mu\left(\bigcup_{j=1}^\infty F_j\right)=\sum_{j=1}^\infty \mu(F_j)$ (*)
Let $\displaystyle F_n = \bigcup ^ \infty _{j=1}E_j \backslash \bigcup ^n _{i=1} E_i$, note that $\displaystyle F_j \subset F_{j+1}$ and $\displaystyle \bigcup ^n_{j=1}F_j$
The backslash is not \ which gives a space in latex, but \backslash
Now, since $\displaystyle \mu$ is continuous from above, I have $\displaystyle \lim _{n \rightarrow \infty } \mu (F_n) = \mu ( \bigcup ^ \infty _{j=1} F_j )$
I guess you're being confused with below and above...
The above continuity deals with an intersection, not an union, and with $\displaystyle F_{j+1}\subset F_j$ (which you actually have)
---------------------------------
To sum up... (not too formally) :
We know that $\displaystyle E_{j+1}\subset E_j$, $\displaystyle \lim_{j\to\infty} \mu(E_j)=\mu\left(\bigcap_{j=1}^\infty E_j\right)$ (continuous from above)
We know that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^n E_j\right)=\sum_{j=1}^n \mu(E_j)$ (finite additivity)
We want to prove that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\sum_{j=1}^\infty \mu(E_j)$
With the information we have, this is equivalent to proving $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\lim_{n\to\infty} \mu\left(\bigcup_{j=1}^n E_j\right)$
Then I recommend you read this : http://www.mathhelpforum.com/math-he...ection-ei.html (the second post).
3. So base on the $\displaystyle F_n$ that I define, I have the following:
1. $\displaystyle \lim _{j \rightarrow \infty } \mu (F_j) = \mu ( \bigcap _{j=1} ^ \infty F_j)$
2. $\displaystyle \bigcup ^ \infty _{j=1} E_j = \bigcap _{j=1} ^ \infty F_j$
3. $\displaystyle F_j = \bigcup ^ \infty _{j=1} E_j \backslash \bigcup _{i=1} ^n E_i$
So I have: $\displaystyle \mu ( \bigcup ^ \infty _{j=1} E_j ) = \mu ( \bigcap _{j=1} ^ \infty F_j ) = \lim _{j \rightarrow \infty } \mu (F_j) = \lim _{ j \rightarrow \infty } ( \bigcup ^ \infty _{j=1} E_j \backslash \bigcup _{i=1} ^n E_i )$
What I'm trying to do is get the last term to be $\displaystyle \sum _{j=1}^ \infty \mu (E_j)$, but I'm suspecting that the F_n that I define is not really help...
Will this get me to the right track? Thanks.
4. Okay, I confused some signs, so the reasoning I was pointing you at was not correct... Sorry
Indeed you defined badly $\displaystyle F_n$. I believe I wrote it above
Let's start it again.
------------------------------
We know that $\displaystyle E_{j+1}\subset E_j, \lim_{j\to\infty} \mu(E_j)=\mu\left(\bigcap_{j=1}^\infty E_j\right)$ (continuous from above)
We know that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^n E_j\right)=\sum_{j=1}^n \mu(E_j)$ (finite additivity)
We want to prove that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\sum_{j=1}^\infty \mu(E_j)$
With the information we have, this is equivalent to proving $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\lim_{n\to\infty} \mu\left(\bigcup_{j=1}^n E_j\right)$, where the Ej are pairwise disjoint << this is important. It uses the finite additivity. Do you understand this point ?
Then, assuming the measure of the whole space is finite (if it's infinite, it may work too but it ain't no fun)...
Let's take a sequence Ej of pairwise disjoint measurable sets.
Define $\displaystyle F_n=\bigcap_{j=1}^n \{E\backslash E_j\}$
- why did I choose the intersection ? Because we need a decreasing sequence for using the continuity from above.
- why did I take the complement ? Because it makes no sense taking the intersection of disjoint sets ! And because of de Moivre's formula, which will help us transform the intersection into an union.
So by the continuity from above applied to the sequence $\displaystyle F_n$, we have $\displaystyle \lim_{n\to\infty} \mu\left(\bigcap_{j=1}^n \{E\backslash E_j\}\right)=\mu\left(\bigcap_{j=1}^\infty\{E\back slash E_j\}\right)$
$\displaystyle \lim_{n\to\infty} \mu(E)-\mu\left(\bigcup_{j=1}^n E_j\right)=\mu(E)-\mu\left(\bigcup_{j=1}^\infty E_j\right)$
and finally $\displaystyle \lim_{n\to\infty}\mu\left(\bigcup_{j=1}^n E_j\right)=\mu\left(\bigcup_{j=1}^\infty E_j\right)$
Does it look somehow clear ? | 2,219 | 6,787 | {"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} | 3.65625 | 4 | CC-MAIN-2018-26 | latest | en | 0.850627 |
https://cas.oslo.no/news/in-real-life-almost-nothing-is-linear-article4451-974.html | 1,679,926,269,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00420.warc.gz | 196,040,663 | 14,352 | ‘Linear is easy and rare; non-linear is common and almost always complicated’, Steffen Oppermann and Aslak Bakke Buan say, in a sentence that feels like it could be about life in general.
The two professors from the Norwegian University of Science and Technology (NTNU) will lead a CAS project in 2022/23.
## What is representation theory?
Oppermann and Buan bring us back to high school in order to explain their CAS project in a way that can be understandable to non-mathematicians: in high school, we learn about linear equations and also linear functions and their graphs, which are simply straight lines on a plane.
‘Linear functions are very useful, but mostly for describing some ideal behavior. Newton taught us that a particle affected by no forces at all will simply move in a straight line’, they say, but reality is harsh: ‘In real life, almost nothing is simple, or linear.’
However, one way to get a clear picture of what is going on is to consider only some snapshots of a moving body at a time. ‘Locally, the behavior could look linear.’
The two scholars say that if they understand the derivative at every point, they have a lot of useful information about the function.
The concept of linearity applies in any dimension. These multi-dimensional linear maps are the subject of linear algebra, which is one of the best understood and most applicable branches of mathematics.
‘The idea of representation theory is to use the well-understood techniques and notions of linear algebra to understand and analyze non-linear structures.’
## Applications for data science
Opperman and Buan explain that in modern mathematics, representation theory is a vast field with many different branches.
Some are directly linked to physical applications, ranging from string theory in physics to
Astronomy, whereas others are more concerned with the intrinsic mathematical notions of representation theory.
‘Our project deals with problems of both kinds, and brings together experts with various perspectives.’
One application that will feature prominently within the project is topological data analysis.
‘The challenge of that subject is to identify topological features within the given data’, they say. For instance, they may ask how many components comprise the given data.
‘It turns out that these questions almost immediately translate to problems in representation theory. However, the questions typically asked by representation theorists don’t completely agree with those asked by data analysts, so we are hoping to learn from each other’s perspectives during our stay at CAS.’
## ‘As good as it gets’
Why did you consider CAS to be a suitable arena for your project?
‘A year at CAS gives a unique possibility to join efforts with leading experts with very varied experience and to work continuously and undisturbed for an entire academic year. The generous funding even makes it feasible for some of the younger experts in the field, travelling with families and children, to join. A program like this is pretty much “as good as it gets” for researchers like us.’ | 599 | 3,091 | {"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} | 3.09375 | 3 | CC-MAIN-2023-14 | latest | en | 0.946957 |
http://blogs.princeton.edu/sas/page/2/ | 1,405,319,645,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776440024.38/warc/CC-MAIN-20140707234040-00076-ip-10-180-212-248.ec2.internal.warc.gz | 16,904,668 | 78,336 | Stochastic Analysis Seminar | Princeton University | Page 2
## Next lecture: one-time room change
The next lecture is on October 17. Please note that we have a one time room change to Sherrerd Hall Room 101 on October 17 as the usual Bendheim room is in use for another event. We will return to the Bendheim center the following week for the rest of the semester.
10. October 2013 by Ramon van Handel
## Lecture 3. Sanov’s theorem
The goal of this lecture is to prove one of the most basic results in large deviations theory. Our motivations are threefold:
1. It is an example of a probabilistic question where entropy naturally appears.
2. The proof we give uses ideas typical in information theory.
3. We will need it later to discuss the transportation-information inequalities (if we get there).
What is large deviations?
The best way to start thinking about large deviations is to consider a basic example. Let be i.i.d. random variables with and . The law of large numbers states that
To say something quantitative about the rate of convergence, we need finer limit theorems. For example, the central limit theorem states that
Therefore, for any fixed , we have
(the probability converges to a value strictly between zero and one). Informally, this implies that the typical size of is of order .
Rather than considering the probability of typical events, large deviations theory allows us to understand the probability of rare events, that is, events whose probabilities are exponentially small. For example, if , then the probability that is at least of order unity (a rare event, as we have just shown that the typical size is of order ) can be computed as
The probability of this rare event decays exponentially at rate . If the random variables have a different distribution, then these tail probabilities still decay exponentially but with a different rate function. The goal of large deviations theory is to compute precisely the rate of decay of the probabilities of such rare events. In the sequel, we will consider a more general version of this problem.
Sanov’s theorem
Let be i.i.d. random variables with values in a finite set and with distribution (random variables in a continuous space will be considered at the end of this lecture). Denote by the set of probabilities on . Let be the empirical distribution of :
The law of large numbers states that a.s. To define a rare event, we fix that does not contain . We are interested in behavior of probabilities of the form , as .
Example. Let be such that . Define for some . Then . Thus the rare events of the type described in the previous section form a special case of the present setting.
We are now in a position to state Sanov’s theorem, which explains precisely at what exponential rate the probabilities decay.
Theorem (Sanov). With the above notations, it holds that
In particular, for “nice” such that the left- and right-hand sides coincide we have the exact rate
In words, Sanov’s theorem states that the exponential rate of decay of the probability of a rare event is controlled by the element of that is closest to the true distribution in the sense of relative entropy.
There are many proofs of Sanov’s theorem (see, for example, the excellent text by Dembo and Zeitouni). Here we will utilize an elegant approach that uses a common device in information theory.
Method of types
It is a trivial observation that each possible value must appear an integer number of times among the samples . This implies, however, that the empirical measure cannot take arbitrary values: evidently it is always the case that , where we define
Each element of is called a type: it contains only the information about how often each value shows up in the sample, discarding the order in which they appear. The key idea behind the proof of Sanov’s theorem is that we can obtain a very good bound on the probability that the empirical measure takes the value for each type .
Type theorem. For every , we have
That is, up to a polynomial factor, the probability of each type behaves like .
In view of the type theorem, the conclusion of Sanov’s theorem is not surprising. The type theorem implies that types such that have exponentially smaller probability than the “optimal” distribution that minimizes the relative entropy in . The probability of the rare event is therefore controlled by the probability of the most likely type. In other words, we have the following intuition, common in large deviation theory: the probability of a rare event is dominated by the most likely of the unlikely outcomes. The proof of Sanov’s theorem makes this intuition precise.
Proof of Sanov’s theorem.
Upper bound. Note that
This yields
[Note that in the finite case, by continuity, the infimum over equals the infimum over as stated in the theorem. The closure becomes more important in the continuous setting.]
Lower bound. Note that is dense in . As is open, we can choose for each , such that . Therefore,
It follows that
[Note that despite that we are in the finite case, it is essential to consider the interior of .]
Of course, all the magic has now shifted to the type theorem itself: why are the probabilities of the types controlled by relative entropy? We will presently see that relative entropy arises naturally in the proof.
Proof of the type theorem. Let us define
Then we can write
It is therefore sufficient to prove that for every
To show this, the key idea is to utilize precisely the same expression for given above, for the case that the distribution that defined the empirical measure is replaced by (which is a type). To this end, let us denote by the empirical measure of i.i.d. random variables with distribution .
Upper bound. We simply estimate using the above expression
Lower bound. It seems intuitively plausible that for every , that is, the probability of the empirical distribution is maximized at the true distribution (“what else could it be?” We will prove it below.) Assuming this fact, we simply estimate
Proof of the claim. It remains to prove the above claim that for every . To this end, note that consists of all vectors such that of the entries take the value , of the entries take the value , etc. The number of such vectors is
It is now straightforward to estimate
Thus the claim is established.
Remark. It is a nice exercise to work out the explicit form of the rate function in the example considered at the beginning of this lecture. The resulting expression yields another basic result in large deviations theory, which is known as Cramèr’s theorem.
General form of Sanov’s theorem
The drawback to the method of types is that it relies heavily on the assumption that take values in a finite state space. In fact, Sanov’s theorem continues to hold in a much more general setting.
Let be a Polish space (think ), and let be i.i.d. random variables taking values in with distribution . Denote by the space of probability measures on endowed with the topology of weak convergence: that is, iff for every bounded continuous function . Now that we have specified the topology, it makes sense to speak of “open” and “closed” subsets of .
Theorem. In the present setting, Sanov’s theorem holds verbatim as stated above.
It turns out that the lower bound in the general Sanov theorem can be easily deduced from the finite state space version. The upper bound can also be deduced, but this is much more tricky (see this note) and a direct proof in the continuous setting using entirely different methods is more natural. [There is in fact a simple information-theoretic proof of the upper bound that is however restricted to sets that are sufficiently convex, which is an unnecessary restriction; see this classic paper by Csiszar.]
We will need the general form of Sanov’s theorem in the development of transportation-information inequalities. Fortunately, however, we will only need the lower bound. We will therefore be content to deduce the general lower bound from the finite state space version that we proved above.
Proof of the lower bound. It evidently suffices to consider the case that is an open set. We use the following topological fact whose proof will be given below: if is open and , then there is a finite (measurable) partition of and such that
Given such a set, the idea is now to reduce to the discrete case using the data processing inequality.
Define the function such that for . Then if and only if the empirical measure of lies in . Thus
As take values in a finite set, and as is open, we obtain from the finite Sanov theorem
where we have used the data processing inequality and in the last inequality. As was arbitrary, taking the supremum over completes the proof.
Proof of the topological fact. Sets of the form
for , , bounded continuous functions, and form a base for the weak convergence topology on . Thus any open subset must contain a set of this form for every (think of the analogous statement in : any open set must contain a ball around any ).
It is now easy to see that each set of this form must contain a set of the form used in the above proof of the lower bound in Sanov’s theorem. Indeed, as is a bounded function, we can find for each a simple function such that . Clearly implies , so we can replace the functions by simple functions. But then forming the partition generated by the sets that define these simple functions, it is evident that if is chosen sufficiently small, then for all implies . The proof is complete.
Remark. It is also possible to work with topologies different than the topology of weak convergence. See, for example, the text by Dembo and Zeitouni for further discussion.
Lecture by Ramon van Handel | Scribed by Quentin Berthet
10. October 2013 by Ramon van Handel
## Next lecture: October 17
Next week (October 10) there will be no stochastic analysis seminar due to conflicting events:
03. October 2013 by Ramon van Handel
## Lecture 2. Basics / law of small numbers
Due to scheduling considerations, we postpone the proof of the entropic central limit theorem. In this lecture, we discuss basic properties of the entropy and illustrate them by proving a simple version of the law of small numbers (Poisson limit theorem). The next lecture will be devoted to Sanov’s theorem. We will return to the entropic central limit theorem in Lecture 4.
Conditional entropy and mutual information
We begin by introducing two definitions related to entropy. The first definition is a notion of entropy under conditioning.
Definition. If and are two discrete random variables with probability mass functions and , then the conditional entropy of given is defined as
where is the conditional probability mass function of given .
Remark. If and are absolutely continuous random variables, the conditional differential entropy is defined analogously (where the probability mass functions are replaced by the corresponding probability densities with respect to Lebesgue measure).
Note that
That is, the conditional entropy is precisely the expectation (with respect to the law of ) of the entropy of the conditional distribution of given .
We now turn to the second definition, the mutual information. It describes the degree of dependence between two random variables.
Definition. The mutual information between two random variables and is defined as
where , and denote the distributions of , and .
Conditional entropy and mutual information are closely related. For example, suppose that has density with respect to the Lebesgue measure, then
In particular, since is always positive (because it is a relative entropy), we have just shown that , that is, conditioning reduces entropy. The same result holds for discrete random variables when we replace by .
Chain rules
Chain rules are formulas that relate the entropy of multiple random variables to the conditional entropies of these random variables. The most basic version is the following.
Chain rule for entropy. . In particular, .
Proof. Note that
Thus,
Taking the expectation on both sides under the distribution gives the desired result.
Corollary. Entropy is sub-additive, that is, .
Proof. Combine the chain rule with .
There is also a chain rule for relative entropy.
Chain rule for relative entropy.
The following identity will be useful later.
Lemma.
Proof. Note that
Data processing and convexity
Two important properties of the relative entropy can be obtained as consequences of the chain rule.
Data processing inequality. Let and be two probability measures on and suppose is measurable. Then , where is the distribution of when .
The data processing inequality tells us that if we process the data (which might come from one of the two distributions and ), then the relative entropy decreases. In other words, it becomes harder to identify the source distribution after processing the data. The same result (with the same proof) holds also if and are transformed by a transition kernel, rather than by a function.
Proof. Denote by and the joint laws of and when and . By the chain rule and nonnegativity of relative entropy
On the other hand, using again the chain rule,
where we used . Putting these together completes the proof.
Convexity of relative entropy. is jointly convex in its arguments, that is, if , , , are probability measures and , then
Proof. Let be a random variable that takes value with probability and with probability . Conditionally on , draw and . Then and . Using the chain rule twice, we obtain
and the right hand side is precisely .
Corollary. The entropy function is concave.
Proof for a finite alphabet. When the alphabet is finite, the corollary can be proven by noting that .
Relative entropy and total variation distance
Consider the hypothesis testing problem of testing the null hypothesis against the alternative hypothesis . A test is a measurable function . Under the constraint , it can be shown that the optimal rate of decay of as a function of the sample size is of the order of . This means that is the measure of how well one can distinguish between and on the basis of data.
We will not prove this fact, but only introduce it to motivate that the relative entropy is, in some sense, like a measure of distance between probability measures. However, it is not a metric since and the triangle inequality does not hold. So in what sense does the relative entropy represent a distance? In fact, it controls several bona fide metrics on the space of probability measures. One example of such metric is the total variation distance.
Definition. Let and be probability measures on . The total variation distance is defined as .
The following are some simple facts about the total variation distance.
1. .
2. If and have probability density functions and with respect to some common probability measure , then . To see this, define . Then
3. .
The following inequality shows that total variance distance is controlled by the relative entropy. This shows that the relative entropy is a strong notion of distance.
Pinsker’s inequality. .
Proof. Without loss of generality, we can assume that and have probability density functions and with respect to some common probability measure on . Let and .
Step 1: Prove this inequality by simple calculation in the case when contains at most elements.
Step 2: Note that and are defined on the space . So Pinsker’s inequality applies to and . Thus,
Law of small numbers
As a first illustration of an application of entropy to probability, let us prove a simple quantitative law of small numbers. An example of the law of small numbers is the well known fact that in distribution as goes to infinity. More generally, if are Bernoulli random variables with , if are weakly dependent, and if none of the dominates the rest, then where . This idea can be quantified easily using relative entropy.
Theorem. If and may be dependent, then
where and .
Proof. Let be independent random variables with . Then . We have
To conclude, it is enough to note that
Remark. If and are independent, then the inequality in the theorem becomes . However, this rate of convergence is not optimal. One can show that under the same condition, , using tools similar to those that will be used later to prove the entropic central limit theorem. Note that it is much harder to prove in the entropic central limit theorem, even without rate of convergence!
Lecture by Mokshay Madiman | Scribed by Che-yu Liu
03. October 2013 by Ramon van Handel
## Lecture 1. Introduction
What is information theory?
The first question that we want to address is: “What is information?” Although there are several ways in which we might think of answering this question, the main rationale behind our approach is to distinguish information from data. We think of information as something abstract that we want to convey, while we think of data as a representation of information, something that is storable/communicable. This is best understood by some examples.
Example. Information is a certain idea, while data is the words we use to describe this idea.
Example. Information is . Possible data describing this information are: , , , followed by a billion zeros.
As we are in a mathematical setting we want to rely on a quantitative approach. The main question that arises naturally is: “How can we measure information?” Making sense of this question requires us to have a model for how data is produced. Throughout this seminar we will consider the probabilistic model which we now introduce.
Definition (Probabilistic model). Data is a random variable taking values on the space (alphabet) having distribution (source distribution). We write .
To be precise, with the above we mean that there exists a probability space and a measurable space with some measurable function such that we have .
Remarks.
1. While this set-up is very similar to what is done in statistics, the focus in information theory is different. In statistics it is assumed that the data comes from one of a family of distributions (statistical model), and the goal is to infer something about the particular distribution generating the data. On the other hand, in information theory the distribution of the data might be known or not, and the goal is to compress or communicate .
2. In the probabilistic model we assume that the data is generated by a certain random source. This is a particular modeling assumption and it is not necessarily an expression of belief in how data are actually produced. This is a reasonable modeling assumption to make and it allows us to draw reasonable conclusions (for example, text data is clearly not randomly produced, but you can still do useful things by making the modeling assumption that the data was produced by a stochastic source).
3. The original motivation behind the development of information theory as based on the probabilistic model came from a practical engineering problem (how to compress/communicate data), and not from the idea of how we measure information (although this aspect was also part of the motivation). The whole field of study was created by the 1948 paper of Claude Shannon.
4. We are going to use the probabilistic model throughout this seminar, but other models exist as well. A popular model used in theoretical computer science is the algorithmic model (which defines the field of algorithmic information theory, as opposed to probabilistic information theory). In this model it is assumed that data is the output of some computer program (running on a Turing machine). While this approach could be thought of as a generalization of the probabilistic model (in fact, one way in which computers can work is to simulate from some probability distribution), many of the basic quantities in algorithmic information theory (like Kolmogorov complexity) are not computable. This is the reason why this field is suitable for theoretical insights, but it is not necessarily suitable for practical purposes.
How do we measure information in the probabilistic model?
In what follows we assume that the information to be conveyed coincides with the data itself (and we now assume that takes values in some countable set ), meaning that there is no universal hidden meaning that we are trying to convey apart from the data itself. For example, assume that the information we want to convey (a particular realization of it) is the text “This is the SAS”. A natural way of measuring the amount of information contained in this data is to look for other representations of this information, and to look for the smallest (in some sense that needs to be specified) representation. As we are in the probabilistic framework, we do not know in advance which data is going to be produced by the random source, so we look for a procedure that takes the random outcome and gives us on average the smallest or most compact representation of the information in that data.
Since the data is random, also the size of a particular realization (encoded data) is random. One way to take into account the randomness is to consider a representation (encoding scheme) that minimizes the expected length/size of the encoded data (and that is uniquely decodable). That is, we measure the amount of information in a given data as
If we set up things in this way, the measure of information is some functional of the source distribution , since is the only quantity governing the data. This functional is called the entropy and it is defined as follows.
Definition (Entropy). If is a discrete distribution, then the entropy of is defined as
where we write for the probability mass function of .
While it can be shown that is the minimal expected length of validly encoded data, we do not proceed this way (the ideas behind this fact are covered in the first couple of lectures of an information theory class). Instead, we will give some intuition on why is a good measure of information.
We first provide some intuition on why the information in the statement should be decreasing as a function of . In fact, recall that presently we assume to know the source distribution . If we know , how informative is a particular outcome form the source distribution? If (i.e., is a point mass at ), being informed that the random outcome is is not informative. On the other hand, if , being informed that the outcome is is extremely informative/significant (something very rare has happened).
The relevant decreasing function turns out to be the following:
In this respect, corresponds to the information that we get from the statement . So the average amount of information in the random outcome is given by
Connection between information theory and statistics
While the connection between information theory and statistics is not surprising as both fields rely on the probabilistic model, this correspondence is very strong and natural. We give some examples.
1. Maximum likelihood estimators (MLE) can be seen as minimal codelength estimators. In a statistical model we assume that , with for some parameter space , and the goal is to find the parameter that generated the data. A popular estimator is the MLE since it is plausible to assume that the parameter that generated the data is the parameter whose corresponding distribution would have given maximal probability to , that is,
Note that we can rewrite the above as
which can be seen to correspond to the minimal number of bits required to represent assuming that it was generated by (codelength). Hence the connection between MLE in statistics and the minimal codelength estimator in information theory. In this setting we assume that we do not know the distribution generating the data and we try to find a good code to encode the data. The problem of finding a good code is in some sense equivalent to the problem of finding the distribution itself, since once you know the distribution you know the best code (in some sense). Also, we mention that many penalized-MLE estimators (where we take into account the complexity of the model by adding a penalty term to the MLE estimator) can be motivated from an information-theoretic point of view in terms of analogue of coding problems; this is the idea behind the “Minimum Description Length” principle.
2. In Hypothesis testing, the optimal error exponents are information-theoretic quantities.
These are not just coincidental connections, but examples of basic relationships between fundamental limits of statistical inference on the one hand, and fundamental limits of communication and compression of data on the other hand.
We now turn to the main topic of this seminar, that is, the connection between information theory and probability theory.
Connection between information theory and probability theory
Since we are using a probabilistic model it is clear that probability theory is the language of information theory. However, it is not so obvious that information theory can say something fundamental about probability theory. In fact, in the past half century or so, it has been realized that information theory captures many fundamental concepts in probability theory. Before turning to one key example of such connection (the entropic central limit theorem) which will serve as motivation for the initial few lectures of the seminar, we introduce some relevant quantities.
Definition (Differential or Boltzmann-Shannon entropy). If , and (i.e., has a density with respect to the Lesbegue measure), then the differential entropy of (equivalently, differential entropy of ) is defined as
with the conventions and .
While we can think of as a measure of disorder (particularly motivated by the setting introduced by Boltzmann in physics), is not a measure of information in the same sense as is. The reason is that in the present context of “continuous” data (recall that we are in and a possible outcome of is a real number) we need infinitely many bits to encode each outcome of , so it is not meaningful to talk of the amount of information in an outcome as this is generally infinity. Nonetheless, the differential entropy represents a crucial quantity in information theory, and shows up for example both when considering communication over channels with continuous noise distributions, and when considering lossy data compression (the only kind of data compression possible with sources like , where one accepts some slight distortion of the data in order to be able to encode it with finitely many bits).
The notion that unifies the continuous entropy with the discrete entropy previously introduced is the relative entropy which we now define.
Definition (Relative entropy). If is a probability measure and is a -finite measure on , then the relative entropy between and is defined as
Typically and have respective densities and with respect to a given reference measure . Then the relative entropy reads
The following examples show how the relative entropy relates and .
1. If is a countable set and is the counting measure, then
2. If and is the Lesbegue measure, then
The following property of relative entropy is the most important inequality in information theory.
Lemma. Let be a probability measure on , and be a sub-probability measure on (i.e., is a nonnegative, countably additive measure with ). Then .
Proof. We only need to consider the case where . Let and . Then we have
where we have applied Jensen’s inequality (which holds as is a probability measure) using that is convex, and used that and that .
As a consequence of this result we can now show that the Gaussian distribution maximizes the entropy under constraints on the first two moments.
Lemma. Let be the class of all probability densities on (with respect to Lebesgue measure) with mean and variance and define
Then for any .
Proof. First of all note that
as is quadratic function and, consequently, only the first two moments are involved in computing its expectation. Hence, we have
We are now ready to present the first example of cross-fertilization where information-theoretic concepts can be used to capture fundamental properties in probability theory. Let us first recall the classical central limit theorem (CLT).
Theorem (CLT). If are i.i.d. real-valued random variables with mean and variance , then
that is,
for nice enough sets .
If we denote by the density of the normalized partial sum introduced in the statement of the theorem above, we note the following.
1. For each we have . This follows immediately from basic properties of expected values.
2. From the previous lemma it follows immediately that
So, the CLT tells us that the sequence converges to the maximizer of the entropy in . In fact, it turns out that the convergence in the central limit theorem can be studied in terms of the entropy and that the CLT is an expression of increasing entropy, as the following entropic central limit theorem describes.
Theorem (Entropic CLT). Let be i.i.d. real-valued random variables with mean and variance , and assume the distribution of has a density (with respect to Lebesgue measure). Under minimal assumptions (specifically, that for some ), we have
or, equivalently,
The entropic central limit theorem is remarkable as usually limit theorems do not come with an associated monotonicity statement. This suggests that the relative entropy is a natural tool to analyze the CLT.
Of course, a natural question that presents itself is whether other limit theorems in probability can be understood from a similar information-theoretic point of view.
Plan for future lectures
In the next lecture or two we will present a full proof of the entropic central limit theorem, and also discuss briefly how other limit theorems can be analogously understood from this information-theoretic point of view. Later, we will look at finer behavior than limit theorems, for instance we may look at how information theory can provide insights into large deviations and concentration inequalities.
Lecture by Mokshay Madiman | Scribed by Patrick Rebeschini
25. September 2013 by Ramon van Handel
## Fall 2013: Information theoretic methods
While information theory has traditionally been based on probabilistic methods, ideas and methods from information theory have recently played an increasingly important role in various areas of probability theory itself (as well as in statistics, combinatorics, and other areas of mathematics). The goal of these informal lectures is to introduce some topics and tools at the intersection of probability theory and information theory. No prior knowledge of information theory will be assumed. Potential topics include: entropic central limit theorems, entropic Poisson approximation, and related information-theoretic and probabilistic inequalities; connections to logarithmic Sobolev inequalities and Stein’s method; entropic inequalities for additive, matroidal, and tree structures, and their applications; transportation cost-information inequalities and their relation to concentration of measure; basic large deviations theory.
Prerequisites: Probability at the level of ORF 526 is assumed.
Time and location: Thursdays, 4:30-6:00, Bendheim Center classroom 103.
The first lecture will be on September 19.
References:
07. September 2013 by Ramon van Handel
## End of Spring 2013 seminar
That was it for the Spring 2013 stochastic analysis seminar! We will be back in the Fall. The topic and location/time slot will be announced in September.
03. May 2013 by Ramon van Handel
## Lectures 7/8. Games on random graphs
The following two lectures by Rene Carmona are on games on random graphs.
Many thanks to Patrick Rebeschini for scribing these lectures!
In the following we discuss some results from the paper “Connectivity and equilibrium in random games” by Daskalakis, Dimakis, and Mossel. We define a random game on a random graph and we characterize the graphs that are likely to exhibit Nash equilibria for this game. We show that if the random graph is drawn from the Erdös-Rényi distribution, then in the high connectivity regime the law of the number of pure Nash equilibria converges toward a Poisson distribution, asymptotically, as the size of the graph is increased.
Let be a simple (that is, undirected and with no self-edges) graph, and for each denote by the set of neighbors of , that is, . We think of each vertex in as a player in the game that we are about to introduce. At the same time, we think of each edge as a strategic interaction between players and .
Definition (Game on a graph). For each let represent the set of strategies for player , assumed to be a finite set. We naturally extend this definition to include families of players: for each , let be the set of strategies for each player in . For each , denote by the reward function for player . A game is a collection .
The above definition describes a game that is static, in the sense that the game is played only once, and local, in the sense that the reward function of each player depends only on its own strategy and on the strategy of the players in its neighbors. We now introduce the notion of pure Nash equilibrium.
Definition (Pure Nash equilibrium). We say that is a pure Nash equilibrium (PNE) if for each we have
A pure Nash equilibrium represents a state where no player can be better off by changing his own strategy if he is the only one who is allowed to do so. In order to investigate the existence of a pure Nash equilibrium it suffices to study the best response function defined below.
Definition (Best response function). Given a reward function for player , we define the best response function for as
Clearly, is a pure Nash equilibrium if and only if for each . We now define the type of random games that we will be interested in; in order to do so, we need to specify the set of strategies and the reward function for each player.
Definition (Random game on a fixed graph). For a graph and an atomless probability measure on , let be the associated random game defined as follows:
1. for each ;
2. is a collection of independent identically distributed random variables with distribution .
Remark. For each game the family is a collection of independent random variables that are uniformly distributed in , and for each , we have almost surely. In fact, note that if and only if and this event has probability since the two random variables appearing on both sides of the inequality sign are independent with the same law and is atomless. As far as the analysis of the existence of pure Nash equilibria is concerned, we could take the present notion of best response functions as the definition of our random game on a fix graph. In fact, note that the choice of in does not play a role in our analysis, and we would obtain the same results by choosing different (atomless) distributions for sampling (independently) the reward function of each player.
Denote by the distribution of a Erdös-Rényi random graph with vertices where each edge is present independently with probability . We now introduce the notion of a random game on a random graph.
Definition (Random game on a ramon graph). For each , and each probability measure on , do the following:
1. choose a graph from ;
2. choose a random game from for the graph .
Henceforth, given a random variable let represent its distribution. Given two measures and on a measurable space, define the total variation distance between and as
where the supremum is taken over measurable functions such that the supremum norm is less than or equal to .
We are now ready to state the main theorem that we will prove in the following (Theorem 1.9 in Daskalakis, Dimakis, and Mossel).
Theorem 1 (High connectivity regime). Let be the number of pure Nash equilibria in the random game on random graph defined above. Let define the high-connectivity regime as
where satisfies the following two properties:
Then, we have
where denotes the conditional probability given the graph and is a Poisson random variable with mean . In particular,
which shows that in this regime a pure Nash equilibrium exists with probability converging to as the size of the network increases.
Remark. Using the terminology of statistical mechanics, the first result in Theorem 1 represents a quenched-type result since it involves the conditional distribution of a system (i.e., the game) given its environment (i.e., the graph). On the other hand, the second result represents an annealed-type result, where the unconditional probability is considered.
In order to prove Theorem 1 we need the following lemma on Poisson approximations. The lemma is adapted from the results of R. Arratia, L. Goldstein, and L. Gordon (“Two moments suffice for Poisson approximations: the Chen-Stein method“, Ann. Probab. 17, 9-25, 1989) and it shows how the total variation distance between the law of a sum of Bernoulli random variables and a Poisson distribution can be bounded by the first and second moments of the Bernoulli random variables. This result is a particular instance of the Stein’s method in probability theory.
Lemma 2 (Arratia, Goldstein, and Gordon, 1989). Let be a collection of Bernoulli random variables with . For each let be such that is independent of . Define
where . Define and . If is a Poisson random variable with mean , then
Proof. We define the following operators that act on each function :
We point out that characterizes in the sense that if and only if is a Poisson random variable with mean ; is an example of Stein’s operator. First of all, we show that for each we have . In fact, if we have
and if we have
For each define and . The following properties hold:
1. ,
2. ,
3. .
In what follows, consider any given function such that . Define the function as for each , and let . From what was seen above we have and we get
The first term is bounded above by while the third term is equal to since is independent of . In order to bound the second term we want to rewrite each term as a telescoping sum. In what follow fix , label the elements of as and define
Noticing that , we have
and we get
Therefore, combining all together we get
Since the total variation distance can be characterized in terms of sets as
from this point on we restrict our analysis to indicator functions, which are easier to deal with than generic functions. For each define , and . The previous result yields
and the proof of the Lemma is concluded if we show that for each . In fact, in what follows we will show that
where the right hand side is clearly upper bounded by . The proof that we are going to present is contained in the Appendix of “Poisson Approximation for Some Statistics Based on Exchangeable Trials” by Barbour and Eagleson.
First of all, note that for each we have and
for each . From this expression it is clear that for each , which suggests that we can restrict our analysis to singletons. For each we have
and from the series expansion of the Poisson probabilities it is easy seen that the function is negative and decreasing if and is positive and decreasing if . Hence, the only positive value taken by the difference function corresponds to the case that can be bounded as follows:
Therefore, for each , we have
Noticing that , then for each , we also get
and we proved that .
We now introduce the notation that will naturally allow us to use Lemma 2 to prove Theorem 1. We label the pure strategy profiles in as , where and as always . Often times, it will be convenient to use the labels to enumerate the vertices of the graph . Accordingly, one can think of the strategy profiles defined in a specific way, for example by positing that is the binary decomposition of . In particular becomes the strategy where each player plays zero, that is, for each , and the strategy where each player plays one, that is, for all . For each define
Clearly the quantity identifies the number of pure Nash equilibria and corresponds to the existence of a pure Nash equilibrium. We recall that both the randomness in the choice of the graph and the randomness in the choice of the game are embedded in the random variables . Note that conditionally on a given graph sampled form we have, for each ,
from which it follows that
That is, the current definition of a game on a fixed graph implies that the expected number of pure Nash equilibria is for any given graph. It follows that also . Notice that Theorem 1 adds more information on the table since it describes the asymptotic distribution of on a particular regime of the Erdös-Rényi random graph.
In the way we set up the stage it seems tempting to apply Lemma 2 to the random variables just defined. However, this approach is not fruitful since, apart from trivial cases, any given random variable has a neighborhood of dependence that coincides with the entire set . To see this, consider any two strategy profiles and . As long as there exists such that , then we can always find a realization of the graph such that does not have any edges attached to it, that is, ; this implies that and, consequently, it implies that and are not independent. Therefore, only and are independent, where and for each . However, Lemma 2 can be fruitfully applied to the random variables when we look at them conditionally on a given graph realization, as the following Lemma demonstrates (Lemma 2.2 of Daskalakis, Dimakis, Mossel).
Lemma 3. Let be a graph. Define
and for each define
where
and is the exclusive-or operation. Then, for each we have that is independent of .
Proof.We first show that is independent of . In order to make the independency structure manifest, we characterize . Recall that is the index set of the pure strategy profiles for which there exists a player having all his neighbors playing . Therefore, each strategy profile corresponding to an index set in is characterized by the fact that each player has at least one neighbor who is playing . Hence, for each we have for all and, consequently, the events
are independent of the events
which proves our claim.
We now generalize this result to show that is independent of , for each . For any , note that the exclusive-or map with respect to preserves the differences in the strategy profiles (and, of course, it also preserves the equalities). That is, if and are such that for some , then also and are such that . Therefore,
holds true if and only if
holds true. Equivalently stated, if and only if , where is the index set such that . Hence, the proof is concluded once we notice that and that is defined as the index set of the pure strategy profiles that are obtained by an exclusive-or map with respect to of a strategy profile in .
For a given graph with , define , where and represents the conditional probability conditionally on the graph . Define as in Lemma 3; then, conditionally on we have that is independent of . Define
where . Define and recall that . If is a Poisson random variable with mean , then Lemma 2 yields
At this point, let us introduce the following two lemmas (Lemmas 2.4 and 2.5 of Daskalakis, Dimakis, Mossel).
Lemma 4. If is sampled from the Erdös-Rényi distribution , we have
Lemma 5. Under the assumptions of Theorem 1 there exists and such that
We now show how the proof of Theorem 1 follows easily from the two lemmas above.
Proof of Theorem 1. Let and as in Lemma 5. Define , and . Clearly, by Lemma 5 we have
Define the event
By the Markov inequality and the previous asymptotic bounds, for each we have
where we used the result of Lemma 4 and the above estimates for and . Since for some , then there clearly exists such that
Hence, we have that for . Let us now define and such that and
Then
which proves the first statement in Theorem 1. In fact, we have
since by definition of , on the event we have
By the properties of conditional expectations we can now prove the convergence in total variation of the unconditional law of to the law of . In fact, for we have
from which it follows that
Since convergence in total variation implies convergence in distribution, the previous result implies that converges in distribution to , which concludes the proof of Theorem 1.
We now provide the proof of Lemma 4, while we refer the reader to Daskalakis, Dimakis, Mossel for the proof of Lemma 5.
Proof of Lemma 4. We begin with the study of . By the symmetry of the model we have
Since , we have . By the symmetry of the Erdös-Rényi distribution we also have that if and have the same number of players playing (equivalently, the same number of players playing ).
Therefore, if we label the vertices of the graph as , we have
where for each the index set is such that the strategy satisfies if and if . Hence, the bound for in the statement of the Lemma is proved if we show that . In fact, by definition of we have
We now study the term . Proceeding as above, by symmetry we have
We now analyze the term . As noticed above, if and only if the graph is such that there exists a player such that . In the case in which such also satisfies the property , then and for each , and it follows that . In fact, the event corresponds to the realizations where both strategy and strategy are pure Nash equilibria, that is, for each . But it can not be that both and are best responses for the player when all player in the neighbor play . Hence, is different from on the event
and on the event
Define and . Note that we have . On the events and we have and we get
Because of the independency structure in the Erdös-Rényi random graph we have
Furthermore, we have
This follows immediately from the definition of pure Nash equilibrium in terms of the best response functions once noticed that on the event there are exactly players (each such that ) such that , while for the remaining players we have . Putting everything together we get
Using the fact that , it clearly follows that .
01. May 2013 by Ramon van Handel
Categories: Random graphs | Comments Off
## Giant component: final remarks
The past three lectures were devoted to the giant component theorem:
Theorem Let be the connected component of that contains .
1. If , then in probability.
2. If , then in probability, for some .
3. If , then in distribution.
We proved only the first (subcritical) and second (supercritical) cases: our presentation was largely inspired by the treatment of Janson, Luczak, and Rucinski and Durrett. We have omitted the critical case, however, as the last two lectures of the semester will be on another topic. The goal of this post is to provide some final remarks and references on the giant component theorem.
Retrospective
At first sight, the “double jump” in the giant component theorem looks quite shocking. In hindsight, however, this does not seem quite so miraculous, as it mirrors an elementary phenomenon that is covered in many introductory probability courses: given a (nice) random walk with initial condition , define the hitting time for some . Then there are three cases:
1. If has negative drift, then . In fact, the random variable has a light (exponential) tail.
2. If has positive drift, then .
3. If has zero drift, then a.s. but . That is, the random variable has a heavy tail.
This “double jump” in the behavior of the hitting probabilities of a random walk is directly analogous to the behavior of the connected components of an Erdös-Rényi graph, and this was indeed the basic idea behind the proofs given in the previous lectures. Of course, it remains a bit of a miracle that the random walk approximation of the exploration process, which only holds for small times, is sufficiently powerful that it describes so completely the behavior of the random graph.
The critical case
In the subcritical case, the size of the largest component is of order because the hitting time of a random walk with negative drift has an exponential tail: that is, we proved
which goes to zero as for .
Similarly, we would expect that we can obtain the size of the largest component in the critical case if we understand the heavy tail behavior of the hitting time of a random walk with zero drift. This is in fact the case. Indeed, when there is zero drift, one can show that
The crude union bound argument used above now does not give the correct answer, but an only slightly better argument is needed. Indeed, note that
Therefore, . With some further work, a corresponding lower bound can also be proved. See the paper by Nachmias and Peres or the notes by van der Hofstad for the details.
It turns out that in the critical case is not only bounded in probability, but in fact converges weakly to some limiting distribution. This distribution, and much more, is beautifully described by Aldous in terms of Brownian excursions. This is an interesting example of the application of stochastic analysis to discrete probability; unfortunately, we do not have the time to cover it.
In a different direction, it turns out that various additional phase transitions appear when we consider a finer scaling, for example, in the “critical window” . For an overview of the various transitions, see, for example, section 11.1 in Alon and Spencer.
Connectivity threshold
Rather than considering the size of the largest component, one could ask when the entire Erdös-Rényi graph is connected. Note that when , the constant in the size of the giant component is always strictly positive, so the graph is not connected. Therefore, in order for the entire graph to be connected, we must let (that is, the edge probability must be superlinear). It turns out that the appropriate scaling for this question is , and another phase transition arises here.
Theorem. Let . If , then the Erdös-Rényi graph is connected with probability tending to as . If , the graph is connected with probability tending to as .
To get some intuition, consider the probability that a vertex is isolated (that is, disconnected from every other vertex):
Thus for , we have
In particular, if , then there must exist an isolated vertex with positive probability as , in which case the graph is not connected (in fact, it is not hard to show that the variance of the number of isolated components is of the same order as its mean, so that the probability that the graph is connected tends to zero). Somewhat miraculously, it turns out that when there are no isolated vertices, then the graph must already be connected, so that we do indeed obtain the sharp transition described in the Theorem above. For a proof of this fact (by a clever combinatorial argument) see, for example, the lecture notes by van der Hofstad. Alternatively, one can use a random walk argument entirely in the spirit of the proofs in the previous lectures to prove that the random graph is connected for : by running simultaneous exploration processes from different vertices as we did in the proof of the supercritical case, one can show that all connected components must intersect when and thus the entire graph must be connected. See section 2.8 in Durrett for such an argument.
28. April 2013 by Ramon van Handel
Categories: Random graphs | Comments Off
## Lecture 6. Giant component (3)
Let us begin with a brief recap from the previous lecture. We consider the Erdös-Rényi random graph in the supercritical case . Recall that denotes the connected component of the graph that contains the vertex . Our goal is to prove the existence of the giant component with size , while the remaining components have size .
Fix sufficiently large (to be chosen in the proof), and define the set
of vertices contained in “large” components. The proof consists of two parts:
• Part 1:
• Part 2: in probability.
Part 1 states that all the sufficiently large components must intersect, forming the giant component. Part 2 counts the number of vertices in the giant component. Part 2 was proved in the previous lecture. The goal of this lecture is to prove Part 1, which completes the proof of the giant component.
Overview
As in the previous lectures, the central idea in the study of the giant component is the exploration process , where
We have seen that , where is a random walk with increments
When , we have . Thus is approximately a random walk with positive drift. The intuitive idea behind the proof of Part 1 is as follows. Initially, the random walk can hit rapidly, in which case the component is small. However, if the random walk drifts away from zero, then with high probability it will never hit zero, in which case the component must keep growing until the random walk approximation is no longer accurate. Thus there do not exist any components of intermediate size: each component is either very small () or very large (we will show , but the precise exponent is not important).
We now want to argue that any pair of large components must necessarily intersect. Consider two disjoint sets and of vertices of size . As each edge is present in the graph with probability , the probability that there is no edge between and is
We therefore expect that any pair of large components must intersect with high probability. The problem with this argument is that we assumed that the sets and are nonrandom, while the random sets themselves depend on the edge structure of the random graph (so the events and are highly correlated). To actually implement this idea, we therefore need a little bit more sophisticated approach.
To make the proof work, we revisit more carefully our earlier random walk argument. The process has positive drift as . Thus the process is still approximately a random walk with positive drift! Applying the above intuition, either dies rapidly (the component is small), or grows linearly in as is illustrated in the following figure:
This means that the exploration process for a component of size will not only grow large () with high probability, but that the exploration process will also possess a large number of active vertices (. To prove that all large components intersect, we will run different exploration processes simultaneously starting from different vertices. We will show that if two of these processes reach a large number of active vertices then there must be an edge between them with high probability, and thus the corresponding components must coincide. This resolves the dependence problem in our naive argument, as the edges between the sets of active vertices have not yet been explored and are therefore independent of the history of the exploration process.
The component size dichotomy
We now begin the proof in earnest. We will first show the dichotomy between large and small components: either the component size is , or the number of active vertices grows linearly up to time . To be precise, we consider the following event:
Our goal is to show that is large.
Define the stopping time
We can write
Now suppose and . Then , as exploration process is alive at time and stays alive until time . We can therefore write
To bound the probabilities inside the sum, we compare to a suitable random walk.
The random walk argument
To bound the probability that , we must introduce a comparison random walk that lies beneath . We use the same construction as was used in the previous lecture. Let
where , are i.i.d. random variables independent of , (the same used in the exploration process), and is the set of the first components of (if , then and thus is undefined; then we simply add variables ).
As in the previous lecture, we have:
• is a random walk with increments.
• whenever and .
Now suppose that and . Then
We therefore obtain for
Thus computing reduces to compute the tail probability of a random walk (or, in less fancy terms, a sum of i.i.d. random variables). That is something we know how to do.
Lemma (Chernoff bound). Let . Then
Proof. Let . Then
The result follows by optimizing over .
Note that . We therefore have by the Chernoff bound
for all (here depends only on and ). In particular, we have
provided is sufficiently large. Thus we can estimate
which goes to zero as provided that is chosen sufficiently large. In particular, the component size dichotomy follows: choosing any , we obtain
Remark: Unlike in the proof of Part 2 in the previous lecture, here we do need to choose sufficiently large for the proof to work. If is too small, then the random walk cannot move sufficiently far away from zero to ensure that it will never return. In particular, even in the supercritical case, the second largest component has size of order .
Large components must intersect
To complete the proof, it remains to show that all large components must intersect. To do this, we will run several exploration processes at once starting from different vertices. If the sets of active vertices of two of these processes grow large, then there must be an edge between them with high probability, and thus the corresponding components intersect. Let us make this argument precise.
In the following, we denote by the exploration process started at . For each such process, we define the corresponding set that we have investigated above:
We have shown above that, provided , we have
We can therefore estimate
Now note that by time , the exploration process has only explored edges where (or ), and similarly for . It follows that
In particular, if are disjoint subsets of vertices, then
On the other hand, implies that must be disjoint at every time . Thus if , there can be no edges between vertices in and at any time (if such an edge exists, then the vertices connected by this edge will eventually be explored by both exploration processes, and then the sets of removed vertices will no longer be disjoint). Therefore,
Thus we finally obtain
and the proof of the giant component theorem is complete.
Many thanks to Quentin Berthet for scribing this lecture!
27. April 2013 by Ramon van Handel
Categories: Random graphs | Comments Off
← Older posts | 11,868 | 57,987 | {"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} | 3.328125 | 3 | CC-MAIN-2014-23 | longest | en | 0.938874 |
https://en.academic.ru/dic.nsf/enwiki/3048941 | 1,606,925,368,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00231.warc.gz | 275,766,426 | 15,689 | # Useful space principle
Useful space principle
The Useful Space Principle, or "USP", was first articulated in a series of six articles in The Bridge World, from November 1980 through April 1981. (The International Bridge Press Association awarded its 1981/1982 award for Best Article or Series on a System or Convention to Jeff Rubens for this series.) The USP is expressed succinctly in [http://www.bridgeworld.com/default.asp?d=bridge_glossary&f=glossu.html The Bridge World glossary] as: "a partnership's assigning meanings to actions so that the remaining bidding space matches the needs of the auction."
