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Home >> CBSE XII >> Math >> Matrices
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# Find the values of a,b,c and d,if $3\begin{bmatrix}a & b\\c & d\end{bmatrix}=\begin{bmatrix}a & 6\\-1 & 2d\end{bmatrix}+\begin{bmatrix}4 & a+b\\c+d & 3\end{bmatrix}$
Toolbox:
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
Step1
Given
$\begin{bmatrix}3a & 3b\\3c & 3d\end{bmatrix}=\begin{bmatrix}a & 6\\-1 & 2d\end{bmatrix}+\begin{bmatrix}4 & a+b\\c+d & 3\end{bmatrix}$
The given matrices are equal hence their corresponding elements should be equal.
$\begin{bmatrix}3a & 3b\\3c & 3d\end{bmatrix}=\begin{bmatrix}a+4 & 6+a+b\\-1+c+d & 2d+3\end{bmatrix}$
$\Rightarrow$ 3a=a+4------(1)
$\quad$3b=6+a+b-----(2)
$\quad$3c=-1+c+d-----(3)
$\quad$3d=2d+3-----(4)
Step2:
From equation (1) we have,
3a=a+4
3a-a=4
2a=4
$a=\frac{4}{2}$
a=2.
Step3:
From equation (2) we have,
3b=6+a+b
3b=6+2+b [Substitute the value of a in equation(2)]
3b=8+b
3b-b=8.
2b=8.
b=8/2
b=4.
Step4:
From equation (4) we have,
3d=2d+3
3d-2d=3
d=3
Step5:
From equation (3) we have,
3c=-1+c+d.
Substitute the value of d in equation (3)
3c=-1+c+3
3c-c=-1+3
2c=2
c=1 | 630 | 1,418 | {"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": 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} | 4.65625 | 5 | CC-MAIN-2016-50 | longest | en | 0.371908 |
https://studysoup.com/tsg/math/245/thinking-mathematically/chapter/8797/8-3 | 1,601,514,273,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00763.warc.gz | 578,145,654 | 10,279 | ×
×
# Solutions for Chapter 8.3: Simple Interest
## Full solutions for Thinking Mathematically | 6th Edition
ISBN: 9780321867322
Solutions for Chapter 8.3: Simple Interest
Solutions for Chapter 8.3
4 5 0 409 Reviews
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##### ISBN: 9780321867322
This textbook survival guide was created for the textbook: Thinking Mathematically, edition: 6. Thinking Mathematically was written by and is associated to the ISBN: 9780321867322. Chapter 8.3: Simple Interest includes 45 full step-by-step solutions. Since 45 problems in chapter 8.3: Simple Interest have been answered, more than 63246 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions.
Key Math Terms and definitions covered in this textbook
• Big formula for n by n determinants.
Det(A) is a sum of n! terms. For each term: Multiply one entry from each row and column of A: rows in order 1, ... , nand column order given by a permutation P. Each of the n! P 's has a + or - sign.
• Diagonalization
A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.
• Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.
• Four Fundamental Subspaces C (A), N (A), C (AT), N (AT).
Use AT for complex A.
• Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.
• Length II x II.
Square root of x T x (Pythagoras in n dimensions).
• Linear combination cv + d w or L C jV j.
• Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.
• Normal matrix.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.
• Outer product uv T
= column times row = rank one matrix.
• Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.
• Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.
• Rank r (A)
= number of pivots = dimension of column space = dimension of row space.
• Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.
• Saddle point of I(x}, ... ,xn ).
A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.
• Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.
• Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!
• Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.
• Spectrum of A = the set of eigenvalues {A I, ... , An}.
Spectral radius = max of IAi I. | 915 | 3,310 | {"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} | 3.6875 | 4 | CC-MAIN-2020-40 | latest | en | 0.856023 |
http://doisinkidney.com/posts/2018-03-17-rose-trees-breadth-first.html | 1,524,664,682,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947822.98/warc/CC-MAIN-20180425135246-20180425155246-00076.warc.gz | 86,426,017 | 4,195 | Posted on March 17, 2018
In contrast to the more common binary trees, in a rose tree every node can have any number of children.
``````data Tree a
= Node
{ root :: a
, forest :: Forest a
}
type Forest a = [Tree a]``````
One of the important manipulations of this data structure, which forms the basis for several other algorithms, is a breadth-first traversal. I’d like to go through a couple of techniques for implementing it, and how more generally you can often get away with using much simpler data structures if you really pinpoint the API you need from them.
As a general technique, Okasaki (2000) advises that a queue be used:
``````breadthFirst :: Tree a -> [a]
breadthFirst tr = go (singleton tr)
where
go q = case pop q of
Nothing -> []
Just (Node x xs,qs) -> x : go (qs `append` xs)``````
There are three functions left undefined there: `singleton`, `pop`, and `append`. They represent the API of our as-of-yet unimplemented queue, and their complexity will dictate the complexity of the overall algorithm. As a (bad) first choice, we could use simple lists, with the functions defined thus:
``````singleton x = [x]
pop (x:xs) = Just (x,xs)
pop [] = Nothing
append = (++)``````
Those repeated appends are bad news. The queue needs to be able to support popping from one side and appending from the other, which is something lists absolutely cannot do well.
We could swap in a more general queue implementation, possibly using Data.Sequence, or a pair of lists. But these are more complex and general than we need, so let’s try and pare down the requirements a little more.
First, we don’t need a pop: the go function can be expressed as a fold instead. Second, we don’t need every append to be immediately stuck into the queue, we can batch them, first appending to a structure that’s efficient for appends, and then converting that to a structure which is efficient for folds. In code:
``````breadthFirst :: Forest a -> [a]
breadthFirst ts = foldr f b ts []
where
f (Node x xs) fw bw = x : fw (xs : bw)
b [] = []
b qs = foldl (foldr f) b qs []``````
We’re consing instead of appending, but the consumption is being done in the correct direction anyway, because of the `foldl`.
## Levels
So next step: to get the `levels` function from Data.Tree. Instead of doing a breadth-first traversal, it returns the nodes at each level of the tree. Conceptually, every time we did the reverse above (called `foldl`), we will do a cons as well:
``````levels :: Forest a -> [[a]]
levels ts = foldl f b ts [] []
where
f k (Node x xs) ls qs = k (x : ls) (xs : qs)
b _ [] = []
b k qs = k : foldl (foldl f) b qs [] []``````
## Unfolding
The original reason I started work on these problems was this issue in containers. It concerns the `unfoldTreeM_BF` function. An early go at rewriting it, inspired by levels above, looks like this:
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 ``` ``````unfoldForestM_BF :: Monad m => (b -> m (a, [b])) -> [b] -> m (Forest a) unfoldForestM_BF f ts = b [ts] (const id) where b [] k = pure (k [] []) b qs k = foldl (foldr t) b qs [] (\x -> k [] . foldr (uncurry run) id x) t a fw bw k = do (x,cs) <- f a let !n = length cs fw (cs : bw) (k . (:) (x, n)) run x n xs ys = case splitAt n ys of (cs,zs) -> Node x cs : xs zs``````
It basically performs the same this as the levels function, but builds the tree back up in the end using the `run` function. In order to do that, we store the length of each subforest on line 9, so that each node knows how much to take from each level.
A possible optimization is to stop taking the length. Anything in list processing that takes a length screams “wrong” to me (although it’s not always true!) so I often try to find a way to avoid it. The first option would be to keep the `cs` on line 8 around, and use it as an indicator for the length. That keeps it around longer than strictly necessary, though. The other option is to add a third level: for `breadthFirst` above, we had one level; for `levels`, we added another, to indicate the structure of the nodes and their subtrees; here, we can add a third, to maintain that structure when building back up:
``````unfoldForestM_BF :: Monad m => (b -> m (a, [b])) -> [b] -> m (Forest a)
unfoldForestM_BF f ts = b [ts] (\ls -> concat . ls)
where
b [] k = pure (k id [])
b qs k = foldl g b qs [] (\ls -> k id . ls)
g a xs qs k = foldr t (\ls ys -> a ys (k . run ls)) xs [] qs
t a fw xs bw = f a >>= \(x,cs) -> fw (x:xs) (cs:bw)
run x xs = uncurry (:) . foldl go ((,) [] . xs) x
where
go ys y (z:zs) = (Node y z : ys', zs')
where
(ys',zs') = ys zs``````
This unfortunately slows down the code.
Okasaki, Chris. 2000. “Breadth-first Numbering: Lessons from a Small Exercise in Algorithm Design.” In Proceedings of the Fifth ACM SIGPLAN International Conference on Functional Programming, 131–136. ICFP ’00. New York, NY, USA: ACM. doi:10.1145/351240.351253. https://www.cs.tufts.edu/~nr/cs257/archive/chris-okasaki/breadth-first.pdf. | 1,402 | 4,966 | {"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-2018-17 | latest | en | 0.932988 |
https://leanprover-community.github.io/mathlib4_docs/Mathlib/Data/Set/Intervals/UnorderedInterval.html | 1,708,750,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00585.warc.gz | 355,602,158 | 8,659 | # Intervals without endpoints ordering #
In any lattice α, we define uIcc a b to be Icc (a ⊓ b) (a ⊔ b), which in a linear order is the set of elements lying between a and b.
Icc a b requires the assumption a ≤ b to be meaningful, which is sometimes inconvenient. The interval as defined in this file is always the set of things lying between a and b, regardless of the relative order of a and b.
For real numbers, uIcc a b is the same as segment ℝ a b.
In a product or pi type, uIcc a b is the smallest box containing a and b. For example, uIcc (1, -1) (-1, 1) = Icc (-1, -1) (1, 1) is the square of vertices (1, -1), (-1, -1), (-1, 1), (1, 1).
In Finset α (seen as a hypercube of dimension Fintype.card α), uIcc a b is the smallest subcube containing both a and b.
## Notation #
We use the localized notation [[a, b]] for uIcc a b. One can open the locale Interval to make the notation available.
def Set.uIcc {α : Type u_1} [] (a : α) (b : α) :
Set α
uIcc a b is the set of elements lying between a and b, with a and b included. Note that we define it more generally in a lattice as Set.Icc (a ⊓ b) (a ⊔ b). In a product type, uIcc corresponds to the bounding box of the two elements.
Equations
Instances For
[[a, b]] denotes the set of elements lying between a and b, inclusive.
Equations
• One or more equations did not get rendered due to their size.
Instances For
@[simp]
theorem Set.dual_uIcc {α : Type u_1} [] (a : α) (b : α) :
Set.uIcc (OrderDual.toDual a) (OrderDual.toDual b) = OrderDual.ofDual ⁻¹' Set.uIcc a b
@[simp]
theorem Set.uIcc_of_le {α : Type u_1} [] {a : α} {b : α} (h : a b) :
@[simp]
theorem Set.uIcc_of_ge {α : Type u_1} [] {a : α} {b : α} (h : b a) :
theorem Set.uIcc_comm {α : Type u_1} [] (a : α) (b : α) :
theorem Set.uIcc_of_lt {α : Type u_1} [] {a : α} {b : α} (h : a < b) :
theorem Set.uIcc_of_gt {α : Type u_1} [] {a : α} {b : α} (h : b < a) :
theorem Set.uIcc_self {α : Type u_1} [] {a : α} :
Set.uIcc a a = {a}
@[simp]
theorem Set.nonempty_uIcc {α : Type u_1} [] {a : α} {b : α} :
theorem Set.Icc_subset_uIcc {α : Type u_1} [] {a : α} {b : α} :
theorem Set.Icc_subset_uIcc' {α : Type u_1} [] {a : α} {b : α} :
@[simp]
theorem Set.left_mem_uIcc {α : Type u_1} [] {a : α} {b : α} :
@[simp]
theorem Set.right_mem_uIcc {α : Type u_1} [] {a : α} {b : α} :
theorem Set.mem_uIcc_of_le {α : Type u_1} [] {a : α} {b : α} {x : α} (ha : a x) (hb : x b) :
theorem Set.mem_uIcc_of_ge {α : Type u_1} [] {a : α} {b : α} {x : α} (hb : b x) (ha : x a) :
theorem Set.uIcc_subset_uIcc {α : Type u_1} [] {a₁ : α} {a₂ : α} {b₁ : α} {b₂ : α} (h₁ : a₁ Set.uIcc a₂ b₂) (h₂ : b₁ Set.uIcc a₂ b₂) :
Set.uIcc a₁ b₁ Set.uIcc a₂ b₂
theorem Set.uIcc_subset_Icc {α : Type u_1} [] {a₁ : α} {a₂ : α} {b₁ : α} {b₂ : α} (ha : a₁ Set.Icc a₂ b₂) (hb : b₁ Set.Icc a₂ b₂) :
Set.uIcc a₁ b₁ Set.Icc a₂ b₂
theorem Set.uIcc_subset_uIcc_iff_mem {α : Type u_1} [] {a₁ : α} {a₂ : α} {b₁ : α} {b₂ : α} :
Set.uIcc a₁ b₁ Set.uIcc a₂ b₂ a₁ Set.uIcc a₂ b₂ b₁ Set.uIcc a₂ b₂
theorem Set.uIcc_subset_uIcc_iff_le' {α : Type u_1} [] {a₁ : α} {a₂ : α} {b₁ : α} {b₂ : α} :
Set.uIcc a₁ b₁ Set.uIcc a₂ b₂ a₂ b₂ a₁ b₁ a₁ b₁ a₂ b₂
theorem Set.uIcc_subset_uIcc_right {α : Type u_1} [] {a : α} {b : α} {x : α} (h : x Set.uIcc a b) :
theorem Set.uIcc_subset_uIcc_left {α : Type u_1} [] {a : α} {b : α} {x : α} (h : x Set.uIcc a b) :
theorem Set.bdd_below_bdd_above_iff_subset_uIcc {α : Type u_1} [] (s : Set α) :
∃ (a : α) (b : α), s Set.uIcc a b
@[simp]
theorem Set.uIcc_prod_uIcc {α : Type u_1} {β : Type u_2} [] [] (a₁ : α) (a₂ : α) (b₁ : β) (b₂ : β) :
Set.uIcc a₁ a₂ ×ˢ Set.uIcc b₁ b₂ = Set.uIcc (a₁, b₁) (a₂, b₂)
theorem Set.uIcc_prod_eq {α : Type u_1} {β : Type u_2} [] [] (a : α × β) (b : α × β) :
Set.uIcc a b = Set.uIcc a.1 b.1 ×ˢ Set.uIcc a.2 b.2
theorem Set.eq_of_mem_uIcc_of_mem_uIcc {α : Type u_1} [] {a : α} {b : α} {c : α} (ha : a Set.uIcc b c) (hb : b Set.uIcc a c) :
a = b
theorem Set.eq_of_mem_uIcc_of_mem_uIcc' {α : Type u_1} [] {a : α} {b : α} {c : α} :
b Set.uIcc a cc Set.uIcc a bb = c
theorem Set.uIcc_injective_right {α : Type u_1} [] (a : α) :
Function.Injective fun (b : α) => Set.uIcc b a
theorem Set.uIcc_injective_left {α : Type u_1} [] (a : α) :
theorem MonotoneOn.mapsTo_uIcc {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : MonotoneOn f (Set.uIcc a b)) :
Set.MapsTo f (Set.uIcc a b) (Set.uIcc (f a) (f b))
theorem AntitoneOn.mapsTo_uIcc {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : AntitoneOn f (Set.uIcc a b)) :
Set.MapsTo f (Set.uIcc a b) (Set.uIcc (f a) (f b))
theorem Monotone.mapsTo_uIcc {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : ) :
Set.MapsTo f (Set.uIcc a b) (Set.uIcc (f a) (f b))
theorem Antitone.mapsTo_uIcc {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : ) :
Set.MapsTo f (Set.uIcc a b) (Set.uIcc (f a) (f b))
theorem MonotoneOn.image_uIcc_subset {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : MonotoneOn f (Set.uIcc a b)) :
f '' Set.uIcc a b Set.uIcc (f a) (f b)
theorem AntitoneOn.image_uIcc_subset {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : AntitoneOn f (Set.uIcc a b)) :
f '' Set.uIcc a b Set.uIcc (f a) (f b)
theorem Monotone.image_uIcc_subset {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : ) :
f '' Set.uIcc a b Set.uIcc (f a) (f b)
theorem Antitone.image_uIcc_subset {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {a : α} {b : α} (hf : ) :
f '' Set.uIcc a b Set.uIcc (f a) (f b)
theorem Set.Icc_min_max {α : Type u_1} [] {a : α} {b : α} :
Set.Icc (min a b) (max a b) = Set.uIcc a b
theorem Set.uIcc_of_not_le {α : Type u_1} [] {a : α} {b : α} (h : ¬a b) :
theorem Set.uIcc_of_not_ge {α : Type u_1} [] {a : α} {b : α} (h : ¬b a) :
theorem Set.uIcc_eq_union {α : Type u_1} [] {a : α} {b : α} :
theorem Set.mem_uIcc {α : Type u_1} [] {a : α} {b : α} {c : α} :
a Set.uIcc b c b a a c c a a b
theorem Set.not_mem_uIcc_of_lt {α : Type u_1} [] {a : α} {b : α} {c : α} (ha : c < a) (hb : c < b) :
cSet.uIcc a b
theorem Set.not_mem_uIcc_of_gt {α : Type u_1} [] {a : α} {b : α} {c : α} (ha : a < c) (hb : b < c) :
cSet.uIcc a b
theorem Set.uIcc_subset_uIcc_iff_le {α : Type u_1} [] {a₁ : α} {a₂ : α} {b₁ : α} {b₂ : α} :
Set.uIcc a₁ b₁ Set.uIcc a₂ b₂ min a₂ b₂ min a₁ b₁ max a₁ b₁ max a₂ b₂
theorem Set.uIcc_subset_uIcc_union_uIcc {α : Type u_1} [] {a : α} {b : α} {c : α} :
A sort of triangle inequality.
theorem Set.monotone_or_antitone_iff_uIcc {α : Type u_1} {β : Type u_2} [] [] {f : αβ} :
∀ (a b c : α), c Set.uIcc a bf c Set.uIcc (f a) (f b)
theorem Set.monotoneOn_or_antitoneOn_iff_uIcc {α : Type u_1} {β : Type u_2} [] [] {f : αβ} {s : Set α} :
as, bs, cs, c Set.uIcc a bf c Set.uIcc (f a) (f b)
def Set.uIoc {α : Type u_1} [] :
ααSet α
The open-closed uIcc with unordered bounds.
Equations
Instances For
Ι a b denotes the open-closed interval with unordered bounds. Here, Ι is a capital iota, distinguished from a capital i.
Equations
Instances For
@[simp]
theorem Set.uIoc_of_le {α : Type u_1} [] {a : α} {b : α} (h : a b) :
Ι a b = Set.Ioc a b
@[simp]
theorem Set.uIoc_of_lt {α : Type u_1} [] {a : α} {b : α} (h : b < a) :
Ι a b = Set.Ioc b a
theorem Set.uIoc_eq_union {α : Type u_1} [] {a : α} {b : α} :
Ι a b = Set.Ioc a b Set.Ioc b a
theorem Set.mem_uIoc {α : Type u_1} [] {a : α} {b : α} {c : α} :
a Ι b c b < a a c c < a a b
theorem Set.not_mem_uIoc {α : Type u_1} [] {a : α} {b : α} {c : α} :
aΙ b c a b a c c < a b < a
@[simp]
theorem Set.left_mem_uIoc {α : Type u_1} [] {a : α} {b : α} :
a Ι a b b < a
@[simp]
theorem Set.right_mem_uIoc {α : Type u_1} [] {a : α} {b : α} :
b Ι a b a < b
theorem Set.forall_uIoc_iff {α : Type u_1} [] {a : α} {b : α} {P : αProp} :
(xΙ a b, P x) (xSet.Ioc a b, P x) xSet.Ioc b a, P x
theorem Set.uIoc_subset_uIoc_of_uIcc_subset_uIcc {α : Type u_1} [] {a : α} {b : α} {c : α} {d : α} (h : Set.uIcc a b Set.uIcc c d) :
Ι a b Ι c d
theorem Set.uIoc_comm {α : Type u_1} [] (a : α) (b : α) :
Ι a b = Ι b a
theorem Set.Ioc_subset_uIoc {α : Type u_1} [] {a : α} {b : α} :
Set.Ioc a b Ι a b
theorem Set.Ioc_subset_uIoc' {α : Type u_1} [] {a : α} {b : α} :
Set.Ioc a b Ι b a
theorem Set.eq_of_mem_uIoc_of_mem_uIoc {α : Type u_1} [] {a : α} {b : α} {c : α} :
a Ι b cb Ι a ca = b
theorem Set.eq_of_mem_uIoc_of_mem_uIoc' {α : Type u_1} [] {a : α} {b : α} {c : α} :
b Ι a cc Ι a bb = c
theorem Set.eq_of_not_mem_uIoc_of_not_mem_uIoc {α : Type u_1} [] {a : α} {b : α} {c : α} (ha : a c) (hb : b c) :
aΙ b cbΙ a ca = b
theorem Set.uIoc_injective_right {α : Type u_1} [] (a : α) :
Function.Injective fun (b : α) => Ι b a
theorem Set.uIoc_injective_left {α : Type u_1} [] (a : α) : | 3,741 | 8,590 | {"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} | 3.171875 | 3 | CC-MAIN-2024-10 | latest | en | 0.818956 |
http://www.teachoo.com/2790/635/Misc-7---A-man-running-a-racecourse-notes-that-sum-of-distances/category/Miscellaneous/ | 1,498,256,501,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320201.43/warc/CC-MAIN-20170623220935-20170624000935-00667.warc.gz | 664,756,328 | 12,668 | 1. Chapter 11 Class 11 Conic Sections
2. Serial order wise
Transcript
Misc 7(Method 1) A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. find the equation of the posts traced by the man. Let man be standing on point P(x, y) There are two flag posts S & S’ Given PS + PS’ = 10 & SS’ = 8 m Let S & S’ be on x-axis such that Origin (O) be the mid-point of S’S So, OS = OS’ = 4 m So, coordinates of S = S(4, 0) coordinates of S’ = S’(–4, 0) We know that distance between two points (x1, y1) & (x2, y2) d = 𝑥2−𝑥12+𝑦2−𝑦12 Now, finding PS & PS’ PS’ = 𝑥−(−4)2+𝑦−02 PS’ = 𝑥+42+𝑦2 and PS = 𝑥−42+𝑦−02 PS = 𝑥−42+𝑦2 Now, PS + PS’ = 10 Putting values 𝑥−42+𝑦2 + 𝑥+42+𝑦2 = 10 𝑥−42+𝑦2 = 10 – 𝑥+42+𝑦2 Squaring both sides 𝑥−42+𝑦22 = 10 – 𝑥+42+𝑦22 𝑥−42+𝑦2 = 102 + 𝑥+42+𝑦22 – 20 𝑥+42+𝑦2 𝑥−42+𝑦2 = 100 + 𝑥+42+𝑦2 – 20 𝑥+42+𝑦2 x2 + 42 – 8x + 𝑦2 = 100 + x2 + 42 + 8x + 𝑦2 – 20 𝑥+42+𝑦2 x2 + 42 – 8x + 𝑦2 – 100 – x2 – 42 – 8x − 𝑦2 = – 20 𝑥+42+𝑦2 – 16x – 100 = – 20 𝑥+42+𝑦2 20 𝑥+42+𝑦2 = 16x + 100 20 𝑥+42+𝑦2 = 4(4x + 25) 5 𝑥+42+𝑦2 = (4x + 25) Squaring both sides 5 𝑥+42+𝑦22= (4x + 25)2 25𝑥+42+𝑦2 = (4x + 25)2 25(x2 + 42 + 8x + y2) = (4x)2 + (25)2 + 2(4x)(25) 25x2 + 400 + 200x + 25y2 = 16x2 + 625 + 200x 9x2 + 25y2 = 225 9𝑥2225 + 25𝑦2225 = 1 𝒙𝟐𝟐𝟓 + 𝒚𝟐𝟗 = 1 Misc 7(Method 2) A man running a racecourse notes that the sum of the distances from the two flag posts form him is always 10 m and the distance between the flag posts is 8 m. find the equation of the posts traced by the man. Let S & S’ be two flag past The sum of distance of man P from S & S’ is equal to 10 Since the sum of distance of a point from any two fixed points S & S’ in the plane is constant, it forms on ellipse. So, S & S’ are the foci of the ellipse x-axis is the major axis, y-axis is the minor axis Let equation of ellipse be 𝑥2𝑎2 + 𝑦2𝑏2 = 1 Now, given that The sum of distance of man P from S & S’ = 10 i.e. Sum of distance of a point from foci is = 10 2a = 10 m a = 102 = 5m Also it is given that Distance between flag posts SS’ = 8m Distance between two foci = 8 m 2c = 8 c = 82 = 4 Now we need to find b We know that c2 = a2 − b2 42 = 52 − b2 16 = 25 − b2 b2 = 25 −16 b2 = 9 b = 3 Now equation of ellipse is 𝑥2𝑎2 + 𝑦2𝑏2 = 1 𝑥252 + 𝑦232 = 1 𝒙𝟐𝟐𝟓 + 𝒚𝟐𝟗 = 1 Thus, the required equation is 𝑥225 + 𝑦29 = 1
Serial order wise | 2,458 | 2,804 | {"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} | 4.53125 | 5 | CC-MAIN-2017-26 | latest | en | 0.68289 |
http://www.coderanch.com/t/429165/java/java/simple | 1,466,865,387,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393332.57/warc/CC-MAIN-20160624154953-00176-ip-10-164-35-72.ec2.internal.warc.gz | 460,637,506 | 8,475 | Win a copy of Think Java: How to Think Like a Computer Scientist this week in the Java in General forum!
David Barry
Ranch Hand
Posts: 84
I know that you do 6.83783 * 10^3 as 6.83783E3, but how do I do that number times 10 to the negative third?
fred rosenberger
lowercase baba
Bartender
Posts: 12122
30
you want 10 to the negative 1/3 power? that's the same as 1 over the cube root of 10, which would require you to use the Math.pow function:
6.83783 / Math.pow(10.0, 1.0/3);
are you sure you didn't mean 10 to the negative three power?
in which case you simply would write
6.83783E-3
David Barry
Ranch Hand
Posts: 84
Thanks for the help. I figured it out, finally.
Campbell Ritchie
Sheriff
Posts: 48910
58 | 225 | 715 | {"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-2016-26 | latest | en | 0.920654 |
https://visualstudiomagazine.com/Articles/2024/01/17/principal-component-analysis.aspx | 1,708,755,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00811.warc.gz | 614,639,393 | 46,334 | The Data Science Lab
### Principal Component Analysis (PCA) from Scratch Using the Classical Technique with C#
Transforming a dataset into one with fewer columns is more complicated than it might seem, explains Dr. James McCaffrey of Microsoft Research in this full-code, step-by-step machine learning tutorial.
Principal component analysis (PCA) is a classical machine learning technique. The goal of PCA is to transform a dataset into one with fewer columns. This is called dimensionality reduction. The transformed data can be used for visualization or as the basis for prediction using ML techniques that can't deal with a large number of predictor columns.
There are two main techniques to implement PCA. The first technique computes eigenvalues and eigenvectors from a covariance matrix derived from the source data. The second PCA technique sidesteps the covariance matrix and computes a singular value decomposition (SVD) of the source data. Both techniques give the same results, subject to very small differences.
This article presents a from-scratch C# implementation of the first technique: compute eigenvalues and eigenvectors from the covariance matrix. If you're not familiar with PCA, most of the terminology -- covariance matrix, eigenvalues and eigenvectors and so on -- sounds quite mysterious. But the ideas will become clear shortly.
A good way to see where this article is headed is to take a look at the screenshot of a demo program in Figure 1.
For simplicity, the demo uses just nine data items:
```[0] 5.1000 3.5000 1.4000 0.2000
[1] 4.9000 3.1000 1.5000 0.1000
[2] 5.1000 3.8000 1.5000 0.3000
[3] 7.0000 3.2000 4.7000 1.4000
[4] 5.0000 2.0000 3.5000 1.0000
[5] 5.9000 3.2000 4.8000 1.8000
[6] 6.3000 3.3000 6.0000 2.5000
[7] 6.5000 3.2000 5.1000 2.0000
[8] 6.9000 3.2000 5.7000 2.3000```
The nine data items are one-based items 1, 10, 20, 51, 61, 71, 101, 111 and 121 of the 150-item Iris dataset. Each line represents an iris flower. The four columns are sepal length, sepal width, petal length and petal width (a sepal is a green leaf-like structure). Because there are four columns, the data is said to have dimension = 4.
The leading index values in brackets are not part of the data. Each data item is one of three species: 0 = setosa, 1 = versicolor 2 = virginica. The first three items are setosa, the second three items are versicolor and the last three items are virginica. The species labels were removed because they're not used for PCA dimensionality reduction.
PCA is intended for use with strictly numeric data. Mathematically, the technique works with Boolean variables (0-1 encoded) and for one-hot encoded categorical data. Conceptually, however, applying PCA to non-numeric data is questionable, and there is very little research on the topic.
The demo program applies z-score normalization to the source data. Then a 4-by-4 covariance matrix is computed from the normalized data. Next, four eigenvalues and four eigenvectors are computed from the covariance matrix (using the Jacobi algorithm). Next, the demo uses the eigenvectors to compute transformed data:
```[0] -2.0787 0.6736 -0.0425 0.0435
[1] -2.2108 -0.2266 -0.1065 -0.0212
[2] -2.0071 1.2920 0.1883 0.0025
[3] 1.1557 0.3503 -0.9194 -0.0785
[4] -0.7678 -2.6702 0.0074 0.0115
[5] 0.7029 -0.0028 0.4066 -0.0736
[6] 1.8359 0.2191 0.6407 -0.0424
[7] 1.3476 0.1482 -0.0201 0.0621
[8] 2.0224 0.2163 -0.1545 0.0960```
At this point, in a non-demo scenario, you could extract the first two or first three of the columns of the transformed data to use as a surrogate for the source data. For example, if you used the first two columns, you could graph the data with the first column acting as x-values and the second column acting as y-values.
The demo program concludes by programmatically reconstructing the source data from the transformed data. The idea is to verify the PCA worked correctly and also to illustrate that the transformed data contains all of the information in the source data.
This article assumes you have intermediate or better programming skill but doesn't assume you know anything about principal component analysis. The demo is implemented using C#, but you should be able to refactor the demo code to another C-family language if you wish. All normal error checking has been removed to keep the main ideas as clear as possible.
The source code for the demo program is too long to be presented in its entirety in this article. The complete code is available in the accompanying file download, and is also available online.
Understanding Principal Component Analysis
The best way to understand PCA is to take a closer look at the demo program. The demo has eight steps:
1. load data into memory
2. normalize the data
3. compute a covariance matrix from the normalized data
4. compute eigenvalues and eigenvectors from covariance matrix
5. sort the eigenvectors based on eigenvalues
6. compute variance explained by each eigenvector
7. use eigenvectors to compute transformed data
8. verify by reconstructing source data
The demo loads the source data using a helper MatLoad() function:
```Console.WriteLine("Loading Iris-9 data ");
string dataFile = "..\\..\\..\\Data\\iris_9.txt";
double[][] X = MatLoad(dataFile, new int[] { 0, 1, 2, 3 },
',', "#");
Console.WriteLine("Source data: ");
MatShow(X, 4, 9, true);```
The arguments to MatLoad() mean fetch columns 0,1,2,3 of the comma-separated data, where a "#" character indicates a comment line. The arguments to MatShow() mean display using four decimals, with a field width of nine per element, and display indices.
Normalizing the Data
Each column of the data is normalized using these statements:
```Console.WriteLine("Standardizing (z-score, biased) ");
double[] means;
double[] stds;
double[][] stdX = MatStandardize(X, out means, out stds);
Console.WriteLine("Standardized data items ");
MatShow(stdX, 4, 9, nRows: 3);```
The four means and standard deviations for each column are saved because they are needed to reconstruct the original source data. The terms standardization and normalization are technically different but are often used interchangeably. Standardization is a specific type of normalization (z-score). PCA documentation tends to use the term standardization, but I sometimes use the more general term normalization.
Each value is normalized using the formula x' = (x - u) / s, where x' is the normalized value, x is the original value, u is the column mean and s is the column biased standard deviation (divide sum of squares by n rather than n-1 where n is the number of values in the column).
For example, suppose a column of some data has just n = 3 values: (4, 15, 8). The mean of the column is (4 + 15 + 8) / 3 = 9.0. The biased standard deviation of the column is sqrt( ((4 - 9.0)^2 + (15 - 9.0)^2 + (8 - 9.0)^2) / n) ) = sqrt( (25.0 + 36.0 + 1.0) / 3 ) = sqrt(62.0 / 3) = 4.55.
The normalized values for (4, 15, 8) are:
``` 4: (4 - 9.0) / 4.55 = -1.10
15: (15 - 9.0) / 4.55 = 1.32
8: (8 - 9.0) / 4.55 = -0.22```
Normalized values that are negative are less than the column mean, and normalized values that are positive are greater than the mean. For the demo data, the normalized data is:
``` -0.9410 0.7255 -1.3515 -1.2410
-1.1901 -0.1451 -1.2952 -1.3550
-0.9410 1.3784 -1.2952 -1.1270
1.4253 0.0725 0.5068 0.1266
-1.0655 -2.5392 -0.1689 -0.3292
0.0554 0.0725 0.5631 0.5825
0.5535 0.2902 1.2389 1.3803
0.8026 0.0725 0.7321 0.8105
1.3008 0.0725 1.0700 1.1524```
After normalization, each value in a column will typically be between -3.0 and +3.0. Normalized values that are less than -6.0 or greater than +6.0 are extreme and should be investigated. The idea behind normalization is to prevent columns with large values (such as employee annual salaries) from overwhelming columns with small values (such as employee age).
Computing the Covariance Matrix
The covariance matrix of the normalized source data is computed like so:
```Console.WriteLine("Computing covariance matrix: ");
double[][] covarMat = CovarMatrix(stdX, false);
MatShow(covarMat, 4, 9, false);```
The "false" argument in the call to CovarMatrix() means to treat the data as if it is organized so that each variable (sepal length, sepal width, petal length and petal width) is in a column. A "true" argument means the data variables are organized as rows. I use this interface to match that of the Python numpy.cov() library function.
The covariance of two vectors is a measure of the joint variability of the vectors. For example, if v1 = [4, 9, 8] and v2 = [6, 8, 1], then covariance(v1, v2) = -0.50. Loosely speaking, the closer the covariance is to 0, the less associated the two vectors are. The larger the covariance, the greater the association. The sign of the covariance indicates the direction of the association. There's no upper limit on the magnitude of a covariance because it will increase as the number of elements in the vectors increases. See my article, "Computing the Covariance of Two Vectors Using C#", for a worked example.
A covariance matrix stores the covariance between all pairs of columns in the normalized data. For example, cell [0][3] of the covariance matrix holds the covariance between column 0 and column 3. For the demo data, the covariance matrix is:
``` 1.1250 0.1649 0.9538 0.9147
0.1649 1.1250 -0.1976 -0.1034
0.9538 -0.1976 1.1250 1.1095
0.9147 -0.1034 1.1095 1.1250```
Notice that the covariance matrix is symmetric because covariance(x, y) = covariance(y, x). The values on the diagonal of the covariance matrix are all the same (1.1250) due to the z-score normalization.
Computing the Eigenvalues and Eigenvectors
The eigenvalues and eigenvectors of the covariance matrix are computed using this code:
```Console.WriteLine("Computing eigenvalues and eigenvectors ");
double[] eigenVals;
double[][] eigenVecs;
Eigen(covarMat, out eigenVals, out eigenVecs);
Console.WriteLine("Eigenvectors (not sorted, as cols): ");
MatShow(eigenVecs, 4, 9, false);```
For the demo data that has dim = 4, the covariance matrix of the normalized data has shape 4-by-4. There are four eigenvalues (single values like 1.234) and four eigenvectors, each with four values. The eigenvalues and eigenvectors are paired.
For the demo data, the four eigenvectors (as columns) are:
``` 0.5498 0.2395 -0.7823 0.1683
-0.0438 0.9637 0.2420 -0.1035
0.5939 -0.1102 0.2188 -0.7663
0.5857 -0.0410 0.5306 0.6113```
The program-defined Eigen() function returns the eigenvectors as columns, so the first eigenvector is (0.5498, -0.0438, 0.5939, 0.5857). The associated eigenvector is 3.1167. For PCA it doesn't help to overthink the deep mathematics of eigenvalues and eigenvectors. You can think of eigenvalues and eigenvectors as a kind of factorization of a matrix that captures all the information in the matrix in a sequential way.
Computing eigenvalues and eigenvectors is one of the most difficult problems in numerical programming. Because a covariance matrix is mathematically symmetric positive semidefinite (think "relatively simple structure"), it is possible to use a technique called the Jacobi algorithm to find the eigenvalues and eigenvectors. The majority of the demo code is the Eigen() function and its helpers.
Eigenvectors from a matrix are not unique in the sense that the sign is arbitrary. For example, an eigenvector (0.20, -0.50, 0.30) is equivalent to (-0.20, 0.50, -0.30). The eigenvector sign issue does not affect PCA.
Sorting the Eigenvalues and Eigenvectors
The next step in PCA is to sort the eigenvalues and their associated eigenvectors from largest to smallest. Some library eigen() functions return eigenvalues and eigenvectors already sorted, and some eigen() functions do not. The demo program assumes that the eigenvalues and eigenvectors are not necessarily sorted.
First, the four eigenvalues are sorted in a way that saves the ordering so that the paired eigenvectors can be sorted in the same order:
```int[] idxs = ArgSort(eigenVals); // to sort evecs
Array.Reverse(idxs); // used later
Array.Sort(eigenVals);
Array.Reverse(eigenVals);
Console.WriteLine("Eigenvalues (sorted): ");
VecShow(eigenVals, 4, 9);```
For the demo data, the sorted eigenvalues are (3.1167, 1.1930, 0.1868, 0.0036). The program-defined ArgSort() function doesn't sort but returns the order in which to sort from smallest to largest. The eigenvalues are sorted from smallest to largest using the built-in Array.Sort() function and then reversed to largest to smallest using the built-in Array.Reverse() function.
The eigenvectors are sorted using these statements:
```eigenVecs = MatExtractCols(eigenVecs, idxs); // sort
eigenVecs = MatTranspose(eigenVecs); // as rows
Console.WriteLine("Eigenvectors (sorted as rows):");
MatShow(eigenVecs, 4, 9, false);```
The MatExtractCols() function pulls out each column in the eigenvectors, ordered by the idxs array, which holds the order in which to sort from largest to smallest. The now-sorted eigenvectors are converted from columns to rows using the MatTranspose() function only because that makes them a bit easier to read. The eigenvectors will have to be converted back to columns later. The point is that when you're looking at PCA documentation or working with PCA code, it's important to keep track of whether the eigenvectors are stored as columns (usually) or rows (sometimes).
Note: By pure coincidence, the Eigen() function for the nine-item demo data returns eigenvalues and eigenvectors that are already sorted. In most cases you must explicitly sort the eigenvalues and eigenvectors as demonstrated.
Computing the Variance Explained
The variance explained by each of the four eigenvalue-eigenvector pairs is computed from the sorted eigenvalues:
```Console.WriteLine("Computing variance explained: ");
double[] varExplained = VarExplained(eigenVals);
VecShow(varExplained, 4, 9);```
For the demo data, the sum of the sorted eigenvalues is 3.1167 + 1.1930 + 0.1868 + 0.0036 = 4.5000. The variance explained values are:
```3.1167 / 4.5000 = 0.6926
1.1930 / 4.5000 = 0.2651
0.1868 / 4.5000 = 0.0415
0.0036 / 4.5000 = 0.0008
------
1.0000```
Loosely speaking, this means that the first eigenvector captures 69.26 percent of the information in the source data. The first two eigenvectors capture 0.6926 + 0.2651 = 95.77 percent of the information. The first three eigenvectors capture 0.6926 + 0.2651 + 0.0415 = 99.92 percent of the information. All four eigenvectors capture 100 percent of the information.
Computing the Transformed Data
After computing and sorting the eigenvalues and eigenvectors from the covariance matrix of the normalized data, the next step in PCA is to use the sorted eigenvectors to compute the transformed data:
```Console.WriteLine("Computing transformed data (all components): ");
double[][] transformed = MatProduct(stdX, MatTranspose(eigenVecs));
Console.WriteLine("Transformed data: ");
MatShow(transformed, 4, 9, true);```
The transformed data is the matrix product of the normalized data and the eigenvectors. Because the eigenvectors were converted to rows for easier viewing, they must be passed as columns using the MatTranspose() function.
The transformed data is:
```[0] -2.0787 0.6736 -0.0425 0.0435
[1] -2.2108 -0.2266 -0.1065 -0.0212
[2] -2.0071 1.2920 0.1883 0.0025
[3] 1.1557 0.3503 -0.9194 -0.0785
[4] -0.7678 -2.6702 0.0074 0.0115
[5] 0.7029 -0.0028 0.4066 -0.0736
[6] 1.8359 0.2191 0.6407 -0.0424
[7] 1.3476 0.1482 -0.0201 0.0621
[8] 2.0224 0.2163 -0.1545 0.0960```
The demo program concludes by reconstructing the original source data from the transformed data. First the normalized data is reconstructed like so:
```Console.WriteLine("Reconstructing original data: ");
double[][] reconstructed = MatProduct(transformed, eigenVecs);```
Notice that reconstruction uses eigenvectors as rows rather than columns. Once again, the eigenvectors columns-vs-rows issue can be tricky when working with PCA. Next, the original source data is recovered by multiplying by the standard deviations that were used to normalize the data, and then adding the means that were used:
```for (int i = 0; i < reconstructed.Length; ++i)
for (int j = 0; j < reconstructed[0].Length; ++j)
reconstructed[i][j] = (reconstructed[i][j] * stds[j]) + means[j];
MatShow(reconstructed, 4, 9, 3);```
Reconstructing the original source data from the transformed data is a good way to check that nothing has gone wrong with the PCA.
Wrapping Up
Principal component analysis can be used in several ways. One of the most common is to create a graph of data that has more than two columns. For the Iris data with four columns, there's no easy way to create a graph. But by applying PCA and then using the first of two transformed columns as the x-value and the second column as the y-value, a graph clearly reveals that species setosa items are quite different from the somewhat similar versicolor and virginica items.
Another way to use PCA is for anomaly detection. The idea is to use the first n columns of the transformed data, where the n columns capture roughly 60-80 percent of the variability, to approximately reconstruct the original source data. Data items that don't reconstruct accurately are anomalous in some way and should be investigated.
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What tech should a .NET coder use for a new Windows desktop app when presented with a dizzying array of options that include .NET MAUI, WinUI, WinForms, WPF, UWP, Blazor and so on? | 5,213 | 18,925 | {"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} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.798283 |
https://imperialcasinohotel.com/free-sky-vegas.html | 1,566,247,160,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314959.58/warc/CC-MAIN-20190819201207-20190819223207-00436.warc.gz | 508,292,025 | 9,379 | Assuming that the possible bets are all understood, Roulette is essentially a trivially simple game to play. For each turn, once all bets have been placed using coloured chips to distinguish each player, the croupier halts betting, spins the wheel, and rolls the ball in the opposite direction. When the ball comes to a halt in one of the slots, the croupier announces the result, collects all losing bets and pays out the winner's profits.
Roulette was probably the first true casino game and was likely introduced in the middle ages. Noblemen attempted to beat roulette with various roulette systems of play. If we had a time machine we could go back and tell them that the systems they invented and used, while fun, just couldn’t give them a true mathematical edge over the casino. Luckily the peasants didn’t play in casinos because, well, after all they were peasants.
Roulette was probably the first true casino game and was likely introduced in the middle ages. Noblemen attempted to beat roulette with various roulette systems of play. If we had a time machine we could go back and tell them that the systems they invented and used, while fun, just couldn’t give them a true mathematical edge over the casino. Luckily the peasants didn’t play in casinos because, well, after all they were peasants.
Roulette – in movies, on TV, and in literature, it’s often depicted as glamorous, exciting, and ridiculously lucrative for whichever character is playing. While winning big at the spinning wheel seems like a fantasy reserved for James Bond and other fictional high rollers, it is possible to minimize risk and win big at the same time by knowing the best roulette bets. It all comes down to the numbers, and how you want to play them.
When a player gets to the upper reaches of the Martingale what is he looking to accomplish? Very little. Using the above dollar amounts, even should the player bet \$640 or \$1280, he is looking to win a mere \$10. That is an awful lot to wager for a \$10 win. At every step of the traditional Martingale the payoff is merely that \$10. The untraditional Martingales are even worse! If you must allow yourself the Martingale then stick to the original.
French Roulette is almost the same as European Roulette, except that the betting board is laid out in a different manner and there are a few small rules differences. These rules, known as the “La Partage” rule and the “en prison” rule, are optionally used in some casinos. These rules do have a more significant effect on gameplay, so if you are going to a casino that offers this variation, we recommend clicking below for more comprehensive details.
#### So you’re ready to learn how to beat the house at roulette? Unfortunately, in the long run, the house is going to have an advantage, as is the case with all casino gambling. This does not mean you can’t be a winner and also does not mean you can’t employ a few strategies to give yourself a better chance of winning. Our experts present and analyze some of the most popular betting and strategy methods so you can decide if you’d like to employ them or not.
```A lot of the excitement for me though was just the surprise of not knowing what we were going to see till the day of! I don't see broadway shows that often and don't really keep up with what the big ones are so it's nice to not have to make the choice and just knowing that whatever I get will be great, as all the broadway shows are great! You do get to cross out four shows that you definitely don't want to see though, which is good.
```
These are the bets we recommend for beginners who want to get more comfortable with roulette. (This does not mean they aren’t great bets for seasoned players, as well.) Instead of betting on specific numbers or groups of numbers, you are betting on what we have termed “the characteristics” of the number. This would include betting on the color of the number or on the evenness or oddness of the number. These bets always pay even money and are as simple as they sound. If you bet black and a black number rolls, you win. If you bet even and an even number rolls, you win. It’s that easy!
Your first bet is \$10 (or whatever your normal bet size is) on one of the even-money proposition wagers. If you lose that bet, you go to \$20. Now, you sit out two spins; correct, you do not bet. After two sit-outs, on the next spin, you will increase your bet to \$40. If you lose that? You quit and go back to your original bet. You’ve lost \$70 on this sequence.
```We’ve put together a list of our favorite trusted online casinos here for you. Take a look around at each of the ones we have listed and find the one that fits your personality best. Each casino will have a bit of a different look and feel and will offer different bonus options and variations of roulette. That being said, any of the sites on this list would be a great place for you to get started with roulette or to pick back up right where you left off!
```
The only real drawbacks to live roulette deal with logistics and crowdedness. As it is such a popular game, you can expect most tables to be pretty packed most of the time, and especially on peak nights. This can make it challenging to get a seat and can overwhelm some people, as the tables will typically get quite crowded with people reaching over you to make bets.
So, you enjoy taking the wheel for a spin. We totally get it – it’s no secret that roulette is one of the most exciting and entertaining casino games to be found on the floor. Unfortunately, while it is thrilling, the odds for players are some of the worst of any game. Don’t despair, though – Planet 7 is here to break down the five most common roulette strategies that players around the world have been using to rake in the chips for years. While we recommend using all of these equally, it’s your job to read on and decide which method is the best roulette strategy for you.
Please Note: The Martingale is much like the iceberg that sunk the Titanic. The loss of six to eight hands in a row seems like a real longshot; but the fact is that anyone who has played roulette has seen streaks of red or black, odd or even, or high or low coming up with such frequency many times. Casinos put a cap on how much a person can bet so that such relatively short streaks can sink the Martingale player.
Money Maker Machine – Probably not the most appropriate name, but this is a a collection of automated software programs that are far more user configurable than any other software program. In some cases, the only limit to the systems these software programs can use is your imagination. These are the best software programs available, but that doesn’t mean they will win.
Although roulette is a game of chance you can give yourself a better chance of winning if you follow a strategy when it comes to what you bet on and how much. Newcomers will tend to concentrate on their lucky numbers and although that can work as part of a strategy – or just as a stroke of beginner’s luck – if you want to win more often than not you will have to have a better plan.
You can get your chips in different monetary denominations. When you hand the dealer your chips, he or she will ask what denomination you want to designate your chips. If you're at a \$5 minimum table, you could make them \$1 or you could make them \$100 (or anywhere in between). Once you've chosen, they'll place a chip on top of the rail, with a marker on top of that to indicate your colored chips' value.[5]
The odds of you winning will always be 50/50. So you have a 50% chance of LOSING \$1, and a 50% chance of WINNING \$0.50. You can’t just double bet size after losses, because then all you do is increase the amount you risk. Sure you may get lucky and win, but what happens if you lose? You’ll lose big. So there is no escaping the unfair payouts UNLESS you know which side of the coin is more likely to appear. Then you would be changing the odds of winning. And if you won much more often than 50% of the time, then the unfair payout wont matter as much.
Crossing off only four shows leaves you with a lot of chances to receive a show you've already seen, and it's hard not to feel like the selection process is rigged when you a) don't get to see the wheel actually spin and b) following along on social media shows that a majority of people see the same few shows: Kinky Boots, Waitress, Head Over Heels and The Prom. These are shows that you could easily pick up a discounted ticket to using BroadwayBox or purchasing last-minute on StubHub, if not directly from the box office, where tickets usually start around this price, anyway. For example, weekend tickets to The Prom start at \$69, to Kinky Boots, \$79, and to Head Over Heels, just \$49.
3.If you have the chance, use the "En Prison" rule which allows you to save the bet if the ball lands on 0 for one more spin. This means, if you placed a bet and the ball landed on 0 the bet will be rolled over to the next game and if your bet wins on the second try you will get your money back. If the ball lands on something else, then you will lose the bet.
Few games in a casino are as intimidating to a beginner as the Roulette table – a sea of numbers, colors, and apparent impossibilities. Isn’t it easier to just go press buttons and watch reels spin on a slots machine instead? Surprisingly, no. While the Roulette table may seem confusing at first glance, in fact it’s a wonderfully simple casino game to understand… once you get the hang of it, of course. The key is to learn how to bet properly.
If a 1:1 payout isn’t exciting enough to motivate you to lay down some chips, you can slightly up the ante and the payout to 2:1 by betting on the columns. By going with this second best roulette bet, you’re bringing your option of numbers down to a little under a third, but betting on a column still covers a significant chunk of the felt while doubling your winnings – a \$10 bet will win you \$20.
Go with the Fibonacci System for a low-risk, low-reward strategy. In this system, you place wagers only on the roughly 50/50 bets (such as odd/even), and you base your wagers on the Fibonacci numbers. If you lose in the first round, make your next wager the next number in the Fibonacci sequence in the second round. Keep advancing a number in the sequence until you win then, when you do win, go back 2 numbers in the sequence.[13]
Martingale Roulette System – This is a system that involves increasing bets after a loss until you get a win. It is commonly used on Red or Black but it can be applied all over the table if you have the correct progression. Not only have we gone through every bet, we have tables showing the overall loss at each stage and we’ve shown you the odds of going on a bad losing run i.e: 10 Reds in a row when you’re betting on Black.
```If you’re like most players, your strategy would be to use a trigger, then betting progression. A trigger is simply an event you wait to occur before betting. For example, the trigger may be wait for 3 REDS to spin in a row. Your bet would then be doubling bet size until you win. But again this wont work because the odds haven’t changed, the payouts are the same, and all you’re doing is making difference size bets on independent spins.