The articles on the USP were the genesis of widely used conventional methods such as Kickback and transfer advances of overcalls. The USP tells bidding theorists that bidding space should be allocated where it is most needed.
A USP Example: Kickback
The Blackwood convention, as originally formulated, violates the USP. Suppose that the agreed trump suit is spades. After the Blackwood "asker" bids 4NT, "teller" can convey four separate messages without bypassing the safety level of 5Spades – four aces or none with 5Clubs, one ace with 5Diams, two aces with 5Hearts and three aces with 5Spades.
But what if the agreed trump suit is clubs? Suppose that asker and teller each have one ace. Then, after 4NT, teller bids 5Diams to show his ace, and the partnership has to play 6Clubs off two aces (or possibly 5NT, which could be worse than 6Clubs, if it has the machinery).
The problem can also occur when the agreed trump suit is diamonds, although it's less likely because there's more space available for responses than when the agreed trump suit is clubs. But if the partnership is using Key Card Blackwood there can be similar problems. Suppose that hearts is agreed, asker has one ace and teller has one ace plus the king and queen of hearts. Asker bids 4NT and teller bids 5Spades to show two key cards plus the trump queen, and the partnership is again too high.
The problem is that Blackwood ignores the USP. The lower in rank the agreed trump suit, the more space that is needed if the partnership is to stay at or below a safety level.
The Kickback ace-asking convention deals with the problem by adjusting the asking bid according to which suit is agreed as trump. The ask is always one step above four of the trump suit. So, if clubs is agreed, the ask is 4Diams; if diamonds is agreed, 4Hearts asks; if hearts, 4Spades; and if spades, 4NT.
The responses to the ask might be similar to Blackwood, but instead of associating a specific suit with a specific number of aces, the responses are in terms of the number of steps above the ask. If spades will be trump, 4NT is the ask, and then 5Clubs, one step, might show zero or four aces, according to partnership agreement. If diamonds will be trump, 4Hearts is the ask, and then 4Spades, one step above the ask, might show zero or four aces.
The effect is to allocate bidding space where it's most useful in the context of the convention. If clubs is agreed and each partner has one ace, asker bids 4Diams and teller bids 4Spades to show one ace. The partnership can now easily sign off in 5Clubs.
There is a cost, of course: the partnership that plays Kickback loses the ability to cue-bid the ace of the suit above trumps. That is, assuming that hearts will be trumps, asker can no longer bid 4Spades to show first round control of spades: that would be the Kickback asking bid.
The solution is to use 4NT to show a first round control in the Kickback asking suit. With diamonds agreed, 4Hearts is the Kickback ask, and 4NT shows the HeartsA or, if credible in the context of the prior bidding, a void.
The agreement that 4NT is a cue-bid still entails a cost, but Kickback users argue that there is a net gain. For example, with clubs agreed, South would bid 4NT to show a first round control in diamonds. This bid not only bypasses the Kickback ask (4Diams), but also prevents North from cue-bidding 4Hearts or 4Spades. Kickback users believe that the gain in space from adjusting the ace-ask outweighs getting in the way of "partner's" cue-bid.
Notice that the Gerber convention, the use of 4Clubs to ask for aces when NT is the likely final strain, is really a special case of Kickback.
Note also that the foregoing is meant only to illustrate the USP. It describes neither additional understandings that Kickback can accommodate, nor the special problems that can arise (for example, the question of which is the agreed trump suit).
The USP at Lower Levels: Transfer Responses to Overcalls
Suppose that North opens a strong NT, North-South are playing Jacoby transfers, and South holds BridgeHandInline|KQ965|6|8752|854. South bids 2Hearts, hoping to pass North's 2Spades. But South would also bid 2Hearts with BridgeHandInline|KQ965|6|8752|A54 (South will force to game) and BridgeHandInline|AKQ65|6|8752|A54 (South will explore slam).
The transfer gives the partnership plenty of space for any continuation it might have in mind. In contrast, the traditional bid of 2Spades as a signoff over 1NT means that the partnership must give up bidding space in order to make forcing bids that start at the three level. It's when South wants to sign off by bidding 2Spades directly that the smallest amount of bidding space is needed, but that bid takes away three steps (2Clubs, 2Diams and 2Hearts). Transfers, whatever costs they entail, tend to conform to the USP.
Now consider competitive bidding. Suppose that West opens 1Spades, North overcalls 2Hearts and East passes. South holds BridgeHandInline|854|6|KQ9653|854. Now:
*If 3Diams is "non"forcing all is well. South describes his hand and leaves the rest to North.
*If 3Diams is forcing South must pass and possibly miss a good diamond contract. The 3Diams bid takes up so much space that, if it's forcing, South cannot show a weak hand with a good suit.
Again after 1Spades – (2Hearts) – P, South holds BridgeHandInline|854|6|KQ9653|KJ4. Now:
*If 3Diams is "non"forcing South must cue bid 2Spades to prepare a rebid in diamonds. The hand is too strong to bid a nonforcing 3Diams. But North's rebid, very often 3Hearts, may well prevent South from showing the diamonds below 3NT.
*If 3Diams is forcing all is well on this hand, and if South has a heart fit and a good hand he can cue bid 2Spades. In this sequence the cue-bid takes up minimal space – but how is that space to be used effectively when South has already shown a heart fit in a strong hand?
Regardless of the agreement on the forcing nature of 3Diams or 3Clubs in this auction, there's a problem caused by the misallocation of bidding space. If 3Diams is forcing, a good diamond suit in a weak hand is problematic. If 3Diams is nonforcing, the ambiguous 2Spades cue-bid may well prompt a rebid by North that preempts South's diamonds.
The USP suggests that in responding to overcalls, a hand with at least invitational strength plus a fit for overcaller's suit make the highest level non-jump bid available. This frees lower bids to be used as natural and forcing, or as transfers – and the transfer buys space to show a weak, a game forcing, or even a slam invitational hand, just as do Jacoby transfers. So doing puts the bidding space where it is most needed – to complete the transfer and possibly to further describe the hand, and to make a natural, forcing new-suit bid below the cue-bid.
Those who play transfer advances of overcalls usually agree that the transfer bids begin with the cue-bid of opener's suit. Bids between the overcall and the cue-bid may be treated as natural and forcing; transfer bids are available to handle weaker hands with their own good suit.
For example, after 1Hearts – (2Clubs) – P, some play this structure:
*2Diams is natural and forcing
*2Hearts, the cue bid, is a transfer to spades with strength to be clarified later
*2Spades is a transfer to clubs – that is, a strong raise of partner's overcall
*2NT is natural and nonforcing
*3Clubs is a limited natural raise
After 1Hearts – (1Spades) – P:
*2Clubs is natural and forcing
*2Diams is natural and forcing
*2Hearts, the cue bid, is a transfer to spades (this is the strong raise)
*2Spades is a limited natural raise
Again, the point of the foregoing is to illustrate how application of the USP can make bidding agreements more effective, not to define an optimal structure for responding to overcalls.
References
The Useful Space Principle, The Bridge World, November 1980 – April 1981.
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https://cassno.com/casino/advanced-blackjack-strategy/ | 1,618,569,158,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00111.warc.gz | 273,378,103 | 9,751 | Blackjack card counting is an advanced technique to win on blackjack. A player must first learn the basic ways of blackjack before learning “counting”. By learning the technique of counting you may have the odds with you.
The essential concept of blackjack counting is {that a} deck filled with 10s and aces can be to your advantage. While the deck of the dealer is stuffed with smaller cards but is to the dealer’s advantage. Merely put, when the deck is in your favor bet to the maximum, when it’s with the dealer’s bet minimum.
This can be as a result of a deck filled with ten increases the chance of the dealer obtaining bust. You on the opposite hand will simply opt to stand. In an opposite manner, a deck crammed with smaller cards makes the dealers likelihood to bust less. If you know the deck is filled with 10s or smaller cards, this could give you the advantage of knowing what strategy to use.
The dealer should then continually hit until having a soft 17. To hit with a hand of 10 & six and knowing the deck is filled with 10s is a unhealthy idea.
If the deck is stuffed with aces, there is high chance that you simply’ll hit blackjack. When the player has blackjack, the pay off odds are three:2. The dealer will only win on the other hand if he also gets a blackjack. That’s why a deck stuffed with aces is usually to the player’s advantage.
Blackjack counting is not done by memorizing each card that has come out of a six deck shoe of cards. If you’re ready to try to to that, I might be in awe: you must be in the planet records book or even even the loony bin.
Blackjack card counting is completed by assigning the different card numbers with totally different purpose values. Perpetually remember to choose a system that’s easy to remember. This can cause fewer mistakes and you’ll lose less money.
As stated above, you will assign a point value for every card and you may have to add the value of the cards that have come out. This is referred to as the running count. Based mostly on the plus/minus strategy, here are the values assigned.
two, three,four,five,half-dozen = +one;
7, eight, nine = 0;
ten, J, Q, K, Ace = -one | 489 | 2,155 | {"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} | 3.328125 | 3 | CC-MAIN-2021-17 | latest | en | 0.966997 |
https://electronics.stackexchange.com/questions/661020/control-a-large-number-of-leds-from-minimal-output-pins | 1,718,349,636,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00752.warc.gz | 194,774,924 | 40,804 | # Control a large number of LEDs from minimal output pins
I am working on a project that uses an Attiny84 to control a large number of LEDs. The number of LEDs far outnumbers the number of I/O pins. What concepts and principles should I be researching into for a solution?
• Essentially your MCU will send the data serially. There are endless ways to do it, not least of which is a basic shift register. Commented Mar 31, 2023 at 10:52
• A LED strip with hundreds of LEDs controlled by one data wire. How many LEDs you need? Commented Mar 31, 2023 at 10:52
• How to control a maximal number of LEDs with minimal output pins is a completely different and unrelated question to controlling a large number of IO with not minimal MCU pins. It has a completely different answer. Commented Mar 31, 2023 at 10:59
• Have a look at the MAX7219, which is specifically designed for this kind of thing, controlling up to 64 LEDs. Or if your design will accomodate it, WS2812 LEDs, which have circuitry already in them. Commented Mar 31, 2023 at 11:13
• If your LEDs all have the same forward voltage you could look into Charlieplexing nl.wikipedia.org/wiki/Charlieplexing
– Bart
Commented Mar 31, 2023 at 11:47
In general, your MCU will communicate serially. There are endless ways of doing this, from the simplest to the most complex.
## With Shift registers
Here's a shift register (74164)
Basically you connect CLOCK and SERIAL_INPUT_A to your CPU (/CLEAR and B to +ve). and you send your data one bit at a time. Most of the families of chips will have enough current to drive LEDs directly through a resistor; usually you use the shift register output to pull the LED to ground. This works well but if you don't clock the bits very fast you can see your LED pattern "ghosting".
To avoid that, you want the data to appear on all the output pins at the same time, and you need a latching one or one with output control, here's the very flexible 74LS595:
Texas datasheet
Here's a tutorial about using it from an Arduino.
## With Special Device
If you want brightness control or very many LEDs, you might use an array and a special chip such as the MAX 7219 or its almost-identical twim 7221. Datasheet.
Here you have very many fewer resistors and because the LEDs are in an array, you can drive 64 of them, and they are commonly used to drive, for example eight-digit 7-segment displays. These chips stack up, so you can use several in row to make a larger array of LEDs. (Indeed, you can buy LED arrays with these chips already in them.)
## With Special LEDs
There are LEDs which take serial inputs, known either as WS2812 or Neopixel. These take a serial interface directly, as each LED has its own chip to decode its signal.
## Special Lighting Protocols DMX-512
For architectural or theatre lighting, the standard is a simple serial protocol called DMX-512.
• "Unbuffered" shift registers ('164) without driver blanking/disable allow severe "ghosting". Commented Mar 31, 2023 at 13:41
• @greybeard Thanks for mentioning it, I've emphasised that in an edtir. Commented Mar 31, 2023 at 13:56 | 754 | 3,099 | {"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} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.923273 |
http://www.homemade-circuits.com/2012/10/designing-grid-tie-inverter-circuit.html | 1,493,437,406,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123270.78/warc/CC-MAIN-20170423031203-00365-ip-10-145-167-34.ec2.internal.warc.gz | 540,198,056 | 120,532 | ## Designing a Grid-Tie Inverter Circuit
A grid tie inverter works quite like a conventional inverter, however the power output from such inverter is fed and tied with the AC mains from the utility grid supply. As long as the mains AC supply is present, the inverter contributes its power to the existing grid mains supply, and stops the process when the grid supply fails.
The concept is indeed very intriguing as it allows each of us to become an utility power contributor. Imagine each house getting involved in this project to generate overwhelming amounts of power to the grid, which in turn provides a passive income source to the involved residences. Since the input is derived from the renewable sources, the income becomes absolutely free of cost.
Making a grid tie inverter at home is considered to be very difficult as the concept involves some strict criteria to be observed, not following may lead to hazardous situations.
The main few things that must be observed are:
The output from the inverter must be perfectly synchronized with the grid AC.
The output voltage amplitude and frequency as mentioned above must all correspond with the grid AC parameters.
The inverter should switch OFF instantly in case the grid voltage fails.
In this post I have tried to present a simple grid-tie inverter circuit which according to me takes care of all the above requirements and delivers the generated AC into the grid safely without creating any hazardous situations.
Let's try to understand the proposed design (exclusively developed by me) with the help of the following points:
Again, as usual our best friend, the IC555 takes the center stage in the entire application. In fact only because of this IC the configuration could become apparently so very simple.
Referring to the circuit diagram, the IC1 and IC2 are basically wired up as a voltage synthesizer or in a more familiar terms a pulse position modulators.
A step down transformer TR1 is used here for supplying the required operating voltage to the IC circuit, and as well as for supplying the synchronization data to the IC, so that it can process the output in accordance with the grid parameters.
Pin#2 and pin#5 of the both the ICs are connected to the point after D1, and via T3 respectively, which provides the frequency count and amplitude data of the grid AC to the ICs respectively.
The above two information provided to the ICs prompts the ICs to modify their outputs at the respective pins in accordance with these information.
The result from the output translates this data into well optimized PWM voltage that's very much synchronized with the grid voltage.
IC1 is used for generating positive PWM, while IC2 produce negative PWMs, both work in tandem creating the required push pull effect over the mosfets.
The above voltages are fed to the respective mosfets, which effectively converts the above pattern into a high current fluctuating DC across the involved step up transformer input winding.
The output of the transformer converts the input into a perfectly synchronized AC, compatible with the existing grid AC.
While connecting the TR2 output with the grid, connect a 100 watt bulb in series with one of the wires. If the bulb glows, means the ACs are out of phase, reverse the connections immediately and now the bulb should stop glowing ensuring proper synchronization of the ACs.
You would also want to see this simplified design
Assumed PWM Waveform (bottom trace) at the Outputs of the ICs
Parts List
All resistors = 2K2
C1 = 1000uF/25V
C2,C4 = 0.47uF
D1,D2 = 1N4007,
D3 = 10AMP,
IC1,2 = 555
MOSFETS = AS PER APPLICATION SPECS.
TR1 = 0-12V, 100mA
TR2 = AS PER APPLICATION SPECS
T3 = BC547
INPUT DC = AS PER APPLICATION SPECS.
WARNING: THE IDEA IS BASED SOLELY ON IMAGINATIVE SIMULATION, VIEWER DISCRETION IS STRICTLY ADVISED.
You would also want to see this simplified design
After receiving a corrective suggestion from one of the readers of this blog Mr. Darren and some contemplation, it revealed that the above circuit had many flaws and it wouldn't actually work practically.
The revised design is shown below, which looks much better and a feasible idea.
Here a single IC 556 has been incorporated for creating the PWM pulses.
One half of the IC has been configured as the high frequency generator for feeding the other half IC which is rigged as a pulse width modulator.
The sample modulating frequency is derived from TR1 which provides the exact frequency data to the IC so that the PWM are perfectly dimensioned in accordance with the mains frequency.
The high frequency makes sure the output is able to chop the above modulation information to precision and provide the mosfets with an exact RMS equivalent of the grid mains.
Finally, the two transistors make sure that the mosfets never conduct together rather only one at a time, as per the mains 50 or 60 Hz oscillations.
Parts List
R1,R2,C1 = select to create around 1 kHz frequency
R3, R4,R5,R6 = 1K
C2 = 1nF
C3 = 100uF/25V
D1 = 10 amp diode
D2, D3, D4, D5 = 1N4007
T1, T2 = as per requirement
T3, T4 = BC547
IC1 = IC 556
TR1, TR2 = as suggested in the previous section design
The above circuit was analyzed by Mr. Selim and he found some interesting flaws in the circuit. The main flaw being the missing negative PWM pulses of the AC half cycles. The second fault was detected with the transistors which did not seem to isolate the switching of the two mosfets as per the fed 50 Hz rate.
The above idea was modified by Mr. Selim, here are the waveform details after the modifications. modifications:
Waveform Image:
CTRL is the 100 Hz signal after the rectifier, OUT is from PWM from both halve waves, Vgs are the gate voltages of the FETs, Vd is the pickup on the secondary winding, which in sync with CTRL/2.
Disregard the frequencies as they are incorrect due low sampling speeds (else it gets too slow on the ipad). At higher sampling freqs (20Mhz) the PWM looks quite impressing.
To fix the duty cycle to 50% at around 9kHz, I had to put a diode in.
Regards,
Selim
For enabling the detection of the negative half cycles, the control input of the IC must be fed with both the half cycles of the AC, this can be achieved by employing a bridge rectifier configuration.
Here's how the finalyzed circuit should look according to me. The transistor base is now connected with a zener diode so that would hopefully enable the transistors to isolate the mosfet conduction such that they conduct alternately in response to the 50 Hz pulses at the base T4.
Hello Swags,
I have tried the zener-diode approach (no-luck), CMOS gates and, much better, op-amps worked best. I've got 90VAC out of 5VDC and 170VAC from 9VDC at 50Hz, I believe it's in sync with the grid ( can't confirm as no oscilloscope). Btw the noise goes if you clamp it with a 0.15u cap. on the secondary coil.
As soon as I put a load on the secondary coil, it's voltage drops to 0VAC with only a slight increase in input DC amps. The Mosfets don't even try to draw more amps. Perhaps some mosfet drivers like IR2113 (see below) could help?
Although in high spirits, I feel that PWM might not be as straight forward as hoped. It definitely is good to control torque on dc motors at low pwm freqs. However when the 50 Hz signal gets chopped at higher freq, it for some reason looses power or the PWMd mosfet can't deliver the needed high amps on the primary coil to keep the 220VAC going under load.
I've found another schematic which is very closely related to yours, except PWM. You might have seen this one before.
The power handling circuit is an H drive with IGBTs (we could use mosfets instead). It looks like it can deliver the power across.
It looks complicated but actually is not too bad, what do you think? I will try to simulate the control circuit and let you how it looks.
Regards,
Selim
Some very interesting modifications and information were provided by Miss Nuvem, one of the dedicated readers of this blog, let's learn them below:
Hello Mr. Swagatam,
I am Miss Nuvem and I'm working in a group that is building some of your circuits during an event about sustentable living in Brazil and Catalonia. You have to visit some day.
I've been simulating your Grid-Tie Inverter Circuit, and I'd like to suggest a couple of modifications to the last design that you had on your post.
First, I was having problems where the PWM out signal (IC1 pin 9) would just blank out and stop oscillating. This was happening whenever the Control voltage at pin 11 would go higher than the Vcc voltage due to the drop across D4. My solution was to add two 1n4007 diodes in series between the rectifier and the control voltage. You might be able to get away with just one diode, but I am using two just to be safe.
Another problem I was having was with the Vgs for T1 and T2 not being very symmetric. T1 was fine, but T2 was not oscillating all the way up to Vcc values because whenever T3 was on, it was putting 0.7V across T4 instead of letting R6 pull up the voltage. I fixed this by putting a 4.7kohm resistor between T3 and T4. I think any value higher than that works, but I used 4.7kohm.
I hope this makes sense. I am attaching an image of the circuit with these modifications and the simulation results that I am getting with LTspice.
We'll be working on this and other circuits for the next week. We will keep you updated.
Warm regards.
Miss Nuvem
1. Hi Swagatam,
what is the perfect TR2 and MOSFETS to use in this circuit?
if i have a 12-0-12 1500watts transformer, how many FETS will i use for iRFZ44n?
Regards,
1. I would suggest you to first check whether the circuit works proposed in the explanation, if it does then you can proceed with the selection of the other sections like mosfets etc.
By the way any type of N mosfet should work here, just make sure they are accurately matched with the transformer primary rating.
2. is this the circuit that can lower your electric bill?
Regards,
3. No this circuit will not lower your electric bill, but in countries where production and selling of excess electricity is allowed, this type of inverters can become a source of passive income.
4. what is the Grid-Tie inverter that i saw in you tube that the meter runs in backward? is it true or just a joke?
5. I haven't tried this circuit practically so have no idea how it would respond...
6. can i use 500mA for TR1?
7. yes, can be used.
8. Hi
You can get H bridge modules with the built in drivers and automatic dead time
Regards
Snuf
9. Hi Mr. Swagatam,
2. hi swagatam this is kurush karanjia here. This circuit of yours can be very help full in the PCU i am trying to make by synchronizing inverter output to the AC mains output and then suppling to loads. So please help and tell me how can i implement this circuit in the way i want.
1. Hi Kurush,
as mentioned in the bottom warning message, the idea has not been tested practically.
So you will have to check it practically using an oscilloscope and other equipment to confirm its working.
If the IC responds just as mentioned in the article then our job is done, we can be sure of the results.
3. thanks swagatam but some changes will have to be made to use it to supply load by synchronizing inverter and mains. so what changes will have to be made?
1. I have assumed the IC to handle a few things which have been mentioned in the article, if that happens correctly rest of the things will automatically shape up.
Synchronizing phase amplitude and frequency are the only things which are important. And also the inverter should switch OFF when mains is not available. All these things have been taken are of in the above circuit.
Unless the circuit is tested practically nothing can be said.
4. here you have shown how to supply the power back to grid after synchronization but i dont wanna supply back to the grid i want to synchronize both the supplies and then supply the load from it
1. OK, by synchronization you mean you want the inverter AC to replace the mains AC immediately when mains AC fails? Right?
2. no by synchronization i mean i want to synchronize the output of inverter and the ac mains and then supply to load when the inverter is not able to supply the load completely..
3. Yes i have tried to design exactly that in the above circuit, the ICs synchronizes the inverter AC output with the grid AC so that both merge together adding up the overall power to the appliances.
4. so please tell what should be the values of T2, mosfets and how much dc input should i use
5. T2 = BC547
mosfet = sk1058
please be cautioned: you are taking a huge risk by making something which has not been confirmed, and it involves potentially dangerous AC currents.
6. ok no problem i will take all possible precautions and will use a mcb before this entire device
And i will be directly using the inverter output so which part of the circuit should i eliminate and where to collect the load
7. and what should be the values of the TR2
8. i think by mistake you have replied to this question in another post but its ok i got what you said but my question now is that i am gonna use the inverter output directly so will need to eliminate some portion of this circuit so what portion should i eliminate...?
9. This circuit cannot be used like a normal inverter because for that we may have to eliminate a few stages and also introduce additional new stages.
You may refer to this circuit for making a simple inverter:
10. i am gonna make the inverter using another circuit and then want to synchronize the output of that inverter with ac mains and then supply to loads
5. i will try making this at home so please tell what should be the values of T2, mosfets and how much dc input should i use
1. The above circuit is of a grid tie inverter, I don't think you are looking for a grid-tie inverter circuit......???
6. The meter will run in the reverse if being fed.
I have seen grid tie inverter and they have a complex a/c synchronization ic circuit and they work in a range of the ac voltage.
Does this circuit also have a working voltage for the grid a/c and also frequency since frequency might drop due to heavy loads being used
1. As per my assumptions, the above circuit derives the frequency and the amplitude information from the existing AC grid input and processes the circuit output accordingly, therefore the circuit should be producing perfectly synchronized AC back to the grid.
2. Thanks but is there any issue w.r.t grid voltage fluctuations
3. The inverter output will follow the grid voltage data accurately so there should not be any issues.
7. Hi Swagatam,
what is the type of capacitors C2,C4?
Regards,
1. They are ordinary ceramic types.
8. Hi Swagatam,
10amps diode is not available in my area. only 6amps. can i use it?
Regards,
1. Use two 6 amps in parallel.
9. Hi Swagatam,
i almost finish with the first circuit but suddenly you modified it, what is wrong with the first circuit?
what is the function of BC547 in the modified circuit?
i would like to know the T1 and T2, is it the same value of FET's?
the T1 Source is connected to T2 Drain?
Regards,
1. I keep modifying my circuits to improve them more.
Both the ICs should produce waveforms in tandem by translating both the halves of the grid AC cycles, therefore BC547 is added so that the bottom IC is able to translate the negative half cycles into the required PWM.
T1 and T2 are same.
"sources" are made common and connected to ground, and the drains are connected to the transformer taps.
2. thanks Swagatam for your quick response...
i just want to clarify the connection of the FET's. because in the diagram my understanding the source and the drain is connected.
i can put the FET's and parallel it into one heatsink?
for testing purposes, can i use a lower wattage for a light bulb?
Regards,
3. Yes in the diagram it's wrongly shown, the source should be made common and connected to ground.
A common heatsink can be used for T1/T2
low wattage bulb can be used for testing, but without oscilloscope you cannot judge the waveforms.
Regards.
4. Hi Swagatam,
it is not sure if the wave form is sine wave even if it is tested with the light bulb?
what will happen if it is not synchronized with the local power?
i already finished with this circuit and i don't have a oscilloscope.
Regards,
5. First test the inverter separately, do not connect the output of the inverter to the grid, instead connect the output to a 100 watt bulb, next connect a DC power source may be a battery 25AH 12Vand to the shown points, check the bulb illumination.
TR1 must be connected to the AC mains as shown for the above testing.
If the bulb illuminates fully then we can proceed further.
WARNING: THE SUGGESTIONS ARE BASED SOLELY ON MY IMAGINATIVE SIMULATION.....
6. can i use DM for testing the output?
7. with DMM you cannot check RMS and amplitude....
10. Hi Swagatam,
i want to clarify the testing procedure.
While connecting the TR2 output with the grid, connect a 100 watt bulb in series with one of the wires. If the bulb glows, means the ACs are out of phase, reverse the connections immediately and now the bulb should stop glowing ensuring proper synchronization of the ACs.
- is TR1 connected to the grid?
- do i need to connect the battery?
Regards,
1. TR1 is connected to the grid so that the 555 circuit is able to assess and the synchronize the waveform accordingly.
Battery is required at the shown DC source points.
Pls read the warning message at the end of the article.
2. so it is not sure if the testing of this circuit will run and safe?
how do i test the wave form? on which point do i connect the probe?
Regards,
3. It seems you are very much new in the field so I will recommend not to build this circuit, it may be dangerous for you.
The output from TR2 can be fed to an oscilloscope to read the waveforms.
11. What kind of simulation tool did you use?
Thank you
1. mind simulation
12. Seems that pin 7 cannot go to + or it will short.
1. OK, thanks, you may add a 1K resistor in series with it.
2. Hi Swagatam,
the 1K resistor is going to the positive of the battery?
pin#4,6,7&8 of IC1,2 will put a 1K resistor going to the +?
Regards,
3. No, the 1K resistors will go from pin#7 to the postive of the respective ICs (pin#4/#8....)
4. Ok i got it... thanks
5. Hi Swagatam,
6. yes both of them should be connected to the 1K resistor.. since they are both joined together...
7. Hi Swagatam,
just for clarification...
pin# 6 and 7 with 1K resistor will connect to pin# 4 and 8? or is it going to the + of the Battery?
Regards,
8. pin#6 and 7 with 1K resistor will connect to pin# 4 and 8....NOT to the positive of the battery.....
Regards.
13. Hi Swagatam,
Been following this post for a while now and cannot beleive the amount of novice`s trying to build this un-tested project with no idea of what the components are doing. Do some research guys before you kill yourselfs.
Looked at the circuit myself and I can understand the half wave entering the chip at pin 5 which will follow the mains frequency but cannot see what you are trying to do with pin 2 as this is also at the same frequency.
You state that the chip is wired as a pulse position modulator which would be fine if it had another higher frequency either injected into pin 2 plus a timing cap or simply ocilating the chip at a higher frequency.
The tall and short of this is where is the higher frequency derived from other than the 50/60Hz that is injected ?.
Regards Darren.
1. Hi Darren,
I have already provided the necessary warning in the article and during the conversations.
Yes pin#2 needs a high frequency for enabling the PWMs at the output.
I'll reconsider the design soon and hopefully correct it.
Thanks!
14. Hi Swagatam,
Having done some more research about the 555 timer chip it seems you would have problems with the duty cycle only operating from 10%-90% which could cause both outputs to be on at once.
I am thinking that a better aproach to this problem would be op-amps.
Using something like a LM324N, 2 op-amps creating a high frequency saw-tooth and fed to the neg of the remaining two op-amps then half waves (50/60Hz) fed to the positive of the op-amps creating two seperate pwm outputs.
Maybe this would be a easier option and only one chip to buy. All off the top of my head at the moment !!!.
Hope this may help. Good Luck !.
Regards, Darren.
1. Hi Darren,
Thanks!
Yes surely opamps can be used also as suggested by you, but since I have already done some hard work in studying IC555, in my new design I employed them yet again, and hopefully eliminated the previous issues in this design.
I have presented the revised design in the above article, I hope this time you won't have any trouble with it :)
Regards.
2. Hi Swagatam,
Good work!
The circuit is shaping up to be a buildable project of which I might even have a go at for my 100w turbine. MANY THANKS!.
Only one error i can see is that T3 and T4 emitters I reckon should go to the low voltage earth point and not the LIVE mains section.
Keep up the good work and thanks again.
Regards Darren.
3. Hi Darren,
Thanks for correcting the diagram yet again, indeed it was a silly mistake from my part, I have corrected it now.
Thank you for making this happen!
15. Hi Swagatam,
May i know if the connection of t1 and t2 is correct? The drain of t2 connected to source of t1?
Regards,
1. Thanks, Good observation!
No it's not correct, the sources should be made common and connected to ground and the drains should go to the taps of the transformer...I'll have to correct the diagram one more time...I'll do it soon hopefully.
16. Hi Swagatam,
to get the 1KHz frequency for the value of R1,R2 and C1. what would be the duty cycle?
Regards,
1. good question! I think it should be preferably 50%, actually I am not very sure, will need to be investigated deeply.
2. Hi Swagatam,
This is what i got in my 555 calculator:
Frequency: 1kHz
Duty Cycle: 50%
Period (T) = 1 Milliseconds
Mark = 500 Microseconds
Space = 500 Microseconds
R1 = 7.143 Kilohms
R1 Preferred Value = 6.8 Kilohms
R2 = 71.429 Kilohms
R2 Preferred Value = 68 Kilohms
what about the testing of this circuit?
Regards,
Mike L
3. Thanks Mike!
That looks neat.
testing may be done just like an ordinary inverter by initially keeping the TR2 output disconnected with grid mains.
Once the waveforms, frequency and power are confirmed using appropriate equipment, the output may be connected to the mains with correct polarity.
17. Hi Swagatam,
what about the testing of the new circuit? is it the same with the first circuit?
Regards,
1. yes it will be the same...
18. Hi Swagatam, great project, I tried the schematic on a breadboard and it worked so far. I changed T2 drain/sorce pins, used IRF740PbF Mosfets and 15VA toroidal TR2. I used a bench power supply as a DC source and went up to 5VDC/0.04A, which produced a 80VAC. Didn't go any higher as the noise levels went up (around 0.8kHz). Also, I did not connect TR2 to TR1 as I have no oscilloscope to confirm wave form and phase. R1 was 6.8k, R2 68k, C1=10n and R7 was 1k (it wasn't mentioned). In your prev. schematics you used capacitors beteween R5 and T1 and (R6 and T2). Could this filter out the 0.8-0.9kHz noise from the transformer TR2? Also, would you suggest that running the timer from a stabilised power source like 78L12 could reduce the noise while feeding in the control frequency through an optical coupler? Could an optical coupler provide both frequency and amplitude data? Thanks for your advise. Br, Selim
1. Hi Salim,
Without an oscilloscope it would be quite dangerous to go head with this circuit...... for reducing the noise you may try increasing the frequency to above 20 kHz by reducing he value of C1.
Regards.
19. Hi Swagatam,
is there any way to check if LM556 is working?
i built this circuit but i got a low output voltage to tr2.
Regards,
Mike L
1. Hi Mike,
Since the output is a PWM, it cannot be checked with a DMM, you will have to incorporate an oscilloscope.
what output are you getting?
Regards.
2. Hi Swagatam,
i only got 6vac...
i will try to find first a oscilloscope to test it again.
Thanks and Regards,
Mike L
3. Hi Mike,
6V at the output of the transformer or at the output of the IC??
4. Hi Swagatam,
6v at the output of tr2.
Regards,
Mike L
5. Hi Mike,
check the voltage at the "PWM out" of the IC.
6. Hi Swagatam,
I still dont have an oscilloscope. Can it be tested with the DMM? What woulb be the output voltage of the ic?
Regards,
7. yes, check the voltage with a DMM at the output of the IC....
8. what is the voltage at the output of the IC?
9. I would like to know it from you....
20. hi swagatam !
great project !
we want to know if is posible to get 240V and 50 hz output
thanks
regards ! !
1. Thanks! since the input is synchronized with the grid voltage, the frequency would be followed accurately..... wattage can be upgraded by using appropriately rated mosfets
21. Those who are asking most of these newbie questions do not know enough to do this safely. Many dangers lurk here, electrocution, fire, damage to connected equipment, etc. Great project, but seriously not one to undertake without a good understanding of whats going on!
22. Hi Swagatam, could you please help me understand the function of T3/4. I know it's important that both FET-gates should not be positive at the same time. Using breadboard, and iCircuit simulator, I always got the same results. Only one FET was operating (10V pulse at gate) and the other one at 0.8V pulse at gate, not enough to open the gate (>2V) . So I tried using 3 NOR gates of a 4001 and it switched sides at 50Hz. Was this the purpose of T3/4 or did you put them in to prevent the FETs T1/2 conducting at the same time only? BTW, can I contact you by email?
Regards,
Selim
ps disregard prev msg
1. Hi Selim,
That's correct, T3/T4 has been introduced so that the mosfets conduct alternately and never together.
23. Thanks, Swagatam.
Somehow T3/4 won't work, but with 4001 it will alternate at 50Hz.
On iCircuit (iPad) and on the breadboard, I get at the gates: T1=+12.2V, and T2=+0.845V PWM-signals, same phase, not alternating between T1/2.
So effectively only one side of the primary winding coil (6V-0-6V) is energised. Leaving T3/4 running on the breadboard, using 2.5VDC/0.05A input, I get around 90VAC at 820-870Hz (wobbling at around 50Hz, I think).
Using NORs from 4001, I can get it alternated in sync between T1/2, now here comes the next question.
On iCircuit, I noticed that the positive wave gets PWM modulated properly where as the negative wave (or the missing bit/flat line between 2 positive waves) is only PWM modulated as a series of short pulses (no width/no power). The output would be distorted, such as larger positive waves vs. smaller negative waves. Is there a way to capture the negative side as well? Maybe another NE555/6 could do this job using the same 1kHz chopper.
Please keep this project up. It seems like it could work well.
Regards
1. Thanks for the observations and suggestions Selim!
OK, If cmos gates work better we can use them instead of the transistors, no issues.
In the first diagram I tried to take care of the negative wave forms also but in the improved version I completely forget about it.
We'll have to incorporate another IC555 for this, and feed the two mosfets with separate signals from the two 555 ICs.
I'll try to present the finalized design soon...
Regards.
2. Hi Swagatam, Any update on promising finalized design diagram with another incorporated IC555?
I was building one when your post came out, so now I'm waiting. Please!
3. Hi Kevin,
The design was updated a long time ago, an additional 555 was not felt necessary as the negative half cycles could be simply derived by using s bridge rectifier network.
please refer to the second last diagram which hopefully satisfies all the conditions.
4. ....sorry, I misspelled you name.
24. Hi Swagatam, I managed to pickup both waves, and PWMed them and then run through 4013 to divide the 100Hz by two. The results look good on iCircuit. Both wave halves are PWMed at 9kHz and alternated on T1/2 evenly. The secondory winding phase in-sync with control input. I'll email you the pics later today. Two 4001 can be used iso 4013, but would require more wiring on the breadboard is needed. I think two 555s could do a much better job. I haven't figured out how to feed a 555 with a negative wave yet.
Regards,
Selim
1. Hi Selim,
Thanks! it would be interesting to see how you derived both the halves of the AC for the required sync. will wait for the images.
25. Hi Swagatam,
I got a little concern about what would happen if this circuit was disconected from the mains but still getting a DC input from say a solar panel/turbine. The way i see it is that TR2 will keep trying to feed TR1 and the whole circuit will go out of control resulting in the end of the plug being live. A normal mains failure will most likely drown TR2 out due to not being able to feed a whole city !!.
Hope i am missing something here.
Best regards and keep up this project.
1. Thank you for the interesting observation!
However if TR2 drowns, TR1 would also stop producing the required data supply to the circuit, and the whole system would stall:-)
But yes your point is noteworthy, and the issue should be checked practically for a reliable solution.
Regards.
26. Hi Swagatam,
What is the value of the zener diode?
Maybe we can put a relay to cut the source coming from tr2.
Regards,
Mike L
1. Hi Mike,
It can be anything from 3V to 8V....we just want to delay the switching fractionally between the transistors.
Yes a relay would be a good idea for correcting the above issues.....
Thanks.
27. Hi Swagatam,
what is the value of zener diode?
R1,R2 and C1? what is the frequency to be use?
Regards,
Mike L
1. Hi Mike,
You will have to calculate them using any online 555 calculator software.
Regards.
2. Is it 1kHz or 20kHz?
3. It has to be a high frequency, the figure does not matter...
4. I am using 1K (R7) resistor and a 5.6 volts zener.
5. That should be OK, the idea is to keep the transistor conductions well isolated, and in an alternate manner.....the zener diode makes sure that the transistor conductions do not overlap (as far as I thing).
28. Hi Swagatam,
what is wrong if the voltage coming out to the PWM out of IC 556 is not stable? i measured it from 10V and then going down to 3V.
Pin#3 (not connected)output is stable to 10V.
Regards,
Mike L
1. Hi Mike,
The output PWM should vary with the control voltage......
29. hi i also trying to do one
can i send it to u
30. http://imageshack.us/photo/my-images/694/img3181oi.jpg/
hi this is the link for my diagram
regards dumidu
1. Good design should work, thanks!
31. can be used for recharge baruti.themigambo @ yahoo.co
32. Hello Sir,
I have a solar panel of 150W, 24V. It is currently hooked up by a charge controller and a lead acid battery. I just need an inverter which directly converts pv supply (24v dc) into 220v ac, without having a battery.
I only need to run an 80w fan on it. Can you please help me in this regard.
Thank you.
Rashid
1. Hello Rashid,
you can make the following circuit, just change the transformer to a 20-0-20V/5 amp.......any nearby type will also do:
can i increase its wattage by adding more mosfets in parallel??
Rashid.
3. Thanks Rashid, yes it can be done.
33. Hi Swagatam
What is the u1 that part no is not clear
Bandula
1. Hi Bandula,
Where's it? can't see it.
34. Hi Sir,
I want final circuit diagram and component list of Grid-Tie Inverter for solar panel for 70-80 watt in home. And give me cost for designing this circuit. Plz reply me sir....
Mohini
1. Hi Mohini,
The second last circuit is the finalized design, but please note that it's only the concept which I have tried to produce, the person who intends to build it should have a sound knowledge regarding electronics and inverters, only then will he or she be able to implement the design correctly for practical feasibility.
If your not an expert in the field please don't even think of making such complex circuits, you will end up wasting a lot of money and time.
2. hi,i have solar panel 58v amorfni 128w 4 pieces do i go in serial 220v then adjust your circuit for it (220vdc) or do i make in paralel 4 x 58v then adjust your circuit for 58v what do you mean by that two cobination?
3. Hi,
the above design is yet to be verified, so I would recommend you to first test the second last design shown above, if it performs as per the expectations, you could go on upgrading it from there....
By the way, putting solar panels in series makes more sense and is always more efficient.
35. Hello Mr. Swagatam,
I am Miss Nuvem and I'm working in a group that is building some of your circuits during an event about sustentable living in Brazil and Catalonia. You have to visit some day.
I've been simulating your Grid-Tie Inverter Circuit, and I'd like to suggest a couple of modifications to the last design that you had on your post.
First, I was having problems where the PWM out signal (IC1 pin 9) would just blank out and stop oscillating. This was happening whenever the Control voltage at pin 11 would go higher than the Vcc voltage due to the drop across D4. My solution was to add two 1n4007 diodes in series between the rectifier and the control voltage. You might be able to get away with just one diode, but I am using two just to be safe.
Another problem I was having was with the Vgs for T1 and T2 not being very symmetric. T1 was fine, but T2 was not oscillating all the way up to Vcc values because whenever T3 was on, it was putting 0.7V across T4 instead of letting R6 pull up the voltage. I fixed this by putting a 4.7kohm resistor between T3 and T4. I think any value higher than that works, but I used 4.7kohm.
I hope this makes sense. I am attaching an image of the circuit with these modifications and the simulation results that I am getting with LTspice.
http://i.imgur.com/jkOoU8k.png
http://i.imgur.com/XpOsQKd.png
http://i.imgur.com/9uv0TGx.jpg
We'll be working on this and other circuits for the next week. We will keep you updated.
Warm regards.
Miss Nuvem
36. Dear Sir
I have to Sun grid tie inverters. One input 10-30 volts and other 22-60.
For my wind turbine I have conclude the best range is 15-40 volts. Is suitable to change them to thiese range? If yes how?.
Best regards
Jose Castanho - Portugal
Email: castanho.jose@gmail.com
1. Dear Jose,
Changing inverter voltage specs would be difficult, you will have to dimension your windmill output as per the inverter requirements, may be by using a buck-boost circuit or other similar concept.
37. Hi Mr Swagatam,
Ryner here, I'm a student doing GTI for my Final Year Project. I would like to raise a more theoretical question: why is there a need to synchronize the output of the GTI with the grid? What would happen if it is not synced?
1. Hi Ryner,
since the involved current is varying from positive to negative every split second, if the two counterparts are not synchronized, would result in possible clashing of their opposite cycles in their AC phases which would in return result either in short circuiting of the power or one of the counterparts dragging the power toward lower levels thereby heating up the inverter devices.
That's why perfect synchronization becomes the ultimate crucial thing in GTI systems.
38. Hi Mr Swagatam,
Ryner here again, sorry to trouble you with another question, but from what I understand, the output of the GTI has to be channeled back to the grid. Does this mean that my output from the GTI should have a slightly higher amplitude compared to that of the grid? (by theory, a potential difference is required for current to flow) On that note, are we transmitting energy in terms of voltage or current? Thank you in advance!
1. Hi Ryner,
Imaging putting batteries in parallel for increasing the backup time, yes, it's the current that is being added here.
If we increase the inverter voltage would load the inverter more than the AC mains input, so both counterparts must bear equal and uniform potential sharing.
39. Hi Mr Swagatam,
Ryner here once again. How exactly do I get the output waveform to synchronize with the grid? Would I be able to do it with a Phase Lock Loop (PLL) Chip? Is there a simpler way to do the synchronization process? Thank You!
1. Hi Ryner,
I have already taken care of and have correctly implemented waveform synchronization in the explained designs,
regarding other options I cannot say much, that will require further investigation.
40. I am doing my final year project with renewable energy, i prefer doing this circuit. I am doing it for a grid of 230V what must be my transformer TR2's ratings? Can i have the full final circuit?
1. The final circuit is already updated in the above article. The concept was designed by me but is not yet tested practically, so pls be cautioned
TR2 wattage will be equal to the required power output from the inverter
2. How will this circuit behave to different voltages from the dc source, however its not a battery being connected to deliver a constant voltage?
3. the DC input could be from a battery or any regulated source. yes we'll have to add a comparator sensing stage for ensuring that the battery supply is cut off as soon as its voltage drops below the input DC ripple voltage from TR1 at pin#11 of the IC
41. Dear sir
In India THD,Freq,voltage all these thing at a time varries, then that how can adjust...if always less then 10 % thd...what is effect on perfomance..of GTI OR ONLOAD
1. Dear Dharamveer, if the DC input to the circuit is maintained at some constant, safe higher level say at 15V for a 12V system, the circuit will automatically take care of all the other factors associated with the mains supply (according to me)
42. dear sir
how can adjust GTI, THD, freq. voltage always varries....thd more then 5%
43. Sir, I need 10 Kw grid tied three phase solar inverter circuit diagram. If you have please send me on "ajinkya.brahma@gmail.com". If you have circuit giagram, please send me as early as possible.
1. Presently i don't have it, I'll try to find it.....
44. hi Swagatam Majumdar
can I connect directly to the solar dc source. can save a lot of costs.
but I have one problem, if larger loads feeds from tie circuit .circuit continues to operate or shut down
thanks
1. Hi Khang,
yes you can connect it directly to a solar panel, however the supply to the ICs must not exceed 15V.
The circuit output will follow the grid voltage proportionately...if the grid voltage drops so will the output from the inverter and vice versa...so with larger loads the output will adjust in accordance with the grid voltage.
45. hi Swagatam Majumdar
iI tested the circuit on the output voltage but only 4vac, f = 1kHz
I use fe core transformer (EI). transformer frequency more than. 100hz. perianth do fire transformer
can i change BC547 to 2N3904 ?
1. BC547 can be changed to 2N3904, .........cant troubleshoot without seeing practically.
46. hi Swagatam Majumdar
can you help me design and calcular an turbin wind capacity 200w.
thanks so much
1. hi khang, pls give complete specification in terms of voltage ad current.
2. Hi Swagatam Majumdar
i need technical specifications
U=24VAC
I=10A
V=0~10m/s (wind speed)
3. do you need the technical specification of the motor??
4. hi Swagatam Majumdar
yes i want it.
choice and calcular :wire size(mm),round wire,magnet.
5. khang, you can refer to this link:
http://www.alibaba.com/product-detail/YC-Series-Single-Phase-24vac-motor_1535220820.html
47. hi Swagatam Majumdar
my problem.
DC input 12v (power supply DC 3A)
transformer 12-0-12
i choice R1=100k,R2=20k,C1=10nF(~1khz)
output(pin 9 IC)=988hz,
current=0.01(DC IN)
output transformer=4VAC,frequency=100hz
Can you give me a guide
thanks
1. Khang, use a 6-0-6V transformer, you will get the required level of output..........but 4v is way too low... something may be wrong in your construction
2. hi Swagatam Majumdar
i send to you a picture result I measured.
3. email is in the "contact" page, pls see it.
48. HI Swagatam Majumdar
I checked the construction .output transformer 200vac.I have a problem ,output frequency is not 50hz .it's >300hz.
can you give me a opinions
1. Hi khang,
I can't get the picture..please post all the details regarding what you are referring to and what exactly have you made so far.
49. ok i understood your point....connect a 1uF/400 cap at the output and then check the frequency, it will show 50Hz.
50. Hi Swagatam Majumdar
Have you received a picture yet.
I have checked output bulbs but no light .it's synchronization phase on grid.
thanks
1. Hi Khang, yes i have seen the pictures sent by you.
I think you should first make the inverter separately, make it to produce the required 220V, once it's confirmed you can proceed with the integrations.
2. Hi Swagatam Majumdar
I sent you that image on meter
3. Hi Khang, yes, I have seen it but could not interpret much...
4. Hi sir
I checked the device by connecting 1 100w bulbs .It's not light . circuits sync phase grid.can I connect on grid?
thank's
5. Hi khang,
something may be wrong in your circuit or you may be using a small battery/trafo..., try using a 12V 25ah battery and a 9-0-9/20amp transformer for getting a valid output
6. Hi sir
I'm using one 3A power supply and 1A tranfo 18-0-18 to check circuit.result check in meter
Thank's
7. for 18-0-18 trafo you will need 24V DC input....... and in that case the mosfet gates will need zener diode clamping for safety
8. hi sir
I'll check back
thank's,
9. Hi Swag
i have a question?
Why create 1kHz instead of 50 hz in output ?
10. khang, 1khz at pin14 of IC 4017 will produce 50Hz on the gates of the mosfets and from the trafo, that's how it works
11. Hi Sir
schematic diagram not ic 4017.
output pin 9 Ic 556 is 1khz.output tranfo 220v 1khz.can't I connect to grid?
12. OK sorry, i confused it with the other circuit...the frequency will is locked with grid frequency so it will be always 50Hz at the trafo output, the 1khz is for generating the PWMs, it's not relevant to the output frequency.
13. hi sir
saving bill
thanks so much.
51. hi Swaggatam, Quick question irfz44n will be ok with 12 0 12..yeah..im going to check if this gonna work. or have you any latest update on grid tie inverter. Thanks
1. Hi Barbe, yes IRF44n will work with 12-0-12 trafo.
Please proceed only if you have understood the concept well, and it's strongly recommended to do it in a stagewise manner
No I do not have any further updates regarding the above circuit, I would appreciate if anybody provides the test results of the same.
2. Sure...the only missing with me is the 68k i hope i can find this afternoon and try to blow in the evening..anyway i let you know once it work smoothly..also i found some diagram from the net but not sure with it...but too many blogs says it works???
3. OK, great!
Try only those which you understand and are able to simulate in mind, otherwise it could be a waste of money and time.
4. I try.. Its busted...any way its fine...im waiting for my scope..its hard to check all of this..!!! I will try your other design and let see we were going to blow that one aswell...i will inform you what ever i found...but this one...my 10amps transformer get hot aswell..its reallly busted.i want to follow only what is in the doagram to check if its actually working but i put all the nessisary precaution..!!!!
52. Hi Miss Nuvem,
Is your Project works...as currently I don't have the scope...with me? Thanks
53. hi Swagatam
I have made such a design above. but I do not see the output of the R5 and R6 using osiloscop. and I have tried to synchronize between the inverter and generator 1 phase using a 30 watt lamp. I see there is a change from the original load dim lights become bright.
when there is no load voltage and frequency phenomenon after the sync is 220 volts to 210 volts and 50Hz be 49.5 Hz. but the mosfet is used even though I use a cooler heat. what should I do to reduce the heat? and what will not be a burden inverter generator?
1. Hi Agung,
Please refer to the following article, I have addressed the heating issue at the end of the article:
54. I want to ask you a schematic, for it is connected to the battery TR2 or mosfet?
55. Great job!... This idea can be extrapolated by the class "D" amplifier implementation with AC grid reference signal employing comercial ASIC!
1. Thanks for the great suggestion! yes it could be feasible.
56. "The inverter should switch OFF instantly in case the grid voltage fails." Why Sir?
1. to make the Grid free from an AC mains potential which could otherwise kill or hurt an electrician who may be possibly at work across a faulty grid line without any caution, assuming the line to be switched OFF and safe.
57. hi swagatam.
i cannot find in your post. what zener diode should use. and how much power it could handle.?
1. Hi kikiloaw,
D10 in the last diagram can be a 6V zener diode, 1/2 watt rated
58. As I understand about the diagram. Any DC voltage can be a source. But depend on how many volts your MOSFET can handle right? And the transformer is depend also of how many power the your DC source can give right?
1. yes that's right....
2. as the mains voltage drops the inverter voltage also drops. when the inverter voltage drop. does it affect the power to be feed to the mains grid? and what is the efficiency of this inverter? and why does the tr2 only 6-0-6 ct?
3. I have only presented the concept only, it needs to be tested practically for all these answers
59. Hi Mr. Swagatam and Miss Nuvem,
Mr Swagatam a great design which was implemented by you. Miss Nuvem, is it possible that I can have the .asc file of your schematic? Because me too tried this in LTPice but output simulation was not as I expected. Thank you
Regards
Lovindu
1. You are welcome Lovindu, I hope Miss Nuvem will respond and fulfill your request.
By the way simulation softwares are never reliable and are no match to a human brain simulation...
60. Dear Sir,
Mr swagatam
good day.
My name is Zeal Njoku
i have been trying to have you in discussion, i am lover of electronics and i have bin going through your circuit designs so i picked interest on them. I must appreciate you for being so king and a generous man and having such time and giving people all this lectures, i say thanks to you.
I have some issues that i will like to discuss with you out of this site, i have some ideas that i brought up to ,my self about most inverter circuit which i want to try but i will like discuss it with you before i will start it then after i found possibility in it after practicing for success it can be post to your site and the explanation and diagram can be uploaded to your site so that all can see.
I will appreciate if you will send to me or type to me in my email address on zealthehightech@yahoo.co.uk so that we can discuss this issue at mail first before it can come up to this site.
Thanks as i waits to see your mail soonest.
zeathehightech@yahoo.co.uk
1. Dear Zeal,
I appreciate your interest, however due to lack of time and many comments pending it would be difficult for me to discuss through email.
So please feel free express your doubts here, I'll try to help
61. dear swagatam,
A great design which was implemented by you,i need a big help from you.I have dsp sine wave solar Hybrid inverter(offgrid) design and all.I want to convert my solar inverter in to bidirectional inverter.That means my inverter output 230v will be fed to ac grid.kindly help me
1. Thanks greens, it can be a little difficult,because to convert a normal inverter into a GTI would require modifying the internal circuitry of the unit and insert an external PWM to it with a duty cycle synchronized with the grid AC
2. Hi Swagatam continueing with Mr Green comment so which internal part of an off grid have to be modified for a GTI?
I half an ups and thought if I could use it's circuit for a GTI.
Thanks
3. Hi Aborg, the mosfet section of the uPS will need to be synchronized by appropriately tailoring its conduction rate with respect to the AC waveform...it will need to be done quite as shown in the second schematic from this article | 11,953 | 48,580 | {"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} | 2.5625 | 3 | CC-MAIN-2017-17 | longest | en | 0.939469 |
http://math.stackexchange.com/questions/172454/is-there-any-graphical-explanation-of-the-derivative-of-sin-x/172466 | 1,469,751,820,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829320.91/warc/CC-MAIN-20160723071029-00156-ip-10-185-27-174.ec2.internal.warc.gz | 161,408,241 | 17,585 | # Is there any graphical explanation of the derivative of $\sin x$? [duplicate]
This question already has an answer here:
I'm trying to understand in a practical/graphical view the derivative of $\sin(x)$ (that results in $\cos(x)$).
Is there any animation or illustration explaining that?
-
## marked as duplicate by Hans Lundmark, Asaf Karagila, TZakrevskiy, Aaron Maroja, Jack D'Aurizio calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 7 '15 at 20:10
I don't know of any animations so you might have to do it yourself. Draw a nice sine curve, draw tangents to it at lots of places, approximate their gradients, and plot those values on another graph. It will look (very satisfyingly if you do it well) much like a cosine curve. – Ragib Zaman Jul 18 '12 at 15:47
I know this isn't graphical, but if you're trying to understand why $\frac{d}{dx}\sin(x) = \cos(x)$ instead of just accepting this as "the rule", then why not expand sine in a Taylor series and differentiate term by term? Then see what you have left. – Derek Allums Jul 18 '12 at 15:52
Please check this demonstration for graphical "proof" of $\sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$. This readily gives the derivative, putting $\alpha=x$ and $\beta = \Delta x$. – Sasha Jul 18 '12 at 15:54
MIT OCW's single variable calculus course has a interactive mathlet explaining the derivates of sines and cosines (and few others) graphically. Please refer to the worked example at here.
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Wondeful! By the way, I think you mean "refer to the Mathlet". Thanks! – Tom Brito Jul 18 '12 at 17:49
Think in polar coordinates.
Draw a circle centered at the origin. Pick an angle $\theta$ on the circle. Draw a ray between the origin and the circle with that angle.
Draw the tangent line on the circle there.
Now, draw a line through the origin parallel to that tangent line. Where does it intersect the circle? At exactly $\cos(\theta)$.
This works because the tangent line on the circle is normal to the ray leading to that point. That means that the line we draw through the origin also intersects that ray at right angles.
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http://functions.wolfram.com/EllipticIntegrals/EllipticF/06/01/07/0002/ | 1,519,013,140,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812327.1/warc/CC-MAIN-20180219032249-20180219052249-00499.warc.gz | 144,319,080 | 10,050 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
EllipticF
http://functions.wolfram.com/08.05.06.0048.01
Input Form
EllipticF[z, m] == (-(1/Sqrt[m])) EllipticK[1/m] (-(-1)^Floor[Arg[z - Subscript[z, 0]]/Pi] + (-1)^Floor[1/2 - Arg[z - Subscript[z, 0]]/Pi]) + EllipticK[m] (-(-1)^Floor[Arg[z - Subscript[z, 0]]/Pi] + (-1)^Floor[1/2 - Arg[z - Subscript[z, 0]]/Pi] - (-1)^(Floor[3/4 - Arg[z - Subscript[z, 0]]/(2 Pi)] + Floor[1/4 + Arg[z - Subscript[z, 0]]/(2 Pi)]) + 2 Round[Re[Subscript[z, 0]]/Pi]) + (z - Subscript[z, 0])/Sqrt[1 - m] + (1/Sqrt[1 - m]) Sum[((I^(k - 1) Binomial[k - 1/2, k - 1])/k!) Sum[((((-1)^q Binomial[k - 1, q])/(2 q + 1)) Sum[Binomial[q, j] (-m)^j (2 - m)^(q - j) 2^(k - j - q - 1) Sum[Binomial[j, i] (2 i - j)^(k - 1) (z - Subscript[z, 0])^k, {i, 0, j}], {j, 0, q}])/(1 - m)^q, {q, 1, k - 1}], {k, 2, Infinity}] /; (z -> Subscript[z, 0]) && Subscript[z, 0] == (3 Pi)/2 + 2 Pi u && Element[u, Integers] && Element[m, Reals] && m > 1
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["EllipticF", "[", RowBox[List["z", ",", "m"]], "]"]], "\[Equal]", RowBox[List[RowBox[List[RowBox[List["-", FractionBox["1", SqrtBox["m"]]]], RowBox[List["EllipticK", "[", FractionBox["1", "m"], "]"]], RowBox[List["(", RowBox[List[RowBox[List["-", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["z", "0"]]], "]"]], "\[Pi]"], "]"]]]]], "+", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "2"], "-", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["z", "0"]]], "]"]], "\[Pi]"]]], "]"]]]]], ")"]]]], " ", "+", RowBox[List[RowBox[List["EllipticK", "[", "m", "]"]], RowBox[List["(", RowBox[List[RowBox[List["-", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["z", "0"]]], "]"]], "\[Pi]"], "]"]]]]], "+", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "2"], "-", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["z", "0"]]], "]"]], "\[Pi]"]]], "]"]]], "-", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List[RowBox[List["Floor", "[", RowBox[List[FractionBox["3", "4"], "-", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["z", "0"]]], "]"]], RowBox[List["2", " ", "\[Pi]"]]]]], "]"]], "+", RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "4"], "+", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["z", "0"]]], "]"]], RowBox[List["2", " ", "\[Pi]"]]]]], "]"]]]]], "+", RowBox[List["2", " ", RowBox[List["Round", "[", FractionBox[RowBox[List["Re", "[", SubscriptBox["z", "0"], "]"]], "\[Pi]"], "]"]]]]]], ")"]]]], "+", FractionBox[RowBox[List["z", "-", SubscriptBox["z", "0"]]], SqrtBox[RowBox[List["1", "-", "m"]]]], "+", RowBox[List[FractionBox["1", SqrtBox[RowBox[List["1", "-", "m"]]]], RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "2"]], "\[Infinity]"], RowBox[List[FractionBox[RowBox[List[SuperscriptBox["\[ImaginaryI]", RowBox[List["k", "-", "1"]]], " ", RowBox[List["Binomial", "[", RowBox[List[RowBox[List["k", "-", FractionBox["1", "2"]]], ",", RowBox[List["k", "-", "1"]]]], "]"]]]], RowBox[List["k", "!"]]], RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["q", "=", "1"]], RowBox[List["k", "-", "1"]]], RowBox[List[FractionBox[RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "q"], " ", RowBox[List["Binomial", "[", RowBox[List[RowBox[List["k", "-", "1"]], ",", "q"]], "]"]]]], RowBox[List[RowBox[List["2", " ", "q"]], "+", "1"]]], SuperscriptBox[RowBox[List["(", RowBox[List["1", "-", "m"]], ")"]], RowBox[List["-", "q"]]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "q"], RowBox[List[RowBox[List["Binomial", "[", RowBox[List["q", ",", "j"]], "]"]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["-", "m"]], ")"]], "j"], " ", SuperscriptBox[RowBox[List["(", RowBox[List["2", "-", "m"]], ")"]], RowBox[List["q", "-", "j"]]], " ", SuperscriptBox["2", RowBox[List["k", "-", "j", "-", "q", "-", "1"]]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["i", "=", "0"]], "j"], RowBox[List[RowBox[List["Binomial", "[", RowBox[List["j", ",", "i"]], "]"]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["2", " ", "i"]], "-", "j"]], ")"]], RowBox[List["k", "-", "1"]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", SubscriptBox["z", "0"]]], ")"]], "k"]]]]]]]]]]]]]]]]]]]]]]], "/;", RowBox[List[RowBox[List["(", RowBox[List["z", "\[Rule]", SubscriptBox["z", "0"]]], ")"]], "\[And]", RowBox[List[SubscriptBox["z", "0"], "\[Equal]", RowBox[List[FractionBox[RowBox[List["3", "\[Pi]"]], "2"], "+", RowBox[List["2", "\[Pi]", " ", "u"]]]]]], "\[And]", RowBox[List["u", "\[Element]", "Integers"]], "\[And]", RowBox[List["m", "\[Element]", "Reals"]], "\[And]", RowBox[List["m", ">", "1"]]]]]]]]
MathML Form
F ( z m ) ( 2 Re ( z 0 ) π + ( - 1 ) 1 2 - arg ( z - z 0 ) π - ( - 1 ) arg ( z - z 0 ) π - ( - 1 ) arg ( z - z 0 ) 2 π + 1 4 + 3 4 - arg ( z - z 0 ) 2 π ) K ( m ) - 1 m ( - ( - 1 ) arg ( z - z 0 ) π + ( - 1 ) 1 2 - arg ( z - z 0 ) π ) K ( 1 m ) + z - z 0 1 - m + 1 1 - m k = 2 k - 1 k ! ( k - 1 2 k - 1 ) TagBox[RowBox[List["(", GridBox[List[List[TagBox[RowBox[List["k", "-", FractionBox["1", "2"]]], Identity, Rule[Editable, True]]], List[TagBox[RowBox[List["k", "-", "1"]], Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] q = 1 k - 1 ( - 1 ) q 2 q + 1 ( k - 1 q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox[RowBox[List["k", "-", "1"]], Identity, Rule[Editable, True]]], List[TagBox["q", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] ( 1 - m ) - q j = 0 q ( q j ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["q", Identity, Rule[Editable, True]]], List[TagBox["j", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] ( - m ) j ( 2 - m ) q - j 2 k - j - q - 1 i = 0 j ( j i ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["j", Identity, Rule[Editable, True]]], List[TagBox["i", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] ( 2 i - j ) k - 1 ( z - z 0 ) k /; ( z "\[Rule]" z 0 ) z 0 3 π 2 + 2 π u u TagBox["\[DoubleStruckCapitalZ]", Function[List[], Integers]] m TagBox["\[DoubleStruckCapitalR]", Function[List[], Reals]] m > 1 F ( z m ) ( 2 Re ( z 0 ) π + ( - 1 ) 1 2 - arg ( z - z 0 ) π - ( - 1 ) arg ( z - z 0 ) π - ( - 1 ) arg ( z - z 0 ) 2 π + 1 4 + 3 4 - arg ( z - z 0 ) 2 π ) K ( m ) - 1 m ( - ( - 1 ) arg ( z - z 0 ) π + ( - 1 ) 1 2 - arg ( z - z 0 ) π ) K ( 1 m ) + z - z 0 1 - m + 1 1 - m k = 2 k - 1 k ! ( k - 1 2 k - 1 ) TagBox[RowBox[List["(", GridBox[List[List[TagBox[RowBox[List["k", "-", FractionBox["1", "2"]]], Identity, Rule[Editable, True]]], List[TagBox[RowBox[List["k", "-", "1"]], Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] q = 1 k - 1 ( - 1 ) q 2 q + 1 ( k - 1 q ) TagBox[RowBox[List["(", GridBox[List[List[TagBox[RowBox[List["k", "-", "1"]], Identity, Rule[Editable, True]]], List[TagBox["q", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] ( 1 - m ) - q j = 0 q ( q j ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["q", Identity, Rule[Editable, True]]], List[TagBox["j", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] ( - m ) j ( 2 - m ) q - j 2 k - j - q - 1 i = 0 j ( j i ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["j", Identity, Rule[Editable, True]]], List[TagBox["i", Identity, Rule[Editable, True]]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]], Rule[Editable, False]] ( 2 i - j ) k - 1 ( z - z 0 ) k /; ( z "\[Rule]" z 0 ) z 0 3 π 2 + 2 π u u TagBox["\[DoubleStruckCapitalZ]", Function[List[], Integers]] m TagBox["\[DoubleStruckCapitalR]", Function[List[], Reals]] m > 1 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["EllipticF", "[", RowBox[List["z_", ",", "m_"]], "]"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[RowBox[List["-", FractionBox[RowBox[List[RowBox[List["EllipticK", "[", FractionBox["1", "m"], "]"]], " ", RowBox[List["(", RowBox[List[RowBox[List["-", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["zz", "0"]]], "]"]], "\[Pi]"], "]"]]]]], "+", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "2"], "-", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["zz", "0"]]], "]"]], "\[Pi]"]]], "]"]]]]], ")"]]]], SqrtBox["m"]]]], "+", RowBox[List[RowBox[List["EllipticK", "[", "m", "]"]], " ", RowBox[List["(", RowBox[List[RowBox[List["-", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["zz", "0"]]], "]"]], "\[Pi]"], "]"]]]]], "+", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "2"], "-", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["zz", "0"]]], "]"]], "\[Pi]"]]], "]"]]], "-", SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], RowBox[List[RowBox[List["Floor", "[", RowBox[List[FractionBox["3", "4"], "-", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["zz", "0"]]], "]"]], RowBox[List["2", " ", "\[Pi]"]]]]], "]"]], "+", RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "4"], "+", FractionBox[RowBox[List["Arg", "[", RowBox[List["z", "-", SubscriptBox["zz", "0"]]], "]"]], RowBox[List["2", " ", "\[Pi]"]]]]], "]"]]]]], "+", RowBox[List["2", " ", RowBox[List["Round", "[", FractionBox[RowBox[List["Re", "[", SubscriptBox["zz", "0"], "]"]], "\[Pi]"], "]"]]]]]], ")"]]]], "+", FractionBox[RowBox[List["z", "-", SubscriptBox["zz", "0"]]], SqrtBox[RowBox[List["1", "-", "m"]]]], "+", FractionBox[RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["k", "=", "2"]], "\[Infinity]"], FractionBox[RowBox[List[RowBox[List["(", RowBox[List[SuperscriptBox["\[ImaginaryI]", RowBox[List["k", "-", "1"]]], " ", RowBox[List["Binomial", "[", RowBox[List[RowBox[List["k", "-", FractionBox["1", "2"]]], ",", RowBox[List["k", "-", "1"]]]], "]"]]]], ")"]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["q", "=", "1"]], RowBox[List["k", "-", "1"]]], FractionBox[RowBox[List[RowBox[List["(", RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List["-", "1"]], ")"]], "q"], " ", RowBox[List["Binomial", "[", RowBox[List[RowBox[List["k", "-", "1"]], ",", "q"]], "]"]]]], ")"]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["1", "-", "m"]], ")"]], RowBox[List["-", "q"]]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["j", "=", "0"]], "q"], RowBox[List[RowBox[List["Binomial", "[", RowBox[List["q", ",", "j"]], "]"]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["-", "m"]], ")"]], "j"], " ", SuperscriptBox[RowBox[List["(", RowBox[List["2", "-", "m"]], ")"]], RowBox[List["q", "-", "j"]]], " ", SuperscriptBox["2", RowBox[List["k", "-", "j", "-", "q", "-", "1"]]], " ", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["i", "=", "0"]], "j"], RowBox[List[RowBox[List["Binomial", "[", RowBox[List["j", ",", "i"]], "]"]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["2", " ", "i"]], "-", "j"]], ")"]], RowBox[List["k", "-", "1"]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["z", "-", SubscriptBox["zz", "0"]]], ")"]], "k"]]]]]]]]]]], RowBox[List[RowBox[List["2", " ", "q"]], "+", "1"]]]]]]], RowBox[List["k", "!"]]]]], SqrtBox[RowBox[List["1", "-", "m"]]]]]], "/;", RowBox[List[RowBox[List["(", RowBox[List["z", "\[Rule]", SubscriptBox["zz", "0"]]], ")"]], "&&", RowBox[List[SubscriptBox["zz", "0"], "\[Equal]", RowBox[List[FractionBox[RowBox[List["3", " ", "\[Pi]"]], "2"], "+", RowBox[List["2", " ", "\[Pi]", " ", "u"]]]]]], "&&", RowBox[List["u", "\[Element]", "Integers"]], "&&", RowBox[List["m", "\[Element]", "Reals"]], "&&", RowBox[List["m", ">", "1"]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2007-05-02 | 4,756 | 13,138 | {"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} | 2.78125 | 3 | CC-MAIN-2018-09 | latest | en | 0.18558 |
https://www.unitconverters.net/flow/milliliter-hour-to-cubic-centimeter-day.htm | 1,726,362,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00409.warc.gz | 982,175,119 | 3,321 | Home / Flow Conversion / Convert Milliliter/hour to Cubic Centimeter/day
# Convert Milliliter/hour to Cubic Centimeter/day
Please provide values below to convert milliliter/hour [mL/h] to cubic centimeter/day, or vice versa.
From: milliliter/hour To: cubic centimeter/day
### Milliliter/hour to Cubic Centimeter/day Conversion Table
Milliliter/hour [mL/h]Cubic Centimeter/day
0.01 mL/h0.24 cubic centimeter/day
0.1 mL/h2.4 cubic centimeter/day
1 mL/h24 cubic centimeter/day
2 mL/h48 cubic centimeter/day
3 mL/h72 cubic centimeter/day
5 mL/h120 cubic centimeter/day
10 mL/h240 cubic centimeter/day
20 mL/h480 cubic centimeter/day
50 mL/h1200 cubic centimeter/day
100 mL/h2400 cubic centimeter/day
1000 mL/h24000 cubic centimeter/day
### How to Convert Milliliter/hour to Cubic Centimeter/day
1 mL/h = 24 cubic centimeter/day
1 cubic centimeter/day = 0.0416666667 mL/h
Example: convert 15 mL/h to cubic centimeter/day:
15 mL/h = 15 × 24 cubic centimeter/day = 360 cubic centimeter/day | 291 | 991 | {"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} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.551815 |
http://www.chestnut.com/en/eg/11/scientific/mathematics/2/3/examples/7/ | 1,477,048,408,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717963.49/warc/CC-MAIN-20161020183837-00027-ip-10-171-6-4.ec2.internal.warc.gz | 359,463,841 | 12,185 | # 11.2.3. Solving Exponential Equations and Their Applications
Determine the solution set of .
• A
• B
• C
• D
### Example
Determine the solution set of .
### Solution
Multiply both sides by 25.
For to be equal to 25,
Therefore, the solution set is
0
correct
0
incorrect
0
skipped | 85 | 289 | {"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} | 2.875 | 3 | CC-MAIN-2016-44 | latest | en | 0.790232 |
https://bob.cs.sonoma.edu/IntroCompOrg-RPi/exercises-25.html | 1,679,364,957,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943589.10/warc/CC-MAIN-20230321002050-20230321032050-00381.warc.gz | 177,176,364 | 7,205 | ## Exercises14.2Programming Exercise
###### 1.
Write a program in assembly language that (a) prompts the user to enter some alphabetic characters, (b) converts any uppercase characters to lowercase, and (c) displays the resulting uppercase string. Your algorithm must allow for the user to enter lowercase characters and leave them as lowercase.
Hint
Consult Table 2.13.1 to see which bit in the ASCII code determines upper/lower case.
Solution
@ lowerCase.s
@ Prompts user to enter alphabetic characters, converts
@ all uppercase to lowercase and shows the result.
@ 2017-09-29: Bob Plantz
@ Define my Raspberry Pi
.cpu cortex-a53
.fpu neon-fp-armv8
.syntax unified @ modern syntax
@ Constant for assembler
.equ nBytes,50 @ amount of memory for string
@ Constant program data
.section .rodata
.align 2
prompt:
.asciz "Enter some alphabetic characters: "
@ The program
.text
.align 2
.global main
.type main, %function
main:
sub sp, sp, 16 @ space for saving regs
@ (keeping 8-byte sp align)
str r4, [sp, 4] @ save r4
str fp, [sp, 8] @ fp
str lr, [sp, 12] @ lr
add fp, sp, 12 @ set our frame pointer
mov r0, nBytes @ get memory from heap
bl malloc
mov r4, r0 @ pointer to new memory
ldr r0, promptAddr @ prompt user
bl writeStr
mov r0, r4 @ get user input
mov r1, nBytes @ limit input size
mov r0, r4 @ convert to lowercase
bl toLower
mov r0, r4 @ echo user input
bl writeStr
mov r0, r4 @ free heap memory
bl free
mov r0, 0 @ return 0;
ldr r4, [sp, 4] @ restore r4
ldr fp, [sp, 8] @ fp
ldr lr, [sp, 12] @ lr
add sp, sp, 16 @ restore sp
bx lr @ return
.word prompt
@ toLower.s
@ Converts all alpha characters to lowercase.
@ Calling sequence:
@ r0 <- address of string to be written
@ bl toLower
@ returns number of characters written
@ 2017-09-29: Bob Plantz
@ Define my Raspberry Pi
.cpu cortex-a53
.fpu neon-fp-armv8
.syntax unified @ modern syntax
@ Useful source code constants
.equ NUL,0
@ The code
.text
.align 2
.global toLower
.type toLower, %function
toLower:
sub sp, sp, 16 @ space for saving regs
str r4, [sp, 0] @ save r4
str r5, [sp, 4] @ r5
str fp, [sp, 8] @ fp
str lr, [sp, 12] @ lr
add fp, sp, 12 @ set our frame pointer
mov r4, r0 @ r4 = string pointer
mov r5, #0 @ r5 = count
whileLoop:
ldrb r3, [r4] @ get a char
cmp r3, NUL @ end of string?
beq allDone @ yes, all done
orr r3, r3, lowerMask @ convert to lowercase
strb r3, [r4] @ update string
add r4, r4, 1 @ increment pointer var
add r5, r5, 1 @ count++ | 884 | 2,901 | {"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} | 2.53125 | 3 | CC-MAIN-2023-14 | latest | en | 0.674559 |
https://www.studypool.com/services/17519/fin-515-managerial-finance-week-4-the-cost-of-capital-discussion-question-2-answer | 1,542,047,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741016.16/warc/CC-MAIN-20181112172845-20181112194845-00380.warc.gz | 1,015,930,260 | 21,176 | # FIN-515- Managerial Finance Week 4_The Cost of Capital_Discussion_Question_2_Answer
Apr 24th, 2015
Studypool Tutor
Price: \$5 USD
Tutor description
FIN-515- Managerial Finance Week 4_The Cost of Capital_Discussion_Question_2_Answer
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The Cost of Capital (graded) What are possible capital components in the WACC equation? How does a company with debt, preferred stock, and common stock in its target capital structure calculate its weighted average cost of capital (WACC)? WACC = wd rd(1-T) + wp rp + wcrs rd=marginal debt capitalrp=marginal preferred stockrs=marginal cost of common equity using retained earnings The weighted average cost of capital (WACC) is calculated using the following formula: WACC = wdrd(1 - T) + wpsrps + wsrs, where wd is the percent of debt, rd is the interest rate on the firm's new debt, T is the firm's marginal tax rate, wps is the percent of preferred stock, rps is the yield investors expect on the preferred stock, ws is the percent of common equity (common stock), and rs is the rate of return investors expect from the firm's common stock. The proportions of debt, preferred stock and common equity typically used in the formula are the target proportions to eliminate slight variability due to slight actual differences from the target proportions. How is the weighted average cost of capital (WACC) used in a capital investment program to select investments?The weighted average cost of capital (WACC) is a tool used to decide whether or not to invest. It represents the minimum rate of return at which a company produces value for its investors. If the companies return is more than the WACC than for each dollar the company invests it is creating value. Conversely if the compa
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https://www.replicon.com/resource/lead-time-calculator/ | 1,713,197,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00722.warc.gz | 901,152,389 | 37,383 | No more losing your sales and revenue
Use Replicon’s Lead Time Calculator to keep up with your Customer Demand!
9 days
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## Introduction to Lead Time Calculator
Lead time is the amount of time that passes between a customer placing an order and the product getting delivered to the customer. This includes three phases:
i) Pre-production time – Time taken to prepare the order for production; like procurement of raw materials, resources, etc.
ii) Production time/ Manufacturing time – Time taken to manufacture the product.
iii) Shipping time – Time taken to ship the product.
Lead Time is calculated based on the organization’s product or services. Irrespective of the organization’s business setup, be it Business to Business(B2B) or Business to Consumer(B2C), below are the three most common formulas used to calculate lead time.
i) Lead Time (Manufacturing Industry) = Pre-production time + Production time/ Manufacturing time + Shipping time
ii) Lead Time (Customer) = Order Received Date – Order Placed Date
iii) Lead Time (Supply Chain Management) = Reordering Delay + Supply Delay
## Lead Time for Customer Orders
This formula is used for calculating the lead time for customer orders. The two main components to calculate lead time using this formula are:
i) Order Placed Date: This is the date on which the order was placed by the customer.
ii) Order Received Date: This is the date on which the order was received by the customer.
The difference between these two dates will give the Lead Time for that particular product.
Let’s see an example using this formula:
A customer placed the order for a customized necklace on 2nd of June . The product was manufactured and shipped. Customer received the order on 12th of June. So the total Lead time is calculated as:
12th of June – 2nd of June = 10 days
## Lead Time in Inventory Management
Inventory control is the process of monitoring a company’s stock. There are two components to be considered when calculating the lead time in inventory management:
i) Supply Delay = Time between placing, and receiving an order
ii) Reordering Delay = Time that has to pass before placing a reorder (after processing)
The formula for calculating Lead Time in Inventory Management:
Lead Time = Supply Delay + Reorder Delay
Let’s see an example of this.
For example, if an order has been placed on a Thursday but the supplier processes orders only on Tuesdays, then there is a reordering delay of six days (Thursday – Tuesday = 6 days). It takes five days to physically ship the order, so the Supply delay is 5 days.
Lead Time = Supply Delay + Reordering Delay
Lead Time = 5 days + 6 Days
## How to Reduce Lead Time
The more the demand, and orders placed, the longer the lead time. There are many strategies to process orders faster and fulfill customer orders. It is important to reduce the supply chain and this can be done in the following ways:
i) Accept smaller orders: this ensures that there is stock available at all times. Instead of one big order, accept smaller orders in regular intervals.
ii) Go Digital: Automate inventory details so stock is calculated online. This saves a lot of time and does not require a constant check on orders and available stock.
iii) Forecast Demands: Understand market demand trends and have up-to-date knowledge of future demands. This helps to keep stock readily available.
iv) Streamline from Order to Delivery: Keep a keen eye on every order from the time it is received till the time it is delivered to the customer. This ensures delays from the manufacturing end.
The benefits of reducing lead time are:
i) Improved customer satisfaction by meeting their demand.
ii) Avoid over or under inventory stock.
iii) It helps in efficient resource management by decreasing wastage of resources.
iv) Improves productivity.
### 1. What is meant by lead time?
Lead time is the total time it takes from the time an order is placed for a product to the time it is delivered to the customer.
### 2. How is lead time different from takt time and cycle time?
Lead time is the total time from the order placed till it is delivered to the customer. Takt time is the rhythm at which a product has to be manufactured to meet customer demand. Cycle time is the actual time it takes to complete a product.
### 3. What is the lead time on an order?
Lead time on an order is the total number of days from when a company places an order till it receives the order.
### 4. How is Lead time measured?
Lead time is measured by date or by the number of days.
By Date: Lead time = Order Shipped Date – Order Received Date
By Days: Receiving order + Manufacturing the product + Delivering Time = Lead Time
### 5. Is lead time the same time as delivery time?
Delivery time is part of the lead time. It is the post-processing phase of lead time. It can be shortened or extended according to the logistic availability. However, lead time is the total time taken from the order placed until it is delivered to the customer. | 1,060 | 5,033 | {"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} | 2.921875 | 3 | CC-MAIN-2024-18 | longest | en | 0.94662 |
https://www.cheenta.com/rational-form-rmo-2019-problem-1-solution/ | 1,624,405,087,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00021.warc.gz | 631,761,122 | 42,263 | # Understand the problem
[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Suppose x is a non zero real number such that both $x^5$ and $20 x + \frac{19}{x}$ are rational numbers. Prove that x is a rational number.
[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Regional Math Olympiad, 2019 Problem 1[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]3/10
[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!
[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]Notice that you can represent higher powers by smaller powers. Suppose $20x + \frac{19}{x} = r$ where r is rational. Multiply both sides by x to get $20 x^2 = r \cdot x - 19$ or $x^2 = r_1 x + r_2$ (that is divide by 20 to have r/20 as the coefficient of x and -19/20 as the x free number. both of these are rationals. therefore we name them $r_1$ and $r_2$ )
Now use the fact that $x^5$ is rational.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]
Suppose $x^5 = r_3$ (some rational number). Then $x^2 \cdot x^2 \cdot x = r_3$ But we know $x^2 = r_1 x + r_2$ Replacing we have $(r_1 \cdot x + r_2 )^2 \cdot x = r_3$ Expanding we have $r_1^2 x^3 + 2r_1r_2 \cdot x^2 + r_2^2 x = r_3$ Again we will replace $x^2$ to have $r_1^2 \cdot (r_1 x + r_2) x + 2r_1r_2 \cdot (r_1 x + r_2) + r_2^2 x = r_3$
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]
Expand $r_1^2 \cdot (r_1 x + r_2) x + 2r_1r_2 \cdot (r_1 x + r_2) + r_2^2 x = r_3$ $r_1^3 x^2 + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3$ Making one final replacement of x^2 and noting that squaring, adding, multiplying rationals gives rationals we have $r_1^3 (r_1 x + r_2) + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3$ $r_1^4 x + r_1^3r_2 + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3$ or we have x as a ratio of rationals.
Hence x is rational. | 1,056 | 2,774 | {"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} | 3.875 | 4 | CC-MAIN-2021-25 | longest | en | 0.584783 |
https://physics.stackexchange.com/questions/541822/covariant-vs-contravariant-vectors | 1,721,736,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00266.warc.gz | 399,346,828 | 41,666 | # Covariant vs contravariant vectors
I understand that, in curvilinear coordinates, one can define a covariant basis and a contravariant basis. It seems to me that any vector can be decomposed in either of those basis, thus one can have covariant components and contravariant components of the same vector, depending on the chosen basis. What is confusing to me, however, is when people talk about covariant and contravariant vectors. Do they just mean the covariant/contravariant components of vectors or there are indeed two distinct types/classes of vectors? If the latter, covariant vectors can only be decomposed into covariant bases and contravariant vectors only in contravariant bases?
## 2 Answers
We don't talk of covariant and contravariant bases. Start with the basis $$\{\mathbf e_i\}$$. Then a general vector can be written $$\mathbf v = v^i \mathbf e_i$$ Now if you double the length of a basis vector, you must halve the component. The components are said to be contravariant, because they change opposite to the basis. In index notation this vector is simply written $$v^i$$, and we call it a contravariant vector meaning that the components are contravariant.
The inner product
$$\mathbf u \cdot \mathbf v = g_{ij}u^iv^j$$ prompts the definition $$u_j = g_{ij}u^i$$ The $$u_j$$ are components of a vector in the dual space. Because the inner product is invariant, the components $$u_j$$ change opposite to contravariant components, which means they change in the same way as the basis vectors. They are called covariant components, and we refer to them as covariant vectors.
Technically contravariant vectors are in one vector space, and covariant vectors are in a different space, the dual space. But there is a clear 1-1 correspondence between the space and its dual, and we tend to think of the contravariant and covariant vectors as different descriptions of the same vector.
• what about the components $u^i$? Aren't those the components of the same vector $\mathbf{u}$? Commented Apr 6, 2020 at 11:24
• I have added a para to clarify. Commented Apr 6, 2020 at 12:10
• God I've been waiting to read something like this for years! Thanks. Commented Apr 6, 2020 at 12:32
• What's the best example of a setting where the mentioned 1-1 correspondence space and its dual is most obscure? The fact that I often equate the two makes it hard to intuitively understand the difference between the two (and thus covariant vs contravariant). Commented Jul 17 at 15:01
You have a basis $${\bf e}_i$$ in some vector space.
The contravariant components of a vector $${\bf v}$$ are given by $${\bf v}=v^i{\bf e_i}$$, as Charles Francis says.
The covariant components of a vector $${\bf v}$$ are given by $$v_i=\mathbf v\cdot\mathbf e_i$$
I think that's a more basic way of thinking about them than going in to their transformation properties - though that is of course true.
Incidentally it's then obvious that $$\mathbf u\cdot\mathbf v=\sum u_i v^i$$ (or $$\sum u^iv_i$$)
I would say (though mathematicians would disagree and will probably downvote this answer as heretical) that a 'physics' vector is neither covariant nor contravariant. It's a pointing arrow. If you want to do anything useful with it you have to write down its components, which can be either covariant of contravariant.
• The problem with the arrow arises, surely, when we try to apply it to grad $\phi$ when working in a non-orthogonal co-ordinates basis, {$\mathbf{e}_i$}. The components of grad $\phi$ on this basis exist but do not give us what we'd like to have, namely {$\frac{\partial \phi}{\partial x_i}$}. [These derivatives are components on the dual basis to {$\mathbf{e}_i$}!] So I'd argue that it's not always $useful$ to think of vectors as arrows. Commented Apr 7, 2020 at 12:05
• If you are standing on a hillside (height $\equiv \phi$) then you know the direction and the magnitude of grad $\phi$ just by releasing a marble and seeing which direction it goes in and how fast it accelerates. You can draw that as an arrow. Without writing down any components and hence without involving a basis. But I agree that this is tricky stuff. Commented Apr 7, 2020 at 12:20
• RogerJBarlow Thanks for replying. It's not that I'm saying that a gradient can't be thought of as an arrow, so much as the arrow thing not being so $suggestive\ of\ useful\ ways\ to\ proceed$ in the case of a gradient, if we're working on a non-orthogonal basis. I don't think we're really at odds. Commented Apr 7, 2020 at 15:42 | 1,154 | 4,506 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-30 | latest | en | 0.930688 |
http://bluebulbprojects.com/MeasureOfThings/results.php?comp=speed&unit=yph&amt=7647.6&sort=pr&p=1 | 1,558,254,787,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00059.warc.gz | 29,549,794 | 6,941 | Bluebulb Projects presents:
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How fast is 7,647.60 yards per hour?
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It's about as fast as Michael Phelps. (at the Beijing Olympics, 2008; 200 m freestyle) (a.k.a. Michael Fred Phelps) (swimmer; 1985-)Setting a world record, Michael Phelps swam the 200 m freestyle in 1:42.96 for an average speed of 7,647.60 yards per hour. Phelps would go on to win nine gold medals individually in the 2008 Olympics - more than all but eight of the competing nations. It's about seven-tenths as fast as a Crocodile. In other words, 7,647.60 yards per hour is 0.720 times the speed of a Crocodile, and the speed of a Crocodile is 1.40 times that amount. (American Crocodile, Crocodylus acutus) (swimming speed)An American crocodile can reach speeds in the water of up to 11,000 yards per hour. On land, larger crocodiles can "gallop" when fleeing danger at speeds of up to 22,000 yards per hour. It's about one-and-a-half times as fast as Walking Pedestrians (in Manhattan). In other words, the speed of Walking Pedestrians (in Manhattan) is 0.670 times 7,647.60 yards per hour. (Manhattan; average speed; 8,978 person-sample)A 2006 Study by the New York City Department of City Planning found that pedestrians in that city walk at an average rate of 5,100 yards per hour. Pedestrians wearing headphones, the study went on to find, walk at a slightly faster 5,600 yards per hour It's about three-tenths as fast as a Bull. In other words, 7,647.60 yards per hour is 0.290 times the speed of a Bull, and the speed of a Bull is 3.40 times that amount. (for animals involved in the Running of the Bulls, a.k.a. Encierro, San Fermin, Pamplona, Spain) (herd average speed)The herd of the annual Encierro in Pamplona, Spain runs at an average speed of 26,000 yards per hour. The Encierro is run annually from July 7th through July 14th and involves 42 bulls, 77 oxen, and an estimated 17,000 runners over the course of the event. It's about one-fourth as fast as Noah Ngeny. In other words, 7,647.60 yards per hour is 0.25630 times the speed of Noah Ngeny, and the speed of Noah Ngeny is 3.9020 times that amount. (in Rieti, Italy; 1999) (sprinter; 1978-)Setting a world record at the Rieti Grand Prix in 1999, Noah Ngeny ran 1,000 m in 2:11.96 for an average speed of 29,830 yards per hour. According to some reports, Ngeny did not begin running competitively until just three years before setting the record. It's about four times as fast as an Iceberg. In other words, the speed of an Iceberg is 0.30 times 7,647.60 yards per hour. (a.k.a. Berg) (Newfoundland iceberg average)Moved by ocean currents and wind, icebergs can drift at speeds of about 2,000 yards per hour. The largest iceberg ever recorded was a found near Baffin Island, Nunavut and was estimated to be nine billion metric tons. It's about one-fifth as fast as Flo-Jo. In other words, 7,647.60 yards per hour is 0.20730 times the speed of Flo-Jo, and the speed of Flo-Jo is 4.8240 times that amount. (at the Seoul Olympics, 1998) (a.k.a. Florence Griffith-Joyner, a.k.a. Florence Delorez Griffith) (swimmer; 1959-1998)Setting a world record in 1988, Flo-Jo ran a 200 m in 0:21.34 for an average speed of 36,900 yards per hour. Known as a 200 m runner, Joyner also set a record time in a 100 m race at in 1987. | 972 | 3,379 | {"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} | 3.21875 | 3 | CC-MAIN-2019-22 | latest | en | 0.944584 |
https://www.mathcelebrity.com/community/threads/your-bill-for-dinner-including-a-7-25-sales-tax-was-49-95-you-want-to-leave-a-15-tip-on-the-co.863/ | 1,685,418,541,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00367.warc.gz | 993,386,955 | 9,109 | # Your bill for dinner, including a 7.25% sales tax, was \$49.95. You want to leave a 15% tip on the co
Discussion in 'Calculator Requests' started by math_celebrity, Feb 7, 2017.
Tags:
Your bill for dinner, including a 7.25% sales tax, was \$49.95. You want to leave a 15% tip on the cost of the dinner before the sales tax. Find the amount of the tip to the nearest dollar.
Find the pretax cost:
49.95/1.0725 = 46.57
Now, add 15% tip to the pretax bill:
46.57(1.15) = \$53.56 | 158 | 482 | {"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} | 3.359375 | 3 | CC-MAIN-2023-23 | longest | en | 0.935987 |
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Prev Next | 1,127 | 4,660 | {"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} | 3.109375 | 3 | CC-MAIN-2018-13 | latest | en | 0.909444 |
https://dsp.stackexchange.com/questions/54977/how-to-solve-signal-mfsk-or-fhss-question-received-signal-noisejamming | 1,718,395,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00539.warc.gz | 194,689,296 | 38,970 | # How to solve signal MFSK or FHSS question (received signal+ noise+jamming)
I'm trying to solve the following: $$$$A \cos(2\pi f t + \theta_1) + B \cos(2\pi f t + \theta_2) = D\cos(?f?\theta)$$$$
I just need to know the correct value of D, the value of frequency and delta is not important since it is non-coherent.
• Link Search for "arbitrary phase shift"... Commented Jan 21, 2019 at 10:20
$$A\cos(2\pi f+\theta_1)+B\cos(2\pi f+\theta_2)=C\cos(2\pi f+\theta_3)$$
where
$$C=|u|\quad\textrm{and}\quad \theta_3=\arg\{u\}$$
with
$$u=Ae^{j\theta_1}+Be^{j\theta_2}$$
The constant $$C$$ can be written as
$$C=\sqrt{A^2+2AB\cos(\theta_1-\theta_2)+B^2}$$ | 258 | 659 | {"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.803266 |
https://www.freecodecamp.org/news/logical-operators-in-php/ | 1,719,270,702,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00613.warc.gz | 697,198,675 | 11,100 | Logical operators play a key role in programming languages. They let you manipulate boolean values and evaluate logical conditions.
In PHP, there are four fundamental logical operators: AND, OR, NOT, and XOR. This guide will help you understand these operators, and I'll explain how they work using code examples and practical use cases.
## The Logical AND Operator (&&)
The AND operator, written like “&&,” evaluates to true only if both of its operands are true. It evaluates to false if any of the operands are false, resulting in a false outcome.
This operator is commonly used to combine multiple conditions in an if statement or a loop. It helps ensure that all conditions are satisfied for the overall condition to be true.
``````
// Checking if two conditions are true using the AND operator
\$mx = 5;
\$my = 10;
if (\$mx > 0 && \$my > 0) {
echo "Both conditions are true!";
} else {
echo "This condition is false.";
}``````
In the above example, the AND operator verifies that both `\$x` and `\$y` are greater than 0.
So, if both conditions are true, the code inside the if block will execute, displaying “Both conditions are true!”. Alternatively, when the else block is triggered, it suggests that one or more of the conditions are false.
Let's now shift our focus to the counterpart of the AND operator – the OR operator.
## The Logical OR Operator (||)
The PHP OR operator, written like “||”, returns true if at least one of its operands is true. It evaluates to false only when both operands are false.
You can use this operator when you want to execute a block of code if any of several conditions are satisfied.
``````
// Checking if at least one condition is true using the OR operator
\$mx = 5;
\$my = 10;
if (\$mx > 0 || \$my > 0) {
echo "At least one condition is true!";
} else {
echo "Both conditions are false.";
}``````
In this example, the OR operator checks if either `\$x` or `\$y` (or both) are greater than 0. If any of the conditions is true, the code inside the if block will execute, displaying “At least one condition is true!”.
Otherwise, the else block will execute, indicating that both conditions are false.
Now let's explore the NOT operator and understand how it works.
## The Logical NOT Operator (!)
The NOT operator, written like "!", is a unary operator that inverses the value of its operand. When the operand is false, it will return true. cvonversely, when the operand is true, it will return false.
This operator is commonly utilized to invert a condition or verify the absence of a specific state.
``````
// Checking if a condition is false using the NOT operator
\$x = 5;
if (!(\$x > 10)) {
echo "Condition is false!";
} else {
echo "Condition is true.";
}``````
In the above example, the NOT operator negates the result of the condition `\$x > 10`. If the condition is false (which it is in this case), the code inside the if block will execute, displaying “Condition is false!”. If the condition is found to be true, the execution of the else block confirms the validity of the condition.
Finally, let's delve deeper into the XOR operator in PHP and gain a better understanding of its usage and behavior.
## The Logical XOR Operator (Exclusive OR)
Although PHP doesn't have a specific XOR operator, we can simulate XOR behavior using a combination of other logical operators. XOR returns true if exactly one of the operands is true, while it returns false if both operands are either true or false.
``````// XOR operator implementation using AND, OR, and NOT operators
\$x = true;
\$y = false;
if ((\$x || \$y) && !(\$x && \$y)) {
echo "Exactly one condition is true (XOR)!";
} else {
echo "Both conditions are either true or false.";
}``````
In this example, we create an XOR behavior by checking if either `\$x` or `\$y` is true (`\$x || \$y`) and ensuring that both conditions are not true at the same time (`!(\$x && \$y)`).
If exactly one condition is true, the code inside the if block will execute, displaying “Exactly one condition is true (XOR)!” Otherwise, the else block will execute, indicating that both conditions are either true or false.
## Wrapping Up
PHP logical operators are powerful tools in PHP that allow us to manipulate boolean values and evaluate logical conditions. Understanding the functionality and usage of logical operators is essential for building reliable and efficient PHP programs.
By mastering these operators, you can create complex conditional statements and make your code more robust and flexible.
In this article, we explored the four fundamental logical operators in PHP: AND, OR, NOT, and XOR.
We provided explanations, code examples, and practical use cases to demonstrate their functionalities. By applying these operators effectively, you can perform intricate logical operations, control the flow of your programs, and make informed decisions based on various conditions.
Remember to practice implementing logical operators in PHP and experiment with different scenarios to deepen your understanding.
Thank you for reading, If you'd like to read more of my articles, you can find them on CodedTag. Stay tuned for my next articles. | 1,138 | 5,167 | {"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} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.876809 |
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Problem Solutions
## The `math` Module
The `math` module is made available to your programs with:
```import math
```
The `math` module contains the following trigonometric functions
`math.acos `( `x` ) → number
arc cosine of x .
`math.asin `( `x` ) → number
arc sine of x .
`math.atan `( `x` ) → number
arc tangent of x .
`math.atan2 `( `y` , `x` ) → number
arc tangent of y / x .
`math.cos `( `x` ) → number
cosine of x .
`math.cosh `( `x` ) → number
hyperbolic cosine of x .
`math.exp `( `x` ) → number
e** x , inverse of log( x ).
`math.hypot `( `x` , `y` ) → number
Euclidean distance, sqrt( x * x + y * y ), length of the hypotenuse of a right triangle with height of y and length of x .
`math.log `( `x` ) → number
natural logarithm (base e) of x , inverse of exp( x ).
`math.log10 `( `x` ) → number
natural logarithm (base 10) of x , inverse of 10** x .
`math.pow `( `x` , `y` ) → number
x ** y .
`math.sin `( `x` ) → number
sine of x .
`math.sinh `( `x` ) → number
hyperbolic sine of x .
`math.sqrt `( `x` ) → number
square root of x . This version returns an error if you ask for sqrt(-1), even though Python understands complex and imaginary numbers. A second module, `cmath`, includes a version of `sqrt`( `x` ) which correctly creates imaginary numbers.