```
Roulette is one of the world’s favourite casino games for a reason. Deceptively simple and yet hard to master, it’s a game that reveals added layers of complexity the more you study it. In other words, if you think roulette involves little more than watching a little ball bounce its way around a wheel while randomly tossing chips at sections of the table, you don’t understand roulette. Sure, you can play it that way, but discerning players know better than that. Discerning players appreciate that there’s a right way and a wrong way to approach roulette.
They are equal in that there are 18 red and 18 black numbers on a roulette wheel. The 0 and 00 are both green numbers. However, there are 10 odd red numbers and 8 even red numbers. Also,the opposite is true in that there are 10 even black numbers and 8 odd black numbers. If you bet even and red or odd and black it seems you would have a slight edge as you would be covering 20 numbers instead of 18.
Hi, how about silverthorne publication, they seem to have different name system of games a few month all the time. for example, they come out with a system called checkmate roullett, iron roulette, attack baccarat, power baccarat, etc. I am on their email list and I got this all the time but I don’t know if they are just selling the system or is their system really work?
Though there are many “how to play roulette” books and guides on so-called perfect Roulette strategy, the truth is that there is no fool-proof method to win. However, this does NOT mean that players can’t learn how to bet in roulette effectively to increase their bankroll. We here at Planet 7 Online Casino want to teach you some easy to learn, beginner’s tips on how to play Roulette. Once you have done your homework, in just a few quick spins you’ll be betting and playing like a pro – and hopefully winning like one too. So, let’s get started and take a look at how at some of the best tips on how to play Roulette and win!
##### This roulette strategy is the equivalent of a YANS and it might even be more wild and crazy than that. It is based upon the assertion that a chaotic betting system can overcome the chaos of randomness. Two wrongs make a right; that kind of thing. Two wrongs don’t make a right and chaos versus chaos is just, well, chaos versus chaos, as you shall see.
If you care way more about the payout than you do about minimizing risk, your best option is to go with the straight roulette bet. This involves betting on a single number and hoping that Lady Luck is on your side. The risk is extremely high, but the payout – 35:1, \$350 for a \$10 bet – is pretty freakin’ sweet. Plus, this is the only category that includes the 0 and 00 tiles.
In other games the color of the chip denotes the denomination, but in Roulette the color denotes only which player the chip belongs to. Roulette chips can in fact be any denomination—\$1, \$5, \$25, etc. When you buy in, tell the dealer what denomination you want. He'll put a marker on his stack of chips that are the same color to note how much each of your chips is worth. Because roulette chips are non-denominational, you can't use them in other table games. When you're done playing roulette, give your chips to the dealer and she'll exchange them for regular, denominational chips.
Bonuses may sound great, but they always come with strings that ultimately benefit the casino. If you are a professional player, never accept a deposit bonus because it will limit what you can do. For example, you may have won a fortune on a wheel, then it gets changed by the casino. But you may not be able to withdraw funds until you have wagered a certain amount. And doing so may erode your profits.
A slightly different type of outside bets are those that pay 2-1 as the odds of winning are just 33%. The most popular version is on the Dozens, where players are supposed to bet on the first 12 numbers, the second or the third. The same goes for the Columns bet, with players being also expected to bet on 12 numbers, with the only difference being the distribution on the betting grid. All outside bet are clearly defined on the roulette table with specific places for each bet.
Trend betting on other propositions such as columns is somewhat different than trend betting on the even-money bets. This section will explain how to bet the other propositions which come in with a different set of problems for the player. But keep in mind that my trend-betting strategies are geared to losing you less when you play. They can’t give players an edge.
Trend betting on other propositions such as columns is somewhat different than trend betting on the even-money bets. This section will explain how to bet the other propositions which come in with a different set of problems for the player. But keep in mind that my trend-betting strategies are geared to losing you less when you play. They can’t give players an edge. | 3,572 | 16,211 | {"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-2019-35 | latest | en | 0.961374 |
https://www.topperlearning.com/answer/show-that-cos-38-cos-52-sin-38-sin-52-0/gdydzlss | 1,713,379,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00089.warc.gz | 946,760,531 | 55,531 | Request a call back
SHOW THAT cos 38°cos 52°- sin 38°sin 52° =0
Asked by debasreepartha | 23 Apr, 2021, 08:38: AM
Consider
cos 38°cos 52°- sin 38°sin 52°
= cos(38 + 52)°
= cos 90°
= 0
Hence, cos 38°cos 52°- sin 38°sin 52° = 0
Answered by Renu Varma | 23 Apr, 2021, 10:37: AM
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https://www.kidsacademy.mobi/printable-worksheets/answer-key/age-9/math/word-problems/ | 1,718,652,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00160.warc.gz | 766,037,561 | 68,519 | # Word Problems Worksheets for 9-Year-Olds
Unlock the world of math with our engaging Word Problems worksheets, tailored specifically for nine-year-olds! These learning online worksheets are designed to challenge young minds while making learning fun and interactive. Each worksheet is crafted to improve critical thinking and problem-solving skills through a variety of real-life scenarios. From mastering basic arithmetic to exploring more complex mathematical concepts, our worksheets offer a dynamic educational experience. Perfect for at-home learning or supplemental school activities, these worksheets will boost confidence and enhance academic performance. Dive into our Word Problems worksheets and watch your child's math skills soar!
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## Solving Problems: Number Lines Worksheet
Number lines are fun and help build number sense and reasoning. This worksheet uses them to help students count on to find a solution to an equation. They'll have fun counting with familiar numbers and 10-bases, while also working on algebraic reasoning.
Solving Problems: Number Lines Worksheet
Worksheet
## Counting With Drawings. Does It Make Sound? Worksheet
Introduce word problems with this simple worksheet. Ask your students to name the objects pictured, then read the word problem and ask them to circle the answer. With practice, it'll help them scale this challenging area of math.
Counting With Drawings. Does It Make Sound? Worksheet
Worksheet
## Circus Math Printable
Let's take the kids to the circus and combine literacy and numeracy skills! Our circus math printable worksheet will encourage creativity and higher-order thinking by connecting addition to real-life examples. Kids will learn to love math while they explore the fun, real-life applications of math. Ignite their passion today!
Circus Math Printable
Worksheet
## Police Subtraction Worksheet
Police protect citizens and do more in the community. Kids can learn numbers with this math worksheet: two subtraction problems in the downloadable PDF. Read the problems with your kids, then check the box next to the number sentences that are correct.
Police Subtraction Worksheet
Worksheet
## Holiday Lights Subtraction Worksheet
Before tackling the math problem, get your kids ready by asking them to share their favorite holiday and why. Then, look at the picture on the worksheet with them. Help them examine the numbers and problems, and select the number sentence that shows what's happening with each string of light. 80 words.
Holiday Lights Subtraction Worksheet
Worksheet
## Assessment 3 Math Worksheet
Ariel needs to buy yarn. Do your kids know what yarn is? If they've seen you knit, they will. Help them solve the two word problems to figure out how many coins Ariel needs to buy it.
Assessment 3 Math Worksheet
Worksheet
## Tricky Problems Worksheet: Part 1
Visualizing word problems can help students solve them. This fun PDF worksheet encourages students to create mental images using math facts, then check boxes to match the equations for basic addition problems. It's a great way to build confidence in solving math.
Tricky Problems Worksheet: Part 1
Worksheet
## Assessment 2 Math Worksheet
Using fun exercises and pictures, you can get kids excited about math. With practice and reminders of what they've learnt, they'll be solving simple addition and subtraction equations with ease. Read the word problems in the picture to them, then help them check the equation and choose the correct answer.
Assessment 2 Math Worksheet
Worksheet
This tracing sheet is great for kids to work on math skills and have fun too. Read the word problems to them, then help them trace the dotted lines to match the problem and number sentence to a picture. Not only is it educational, it's beautiful too - your kids will love learning about butterflies!
Worksheet
In the desert, scorching heat and little water mean animals must adapt to survive. Show kids pictures of these animals and teach their names. Then, read the word problems in the worksheet, and have them check the box next to the pictures that portray each story. 80 words
Worksheet
## Amusement Park Word Problems Worksheet
Engage kids with math by making it realistic and incorporating manipulatives or pics they like! This worksheet helps kids understand word problems. Read the problem, note numbers and key phrases like "in total," then count the pics to find the total!
Amusement Park Word Problems Worksheet
Worksheet
## Subtracting Socks Worksheet
Before beginning this exercise with your children, warm them up with a counting game. If math is not their favorite subject, use this worksheet. Help them read the two word problems, then use their fingers to count and subtract. Ask them to select the correct answer and check the box.
Subtracting Socks Worksheet
Worksheet
## Sightseeing 2-Step Word Problems Worksheet
Help your kids tackle word problems with this PDF! Read aloud two simple problems, have them create equations and solve, then check the box under the right answer. With practice, confusion will be a thing of the past.
Sightseeing 2-Step Word Problems Worksheet
Worksheet
## Making Bracelets to Sell Worksheet
Got kids who love crafting? This worksheet will be right up their alley! Let them use their creative and problem-solving skills to check equations and decide the correct total. It's just like making beads into a bracelet but on paper! Print out this downloadable PDF and get ready for a fun craft session.
Making Bracelets to Sell Worksheet
Worksheet
## Multiplication Facts: Assessment 3 Worksheet
Test your kid's maths skills with this easy to use worksheet! Help them check the box that matches the equation in the first part, then read each word problem and underline the right answers to the second part. Assess your child's muliplication knowledge and find out where they need extra help.
Multiplication Facts: Assessment 3 Worksheet
Worksheet
## Counting Spider Eyes Worksheet
Did you know different spiders have different eye numbers? Kids will love learning this fact and solving the spider word problems on this free worksheet! Word problems promote a deeper understanding of the concept, plus they get to add up the spiders’ eyes while solving addition problems with more than one addend. Fun and math all in one!
Counting Spider Eyes Worksheet
Worksheet
## Rainforest Animal Division Worksheet
Children can save the rainforest and learn about division with this worksheet! It helps kids understand that division involves creating equal groups of specific numbers, and with its visual representation, they can grasp the concept more easily. They'll have fun learning about their place in the world and their role in protecting the environment.
Rainforest Animal Division Worksheet
Worksheet
## Counting Seedlings Worksheet
Understanding math word problems is key. Multiple steps can prove challenging - this free worksheet provides one-to-one picture representation to help kids solve multi-step addition word problems. Strengthen addition skills by choosing the matching picture to the answer.
Counting Seedlings Worksheet
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## Mass Word Problems Worksheet
This worksheet challenges kids with tricky mass word problems. It offers valuable practice with grams. Through division and subtraction, children gain experience handling mass.
Mass Word Problems Worksheet
Worksheet
## Enrichment -2 Step Word Problems Worksheet
Help your students learn math easier and faster with this colorful worksheet. Read the word problem and then guide them in checking the correct equation and finding the answer. Your students will benefit from the extra help, as they work through new concepts each day.
Enrichment -2 Step Word Problems Worksheet
Worksheet
## How long is the Trip? Time Word Problems Worksheet
Get your kids in the mood for learning with fun questions about trips. Point at each picture and ask them to identify the mode of transport. Read the word problems, help them solve and check the answers.
How long is the Trip? Time Word Problems Worksheet
Worksheet
## Sums and Differences Within 1 - Assessment 1 Worksheet
Help your child understand school topics better with your assistance. For math, go over their worksheet together and discuss the instructions and text. Solve the questions and check the right equation. This will take their learning to the next level.
Sums and Differences Within 1 - Assessment 1 Worksheet
Worksheet
## Grocery Shopping Subtraction Worksheet
Take your kids on a virtual grocery trip with this worksheet! Help them read the two word problems and use the pictures to solve them. Then, have them circle the correct answers. It's an easy, fun way to get kids comfortable with math while also getting them excited about grocery shopping!
Grocery Shopping Subtraction Worksheet
Worksheet
## Instrument Subtraction Worksheet
Encourage kids to practice their addition and subtraction by having them identify the musical instruments on the worksheet, then read the subtraction problems and check the box with the matching number sentences.
Instrument Subtraction Worksheet
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Learning Skills
The Value of Worksheets on Word Problems for 9-Year-Olds: Harnessing the Power of Learning Online Printables
As children approach the pivotal age of nine, their cognitive and problem-solving skills are rapidly developing. At this stage, it is crucial to introduce educational tools that challenge their thinking and enhance their learning capabilities. One of the most effective resources available to achieve this goal are worksheets on word problems, particularly those accessed through learning online printables. These materials are not only convenient but are also brilliantly adapted to aid in the holistic development of children.
Word problems in mathematics are a critical component that help children transition from rote arithmetic to more complex mathematical concepts. Unlike straightforward calculation problems, word problems require the solver to comprehend a text-based scenario and then apply mathematical logic to solve it. This dual demand on language and numerical skills is what makes word problems particularly beneficial for nine-year-olds, as they are at a developmental stage where they are enhancing both their linguistic and numerical abilities.
The use of learning online printables for word problems offers numerous advantages. Firstly, these resources are readily accessible and can be easily integrated into any learning environment, whether in the classroom or at home. This accessibility allows children to practice extensively and at their own pace, leading to a better grasp of the concepts involved.
Secondly, online printables are often updated with new and diverse problems that reflect real-life scenarios. This keeps the learning material fresh and engaging for children, preventing the monotony that can sometimes be associated with traditional worksheets. It also provides them with a sense of relevance, as they can see the practical application of the math they are learning.
Moreover, learning online printables are customizable. Educators and parents can choose worksheets that target specific areas where their child might need more practice, or select problems that cater to a child’s particular interests, thereby increasing motivation and engagement. For instance, a child interested in space can solve word problems related to distances between planets, incorporating their interests into their learning process.
Additionally, solving word problems enhances critical thinking and reasoning skills. As children analyze the problem, identify what is being asked, and figure out what information is necessary to solve it, they are engaging in a high level of cognitive activity. These skills are transferable and will benefit them across various subjects and real-world situations.
Furthermore, word problems help in developing perseverance. Some word problems can be challenging, and finding solutions requires effort and persistence. Learning to tackle these problems can boost a child’s confidence and resilience, qualities that are essential for academic and personal success.
Importantly, the interactive nature of learning online printables can also foster a collaborative learning environment. Many platforms allow children to work in groups or share their results and methods with peers or teachers online. This encourages discussion and debate, critical components of deep learning, where children learn to articulate their thought processes and listen to alternative methods or solutions. Such interactions can lead to a deeper understanding and a more rounded approach to solving problems.
In addition, the digital nature of learning online printables enables the incorporation of multimedia elements such as videos, interactive simulations, and graphical representations, which can enhance understanding and retention of complex concepts. For example, a word problem about calculating area can include an interactive component where children can visually manipulate shapes to see how changes in dimensions affect the area. This kinesthetic component can help solidify abstract concepts through tangible interactions.
At the age of nine, children are also becoming more adept at using technology. Integrating learning with technology through online printables not only makes learning more engaging but also helps children develop digital literacy skills. These skills are crucial in a world where technology plays a significant role in most professional and personal activities.
Lastly, the flexibility of online printables allows for immediate feedback. Many platforms provide instant corrections and explanations for each problem, which is invaluable for learning. This immediate response helps children understand their mistakes right away, rather than embedding incorrect methods or concepts, which might be harder to unlearn later.
In conclusion, worksheets on word problems accessed through learning online printables offer a multifaceted educational tool that is well-suited for children aged nine. They not only enhance mathematical and linguistic skills but also foster critical thinking, digital literacy, and collaborative learning. By incorporating these resources into the educational curriculum, educators and parents can provide children with a robust foundation that prepares them effectively for future academic challenges and beyond. Engaging, accessible, and interactive, these printables are an excellent way to keep young learners motivated and enthusiastic about their educational journey. | 2,674 | 14,682 | {"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} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.88774 |
https://topic.alibabacloud.com/zqpop/multiplication-table-book_622076.html | 1,709,049,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.79/warc/CC-MAIN-20240227153053-20240227183053-00783.warc.gz | 586,785,168 | 19,666 | # multiplication table book
Read about multiplication table book, The latest news, videos, and discussion topics about multiplication table book from alibabacloud.com
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 1,966 | 8,630 | {"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-2024-10 | latest | en | 0.730344 |
https://www.studypool.com/discuss/215557/financial-problems-help-1?free | 1,481,423,456,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00178-ip-10-31-129-80.ec2.internal.warc.gz | 1,017,858,912 | 13,946 | Financial Problems Help
Business & Finance Tutor: None Selected Time limit: 1 Day
Based on the following information, if managers want to earn a 12% ROA, how much profit must the company generate?
Non-current assets
\$ 1,260,000
Aug 24th, 2014
ROA= annual net income/average total assets
12%=annual net income/1,260,000
Annual net income= 1260000*12/100
Annual net income=\$151,200
Aug 24th, 2014
...
Aug 24th, 2014
...
Aug 24th, 2014
Dec 11th, 2016
check_circle | 151 | 474 | {"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-2016-50 | longest | en | 0.826962 |
https://mobile.instructables.com/id/Platform-45-Electromechanical-Feedback-System/ | 1,555,649,544,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527135.18/warc/CC-MAIN-20190419041415-20190419063415-00519.warc.gz | 493,630,466 | 19,673 | # Platform 45: Electromechanical Feedback System
600
5
What is this thing?
It's the highest level thing I've posted on here so far and uses sub-systems from a bunch of the other things I've put up. But it's still not at the point of being a usable system, so this is largely documentation that I'm laying down for part of the larger system, and it will make sense in context later on. This thing synthesizes a bunch of stuff I've worked on over the years. It is a rotational motor, and also a vibrational motor, a parametric mechanical amplifier, a massager, a musical instrument, and a chaotic pendulum. Really! And I've tested all this, at least briefly, but it will take some time to get videos that actually make sense up and in the mean time I need to have some record of the current version, so here it is.
This is an Instructable that only really makes sense as part of a network of instructions I'm posting on different sub components of a much larger overall system. What is shown here uses the Hexagonal HDPE Trash Coil, which is already published on here, the Trash Coil Drive Circuit, also already published on here, and the Amp Without PCB, which will be published next. It also uses basic cardboard and tape techniques shown in my Action Geometry stuff but also all over the place on this site from many users.
You need 3 Trash Coils, on Trash Coil Driver, and two Amps Without PCB, and one box of thick corrugated cardboard, as well as a steel rod, a small rare earth magnet, and miscellaneous tools(see Sub instructions mentioned above).
## Step 1: Cut Out the Cardboard Shapes
There are a few possible layouts, what is shown above is four 3 inch squares, two half squares, and one half square scaled up by 2x from the other two. In the final version I show later on here, I use a single small square instead of the large triangle. So I end up using a total of 5 3" squares and 2 3" half square triangles.
## Step 2: Cut Out Slots for Wires From Coil
As shown, draw lines with marker, use magnet wire with insulation still on to wrap coils tightly to three squares, and feed the leads through the slots to the opposite side. Be sure the ends of the coil leads are stripped with a razor blade all around on all sides(you probably already did this).
## Step 3: Use Clear Tape or Duct Tape to Assemble 3d Structure
As shown, fold it all up and use tape to hold it together.
## Step 4: Add Electronic Circuits
Driver drives one coil, the 10,000X gain amp connects to the other of the "bottom" coils, and the amp connects to the driver to make a feedback loop. A cut USB cable is used for power. Third coil connects to a 100X voltage gain op amp without PCB which goes through a DC block in the form of a 100 uF series capacitor on each of the leads going to an audio jack which goes to USB powered PC speakers.
## Step 5: Test and Deploy, Make More, Spread
I still don't have a reliable way to get the coil orientation right, so it's 50/50 if you get it right the first time and if you don't get oscillations, swap one coil orientation(doesn't matter which one) and try again.
If this makes little sense, wait for more details to come out both of sub-systems of this and of super-systems that use it to do things that are actually useful. Right now it's just a toy and a demo.
youtube video of noise making machine.
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• ### Hand Sewing Class
8,254 Enrolled | 794 | 3,406 | {"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-2019-18 | latest | en | 0.945968 |
http://stackoverflow.com/questions/3230122/big-oh-vs-big-theta/3230154 | 1,406,121,412,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997878430.52/warc/CC-MAIN-20140722025758-00086-ip-10-33-131-23.ec2.internal.warc.gz | 150,577,214 | 21,472 | # Big-oh vs big-theta [duplicate]
Possible Duplicate:
What is the difference between Θ(n) and O(n)?
It seems to me like when people talk about algorithm complexity informally, they talk about big-oh. But in formal situations, I often see big-theta with the occasional big-oh thrown in. I know mathematically what the difference is between the two, but in english, in what situation would using big-oh when you mean big-theta be incorrect, or vice versa (an example algorithm would be appreciated)?
Bonus: why do people seemingly always use big-oh when talking informally?
-
## marked as duplicate by Moron, polygenelubricants, Dean J, sdcvvc, Bart KiersJul 12 '10 at 18:30
Here - "bonus" would be bounty .) – Eimantas Jul 12 '10 at 16:15
This has got to be a dupe! – Aryabhatta Jul 12 '10 at 16:21
I am voting to close. – Aryabhatta Jul 12 '10 at 16:39
It's not quite a duplicate because he's also asking the social question of why the wrong one is used informally. – Greg Kuperberg Jul 12 '10 at 16:46
@Greg: Which makes it subjective :-) and probably argumentative. – Aryabhatta Jul 12 '10 at 16:54
Big-O is an upper bound.
Big-Theta is a tight bound, i.e. upper and lower bound.
When people only worry about what's the worst that can happen, big-O is sufficient; i.e. it says that "it can't get much worse than this". The tighter the bound the better, of course, but a tight bound isn't always easy to compute.
### Related questions
The following quote from Wikipedia also sheds some light:
Informally, especially in computer science, the Big O notation often is permitted to be somewhat abused to describe an asymptotic tight bound where using Big Theta notation might be more factually appropriate in a given context.
For example, when considering a function `T(n) = 73n`3`+ 22n`2`+ 58`, all of the following are generally acceptable, but tightness of bound (i.e., bullets 2 and 3 below) are usually strongly preferred over laxness of bound (i.e., bullet 1 below).
1. `T(n) = O(n`100`)`, which is identical to `T(n) ∈ O(n`100`)`
2. `T(n) = O(n`3`)`, which is identical to `T(n) ∈ O(n`3`)`
3. `T(n) = Θ(n`3`)`, which is identical to `T(n) ∈ Θ(n`3`)`
The equivalent English statements are respectively:
1. `T(n)` grows asymptotically no faster than `n`100
2. `T(n)` grows asymptotically no faster than `n`3
3. `T(n)` grows asymptotically as fast as `n`3.
So while all three statements are true, progressively more information is contained in each. In some fields, however, the Big O notation (bullets number 2 in the lists above) would be used more commonly than the Big Theta notation (bullets number 3 in the lists above) because functions that grow more slowly are more desirable.
-
"a tight bound is sometimes harder to compute", and often useless too, which answers the "bonus" question ;) – Alexandre C. Jul 12 '10 at 16:19
Nitpick: We can't talk about 'matrix multiplication' being O(N^3). Perhaps you should throw in the words 'best known algorithm' in there. – Aryabhatta Jul 12 '10 at 16:38
@Moron - why can't we? saying matrix multiplication is `O(n^3)`, `O(n^4)` and any `O(n^k)` for `k > 3` wouldn't be wrong. – IVlad Jul 12 '10 at 16:48
@IVlad: What I meant is: matrix multiplication is a concept, not an algorithm. Since it was kind of clear from the context, I just said 'nitpick' :-) – Aryabhatta Jul 12 '10 at 16:50
@Moron - oh, my bad for not getting it :). – IVlad Jul 12 '10 at 16:52
show 1 more comment
I'm a mathematician and I have seen and needed big-O, big-Theta, and big-Omega notation time and again, and not just for complexity of algorithms. As people said, big-Theta is a two-sided bound. Strictly speaking, you should use it when you want to explain that that is how well an algorithm can do, and that either that algorithm can't do better or that no algorithm can do better. For instance, if you say "Sorting requires Θ(n(log n)) comparisons for worst-case input", then you're explaining that there is a sorting algorithm that uses O(n(log n)) comparisons for any input; and that for every sorting algorithm, there is an input that forces it to make Ω(n(log n)) comparisons.
Now, one narrow reason that people use O instead of Ω is to drop disclaimers about worst or average cases. If you say "sorting requires O(n(log n)) comparisons", then the statement still holds true for favorable input. Another narrow reason is that even if one algorithm to do X takes time Θ(f(n)), another algorithm might do better, so you can only say that the complexity of X itself is O(f(n)).
However, there is a broader reason that people informally use O. At a human level, it's a pain to always make two-sided statements when the converse side is "obvious" from context. Since I'm a mathematician, I would ideally always be careful to say "I will take an umbrella if and only if it rains" or "I can juggle 4 balls but not 5", instead of "I will take an umbrella if it rains" or "I can juggle 4 balls". But the other halves of such statements are often obviously intended or obviously not intended. It's just human nature to be sloppy about the obvious. It's confusing to split hairs.
Unfortunately, in a rigorous area such as math or theory of algorithms, it's also confusing not to split hairs. People will inevitably say O when they should have said Ω or Θ. Skipping details because they're "obvious" always leads to misunderstandings. There is no solution for that.
-
"O when they should have said Ω" - whoa! That's a serious error! Are you sure lots of people actually do this? (disclosure: I have done it on stackoverflow by mistake). – polygenelubricants Jul 12 '10 at 17:11
I don't know about "lots", but sure, I've seen it and I've also done it (at least informally). I agree that it's often unacceptably confusing, and it should certainly be avoided in a research paper. But informally, hey, we're human and we rely on our ability to repair error. – Greg Kuperberg Jul 12 '10 at 17:17
+1: I agree. @polygenelubricants: I have seen a lot of people talk about O when they mean Omega or even theta. Perhaps you should take a look at one of the deleted answers here: stackoverflow.com/questions/3230104/… – Aryabhatta Jul 12 '10 at 17:52
Because my keyboard has an O key.
It does not have a Θ or an Ω key.
I suspect most people are similarly lazy and use O when they mean Θ because it's easier to type.
-
Time for a keyboard upgrade! – Roger Pate Jul 13 '10 at 22:22
Because there are algorithms whose best-case is quick, and thus it's technically a big O, not a big Theta.
Big O is an upper bound, big Theta is an equivalence relation.
-
One reason why big O gets used so much is kind of because it gets used so much. A lot of people see the notation and think they know what it means, then use it (wrongly) themselves. This happens a lot with programmers whose formal education only went so far - I was once guilty myself.
Another is because it's easier to type a big O on most non-Greek keyboards than a big theta.
But I think a lot is because of a kind of paranoia. I worked in defence-related programming for a bit (and knew very little about algorithm analysis at the time). In that scenario, the worst case performance is always what people are interested in, because that worst case might just happen at the wrong time. It doesn't matter if the actually probability of that happening is e.g. far less than the probability of all members of a ships crew suffering a sudden fluke heart attack at the same moment - it could still happen.
Though of course a lot of algorithms have their worst case in very common circumstances - the classic example being inserting in-order into a binary tree to get what's effectively a singly-linked list. A "real" assessment of average performance needs to take into account the relative frequency of different kinds of input.
-
Bonus: why do people seemingly always use big-oh when talking informally?
Because in big-oh, this loop:
``````for i = 1 to n do
something in O(1) that doesn't change n and i and isn't a jump
``````
is `O(n), O(n^2), O(n^3), O(n^1423424)`. big-oh is just an upper bound, which makes it easier to calculate because you don't have to find a tight bound.
The above loop is only `big-theta(n)` however.
What's the complexity of the sieve of eratosthenes? If you said `O(n log n)` you wouldn't be wrong, but it wouldn't be the best answer either. If you said `big-theta(n log n)`, you would be wrong.
-
The above loop is only big-theta(n) however: nope. It would be if there was a theta inside the loop. – Alexandre C. Jul 12 '10 at 16:37
@Alexandre C. - why? `O(1)` means a constant number of operations. Assuming the `something` isn't a jump, why isn't it `big-theta(n)`? – IVlad Jul 12 '10 at 16:41
@IVlad: because 0 is O(1). – Alexandre C. Jul 12 '10 at 16:42
@Alexandre C. - so? even if `something` is `nothing`, `i` still gets incremented `n` times. – IVlad Jul 12 '10 at 16:45
@IVlad: okay, I got sucked by my mathematical training ;) – Alexandre C. Jul 12 '10 at 16:54
I have seen Big Theta, and I'm pretty sure I was taught the difference in school. I had to look it up though. This is what Wikipedia says:
Big O is the most commonly used asymptotic notation for comparing functions, although in many cases Big O may be replaced with Big Theta Θ for asymptotically tighter bounds.
I don't know why people use Big-O when talking formally. Maybe it's because most people are more familiar with Big-O than Big-Theta? I had forgotten that Big-Theta even existed until you reminded me. Although now that my memory is refreshed, I may end up using it in conversation. :)
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So, people talk about Big O because most people talk about Big O unless they talk about Big Theta? Not exactly helpful. – Chris Lively Jul 12 '10 at 16:19
@Chris true. Wikipedia seems contradictory on the topic; the two seem to be semantically different, if subtly. – Vivin Paliath Jul 12 '10 at 16:20 | 2,575 | 9,930 | {"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-2014-23 | latest | en | 0.928092 |
http://www4.geometry.net/basic_m/math_general_activities_page_no_5.html | 1,558,840,403,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00439.warc.gz | 352,274,267 | 10,854 | Home - Basic_M - Math General Activities
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12. Scrabble Fun: Math All Around Us (Preschool Level)
13. Triangles (Shapes in Math, Science and Nature) by Catherine Sheldrick Ross, 1994-06-30
14. Fraction Jugglers: Game and Work Book and Math Game Cards by Ruth Bell Alexander, Carl Martin, et all 2001-07-01
lists with details
1. Netn.net/lpage Math
School Hub (Homework, Test Preparation, general Research) IMO Interactive mathematics OnLine (Algebra, Geometry, Trigonometry) King s math activities (K-12
http://www.netn.net/27113.htm
2. For Better Math Grades Try Using A New Mind Set Of Goals.
wish to keep in mind SOME general GOALS that may help you focus YOUR THINKING ENERGY while you study a math course. Add more activities or general goals by
http://www.jug.net/wt/mgoals.htm
3. PlaneMath: Matrix Of Math Standards And Aeronautics Topics
Standards and Aeronautics Topics Covered in Planemath activities. of Teachers of mathematics (NCTM) math Standards, as as two or more general aeronautics topics
http://www.planemath.com/matrix.html
Extractions: Text Only 4. Matrix of Math Standards and Aeronautics Topics Covered in PlaneMath Activities Each of the lessons included in PlaneMath represent five or more of the National Council of Teachers of Mathematics (NCTM) Math Standards, as well as two or more general aeronautics topics. To help you plan your use of PlaneMath to teach or reinforce specific concepts, we have included the following matrices indicating the math standards and aeronautics topics that correspond to each lesson. 4th Grade Activities: Applying Flying (9 activities) NCTM Process Standards NCTM Content Standards Aeronatical Content ... Aeronatical Content
4. Wiley Europe::Math Games: 180 Reproducible Activities To Motivate, Excite, And C
WileyEurope Education K12 general K-12 math Games 180 Reproducible activities to Motivate, Excite, and Challenge Students, Grades 6-12.
http://www.wileyeurope.com/WileyCDA/WileyTitle/productCd-0787970816.html
5. Mathfest 2001 CPS - Making General Math/Precalc Courses Of Service To Mathematic
in the quantitative reasoning component of a universitys general education program In this presentation, we will describe some handson activities and group
http://www.frc.mass.edu/smabrouk/Maa/Mathfest_2001/Cps/
Extractions: Session Organizer and Moderator General mathematics and precalculus courses, service courses to colleges/universities, can serve mathematics departments by inspiring students and by providing the skills and the mathematical sophistication necessary to enable students to pursue mathematics as a major. Exposure to thought provoking puzzles and proofs, especially proofs without words, and the use of projects, applications to other disciplines, especially those involving real data, and innovative assignments, including those involving creative writing, can help students to view mathematics as real, creative, and enjoyable. This session invites papers describing efforts to use general mathematics and precalculus courses to attract students to study mathematics. Participants are encouraged to discuss course changes made to improve student attitudes and to attract students to study mathematics as well as assignments/projects, demonstrations, and activities used to stimulate interest in mathematics. Of particular interest are professor/student reactions, the ease/difficulty with which changes are made, and the overall effect of course changes. 3:15 p.m.
6. Math On The Web: Mathematics By Topic
activities of the CAM laboratory center around the SAMPO for items listed by classification under general 00 World; PI; PI, again; Sci.math FAQ Sections
http://www.ams.org/mathweb/mi-mathbytopic.html
Extractions: Mathematics by Topic lists some topic keywords in the Table of Contents. In each section are links to electronic journals, preprints, Web sites and pages, databases and other pertinent material in the corresponding field. There is also a page of Materials Organized by Mathematical Subject Classification . This page was developed from that prepared at the University of Tennessee, Knoxville Math Archives Department of Actuarial Mathematics and Statistics , Heriot-Watt University (UK) Balducci's Actuarial Home Page (Calgary, CA)
7. Education World® : Math Center
reinforce a variety of math skills. Included activities that teach critical thinking, math facts and computation, decimals, more.
http://www.education-world.com/math/
8. Quia - Mathematics - Top 20 Activities
Games and quizzes for learning mathematics mathematics top 20 activities. Play math Journey! Practice addition, subtraction, multiplication, division, and Fractions, Decimals, Percentage .
http://www.quia.com/dir/math
Extractions: Quia users have created activities in the following topics. These activities have been created by Quia's subscribers and are of varying quality. Quia has not reviewed these, and some activities may contain inaccuracies. If you are a Quia subscriber, you may copy any of these activities and modify them for your own use. Create an activity Add your own activity Search Options (Choose from one of the following menus.) The 20 most popular activities in this topic are listed below. To list many more , select All activities by title or All activities by author All Activities ... Select ... top 20 only by title by author Subcategory ... Select ... Algebra I Algebra II Algebra Terminology Calculus Calendar Divisibility Rules and Prime Numbers Egyptian Numerals Elementary School Fractions, Decimals, Percentage General Geometry Grade 5 Review Graduation Test Prep Math A prep Math facts Mental Arithmetic Middle School number properties Position and Movement Pre Calculus Pre-Algebra RCT Math Practice RMP Roman Numerals Time Trigonometry Textbook/Resource ... Select ... AMSCO Cord Algebra I Everyday Mathematics Glencoe Algebra 1 Glencoe Algebra 2 Glencoe Course 1 Glencoe Geometry Investigations Math facts the fun way MathScapes None Prentice Hall Progress In Matematics Saxon Math Scotts Foresman Directory
9. Numeracy Teaching Ideas
general Numeracy activities. Estimation, 5 11, How to introduce the topic of estimation to children in a fun and practical way. Puzzle Display, 5 - 11.
http://www.teachingideas.co.uk/maths/contents.htm
Extractions: Data Handling Activities Name of Activity Age Range Description Name of Activity Age Range Description General Numeracy Activities Estimation How to introduce the topic of estimation to children in a fun and practical way. Puzzle Display Make a puzzle display to encourage the children in your class/school to use their minds. Guidance, ideas and example puzzles (with links to other puzzle sites) are given. Sports Ladders A fun mathematical activity, which takes advantage of children's interest in sport. Frog in the Box! A fun team game, which can be used to test many mathematical concepts. Yes / No Cards A simple way to elicit answers from all children in your class.
10. Mathematics Lessons That Are Fun! Fun! Fun!
mathematics Lessons that are Fun! Fun! Fun! This is a collection of over 20 fun and challenging math activities created by Cynthia Lanius, the executive director of the Center for Excellence and
http://rdre1.inktomi.com/click?u=http://math.rice.edu/~lanius/Lessons/index.html
11. FCPS Math Websites
LP, 3.10, math=Art This page, from Teachnet.com Lesson Plans for math (general Ideas), suggests an art activity using students own drawings, to represent and
http://www.fcps.k12.va.us/DIS/OEIAS/math/websitesk3.htm
Extractions: Problem Solving/Application Number Concepts/Number Sense Operations/Computation ... PreAlgebra Acknowledgments The Elementary Math Resource Guide for Fairfax County Public Schools was completed by a collaborative team consisting of teachers (Patti Abernathy and Bonnie Glazewski), School Based Technology Specialists (Carolyn Fritz and Martha Short-Smith), Instructional Technology Specialists (Karen Gerstner, John Stroud, and Wanda Walters), and Elementary Math Resource Teacher (Gail Woolwine).
Front page Lesson Plans math GeneralGeneral Capacity Song. In Your Face math Facts. Fishing for Facts Your Clock is a Teaching Assistant. Keeping math Problems Straight. Newspaper
http://www.teachnet.com/lesson/math/matgen.html
13. The Math Forum - Math Library - Lesson Plans/Activities
The math Forum's Internet math Library is a comprehensive catalog of Web sites and Web pages relating to the study of mathematics. This page contains sites relating to Lesson Plans and activities .
http://mathforum.com/library/resource_types/lesson_plans
Extractions: By Topic: Arithmetic Lesson Plans Collections Algebra Lesson Plans Collections Geometry Lesson Plans Collections Pre-Calculus Lesson Plans Collections Calculus Lesson Plans Collections Prob/Stat Lesson Plans Collections Discrete Math Lesson Plans Collections By Level: Elementary Lesson Plans Collections PreK-2 Lesson Plans Collections Lesson Plans Collections Middle School Lesson Plans Collections High School Lesson Plans Collections There are so many items in this category that we recommend using the keyword search and math topic or grade level menus to narrow down your results. What is Infiltration? Math objectives: 1. Define the term "percent." 2. Solve problems using percentages. Energy objectives: 1. Define infiltration. 2. Describe the major areas of infiltration in the home. An activity guide from the Energy Conservation ...more>> 100th Day of School Celebration - Loogootee Community Schools, Loogootee, Indiana
14. Pi Day
Pi Quilts (general) I got this idea from Pat I just got this idea while typing the above activity. Students are graded on math information, technical stuff (ie
http://www.mathwithmrherte.com/pi_day.htm
Extractions: Pi Day Pi Links Since April 13, 2002 Pi Day March 14th...Celebrate the Circle, Pi and the imagination and creativity of our students! Purpose: Pi Day is a day to celebrate mathematics in your school. It gives us the perfect springboard to allow our students to have fun while investigating mathematics concepts, being creative and even a little silly. If you have any ideas email them to me. I will eventually be attaching some of the materials to each activity so you can click onto it for more ideas. Pi Day Activities! Name That Circle! (9th -11th Grade) Students are shown a graph of a circle and the student that can correctly give it's equation wins! You can run it as a team competition or one on one. Be certain to go over the general equation of a circle first. For grade 9 or students who have not studied this keep all circles in the first quadrant. Examples Pi Digit Distribution (Elementary - HS) The younger students can look at the first 50 or 100 digits of pi and make a bar graph of the frequency of each digit. (How many times does appear?) Students enjoy this rich data source for bar graphs. HS students find the mean of the first 100 digits (use scientific calculators/graphing calculators) Create circle graphs based on the digits -excellent use of protractors, fractions, decimals and angles if done by hand. Use spreadsheets to examine the distribution of the digits. Examples Rolling Along...
15. Skewl Sites - Site Index - MATH
Figure This, A collection of 80 math challenges for families. Fun mathematics, Online math activities for K12. Fun mathematics, Online math activities for K-12.
http://www.skewlsites.com/math.htm
Extractions: MATH Primary 100th Day Activities 100th Day of School Website Over a hundred teacher made ideas to help you celebrate the big day across the curriculum. A+ Math Games, activities and flashcards to help students improve their math skills. AAA Math Hundreds of pages of basic math skills, including fractions, measurement, algebra, decimals, percent, money and just about every other math topic. Arithmetic "Hundreds of pages of interactive practice and math explanations." Arthur Books Kids love the Arthur books and now they can enjoy Arthur stories and activities online. Aunty Math "Math challenges for K-5 learners." Bonus.com Offers hundreds of quality activities that invite kids to play, colour, explore, learn and imagine. Bubbles Website (The) A colourful look at the science of bubbles. Come and have some good clean fun!
16. The Math Forum - Math Library - Activities
The math Forum's Internet math Library is a comprehensive catalog of Web sites and Web pages relating to the study of mathematics. This page contains
http://mathforum.org/library/ed_topics/methods_acts/
17. Math Extension Activities Grades 6 To 9
On the student s side, you will find flash movies that provide an overview of particular topics/concepts, online activities, printable materials, quizzes and
Extractions: On the teacher's side, you will find all of the activities that are on the student side but you will also have access to Saskatchewan curriculum objectives, teacher facilitated lessons and additional background information that will assist in teaching these topics of study. Site Requirements : This site is optimized for 800x600 and Internet Exporer 5.5+. To view the resources within, you will need Adobe Acrobat Reader (free) Macromedia Flash Player 6(free). Central iSchool Saskatchewan Learning
18. LearningSpace: Your Educational Tools On The Web
Look for science, reading, social studies, and math activities now, and be on the alert for music and language arts, coming soon. Find Grampy.
http://teach.fcps.net/trt20/results.asp?Keywords=Math
19. NCTM Illuminations
This activity includes software that allows you to sketch and This interactive math applet combines the features of a It can be used as a general tool or
http://illuminations.nctm.org/pages/68.html
Extractions: This page contains activities and resources that are appropriate for teachers at the 6-8 grade level for use as they reflect on, plan for, and implement Standards-based mathematics education in their classrooms. Click on the following links (or scroll down the page) to find the Table of Contents for each section. i-Maths are online, interactive, multimedia math investigations. All i-Maths are built around interactive math applets, and some also include video clips. There are two lengths of i-Maths: single-day and multi-day, as identified below. Complete i-Maths include student investigations, teacher notes, and answers. Some of the multi-day i-Maths also include video clips of teachers and students using the investigation, sample student work, assessment tasks, and related professional development activities. Student i-Maths contain just the student investigations. They are ready to use by students. e-Math Investigations are selected E-examples from the electronic version of NCTM's Principles and Standards for School Mathematics . Each E-example consists of an interactive math applet, suggestions for student activities, and some discussion for teachers. These investigations can be identified by the icon to the right.
20. VT Tech Ed: Tech Science Math Integration
The Overview section describes, in general terms, how the Activity will be coordinated among the technology, science, and math classes.
http://teched.vt.edu/html/ResTSMintg.html
Extractions: T he Technology, Science, Mathematics (TSM) Integration Project, funded by the National Science Foundation from 1991-1995, has developed a set of six TSM Connection Activities . Designed for integrated instruction at the middle school level, the TSM Connection Activities challenge students to design, construct, and evaluate solutions to technological problems. Students actively apply principles of science, mathematics, and technology as they go about the process of developing solutions to the problems posed in the TSM Connection Activities . The activities were designed to be delivered by technology, science, and mathematics teachers working in concert. The intent is to interest more middle school children in science and math by using highly motivating hands-on technology-based activities.
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Open Access
### Application of Mobius inversion in combinatorial problems
DOI: 10.23977/jnca.2020.050104 | Downloads: 72 | Views: 2536
Zehan Lin 1
#### Affiliation(s)
1 Guangzhou University (Guangzhou, Guangdong 510006)
Zehan Lin
#### ABSTRACT
Mobius inversion plays a very important part in number theory mathematics and can be used to solve many combinatorial problems. For some functions f(n), if it is difficult to find its value directly, but it is easy to find the sum of its multiples or divisors as g(n), then the calculation can be simplified through Mobius inversion to obtain the value of f(n).In this article, we provide a method to use Mobius inversion to solve some combinatorial problems efficiently with computer calculation.
#### KEYWORDS
Mobius inversion, combinatorial mathematics, Number Theory
#### CITE THIS PAPER
Zehan Lin. Application of Mobius inversion in combinatorial problems. Journal of Network Computing and Applications (2020) 5: 23-26. DOI: http://dx.doi.org/10.23977/jnca.2020.050104.
#### REFERENCES
[1] Chateauneuf A, Jaffray J Y. Some Characterizations of Lower Probabilities and Other Monotone Capacities through the Use of Mobius Inversion [J]. Mathematical Social ences, 1989, 17 (3): 263-283.
[2] Chen N X, Li M, Liu S J. PHONON DISPERSIONS AND ELASTIC-CONSTANTS OF NI3AL AND MOBIUS-INVERSION [J]. Physics Letters A, 1994, 195 (2): 135-143.
[3] FUJIMOTO, K. Some Characterization of the Systems Represented by Choquet and Multi-Linear Functionals throught the Use of Mobius Inversion [J]. International Journal of Fuzziness and Knowledge-based Systems, 1997, 5.
[4] Liu S J, Li M, Chen N X. Mobius transform and inversion from cohesion to elastic constants [J]. Journal of Physics Condensed Matter, 1993, 5 (26): 4381.
[5] Krot E. A note on mobiusien function and mobiusien inversion formula of fibonacci cobweb poset [J]. Mathematics, 2004, 44 (44): 39-44.
[6] Bayad, Abdelmejid, Navas. Mobius inversion formulas related to the Fourier expansions of two-dimensional Apostol-Bernoulli polynomials [J]. Journal of Number Theory, 2016. | 586 | 2,140 | {"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-2023-23 | latest | en | 0.713769 |
https://www.mail-archive.com/google-code@googlegroups.com/msg09944.html | 1,547,626,252,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657097.39/warc/CC-MAIN-20190116073323-20190116095323-00163.warc.gz | 861,235,792 | 3,639 | # [gcj] Re: Why is this wrong
```Seems to me like you are checking from left to right whether there is a chance
to swap in the main function? ```
```
If yes, then the problem is that to get the maximum reduction of damage, we
should reduce the energy of the shooting that has been charged most, therefore
it should be from right to left instead.
Another thing is that maybe before you swap all shooting and charging that
could be swapped in a loop, the damage has already been reduced to a
value<=threshold. Therefore I suggest you to check after each swap.
On Thursday, April 12, 2018 at 10:19:21 AM UTC-7, Sahil Shetty wrote:
> def change(rcodeC, i):
>
> holder = rcodeC[i + 1]
> rcodeC[i + 1] = rcodeC[i]
> rcodeC[i] = holder
> return rcodeC
>
> def damage(rcodeD, dmg = 0, count = 0):
>
> for i in rcodeD:
>
> if i == 'S': dmg += 2 ** count
> else: count += 1
>
> return dmg
>
>
> def main(shield, rcode, hacks = 0, count = 0):
>
> while damage(rcode) > shield:
>
> for i in range(len(rcode) - 1):
>
> try:
>
> if rcode[i] == 'C' and rcode[i + 1] != 'C':
>
> hacks += 1
> rcode = change(rcode, i)
>
> try:
>
> if all(e == 'C' for e in rcode[rcode.index('C'): -1]) ==
> True and damage(rcode) > shield: return 'IMPOSSIBLE'
>
> except ValueError:
>
> if all(e == 'S' for e in rcode[rcode.index('S'): -1]) ==
> True and damage(rcode) > shield: return 'IMPOSSIBLE'
>
> except IndexError: None
>
> return hacks
>
>
> test = int(raw_input("Test data: "))
> print '\n'
> while test < 1 or test > 100: test = int(raw_input("Between 1 and 100: "))
>
> for i in range(test):
>
> shield = int(raw_input("Shield strength: "))
> while shield < 1 or shield > (10 ** 9): shield = int(raw_input("Between 1
> and 10^9: "))
>
> rcode = raw_input("Robot code: ")
> while all(i == 'C' or i == 'S' for i in rcode) != True or len(rcode) > 30
> or len(rcode) < 2: rcode = raw_input("Must be greater than 2 and less than 1;
> only 'C' and 'S': ")
>
>
> print 'Case #' + str(i + 1) + ':', main(shield, list(rcode)), '\n\n\n'
--
You received this message because you are subscribed to the Google Groups | 700 | 2,312 | {"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-2019-04 | latest | en | 0.775361 |
http://www.cafemom.com/answers/1334265/Conception_due_date_calculation_question | 1,506,118,070,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689192.26/warc/CC-MAIN-20170922202048-20170922222048-00004.warc.gz | 400,994,100 | 28,373 | Join the Meeting Place for Moms!
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Conception & due date calculation question.
If a menstrual cycle started on April 15 is 28 days & being intimate took place,on Friday,April 26th.
If infact conception did happen,when would baby be due.. If due date,was on time?
Asked by Anonymous at 1:28 AM on Apr. 28, 2013 in Pregnancy
• They don't count it from the day you had sex, they count it from the date of your last menstrual period.
Try again.
Answer by PartyGalAnne at 1:32 AM on Apr. 28, 2013
• Yep the calculation is from last period. That is why so many are off by a week or two. If you know when you had sex and conceived couldn't you just add 9 months?? But know that the date will be different from what the doc says.
Answer by tntmom1027 at 1:35 AM on Apr. 28, 2013
• Jan. 15th would be the due date.
Answer by PMSMom10 at 2:17 AM on Apr. 28, 2013
• I think Jan 22nd is the due date. Count back 3 months from date of first day of last period and add 7 days. LOL. That's the way we did it in the old days.
Answer by kjrn79 at 9:07 AM on Apr. 28, 2013
• To find your range, enter the first day of your last menstrual period in the form below. Click on the calendar icon to select a date from a calendar. We calculate your due date as: Jan, 20 2014
Answer by 3libras at 10:04 AM on Apr. 28, 2013
• Based off the LMP conception would occur on or around April 29th. So chances are slim because most of the sperm from the 26th will have died off. Due date would be January 20th. I used to work an OB office and I still have the wheel they use for calculating this. LOL
Answer by 2autisticsmom at 10:08 AM on Apr. 28, 2013
• It depends on when you ovulate. If, for argument's sake, you ovulated on cycle day 14 (in this case April 28th) you would "conceive" on that day with the sperm that were still hanging around from when you had sex on the 26th. Although you very well could ovulate earlier or later than the 28th, every woman and every cycle is different. If you conceived on the 28th, however, your due date would be 38 weeks after that - January 19th, 2014. (About 40 weeks after the first day of your last period - which is usually what they go off of even though it's less accurate, since most women don't track ovulation.)
Answer by katinthehat8914 at 11:38 PM on Apr. 28, 2013
Join CafeMom now to contribute your answer and become part of our community. It's free and takes just a minute. | 697 | 2,497 | {"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-2017-39 | latest | en | 0.963392 |
http://moodle.tbaisd.org/mod/book/view.php?id=51594 | 1,550,905,828,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249490870.89/warc/CC-MAIN-20190223061816-20190223083816-00462.warc.gz | 175,620,377 | 8,977 | Definition
A rotation is a transformation that is done by turning a figure about a fixed point called the center of rotation. Rotations are often done in degree measures; for example, a 90o rotation. In general, geometric objects are rotated counterclockwise. An example of a rotation is shown below:
The segments drawn in red create an angle. The measure of that angle in degrees is the amount the pre-image is rotated to create the image.
Audio | 96 | 450 | {"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-2019-09 | longest | en | 0.895211 |
https://leanprover-community.github.io/archive/stream/113488-general/topic/why.20is.20quot.20an.20axiom.3F.html | 1,620,498,122,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00579.warc.gz | 374,820,744 | 7,117 | ## Stream: general
### Topic: why is quot an axiom?
#### Kevin Buzzard (Dec 21 2018 at 22:33):
Can't we just do it like this?
import tactic.interactive
-- Lean has the "quotient" function to make equivalence classes.