`math.tan `( `x` ) → number
tangent of x .
`math.tanh `( `x` ) → number
hyperbolic tangent of x .
Additionally, the following constants are also provided.
`math.pi`
the value of pi, 3.1415926535897931
`math.e`
the value of e, 2.7182818284590451, used for the `exp`( `x` ) and `log`( `x` ) functions.
The math module contains the following other functions for dealing with floating point numbers.
`math.ceil `( `x` ) → number
next larger whole number. `math.ceil(5.1) == 6`, `math.ceil(-5.1) == -5.0`.
`math.fabs `( `x` ) → number
absolute value of the real x .
`math.floor `( `x` ) → number
next smaller whole number. `math.floor(5.9) == 5`, `math.floor(-5.9) == -6.0`.
`math.fmod `( `x` , `y` ) → number
floating point remainder after division of x / y . This depends on the platform C library and may return a different result than the Python `x % y`.
`math.modf `( `x` ) → ( number, number )
creates a tuple with the fractional and integer parts of x . Both results carry the sign of x so that x can be reconstructed by adding them.
`math.frexp `( `x` ) → ( number, number )
this function unwinds the usual base-2 floating point representation. A floating point number is m *2** e , where m is always a fraction between 1/2 and 1, and e is an integer power of 2. This function returns a tuple with m and e . The inverse is `ldexp(m,e)`.
`math.ldexp `( `m` , `e` ) → number
m *2** e , the inverse of `frexp(x)`.
Published under the terms of the Open Publication License Design by Interspire | 943 | 3,281 | {"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} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.493085 |
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*********** | 336 | 1,449 | {"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} | 2.53125 | 3 | CC-MAIN-2017-51 | longest | en | 0.846358 |
https://community.wanikani.com/t/spoiler-edition-descent-of-the-durtle-into-madness/34542?page=8 | 1,547,915,838,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583671342.16/warc/CC-MAIN-20190119160425-20190119182425-00260.warc.gz | 465,255,266 | 7,850 | # (SPOILER EDITION) Descent of the Durtle into Madness
#141
But how do you get from 3 to E and 5 to G? And sometimes you shift in different directions.
Also O is not 0.
#142
well… 3 looks like an E and a 5 like a G, also 5 is ご
53NKI I
And I shift right or left… whatever I feel like. I know… it’s too farfetched
#143
### Letters for numbers
• 0 = O or D or space
• 1 = I or L
• 2 = Z or e
• 3 = E or
• 4 = h or A
• 5 = S
• 6 = b or G
• 7 = T or j
• 8 = X
• 9 = g or J
Even worse
#144
I’m with @emucat in still trying to decode the previous codes just to try and verify if they are just a means to get to the other pages we found with comments or determine if they are still part of the puzzle.
I found a pattern that partially links up between the codes and the pages found, but it doesn’t 100% fit.
Recap of the pattern:
AABQD010L76223CEGD72FI5DPBOJJMO+
B51BIQPGS206HD
00LB269G
D02315CPQAWDF0B~IL7KB1FG+A
I decided to count the number of characters in each encrypted row and you get:
32
14
8
26
I then cross-referenced with the Tofugu articles that were discovered and found a similarity between the codes and the URL of each article. Now remember, below our four encrypted clues it said “Caution! Space is also required.”, so each hyphen below (used in place of spaces in the URL) is included with the count:
32 how-to-install-japanese-keyboard
14 genki-textbook
8
26 beginner-japanese-textbook
So 3 of the 4 pages with comments fit the pattern, but the page that does not fit has the comment “Durt Durt! Nice try!” which makes me think it was a false trail we may have found through some other means.
The page that does not fit is:
36 dictionary-of-basic-japanese-grammar
And we don’t have anything matching 8 characters. So maybe there is one article out there we are still missing.
Thoughts?
#145
Coincidentally I was looking at it the same way, but didn’t look further once I came across the first length mismatch.
If you want to pull that thread some more, here’s all the 22 article URL’s where the last segment in the path is eight characters in length:
Summary
#146
Maybe the furigana one? Pretty sure none of the others match up with the theme.
#147
This doesn’t fill me with joy, considering that’s the page we found the codes on in the first place.
Question is, is there any way to reverse-engineer a black box algorithm that renders “how-to-install-japanese-keyboard” as “AABQD010L76223CEGD72FI5DPBOJJMO+” and then see if it works with the other codes too?
“hinative”, “skritter” and “tonoharu” are all reviews. But yeah, not reviews of the books we’re looking at.
#148
Working on analyzing that text. Considering rotating keys are used to encode this cipher…
#149
That looks good!
#150
So for a simple XOR cipher, you can recover the key if you have both plaintext and ciphertext by xor-ing them back together. If it’s that, AND if the key is the same for all 4 items, we can get the unknown ones.
K = C1 xor P1
then,
Pn = K xor Cn
complication: how the xor works depends (once again) on the encoding scheme of the characters. ASCII? Unicode? etc.
#151
What are we XORing, though? ASCII values?
#152
That’s not a complication because the Latin range of both schemes is the same. Naturally, that ensures backwards compatibility of documents encoded in ASCII when they are re-encoded in Unicode. Or put more specifically, Captial A is 0x41 and U+0041 in ASCII and Unicode, respectively.
#153
Still, I have done a lot trying to XOR against this text before we discovered the book cipher elements, to no avail. That’s why I’m exploring Vigenere most recently. In part based on a hunch that if the puzzle IS NOT using cryptographic techniques that require computer science like XOR, then the puzzle is friendlier to a larger user base.
EDIT: But I like XOR because it gives a more plausible explanation for the appearance of the symbols in the ciphertext like ~ and +. Sigh so much to speculate
EDIT: If someone likes to play, this site has lot’s of tools to attack https://www.dcode.fr/vigenere-cipher
I used this one more for the XOR, it’s SOOO stupidly good versatile, but a less friendly to non techies. This link will load it with the recipe that tests the entropy of the ciphertext. Don’t be misled by the scale when it shows that this text doesn’t look like properly encrypted text. Take that to mean, if it’s encrypted it’s “weak” as in not a proper industry-grade level of encryption. https://gchq.github.io/CyberChef/#recipe=Entropy()&input=QUFCUUQwMTBMNzYyMjNDRUdENzJGSTVEUEJPSkpNTytCNTFCSVFQR1MyMDZIRDAwTEIyNjlHRDAyMzE1Q1BRQVdERjBCfklMN0tCMUZHK0EK
#154
… How did I not see that?
#155
Well, I went through the source code for all the likely ones and saw nothing, so hmm.
I’ll keep looking but I don’t think this is it.
#156
How are we doing? Do we have more words? Leads?
#157
I’m personaly working on the Adventures in Japanese page but I’m not finding anything.
#158
I tried that one and I got:
eymtof
#159
Aye, that’s what I got too. Or “tamscl” if I include the heading.
#160
we should add it to the top post so nobody else tries | 1,397 | 5,116 | {"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} | 2.578125 | 3 | CC-MAIN-2019-04 | latest | en | 0.934374 |
https://www.coursehero.com/file/46301307/IMG-0398jpg/ | 1,571,561,501,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00318.warc.gz | 836,065,984 | 35,830 | IMG_0398.jpg - Correlational Research demonstrates a correlation relationship between 2 or more variables Positive variable In variable 2 T variable 7
# IMG_0398.jpg - Correlational Research demonstrates a...
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Unformatted text preview: Correlational Research demonstrates a correlation ( relationship ) between 2 or more variables. + Positive: variable In variable 2 T variable 7 variable 2 V -1 Negative : variable 7 & variable 2 1 O Zero : No relationship between variables correlation coefficient ( R ): measures strength of relationship between 2 or more variables ex -. 4 (. 6 ) +. 3 (.7 ) CORRELATION CAUSATION closer to - 1 Which correlation coefficient is stronger ?...
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• Spring '16
• Addition, Correlation and dependence, Pearson product-moment correlation coefficient, Covariance and correlation, Correlational Research
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Answers in as fast as 15 minutes | 444 | 2,008 | {"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} | 2.59375 | 3 | CC-MAIN-2019-43 | latest | en | 0.855657 |
https://discuss.fogcreek.com/techInterview/default.asp?cmd=show&ixPost=1905 | 1,603,924,211,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902038.86/warc/CC-MAIN-20201028221148-20201029011148-00199.warc.gz | 286,220,279 | 4,412 | Hungry Elephant this one is a easy puzzle there are 2 cities A and B, dist. between them is 1000 Km we have 3000 bananas in city A and a elephant can carry max 1000 bananas at any given time and it needs to eat a banana every 1 Km How Many Max. No. of bananas u can tranfer to city B??? note : the elephant cannot go without bananas to at. BondOfUK Wednesday, April 7, 2004 500. Go 250km drop off 500, go back. Go 250km drop off 500, go back. Go 250km, now you have 1750 at 250km. Go another 250km to 500km, drop off 500, go back. Pick up remaining 750, go to 500km. Now you have 1000 bananas at the 500km mark. The trip to the end leaves you with 500 bananas. -Daws Daws Wednesday, April 7, 2004 533. Pick up 1000, go 200 km, drop 600, return. Pick up 1000, go 200 km, drop 600, return. Pick up 1000, go 200 km, drop 800. Now you have 2000 bananas at 200 km. Pick up 1000, go 333 km, drop 334, return. Pick up 1000, go 333 km, drop 667. Now you have 1001 bananas at 533 km. Pick up 1000, go 467 km, drop 533 banans. You now have 533 bananas in city B and one banana rotting away at 533 km. Vigor Thursday, April 8, 2004 very good thats the correct Answer 533 BondofUk Saturday, April 10, 2004 I'll sketch my thought process. First, I realized that if you take 1000 bananas and walk 1000 kilometres, you arrive with no banana at all (and the elephant is stuck in city B). So the elephant has to travel shorter distances (but more often). So, I asked myself, why not carry the bananas kilometre by kilometre. It needs to walk five times one kilometre to move the bananas one kilometre forward (which costs five bananas). At 200 km, you have only 2000 bananas left, so you only need to walk three times to get the bananas one km farther. I arrived with 533 bananas in B (an optimal solution), but the poor elephant! It must feel dizzy of turning around so often. That's not what I thought at that moment. I was happy that I had found a solution (which I believed to be correct (which it was)) and went to bed. The next day, I realized that it's not necessary to make the elephant turn around so often. Transporting the bananas two kilometres is the same as transporting them one kilometre twice (as long as you don't cross the 2000 banana line or the 1000 banana line). I have posted the simplified solution above. The logical explanation for the solution is this one: As long as you have to walk every kilometre five times, each kilometre costs you 5 bananas, so you want to stop that as soon as possible. Which is when you have only 2000 bananas left, which is after 1000 km / 5 = 200 km. Daws' elephant walks the first 250 km five times, wasting some bananas here. (Daws, I hope you aren't offended. I just try to explain why your solution is suboptimal.) So with only 2000 bananas left, every kilometre costs 3 bananas, which we want to do until you have only 1000 bananas left, and no farther, because with only one way to walk, transportation becomes even cheaper. Now 1000/3 is not an integer, so we lose one banana due to rounding. That last part sound really odd. People in B will ask: "Why have you left one banana on the road?" - "I couldn't do anything about it; rounding error." - "WHAT?" Vigor Sunday, April 11, 2004 Sorry i feel the answer is 534 B : Bananas now when we reach 533KM we have 1001 B so pick 1000 B and go 1/3 KM drop 1000 B and come back 1/3 KM and pick 1 B and go 1/3 KM so now we have 1000 B at 533 + 1/3 KM and we dont have to give a B for travelling 2/3 KM at the end so we reach with 534 B cool :-) BondOfUk Wednesday, April 14, 2004 The puzzle said that the elephant cannot go without bananas to eat. However, that statement is ambiguous. It needs to specify whether the elephant eats the banana before going 1km, or eats after going 1km, or eats somewhere along the 1km. Does it eat the whole banana in one gulp, or takes little bites. If one is to find a precise and clear-cut solution, the statements in the problem must be precise. Unless you intend to play tricks on words. JHY Wednesday, April 14, 2004 Answer is 833 Bananas 1. Go 1 Km and leave 999 repeat this for another 2 times. 2. Now u were left with 2997 bananas 3. Repeat this process untill u will be left with less than 2000 bananas. () 4. Now go 1 KM in two sets (1000 + 1000) as u were left with below 2000 bananas 5. Repeat this process untill u will be left with less than 1000 bananas. () 6. Now keep going the city B at the end you will be left with 833 banas ----------------------- Code : ORACLE ----------------------- DECLARE b NUMBER := 3000; BEGIN dbms_output.put_line(' b : '|| b); FOR i IN 1..1000 LOOP IF b > 2000 THEN b := b - 3; ELSIF b > 1000 THEN b := b - 2; ELSE b := b - 1; END IF; END loop; dbms_output.put_line(' b : '|| b); END; Babu Tuesday, July 20, 2004 Fog Creek Home | 1,370 | 4,851 | {"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} | 3.875 | 4 | CC-MAIN-2020-45 | latest | en | 0.919019 |
https://www.physicsforums.com/threads/volume-of-water-flow.283117/ | 1,547,816,204,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660070.15/warc/CC-MAIN-20190118110804-20190118132804-00366.warc.gz | 876,807,443 | 12,382 | Volume of water flow
1. Jan 5, 2009
berniebs
A pipe broke on the roof of a public building here in Chicago. Although I am an electrical engineer and very rusty on hydraulics and water flow, I have been asked to estimate the amount of water dumped on the roof of this building, as follows:
Diameter of the burst pipe: 3/4 in.
Water temperature: 130 degr F
Pressure: 15 lbs
Water flowed for 45 minutes before being shut off. I assume the break allowed water to flow freely from a 3/4 in diameter opening.
I would appreciate it if someone can send me the formula for this calculation. I have searched a number of PF postings, without luck.
Thanks,
berniebs
berniebs@comcast.net
2. Jan 6, 2009
tiny-tim
Welcome to PF!
Hi berniebs! Welcome to PF!
Bernoulli's equation along a streamline (essentially, KE = work done ), assuming zero speed at zero pressure:
pressure + (density)(speed)2/2 = 0
and of course flow rate = speed x area.
(For details to impress your colleagues with, see the PF Library ) | 260 | 1,008 | {"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} | 2.75 | 3 | CC-MAIN-2019-04 | latest | en | 0.934514 |
https://www.teacherspayteachers.com/Product/Roll-a-Product-Multiplication-Game-2262323 | 1,487,574,139,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170425.26/warc/CC-MAIN-20170219104610-00469-ip-10-171-10-108.ec2.internal.warc.gz | 917,312,308 | 24,519 | # Roll a Product - Multiplication Game
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
0.37 MB | 9 pages
### PRODUCT DESCRIPTION
Number of Players: 2-5
Object of the Game: Be the first to color a box on the top row
Materials:
Dice
Crayon
This is a quick game that helps students practice a set of facts. Sheets for 2 through 9 are included. Each group chooses a number to work on for multiplication. They take turns rolling dice and multiplying the number by the practice sheet number. The repetition helps students practice their facts in a fun way.
Example:
Laura and Jake want to work on their 4 Facts, so they grab the 4 Product Practice Sheet. Laura goes first and she rolls a 10-sided die while Jake holds the sheet (so Laura can't see). Laura's die lands on 4. Laura knows that 4 x 4 is 16, which she tells her partner. Jake agrees and hands her the paper to color. Then, they switch rolls. Jake rolls the die and Laura holds the sheet.
This continues until on of the players colors the top box. Anyone can win, even if the top box is the ONLY box they've colored in the whole column.
Have fun! :)
Total Pages
9
N/A
Teaching Duration
30 Minutes
### Average Ratings
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
1 rating
\$2.00
User Rating: 4.0/4.0
(1 Followers)
\$2.00 | 381 | 1,409 | {"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} | 2.53125 | 3 | CC-MAIN-2017-09 | longest | en | 0.88691 |
https://numberworld.info/42665274 | 1,621,148,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989690.55/warc/CC-MAIN-20210516044552-20210516074552-00420.warc.gz | 449,968,689 | 4,040 | # Number 42665274
### Properties of number 42665274
Cross Sum:
Factorization:
2 * 3 * 3 * 17 * 139429
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
28b053a
Base 32:
18m19q
sin(42665274)
0.9994094169712
cos(42665274)
-0.03436302183001
tan(42665274)
-29.083862936012
ln(42665274)
17.568895892058
lg(42665274)
7.6300745390535
sqrt(42665274)
6531.866042717
Square(42665274)
### Number Look Up
Look Up
42665274 which is pronounced (forty-two million six hundred sixty-five thousand two hundred seventy-four) is a very impressive number. The cross sum of 42665274 is 36. If you factorisate 42665274 you will get these result 2 * 3 * 3 * 17 * 139429. The figure 42665274 has 24 divisors ( 1, 2, 3, 6, 9, 17, 18, 34, 51, 102, 153, 306, 139429, 278858, 418287, 836574, 1254861, 2370293, 2509722, 4740586, 7110879, 14221758, 21332637, 42665274 ) whith a sum of 97879860. 42665274 is not a prime number. The number 42665274 is not a fibonacci number. The number 42665274 is not a Bell Number. The number 42665274 is not a Catalan Number. The convertion of 42665274 to base 2 (Binary) is 10100010110000010100111010. The convertion of 42665274 to base 3 (Ternary) is 2222021121202100. The convertion of 42665274 to base 4 (Quaternary) is 2202300110322. The convertion of 42665274 to base 5 (Quintal) is 41410242044. The convertion of 42665274 to base 8 (Octal) is 242602472. The convertion of 42665274 to base 16 (Hexadecimal) is 28b053a. The convertion of 42665274 to base 32 is 18m19q. The sine of the figure 42665274 is 0.9994094169712. The cosine of 42665274 is -0.03436302183001. The tangent of the figure 42665274 is -29.083862936012. The square root of 42665274 is 6531.866042717.
If you square 42665274 you will get the following result 1820325605495076. The natural logarithm of 42665274 is 17.568895892058 and the decimal logarithm is 7.6300745390535. I hope that you now know that 42665274 is unique figure! | 757 | 2,073 | {"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} | 2.859375 | 3 | CC-MAIN-2021-21 | latest | en | 0.719711 |
www.inthefashionjungle.com | 1,709,377,121,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475806.52/warc/CC-MAIN-20240302084508-20240302114508-00186.warc.gz | 825,412,372 | 23,078 | Have you ever experienced swimming in a pool filled with soft water beads? Water beads are polymer-made multi-colored small balls that grow a hundred times their original size when submerged in water.
These beads come in all sorts of colors. They can be used to teach children about absorption and improve scientific thinking abilities such as observations and predictions in them.
As these are very tiny balls so the question that arises in our minds is how many water beads can fill a pool. Well, about 600,000 water beads are required to fill an average size water pool. Large pools need more. A 168-gallon pool is filled with 2,688 cups of water.
So, if we divide this by 72 cups that is the amount one 4 Orbeez bag supplies. Therefore, we need 37 bags of Orbeez to fill this pool and enjoy it to the fullest.
## How many water beads fill a gallon?
According to the information taken from the Orbeez website, 100 dried water beads produce one cup of 240 milliliters of fully hydrated Orbeez. As there are 16 cups in a gallon, 1600 water beads fill a gallon.
Similarly, for a 42-gallon tub, 67,200 water beads are needed to fill a standard-sized bathtub.
## How many water beads do I need?
First, add 1-2 cups of water into the container, then add 1-2 tablespoons of water beads into it. After that allow 6-8 hours for the water beads to absorb the water in the container.
Smaller water beads take less time to grow and will be ready for play and enjoyment in less than a day. Whereas, larger beads take up to 36 hours to be prepared.
## How long do water beads last for?
Water beads can be conserved almost indefinitely if they are kept in an airtight environment with low humidity. Usually, Water beads last for about two weeks. After this period, they shrink up and dehydrate themselves.
But if these are exposed to light in a vase for decorative purposes, then they last for approximately 1-2 years. If mixed in the soil, they stay for a long period of 7-9 years as a source to retain moisture.
## How much water do I add to 1000 Orbeez?
Water beads and Orbeez are the same. Orbeez is simply the registered brand for the product and water beads are the all-encompassing version.
Both are made of the same material, a polymer which is super absorbent and biodegradable and increases in size as it absorbs water.
Each Orbeez color seed pack is served with 1000 multi-colored Orbeez seeds to grow. To hydrate your Orbeez seeds, you will have to pour some of the seeds into a bowl of water and watch them magically rise right before your eyes.
In about four hours, your active seeds will become juicy and grown Orbeez. For 100 Orbeez, it is recommended to use 1 cup of water that contains 240 mL of warm water.
So if we are given 1000 Orbeez, then 10 cups of water are required. It is very important to use a sufficient quantity of water as Orbeez soak up the water as they assemble and expand in the bowl.
Moreover, if you wish to keep this lovely Orbeez for a long time then add a pinch of salt to the water. You can preserve this amount of Orbeez for multiple purposes. These can be used for party decorations, can be planted, and used as candle holders and flower holders.
## Conclusion:
Water beads are made of a combination of water and a water-absorbing polymer and can last for approximately 1-2 years. When dry water beads are drenched in water, they fill up and enhance just like a sponge.
Nearly 600,000 water beads and 37 bags of Orbeez fill a medium-sized pool. Almost 1600 water beads fill a gallon and 67,200 are required for a bathtub.
Moreover, 1-2 tablespoons of water beads are needed to get soaked in 1-2 cups of water and 10 cups of water are essential for 1000 Orbeez.
Check out our article on whether you can sell perler beads? | 885 | 3,777 | {"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} | 2.796875 | 3 | CC-MAIN-2024-10 | longest | en | 0.945677 |
https://www.paraisos-fiscales.info/answers/628721-if-x-nbsp-3-and-y-2-find-z-when-z-nbsp-nbsp-2x-y-2help-please-im-lost-and | 1,708,525,604,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00234.warc.gz | 965,167,380 | 4,760 | # If x= -3 and y= -2, find z when z = 2x y^2Help please :) Im lost and so confused
z=2x(y^2)
z=-6(-4)
z=24
## Related Questions
100 times 1 is 100
Anything times 1 is itself
100
Step-by-step explanation:100x 1 is all was going to be the same number
Brad puts an equal amount of money in savings account once a month.He started with \$25. The next month he had \$35 in his account. How much money will brad have in his account after 6 months?
35-25= 10
Brad saves ten dollars a month
6*10= 60
35+60= 95
35+60=\$95 in his account after 6 months.
Write a proportion that you can use to convert 60 inches to centimeters.
1 inch = 2.52 centimeters
55.8 ÷ (-3.1) do i estimate
Step-by-step explanation:
5. Lenny is moving tables in the schoolcafeteria. He places all the tables in a
7 X 4 array. How many tables are in
the cafeteria? | 264 | 842 | {"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} | 3.921875 | 4 | CC-MAIN-2024-10 | latest | en | 0.904654 |
https://www.doubtnut.com/question-answer/the-number-of-solutions-of-the-equation-3-cos-x24-2sin8x-in-0-9pi-is-equal-to-645080866 | 1,656,287,814,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00551.warc.gz | 792,335,429 | 103,945 | Home
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Updated On: 20-06-2022
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Transcript
question that was given to us that is the number of solutions of the equation that is 3 + cos x whole square is equals to 4 minus 2 Sin power 8X in that is interval close bracket 0295 open bracket is equals to have to find out the number of solutions it was given that 3 + cos x that is whole square is equal to that is 4 - 2 Sin power medicine powertex sin power 8X in the interval that is given it is 0295 open bracket so now we can say that we know that the range of cos x is between that the range of cos x is between that is greater than equal to minus one but less than equal to 1 so the range of 3 + cosx that should be greater than equals to that is so minus 1 + 3 that is minus 2 and 3 plus one that is 4 and the range of that is 3 + cos x whole square 3 + cos x whole square that should be close to that is greater than equal
to minus 2 whole square which is more in less than equal to 60 similarly we can we know that the range of sin x that the range of sin x is also between that is -1 to 1 so we can say that the sin8x sin power take that is greater than equal to zero but less than equal to one because its whole square Suite will be 10021 only and now the value of that is minus sin power 8X so now the this was - so the equality sign change is we get that is equals to the - 120 that is minus sin powertex is between that is equality science changes and NB x minus 1 so we get - is greater than equal to minus is equal to or greater than equal to minus one but less than equal to zero so now if we multiply it by that is minus of to now we will X to we get that is equals to G distributor -2 this will become
minus 2 root 2 x 2 so we get -2 here that is minus 2 Sin power a text that is less than 0 9 we will we will have to add the four sorry for -2 that is to Saudi 4 - 2 Sin power 8 x is greater than equal to 2 but less than equal to 4 from where we get these two things sunao VP draw is this on the number line we get that is this is relying so this is the point that is let this valuable that is to this is very 4 and this will be the 16 so the so the cause excess between that is 4260 and similarly that is this value that is sin x is between 2 to 4 and both are close bracket so they both have Sodhi Sodhi left inside and right inside is equals to left and right inside is
equals to show their common point is between that is equals to the four so from here we get that that 3 + cos x whole square is equals to 4 that the value of 3 + cos x whole square should also equals to 4 and that is 4 - 2 Sin power a text should also = 24 because the common point is on the four value we get if we taking under root on both side we get 3 + cosx that is equals to plus minus 2 and here we get that is 4 and 4 cancel out we get minus 2 sin a text that is equal to zero from here we can say that sin power 8 x is equals to zero and finally we get that is sin x is equals to zero to get the value that is sin x is equals to zero from here similarly hair we get that is we get two values one is that is 3 + cos x 3 + cos x is equals to plus two second values
that is 3 + cos x is equals to minus 2 Sohail Kashmiri gate that is cos x is equals to minus 2 and minus 3 x minus of 5 and from here that is cos x = 2 - 3 which is -1 but we know that but we know that the range of cos x is written -121 Saudi cos x = 2 - 5 it is not possible so the value of cos x is equals to get is -1 now we get the value of sin x = 20 and cos x is equal to minus we get that is causing is equal to -1 and the other values that is sin x = 20 this value is only possible this value is only possible these values only possible at only possible that ought at odd multiples of 5 you know that these values will be only possible at odd multiples of that its example that is by at 3555 and so on that
is equals to the 2n + 1 by 2n + 15 where that is any request 2012 and upto Infinity so that if we put zero we get that is why we put one we get 3542 big 855 so this is the founder and now the range that was given the question is that is zero 2 and 5 range is given that is between close bracket 0295 open bracket chunavi find the values we get the value that is our that is if we put 0 we get by if we put one we get 35 Puttu we get 55 if we put that is 4 he put three we get 7 by heavy before we get 9 Pie 5 wicket 11 5G ranges between 0295 so we can consider 11/5 and similarly there is close bracket this means 95 not included so we can't take 95 so we get 4 value that is pi 3 5 5 5 and
7 so there are 4 solutions there are four solutions of this equation we can see that there are four solutions of this equation sober answer from the options is that is first option that is ko thank you | 1,317 | 5,062 | {"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} | 4.21875 | 4 | CC-MAIN-2022-27 | latest | en | 0.966402 |
https://www.knowpia.com/knowpedia/Cosmic_time | 1,656,753,263,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00499.warc.gz | 903,243,767 | 15,671 | BREAKING NEWS
Cosmic time
## Summary
Cosmic time, or cosmological time, is the time coordinate commonly used in the Big Bang models of physical cosmology.[1][2][3] Such time coordinate may be defined for a homogeneous, expanding universe so that the universe has the same density everywhere at each moment in time (the fact that this is possible means that the universe is, by definition, homogeneous). The clocks measuring cosmic time should move along the Hubble flow.
Cosmic time ${\displaystyle t}$[4][5] is a measure of time by a physical clock with zero peculiar velocity in the absence of matter over-/under-densities (to prevent time dilation due to relativistic effects or confusions caused by expansion of the universe). Unlike other measures of time such as temperature, redshift, particle horizon, or Hubble horizon, the cosmic time (similar and complementary to the comoving coordinates) is blind to the expansion of the universe.
There are two main ways for establishing a reference point for the cosmic time. The most trivial way is to take the present time as the cosmic reference point (sometimes referred to as the lookback time).
Alternatively, the Big Bang may be taken as reference to define ${\displaystyle t}$ as the age of the universe, also known as time since the big bang. The current physical cosmology estimates the present age as 13.8 billion years.[6] The ${\displaystyle t=0}$ doesn't necessarily have to correspond to a physical event (such as the cosmological singularity) but rather it refers to the point at which the scale factor would vanish for a standard cosmological model such as ΛCDM. For instance, in the case of inflation, i.e. a non-standard cosmology, the hypothetical moment of big bang is still determined using the benchmark cosmological models which may coincide with the end of the inflationary epoch. For technical purposes, concepts such as the average temperature of the universe (in units of eV) or the particle horizon are used when the early universe is the objective of a study since understanding the interaction among particles is more relevant than their time coordinate or age.
Cosmic time is the standard time coordinate for specifying the Friedmann–Lemaître–Robertson–Walker solutions of Einstein's equations.
## Notes
1. ^ In mathematical terms, a cosmic time on spacetime ${\displaystyle M}$ is a fibration ${\displaystyle t\colon M\to R}$ . This fibration, having the parameter ${\displaystyle t}$ , is made of three-dimensional manifolds ${\displaystyle S_{t}}$ .
2. ^ On the physical basis of cosmic time by S.E. Rugh and H. Zinkernagel
3. ^ D'Inverno, Ray (1992). Introducing Einstein's Relativity. Oxford University Press. p. 312. ISBN 0-19-859686-3.
4. ^ Dodelson, Scott (2003). Modern Cosmology. Academic Press. pp. 29. ISBN 9780122191411.
5. ^ Bonometto, Silvio (2002). Modern Cosmology. Bristol and Philadelphia: Institute of Physics Publishing. pp. 2. ISBN 9780750308106.
6. ^ How Old is the Universe?
## References
• Dodelson, Scott (2003). Modern Cosmology. Academic Press. ISBN 978-0-12-219141-1.
• Bonometto, Silvio (2002). Modern Cosmology. Bristol and Philadelphia: Institute of Physics Publishing. ISBN 978-0750308106. | 763 | 3,213 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.928811 |
http://hackage.haskell.org/package/matrix-lens | 1,600,442,685,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00539.warc.gz | 63,845,896 | 8,513 | matrix-lens: Optics for the "matrix" package
[ bsd3, library, math ] [ Propose Tags ]
Versions [faq] 0.1.0.0 ChangeLog.md base (>=4.7 && <5), lens (>=4.19.2 && <4.20), matrix (>=0.3.6 && <0.4), vector (>=0.12.1 && <0.13) [details] BSD-3-Clause 2020 Interos, Inc. Interos, Inc. jevans@interos.ai Math https://github.com/interosinc/matrix-lens#readme https://github.com/interosinc/matrix-lens/issues head: git clone https://github.com/interosinc/matrix-lens by lgastako at 2020-07-07T06:39:57Z NixOS:0.1.0.0 38 total (4 in the last 30 days) (no votes yet) [estimated by Bayesian average] λ λ λ Docs available Last success reported on 2020-07-07
Modules
[Index] [Quick Jump]
Flags
NameDescriptionDefaultType
developer
Developer mode -- stricter handling of compiler warnings.
DisabledManual
Use -f <flag> to enable a flag, or -f -<flag> to disable that flag. More info
Maintainer's Corner
For package maintainers and hackage trustees
[back to package description]
matrix-lens
Optics for the matrix package.
That’s how it is with people. Nobody cares how it works as long as it works.
• Councillor Hamann
Sparked by this reddit post.
Examples
The examples below make use of the following three matrices:
exampleInt :: Matrix Int
exampleInt = Matrix.fromLists
[ [1, 2, 3]
, [4, 5, 6]
, [7, 8, 9]
]
exampleInvertible :: Matrix (Ratio Int)
exampleInvertible = Matrix.fromLists
[ [ -3, 1 ]
, [ 5, 0 ]
]
exampleNotSquare :: Matrix Int
exampleNotSquare = Matrix.fromLists
[ [10, 20, 30]
, [40, 50, 60]
, [70, 80, 90]
, [100, 110, 120]
]
Accessing individual elements:
λ> exampleNotSquare ^. elemAt (4, 3)
120
λ> exampleInt & elemAt (2, 2) *~ 10
┌ ┐
│ 1 2 3 │
│ 4 50 6 │
│ 7 8 9 │
└ ┘
Accessing individual columns:
λ> exampleInt ^. col 1
[1,4,7]
λ> exampleInt & col 2 . each *~ 10
┌ ┐
│ 1 20 3 │
│ 4 50 6 │
│ 7 80 9 │
└ ┘
Accessing individual rows:
λ> exampleInt ^. row 1
[1,2,3]
λ> exampleInt & row 2 . each *~ 100
┌ ┐
│ 1 2 3 │
│ 400 500 600 │
│ 7 8 9 │
└ ┘
Manipulating all columns as a list:
λ> exampleInt ^. cols
[[1,4,7],[2,5,8],[3,6,9]]
λ> exampleInt & cols %~ reverse
┌ ┐
│ 3 2 1 │
│ 6 5 4 │
│ 9 8 7 │
└ ┘
Accessing all rows as a list:
λ> exampleInt ^. rows
[[1,2,3],[4,5,6],[7,8,9]]
λ> exampleInt & rows %~ map reverse
┌ ┐
│ 3 2 1 │
│ 6 5 4 │
│ 9 8 7 │
└ ┘
λ> exampleInt & partsOf (dropping 1 (rows . each)) %~ reverse
┌ ┐
│ 1 2 3 │
│ 7 8 9 │
│ 4 5 6 │
└ ┘
In addition to the above there are also switching and sliding Isos for both rows and columns which allow you to swap two arbitrary rows or columns or slide a row or column through the matrix to a different row or column (moving all intervening rows or columns over in the direction of the source row or column):
λ> exampleNotSquare ^. switchingRows 1 4
┌ ┐
│ 100 110 120 │
│ 40 50 60 │
│ 70 80 90 │
│ 10 20 30 │
└ ┘
λ> exampleNotSquare ^. slidingRows 1 4
┌ ┐
│ 40 50 60 │
│ 70 80 90 │
│ 100 110 120 │
│ 10 20 30 │
└ ┘
..and similary for switchingCols and switchingRows.
An Iso exists for accessing the matrix with a given row scaled:
λ> exampleInt ^. scalingRow 1 10
┌ ┐
│ 10 20 30 │
│ 4 5 6 │
│ 7 8 9 │
└ ┘
λ> exampleInt & scalingRow 1 10 . flattened *~ 2
┌ ┐
│ -200 -400 -600 │
│ 8 10 12 │
│ 14 16 18 │
└ ┘
Any valid sub matrix can be accessed via the sub lens:
λ> exampleNotSquare ^. sub (2, 1) (3, 2)
┌ ┐
│ 40 50 │
│ 70 80 │
└ ┘
λ> exampleNotSquare & sub (2, 1) (3, 2) . rows %~ reverse
┌ ┐
│ 10 20 30 │
│ 70 80 60 │
│ 40 50 90 │
│ 100 110 120 │
└ ┘
The transposition of the matrix can be accessed via the transposed Iso:
λ> exampleInt ^. transposed
┌ ┐
│ 1 4 7 │
│ 2 5 8 │
│ 3 6 9 │
└ ┘
λ> exampleInt & transposed . taking 4 flattened *~ 10
┌ ┐
│ 10 20 3 │
│ 40 5 6 │
│ 70 8 9 │
└ ┘
You can also traverse the flattened matrix:
λ> exampleInt ^.. flattened
[1,2,3,4,5,6,7,8,9]
which is more useful for making modifications:
λ> exampleInt & flattened . filtered even *~ 10
┌ ┐
│ 1 20 3 │
│ 40 5 60 │
│ 7 80 9 │
└ ┘
λ> exampleInt & dropping 4 flattened *~ 10
┌ ┐
│ 1 2 3 │
│ 4 50 60 │
│ 70 80 90 │
└ ┘
Accessing the diagonal:
λ> exampleInt ^. diag
[1,5,9]
λ> exampleInt & diag %~ reverse
┌ ┐
│ 9 2 3 │
│ 4 5 6 │
│ 7 8 1 │
└ ┘
λ> exampleInt & diag . each *~ 10
┌ ┐
│ 10 2 3 │
│ 4 50 6 │
│ 7 8 90 │
└ ┘
Accessing inverse matrix is possible via the inverted optic. Since not all matrices have inverses inverted is a prism:
λ> exampleInvertible ^? inverted
Just ┌ ┐
│ 0 % 1 1 % 5 │
│ 1 % 1 3 % 5 │
└ ┘
λ> exampleInvertible & inverted . flattened *~ 2
┌ ┐
│ (-3) % 2 1 % 2 │
│ 5 % 2 0 % 1 │
└ ┘
Minor matrices can be accessed by specifying the (r, c) to be removed:
λ> exampleInt ^. minor (1, 2)
┌ ┐
│ 4 6 │
│ 7 9 │
└ ┘
λ> exampleInt & minor (1, 2) . flattened *~ 10
┌ ┐
│ 1 2 3 │
│ 40 5 60 │
│ 70 8 90 │
└ ┘
An Iso exists for accessing a scaled version of a matrix:
λ> exampleInt ^. scaled 10
┌ ┐
│ 10 20 30 │
│ 40 50 60 │
│ 70 80 90 │
└ ┘
λ> exampleInt & minor (1, 1) . scaled 10 . flattened +~ 1
┌ ┐
│ 1 2 3 │
│ 4 -510 -610 │
│ 7 -810 -910 │
└ ┘
Getters for the matrix determinant and size are also provided:
λ> exampleInt ^. determinant
Just 0
λ> exampleInvertible ^. determinant
Just ((-5) % 1)
λ> exampleNotSquare ^. determinant
Nothing
λ> exampleInt ^. size
(3,3)
λ> exampleInvertible ^. size
(2,2)
λ> exampleNotSquare ^. size
(4,3) | 2,324 | 5,879 | {"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} | 2.578125 | 3 | CC-MAIN-2020-40 | latest | en | 0.567949 |
http://openwetware.org/wiki/Makeschematic.py | 1,481,013,288,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541886.85/warc/CC-MAIN-20161202170901-00070-ip-10-31-129-80.ec2.internal.warc.gz | 207,622,365 | 5,920 | # Makeschematic.py
```import sys, pygame, re
from pygame.draw import line
#globals
unitlength = 20
def getcolend(acol, arrow):
if arrow%2 == 0:
return 62 + acol*4*unitlength+unitlength/2
else:
return 62 + acol*4*unitlength+unitlength*7/2
def getrowloc(arow):
if arow%2 == 0:
return 75+2*unitlength*arow+unitlength/2
else:
return 75+2*unitlength*arow-unitlength/2
def drawat(x,y,z, screener, cola1): #draw from (x,y) to (x,z)
yc = getrowloc(x)
if y < z:
xca = 62+4*unitlength*y+unitlength/2
xcb = 62+4*unitlength*z+7*unitlength/2
line(screener, cola1, (xca, yc), (xcb, yc),1)
else:
xca = 62+4*unitlength*z+unitlength/2
xcb = 62+4*unitlength*y+7*unitlength/2
line(screener, cola1, (xca, yc), (xcb, yc),1)
def drawup(col, row1, row2, screener, oliga, cola2,fonz): #draw from (y,x) to (z,x)
yc1 = getrowloc(row1)
yc2 = getrowloc(row2)
xc = getcolend(col, row1)
line(screener, cola2, (xc, yc1), (xc, yc2),1)
if col%2 == 0:
xc = xc-10
screener.blit(fonz.render(str(oliga),1,cola2),(xc,yc1))
screener.blit(fonz.render(str(oliga),1,cola2),(xc,yc2))
drawat(row1, col, col, screener, cola2) #possibly draws over previously drawn segments
drawat(row2, col, col, screener, cola2)
def maked(inf, outf):
pygame.init()
font18=pygame.font.Font("arial.TTF",18)
font12=pygame.font.Font("arial.TTF",12)
black = 0, 0, 0
white = 250, 250, 250
size = width, height = 1100, 3000
screen = pygame.display.set_mode(size)
screen.fill(white)
#numrows should be even for now
numrows = 72
for x in range(11):
line(screen, (175,175,175), (62+4*unitlength*(x+1), 0), (62+4*unitlength*(x+1), 3000), 1)
for x in range(12):
screen.blit(font18.render(str(x),1,black),(62+unitlength*(4*x+2)-5,32))
#rects for making arcs
arcthelad = pygame.Rect(0, 0, 2*unitlength, 2*unitlength)
for x in range(numrows/2):
screen.blit(font18.render(str(2*x),1,black),(31,75+unitlength*4*x-9))
line(screen, (0,0,0), (62, 75+unitlength*4*x), (62+12*4*unitlength, 75+4*unitlength*x),1)
pygame.draw.arc(screen, (250,0,0), arcthelad, -1.57, 1.57, 1)
screen.blit(font18.render(str(2*x+1),1,black),(31,75+(4*x+2)*unitlength-9))
line(screen, (0,0,0), (62+12*4*unitlength, 75+(4*x+2)*unitlength), (62, 75+(4*x+2)*unitlength),1)
pygame.draw.arc(screen, (250,0,0), arcthelad2, 1.57, 4.71, 1)
num = re.compile('\d+')
oligo = -1
row = 0
col = 0
xc = 0
yc = 0
oligocolor = 0,0,0
for lin in open(inf, 'r'):
haha = num.findall(lin)
haha[0] = int(haha[0])
haha[1] = int(haha[1])
haha[2] = int(haha[2])
oligocolor = (haha[0])%100, (haha[0] * 2)%100, (haha[0] * 3)%100 # darker colors
if haha[0] != oligo: #arrows
if (oligo != -1):
yc = getrowloc(row)
xc = getcolend(col, row)
if row%2 == 0:
pygame.draw.polygon(screen, oligocolor, ((xc,yc+unitlength/5),(xc,yc-unitlength/5),(xc-unitlength/2,yc)), 0)
screen.blit(font18.render(str(oligo),1,oligocolor),(xc,yc))
elif row%2 == 1:
pygame.draw.polygon(screen, oligocolor, ((xc,yc+unitlength/5),(xc,yc-unitlength/5),(xc+unitlength/2,yc)), 0)
screen.blit(font18.render(str(oligo),1,oligocolor),(xc,yc))
oligo = haha[0]
row = haha[1]
col = haha[2]
pygame.draw.circle(screen, oligocolor, (getcolend(col, row+1), getrowloc(row)), 2, 0)
else:
if haha[1] == row:
drawat(row, col, haha[2], screen, oligocolor)
else:
drawup(col, row, haha[1], screen, oligo, oligocolor, font12)
row = haha[1]
col = haha[2]
#didn't take care of last oligo in above
yc = getrowloc(row)
xc = getcolend(col, row)
if row%2 == 0:
pygame.draw.polygon(screen, oligocolor, ((xc,yc+unitlength/5),(xc,yc-unitlength/5),(xc-unitlength/2,yc)), 0)
screen.blit(font18.render(str(oligo),1,oligocolor),(xc,yc))
elif row%2 == 1:
pygame.draw.polygon(screen, oligocolor, ((xc,yc+unitlength/5),(xc,yc-unitlength/5),(xc+unitlength/2,yc)), 0)
screen.blit(font18.render(str(oligo),1,oligocolor),(xc,yc))
###
pygame.image.save(screen, outf)
pygame.quit() | 1,501 | 3,817 | {"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} | 2.84375 | 3 | CC-MAIN-2016-50 | latest | en | 0.386911 |
http://forum.arduino.cc/index.php?topic=110647.0 | 1,472,520,414,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982967939.90/warc/CC-MAIN-20160823200927-00171-ip-10-153-172-175.ec2.internal.warc.gz | 89,613,566 | 11,517 | Go Down
### Topic: How to Measure Small or Low Resistance (less than 1 Ohm). (Read 6561 times)previous topic - next topic
#### jaydie
##### Jun 19, 2012, 02:49 am
RefV variable contains a pre measured value, the measurement was performed by closing the circuit without R1 so basically it is the Voltage source or the VT(Voltage Total).
On my code below, the first thing it does is close the relay to add the resistor in the circuit then perform a measurement. The measure value will be stored at dRes. dRes now represent V2 in the circuit, then by doing RefV-dRes v1 is calculated. Since R2 is known in theory I can calculate for R1. With my code I was able to calculate for R1 if it is 100 OHM or more (tried and tested with 100, 470 and 1k Ohms), but I was hoping I could capture up to .2 Ohms if possible. I was trying it with my 2.4, 5.6 and 10 Ohms resistor and I can't seem to get an accurate measurement or it won't show measurement at all.
Code: [Select]
` digitalWrite(13, HIGH);//Turn the relay on and add the resistor to be measured in the circuit delay(500);//Stabilize the relay contacts test=analogRead(A0); dRes=map(test,0 ,1023, 0, 6600);//Map A/D digital data. 6V max voltage divider is two 1.5k OHM resistor with 1% tolerance dRes=(dRes/1000); digitalWrite(13, LOW); delay(10); float v1; float v2; float R2=3000;// Load resistors value-This is the voltage divider for the analog input composed of two 1.5K resistor float I; v1=RefV-dRes;//Get voltage drop on relay. RefV is the reference voltage. v2=dRes; I=(v2/R2);//Compute for Current-I=I1=I2=..... //By knowing the I, R1 can be calculated ContR=(v1/I);// ContR is the R1 in the circuit so R1=V1/I, R1 also represent the resistor to be measured dContR=ContR; //Transfere value to another variable (for LCD Display)`
I'm hoping to get away with simple circuit or without using amplification if possible.
Thanks.
#1
##### Jun 19, 2012, 03:38 am
Quote
but I was hoping I could capture up to .2 Ohms if possible.
The Arduino hardware is nowhere near good enough to read that sort of resolution, you have about .1% native accuracy but that doesn't include the non-linearities in the ADC. You're looking for 1 in 30000 accuracy.
Quote
can't seem to get an accurate measurement or it won't show measurement at all.
With a 5v reference the ADC has a resolution of ~5mV, I suspect the voltage across those small resistors is not enough to register.
Also if reading .1R in 3000R the other components have to be very accurate, resistors, power supply, voltage reference etc.
You need to amplify the signal, maybe using a current-sensing amp designed for current shunts would work OK, or a diff-amp.
______
Rob
Rob Gray aka the GRAYnomad www.robgray.com
#### Magician
#2
##### Jun 19, 2012, 04:34 am
Change your hardware setup, measure voltage across R1 :
GND ------ R1 -------R2 --------- +5V
Know, that arduino could read a voltage 100 mV more or less accurate, to measure 0.2 OHm you should supply 0.5 A into resistor. Check if resistor able to handle this current (and PSU). Than you choose R2 = 4.9V / 0.5A = 9.8 OHm.
To decrease a current, or measure even smaller R, you need OPA, which could bring 1000 000! times difference, make it possible to measure as low R as 0.000 0002 OHm, or use uA current,
#### jaydie
#3
##### Jun 19, 2012, 05:36 pm
Thank you so much for a quick respond. Your inputs are noted and highly appreciated. You have pointed me on the right direction and saved me a lot of time to research now that I know where to start.
Jaydie
#### MarkT
#4
##### Jun 19, 2012, 10:13 pm
Standard way is to wire the resistor as a 4-terminal device (I think the term is a Kelvin resistor) and measure the voltage of both ends using the sense wires and take the difference. No current must flow along the sense wires, you use the other two connections for that.
If the voltages are going to be small a rail-to-rail instrumentation op-amp is the best way to boost the differential signal level up to the 5V range.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
Go Up
Please enter a valid email to subscribe | 1,151 | 4,201 | {"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} | 3.15625 | 3 | CC-MAIN-2016-36 | longest | en | 0.858256 |
https://www.cleancss.com/convert-units/units-of-power/kwhr/ftlb | 1,722,702,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00896.warc.gz | 562,968,091 | 18,372 | # Convert KWHR to FTLB
1
Kilowatt Hours
=
2655219.7194318
Foot Pounds
How To Calculate Kilowatt Hours To Foot Pounds
To convert kilowatt hours to foot pounds you simply multiply your kilowatt hours by 2655219.7194318. The formula would look like this:
Yftlb = Xkwhr * 2655219.7194318
#### 1 Kilowatt Hours equals
Joules 3.6e+06 Kilojoules 3600 Megajoules 3.6 Calories 860.421 Newton Meters 3.6e+06 Foot Pounds 2.65522e+06 Watt Hours 1000 Megawatt Hours 0.001 Mega Electron Volts 2.25e+22
#### Kilowatt Hours To Foot Pounds Conversion Table
From To
1 kwhr2655219.7194318 ftlb
2 kwhr5310439.4388636 ftlb
3 kwhr7965659.1582953 ftlb
4 kwhr10620878.877727 ftlb
5 kwhr13276098.597159 ftlb
6 kwhr15931318.316591 ftlb
7 kwhr18586538.036022 ftlb
8 kwhr21241757.755454 ftlb
9 kwhr23896977.474886 ftlb
10 kwhr26552197.194318 ftlb
11 kwhr29207416.91375 ftlb
12 kwhr31862636.633181 ftlb
13 kwhr34517856.352613 ftlb
14 kwhr37173076.072045 ftlb
15 kwhr39828295.791477 ftlb
16 kwhr42483515.510909 ftlb
17 kwhr45138735.23034 ftlb
18 kwhr47793954.949772 ftlb
19 kwhr50449174.669204 ftlb
20 kwhr53104394.388636 ftlb
30 kwhr79656591.582953 ftlb
40 kwhr106208788.77727 ftlb
50 kwhr132760985.97159 ftlb
60 kwhr159313183.16591 ftlb
70 kwhr185865380.36022 ftlb
80 kwhr212417577.55454 ftlb
90 kwhr238969774.74886 ftlb
100 kwhr265521971.94318 ftlb | 561 | 1,329 | {"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} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.39798 |
https://rustgym.com/leetcode/1685 | 1,669,572,452,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00423.warc.gz | 552,618,065 | 3,324 | 1685. Sum of Absolute Differences in a Sorted Array
You are given an integer array `nums` sorted in non-decreasing order.
Build and return an integer array `result` with the same length as `nums` such that `result[i]` is equal to the summation of absolute differences between `nums[i]` and all the other elements in the array.
In other words, `result[i]` is equal to `sum(|nums[i]-nums[j]|)` where `0 <= j < nums.length` and `j != i` (0-indexed).
Example 1:
```Input: nums = [2,3,5]
Output: [4,3,5]
Explanation: Assuming the arrays are 0-indexed, then
result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,
result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,
result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.
```
Example 2:
```Input: nums = [1,4,6,8,10]
Output: [24,15,13,15,21]
```
Constraints:
• `2 <= nums.length <= 105`
• `1 <= nums[i] <= nums[i + 1] <= 104`
1685. Sum of Absolute Differences in a Sorted Array
``````struct Solution;
impl Solution {
fn get_sum_absolute_differences(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len();
let mut prev = nums[0];
let mut prev_sum = 0;
let mut res = vec![0; n];
for i in 0..n {
let diff = nums[i] - prev;
prev_sum += diff * i as i32;
res[i] += prev_sum;
prev = nums[i];
}
prev = nums[n - 1];
prev_sum = 0;
for i in (0..n).rev() {
let diff = (nums[i] - prev).abs();
prev_sum += diff * (n - 1 - i) as i32;
res[i] += prev_sum;
prev = nums[i];
}
res
}
}
#[test]
fn test() {
let nums = vec![2, 3, 5];
let res = vec![4, 3, 5];
assert_eq!(Solution::get_sum_absolute_differences(nums), res);
let nums = vec![1, 4, 6, 8, 10];
let res = vec![24, 15, 13, 15, 21];
assert_eq!(Solution::get_sum_absolute_differences(nums), res);
}
`````` | 614 | 1,678 | {"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} | 3.5625 | 4 | CC-MAIN-2022-49 | latest | en | 0.654145 |
https://engineerboards.com/profile/33166-zach-stone-pe/ | 1,603,601,739,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107887810.47/warc/CC-MAIN-20201025041701-20201025071701-00243.warc.gz | 323,077,063 | 17,877 | Engineer Boards
# Zach Stone, P.E.
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1. ## April/October 2020 Power PE Exam Prep
Keep your head up everyone! Don't forget to take a break when you need to. I highly recommend taking advantage of the extra month or two you have before your CBT exam date compared to the October exam date to deep dive on areas you are most uncomfortable with.
2. ## 2020/2021 Power CBT Exam Thread
FYI for anyone that hasn't already seen it: First Impressions of the New NCEES® Electrical Power Reference Handbook for the Computer Based Testing (CBT) Exam.
3. ## NCEES PE electrical and Computer practice exam 512
I'm assuming this is the NCEES practice problem where there are two conductors run in parallel because this question comes up often. What is the equivalent resistance of two identical resistors in parallel? if: R1 = R2 = R Then: Req = R//R Req = (R*R)/(R+R) Req = (R*R)/(2R) Req = R/2 The value is halved. The same happens with conductors run in parallel. Each of the NCEES practice exam problems have been discussed to great length on engineer boards with multiple threads. You can find them easily by using the search function instead of making a new threa
4. ## NCEES PE Power Prep Exam Problem 515
The problem is not asking for the output voltage of the half wave rectifier. It's asking for the reverse minimum voltage rating of the diode. The voltage rating of the diode is how much voltage it can be subjected to without failing or becoming damaged. You have to calculate what the maximum voltage is across the diode during the entire period of the input AC voltage signal (from zero to +Vpeak, back to zero, to -Vpeak, and back to zero). In order to use a diode in this circuit, the diode will have to be rated for a "minimum" of this value, meaning as long as the voltage rating
5. ## Voltage Drop percentage - percentage of sending/source voltage (Shorebrook Question 49)
The NEC handbook edition gives some good examples using the magnitude approximation method only (by subtracting magnitudes instead of complex numbers).
6. ## Voltage Drop percentage - percentage of sending/source voltage (Shorebrook Question 49)
This is incorrect. Voltage drop percent by definition in the ratio of the amount dropped across the conductor to the amount supplied by the source.
7. ## EB Mafia
I would love to. Unfortunately, we are currently in the middle of this semester's live class program while trying to incorporate a lot of changes for the new CBT version of the power PE exam has me really putting in the hours. I've never played before, I'd be happy to take a rain check for anytime after the (now cancelled) Oct 23rd PE exam
8. ## Battery reference material
Not a bad book at all for the price but it is not as in depth as the Linden handbook.
9. ## EB Mafia
Never played before, but sounds fun!
10. ## EB Mafia
Wow this is a big thread 😮
11. ## Just got an NCEES email saying October PE exams are canceled.
Yes, I contacted NCEES per @Tim @ NCEES's instructions. They said they are working on revising issues.
12. ## 2-wattmeter load power factor angle
If you're looking for a breakdown of the phasor diagram relationships these two videos sort it out:
13. ## Just got an NCEES email saying October PE exams are canceled.
Thanks Tim, will do.
14. ## Just got an NCEES email saying October PE exams are canceled.
Hi @Tim @ NCEES, I've found a few mistakes in the new NCEES® Power PE Reference Manual for the CBT exam. One of them is a formula mistake with the variable in the wrong place that will result in a wrong answer for anyone that uses it. Who can I can notify at NCEES to help bring this to the right persons attention?
Good catch.
×
• Pages | 1,010 | 4,299 | {"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} | 3.484375 | 3 | CC-MAIN-2020-45 | latest | en | 0.862168 |
https://www.doorsteptutor.com/Exams/NCO/Class-2/Questions/Topic-Mental-Ability-2/Subtopic-Basic-Arithmetic-1/Part-23.html | 1,527,408,796,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00477.warc.gz | 730,561,879 | 13,405 | # Mental Ability-Basic Arithmetic (NCO- Cyber Olympiad (SOF) Class 2): Questions 182 - 190 of 338
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## Question number: 182
» Mental Ability » Basic Arithmetic
MCQ▾
### Question
I have 12 baseball cards. Here are 6. The rest are in my book. How many are there in my book?
### Choices
Choice (4) Response
a.
12
b.
17
c.
8
d.
6
## Question number: 183
» Mental Ability » Basic Arithmetic
MCQ▾
### Question
Start at 9. Count on 3. What number do you reach?
### Choices
Choice (4) Response
a.
13
b.
14
c.
12
d.
10
## Question number: 184
» Mental Ability » Basic Arithmetic
MCQ▾
### Choices
Choice (4) Response
a.
It is less than
b.
It is the same as
c.
It is greater than
d.
It is the same as
## Question number: 185
» Mental Ability » Basic Arithmetic
MCQ▾
### Question
I have 22 baseball cards. Here are 16. The rest are in my book. How many are there in my book?
### Choices
Choice (4) Response
a.
6
b.
17
c.
8
d.
12
## Question number: 186
» Mental Ability » Basic Arithmetic
MCQ▾
230 is ________
### Choices
Choice (4) Response
a.
Less than
b.
Greater than
c.
Equal to
d.
None of the above
## Question number: 187
» Mental Ability » Basic Arithmetic
MCQ▾
120 is ________
### Choices
Choice (4) Response
a.
Greater than
b.
Less than
c.
Equal to
d.
All of the above
## Question number: 188
» Mental Ability » Basic Arithmetic
MCQ▾
120 is ________
### Choices
Choice (4) Response
a.
Less than
b.
Greater than
c.
Equal to
d.
None of the above
## Question number: 189
» Mental Ability » Basic Arithmetic
MCQ▾
### Question
What 2 numbers could balance this scale?
### Choices
Choice (4) Response
a.
38 and 4
b.
5 and 27
c.
39 and 10
d.
26 and 7
## Question number: 190
» Mental Ability » Basic Arithmetic
MCQ▾
If , then
### Choices
Choice (4) Response
a.
15
b.
14
c.
12
d.
13
f Page | 633 | 2,154 | {"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} | 2.546875 | 3 | CC-MAIN-2018-22 | latest | en | 0.744022 |
https://www.mumsnet.com/Talk/education/62490-do-they-learn-times-tables-by-chanting-still | 1,544,840,153,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00336.warc.gz | 962,149,749 | 27,725 | # Talk
## do they learn times tables by chanting still?
(10 Posts)
Cod Fri 04-Mar-05 08:29:44
Message withdrawn
soapbox Fri 04-Mar-05 08:32:59
Cod - DD's been doing them this year and seems to have learned them by 'counting in' different numbers. So started in Y1 counting 2,4,6,8... then 5,10,15... Now this year 3,6,9... 4,8,12...
Once they are reasonably proficient at the counting in different bases they then add on the 2*3 = 6 etc etc.
I don't think they have ever chanted in the way we would have done as children. Not least because the whole class is at completely different levels in maths and don't seem to do much whole class work in this subject!
Miaou Fri 04-Mar-05 08:33:51
My kids do chanting, clapping their hands at the same time to create a rhythm. works really well if you ask me.
soapbox Fri 04-Mar-05 08:33:57
We also only used to go up to x * 12, they go up to x*15 - which taxes my brain sometimes
roisin Fri 04-Mar-05 08:38:43
Cod at our school they do lots of counting in 2s, counting in 3s etc. in yr1. But then in yr2 they do start chanting tables x2, x5 and x10. (I only know this because ds1's yr2 report said he often declined to join in!) And progressing in yr3. They are expected to learn them by rote.
Tbh ds1 hasn't really got this yet. He can 'work them out' so quickly he doesn't see the point of learning them by rote.
I hope he picks them up at school soon, otherwise we are going to have to start working on them at home, which doesn't really appeal
Miaou Fri 04-Mar-05 08:54:34
At dd1's old school, they used to recite the tables (2 nothings are nothing, 2 ones are two etc), and be timed doing them. They had a chart called "beat the teacher" to see if they could do it faster than the teacher. However, this soon got changed to "Beat dd1"! She thrives on this style of learning, obviously!
lunavix Fri 04-Mar-05 08:56:59
They go up to *15????
Dear god
I only just managed up to *12! What are they going to be doing by the time ds goes to school, *18??
Cod Fri 04-Mar-05 09:13:37
Message withdrawn
Cod Fri 04-Mar-05 09:14:24
Message withdrawn
Cam Fri 04-Mar-05 11:06:47
At dd's school they go up to x10, this makes sense to me now that we have decimilisation. Surely up to x12 was because of old money?
Anyway, dd's class don't seem to chant in the way we did (and boy did we - I have all the times tables fixed in my head forever) but IMO it is the only way to learn them. We have a tape in the car and I randomly ask her what's x times x? She is confident in 2,3,4,5,6,7 and 10 but is still learning 8 and 9 (year 3).
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Registering is free, easy, and means you can join in the discussion, watch threads, get discounts, win prizes and lots more. | 789 | 2,734 | {"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} | 2.8125 | 3 | CC-MAIN-2018-51 | latest | en | 0.973531 |
https://stonespounds.com/505-6-stones-in-stones-and-pounds | 1,606,424,798,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00434.warc.gz | 507,302,472 | 4,646 | # 505.6 stones in stones and pounds
## Result
505.6 stones equals 505 stones and 8.4 pounds
You can also convert 505.6 stones to pounds.
## Converter
Five hundred five point six stones is equal to five hundred five stones and eight point four pounds (505.6st = 505st 8.4lb). | 78 | 279 | {"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} | 3.0625 | 3 | CC-MAIN-2020-50 | latest | en | 0.843124 |
http://www.governmentadda.com/quant-quiz-for-sbi-po-ibps-clerk-railway-ibps-so-canara-bank-po-rbi-other-exams-9/ | 1,553,065,717,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202303.66/warc/CC-MAIN-20190320064940-20190320090940-00429.warc.gz | 279,097,009 | 29,096 | Wednesday , March 20 2019
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# Quant Quiz For SBI PO | IBPS Clerk | Railway | IBPS SO | Canara Bank PO | RBI & Other Exams
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Directions (Q. 1–5): What approximate value should come in the place of question mark (?) in the following questions?
Q1. 12.98 × 139.90 = ?
1) 1720 2) 1620 3) 1680 4) 1820 5) 1860
Q2. 139.89% of 1580 = ?
1) 2212 2) 2342 3) 2416 4) 2314 5) 2312
Q3. 8195.01 ÷ 744.9 + ? × 11.9 = 7847.01
1) 683 2) 753 3) 783 4) 693 5) 653
Q4. 49.89 × 42.9 = ?
1) 2050 2) 2150 3) 2460 4) 2250 5) 2480
Q5. 14.9% of 124 = ?
1) 10.6 2) 27.6 3) 18.6 4) 24.6 5) 28.6
Directions (Q. 6 – 10): In the following number series one number is wrong. Find out the wrong number.
Q6. 27, 28, 30, 35, 42, 58
1) 30 2) 28 3) 58 4) 35 5) 42
Q7. 6, 12, 48, 288, 2304, 2890
1) 2890 2) 288 3) 2304 4) 12 5) 48
Q8. 31, 38, 50, 70, 95, 126
1) 38 2) 126 3) 50 4) 95 5) 70
Q9. 10, 10, 15, 30, 75, 100
1) 100 2) 15 3) 10 4) 75 5) 30
Q10. 21, 22, 26, 35, 50, 76
1) 22 2) 26 3) 76 4) 35 5) 50
GovernmentAdda Recommends Online Tyari Mock Test RRB Group D Speed Test Attempt Now Attempt Now Railway RRB Pshyco Complete Package Get It Now SSC CHSL, CGL Mock Test (1 free+30 Paid) Attempt Now | 802 | 2,128 | {"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} | 3.8125 | 4 | CC-MAIN-2019-13 | latest | en | 0.352619 |
https://www.jiskha.com/display.cgi?id=1267811712 | 1,516,377,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888041.33/warc/CC-MAIN-20180119144931-20180119164931-00409.warc.gz | 942,492,518 | 4,145 | # Math
posted by .
Match the trigonometric expression:
cos^2 [(pi/2) - x] / cos x
Does csc x match with this?
• Math -
cos(pi/2-x) = sinx
(e.g. cos 70 = sin 20)
[cos^2 (pi/2-x)]/cosx = sin^2 x/cosx
= (sinx/cosx)sinx
= tanxsinx
Don't know what you mean by "match with this"
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More Similar Questions | 885 | 2,548 | {"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} | 4 | 4 | CC-MAIN-2018-05 | latest | en | 0.646193 |
http://math.stackexchange.com/questions/218085/finding-reflection-of-a-matrix | 1,469,723,468,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828286.80/warc/CC-MAIN-20160723071028-00325-ip-10-185-27-174.ec2.internal.warc.gz | 155,881,577 | 17,523 | # Finding reflection of a matrix
To 2 decimal places, what is the value of the lower-right entry in the reflection matrix $Q_a$if a = 1.05?
Not even sure where to begin, is there a formula?
This is what I could find in my textbook
-
"in the reflection matrix $Q_a$" - what is the definition of $Q_a$ ? – Belgi Oct 21 '12 at 15:37
Thats it the question I was given. – JackyBoi Oct 21 '12 at 15:45
@Belgi this has to do with transformations – JackyBoi Oct 21 '12 at 15:46
There is no universal definition of $Q_a$. You're going to have consult your textbook, class notes, or your instructor to determine the meaning of this symbol before proceeding. – Michael Joyce Oct 21 '12 at 16:02
@MichaelJoyce I have edited my question please take a look. – JackyBoi Oct 22 '12 at 1:25
Okay, well the lower right entry of the matrix $q_a$ is $-\cos(2a)$. So plug in $a = 1.05$ and use a calculator to get an approximate value for $-\cos(2.10) \approx 0.50485$. Be sure to use 'radian' mode on your calculator. (Always assume angle measures for trigonometric functions are in radians unless explicitly told otherwise.) | 320 | 1,110 | {"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} | 2.9375 | 3 | CC-MAIN-2016-30 | latest | en | 0.860057 |
http://alchemipedia.blogspot.com/2009/08/centipede-game_11.html | 1,601,307,167,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00754.warc.gz | 5,705,018 | 103,565 | Tuesday
Centipede Game (Game Theory) Rosenthal 1981
Centipede Game Important Facts:
• Is an extensive form game in the field of mathematical game theory.
• This game was first introduced by Rosenthal (1981).
• Two players take turns choosing either to take a slightly larger share of a slowly increasing pot, or to pass the pot to the other player.
• Payoffs are arranged so that if one passes the pot to one's opponent and the opponent takes the pot on the next round, one receives slightly less than if one had taken the pot on the previous round.
• The traditional centipede game had a limit of 100 rounds (this is how the centipede name came about).
• The Nash equilibrium indicates that the first player take the pot on the very first round of the game.
• In real world testing relatively few players will take the pot on the first round.
• The Centipede game is commonly used in teaching to introduce the concept of backward induction and the iterated elimination of dominated strategies.
Centipede Game Significance
• Self-interest or distrust can interfere with cooperation to create a game situation where both players may do worse than if they had blindly cooperated. | 249 | 1,179 | {"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} | 2.65625 | 3 | CC-MAIN-2020-40 | latest | en | 0.947559 |
http://en.wikibooks.org/wiki/A-level_Physics/Electrons,_Waves_and_Photons/D.C._circuits | 1,432,503,892,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928078.25/warc/CC-MAIN-20150521113208-00108-ip-10-180-206-219.ec2.internal.warc.gz | 81,751,762 | 11,042 | # A-level Physics/Electrons, Waves and Photons/D.C. circuits
A direct current (DC) circuit usually has a steady and constant voltage supplied to it. A direct current does not have a continually changing polarity, unlike an alternating current (AC), but instead a constant direction and rate of flow. DC is generally provided by batteries or via a transformer, rather than generators.
## Circuit diagrams
Below are the symbols and names for all of the components that you are required to know:
## Series circuits
When resistors are set up in series, the formula to work out the total resistance is:
$R_T = r_1+r_2+r_n...etc$
Where $r_1, r_2$ etc., are the resistance of each resistor in series.
## Parallel circuits
When resistors are set up in parallel, the formula to work out the total resistance is:
$\frac{1}{r_t}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_n} ...etc$
Where $r_1, r_2$ etc., are the resistance of each resistor in parallel.
## Internal resistance
A electrical source has its own resistance, known as Internal Resistance. This is caused by the electrons in the source having to flow through wires within it, or in the case of a chemical battery, the charge may have to flow through the electrolytes and electrodes that make up the cell.
By considering a battery of EMF E, in series with a resistor of resistance R we can calculate the internal resistance r:
$E=I(R+r)$ (See Series Circuits above)
Combining with V=IR:
$V=E-Ir$
The quantity Ir is called the lost volts. The lost volts shows us the energy transferred to the internal resistance of the source, so if you short circuit a battery, I is very high and the battery gets warm.
## Potential dividers
A potential (or voltage) divider is made up of two resistors. The output voltage from a potential divider will be a proportion of the input voltage and is determined by the resistor values.
The values of a battery with voltage V1 passing through two resistors in series of resistance R1 and R2, with an output circuit in parallel with Resistor R1 with output voltage V2 are related by the equation: $V1=(R1/(R1+R2))$ × $V2$
## Kirchhoff's laws
First Law states "The sum of the current (A) entering a junction is equal to the sum of the current (A) leaving the junction". This is a consequence of conservation of charge.
Second law states that the EMF is equal to the voltage of the circuit. This is a consequence of conservation of energy.
## Use of other components
Thermistors can be placed in circuits when temperature plays a role. As the temperature increases, the resistance of the device decreases. This does not obey the Ohms law. Light dependent resistors are resistors that decrease their resistance when exposed to light. | 634 | 2,730 | {"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} | 4.15625 | 4 | CC-MAIN-2015-22 | latest | en | 0.919302 |
http://math.stackexchange.com/questions/821058/quadratic-formula-question | 1,469,529,151,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00176-ip-10-185-27-174.ec2.internal.warc.gz | 164,540,032 | 18,028 | Concerning the quadratic formula. What does it mean if $b^2-4ac>0$, $b^2-4ac<0$, and $b^2-4ac=0$?
-
What do you exactly mean? Those expresions are inequalities, and their content is the discriminant of a quadratic expresion $ax^{2}+bx+c=0$. Can you explain your question please? You could add another flag too, just like "homework" or "algebra-precalculus". – Solid Snake Jun 4 '14 at 23:29
Due to the nature of the quadratic formula I removed the square root signs in the three inequalities. – Américo Tavares Jun 4 '14 at 23:36
$\Delta={ b^2-4ac}>0$ means that the equation has two real solutions.
$\Delta= { b^2-4ac}<0$ means that the equation has no real solutions, but two complex solutions.
$\Delta={ b^2-4ac}=0$ means that the equation has one solution.
-
And when it's $0$, the solution is real. I think it's worth pointing out that all of this can be determined by a careful look directly at the quadratic formula and thinking about what happens in each of these cases. – jpmc26 Jun 5 '14 at 0:37
Yes, you're right!!! – Mary Star Jun 5 '14 at 0:39
Since one has the term $\sqrt{b^2-4ac}$ in solution of the quadratic formula, if $b^2-4ac>0$, then the equation has two real solutions (since $\sqrt{b^2-4ac}$ is real, and $b$ and $2a$ are also real). When $b^2-4ac<0$, then the quadratic equation has two complex solutions (since $\sqrt{b^2-4ac}$ is complex imaginary). If $b^2-4ac=0$, then $\sqrt{b^2-4ac}=0$, implying that the solution is $x=\dfrac{-b}{2a}$ (can you see why?).
- | 465 | 1,495 | {"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} | 4.0625 | 4 | CC-MAIN-2016-30 | latest | en | 0.893358 |
https://www.enotes.com/homework-help/can-any-one-theoretically-prove-that-317434 | 1,516,118,194,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00714.warc.gz | 902,298,393 | 10,129 | # CAN ANY ONE THEORETICALLY PROVE THAT - * - IS +?????I actually noe the answer . i am just expecting some new methods. many people give logicall proofs. please give me theoritical proofs...
beckden | Certified Educator
We are going to use the following definitons
0 is the additive identity and 0 + a = a + 0 = a
For any number a there exists an additive inverse denoted by -a, which has the property that a + -a = -a + a = 0
(c) Multiplicative identity property
1 is the multiplicative identity and 1 * a = a * 1 = a
(d) Multiplicative property of 0
For any a 0 * a = a * 0 = 0
(d) Distributive property
For any three numbers a, b and c, a(b + c) = ab + ac
For any a, b and c if a = b then a + c = b + c
(f) Transitive property
For any a, b, and c. If a = c and b = c then a = b.
Lemma : -a = (-1)a
a + -a = 0 Inverse property of addition
a* 0 = 0 Multiplicative property of zero
1 + -1 = 0 Inverse property of addition
a (1 + -1) = 0 because 0 = 1 + -1
a + (-1)a = 0 because of the distributive property
a + -a + (-1)a = -a because of the additive property of equality
0 + (-1)a = -a because of the inverse property of equality
(-1)a = -a because of the identity property of addition
By definiton
b + -b = 0 Because of the definiton of additive inverse
-a * 0 = 0 By the multiplicative property of 0
-a (b + -b) = 0 By the transitive property
-a(b) + -a(-b) = 0 By the distributive property.
ab + -ab + (-a)(-b) = ab By the additive property of equality
0 + (-a)(-b) = ab By the additive inverse property.
(-a)(-b) = ab By the additive identity property | 507 | 1,616 | {"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} | 4.28125 | 4 | CC-MAIN-2018-05 | latest | en | 0.82007 |
https://www.physicsforums.com/threads/heat-equation-periodic-heating-of-a-surface.801470/ | 1,527,358,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867841.63/warc/CC-MAIN-20180526170654-20180526190654-00177.warc.gz | 838,721,493 | 18,526 | # Homework Help: Heat equation, periodic heating of a surface
1. Mar 5, 2015
### bobred
1. The problem statement, all variables and given/known data
The temperature variation at the surface is described by a Fourier
series
$$\theta(t)=\sum^\infty_{n=-\infty}\theta_n e^{2\pi i n t /T}$$
find an expression for the complex Fourier
series of the temperature at depth $d$ below the surface
2. Relevant equations
Solution of the diffusion equation
$$\theta(x,t)=\cos\left(\phi+\omega t-\sqrt{\dfrac{\omega}{2D}}x\right)\exp\left(-\sqrt{\dfrac{\omega}{2D}}x\right)$$
3. The attempt at a solution
A the surface $x=0$ so
$$\theta(0,t)=\cos\left(\omega t + \phi\right)$$
To find the coefficients $\theta_n$ I'm guessing I use the Fourier formula
$$\theta_n=\frac{1}{T}\int_0^T dt \, \theta(0,t) \exp\left( -2\pi i n t/T \right)$$
2. Mar 6, 2015
### Staff: Mentor
I don't think you can calculate the $\theta_n$. They are some given (but unknown) constants, and you have to modify your solution to fit this constraint.
You can start with an easy case: imagine $\theta_1=1$ and $\theta_n=0$ for all other n. That makes the temperature at x=0 a sine. Can you find the temperature at depth d?
What happens with $\theta_2=3$ and $\theta_n=0$ for all other n? What happens in the general case?
3. Mar 6, 2015
### bobred
Typo, $\theta(x,t)$ should be
$\theta(x,t)=A\cos\left(\phi+\omega t-\sqrt{\dfrac{\omega}{2D}}x\right)\exp\left(-\sqrt{\dfrac{\omega}{2D}}x\right)$ (c)
'The temperature is a superposition of solutions found in (c).
On the surface x=0...use this to determine the coefficients in the above and complete.'
Assuming I have got (c) correct!
James
4. Mar 7, 2015
### bobred
Appologies, $A=7.5^{\circ}$C, $\omega=\frac{\pi}{43200} s^{-1}$, $\phi=\frac{4}{3}$ and $D=5\times10^{-7} m^2/s$
5. Mar 7, 2015
### Staff: Mentor
Where do those numbers come from? You cannot fix them like that for the problem.
A superposition of those solutions is the right approach.
6. Mar 7, 2015
### bobred
These were given earlier in the question and you are right have no baring on this.
So to get the coefficients I should set x=0 and use the Fourier formula to find them?
7. Mar 7, 2015
Yes.
8. Mar 7, 2015
### bobred
I have done some calculations but getting trivial answers.
I am taking
$$\theta(0,t)=\cos\left(2\pi n \ t/T + \phi\right)$$
and converting it to its complex exponential form and inserting into
$$\theta_n=\frac{1}{T}\int_0^T dt \, \theta(0,t) \exp\left( -2\pi i n t/T \right)$$
Should phi be included?
Last edited: Mar 7, 2015
9. Mar 7, 2015
### pasmith
You probably want to use the solution of the diffusion equation in the form $$Ce^{i\omega t}\exp\left(-\sqrt{\frac{\omega}{2D}}x - i\sqrt{\frac{\omega}{2D}}x \right)$$ with $x$ measuring distance below the surface.
10. Mar 7, 2015
### Staff: Mentor
You'll need phi if θn can be complex.
I still don't see what "inserting into [long equation]" is supposed to mean, but I agree that you don't need long calculations.
11. Mar 7, 2015
### bobred
By inserting into, I mean to work out the coefficients.
I'm a bit lost at the moment.
12. Mar 7, 2015
### pasmith
You should treat the coefficients $\theta_n$ as known. You are asked for "an expression for the complex Fourier
series of the temperature at depth d below the surface" so you are looking for an expression of the form $$\theta(d,t) = \sum_{n=-\infty}^\infty e^{2n\pi it/T}f_n(d)$$ where $e^{2n\pi it/T}f_n(x)$ is a solution of the diffusion equation with $f_n(0) = \theta_n$.
13. Mar 7, 2015
### bobred
So would
$$f_n(0)=\cos\left( 2\pi n t /T + \phi \right)$$ | 1,160 | 3,612 | {"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": 2, "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} | 3.625 | 4 | CC-MAIN-2018-22 | latest | en | 0.764538 |
https://qualifiedgraduatetutors.com/homework-21-a-parameter-is-to-a-population-as-a-statistic-is-to-a-__________a-independent-variableb-samplec-z-scored-correlation2-what-is-the-scale/ | 1,709,141,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00153.warc.gz | 470,281,484 | 19,464 | ### Our Services
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# Homework 2 1. A parameter is to a population as a statistic is to a __________ a. Independent variable b. Sample c. Z-score d. Correlation 2. What is the scale
Homework 2
1. A parameter is to a population as a statistic is to a __________
a. Independent variable
b. Sample
c. Z-score
d. Correlation
2. What is the scale of measure for the variable hours worked?
a. Nominal
b. Ordinal
c. Interval
d. Ratio
3. Create a frequency distribution for the following data set; 3, 5, 5, 5, 6, 6, 7, 7, 7, 8, 10
4. Create a histogram from the data in 3.
5. Create a frequency polygon from the data in 3.
6. The following data set is a sample of voter ages in the recent Presidential election; 21, 21, 21, 18, 19, 20, 20, 25, 26, 27, 27, 32, 55, 55, 67, 67, 67, 44, 45, 86, 96, 56, 67, 37, 39
7. Create a stem and leaf plot of the above data set
9. Create a group histogram for the above data set
10. Create a frequency polygon for this data set
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### Curriculum Vitae
Curriculum Vitae Manoj Kumar Address: 493/9B, Saket Nagar,Bhopal – 462024 ( M.P. ). E-mail- [email protected] Mobile No.8827714636 CAREER OBJECTIVE- To obtain a position where my teaching knowledge and soft skills would be well utilized and a opportunity for value based growth and career advancement while contributing handsomely to the most | 2,076 | 8,361 | {"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} | 3.640625 | 4 | CC-MAIN-2024-10 | latest | en | 0.852791 |
magicofsecurity.com | 1,726,450,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00477.warc.gz | 329,549,655 | 17,981 | Partial passwords is an authentication system where users enter a random 2-3 characters of their password. While it makes random guesses easier, attackers must eavesdrop on more than one logons.
How to implement partial passwords correctly? The problem made us think and eventually realise why online banking systems using authentication with partial passwords ask for two or three letters only and they also limit the maximum length of passwords.
Note: Actually – the above is only true for well-designed systems.
# Problem
If a system asks users for two letters, the database of the system will store all possible pairs of letters that appear in a user password – all “obfuscated” (e.g., with a cryptographic hash function). We say obfuscated because brute-force attack on data with about 10 bits of entropy (the number of all possible options being 1,000) is trivial.
Those systems can not ask for more than two letters because more letters would require a lot more database space. A two-letter partial password system (with eight characters’ passwords) requires 56 strings (hash values), while a three-letter system would require 336 strings – easily taking more than 5kB of data for a single password.
It appears that most companies indeed store all possible combination of letters a user may enter in a database. The second option is to store the password unencrypted or encrypted with a symmetric algorithm (e.g. 3DES, AES). Passwords cannot be stored as hash values as the system needs to have access to plaint text passwords to compare the letters entered by users.
# Better Solution
We have suggested a solution based on Shamir secret sharing scheme. This saves database space and it is also fast as only a polynomial computation (in the simplest form) is needed for verification.
## A. Global Parameters
At the beginning, someone has to define global parameters of the system. Well, there is actually just one – how many letters will users have to select. Let us call the parameter N. The maximum length of password (L) is important only from the database point of view – you will need to store a 32b long number for each character.
1. User chooses his/her password P. It consists of letters p1, p2, p3, … pk.
2. The system will generate at least 32 bits’ long secret key – K – unique for each user. 32 bits is enough if N is 3, for larger N (the number of letters users have to correctly select each time) you may want to increase the length of K.
3. The system will also generate N-1 32 bits’ long random numbers R1, R2, … R(N-1)
4. The next step is a computation of k points (k being the length of the password) on a polynomial: y=K+R1*x + R2*x^2 + … + R(N-1)*x^(N-1), for x = 1, 2, … k. Let us denote the results as y1, y2, …, y(N-1).
5. Values s1=(y1-p1), s2=(y2-p2), … sk=(yk-pk) is stored in the database. Each number takes 32 bits. One will also need to store K, or the hash of K.
## C. Authentication
The next part of the system is user authentication, which is very simple and fast.
1. The system selects N positions in the password – i1, i2, … iN.
2. A user selects N letters from her/his password at specified positions so that we have pairs (p’1, i1), (p’2, i2), …, (p’N, iN).
3. The system recovers yi values for indices i selected in step 1 – simply just adding stored values (see step 5 above) to values p’i entered by the users.
4. Now we have to solve the polynomial equation to obtain K’. The equation for that looks horribly but it is quick to compute and can even be partially pre-computed as it uses indices (positions of letters): K’ = \sum_i [ yi * [ (\PI_j (j) ) / (\PI_j (i-j)) ] ], where i and j run over i1, i2, …, iN (step 1), and j skips the actual selected i.
Example: let’s say that user selected 2nd, 3rd, and 4th letter, the solution will be: K’=y2*( 3*4/[(2-3)*(2-4)] )+y3*( 2*4/[(3-2)*(3-4)] ) + y4*( 2*3/[(4-2)*(4-3)] )
5. The last step is to compare K and K’. If they are equal, user entered correct values and is logged in.
One note here – the secret is not K, but values yi reconstructed when user enters his/her letters.
# Pros and Cons
The highlights of this solution are:
1. The number of letters for authentication and the length of passwords is not really limited.
2. The database space that is needed to store all necessary information is linear with the length of a password, not quadratic.
3. Secrets are not stored in plain-text and need certain computation.
4. All the data can be still encrypted for storage if needed.
5. Faster to compute – verification is several multiplications of 32 bit numbers that is much much faster than computing a hash value.
The low points are:
1. It is not a straight forward solution.
2. The security is still not increased beyond the difficulty of finding the correct K letters.
# Improvements
The discussion actually continued and a stranger – Tom – showed that storing the constant K in plaintext allows for an attack. I have initially suggested that one can store this constant in plaintext or a hash value of it. Currently, we have to recommend using the hash value.
For those interested in detail, here is an excerpt from Tom’s email. (It assumes password letters come from a set of 100 characters – hence the magic constant of 100.)
For match (p1p2..pN) I think you can compute the whole password in “constant” time O(100*k), by using the function again for each next unknown password character pN+1..pk, as an equation with 1 variable. E.g. we’ve got a match (p1p2…pN), then to reveal pN+1 we solve the equation for set 1, 2, …, N-1, N+1 with just discovered (y1, y2, …, yN-1), and with an unknown yN+1:
K’ = \sum_i [ yi * [ (\PI_j (j) ) / (\PI_j (i-j)) ] ] would be like
K’ = yN+1 * [ (\PI_j (j) ) / (\PI_j (N+1-j)) ] ] + C
where C is constant (computed).
If K is in plain text, then it is a simple equation with 1 variable and can be solved in O(1). If K is stored as a hash, then iterate through every character again and compare hash(K’) with stored hash(K). Note, that in this case, computing each next character requires O(100) operations, in total of O(100*k) rather than O(100^k).
10^12/16!*10! was suggested as if you brute force N=6 characters with plain K (if using the trick with modulo it is be better to brute force last N chars than first N, due to the lower charset for p16 than p1). | 1,634 | 6,329 | {"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} | 2.71875 | 3 | CC-MAIN-2024-38 | latest | en | 0.909742 |
https://www.computerbitsdaily.com/2023/05/python-program-to-find-average-of-n-numbers.html | 1,701,640,614,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00613.warc.gz | 801,105,526 | 57,592 | Python Program to Calculate the Average of N Numbers: A Step-by-Step Guide with Examples
How to Calculate the Average of N Numbers in Python
Calculating the average of N numbers is a common task in programming. In this blog post, we will discuss how to write a Python program to read N numbers from users and calculate the average of those N numbers.
Table of Contents
• Introduction
• Prerequisites
• Steps to Calculate the Average of N Numbers in Python
• Example Program
• FAQ
Introduction
Calculating the average of N numbers is a simple task that involves taking the sum of the numbers and dividing it by the total number of numbers. In Python, we can use loops and conditional statements to read N numbers from users and calculate their average.
Prerequisites
Before we start writing the program, we need to have a basic understanding of Python programming language. We also need to know how to use loops and conditional statements in Python.
Steps to Calculate the Average of N Numbers in Python
- Here are the steps to calculate the average of N numbers in Python:
- First, we need to ask the user to enter the total number of numbers they want to calculate the average for. We can use the input() function to get the user input.
- Next, we need to use a loop to read N numbers from the user. We can use a for loop or a while loop to achieve this.
- After reading the N numbers, we need to calculate their sum. We can use a variable to keep track of the sum of the numbers.
- Once we have the sum of the numbers, we can calculate the average by dividing the sum by the total number of numbers.
- Finally, we can print the average of the N numbers to the user.
Example Program
- Here is an example program that reads N numbers from the user and calculates their average:
# Python program to calculate the average of N numbers
# Step 1: Ask the user to enter the total number of numbers
n = int(input("Enter the total number of numbers: "))
# Step 2: Use a loop to read N numbers from the user
sum = 0
for i in range(n):
num = int(input("Enter a number: "))
sum += num
# Step 3: Calculate the average of the N numbers
average = sum / n
# Step 4: Print the average to the user
print("The average of the", n, "numbers is", average)
Output:
Enter the total number of numbers: 5
Enter a number: 5
Enter a number: 5
Enter a number: 5
Enter a number: 5
Enter a number: 5
The average of the 5 numbers is 5.0
Calculating the average of N numbers is a simple task that can be achieved using loops and conditional statements in Python. In this blog post, we discussed how to write a Python program to read N numbers from users and calculate their average. We hope this blog post was helpful to you.
FAQ : Calculate the Average of N Numbers Using Python
What is the meaning of "n" in this program?
In this program, "n" refers to the total number of numbers that the user wants to enter to calculate the average.
What is the formula to calculate the average of n numbers?
The formula to calculate the average of n numbers is to add all the numbers and divide the sum by the total number of numbers. Mathematically, it can be represented as:
Average = (Sum of numbers) / (Total number of numbers)
What is the easiest way to calculate the average of n numbers in Python?
The easiest way to calculate the average of n numbers in Python is by using a for loop. First, we define the total number of inputs we want to enter. Then, we take the numbers and calculate the total sum of those numbers using the for loop. Finally, we calculate the average of those numbers using a formula and print the average value.
Can we use a while loop to calculate the average of n numbers in Python?
Yes, we can use a while loop to calculate the average of n numbers in Python. We can use a while loop to read n numbers from the user and calculate their sum. Once we have the sum of the numbers, we can calculate the average by dividing the sum by the total number of numbers.
Is there any built-in function in Python to calculate the average of n numbers?
Yes, there is a built-in function in Python to calculate the average of n numbers. We can use the mean() method of the statistics module to directly calculate the average of the elements of the list. We will pass the given list of numbers as input to the mean() method and it will return the average of numbers. | 979 | 4,391 | {"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} | 4.125 | 4 | CC-MAIN-2023-50 | latest | en | 0.895757 |
https://processing.org/reference/rightshift | 1,723,262,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00844.warc.gz | 345,446,215 | 9,229 | ### >> (right shift)
#### Description
Shifts bits to the right. The number to the left of the operator is shifted the number of places specified by the number to the right. Each shift to the right halves the number, therefore each right shift divides the original number by 2. Use the right shift for fast divisions or to extract an individual number from a packed number. Right shifting only works with integers or numbers which automatically convert to an integer such at byte and char.
Bit shifting is helpful when using the color data type. A right shift can extract red, green, blue, and alpha values from a color. A left shift can be used to quickly reassemble a color value (more quickly than the color() function).
#### Examples
• ``````int m = 8 >> 3; // In binary: 1000 to 1
println(m); // Prints "1"
int n = 256 >> 6; // In binary: 100000000 to 100
println(n); // Prints "4"
int o = 16 >> 3; // In binary: 10000 to 10
println(o); // Prints "2"
int p = 26 >> 1; // In binary: 11010 to 1101
println(p); // Prints "13"
``````
• ``````// Using "right shift" as a faster technique than red(), green(), and blue()
color argb = color(204, 204, 51, 255);
int a = (argb >> 24) & 0xFF;
int r = (argb >> 16) & 0xFF; // Faster way of getting red(argb)
int g = (argb >> 8) & 0xFF; // Faster way of getting green(argb)
int b = argb & 0xFF; // Faster way of getting blue(argb)
fill(r, g, b, a);
rect(30, 20, 55, 55);
``````
#### Syntax
• `value >> n`
#### Parameters
• `value`int: the value to shift
• `n`int: the number of places to shift right | 470 | 1,574 | {"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} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.566927 |
https://www.theburningofrome.com/users-questions/how-many-feet-are-in-a-mr/ | 1,695,825,506,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00499.warc.gz | 1,146,430,875 | 39,647 | # How many feet are in a MR?
## How many feet are in a MR?
Meters to feet conversion table
Meters (m) Feet (ft)
1 m 3.28084 ft
2 m 6.56168 ft
3 m 9.84252 ft
4 m 13.12336 ft
## How do you convert feet to linear meters?
The length in meters is equal to the feet multiplied by 0.3048.
How many feet is a ma?
Meters to Feet table
Meters Feet
1 m 3.28 ft
2 m 6.56 ft
3 m 9.84 ft
4 m 13.12 ft
### How do you calculate feet to meters?
Multiply or divide your measurement by a conversion factor. Because there are 3.28 feet in a meter, take your measurement (in feet) and divide it by 3.28 to convert to meters. You can also multiply your measurement in feet by 0.3048 to get the exact same answer because there are 0.3048 meters in a foot.
### How many meters is 5 foot 5?
Feet to meters chart
Feet & Inches Feet Meters
5 feet 5 inches 5.42 feet 1.65 m
5 feet 6 inches 5.5 feet 1.68 m
5 feet 7 inches 5.58 feet 1.7 m
5 feet 8 inches 5.67 feet 1.73 m
Is a meter exactly 3 feet?
A meter is equal to approximately 3.28084 feet.
#### How do you convert feet into linear feet?
To measure linear footage, start by measuring the length in inches. Then divide the total inches by 12. The length is the linear footage, so no fancy linear foot calculator is required. To convert linear feet to feet, there’s no math.
#### What is linear foot?
Technically, a linear foot is a measurement that is 12 inches long (so, one foot) and that is measured in a straight line, which is why it’s called linear.
What is the difference between Metre and feet?
Meter to Feet Conversion For converting meter to feet firstly we should know the difference between their lengths. That is one meter is equal to 3.28 feet and one foot is equal to 12 inches as per rule. So, to convert meter to feet just simply multiply the number of meter to the value of feet per meter. = 19.91 ft.
## What is longer meter or feet?
A meter is approximately equal to 3.28084 feet.
## How many meters is 5 feet?
Quick Lookup Feet To Metres Common Conversions
ft & in m
5′ 1″ 1.55
5′ 2″ 1.57
5′ 3″ 1.60
5′ 4″ 1.63 | 612 | 2,083 | {"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} | 3.71875 | 4 | CC-MAIN-2023-40 | longest | en | 0.912379 |
https://www.codeproject.com/Articles/871918/Magic-Cube?msg=4994757 | 1,624,412,836,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488528979.69/warc/CC-MAIN-20210623011557-20210623041557-00249.warc.gz | 619,869,615 | 24,468 | 14,933,981 members
Articles / Programming Languages / Java
Article
Posted 2 Feb 2015
29K views
7 bookmarked
# Magic Cube
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Code to generate Magic Cubes of odd and doubly even orders
## Introduction
In mathematics, a magic cube is the 3-dimensional equivalent of a magic square, that is, a number of integers arranged in a n x n x n pattern such that the sum of the numbers on each row, each column, each pillar and the four main space diagonals is equal to a single number, the so-called magic constant of the cube.
## Terminology
• Order - The value of n in n x n x n cube.
• Magic constant - The unique sum of any row,column,pillar or triagonal. Its value is equal to n(n3+1)/2.
## Extended Siamese Method for Odd Magic Cubes
Siamese method is one method to construct magic squares of odd orders. This method can be extended to 3 dimension to construct magic cubes. Assumtion is that reader has understading of the Siamese method.
We can represent any cube using index notation e.g. [1,2,3] which represents layer-1,row-2 and column-3 of the cube. The index starts from [0,0,0] and the maximum value of the index would be [n-1,n-1,n-1].
Any move will be wrapped around (like pac man or train tunnel in Matrix). -1 value is equivalent to n-1 and n is equivalent to 0.
1. Place number 1 in middle of the top layer [0,(n-1)/2,(n-1)/2].
2. Let us represent current location as [l,r,c].
3. Move to [l-1,r,c-1].
4. If [l-1,r,c-1] is already occupied move to [l-1,r-1,c].
5. If [l-1,r-1,c] is also occupied move to [l+1,r,c].
6. Fill the current cell with next number and keep iterating from step #2 till all the cells are filled with numbers from 1 to n3.
## Doubly Even Magic Cube
The logic to create magic cube can be extended from following magic square
1 15 14 4 12 6 7 9 8 10 11 5 13 3 2 16
The above magic square can be created in following manner.
Start filling the square from top left hand corner. If number lies in red colored box fill the number else fill with 17-number (or n2+1-number).
Let us call the red=parity 0 and blue=parity 1.
Logic to calculate parity of index [r,c] of a magic square
Java
```int nBy4 = n / 4;
boolean parity = false;
if (r >= nBy4 && r < 3 * nBy4) {
parity = true;
}
if (c >= nBy4 && c < 3 * nBy4) {
parity = !parity;
}```
On similar lines a magic cube can be constructed.
## Singly Even Magic Cube
Here the logic becomes little complex and hence leaving it out of scope of this article. Medjig method can be used to construct these magic cubes. In the demo source I have also inculded code to generate magic square of any order (including singly even magic squares). Same logic can be exended to generate cubes.
## The code
The code for this is very simple.
Java
```public static int[][][] getMagicCube(int n) {
if (n % 2 == 1) {
int[][][] magicCube = new int[n][n][n];
// layer index, row index, column index
int l, r, c;
l = 0;
r = n / 2;
c = n / 2;
// If you want to port this code in C++
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < magicCube.length; k++) {
magicCube[i][j][k] = 0;
}
}
}
int last = (int) Math.pow(n, 3);
for (int i = 0; i < last; i++) {
magicCube[l][r][c] = i + 1;
l--;
l = normalize(n, l);
c--;
c = normalize(n, c);
if (magicCube[l][r][c] != 0) {
r--;
r = normalize(n, r);
c++;
c = normalize(n, c);
if (magicCube[l][r][c] != 0) {
r++;
r = normalize(n, r);
l += 2;
l = normalize(n, l);
}
}
}
return magicCube;
} else if (n % 4 == 0) {
int[][][] magicCube = new int[n][n][n];
int lastPlusOne = (int) Math.pow(n, 3) + 1;
int number = 1;
int nBy4 = n / 4;
for (int l = 0; l < n; l++) {
for (int r = 0; r < n; r++) {
for (int c = 0; c < magicCube.length; c++) {
boolean parity = false;
if (l >= nBy4 && l < 3 * nBy4) {
parity = true;
}
if (r >= nBy4 && r < 3 * nBy4) {
parity = !parity;
}
if (c >= nBy4 && c < 3 * nBy4) {
parity = !parity;
}
magicCube[l][r][c] = (parity) ? lastPlusOne - number: number;
number++;
}
}
}
return magicCube;
}
throw new RuntimeException(
"Singly even magic cubes are not returned by this method.");
}
// index should be between 0 to n-1
private static int normalize(int n, int index) {
while (index < 0) {
index = index + n;
}
while (index > n - 1) {
index = index - n;
}
return index;
}```
This concept can be extended to cubes of higher dimensions (hyper cube). You can visit my website where you can generate a magic cube of of any given order and dimension.
## Sample of cube of order 5
Generated from code
Layer 1
109 77 75 43 11 97 70 38 6 104 65 33 1 124 92 28 21 119 87 60 16 114 82 55 48
Layer 2
15 108 76 74 42 103 96 69 37 10 91 64 32 5 123 59 27 25 118 86 47 20 113 81 54
Layer 3
41 14 107 80 73 9 102 100 68 36 122 95 63 31 4 90 58 26 24 117 53 46 19 112 85
Layer 4
72 45 13 106 79 40 8 101 99 67 3 121 94 62 35 116 89 57 30 23 84 52 50 18 111
Layer 5
78 71 44 12 110 66 39 7 105 98 34 2 125 93 61 22 120 88 56 29 115 83 51 49 17
## Share
India
No Biography provided | 1,673 | 4,958 | {"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} | 3.71875 | 4 | CC-MAIN-2021-25 | latest | en | 0.784924 |
https://wildmath.org/multiplying-decimals-by-whole-numbers-worksheet/ | 1,581,963,910,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00043.warc.gz | 626,006,191 | 11,542 | ## Wildmath
Math Worksheets & Free Printables
# Multiplying Decimals by whole Numbers Worksheet
Posted at February 6, 2020 10:37 by admin in Wildmath
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We try to provide the best Pic in this article because we know that this may be one of the best resource for any 4th grade math multiplication worksheets ideas. Don’t you come here to learn some new New 4th... | 486 | 2,336 | {"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} | 2.625 | 3 | CC-MAIN-2020-10 | latest | en | 0.87345 |
https://stickinthemudpodcast.com/betting/what-is-the-probability-of-rolling-a-total-of-9-with-three-dice.html | 1,675,300,713,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00409.warc.gz | 565,534,862 | 20,430 | # What is the probability of rolling a total of 9 with three dice?