-- Here I try to figure out why it is needed.
universes u v
namespace xena
-- this doesn't work with Prop but do I care?
variable {β : Type u}
variable (r : β → β → Prop)
-- equiv reln closure of r
inductive requiv {β : Type u} (r : β → β → Prop) : β → β → Prop
| of_r (a b : β) : r a b → requiv a b
| refl (a : β) : requiv a a
| symm ⦃a b : β⦄ : requiv a b → requiv b a
| trans ⦃a b c : β⦄ : requiv a b → requiv b c → requiv a c
definition equiv_class (r : β → β → Prop) (b : β) : set β := {c : β | requiv r b c}
definition quot {β : Type u} (r : β → β → Prop) : Type u :=
{eq_cl : set β // ∃ a : β, equiv_class r a = eq_cl}
namespace quot
definition mk : Π {α : Type u} (r : α → α → Prop), α → quot r :=
λ α r a, ⟨equiv_class r a,a,rfl⟩
theorem ind : ∀ {α : Type u} {r : α → α → Prop} {β : quot r → Prop},
(∀ (a : α), β (mk r a)) → ∀ (q : quot r), β q :=
λ α r β h q,
begin
rcases q with ⟨C,a,Ha⟩,
convert h a,
rw Ha,
end
noncomputable definition lift :
Π {α : Type u} {r : α → α → Prop} {β : Sort v} (f : α → β),
(∀ (a b : α), r a b → f a = f b) → quot r → β :=
λ α r β f h q,begin
apply f,
rcases q with ⟨C,HC⟩,
-- cases HC with a Ha, MEH
let a := classical.some HC,
have Ha : equiv_class r a = C := classical.some_spec HC,
exact a,
end
definition sound : ∀ {α : Type u} {r : α → α → Prop} {a b : α}, r a b → quot.mk r a = quot.mk r b :=
λ α r a b h,begin
unfold mk,
suffices : equiv_class r a = equiv_class r b,
simp [this],
ext,
split,
{ intro Hx,
show requiv r b x,
apply requiv.trans _ Hx,
apply requiv.symm,
apply requiv.of_r,
assumption,
},
{ intro Hx,
show requiv r a x,
apply requiv.trans _ Hx,
apply requiv.of_r,
assumption,
}
end
end quot
end xena
I think I do all the basic theory of quot. I've been teaching quotients (in ZFC) in my class and I was trying to figure out how to explain to mathematicians why all this quot.sound stuff all had to be dealt with via extra axioms, but I've just done it all myself. The two sacrifices I had to make were: (1) it doesn't work for props (but who takes a quotient on proofs of a prop?) and (2) lift (the map which mathematicians think of as a "descent", quite the other direction to the computer science word) is noncomputable. What does quot.sound offer that my set-up doesn't but which I want or need?
#### Kevin Buzzard (Dec 21 2018 at 22:33):
I just make the quotient type as a bunch of equivalence classes.
#### Kevin Buzzard (Dec 21 2018 at 22:34):
A subtype of set β consisting of the equiv classes.
#### Reid Barton (Dec 21 2018 at 22:40):
The differences are that lift is computable, and lift f _ (mk r x) = f x(which it looks like you haven't proved, but I'm sure you can) hold definitionally (this is quot.lift_beta).
#### Kevin Buzzard (Dec 21 2018 at 23:01):
Actually I'm struggling with the theorem about lift:
-- the computation principle
variables (β : Type v) (f : α → β) (a : α)
variable (h : ∀ a b , r a b → f a = f b)
theorem thm : lift f h (mk r a) = f a := begin
rcases (mk r a) with ⟨C,HC⟩, -- HC has nothing to do with a now.
let b := classical.some HC,
have Hb : equiv_class r b = C := classical.some_spec HC,
/-
α : Type u,
r : α → α → Prop,
β : Type v,
f : α → β,
a : α,
h : ∀ (a b : α), r a b → f a = f b,
C : set α,
HC : ∃ (a : α), equiv_class r a = C,
b : α := classical.some HC,
Hb : equiv_class r b = C
⊢ lift f h ⟨C, HC⟩ = f a
-/
show f b = f a,
apply h b a,
-- but I don't know a ∈ C
sorry
end
#### Chris Hughes (Dec 22 2018 at 00:14):
Also, your proof of sound uses quot.sound
#### Kevin Buzzard (Dec 22 2018 at 11:57):
Oh does it? I suppose that's cheating :-)
#### Mario Carneiro (Dec 22 2018 at 12:25):
You use set extensionality in the proof, which is an axiom in ZFC and is derived from propext and quot.sound in lean
#### Kevin Buzzard (Dec 24 2018 at 09:35):
OK so I proved xena.quot.thm. I think definitional equality is overrated, and if I have my maths hat on I'd say the same about computability. I am concerned about Chris' comment though. I am doing this to try and figure out how to explain to mathematicians why Lean wants quotients to be an extra axiom. But set extensionality is implied by function extensionality, right? Would another approach have been to make funext and/or propext axioms and then get quotients using my method? At least if we decide not to care about computability and definitional equality and be 100% mathematician.
#### Mario Carneiro (Dec 24 2018 at 09:53):
yes, that's called zfc
#### Mario Carneiro (Dec 24 2018 at 09:53):
you make set.ext an axiom
#### Mario Carneiro (Dec 24 2018 at 09:53):
and derive funext and the others by constructing everything from sets
#### Mario Carneiro (Dec 24 2018 at 09:54):
if you want to convince yourself, make a copy of set.ext as an actual axiom, and then derive all that and check that you didn't use propext or quot.sound in #print axioms
#### Kevin Buzzard (Feb 15 2019 at 23:58):
I now understand this quotient stuff better. I always understand stuff better with an example, and whilst equivalence classes of continuous non-archimedean valuations on a topological ring might not be the example best suited to everyone, working a lot on this example now I'm back on the perfectoid project has made me see the big picture much better.
Here's something I don't understand going back to the conversation above though. I can see why lift f _ (mk r x) = f x isn't going to be proved by rfl in my set-up above (quotients as equivalence classes), but now I don't understand why it can be proved with rfl if we use Lean's inbuilt quotient. If TPIL is to believed, the whole quot set-up in core is just made by writing down a load of axioms. How can rfl prove the computation law for quot?
#### Mario Carneiro (Feb 16 2019 at 00:00):
because it's an axiom
#### Mario Carneiro (Feb 16 2019 at 00:00):
it's literally baked into lean that this happens
#### Kevin Buzzard (Feb 16 2019 at 00:01):
I don't see the axiom in TPIL
#### Mario Carneiro (Feb 16 2019 at 00:01):
that's why quot has special magic powers you can't achieve even with isomorphic constructions
#### Kevin Buzzard (Feb 16 2019 at 00:01):
but this gives me an idea -- why don't I make the computation law I want to be true by rfl an axiom?
#### Kevin Buzzard (Feb 16 2019 at 00:02):
I can prove it after all, so it's true, so it's safe to make it an axiom, and then I can prove it with rfl?
not necessarily
#### Mario Carneiro (Feb 16 2019 at 00:02):
just because it's true doesn't mean it's a good reduction rule - you've seen that with simp
#### Mario Carneiro (Feb 16 2019 at 00:03):
and I mean axiom in the conventional sense of "a thing that lets you prove things you couldn't before", not the lean sense of "a constant with no computation rules that is called axiom or constant"
#### Kevin Buzzard (Feb 16 2019 at 00:03):
Oh. You mean "it's in the C++ code"?
basically
#### Mario Carneiro (Feb 16 2019 at 00:05):
one of the things I want to experiment with in the fork is user computation rules, which would allow you to install your own computation rules just like you can add your own axioms
#### Johan Commelin (Feb 16 2019 at 05:10):
@Mario Carneiro But in our case of a "fake" quotient, I guess it would be a good reduction rule. Is that true?
#### Mario Carneiro (Feb 16 2019 at 07:26):
It's hard to say. Part of the problem with having computation rules on a type that's not a constant (like quot) is that there are already computation rules, if not very useful ones, and adding a reduction rule on top of that can cause failure of confluence
#### Mario Carneiro (Feb 16 2019 at 07:27):
For a structure, that will be things like a projection on a structure instance computing to the value
Last updated: May 08 2021 at 17:33 UTC | 2,380 | 7,967 | {"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} | 3.375 | 3 | CC-MAIN-2021-21 | longest | en | 0.652289 |
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# Top 10 how much is 31 000 a year per hour That Will Change Your Life
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Below is information and knowledge on the topic how much is 31 000 a year per hour gather and compiled by the nhomkinhnamphat.com team. Along with other related topics like: How much is \$39,000 a year per hour, 31 000 a year is how much a month, \$31,000 a year is how much biweekly, 40 000 a year is how much an hour, 31 000 a year is how much a week, \$38 000 a year is how much an hour, \$31,000 a year is how much an hour 40 hours a week, \$31 an hour is how much a year.
# \$31,000 a year is how much an hour?
## ››\$31,000 salary working 2,000 hours
As a simple baseline calculation, let’s say you take 2 weeks
off each year as unpaid vacation time. Then you would be working
50 weeks of the year, and if you work a typical 40 hours a week,
you have a total of 2,000 hours of work each year. In this case,
you can quickly compute the hourly wage by dividing the annual
salary by 2000. Your yearly salary of \$31,000 is then
equivalent to an average hourly wage of
\$15.50 per hour.
Want to reverse the calculation?
15.50 an hour is how much a year?
## ››What if you get 2 weeks of paid vacation, or you take no vacation time?
Now let’s consider the case where you get paid \$31,000 a
year, but you get 2 weeks of paid vacation. You get the same
result if you work all year with no vacation time. In the
previous case, we assumed 2 weeks of unpaid vacation, so your
total year consisted of 50 weeks. But if you get paid for 2
extra weeks of vacation (at your regular hourly rate), or you
actually work for those 2 extra weeks, then your total year now
consists of 52 weeks. Assuming 40 hours a week, that equals
2,080 hours in a year. Your annual salary of \$31,000 would
end up being about
\$14.9 per hour.
## ››What’s the total number of working days in 2022?
If you wanted to be even more accurate, you can count the
exact number of working days this year. 2022 starts on
a Saturday (January 1, 2022) and ends on
a Saturday (December 31, 2022). It has a total
of 365 days in the year including both weekdays and
weekends. There are 105 weekend days (counting every
Saturday and Sunday in the year), and 260 weekdays
(Monday through Friday). So if you worked a normal 8 hour day
on every weekday, and didn’t work any overtime on the weekends,
you would have worked a total of 2,080 hours
over the 2022 year. You can then convert
your annual salary to an hourly wage of roughly
\$14.9 per hour.
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## ››What about the holidays?
Most companies give employees time off for various holidays,
so that should really be included in the calculation. The problem
is that different countries can vary in their national holidays,
and your company may give certain days off but not others.
On average, in the U.S. there are 7 major holidays: New Years Day,
Memorial Day, Independence Day, Labor Day, Thanksgiving,
the day after Thanksgiving (since this is always a Friday),
and Christmas. Many businesses also have an additional floating
holiday that is sometimes used for Christmas Eve, New Years Eve,
or one of the days near July 4th. Often this is based on whether
one of those days falls on a Tuesday or Thursday. Adding that
extra holiday can give employees an extra long weekend. Some
schools can also be off on Presidents Day, and banks may have
their own official holidays. Retail businesses often have
different rules for what’s considered a typical “business day”.
Federal holidays may include Veterans Day, Columbus Day,
Martin Luther King Day, and Washington’s birthday. Of course,
holidays may also fall on a weekend day, in which case you might
get the previous Friday or the following Monday off. Given all
these variations, here’s a table showing how the number of
holidays affects your hourly rate:
\$31,000 annual salary in 2022 # holidays # working days hourly rate 6 254 \$15.26 per hour 7 253 \$15.32 per hour 8 252 \$15.38 per hour 9 251 \$15.44 per hour 10 250 \$15.50 per hour 11 249 \$15.56 per hour 12 248 \$15.63 per hour 13 247 \$15.69 per hour 14 246 \$15.75 per hour
## ››How much do I make each month?
Since there are 12 months in a year, you can estimate
the average monthly earnings from your \$31,000 salary as
\$2,583.33 per month.
Of course, some months are longer than others, so this is
just a rough average. If you get paid biweekly, you may get
2 checks in one month and 3 checks in another, so some calculators
look at a month as a 4-week period, with 13 of these periods in a
year. In that case, you would be getting
\$2,384.62 per 4-week period.
## ››What about each week?
Assuming 52 weeks in a year, you would make
\$596.15 per week.
You can check how many weeks in a year to get the exact number.
## ››How much money would I make in a day?
First of all, if you’re working a regular 8-hour day, then
you can simply take any of the hourly rates listed above and
multiply it by 8 to get your daily rate. For example, if
you worked a total of 2,000 hours in the year, then your
hourly rate is \$15.50 which means your daily rate is
\$124 per day.
## ››Convert annual salary to hourly wage
ConvertUnits.com
provides a tool you can use to calculate the
equivalent hourly wage based on your annual salary.
You can factor in paid vacation time and holidays to
figure out the total number of working days in a year.
The salary calculator will also give you information on
your weekly income and monthly totals. Remember that
a full salary with benefits can include health insurance
and retirement benefits that add more value to your total
annual salary compared to similar hourly rates. You may
also want to factor in overtime pay and the effects of any
income taxes on your take home pay. Type in your own numbers and
convert annual salary to hourly wage
for your job to find out how much you’re worth!
## Extra Information About how much is 31 000 a year per hour That You May Find Interested
If the information we provide above is not enough, you may find more below here.
### \$31,000 a year is how much an hour? – Convert Units
• Author: convertunits.com
• Rating: 5⭐ (19378 rating)
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• Sumary: Convert annual salary of \$31,000 to the equivalent hourly wage to calculate take home pay after holidays and vacation time.
• Matching Result: In this case, you can quickly compute the hourly wage by dividing the annual salary by 2000. Your yearly salary of \$31,000 is then equivalent to an average …
• Intro: \$31,000 a year is how much an hour? ›› \$31,000 salary working 2,000 hours As a simple baseline calculation, let’s say you take 2 weeks off each year as unpaid vacation time. Then you would be working 50 weeks of the year, and if you work a typical 40 hours…
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### What is a \$31,000 Salary on a Per-Hour Basis?
• Author: calculateme.com
• Rating: 5⭐ (19378 rating)
• Highest Rate: 5⭐
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• Sumary: Convert \$31k a year to hourly wage. Use this easy calculator to convert an annual salary to its equivalent as an hourly wage.
• Matching Result: It depends on how many hours you work, but assuming a 40 hour work week, and working 50 weeks a year, then a \$31,000 yearly salary is about \$15.50 per hour.
• Intro: What is a \$31,000 Salary on a Per-Hour Basis? \$31,000 per year = \$15.50 per hour (\$620 per week) \$31,000 a year is what per hour? It depends on how many hours you work, but assuming a 40 hour work week, and working 50 weeks a year, then a \$31,000…
### \$31000 Per Year to Hourly Salary – Finatopia
• Author: finatopia.com
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• Sumary: 31,000 a year is how much an hour? Salary Calculator for \$31000 per year to hourly.
• Matching Result: If you work 40 hours per week and 50 weeks per year, you will make an hourly salary of \$15.50 if you make 31,000 per year. Calculate an hourly salary if you …
• Intro: 31,000 a year is how much an hour? Salary Calculator for \$31000 per year to hourly. If you work 40 hours per week and 50 weeks per year, you will make an hourly salary of \$15.50 if you make 31,000 per year. Calculate an hourly salary if you make 31k…
### \$31,000 a Year is How Much an Hour? – Online Calculator
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• Sumary: \$31,000 a year is how much an hour? – \$31,000 in annual salary equals \$14.90 in hourly rate. Assume you work 40 hours per week, 52 weeks per year with a total working hours of 2,080, divide \$31,000 by 2,080 hours, and you get \$14.90 per hour.
• Matching Result: a year is how much an hour? – \$31000 in annual salary equals \$14.90 in hourly rate. Assume you work 40 hours per week, 52 weeks per year with a total …
• Intro: \$31,000 a Year is How Much an Hour? Online Calculators > Financial Calculators \$31,000 a year is how much an hour? – \$31,000 in annual salary equals \$14.90 in hourly rate. Assume you work 40 hours per week, 52 weeks per year with a total working hours of 2,080, divide…
### \$31000 a Year is How Much an Hour? (Before and After Taxes)
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• Sumary: So you’re going to make \$31,000 a year. But \$31,000 a year is how much an hour, daily, weekly, biweekly, monthly, and even after taxes?
• Matching Result: If you make \$31,000 a year, your hourly wage would be \$14.90. This is calculated by dividing the \$31,000 yearly salary by 2,080 total annual hours worked. There …
• Intro: \$31,000 a Year is How Much an Hour? (Before and After Taxes) So you found out you will soon be getting a new job or raise and will be making \$31,000 a year. But \$31,000 a year is how much an hour? Below you will find how much \$31,000 is…
### \$31,000 Per Year is What Per Hour? – DollarTimes
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• Sumary: What hourly wage is equivalent to \$31,000/yr? Convert \$31,000 per year to an hourly wage.
• Matching Result: Starting with \$31,000 a year; 31 / 2 = 15.5; So you make approximately \$15.50 an hour. Median Annual Salary by State. Source: Bureau of …
• Intro: \$31,000 Per Year is What Per Hour? What hourly wage is equivalent to \$31,000/yr? Convert \$31,000 per year to an hourly wage. Results You work about 1,927 hours a year. Your salary of \$31,000 a year equals about \$16.09 an hour. Tip: An easy way to estimate your hourly wage…
### 31,000 a year is how much an hour? – Osh.net
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• Sumary: Convert between annual and hourly salaries
• Matching Result: \$31,000 a year is how much an hour? Convert between annual and hourly salaries. Type into the calculator above or browse the chart below.
• Intro: 31,000 a year is how much an hour? 31,000 a year is how much an hour? Convert between annual and hourly salaries Salary \$ Hours Worked per Week hours Weeks Worked per Year weeks \$31,000 a year is how much an hour? Convert between annual and hourly salaries. Type into…
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### \$31,000 a year is how much an hour? – Zippia
• Author: zippia.com
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• Sumary: \$31,000 is \$15.50 an hour. \$15.50 is the hourly wage a person who earns a \$31,000 salary will make if they work 2,000 hours in a year for an average of 40 hours per week, with two weeks of total holidays.
• Matching Result: Mar 2, 2022 · 1 answer
• Intro: \$31,000 a year is how much an hour?This question is about careers.By Zippia Expert – Mar. 3, 2022\$31,000 is \$15.50 an hour. \$15.50 is the hourly wage a person who earns a \$31,000 salary will make if they work 2,000 hours in a year for an average of 40 hours…
## Frequently Asked Questions About how much is 31 000 a year per hour
If you have questions that need to be answered about the topic how much is 31 000 a year per hour, then this section may help you solve it.
### Is \$31,000 per year a decent wage?
No, \$0.000 is not a great salary for a single person, but depending on location and costs, it may be livable. The average personal income in the United States is \$3,214 annually, which is more than twice the \$0.000 threshold
7,440 a year
### What is 31000 hours annually?
This number is based on 40 hours of work per week and assumes it’s a full-time job (8 hours per day) with paid vacation time. A yearly salary of 1,000 is b>4.90 per hour/b>.
5.48% per hour
### What should my salary be when I’m 25?
As a general rule, earnings tend to increase in your 20s and 30s as you start to climb the ladder. For Americans ages 25 to 34, the median salary is 60 per week or 9,920 per year, a significant increase from the median salary for 20- to 24-year-olds.
### Making \$30k, can you afford a house?
A second rule to abide by is that your home should not cost more than 2.5 to 3 times your annual salary, which means that if you earn 0 per year, your maximum budget should be 0 per year. Using the 28% rule, you could afford a monthly mortgage payment of 00 on a yearly income of 0 if you were to follow this rule.
### What would a fair hourly rate be?
Why not be way above average and find a job that pays 0 more than the average hourly salary? According to the National Compensation Survey, the national mean salary in the United States is 6,310, or 7 per hour. Therefore, in order to be above average, you must earn more than 8 per hour.
### How much does 0.00 hours cost yearly?
If you earn 0 per hour, your annual salary would be \$1,600. This figure is calculated by multiplying your base salary by the number of hours, weeks, and months in a year, assuming you work 40 hours per week.
### What is two hours per year?
With a wage of \$2 per hour, the gross annual salary is \$5,760.
### How much does 17.50 hours of work pay?
7.50 per hour equates to 5,000 per year in income. If you work 40 hours per week on average for 50 weeks per year, you will make this amount of money in a year.
### What is the annual cost of 22.50 dollars an hour?
The calculation assumes 50 work weeks with an average of two weeks off in a year. If you earn an hourly wage of 2.50 an hour and work for an average of 40 hours a week for 50 weeks a year, you will earn a yearly salary of 5,000.
### What is 5 000 dollars an hour?
If a person works an average of 40 hours per week and earns \$5,000 per year, their hourly wage would be \$1.63.
### How much does 0 an hour cost annually?
An hourly wage of zero results in an annual income of about 3,200.
Rate this post | 4,024 | 14,867 | {"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} | 4.03125 | 4 | CC-MAIN-2023-06 | latest | en | 0.943785 |
https://testinar.com/article/real_numbers_and_integers__arrange,_order,_and_comparing_integers_course | 1,722,737,016,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00288.warc.gz | 447,537,810 | 13,389 | ## How to Arrange, Order, and Compare Integers
Read,3 minutes
The term integers represent the meaning “intact” or “whole”. So, you can generally refer an integer as a whole number, except integers can be negative also!
What is an Integer?
An integer is a whole number (not any decimal or fraction numbers) which can be zero, positive or negative numbers. Some examples of integers can be $$6 \ , \ 2 \ , \ 0 \ , \ -7 \ , \ -19,$$ etc. Moreover, we can represent integers by the denotation $$Z$$ which comprises of:
Positive Integers: As the name suggests, any integer that is greater than zero is termed as a positive integer.
Negative Integer: As from the name, any integer that is less than zero is termed as a negative integer.
Zero: Zero is neither a positive integer or a negative integer. It is just a whole number.
So, we can write $$Z = \{…… \ ,-3 \ , \ -2 \ , \ -1 \ , \ 0 \ , \ 1 \ , \ 2 \ , \ 3 \ , \ ……\}$$
Also, we can place all integers on a number line where the negative ones are placed on the left of $$“0”$$ and the positive ones on the right. Moreover, we can perform the 4 basic mathematic properties with integers. They are:
• Addition
• Subtraction
• Multiplication
• Division
We often see that negative integers are always written as $$-7 \ , \ -19$$, and so on. But it’s not generally considered necessary to write positive integers like $$+7 \ , \ +19$$, and so on. So, when we write just $$7,$$ we mean $$+7$$.
Another thing to note is, that an absolute value of any integer is always positive. So, $$\lvert -3 \rvert=3$$ and $$\lvert 3 \rvert$$ is also $$3$$.
#### How to Arrange/Order Integers and Numbers
To arrange/order integers and numbers, follow these steps:
• First, you should identify the negative numbers from the set. Remember, the farthest the negative number is from zero, the smaller it gets. In other words, the bigger the negative integer, the smaller it is.
• If the set contains a zero, then it should be written and ranked above all negative numbers.
• Finally, place the positive numbers higher than zero. A greater positive number always has a greater value.
Example: $$-11< -6< -2<0<1<4<6<13$$ (This is an ascending order of all integers in the set).
#### How to Compare Integers
Comparing integers is really very simple. The only thing to keep in mind is that the larger the negative integer, the lower is its value. Also, the greater the positive integer, the higher will be its value.
Example:
• $$-12 < -5$$
• $$5 > -7$$
• $$-3 > -8$$
### Exercises for Arrange, Order, and Compare Integers
1)$$-12, \ -7, \ 12, \ -9, \ -3, \ 4$$ $$\Rightarrow \$$
2) $$16, \ 27, \ 4, \ -12, \ -10, \ 2$$ $$\Rightarrow \$$
3) $$8, \ 34, \ -18, \ 32, \ 20, \ 6$$ $$\Rightarrow \$$
4) $$8, \ 10, \ -17, \ 20, \ 5, \ -10$$ $$\Rightarrow \$$
5) $$-6, \ -3, \ 24, \ 27, \ 19, \ -24$$ $$\Rightarrow \$$
6) $$1, \ -1, \ -3, \ 13, \ 10, \ -8$$ $$\Rightarrow \$$
7) $$12, \ -6, \ -14, \ 20, \ 13, \ -2$$ $$\Rightarrow \$$
8) $$-3, \ 27, \ 21, \ 17, \ -8, \ 24$$ $$\Rightarrow \$$
9) $$9, \ 13, \ 11, \ -11, \ -9, \ -5$$ $$\Rightarrow \$$
10) $$5, \ 1, \ -15, \ -18, \ 20, \ -6$$ $$\Rightarrow \$$
1) $$-12, \ -7, \ 12, \ -9, \ -3, \ 4$$ $$\Rightarrow \ \color{red}{12, \ 4, \ -3, \ -7, \ -9, \ -12}$$
Solution:
Step 1: Find the smallest negative integer (farthest from zero): $$-12$$ and the largest positive integer: $$33$$
Step 2: Order the numbers from the lagest one to the smallest one: $$12, \ 4, \ -3, \ -7, \ -9, \ -12$$
2) $$16, \ 27, \ 4, \ -12, \ -10, \ 2$$ $$\Rightarrow \ \color{red}{27, \ 16, \ 4, \ 2, \ -10, \ -12}$$
3) $$8, \ 34, \ -18, \ 32, \ 20, \ 6$$ $$\Rightarrow \ \color{red}{34, \ 32, \ 20, \ 8, \ 6, \ -18}$$
4) $$8, \ 10, \ -17, \ 20, \ 5, \ -10$$ $$\Rightarrow \ \color{red}{20, \ 10, \ 8, \ 5, \ -10, \ -17}$$
5) $$-6, \ -3, \ 24, \ 27, \ 19, \ -24$$ $$\Rightarrow \ \color{red}{27, \ 24, \ 19, \ -3, \ -6, \ -24}$$
6) $$1, \ -1, \ -3, \ 13, \ 10, \ -8$$ $$\Rightarrow \ \color{red}{13, \ 10, \ 1, \ -1, \ -3, \ -8}$$
7) $$12, \ -6, \ -14, \ 20, \ 13, \ -2$$ $$\Rightarrow \ \color{red}{20, \ 13, \ 12, \ -2, \ -6, \ -14}$$
8) $$-3, \ 27, \ 21, \ 17, \ -8, \ 24$$ $$\Rightarrow \ \color{red}{27, \ 24, \ 21, \ 17, \ -3, \ -8}$$
9) $$9, \ 13, \ 11, \ -11, \ -9, \ -5$$ $$\Rightarrow \ \color{red}{13, \ 11, \ 9, \ -5, \ -9, \ -11}$$
10) $$5, \ 1, \ -15, \ -18, \ 20, \ -6$$ $$\Rightarrow \ \color{red}{20, \ 5, \ 1, \ -6, \ -15, \ -18}$$
## Arrange, Order, and Compare Integers Practice Quiz
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# Exploring Algebra: Symbolism, Equations, and Complexity
Created by
@SolicitousLimerick
## Questions and Answers
### What is the key difference between an equation and an inequality?
• Equations have multiple solutions, while inequalities have only one solution.
• Equations cannot be solved, while inequalities can be solved.
• Equations use signs like less than or greater than, while inequalities use the equals sign.
• Equations involve equal relationships, while inequalities involve unequal relationships. (correct)
• ### How can algebraic equations be solved effectively?
• By introducing new variables
• By using only addition operations
• By factoring out all terms
• By isolating variables and applying inverse operations (correct)
• ### What do multiplicative properties in algebra assert?
• Multiplying any number by zero results in one
• Multiplying any number by one maintains the original value (correct)
• Multiplication is not associative
• Multiplying two numbers in any order gives different products
• ### In what situations do inequalities commonly arise?
<p>In life situations where quantitative relationships involve comparisons</p> Signup and view all the answers
### What are some common methods for handling inequalities?
<p>Graphing lines, plotting points, and evaluating test cases</p> Signup and view all the answers
## Exploring Mathematical Foundations: Algebraic Perspectives
Algebra is a branch of mathematics focused on abstract representations, symbolic expressions, and problem-solving techniques using various rules and properties. It forms the backbone of modern mathematics, serving as both a fundamental building block and a practical tool across science, engineering, economics, and other disciplines. Let's dive into this fascinating realm together, examining how algebra connects different concepts and provides solutions to problems through its distinct approaches.
### Symbolism and Expressions
At the heart of algebra lies our ability to represent quantities by symbols such as (x), (y), and (z), which stand for arbitrary numbers that we later define based on specific contexts or conditions. These variables enable us to formulate generalizations, create models, and solve equations with relative ease. By manipulating these expressions according to precise principles like order of operations, commutativity, associativity, and distributivity, we can simplify complex statements down to their core components before solving them.
The four basic arithmetic operations—addition, subtraction, multiplication, and division (ASMD) —form the basis for constructing more advanced topics in algebra, and they obey certain laws. For example, addition properties state that any number added to zero results in itself; adding two equal numbers yields another equal result. Similarly, multiplicative properties assert that multiplying any number by one maintains the original value; two numbers multiplied together will give the product of their respective factors regardless of their order. Understanding and applying these principles allows us to perform elegant calculations and effectively communicate mathematical ideas.
### Equations and Inequalities
An equation is a statement expressing equality between two expressions using an equals sign ((=)). In contrast, an inequality indicates an unequal relationship between expressions, represented using signs like less than ((\lt)), greater than ((\gt)), less than or equal to ((\le)), and greater than or equal to ((\ge)). Solving algebraic equations involves determining values for unknown variables so that the resulting expression becomes true when substituted in place of those variables within the given equation. Factoring out common terms, applying inverse operations, isolating variables, and using identities from geometry help us tackle these challenges efficiently.
Inequalities arise naturally in life situations, where quantitative relationships do not necessarily entail strict equivalence, but rather comparisons among magnitudes. Commonly used methods for handling inequalities include graphing lines, plotting points, identifying patterns, and evaluating test cases. We also employ transformational properties of inequalities that allow us to change their appearance while maintaining their meaning, providing valuable insights into the structure of numerical data.
### Linear Systems and Complexity
A linear system consists of simultaneous equations involving only ASMD operations, representing scenarios such as systems of balance, inventory management, and supply chain optimization. This type of system has straightforward solutions via techniques like elimination, substitution, and matrices, opening doors to countless applications outside mathematics proper. Extensions of linear algebra to higher dimensions introduce even more versatile tools and opportunities for exploring real-world phenomena.
As complexity increases exponentially along with the size and nature of problems addressed in algebra, students must learn strategies for recognizing structure amidst chaos. Graph theory, computability theory, and combinatorial analysis offer ready resources for dealing with intricate issues confronted daily by mathematicians working at cutting-edge frontiers. Nonetheless, simplicity remains key throughout algebra: seeking brevity and clarity often leads to powerful generalizations and remarkable outcomes.
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## Description
Delve into the fascinating realm of algebra, where abstract symbols, equations, and linear systems form the backbone of modern mathematics. Learn about symbolism and expressions, solving equations and inequalities, and navigating complexity through linear systems and advanced techniques. Discover how algebra connects different concepts and provides solutions to complex problems in various fields.
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Calculus MA 224 Chapter 7 Section 1: Functions of Several Variables Function of Two Variables : A function f of the two independent variables x and y is a rule that assigns to each ordered pair (x,y) in a given set D (the domain of f) exactly one real number, denoted by f(x,y). Domain Convention : Assume that the domain of f is the set of all (x,y) for which the expression f(x,y) is defined. Ex. 1 Suppose f(x,y) = Find domain of f Compute f(1,-2) Solution: Domain= x-y0 or xy f(1,-2)= Graphs of Functions of Two Variables : The graph of a function of two variables f(x,y) is the set of all triples (x,y,z) such that (x,y) is in the domain of f and z=f(x,y).
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Unformatted text preview: To do this, you must construct a three-dimensional graph. the xy plane is laid horizontally and z is perpendicular to this plane. To graph a function f(x,y) of the two independent variables x and y, it is customary to introduce the letter z to stand for the dependent variable and to write z=f(x,y). The ordered pairs (x,y) in the domain of f are thought of as points in the xy plane, and the function f assigns a "height" z to each such point. Thus, if f(1,2)=4, you would express this fact geometrically by plotting the point (1,2,4) in a three-dimensional space....
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# Anirban Pal
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You have a two vectors, a and b. They are monotonic and the same length. Given a value, va, where va is between a(1) and a(end... | 1,134 | 4,350 | {"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-2020-45 | latest | en | 0.725211 |
https://se.mathworks.com/matlabcentral/answers/1722895-how-to-filter-the-following-vector-from-multiple-vectors?s_tid=prof_contriblnk | 1,680,160,441,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00655.warc.gz | 572,540,647 | 29,715 | # How to filter the following vector from multiple vectors?
9 views (last 30 days)
M on 19 May 2022
Edited: Jon on 19 May 2022
How to filter the following vector from multiple vectors?
I want to do the following for each matrix stored in M
suppose each vector is named as the following in Matrix M: [deltaX
deltaY
1]
I want to filter the vector that satisfy the following conditions:
if abs(deltaY) is less than threshold for example 32
then select that vectors/vector
after that if there are more than selected vectors, filter them as the following:
select the vector that has the min deltaX
*The out should be 23 separated vectors each corrosponding to its matrix
Jan on 19 May 2022
Edited: Jan on 19 May 2022
What does this mean: "suppose the each vector is named as the following in Matrix MM"?
Would it be impolite if I reply although I'm neither Matt J nor Star Strider?! Please avoid addressing specific persons in the forum except if you have really good reasons.
Jon on 19 May 2022
Edited: Jon on 19 May 2022
I think this does what you describe
% filter matrix mm
% parameters
threshold = 32
% extract the matrix out of the cell array MM
mm = MM{1}
% find all of the columns where the value in the second row is over
% threshold
result1 = mm(:,abs(mm(2,:))>threshold)
% now pick the one with the smallest value of "delta x" (first row)
[~,idx] = min(result1(1,:));
result2 = result1(:,idx)
For the calculation of result2, which pick the one with the minimum value in the first row, I wasn't sure from your description if you wanted the minimum of the actual values, or the minimum absolute value. I picked the minimum, so it is one with a negative value. You can modify the code accordingly.
Jon on 19 May 2022
Edited: Jon on 19 May 2022
% filter matrices in M.mat
% parameters
threshold = 32
% load the cell array of matrices to be filtered
% loop through matrices filtering each one
% accumulate results (selected column) in a matrix where each column is the
% selected column from the corresponding matrix in M
numFilter = numel(M); % number of matrices to be filtered
R = zeros(3,numFilter); % preallocate array to hold results
for k = 1:numFilter
% extract the kth matrix to be filtered
m = M{k};
% find all of the columns where the value in the second row is less than
% threshold
result1 = m(:,abs(m(2,:))<threshold);
% now pick the one with the smallest value of "delta x" (first row)
[~,idx] = min(result1(1,:));
% store the result
R(:,k) = result1(:,idx);
end
Jan on 19 May 2022
Maybe:
V = MM(:, abs(MM(2, :)) < 32);
[~, index] = min(V(1, :));
W = V(:, index) | 688 | 2,581 | {"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-2023-14 | latest | en | 0.918963 |
https://ironion.com/blog/2015/12/22/bonfire-chunky-monkey-solution/ | 1,618,417,980,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077843.17/warc/CC-MAIN-20210414155517-20210414185517-00634.warc.gz | 445,935,989 | 19,316 | Bonfire: Chunky Monkey Solution
Ah, the Chunky Monkey. I’ve been, ever so gratefully, swamped with work (hooray!) But this Bonfire Solution has been a long time coming. I’ve been eyeing this one since I saw the name of it so let’s take a look at it. Unfortunately, it has very little to do with the Ben & Jerry’s ice cream flavor and a whole lot to do with Javascript arrays.
Here’s what Free Code Camp has provided us in this Bonfire:
```function chunk(arr, size) {
// Break it up.
return arr;
}
chunk(["a", "b", "c", "d"], 2);```
FCC wants us to split the array and group them into a multidimensional array based upon the size argument in the chunk() function call. Let’s get started.
Approaching the Problem
1. First thing’s first, if I’m not already crystal clear on it, I’m going to have to define what a multidimensional array is and how JS makes things confusing and probably start an empty one to fill in later with my return values.
2. I’ll have to approach the array splitting in one of two ways. I can use split() or I can use splice(). Whichever one I end up with, it’s going to have to be approached in the form of a loop and be based on whatever number falls into that second function argument.
The Different Flavors of Arrays
Arrays come in different types, and in various programming languages you might see support for more array types or less. Now, we’ve been working with arrays a lot with these Bonfires, and in this one FCC wants us to create a multidimensional array. I don’t spend days pondering the intricacies of JS arrays, so I might not have the most definitive answers for you. You can check out the MDN page on arrays for more information. For the rest of us just learning, here’s a simple rundown:
Associative Arrays – Are arrays that use strings instead of numbers to act as a “key” in the key value pair. For example:
```// create an empty array called myCar.
var myCar = [ ];
// populate the myCar array with key value pairs.
myCar['color'] = 'red';
myCar['type'] = 'sedan';
myCar['make'] = 'Honda';
myCar['year'] = 2010;```
In the above code, we created an associative array called myCar. Then we go on to populate it with values using keys to identify them. For example in the declaration: myCar[‘color’] = ‘red’, ‘color’ is considered the key and ‘red’ is the value. So when you need to reference the value red, you would use the key to call it up:
`console.log (myCar['color']);`
Seeing an associative array at work, it’s hard not to see how similar in function it is to a regular old object. Which is why some programmers recommend that instead of using JS’ associative arrays which can be confusing and aren’t seen all that often, you just create an object instead. It’s easier to understand, and you can do more with it and more efficiently. Here’s the same thing as what you saw above, except we use objects instead:
```// create an object using object literal notation and call it myCar then populate it with properties.
var myCar = {
color : 'red',
type : 'sedan',
make : 'Honda',
year : 2010
}```
And to get at the color for the myCar object, you would also use the key to call it up:
`console.log (myCar['color']);`
So that’s Associative Arrays, long story short, objects are easier and faster to work with. But what about Multidimensional Arrays?
Multidimensional Arrays – Are arrays within arrays. So technically, a multidimensional array can hold other arrays inside of itself. For example:
```var myDesserts = [
['carrot cake', 'chocolate cake', 'cheesecake'],
['apple pie', 'pumpkin pie', 'plum pie'],
];```
In the above code, we’ve created an array called myDesserts. Within the myDesserts array, it contains three elements. Each of those elements is an array that contains three values each. So if I wanted to access ‘plum pie’ in the above code, I’d have to call it up like this:
`console.log (myDesserts[1][2]);`
In the above code, I’m referencing the myDesserts multidimensional array, then telling it to go into the second array (we reference the number 1 to find the second array because JS starts counting at 0) and then give me the third value in that second array (we reference 2 to get the third value because 0 is ‘apple pie’ since JS starts counting at 0), which should be plum pie.
So now that we know what these two kinds of arrays are, let’s get back to our Bonfire, which is actually considerably simpler than all this talk about arrays.
Building the Solution with a While Loop
I’m going to use a While Loop for this chunky monkey challenge, but a For Loop will work just as well–it’s only a little more verbose and not really necessary. First thing I’m going to do is create my empty While Loop and my empty multidimensional array to push my values into:
```function chunk (arr, size) {
var chunked = [];
while () {
}
return chunked;
}```
The above code is just a skeleton of what I’m planning to do. But I’ve declared an empty array called chunked. Then set up an empty while loop that doesn’t do anything useful yet. And finally, I’m returning the chunked array. All stuff we should be really familiar with by now. Next thing I’m going to do is evaluate the length of arr. So that it will essentially run the loop and take care of all the values that arr might have fed into the function. This is what that will look like:
```function chunk (arr, size) {
var chunked = [];
while (arr.length) {
}
return chunked;
}```
Finally, I’m going to use a bunch of methods to perform two major things. I want to add values into an array where each array contains two values within it. This means we need a and b in one array and c and d in another, both of those arrays will have to go into my chunked array. What I’m going to use is a method called splice(). Hold onto your hats, we’re heading into another sidetrack.
The Splice() Method
Splice() is a handy method in Javascript that allows you to remove elements from an array and, if it pleases you, add elements back in. It can even return the removed elements. It has a couple of efficiency boosts over split() because with one splice(), you can both add, remove and return. Here’s splice() in action:
```var myJams = ['Cranberry', 'Strawberry', 'Grape', 'Apple'];
myJams.splice(1, 3, 'Crabapple', 'Orange');
console.log(myJams);```
In the above code, I declared an array called myJams, which originally contains four types of jams: Cranberry, Strawberry, Grape and Apple. On the next line, I use the splice() method to select the start and end of where I want to do my splicing (start at position 1, end at position 3) and I designated two new values to add to the array: Crabapple and Orange.
The result when logged out will be: Cranberry, Crabapple, Orange. What splice did was remove the old values in myJams starting at position 1 (Strawberry) and ending on position 3 (Apple). Then it put Crabapple and Orange at the end of myJams. The two numbers in the arguments will give you the range of splicing, then what comes after is what you want to add in. So, I could very well have something like this instead:
```var myJams = ['Cranberry', 'Strawberry', 'Grape', 'Apple'];
myJams.splice(1, 2, 'Crabapple', 'Orange', 'Plum', 'Banana', 'Pomegranate');
console.log(myJams);```
And the log would reveal this list for myJams: Cranberry, Crabapple, Orange, Plum, Banana, Pomegranate, Apple. This is because I started at 1, but ended at 2 and pushed the remaining values into the place left open from where I removed those values. Therefore, instead of seeing the new values added at the end, they get placed into the middle spaces where the old values used to be.
Using splice, we can trim away the values from the array we don’t need and replace it with the values declared as the arr argument.
Solving the Chunky Monkey Challenge
Here’s the solution:
```function chunk (arr, size) {
var chunked = [];
while (arr.length) {
chunked.push(arr.splice(0, size));
}
return chunked;
}
chunk(["a", "b", "c", "d"], 2);```
We’ve already gone through the easy stuff, so let’s focus on the magic inside of the while loop. We’re checking to make sure that we loop through all values within the arr argument. So for each of those values, we go in and push to the chunked array each value in arr. For each of the arr values, we want to splice them. We start splicing at 0 and splice it using the value for size (which is 2 in this case). So each time we run through the splice, it knows that we want it all spliced, and to split things up into groups of two. This works well enough if you want groups of 3, 4, 5, etc. So long as you have the values and an appropriate size argument for each. Even if you don’t have enough values to feed into the function, it would still group as many as it can into groups and push the rest into an array. Give it a try with ten values, then push them into groups of 4 to see how it works. In the mean time, we’ve solved Free Code Camp’s Chunky Monkey Bonfire.
I had to do a lot of looking up and experimentation to get at the solution for this one, so check out some of the Resources I referenced for additional help and information.
Resources
Microsoft DN, Splice Method
For some reason, I found Microsoft DN’s explanation of splice() to be the easiest to understand so here it is.
MDN, Splice Method
Another resource for splice(), more detailed this time.
SharkySoft, Multidimensional Arrays Explanation
SharkySoft’s ultra-minimalist approach to design and development has put this particular page in a warm place within my heart. It’s also a great explanation of JS Multidimensional Arrays.
6 comments on “Bonfire: Chunky Monkey Solution”
1. Mark on
I had found another solution in the usual place which actually gave 3 solutions ranging from beginning to intermediate to advanced. I think you know what I’m referring to. In each case I always ran into at least one line of code or explanation of code that still didn’t connect the dots for me. But twice now I have looked at your explanations and they all make perfect sense.
I also noticed that on that other page the advanced solution always used the least amount of code.
Thanks so much for your passion not only as a programmer but as a teacher.
2. Nicolas Pham-Dinh on
I want to clarify the role of the second parameter in splice(). Correct me if I am wrong, but the second parameter is not the ending position (or index), as it is in slice(), but rather the number, or range, of values we want to remove. For example:
var array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
array.splice(0,3); // [0, 1, 2]
array.splice(1,3); // [1, 2, 3]
array.splice(2,3); // [2, 3, 4]
etc.
• Khanh on
Hi Nicolas, thanks for leaving a comment!
If I understand your question correctly, then yes, the second parameter in splice() is the amount we want to remove. The second parameter in the solution in the post is pulling the parameter from the function chunk(). That parameter gets fed when chunk() is called.
3. Hoa on
hi,
Thanks for your super precise explanation. I just have some small questions.
I’m sorry, I’m super noop in JS, so please forgive me if my questions are too dump. But according to what I’ve learnt, the while loop should be written in something like this:
var n = 0;
var x = 0;
while (n < 3) {
n++;
x += n;
}
So if we declare "while (arr.length)", would it equal while(5) (is this case the length of my arr is 5), and would it become infinite loop? is there any reason for you to do it that way? because I tried to rewrite your but i became so weird. I hoped you would bother to give me the answer and thanks so much for your time. | 2,841 | 11,582 | {"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-2021-17 | longest | en | 0.875826 |
http://spolocnostskeptikov.sk/mj60yyd/d7vws.php?1fc491=application-of-fourier-transform-to-partial-differential-equations | 1,718,687,338,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00077.warc.gz | 28,634,419 | 8,883 | Visit to download.. The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes of vibration (frequencies), the Laplace transform resolves a function into its moments. Since the beginning Fourier himself was interested to find a powerful tool to be used in solving differential equations. Once we have calculated the Fourier transform ~ of a function , we can easily find the Fourier transforms of some functions similar to . Sections (1) and (2) … Like the Fourier transform, the Laplace transform is used for solving differential and integral equations. However, the study of PDEs is a study in its own right. 6. Researchers from Caltech's DOLCIT group have open-sourced Fourier Neural Operator (FNO), a deep-learning method for solving partial differential equations … Featured on Meta “Question closed” notifications experiment results and graduation In physics and engineering it is used for analysis of The first topic, boundary value problems, occur in pretty much every partial differential equation. This second edition is expanded to provide a broader perspective on the applicability and use of transform methods. All the problems are taken from the edx Course: MITx - 18.03Fx: Differential Equations Fourier Series and Partial Differential Equations.The article will be posted in two parts (two separate blongs) The second topic, Fourier series, is what makes one of the basic solution techniques work. This paper aims to demonstrate the applicability of the L 2-integral transform to Partial Differential Equations (PDEs). Also, like the Fourier sine/cosine series we’ll not worry about whether or not the series will actually converge to $$f\left( x \right)$$ or not at this point. Review : Systems of Equations – The traditional starting point for a linear algebra class. We can use Fourier Transforms to show this rather elegantly, applying a partial FT (x ! In this section, we have derived the analytical solutions of some fractional partial differential equations using the method of fractional Fourier transform. Hajer Bahouri • Jean-Yves Chemin • Raphael Danchin Fourier Analysis and Nonlinear Partial Differential Equations ~ Springer Making use of Fourier transform • Differential equations transform to algebraic equations that are often much easier to solve • Convolution simplifies to multiplication, that is why Fourier transform is very powerful in system theory • Both f(x) and F(ω) have an "intuitive" meaning Fourier Transform – p.14/22. The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes of vibration (frequencies), the Laplace transform resolves a function into its moments. S. A. Orszag, Spectral methods for problems in complex geometrics. We will only discuss the equations of the form 10.3 Fourier solution of the wave equation One is used to thinking of solutions to the wave equation being sinusoidal, but they don’t have to be. 1 INTRODUCTION. Faced with the problem of cover-ing a reasonably broad spectrum of material in such a short time, I had to be selective in the choice of topics. In Numerical Methods for Partial Differential Equations, pp. Systems of Differential Equations. For now we’ll just assume that it will converge and we’ll discuss the convergence of the Fourier series in a later PARTIAL DIFFERENTIAL EQUATIONS JAMES BROOMFIELD Abstract. How to Solve Poisson's Equation Using Fourier Transforms. Summary This chapter contains sections titled: Fourier Sine and Cosine Transforms Examples Convolution Theorems Complex Fourier Transforms Fourier Transforms in … UNIT III APPLICATIONS OF PARTIAL DIFFERENTIAL 9+3 Classification of PDE – Method of separation of variables - Solutions of one dimensional wave equation – One dimensional equation of heat conduction – Steady state solution of two dimensional equation of heat conduction (excluding insulated edges). 4. Partial Differential Equations (PDEs) Chapter 11 and Chapter 12 are directly related to each other in that Fourier analysis has its most important applications in modeling and solving partial differential equations (PDEs) related to boundary and initial value problems of mechanics, heat flow, electrostatics, and other fields. Applications of fractional Fourier transform to the fractional partial differential equations. Like the Fourier transform, the Laplace transform is used for solving differential and integral equations. The Fourier transform can be used for sampling, imaging, processing, ect. Fractional heat-diffusion equation k, but keeping t as is). The Fourier transform, the natural extension of a Fourier series expansion is then investigated. 4 SOLUTION OF LAPLACE EQUATIONS . And even in probability theory the Fourier transform is the characteristic function which is far more fundamental than the … 3 SOLUTION OF THE HEAT EQUATION. Table of Laplace Transforms – This is a small table of Laplace Transforms that we’ll be using here. 47.Lecture 47 : Solution of Partial Differential Equations using Fourier Cosine Transform and Fourier Sine Transform; 48.Lecture 48 : Solution of Partial Differential Equations using Fourier Transform - I; 49.Lecture 49 : Solution of Partial Differential Equations using Fourier Transform - II Poisson's equation is an important partial differential equation that has broad applications in physics and engineering. The Fourier transform can be used to also solve differential equations, in fact, more so. This text serves as an introduction to the modern theory of analysis and differential equations with applications in mathematical physics and engineering sciences. INTRODUCTORY APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS With Emphasis on Wave Propagation and Diffusion This is the ideal text for students and professionals who have some familiarity with partial differential equations, and who now wish to consolidate and expand their knowledge. Transform Methods for Solving Partial Differential Equations, Second Edition by Dean G. Duffy (Chapman & Hall/CRC) illustrates the use of Laplace, Fourier, and Hankel transforms to solve partial differential equations encountered in science and engineering. APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS . So, a Fourier series is, in some way a combination of the Fourier sine and Fourier cosine series. Having outgrown from a series of half-semester courses given at University of Oulu, this book consists of four self-contained parts. The finite Fourier transform method which gives the exact boundary temperature within the computer accuracy is shown to be an extremely powerful mathematical tool for the analysis of boundary value problems of partial differential equations with applications in physics. Therefore, it is of no surprise that we discuss in this page, the application of Fourier series differential equations. Applications of Fourier transform to PDEs. Partial Differential Equations ..... 439 Introduction ... application for Laplace transforms. Of special interest is sec-tion (6), which contains an application of the L2-transform to a PDE of expo-nential squared order, but not of exponential order. The following calculation rules show examples how you can do this. problems, partial differential equations, integro differential equations and integral equations are also included in this course. Anna University MA8353 Transforms And Partial Differential Equations 2017 Regulation MCQ, Question Banks with Answer and Syllabus. A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. We will present a general overview of the Laplace transform, a proof of the inversion formula, and examples to illustrate the usefulness of this technique in solving PDE’s. Browse other questions tagged partial-differential-equations matlab fourier-transform or ask your own question. 273-305. 9.3.3 Fourier transform method for solution of partial differential equations:-Cont’d At this point, we need to transform the specified c ondition in Equation (9.12) by the Fourier transform defined in Equation (a), or by the following expression: T T x T x e dx f x e i x dx g It is analogous to a Taylor series, which represents functions as possibly infinite sums of monomial terms. cation of Mathematics to the applications of Fourier analysis-by which I mean the study of convolution operators as well as the Fourier transform itself-to partial differential equations. The purpose of this seminar paper is to introduce the Fourier transform methods for partial differential equations. 1 INTRODUCTION . APPLICATIONS OF THE L2-TRANSFORM TO PARTIAL DIFFERENTIAL EQUATIONS TODD GAUGLER Abstract. In this chapter we will introduce two topics that are integral to basic partial differential equations solution methods. The introduction contains all the possible efforts to facilitate the understanding of Fourier transform methods for which a qualitative theory is available and also some illustrative examples was given. Heat equation; Schrödinger equation ; Laplace equation in half-plane; Laplace equation in half-plane. In mathematics, a Fourier transform (FT) is a mathematical transform that decomposes functions depending on space or time into functions depending on spatial or temporal frequency, such as the expression of a musical chord in terms of the volumes and frequencies of its constituent notes. 2 SOLUTION OF WAVE EQUATION. Wiley, New York (1986). But just before we state the calculation rules, we recall a definition from chapter 2, namely the power of a vector to a multiindex, because it is needed in the last calculation rule. M. Pickering, An Introduction to Fast Fourier Transform Methods for Partial Differential Equations with Applications. 4.1. This paper is an overview of the Laplace transform and its appli- cations to partial di erential equations. 5. The course begins by characterising different partial differential equations (PDEs), and exploring similarity solutions and the method of characteristics to solve them. Partial differential equations also occupy a large sector of pure ... (formally this is done by a Fourier transform), converts a constant-coefficient PDE into a polynomial of the same degree, with the terms of the highest degree (a homogeneous polynomial, here a quadratic form) being most significant for the classification. This is the 2nd part of the article on a few applications of Fourier Series in solving differential equations.All the problems are taken from the edx Course: MITx - 18.03Fx: Differential Equations Fourier Series and Partial Differential Equations.The article will be posted in two parts (two separate blongs) We shall see how to solve the following ODEs / PDEs using Fourier series: In this article, a few applications of Fourier Series in solving differential equations will be described. Academic Press, New York (1979). Point for a linear algebra class this section, we can use Fourier Transforms to show rather... Equation Since the beginning Fourier himself was interested to find a powerful tool to be used solving. Numerical methods for partial differential equations, integro differential equations to be used for solving differential and equations... Transform to partial Differential equations ( PDEs ), Spectral methods for partial differential equations and integral.... 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This paper is an important partial differential equations solution methods this page, the transform... A broader perspective on the applicability and use of transform methods this seminar paper is an overview of the to... This seminar paper is an overview of the Laplace transform is used sampling. Represents functions as possibly infinite sums of monomial terms applications of the Laplace transform is used for solving equations. Surprise that we ’ ll be using here a partial FT ( x s. A. Orszag, Spectral for! Solving differential and integral equations and integral equations topics that are integral to partial. Do this application of fourier transform to partial differential equations topics that are integral to basic partial differential equations 2017 Regulation MCQ, question Banks Answer! Of some fractional partial differential equation complex geometrics and its appli- cations to partial differential equations fourier-transform or ask own. Boundary value problems, occur in pretty much every partial differential equations solution methods broader perspective on the and... Much every partial differential equations solution methods integral to basic partial differential equations TODD GAUGLER.... Use Fourier Transforms to show this rather elegantly, applying a partial FT ( x represents! To partial Differential equations ( PDEs ) basic solution techniques work find the Fourier transform be!, processing, ect, occur in pretty much every partial differential equation that has broad applications in and. Of this seminar paper is an overview of the L 2-integral transform the! Using here, in fact, more so Fourier himself was interested to find a powerful tool to used! Boundary value problems, partial differential equations and integral equations are also included in this course it is of surprise. Calculated the Fourier Transforms to show this rather elegantly, applying a partial FT ( x and.., Fourier series expansion is then investigated ) … 4 one of the basic solution techniques work of Laplace –. Integro differential equations and its appli- cations to partial di erential equations, so... Is of no surprise that we ’ ll be using here ) and ( )! Infinite sums of monomial application of fourier transform to partial differential equations this page, the application of Fourier expansion! Equations ( PDEs ) topic, Fourier series expansion is then investigated Answer Syllabus! Value problems, occur in pretty much every partial differential equation easily the. For partial differential equation that has broad applications in physics and engineering, applying a partial FT x... The following calculation rules show examples how you can do this s. A. Orszag, Spectral methods for partial equations. In this chapter we will introduce two topics that are integral to basic differential. 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To show this rather elegantly, applying a partial FT ( x the Laplace transform is used for,. S. A. Orszag, Spectral methods for partial differential equations the first topic, value... | 4,276 | 21,983 | {"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} | 3.03125 | 3 | CC-MAIN-2024-26 | latest | en | 0.927354 |
http://mathhelpforum.com/statistics/43861-organized-counting-problem-help.html | 1,524,360,359,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945484.57/warc/CC-MAIN-20180422002521-20180422022521-00460.warc.gz | 190,132,324 | 9,523 | # Thread: Organized Counting Problem Help
1. ## Organized Counting Problem Help
Problem:
A Canadian postal code uses six characters. The first, third, and fifth are letters, while the second, forth, and sixth are digits. A U.S.A. zip code contains five characters, all digits.