Contents
## Which is more likely rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled?
Which is more likely, rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled? Answer: 9 = 3+6 = 4+5 = 5+4 = 6+3, — 4 ways to get total of 9 points rolling two dice, so the probability that the outcome is 9 points is: 4/(6*6) = 1/9 = 0.111.
## What is the probability of rolling a sum of 9 with two dice?
The probability of getting 9 as the sum when 2 dice are thrown is 1/9.
## What is the probability of getting a sum of at least 9?
❖ “At least one” is equivalent to “one or more.” To find the probability of at least one of something, calculate the probability of none and then subtract that result from 1. That is, P(at least one) = 1 – P(none).
## How many outcomes have a sum of at least 9?
(ii) two numbers appearing on them whose sum is 9. Therefore, total number of possible outcomes = 36.
THIS IS IMPORTANT: How far is Hardrock Casino from Clearwater Florida?
## What is the probability of rolling 3 dice?
Two (6-sided) dice roll probability table
Roll a… Probability
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
= 91/216.
## What is the expected sum of the numbers that appear when three fair dice are rolled?
E(S) = E(s1+s2+s3) = E(s1)+E(s2)+E(s3). The expectation of the sum is the sum of the expectation values for the three dice. But since they are all fair dice, they have the same expectation values. Hence E(S) = 3 E(s1). | 466 | 1,619 | {"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} | 4.625 | 5 | CC-MAIN-2023-06 | latest | en | 0.917076 |
http://www.wyzant.com/resources/answers/kinetic_energy | 1,408,675,654,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500822407.51/warc/CC-MAIN-20140820021342-00267-ip-10-180-136-8.ec2.internal.warc.gz | 697,256,191 | 10,638 | Search 75,854 tutors
A 5g apple is thrown straight upward with an initial velocity of 20 m/s @ a point of 1.5m height from ground. Ignoring air friction, find: a. Total mechanical energy of the...
A boy pulls a 10.0 kg book rack at rest on a smooth floor with a constant horizontal force of 1.2N starting from rest. Compute the change in kinetic energy of the box after the...
if the average kinetic energy of a substance changes, a. its temperture changes b. it may change phase c. thermal energy has been transferred into or...
Suppose at P point, a body loses its entire K.E which it had at O point and at Q point body losses 1/3 of its K.E which is at has at O. Find out the height h.
A 15-g CD with a radius of 7.5 cm rotates with an angular speed of 40 rad/s. What is its kinetic energy? | 214 | 807 | {"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} | 2.609375 | 3 | CC-MAIN-2014-35 | longest | en | 0.911409 |
https://topnursingessay.com/topic-4-exercises/ | 1,627,943,057,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154385.24/warc/CC-MAIN-20210802203434-20210802233434-00239.warc.gz | 589,147,118 | 11,986 | # Topic 4 exercises
• Chapter 10, numbers 10.9, 10.10, 10.11, and 10.12
• Chapter 11, numbers 11.11, 11.19, and 11.20
• Chapter 12, numbers 12.7, 12.8, and 12.10
10.9 The normal range for a widely accepted measure of body size, the body mass index (BMI), ranges from 18.5 to 25. Using the midrange BMI score of 21.75 as the null hypothesized value for the population mean, test this hypothesis at the .01 level of significance given a random sample of 30 weight-watcher participants who show a mean BMI = 22.2 and a standard deviation of 3.1.
10.10 Let’s assume that, over the years, a paper and pencil test of anxiety yields a mean score of 35 for all incoming college freshmen. We wish to determine whether the scores of a random sample of 20 new freshmen, with a mean of 30 and a standard deviation of 10, can be viewed as coming from this population. Test at the .05 level of significance.
10.11 According to the California Educational Code (http://www.cde.ca.gov/ls/fa/sf/pegui-demidhi.asp), students in grades 7 through 12 should receive 400 minutes of physical education every 10 school days. A random sample of 48 students has a mean of 385 minutes and a standard deviation of 53 minutes. Test the hypothesis at the .05 level of significance that the sampled population satisfies the requirement.
10.12 According to a 2009 survey based on the United States census (http://www.census.gov/prod/2011pubs/acs-15.pdf), the daily one-way commute time of U.S. workers averages 25 minutes with, we’ll assume, a standard deviation of 13 minutes.
An investigator wishes to determine whether the national average describes the mean commute time for all workers in the Chicago area. Commute times are obtained for a random sample of 169 workers from this area, and the mean time is found to be 22.5 minutes. Test the null hypothesis at the .05 level of significance.
11.11 Give two reasons why the research hypothesis is not tested directly.
11.19 How should a projected hypothesis test be modified if you’re particularly concerned about (a) the type I error? (b) the type II error?
11.20 Consult the power curves in Figure 11.7 to estimate the approximate detection rate, rounded to the nearest tenth, for each of the following situations: (a) a four-point effect, with a sample size of 13 (b) a ten-point effect, with a sample size of 29 (c) a seven-point effect with a sample size of 18 (Interpolate) (I ATTACHED FIGURE 11.7)
*12.7 In Question 10.5 on page 191, it was concluded that, the mean salary among the population of female members of the American Psychological Association is less than that (\$82,500) for all comparable members who have a doctorate and teach full time. (a) Given a population standard deviation of \$6,000 and a sample mean salary of \$80,100 for a random sample of 100 female members, construct a 99 percent confidence interval for the mean salary for all female members (b) Given this confidence interval, is there any consistent evidence that the mean salary for all female members falls below \$82,500, the mean salary for all members?
12.8 In Review Question 11.12 on page 218, instead of testing a hypothesis, you might prefer to construct a confidence interval for the mean weight of all 2-pound boxes ofcandy during a recent production shift. (a) Given a population standard deviation of .30 ounce and a sample mean weight of 33.09 ounces for a random sample of 36 candy boxes, construct a 95 percent con-fidence interval (b) Interpret this interval, given the manufacturer’s desire to produce boxes of candy that, on the average, exceed 32 ounces.
12.10 Imagine that one of the following 95 percent confidence intervals estimates the effect of vitamin C on IQ scores:
95% CONFIDENCE INTERVAL LOWER LIMIT UPPER LIMIT
1100102
29599
3102106
490 111
59198
(a) Which one most strongly supports the conclusion that vitamin C increases IQ scores?
(b) Which one implies the largest sample size?
(c) Which one most strongly supports the conclusion that vitamin C decreases IQ scores?
(d) Which one would most likely stimulate the investigator to conduct an additional experiment using larger sample sizes? | 991 | 4,153 | {"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} | 3.453125 | 3 | CC-MAIN-2021-31 | latest | en | 0.887806 |
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Broad Topics > Stage 5 Mechanics mapping document > Work, energy and power
### Go Spaceship Go
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Show that even a very powerful spaceship would eventually run out of overtaking power
### The Ultra Particle
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Explore the energy of this incredibly energetic particle which struck Earth on October 15th 1991
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Explore the power of aeroplanes, spaceships and horses.
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Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon?
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Play with the energetic forces in this sling shot | 390 | 1,701 | {"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} | 2.5625 | 3 | CC-MAIN-2017-43 | latest | en | 0.857741 |
https://jp.mathworks.com/matlabcentral/answers/519867-epsilon-neighborhood-for-polar-coordinates | 1,726,818,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00342.warc.gz | 308,590,535 | 30,509 | # epsilon-neighborhood for polar coordinates
11 ビュー (過去 30 日間)
Christian Metz 2020 年 4 月 22 日
Hi,
I am having a density-based radar database that I try to cluster with a grid-based DBSCAN algorithm.
The part of the algorithm my question focuses on is "the core point condition". Which is basically the following. To determine if a point is a core point, it needs to have more than a certain number of neighbours (MinPoints) in its epsilon-neighbourhood.
The epsilon neighbourhood can be determined as follows:
1. Calculate the distance from a point P to all the other points in the database.
2. Determine the epsilon-neighbourhood: All the points that are within a max. distance of eps are part of the neighbourhood. Meaning epsilon is the search radius.
3. The core point condition: A point is a core point if in its epsilon-neighbourhood at least MinPoints are.
My question is now, how to determine the epsilon-neighbourhood for polar points with azimuth (azi) and range (r). I do not completely know how to calculate the distance and then from there how to determine the epsilon neighbourhood.
---------------------------------------------------
I use among others the "Phased-Array Toolbox" and the "clusterDBSCAN" algorithm has to possiblity to also use a 1-by-P row vector. Where P is the number of features in the input data. Does someone know how this neighborhood search works with a vector?
Because in the end I want to have a range dependend epsilon(x,y)-vector that varients in azimuth direction with the range.
-------------------------------------------------------
Edit:
I implemented my own algorithm. My problem is that I do not know how to implement the ellipse equation to determine the size of the epsilon-neighbourhood. To basically how to implement the regionQuery function for an ellipse search radius. Below you find my temporary implementation were I only look at the points on the same range or azimuth angle. So it is basically a cross, but not a circual search area. P is the temporary point which has to be checked for the core point criterion.
function Neighbors=RegionQuery(P) %Determine range/DeltaR to know which
%Element of the w(c) vector
%should be used and then
%also determine the range.
RANGEPOINT = Data(P,2)-min(Data(:,2))/Q_range+1;
% Find all the points with the same Range:
All_Azi = find(Data(:,2) == Data(P,2));
% Find all the points in the Eps neighborhood
Neighbors_azi = All_Azi(find(D_azi(P,All_Azi) <= epsilon_azi(RANGEPOINT)));
% Find all the points with the same Azi:
All_Range = find(Data(:,1) == Data(P,1));
% Find all the points in the Eps neighborhood
Neighbors_range = All_Range(find(D_range(P,All_Range) <= epsilon_range));
Neighbors = [Neighbors_azi; Neighbors_range];
Neighbors = sort(unique(Neighbors)); % delete dublicates and sort it in raising order
end
The implementation above is as said only looking linear but not elliptical. Can you help me with that? I created a short data set of points in polar coordinates in degree and meter:
polar_corepoint = [0,6];
polar_borderpoints = [10,6;5,6.5;-5,5.5; 0,8; 0,4 ];
polar_noise = [25,6;-15,4;15,9;45,8.5;-25,8];
eps_azi = 10; % epsilon radius in azi direction in degree
eps_range = 2; % epsilon radius in range direction in meter
I want to get the neighbourhood of the corepoint that only includes the boarderpoints and not the noise. So basically the size of the neighborhood should be in the end = 6 (core point + borderpoints).
Do you know how to implement that epsilon search? It does not matter which distance metric is used, I personally have used the Euclidean until now (pdist2(data,data)).
Thanks for you help
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### 回答 (1 件)
Honglei Chen 2020 年 4 月 27 日
I'll explain how clusterDBSCAN works in this situation.
In your case, range and angle will be two features so you can specify epsilon independently for the two features. The based on these two epsilon values, the algorithm will draw a ellipse to see if a point is within the neighborhood.
HTH
##### 3 件のコメント1 件の古いコメントを表示1 件の古いコメントを非表示
Honglei Chen 2020 年 4 月 29 日
Are you trying to implement your own? You can think of the ellipse equation, like (x1/epsilon1)^2+(x2/epsilon2)^2=1.
HTH
Christian Metz 2020 年 5 月 1 日
Yes, I implemented my own. My problem is that I do not know how to implement the ellipse equation to determine the size of the epsilon-neighbourhood. Can you help me with that? I created a short data set of points in cartesian coordinates:
corepoint = [5 4];
borderpoints = [5 3; 6 4.5; 4 4.5; 7 4; 3 4];
noise = [1 6; 2 2; 5 2; 8 3];
eps_x = 2; % epsilon radius in x direction
eps_y = 1; % epsilon radius in y direction
I want to get the neighbourhood of the corepoint that only includes the boarderpoints and not the noise. So basically the size of the neighborhood should be in the end = 6 (core point + borderpoints).
Do you know how to implement that epsilon search? It does not matter which distance metric is used, I personally have used the Euclidean until now (pdist2(data,data)).
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Translated by | 1,324 | 5,232 | {"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} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.916802 |
https://infinitylearn.com/surge/question/mathematics/state-true-or-falsethe-midpoint-of-the-hypotenuse-of-a-righ/ | 1,716,957,734,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00744.warc.gz | 257,084,557 | 17,062 | State true or false:The midpoint of the hypotenuse of a right triangle is equidistant from its vertices.
# State true or false:The midpoint of the hypotenuse of a right triangle is equidistant from its vertices.
1. A
True
2. B
False
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### Solution:
Given, a right angled triangle.
The midpoint of the hypotenuse is the circumcenter of the right angled triangle.
Hence, the vertices are equidistant from the circumcenter.
So, the midpoint of the hypotenuse of a right triangle is equidistant from its vertices.
Hence, the given statement is true.
So, option (1) is correct.
## Related content
Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics
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https://www.askiitians.com/forums/Differential-Calculus/25/6375/applications-of-derivatives.htm | 1,723,772,494,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00344.warc.gz | 516,823,494 | 43,260 | # the number of real values of x where the function f(x) = cosx+cos{(√2)x} attains its maximum is(a) 0(b) 1(c) 2(d) infinite
148 Points
14 years ago
Dear raman malik.
f(x) = cosx+cos{(√2)x}
f'(x)=-sin x -√2 sin √2x =0
it gives x= 2m ∏/(√2 +1 ) and (2m +1 ) ∏ /(√2 -1) where m is any natural number
clearly both values are not multiple of 2∏
only for m=0
and f(0) = 1+1 =2 is its maximum value
so only one solution option b is correct
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
All the best.
Regards, | 222 | 680 | {"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} | 3.40625 | 3 | CC-MAIN-2024-33 | latest | en | 0.842792 |
https://ageekoutside.com/how-many-cups-is-3-egg-whites/ | 1,659,997,855,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00499.warc.gz | 120,477,689 | 10,220 | Contents
# How many cups is 3 egg whites?
## How many cups are in one egg white?
1 Large Egg White = 2 tablespoons. 8 to 10 Large Egg Whites = 1 cup.
## How many cups is 3 eggs?
Three large eggs equals approximately one-half cup plus two tablespoons volume.
## How Cups is 4 egg whites?
While each egg will vary in size, one egg white from a large egg will be about 1 ounce. Four egg whites is about 4 ounces or a half cup.
## How much liquid is 1 egg white?
To substitute for egg whites fresh from the shell, 2 tablespoons (30 mL) = 1 large egg white.
## How many cups are in 6 egg white?
To Make 1 CupEgg SizeWholeWhitesX-Large46Large57Medium58Small69
## How many cups is 4 eggs?
1 cup
4 large eggs equal 1 cup, or 1 egg equals 1/4 cup. Tip: Unless otherwise specified, large eggs should always be used in recipes.
## How many grams is 3 egg whites?
Conversion ChartsIn Shell57 gramsWithout Shell50 gramsWhite Only30 gramsYolk Only18 gramsLiquid Egg Products
## How many eggs are in one cup?
ANSWER: 4 large eggs equal 1 cup, or 1 egg equals 1/4 cup.
## How many eggs are equivalent to egg white?
Two egg whites
Consider egg whites: When you’re making store-bought cake mixes, you can get away with using just egg whites as your substitute for whole eggs because the mixes usually include other ingredients that help with tenderness and texture. Two egg whites—or 1/4 cup fat-free egg substitute—can replace 1 whole egg.
## How many grams is 3 eggs?
3 eggs – Bob Evans Farms Inc.Nutrition FactsFor a Serving Size of 3 each (170g)How many calories are in Scrambled Eggs, 3 eggs? Amount of calories in Scrambled Eggs, 3 eggs: Calories 253Calories from Fat 144 (56.9%)% Daily Value * | 445 | 1,703 | {"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} | 3.125 | 3 | CC-MAIN-2022-33 | latest | en | 0.914343 |
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This post follows the previous two posts on Categories. (First and second). It explains how one uses the universal properties to prove nontrivial results. Since $\LaTeX$ rendering in WordPress is not so efficient, I decided to upload the notes in my Notes page, and can also be accessed via this link.
This post is a continuation of the previous post – A mild introduction to Categories. Before I jump into the technical details of Categorical stuff like Functors, I would like to introduce the notion of a universal mapping property (henceforth referred to as UMP).
I won’t bog you down with the details of jargon that one has to absorb before getting the definition. Those interested in the definition of a UMP may refer the Wikipedia article here. Loosely speaking, a UMP of an object in a category is a property that characterizes it completely. In this post, I shall only present the universal properties of some familiar objects. We start with the property of a kernel of a group homomorphism.
Kernel of a group homomorphism: Let Grp be the category of groups. (One may similarly define the UMP of the kernel of a ring, module, field homomorphism). Let $A$ and $B$ be groups and $f: A \to B$ be a group homomorphism. To familiarize ourselves with the categorical notation, we continue this example in categorical language- Let $A, B \in \text{Obj}$ (Grp) and $f \in \text{Mor} (A, B)$. There exists an object $K\in \text{Obj}$ (Grp) and a morphism $\iota \in \text{Mor} (K, A)$ such that the pair $(K, \iota)$ satisfies the following properties:
• $f \circ \iota = 0$ as a morphism (group homomorphism) from $K$ to $B$.
• Whenever $latex (K’, \iota’ \in \text{Mor}(K’, A))$ is such that $f \circ \iota' = 0$, then there is a unique morphism $\Phi \in \text{Mor}(K', K)$ such that $\iota \circ \Phi = \iota'$.
In group-theoretic parlance, kernel (of a homomorphism) is a group which has an embedding into the preimage and whose image under the homomorphism is trivial. The group is unique in that any other group that has these properties must factor uniquely via the kernel. This property of the kernel, in fact, characterizes it.
We begin with examples that create new rings from old ones:
Direct sum of $R$-Modules: Given $R$-modules $M_1, M_2$, the direct product of $M_1$ and $M_2$ is an $R$-module $M_1\oplus M_2$ along with (inclusion) maps $\iota_i: M_i \to M_1\oplus M_2\; (i=1,2)$ such that given any $R$-module $N$ and any $R$-module homomorphisms $\phi_i : M_i \to N$, there is a unique homomorphism $\Phi: M_1 \oplus M_2 \to N$ such that $\Phi \circ \iota_i = \phi_i$.
Note: This can be generalized to an arbitrary family of submodules of an $R$-module. The existence follows from by taking all but finitely many nonzero elements of each submodule.
Polynomial ring in one variable over a ring: Given rings $R, S$, a ring-homomorphism $\phi: R \to S$ and an element $a\in S$, there exists a unique ring-homomorphism $\Phi: R[X] \to S$ extending $\phi$ such that $\Phi(X)=a$.
Quotient Ring: Given a ring $R$ and its ideal $I$ (possibly trivial or the whole ring), the quotient ring is a ring $Q$ along with ring homomorphism $\pi: R \to Q$ such that
• $\pi(I)=0$,
• For every ring $S$ and homomorphism $\phi:R\to S$ with $\phi(I)=0$, there is a unique ring homomorphism $\Phi: Q\to S$ such that $\phi = \pi \circ \Phi$.
The UMP of a free module (or free group) is in some sense, more fundamental.
Free $R$-module: Let $S$ be a set and $R$ be a ring. A free $R$-module over $S$ is an $R$-module $F(S)$ along with an (injective) set map $\iota: S \to F(S)$ such that given any $R$-module $M$ and a set map $\phi: S \to M$, there is a unique $R$-module homomorphism $\Phi: F(S) \to M$ such that $\Phi \circ \iota = \phi$.
The UMP of a ring of fractions of a ring is often useful.
Ring of fractions of a ring: Given a domain $R$ and a multiplicatively closed subset $S$ of $R$, (that contains 1 but no zerodivisors of $R$), the ring of fractions of $R$ is a ring (denoted by) $S^{-1}R$ along with a map $\iota: R \to S^{-1}R$ such that given a ring $Q$ and ring homomorphism $\phi: R \to Q$ such that $\phi(S)$ consists of units (of $Q$), then there is a unique homomorphism $\Phi: S^{-1}R \to Q$ such that $\Phi\circ\iota=\phi$.
Note: If we take $R$ to be a domain and $S=R\backslash \{0\}$ then we get the field of fractions of $R$.
We give two examples to show that UMPs exist for non-algebraic objects too. (whatever that means!)
Quotient of a set: Let $X$ be a set with an equivalence relation $\tilde{}$ on it. The quotient of $X$ with respect to $\tilde{}$ is a set $\overline{X}$ with a map $\pi: X \to \overline{X}$ such that given any set $Z$ and a map $\phi: X\to Z$ with $\phi(x) = \phi(x')$ whenever $x$ $\tilde{}$ $x'$, then there is a unique map $\Phi: \overline{X}\to Z$ such that $\Phi\circ\pi=\phi$.
Product of sets: Let $X_i\; (i\in I)$ be an arbitrary family of sets. We define the product set of $X_i$‘sto be a set $\Pi_{i\in I} X_i$ alongwith (projection) maps $\displaystyle\pi_i : \Pi_{i\in I} X_i \to X_i$ such that for any __ (complete this!)
The above two properties – quotient and product of sets generalize immediately and can be used to define the UMP for Quotient topology and Box topology in the category Top of topological spaces. To state the UMP of the Product topology, one uses an analog of Direct Sum of Modules stated above.
Exercise: Deduce the UMP for cokernel of a module homomorphism.
Uniqueness of a UMP: The universal property of an object need not be unique. A trivial example to see this is as follows: The identity element of a group (denote it by $e$) satisfies the two following properties:
1. Given any group $G$, there is a unique homomorphism $f:\{e\} \to G$.
2. Given any group $G$, there is a unique homomorphism $f: G \to \{e\}$.
Future post:
• A small application of UMP.
Galois, Abel, Cauchy, Cayley and other stalwarts of their era developed group theory. Kummer, Dedekind, Noether were among those who enriched our understanding of rings and ideals, especially over $\mathbb Z$ and over $\mathbb Q$. This marked the beginning of what is coined as “Abstract Algebra” in many undergraduate texts and curricula. But as Herstein puts it, abstractness, being a relative term, this algebra of rings, fields, modules and vector-spaces might not be so abstract, as far as today’s development in Algebra is concerned. Perhaps I might call today’s research in Algebraic Geometry, the pinnacle of abstraction. Indeed, Grothendieck took the subject to dizzying heights of abstraction. There was another abstractness introduced by founders of homological algebra including McLane and Eilenberg. “Category Theory” is sometimes notoriously humoured as “Abstract Nonsense”.
What is a Category?
A category, roughly speaking, consists of objects and relations between objects. To be more formal, I would say a category is a collection of objects (Obj) and for any two objects $A, B \in$ Obj, a (possibly empty) set (Mor($A, B$)) called as morphisms from $A$ to $B$ (which can be thought of as the set of all maps from $A$ to $B$); and for any three objects $A, B, C \in$ Obj, a law of composition (i.e. a map)
Mor($A, B$) $\times$ Mor($B, C$) $\to$ Mor($A, C$)
satisfying a few properties that we may want them to satisfy. Before I start-off with some examples, let me tell a fundamental category that can stand as a guiding example in better understanding properties of categories. Set is a category whose objects are sets (not classes of sets; for I don’t want to start off paradoxes again :P) and morphisms are functions (or plain set-maps).
The properties a category $\mathfrak C$(Obj, Mor) is expected to satisfy are obvious —
1. Two sets Mor($A, B$) and Mor($A', B'$) are disjoint unless $A=A'$ and $B=B'$, in which case they are equal.
2. For each object $A$ of Obj, there is a morphism $\text{id}_A \in$ Mor($A, A$) which acts as left and right identity for the elements of Mor($A, B$) and Mor($B, A$) respectively, for all objects $B\in$ Obj($\mathfrak C$).
3. The law of composition is associative whenever defined (whatever that means!)
(In (2) above, note that Mor($A,A$) may contain more than one object; there are so many bijections from a set onto itself.)
Notice that these properties are satisfied by the aforementioned category Set quite trivially. Categories are abundant in Mathematics. Here are a few to begin with:
(PS: Fat rigorous books in Maths make comments like: Henceforth, by abuse of notation, we will denote a category by its objects. I refrain from making such obvious remarks.)
(PPS: Ignore those examples below which you don’t follow.)
• PSet, (meaning, pointed sets) with nonempty sets with a fixed element (special point) of every set as objects, and morphisms as set maps with the property that special points are mapped to special points,
• Grp, or the category with groups as objects and group-homomorphisms as morphisms,
• R-Mod, or the category of R-modules with R-module-homomorphisms as morphisms,
• FD$k$Vec, or finite-dimensional vector spaces over a field $k$ with $k$-linear homomorphisms as morphisms,
• PPTop, or Pointed path-connected topological spaces with a special point and with continuous functions (respecting special-points) serving as morphisms,
• $\mathcal C^0(\mathbb R)$ where the objects are open sets in $\mathbb R$ and morphisms are continuous functions on these open sets.
• Top$\mathbb X$ with objects as open sets of a topological space $\mathbb X$ and inclusion maps as morphisms. Note that here, Mor($U, V$) is empty if $U \not\subseteq V$ and consists of one object, the inclusion map from $U$ to $V$ otherwise.
• Hol, where objects are regions (open & path-connected subsets) of $\mathbb C$ and morphisms are holomorphic functions on them,
• $k$Var Let $k$ be an algebraically closed field. (Affine or Projective) algebraic varieties constitute objects in this category and (variety) morphisms are morphisms here. (PS: If you know this category, then you know most of the categories above :P)
We also have something known as Subcategories; a subcategory is a category whose objects and morphisms are the objects and morphisms of another (bigger) category. A few examples to illustrate this —
• PSet is a subcategory of Set.
• Ab, the category of abelian groups is a subcategory of Grp,
• Rng = Category of rings (possibly without identity), Ring = Category of rings with identity and CRing = Category of commutative rings, Field = Category of fields. Then Field $\subset$ CRing $\subset$ Ring $\subset$ Rng (PS: This fancy notation is by Jacobson.)
Why study Category Theory?
Categories were introduced to make rigorous the meaning of the word natural used in many contexts. As an example, consider this. Given a $k$-vector space $V$ and a basis of $V$, we have an explicit map from $V$ to its dual, $V^*$, but there is no such explicit map without a basis. However, we have a natural map from $V$ to its bidual, $V^{**}$, namely, $v \mapsto (F_v: V^* \to k)$; $F_v(f) = f(v)$.
A category, say of groups, has group-homomorphisms that take a group to another retaining the juice or information of the original group. We can then study a tough group by looking at its image in a relatively easier group and can deduce information about the tough group. In a somewhat similar fashion, we have maps (called as functors) that take objects and morphisms of one category to another. Thus, study of a tougher category can be simplified by looking at corresponding objects (and morphisms) in the other category.
For instance, a geometric assertion on affine varieties can be transferred to a statement in commutative algebra with the use of a functor that associates to each affine variety, its coordinate ring. Similarly, one observes that homeomorphism of path-connected topological spaces implies group-isomorphism of their fundamental groups. Thus, if the fundamental groups are different, then the two spaces cannot be homeomorphic.
More on Functors in a future post!
Future posts:
• Functors
• Products & coproducts
• Limits & colimits
• Yoneda Lemma
Abhishek Parab
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Rachit March 4, 2013 00:45
Rhie Chow interpolation
Hi everyone,
I have been trying to write a MATLAB code for 2D flow of the SIMPLE algorithm for incompressible flows on a collocated grid. I do understand that the Rhie Chow interpolation is used against the normal interpolation to calculate the velocity on the faces, in order to avoid checkerboarding effect. However, from what I gathered from the literature, there is no reference to where is the Rhie Chow interpolation used. In the calculation face velocities in the discretization of the continuity equation? Or is it used to calculate the face velocities in the calculation of convective flux in the momentum equation as well? If that is so then what is the point of using schemes like Upwind, Central Difference, QUICK, etc?
I would be extremely grateful, if someone could help me out on this.
Rachit
dsanthosh2 March 4, 2013 01:54
you have to use Rhie and Chow interpolation in the continuity equation. Not in the momemtume equation.
RodriguezFatz March 4, 2013 05:25
Quote:
Originally Posted by dsanthosh2 (Post 411259) you have to use Rhie and Chow interpolation in the continuity equation. Not in the momemtume equation.
But how do you get the flux at the cell faces for the convective term in the momentum euqation?
Quote:
Originally Posted by Rachit (Post 411252) Or is it used to calculate the face velocities in the calculation of convective flux in the momentum equation as well? If that is so then what is the point of using schemes like Upwind, Central Difference, QUICK, etc?
I think it is used in both cases, continuity and momentum. Remember, that you don't really directly solve the continuity equation in schemes like SIMPLE...
The momentum term is not linear. Thus, you have to interpolate the flux on the cell face (as the "transporting" medium) and you have to interpolate the velocity on the cell faces (as the "independent variable"). If you have "rho*u*u", then "rho*u" is the flux through the face, and one single "u" is the independent variable. That's the one, you create your linear system of equations for. And this is also the one, you can decide the way of interpolation for (as you described).
If you have "rho*v*u" and you are creating the momentum equation for "u", then "rho*v" is the flux through the face, and "u" is the one you solve with your matrix...
Did that become clearer?
andy_ March 4, 2013 06:14
Quote:
Originally Posted by Rachit (Post 411252) Hi everyone, I have been trying to write a MATLAB code for 2D flow of the SIMPLE algorithm for incompressible flows on a collocated grid. I do understand that the Rhie Chow interpolation is used against the normal interpolation to calculate the velocity on the faces, in order to avoid checkerboarding effect. However, from what I gathered from the literature, there is no reference to where is the Rhie Chow interpolation used. In the calculation face velocities in the discretization of the continuity equation? Or is it used to calculate the face velocities in the calculation of convective flux in the momentum equation as well? If that is so then what is the point of using schemes like Upwind, Central Difference, QUICK, etc? I would be extremely grateful, if someone could help me out on this. Thanks in advance, Rachit
Calling the addition of pressure gradient terms to the mass flux "interpolation" is rather misleading because there are no pressure terms in the mass flux. Nonetheless adding pressure smoothing terms to the mass flux is a requirement for many schemes.
The pressure smoothing terms are usually needed in the mass fluxes in the continuity equation to oppose pressure/velocity decoupling. For the mass fluxes in the other transport equations the pressure smoothing terms are optional but are usually included so that what is a mass flux is consistent and various numerical fixes that rely on the mass fluxes in a cell summing to zero hold.
The flux of momentum involves squared velocities. One velocity can be involved in determining the mass flux and the other velocity that is being transported by the mass flux can be evaluated using upwind, central difference, QUICK or whatever.
RodriguezFatz March 4, 2013 06:21
Quote:
Originally Posted by andy_ (Post 411312) One velocity can be involved in determining the mass flux and the other velocity that is being transported by the mass flux can be evaluated using upwind, central difference, QUICK or whatever.
That's what I was trying to explain. Better read andy's post - it's easier to understand.
andy_ March 4, 2013 06:28
Quote:
Originally Posted by RodriguezFatz (Post 411313) That's what I was trying to explain. Better read andy's post - it's easier to understand.
Had I seen that you had posted I would not have replied. My technique of putting posts to reply to in tabs and then getting on with other things while replying perhaps needs modifying.
Rachit March 4, 2013 09:57
Dear Santo, Rodriguez and Andy,
Thank you for your replies. I shall work it out this way in my code. Thanks once again.
Regards,
Rachit
Rachit March 5, 2013 10:11
Hi,
Looks like my confusion pertaining the Rhie Chow interpolation hasn't ended yet. :( From what's there in Versteeg and Malalasakera, to calculate the velocity at face e, I will need to the pressure at EE. What if my EE is outside the domain? Is there any way find the value of PEE without using the concept of ghost cells?
Also, to calculate the pressure smoothing term I need to aP which is the coefficient of variable u at point P. However, the coefficient is itself made up of convective flux terms (something for which I am using the Rhie Chow interpolation to calculate). What do I do then? Do I take the flux values from the previous step or one has to do something else?
Rachit
andy_ March 5, 2013 11:38
> What if my EE is outside the domain?
What to do about the pressure smoothing on the boundary is up to you. For example, on a solid boundary the mass flow is zero but the pressure gradients will generally be non zero. So do you use a zero mass flux or include the pressure smoothing term? If the former, global mass conservation will be physically reasonable but there will be a jump caused by smoothing on one face but not the other and in the presence of strong pressure gradients from body forces like swirl it can be large and unreasonable. If the latter, you will have unreasonable mass fluxes on the solution boundary and issues such as coming up with ways to evaluate pressure gradients at the boundary.
> Do I take the flux values from the previous step or one has to do something else?
Storing the mass fluxes along with the solution variables is a common thing to do for schemes using Rhie and Chow type pressure smoothing.
FMDenaro March 5, 2013 12:56
Justo to give you my opinion ...
First, start considering everywhere in the domain the Hodge decomposition
Vn+1 + Grad phi = V*
1) in the interior, after computing V* somehow, you will search for a "pressure" gradient such that it ensures Div Vn+1 = 0 -> Div Grad phi = Div V*
2) on the boundaries, you only have to prescribe something about Vn+1 and use the Hodge decomposition to ensures the correct mass flux by solving the modified pressure equation.
On co-located grids, you can use the approximate projection method (APM) to avoid checkerboard modes in the pressure, the RC interpolation can be applied on the flux defined by the Hodge decomposition.
However, the RC interpolation has some degrading in the accuracy...
If you want, some more deatils about possible remedy for checkerboard modes are addressed at:
http://onlinelibrary.wiley.com/doi/1....1368/abstract
http://onlinelibrary.wiley.com/doi/1....1368/abstract
malaboss March 4, 2014 22:26
Hi,
I am actually also trying to understand how is done the Rhie Chow interpolation in OpenFOAM and there are several things that still remain unclear. I understand that we interpolate pressure on the faces of the cell and then use this face pressure to construct the central difference scheme for pressure, which couples the adjacent pressures and avoid the checkerboard effect. This is refered as the pressure smoothing by Philipp and Andy.
1) I don't get were it is formulated in OpenFOAM. I see some descriptions of the RC interpolation with the usage of the H and A Matrix, but I don't really understand the link between theses matrices and the actual RC interpolation.
2) In the literature, some improvements of the RC interpolation are mentioned. Then, I am not sure which version has been implemented in OpenFOAM. Do you have some info about it ?
3) In order to better understand the effect of this anti-chekboarding interpolation, do you know if it is possible to disable the Rhie Chow interpolation in the OpenFOAM and how to do it ?
sbaffini March 5, 2014 03:01
In my opinion, a very detailed and clear set of notes on unstructured FV methods for CFD is the one by Prof. Murthy, former Fluent developer:
https://engineering.purdue.edu/ME608...ass-notes.html
where you can find most of the things you need about a CFD solver.
For what concerns OpenFOAM specifically, consider the following article:
Tukovic, Jasak: A moving mesh finite volume interface tracking method for surface tension dominated interfacial fluid flow. Comp. Fluids 55 (2012), pp. 70-84
Here, the OpenFOAM implementation and modifications are discussed at length.
However, consider that most of RC has nothing to do with the pressure equation, where it simply reduces to use an Approximate Projection Method (APM), as discussed by F.M. Denaro above. More roughly speaking this just means using a compact, straight, Laplacian operator in the pressure equation. Moreover, this is actually your only option for unstructured FV codes.
Where RC really comes into play is in the determination of the mass flux, which in order to be consistent with the above pressure equation has to include a pressure part. Still, this part is only a perturbation vanishing with the grid spacing going to 0.
malaboss March 6, 2014 17:00
Hi and thanks for the answer.
I found in Tukovic and Jasak's paper what I was looking for.
I will sum it up to make sure that I really get it and to save time to other people interested in it :
OpenFOAM uses collocated grid in order to handle complex geometries. One known problem about the collocated grids is the checkboarder effect, that is to say, the decoupling between adjacent cells for pressure and velocity which will not be able to damp possible oscillations of the fields, leading to an unphysical result.
So why do we have a decoupling between adjacent cells ?
Let's consider the 3 same cubic cells (W P and E separated by faces which locations are denoted by w and e) and a 1D case, and a constant density case.
We do not want to calculate grad(P) with a 1st order scheme but a central difference scheme.
Momentum equation applied to the 3 cells
I do not discretize grad(p) for the moment because this is where lies the RC interpolation
Continuity equation with FVM applied to the cell P.
rho ue Se - rho uw Sw=0
We need to know the velocity at the faces e and w, but we have it at the center.
So we approximate ue by (uE+uP)/2 and uw by (uW+uP)/2
As a result
And the continuity equation reduces to
1/2 * [(H/A)_E + (H/A)_P - grad(p)_P/A_P - grad(p)_E/A_E ] - 1/2 * [(H/A)_W + (H/A)_P - grad(p)_P/A_P - grad(p)_W/A_W ] = 0
This is an approximation for
1/2 * [(H/A)_e - grad(p)_e/A_e ] - 1/2 * [(H/A)_w - grad(p)_w/A_w ] = 0
The goal of the RC interpolation is to involve Pressure at adjacent cells in the continuity equation.
What is not RC interpolation
Calculate grap(P) at cell centers or neighboring cells first.
Then the continuity equations without the H/A terms reduces to
= 1/(4*deltaX) * [- (P_E-P_W)- (P_P-P_EE) +(P_E-P_W) +(P_P-P_WW) ]
= 1/(4*deltaX) * [P_EE -P_WW ]
Here only P_EE and P_WW are coupled leading to unphysical oscillations
What is RC interpolation
Calculate grap(P) at cell faces directly
Then the continuity equations without the H/A terms reduces to
= 1/(4*deltaX) * [- (P_P-P_E) - (P_P-P_W)]
= 1/(4*deltaX) * [P_E - 2P_P +P_W ]
Here, adjacent cells are coupled !
Remark
In the continuity equation, the term (H/A)_P will also disappear. I don't know which treatment is done by RC interpolation for this.
1/2 * [(H/A)_E + (H/A)_P - grad(p)_P/A_P - grad(p)_E/A_E ] - 1/2 * [(H/A)_W + (H/A)_P - grad(p)_P/A_P - grad(p)_W/A_W ] = 0
Answer to question 1 : How is formulated the RC interpolation in OpenFOAM ?
The particular gradient calculation at the faces for p is done in the formulation of the laplacian of p which calculates a flux of gradient p, in the pressure correction equation. This laplacian of p is equivalent to div(snGrad(p)). snGrad(P) is the Rhie Chow interpolation
Answer to question 2 : Has the last improvement of RC interpolation been implemented in OpenFOAM ?
The basic Rhie chow interpolation seems implemented this way. If the implementation of rhie chow interpolation is really what I described then there is room for improvement.
Answer to question 3 : Can we remove the Rhie Chow interpolation from OpenFOAM ?
Yes, we can remove RC interpolation from OpenFOAM by just replacing laplacian(p) by div(grad(p))
Hope that helps someone !
sbaffini March 7, 2014 02:52
I am not an expert in OpenFOAM, but i would consider the following two points:
1) Improvements in the RC procedure, as i understand them, have been around for a while and i would be impressed if the OpenFOAM formultation would still be based on the original one. I'm referring here to the time-step/under-relaxation dependence. I don't know of other relevant improvements. These are also clearly mentioned in the Tukovic paper.
2) OpenFOAM works with classes for implicit (fvm) and explicit (fvc) operators which are clearly distinguished. With the former you can build up a discretization matrix, with the latter you can only work on the source term side. This is fundamental to understand and also clarifies what approach you can actually use for the Poisson equation.
In OpenFOAM, according to my knowledge, you can only use the laplacian as an impicit discretization and not the extended one. There are different reasons for this, like the inherent difficulty to track the extended stencil for each cell, the possible bad properties of such a wide band matrix, the nearly impossible task to obtain the discretization coefficients for such a discrete operator. I am not aware of any unstructured code doing differently.
When looked in this perspective, RC is just a justification for such a compact Laplacian and the resulting derivation of the correct mass fluxes to use in the remaining equations.
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# Problem 17. Find all elements less than 0 or greater than 10 and replace them with NaN
Solution 2279440
Submitted on 14 May 2020 by silki baghla
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [ 5 17 -20 99 3.4 2 8 -6 ]; y_correct = [ 5 NaN NaN NaN 3.4 2 8 NaN ]; assert(isequalwithequalnans(cleanUp(x),y_correct))
x = 5.0000 NaN NaN NaN 3.4000 2.0000 8.0000 NaN
2 Pass
x = [ -2.80 -6.50 -12.60 4.00 2.20 0.20 -10.60 9.00]; y_correct = [ NaN NaN NaN 4.00 2.20 0.20 NaN 9.00] assert(isequalwithequalnans(cleanUp(x),y_correct))
y_correct = NaN NaN NaN 4.0000 2.2000 0.2000 NaN 9.0000 x = NaN NaN NaN 4.0000 2.2000 0.2000 NaN 9.0000
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Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 328 | 896 | {"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} | 2.71875 | 3 | CC-MAIN-2021-04 | latest | en | 0.547281 |
https://www.giassa.net/?page_id=207 | 1,618,390,754,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00159.warc.gz | 887,049,061 | 19,616 | # I – Nearest Neighbour Interpolation
Nearest neighbour interpolation is the simplest approach to interpolation. Rather than calculate an average value by some weighting criteria or generate an intermediate value based on complicated rules, this method simply determines the “nearest” neighbouring pixel, and assumes the intensity value of it.
First, let’s consider this in a 1-dimensional case. Observe the plot below.
y=f(x)
Let’s say that we wanted to insert more data points between the points x1 (=2) and x2 (=3), that approximate the values between them, which range between f(x1) (=4) and f(x2) (=6). Using nearest neighbour interpolation, our result would look like:
y=f(x) using 1D Nearest Neighbour Interpolation
We can see above that for each data point, xi, between our original data points, x1 and x2, we assign them a value f(xi) based on which of the original data points was closer along the horizontal axis.
Now, extending this to 2D, assume that we want to re-size a tiny, 2×2 pixel image, X, as shown below, so that it “fits” in the larger 9×9 image grid, Y, to the right of it.
Example of Upsampling an Image by a Non-Integral Factor
As shown above, when we resize by a non-integral factor (as outlined in the beginnging of this section on interpolation) pixels cannot simply be cloned by column/row – we need to interpolate them. The squares (representing pixels) forming a vertical and horizontal line through the rightmost image, for example, cannot contain different color values. They can only contain a single color value.
To visualize nearest neighbour interpolation, consider the diagram below. The data points in the set X represent pixels from the original source image, while the data points in the set Y represent pixels in our target output image.
Coordinate System
So, for each pixel in the output image Y, we must calculate the nearest neighbouring pixel in our source image X. Furthermore, we should only need to rely on 4 specific data points: X(A,B), X(A+1,B), X(A,B+1), and X(A+1,B+1). We can handle this decision very quickly with a few IF statements, as outlined in the pseudocode below.
This is very straightforward, but how do we properly index our images when we are dealing with an image larger than 2×2 pixels? We need a transformation that will handle this. Consider a small image which is ‘m’ pixels wide by ‘n’ pixels high, which we want to re-size to ‘p’ pixels wide by ‘q’ pixels high, assuming that p>m and q>n. Now, we need two scaling constants:
• s1 = p/m
• s2 = q/n
Now, we simply loop through all the pixels in the target/output image, addressing the source pixels to copy from by scaling our control variables by s1 and s2, and rounding the resulting scaled index values. The following example shows this in action on a 2×2 pixel “checkerboard” pattern image.
As shown in the images below, the 2×2 checkerboard image was upsampled to a 640×480 pixel image without any changes at all. Furthermore, due to the simplicity of this algorithm, the operation takes very little time to complete. Note the units on the axes for the images below. They look the same, but the one on the left is actually MUCH smaller than the one on the right.
Upsampled Checkerboard Pattern
The major drawback to this algorithm is that, despite its speed and simplicity, it tends to generate images of poor quality. Although the checkerboard pattern was upsampled flawlessly, images such as photographs come out “blocky”, though with little to no noticable loss of sharpness, as demonstrated below. The key point here is that this is a simple and fast algorithm. Included below are two more examples that demonstrate the drawbacks of this algorithm.
Small image of an ‘X’
Small Image of Trees
Now, after applying the nearest neighbour algorithm, we get the following results:
Upsampled with Aliasing
As we can see in the results above (you can click on it to get a better view) the results are of poor quality. Although the sharpness of the original image is retained, we notice how the image on the left of an ‘X’ has “jaggies” (ie: jagged edges) and the image on the right looks pixelated and distorted. This is referred to as aliasing, and there are several ways to deal with it. Two of the most straightforward ways are using a better interpolation method, as covered on the proceeding subsection on interpolation, or the use of spatial domain image filtering, which is covered in the sections on filtering. Finally, included below is the code used to generate the above results, along with the nearest neighbour algorithm MATLAB code rewritten as a MATLAB M-function for convenience.
# Sample Code:
## 8 thoughts on “I – Nearest Neighbour Interpolation”
1. Hi. I’m trying to implement my own nearest neighbour resizing algorithm. Your article is really helpfull. I just don’t understand why are you substracting 1 from j and k in your code, although in text you are only dividing p/m and q/n. This seems more logical to me, but I do have problems with my code, because I’m getting coordinates (0,0) in my code, which are of course invalid.
Thanks for your help!
• Hi there.
The subtracting of (1) from J and K is basically to deal with how arrays are indexed in MATLAB. eg: In C/C++, an array of ‘N’ integers is addressed on [0,N-1], while in MATLAB, it is addressed on [1,N]. If you don’t include this piece of code, you will attempt to access non-existent array members.
Cheers!
2. I’m very grateful for your answer. But I just can’t see how substraction of j and k is related to array indexing – we are actually looking for ratio in that equation, and your technique always seems to return a bit too big ratio. I also don’t understand why are you using dot operator in front of division operator? I’m real beginner in matlab and I’m trying to relearn math I forgot many years ago, so I might be missing something 🙂
I found a working solution here – in second post:
http://stackoverflow.com/questions/1550878/nearest-neighbor-interpolation-algorithm-in-matlab
This seems to avoid access to nonexistent pixels and it seems to be cleaner, easier to understand solution:
Ratio in this case is always correct. Because we substract 0.5 from j and add 0.5back after division, we never get any nonexisting coordinates. This seems a cleaner solution to me, but it might be wrong in some way.
x_output = round((j-0.5)/resizing_factor+0.5);
What do you think?
Thanks.
• I think the best way to see why I’m using the subtraction of 1 for my scaling coefficients would be to just remove the semi-colon at the end of a few lines of my code and let MATLAB display the results while generating them. It should should how this effectively scales the image coordinates so the array indexes are never out of bounds.
3. Hi,
Your code is very neat and easy to understand. However, it is only useful when we want to produce an output image which is larger than input image. How does your code change if we want to produce an output image which is smaller than input image.
Consider the following example:
Input image resolution: 256×256
We want to have the output image with 1024×1024 resolution.
• Not trying to nitpick, but I think you mean when the input image resolution is 1024×1024, and the output resolution is 256×256. In other parts of the tutorial, I note that you can either use decimation (ie: discarding columns and rows), or, you can first upscale and then downscale by an integer ratio of the two for better results, at the expense of increased cycles. Sorry for the late reply
4. You have no idea how much I love you right now. This has seriously helped me out big time with my coursework. Thank you!
5. Hi,
Firstly thank you for perfect explanation.I apply affine transformation an image and my output image has a missing pixel.So I want to apply my output image nearest neighbour interpolation.However, when apply an nearest n. interpolation ,the output image has a just black white lines.How can implement nearest neighbour interpolation to transformed image?
I use affine transformation matrix just like below:
Output image: (O)
[X1
X2]
Input image:(I)
[1 x1 y1 0 0 0
0 0 0 1 x1 y1]
6 Parameters: (P)
[a
b
c
d
e
f]
[O] = [I].[P]
Could you explain briefly,please? | 1,888 | 8,246 | {"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} | 3.84375 | 4 | CC-MAIN-2021-17 | longest | en | 0.893745 |
http://www.apmaths.uwo.ca/~arich/IntegrationProblems/MathematicaSyntaxFiles/MathematicaSyntaxFiles.html | 1,532,257,261,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00523.warc.gz | 403,974,792 | 5,889 | # Rubi Integration Test Suite
## Expressed in Mathematica Syntax
This page provides links to text files containing the Rubi Integration Test Suite expressed in Mathematica syntax. Each problem is stored as a list of the following items:
1. the integrand;
2. the integration variable;
3. the number of rule applications required to integrate the integrand;
4. the optimal antiderivative; and
5. on a few problems, a nonoptimal, but acceptable, antiderivative.
The optimal antiderivatives in the test suite are just the simplest ones found so far. If you find significantly simpler antiderivatives and want to contribute them to the test suite, please email them to Albert Rich.
Mathematica is required to load these package files; however, they can also be viewed as text files. To view or download the test suite files, click on one of the following links:
Integration problems from independent test suites:
Integration problems involving rational and algebraic functions:
Problems involving powers of linear binomials:
Problems involving powers of quadratic binomials:
Problems involving powers of general binomials:
Problems involving powers of improper binomials:
Problems involving powers of quadratic trinomials:
Problems involving powers of quartic trinomials:
Problems involving powers of general trinomials:
Problems involving powers of improper trinomials:
Miscellaneous rational and algebraic function problems:
Integration problems involving exponentials:
Integration problems involving logarithms:
Integration problems involving trig functions:
Sine function problems:
• Download (a+b sin(e+f x))m (c+d sin(e+f x))n (A+B sin(e+f x)+C sin(e+f x)2)
Cosine function problems:
• Download (a+b cos(e+f x))m (c+d cos(e+f x))n (A+B cos(e+f x)+C cos(e+f x)2)
Tangent function problems:
• Download (a+b tan(e+f x))m (c+d tan(e+f x))n (A+B tan(e+f x)+C tan(e+f x)2)
Cotangent function problems:
Secant function problems:
• Download (a+b sec(e+f x))m (d sec(e+f x))n (A+B sec(e+f x)+C sec(e+f x)2)
Cosecant function problems:
• Download (a+b csc(e+f x))m (d csc(e+f x))n (A+B csc(e+f x)+C csc(e+f x)2)
Miscellaneous trig function problems:
Integration problems involving inverse trig functions:
Inverse sine function problems:
Inverse cosine function problems:
Inverse tangent function problems:
Inverse cotangent function problems:
Inverse secant function problems:
Inverse cosecant function problems:
Integration problems involving hyperbolic functions:
Hyperbolic sine function problems:
Hyperbolic cosine function problems:
Hyperbolic tangent function problems:
Hyperbolic cotangent function problems:
Hyperbolic secant function problems:
Hyperbolic cosecant function problems:
Miscellaneous hyperbolic function problems:
Integration problems involving inverse hyperbolic functions:
Inverse hyperbolic sine function problems:
Inverse hyperbolic cosine function problems:
Inverse hyperbolic tangent function problems:
Inverse hyperbolic cotangent function problems:
Inverse hyperbolic secant function problems:
Inverse hyperbolic cosecant function problems:
Integration problems involving special functions: | 702 | 3,165 | {"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} | 2.796875 | 3 | CC-MAIN-2018-30 | latest | en | 0.785761 |
http://mathhelpforum.com/calculus/15703-one-more-de-question.html | 1,529,951,648,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00497.warc.gz | 206,086,971 | 8,633 | # Thread: One more DE question
1. ## One more DE question
Im a bit hazy on my log rules....and this problem isnt helping...
Original problem: ((e^y+1)^2)e^-ydx+ ((e^x+1)^3)e^-xdy=0
Where i've gotten: -((e^x+1)^-3)e^xdx= ((e^y+1)^-2))e^y dy
im not sure how to get rid of those e^x and the e^y.....probaly some log rule im missing, Thanks for any help!
2. Hello, neven87!
Problem: .$\displaystyle (e^y+1)^2e^{-y}dx + (e^x+1)^3e^{-x}dy\:=\:0$
Answer: .$\displaystyle (e^x+1)^{-2} + 2(e^y+1)^{-1}\:=\:C$
Re-arrange the equation: .$\displaystyle \frac{(e^y + 1)^2}{e^y}\ dx + \frac{(e^x + 1)^3}{e^x}\ dy \:=\:0\quad\Rightarrow\quad\frac{(e^x + 1)^3}{e^x}\,dy \;=\;-\frac{(e^y + 1)^2}{e^y}$
Separate variables: .$\displaystyle \frac{e^y}{(e^y+1)^2}\,dy \;=\;-\frac{e^x}{(e^x+1)^3}\,dx\quad\Rightarrow\quad (e^y+1)^{-2}(e^y\,dy) \;=\;-(e^x+1)^{-3}(e^x\,dx)$
Integrate: .$\displaystyle \int(e^y+1)^{-2}(e^y\,dy) \;=\;-\int(e^x+1)^{-3}(e^x\,dx)$
Let $\displaystyle u = e^y + 1\quad\Rightarrow\quad du = e^y\,dy$
Let $\displaystyle v = e^x + 1\quad\Rightarrow\quad dv= e^x\,dx$
Substitute: .$\displaystyle \int u^{-2}du \;=\;-\int v^{-3}\,dv\quad\Rightarrow\quad -u^{-1}\;=\;\frac{v^{-2}}{2} + c$
. . $\displaystyle \frac{v^{-2}}{2} + u^{-1}\;=\;c\quad\Rightarrow\quad v^{-2} + 2u^{-1} \;=\;C$
Back-substitute: .$\displaystyle (e^x + 1)^{-2} + 2(e^y + 1)^{-1} \;=\;C$ | 635 | 1,372 | {"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} | 3.875 | 4 | CC-MAIN-2018-26 | latest | en | 0.211772 |
https://ezeenotes.in/a-particle-moves-a-distance-x-in-time-t-according-to-equation-x-t5-1-the-acceleration-of-particle-is-proportional-to/ | 1,708,496,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00666.warc.gz | 258,733,166 | 20,166 | # A particle moves a distance x in time t according to equation x = (t+5)-1 . The acceleration of particle is proportional to
Question : A particle moves a distance x in time t according to equation x = (t+5)-1 . The acceleration of particle is proportional to
(a) (Velocity)2/3
(b) (Velocity)3/2
(c) (distance)2
(d) (distance)-2
Answer : The Right option is (b)
Solution :
Tags: No tags | 113 | 395 | {"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} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.57878 |
http://underpop.online.fr/j/java/algorithims-in-java-1-4/ch05lev1sec6.htm | 1,519,181,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813322.19/warc/CC-MAIN-20180221024420-20180221044420-00264.warc.gz | 375,526,840 | 5,000 | Previous Next
### Tree Traversal
Before considering algorithms that construct binary trees and trees, we consider algorithms for the most basic tree-processing function: tree traversal: Given a (reference to) a tree, we want to process every node in the tree systematically. In a linked list, we move from one node to the next by following the single link; for trees, however, we have decisions to make, because there may be multiple links to follow.
We begin by considering the process for binary trees. For linked lists, we had two basic options (see Program 5.5): process the node and then follow the link (in which case we would visit the nodes in order), or follow the link and then process the node (in which case we would visit the nodes in reverse order). For binary trees, we have two links, and we therefore have three basic orders in which we might visit the nodes:
• Preorder, where we visit the node, then visit the left and right subtrees
• Inorder, where we visit the left subtree, then visit the node, then visit the right subtree
• Postorder, where we visit the left and right subtrees, then visit the node
We can implement these methods easily with a recursive program, as shown in Program 5.14, which is a direct generalization of the linked-list–traversal program in Program 5.5. To implement traversals in the other orders, we permute the method invocations in Program 5.14 in the appropriate manner. Screenshot shows the order in which we visit the nodes in a sample tree for each order. Screenshot shows the sequence of method invocations that is executed when we invoke Program 5.14 on the sample tree in Screenshot.
##### Screenshot Preorder-traversal method invocations
This sequence of method invocations constitutes preorder traversal for the example tree in Screenshot.
##### Screenshot Tree-traversal orders
These sequences indicate the order in which we visit nodes for preorder (left), inorder (center), and postorder (right) tree traversal.
We have already encountered the same basic recursive processes on which the different tree-traversal methods are based, in divide-and-conquer recursive programs (see Figures 5.8 and 5.11), and in arithmetic expressions. For example, doing preorder traversal corresponds to drawing the marks on the ruler first, then making the recursive calls (see Screenshot); doing inorder traversal corresponds to moving the biggest disk in the towers of Hanoi solution in between recursive calls that move all of the others; doing postorder traversal corresponds to evaluating postfix expressions, and so forth. These correspondences give us immediate insight into the mechanisms behind tree traversal. For example, we know that every other node in an inorder traversal is an external node, for the same reason that every other move in the towers of Hanoi problem involves the small disk.
### Recursive tree traversal
This recursive method takes a link to a tree as an argument and calls visit with each of the nodes in the tree as argument. As is, the code implements a preorder traversal; if we move the call to visit between the recursive calls, we have an inorder traversal; and if we move the call to visit after the recursive calls, we have a postorder traversal.
```private void traverseR(Node h) { if (h == null) return; h.item.visit(); traverseR(h.l); traverseR(h.r); } void traverse() { traverseR(root); }
```
It is also useful to consider nonrecursive implementations that use an explicit pushdown stack. For simplicity, we begin by considering an abstract stack that can hold items or trees, initialized with the tree to be traversed. Then, we enter into a loop, where we pop and process the top entry on the stack, continuing until the stack is empty. If the popped entity is an item, we visit it; if the popped entity is a tree, then we perform a sequence of push operations that depends on the desired ordering:
• For preorder, we push the right subtree, then the left subtree, and then the node.
• For inorder, we push the right subtree, then the node, and then the left subtree.
• For postorder, we push the node, then the right subtree, and then the left subtree.
We do not push null trees onto the stack. Screenshot shows the stack contents as we use each of these three methods to traverse the sample tree in Screenshot. We can easily verify by induction that this method produces the same output as the recursive one for any binary tree.
##### Screenshot Stack contents for tree-traversal algorithms
These sequences indicate the stack contents for preorder (left), inorder (center), and postorder (right) tree traversal (see Screenshot), for an idealized model of the computation, similar to the one that we used in Screenshot, where we put the item and its two subtrees on the stack, in the indicated order.
### Preorder traversal (nonrecursive)
This nonrecursive stack-based method is functionally equivalent to its recursive counterpart, Program 5.14.
```private void traverseS(Node h) { NodeStack s = new NodeStack(max); s.push(h); while (!s.empty()) { h = s.pop(); h.item.visit(); if (h.r != null) s.push(h.r); if (h.l != null) s.push(h.l); } } void traverseS() { traverseS(root); }
```
The scheme described in the previous paragraph is a conceptual one that encompasses the three traversal methods, but the implementations that we use in practice are slightly simpler. For example, for preorder, we do not need to push nodes onto the stack (we visit the root of each tree that we pop), and we therefore can use a simple stack that contains only one type of item (tree link), as in the nonrecursive implementation in Program 5.15. The system stack that supports the recursive program contains return addresses and argument values, rather than items or nodes, but the actual sequence in which we do the computations (visit the nodes) is the same for the recursive and the stack-based methods.
A fourth natural traversal strategy is simply to visit the nodes in a tree as they appear on the page, reading down from top to bottom and from left to right. This method is called level-order traversal because all the nodes on each level appear together, in order. Screenshot shows how the nodes of the tree in Screenshot are visited in level order.
##### Screenshot Level-order traversal
This sequence depicts the result of visiting nodes in order from top to bottom and left to right in the tree.
Remarkably, we can achieve level-order traversal by substituting a queue for the stack in Program 5.15, as shown in Program 5.16. For preorder, we use a LIFO data structure; for level order, we use a FIFO data structure. Otherwise, there is no difference at all between the two programs. These programs merit careful study, because they represent approaches to organizing work remaining to be done that differ in an essential way. In particular, level order does not correspond to a recursive implementation that relates to the recursive structure of the tree.
Preorder, postorder, and level order are well defined for forests as well. To make the definitions consistent, think of a forest as a tree with an imaginary root. Then, the preorder rule is "visit the root, then visit each of the subtrees," the postorder rule is "visit each of the subtrees, then visit the root." The level-order rule is the same as for binary trees. Direct implementations of these methods are straightforward generalizations of the stack-based preorder traversal programs (Programs 5.14 and 5.15) and the queue-based level-order traversal program (Program 5.16) for binary trees that we just considered. We omit consideration of implementations because we consider a more general procedure in .
### Level-order traversal
Switching the underlying data structure in preorder traversal (see Program 5.15) from a stack to a queue transforms the traversal into a level-order one.
```private void traverseQ(Node h) { NodeQueue q = new NodeQueue(max); q.put(h); while (!q.empty()) { h = q.get(); h.item.visit(); if (h.l != null) q.put(h.l); if (h.r != null) q.put(h.r); } } void traverseQ() { traverseQ(root); }
```
#### Exercises
5.79 Give preorder, inorder, postorder, and level-order traversals of the following binary trees:
5.80 Show the contents of the queue during the level order traversal (Program 5.16) depicted in Screenshot, in the style of Screenshot.
Show that preorder for a forest is the same as preorder for the corresponding binary tree (see Property 5.4), and that postorder for a forest is the same as inorder for the binary tree.
Give a nonrecursive implementation of inorder traversal.
5.83 Give a nonrecursive implementation of postorder traversal.
5.84 Write a program that takes as input the preorder and inorder traversals of a binary tree and that produces as output the level-order traversal of the tree.
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# GEORGIOS BEKAS
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Solved | 1,514 | 5,568 | {"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} | 3.015625 | 3 | CC-MAIN-2020-34 | latest | en | 0.888233 |
https://www.english-arabic.org/english-to-arabic-meaning-decibel | 1,642,483,558,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00237.warc.gz | 807,669,518 | 17,955 | # English to Arabic Meaning :: decibel
Decibel :
ديسيبل وحدة قياس
ديسيبل وحدة قياس - ديسيبل وحدة قياسديسيبل وحدة قياسديسيبل
#### Show English Meaning (+)
Noun(1) A logarithmic unit of sound intensity(2) 10 times the logarithm of the ratio of the sound intensity to some reference intensity
#### Show Examples (+)
(1) The 215 - decibel sound waves can travel 300 miles through the ocean.(2) The device can create a sound at 120 decibels , loud enough to disable enemy combatants.(3) Sound pressure against the ears is measured in decibels .(4) Mr. Russell apparently had not overheard the conversation, as the din in the room had risen a few decibels .(5) Humans feel pain when they hear sounds of 120 decibels , a level typically reached next to the speakers at a rock concert.(6) When the woman had the noise level tested, it measured 67 to 72 decibels .(7) According to the results, the noise on Sunday was 58 decibels, compared to 80 decibels on Friday afternoon.(8) Hearing protection is recommended when sound exceeds 85 decibels .(9) She glared at me, her voice rising up ten decibels .(10) Failing to get the desired answer, she raised her voice several decibels(11) Because the range of sound pressures that can be heard is so large, a logarithmic scale of decibels is used to measure sound intensity.(12) Efficiency of the speakers determines the distance sound will travel and is measured in terms of decibels , the higher the better.(13) In the Stotts' front garden, they found sound levels reached 97.8 decibels , which can cause serious damage to the ears.(14) Vibrations are measured in decibels (d.b.g.) which is the level of vibration.(15) Intensity is measured in decibels , most commonly using the decibel u251cu00f6u251cu00e7u251cu2510Au251cu00f6u251cu00e7u251cu00fb scale (dbA).(16) An aircraft taking off produces about 140 decibels of noise and motorways a further 75 decibels , both above levels deemed unacceptable by some health experts.
Synonyms
M
1. decibel ::
ديسيبل وحدة قياس
Antonyms
1. quiet ::
هادئ
2. silence ::
الصمت
3. still ::
ما يزال
Different Forms
decibel, decibels
Word Example from TV Shows
The best way to learn proper English is to read news report, and watch news on TV. Watching TV shows is a great way to learn casual English, slang words, understand culture reference and humor. If you have already watched these shows then you may recall the words used in the following dialogs.
...you'll keep the DECIBEL level to a minimum.
#### The Big Bang Theory Season 3, Episode 3
English to Arabic Dictionary: decibel
Meaning and definitions of decibel, translation in Arabic language for decibel with similar and opposite words. Also find spoken pronunciation of decibel in Arabic and in English language.
Tags for the entry 'decibel'
What decibel means in Arabic, decibel meaning in Arabic, decibel definition, examples and pronunciation of decibel in Arabic language.
## English-Arabic.Org | English to Arabic Dictionary
This is not just an ordinary English to Arabic dictionary & Arabic to English dictionary. This dictionary has the largest database for word meaning. It does not only give you English toArabic and Arabic to English word meaning, it provides English to English word meaning along with Antonyms, Synonyms, Examples, Related words and Examples from your favorite TV Shows. This dictionary helps you to search quickly for Arabic to English translation, English to Arabic translation. It has more than 500,000 word meaning and is still growing. This English to Arabic dictionary also provides you an Android application for your offline use. The dictionary has mainly three features : translate English words to Arabic translate Arabic words to English, copy & paste any paragraph in the Reat Text box then tap on any word to get instant word meaning. This website also provides you English Grammar, TOEFL and most common words. | 923 | 3,899 | {"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} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.769089 |
https://www.education.com/activity/article/create-cube-find-surface-area/ | 1,685,546,184,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00787.warc.gz | 825,482,579 | 22,380 | # Create a Cube and Find Its Surface Area
### What You Need:
• Cube template
• Ruler
• Pencil
• Glue or tape
• Markers, crayons, and/or colored pencils
• Scissors
### What You Do:
1. Print out the cube template.
2. Have your child decorate the cube using markers, crayons, and/or colored pencils. She can also add glitter glue, sequins, etc.
3. Have your child carefully cut the pattern out along the outer edge only.
4. Fold along the remaining lines.
5. Attach the small flaps to the squares with glue or tape to form the cube. If using glue you may need to give it time to dry before handling.
6. Have your child measure the length and width of each side. Find the area of each side by multiplying the length and width together. The length should be 2 inches and the width 2 inches also. Multiply 2 X 2 = 4 inches. Then add the area of the six sides together: 4 + 4 + 4 + 4 + 4 + 4 = 24 inches squared is the surface area.
7. After the activity is complete, have your child calculate the surface area of other cube or rectangular shaped objects around the house. For example, shoe box, cable box, digital camera, etc.
Create new collection
0
0 items | 293 | 1,159 | {"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} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.821384 |
http://mathhelpforum.com/algebra/39856-expanding-triple-brackets-easy.html | 1,498,755,529,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128329372.0/warc/CC-MAIN-20170629154125-20170629174125-00409.warc.gz | 262,633,641 | 11,030 | # Thread: Expanding triple brackets (easy?)
1. ## Expanding triple brackets (easy?)
I'm gonna feel really dumb asking this but how do you expand triple brackets?
I can do single and double brackets no problem with the FOIL method, but what multiplies with what when you have triple brackets?
Also I was hoping you could help me on a question also here it is:
$
x^3 - 2x^2 - 11x + 12 = 0$
I've deduced that $(fx = 1)$ because $1^3 - 2 (1^2) - 11 (1) + 12 = 0
$
But what do I do next to factorise?
2. Hello,
Originally Posted by Nick87
I'm gonna feel really dumb asking this but how do you expand triple brackets?
I can do single and double brackets no problem with the FOIL method, but what multiplies with what when you have triple brackets?
Also I was hoping you could help me on a question also here it is:
$
x^3 - 2x^2 - 11x + 12 = 0$
I've deduced that $(fx = 1)$ because $1^3 - 2 (1^2) - 11 (1) + 12 = 0
$
But what do I do next to factorise?
If $\alpha$ is a root of a polynomial $P(x)$ of degree n, then you can write : $P(x)=(x-\alpha)Q(x)$, with $Q(x)$ a polynomial of degree n-1.
Here, you can write that $x^3-2x^2-11x+12=(x-1)Q(x)$, where $Q(x)$ is of degree 2 --------> $Q(x)=ax^2+bx+c$
Try to find a,b & c
3. $x^3-2x^2-11x+12)\div(x-1)=x^2-x-12$
Now factor the quadratic $x^2-x-12=0$
$(x-4)(x+3)=0$
$x=4 \ or \ x=-3$
Solutions {-3, 1, 4}
Give an example of your triple bracket problem.
4. Example of my triple bracket problem?
I just need to PROVE that $(x-1)(x-3)(x+2)$ = $x^3 - 2x^2 - 5x + 6$
I know that it does because of the -3 and 2 being factors making the -2, -5 and +6 but i need to show expansion methods.
By the way thanks Moo and Masters! Muchly helpful! :P
5. First, expand the last two binomials using FOIL, since you know that method.
$(x-1)(x^2-x-6)$
Then, use the distributive property to multiply the binomial and the trinomial:
$x(x^2-x-6)-1(x^2-x-6)$
Continuing:
$x^3-x^2-6x-x^2+x+6$
Combine terms:
$x^3-2x^2-5x+6$
6. Thank you once again. My problem is solved! | 692 | 2,028 | {"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": 23, "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} | 4.375 | 4 | CC-MAIN-2017-26 | longest | en | 0.8744 |
http://www.padtinc.com/blog/page/43/ | 1,568,850,539,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573385.29/warc/CC-MAIN-20190918234431-20190919020431-00086.warc.gz | 303,068,308 | 55,740 | ## To Use Large Deflection or Not, That Is the Question
It seems like I’ve been explaining large deflection effects a lot recently. Between co-teaching an engineering class at nearby Arizona State University and also having a couple of customer issues regarding the concept, large deflection in structural analyses has been on my mind.
Before I explain any further, the thing you should note if you are an ANSYS Mechanical simulation user is this: If you don’t know if you need large deflection or not, you should turn it on. There is really no way to know for certain if it’s needed or not unless you perform a comparison study with and without it.
So, what are large deflection effects? In simple terms the inclusion of large deflection means that ANSYS accounts for changes in stiffness due to changes in shape of the parts you are simulating. The classic case to consider is the loaded fishing rod.
In its undeflected state, the fishing rod is very flexible at the tip. With a heavy fish on the end of the line, the rod deflects downward and it is then easy to observe that the stiffness of the rod has increased. In other words, when the rod is lightly loaded, a small amount of force will cause a certain downward deflection at the top. When the rod is heavily loaded however, a much larger amount of force will be needed to cause the tip to deflect downward by the same amount.
This change in the force amount required to achieve the same change in displacement implies that we do not have a linear relationship between force and displacement.
Consider Hooke’s law, also known as the spring equation:
F = Kx
Where F is the force applied, K is the stiffness of the structure, and x is the deflection. In a linear system, doubling the force results in double the displacement. In our fishing rod case, though, we have a nonlinear system. We might need to triple the force to double the displacement, depending on how much the rod is loaded relative to its size and other properties, and then to double the displacement again we might need to apply four times that force, just using numbers out of my head as examples.
So, in the case of the fishing rod, Hooke’s law in a linear form does not apply. In order to capture the nonlinear effect we need a way for the stiffness to change as the shape of the rod changes. In our finite element solution in ANSYS, it means that we want to recalculate the stiffness as the structure deflects.
This recalculation of the stiffness as the structure deflects is activated by turning on large deflection effects. Without large deflection turned on, we are constrained to using the linear equation, and no matter how much the structure deflects we are still using the original stiffness.
So, why not just have large deflection on by default and use it all the time? My understanding is that since large deflection adds computation expense to have it on, it’s off by default. It’s the same as for a lot of advanced usage, such as frictionless or frictional contact vs. the default bonded (simpler) behavior. In other words, turning on large deflection will trigger a nonlinear solution, meaning multiple passes through the solver using the Newton Raphson method instead of the single pass needed for a linear problem.
Here is an example of a simplified fishing rod. The image shows the undeflected rod (top), which is held fixed on the left side and has a downward force load applied on the right end. The bottom image shows the final deflected shape, with large deflection effects included. The deflection at the tip in this case is 34 inches.
In comparison running the same load with large deflection turned off resulted in a tip deflection of 40 inches. Thus, the calculated tip deflection is 15% less with large deflection turned on, since we are now accounting for change in stiffness with change in shape as the rod deflects.
Below we have a force (horizontal axis) vs. deflection (vertical axis) plot for a nonlinear simulation of a fishing rod with large deflection turned on. The fact that the curve is not a straight line confirms that this is a nonlinear problem, with the stiffness (slope of the curve) not constant. We can also see that as the force gets higher, the slope of the curve is more horizontal, meaning that more force is needed for each incremental amount of displacement. This matches our observations of the fishing rod behavior.
So, getting back to our original point, it’s often the case that we don’t know if we need to include large deflection effects or not. When in doubt, run cases with and without. If you don’t see a change in your key results, you can probably do without large deflection.
Here is an example using an idealized compressor vane. In this case, the deflections and stresses with and without large deflection effects are nearly the same (the stress difference is about 0.2%).
Large Deflection On:
Small Deflection:
Bottom line: when in doubt, try it out, with and without large deflection. In ANSYS Mechanical, Large Deflection effects are turned on or off in the details of the Analysis Settings branch.
It’s worth noting that turning on large deflection in ANSYS actually activates four different behaviors, known as large deflection which include large rotation, large strain, stress stiffening, and spin softening. All of these involve change in stiffness due to deformation in one way or another.
If you like this kind of info, or find it useful, we cover topics like this in our training classes. For more info, check out our training pages at http://www.padtinc.com/support/software/training.html.
## 3D Printing – A 2D Explainer from Shapeways
What is this 3D Printing anyway? It doesn’t take long for someone new to the technology to see the wide range of applications and implications it brings to the table. But what how does it actually work. Our friends at Shapeways have put together a great infographic that explains things well.
Take a look and share:
If you scrolled down this far, you may be asking, “Why is PADT sharing Shapeways material? Are they not competitors?” Well, to be honest, we recommend Shapeways to people all the time. Our Additive Manufacturing business is about producing engineering prototypes, tooling, and end-use products for manufacturing companies. When a hobbiest or artist comes to ask us for a prototype, we often recommend that they go visit Shapeways.
We also recommend that people who are interested in all the non-engineering applications for 3D Printing check out their marketplace. The things that people have come up with is just amazing and shows the unbounded potential of this technology.
## Presentation: Leveraging Simulation for Product Development of IoT Devices
The local SEMI chapter here in Arizona held a breakfast meeting on Monetizing Internet of Things (IoT) and PADT was pleased to be one of the presenters. Always a smart group, this was a chance to sit with people making the sensors, chips, and software that enable the IoT and dig deep in to where things are and where they need to be.
The event was hosted by one of our favorite customers, and neighbor right across the street, Freescale Semiconductor. Speakers included IoT experts from Freescale, Intel, Medtronics, ASU, and SEMICO Research.
Not surprisingly I talked about how Simulation can play a successful role in product development of IoT devices.
You can download a copy of the presentation here: PADT-SEMI-IOT-Simulation-1.pdf
UPDATE (11/9/2015): Great write-up by Don Dingee on this event in the SemiWiki. Click here to read it. It includes a great summary of the other speakers.
You can also see more details on how people use Simulation for this application on the ANSYS, Inc. website here. We also like this video from ANSYS that shows some great applications and how ANSYS is used with them:
A couple of common themes resonated across the speakers:
1. Price and size need to come down on the chips used in IoT (this was a semiconductor group, so this is a big part of their focus)
2. Lowering power usage and increasing power density in batteries is a key driver
3. The biggest issue in IoT is privacy and security. Keeping your data private and keeping people from hacking in to IoT devices.
4. Another big problem is dealing with all the data collected by IoT devices. How to make it useful and how to store it all. One answer is reducing the data on the device, another is only keeping track of what changes.
5. It is early, standards are needed but they are still forming.
If you look at this list, the first two problems are addressable with simulation:
PADT has a growing amount of experience with helping customers simulate and design IoT devices as well as the chips, sensors, and antenna that go in to IoT devices. To learn more, shoot us an email at info@padtinc.com or call 480.813.4884.
## Manufacturing Open House Highlights – October 2015
Here at PADT we help people who make products, stuff that gets manufactured. So we focused our open house yesterday on advanced manufacturing and invited the community to come out and network, learn, and share. Even though it was a busy week for technology events in Arizona, we had a great turnout on a surprisingly cloudy Wednesday evening.
October is Manufacturing month and this open house was part of the Arizona Commerce Authority’s coordinated events to highlight manufacturing in Arizona. You can learn more about other events in the state here.
This event was a bit more casual and less structured then past PADT open houses, letting attendees spend more time one-on-one with various experts and dig deep in to technologies like metal 3D Printing, urethane casting, topological optimization, and scanning.
What struck all of us here was the keen interest in and knowledge about the various tools we were showing across a wide range of attendees. From students with home built 3D Printers to managers from local aerospace companies that are on the forefront of Additive Manufacturing, the questions that were asks and comments that were made with insightful and show a transition of this technology from hype to real world application.
Below are some more quick snapshot taken during the event.
A big thanks to everyone who made it out and we hope to see more of you next time. If you have any questions about the application of advanced manufacturing technologies to your products, don’t hesitate to reach out to us at info@padtinc.com or 480.813.4884. As always, visit www.PADTINC.com to learn more.
## Free Training and Evaluation for ANSYS AIM
PADT is hosting a series of free training classes to introduce users to ANSYS AIM. We have pasted the invitation below. You can register here. We are very excited about this new tool from ANSYS, Inc. and are eager to share it with everyone. Look for more AIM information on this blog in the near future.
## Free Training and Evaluation for ANSYS® AIM™. Register Today – Seats Are Limited.
Discover how to design your next product
better… and faster
## ANSYS AIM: Integrated Multiphysics Simulation Environment for All Engineers
### Free Training and Evaluation for ANSYS® AIM™ – An Integrated Multi-physics Simulation Environment for All Engineers
As a special offer, PADT Inc. is offering FREE “Jump Start” training and hands-on evaluation for ANSYS® AIM™. Design engineers, method engineers and managers seeking to learn the latest simulation software, boost adoption and usability for the occasional user, or extend their existing CAD-based tool’s limited functionality will benefit from this no-obligation course.
Register Today – Seats are limited and will be filled on a first-come, first-served basis. On completion of the class, you’ll be qualified to receive and use a FREE 30-day ANSYS AIM download for evaluation.
All classes will be held from 9:00 a.m. – 4:00 p.m. local time and include a complimentary lunch.
PADT’s support team of ANSYS experts will help attendees understand where ANSYS AIM fits in to their organization and workflow. The class will address both situations and how ANSYS AIM provides the integration of CAD based systems and the ease of use of a modern tool in a product that steps the occasional user through the process without limiting functionality.
Watch this short video to learn more about the capabilities and benefits of ANSYS® AIM™ for the simulation of 3-D physics and multiphysics
## Beyond the Hype – Additive Manufacturing and 3D Printing Worldwide, A Summary of Terry Wholers’ Thoughts
Terry Wholers is the founder and principal consultant of Wohlers Associates Inc., an independent consulting firm that was launched 28 years ago. Wohlers and his team have provided consulting work to over 240 organizations in 24 countries as well as to 150 companies in the investment community. He has authored over 400 books, articles, and technical papers. Terry has twice served as a presenter at the White House. For the past 20 years hes has been the principal author for the Wohlers Report which is an annual worldwide publication focused on Additive Manufacturing and 3D Printing. In 2007 more than a 1,000 industry professionals from around the world selected Terry as the most influential person in Rapid Prototyping Development and Additive Manufacturing.
PADT was fortunate enough to sponsor, with the local SME group, an event in Fort Collins, Colorado where Terry came and shared his views on the industry. What follows is a summary of what we learned. They are basically notes and observations. Please contact us for any clarification or details:
Terry Wohlers started his talk by asking: How many people have heard of 3D printing?
He noted that these days it was pretty much everyone and if you haven’t then you must be living in a cave. It is like everyone can’t get enough of it.
There has been a lot of growth. In the last 5 years the industry has quadrupled. Last year it was a 4.1 billion industry and this year 5.5 billion. Terry doesn’t own any stock in any of the different 3D printing companies. He cautioned everyone to not confuse the share prices with the growth and the expansion within this industry.
After this introduction, Terry stated that there were really two things in the industry that really excited him. 3D Printing for Manufacturing and for Production Parts.
## 3D Printing in Manufacturing.
The first area to watch is the use of this technology for manufacturing applications. The team looking at the sales data drew a line in the sand for the low cost hobbyist printers at \$5,000. There were 140,000 of them sold last year compared to under 13,000 above \$5k. However, they don’t cost much so the money is still in the industrial machines. Here are the revenues for 2014:
Industrial: 1.12 Billion, or 86.6%.
Hobbyist: 173.3 Million, or 13.4%
There are FDM clones everywhere. 300 or more brands. There is a lot of open source software out there to develop your own FDM printer.
One thing to watch in the industry is expiring patents. This opens up competition and lowers prices and sometimes brings better machines to market. Right now, the SLS patent expired in June of last year so we are seeing new Selective Laser Sintering devices coming to market.
An exciting example of using 3D printing in manufacturing is the landing gear created by Stratasys. It was built and assembled with a Stratasys FDM printer and used for a fit check. Very Cool!
In medical, some great examples of tooling are jigs, fixtures, drill press, and custom cutting guide for knee replacement. You can take scanned data and create a custom cutting guide for replacing your knee. Tens of thousands of those have been done.
Lots of work is being done on test fixtures as well.
In tooling, with additive manufacturing you can do things that are highly complex. Instead of just straight gun drilled cooling channels you can make the cooling channels conform to the purpose of the part. You can reduce 30-300% cycle time by improving the cooling channels for injection molding dies. It turns out that Lego is printing their molds! They are using conformal cooling to increase their cycle times.
On the aerospace side of things, end use parts are literally taking off. Airbus is flying today 45,000 to 60,000 Ultem plastic parts. Both passenger and non-passenger planes have Ultem parts on them.
## 3D Printing for Final Production Parts
The second area to watch is the next frontier, and that is what excites him. You can do structural ribs in 3D printed parts. You need to make sure there are places in your parts to remove the support material used if you are going to use structural ribs. Design is absolutely critical. When he was at Solidworks world in Orlando a few years ago, there was a 3D printed bird that was flapping its wings.
This is a part of that bird that was being flown.
Two weeks ago Terry did a four day course at NASA on Design for Additive Manufacturing. The importance of the subject now is that companies and organizations are paying a lot of money to host people to teach them how to design for additive manufacturing. It was a great learning experience and NASA has already signed up for a second course that is focused on metals. NASA 3D printed a turbopump with 45%fewer parts that runs at 90,000 rpm, and creates 2,000 hp. This turbopump manufactured with conventional methods costs \$220,000 for one, they can 3D print 2 of them in Inconel for \$20,000.
A big part of Design for Additive Manufacturing is using the correct thinking but also using the right tools. There is a lack of both. We are taught to design for the conventional method of manufacturing. Now we have to undo some of that and think, hey there can be a better way to design this part.
One of those ways is Topology Optimization (let mathematics decide where to place the support structure so there is a increased strength to weight ratio). Another is the use of lattice structure (mesh and cellular). Ever since the beginning of time, man would make parts out of a solid material. Well now you can have a thin skin and a lattice structure on the interior to produce something superior in some cases.
We need these kind of tools integrated into the different CAD software’s so that we can design better parts. This bracket is flying on a Airbus. This cabinet bracket is made out of titanium and is flying on the A35 Airbus. It was designed for 2.3 tons and actually holds up to 12.5-14 tons depending on the test. Peter Zander at Airbus believes that in 2 years they will be printing 30 tons of metal per month!
GE Aviation is building fuel nozzles for the new leap engine. The new design is 25% lighter and five times more durable than the previous design that took 20 different parts to assemble to make one fuel nozzle. The will be printing 40,000 fuel nozzles per year.
Consumer Products:
It is going to be very big. Terry thinks this is going to be a sweet spot in the industry. Once example is this guitar called the Hive Bass. It is built out of Nylon and would cost you \$3,500. You can have a custom guitar made for that price.
There is a Belgium company that creates custom frames for eyewear.
There is also a lot of Jewelry available for consumers along with many other products.
For metal part production there are many steps needed to finish the part. About 9 steps that Terry counted so it can be a long process.
Myth: Additive Manufacturing is fast! Well that depends on Polymers versus Metals and the size and complexity of the parts. Airbus had one build that took 14 days to print with their metal printer! GE mentioned that they have to print the same part twice before they get it right because they will have to reorient the part or change the build parameters to get the best quality build possible.
According to some estimates the global manufacturing economy is in the range of \$13 trillion. If this technology were to penetrate 2% of it then that is over a quarter of a trillion dollars. 5% is approaching two thirds of a trillion!
Terry finished by asking: How many of you think this will be North of the 5% estimate?
We want to thank Terry for giving such an informative talk, and New Belgium Brewing for hosting. The networking afterwords was fantastic.
If you would like to stay up to date on 3D Printing, we recommend the Wohlers Report. It is our primary reference document here at PADT.
## NICE Desktop Cloud Visualization
In a previous post I argued that engineers do magic (read it here). And to help them do their magic better PADT Inc. introduced CoresOnDemand.com.
Among the magical skills engineers use in their daily awesomeness is their ability to bend the time fabric of the universe and perform tasks in almost impossible deadlines. It’s as if engineers work long hours and even work from home, while commuting and even at the coffee shop. Wait, is that what they actually do?
Among a myriad of tools that facilitate remote access and desktop redirection available, one stands out with distinction. NICE-Software developed a tool called Desktop Cloud Visualization (DCV for short). DCV has numerous advantages that we will get into shortly. The videos below give a general idea of what can be achieved with NICE-DCV.
Here is a video from the people at NICE:
And here is one of two PADT Employees using an iPhone to check their CFD results:
# Advantages of Nice-DCV
## Physical location of cluster/workstation or the engineers becomes irrelevant
Because engineers have fast, efficient and secure access to their workstations and clusters, they no longer need to be in the same office or on the same network segment to utilize the available compute resources. They can utilize NICE-DCV to create a fast, efficient and encrypted connection to their resources to submit, monitor and process results. The DCV clients are supported on Windows, Linux & IOS and even have a stand-alone Windows client that can be run on shared or public computers. In a recent live test, one of our engineers was travelling on a shuttle bus to a tiny ski town in Colorado, he was able to connect over the courtesy Wifi, check the status of his jobs and visualize some of the results.
## The need for a powerful laptop or remote workstation to enable offsite work is no longer the only solution
There is no need for offsite engineers lug around a giant laptop in order to efficiently launch and modify their designs or perform simulation runs. Users launch the DCV client, connect to their workstation or cluster and are immediately given access to their desktop. No need to copy files, borrow licenses or transfer data. Engineers don’t need to create copies of files and carry them around on the laptops or on external storage which is an unnecessary security risk.
## “If it ain’t broken don’t fix it!”
Every engineer uses ANSYS in his own special way. Some prefer the good old command line for everything even when a flashy GUI option is available. Others are comfortable using the Windows like GUI interface and would
## Opens the door for GUI-only users to utilize large cluster resources without a steep learning curve or specialized tools.
Nice-DCV makes the use of ANSYS on large HPC clusters within reach for everyone. Engineers can log into pre-configured environments with all of the variables needed for parallel ANSYS runs already defined. Users can use can have their favorite ANSYS software added to the desktop as shortcuts or system admins can write small scripts or programs that serve as an answer file for custom job scripts.
## From 0-60 in about…10 Minutes
For an engineer with the smallest amount of system administration skills it takes about 10 minutes to install the Nice-DCV server and launch the first connection. It’s surprisingly simple and straightforward on both the server and the client side. The benefits of Nice-DCV can be immediately realized in both simplified cluster administration and peace of mind for both the engineers and the system admins.
## PADT’s CoresOnDemand and Nice-DCV
The CoresOnDemand service that PADT introduced last year utilizes the Nice-DCV tool to simplify and enhance the user experience. If you are interested in a live demo on Nice-DCV or the CoresOnDemand environment contact us either by phone: 480-813-4884 or by email cod@padtinc.com. For more information please visit: CoresOnCemand.com
(Note: some of the social media posts had a typo in the title, that was my fault (Eric) not Ahmed’s…)
## ReBlog: An Insider’s View on 3D Printing in Aerospace
In all the hype and hoopla around 3D Printing there are teams around the world that are quietly making a difference in manufacturing – making real parts and figuring out the processes, testing, and protocols needed to realize the dream of additive manufacturing. One such team is at Honeywell Aerospace, and we are proud to be one of their vendors.
They just published a great blog on where they are and what they have achieved and we recommend you give it a read. Very informative.
### An Insider’s View on 3D Printing in Aerospace
If you would like to learn how you can use this same technology to move your manufacturing process forward, fill out our simple form here, call us at 480.813.4884, or send an email to info@padtinc.com.
## Special Event: Beyond the Hype – Additive Manufacturing and 3D Printing Worldwide
We are pleased and honored to announce a special event that PADT is sponsoring with the Colorado Society of Manufacturing Engineers. Terry Wholers, a leading voice in the additive manufacturing space, is giving a presentation on the current state of all things AM. The event is being held at the world famous New Belgium Brewery in Fort Collins, Colorado on September 15, 2015.
## Additive Manufacturing and 3D Printing Worldwide
Cut through the hype and hear about real world applications for additive manufacturing and 3D printing:
• Where is the industry growth?
• Which types of polymers and metals are used in 3D printing?
• What are practical uses for the technology in the engineering environment?
• What are current industry implementations for AM/3D printing?
• How is it being implemented in industry today?
• What kind of parts can be manufactured for final products?
• How important is the design process?
• What are the most common myths and misconceptions?
• What does the future hold?
If your company is thinking about how to practically introduce AM into your design/workflow/manufacturing process, this presentation is for you. Ask questions, discuss business opportunities, and speak in depth about the future.
Mr. Wohlers will highlight recent developments and growth trends that point to where the industry is headed and what the future holds. New products and services are being introduced at an astounding rate. Mr. Wohlers will sort through the maze of choices and opportunities associated with the methods used for rapid product development and additive manufacturing (AM).
Here is the agenda:
4:00 – Brew Tour of New Belgium Brewery –Must RSVP– only 16 spots available!
5:00 – Packaging tour of New Belgium Brewery –Must RSVP– only 48 spots available!
5:30 – 6:30 – Meet, Greet and Network
6:30 – Light Dinner Buffet – Must RSVP for buffet and presentation
7:00 – Presentation by Mr. Terry Wohlers
8:00 – Networking (meet Terry)
9:00 – End of event
Reservations are accepted through Friday, September 11th, so please register now!
Cost of the event is \$20 but is FREE if you mention PADT when you register.
To register please email reservations@sme354.org or
call Chuck Otoupalik at 303-678-8414
Terry Wohlers, founder of Wohlers Associates, Inc., a 28-year old independent consulting firm. Wohlers and his team have provided consulting assistance to more than 240 organizations in 24 countries, as well as to 150+ companies in the investment community. He has authored 400 books, articles, and technical papers and has given 125 keynote presentations on five continents. Wohlers has twice served as a featured speaker at events held at the White House. He is a principal author of the Wohlers Report, the undisputed industry-leading study on additive manufacturing and 3D printing for 20 consecutive years.
Here is what industry experts have to say about Mr. Wohlers and his company’s industry report:
“Why waste time and money when you can get a worldwide overview of additive manufacturing from Wohlers Associates-experts that have focused on AM for 26 years. The Wohlers Report is worth every dollar.”
Peter Sander, Vice President, Airbus Germany
“Now in its 18th year of publication (that’s right all you 3-D printing arrivistes, this stuff has been around for a while), the annual report describes a healthy and growing market for 3-D printing equipment, services, materials and processes, albeit one where the value continues to accrue to industrial applications.”
Michael Copeland, Senior Editor, WIRED
“The amount of information in the report is almost overwhelming. I am awed by its depth and breadth. What’s more, the information is not available anywhere else. For example, we hear how China is changing, but few people fully understand the transformation that’s underway. The report gives insight and clarification on China and it covers the rest of the world with the same careful analysis. It also provides insight into new products and applications that you normally would not hear about, such as light-weight structures, nanomanufacturing, growing organs, gaming, and new types of protective gear. One of the secrets of your success is the extensive travel worldwide, coupled with the information you seek from experts globally. Thank you for making something so remarkable available each year.”
Boris Fritz, Northrop Grumman
“If you need to know anything about where this technology is today or where it is going tomorrow, Wohlers Report is your guide.”
Anthony J. Lockwood, former editorial director, Desktop Engineering
## Press Release: Southern California Expansion Grows PADT’s ANSYS Product Development Software Distribution Business
Palm trees and movie stars. Endless beaches and deserts that fade to the horizon. Aerospace companies, world class universities, med device developers, and toy manufacturers. Oil, freeways, and big construction. Southern California. A place larger and more diverse than most countries in the world. PADT has done work in the area since our first weeks in business. As our business continued to grow, our customers started asking when we were opening up a local office, but the time never seemed right. Until now.
PADT is pleased to announce that we will be loading furniture and computers in a truck and head on the I-10 to Torrance, California where we will open up a new office. ANSYS, Inc. has expanded our sales territory to include small and medium sized new accounts in the Southern California area. The focus of this new office will be building that business.
You can read the official details in the press release below, or the PDF here. As usual, we want to share some more informal information with our blog readers.
The office will be started with an engineer and a salesperson who have been with us for a while, and another pair that we are hiring locally. This combination of company experience and local knowledge should get us going quickly. Over time, the plan is to grow the Torrance office, and add at least two more. Long term we would like to have between 3 and 10 employees per office in Southern California.
Our team will conduct training and seminars from this office and use it as a base to spread the word on simulation driven product development across Southern California. The initial focus for sales will be on small and medium sized businesses that are currently not using ANSYS products, that want to work with a technical sales and support team who can provide more than the software tool – customers who want a partner who can also help them apply the tools effectively. The dense hotbeds of engineering along the coast will be an obvious area of concentration. We also aim to represent the value of ANSYS products in less visited areas of the region, including the high deserts, “in-between” towns, and inland locations beyond LA, Orange County, and San Diego.
The good news is that we are not starting from scratch. This first office is right down the street from the California campus of PADT’s largest and oldest customer. We also have over one hundred customers who have used PADT for simulation services, training, rapid prototyping, and product development, and we will be reaching out to them shortly to start building our local network even further. And then, our new employees who we will hire locally will be contacting their network as well.
Before the end of the summer we hope to have a grand opening event, as well as several seminars that will continue through the end of the year. If you live in the area and want to be invited, visit here to register as someone who want to be on the California contact list.
This blog and social media will be used to post our progress. The entire sales and technical team is looking forward to meeting everyone in the area in the coming months.
If you have any questions or suggestions for us, please contact us. Our standard number 480.813.4884 works for all of our offices.
Below is a copy of the press release, or you can view the “official” version here.
Press Release:
Southern California Expansion Grows PADT’s ANSYS Product Development Software Distribution Business
PADT opens Torrance office to provide consultant-focused ANSYS Product Sales and Support for small and medium sized engineering businesses in the region
Tempe, Ariz., August 24, 2015 —Phoenix Analysis & Design Technologies, Inc. (PADT) the Southwest’s largest provider of Numerical Simulation, Product Development, and 3D Printing services and products, today announced the addition of Southern California to its ANSYS, Inc. Product Sales and Support territory. PADT is a long time ANSYS Channel Partner who has built a reputation for outstanding technical abilities and customer support in Arizona, New Mexico, Colorado, Utah, and Nevada. The company is now taking the same customer focused approach to selling and supporting the world’s leading product development simulation tools from ANSYS to new customers in Southern California.
“We are honored by ANSYS’ trust in PADT and are eager to start working more closely with their team in Southern California,” said Bob Calvin, PADT’s manager of Simulation Sales. “We have been doing business in this area since PADT was founded 21 years ago. Expanding our offering to include ANSYS products and support is something that makes sense for users, ANSYS and PADT.”
Located in Torrance California, PADT’s new office will be staffed by two sales people and two application engineers. Aggressive growth will follow.
“We selected Torrance for our new Southern California office because it’s centrally located, easily accessible and right down the street from the California campus of our largest customer,” said Ward Rand, co-owner, PADT. “Having staff with real world industry experience located nearby will strengthen our ability to drive our customer’s product development process, resulting in higher quality products, improved performance and lower costs.”
PADT will open additional offices across the Southern California region in the coming two years with the long term goal of three total offices with three to ten employees each. The location of these offices, just like the initial Torrance facility, will be chosen to provide service where the demand is greatest.
The ANSYS Channel Partner program is unique in the industry because it allows customers the option to purchase software and support from ANSYS directly, or from highly technical local consulting companies like PADT. Since Southern California has not had an ANSYS Channel Partner for thirteen years, PADT’s engineering experience and ANSYS product expertise will be a tremendous help to small and medium sized companies seeking to discover the power of ANSYS products, and efficiently implement Simulation Driven Product Development (SDPD).
Events, both on-line and face-to-face, will be announced in the coming months to celebrate the arrival of PADT in the area. Those interested in following PADT’s progress, can subscribe to any of the company’s social media outlets, PADT California emails, or visit the new PADT California web page (www.padtinc.com/socal). Anyone needing immediate information can contact PADT at info@padtinc.com or call 480.813.4884.
About Phoenix Analysis and Design Technologies
Phoenix Analysis and Design Technologies, Inc. (PADT) is an engineering product and services company that focuses on helping customers who develop physical products by providing Numerical Simulation, Product Development, and Rapid Prototyping solutions. PADT’s worldwide reputation for technical excellence and experienced staff is based on its proven record of building long term win-win partnerships with vendors and customers. Since its establishment in 1994, companies have relied on PADT because “We Make Innovation Work.” With over 75 employees, PADT services customers from its headquarters at the Arizona State University Research Park in Tempe, Arizona, and from offices in Littleton, Colorado, Albuquerque, New Mexico, and Murray, Utah, as well as through staff members located around the country. More information on PADT can be found at http://www.PADTINC.com.
# # #
Company contact:
Eric Miller
480.813.4884
Media contact:
Linda Capcara
TechTHiNQ
480-229-7090
linda.capcara@techthinq.com
# Engineers Do Magic
In the world of simulation there are two facts of life. First, the deadline of “yesterday would be good” is not too uncommon. Funding deadlines, product roll-out dates, as well as unexpected project requirements are all reliable sources for last minute changes. Engineers are required to do quality work and deliver reliable results in limited time and resources. In essence perform sorcery.
Second, the size and complexity of models can vary wildly. Anything from fasteners and gaskets to complete systems or structures can be in the pipeline. Engineers can be looking at any combination of hundreds of variables that impact the resources required for a successful simulation.
Required CPU cores, RAM per core, interconnect speeds, available disk space, operating system and ANSYS version all vary depending on the model files, simulation type, size, run-time and target date for the results.