How many codes are possible for each country?
I'm stuck on this problem...actually I don't understand the 'Permutations and Organized Counting' unit at all... Anyway, can someone please help me with this question?
2. Originally Posted by Morphayne
Problem:
A Canadian postal code uses six characters. The first, third, and fifth are letters, while the second, forth, and sixth are digits. A U.S.A. zip code contains five characters, all digits.
How many codes are possible for each country?
I'm stuck on this problem...actually I don't understand the 'Permutations and Organized Counting' unit at all... Anyway, can someone please help me with this question?
In general, when dealing with this unit, ask your self 2 questions. Does order matter, and do you pick with replacement or not (i.e. do the possibilities change as you pick from the "pile")?
Order matters and you do pick with replacement. | 265 | 1,182 | {"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} | 2.703125 | 3 | CC-MAIN-2018-17 | latest | en | 0.931127 |
https://www.studypool.com/discuss/531277/physics-homework-question-8?free | 1,480,702,483,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00294-ip-10-31-129-80.ec2.internal.warc.gz | 1,017,028,617 | 15,175 | ##### Physics Homework question 8
Physics Tutor: None Selected Time limit: 1 Day
The magnitude of the magnetic field at point
P for a certain electromagnetic wave is 2.12 μT. What is the magnitude of the electric field for that wave at ? ()
The magnitude of the magnetic field at point for a certain electromagnetic wave is 2.12 μT. What is the magnitude of the electric field for that wave at ? ()
May 12th, 2015
636 N/C
Electric field / Magnetic field = c
E = 2.12 * 10^(-6) * 3 * 10^8
E = 636N/C
May 12th, 2015
...
May 12th, 2015
...
May 12th, 2015
Dec 2nd, 2016
check_circle | 189 | 590 | {"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} | 3.390625 | 3 | CC-MAIN-2016-50 | longest | en | 0.808444 |
https://fresherbell.com/quizdiscuss/quantitative-aptitude/3889-12952-3854002 | 1,719,025,494,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00132.warc.gz | 227,345,908 | 11,672 | # Quiz Discussion
3889 + 12.952 - ? = 3854.002
Course Name: Quantitative Aptitude
• 1] 47.095
• 2] 47.752
• 3] 47.932
• 4] 47.95
##### Solution
No Solution Present Yet
#### Top 5 Similar Quiz - Based On AI&ML
Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api
# Quiz
1
Discuss
The sum of is
• 1]
• 2]
• 3]
• 4]
##### Solution
2
Discuss
Which part contains the fractions in ascending order ?
• 1]
$$\frac{{11}}{{14}},\frac{{16}}{{19}},\frac{{19}}{{21}}$$
• 2]
$$\frac{{16}}{{19}},\frac{{11}}{{14}},\frac{{19}}{{21}}$$
• 3]
$$\frac{{16}}{{19}},\frac{{19}}{{21}},\frac{{11}}{{14}}$$
• 4]
$$\frac{{19}}{{21}},\frac{{11}}{{14}},\frac{{16}}{{19}}$$
##### Solution
3
Discuss
[(?)2 + (18)2] ÷ 125 = 3.56
• 1] 11
• 2] 12
• 3] 14
• 4] 15
• 5] None of these
##### Solution
4
Discuss
The rational number for recurring decimal 0.125125.... is:
• 1]
63/487
• 2]
119/993
• 3]
125/999
• 4]
None of these
##### Solution
5
Discuss
The expression (11.98 × 11.98 + 11.98 × X + 0.02 × 0.02) will be a perfect square for X equal to:
• 1] 0.02
• 2] 0.2
• 3] 0.04
• 4] 0.4
##### Solution
6
Discuss
Which of the following is equal to 3.14 x 106 ?
• 1] 314
• 2] 3140
• 3] 3140000
• 4] None of these
##### Solution
7
Discuss
$$\frac{{4.2 \times 4.2 - 1.9 \times 1.9}}{{2.3 \times 6.1}}$$ is equal to:
• 1] 0.5
• 2] 1
• 3] 20
• 4] 22
##### Solution
8
Discuss
The value of $$\left( {\frac{{0.943 \times 0.943 - 0.943 \times 0.057 + 0.057 \times 0.057}}{{0.943 \times 0.943 \times 0.943 + 0.057 \times 0.057 \times 0.057}}} \right)$$ is :
• 1] 0.32
• 2] 0.886
• 3] 1.1286
• 4] None of these
##### Solution
9
Discuss
How many digits will be there to the right of the decimal point in the product of 95.75 and .02554 ?
• 1] 5
• 2] 6
• 3] 7
• 4] None of these
##### Solution
10
Discuss
Which of the following fractions is greater than 3/4 and less than 5/6 ?
• 1]
1/2
• 2]
2/3
• 3]
4/5
• 4]
9/10
# Quiz | 843 | 1,978 | {"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} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.545078 |
http://nrich.maths.org/public/leg.php?code=-68&cl=2&cldcmpid=25&setlocale=en_US | 1,503,182,037,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105927.27/warc/CC-MAIN-20170819220657-20170820000657-00220.warc.gz | 316,751,735 | 8,961 | # Search by Topic
#### Resources tagged with Visualising similar to Hoops/rope:
Filter by: Content type:
Stage:
Challenge level:
### Making Maths: Rolypoly
##### Stage: 1 and 2 Challenge Level:
Paint a stripe on a cardboard roll. Can you predict what will happen when it is rolled across a sheet of paper?
### Making Tangrams
##### Stage: 2 Challenge Level:
Here's a simple way to make a Tangram without any measuring or ruling lines.
### Reef and Granny
##### Stage: 2 Challenge Level:
Have a look at what happens when you pull a reef knot and a granny knot tight. Which do you think is best for securing things together? Why?
### Midpoint Triangle
##### Stage: 2 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Move Those Halves
##### Stage: 2 Challenge Level:
For this task, you'll need an A4 sheet and two A5 transparent sheets. Decide on a way of arranging the A5 sheets on top of the A4 sheet and explore ...
### Redblue
##### Stage: 2 Challenge Level:
Investigate the number of paths you can take from one vertex to another in these 3D shapes. Is it possible to take an odd number and an even number of paths to the same vertex?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### World of Tan 11 - the Past, Present and Future
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the telescope and microscope?
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### World of Tan 1 - Granma T
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Granma T?
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### Folding Flowers 1
##### Stage: 2 Challenge Level:
Can you visualise what shape this piece of paper will make when it is folded?
### Fractional Triangles
##### Stage: 2 Challenge Level:
Use the lines on this figure to show how the square can be divided into 2 halves, 3 thirds, 6 sixths and 9 ninths.
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### Folding, Cutting and Punching
##### Stage: 2 Challenge Level:
Exploring and predicting folding, cutting and punching holes and making spirals.
### Map Folding
##### Stage: 2 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### Single Track
##### Stage: 2 Challenge Level:
What is the best way to shunt these carriages so that each train can continue its journey?
### World of Tan 7 - Gat Marn
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this plaque design?
### Jomista Mat
##### Stage: 2 Challenge Level:
Looking at the picture of this Jomista Mat, can you decribe what you see? Why not try and make one yourself?
### World of Tan 3 - Mai Ling
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### More Building with Cubes
##### Stage: 2 Challenge Level:
Here are more buildings to picture in your mind's eye. Watch out - they become quite complicated!
### Green Cube, Yellow Cube
##### Stage: 2 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 cube that is green all over AND a 2 x 2 cube that is yellow all over?
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### World of Tan 5 - Rocket
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the rocket?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Shunting Puzzle
##### Stage: 2 Challenge Level:
Can you shunt the trucks so that the Cattle truck and the Sheep truck change places and the Engine is back on the main line?
### Little Boxes
##### Stage: 2 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### Waiting for Blast Off
##### Stage: 2 Challenge Level:
10 space travellers are waiting to board their spaceships. There are two rows of seats in the waiting room. Using the rules, where are they all sitting? Can you find all the possible ways?
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
### World of Tan 12 - All in a Fluff
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these rabbits?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### World of Tan 21 - Almost There Now
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### World of Tan 22 - an Appealing Stroll
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### World of Tan 24 - Clocks
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### Pyramid Numbers
##### Stage: 2 Challenge Level:
What are the next three numbers in this sequence? Can you explain why are they called pyramid numbers?
### World of Tan 20 - Fractions
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### Regular Rings 2
##### Stage: 2 Challenge Level:
What shape is made when you fold using this crease pattern? Can you make a ring design?
### Open Boxes
##### Stage: 2 Challenge Level:
Can you work out how many cubes were used to make this open box? What size of open box could you make if you had 112 cubes?
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Triangular Faces
##### Stage: 2 Challenge Level:
This problem invites you to build 3D shapes using two different triangles. Can you make the shapes from the pictures?
### Display Boards
##### Stage: 2 Challenge Level:
Design an arrangement of display boards in the school hall which fits the requirements of different people.
### World of Tan 19 - Working Men
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this shape. How would you describe it?
### Folding Flowers 2
##### Stage: 2 Challenge Level:
Make a flower design using the same shape made out of different sizes of paper.
### Regular Rings 1
##### Stage: 2 Challenge Level:
Can you work out what shape is made by folding in this way? Why not create some patterns using this shape but in different sizes?
### World of Tan 9 - Animals
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone? | 1,875 | 7,996 | {"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-2017-34 | latest | en | 0.823571 |
https://discuss.codechef.com/t/hldots-editorial/9768 | 1,653,090,696,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00609.warc.gz | 272,073,850 | 8,368 | # HLDOTS - editorial
Author: Sergey Kulik
Tester: Roman Furko
Editorialist: Balajiganapathi Senthilnathan
Medium
### PREREQUISITES:
Dynamic programming
### PROBLEM:
A heavy light decomposition of a tree is where you label exactly one edge going out of each node (i.e. away from the root) as heavy and others as light. A correct heavy light decomposition is where the number of light edges from root to any node is at most \lfloor{log_2} n \rfloor. Given a tree, find the number of correct heavy light decomposition it can have.
### QUICK EXPLANATION:
This can solved by using dynamic programming with the state(node index, number of light edges seen from root so far). To calculate dp[x][cnt] we loop through each outgoing edge of x and mark it as heavy in turn. We recursively solve for the children and only increment cnt if that edge is light. If at any stage the number of light edges is \gt log_2 n we return 0. If we reach a leaf then we return 1. See explanation for how to calculate this optimally.
### Explanation
Since this is a tree, it is not much of a leap to guess that we need some sort of recursive solution. Suppose we start recursion from the root and solve recursively for children, what extra information do we need? And what do we do for each node? From the statement, it is clear that the action we must perform for each node is: choose each of the edges as a heavy edge and calculate the number of ways we can make a correct heavy light decomposition with that edge marked heavy. Sum up these ways for all the edges.
The only other constraint is that the number of light edges from root to any node must be \le log_2 n. So, besides the node index itself, we also need the number of light edges traversed so far. This gives us the state of our recursion: (x, cnt) where x is the current node index and cnt is the number of light edges traversed from the root to x.
Suppose the function solve does this, the final answer will be solve(1, 0).
How to calculate solve(x, cnt)?
x has say m children c_1, c_2 ... c_m. We make each of the m edges heavy in turn and increment cnt while calling recursively for all children except the one we made heavy. Now since the number of correct heavy light decomposition for one child is independent of the other children, we multiply the ways we get for each children. We sum this up for all possible choices of heavy edges to get the answer for solve(x, cnt). Since a state may repeat, we do memoization to avoid re calculations.
So,
solve(x, cnt) = \sum_{i = 1}^{m} (\prod_{j \ne i} solve(c_j, cnt + 1) * solve(c_i, cnt))
A straight forward implementation of this formula will TLE for subtask 3 as it will take O(m^2) = O(n^2) (Consider a tree where the root has n - 1 children).
To optimize let us see the inner product in detail.
We have \prod_{j \ne i} solve(c_j, cnt + 1) * solve(c_i, cnt).
We can split it as
\prod_{j = 1}^{i - 1} solve(c_j, cnt + 1) * \prod_{j = i + 1}^{m} solve(c_j, cnt + 1) * solve(c_i, cnt).
Why is this helpful? Well, we can see that we need two types of products: from left side till i - 1 and from right side till i + 1. We can precalculate this.
Let us define two arrays left and right.
left[i] = \prod_{j = 1}^{i} solve(c_j, cnt + 1)
right[i] = \prod_{j = i}^{m} solve(c_j, cnt + 1)
Now note that we can calculate this array in O(m):
left[0] = solve(c_0, cnt + 1)
left[j] = left[j - 1] * solve(c_j, cnt + 1)
and similarly for right.
Putting these two together:
solve(x, cnt) = \sum_{i = 1}^{m} (left[i - 1] * solve(c_i, cnt) * right[i + 1])
Remember that all calculations are modulo 19101995.
### Complexity
How many possible states are there? There are n possible values for x and since we stop when cnt is log_2 n, the total number of states is O(n log n). Work done in each state is equal to the number of edges from that node. So, overall we can add them up and they will be equal to n - 1 - the number of edges. So, overall complexity is still O(nlogn)
1 Like
# WEAK TESTS
written in bold, italic and double heading style, because my code is O(N^2\log{N}) for stars (it easily runs over 10 seconds) and it gets AC with 0.15 seconds worst case. Seriously, no star? That’s one of the basic tests for beating slow codes when the input is a tree.
Thanks for the good question.
We would have solved this problem much more easily by making use of modulo inversion if given mod was some large prime say 1e9+7
1 Like
Can anyone please explain it in another way?
Next time write some testcases for N^2 solutions beforehand…
I just managed to fit in O(n(logn)^2) solution using segment trees
I am just wondering we had added test cases of star during the contest itself.
Yes, your code got TLE on the new added test case. You have got 55 points for the task.
I had 100 points 1.5 hours after the start of the contest.
Mind you, it’s not a good idea to add tests in the middle of / at the end of the contest.
1 Like
@gdisastery1: I think no \mathcal{O}(n^2) passed in the problem.
@dpraveen Please read Xellos’ entry - I was in the same situation. I was immersed in another problem and without any notification system there was no way to figure out that my solution to this problem has been rejudged as WA ;( Luckily, I saw this at the very end of the contest.
I think the notification system needs to be improved regarding these kind of scenarios. I will ask codechef developers regarding this.
IMHO, it was a good idea to add tests in the middle of contest because we thought that letting \mathcal{O}(n^2) solutions pass is injustice to people who have used expected algorithm to pass the problem.
Nope it’s just ethically wrong and never practiced in any kind of contests where the prize is high (see IOI especially).
It’s better than not adding these tests in that ppl will learn more on it, but it’s not fair to anyone who thought they have a full score on the problem and ignored it or something. 1.5 hours is plenty of time to lose interest in trying anymore and go do something else. It’s generally bad practice to change results in the middle of a contest.
2 Likes
I second @xellos0 's view. Once the contest has started, no test files should be added or removed. Definitely not after half the contest. A similar thing happened at IOI 2014; some test files were dropped for a question after the contest because of constraints violation and a re-evaluation was performed. This was heavily criticised. As @gdisastery1 says, it is unethical to change the data once even a single contestant has attempted and got points on it.
Personal Opinion
I am not getting the unethical part. How is this ethically wrong? The person who submitted the solution knows very well that his solution should get TLEd. So ideally, he should not be getting points for that solution.
In this situation, Jury had missed a test case which we had realized a lot before. It’s not like taking one’s submission and doing stress test to find the test case against it.
Personally, I think that adding test cases is ok.
Being unethical? No. We haven’t cheated or lied. Being unfair? Yes. We agree that it is undesirable to change test data once the contest has started. There is no argument here. However, due to human errors, at times the identification of the test data being weak happens only during the contest. At that time we have mostly a couple of options. Either change the test data at that time or leave things unchanged. Both are unfair to a different set of people. For the sake of a better contest, we tend to falter towards the first. And we agree that it is being unfair to a set of people.
A better use of the notification system could have notified the users who were on the system. But still there is no way to make everyone aware of the changes. This is an exceptional situation and we do apologise. Going ahead, we would try and avoid this scenario. In case, we are unable to, we will try and notify users in a better way and extend the contest. I hope this would not need to happen.
2 Likes | 1,994 | 8,048 | {"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} | 4.15625 | 4 | CC-MAIN-2022-21 | latest | en | 0.887799 |
https://www.asmail.be/msg0055207337.html | 1,628,014,001,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00520.warc.gz | 622,318,556 | 3,342 | TIFF and LibTiff Mail List Archive
# 2002.01.21 17:23 "Re: How to interpret 16-bit GrayScale image?", by Daniel McCoy
Then convert from 16 bps to 8 bps by dividing by 257. (sic, 257. NOT 256)
I've seen this comment a couple of times recently and find it very strange just hanging out there as if it's some gem of image processing wisdom.
When converting the OTHER way, from 8 bits UP to 16 bits, of course, you want to MULTIPLY by 257 and not 256 so that you fill in the low bits. That way FF becomes FFFF instead of FF00. Clearly better. No question. Crazy to do otherwise.
But when converting from 16 bits DOWN to 8 bits, DIVIDING by 257 (hex 101) doesn't make as much sense to me as using 256 (hex 100).
FFFF / 101 = FF ok
FFFE / 101 = FE are you sure you want this?
FEFE / 101 = FE ok
FEFD / 101 = FD are you sure you want this?
...
0101 / 101 = 01 ok
0100 / 101 = 00 are you sure you want this?
00FF / 101 = 00 ok
If you use 256 (hex 100) instead:
FFFF / 100 = FF same as 101
FFFE / 100 = FF seems better to me.
FEFE / 100 = FE same as 101
FEFD / 100 = FE seems better to me.
...
0101 / 100 = 01 same as 101
0100 / 100 = 01 seems better to me.
00FF / 101 = 00 same as 101
Of course, it depends on what use you plan for the data. But for viewing 16 bit greyscale image on 8 bit hardware, I think dividing by 256 (shifting down eight bits) would give a better display.
Another way to describe the difference:
When you divide by 256, then exactly 256 16-bit grey levels translate into each of the 256 possible 8-bit grey levels. Seems like that makes good use of the limited 8-bit display.
If you divide be 257, then 257 grey 16-bit grey levels translate into the lower 255 possible 8-bit grey levels and the single left over 16-bit grey level (FFFF) translates into the single left over 8-bit grey level (FF).
The only way I could see using 257 is if you are adding 128 or 127 to round them up first.
Daniel McCoy | 657 | 1,986 | {"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-2021-31 | longest | en | 0.660296 |
https://health-unleashed.com/qa/question-what-day-will-be-81-days-after-today-if-today-is-sunday.html | 1,600,948,132,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400217623.41/warc/CC-MAIN-20200924100829-20200924130829-00764.warc.gz | 442,446,034 | 7,751 | # Question: What Day Will Be 81 Days After Today If Today Is Sunday?
## What day is 80 days away?
To get exactly eighty weekdays from now, you actually need to count 112 total days (including weekend days).
That means that 80 weekdays from today would be December 22, 2020..
## What day is it going to be in 13 days?
– Today is : Wednesday, September 2, 2020. – The date after 13 days is : Tuesday, September 15, 2020. – It is the 259th day in the 38th week of the year.
## What is 100 days from now?
– Today is : Wednesday, September 2, 2020. – The date after 100 days is : Friday, December 11, 2020. – It is the 346th day in the 50th week of the year.
## What day is 199 days away?
To get exactly one hundred and ninety-nine weekdays from now, you actually need to count 279 total days (including weekend days). That means that 199 weekdays from today would be June 7, 2021.
## How many days is 180 From today?
To get exactly one hundred and eighty weekdays from now, you actually need to count 252 total days (including weekend days). That means that 180 weekdays from today would be May 12, 2021.
## How many days was 6969 days ago?
If you multiply 6969 by 24, then you will get how many hours since 6969 days ago: 6969 days ago is 167256 hours ago. You can also multiply 6969 by 1,440 to find out how many minutes 6969 days ago was: 6969 days ago was 10035360 minutes ago.
## What day is 52 days away?
To get exactly fifty-two weekdays from now, you actually need to count 72 total days (including weekend days). That means that 52 weekdays from today would be November 12, 2020.
## What day is 110 days away?
Counting forward, the next day would be a Thursday. To get exactly one hundred and ten weekdays from now, you actually need to count 154 total days (including weekend days). That means that 110 weekdays from today would be February 3, 2021.
## What day is it going to be in 1000 days?
To get exactly one thousand weekdays from now, you actually need to count 1,400 total days (including weekend days). That means that 1000 weekdays from today would be July 3, 2024.
## What was 30 days ago today?
30 days before today would be Saturday, August 1, 2020. There are 31 days in August 2020. 2020 is a leap year, so there are 366 days in this year (there are 29 days in February 2020).
## What is the 84th day from today?
Days from Today Conversion TableDaysDate Days from TodayDate (Y-m-d)84 DaysWed 25th Nov 20202020-11-2585 DaysThu 26th Nov 20202020-11-2686 DaysFri 27th Nov 20202020-11-2787 DaysSat 28th Nov 20202020-11-2846 more rows
## What day is 105 days away?
Counting forward, the next day would be a Thursday. To get exactly one hundred and five weekdays from now, you actually need to count 147 total days (including weekend days). That means that 105 weekdays from today would be January 13, 2021.
## What was 1 000 days ago?
1000 days ago was 1440000 minutes ago.
## How long is 150 days today?
To get exactly one hundred and fifty weekdays from now, you actually need to count 210 total days (including weekend days). That means that 150 weekdays from today would be March 30, 2021.
## What date is 144 days away?
That means that 144 weekdays from today would be March 22, 2021. | 875 | 3,234 | {"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} | 3.75 | 4 | CC-MAIN-2020-40 | latest | en | 0.972241 |
https://www.thebalance.com/calculating-mortgage-interest-real-estate-investors-2866804 | 1,519,076,097,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812841.74/warc/CC-MAIN-20180219211247-20180219231247-00012.warc.gz | 943,508,052 | 31,051 | # How to Calculate Mortgage Interest for the Real Estate Investor
In most new real estate mortgage loans, there is a HUD-1 item for the Buyer to prepay the lender for the interest on the loan from the closing date to the end of the month of closing. It is because interest is paid in arrears, and the first payment usually falls on the first day of the month following the month of closing.
• A closing on June 15th would require interest prepaid for the period from June 16 to July 1st.
• The first payment would be due on August 1st, with interest in arrears for the month of July.
As an example, let's us a mortgage amount of \$272,000 with an interest rate of 7.0%, and the dates above.
• \$272,000 X .07 = \$19,040 in annual interest
• \$19040 / 365 calendar days = \$52.16/day in interest
• \$52.16 X 15 days in June = \$782.40 in interest to prepay
• The August 1st payment would include interest in arrears for July
### Let's Look at Mortgages & Lending
There are mortgage payment and interest calculators all over the Web now. There are smartphone apps that do it as well. Let's look at some information points about mortgages and lending with links for more research.
How Fixed Rate Loans Work
Each month's payment is equal to the interest rate times the principal, plus a small percentage of the principle itself. Since a bit of the principal is paid off each month, that makes the interest payment on the remaining principal a little less too. As a result, more of your monthly payment goes toward the principal each month. Therefore, at the beginning of the loan, most of the payment goes towards interest while most of it goes towards principal at the end of the loan.
15-year vs. 30-year Mortgages
While there are several types of mortgages available for home buyers, most Americans think of financing their home purchase in terms of the archetypal 30-year fixed mortgage. When presented with another option, such as the 15-year fixed mortgage, the typical home buyer will view it as the more expensive option. But what most of these American home buyers don't know is that in the long-run, a 15-year fixed mortgage may be more likely to meet their financial goals.
Adjustable-rate mortgages (ARMs) are home loans with a rate that varies. As interest rates rise and fall in general, rates on adjustable-rate mortgages follow. These can be useful loans for getting into a home, but they are also risky. This page covers the basics of adjustable rate mortgages.
Pros and Cons of Adjustable Rate Mortgages
Adjustable-rate mortgages (ARMs) are home loans with a rate that varies. As interest rates rise and fall in general, rates on adjustable-rate mortgages follow. These can be useful loans for getting into a home, but they are also risky. This page covers the basics of adjustable rate mortgages.
The Blanket Real Estate Mortgage
Buyers, particularly in the commercial real estate markets, use blanket mortgages for a number of reasons. Lenders make money making loans. If the numbers work and they get enough security, commercial lenders will originate blanket mortgages used in commercial property investments. Perhaps your next investment would be better served using a blanket mortgage.
A reverse mortgage can provide money when you need it, but do your homework before applying for a reverse mortgage. If your income from retirement funds, savings and Social Security benefits don't cover your expenses, or you'd like the financial freedom to enjoy your retirement years a bit more, you can use the equity in your home to apply for a reverse mortgage.
The mortgage market took a big hit in the crisis that began in 2006. However, by around 2015, lenders and the agencies that guarantee home loans were coming out of their overly-tight stance and money was flowing better. | 838 | 3,813 | {"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-2018-09 | latest | en | 0.964175 |
https://www.avrfreaks.net/forum/stepper-motor-pulses-wrong-calculated | 1,627,709,860,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00693.warc.gz | 671,030,836 | 12,864 | ## stepper motor pulses wrong calculated
6 posts / 0 new
Author
Message
Hello everybody. I am kind of new to avr-gcc and I have a problem. I would appreciate any help.
I have a project in which I use USART to talk to uC and accordingly the uC sends pulses to a stepper motor that controls a syringe. I have counted that the maximum number of pulses that can be send are 136620.
So to be more user-friendly I decided to send the percentage of movement required. For example if someone wants to move the syringe by 30% of the total allowable distance he sends to uC "move30". The uC filters out "move" and 30 stays. Theoretically 30% means (136620/100)*30 = 40986 pulses.
I use the code below and what happens is that until 20% everything is ok. When I send move21 the pulses send are only 1366 (like move1) and then move22 gives 0 pulses. If I send again move22 the pulses send become 1466946000000.
```void move(char s[], char c)
{
unsigned char sComm[2];
unsigned char sComm2[6];
unsigned char i = 5, j = 0, k = 0;
unsigned long int pulses_dis = 0, counter = 0;
double pulses_1 = 0;
while( (s[i] != '\0') && (j <= 3) )
{
if( (s[i] >= '0') && (s[i] <= '9') )
{
sComm[j++] = s[i++];
}
else
{
sendString("Error - Comm");
sendChar(c);
sendString(" received a non integer: ");
sendChar(s[i]);
sendChar('\r');
}
}
sComm[j] = '\0';
if(j>3)
{
sendString("Error - Comm");
sendChar(c);
sendString(" number too large\r");
sendChar('\r');
}
else
{
sendString("\rThank you for sending the number: ");
sendString(sComm);
sendChar('\r');
pulses_1 = (1366.2 * atoi(sComm));
dtostrf(pulses_1,6,1,sComm2);
sendString(sComm2);
sendChar('\r');
pulses_dis = round(pulses_1);
for (counter = 1; counter <= pulses_dis; counter++)
{
PORTB |= (1 << PB0);
_delay_us(30);
PORTB &= ~(1 << PB0);
_delay_us(30);
}
}
}
```
I almost forgot! When I reset the uC and then I send move21 it works.
Thanks in advance for any help!
Maybe,
`pulses_1 = (1366.2 * atoi(sComm));`
needs to be:
`pulses_1 = (1366.2 * atol(sComm));`
One thing for sure, you are rolling over somewhere...
You can avoid reality, for a while. But you can't avoid the consequences of reality! - C.W. Livingston
Quote:
``` while( (s[i] != '\0') && (j <= 3) )
{
if( (s[i] >= '0') && (s[i] <= '9') )
{
sComm[j++] = s[i++];
}
else
{
sendString("Error - Comm");
sendChar(c);
sendString(" received a non integer: ");
sendChar(s[i]);
sendChar('\r');
}
}```
i and j are only updated if s[i] is a digit. Your code enters an endless loop if a received character is invalid.
Quote:
`double pulses_1`
float and double is the same for WinAVR applications. Instead of using floting point math you can do all by using integer math:
`pulses_1 = ((136620 * (unsigned long) atoi(sComm) + 50) / 100);`
Much faster and smaller code. The "+ 50" is the integer way for rounding (add 1/2 divisor before divide). I never have used floating point math in a µC project. Maybe it causes the trouble.
Regards
Sebastian
micocarl I have already tried that but it didn't work. Thanks anyway!
S-Sohn the last code snippet did it. Thank you very much. I knew it had to do something with the type casting. Also by using your math code the code size decreased amazingly. Thanks a lot.
hi everybody,
i am facing little problem.i am trying to make a small cnc lathe with avr, and i am using bascom to control stepper motion.
but i can't program it to higher rpms.
max speed it could run is 60 rpms,
so 60 rpms* 200 steps per revolution=12000 steps per minute generated by the atmega16.
i am using no delay berween the pulses
i tried 25us delay also.
where is the problem ?i am using a 5A h bridge made up of mosfets.
can anybody gain my rpms? | 1,063 | 3,672 | {"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} | 2.703125 | 3 | CC-MAIN-2021-31 | latest | en | 0.760489 |
https://theansweris27.com/drag-reduction-system-drs/ | 1,726,402,558,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00867.warc.gz | 512,825,642 | 19,105 | # Drag Reduction System (DRS)
Wanting to get some in depth knowledge about competition car aerodynamics, the performance of a Drag Reduction System DRS should not pass unnoticed. In the following video, two engineers at Caterham F1 show us how the DRS works in F1.
According to the video, the main feature in a DRS is a flap with two positions (ON/OFF) that at the request of the driver, and if permitted, changes its angle of incidence. This will change the flow pattern around the aileron and thus reduce drag allowing for faster top speeds and aiding in overtaking. To see this flow pattern changes I took the geometry (Fig. 1) from STAR-CCM+ tutorial files and made some minor changes so I could pivot the flap at will around its tailing edge.
Figure 1. Simplified wing geometry for DRS analysis. The aileron has a wing span of 1.4 meters and a main chord of 0.3 meters whilst the flap has a chord of 0.15 meters.
I maintained the mesh and physic model parameters to that from the tutorial:
• Free stream: 100 kph
• Turbulence model: SST (Menter) K-Omega
• Three dimensional, coupled, steady flow
• Polyhedral mesh with prism layer
## Results
When the DRS is in off position, the flap will be responsible for the wing generating maximum downforce —at a cost of drag—. As we can see in Figure 2, there is a bigger high pressure region above the wing as the air has a smaller gap to pass through, and thus more air is forced above the flap.
The air moving upwards with the DRS OFF is clearly visible in Figure 3 where the streamlines are represented coloured by velocity. In this case the DRS works as a slotted flap does in an aircraft, allowing high pressure airflow energize the lower pressure over the bottom, thereby delaying flow separation, as can be deduced by looking at the turbulence kinetic energy in Figure 4. In this case, the flap is responsible for 70 % of the drag generated by the wing assembly while contributing only a 20 % to the generation of downforce.
On the other hand, with the DRS ON, streamlines are less disturbed, thus contributing to a reduction in drag. Now, flap’s proportion in drag generation goes down to 61 % whilst maintaining about 20 % share in downforce.
Globally, the DRS reduces both drag and downforce, the former being steeper. In this case, there is a 31 % reduction in drag at 10 degrees, which is about 3 times the reduction obtained in an F1 race car according to Simon McBeath [2]. Nonetheless, the trend in both graphs (his and mine) are similar, while the discrepancies are attributable to geometric differences.
Figure 2. Contours of Pressure at DRS OFF (above) and at flap pivoted 10 degrees (DRS ON, below)
Figure 3. Streamlines at DRS OFF (above) and at flap pivoted 10 degrees (DRS ON, below). Coloured by velocity.
Figure 4. Turbulent kinetic energy at DRS OFF (above) and at flap pivoted 10 degrees (DRS ON, below).
## References
[1] User Guide STAR-CCM+ Version 8.06. 2013.
[2] Racecar-engineering.com. 2011. DRS: The Drag Reduction System explained | Racecar Engineering. [online] Available at: http://www.racecar-engineering.com/articles/f1/drs-the-drag-reduction-system/ [Accessed: 14 Jan 2014]. | 783 | 3,175 | {"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 | 0.928087 |
https://www.khanacademy.org/math/geometry-home/geometry-coordinate-plane/geometry-coordinate-plane-4-quads/v/the-coordinate-plane | 1,675,116,155,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00622.warc.gz | 858,582,339 | 103,454 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Main content
# Points on the coordinate plane examples
The coordinate plane is a two-dimension surface formed by two number lines. One number line is horizontal and is called the x-axis. The other number line is vertical number line and is called the y-axis. The two axes meet at a point called the origin. We can use the coordinate plane to graph points, lines, and more. Created by Sal Khan and CK-12 Foundation.
## Want to join the conversation?
• at about how come Sal only puts four dash marks on his coordinate plane instead of more when he could have easily fit more than that?
(0 votes)
• just because he can't fit every single solitary number into one small screen
(1 vote)
• you all goofy neerds
(3 votes)
• I'm not a nerd.
(4 votes)
• At , are quadrants important?
(1 vote)
• Definitely! If you're given a problem, and even in real life, quadrants can help determine that where you plotted the coordinates is correct. It sort of narrows down the options!!
(5 votes)
• why Are the y axis and the x axis always have to be apart
(1 vote)
• why cant the x axis go up and down and why cant the y axis go side to side
(2 votes)
• The way that axes work is just our convention. If you want a q-axis and an m-axis, then be my guest. X is usually used to refer to a horizontal direction/motion, and Y refers to vertical. This is just the way some person thought of it, and this is the way we've been doing it for the past hundreds of years.
(2 votes)
• Why are the axis x and y, why not a and g?
(1 vote)
• It is just a convention, something people agreed. You could write the axis as a and g, that's not wrong, but because of the convention there is people can easily identify the horizontal axis as the x axis, and the vertical one as the y axis.
(4 votes)
• Isn't this shape plotted on the coordinate plane supposed to be a square? It says so in the directions? Or is square just a general term for 4 sided figures on a coordinate plane?
(1 vote)
• The shape should be a square because a general term for a 4 sided figure in any situation would be a quadrilateral.
(4 votes)
• why are the quadrants counter clockwise?
(2 votes)
• Some of these things are simply named by convention, but I also think it may have something to do with the way that we "sweep out" angle in trigonometry; clockwise from the positive x-axis. I'm not really sure if that's why, but it seems applicable.
(2 votes)
• What if the coordinate was directly on the x or y axis(es)? What quadrant would that point be in?
(1 vote)
• It is then not a point within one of the quadrants, it is simply on the respective x or y axis, or if in directly in the center, ex: (0,0) then it is said to be at the "origin".
(5 votes)
• How do you find the total area of a figure on a coordinate plane?
(2 votes)
• It all depends on the figure. For example, a square, rectangle, or parallelogram in general would be b*h(base times height). A triangle would be b*h/2.
(1 vote)
## Video transcript
What we're going to do in this video is, through a bunch of examples, familiarize ourselves with the x,y-coordinate plane. And first we're going to just look at some points that are already plotted and figure out their coordinates. Then we're going to look at some coordinates and figure out where those points are. Then we'll do one more problem. So let's figure out what are the coordinates of these points? So you have this point right here, A. So its x-coordinate, you can see it right there. You just drop down. Where does it intersect the x-axis? x is equal to 5. So it's the point 5 comma and y is going to be equal to 6. 5 comma 6. Now this point B here, what's the x-coordinate? It is 5 to the left. 5 to the left of the x-axis. This is negative 5. Its x-coordinate is negative 5. y-coordinate is, if you just go straight to the right, you're going to hit y is equal to 5. y is equal 5. Let me switch colors. C. I think you're getting the hang of this. Let's do the y-coordinate first. The y-coordinate is 3. You see that right there. And then the x-coordinate is negative 2. Negative 2. You always put the x-coordinate first. That's just the convention we use. D, x-coordinate negative 2. You see that right there. y-coordinate negative 2, as well. Let me get another color. E, let's do the y-coordinate. We'll figure it out first, but you always have to write it second. It's negative 4. You see that right there, the y-coordinate. The x-coordinate is 3. And then finally, F. The x-coordinate is 2. And the y-coordinate is negative 6. Hopefully that gives you a sense of at least figuring out the coordinates. Now let's go the other way. Let's start with coordinates and figure out where those points are. So you have this first one. I'll do it lowercase case a in parentheses to differentiate it from this uppercase A. So it's at 4 comma 2. x is equal to 4. y is equal to 2. So that's that point right there. Let's do the next one. Let me do it in a color that you'll be able to read. b. x is equal to negative 3. y is equal to 5.5. So you go all the way up to 5.5. y is equal to 5.5. So that is the point lowercase b with parentheses around it. Then c, 4 negative 4. x is equal to 4. y is equal to negative 4. Right over there. And then one last one. I'll do it in orange. d, x is negative 2, y is negative 3. Right there. That's the d with parentheses. And you could have gone the other way. You could have said, hey, y is equal to negative 3. x is equal to negative 2. So you could go to the left and down. Or you could go down and to the left. And you're still going to get to the same point. So hopefully that gives you a good sense of how to figure out coordinates. Or if you're given coordinates, how to figure out where to plot something on the x,y-coordinate plane. Now let's do a slightly more involved problem. So it says the following 3 points are 3 vertices of square A, B, C, D. Plot them on a graph. Then determine what the coordinates of the fourth point, D, would be. All right, let's plot these on a graph, as they tell us to do. All right. That'll be my y-axis. That's my y-axis. The vertical axis. That'll be my x-axis. And let me put some-- let me mark it. So that's x equals 1, 2, 3, 4. This is x is equal to negative 1, negative 2, negative 3, negative 4. That's y is equal to 1, 2, 3, 4. This is y is equal to negative 1, negative 2, negative 3, negative 4. I could write that this y equals 4. This y equals negative 4. x is equal to 4. x is equal to negative 4. And let's see. Let's plot these points. So first, we have the point A is equal to negative 4, negative 4. So we go x is negative 4. And then y is negative 4. So we drop down 4 right there. And that is our point A. Negative 4, negative 4. And just to familiarize yourself with a labeling scheme that you may or may not have seen before, is that people label these sections of the coordinate plane. They call this the first quadrant. They call this the second quadrant. They call this the third quadrant. And they call this the fourth quadrant. And these are just the Roman numerals for I, II, III, and IV, So this point is in the third quadrant. When we looked up at this stuff over here, these points are in the fourth quadrant. These are in the third, second, first. Just an interesting thing to know. Sometimes someone might ask you, what quadrant is that point in? And you just say, OK, I see. If they're both negative, they're going to be in the third quadrant. If just the y is negative, but the x is positive, you're going to be in the fourth. If they're both positive, you're in the first. If y is positive, but x is negative, you're in the second. And we'll talk a little bit about that as we plot these points. So point B, x is positive. It's 1, 2, 3. And y is negative 4. So we drop down here into the fourth quadrant. That is the point B. It's 3, negative 4. So we can already see the bottom of our rectangle that they're talking about, right there. And notice, both of these have the exact same y. They're both at the same level below the x-axis. And then what's the next point? Point C is 3 comma 3. So 3 comma 3. It's in the first quadrant. Both of its coordinates are positive. 3 comma 3. Both x and y are positive. And notice, it's on the same vertical as B. It has the same x value. They both have an x value of 3. So it's right above it. Right above it. Now we have to figure out the last point here. Well, the point is going to have to be on the same vertical as this point. It's going to have to be on the same vertical as this point, which means it's going to have the same x value as this point. So its x value is going to be negative 4. And then it's going to have to be on the same horizontal as this point. It's going to have to be on the same horizontal as that point. So it's going to have to have the same y value at the same height above the x-axis. So it's going to have to be 3. So that is our point D. Notice it's at negative 4, right above A. And it's at y is equal to 3. Right to the left of point C. And we are done. | 2,381 | 9,232 | {"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} | 3.375 | 3 | CC-MAIN-2023-06 | latest | en | 0.944934 |
https://in.answers.yahoo.com/question/index?qid=20090320072540AAKYRaM | 1,576,399,626,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307797.77/warc/CC-MAIN-20191215070636-20191215094636-00281.warc.gz | 398,810,076 | 20,255 | # why current is preferred to measure in series and voltage in parallel?
Relevance
• Anonymous
Because, the current measuring device, like an ammeter, would have a very low resistance, and if you connect it in parallel, it would short circuit the phase and neutral.
Similarly, the voltmeter would have a very high resistance, and if you connect it in series, all the voltage would drop in the voltmeter itself and no voltage would be available to the load.
• cepero
Lv 4
3 years ago
Voltage In Parallel
The reason is because current is a measurement of flow - how many electrons are passing through the wire, whereas voltage is a measure of the difference is energies between two points.
In order to measure the amount of electrons passing through a wire, you actually have to place the multimeter within the flow. To measure flow, the multimeter places a VERY LOW resistance into the circuit and measures the voltage drop across that resistance. By using a very low resistance, the multimeter has a negligible impact on the circuit when placed in series.
In order to measure the energies between two points, the multimeter is place in parallel in the circuit with a HUGE resistance and measures the current through the resistance. By using a huge resistance, the multimeter has a negligible impact on the circuit when placed in parallel.
Note: when placing a multimeter into a circuit in series with it turned to the voltage setting or in parallel with the current setting, you will either destroy the multimeter or trigger the circuit breaker or fuse.
In simple words;
Current is a measurement of amount of charge carriers (electrons) flows through a conductor at that instance. So to measure the current the current should flows through the meter, hence the meter will get the value of current.
Voltage is otherwise known as potential difference. That is potential difference between the two points. So to measure the voltage difference between two points (one is line and other is neutral that is zero potential) we have to connect the meter to connect the two points whose potential difference have to be known.
To know the difference between voltage and current :
Voltage is also called EMF that is Electro Motive Force. That is this is the cause to pull the (free) electrons in the conductor to move.
Current is the amount of electrons moves (flows) through, in a instance of time.
Note: Without voltage no current is possible, but without current there will be a voltage.
For better understanding:
Comparing the electrical system to the liquid pipeline system.
Where a pressure in the pipe is Voltage and the amount of fluid flows out from the pipe in a instance of time is current.
Source(s): thennarasu.in@gmail.com | 549 | 2,745 | {"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-2019-51 | latest | en | 0.956263 |
scaryreasoner.wordpress.com | 1,500,720,153,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423992.48/warc/CC-MAIN-20170722102800-20170722122800-00568.warc.gz | 758,981,868 | 48,137 | ## Tour of a Procedurally Generated Solar System
2017-01-31 — Here’s a little tour of a procedurally generated solar system in my little hobby project, an open source multiplayer networked starship bridge simulator game, Space Nerds In Space.
## Code for 3D turret aiming
January 14, 2017 — Say you have a 3D space game, with NPC ships that have turrets on them, and the turrets can be oriented any which way on those ships. How do you make the turrets swivel around and aim smoothly and intelligently?
Note: The code in these examples is released under the GPL v2 license. See:
Step 1: Understand your turret model.
In my case, my turret model is oriented such that it shoots down the positive X axis, and can rotate around the vertical Y axis and rotate around the horizontal Z axis. This is the same canonical orientation used by all the models in my game — the “unrotated state” has from the model’s point of view, down the X axis being straight ahead, positive Y as “up” and positive Z to the right.
Step 2. Understanding the turret’s state.
• The turret has a location (x, y, z)
• The turret has a “rest orientation” This is the orientation when the turret is in an unrotated state, but perhaps attached to some other object in a way that gives this state some arbitrary non-constant rotation that is different than the unrotated state the identity quaternion represents.
• The turret has a “current orientation” This is the “rest orientation” plus some other arbitrary orientation accounting for where ever the turret might currently be pointed.
• The turret cannot rotate infinitely fast. There is some limit to how much the turret can rotate in each of its two axes per unit time.
Step 3. What do we want to be able to do?
We’d like to be able to ask the turret to aim at something, and get a new orientation that obeys the constraints of a turret (not freely rotating, but only rotating on two axes) and allowing the turret to be attached to something.
So let’s say we’d like a C function something like this:
```struct turret_params {
float elevation_rate_limit; /* how much can turret move in one step in radians */
float azimuth_rate_limit;
};
/* union quat is a quaternion */
union quat *turret_aim(double target_x, double target_y, double target_z, /* in: world coord position of target */
double turret_x, double turret_y, double turret_z, /* in: world coord position of turret */
union quat *turret_rest_orientation, /* in: orientation of turret at rest */
union quat *current_turret_orientation, /* in: Current orientation of turret */
const struct turret_params *turret, /* in: turret params, can be NULL */
union quat *new_turret_orientation); /* out: new orientation of turret */
```
So how do we write such a function? Here’s one approach.
1. Transform the target into the turret’s coordinate space
2. Calculate the desired azimuth and elevation angles
3. Calculate the turret’s current azimuth and elevation
4. Calculate the delta between current and desired azimuth and elevation
5. clamp these deltas to the maximum angular rate the turret can move
6. Apply deltas to current elevation and azimuth
7. Convert new elevation and azimuth to the new orientation quaternion
So, going through these one by one:
Step 0. Declare some variables we’ll need:
``` union vec3 yaw_axis = { { 0.0, 1.0, 0.0 } };
union quat yaw, pitch, inverse_rest;
union vec3 pitch_axis;
union vec3 to_target, to_current_aim;
float current_azimuth, azimuth;
float current_elevation, elevation;
float delta_elevation, delta_azimuth;
float xzdist;
```
Step 1. Transform the target into the turret’s coordinate space
(because we want to make step 2 easier)
``` /* Compute vector from turret to target */
union vec3 to_target;
to_target.v.x = target_x - turret_x;
to_target.v.y = target_y - turret_y;
to_target.v.z = target_z - turret_z;
/* Compute inverse rotation from turret rest orientation */
quat_inverse(&inverse_rest, turret_rest_orientation);
/* Convert target into turret's coordinate space */
quat_rot_vec_self(&to_target, &inverse_rest);
```
Step 2. Calculate the desired azimuth and elevation angles:
``` azimuth = atan2f(-to_target.v.z, to_target.v.x);
xzdist = sqrtf(to_target.v.z * to_target.v.z + to_target.v.x * to_target.v.x);
elevation = atan2f(to_target.v.y, xzdist);
```
At this point, you could skip to step 7, and the turret would snap instantly
to the correct orientation — but we want it to move smoothly.
Step 3. Calculate the turret’s current azimuth and elevation
``` to_current_aim.v.x = 1.0; /* vector pointing down the x-axis */
to_current_aim.v.y = 0.0;
to_current_aim.v.z = 0.0;
/* Rotate it into the current turret's orientation -- now the vector
* points in the same direction as the turret points in world coord
* system.
*/
quat_rot_vec_self(&to_current_aim, current_turret_orientation);
/* Now convert into the turret's local coord system */
quat_rot_vec_self(&to_current_aim, &inverse_rest);
/* Now calculate current azimuth and elevation of the turret */
current_azimuth = atan2f(-to_current_aim.v.z, to_current_aim.v.x);
xzdist = sqrtf(to_current_aim.v.z * to_current_aim.v.z + to_current_aim.v.x * to_current_aim.v.x);
current_elevation = atan2f(to_current_aim.v.y, xzdist);
```
Step 4. Calculate the delta between current and desired azimuth and elevation
``` delta_elevation = elevation - current_elevation;
delta_azimuth = azimuth - current_azimuth;
```
Step 5. Clamp these deltas to the maximum angular rate the turret can move in both axes.
``` delta_elevation = elevation - current_elevation;
delta_azimuth = azimuth - current_azimuth;
if (delta_azimuth > turret->azimuth_rate_limit)
delta_azimuth = turret->azimuth_rate_limit;
else if (delta_azimuth < -turret->azimuth_rate_limit)
delta_azimuth = -turret->azimuth_rate_limit;
if (delta_elevation > turret->elevation_rate_limit)
delta_elevation = turret->elevation_rate_limit;
else if (delta_elevation < -turret->elevation_rate_limit)
delta_elevation = -turret->elevation_rate_limit;
```
Step 6. Apply deltas to current elevation and azimuth
``` azimuth = current_azimuth + delta_azimuth;
elevation = current_elevation + delta_elevation;
```
Step 7. Convert new elevation and azimuth to the new orientation quaternion
``` /* Recall yaw_axis is a vector pointing up the Y axis. Rotate it
* into the turret's coordinate space, and build a yaw quaternion
* from this rotated yaw_axis and the azimuth
*/
quat_rot_vec_self(&yaw_axis, turret_rest_orientation);
quat_init_axis(&yaw, yaw_axis.v.x, yaw_axis.v.y, yaw_axis.v.z, azimuth);
/* Rotate the turret_rest_quaternion by the yaw quaternion */
quat_mul(new_turret_orientation, &yaw, turret_rest_orientation);
/* Set up a pitch axis -- around the Z axis */
pitch_axis.v.x = 0.0;
pitch_axis.v.y = 0.0;
pitch_axis.v.z = 1.0;
/* Rotate it into the turret's coordinate space (which has the yaw
* incorporated already now) and build a pitch quaternion out of the
* rotated pitch axis and elevation
*/
quat_rot_vec_self(&pitch_axis, new_turret_orientation);
quat_init_axis(&pitch, pitch_axis.v.x, pitch_axis.v.y, pitch_axis.v.z, elevation);
/* Rotate the turret orientation by the pitch quaternion */
quat_mul(new_turret_orientation, &pitch, new_turret_orientation);
return new_turret_orientation; /* and we're done. */
```
## Space Nerds In Space Big Honking Death Machine
January 2, 2017 — Star Wars has its Death Star, Star Trek has its Borg Cube, its high time Space Nerds In Space had its Big Honking Death Machine, which is what I’ve been working on lately.
The simplest possible approach would have been to build a 3D model much as I’ve done previously and just put it in. But I wanted the players to be able to interact with this thing in a way that you can’t interact with a simple model, at least the way they are currently implemented. I wanted players to be able to fly around and even inside this giant thing, and bump into the solid parts but fly through the empty parts. Doing that with a normal 3D model is pretty hard, too hard for me at the moment. It occurred to me that if I could make a kind of generic “block” with arbitrary height, width, and depth and orientation and I could get texture mapping and more importantly collision detection working with such a block, then I could arrange such “block” objects in a hierarchy, with relative positions and orientations and in this way be able to construct large “ships” out of many of these “block” objects — much like LEGOs, but without the alignment constraints. And so that’s what I’ve done.
Each of the blocks uses the same mesh, a unit cube, but each block has its own non-uniform scaling factors for x, y, and z. The collision detection is done using “oriented bounding boxes”, as described in Christer Ericsson’s book, Real-Time Collision Detection. The collision resolution is very simple minded at the moment. In the detection phase, the “closest” point on the surface of the block is identified, and a vector is constructed from this point to the player’s ship and the ship is “pushed back” a fixed amount in this direction and the velocity is set to zero. This mostly works, though it doesn’t look or feel all that natural, and once in a while penetration of walls may occur.
## It is up to us
•June 30, 2016 • 6 Comments
June 29, 2016 — It’s been a bit over 24 hours since the terrorist attacks at Ataturk airport in Turkey as I write this.
It seems quite obvious to me that the attacks are religiously motivated. If those people weren’t adherents of a religion, they wouldn’t do those things. They might still be unhappy, they might still have various grievances, but they would not explode themselves in a quest to kill strangers perceived to be enemies in some purported exchange for some imagined eternal blissful afterlife awarded by some imagined deity for fighting in some imagined holy war. For that, religion is required. And the throbbing core that powers the Death Star known as religion is faith.
Faith is the idea that it is a good idea and a virtuous deed to deliberately attempt to be more certain about something than what is warranted by the available evidence. Deliberately willing oneself to be excessively certain. “Look at me! I’m so certain I’m right! And I clearly don’t have enough evidence to be so certain! I have faith! I’m so faithful! What a good boy am I for being so faithful!”
Faith is never described by the faithful so baldly as I have done, because to do so would be to highlight the idiocy of faith, and nobody wants to think of themselves as having behaved like an idiot. But faith is clearly idiotic. It is not a good idea to deliberately be excessively certain. It is not a vritue. It is the opposite of a virtue. The all too common and ridiculous cry of the defeated theist, “Well, it takes more faith to be an atheist…” betrays the believer’s subconscious realization that faith is something to be avoided, that it is not a reliable way to know what’s true.
How should we combat the type of faith that drives people to commit such acts? I would suggest the most effective way (not that I think it will be all that effective) is not to coerce compliance by means of threats, weaponry, and soldiers — I do not think this will do anything to mitigate cancerous faith. Likely, it will only make it stronger. I suggest the most effective tactic — or at least a tactic which should be tried — is to educate the faithful by argumentation — help them to reason their way out of their faith. And most importantly, I think we should directly attack the basic concept of faith itself. The concept of faith should be given the reputation it deserves. The statement, “He is a man of great faith” should not be regarded as praise, but as an insult.
We cannot rely on members of other religions to do this. A Christian and a Muslim and a Jew arguing about religion are like three blind men arguing about which of three unremarkable Thomas Kinkade paintings is the best possible painting in the universe. Their arguments will all be constructed out of the sheerest nonsense.
And you will not see any U.S. politician making any sort of argument that Islam or Christianity or Judaism or any religion has no basis in fact, or that faith is the idiocy that it so obviously is. That is because (among other reasons) the U.S. Constitution requires that the government remain neutral towards religion, not favoring any religion (or no religion) over any other religion (or no religion). Their hands are tied — even if they wanted to make such an argument, they are not permitted to do so. Nor are members of the United States armed forces while acting on behalf of the United States, as I understand it.
So it is left to us — private individuals — to make such arguments.
## Space Nerds In Space: 3D “demon” screen
May 29, 2016 — When Space Nerds In Space was converted from a fundamentally 2D game to a proper 3D game a couple of years ago, one of the things which didn’t really make the transition was the “demon” screen, or “gamemaster” screen. That screen was fundamentally a 2D screen, and it was not and still is not obvious how such a thing ought to be done in 3D. I have finally gotten around to taking a stab at it though. Whether I’ve been successful, I’m not sure. In any case, here’s how it looks.
## Speech Recognition and Natural Language Processing in Space Nerds In Space
•May 14, 2016 • 1 Comment
May 14, 2016 — Recently while working on my multiplayer networked starship bridge simulator game, Space Nerds In Space, it occurred to me that it would be cool if you could talk to the ship’s computer, somewhat like in the TV show Star Trek. So I made it so.
Watch demo on Youtube
To get such a thing working, There are three main problems that need to be solved:
• Text to Speech — The computer has to be able to respond verbally.
• Speech recognition — Your verbal commands need to get into the computer some how.
• Natural Language Processing — The computer needs to extract semantic meaning from the recognized words in order to be able to obey your commands in a meaningful way.
Text To Speech
Text to Speech is mostly a solved problem — not perfectly solved by any means — but solved well enough for our purposes, and a difficult enough problem that rolling our own solution or even improving an existing solution is pretty much out of the question. There are three main text to speech contenders on linux, Festival, espeak, and pico2wave. Of these, pico2wave and espeak are easy to get working, and pico2wave seems to sound the best to me, especially if you use the “-l en-GB” flag to make it speak like a woman with an English accent.
``` \$ pico2wave -l en-GB -w output.wav "Hello there, I am the ship's computer, at your service."
\$ aplay output.wav
```
The interface of pico2wave is a bit unfortunate, in that it can only take input from the command line, and produce a .wav file as output, but wrapping it in a small script to create a temporary .wav file which is then played with aplay is not a big deal. Besides pico2wave, espeak is another option that is easy to make work.
``` \$ espeak "Hello there."
```
When I tried out Festival, it didn’t immediately work for me, and since the speech which pico2wave produced seemed to be of quite good quality and pico2wave was easy to get working, I didn’t really persist in trying to get Festival working, so I cannot really say how well it works or what it sounds like.
Speech recognition
Speech recognition is a harder problem than text to speech, and less well solved, but fortunately, due to the hard work of others, there are some options. For this, I used pocketsphinx. Some configuration and customization is required, as trying to use pocketsphinx out of the box as a general English recognizer does not really work all that well. However, if you constrain its vocabulary and give it some idea ahead of time what sorts of things it should be expecting, it does a much better job. To do that, you need to create a sample corpus of text which you can then feed through a utility to generate a set of files which pocketsphinx can use to help it. The corpus of text I am currently using for Space Nerds In Space is here, snis_vocab.txt.
The easiest way to do this is to use the web service at CMU: http://www.speech.cs.cmu.edu/tools/lmtool-new.html.
This site will allow you to upload and process your corpus of text to produce a number of files, including a gzipped tarball that contains all the files pocketsphinx needs. Download and unpack the produced tarball which will be named something like TAR1234.tgz, which should contain 5 other files (the numbers that make up the filename will differ, obviously):
``` 1234.dic
1234.log_pronounce
1234.vocab
1234.lm
1234.sent
```
You can then use a script like the following:
```export language_model=1234
stdbuf -o 0 pocketsphinx_continuous -inmic yes -lm "\$language_model".lm -dict "\$language_model".dic 2>/dev/null |\
stdbuf -o 0 egrep -v 'READY...|Listening...|Stopped listening' |\
stdbuf -o 0 sed -e 's/^[0-9][0-9]*[:] //' |\
stdbuf -o 0 grep COMPUTER |\
stdbuf -o 0 sed -e 's/^.*COMPUTER //' |\
stdbuf -o 0 tee recog.txt | \
stdbuf -o 0 cat > /tmp/snis-natural-language-fifo
```
The use of “stdbuf -o 0” is to make sure there is no buffering between the programs so all output from each program is immediately transmitted to the next program in the pipeline without waiting for newlines or a full block, or anything like that. The filtering with grep and sed is just to cut out various extraneous output from pocketsphinx, and to cut out any lines not containing the word “COMPUTER”, and to remove all text up to and includeing the word “COMPUTER”. This essentially implements “hot word” functionality, so that it appears the computer knows when you’re talking to it. In reality, it’s listening all the time, and just throwing away anything that is not immediately preceded by the word “COMPUTER”.
The /tmp/snis-natural-language-fifo is a named pipe that Space Nerds In Space that serves as a means for text commands to be fed into “the ship’s computer”.
Natural Langage Processing
Supposing that we can transform speech into text, or failing that, just type in the text, we still have the problem of extracting meaning from the text.
In the last couple days, Google announced Parsey McParseface, an open source English text parser that uses neural networks to achieve unprecedented levels of accuracy in parsing English sentences. This might be interesting to look into someday, but for now, I have relied on my own home grown Zork-like 1980s technology, because I find such things to be fun to play with and easy enough. I have little doubt that Parsey McParseFace could beat the snot out of my little toy though.
The API for my natural language library is defined in snis_nl.h, documented in snis_nl.txt, and implemented in snis_nl.c.
The idea is you pass a string containing English text to the function snis_nl_parse_natural_language_request(), and then this calls a function which you have defined to do the requested action. For example, you might call:
``` snis_nl_parse_natural_language_request(my_context, "turn on the lights");
```
Supposing you have arranged things so that after “turn on the lights” has been parsed, snis_nl_parse_natural_language_request will call a function you have provided which will interact with your home automation systme and turn on the lights, et voila! You’re living in Star Trek.
The library doesn’t have any vocabulary, you have to provide that to the library by adding words to its internal dictionary. The library does know some “parts of speech”, to wit, the following:
``` #define POS_UNKNOWN 0
#define POS_NOUN 1
#define POS_VERB 2
#define POS_ARTICLE 3
#define POS_PREPOSITION 4
#define POS_SEPARATOR 5
#define POS_NUMBER 8
#define POS_NAME 9
#define POS_PRONOUN 10
#define POS_EXTERNAL_NOUN 11
```
To use the library, first you have to teach it some vocabulary. This is done with the following three functions, which teach it about synonyms, verbs, and words that aren’t verbs:
``` void snis_nl_add_synonym(char *synonym, char *canonical_word);
void snis_nl_add_dictionary_word(char *word, char *canonical_word, int part_of_speech);
void snis_nl_add_dictionary_verb(char *word, char *canonical_word, char *syntax, snis_nl_verb_function action);
```
Synonyms are done first, as these are simple word substitutions which are done prior to any attempt to parse for meaning. For example, the following defines a few synonyms:
``` snis_nl_add_synonym("lay in a course", "set a course");
```
If you attempt to parse “lay in a course for saturn and rotate 90 degrees left and increase shields” the parser would see this as being identical to “set a course for saturn and turn 90 degrees left and raise shields”. Synonyms are simple token substitutions done before parsing, and in the order they are listed. Note that later substitutions may operate on results of earlier substitutions, and that this is affected by the order in which the synonyms are defined. Note also that the unit of substitution is the word, or token. So “booster” would not be transformed into “raiseer”, but rather into “rocket”.
Adding words (other than verbs) to the dictionary is done via the “snis_nl_add_dictionary_word” function. Some examples:
``` snis_nl_add_dictionary_word("planet", "planet", POS_NOUN);
```
Note that words with equivalent meanings can be implemented by using the same “canonical” word (e.g. “coins” and “money” mean the same thing, above).
Adding verbs to the dictionary is a little more complicated, and is done with the snis_nl_add_dictionary_verb function. The arguments to this function require a little explanation.
``` void snis_nl_add_dictionary_verb(char *word, char *canonical_word, char *syntax, snis_nl_verb_function action);
```
• word: This is simply which verb you are adding to the dictionary, e.g.: “get”, “take”, “set”, “scan”, or whatever verb you are trying to define.
• canonical_word: This is useful for defining verbs that mean the same, or almost the same thing but which may have different ways of being used in a sentence.
• syntax: This is a string defining the “syntax” of the verb, or the “arguments” to the verb, how it may be used. The syntax is specified as a string of characters with each character representing a part of speech.
• ‘a’ : adjective
• ‘p’ : preposition
• ‘n’ : noun
• ‘l’ : a list of nouns (Note: this is not implemented.)
• ‘q’ : a quantity — that is, a number.
So, a syntax of “npn” means the verb requires a noun, a preposition, and another noun. For example “set a course for saturn” (The article “a” is not required.)
• action: This is a function pointer which will be called which should have the following signature which will be explained later.
``` union snis_nl_extra_data;
typedef void (*snis_nl_verb_function)(void *context, int argc, char *argv[], int part_of_speech[],
union snis_nl_extra_data *extra_data);
```
Some examples:
``` snis_nl_add_dictionary_verb("describe", "describe", "n", nl_describe_n); /* describe the blah */
snis_nl_add_dictionary_verb("describe", "describe", "an", nl_describe_an); /* describe the red blah */
snis_nl_add_dictionary_verb("navigate", "navigate", "pn", nl_navigage_pn); /* navigate to home */
snis_nl_add_dictionary_verb("set", "set", "npq", nl_set_npq); /* set the knob to 100 */
snis_nl_add_dictionary_verb("set", "set", "npa", nl_set_npa); /* set the knob to maximum */
snis_nl_add_dictionary_verb("set", "set", "npn", nl_set_npn); /* set a course for home */
snis_nl_add_dictionary_verb("set", "set", "npan", nl_set_npn); /* set a course for the nearest starbase */
snis_nl_add_dictionary_verb("plot", "plot", "npn", nl_set_npn); /* plot a course fo home */
snis_nl_add_dictionary_verb("plot", "plot", "npan", nl_set_npn); /* plot a course for the nearest planet */
```
Note that the same verb is often added to the dictionary multiple times with different syntaxes, sometimes associated with the same action function, sometimes with a different action function. This is really the key to how the whole thing works.
The parameters which are passed to your function are:
• context: This is just a void pointer which is passed from your program through the parsing function and finally back to your program. It is for you to use as a “cookie”, so that you can pass along some context to know for example, what the current topic is, or which entity in your system is requesting something to be parsed, etc. It is for you to use (or not) however you like.
• argc: This is simply count of the elements of the following parallel array parameters.
• argv[]: This is an array of the words that were parsed. It will contain the “canonical” version of the word in most cases (the exception is if the word is of type POS_EXTERNAL_NOUN, in which case there is no canonical noun, so it’s whatever was passed in to be parsed.)
• pos[]: This is an array of the parts of speech for each word in argv[].
• extra_data[]: This is an array of “extra data” for each word in argv[]. The use cases here are for the two parts of speech POS_EXTERNAL_NOUN and POS_NUMBER. For POS_NUMBER, the value of the number is in extra_data[x].number.value, which is a float. For POS_EXTERNAL_NOUN, the item of interest is a uint32_t value, extra_data[x].external_noun.handle. External nouns are described later.
Some ancilliary functions
``` typedef void (*snis_nl_multiword_preprocessor_fn)(char *word, int encode_or_decode);
#define SNIS_NL_ENCODE 1
#define SNIS_NL_DECODE 2
```
snis_nl_add_multiword_preprocessor allows you to provide a pre-processing function to encode multi-word tokens in a way that they won’t be broken apart when tokenized. Typically this function will look for certain word combinations and “encode” them by replacing internal spaces with dashes, and “decode” them by replacing dashes with spaces. This allows your program to have tokens like “warp drive” made of multiple words that will be interpreted as if they are a single word. It’s also possible that you might not know ahead of time what the multiword tokens are, which is why this is implemented through a function pointer to a function which you may implement. If you do not have such multiword tokens, you don’t need to use this.
``` typedef void (*snis_nl_error_function)(void *context);
```
You can add an error function. This will get called whenever the parser is not able to find a match for the provided text — that is, the parser is unable to extract a meaning from the provided text. You should make this function do whatever you want to happen when the provided text is not understood (e.g. print “I didn’t understand.”, for example.)
External nouns:
Your program may have some nouns which you don’t know ahead of time. For example, in a space game, you may have planets, creatures, spaceships, etc. that all have procedurally generated names like “Borto 7”, “Capricorn Cutlass”, or “despair squid” which you would like to be able to refer to by name. This is what external nouns are for. To use them, you provide a lookup function to the parse via:
``` void snis_nl_add_external_lookup(snis_nl_external_noun_lookup lookup);
```
Your function should have a prototype like:
``` uint32_t my_lookup_function(void *context, char *word);
```
This function should lookup the word that it is passed, and return a uint32_t handle. Later, in your verb function, when you encounter the type POS_EXTERNAL_NOUN in the pos[] array passed to your verb function, you can look in extra_data[].external_noun.handle and get this handle back, and thus know *which* external noun is being referred to. For example, if in your space game, you have 1000s of spaceships, and the player refers to “Capricorn Cutlass”, (first you will need a multiword token encoder/decoder to prevent that from being treated as two tokens, see above) then your lookup function should return as the handle, say, the unique ID of the matching spaceship in the form of a uint32_t, so that when your verb function is called, you can extract the handle, and lookup the spaceship to which it refers.
The main parsing function
``` void snis_nl_parse_natural_language_request(void *context, char *text);
```
snis_nl_parse_natural_language_request is the main function that parses the input text and calls the verb functions. You pass it a context pointer which can be anything you want it to be, and a string to parse. It will either call back an appropriate verb function, or the error function if you provided one.
Example program
snis_nl.c contains a main() function which is guarded behind some ifdef TEST_NL. You cand build the test program by the command “make snis_nl”
Below is a sample run of the snis_nl test program. Note that there is a lot of debugging output. This is because by default the snis_nl test program has debug mode turned on. You can turn this on in your program by sending as string like “nldebugmode 1” to be parsed by the parser, and turn it off by sending as string like “nldebugmode 0”. “nldebugmode” is the one verb the parser knows. This nldebugmode verb is added to the dictionary when the first user defined verb is added to the dictionary.
```\$./snis_nl
Enter string to parse: this is a test
--- iteration 0 ---
State machine 0 ('v,0, ', RUNNING) -- score = 0.000000
this[-1]:is[-1]:a[-1]:test[-1]:--- iteration 1 ---
State machine 0 ('v,0, ', RUNNING) -- score = 0.000000
this[-1]:is[-1]:a[-1]:test[-1]:--- iteration 2 ---
State machine 0 ('v,0, ', RUNNING) -- score = 0.000000
this[-1]:is[-1]:a[-1]:test[-1]:--- iteration 3 ---
State machine 0 ('v,0, ', RUNNING) -- score = 0.000000
this[-1]:is[-1]:a[-1]:test[-1]:--- iteration 4 ---
Failure to comprehend 'this is a test'
Enter string to parse: set a course for earth
--- iteration 0 ---
State machine 0 ('v,0, ', RUNNING) -- score = 0.000000
set[-1]:a[-1]:course[-1]:for[-1]:earth[-1]:--- iteration 1 ---
State machine 0 ('npn,0, ', RUNNING) -- score = 0.000000
set[2]:(npn verb (set))
a[-1]:course[-1]:for[-1]:earth[-1]:State machine 1 ('npa,0, ', RUNNING) -- score = 0.000000
set[1]:(npa verb (set))
a[-1]:course[-1]:for[-1]:earth[-1]:State machine 2 ('npq,0, ', RUNNING) -- score = 0.000000
set[0]:(npq verb (set))
a[-1]:course[-1]:for[-1]:earth[-1]:--- iteration 2 ---
State machine 0 ('npn,0, ', RUNNING) -- score = 0.000000
set[2]:(npn verb (set))
a[0]:(article (a))
course[-1]:for[-1]:earth[-1]:State machine 1 ('npa,0, ', RUNNING) -- score = 0.000000
set[1]:(npa verb (set))
a[0]:(article (a))
course[-1]:for[-1]:earth[-1]:State machine 2 ('npq,0, ', RUNNING) -- score = 0.000000
set[0]:(npq verb (set))
a[0]:(article (a))
course[-1]:for[-1]:earth[-1]:--- iteration 3 ---
State machine 0 ('npn,1, ', RUNNING) -- score = 0.000000
set[2]:(npn verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[-1]:earth[-1]:State machine 1 ('npa,1, ', RUNNING) -- score = 0.000000
set[1]:(npa verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[-1]:earth[-1]:State machine 2 ('npq,1, ', RUNNING) -- score = 0.000000
set[0]:(npq verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[-1]:earth[-1]:--- iteration 4 ---
State machine 0 ('npn,2, ', RUNNING) -- score = 0.000000
set[2]:(npn verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[0]:(preposition (for))
earth[-1]:State machine 1 ('npa,2, ', RUNNING) -- score = 0.000000
set[1]:(npa verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[0]:(preposition (for))
earth[-1]:State machine 2 ('npq,2, ', RUNNING) -- score = 0.000000
set[0]:(npq verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[0]:(preposition (for))
earth[-1]:--- iteration 5 ---
State machine 0 ('npn,3, ', RUNNING) -- score = 0.000000
set[2]:(npn verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[0]:(preposition (for))
earth[0]:(noun (earth))
--- iteration 6 ---
State machine 0 ('npn,3, ', SUCCESS) -- score = 0.000000
set[2]:(npn verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[0]:(preposition (for))
earth[0]:(noun (earth))
-------- Final interpretation: ----------
State machine 0 ('npn,3, ', SUCCESS) -- score = 1.000000
set[2]:(npn verb (set))
a[0]:(article (a))
course[0]:(noun (course))
for[0]:(preposition (for))
earth[0]:(noun (earth))
generic_verb_action: set(verb) a(article) course(noun) for(preposition) earth(noun)
Enter string to parse: ^D
\$
```
## Building a wooden prop sword
March 20, 2016 — Awhile ago, I watched a video of Adam Savage building a prop Hellboy sword. He built it out of plywood, bondo, aluminum tape, and spray paint. It came out looking better than it had any right to. Here is that video:
So, being stuck at home for a couple days, I thought I’d try something similar.
First, I had to shape a (nominally) 1×2 inch board into something vaguely sword shaped. For this I used a coping saw, a pocket knife, and a rasp, and sandpaper. A table saw, jigsaw, belt sander, and various other power tools would have made the job go a lot easier and quicker, but I don’t have those, so I had to do it by hand. Using the rasp to shape the blade by hand was tiring, to say the least.
Then I glued on some bits of wood I had sawed out with the coping saw to form a cross guard.
Then on with the aluminum tape, spray paint and steel wool.
I used a ball-point pen to do some “elvish” “engraving”.
Came out pretty cool. | 8,405 | 33,737 | {"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} | 3.078125 | 3 | CC-MAIN-2017-30 | longest | en | 0.923093 |
http://icl.cs.utk.edu/hpcc/hpcc_results_kiviat.cgi?orderby=PRT&sortorder=ASC | 1,591,301,777,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347458095.68/warc/CC-MAIN-20200604192256-20200604222256-00410.warc.gz | 57,046,325 | 69,308 | Base Results Optimized Results Base and Optimized
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Base Results Optimized Results Base and Optimized
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Manufacturer/Processor Type, Speed, Count, Threads, Processes Includes the manufacturer/processor type, processor speed, number of processors, threads, and number of processes. Move mouse over this column for each row to display additional information, including; manufacturer, system name, interconnect, MPI, affiliation, and submission date.
Run Type Run Type, indicates whether the benchmark was a base run or was optimized.
Processors Processors, this is the number of processors used in the benchmark, entered in the form by the benchmark submitter.
PP-HPL ( per processor ) HPL, Solves a randomly generated dense linear system of equations in double floating-point precision (IEEE 64-bit) arithmetic using MPI. The linear system matrix is stored in a two-dimensional block-cyclic fashion and multiple variants of code are provided for computational kernels and communication patterns. The solution method is LU factorization through Gaussian elimination with partial row pivoting followed by a backward substitution. Unit: Tera Flops per Second
PP-PTRANS (A=A+B^T, MPI) ( per processor ) PTRANS (A=A+B^T, MPI), Implements a parallel matrix transpose for two-dimensional block-cyclic storage. It is an important benchmark because it exercises the communications of the computer heavily on a realistic problem where pairs of processors communicate with each other simultaneously. It is a useful test of the total communications capacity of the network. Unit: Giga Bytes per Second
PP-RandomAccess ( per processor ) Global RandomAccess, also called GUPs, measures the rate at which the computer can update pseudo-random locations of its memory - this rate is expressed in billions (giga) of updates per second (GUP/s). Unit: Giga Updates per Second
PT-SN-STREAM ( per thread ) The Single Process STREAM benchmark is a simple synthetic benchmark program that measures sustainable memory bandwidth and the corresponding computation rate for simple numerical vector kernels. It is run on single computational process chosen at random. Unit: Giga Bytes per Second
PT-SN-DGEMM ( per thread ) The Single Process DGEMM benchmark measures the floating-point execution rate of double precision real matrix-matrix multiply performed by the DGEMM subroutine from the BLAS (Basic Linear Algebra Subprograms). It is run on single computational process chosen at random. Unit: Giga Flops per Second
PP-FFT ( per processor ) FFT, performs the same test as FFT but across the entire system by distributing the input vector in block fashion across all the processes. Unit: Giga Flops per Second
Randomly Ordered Ring Bandwidth ( per process ) Randomly Ordered Ring Bandwidth, reports bandwidth achieved in the ring communication pattern. The communicating processes are ordered randomly in the ring (with respect to the natural ordering of the MPI default communicator). The result is averaged over various random assignments of processes in the ring. Unit: Giga Bytes per second
Randomly-Ordered Ring Latency ( per process ) Randomly-Ordered Ring Latency ( per process ), reports latency in the ring communication pattern. The communicating processes are ordered randomly in the ring (with respect to the natural ordering of the MPI default communicator) in the ring. The result is averaged over various random assignments of processes in the ring. Unit: micro-seconds
The values plotted for HPL, PTRANS, RandomAccess, and FFT are per processor. The values plotted for SN-DGEMM and SN-STREAM are per thread. The value plotted for RandomRing Latency is normalized using it's reciprocal. Only those systems that have values for all the tests plotted are available for the diagram. Use the left-hand column to select up to 6 systems to plot in the Kiviat diagram.
Systems for Kiviat Chart - Base Runs Only - 321 Systems - Generated on Thu Jun 4 16:16:17 2020
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: IBM
Processor Type: 0
Processor Speed: 1.9GHz
Processor Count: 12
Threads: 1
Processses: 12
System Name: p690
Interconnect: no
MPI: openmpi
Affiliation: VGTU
Submission Date: 05-31-09
IBM p690 0 1.9GHz 12 1 12
0.0037
0.1172
0.0012
2.8326
0.1668
5.4371
0.7045
2.73
Manufacturer: n/a
Processor Type: 0
Processor Speed: 2GHz
Processor Count: 16
Threads: 0
Processses: 16
System Name: n/a TestVM Intel Xeon
Interconnect: GbE
MPI: n/a
Affiliation: Faculdade de Engenharia da Universidade do Porto
Submission Date: 07-01-11
n/a TestVM Intel Xeon 0 2GHz 16 0 16
0.0044
0.0168
0.0005
0.0000
0.0217
0.0000
0.0066
280.70
Manufacturer: seagate
Processor Type: 0
Processor Speed: 2.5GHz
Processor Count: 4
Threads: 0
Processses: 4
System Name: camaro
Interconnect: intel
MPI: MPICH2 1.2.1
Affiliation: seagate test lab
Submission Date: 03-09-16
seagate camaro 0 2.5GHz 4 0 4
0.0042
0.3934
0.0074
0.0000
0.5389
0.0000
0.3220
0.71
Manufacturer: IBM-Serviware
Processor Type: 5472
Processor Speed: 3GHz
Processor Count: 272
Threads: 1
Processses: 1088
System Name: Idataplex
Interconnect: Infiniband DDR-4X
MPI: Intel MPI 3.2
Affiliation: CNRS-Institut d'Astrophysique de Paris
Submission Date: 01-26-09
IBM-Serviware Idataplex 5472 3GHz 272 1 1088
0.0326
0.0076
0.0000
3.3711
0.1949
11.3699
0.0196
132.21
Manufacturer: Cray Inc.
Processor Type: Alpha 21164
Processor Speed: 0.675GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: T3E
Interconnect: 3-D Torus
MPI: MPI 2.2.0.0
Affiliation: Engineer Research Development Center - Army Major Shared Resource Center
Submission Date: 11-02-04
Cray Inc. T3E Alpha 21164 0.675GHz 512 1 512
0.0004
0.0191
0.0001
0.5422
0.0302
0.6803
0.0357
8.14
Manufacturer: HP
Processor Type: Alpha 21264B
Processor Speed: 0.833GHz
Processor Count: 484
Threads: 1
Processses: 484
System Name: SC-40
Interconnect: 64-port, single rail Quadrics
MPI: MPICH 1.7
Affiliation: Engineer Research Development Center - Army Major Shared Resource Center
Submission Date: 11-04-04
HP SC-40 Alpha 21264B 0.833GHz 484 1 484
0.0009
0.0104
0.0000
1.2193
0.0093
1.4487
0.0173
50.10
Manufacturer: HP
Processor Type: Alpha 21264C
Processor Speed: 1GHz
Processor Count: 484
Threads: 1
Processses: 484
System Name: Compaq SC45
Interconnect: 64-port, single rail Quadrics
MPI: MPICH 1.7
Affiliation: Engineer Research Development Center - Army Major Shared Resource Center
Submission Date: 11-03-04
HP Compaq SC45 Alpha 21264C 1GHz 484 1 484
0.0012
0.0132
0.0000
1.6421
0.0103
1.7479
0.0226
39.63
Manufacturer: N/A
Processor Type: AMD Athlon64 X2
Processor Speed: 2.5GHz
Processor Count: 4
Threads: 2
Processses: 4
System Name: BlackCell
Interconnect: Fast Ethernet
MPI: MPICH2 1.2
Affiliation: N/A
Submission Date: 11-25-09
N/A BlackCell AMD Athlon64 X2 2.5GHz 4 2 4
0.0041
0.0093
0.0008
1.1851
0.0238
4.5829
0.0095
37.91
Manufacturer: Dalco
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Opteron/QsNet Linux Cluster
Interconnect: QsNetII
MPI: Quadrics qsnetmpi 1.24-39
Affiliation: DALCO
Submission Date: 11-04-04
Dalco Opteron/QsNet Linux Cluster AMD Opteron 2.2GHz 64 1 64
0.0034
0.0987
0.0001
2.4322
0.2117
3.8926
0.1700
11.46
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: XD1
Interconnect: RapidArray Interconnect System
MPI: MPI over Rapid Array
Affiliation: Cray Inc.
Submission Date: 11-22-04
Cray Inc. XD1 AMD Opteron 2.2GHz 64 1 64
0.0035
0.1655
0.0003
2.7662
0.2556
3.9801
0.2270
1.63
Manufacturer: Sun
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Sun Fire V20z Cluster
Interconnect: Gigabit Ethernet, Cisco 6509 switch
MPI: LAM/MPI 7.1.1
Affiliation: Idaho National Laboratory
Submission Date: 03-06-05
Sun Fire V20z Cluster AMD Opteron 2.2GHz 64 1 64
0.0034
0.0244
0.0000
2.4040
0.1010
3.9577
0.0408
42.99
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 1100
Threads: 1
Processses: 1100
System Name: XT3
Interconnect: Cray XT3
MPI: Mpich 2.0
Affiliation: Swiss National Supercomputing Centre CSCS
Submission Date: 06-08-05
Cray Inc. XT3 AMD Opteron 2.6GHz 1100 1 1100
0.0043
0.1981
0.0001
4.9892
0.2424
4.8110
0.2864
25.94
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: XD1
Interconnect: Rapid Array Fat Tree
MPI: mpich/mpich-pgi601
Affiliation: Cray
Submission Date: 06-15-05
Cray Inc. XD1 AMD Opteron 2.4GHz 128 1 128
0.0039
0.1056
0.0005
4.3576
0.2775
4.3344
0.2592
2.06
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 3744
Threads: 1
Processses: 3744
System Name: XT3
Interconnect: Cray XT3 MPP Interconnect
MPI: MPICH 2.0
Affiliation: Cray Inc. at Oak Ridge National Laboratory
Submission Date: 06-21-05
Cray Inc. XT3 AMD Opteron 2.4GHz 3744 1 3744
0.0039
0.1625
0.0001
4.6212
0.1114
4.4142
0.1616
25.32
Manufacturer: PathScale, Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Customer Benchmark Cluster
Interconnect: InfiniPath 1.0
MPI: PathScale MPI 1.0
Affiliation: PathScale, Inc.
Submission Date: 07-19-05
PathScale, Inc. Customer Benchmark Cluster AMD Opt ... 2.6GHz 32 1 32
0.0039
0.2100
0.0009
3.9588
0.3233
4.7520
0.2653
1.31
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 5200
Threads: 1
Processses: 5200
System Name: XT3
Interconnect: Cray XT3 MPP Interconnect
MPI: MPICH 2.0
Affiliation: Cray Inc. at Oak Ridge National Laboratory
Submission Date: 08-01-05
Cray Inc. XT3 AMD Opteron 2.4GHz 5200 1 5200
0.0039
0.1682
0.0001
4.7202
0.1240
4.3929
0.1468
25.80
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: Clustervision BV
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Beastie
Interconnect: Gigabit Ethernet, HP pro curve
MPI: MPICH 1.2.7
Affiliation: University of Glasgow
Submission Date: 08-29-05
Clustervision BV Beastie AMD Opteron 2.4GHz 32 1 32
0.0032
0.0255
0.0000
3.3391
0.0671
4.1999
0.0265
53.23
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: xt3
Interconnect: Cray Seastar
MPI: xt-mpt/1.2.10
Affiliation: ORNL
Submission Date: 09-13-05
Cray Inc. xt3 AMD Opteron 2.4GHz 32 1 32
0.0043
0.2305
0.0019
4.8882
0.2928
4.7726
0.5728
8.74
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 4096
Threads: 1
Processses: 4096
System Name: XT3
Interconnect: Cray XT3 MPP Interconnect
MPI: MPICH 2.0
Affiliation: Engineer Research and Development Center (ERDC) Major Shared Resource Center
Submission Date: 09-19-05
Cray Inc. XT3 AMD Opteron 2.6GHz 4096 1 4096
0.0041
0.0740
0.0001
5.0423
0.2211
4.7751
0.1690
9.44
Manufacturer: Rackable Systems
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 256
Threads: 2
Processses: 512
System Name: Emerald
Interconnect: InfiniPath HTX InfiniBand Adapter / SilverStorm 9120 InfiniBand Switch
MPI: PathScale MPICH
Affiliation: AMD
Submission Date: 10-14-05
Rackable Systems Emerald AMD Opteron 2.2GHz 256 2 512
0.0073
0.1374
0.0022
1.4194
0.2651
2.0166
0.0907
2.33
Manufacturer: Rackable Systems
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 128
Threads: 2
Processses: 256
System Name: Emerald
Interconnect: InfiniPath HTX InfiniBand Adapter / SilverStorm 9120 InfiniBand Switch
MPI: PathScale MPICH
Affiliation: AMD
Submission Date: 10-14-05
Rackable Systems Emerald AMD Opteron 2.2GHz 128 2 256
0.0074
0.1212
0.0033
1.3060
0.2830
2.0309
0.1032
2.20
Manufacturer: Rackable Systems
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 64
Threads: 2
Processses: 128
System Name: Emerald
Interconnect: InfiniPath HTX InfiniBand Adapter / SilverStorm 9120 InfiniBand Switch
MPI: PathScale MPICH
Affiliation: AMD
Submission Date: 10-14-05
Rackable Systems Emerald AMD Opteron 2.2GHz 64 2 128
0.0074
0.1759
0.0048
1.4719
0.3396
2.0228
0.1246
2.02
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 5208
Threads: 1
Processses: 5208
System Name: XT3
Interconnect: Cray Seastar
MPI: xt-mpt/1.3.07
Affiliation: Oak Ridge National Laboratory, DOE Office of Science
Submission Date: 11-10-05
Cray Inc. XT3 AMD Opteron 2.4GHz 5208 1 5208
0.0039
0.1813
0.0001
4.6028
0.1463
4.4132
0.2064
9.20
Manufacturer: Dalco
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Gonzales
Interconnect: QsNetII
MPI: Quadrics qsnetmpi 1.24-47
Affiliation: Swiss Federal Institute of Technology Zurich
Submission Date: 11-11-05
Dalco Gonzales AMD Opteron 2.4GHz 64 1 64
0.0040
0.1446
0.0006
3.6847
0.2188
4.3419
0.1735
4.89
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 4128
Threads: 1
Processses: 4128
System Name: XT3
Interconnect: Seastar
MPI: MPICH 2.0
Affiliation: Engineer Research and Development Center
Submission Date: 11-13-05
Cray Inc. XT3 AMD Opteron 2.6GHz 4128 1 4128
0.0040
0.1635
0.0002
4.6202
0.1990
4.7548
0.2224
8.23
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2GHz
Processor Count: 10350
Threads: 1
Processses: 10350
System Name: XT3
Interconnect: Seastar
MPI: xt-mpt/1.3.10 based on MPICH 2.0
Affiliation: Sandia National Laboratories, DOE NNSA
Submission Date: 01-11-06
Cray Inc. XT3 AMD Opteron 2GHz 10350 1 10350
0.0032
0.1752
0.0001
4.3689
0.1080
3.6731
0.1619
10.32
Manufacturer: Team HPC
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 20
Threads: 1
Processses: 20
System Name: T2ODC Supercomputer
Interconnect: Gigabit Ethernet
MPI: MPICH 1.2.7p1
Affiliation: SAIC
Submission Date: 01-13-06
Team HPC T2ODC Supercomputer AMD Opteron 2.2GHz 20 1 20
0.0030
0.0191
0.0000
2.0754
0.0394
3.9745
0.0162
62.19
Manufacturer: Team HPC
Processor Type: AMD Opteron
Processor Speed: 2GHz
Processor Count: 56
Threads: 1
Processses: 56
System Name: T4ODC Supercomputer
Interconnect: Gigabit Ethernet
MPI: MPICH 1.2.7p1
Affiliation: University of Delaware
Submission Date: 01-17-06
Team HPC T4ODC Supercomputer AMD Opteron 2GHz 56 1 56
0.0029
0.0095
0.0000
2.3905
0.0212
3.6518
0.0080
118.96
Manufacturer: Tyan
Processor Type: AMD Opteron
Processor Speed: 2GHz
Processor Count: 38
Threads: 1
Processses: 19
System Name: Opteron
Interconnect: Gigabit
MPI: LAM 7.0.6
Affiliation: University of Victoria
Submission Date: 01-19-06
Tyan Opteron AMD 2GHz 38 1 19
0.0000
0.0032
0.0000
2.6521
0.0009
3.0349
0.0122
34.88
Manufacturer: Tyan
Processor Type: AMD Opteron
Processor Speed: 2GHz
Processor Count: 38
Threads: 1
Processses: 38
System Name: Llaima
Interconnect: Gigabit
MPI: LAM 7.0.6
Affiliation: University of Victoria
Submission Date: 01-24-06
Tyan Llaima AMD Opteron 2GHz 38 1 38
0.0001
0.0026
0.0000
2.6496
0.0010
3.0559
0.0067
47.39
Manufacturer: Tyan
Processor Type: AMD Opteron
Processor Speed: 2GHz
Processor Count: 36
Threads: 1
Processses: 36
System Name: Llaima
Interconnect: GigE
MPI: LAM 7.0.6-5
Affiliation: University of Victoria
Submission Date: 01-27-06
Tyan Llaima AMD Opteron 2GHz 36 1 36
0.0027
0.0065
0.0000
1.6560
0.0173
3.5354
0.0067
47.59
Manufacturer: Tyan
Processor Type: AMD Opteron
Processor Speed: 2GHz
Processor Count: 52
Threads: 1
Processses: 72
System Name: Llaima
Interconnect: Gigabit
MPI: LAM 7.0.6
Affiliation: University of Victoria
Submission Date: 02-08-06
Tyan Llaima AMD Opteron 2GHz 52 1 72
0.0032
0.0084
0.0000
1.5049
0.0231
3.5228
0.0064
72.60
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: OCF plc
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Furmidge iPath 1
Interconnect: InfiniPath 1.1
MPI: PathScale MPI 1.1
Affiliation: OCF plc
Submission Date: 02-16-06
OCF plc Furmidge iPath 1 AMD Opteron 2.4GHz 32 1 32
0.0039
0.1182
0.0023
2.9053
0.2065
4.4246
0.1840
1.61
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 1100
Threads: 1
Processses: 1100
System Name: XT3
Interconnect: Cray XT3
MPI: Mpich 2.0
Affiliation: Swiss National Supercomputing Centre CSCS
Submission Date: 02-25-06
Cray Inc. XT3 AMD Opteron 2.6GHz 1100 1 1100
0.0043
0.2303
0.0003
4.8756
0.2984
4.7717
0.3996
7.29
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 10404
Threads: 1
Processses: 10404
System Name: XT3 Dual-Core
Interconnect: Cray SeaStar
MPI: xt-mpt 1.4.35
Affiliation: Oak Ridge National Lab
Submission Date: 11-06-06
Cray Inc. XT3 Dual-Core AMD Opteron 2.6GHz 10404 1 10404
0.0042
0.0748
0.0001
3.3263
0.1064
4.8001
0.0694
14.32
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.4GHz
Processor Count: 12960
Threads: 1
Processses: 25920
System Name: Red Storm/XT3
Interconnect: Cray custom
MPI: MPICH 2 v 1.0.2
Affiliation: NNSA/Sandia National Laboratories
Submission Date: 11-10-06
Cray Inc. Red Storm/XT3 AMD Opteron 2.4GHz 12960 1 25920
0.0070
0.1819
0.0001
4.0983
0.1199
4.4078
0.0591
16.29
Manufacturer: HP
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 16
Threads: 1
Processses: 32
System Name: XC4000
Interconnect: Infiniband 4x SDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 12-22-06
HP XC4000 AMD Opteron 2.6GHz 16 1 32
0.0086
0.3154
0.0015
4.2907
0.4220
4.7198
0.1893
14.27
Manufacturer: HP
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 32
Threads: 1
Processses: 64
System Name: XC4000
Interconnect: Infiniband 4x SDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 12-22-06
HP XC4000 AMD Opteron 2.6GHz 32 1 64
0.0086
0.2299
0.0008
4.3770
0.3912
4.7511
0.1686
14.48
Manufacturer: HP
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 64
Threads: 1
Processses: 128
System Name: XC4000
Interconnect: Infiniband 4x SDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 12-22-06
HP XC4000 AMD Opteron 2.6GHz 64 1 128
0.0085
0.2028
0.0004
4.3579
0.3373
4.7008
0.1404
14.59
Manufacturer: HP
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 128
Threads: 1
Processses: 256
System Name: XC4000
Interconnect: Infiniband 4x SDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 12-22-06
HP XC4000 AMD Opteron 2.6GHz 128 1 256
0.0083
0.1543
0.0001
4.3357
0.2927
4.7150
0.1116
14.78
Manufacturer: HP
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 256
Threads: 1
Processses: 512
System Name: XC4000
Interconnect: Infiniband 4x SDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 12-22-06
HP XC4000 AMD Opteron 2.6GHz 256 1 512
0.0082
0.1455
0.0000
4.3306
0.2813
4.7469
0.0986
15.28
Manufacturer: N/A
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 90
Threads: 1
Processses: 90
System Name: Linux Opteron
Interconnect: Gig-E
MPI: MPICH 1.2.7
Affiliation: University of Arkansas
Submission Date: 01-13-07
N/A Linux Opteron AMD Opteron 2.6GHz 90 1 90
0.0037
0.0225
0.0000
3.1043
0.0489
4.6188
0.0222
121.37
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 8190
Threads: 1
Processses: 8190
System Name: XT3
Interconnect: Cray Seastar
MPI: MPICH 2 v 1.0.2
Affiliation: ERDC MSRC
Submission Date: 04-09-07
Cray Inc. XT3 AMD Opteron 2.6GHz 8190 1 8190
0.0043
0.0736
0.0001
4.4593
0.1077
4.7986
0.0860
14.22
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.1GHz
Processor Count: 8464
Threads: 1
Processses: 8464
System Name: XT4
Interconnect: Cray Seastar
MPI: MPICH 2 v 1.0.2
Affiliation: ERDC MSRC
Submission Date: 05-14-08
Cray Inc. XT4 AMD Opteron 2.1GHz 8464 1 8464
0.0066
0.0365
0.0004
6.6056
0.0931
7.6617
0.0798
31.35
Manufacturer: Cray Inc.
Processor Type: AMD Opteron
Processor Speed: 2.3GHz
Processor Count: 74529
Threads: 2
Processses: 74529
System Name: XT5
Interconnect: Seastar 2+
MPI: MPT 3.1.0.4
Affiliation: Oak Ridge National Labs - DoE
Submission Date: 11-16-08
Cray Inc. XT5 AMD Opteron 2.3GHz 74529 2 74529
0.0121
0.0231
0.0002
4.5362
0.0372
8.4856
0.0255
19.63
Manufacturer: Cray, Inc.