Engineers usually do magic. But sometimes limited time or resources that are out of reach can delay on-time delivery of project tasks.
# At PADT, We Can Help
PADT Inc. has been nostrils deep in engineering services and simulation products for over 20 years. We know engineering, we know how to simulate engineering and we know ANSYS very well. To address the challenges our customers are facing, in 2015 PADT introduced CoresOnDemand to the engineering community.
CoresOnDemand offers the combination of our proven CUBE cluster, ANSYS simulation tools and the PADT experience and support as an on demand simulation resource. By focusing on the specific needs of ANSYS users, CoresOnDemand was built to deliver performance and flexibility for the full range of applications. Specifics about the clusters and their configurations can be found at CoresOnDemand.com.
CoresOnDemand is a high performance computing environment purpose built to help customers address numerical simulation needs that require compute power that isn’t available or that is needed on a temporary basis.
# Call Us We’re Nice
CoresOnDemand is a new service in the world of on-demand computing. Prospective customers just need to give us a call or send us an inquiry here to get all of their questions answered. The engineers behind CoresOnDemand have a deep understanding of the ANSYS tools and distributed computing and are able to asses and properly size a compute environment that matches the needed resources.
Call us we’re nice!
# Two Halves of the Nutshell
The process for executing a lease on a CoresOnDemand cluster is quite straight forward. There are two parts to a lease:
## PART 1: How many cores & how long is the lease for?
By working with the PADT engineers – and possibly benchmarking their models – customers can set a realistic estimate on how many cores are required and how long their models need to run on the CoresOnDemand clusters. Normally, leases are in one-week blocks with incentives for longer or regular lease requirements.
Clusters are leased in one-week blocks, but we’re flexible.
## Part 2: How will ANSYS be licensed?
An ANSYS license is required in order to run on the CoresOnDemand environment. A license lease can be generated by contacting any ANSYS channel partner. PADT can generate license leases in Arizona, Colorado, New Mexico, Utah & Nevada. Licenses can also be borrowed from the customer’s existing license pool.
An ANSYS license may be leased from an ANSYS channel partner or borrowed from customer’s existing license pool.
# Using the Cluster
Once the CoresOnDemand team has completed the cluster setup and user creation (takes a couple of hours for most cases), customers can login and begin using the cluster. The CoresOnDemand clusters allow customers to use the connection method they are comfortable with. All connections to CoresOnDemand are encrypted and are protected by a firewall and an isolated network environment.
## Step 1: Transfer files to the cluster:
Files can be transferred to the cluster using Secure Copy Protocol which creates an encrypted tunnel for copying files. A graphical tool is also available for Windows users (& it’s freeJ). Also, larger files can be loaded to the cluster manually by sending a DVD, Blu-ray disk or external storage device to PADT. The CoresOnDemand team will mount the volume and can assist in the copying of data.
## Step 2: Connect to the cluster and start jobs
Customers can connect to the cluster through an SSH connection. This is the most basic interface where users can launch interactive or batch processing jobs on the cluster. SSH is secure, fast and very stable. The downside of SSH is that is has limited graphical capabilities.
Another option is to use the Nice Software Desktop Cloud Visualization (DCV) interface. DCV provides enhanced interactive 2D/3D access over a standard network. It enables users to access the cluster from anywhere on virtually any device with a screen and an internet connection. The main advantage of DCV is the ability to start interactive ANSYS jobs and monitor them without the need for a continuous connection. For example, a user can connect from his laptop to launch the job and later use his iPad to monitor the progress.
Figure 1. 12 Million cell model simulated on CoresOnDemand
The CoresOnDemand environment also has the Torque resource manager implemented where customers can submit multiple jobs to a job queue and run them in sequence without any manual intervention.
Customers can use SCP or ship external storage to get data on the cluster. SSH or DCV can be used to access the cluster. Batch, interactive or Torque scheduler can be used to submit and monitor jobs.
# All Done?
Once the simulation runs are completed customers usually choose one of two methods to transfer data back. First is to download the results over the internet using SCP (mentioned earlier) or have external media shipped back (External media can be encrypted if needed).
After the customer receives the data and confirms that all useful data was recovered from the cluster, CoresOnDemand engineers re-image the cluster to remove all user data, user accounts and logs. This marks the end of the lease engagement and customers can rest assured that CoresOnDemand is available to help…and it’s pretty fast too.
At the end of the lease customers can download their data or have it shipped on external media. The cluster is later re-imaged and all user data, accounts & logs are also deleted in preparation for the next customer.
## ANSYS Launches Free Student Version
This week ANSYS, Inc. made a fantastic announcement that has been in the works for a while, and that we think will greatly benefit the simulation community: A free ANSYS Student product. This is an introductory product that is focused on students who are learning the fundamentals of simulation who also want to learn the full power and capability of the ANSYS product suite. It includes ANSYS® Multiphysics™ , ANSYS® CFD™ , ANSYS® Autodyn®, ANSYS® Workbench™, ANSYS® DesignModeler™and ANSYS®DesignXplorer™
Yes you read that right, all of the flagship products for free. No features or capabilities are turned off. It is the exact same software as the commercial product, but the size of problems that you can solve is limited. It runs on MS Windows. Perfect for students.
PADT is excited about this because it gives students access to the ability to learn FEA and CFD simulation with the world’s most popular and capable simulation tool, without running in to brick walls. Want to do a flat plate with a hole in it? No Problem. Want to model fluid-solid-interaction on a flexible membrane valve? No Problem. Want to model explosive forming? No Problem. Want to model combustion with complex turbulence? No problem.
All in the same interface as students will use when they enter the work force or do research at University.
This is great news and we can’t wait to see what schools and students do with this access.
## How to Get It – The New Academic Web Pages
The previous Student Portal is being replaced with an Academic Web area on the ansys.com site: ansys.com/academic.
Go to the ANSYS Student site to learn more about ANSYS Student and how to download your copy. These same pages will have resources to help you learn and understand the product.
## The “Pictures”
Let me state categorically that PADT was not consulted on the image that ANSYS, Inc. used for the “student” user that was so happy to find out that there is now a free version of the ANSYS software suite. Here is their picture:
We would have preferred something like this:
Just kidding. We were happy to see this product come out and thought the picture was hilarious. In all seriousness, we will also plug the recent #ilooklikeanengineer twitter hash tag , highlighting the diversity of female engineers. that was awesome and we would love to see more chances for engineers to show their true selves.
## 3D Printing the 4th Dimension – GISHWHES 2015 Scavenger Hunt
GISHWHES is a huge international scavenger hunt. Every year teams around the globe comb through the list of 215 tasks and pick as many as possible that their team can do. Last year they introduced 3D Printing as a task, and we helped a team 3D Print a quill pen. That was a lot of fun, so when this year’s list included an item on 3D printing, we jumped at the chance to be involved.
The item was:
110: VIDEO. Use a cutting edge 3D printer to 3D print your representation of the 4th dimension.62 POINTS
Being engineers we said “4th Dimension? Time.” Then it became a choice between the way mass distorts the space-time continuum or some sort of clock’ish thing. The distortion thing seemed difficult so we focused on a clock. Being that we were constrained on budget and time we decided to do a sundial.
The result can be seen here in this YouTube video.
It was a fun project and the team spent a bit of time in the 112F sunshine trying it out. We can’t wait to see what we will get to do for the 2016 scavenger hunt.
# Making the Model
A couple of people have asked if we downloaded the solid model for the sundial or if we made it. We actually made it. After a little bit of research we found that making a simple horizontal sundial like this one is very easy. Here are the steps we took:
## Get Geometry Values
So it turns out that the angle of each hour line is determined by the latitude of where the dial will go. The angle of the pointy thing, called a gnomon, is also the latitude. So for Tempe, AZ that is 33.4294°.That gets applied to the equation:
angle(h) = arctan(sin(L*tan(15° · h))
h = integer of the hour, 6 am to 6 pm
L = latitude
I plopped that into Excel:
and got the following:
Latitude 33.4294 Hour Angle 6 90.00 7 64.06 8 43.66 9 28.85 10 17.64 11 8.40 12 0.00
## Build the Solid Model
The next step is to build the model. I used SolidEdge because I know it real well and was able to knock it out quickly. But all CAD tools would be the same:
1. Pick a center point.
2. Add lines as rays from that using the angles in the table above for each hour.
3. Design the shape of your sundial to look cool. I did a simple circle .
4. Mark the hours using the sketch. I raised up thin rectangles.
5. Model the gnomon using the latitude as the angle. Make this as fancy or simple as you want.
7. Label the hours if you want.
8. Save to STL
Here is what my sketch looked like:
And the final solid model looked like this:
We sent this to the printer as shown in the video, and got a sundial.
## Major Enhancements in FLOWNEX 2015: Combustors, Importers, and Pipes
Simulation has revolutionized flow and heat transfer dependent systems over the past decades by minimizing costly physical testing and accelerating time to operation around the world. But for many companies, such simulation has largely focused on components and proved to be very time consuming. The technology advancements delivered by Flownex SE now offer a fast, reliable, and accurate total system and subsystem approach to simulation.
With the release of FLOWNEX 2015, users now have access to advanced combustor system level modeling and they can interact with more system and component simulation tools. This is on top of the already considerable capabilities found in the tool
## Gas Turbine Combustor Heat Transfer Library
During the Preliminary design phase or when considering modifications to existing combustor designs it’s essential to make realistic predictions of mass flow splits through the various air admission holes, total pressure losses liner temperatures along the length of the combustor etc.
Although very powerful, 3D CFD solutions of combustors are specialized, time consuming processes and therefore are seldom exclusively used during initial sizing of a combustor.
It has been demonstrated that 1D/2D network tools, like Flownex, are capable of predicting with reasonable accuracy the same trends as more detailed numerical models.
The advantage, however, is Flownex’s rapid execution, which allows design modifications and parametric studies to be conducted more simply than ever before. The ease of use and incredible speed of Flownex allows 1000s of preliminary designs to be evaluated under all modes of operation for steady state and dynamic cases. Furthermore, the data obtained from the one-dimensional analysis can be used as boundary conditions for a more detailed three-dimensional model, ultimately supplementing a typical combustor design work flow.
While the simulation of combustor systems was previously possible in the Flownex environment, much of the work of implementing industry standard heat transfer correlations was left to the user through scripting .Now in Flownex SE 2015 it’s all been built in to the tool, while maintaining the flexibility required to model any combustor configuration.
New components include
• Film convection component
• Fluid radiation component
• Jet impingement heat transfer component
To sum up Flownex allows more accurate initial designs, less time is spent on advanced 3D combustor simulations and rig tests, thus reducing development time and cost.
Here is a Video that shows off these features:
## Added importers and integration features
### AFT Fathom/Impulse/Arrow importer
An importer was added to import the file formats of AFT products. The importer imports all the diameters, loss factors heights, etc. so 90% of the effort is done, and in some cases the networks solve without any modifications.
ROHR2 Integration (pipe stress analysis software)
Flownex has the ability to calculate forces during dynamic simulations. This is very useful in pipe stress analysis for surge or water hammer cases. The ability to import complete geometries from ROHR2 and export results in the format that ROHR2 expects natively has been added. This means a user can perform these combined analysis now with ROHR2 with the minimum of effort.
Fluid Importers
An Importer was added to import liquid and gas properties from CoolProp an open source fluid property library. The existing Aspen/Hysys fluid importer was changed to be a generic Cape-Open compliant importer. This means that fluid properties can now be imported from any Cape-Open compliant server software.
## Donny Don’t – Thin Sweep Meshing
It’s not a series of articles until there’s at least 3, so here’s the second article in my series of ‘what not to do’ in ANSYS…
Just in case you’re not familiar with thin sweep meshing, here’s an older article that goes over the basics. Long story short, the thing sweep mesher allows you to use multiple source faces to generate a hex mesh. It does this by essentially ‘destroying’ the backside topology. Here’s a dummy board with imprints on the top and bottom surface:
If I use the automatic thin sweep mesher, I let the mesher pick which topology to use as the source mesh, and which topology to ‘destroy’. A picture might make this easier to understand…
As you can see, the bottom (right picture) topology now lines up with the mesh, but when I look at the top (left picture) the topology does not line up with the mesh. If I want to apply boundary conditions to the top of the board (left picture), I will get some very odd behavior:
I’ve fixed three sides of the board (why 3? because I meant to do 4 but missed one and was too lazy to go back and re-run the analysis to explain for some of future deflection plots…sorry, that’s what you get in a free publication) and then applied a pressure to all of those faces. When I look at the results:
Only one spot on the surface has been loaded. If you go back to the mesh-with-lines picture, you’ll see that there is only a single element face fully contained in the outline of the red lines. That is the face that gets loaded. Looking at the input deck, we can see that the only surface effect element (how pressure loads are applied to the underlying solid) is on the one fully-contained element face:
If I go back and change my thin sweep to use the top surface topology, things make sense:
The top left image shows the thin sweep source definition. Top right shows the new mesh where the top topology is kept. Bottom left shows the same boundary conditions. Bottom right shows the deformation contour.
The same problem occurs if you have contact between the top and bottom of a thin-meshed part. I’ll switch the model above to a modal analysis and include parts on the top and bottom, with contact regions already imprinted.
I’ll leave the thin sweeping meshing control in place and fix three sides of the board (see previous laziness disclosure). I hit solve and nothing happens:
Ah, the dreaded empty contact message. I’ll set the variable to run just to see what’s going on. Pro Tip: If you don’t want to use that variable then you would have to write out the input deck, it will stop writing once it gets to the empty contact set. Then go back and correlate the contact pair ID with the naming convection in the Connections branch.
The model solves and I get a bunch of 0-Hz (or near-0) modes, indicating rigid body motion:
Looking at some of those modes, I can see that the components on one side of my board are not connected:
The missing contacts are on the bottom of the board, where there are three surface mounted components (makes sense…I get 18 rigid body modes, or 6 modes per body). The first ‘correct’ mode is in the bottom right image above, where it’s a flapping motion of a top-mounted component.
So…why don’t we get any contact defined on the bottom surface? It’s because of the thin meshing. The faces that were used to define the contact pair were ‘destroyed’ by the meshing:
Great…so what’s the take-away from this? Thin sweep meshing is great, but if you need to apply loads, constraints, define contact…basically interact with ANYTHING on both sides of the part, you may want to use a different meshing technique. You’ve got several different options…
1. Use the tet mesher. Hey, 2001 called and wants its model size limits back. The HPC capabilities of ANSYS make it pretty painless to create larger models and use additional cores and GPUs (if you have a solve-capable GPU). I used to be worried if my model size was above 200k nodes when I first started using ANSYS…now I don’t flinch until it’s over 1.5M
Look ma, no 0-Hz modes!
2. Use the multi-zone mesher. With each release the mutli-zone mesher has gotten better, but for most practical applications you need to manually specify the source faces and possibly define a smaller mesh size in order to handle all the surface blocking features.
Look pa, no 0-Hz modes!Full disclosure…the multi-zone mesher did an adequate job but didn’t exactly capture all of the details of my contact patches. It did well enough with a body sizing and manual source definition in order to ‘mostly’ bond each component to the board.
3. Use the hex-dominant mesher. Wow, that was hard for me to say. I’m a bit of a meshing snob, and the hex dominant mesher was immature when it was released way back when. There were a few instances when it was good, but for the most part, it typically created a good surface mesh and a nightmare volume mesh. People have been telling me to give it another shot, and for the most part…they’re right. It’s much, much better. However, for this model, it has a hard time because of the aspect ratio. I get the following message when I apply a hex dominant control:
4. The warning is right…the mesh looks decent on the surface but upon further investigation I get some skewed tets/pyramids. If I reduce the element size I can significantly reduce the amount of poorly formed elements:
5. That’s going on the refrigerator door tonight!
And…no 0-Hz modes!
• Lastly…go back to DesignModeler or SpaceClaim and slice/dice the model and use a multi-body part.
3 operations, ~2 minutes of work (I was eating at the same time)
Modify the connection group to search/sort across parts
That’s a purdy mesh! (Note: most of the lower-quality elements, .5 and under, are because there are 2-elements through thickness, reducing the element size or using a single element thru-thickness would fix that right up)
And…no 0-Hz modes.
Phew…this was a long one. Sorry about that. Get me talking about meshing and look what happens. Again, the take-away from all of this should be that the thin sweeper is a great tool. Just be aware of its limitations and you’ll be able to avoid some of these ‘odd’ behaviors (it’s not all that odd when you understand what happens behind the scenes). | 12,751 | 59,259 | {"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} | 3.625 | 4 | CC-MAIN-2019-39 | longest | en | 0.944132 |
https://socratic.org/questions/how-do-you-find-int-x-3-sqrt-2x-2-4-dx-using-trigonometric-substitution | 1,726,440,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00185.warc.gz | 497,930,674 | 6,054 | # How do you findint x^3/sqrt(2x^2 + 4) dx using trigonometric substitution?
Jun 4, 2018
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{\left({x}^{2} - 4\right) \sqrt{2 {x}^{2} + 4}}{6} + C$
#### Explanation:
You do not really need to use trigonometric substitutions.
As:
$\frac{d}{\mathrm{dx}} \left(\sqrt{2 {x}^{2} + 4}\right) = \frac{2 x}{\sqrt{2 {x}^{2} + 4}}$
we can integrate by parts:
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \int {x}^{2} / 2 \frac{d}{\mathrm{dx}} \left(\sqrt{2 {x}^{2} + 4}\right) \mathrm{dx}$
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} \sqrt{2 {x}^{2} + 4}}{2} - \int \frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right) \sqrt{2 {x}^{2} + 4} \mathrm{dx}$
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} \sqrt{2 {x}^{2} + 4}}{2} - \int x \sqrt{2 {x}^{2} + 4} \mathrm{dx}$
Solve the resulting integral by substitution:
$t = 2 {x}^{2} + 4$
$\mathrm{dt} = 4 x \mathrm{dx}$
$\int x \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{1}{4} \int \sqrt{t} \mathrm{dt}$
$\int x \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{1}{4} {t}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C = \frac{1}{6} t \sqrt{t} + C$
and undoing the substitution:
$\int x \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} + 2}{3} \sqrt{2 {x}^{2} + 4} + C$
Putting the results together:
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} \sqrt{2 {x}^{2} + 4}}{2} - \frac{{x}^{2} + 2}{3} \sqrt{2 {x}^{2} + 4} + C$
and simplifying:
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \left(\frac{{x}^{2}}{2} - \frac{{x}^{2} + 2}{3}\right) \sqrt{2 {x}^{2} + 4} + C$
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \left(3 {x}^{2} - 2 {x}^{2} - 4\right) \frac{\sqrt{2 {x}^{2} + 4}}{6} + C$
$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{\left({x}^{2} - 4\right) \sqrt{2 {x}^{2} + 4}}{6} + C$ | 952 | 1,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.298263 |
https://www.physicsforums.com/threads/equilibrium-of-spring.343192/ | 1,532,132,544,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592001.81/warc/CC-MAIN-20180720232914-20180721012914-00291.warc.gz | 952,082,058 | 13,882 | # Homework Help: Equilibrium of spring
1. Oct 5, 2009
### songoku
1. The problem statement, all variables and given/known data
A light spring with a force constant k = 160 N/m rest vertically on the bottom of a large beaker of water (a). A 5 kg block of wood (density = 650 kg/m^3) is connected to the spring, and the block-spring system is allowed to come to static equilibrium (b). What is the elongation $$\Delta L$$ of the spring?
2. Relevant equations
F = kx
3. The attempt at a solution
I don't have idea to start. The spring rests on the bottom so when the mass is connected, I think the spring should be compressed rather than become longer. Maybe we should consider external force (I doubt this myself because there is no such thing in the question).
And why does the question mention density? The only thing I can come up with is to find volume of the block and I absolutely clueless what to do with the volume....
Thanks
2. Oct 5, 2009
### kuruman
What keeps the wood from floating to the surface of the water? Draw a free body diagram of the block.
3. Oct 6, 2009
### songoku
Hi kuruman
Maybe I get it. From the free body diagram, there are 3 forces that acts on the block which are weight, upthrust, and the restoring force from the spring.
$$\rho *g*V + kx = mg$$
From the above equation, I got x = - 16.5 cm. I want to ask why I got negative value?
Maybe I should use : restoring force = - kx, instead of kx ?
Thanks
4. Oct 6, 2009
### kuruman
If up is positive and down is negative and the sum of all the forces must be zero,
+ρgV - kx - mg = 0
The buoyant force is up, gravity is down and the spring force is down because the spring stretches up to keep the block from floating to the surface.
5. Oct 7, 2009
### songoku
Hi kuruman
Oh I see. The direction of the restoring force is downward
Thanks a lot for your help | 492 | 1,860 | {"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} | 3.515625 | 4 | CC-MAIN-2018-30 | latest | en | 0.921014 |
https://9jainformed.com/2021/02/27/when-will-rapture-take-place/ | 1,726,585,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00891.warc.gz | 58,203,403 | 99,097 | You dont have javascript enabled! Please enable it! How Many Years Left Before Rapture Takes Place - Chris Oyakhilome
# In Case You Missed It: This Is How Many Years Left Before Rapture Takes Place – Chris Oyakhilome
## How Many Years Left Before Rapture Takes Place – Chris Oyakhilome
The popular Nigerian preacher and teacher of the gospel, Chris Oyakhilome has given his own prediction on the year rapture will happen. In Case You Missed It: This Is How Many Years Left Before Rapture Takes Place – Chris Oyakhilome.
#### -When Exactly will rapture Take place?
This post is to call your mind back or let you know, in case you missed the news, about the rapture prediction and years breakdown made by the Founder of Loveworld Ministry a.k.a Christ Embassy, Pastor Chris Oyakhilome. He made this prediction during a Sunday Service teaching before his congregation.
(Image credit:Â BBC)
He predicted the years left before the son of perdition which also means the man of sin will be revealed. But before that, the rapture must have taken place. Pastor Christ gave his prediction, using a chart. He showed his members the chart. He said the time is not far anymore.
From his chart, he calculated the rapture time, drawing an inference, using the range of time when Jesus was born which was within 4BC to 2BC, and the very time he walked out of the temple which was around 30AD.
In his word; “if we take an average of 30AD and add it to the year 2000, it gives us 2030”. The year 2000 was the year he died on the cross.
“From 2020 to 2030 is 10 years. This means that it will not exceed 10 years before the man of sin will be revealed”. Said Pastor Chris.
Pastor Chris made his rapture prediction given 3 options.
I have already said that the man of sin or perdition will not be revealed unless rapture takes place. Pastor Christ has said that it is remaining 10 years before the man of sin will be revealed. He also said it is along with this range of time that Christ will return.
“At The Last Trump”. What Bible Says about “Trump” and the End Time
## How Many Years Left Before Rapture Takes Place – Chris Oyakhilome
In his words, he said;Â “we have 7 years either after or within or across. If it happens after 7 years, it means there is only 17 years left before Christ return. 7 years for tribulation, 10 years before Anti-Christ will come”.
Pastor Chris also said the 10 years is not static. The man of sin may show up in 3 years or in six years. “If this is the case, rapture may take place in the next 3 years or 6 years”.
In his teaching, Pastor Chris further claimed that the Scientists are aware of what lies ahead. That is the major reason why they are insisting that by the year 2030, everybody in the world must be vaccinated and chipped.
(Image credit:Â BBC)
From what Pastor Chris predicted, the children of God should get themselves prepared and repent from all their sins because it will not be more than 10 years again before the rapture will take place. It could happen in the next 3 years. It could take place in the next 6 years. But it will not exceed the year 2030.
Meanwhile, the Bible made it clear that Christ’s second coming and the time for rapture will no man know. But it listed out the signs that will signal His second coming. Nevertheless, almost all the signs have been fulfilled. Pastor Chris could be right. He could be speaking as revealed.
News Source:Â BBCÂ Pigin News
## One thought on “In Case You Missed It: This Is How Many Years Left Before Rapture Takes Place – Chris Oyakhilome”
1. Prediction or no prediction, Christ must surely return. So, be prepared. | 854 | 3,623 | {"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} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.955964 |
https://www.onlinemathlearning.com/sin-cos-equation.html | 1,540,015,491,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512592.60/warc/CC-MAIN-20181020055317-20181020080817-00166.warc.gz | 1,016,126,710 | 11,292 | Find the Equation of a Sine or Cosine Graph
Related Topics:
More Lessons on Finding An Equation for Sine or Cosine Graphs
More Algebra 2 Lessons
More Trigonometric Lessons
Videos, worksheets, games and activities to help Algebra 2 students learn how to find the equation of a given sine or cosine graph.
We have included a tool that with plot the sine graph f(x) = A sin(B(x-h))+ k, given the values A, B, h and k. Use it to check your answers.
The following diagram shows how to find the equation of a sine graph. Scroll down the page for examples and solutions.
Find an Equation for the Sine or Cosine Wave
When finding the equation for a trig function, try to identify if it is a sine or cosine graph.
To find the equation of sine waves given the graph
1. Find the amplitude which is half the distance between the maximum and minimum.
2. Find the period of the function which is the horizontal distance for the function to repeat. If the period is more than 2π then B is a fraction; use the formula period = 2π/B to find the exact value.
3. Find any phase shift, h.
How to determine the equation of a sine and cosine graph?
The general equation of a sine graph is y = A sin(B(x - D)) + C
The general equation of a cosine graph is y = A cos(B(x - D)) + C
Example:
Given a transformed graph of sine or cosine, determine a possible equation.
Writing equation of sin and cos graph
Example:
Find the equation of the given graph in terms of sine and cosine. Trigonometry: Finding an Equation from a Graph Determining the equation of a trigonometric function
Determining the amplitude and period of sine and cosine functions. How to come up with the equation of a sin/cos function when given the graph? How to Get Equation of a Sine Curve? How to identify the graph of a stretched cosine curve?
This video demonstrates how to stretch and shrink the sine and cosine curves. Finding the equation of Sine or Cosine given its graph
Sine Graph Calculator
Enter in the values for f(x) = Asin(B(x-h))+k into the sine graph calculator to check your answer.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 550 | 2,454 | {"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} | 4.21875 | 4 | CC-MAIN-2018-43 | longest | en | 0.857347 |
http://virtualdub.org/blog/archives/archive_2009-m02.php?month=6&year=2020 | 1,596,878,249,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737319.74/warc/CC-MAIN-20200808080642-20200808110642-00523.warc.gz | 113,591,560 | 7,770 | v1.10.4 (stable)
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## § ¶Fast noise
A user on my forum asked how to generate video noise, and I was intrigued enough to do a bit of tinkering.
The first issue to deal with is what random number generator to use. The easiest thing to do is just to use the C runtime library rand() function... until you realize how slow it is. The first reason is that it's commonly a linear congruential generator of the form:
x' = (x * A + B) mod 2^N
The multiply already isn't great, especially since it's usually a 32-bit multiply, but what makes this worse is that the seed used by rand() is shared state, which means that on a multithreaded CRT you're also taking a hit for thread-local storage. You can bypass the TLS issue easily enough just by reimplementing rand() locally, but then you run into the problem of how to vectorize this generator. 32-bit multiplication and addition are needed for this generator, and that's a pain in MMX/SSE2/3/4. And a 16-bit generator has way too short of a period.
In these cases, I prefer to use a linear feedback shift register, because they only require very simple logical operations. Here's a 31-bit generator:
x' = (x << 1) + (((x >> 27) ^ (x >> 30)) & 1);
That is, you XOR two bits in the register and shift the result in at one end. With appropriately chosen taps, this gives you a period of 2^N-1 for a register of size N. Just make sure to avoid zero, since the generator will get stuck there. LFSRs are a bit deficient in some properties typically desired for a random number generator, but that's less of an issue for noise generation than for, say, statistical sampling.
There is a problem with LFSRs, though, which is that if you use the straight formulation that is normally given for more than single bit output, you'll end up with a pretty crappy RNG in practice, because you get strings of values that are obviously just doubles or halves of the previous value due to the single bit shift. This looks like a lot of gradients if you're dealing with chunky images and a terrible "magic eye" image if you're filling bitplanes. Fortunately, it's fairly easy to generate more bits at a time, because you just do the generation in parallel:
x' = (x << 16) + (((x >> 12) ^ (x >> 15)) & 0xffff);
The straightforward way to vectorize this with SSE2 would be to run four generators in parallel in a 128-bit vector, and then extract the low 16 bits of each generator. The problem with doing this is that you spend a decent amount of time doing the masking and packing. A better way is to instead split the high and low halves of each generator into different registers, thus making both the extraction and 16 bit shift trivial:
` ;xmm0 = high 16 bits of 8 generators ;xmm1 = low 16 bits of 8 generators paddw xmm0, xmm0 ;xmm0 = hi << 1 movaps xmm2, xmm1 ;xmm2 = lo psrlw xmm1, 12 ;xmm1 = lo >> 12 movaps xmm3, xmm0 ;xmm3 = hi << 1 pxor xmm3, xmm1 ;xmm3 = (hi << 1) ^ (lo >> 12) psllw xmm0, 3 ;xmm0 = hi << 4 psrlw xmm1, 3 ;xmm1 = lo >> 15 pxor xmm1, xmm0 ;xmm1 = (hi << 4) ^ (lo >> 15) movaps xmm0, xmm2 ;xmm0 = lo pxor xmm1, xmm3 ;xmm1 = (hi << 1) ^ (lo >> 12) ^ (hi << 4) ^ (lo >> 15)`
This generates eight 16-bit random values in 10 instructions using four registers, two being temporaries. It's also possible to generate 8-bit values from 16 parallel generators in a similar fashion, although then you'd need four registers just for the shift registers. You might wonder why I used a 31-bit generator instead of a 32-bit generator, since that would eliminate one of the shifts. That's because a 32-bit LFSR annoyingly requires four taps to hit maximum period instead of just two.
The last and trickiest part is how to preroll the eight generators so that they're all reading from different parts of the sequence. If they're too close together, they'll form an obvious pattern in the noise image. I'm guessing this may be better known in hardware engineering circles, but I had to figure this one out. Basically, you can model an N-bit LFSR as a multiplication by a NxN modulo-2 matrix:
x' = x * M
Note that in this case, scalar "multiplication" is a logical AND, and "addition" is an XOR.
Since multiplying by M advances the state by one step, multiplying by M twice advances it by two, and since matrix multiplication is associative, you can instead multiply by M^2:
x' = (x * M) = x * (M * M) = x * M^2
From here, you can then compute powers of two of M, and multiply together combinations of those results in order to quickly compute M to any desired power. This basically gives a closed-form solution to directly compute the state of the LFSR at any position, without having to iteratively step it up to 2^31 times. Not only does this make it easy to preroll the eight generators to adjacent portions of the sequence, but it also means that the noise generator can be made fully deterministic and support arbitrary seeking and decoding order.
You might have noticed that I used post-multiplication as the convention above. The reason is that this convention is often used along with a row-major storage ordering in memory, which is advantageous from a computation perspective as it turns vector-matrix multiplication into a series of parallel multiplies and adds instead of dot products. This is true as well for the binary matrices here, and in fact my vector-matrix multiply routine looks like this:
`uint32 Mult(uint32 v, const BinMat32& m) { uint32 r = 0;`
` for(int i=0; i<32; ++i) { if (v & (1 << i)) r ^= m.m[i]; }`
` return r;}`
Matrix-matrix multiplication is then just this applied to each row of the left-hand matrix. This isn't so fast that you'd do it in the inner loop, of course, but it's a lot faster than nested 31x31 loops and if done once a frame it's barely going to show a blip on a profile. The base matrix and the power of two matrices are properties of the LFSR and can be precomputed.
I implemented this as a VirtualDub filter, and it does generate noise very quickly. It still doesn't look like natural noise, though, because the spectrum isn't right. I tried the trick of adding groups of samples to approximate a normal distribution with a binomial, and it still doesn't look right due to the spatial spectrum. At that point I lost interest, but I think the next thing to try would have been to generate several octaves of noise and blend them together, which in my experience generates much more interesting noise.
## § ¶Video operations without pixel shaders
May be this is not the right place to ask, but can I implement brightness/contrast/hue/etc. correction in Direct3D 9 without using pixel shaders? I mean - what's the way this kind of thing is done? Where should I start from?
Well... yes, you can do some of those, but it's a pain.
If you don't have pixel shaders, then you don't have a lot of options. You basically have two choices for getting images on the screen, StretchRect() or drawing polygons. StretchRect() doesn't give you any control other than filtering, so that means drawing polygons, and without pixel shaders, that means the dreaded fixed function pipeline of Direct3D 7. The fixed function pixel pipeline is a cascade of parallel color and alpha stages, with diffuse and specular going in the head, a texture being injected to each color/alpha stage, and the output going to the framebuffer.
Now, if you're assuming lack of pixel shader support, that means you're looking at cards like an TNT/TNT2, GeForce 1/2, or Radeon. The NVIDIA cards have two textures and two blend stages, while the Radeon has three. Furthermore, none of these cards have dependent texture read support, so texture-based table lookups are out and you're going to need very simple algorithms. Brightness/contrast adjustments are doable in two stages as a multiply and an add or subtract, or one if you use a multiply-add operation. You can squeeze in simple color balance too by using separate scale and bias values for each channel. Saturation can be done as a linear blend between luminance and the original color, i.e. lerp(dot(luma_basis, color), color, factor), but can be tricky to set up with the weird DX7 signed dot product. I think it might be doable in one pass if you're not doing anything else and use the framebuffer blend hardware.
Hue shifts are annoying to do if you want the standard HSV/HSL way of rotating colors around a hexagonal prism -- I think it'd be tough to do in fixed function without at least a lot of passes. Simple rotation in chroma would be easier, but you need a minimum of two dot products to do that and that practically means multipassing.
There are two other ways that some of these operations can be done. If YCbCr-to-RGB conversion is also being done through the rasterizer hardware, then some of these operations can be folded into that operation via the color matrix coefficients, which is desirable for both performance and accuracy reasons. In full-screen mode, it is also possible to use gamma ramp tables, although you run into problems if you have any on-screen UI that you don't want whacked by the post-process transform.
Finally, this isn't going to help if you must target Direct3D, but it's worth noting that the NVIDIA cards of that era are considerably more powerful if driven from OpenGL with NV extensions instead of Direct3D. On a GeForce 2, you get to play with two full register combiners along with a lerp-capable final combiner instead of two pathetic texture blend stages.
Overall, using fixed function rasterizing hardware to process video is difficult and restrictive. Creativity is a requirement as even simple operations like subtraction can require ingenuity. You can gain additional flexibility by multipassing in the framebuffer or off-screen surfaces using the post-blender, but you pay dearly in terms of fill rate and possibly accuracy. I did this in order to implement bicubic stretching in fixed function, and when you're doing somewhere between 5-9 passes with blending on and all texture stages active the GPU may not have enough fill rate to run at full video frame rate. | 2,535 | 10,437 | {"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} | 2.625 | 3 | CC-MAIN-2020-34 | latest | en | 0.92267 |
http://www.jiskha.com/display.cgi?id=1319916516 | 1,462,514,246,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861727712.42/warc/CC-MAIN-20160428164207-00202-ip-10-239-7-51.ec2.internal.warc.gz | 601,764,045 | 3,678 | Friday
May 6, 2016
# Homework Help: Trig
Posted by Julianne on Saturday, October 29, 2011 at 3:28pm.
Find the measure of the central angle, in radians, given the area of the sector as
22.6 cm2 and a radius equal to 3.7 cm
• Trig - Damon, Saturday, October 29, 2011 at 4:39pm
total area = pi r^2 = pi(3.7*2) = 43
22.6/43 = .5255
.5255 *2 pi = 3.3 radians | 138 | 357 | {"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} | 3.484375 | 3 | CC-MAIN-2016-18 | longest | en | 0.932196 |
https://openscience.fr/Permutation-groups-with-two-orbits-having-constant-movement | 1,725,847,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00423.warc.gz | 411,473,262 | 9,765 | Mathematics > Home > Advances in Pure and Applied Mathematics > Forthcoming papers > Article
# [FORTHCOMING] Permutation groups with two orbits having constant movement
## [FORTHCOMING] Groupe de permutations avec deux orbites à mouvement constant
Mehdi Rezaei
Buein Zahra Technical University
Iran
Mehdi Alaeiyan
Iran University of Science and Technology
Iran
Validated on 7 August 2024 DOI : TBA
### Mots-clés
Let $G$ be a permutation group on a set $\Omega$ with no fixed points in $\Omega$, and let $m$ be a positive integer. If for each subset $\Gamma$ of $\Omega$ the size $\Gamma^{g}-\Gamma|$ is bounded, for $g\in G$, the movement of $g$ is defined as move $(g):=\max{|\Gamma^{g}-\Gamma|}$ over all subsets $\Gamma$ of $\Omega$, and move $(G)$ is defined as the maximum of move $(g)$ over all non-identity elements of $g\in G$. Suppose that $G$ is not a 2-group. It was shown by Praeger that $|\Omega|\leqslant\lceil\frac{2mp}{p-1}\rceil+t-1$, where $t$ is the number of $G$-orbits on $\Omega$ and $p$ is the least odd prime dividing $|G|$. In this paper, we classify all permutation groups with maximum possible degree $|\Omega|=\lceil\frac{2mp}{p-1}\rceil+t-1$ for $t=2$, in which every non-identity element has constant movement $m$.
Let $G$ be a permutation group on a set $\Omega$ with no fixed points in $\Omega$, and let $m$ be a positive integer. If for each subset $\Gamma$ of $\Omega$ the size $\Gamma^{g}-\Gamma|$ is bounded, for $g\in G$, the movement of $g$ is defined as move $(g):=\max{|\Gamma^{g}-\Gamma|}$ over all subsets $\Gamma$ of $\Omega$, and move $(G)$ is defined as the maximum of move $(g)$ over all non-identity elements of $g\in G$. Suppose that $G$ is not a 2-group. It was shown by Praeger that $|\Omega|\leqslant\lceil\frac{2mp}{p-1}\rceil+t-1$, where $t$ is the number of $G$-orbits on $\Omega$ and $p$ is the least odd prime dividing $|G|$. In this paper, we classify all permutation groups with maximum possible degree $|\Omega|=\lceil\frac{2mp}{p-1}\rceil+t-1$ for $t=2$, in which every non-identity element has constant movement $m$. | 636 | 2,095 | {"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} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.751619 |
https://www.catalyzex.com/paper/arxiv:1910.11217 | 1,652,864,046,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00168.warc.gz | 779,644,866 | 11,279 | Get our free extension to see links to code for papers anywhere online!
# Katyusha Acceleration for Convex Finite-Sum Compositional Optimization
Oct 24, 2019
Yibo Xu, Yangyang Xu
Structured problems arise in many applications. To solve these problems, it is important to leverage the structure information. This paper focuses on convex problems with a finite-sum compositional structure. Finite-sum problems appear as the sample average approximation of a stochastic optimization problem and also arise in machine learning with a huge amount of training data. One popularly used numerical approach for finite-sum problems is the stochastic gradient method (SGM). However, the additional compositional structure prohibits easy access to unbiased stochastic approximation of the gradient, so directly applying the SGM to a finite-sum compositional optimization problem (COP) is often inefficient. We design new algorithms for solving strongly-convex and also convex two-level finite-sum COPs. Our design incorporates the Katyusha acceleration technique and adopts the mini-batch sampling from both outer-level and inner-level finite-sum. We first analyze the algorithm for strongly-convex finite-sum COPs. Similar to a few existing works, we obtain linear convergence rate in terms of the expected objective error, and from the convergence rate result, we then establish complexity results of the algorithm to produce an $\varepsilon$-solution. Our complexity results have the same dependence on the number of component functions as existing works. However, due to the use of Katyusha acceleration, our results have better dependence on the condition number $\kappa$ and improve to $\kappa^{2.5}$ from the best-known $\kappa^3$. Finally, we analyze the algorithm for convex finite-sum COPs, which uses as a subroutine the algorithm for strongly-convex finite-sum COPs. Again, we obtain better complexity results than existing works in terms of the dependence on $\varepsilon$, improving to $\varepsilon^{-2.5}$ from the best-known $\varepsilon^{-3}$. | 414 | 2,051 | {"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} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.890327 |
https://picturethismaths.wordpress.com/2016/02/04/spectral-sequences/ | 1,500,914,051,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424889.43/warc/CC-MAIN-20170724162257-20170724182257-00670.warc.gz | 691,694,606 | 37,410 | ## Spectral sequences
Last post Anna talked about representing a large tensor by a collection of smaller ones, which gave me the inspiration to write this post.
A lot of the time in topology (or at least in the papers I read) it’s convenient to use a tool called a spectral sequence to arrive at a result. In general, the spectral sequence breaks down the computation you want to perform in to a lot of smaller computations. This can seem like more work but in truth it can sometimes be impossible to perform your original computation in any other way.
A spectral sequence can be thought of as a book with a grid on each page, as shown above. On each vertex of the grid lies a group (shown as black dots in the picture) and there are maps going between certain groups (shown as arrows in the picture) satisfying certain properties. Given the first page there is a formula to give you the second page, and so on. Therefore if this was a book, the first page would determine the whole story. But unlike a story, this book need never end. If a group at a certain grid point stabilises (stays the same from a certain page forevermore), then the group at that grid point is called the limiting group at that point, and these are the groups that can give us answers to computations.
In particular you can use a spectral sequence to calculate the homology of a space, using only knowledge about a filtration of the space: a string of subspaces of that space which fit inside each other. So if you had a space built from dots, lines, triangles etc. then you can calculate it’s homology using knowledge of the interactions between the space with only the dots, only the lines, only the triangles etc. (it’s actually something called relative homology but we wont get into that). Quite cool eh? For example
Can be broken up into these three spaces, each one fitting inside the next:
(I just want to mention here that this space has very boring homology and in reality the spaces and filtrations we use are a lot more complicated!)
I’ll leave you with a picture of the first page of a spectral sequence that I drew on my blackboard today.
### 3 thoughts on “Spectral sequences”
1. Patrick February 4, 2016 / 11:07 pm
So probably “The Neverending Story 2” is the E_2 page. Is this where the franchise collapsed?
Liked by 2 people
• Rachael February 5, 2016 / 2:20 pm
The pictures are really something else! All maths textbooks should be this expressive, though maybe not as nightmare inducing 😛 thanks for the link, I’m trying to collect some older drawings of topological concepts for a historical post 🙂
Liked by 1 person | 591 | 2,626 | {"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} | 3.296875 | 3 | CC-MAIN-2017-30 | longest | en | 0.932508 |
http://www.greenspun.com/bboard/q-and-a-fetch-msg.tcl?msg_id=0008GT | 1,526,928,636,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00249.warc.gz | 399,469,470 | 4,274 | ### Merklinger: Focusing the view camera
greenspun.com : LUSENET : Large format photography : One Thread
Hello,
I just acquired the book "focusing the view camera" by Harold Merklinger, and read about his "hinge rule". My question: has anybody practical experience with Merklinger's system? Does it work, and do you have any tips to give? How convenient is it to estimate the distance from the camera to the plane of sharp focus? I still find it somewhat difficult to imagine how this should be done practically.
Lukas Werth
-- Lukas Werth (werthlvk@zedat.fu-berlin.de), August 13, 1998
Lukas,
I really love Merklinger's book, and his other one, too, The Ins and Outs of Focus. Other explanations of the Scheimpflug rule I have seen do not account for how tilts affect focus, but Merklinger does. I still have a hell of a time with tilts but Merklinger's procedure makes it way easier - you set your tilt according to desired plane of focus and then rack the standard back and forth until it lies in the desired angle.
-- Erik Ryberg (ryberg@seanet.com), August 13, 1998.
I haven't seen the books, but am reading the articles on his web site: http://fox.nstn.ca/~hmmerk/, including the animations showing how movements change the plane of sharpest focus. Brilliant. It has stared me thinking about knocking up some "angle templates" for my camera.
In the articles, the critical dimension is "J", the distance from the lens to the required plane of sharp focus, measured parallel to the film. If you want the ground in focus, and the back is vertical, and the lens is six foot above the ground, then the distance is six feet. If you are sideways on to a wall, looking up, J can still be measured, and the lens angle calculated, but translating that angle into a combined tilt and swing is much harder.
-- Alan Gibson (gibson.al@mail.dec.com), August 13, 1998.
I know this post is a few months old but I wanted to add what I felt was most interesting to learn from this book. In most average situations, not including close-up studio or still life, it is not necessary to use alot of lense tilt. Most lens tilts will be 5 degrees or less, maybe a little more for longer lenses. Before reading this I would struggle with my lens tilts because it seemed like I should be using more tilt. It is probably from looking at the camera advertisements where they show off the camera with a 45 degree tilt. Well, thats it in a nutshell, most pictures will benifit from at least a small amount of tilt and if you are tilting more than 10 degrees then you are probably tilting too much except for close-up work.
-- Jeff White (zonie@computer-concepts.com), November 10, 1998.
I have not read Merklingers books but I did read his article in View Camera a while back. I'm no engineer or mathematician and it seemed IMHO, rather verbose to me. I prefer Jim Stones explanation of focus, tilt and swing in his book on the view camera, combined with an article in Darkroom & Creative Camera Mar/Apr '96, about focus spread. It really isn't that hard. I think maybe - call me crazy - that people starting out would benefit from some hands on with an 8 X 10 as everything is magnified - the depth of field is shallower, the lenses longer, the need for focus control greater, AND, it's EASIER to see what's happening on that big ground glass.
-- Sean yates (yatescats@yahoo.com), November 10, 1998.
I agree that Jim Stones article is much better for understanding depth of field, focus and swings and tilts. It is the approach that I use when working in LF. Both of Merklinger's books are too long, over 100 pages each and could have been presented in 4 pages as effectively. He puts out a fairly easy to understand idea and then hits you with pages and pages to back it up. They are good books if you really want to understand what is going on with focus but I am not going to do all that math in the field. If you understand the hinge distance identified as "J" he gives you a couple of charts that list how much tilt or swing would be needed for the particular lens that you are using. It was intersting for me to see that the amount of correction necessary was usually a small amount.
-- Jeff White (zonie@computer-concepts.com), November 10, 1998.
I too have tried to wade through Merklinger's book and found it both interesting but rather tedious. I have applied a simple principle that I once read in an old Zone VI catalog regarding focusing: "Focus on the far, and tilt for the near." If you want to emphasize the near you use back tilt. With base tilt (like most field cameras) then the focus will have to be adjusted slightly after tilting. For me, this simple principle seems to have worked. Wondering if others have done the same.
-- John Wiemer (Wiemerjo@slcc.edu), November 12, 1998.
Short answer: Yes, it does work. Suggestion/tip....Reread the book several times, do the calcs on the samples, compare the chart answers, and get out in the field and use it. Harry
-- Harry L. Martin (hmartin@access1.net), October 20, 1999.
I read the book-twice-.VV=Very verbose. I just got back from a Howard Bond course on the view camera. So simple. The work we did involved tilts and swings so minute that it was difficult to accurately make slight changes. For most non-studio work lens and/or back changes are very small. The most important part I think is a detailed inspection of the GG with a loupe. George
-- George Nedleman (gnln@thegrid.net), October 20, 1999. | 1,299 | 5,472 | {"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} | 2.640625 | 3 | CC-MAIN-2018-22 | longest | en | 0.948633 |
https://www.educative.io/answers/standardization-vs-min-max-normalization | 1,660,095,803,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00245.warc.gz | 699,750,465 | 72,402 | Related Tags
python
# Standardization vs. min-max normalization
## Normalization (Min-Max Scalar)
Normalization makes sure all elements lie within zero and one. It is useful to normalize our data, given that the distribution of data is unknown. Moreover, Normalization cannot be used if the distribution is not a bell curve (like Gaussian distributions). Normalization is useful in models such as k-nearest neighbors and artificial neural networks, or anywhere where the data we are using has varying scales or precision (this will be more clear in the example below). For example, consider a dataset that has two features (salary and years worked). Both of these features will have high differences between their values. For example, salary might be a thousand times bigger than years worked – normalization makes sure the entire data in the same range.
Equation:
$\frac{X - X_{min}}{X_{max} - X_{min})}$
Not every dataset needs normalization in Machine Learning. It is only required when data features have varying ranges.
## Standardization
Standardizing a vector mostly means scaling a vector to the mean so that the values are closer to the mean than with a standard deviation of one. Standardization is important when data variability is in question. Many variables are measured using different units and scales. For instance, a variable ranging from 0 to 100 will outweigh the effect of a variable that has values between 0 and 1. This may result in a bias, which is why it is important to transform data to comparable scales. Moreover, it also assumes that data is Gaussian (follows a bell curve) in nature. Standardization is also beneficial when your algorithm predicts using a Gaussian distribution, such as linear regression, logistic regression, or linear discriminant analysis.
## Code for standardization
StandardScaler utility class from scikit-learn can be used to remove the mean and scale the data to unit variance.
from sklearn.preprocessing import StandardScaler
data = [[-1, -2], [0.5, 2], [0, 5], [12, 18]]
scaler = StandardScaler()
scaler.fit(data)
print(scaler.transform(data))
## Code for normalization (Min-Max Scalar)
MinMaxScalar from scikit-learn can be applied to a dataset, as shown below:
from sklearn.preprocessing import MinMaxScaler
data = [[-1, -2], [0.5, 2], [0, 5], [12, 18]]
scaler = MinMaxScaler()
scaler.fit(data)
MinMaxScaler()
print(scaler.transform(data))
RELATED TAGS
python
CONTRIBUTOR | 539 | 2,451 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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} | 2.71875 | 3 | CC-MAIN-2022-33 | latest | en | 0.901278 |
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