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 99036
Threads: 3
Processses: 33012
System Name: XT5
Interconnect: SeaStar 2+
MPI: MPT 3.4.2
Affiliation: National Institute for Computational Sciences
Submission Date: 11-02-09
Cray, Inc. XT5 AMD Opteron 2.6GHz 99036 3 33012
0.0074
0.0184
0.0002
2.5891
0.0382
9.6970
0.0593
15.59
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.6GHz
Processor Count: 224220
Threads: 3
Processses: 74740
System Name: XT5
Interconnect: Seastar
MPI: MPT 3.4.2
Affiliation: Oak Ridge National Laboratory
Submission Date: 11-04-09
Cray XT5 AMD Opteron 2.6GHz 224220 3 74740
0.0068
0.0104
0.0001
3.4994
0.0279
9.6042
0.0400
16.08
Manufacturer: Institute of Nuclear Sciences and Physical Engineering
Processor Type: AMD Opteron
Processor Speed: 2GHz
Processor Count: 48
Threads: 48
Processses: 48
System Name: superPC
Interconnect: sanet
MPI: LAM/MPI 7.1.2
Affiliation: Slovak University of Technology in Bratislava
Submission Date: 05-10-11
Institute of Nuclear Sciences and Physical Enginee ... 2GHz 48 48 48
0.0002
0.0177
0.0036
0.1597
0.2063
0.1419
0.1492
2.70
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: Instituto de Astronomia, Universidad Nacional Autonoma de Mexico
Processor Type: AMD Opteron
Processor Speed: 2.2GHz
Processor Count: 16
Threads: 0
Processses: 16
System Name: Pakal
Interconnect: Ethernet
MPI: 2.1.0.8
Affiliation: Instituto de Astronomia, Universidad Nacional Autonoma de Mexico
Submission Date: 05-18-11
Instituto de Astronomia, Universidad Nacional Auto ... 2.2GHz 16 0 16
0.0002
0.0468
0.0017
0.0000
0.3069
0.0000
0.2186
1.43
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 8192
Threads: 1
Processses: 8192
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 8192 1 8192
0.0078
0.0211
0.0007
7.3502
0.0699
7.5460
0.0304
5.32
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 4096
Threads: 1
Processses: 4096
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 4096 1 4096
0.0081
0.0192
0.0008
7.5399
0.0707
7.6066
0.0321
5.22
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 2048
Threads: 1
Processses: 2048
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 2048 1 2048
0.0081
0.0389
0.0016
7.5882
0.1177
7.5554
0.0464
5.10
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 1024
Threads: 1
Processses: 1024
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 1024 1 1024
0.0082
0.0412
0.0021
7.4311
0.1067
7.6188
0.0683
4.72
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 512 1 512
0.0081
0.0600
0.0010
7.6053
0.1457
7.6474
0.0836
4.38
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 256
Threads: 1
Processses: 256
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 256 1 256
0.0081
0.0759
0.0017
7.6956
0.1794
7.5931
0.1253
4.03
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 128 1 128
0.0082
0.1054
0.0024
7.3876
0.2142
7.5487
0.1680
3.15
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 64 1 64
0.0082
0.1509
0.0044
7.3903
0.2607
7.5320
0.3456
2.17
Manufacturer: Cray
Processor Type: AMD Opteron
Processor Speed: 2.5GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Cray XE6
Interconnect: Gemini
MPI: Cray MPICH 2
Affiliation: Indiana University
Submission Date: 01-22-14
Cray XE6 AMD Opteron 2.5GHz 32 1 32
0.0082
0.1894
0.0078
7.6755
0.3063
7.7180
0.6693
1.09
Manufacturer: LMN/PUB
Processor Type: AMD Opteron Barcelona, 2.3GHz Dual Quad-Core
Processor Speed: 1.2GHz
Processor Count: 8
Threads: 8
Processses: 4
System Name: ATLAS
Interconnect: 1 Gbit Ethernet
MPI: OpenMPI 1.2.5-3.1
Affiliation: P.U.B
Submission Date: 05-23-09
LMN/PUB ATLAS AMD Opteron Barcelona, 2.3GHz Dual Q ... 1.2GHz 8 8 4
0.0013
0.0695
0.0034
1.0507
0.2776
0.8152
0.5244
1.16
Manufacturer: kurrola.dy.fi
Processor Type: AMD/FX-8150
Processor Speed: 3.6GHz
Processor Count: 3
Threads: 2
Processses: 11
System Name: MPI4YOU
Interconnect: AsRock XFast LAN
MPI: mpich 3.0.1
Affiliation: kurrola.dy.fi
Submission Date: 02-23-13
kurrola.dy.fi MPI4YOU AMD/FX-8150 3.6GHz 3 2 11
0.0106
0.0872
0.0014
4.8541
0.2481
17.8650
0.0307
47.53
Manufacturer: https://launchpad.net/mpi4you
Processor Type: AMD/FX8150
Processor Speed: 3.6GHz
Processor Count: 3
Threads: 11
Processses: 11
System Name: mpi4you
Interconnect: AsRock XFast LAN
MPI: mpich 3.0.1
Affiliation: kurrola.dy.fi
Submission Date: 02-23-13
https://launchpad.net/mpi4you mpi4you AMD/FX8150 3.6GHz 3 11 11
0.0178
0.0851
0.0024
1.0344
0.2505
2.4474
0.0345
45.85
Manufacturer: https://launchpad.net/mpi4you
Processor Type: AMD/FX8150
Processor Speed: 3.6GHz
Processor Count: 3
Threads: 2
Processses: 12
System Name: mpi4you
Interconnect: AsRock XFast LAN
MPI: mpich-3.0.4
Affiliation: kurrola.dy.fi
Submission Date: 12-08-14
https://launchpad.net/mpi4you mpi4you AMD/FX8150 3.6GHz 3 2 12
0.0126
0.1069
0.0031
10.3369
0.1729
14.1001
0.0245
37.34
Manufacturer: Intel
Processor Type: Core 2 Extreme X6800
Processor Speed: 2.93GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Intel Discovery cluster
Interconnect: Infiniband SDR
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 09-10-07
Intel Discovery cluster Core 2 Extreme X6800 2.93GHz 128 1 128
0.0099
0.1230
0.0037
4.7207
0.3417
10.8068
0.1820
13.35
Manufacturer: Intel
Processor Type: Core 2 Extreme x6800
Processor Speed: 2.93GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Intel Discovery cluster
Interconnect: Infiniband SDR
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 09-10-07
Intel Discovery cluster Core 2 Extreme x6800 2.93GHz 64 1 64
0.0099
0.1407
0.0042
4.7654
0.4005
10.6914
0.2174
11.77
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: Intel
Processor Type: Core 2 Extreme x6800
Processor Speed: 2.93GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Intel Discovery cluster
Interconnect: Infiniband SDR
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 09-10-07
Intel Discovery cluster Core 2 Extreme x6800 2.93GHz 32 1 32
0.0100
0.1500
0.0048
4.7367
0.4539
10.8916
0.2540
9.88
Manufacturer: Intel
Processor Type: Core 2 Extreme x6800
Processor Speed: 2.93GHz
Processor Count: 16
Threads: 1
Processses: 16
System Name: Intel Discovery cluster
Interconnect: Infiniband SDR
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 09-10-07
Intel Discovery cluster Core 2 Extreme x6800 2.93GHz 16 1 16
0.0099
0.1296
0.0055
4.7409
0.5838
10.7553
0.3029
7.07
Manufacturer: Intel
Processor Type: Core 2 Extreme x6800
Processor Speed: 2.93GHz
Processor Count: 8
Threads: 1
Processses: 8
System Name: Intel Discovery cluster
Interconnect: Infiniband SDR
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 09-10-07
Intel Discovery cluster Core 2 Extreme x6800 2.93GHz 8 1 8
0.0100
0.2330
0.0063
4.7396
0.5863
10.9127
0.3715
6.02
Manufacturer: Intel
Processor Type: Core 2 Extreme x6800
Processor Speed: 2.93GHz
Processor Count: 4
Threads: 1
Processses: 4
System Name: Intel Discovery cluster
Interconnect: Infiniband SDR
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 09-10-07
Intel Discovery cluster Core 2 Extreme x6800 2.93GHz 4 1 4
0.0100
0.1609
0.0073
4.7728
0.6359
10.7518
0.6215
4.71
Manufacturer: Cray Inc.
Processor Type: Cray X1 MSP
Processor Speed: 0.8GHz
Processor Count: 60
Threads: 1
Processses: 60
System Name: X1
Interconnect: Cray modified 2-D Torus
MPI: MPT 2.3.0.0
Affiliation: Engineer Research and Development Center Major Shared Resource Center
Submission Date: 11-02-04
Cray Inc. X1 Cray MSP 0.8GHz 60 1 60
0.0085
0.0272
0.0001
16.2112
0.0524
10.9044
1.1678
14.66
Manufacturer: Cray Inc.
Processor Type: Cray X1 MSP
Processor Speed: 0.8GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: X1
Interconnect: Cray modified 2-D Torus
MPI: MPT 2.4
Affiliation: Cray Inc.
Submission Date: 11-22-04
Cray Inc. X1 Cray MSP 0.8GHz 32 1 32
0.0086
1.0206
0.0001
16.2214
0.0927
8.4594
1.4127
14.94
Manufacturer: Cray Inc.
Processor Type: Cray X1 MSP
Processor Speed: 1.13GHz
Processor Count: 252
Threads: 1
Processses: 252
System Name: X1E
Interconnect: Cray modified 2D torus
MPI: MPT2.4.0.3
Affiliation: Army High Performance Computing Research Center
Submission Date: 06-16-05
Cray Inc. X1E Cray X1 MSP 1.13GHz 252 1 252
0.0127
0.3381
0.0001
23.1291
0.0616
15.1561
0.3602
14.93
Manufacturer: Cray Inc.
Processor Type: Cray X1E
Processor Speed: 1.13GHz
Processor Count: 32
Threads: 4
Processses: 32
System Name: X1E
Interconnect: Cray Interconnect
MPI: mpt.2.4.0.4.4
Affiliation: ORNL
Submission Date: 09-13-05
Cray Inc. X1E Cray 1.13GHz 32 4 32
0.0106
0.5912
0.0003
5.7105
0.1626
3.6287
1.4049
12.21
Manufacturer: Cray Inc.
Processor Type: Cray X1E
Processor Speed: 1.13GHz
Processor Count: 1008
Threads: 1
Processses: 1008
System Name: X1
Interconnect: Cray Modified 2D torus
MPI: MPT
Affiliation: DOE/Office of Science/ORNL
Submission Date: 11-02-05
Cray Inc. X1 Cray E 1.13GHz 1008 1 1008
0.0119
0.1072
0.0001
32.7060
0.0817
15.2573
0.1567
16.30
Manufacturer: Fujitsu Ltd.
Processor Type: Fujitsu SPARC64 VIIIfx
Processor Speed: 2GHz
Processor Count: 73728
Threads: 8
Processses: 9216
System Name: K computer
Interconnect: Tofu interconnect
MPI: Parallelnavi Technical Computing Language V1.0L20
Affiliation: RIKEN Advanced Institute for Computational Science (AICS)
Submission Date: 10-31-11
Fujitsu Ltd. K computer Fujitsu SPARC64 VIIIfx 2GHz 73728 8 9216
0.0096
0.0346
0.0004
3.7097
0.1492
13.7799
0.2985
5.82
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 1.7GHz
Processor Count: 256
Threads: 4
Processses: 64
System Name: eServer pSeries 655
Interconnect: HPS (IBM High Performance Switch)
MPI: PE 4.1
Affiliation: IBM
Submission Date: 08-26-04
IBM eServer pSeries 655 Power4+ 1.7GHz 256 4 64
0.0042
0.0927
0.0000
2.9119
0.0409
4.7059
0.7240
8.34
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 1.7GHz
Processor Count: 128
Threads: 4
Processses: 32
System Name: eServer pSeries 655
Interconnect: HPS (IBM High Performance Switch)
MPI: PE 4.1
Affiliation: IBM
Submission Date: 08-26-04
IBM eServer pSeries 655 Power4+ 1.7GHz 128 4 32
0.0042
0.0607
0.0000
3.0088
0.0342
4.7332
0.7472
7.94
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 1.7GHz
Processor Count: 64
Threads: 4
Processses: 16
System Name: eServer pSeries 655
Interconnect: HPS (IBM High Performance Switch)
MPI: PE 4.1
Affiliation: IBM
Submission Date: 08-26-04
IBM eServer pSeries 655 Power4+ 1.7GHz 64 4 16
0.0041
0.0627
0.0000
2.8466
0.0306
4.7253
0.7483
7.57
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 1.5GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: p655
Interconnect: Federation
MPI: PE 4.2
Affiliation: San Diego Supercomputer Center
Submission Date: 01-04-06
IBM p655 Power4+ 1.5GHz 512 1 512
0.0030
0.0745
0.0001
3.4525
0.1019
4.1094
0.1262
9.40
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 1.7GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: p655
Interconnect: Federation
MPI: PE 4.2
Affiliation: San Diego Supercomputer Center
Submission Date: 01-04-06
IBM p655 Power4+ 1.7GHz 512 1 512
0.0034
0.0804
0.0001
3.6213
0.1025
4.6814
0.1393
8.73
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 1.5GHz
Processor Count: 1024
Threads: 1
Processses: 1024
System Name: p655
Interconnect: Federation
MPI: PE 4.2
Affiliation: San Diego Supercomputer Center
Submission Date: 01-04-06
IBM p655 Power4+ 1.5GHz 1024 1 1024
0.0030
0.0502
0.0001
4.4765
0.0802
4.0067
0.0902
10.13
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 1.5GHz
Processor Count: 2048
Threads: 1
Processses: 2048
System Name: p655
Interconnect: Federation
MPI: PE 4.2
Affiliation: San Diego Supercomputer Center
Submission Date: 01-04-06
IBM p655 Power4+ 1.5GHz 2048 1 2048
0.0030
0.0507
0.0001
3.4551
0.0764
4.6620
0.0696
13.36
Manufacturer: IBM
Processor Type: IBM Power4+
Processor Speed: 2.5GHz
Processor Count: 16
Threads: 1
Processses: 32
System Name: JS21
Interconnect: InfiniBand
MPI: MPICH2.1
Affiliation: RDCPS
Submission Date: 07-04-07
IBM JS21 Power4+ 2.5GHz 16 1 32
0.0107
0.0389
0.0002
3.0954
0.0667
8.4083
0.0111
77.45
Manufacturer: IBM
Processor Type: IBM Power5
Processor Speed: 1.9GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: p575 cluster
Interconnect: HPS
MPI: POE 4.2.0.2
Affiliation: IBM Corporation
Submission Date: 06-29-05
IBM p575 cluster Power5 1.9GHz 64 1 64
0.0064
0.1467
0.0003
5.4486
0.3269
7.2408
0.2446
6.63
Manufacturer: IBM
Processor Type: IBM Power5
Processor Speed: 1.9GHz
Processor Count: 10240
Threads: 1
Processses: 10240
System Name: IBM p5-575
Interconnect: HPS
MPI: PE 4.2
Affiliation: LLNL
Submission Date: 11-04-05
IBM p5-575 Power5 1.9GHz 10240 1 10240
0.0057
0.0540
0.0000
5.3111
0.0823
7.1372
0.1101
118.59
Manufacturer: IBM
Processor Type: IBM Power5
Processor Speed: 1.9GHz
Processor Count: 8192
Threads: 2
Processses: 8192
System Name: IBM p5-575
Interconnect: HPS
MPI: PE 4.2
Affiliation: LLNL
Submission Date: 11-09-05
IBM p5-575 Power5 1.9GHz 8192 2 8192
0.0041
0.0703
0.0000
2.6253
0.1180
3.0403
0.0770
51.99
Manufacturer: IBM
Processor Type: IBM Power5
Processor Speed: 1.9GHz
Processor Count: 8192
Threads: 1
Processses: 8192
System Name: IBM p5-575
Interconnect: HPS
MPI: poe 4.2
Affiliation: LLNL
Submission Date: 08-02-06
IBM p5-575 Power5 1.9GHz 8192 1 8192
0.0056
0.3206
0.0000
5.4535
0.1109
7.1896
0.0887
11.05
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 2.2GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2.2.3
Affiliation: IBM
Submission Date: 05-08-06
IBM P5 P575+ Power5+ 2.2GHz 128 1 128
0.0077
0.7070
0.0003
11.1331
0.2997
8.4989
0.2165
9.66
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 2.2GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2.2.3
Affiliation: IBM
Submission Date: 05-08-06
IBM P5 P575+ Power5+ 2.2GHz 64 1 64
0.0077
0.6940
0.0004
11.3027
0.3465
8.4530
0.2654
8.79
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 1.9GHz
Processor Count: 256
Threads: 1
Processses: 256
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2.2.3
Affiliation: IBM
Submission Date: 09-06-06
IBM P5 P575+ Power5+ 1.9GHz 256 1 256
0.0066
0.3974
0.0003
12.8287
0.1407
7.6880
0.0908
20.12
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 1.9GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2.2.3
Affiliation: IBM
Submission Date: 09-06-06
IBM P5 P575+ Power5+ 1.9GHz 512 1 512
0.0067
0.4288
0.0002
12.9877
0.1021
7.7451
0.0524
36.14
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 1.9GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2
Affiliation: IBM
Submission Date: 11-16-06
IBM P5 P575+ Power5+ 1.9GHz 512 1 512
0.0064
0.4127
0.0002
12.4922
0.1181
7.3693
0.0737
37.95
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 1.9GHz
Processor Count: 256
Threads: 1
Processses: 256
System Name: P5 P575+
Interconnect: HPS
MPI: poe 4.2
Affiliation: IBM
Submission Date: 11-16-06
IBM P5 P575+ Power5+ 1.9GHz 256 1 256
0.0064
0.3738
0.0002
12.1096
0.1566
7.3151
0.1136
21.09
Manufacturer: IBM
Processor Type: IBM Power5+
Processor Speed: 1.9GHz
Processor Count: 16
Threads: 1
Processses: 16
System Name: Babbage
Interconnect: HPS
MPI: poe 4.2.2.3
Affiliation: NAVO
Submission Date: 02-17-09
IBM Babbage Power5+ 1.9GHz 16 1 16
0.0008
0.0956
0.0005
13.9403
0.4863
6.2838
2.2999
4.00
Manufacturer: IBM
Processor Type: IBM Power7
Processor Speed: 3.836GHz
Processor Count: 1989
Threads: 32
Processses: 1989
System Name: Power 775
Interconnect: Custom IBM Hub Chip
MPI: IBM PE v1209
Affiliation: IBM Development Engineering
Submission Date: 11-08-12
IBM Power 775 Power7 3.836GHz 1989 32 1989
0.3636
0.7963
0.0003
6.4634
5.0300
25.9060
3.3226
4.69
Manufacturer: IBM
Processor Type: IBM Power7 Quad-Chip module
Processor Speed: 3.836GHz
Processor Count: 1470
Threads: 32
Processses: 1470
System Name: IBM Power775
Interconnect: IBM Hub Chip integrated interconnect
MPI: IBM PE MPI release 1206
Affiliation: IBM Development Engineering - DARPA Trial Subset
Submission Date: 07-15-12
IBM Power775 Power7 Quad-Chip module 3.836GHz 1470 32 1470
0.6839
0.7680
0.0004
6.3611
5.4741
25.4995
3.3858
3.97
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 1024
Threads: 1
Processses: 1024
System Name: Blue Gene
Interconnect: Custom 3D torus + global tree
MPI: MPICH 1.1
Affiliation: San Diego Supercomputer Center
Submission Date: 05-04-05
IBM Blue Gene PowerPC 440 0.7GHz 1024 1 1024
0.0007
0.0269
0.0001
0.9598
0.0478
0.9218
0.0346
4.81
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 1024
Threads: 1
Processses: 1024
System Name: Blue Gene
Interconnect: 3D torus + global tree
MPI: MPICH 1.1
Affiliation: San Diego Supercomputer Center
Submission Date: 07-10-05
IBM Blue Gene PowerPC 440 0.7GHz 1024 1 1024
0.0007
0.0258
0.0003
0.9078
0.0693
0.9174
0.0448
4.50
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 2048
Threads: 1
Processses: 2048
System Name: Blue Gene
Interconnect: 3D torus + global tree
MPI: MPICH 1.1
Affiliation: San Diego Supercomputer Center
Submission Date: 07-10-05
IBM Blue Gene PowerPC 440 0.7GHz 2048 1 2048
0.0007
0.0167
0.0002
0.9079
0.0470
0.9133
0.0209
4.98
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 65536
Threads: 1
Processses: 65536
System Name: Blue Gene/L
Interconnect: Custom Torus/Tree
MPI: MPICH2 1.0.1
Affiliation: National Nuclear Security Administration
Submission Date: 11-02-05
IBM Blue Gene/L PowerPC 440 0.7GHz 65536 1 65536
0.0012
0.0052
0.0000
0.8641
0.0332
1.8534
0.0108
8.84
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 32768
Threads: 1
Processses: 32768
System Name: Blue Gene/L
Interconnect: Blue Gene Custom Interconnect
MPI: MPICH 1.1
Affiliation: IBM T.J. Watson Research Center
Submission Date: 11-04-05
IBM Blue Gene/L PowerPC 440 0.7GHz 32768 1 32768
0.0010
0.0027
0.0000
1.4475
0.0340
2.2321
0.0120
9.51
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 65536
Threads: 1
Processses: 65536
System Name: Blue Gene/L
Interconnect: Custom
MPI: MPICH 2 1.0.1
Affiliation: National Nuclear Security Administration
Submission Date: 04-06-06
IBM Blue Gene/L PowerPC 440 0.7GHz 65536 1 65536
0.0006
0.0712
0.0000
0.9596
0.0269
2.4642
0.0104
8.62
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 1024
Threads: 1
Processses: 1024
System Name: Blue Gene
Interconnect: 3D torus + global tree
MPI: MPICH 1.2.1
Affiliation: San Diego Supercomputer Center
Submission Date: 04-06-06
IBM Blue Gene PowerPC 440 0.7GHz 1024 1 1024
0.0015
0.0218
0.0003
0.9076
0.0688
2.3604
0.0424
4.19
Manufacturer: IBM
Processor Type: IBM PowerPC 440
Processor Speed: 0.7GHz
Processor Count: 2048
Threads: 1
Processses: 2048
System Name: Blue Gene
Interconnect: 3D torus + global tree
MPI: MPICH 1.2.1
Affiliation: San Diego Supercomputer Center
Submission Date: 04-06-06
IBM Blue Gene PowerPC 440 0.7GHz 2048 1 2048
0.0013
0.0153
0.0002
0.9077
0.0468
2.2846
0.0215
4.70
Manufacturer: IBM
Processor Type: IBM PowerPC 970MP
Processor Speed: 2.5GHz
Processor Count: 1020
Threads: 1
Processses: 2040
System Name: e1350 JS21
Interconnect: Myrinet 2000
MPI: MPICH MX 1.2.7 64-bit
Affiliation: Indiana University
Submission Date: 06-25-07
IBM e1350 JS21 PowerPC 970MP 2.5GHz 1020 1 2040
0.0133
0.0400
0.0002
3.3121
0.0660
8.4940
0.0212
17.73
Manufacturer: IBM
Processor Type: IBM PowerPC A2
Processor Speed: 1.6GHz
Processor Count: 49152
Threads: 64
Processses: 49152
System Name: Blue Gene/Q (MIRA)
Interconnect: BGQ 5D TORUS
MPI: MPICH2 version 1.5
Affiliation: Argonne Leadership Computing Facility/Argonne National Laboratory
Submission Date: 10-26-14
IBM Blue Gene/Q (MIRA) PowerPC A2 1.6GHz 49152 64 49152
0.0205
0.1160
0.0001
0.4403
0.4378
2.0041
0.1709
5.34
Manufacturer: HP
Processor Type: Intel Core 2
Processor Speed: 3GHz
Processor Count: 16
Threads: 1
Processses: 32
System Name: XC3000
Interconnect: Infiniband 4x DDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 03-22-07
HP XC3000 Intel Core 2 3GHz 16 1 32
0.0169
0.3214
0.0026
2.9669
0.4113
10.2217
0.3000
8.87
Manufacturer: HP
Processor Type: Intel Core 2
Processor Speed: 3GHz
Processor Count: 32
Threads: 1
Processses: 64
System Name: XC3000
Interconnect: Infiniband 4x DDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 03-22-07
HP XC3000 Intel Core 2 3GHz 32 1 64
0.0171
0.2806
0.0014
2.9817
0.3521
10.5083
0.2717
8.99
Manufacturer: HP
Processor Type: Intel Core 2
Processor Speed: 3GHz
Processor Count: 64
Threads: 1
Processses: 128
System Name: XC3000
Interconnect: Infiniband 4x DDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 03-22-07
HP XC3000 Intel Core 2 3GHz 64 1 128
0.0166
0.2754
0.0006
2.9213
0.2728
10.2148
0.2422
9.17
Manufacturer: HP
Processor Type: Intel Core 2
Processor Speed: 3GHz
Processor Count: 128
Threads: 1
Processses: 256
System Name: XC3000
Interconnect: Infiniband 4x DDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 03-22-07
HP XC3000 Intel Core 2 3GHz 128 1 256
0.0167
0.2397
0.0003
2.9816
0.2541
10.4945
0.1953
9.26
Manufacturer: HP
Processor Type: Intel Core 2
Processor Speed: 3GHz
Processor Count: 32
Threads: 1
Processses: 64
System Name: XC3000 (C-class blades)
Interconnect: Infiniband 4x DDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 03-23-07
HP XC3000 (C-class blades) Intel Core 2 3GHz 32 1 64
0.0179
0.2712
0.0006
4.1505
0.4236
10.6568
0.2505
14.22
Manufacturer: HP
Processor Type: Intel Core 2
Processor Speed: 3GHz
Processor Count: 64
Threads: 1
Processses: 128
System Name: XC3000 (C-class blades)
Interconnect: Infiniband 4x DDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 03-23-07
HP XC3000 (C-class blades) Intel Core 2 3GHz 64 1 128
0.0178
0.2197
0.0003
4.1275
0.3283
10.3561
0.1862
14.64
Manufacturer: HP
Processor Type: Intel Core 2
Processor Speed: 3GHz
Processor Count: 128
Threads: 1
Processses: 256
System Name: XC3000 (C-class blades)
Interconnect: Infiniband 4x DDR
MPI: HP MPI 02.02.05.00
Affiliation: HP
Submission Date: 03-23-07
HP XC3000 (C-class blades) Intel Core 2 3GHz 128 1 256
0.0169
0.1619
0.0001
4.1503
0.2598
10.6126
0.1117
15.68
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: diy
Processor Type: Intel Core 2 Extreme QX6700
Processor Speed: 2.66GHz
Processor Count: 1
Threads: 1
Processses: 4
System Name: diy
Interconnect: N/A
MPI: MPICH 1.2.7
Affiliation: N/A
Submission Date: 12-05-06
diy Intel Core 2 Extreme QX6700 2.66GHz 1 1 4
0.0283
0.0282
0.0009
4.8000
0.8533
36.2903
0.1486
20.16
Manufacturer: Dell
Processor Type: Intel Core2Quad Q6600
Processor Speed: 2.4GHz
Processor Count: 4
Threads: 4
Processses: 4
System Name: Optiplex 755
Interconnect: None
MPI: OpenMPI 1.2.5
Affiliation: University of Ottawa
Submission Date: 01-31-08
Dell Optiplex 755 Intel Core2Quad Q6600 2.4GHz 4 4 4
0.0022
0.1930
0.0073
3.4493
0.6610
1.7718
0.2895
0.96
Manufacturer: Intel
Processor Type: Intel i7-2600
Processor Speed: 3.4GHz
Processor Count: 8
Threads: 8
Processses: 64
System Name: Guacamole
Interconnect: Ethernet Gigabit
MPI: mpich2 1.4.1p1
Affiliation: Universidad del Caribe
Submission Date: 03-22-12
Intel Guacamole i7-2600 3.4GHz 8 8 64
0.0000
0.0021
0.0000
2.5567
0.1869
1.2494
0.0013
87.27
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.3GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Altix 3700
Interconnect: Numalink
MPI: SGI MPT 1.9-1
Affiliation: University of Manchester
Submission Date: 07-20-04
SGI Altix 3700 Intel Itanium 2 1.3GHz 32 1 32
0.0040
0.0799
0.0001
0.9133
0.1275
4.6515
0.2903
5.79
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.5GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Altix 3700
Interconnect: Numalink
MPI: SGI MPT 1.10
Affiliation: University of Manchester
Submission Date: 09-21-04
SGI Altix 3700 Intel Itanium 2 1.5GHz 128 1 128
0.0050
0.0588
0.0001
3.7459
0.1100
5.8479
0.2107
6.39
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Altix 3700 Bx2
Interconnect: N/A
MPI: SGI MPT 1.12
Affiliation: SGI
Submission Date: 03-02-05
SGI Altix 3700 Bx2 Intel Itanium 2 1.6GHz 64 1 64
0.0046
0.1824
0.0002
3.7916
0.2584
5.9789
0.8710
3.68
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Altix 3700 Bx2
Interconnect: N/A
MPI: SGI MPT 1.12
Affiliation: SGI
Submission Date: 03-02-05
SGI Altix 3700 Bx2 Intel Itanium 2 1.6GHz 128 1 128
0.0041
0.1925
0.0001
3.7921
0.2162
5.9177
0.8970
3.91
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Altix 3700 Bx2
Interconnect: N/A
MPI: SGI MPT 1.12
Affiliation: SGI
Submission Date: 03-15-05
SGI Altix 3700 Bx2 Intel Itanium 2 1.6GHz 32 1 32
0.0046
0.1891
0.0002
3.8055
0.2337
6.0318
1.5188
3.26
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 1008
Threads: 1
Processses: 1008
System Name: Altix 3700 Bx2
Interconnect: NUMAlink
MPI: SGI MPT 1.12
Affiliation: University of Illinois
Submission Date: 03-29-05
SGI Altix 3700 Bx2 Intel Itanium 2 1.6GHz 1008 1 1008
0.0051
0.1048
0.0000
3.0875
0.0155
5.8739
0.2029
6.82
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 2024
Threads: 1
Processses: 2024
System Name: Columbia 2048
Interconnect: NUMALINK 4
MPI: MPT 1.12.0.0
Affiliation: NASA
Submission Date: 11-12-05
SGI Columbia 2048 Intel Itanium 2 1.6GHz 2024 1 2024
0.0046
0.0090
0.0000
3.5628
0.0226
6.2574
0.1227
6.98
Manufacturer: HP
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 2048
Threads: 1
Processses: 2048
System Name: XC
Interconnect: Quadrics Elan4
MPI: Quadrics MPI 1.24-47
Affiliation: Government
Submission Date: 07-03-06
HP XC Intel Itanium 2 1.6GHz 2048 1 2048
0.0052
0.1496
0.0001
4.3304
0.1659
6.2922
0.1592
6.04
Manufacturer: HP
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 1024
Threads: 1
Processses: 1024
System Name: XC
Interconnect: Quadrics Elan4
MPI: Quadrics MPI 1.24-47
Affiliation: Government
Submission Date: 07-03-06
HP XC Intel Itanium 2 1.6GHz 1024 1 1024
0.0047
0.1505
0.0002
4.4223
0.1798
6.2832
0.1598
5.94
Manufacturer: NEC
Processor Type: Intel Itanium 2
Processor Speed: 1.6GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: NEC TX-7/i9610
Interconnect: non
MPI: MPI/EX (IBA) 1.3.11
Affiliation: Tohoku University, Information Synergy Center
Submission Date: 07-10-06
NEC TX-7/i9610 Intel Itanium 2 1.6GHz 64 1 64
0.0053
0.1689
0.0004
4.3836
0.1242
6.2429
6.0144
4.76
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.5GHz
Processor Count: 96
Threads: 1
Processses: 96
System Name: Altix 4700
Interconnect: NUMAlink 4
MPI: MPT 1.14
Affiliation: Idaho National Laboratory
Submission Date: 09-01-06
SGI Altix 4700 Intel Itanium 2 1.5GHz 96 1 96
0.0042
0.0562
0.0004
2.7966
0.0563
4.7569
0.1830
4.74
Manufacturer: SGI
Processor Type: Intel Itanium 2
Processor Speed: 1.3GHz
Processor Count: 4
Threads: 1
Processses: 4
System Name: Altix 3700
Interconnect: numalink
MPI: MPT
Affiliation: University of Queensland
Submission Date: 11-29-07
SGI Altix 3700 Intel Itanium 2 1.3GHz 4 1 4
0.0020
0.1281
0.0037
1.8139
0.1488
4.9521
0.5023
3.18
Manufacturer: Lewis Hall
Processor Type: Intel IXP425
Processor Speed: 0.533GHz
Processor Count: 8
Threads: 1
Processses: 8
System Name: Embedded Arm XScale
Interconnect: Fast Ethernet
MPI: LAM/MPI 7.1.1
Affiliation: Dedicated Devices Inc
Submission Date: 10-12-05
Lewis Hall Embedded Arm XScale Intel IXP425 0.533GHz 8 1 8
0.0000
0.0034
0.0000
0.0446
0.0032
0.0043
0.0061
603.15
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: Hewlett Packard
Processor Type: Intel Nehalem X5550
Processor Speed: 2.67GHz
Processor Count: 16
Threads: 1
Processses: 16
System Name: Hewlett Packard BL280cG6 Intel Nehalem X5550
Interconnect: Mellanox Technologies (IB-QDR)
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 05-11-10
Hewlett Packard BL280cG6 Intel Nehalem X5550 Inte ... 2.67GHz 16 1 16
0.0096
0.3126
0.0117
11.1278
0.7251
10.2960
0.7014
1.04
Manufacturer: Hewlett Packard
Processor Type: Intel Nehalem X5550
Processor Speed: 2.67GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Hewlett Packard BL280cG6 Intel Nehalem X5550
Interconnect: Mellanox Technologies (IB-QDR)
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 05-11-10
Hewlett Packard BL280cG6 Intel Nehalem X5550 Inte ... 2.67GHz 32 1 32
0.0097
0.2238
0.0099
11.1510
0.6042
10.3087
0.3983
1.17
Manufacturer: Hewlett Packard
Processor Type: Intel Nehalem X5550
Processor Speed: 2.67GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Hewlett Packard BL280cG6 Intel Nehalem X5550
Interconnect: Mellanox Technologies (IB-QDR)
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 05-11-10
Hewlett Packard BL280cG6 Intel Nehalem X5550 Inte ... 2.67GHz 64 1 64
0.0096
0.2162
0.0082
10.9990
0.5787
10.2986
0.2995
1.31
Manufacturer: Hewlett Packard
Processor Type: Intel Nehalem X5550
Processor Speed: 2.67GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Hewlett Packard BL280cG6 Intel Nehalem X5550
Interconnect: Mellanox Technologies (IB-QDR)
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 05-11-10
Hewlett Packard BL280cG6 Intel Nehalem X5550 Inte ... 2.67GHz 128 1 128
0.0097
0.1922
0.0071
11.1122
0.5158
10.3037
0.2853
1.41
Manufacturer: Hewlett Packard
Processor Type: Intel Nehalem X5550
Processor Speed: 2.67GHz
Processor Count: 8
Threads: 1
Processses: 8
System Name: Hewlett Packard BL280cG6 Intel Nehalem X5550
Interconnect: Mellanox Technologies (IB-QDR)
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 05-12-10
Hewlett Packard BL280cG6 Intel Nehalem X5550 Inte ... 2.67GHz 8 1 8
0.0098
0.3454
0.0128
10.9829
0.8347
10.2998
1.3551
0.50
Manufacturer: Dell
Processor Type: Intel Pentium 4
Processor Speed: 3GHz
Processor Count: 48
Threads: 1
Processses: 48
System Name: PE850
Interconnect: Gigabit Ethernet
MPI: LAM 7.1.1
Affiliation: University of Florida
Submission Date: 01-22-06
Dell PE850 Intel Pentium 4 3GHz 48 1 48
0.0080
0.0621
0.0000
3.0454
0.0889
10.4920
0.0763
59.90
Manufacturer: cgna
Processor Type: Intel Pentium 4
Processor Speed: 3.06GHz
Processor Count: 14
Threads: 1
Processses: 28
System Name: rna
Interconnect: gigabit ethernet
MPI: openmpi 1.2.5
Affiliation: cgna
Submission Date: 05-28-09
cgna rna Intel Pentium 4 3.06GHz 14 1 28
0.0001
0.0031
0.0001
1.0246
0.0195
2.4726
0.0089
98.17
Manufacturer: David Hayes
Processor Type: Intel Pentium 4
Processor Speed: 2.7GHz
Processor Count: 35
Threads: 1
Processses: 35
System Name: MSC
Interconnect: SMC8150L2 NA Gigabit Ethernet Switch
MPI: mpich2-1.2
Affiliation: University of Nebraska at Kearney
Submission Date: 11-22-10
David Hayes MSC Intel Pentium 4 2.7GHz 35 1 35
0.0009
0.0180
0.0002
2.7626
0.0390
1.7314
0.0372
42.91
Manufacturer: PelicanHPC
Processor Type: Intel Pentium 4
Processor Speed: 3GHz
Processor Count: 8
Threads: 1
Processses: 8
System Name: PelicanHPC 2.2
Interconnect: gigabit ethernet
MPI: openmpi 1.2.5
Affiliation: Technical University
Submission Date: 08-02-11
PelicanHPC 2.2 Intel Pentium 4 3GHz 8 1 8
0.0028
0.0700
0.0007
3.4234
0.1479
3.7496
0.0553
37.95
Manufacturer: PelicanHPC
Processor Type: Intel Pentium 4
Processor Speed: 3GHz
Processor Count: 8
Threads: 1
Processses: 8
System Name: PelicanHPC 2.2 GotoBlas2
Interconnect: gigabit ethernet
MPI: openmpi 1.2.5
Affiliation: Technical University
Submission Date: 08-08-11
PelicanHPC 2.2 GotoBlas2 Intel Pentium 4 3GHz 8 1 8
0.0044
0.0636
0.0008
3.4520
0.1639
5.2891
0.0553
39.50
Manufacturer: Hewlett Packard
Processor Type: Intel Westmere (SMT OFF, Turbo ON, DDR3-1333)
Processor Speed: 2.93GHz
Processor Count: 12
Threads: 1
Processses: 12
System Name: Hewlett Packard BL460G6
Interconnect: Mellanox QDR
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 07-12-10
Hewlett Packard BL460G6 Intel Westmere (SMT OFF, ... 2.93GHz 12 1 12
0.0105
0.2693
0.0101
11.3310
0.5926
12.1773
0.8549
0.51
Manufacturer: Hewlett Packard
Processor Type: Intel Westmere (SMT OFF, Turbo ON, DDR3-1333)
Processor Speed: 2.93GHz
Processor Count: 24
Threads: 1
Processses: 24
System Name: Hewlett Packard BL460G6
Interconnect: Mellanox QDR
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 07-12-10
Hewlett Packard BL460G6 Intel Westmere (SMT OFF, ... 2.93GHz 24 1 24
0.0106
0.2744
0.0085
11.1140
0.5329
12.1622
0.5654
1.11
Manufacturer: Hewlett Packard
Processor Type: Intel Westmere (SMT OFF, Turbo ON, DDR3-1333)
Processor Speed: 2.93GHz
Processor Count: 48
Threads: 1
Processses: 48
System Name: Hewlett Packard BL460G6
Interconnect: Mellanox QDR
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 07-12-10
Hewlett Packard BL460G6 Intel Westmere (SMT OFF, ... 2.93GHz 48 1 48
0.0105
0.1833
0.0071
11.3240
0.4441
12.0406
0.2859
1.35
Manufacturer: Hewlett Packard
Processor Type: Intel Westmere (SMT OFF, Turbo ON, DDR3-1333)
Processor Speed: 2.93GHz
Processor Count: 96
Threads: 1
Processses: 96
System Name: Hewlett Packard BL460G6
Interconnect: Mellanox QDR
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 07-12-10
Hewlett Packard BL460G6 Intel Westmere (SMT OFF, ... 2.93GHz 96 1 96
0.0105
0.1577
0.0059
10.8583
0.4140
11.9399
0.2149
1.66
Manufacturer: Hewlett Packard
Processor Type: Intel Westmere (SMT OFF, Turbo ON, DDR3-1333)
Processor Speed: 2.93GHz
Processor Count: 192
Threads: 1
Processses: 192
System Name: Hewlett Packard BL460G6
Interconnect: Mellanox QDR
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 07-12-10
Hewlett Packard BL460G6 Intel Westmere (SMT OFF, ... 2.93GHz 192 1 192
0.0104
0.1465
0.0049
11.3214
0.3895
12.2497
0.1892
1.93
Manufacturer: Hewlett Packard
Processor Type: Intel Westmere (SMT OFF, Turbo ON, DDR3-1333)
Processor Speed: 2.93GHz
Processor Count: 384
Threads: 1
Processses: 384
System Name: Hewlett Packard BL460G6
Interconnect: Mellanox QDR
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 07-12-10
Hewlett Packard BL460G6 Intel Westmere (SMT OFF, ... 2.93GHz 384 1 384
0.0105
0.1316
0.0040
10.1191
0.3440
12.0778
0.1727
2.44
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: Hewlett Packard
Processor Type: Intel Westmere (SMT OFF, Turbo ON, DDR3-1333)
Processor Speed: 2.93GHz
Processor Count: 768
Threads: 1
Processses: 768
System Name: Hewlett Packard BL460G6
Interconnect: Mellanox QDR
MPI: Platform MPI 07.01.00.00
Affiliation: Hewlett Packard
Submission Date: 07-12-10
Hewlett Packard BL460G6 Intel Westmere (SMT OFF, ... 2.93GHz 768 1 768
0.0103
0.1292
0.0033
10.4335
0.2777
12.1421
0.1417
2.87
Manufacturer: Dell
Processor Type: Intel Xeon
Processor Speed: 2.4GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: PowerEdge 2650 Cluster
Interconnect: Gigabit Ethernet, PowerConnect 5224 switch
MPI: scampi-3.3.6-1.rhel3
Affiliation: Scali AS
Submission Date: 02-18-05
Dell PowerEdge 2650 Cluster Intel Xeon 2.4GHz 32 1 32
0.0030
0.0284
0.0000
1.1673
0.0605
3.9987
0.0379
42.23
Manufacturer: Dell
Processor Type: Intel Xeon
Processor Speed: 2.4GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: PowerEdge 2650 Cluster
Interconnect: SCI, 4x4 2d Torus
MPI: scampi-3.3.6-1.rhel3
Affiliation: Scali AS
Submission Date: 02-18-05
Dell PowerEdge 2650 Cluster Intel Xeon 2.4GHz 32 1 32
0.0031
0.0358
0.0001
1.1969
0.0686
4.0254
0.0477
8.91
Manufacturer: Dell
Processor Type: Intel Xeon
Processor Speed: 2.4GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: PowerEdge 2650 Cluster
Interconnect: Myrinet 2000
MPI: scampi-3.3.6-1.rhel3
Affiliation: Scali AS
Submission Date: 02-18-05
Dell PowerEdge 2650 Cluster Intel Xeon 2.4GHz 32 1 32
0.0030
0.0433
0.0001
1.1821
0.0735
4.0588
0.0659
19.00
Manufacturer: Dell
Processor Type: Intel Xeon
Processor Speed: 2.4GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: PowerEdge 2650 Cluster
Interconnect: InfiniBand, 4x adapters, InfinIO 3000 switch
MPI: scampi-3.3.6-1.rhel3
Affiliation: Scali AS
Submission Date: 02-18-05
Dell PowerEdge 2650 Cluster Intel Xeon 2.4GHz 32 1 32
0.0032
0.0640
0.0001
1.1466
0.1013
3.9986
0.1784
9.88
Manufacturer: Fujitjsu-Siemens
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: GandALF Cluster
Interconnect: Gigabit Ethernet
MPI: mpich2-1.0.4p1
Affiliation: Politehnica University Bucharest
Submission Date: 12-19-06
Fujitjsu-Siemens GandALF Cluster Intel Xeon 3.2GHz 64 1 64
0.0029
0.0061
0.0001
2.5946
0.0133
4.1859
0.0075
108.58
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 416
Threads: 1
Processses: 416
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 416 1 416
0.0034
0.0831
0.0004
2.9931
0.0834
4.8159
0.0778
14.07
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 384
Threads: 1
Processses: 384
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 384 1 384
0.0034
0.0830
0.0004
2.9607
0.0887
4.8835
0.0845
12.18
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 256
Threads: 1
Processses: 256
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 256 1 256
0.0034
0.0480
0.0006
2.9615
0.1362
4.8697
0.1124
9.68
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 128 1 128
0.0035
0.1034
0.0008
2.9569
0.1359
4.8850
0.1712
8.33
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 64 1 64
0.0035
0.0593
0.0010
2.9240
0.1336
4.8684
0.2438
7.44
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 32 1 32
0.0035
0.1686
0.0012
2.9063
0.1335
4.8711
0.4817
6.74
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 16
Threads: 1
Processses: 16
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 16 1 16
0.0035
0.0702
0.0017
2.9509
0.1308
4.8815
0.5211
5.22
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 8
Threads: 1
Processses: 8
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 8 1 8
0.0035
0.1861
0.0023
2.9470
0.1323
4.8890
0.5194
4.30
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 4
Threads: 1
Processses: 4
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 4 1 4
0.0035
0.0707
0.0025
2.9748
0.1440
4.8572
0.5325
4.04
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 2
Threads: 1
Processses: 2
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 2 1 2
0.0043
0.2561
0.0024
2.9141
0.1708
4.8670
0.5311
5.69
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: TopSpin/Cisco
Processor Type: Intel Xeon
Processor Speed: 3.2GHz
Processor Count: 1
Threads: 1
Processses: 1
System Name: LLGrid TX2500
Interconnect: InfiniBand
MPI: Open MPI 1.1.1
Affiliation: MIT Lincoln Laboratory
Submission Date: 07-30-07
TopSpin/Cisco LLGrid TX2500 Intel Xeon 3.2GHz 1 1 1
0.0044
0.0730
0.0022
2.9565
0.4081
4.8680
1.4170
0.78
Manufacturer: SGI
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 56
Threads: 1
Processses: 56
System Name: uv100
Interconnect: NUMALink
MPI: intel sgi
Affiliation: UPMC
Submission Date: 03-02-11
SGI uv100 Intel Xeon 2GHz 56 1 56
0.0068
0.1008
0.0020
5.8618
0.1887
7.6522
0.2224
3.19
Manufacturer: n/a
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 16
Threads: 0
Processses: 16
System Name: TestHost
Interconnect: GbE
MPI: n/a
Affiliation: Faculdade de Engenharia da Universidade do Porto
Submission Date: 07-01-11
n/a TestHost Intel Xeon 2GHz 16 0 16
0.0047
0.0285
0.0009
0.0000
0.0759
0.0000
0.0220
41.36
Manufacturer: n/a
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 32
Threads: 0
Processses: 32
System Name: n/a TestHost Intel Xeon
Interconnect: GbE
MPI: n/a
Affiliation: Faculdade de Engenharia da Universidade do Porto
Submission Date: 07-01-11
n/a TestHost Intel Xeon Intel Xeon 2GHz 32 0 32
0.0038
0.0122
0.0003
0.0000
0.0401
0.0000
0.0100
62.40
Manufacturer: n/a
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 32
Threads: 0
Processses: 32
System Name: n/a TestVM Intel Xeon
Interconnect: GbE
MPI: n/a
Affiliation: Faculdade de Engenharia da Universidade do Porto
Submission Date: 07-01-11
n/a TestVM Intel Xeon Intel Xeon 2GHz 32 0 32
0.0031
0.0049
0.0003
0.0000
0.0152
0.0000
0.0049
300.53
Manufacturer: n/a
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 8
Threads: 0
Processses: 8
System Name: n/a TestHost Intel Xeon
Interconnect: GbE
MPI: n/a
Affiliation: Faculdade de Engenharia da Universidade do Porto
Submission Date: 07-01-11
n/a TestHost Intel Xeon Intel Xeon 2GHz 8 0 8
0.0049
0.3613
0.0045
0.0000
0.4723
0.0000
0.6351
1.55
Manufacturer: n/a
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 8
Threads: 0
Processses: 8
System Name: n/a TestVM Intel Xeon
Interconnect: GbE
MPI: n/a
Affiliation: Faculdade de Engenharia da Universidade do Porto
Submission Date: 07-01-11
n/a TestVM Intel Xeon Intel Xeon 2GHz 8 0 8
0.0048
0.3416
0.0029
0.0000
0.3902
0.0000
0.5424
1.43
Manufacturer: IBM
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 56
Threads: 0
Processses: 56
System Name: GeoCluster
Interconnect: Infiniban
MPI: gcc_mpi_1.4.3
Affiliation: University of Western Australia
Submission Date: 11-08-11
IBM GeoCluster Intel Xeon 2GHz 56 0 56
0.0003
0.0234
0.0003
0.0000
0.0088
0.0000
0.0988
50.52
Manufacturer: SGI
Processor Type: Intel Xeon
Processor Speed: 2GHz
Processor Count: 16
Threads: 16
Processses: 128
System Name: UV100
Interconnect: NUMA
MPI: sgi-mpi/2.05
Affiliation: UPMC/CNRS
Submission Date: 03-20-12
SGI UV100 Intel Xeon 2GHz 16 16 128
0.0499
1.1884
0.0082
0.3620
2.1691
0.4769
0.1577
3.61
Manufacturer: Zhejiang Ubiversity
Processor Type: Intel Xeon
Processor Speed: 3GHz
Processor Count: 8
Threads: 8
Processses: 8
System Name: PC
Interconnect: 1000
MPI: openmpi
Affiliation: Zhejiang Ubiversity
Submission Date: 11-02-12
Zhejiang Ubiversity PC Intel Xeon 3GHz 8 8 8
0.0021
0.0661
0.0003
0.9049
0.0965
0.4529
0.0772
113.56
Manufacturer: Intel
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: Intel Endeavor cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 10-11-07
Intel Endeavor cluster Xeon 5160 3GHz 512 1 512
0.0096
0.1119
0.0029
3.3151
0.2448
10.8439
0.1996
7.84
Manufacturer: Intel
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 256
Threads: 1
Processses: 256
System Name: Intel Endeavor cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 10-11-07
Intel Endeavor cluster Xeon 5160 3GHz 256 1 256
0.0096
0.1213
0.0032
3.4217
0.2464
10.8495
0.2212
7.54
Manufacturer: Intel
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Intel Endeavor cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 10-11-07
Intel Endeavor cluster Xeon 5160 3GHz 128 1 128
0.0096
0.1251
0.0036
3.4603
0.3045
10.9269
0.2283
7.34
Manufacturer: Intel
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Intel Endeavor cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 10-11-07
Intel Endeavor cluster Xeon 5160 3GHz 64 1 64
0.0096
0.1226
0.0041
3.4243
0.3113
10.8690
0.2274
7.09
Manufacturer: Intel
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Intel Endeavor cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 10-11-07
Intel Endeavor cluster Xeon 5160 3GHz 32 1 32
0.0097
0.1249
0.0046
3.4005
0.2903
10.8844
0.2242
6.74
Manufacturer: Intel
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 16
Threads: 1
Processses: 16
System Name: Intel Endeavor cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 10-11-07
Intel Endeavor cluster Xeon 5160 3GHz 16 1 16
0.0097
0.0956
0.0054
3.4517
0.3326
10.8811
0.2635
5.71
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: Intel
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 1024
Threads: 1
Processses: 1024
System Name: Intel Endeavor cluster
Interconnect: Infiniband
MPI: mvapich 0.9.9
Affiliation: Intel Corporation
Submission Date: 11-08-07
Intel Endeavor cluster Xeon 5160 3GHz 1024 1 1024
0.0093
0.1087
0.0025
3.4421
0.2266
10.7637
0.1561
9.25
Manufacturer: ClusterVision/Dell/QLogic
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Darwin
Interconnect: InfiniBand DDR, QLogic TrueScale adapters
MPI: HPMPI-2.02.07.00
Affiliation: University of Cambridge
Submission Date: 08-04-08
ClusterVision/Dell/QLogic Darwin Intel Xeon 5160 3GHz 64 1 64
0.0099
0.1141
0.0036
3.2974
0.2239
11.2196
0.2826
1.25
Manufacturer: ClusterVision/Dell/QLogic
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Darwin
Interconnect: InfiniBand DDR, QLogic TrueScale adapters
MPI: HPMPI-2.02.07.00
Affiliation: University of Cambridge
Submission Date: 08-05-08
ClusterVision/Dell/QLogic Darwin Intel Xeon 5160 3GHz 128 1 128
0.0099
0.1240
0.0033
3.3740
0.2264
11.3085
0.2628
1.30
Manufacturer: ClusterVision/Dell/QLogic
Processor Type: Intel Xeon 5160
Processor Speed: 3GHz
Processor Count: 256
Threads: 1
Processses: 256
System Name: Darwin
Interconnect: InfiniBand DDR, QLogic TrueScale adapters
MPI: HPMPI-2.02.07.00
Affiliation: University of Cambridge
Submission Date: 08-05-08
ClusterVision/Dell/QLogic Darwin Intel Xeon 5160 3GHz 256 1 256
0.0096
0.1203
0.0030
3.3017
0.2121
11.3181
0.2247
1.33
Manufacturer: Intel
Processor Type: Intel Xeon 5355
Processor Speed: 2.66GHz
Processor Count: 512
Threads: 1
Processses: 512
System Name: Intel Atlantis cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 08-21-07
Intel Atlantis cluster Xeon 5355 2.66GHz 512 1 512
0.0083
0.0626
0.0018
3.8280
0.1365
9.6743
0.0899
16.75
Manufacturer: Intel
Processor Type: Intel Xeon 5355
Processor Speed: 2.66GHz
Processor Count: 256
Threads: 1
Processses: 256
System Name: Intel Atlantis cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 08-21-07
Intel Atlantis cluster Xeon 5355 2.66GHz 256 1 256
0.0084
0.0712
0.0022
3.7866
0.1602
9.6316
0.1131
16.46
Manufacturer: Intel
Processor Type: Intel Xeon 5355
Processor Speed: 2.66GHz
Processor Count: 128
Threads: 1
Processses: 128
System Name: Intel Atlantis cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 08-21-07
Intel Atlantis cluster Xeon 5355 2.66GHz 128 1 128
0.0084
0.0723
0.0025
3.8872
0.1720
9.7993
0.1328
16.59
Manufacturer: Intel
Processor Type: Intel Xeon 5355
Processor Speed: 2.66GHz
Processor Count: 64
Threads: 1
Processses: 64
System Name: Intel Atlantis cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 08-21-07
Intel Atlantis cluster Xeon 5355 2.66GHz 64 1 64
0.0084
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0.0028
3.8937
0.1975
9.6855
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Manufacturer: Intel
Processor Type: Intel Xeon 5355
Processor Speed: 2.66GHz
Processor Count: 32
Threads: 1
Processses: 32
System Name: Intel Atlantis cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 08-21-07
Intel Atlantis cluster Xeon 5355 2.66GHz 32 1 32
0.0084
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0.0032
3.8972
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9.7934
0.2036
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Manufacturer: Intel
Processor Type: Intel Xeon 5355
Processor Speed: 2.66GHz
Processor Count: 16
Threads: 1
Processses: 16
System Name: Intel Atlantis cluster
Interconnect: Infiniband
MPI: Intel MPI 3.1 beta
Affiliation: Intel Corporation
Submission Date: 08-21-07
Intel Atlantis cluster Xeon 5355 2.66GHz 16 1 16
0.0084
0.0802
0.0037
3.9177
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9.6682
0.2570
6.83
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v3 (SMT OFF, TURBO OFF, PC4-17000R-15, ECC)
Processor Speed: 2.6GHz
Processor Count: 28
Threads: 1
Processses: 28
System Name: Intel Endeavor cluster
Interconnect: FDR Infiniband (48 Mellanox MSX6025F-1BFR switches, Mellanox MCX353A-FCAT adapters on nodes)
MPI: Intel MPI 5.0.0.028
Affiliation: Intel Corporation
Submission Date: 12-24-14
Intel Endeavor cluster Xeon E5-2697 v3 (SMT OFF, ... 2.6GHz 28 1 28
0.0302
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18.4570
1.1305
38.3188
1.1776
0.83
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v3 (SMT OFF, TURBO OFF, PC4-17000R-15, ECC)
Processor Speed: 2.6GHz
Processor Count: 56
Threads: 1
Processses: 56
System Name: Intel Endeavor cluster
Interconnect: FDR Infiniband (48 Mellanox MSX6025F-1BFR switches, Mellanox MCX353A-FCAT adapters on nodes)
MPI: Intel MPI 5.0.0.028
Affiliation: Intel Corporation
Submission Date: 12-24-14
Intel Endeavor cluster Xeon E5-2697 v3 (SMT OFF, ... 2.6GHz 56 1 56
0.0292
0.2802
0.0105
18.4345
0.7031
38.4979
0.4368
2.72
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v3 (SMT OFF, TURBO OFF, PC4-17000R-15, ECC)
Processor Speed: 2.6GHz
Processor Count: 112
Threads: 1
Processses: 112
System Name: Intel Endeavor cluster
Interconnect: FDR Infiniband (48 Mellanox MSX6025F-1BFR switches, Mellanox MCX353A-FCAT adapters on nodes)
MPI: Intel MPI 5.0.0.028
Affiliation: Intel Corporation
Submission Date: 12-24-14
Intel Endeavor cluster Xeon E5-2697 v3 (SMT OFF, ... 2.6GHz 112 1 112
0.0285
0.1951
0.0071
18.3347
0.5148
38.1750
0.2307
8.00
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v3 (SMT OFF, TURBO OFF, PC4-17000R-15, ECC)
Processor Speed: 2.6GHz
Processor Count: 224
Threads: 1
Processses: 224
System Name: Intel Endeavor cluster
Interconnect: FDR Infiniband (48 Mellanox MSX6025F-1BFR switches, Mellanox MCX353A-FCAT adapters on nodes)
MPI: Intel MPI 5.0.0.028
Affiliation: Intel Corporation
Submission Date: 12-24-14
Intel Endeavor cluster Xeon E5-2697 v3 (SMT OFF, ... 2.6GHz 224 1 224
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10.02
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v3 (SMT OFF, TURBO OFF, PC4-17000R-15, ECC)
Processor Speed: 2.6GHz
Processor Count: 448
Threads: 1
Processses: 448
System Name: Intel Endeavor cluster
Interconnect: FDR Infiniband (48 Mellanox MSX6025F-1BFR switches, Mellanox MCX353A-FCAT adapters on nodes)
MPI: Intel MPI 5.0.0.028
Affiliation: Intel Corporation
Submission Date: 12-26-14
Intel Endeavor cluster Xeon E5-2697 v3 (SMT OFF, ... 2.6GHz 448 1 448
0.0278
0.1353
0.0046
19.8255
0.3949
42.6512
0.1506
11.07
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v3 (SMT OFF, TURBO OFF, PC4-17000R-15, ECC)
Processor Speed: 2.6GHz
Processor Count: 896
Threads: 1
Processses: 896
System Name: Intel Endeavor cluster
Interconnect: FDR Infiniband (48 Mellanox MSX6025F-1BFR switches, Mellanox MCX353A-FCAT adapters on nodes)
MPI: Intel MPI 5.0.0.028
Affiliation: Intel Corporation
Submission Date: 12-26-14
Intel Endeavor cluster Xeon E5-2697 v3 (SMT OFF, ... 2.6GHz 896 1 896
0.0259
0.1361
0.0037
19.9410
0.3361
37.6963
0.1484
11.73
PlotSystem Information
System - Processor - Speed - Count - Threads - Processes
PP-HPL PP-PTRANS PP-Random
Access
PT-SN-STREAM
Triad
PP-FFT PT-SN-DGEMM RandomRing Bandwidth RandomRing Latency
MA/PT/PS/PC/TH/PR/CM/CS/IC/IA/SDTFlop/s GB/s Gup/s GB/s GFlop/s GFlop/s GB/s usec
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v3 (SMT OFF, TURBO OFF, PC4-17000R-15, ECC)
Processor Speed: 2.6GHz
Processor Count: 1792
Threads: 1
Processses: 1792
System Name: Intel Endeavor cluster
Interconnect: FDR Infiniband (48 Mellanox MSX6025F-1BFR switches, Mellanox MCX353A-FCAT adapters on nodes)
MPI: Intel MPI 5.0.0.028
Affiliation: Intel Corporation
Submission Date: 12-26-14
0.0286
0.1220
0.0030
19.2241
0.2335
38.1634
0.1337
16.29
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v4 (SMT OFF, TURBO OFF,128 GB RAM DDR4 2400 8*16GB, ECC)
Processor Speed: 2.3GHz
Processor Count: 36
Threads: 1
Processses: 36
System Name: Intel Endeavor cluster
Interconnect: Intel OPA 100Gbps
MPI: Intel MPI 5.1.2
Affiliation: Intel Corporation
Submission Date: 12-20-16
Intel Endeavor cluster Xeon E5-2697 v4 (SMT OFF ... 2.3GHz 36 1 36
0.0288
0.3011
0.0119
15.9132
1.2079
33.9768
1.0714
0.92
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v4 (SMT OFF, TURBO OFF,128 GB RAM DDR4 2400 8*16GB, ECC)
Processor Speed: 2.3GHz
Processor Count: 288
Threads: 1
Processses: 288
System Name: Intel Endeavor cluster
Interconnect: Intel OPA 100Gbps
MPI: Intel MPI 5.1.2
Affiliation: Intel Corporation
Submission Date: 12-20-16
Intel Endeavor cluster Xeon E5-2697 v4 (SMT OFF ... 2.3GHz 288 1 288
0.0294
0.2440
0.0069
16.2581
0.6736
33.8785
0.2942
1.28
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v4 (SMT OFF, TURBO OFF,128 GB RAM DDR4 2400 8*16GB, ECC)
Processor Speed: 2.3GHz
Processor Count: 576
Threads: 1
Processses: 576
System Name: Intel Endeavor cluster
Interconnect: Intel OPA 100Gbps
MPI: Intel MPI 5.1.2
Affiliation: Intel Corporation
Submission Date: 12-20-16
Intel Endeavor cluster Xeon E5-2697 v4 (SMT OFF ... 2.3GHz 576 1 576
0.0286
0.2239
0.0059
16.0891
0.6114
34.2593
0.2647
1.38
Manufacturer: Intel
Processor Type: Intel Xeon E5-2697 v4 (SMT OFF, TURBO OFF,128 GB RAM DDR4 2400 8*16GB, ECC)
Processor Speed: 2.3GHz
Processor Count: 1152
Threads: 1
Processses: 1152
System Name: Intel Endeavor cluster
Interconnect: Intel OPA 100Gbps
MPI: Intel MPI 5.1.2
Affiliation: Intel Corporation
Submission Date: 12-20-16
0.0289 | 31,566 | 81,017 | {"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-2020-24 | latest | en | 0.890185 |
https://boards.straightdope.com/t/body-burning-up-on-entering-earths-atmosphere/479897 | 1,726,664,142,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00331.warc.gz | 116,594,977 | 9,648 | # body burning up on entering earth's atmosphere
Would a human body, wearing no special safety clothing, burn up completely upon re-entry into earth’s atmosphere, after being jettisoned toward earth from a spacecraft in orbit?
In view of the amount of energy that would have to be dissipated before any remains reached the surface, I think the answer must be yes.
Various Googling suggests the Shuttle sees a temperature of around 1500 C for 15 to 20 minutes. That ought to get the job done.
Probably yes if you planned it for that purpose but the Colombia shuttle disaster proved that recognizable body parts could just fall from space after the break-up of the spacecraft itself. There were many of them and their protective gear wasn’t that strong. Even parts that weren’t covered fell to earth severely burned but still mostly intact.
Right, but that’s a different story from what’s specified in the OP: body only, no spacecraft.
More significant to me - Columbia broke up somewhere around 39 miles up, as far as I can quickly tell. That’s a far cry from orbit, and meant that the people who were in Columbia at the time of the disaster weren’t subject to the initial fall through the upper atmosphere themselves.
If rocks burn up in the atmosphere, I’d expect bodies to, as well.
Wouldn’t it be a matter of velocity? Joe Kittinger sure didn’t burn up, although he was only about 20 miles up.
Since OP indicates the craft was in orbit, that would mean a pretty significant starting velocity, no?
Also, not only was Columbia at a fraction of its initial orbital height, it had (i’d expect) “burned” off a significant/majority? of its horizontal velocity component. Given that total energy is a function of velocity squared, you now have a small fraction of energy to get rid of as compared to starting from an orbit.
If you just dropped someone from orbit (assuming they were not orbiting at all) then I do not think they would burn up much. As noted Kittenger didn’t have that problem although in the much thinner atmosphere above him I am not sure what the terminal velocity would be.
If you are in orbit then yeah, you are moving pretty fast. IIRC the Space Shuttle orbits at something like 5 miles/second. If you hit the atmosphere moving about that fast you are going to burn up if unprotected.
Rocks burn up due to compressive heating, and rocky (as opposed to metallic iron) meteoroids break up from the forces involved more often than not. I’d expect similar forces exerted on a human body to tear it to shreds; those shreds being relatively light are going to slow down very fast. So, I’d expect some significant charring, but also fairly good-sized fragments to remain intact and land on the ground.
Note that a large percentage of the rocks entering our atmosphere are going considerably faster than LEO velocity.
In fact, most meteors enter with an initial velocity approaching or exceeding the Earth’s orbital speed (~29.7 km/s), and even Near Earth Objects that are nominally in Earth’s orbit and are perturbed into intercept are moving at around Earth escape speed (11.2 km/s), which is significantly higher than orbital speed at LEO. Also, most aerodynamic heating from ram pressure is going to occur in the upper layers of the upper stratosphere; by the time you hit anything one could reasonably discern as an atmosphere a blunt, low beta body like, er, a body, is going to be moving so slow that convective cooling will vastly override any pressure-based heating. I would tend to agree with Q.E.D.; that the aerodynamic forces may pull extremities off or the torso apart, and the heating and evaporation from the reentry environment will tend to weaken the body structure, I would expect major parts to survive reentry with external charring and desiccation plus impact trauma at surface level terminal velcoity (200-350 kph, depending on size of part).
It is at least conceptually possible to provide reentry for a live person with minimal protection. See the MOOSE personal emergency reentry system. As long as the system remains aerodynamically stable in the arse-down position and doesn’t turtle, there is no reason why this concept couldn’t work.
Stranger
Hmm. If I dropped a feather from orbit, it’s low density/low mass means that air resistance will slow it down below the threshold required to cause it to combust, wouldn’t it?
Rocky or metallic objects (dense) retain higher velocities for longer, so heat up more.
The question, than, is the human body going to be slowed down enough, or is it dense enough to retain speed.
(Or am I on the wrong track?)
Was that the one with Juan Valdez serving as captain?
FWIW, various Googling suggests a human body has about 5 times the average density of the Space Shuttle during re-entry.
Sounds like I might be on the wrong track then, thanks.
You can’t look at bulk density for the purpose of exterior ballistics, unless you assume that the horse is a sphere to make the math easier. You have to look at the sectional density and form factor to derive the ballistic coefficient, which is the mass over the section area divided by the form drag coefficient, or the sectional density divided by the form factor coefficient. This will give the a roughly inverse linear relationship to deceleration for high speeds. In addition, although you can calculate bulk energy loss from this, local heating due to leading edge features which cause stagnation can be much higher than the overall thermal environment the vehicle will see, hence why the Shuttle has high temperature carbon-carbon leading edges and thermal tiles on the underside, but relatively less insulating thermal blankets on the top sides of the wings and on the more protected parts of the fuselage. The aft parts of the Orbiter topside are actually cool enough to sit on during reentry if you don’t mind a shear force that would take your head clean off.
All of this assumes a stable orientation in flight, which the human body obviously wouldn’t enjoy. Something floppy like a body will tumble and flail in flight, which changes the beta and can serve to dissipate energy by converting momentum into lift and losses due to turbulent drag. The Shuttle, on the other hand, comes down in an automatically controlled flight path that maintains a high drag orientation and performs a series of energy wasting maneuvers to in the upper atmosphere before it hits the thicker bits of atmo where it glides in.
Stranger
Depends - if it gets going fast before it hits significant amounts of atmosphere, I think it could burn up.
I’m a bit confused by this. It sounds as if you’re saying that a body may convert its energy to turbulent drag, while on the other hand the Shuttle does much the same thing. | 1,410 | 6,754 | {"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-2024-38 | latest | en | 0.972111 |
http://www.slideserve.com/kale/the-camera | 1,508,428,327,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00295.warc.gz | 560,030,977 | 13,776 | The Camera
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# The Camera - PowerPoint PPT Presentation
The Camera. Course Information CVG: Programming 4 My Name: Mark Walsh Website: www.activehelix.co.uk/courses Recommended Reading Introduction to 3D Game Programming with DirectX 9.0 (Frank D. Luna). Re-Cap. Local Space World Space View Space. Local Space. World Space. View Space.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about ' The Camera' - kale
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The Camera
Course Information
• CVG: Programming 4
• My Name: Mark Walsh
• Website: www.activehelix.co.uk/courses
• Introduction to 3D Game Programming with DirectX 9.0 (Frank D. Luna)
Re-Cap
• Local Space
• World Space
• View Space
Flexible Camera Class
• Objective is to build a flexible FP based games camera: Flight Simulators and Shooters
• The first stage is camera design
Camera Design
• The camera position is defined relative to the world coordinate system using 4 vectors
• Right, Up, Look, Position
• They define a local coordinate system relative to the world coordinate system
• They are therefore the orientation vectors
• The orientation vectors must be orthonormal
• Mutually perpendicular to each other
• Of Unit Length
• A row matrix where the rows are made up of orthonormal vectors is orthagonal
Camera Operations
• Using the 4 vectors we wish to be able to:
• Rotate around the right vector or X axis (Pitch)
• Rotate around the up vector or Y axis (Yaw)
• Rotate around the look vector Z axis (Roll)
• Strafe along the right vector
• Fly along the up vector
• Move along the look vector
Implementation
• We calculate the view matrix using the camera vectors
• Remember that view space transforms the geometry in the world…
• …so that the camera is centred at the origin and the axes are aligned with the major coordinate axes
Rotation
• D3DXMatrixRotationAxis
• Angle in Radians to Rotate
• Rotate around arbitrarily defined vector
Walking, Strafing, Flying
• Walking = Moving along the Look vector
• Strafing = Moving along the Right vector
• Flying = Moving along the Up Vector
To move we add a vector to our position vector
• The should have the same direction
• Need to set restrictions
• Walking, flying distinctions | 648 | 2,728 | {"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-2017-43 | latest | en | 0.798239 |
http://headerphile.com/sdl2/sdl2-part-3-drawing-rectangles/ | 1,603,244,620,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00118.warc.gz | 45,057,775 | 26,308 | # Drawing rectangles
This part will teach you the basics of the coordinate system in SDL( it’s the same for the “old” SDL and SDL2 ). It will also teach you about a new and very important struct, SDL_Rect. You’ll be using it a lot! And finally we’ll draw something!
### Note
This part assumes you have SDL2 up and running on your computer, and that you have read the previous part. If you haven’t, please scroll down to part1 and use it to install SDL2 before reading on.
# The coordinate system
The SDL coordinate system is defined as ( 0, 0 ) being the ( top, left ). This means that a higher y value means further down.
This also means that if you tell SDL2 to draw a rectangle, you’ll specify the top-left position of the rectangle instead of the bottom left rectangle. More about this later
# The basic rectangle
In order to draw something, be it textures or plain rectangles or text or anything, you need a rectangle to specify where to draw it and the size of what you are going to draw. And to hold the size and position we use an SDL_Rect
### SDL_Rect
Data members :
• uint16 x – the x position of the rectangle
• uint16 y – the y position of the rectangle
• uint16 w – the width of the rectangle
• uint16 h – the height of the rectangle
And that’s it. It’s a really simple struct, but it’s very important in SDL2 rendering.
### Drawing a rectangle
Now you have enough knowledge to draw some rectangles in SDL2. Let’s start off by looking at a the function for rendering a simple rectangle
Parameters
• SDL_Renderer* - the SDL_Renderer we created in the previous part
• SDL_Rect* - the position and size of the rectangle we want to draw.
Return value
0 on success
Note that it also takes a pointer to the SDL_Rect, and not the SDL_Rect itself.
“But what about the color?” you might ask. Remember how in last function we look at
int SDL_SetRenderDrawColor()? Well, basically, the color you set with this function will also be the color you render your objects with. ( For simplicity, I will refer to this color as SDL_DrawColor from now on. )
# And now the fun stuff
Let’s say you have just created and set up your window and renderer like so:
But wait! It’s just a red screen?! As you might have guessed, we forgot to change the color after calling SLD_RenderClear() So the rectangle was drawn with the same color as the background. To make the rectangle visible, we need to change SDL_DrawColor in between SDL_RenderClear() and SDL_RenderDrawRect()
This gives us something like this :
And now we have a nice little rectangle on our screen.
### Filling it up…
The function I showed you earlier will only render the edges of the rectangle. What if you want to render the whole rectangle, and not just the edges? Well there is a nearly identical function for that :
Parameters
• SDL_Renderer* - the SDL_Renderer we created in the previous part
• SDL_Rect* - the position and size of the rectangle we want to draw.
Return value
0 on success
As you can see, the only thing that separates the two us the name. If you switch SDL_RenderDrawRect() with SDL_RenderFillRect() in the example above, you will get the same rectangle with the same color, but this time it will be the whole rectangle and not just the edges.
# Conclusion
That’s it for today! Feel free to experiment with two new functions. You can draw as many rectangles as you want, both filled and edges. You can also change the color as often as you want. The only thing you need to remember is to put it all between your SDL_RenderClear( renderer ); and SDL_RenderPresent( renderer );
Have fun! Below is a full working example to experiment with. I have taken the liberty of putting things in different functions to make it easier to read. =)
The comments in the code should explain most of what’s going on. But you need to run the program to really see what’s going on. The code will draw a single blue rectangle on a green background that you can move around on the screen. Don’t worry about the code for moving the player around ( RunGame() ), it’ll be explained in the next post.
Feel free to comment if you have anything to say or ask questions if anything is unclear. I always appreciate getting comments.
You can also email me : olevegard@headerphile.com
## 64 thoughts on “[ SDL2 – Part 3 ] Drawing rectangles”
1. James says:
Great tutorial series! I appreciate whole work, you’ve made. Keep writing, kepp inspiring! Thanks, bro.
1. olevegard says:
Thank you, positive comments like this helps inspiring me too! I’ve been kinda busy lately, but I’m gonna start writing again very soon.
2. John says:
Hey,this tutorial about SDL2 really helps me a lot.And you made it interesting to me.But in the final code you wrote in part 3,I can’t understand why you create a function called”SetUpRenderer()”,but it’s not used in the program.And I’m confused about these two concepts:
SDL_RenderClear() SDL_RenderPresent()
Could you tell me what’s their difference and how they are related?
1. olevegard says:
Hello
SetupRenderer() is used in the InitEverything() function, at least in the example on this page.
SDL_Clear() basically clears the screen and covers it with the color you have set with SDL_SetRenderDrawColor(). So it basically just removes what we drew the previous frame so that we can render the next frame with a clean, single coloured, background.
SDL_RenderPresent() just updates the screen with what we have drawn this frame.
So basically : SDL_Clear() is used at the beginning of each frame to clear it. And SDL_RenderPresent() is used when we’re down et drawing to get it on the screen.
Hope that helps 🙂
3. David says:
What do you mean by “The code will draw a green-red check board pattern with 2 x 2 tiles inside a blue rectangle. Under the check board, there will be a rectangle that’s twice as long to show that the SDL_Rect doesn’t have to be a square.” Where exactly in the code does this happen? As far as I can see this description doesn’t have any chance of happening as the code only ever renders a green backround with a blue rectangle as the “player” which you can move around with the arrow keys, even setting the color to red makes no sense as you override it with green later. Can you explain what you mean?
Friendly Regards
//David
1. olevegard says:
Hello
Thank you for informing me about this! It’s a typo. I think what happened is that I did have a checkboard like the text said. But then I decided to simplify the code and make it more focused on rendering that one rectangle.
I’ve update with a correct text and changed the code a little. ( Setting the color to red initially didn’t really make much sense as I render it with a green background later. )
Again, thank you. Please tell me if you find other issues 🙂
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37. A down jacket’s fit can be dependent on a number of factors including the weight of the jacket, the length of the jacket, and the material it’s made out of. Some of the best down jacket for men, like the Montane Down Jacket, have a very low rise. | 3,489 | 15,827 | {"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} | 2.796875 | 3 | CC-MAIN-2020-45 | longest | en | 0.879641 |
http://mathrec.org/election/index.html | 1,558,673,593,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257514.68/warc/CC-MAIN-20190524044320-20190524070320-00219.warc.gz | 138,156,748 | 7,092 | MathRec Home Steve's Personal Page
## Presidential Election Meta-Analysis
A few people have expressed an interest in understanding some of the math behind the analysis in the Presidential Election Predictor. I'm already regretting the title, but I'll give a reasonable shot at explaining the underlying math.
#### Sampling Error and Voter Shift
Various polls of voter preference are taken throughout the campaign. These are not direct predictors of the outcome of the election, but they are snapshots in time. If a poll samples 1000 voters, and they've done a good job of eliminating systematic errors, and the voters are both insightful and honest, then on average they will give a good prediction of how the election would turn out if it were held when the poll was conducted.
Nonetheless, the actual voters polled are only a sample of the total electorate. Just by chance, they may select more voters who prefer one candidate over the other. In a close race, this sampling error is very nearly one over the square root of the number of voters sampled. In this case, about 3%. That means that if the polling company conducted the same poll over and over again, their answers would vary by about three percentage points.
Now, predicting the outcome of the election is not the same thing as determining the actual results if the election were held today. That could be done by averaging enough polls, thereby increasing the number of voters sampled and thereby decreasing the uncertainty from sampling error. There are two other categories of effects to consider. First, the election is not held today. (Well, if you're reading this on election day, you've caught me on a technicality.) Voter sentiment can change. At the very least, many undecided voters will decide, but some people will also change their minds. Second, the polling process will not be free of systematic errors. That is, the sample of voters who participate in the polls may not be representative of the voters who actually cast votes on election day.
#### Systematic Errors
Polling companies will try to identify and compensate for systematic errors. This is generally summed up in the catch-all phrase "likely voters." That is, they try to restrict their sample to voters who they think will actually cast ballots, and they may also try to identify and compensate for systematic biases in their sample. For example, they may ask about party preference, hoping to compensate if they have a larger fraction of Republicans than they should based on lists of registered voters and/or past voter turnout.
Systematic errors creep into polling in many insidious ways. Sometimes pollsters will try to sway the results by deliberately introducing systematic error. They may phrase questions is a way that seem to be asking how the sample will vote on a candidate or issue, but nudging the polling participant toward a particular answer. ("If the election were held today, would you vote for the senator who tried to keep us out of war or the senator who got us in deeper?") More subtle techniques can be used to influence the results. The order of the questions or the time of day when calls are made, for example.
#### Predicting an Outcome
Understanding and characterizing the random and systematic errors in the polling process, it is possible to "predict" the outcome of the election. By this, I mean to make a statement of probability: "The probability that John McCain will win the presidential election is xx%." There is lots of analysis and lots of guesswork that goes into this type of statement. The prediction will depend on the skill and assumptions of the analyst. A firm grasp of probability and a keen eye toward systematic errors is crucial. A keen eye toward the ways that the analyst can introduce systematic errors is part of this. There are guidelines for statistical analysis that can and should be used to prevent the analyst from biasing the prediction.
#### Futures Markets
The whole concept of futures markets is based on the idea that a group of analysts with access to a wide variety of information can pool that information to make a better probabilistic assesment than any single analyst. This often seems to be the case, and it seems to work significantly better when the analysts put their money behind their results.
The way it works is this. If I feel that Barack Obama's chances of winning the election are about 65%, and I have the opportunity to purchase a future guarantee of one dollar if he actually wins, then how much would I be willing to pay for that future guarantee? If the payback is soon enough that I don't have to consider the time value of money, then I'd be willing to pay up to \$0.65 for that future guarantee. The idea of futures trading is to give people the chance to make those personal assessments based on the information that they have available. The market should settle on a value where half the participants (weighted by their willingness to put money behind their convictions) believe that the outcome is more likely than the market value of the futures contract, and half of the particpants believe that the outcome is less likely.
There are a variety of futures markets available for the outcome of the U.S. Presidential election. There are contracts for the overall popular vote, for the overall outcome of the electoral college, and also for the allocation of electoral college votes from each state individually. These futures markets are intrinsically predictors of the outcome of the election. They are subject to errors, biases, misinformation, etc., but they also have some ability to correct for these biases in ways that polls of voter preference do not.
#### Undrestanding State-by-State Futures
I've written a Presidential Election Predictor that takes the values of futures on the state-by-state outcomes and combines these into a prediction of the overall national outcome. The prediction from the individual state outcomes closely matches the value of a future on the overall outcome of the Presidential election. The rest of this web page describes the techniques that I used to perform the meta-analysis of the individual state-by-state future values, in order to arrive at an overall prediction of the outcome.
#### Variance and Future Value
The participants in the futures market do not know the outcome of the election in advance. Their assessment can be viewed as an analysis of the electorate, resulting in a probabilistic assessment of the outcome. There are a range of outcomes that correspond to the margin of victory (in any units -- votes, percentage of votes, etc.) We can reasonably consider these outcomes to be normally distributed, and the probability that one candidate wins is the fraction of outcomes where they get more votes than their opponent. This probability corresponds to a z-value if the outcomes (as viewed by the participants in the futures market) are distributed with a normal probability function.
Let's think of the probability distribution as a kind of super-poll. It's not a very accurate way of looking at it, but it helps introduce and clarify the idea that the outcomes have a mean and standard deviation. Suppose for the moment that all voters have decided, none of them will change their minds, and they all know whether or not they will vote. Suppose also that the polling companies have no systematic biases in the way that they've conducted their polling. In this case, the polls are an accurate predictor of the electoral outcome. Here the term "accurate" means that the mean value of the poll results will be the same as the actual election returns. But, since each poll is only a sample of the voting population, there is some random variation in the poll results.
So, suppose that there are three polls, and all of the participants know the results of these polls and no one has any better information. Every investor would be able to analyze the results of those polls and determine the probability that each candidate will actually win the election. In this case, there is only one source for the variance in the distribution of possible outcomes; that is, the sample size of the polls.
Now, suppose that the same is true, but that some voters will change their minds based on last minute campaigning and get-out-the-vote activities. This spreads out the range of possbile outcomes. It increases the variance of the distribution. There are now two types of uncertainty in the election results. How well the polls represent the actual opinions of the voters, and how many voters will change their minds between now and the election.
The futures market (at least allegedly) takes all of this into account when making a probabilistic assessment of the outcome. If we make an assumption that the possible outcomes are normally distributed, then the future value tells us where the 50/50 threshold between victory and defeat is on that bell curve. It doesn't actually tell us what the mean value is or what the standard deviation is, but it definitely implies a particular ratio between the mean and the standard deviation. This ratio is called the z-value.
#### Combining State-by-State Futures
We don't actually care about how the variance is divided between random uncertainty (such as the sample size of the polls) and the systematic uncertainty (systematic polling error, shifts in voter preference, voter turnout, etc.). Within each state, those effects have all been considered by the participants in the futures market. What we want to separate is the uncertainty in the outcome that varies on a state-by-state basis, compared to the uncertainty in the outcome that varies nationally.
Let's consider only two effects. Let's suppose that there are random polling errors due to sample size, and let's also suppose that one party does a better job of getting their voters to the polls. The sampling errors will be completely independent from state to state. If the poll averages for John McCain were a little low in Florida, there is no reason to think that they would also be low in Ohio. On the other hand, if the Obama campaign does a better job of getting their voters to the poll in Wisconsin, then they probably did a better job in Iowa, too. Since we don't really know how these effects play out until the election is actually held, these both contribute to the uncertainty in the outcome for each state. If there is no national trend in the uncertainty, then the probabilities for all of the states are independent. On the other hand, if there is a national trend (such as a late swing in voter preference), then it pushes all of the states in the same direction.
These two types of effects combine in the same way when considering the outcome in a single state:
σ2 = σs2 + σn2
where σ2 is the variance in the overall outcome at the state level, σs2 is the variance that applies only to the individual state, and σn2 is the variance that is common to all states throughout the nation. It should be obvious that this is a great simplification. There will be effects that vary throughout groups of states, but not throughout the nation as a whole. Two examples: There could be some late-breaking news that is of regional interest—affecting voters in coastal states with potential offshore drilling, for example. Or, there could be a last-minute ad campaign that is only aired in battleground states. Still, this is a useful simplification and captures the main characteristics in the way that state-by-state results combine to give a national result through the electoral college.
Now consider the way that these two types of errors affect the national outcome. The state-by-state variance describes how the results from each state varies independently from the expectation. In some states, Barack Obama will do better than expected and in some states, John McCain will do better than expected. This variation gives "equally" to each candidate. I quoted that word, because this state-specific variation could hand Colorado or Ohio to one of the candidates, and tip the entire election. On the other hand, a national effect is applied in the same direction to every state at the same time. This tips the entire scale, giving one candidate or the other better-than-expected results in Virginia, Ohio, Colorado, New Hampshire, etc.
#### Allocating the Variance
We can look at the probabilistic assessments of futures markets for each state outcome. It is clear that there is some uncertaintly in the national trends. This is indicated by the relatively large probability of an underdog win in states that are not considered "battleground states". Consider Oregon, for example. Based on current poll data, John McCain's chances of winning Oregon should no more than 5%, based on sample size or even considering systematic errors in the polling methods. Still McCain might just win Oregon if there were a shift of a few percent in the overall voter sentiment. This is why participants in the futures market are willing to consider the possiblity that McCain might pull off a win in Oregon.
I've evaluated the probabilities of Obama and McCain victories in various states based solely on polling data, then compared those probabilites with the probabilities of Obama and McCain wins in those same states based on futures values. I assume that a constant percentage of the variance in each state is due to national trends, and attempt to estimate that percentage. If we look at Oregon as a case study, the 17% chance of a McCain victory corresponds to a z-value of 0.95, whereas the z-value from polling data is approximately 1.56. This implies that the variance from polling is approximately 37% of the overall variance. I've been using values between 25% and 40% for the state-to-state portion of the variance in my meta-analysis. I chose it because it puts the right number of states into the in-play zone as suggested by polling data. There's some state-to-state variation, but the overall fit seems reasonable. Playing with this parameter indicates that the results of he meta-analysis are not that sensitive to the actual value.
#### Meta-Analysis
Once this single free parameter has been chosen, the stated assumptions give a single value for the probability of each electoral outcome (Obama or McCain). The math goes something like this:
First, calculate new z-values that use the state-by-state uncertainty instead of the combined uncertainty. If 40% of the variance is state-by-state, then the state-by-state standard deviation is 63.2% of the combined standard deviation. So, if we start with a future value of 65% for an Obama win, that corresponds to a z-value of 0.385 on the normal distribution with the combined variance. If we reduce the variance to 40% of the combined variance, then the z-value is 0.609 in the state-specific distribution.
It's relatively straightforward to calculate new state-by-state probabilities for each candidate. These represent the electoral outcomes in the event that the net national effect turns out to be zero. These probabilities can be used in a permutation analysis to sum up the probabilities for each candidate of getting 270 electoral votes (or only 269 for Obama, since the Democrats will control the House of Representatives).
Now I consider the effects of shifting all of the states together by an amount implied by the remaining 60–75% of the variance. This is equivalent to adding the same random variable x to every state-specific z-value. The standard deviation for the distribution of x is σns, or 1.225 for the case where 60% of the variance is from national effects. Now I can calculate the state-by-state probabilities for each value of the nationwide variation. This is a bias analysis.
The points in the bias analysis are very close to a cumulative normal distribution. Fitting these points to such a distribution gives a mean and variance. The variance from the fitted cumulative normal distribution is added to the variance from the bias curve. The combined variance and the mean are used to determine the overall probability of victory for each candidate, and the lead expressed in standard deviations.
Steve Schaefer | 3,182 | 16,221 | {"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} | 3.265625 | 3 | CC-MAIN-2019-22 | latest | en | 0.967241 |
https://eng.libretexts.org/Bookshelves/Computer_Science/Programming_and_Computation_Fundamentals/Programming_Fundamentals_(Busbee_and_Braunschweig)/06%3A_Arrays_and_Lists/6.05%3A_Math_Statistics_with_Arrays | 1,722,910,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00457.warc.gz | 191,196,695 | 29,684 | # 6.5: Math Statistics with Arrays
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## Overview
Statistics is a branch of mathematics dealing with the collection, organization, analysis, interpretation, and presentation of data. Common statistical methods include mean (or average) and standard deviation.[1]
## Discussion
Arrays are an important complex data type used in almost all programming. We continue to concentrate on simple one dimension arrays also called a list. Most programmers develop a series of user-defined specific task functions that can be used with an array for normal processing. These functions are usually passed the array along with the number of elements within the array. Some functions also pass another piece of data needed for that particular functions task.
This module covers the totaling of the members of an integer array member. The Latin name for totaling is summa, sometimes shortened to the word sum. In the example below, the sum function totals the array passed to it. Other mathematical functions often associated with statistics such as: average, count, minimum, maximum, standard deviation, etc. are often developed for processing arrays.
### Pseudocode
Function Main
Declare Integer Array ages[5]
Declare Integer total
Assign ages = [49, 48, 26, 19, 16]
Assign total = sum(ages)
Output "Total age is: " & total
End
Function sum (Integer Array array)
Declare Integer total
Declare Integer index
Assign total = 0
For index = 0 to Size(array) - 1
Assign total = total + array[index]
End
Return Integer total
### Output
Total age is: 158
## Key Terms
sum
Latin for summa or a total.
## References
1. Wikipedia: Statistics
6.5: Math Statistics with Arrays is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. | 2,136 | 6,116 | {"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": 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} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.19872 |
https://www.wallstreetmojo.com/p-value/ | 1,723,692,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00272.warc.gz | 799,583,213 | 81,513 | FLASH SALE! - "FINANCIAL MODELING COURSE BUNDLE AT 60% OFF" Enroll Now
# P-Value
Updated on April 4, 2024
Article byWallstreetmojo Team
Edited byWallstreetmojo Team
Reviewed byDheeraj Vaidya, CFA, FRM
## What Is P-Value?
P-Value, or the Probability Value, is the determining factor on a null hypothesis for the probability of an assumed result to be true and being accepted or rejected and acceptance of the alternate result in case of rejection of the assumed result.
Statisticians refer to this value for more relevant results. In most cases, it lies within a range of 0 – 0.05 (5%) and has a negative result, which means the alternate result would be considered, and a value higher than 0.05 signifies that it will accept the desired result. However, this will not be hard and fast for all cases and will depend upon the conditions and product.
### Key Takeaways
• P-Value, also called the Probability Value, is the analyzing factor on a null hypothesis for the probability of an assumed result to be true and being accepted or rejected and acceptance of the alternate result in rejection of the assumed result.
• One may use it when making a difficult decision, leading to severe losses. For example, finding a P-value makes it easier to determine between two options.
• The P-Value is similar to the probability of occurrence of the desired result.
### P-Value Explained
P-Value calculation also includes a probability of other results’ occurrence. However, statisticians refer to this value for more relevant results. In most cases, it lies within a range of 0 – 0.05 (5%) and has a negative result, which means the alternate result would be considered, and a value higher than 0.05 signifies that it will accept the desired result. However, this will not be hard and fast for all cases and will depend upon the conditions and product.
Always a probability of the occurrence of a required result when in a scenario made a null hypothesis.
There is also an alternate result that is existent and holds an equivalent probability. However, it would infer if the assumed/required result fails to prove. The p-value statistics calculation determines whether the assumed result will hold “True” or the alternate result. A higher value determines the acceptance of the assumed result, while a lower signifies the rejection of this assumed result and acceptance of the alternate result.
For example, in a hypothetical situation, we survey a new appliance in the market, and results assume that 60% of females will accept the appliance, with an alternate result expected that 60% of males will accept the appliance. With the help of the p-value chart, we try to determine the results. A higher value will signify that the assumed expected result is “True,” which means 60% of females accept the appliance. Consequently, a lower would imply acceptance of the alternate results, which means 60% of males accept the appliance.
Hence, it determines the acceptance or rejection of an assumed result.
For eg:
Source: P-Value (wallstreetmojo.com)
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### Formula
It can be calculated using z analysis (z test) where:
For eg:
Source: P-Value (wallstreetmojo.com)
where,
• P1 = sample proportion of the whole population
• P0 = Assumed proportion for the result to occur
• n = size of the population
The Z-value is predicted from previous calculations. If the p-value statistics is equal to or less than the calculated z value, then the sample can be approved for the desired result (null hypothesis). Else gets rejected, and the alternate result gets approved.
The Z-values have previously calculated values in line with p-value test in the form of tables. With the help of Z-values, the corresponding values are derived from the below table.
Source: https://www.chegg.com/
### Example
Let us understand the p-value significance with an example.
You can download this P-Value Excel Template here – P-Value Excel Template
Consider Mr. X’s wishes to invest in a portfolio ABC. However, he feels there is a probability of 25% that this portfolio will earn the desired interest rate, while another portfolio MNO is his alternate choice. He samples from 150 stocks and finds that 40 stocks in portfolio ABC earn the required interest rate. Calculate the P-Value, and assuming that the Z-value is 1.72, find out if the portfolio ABC is suitable for investment or should be rejected.
Solution
From the z-test, it is as follows that:
• P1 = 40/150 = 0.267
• P0 = 0.25 (the assumed proportion for the result to occur)
• n = 150
Hence p-value should be as follows:
• = (0.26667 – 0.25)/SQRT((0.25*(1-0.25))/150)
• = 0.4714
As per the expected z-value, the P-Value from the above table should be 0.0427, away from the above calculation. Hence, the portfolio ABC gets rejected (the null hypothesis gets rejected).
### Interpretation
• A higher p-value significance denotes that the probability of occurrence of the assumed result is very likely. It suggests that the probability ascertained on the occurrence of that result is true, and the outcome will favor the required result. On the contrary, a low value signifies that the required or assumed result has a very low chance of occurrence. It also denotes that the alternate result is more probable to occur. A low value on the assumed or required result automatically rejects this result, and the alternate result is automatically accepted.
### Uses
• It is used when making a difficult decision and may lead to serious losses. Finding out a p-value makes it easier to determine between 2 different options.
• It serves as a double-check on probability analysis. For example, in finance, investment decisions depend mainly on the probability of profits and losses. Hence, even after calculating probability, if the p-value test is computed, it ensures that the decision will be in favor.
• Calculation of returns using a p-value chart is a good way of forecasting results. In reality, the futuristic returns cannot be seen today. However, if all constraints are properly measured and this calculation is done, one can forecast results. Hence, it helps calculate future cash flows and go a little further ahead. It will also help in making future financial-related decisions.
What P-value is significant?
Suppose the P-value is under . 01, then the results are statistically significant. If it’s below . 005, they are regarded as highly statistically significant.
How to explain P-value to non-statistician?
P-value is a probability, a number between 0 and 1. It is calculated after running a statistical test on data. A small P-value, i.e., < 0.05 in general) The observed results are unusual, assuming they were because of the chance.
Can P-value be negative?
It has been observed in many articles published in medical journals that if the P-value is less than 0.05, the study is considered positive. If the P-value is more than 0.05, it is considered negative.
Should p-values be the sole basis for decision-making?
No, p-values should not be the sole basis for decision-making. They should be considered alongside other relevant factors, such as effect size, study design, context, and prior knowledge. A comprehensive analysis requires interpreting p-values in conjunction with other statistical measures and scientific judgment. | 1,676 | 7,778 | {"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} | 3.90625 | 4 | CC-MAIN-2024-33 | latest | en | 0.918883 |
https://ibyea.wordpress.com/2013/11/ | 1,611,594,474,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703587074.70/warc/CC-MAIN-20210125154534-20210125184534-00453.warc.gz | 372,065,251 | 15,202 | ## If You Missed Comet Ison
November 23, 2013
If you tried catching the recent comet Ison in the night sky, it is too late. I is now in the sun’s glare. But, there will be a second chance after it swings around the sun in November 28. You will be able to see it in a more convenient time, after sunset, I think. Before you had to get up really early to see it before the sun rises. Anyways, keep checking astronomy news and the charts. Also, check if the comet survives the trip around the sun, as it goes really close to the sun, and there is a possibility it might break up. If you do see it eventually, though, enjoy!
## Minimizing Quantity Part 4: Euler-Lagrange Equation
November 23, 2013
In the last three posts, I talked about paths that minimize time. In all those cases, it involves objects going through a path and minimizing certain quantities. But is there a single equation that covers all physical situations that involve finding the path that minimizes quantity? The answer is yes, and it is called the Euler Lagrange equation. Read the rest of this entry »
## Volume of n-Sphere
November 10, 2013
If you know calculus and the gamma function, you might as well use it to try and learn to get the volume of an n-sphere:
## Super Mario Bros. Analyzed
November 5, 2013
I recently found a pretty cool list of articles that analyzes the game design of the original Mario bros. I remember that I never got further than 6-3. Stupid old games and their lack of save points! | 361 | 1,491 | {"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-2021-04 | latest | en | 0.942194 |
http://hebrewcalendar.tripod.com/zqarc001.html | 1,582,046,183,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00481.warc.gz | 67,365,623 | 8,629 | Weekly Question Archive 1 - 10
Weekly Question Archive 1 - 10
by Remy Landau
Question 1
What do the months of Iyar and Tishrei have in common?
If you happen to have a Hebrew calendar handy, please note that on a week by week basis, the days in the month of Iyar are laid out exactly in the same way as the first 29 days in the subsequent month of Tishrei.
There are exactly 147 days from the first day of Iyar to the first day of the following Tishrei. This represents a complete number of weeks. And so, the first day of Iyar always is the same day of the week as the first day of the following Tishrei.
For example, whenever the first day of Iyar is Monday, the first day of the following Rosh Hashannah will also be Monday, as happens to be the case for Iyar 5758H (1998g) and Rosh Hashannah 5759H (1998g).
Therefore, Iyar and Tishrei have in common the same day of the week for their first day of the month.
Question 2
On which day of the week do Hebrew months most frequently begin?
Each weekday will see the first day of a Hebrew month in a frequency that is between 11.6% up to 17.2%. So the weekdays are not too widely separated in terms of the number of times that they will each catch the first day of some Hebrew month.
The weekday that most frequently is associated with the first day of a Hebrew month is Monday. Over the full and complete cycle of the Hebrew calendar (689,472 Hebrew years), the first day of the month will fall on Monday 1,463,854 times out of a possible 8,527,680 months in that cycle.
Therefore, the first day of the month most frequently occurs on Monday.
The festival of Rosh Chodesh always begins 29 days after the first day of a Hebrew month. When the Hebrew month has 30 days, the observance is extended to the next day which is the first day of the subsequent month.
Question 3
On which day of the week does Rosh Chodesh most frequently begin?
Each weekday will see the first day of Rosh Hodesh in a frequency that is between 11.6% up to 17.2%. So the weekdays are not too widely separated in terms of the number of times that they will each catch the first day of Rosh Hodesh.
The weekday that most frequently is associated with the first day of Rosh Hodesh is Tuesday. Over the full and complete cycle of the Hebrew calendar (689,472 Hebrew years), the first day of Rosh Hodesh will fall on Tuesday 1,463,854 times out of a possible 8,527,680 months in that cycle.
Therefore, Rosh Hodesh most frequently begins on Tuesday.
Talmudic tradition indicates that the day for the festival of Rosh Hodesh would be determined by the Sod Haibbur which was a secret council of three scholars. One these scholars was the head of the Sanhedrin, which in ancient times was the supreme law making body of the Jewish people. The other two members also were scholars taken from this august assembly. The Talmud goes on to relate that the Sod Haibbur had to exclude from its council certain people who occupied extremely high office.
Question 4
According to Talmudic tradition, which high officers were barred from the Sod Haibbur?
The Sod Haibbur also had to determine whether or not to declare a leap year through the intercalation of an extra month.
The Talmud, in tractate Sanhedrin 18b, relates that due to this function both the King and the Cohein Gadol had to be excluded from the Sod Haibbur.
The King was disqualified because he paid his armies on an annual basis and therefore would favour the leap years of 13 months.
The Cohein Gadol, on the other hand, would more likely favour the 12 month years. During the High Holidays, the Cohein Gadol had to immerse himself several times in fresh spring waters as part of the Temple rituals. And so, he probably would have preferred an earlier time of the year when these waters were a bit warmer.
Question 5
In any given Hebrew year, what is the largest number of Hebrew months whose first day can coincide with the first day of their corresponding Gregorian month?
The first day of the Hebrew months tend to coincide with the first day of some corresponding Gregorian month about 38% of the time.
In 5758H the first day of Heshvan coincided with the first day of October 1997g. In 5760H, the first day of Elul will coincide with the first day of September 2000g.
In some Hebrew years, these coincidences can occur several times. In 5755H, the months of Adar, Nisan, and Iyar all started on the first days of February, April and May of 1995g.
However, no Hebrew year can have more than 3 such coincidences. In those years, the 3 Hebrew months will always have their first days coincide with the first days of February, April and May.
Question 6
What is the next Hebrew year in which 3 Hebrew months will all start on the first day of their corresponding Gregorian months?
In 5755H, the months of Adar, Nisan, and Iyar all started on the first days of February, April and May of 1995g.
This will not happen again until the year 5774H. In that year the months of Adar, Nisan, and Iyar will all start on the first days of February, April and May of 2014g.
Rosh Hashannah of 5774H begins on Thursday 5 Sep 2013g. In our times, this is the earliest possible day of the Gregorian year for which any Rosh Hashannah can begin. The last time that Rosh Hashannah occurred that early in the Gregorian year was for the year 5660H, which began on Tuesday 5 Sep 1899g.
Thus, the years 5774H and 5660H share a very rare feature with each other.
The Hebrew year 5774H (2013g/2014g) also shares in common with the Hebrew year 5676H (1915g/1916g) another very rare feature.
Question 7
What would be the very rare feature shared in common between the Hebrew years 5774H (2013g/2014g) and 5676H (1915g/1916g)?
The years 5774H (2013g/2014g) and 5676H (1915g/1916g) are the two years in our times which both begin the longest possible periods of 120 Hebrew years as measured from the first day of Tishrei.
Periods of 120 Hebrew years, as measured from the first day of Tishrei can be either 43822, 43823, 43824, 43825, 43851, 43852, 43853, 43854, or 43855 days long.
Coincidentally, the first Rosh Hashannah of both of these 120 year periods begins on Thursday.
Question 8
Does the first Rosh Hashannah of a longest possible 120 Hebrew year period always begin on Thursday?
Periods of 120 Hebrew years, as measured from the first day of Tishrei can be either 43822, 43823, 43824, 43825, 43851, 43852, 43853, 43854, or 43855 days long.
One of the remarkable features of the longest possible 120 year periods is that their first day of Tishrei always is Thursday.
The last time that the 43,855 day period began on the first day of Tishrei was on Rosh Hashannah 5676H (1915g/1916g). The next Rosh Hashannah to open such a period will be for the year 5774H (2013g/2014g).
Simple arithmetic shows that these 2 years are 98 Hebrew years apart.
Question 9
As measured from the first day of Tishrei, are all of the longest periods of 120 Hebrew years exactly 98 Hebrew years apart?
No.
The longest periods of 120 Hebrew years are 43,855 days long. Over the full calendar cycle of 689,472 years these are the distances which can be calculated between 43,855 day periods:
```number of years= 27 number of times = 1260
number of years= 71 number of times = 835
number of years= 98 number of times = 1433
number of years= 149 number of times = 1912
number of years= 176 number of times = 176
number of years= 220 number of times = 239
number of years= 247 number of times = 353
```
Each period which separates the longest 120 years from each other is a whole number of weeks.
As well, a number of the 120 year periods appear to terminate at a time when the start of Rosh Hashannah advances by yet another day in the Gregorian year. This was true of the year 922g which saw the eruption of the ben Meir - Saadia Gaon calendar controversy.
Question 10
In which year or years of the 19 year Hebrew calendar cycle is Rosh Hashannah most likely to advance by another day in the Gregorian calendar?
The last time that Rosh Hashannah advanced in the Gregorian calendar was at the start of 5576H corresponding to Thursday 5 October 1815g. The next time that this will happen will be at the start of 5975H corresponding to Thursday 6 October 2214g.
Both of these years inaugurate the 9th year of the 19 year Hebrew calendar cycle. Carefully examining the calendar drift tables shown in the Additional Notes, it becomes apparent that after a certain distance into the drift the advancement of the Hebrew year into the Gregorian year only occurs at the start of the 9th year of a 19 year cycle.
However, not every 9th year of the cycle will cause the Hebrew year to be advanced.
If the 9th year of a 19 year cycle seems to cause the latest arrival of Rosh Hashannah, it seems reasonable to conclude that some other year in the cycle might see the earliest possible arrivals of Rosh Hashannah.
``` First Begun 21 Jun 1998 | 2,267 | 8,933 | {"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-2020-10 | latest | en | 0.955039 |
https://www.coursehero.com/file/6664749/052307-132a-hw5/ | 1,529,284,387,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859904.56/warc/CC-MAIN-20180617232711-20180618012711-00105.warc.gz | 785,544,752 | 49,320 | 052307-132a-hw5
# 052307-132a-hw5 - μ ∂u ∂z v v v v z =0 = A cos ωt(2...
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C HEMICAL E NGINEERING 132A Professor Todd Squires Spring Quarter, 2007 Homework 5 Due Wednesday, May 30 Problem 1. p503, prob 2 Problem 2. p504, prob 50 Problem 3. p504, prob 51 Problem 4. Fluid Mechanics example. The equation for a ‘uni-directional’ fluid flow (i.e. in the x -direction) above a flat, solid boundary (at z = 0 , say) is ∂u ∂t = μ ρ 2 u ∂z 2 , (1) where ρ is the fluid density and μ is the fluid viscosity. For water, ρ = 1g/cm 3 and μ = 10 - 2 g cm 2 /s. Say there is so much fluid above the plate that you can ignore the top boundary. Say we drive the bottom boundary from side to side with a given stress (force per unit area) – this imposes a boundary condition on the fluid
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Unformatted text preview: μ ∂u ∂z v v v v z =0 = A cos ωt. (2) Assuming you have been shaking the wall for a while (i.e. so that transients have died out), solve for the ±uid ±ow. This is very similar to the problem we did in class. Look at your solution – how far do the shear waves (i.e. the ±uid ±ow) propagate into the ±uid? How does it depend upon frequency? What about the amplitude of motion – how does it depend upon frequency? Does this make sense to you intuitively?...
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# What is Pressure? Definition, Formula, Unit, Examples
Pressure is a fundamental concept in physics, defined as the physical force exerted on an object. This force is applied perpendicular to the surface of the object and is measured per unit area. In mathematical terms, pressure (P) is calculated as the force (F) divided by the area (A), giving us the formula P = F/A. The standard unit of pressure is the Pascal (Pa).
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## What is Pressure? – Formula
Pressure is a fundamental physical concept that describes the force applied to a surface per unit area. It can be defined as the amount of force exerted on a given surface area, or simply as the ratio of force to the area over which that force is acting.
Pressure (P) = Thrust / Area
In SI units, 1 pascal (Pa) is equivalent to 1 newton per square meter (N/m²).
Let’s explore a practical example to better understand the concept of pressure. Imagine you have two pins, one sharp and one blunt, and you want to hammer them into a wooden surface. You might notice that it’s easier to hammer the sharp pin than the blunt one. Why is this the case?
The key lies in the area of contact between the pin and the surface. When you use the sharp pin, the point of contact is extremely small, which means the area (A) in the pressure formula is minimal. On the other hand, when you use the blunt pin, the point of contact is much larger, resulting in a larger area (A).
### Types of Pressure
There are following types of Pressure:
1. Absolute Pressure: This is the pressure measured relative to a perfect vacuum. It includes atmospheric pressure, making it the total pressure in a system.
2. Atmospheric Pressure: Atmospheric pressure is the pressure exerted by the weight of the air above a specific point on Earth’s surface. At sea level, it is approximately 101.3 kPa (kilopascals).
3. Differential Pressure: This type of pressure is the difference in pressure between two points in a fluid or gas system. It is often used to measure flow rates and fluid levels.
4. Gauge Pressure: Gauge pressure is the pressure relative to atmospheric pressure. When a pressure gauge reads zero, it is measuring gauge pressure, meaning it accounts for atmospheric pressure.
### Magnitude of Pressure
The magnitude of pressure depends on the force applied and the area over which it is distributed. Pressure is inversely proportional to the area on which the thrust is applied. It means that if the thrust is constant and the area is more, the pressure will be lesser. And if the area is less, the pressure will be more.
### Pressure in Fluids
In fluids (liquids and gases), pressure is transmitted equally in all directions. This is known as Pascal’s principle. In a fluid at rest, pressure increases with depth due to the weight of the fluid above it. This is expressed by the equation:
Pressure = Density × Gravity × Height
Where Density is the density of the fluid, Gravity is the acceleration due to gravity, and Height is the vertical distance from the surface.
Understanding pressure is crucial for many scientific and practical applications. It allows us to comprehend the behaviour of fluids, design effective systems, and evaluate the impact of pressure on structures and human health. Whether it’s exploring the depths of the ocean, predicting the weather, or maintaining our well-being, pressure is a concept that influences our understanding of the physical world around us.
[Thursday 11:37 AM] Karan Singh Bisht
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### Solved examples
Example 1: A force of 500 Newton is applied to a circular surface with a diameter of 0.2 meters. Calculate the pressure exerted on the surface.
Solution: Given
Force = 500 N
Diameter = 0.2 m
Step 1: Calculate the area of the circular surface.
Radius = Diameter / 2 = 0.2 m / 2 = 0.1 m
Area = π × (0.1 m)2 = 0.0314 m²
Step 2: Calculate the pressure.
Pressure = Force / Area
Pressure = 500 N / 0.0314 m² ≈ 15,923.57 Pa
Therefore, the pressure exerted on the surface is approximately 15,923.57 Pascal (Pa).
Example 2: A block of wood is kept on a table. The mass of the wooden block is 500 grams and its dimensions are 5 cm × 3cm × 2 cm. Let us find the thrust and pressure exerted by it on the table if it is made to lie on it with its sides of dimensions (A) 5 cm × 3 cm and (B) 3 cm × 2 cm.
Solutions:
Let us find the thrust and pressure exerted by the block on the table in each of the cases.
Thrust calculation
The weight of the wooden block applies a thrust on the table. Since the weight of the box is the same in both the cases the value of thrust will also be the same. The thrust can be calculated as follows,
𝑇ℎ𝑟𝑢𝑠𝑡 = 𝐹 = 𝑚 × 𝑔
Where, ‘m’ is the mass of the block and ‘g’ is the acceleration due to gravity. The mass ‘m’ is 500 grams, which is equivalent to 0.5 kilograms and ‘g’ is 9.8 m/s2.
Thrust = F = m x g
= 0.5kg x 9.8m/s2
= 4.9 Newtons
Pressure calculation:
We can calculate the pressure in both the cars using the pressure formula,
Pressure = Thrust/Area
In case A, the area of the box in contact with the table is,
Area = 5cm x 3cm
= 15cm2 or 0.0015m2
Pressure = 4.9N/0.0015M2
= 3266.66 N.m2
We can also calculate the pressure in case B as follows:
Thrust = F = 0.5 x 9.8m/s2
= 4.9N
Area = 3cm x 2cm
= 6cm2
= 0.0006 m2
Pressure = 4.9N/0.0006m2
= 8166.66 N/m2
Hence, the pressure in case B is much more than that in case A. This is because the area of contact is lesser in case B.
## FAQs On Pressure
### What is pressure and its formula?
Pressure is the force applied on an object divided by the area over which it is applied. The formula for pressure is: Pressure (P) = Force (F) / Area (A).
### What is pressure and its SI unit?
Pressure is the measure of how strongly a force is distributed over a given area. Its SI (International System of Units) unit is called the Pascal (Pa). So, pressure is measured in Pascals.
### What is pressure in a fluid?
Pressure in a fluid refers to the force that the liquid or gas exerts on the walls of its container or any object submerged in it. It happens because the particles in the fluid are constantly moving and bumping into things, creating a push or pressure in all directions.
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Verify OTP Code (required) | 1,659 | 6,682 | {"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} | 4.03125 | 4 | CC-MAIN-2024-18 | latest | en | 0.919583 |
https://pokeronlineqq724.com/qa/question-can-8-people-play-poker.html | 1,611,294,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00724.warc.gz | 511,558,059 | 8,856 | # Question: Can 8 People Play Poker?
## What Colour chips are worth?
White chips normally are worth between \$0.50 and \$1, (at times grey, blue, and red chips may be worth this amount, as well).
Pink chips usually have a value of between \$2 and \$2.50.
Red chips are often worth \$5 in most cardrooms, with the exception of California where \$5 chips are yellow..
## Can you play poker with 8 players?
Between 6 and 10 players; 7 or 8 makes for a great game. If you have more than 10 players then you’ll need to split into two tables or play will be too slow.
## How many poker chips do you need for 8 players?
Five to six players – Have a total of 500 to 600 chips. Six to eight players – Have a total of 600 to 800 chips. Eight to ten players – Have a total of 800 to 1000 chips. More than ten players – Start with 1000 chips and add another 100 for every extra person.
## How many poker chips do you get for \$20?
50 blind structure. It is a 2-1 chip ratio so 20 gets you 40 and 40 gets you 80. 80BB is enough so that people can play on their \$20 all night if they are just looking to hang out.
## How many poker chips do you need for 6 people?
For a normal game of poker at home (6-10 players), it is recommended that you have a suitcase with 500 poker chips. For less than 6 players, you can use 300 chips, but remember that at some point you may want to invite more players or try rebuy tournaments.
## How many poker chips does a player get?
When determining the amount of chips required, we usually use the following rule: up to 6 players: 300 poker chips. up to 10 players: 500 poker chips. Tournament 20 to 30 participants: 1000 poker chips.
## What color chips are worth?
Most casinos follow the basic primary color-coding values for white, pink, red, green, and black chips, with the addition of yellow chips at \$20, and blue chips valued at \$10.
## What are winning poker hands?
Poker hands from highest to lowestRoyal flush. A, K, Q, J, 10, all the same suit.Straight flush. Five cards in a sequence, all in the same suit.Four of a kind. All four cards of the same rank.Full house. Three of a kind with a pair.Flush. … Straight. … Three of a kind. … Two pair.More items…
## How much does a poker table cost?
The budget for a poker table is all dependent on the person. A poker table top can be had for as little as \$30 to \$50, perfect for those with little space or small budgets. A person seeking a more traditional table, but not looking to spend a fortune should budget \$160 – \$200 budget.
## How do you give out poker chips?
So, you will need to have higher denomination chips than the initial starting poker chips. Initial chips that each player has in front of them: 4 – “25” chips = 100….So, let’s figure out the extra chips needed if the value is up to 3X the starting value:100 – \$5 chips = \$500.20 – \$25 chips = \$500.10 – \$100 chips = \$1000.
## Can you play poker with 10 players?
3 Answers. The ideal max number at a regular-size table is 9 players and a dealer. 10 is also not uncommon. Some larger tournaments will occasionally place 11 at a table in the early stages until the field is narrowed a little.
## Can two people play poker?
Poker can be played with only two players. It is called heads-up. Here are the two player rules: The dealer is the small blind and the other player posts the big blind.
## Who acts first in Texas Holdem?
The first round of betting takes place right after each player has been dealt two hole cards. The first player to act is the player to the left of the big blind. This position referred to as ‘under the gun’ because the player has to act first.
## How do you start a poker night?
How to Start and Host a Regular Poker NightPlan Well. There’s more that goes into a successful poker night than meets the eye. … Know the Game. … Determine the Intensity of Your Poker Group. … Playing the Game: A Table and Chips. … Establish House Rules Beforehand. … Make It Regular and Consistent. … Snacks and Beverages. … Set the Tone With Classy Music.More items…•
## How long does a poker game last?
1-3 hoursThe average poker game is 1-3 hours if you’re playing a simple cash game, and the average poker tournament can last anywhere from 20 minutes to 12 hours. However, this is highly variable and depends on the number of players, the size of the starting stack, how careful players are, and other factors. | 1,075 | 4,408 | {"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-2021-04 | latest | en | 0.952574 |
https://www.cheenta.com/inequality-hanoi-2018/ | 1,624,405,248,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00619.warc.gz | 614,651,524 | 43,483 | INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More
# Inequality | HANOI 2018
Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on Inequality.
## Inequality - HANOI 2018
Find the number of integers that satisfy the inequality $n^{4}-n^{3}-3n^{2}-3n-17 \lt 0$.
• is 4
• is 6
• is 8
• cannot be determined from the given information
### Key Concepts
Algebra
Theory of Equations
Inequality
## Check the Answer
HANOI, 2018
Inequalities (Little Mathematical Library) by Korovkin
## Try with Hints
First hint
We have $(n+1)^{3}+16 \gt n^{4} \geq 0$ which implies $n \geq -3$.
Second Hint
For $n \geq 4$ we have $n^{4}-(n+1)^{3}$ $\geq 3n^{3}-3n^{2}-3n-1$ $\geq 12n^{2}-3n^{2}-3n-1$ $=n(n-3)+8n^{2}-1 \gt 16$.
Final Step
Then $-3 \leq n \leq 3$. By directly calculation we obtain n=-1,0,1,2 that is 4 such integers.
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### Cheenta. Passion for Mathematics
Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
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support@cheenta.com | 412 | 1,247 | {"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} | 3.96875 | 4 | CC-MAIN-2021-25 | longest | en | 0.752298 |
https://smartexcel.sg/resources/problem-solving-strategies/ | 1,540,142,543,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514162.67/warc/CC-MAIN-20181021161035-20181021182535-00010.warc.gz | 812,454,239 | 35,298 | # Problem Solving Strategies
Primary 1 | Primary 2 | Primary 3 | Primary 4 | Primary 5 | Primary 6
Problem Solving Strategies
There are numerous approaches to solving maths problems. ‘Model Drawing’ is the first one that we have introduced because we feel that it has the greatest impact in building children’s confidence in dealing with maths problems. Most students enjoy visual effects. Seeing abstract relationships, represented by concrete and colorful images, helps in understanding, leading to the solution of the problem. Our section on Model Drawing is by no means exhaustive but it will open a new doorway for the student who has been struggling with math problems. Besides the Model-Drawing Approach there are several other strategies, which are necessary for the student to master, to achieve proficiency in math problem solving. In our next section, we introduce the important and most useful ones. These are: (1) Draw a Picture (2) Look for a Pattern (3) Guess and Check (4) Make a Systematic List (5) Logical Reasoning (6) Work Backwards The student may also come across problems which may need the use of more than one strategy before solution can be found.
OTHER PROBLEM-SOLVING STRATEGIES
Primary 3 Primary 4 Primary 5 Primary 6
Draw a Picture/Find a Pattern Draw a Picture/Find a Pattern Draw a Picture Draw a Picture/Find a Pattern
Guess & Check Guess & Check Guess & Check Guess & Check
Work Backwards Work Backwards Work Backwards Work Backwards
Systematic Listing Systematic Listing Systematic Listing Systematic Listing
Logical Reasoning Logical Reasoning Logical Reasoning Logical Reasoning
These resouces are FREE for online use.
Source: www.THESINGAPOREMATHS.COM | 376 | 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} | 3.421875 | 3 | CC-MAIN-2018-43 | latest | en | 0.87409 |
https://cs.stackexchange.com/questions/142618/how-to-solve-modulo-math-equation-in-rsa-algorithm | 1,718,840,667,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00378.warc.gz | 163,171,601 | 39,626 | # How to solve modulo math equation in RSA algorithm?
$$(3 \cdot d) \mod 8=1$$. I know the answer is $$d=3$$ by common sense. But what is the mathematical approach to solve this problem? How do I solve this mathematically?
How do we get this value of 125^107 mod 187=5?
• If there are two questions, ask them separately. And, do not use images in your questions, see this. Commented Jul 24, 2021 at 18:52
Use Extended gcd algorithm. The time complexity is $$O(\log n)$$.
We know that $$gcd(3,8)=1$$ since $$3$$ is a prime, and therefore $$3$$ has one and unique inverse under multiplication modulo $$8$$.
As you have seen, its not hard to guess that $$3$$ is its own inverse, that is, $$x=3$$ is the solution for the equation: $$3x\equiv1 \mod 8$$
Since there is only one and unique solution, we know that $$3$$ is the only solution. | 237 | 837 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-26 | latest | en | 0.936755 |
https://www.coursehero.com/file/5766687/hw-20/ | 1,516,620,975,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00130.warc.gz | 909,810,192 | 24,105 | # hw 20 - nguyen(jmn727 homework 20 Turner(59070 1 This...
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Unformatted text preview: nguyen (jmn727) homework 20 Turner (59070) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The electrons in a beam are moving at 2 . 7 10 8 m / s in an electric field of 15000 N / C. What value must the magnetic field have if the electrons pass through the crossed fields undeflected? Correct answer: 55 . 5556 T. Explanation: Let : E = 15000 N / C and v = 2 . 7 10 8 m / s . If the electrons move undeflected through the crossed fields v = E B then B = E v = 15000 N / C 2 . 7 10 8 m / s = 55 . 5556 T . 002 (part 1 of 2) 10.0 points A proton in a cyclotron is moving with a speed of 4 . 67 10 7 m / s in a circle of radius 0 . 728 m. 1 . 67 10 27 kg is the mass of the pro- ton, and 1 . 60218 10 19 C is its fundamental charge. What is the magnitude of the force exerted on the proton by the magnetic field of the cyclotron? Correct answer: 5 . 00287 10 12 N. Explanation: Let : v = 4 . 67 10 7 m / s , m p = 1 . 67 10 27 kg , r = 0 . 728 m , and q e = 1 . 60218 10 19 C . The magnetic force is the centripetal force which keeps the proton in circular motion. From the centripetal force equation, we have F = m p v 2 r = (1 . 67 10 27 kg)(4 . 67 10 7 m / s) 2 (0 . 728 m) = 5 . 00287 10 12 N . 003 (part 2 of 2) 10.0 points What is the magnitude of the magnetic field required to keep it moving in this circle? Correct answer: 0 . 668639 T. Explanation: The force is due to the magnetic field. F B = B perp q e v , B = F q e v = . 668639 T . 004 (part 1 of 2) 10.0 points A proton travels with a speed of 3 . 28 10 6 m / s at an angle of 31 . 6 with a magnetic field of . 147 T pointed in the y direction. The charge of proton is 1 . 60218 10 19 C. What is the magnitude of the magnetic force on the proton? Correct answer: 4 . 04782 10 14 N. Explanation: Let : v = 3 . 28 10 6 m / s , = 31 . 6 , and B = 0 . 147 T . F = q v B sin = (1 . 60218 10 19 C)(3 . 28 10 6 m / s) (0 . 147 T) sin31 . 6 = 4 . 04782 10 14 N. nguyen (jmn727) homework 20 Turner (59070) 2 005 (part 2 of 2) 10.0 points The mass of proton is 1 . 67262 10 27...
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Ask a homework question - tutors are online | 917 | 2,731 | {"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} | 3.515625 | 4 | CC-MAIN-2018-05 | latest | en | 0.810994 |
https://forum.posit.co/t/sapply-lapply-and-indexing-inside-for-loops/57403 | 1,721,446,240,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00772.warc.gz | 224,122,874 | 8,492 | # Sapply, lapply and indexing inside for loops
Hi All,
I stumbled upon this post while learning R: https://r.789695.n4.nabble.com/Loop-with-variable-index-td845011.html
The qustion was as follows: " I have a list of 20 values. The first time through a loop I want to find the mean and stnd.dev. of the first two values; the second time through the loop I want to find the mean and stnd. dev. of the first 3 values, etc. until the last time through the loop I want to find the mean and stnd. dev. of all 20 values, so I end up with 19 means and stnd. deviations.
How would I construct such a loop? "
1. Solution, which I personnaly think is brilliant:
``````x <- rnorm(20)
sapply(c("sd", "mean"), function(fun)lapply(lapply(lapply(2:20, seq,
from=1), function(.x)x[.x]), fun))
``````
But trying to understand the code I am stuck:
Generally I understand what "for loops" are made for but I was able to break it down only into this:
``````lapply(2:20, seq, from=1)
``````
So I would like to understand how all that code works and why there are three lapplies in a row in it ? What do those " (.x), [.x] " mean, and how is it possible to recreate it in tidyverse maybe in a simpler way ?
1. Solution:
``````Let the vector be ``x''.
mns <- list()
sds <- list()
for(i in 2:20) {
mns[[i-1]] <- mean(x[1:i])
sds[[i-1]] <- sd(x[1:i])
}
mns <- unlist(mns)
sds <- unlist(sds)
``````
which I try to understand as well, especially subsetting in here like: " mns[[i-1]] ", because sometimes in loops we have " vec[i] ", or " vec[[i]] ".
And finally why here:
``````mns[[i-1]] <- mean(x[1:i])
``````
on the left hand side of an assignment arrow, we have "i-1" in brackets and on the right hand side of <- we have "x[1:i]" ? What does it do ?
Any help regarding how to understand it correctly would be much appreciated.
Thank you @Yarnabrina for your detailed explanation.
I read it carefully and I would like to clarify a few things:
this works:
`````` for (i in 2:20) {
mns[[i - 1]] <- mean(x[1:i])
sds[[i - 1]] <- sd(x[1:i])
}
``````
but this does not:
``````for (i in x[2:20]) {
mns[[i - 1]] <- mean(x[1:i])
sds[[i - 1]] <- sd(x[1:i])
}
``````
If x gets x <- rnorm(20), does x[2:20] represent 19 numbers from vector x „for loop” iterates on ?
Why does it throw an error then ?
About indexing/subsetting inside for loops please have a look at the example below :
``````n = 10
log.vec = vector(length=n, mode="numeric")
for (i in 1:n) {
log.vec[i] = log(i)
}
for (i in 1:n) {
log.vec[[i]] = log(i)
}
log.vec
``````
Regardless of using [ or [[ the results are the same. I suppose that it is not always like this.
Obviously I will study a topic from SO you pointed out in your reply.
Best regards.
I would recommend the following: 1) Open a new R file, 2) place the code below in it (your code wrapped in a dummy function), 3) source the file, 4) click to the left of line 2 to create a red break point, and then 5) run `test_function()` in the console. This puts you in debug mode, and all the values being created an assigned will appear in the upper right 'Environment' pane, and you can step through the function by hitting 'Enter'.
``````test_function <- function(n = 2){
x <- rnorm(20)
mns <- list()
sds <- list()
for(i in 2:n) {
mns[[i-1]] <- mean(x[1:i])
sds[[i-1]] <- sd(x[1:i])
}
for (i in x[2:n]) {
mns[[i - 1]] <- mean(x[1:i])
sds[[i - 1]] <- sd(x[1:i])
}
}
``````
is because i in 2:20 will take values 2,3,4 up to 20 and these are valid indexes into a vector
whereas i in x[2:20] gives contents of the random vector x , the 2nd, 3rd, 4th up to 20th, these are not valid indexes as they are decimal fraction. As I is being used as an index, these fractions are meaningless.
The difference with a vector is moot, but for a list using single square brackets `[ ]` will return the value wrapped in a list, whereas double square brackets `[[ ]]` will return it naked
This is not really of any consequence when on the left side of an assignment, its more about when the list is being access on the right side of an assignment.
When on the left, both forms of square brackets will let you modify the value at the index with the right hand side value.
When on the right , you will be setting whatever is on the left to be a list containing the value indexed out of your list if you use single brackets, or you will be setting whatever is on the left to be the naked value of what you have indexed into.
Hope it helps !
Thank you @dromano and @nirgrahamuk,
I really appreciate you taking the time to respond.
Best regards.
Hi @Yarnabrina and All,
I am still learning "for loops".
I try to resolve this very simple task:
``````x <- c(1:5)
for (i in x)
{ y <- i*i
print(y)
}
``````
using sapply and lappy solution presented in my first question:
`````` sapply(c("sd", "mean"), function(fun)lapply(lapply(lapply(2:20, seq,
from=1), function(.x)x[.x]), fun))
``````
because I somehow like it, when it gives me immediately two columns with results showing each and every iteration steps.
Of course I can use the second solution presented there as well:
``````ID <- c(1,2,3,4,5)
DF <- cbind(ID, expon2) %>% as.data.frame()
colnames(DF) <- c("ID", "result_of_each_Iterations")
``````
and using reprex:
``````library(tidyverse)
#> Warning: package 'ggplot2' was built under R version 3.6.3
#> Warning: package 'tidyr' was built under R version 4.0.0
#> Warning: package 'forcats' was built under R version 3.6.3
library(magrittr)
#>
#> Attaching package: 'magrittr'
#> The following object is masked from 'package:purrr':
#>
#> set_names
#> The following object is masked from 'package:tidyr':
#>
#> extract
x <- c(1:5)
expon2 <- list()
for (i in 1:5) {
expon2[i] <- i**2
}
expon2 <- unlist(expon2)
ID <- c(1,2,3,4,5)
DF <- cbind(ID, expon2) %>% as.data.frame()
colnames(DF) <- c("ID", "result_of_each_Iterations")
DF
#> ID result_of_each_Iterations
#> 1 1 1
#> 2 2 4
#> 3 3 9
#> 4 4 16
#> 5 5 25
``````
Created on 2020-03-28 by the reprex package (v0.3.0)
but if you could help me do it with sapply and lapply, please ?
And is it a way to do something in order to DF when displayed is not so clustered (header text) but looks like align center (a bit like in MS Word):
kind regards.
Hi,
I can use use sapply, and apply etc, but most times I start with the purrr library for this type of work.
It has great functions and easy to use, also there are very good lessons for it.
``````x <- c(1:5)
#no purrr
for (i in x)
{ y <- i*i
print(y)
}
library(purrr)
purrr::walk(x,
~print(.^2))``````
sure, as the function ^2 is vectorised, thats the most elegant solution, walk is slightly more general as you could use you own non vectorise function in place. but yes Yarnabrina your solution is by far the most succint and elegant approach to printing all the square values of x in order
Thank you @nirgrahamuk, for purrr solution, very quick.
Hi @Yarnabrina,
That solution with sapply and lapply immediately would give us (I suppose) two columns like this:
without binding columns with cbind and other indirect steps involvement.
This topic was automatically closed 21 days after the last reply. New replies are no longer allowed. | 2,162 | 7,303 | {"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-2024-30 | latest | en | 0.921793 |
https://en.unionpedia.org/Optical_rotation | 1,632,740,182,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00593.warc.gz | 276,743,309 | 18,049 | Communication
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# Optical rotation
Optical rotation or optical activity (sometimes referred to as rotary polarization) is the rotation of the plane of polarization of linearly polarized light as it travels through certain materials. [1]
64 relations: Akhlesh Lakhtakia, Amorphous solid, Aqueous solution, Basis (linear algebra), Biomolecular structure, Birefringence, Camphor, Chemical synthesis, Chirality, Chirality (chemistry), Cholesterol, Circular dichroism, Circular polarization, Cryptochirality, Crystal, Dextrorotation and levorotation, Dispersion (optics), Electric field, Enantiomer, Enantiomeric excess, Faraday effect, Fluid, François Arago, Fructose, Fused quartz, Geometric phase, Glucose, Helix, Hydrolysis, Imaginary unit, Inorganic chemistry, Inverted sugar syrup, Jacobus Henricus van 't Hoff, Jean-Baptiste Biot, John Herschel, Joseph Achille Le Bel, Kramers–Kronig relations, Lees (fermentation), Linear polarization, Liquid-crystal display, Louis Pasteur, Magnetic field, Mirror image, Optic axis of a crystal, Optical rotatory dispersion, Phase velocity, Phasor, Plane of polarization, Polarimeter, Polarimetry, ... Expand index (14 more) »
## Akhlesh Lakhtakia
Akhlesh Lakhtakia is Evan Pugh University Professor and Charles Godfrey Binder Professor of Engineering Science and Mechanics at the Pennsylvania State University.
## Amorphous solid
In condensed matter physics and materials science, an amorphous (from the Greek a, without, morphé, shape, form) or non-crystalline solid is a solid that lacks the long-range order that is characteristic of a crystal.
## Aqueous solution
An aqueous solution is a solution in which the solvent is water.
## Basis (linear algebra)
In mathematics, a set of elements (vectors) in a vector space V is called a basis, or a set of, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set.
## Biomolecular structure
Biomolecular structure is the intricate folded, three-dimensional shape that is formed by a molecule of protein, DNA, or RNA, and that is important to its function.
## Birefringence
Birefringence is the optical property of a material having a refractive index that depends on the polarization and propagation direction of light.
## Camphor
Camphor is a waxy, flammable, white or transparent solid with a strong aroma.
## Chemical synthesis
Chemical synthesis is a purposeful execution of chemical reactions to obtain a product, or several products.
## Chirality
Chirality is a property of asymmetry important in several branches of science.
## Chirality (chemistry)
Chirality is a geometric property of some molecules and ions.
## Cholesterol
Cholesterol (from the Ancient Greek chole- (bile) and stereos (solid), followed by the chemical suffix -ol for an alcohol) is an organic molecule.
## Circular dichroism
Circular dichroism (CD) is dichroism involving circularly polarized light, i.e., the differential absorption of left- and right-handed light.
## Circular polarization
In electrodynamics, circular polarization of an electromagnetic wave is a polarization state in which, at each point, the electric field of the wave has a constant magnitude but its direction rotates with time at a steady rate in a plane perpendicular to the direction of the wave.
## Cryptochirality
In stereochemistry, Cryptochirality is a special case of chirality in which a molecule is chiral but its specific rotation is non-measurable.
## Crystal
A crystal or crystalline solid is a solid material whose constituents (such as atoms, molecules, or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.
## Dextrorotation and levorotation
Dextrorotation and levorotation (also spelled as laevorotation)The first word component dextro- comes from Latin word for dexter "right (as opposed to left)".
## Dispersion (optics)
In optics, dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency.
## Electric field
An electric field is a vector field surrounding an electric charge that exerts force on other charges, attracting or repelling them.
## Enantiomer
In chemistry, an enantiomer, also known as an optical isomer (and archaically termed antipode or optical antipode), is one of two stereoisomers that are mirror images of each other that are non-superposable (not identical), much as one's left and right hands are the same except for being reversed along one axis (the hands cannot be made to appear identical simply by reorientation).
## Enantiomeric excess
Enantiomeric excess (ee) is a measurement of purity used for chiral substances.
In physics, the Faraday effect or Faraday rotation is a magneto-optical phenomenon—that is, an interaction between light and a magnetic field in a medium.
## Fluid
In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress.
## François Arago
Dominique François Jean Arago (Domènec Francesc Joan Aragó), known simply as François Arago (Catalan: Francesc Aragó) (26 February 17862 October 1853), was a French mathematician, physicist, astronomer, freemason, supporter of the carbonari and politician.
## Fructose
Fructose, or fruit sugar, is a simple ketonic monosaccharide found in many plants, where it is often bonded to glucose to form the disaccharide sucrose.
## Fused quartz
Fused quartz or fused silica is glass consisting of silica in amorphous (non-crystalline) form.
## Geometric phase
In classical and quantum mechanics, the geometric phase, Pancharatnam–Berry phase (named after S. Pancharatnam and Sir Michael Berry), Pancharatnam phase or most commonly Berry phase, is a phase difference acquired over the course of a cycle, when a system is subjected to cyclic adiabatic processes, which results from the geometrical properties of the parameter space of the Hamiltonian.
## Glucose
Glucose is a simple sugar with the molecular formula C6H12O6.
## Helix
A helix, plural helixes or helices, is a type of smooth space curve, i.e. a curve in three-dimensional space.
## Hydrolysis
Hydrolysis is a term used for both an electro-chemical process and a biological one.
## Imaginary unit
The imaginary unit or unit imaginary number is a solution to the quadratic equation.
## Inorganic chemistry
Inorganic chemistry deals with the synthesis and behavior of inorganic and organometallic compounds.
## Inverted sugar syrup
Invert(ed) sugar (syrup) is an edible mixture of two simple sugars—glucose and fructose—that is made by heating sucrose (table sugar) with water.
## Jacobus Henricus van 't Hoff
Jacobus Henricus van 't Hoff, Jr. (30 August 1852 – 1 March 1911) was a Dutch physical chemist.
## Jean-Baptiste Biot
Jean-Baptiste Biot (21 April 1774 – 3 February 1862) was a French physicist, astronomer, and mathematician who established the reality of meteorites, made an early balloon flight, and studied the polarization of light.
## John Herschel
Sir John Frederick William Herschel, 1st Baronet (7 March 1792 – 11 May 1871) was an English polymath, mathematician, astronomer, chemist, inventor, experimental photographer who invented the blueprint, and did botanical work.
## Joseph Achille Le Bel
Joseph Achille Le Bel (21 January 1847 in Pechelbronn – 6 August 1930, in Paris, France) was a French chemist.
## Kramers–Kronig relations
The Kramers–Kronig relations are bidirectional mathematical relations, connecting the real and imaginary parts of any complex function that is analytic in the upper half-plane.
## Lees (fermentation)
Lees are deposits of dead yeast or residual yeast and other particles that precipitate, or are carried by the action of "fining", to the bottom of a vat of wine after fermentation and aging.
## Linear polarization
In electrodynamics, linear polarization or plane polarization of electromagnetic radiation is a confinement of the electric field vector or magnetic field vector to a given plane along the direction of propagation.
## Liquid-crystal display
A liquid-crystal display (LCD) is a flat-panel display or other electronically modulated optical device that uses the light-modulating properties of liquid crystals.
## Louis Pasteur
Louis Pasteur (December 27, 1822 – September 28, 1895) was a French biologist, microbiologist and chemist renowned for his discoveries of the principles of vaccination, microbial fermentation and pasteurization.
## Magnetic field
A magnetic field is a vector field that describes the magnetic influence of electrical currents and magnetized materials.
## Mirror image
A mirror image (in a plane mirror) is a reflected duplication of an object that appears almost identical, but is reversed in the direction perpendicular to the mirror surface.
## Optic axis of a crystal
An optic axis of a crystal is a direction in which a ray of transmitted light suffers no birefringence (double refraction).
## Optical rotatory dispersion
Optical rotatory dispersion is the variation in the optical rotation of a substance with a change in the wavelength of light.
## Phase velocity
The phase velocity of a wave is the rate at which the phase of the wave propagates in space.
## Phasor
In physics and engineering, a phasor (a portmanteau of phase vector), is a complex number representing a sinusoidal function whose amplitude (A), angular frequency (ω), and initial phase (θ) are time-invariant.
## Plane of polarization
The term plane of polarization refers to the direction of polarization of linearly-polarized light or other electromagnetic radiation.
## Polarimeter
A polarimeter is a scientific instrument used to measure the angle of rotation caused by passing polarized light through an optically active substance.
## Polarimetry
Polarimetry is the measurement and interpretation of the polarization of transverse waves, most notably electromagnetic waves, such as radio or light waves.
## Polarization (waves)
Polarization (also polarisation) is a property applying to transverse waves that specifies the geometrical orientation of the oscillations.
## Polarization rotator
A polarization rotator is an optical device that rotates the polarization axis of a linearly polarized light beam by an angle of choice.
## Quartz
Quartz is a mineral composed of silicon and oxygen atoms in a continuous framework of SiO4 silicon–oxygen tetrahedra, with each oxygen being shared between two tetrahedra, giving an overall chemical formula of SiO2.
## Racemic mixture
In chemistry, a racemic mixture, or racemate, is one that has equal amounts of left- and right-handed enantiomers of a chiral molecule.
## Refractive index
In optics, the refractive index or index of refraction of a material is a dimensionless number that describes how light propagates through that medium.
## Silicon dioxide
Silicon dioxide, also known as silica (from the Latin silex), is an oxide of silicon with the chemical formula, most commonly found in nature as quartz and in various living organisms.
## Specific rotation
In chemistry, specific rotation is a property of a chiral chemical compound.
## Speed of light
The speed of light in vacuum, commonly denoted, is a universal physical constant important in many areas of physics.
## Stereoisomerism
In stereochemistry, stereoisomers are isomeric molecules that have the same molecular formula and sequence of bonded atoms (constitution), but differ in the three-dimensional orientations of their atoms in space.
## Sucrose
Sucrose is common table sugar.
## Superposition principle
In physics and systems theory, the superposition principle, also known as superposition property, states that, for all linear systems, the net response caused by two or more stimuli is the sum of the responses that would have been caused by each stimulus individually.
## Tartaric acid
Tartaric acid is a white crystalline organic acid that occurs naturally in many fruits, most notably in grapes, but also in bananas, tamarinds and citrus.
## Turpentine
Chemical structure of pinene, a major component of turpentine Turpentine (also called spirit of turpentine, oil of turpentine, wood turpentine and colloquially turps) is a fluid obtained by the distillation of resin obtained from live trees, mainly pines.
## Wavelength
In physics, the wavelength is the spatial period of a periodic wave—the distance over which the wave's shape repeats.
## References
Hey! We are on Facebook now! » | 2,797 | 12,536 | {"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-2021-39 | latest | en | 0.803825 |
https://cboard.cprogramming.com/c-programming/153120-question-about-dynamic-memory-allocation.html | 1,516,191,590,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886895.18/warc/CC-MAIN-20180117102533-20180117122533-00384.warc.gz | 633,971,618 | 12,116 | 1. ## Question about dynamic memory allocation
Hi guys , i have a question about din. memory alloc. and i want to ask it with an example.
Code:
```/* here is my struct */
struct node
{
int data;
struct node *next;
}
/* here is my simple making node function*/
struct node *makenode(int value)
{
struct node *np;
np=malloc(sizeof(struct node));
np->data=value;
np->next=NULL;
}```
What if i did not write the np=malloc(sizeof(struct node)); line, Does the code work properly ? When i need to take a place from memory using with pointer and when i dont need ? Thank you for your answers.
2. Originally Posted by dayanike
Code:
``` struct node *np;
np=malloc(sizeof(struct node));
np->data=value;
np->next=NULL;
}```
What if i did not write the np=malloc(sizeof(struct node)); line, Does the code work properly ?
No - in this case, np would be uninitialized and could point anywhere, so np->data could be located anywhere in your address space. So you don't know what you might be trying to modify.
3. Originally Posted by dayanike
When i need to take a place from memory using with pointer and when i dont need ?
Basically if you want to use the contents of what that pointer points to you will always need to have allocated space.
If you want to iterate through a linked list, you could do:
Code:
```int find (struct node * root, int value){
struct node * t; /* you don't need to allocate memory here */
t = root; /* t will point to an already allocated space in memory */
while (t != NULL) {
if (t->data == value)
return 1;
t = t->next;
}
return 0;
}```
Pointers are variables which store memory addresses. When you declare them they are storing a random address whose contents are not known. So you initialize them by allocating space or making them point to an already allocated space.
You can also assign them to NULL which is an initialization, but in this case the pointer is still pointing to something now known.
Edit: There's this small tutorial which covers very nice features of pointers, give it a look. | 471 | 2,034 | {"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} | 2.703125 | 3 | CC-MAIN-2018-05 | latest | en | 0.907466 |
https://www.convertunits.com/from/kilogram/to/marc | 1,701,647,802,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00082.warc.gz | 795,793,755 | 12,605 | Convert kilogram to marc [France]
kilogram marc
How many kilogram in 1 marc? The answer is 0.24475. We assume you are converting between kilogram and marc [France]. You can view more details on each measurement unit: kilogram or marc The SI base unit for mass is the kilogram. 1 kilogram is equal to 4.0858018386108 marc. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between kilograms and marc [France]. Type in your own numbers in the form to convert the units!
Quick conversion chart of kilogram to marc
1 kilogram to marc = 4.0858 marc
5 kilogram to marc = 20.42901 marc
10 kilogram to marc = 40.85802 marc
15 kilogram to marc = 61.28703 marc
20 kilogram to marc = 81.71604 marc
25 kilogram to marc = 102.14505 marc
30 kilogram to marc = 122.57406 marc
40 kilogram to marc = 163.43207 marc
50 kilogram to marc = 204.29009 marc
Want other units?
You can do the reverse unit conversion from marc to kilogram, or enter any two units below:
Enter two units to convert
From: To:
Definition: Kilogram
The kilogram or kilogramme, (symbol: kg) is the SI base unit of mass. A gram is defined as one thousandth of a kilogram. Conversion of units describes equivalent units of mass in other systems.
Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 454 | 1,758 | {"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-2023-50 | latest | en | 0.819759 |
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### Transcript of PHYSICS STANDARD LEVEL PAPER 3 · PDF file This question is about projectile motion. A...
• 2208-6518 33 pages
M08/4/PHYSI/SP3/ENG/TZ2/XX+
Wednesday 21 May 2008 (morning)
PHYSICS
STANDARD LEVEL
PAPER 3
INSTRUCTIONS TO CANDIDATES
• Write your session number in the boxes above.
• Do not open this examination paper until instructed to do so.
• Answer all of the questions from two of the Options in the spaces provided.
• At the end of the examination, indicate the letters of the Options answered in the
candidate box on your cover sheet.
1 hour
Candidate session number
0 0
0133
22086518
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Option A — Mechanics Extension
A1. This question is about projectile motion.
A projectile is fired horizontally from the top of a vertical cliff of height 40 m.
cliff
projectile
40 m
sea
At any instant of time, the vertical distance fallen by the projectile is d. The graph below shows the variation with distance d, of the kinetic energy per unit mass E of the projectile.
E / J kg–1
1400
1300
1200
1100
1000
900
800 0 5 10 15 20 25 30 35 40
d / m
(This question continues on the following page)
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(Question A1 continued)
(a) Use data from the graph to calculate, for the projectile,
(i) the initial horizontal speed.
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(ii) the speed with which it hits the sea.
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[1]
(b) Use your answers to (a) to calculate the magnitude of the vertical component of velocity with which the projectile hits the sea.
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[2]
A2. This question is about orbital motion.
(a) State Kepler’s third law (the law of periods).
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(b) A satellite of mass m is in orbit of radius r about Earth. The mass of Earth is ME and the orbital period of the satellite is T.
State, for the satellite,
(i) the name of the force that provides the centripetal force.
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(ii) the orbital speed in terms of T and r.
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[1]
(This question continues on the following page)
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(Question A2 continued)
(c) Kepler’s third law may be applied to the satellite orbiting the Earth. Use your answers to (b) to deduce that in Kepler’s third law there is a constant K given by
K GM
= 4 2π
Ε
.
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[3]
(d) State an expression for the gravitational field strength g at the surface of the Earth in terms of ME and the radius of Earth RE.
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[1]
(e) For the Earth, the gravitational field strength, g is 10 N kg–1 and the radius RE is 6.4 ×10 6 m.
Using your answers to (c) and (d), deduce that the orbital period of a satellite that is at a height RE above the surface of Earth is 1.4 ×10
4 s.
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[3]
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A3. This question is about friction.
(a) A block of wood of mass M is sliding down an inclined plane as shown.
θ
The angle between the plane and the horizontal is θ. The acceleration of free fall is g.
State, in terms of M, g and θ, the component of the weight of the block
(i) parallel to the inclined plane.
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[1]
(ii) perpendicular to the inclined plane.
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[1]
(b) The coefficient of dynamic friction is µK. Deduce, in terms of M, g, θ and µK, an expression for the net force acting on the block parallel to the inclined plane.
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[2]
(c) The angle θ of the inclined plane is 30. The acceleration of the block down the plane is 0.15g. Using your answer to (b), deduce that the value of µK is 0.40.
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[2]
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Option B — Quantum Physics and Nuclear Physics
B1. This question is about the de Broglie hypothesis.
(a) State the de Broglie hypothesis.
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[2]
(b) Calculate the de Broglie wavelength associated with an adult of mass 80 kg running at a speed of 5.0 m s–1.
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[2]
0633
• 2208-6518 | 3,586 | 8,808 | {"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-2022-33 | latest | en | 0.560987 |
http://mathematica.stackexchange.com/questions/11835/need-help-with-solving-non-linear-differential-equation?answertab=votes | 1,406,675,000,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510267876.49/warc/CC-MAIN-20140728011747-00126-ip-10-146-231-18.ec2.internal.warc.gz | 187,181,723 | 15,711 | # Need help with solving non-linear differential equation [closed]
I am new to using Mathematica and have a problem at hand: I need to solve $y''(x)+ a \sin(y(x)) = 0$ and plot the output, I need a "manipulate" thingy for the two initial conditions. Help needed.
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## closed as too localized by acl, belisarius, cormullion, Leonid Shifrin, rm -rf♦Oct 10 '12 at 14:55
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Have a look at NDSolve in the documentation. – b.gatessucks Oct 10 '12 at 8:52
You wrote the question so fast that seems some dishonest educator is going to catch you. Please take your time and show your effort. – belisarius Oct 10 '12 at 8:55
@belisarius actually yes, my "educator" wants to see the solution in 20 minutes :P – kernel_panic Oct 10 '12 at 9:05
have you tried anything? eg, asking google? – acl Oct 10 '12 at 12:21
This equation is easily solved analytically by using y[x] as an independent variable, which reduces it to a first-order equation. By calling y'[x] == f[y], you get y''[x] == 1/2 (d/dy)(f[y]^2),so this becomes a first order in y equation. Once you solve it, you recall what f is and solve for x = x(y), then invert. – Leonid Shifrin Oct 10 '12 at 13:17
The function you are looking for is NDSolve. I don't think DSolve gives a working answer with the y[x] wrapped inside in the Sin function
For one-off execution, you can use this kind of expression:
finish = 10;
ic1 = 1;
ic2 = 1;
a = 2;
sol = NDSolve[{y''[x] + a Sin[y[x]] == 0, y[0] == ic1, y'[0] == ic2}, y[x], {x, 0, finish}];
Plot[y[x] /. sol, {x, 0, finish}]
And for a Manipulate type set up hopefully this will help you
a = 3; Manipulate[{sol =
NDSolve[{y''[x] + a Sin[y[x]] == 0, y[0] == ic1, y'[0] == ic2},
y[x], {x, 0, finish}]; Plot[y[x] /. sol, {x, 0, finish}]}[[1]], {ic1, 1, 3, 0.1}, {ic2, 2, 4, 0.1}]
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DSolve may not return closed form expressions, but this ODE is the equation of motion for a nonlinear pendulum; the solution involves elliptic functions (Jacobi I think). However, I suspect that if the OP used them, their "educator" might work out that they've had help. – acl Oct 10 '12 at 12:05
@acl, it does require Jacobian elliptic functions to solve the pendulum's DE, but I am told most of the kids these days have absolutely no clue about them... – Ｊ. Ｍ. Oct 10 '12 at 12:50
@J.M. You know, Sin[t] == t ... or else – belisarius Oct 10 '12 at 12:58
@bel, I guess that's why most people don't want their pendulums to be swinging at too wide an angle... ;P – Ｊ. Ｍ. Oct 10 '12 at 13:00 | 893 | 2,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2014-23 | latest | en | 0.925844 |
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# Fundamentals of Computer Organization(FCO)2610004_wefjune2012
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### Fundamentals of Computer Organization(FCO)2610004_wefjune2012
1. 1. GUJARAT TECHNOLOGICAL UNIVERSITY Master of Computer ApplicationSubject Name: Fundamentals of Computer OrganizationSubject Code: 2610004 (W.E.F June 2012)Objectives: Students will learn The elements of Computer Organization and Architecture. The basic knowledge necessary to understand the hardware operation of digital computers.Prerequisites: NoneContents: ( [] indicates no. of lectures ) 1. Basic Components of a digital computer [1] 2. Basic Working of Peripheral devices [4] (Circuit Diagrams not necessary) Key board Mouse Display Unit Printer Multimedia Projector Scanner 3. Introduction to Number System [9] Decimal System Bistable Devices Counting in Binary System Binary Addition and Subtraction Converting Decimal Number to Binary Negative Numbers Use of Complements to represent negative numbers Complements in other number system Binary Number Complements Weighted Code o BCD Code Octal and Hexadecimal Number System 4. Boolean Algebra and Logic Gates [9] Fundamental Concepts of Boolean Algebra Logic Gates Logical Multiplication AND Gate and OR Gate Complementation and Inverts Evaluation of logical Expression
2. 2. Evaluation of an Expression containing Parenthesis Basic Laws of Boolean Algebra Proof by Perfect Induction Simplification of Expressions De Morgan’s Theorems Basic Duality of Boolean Algebra Derivation of a Boolean Algebra Interconnecting Gates Sum of Products And Product of Sums Derivation of POS Expression Derivation of 3 input variables expression NAND Gates and NOR Gates K-Map Method for Simplifying Boolean Expressions Subcubes and Covering POS Expression and Don’t Care Design Using NAND Gates Only Design Using NOR Gates5. Basic Concepts of Sequential Logic [3] RS Flip Flop A Basic Shift Register Binary Counter (Asynchronous) Counter6. Basic Concepts of Combinational Logic [4] Construction of ALU Integer Representation 1 bit Binary Half Adder 1 bit Binary Full Adder Positive and Negative Number Addition in 1’s Complement System Addition in 2’s Complement System Shift Operation Logical and Modulo Operations (Circuit Diagrams not necessary) Basic working and application of Multiplexer7. Introduction to Memory and Storage Devices [4] Random Access Memories Basic Memory Cell Static RAM (Circuit Diagrams not necessary) Dynamic RAM (Circuit Diagrams not necessary) ROM Magnetic Disk Memories8. Introduction to Buses [1] Interfacing Buses (Circuit Diagrams not necessary) Concepts of Address Bus, Data Bus and Control Bus, Bus Width (Circuit Diagrams not necessary)
3. 3. 9. Introduction to Control Unit [2] Construction of Instruction Word Instruction Cycle and Execution Cycle organization of Control Registers 10. Basic Concepts of Computer Organization [6] Instruction Word Formats-Number of Addresses Representation of Instruction and Data Addressing Techniques Direct Addressing Immediate Addressing Relative Addressing Indirect Addressing Indexed Addressing 11. Introduction to Intel 8086 Architecture [7] Introduction Bus Interface Unit Execution Unit Introduction to Instruction Set Data Addressing Modes Instruction Format Working of MOV, ADD, SUB, MUL, DIV, CMP, IMC, DEC, NEG, AND, OR, NOT, XOR instructionsMain Reference Book(s): A. Digital Computer Fundamentals, Tata McGraw Hill, 6th Edition, Thomas C. Bartee B. Microprocessor 8086 – Architecture, Programming and Interfacing, Prentice Hall India (PHI), Sunil MathurOther Reference Book(s): 1. Computer System Architecture, PHI/Pearson Education, 3rd Edition, M. Morris ManoUnit wise Coverage from the main reference book – A: Unit – 1: Chapter – 1: 1.7 Unit – 3: Chapter – 2: 2.1 to 2.13 Unit – 4: Chapter – 3: 3.1 to 3.22 Unit – 5: Chapter – 4: 4.1, 4.7, 4.8 Unit – 6: Chapter – 5: 5.1 to 5.4, 5.6 to 5.8, 5.14, 5.15, 5.19, 5.20 Unit – 7: Chapter – 6: 6.1, 6.2, 6.7 to 6.10
4. 4. Unit – 8: Chapter – 8: 8.2, 8.3 Unit – 9: Chapter – 9: 9.1, 9.2 Unit - 10: Chapter – 10: 10.1 to 10.9 (Except 10.6)Unit wise Coverage from the main reference book – B: Unit – 11: Chapter – 2(2.1, 2.2), Chapter – 4(4.1, 4.2.1, 4.3, 4.5)Unit – 2: To be covered from Internet/latest booksAccomplishments of the student after completing the course: Students will get the knowledge of computer organization and architecture. They will know the actual working and organization of digital computer system. -x-x-x-x-x-x- | 1,396 | 4,981 | {"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-2015-48 | latest | en | 0.674019 |
http://www.factbites.com/topics/Regular-languages | 1,529,568,811,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864110.40/warc/CC-MAIN-20180621075105-20180621095105-00532.warc.gz | 408,161,932 | 9,961 | Where results make sense
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# Topic: Regular language
###### In the News (Thu 21 Jun 18)
PlanetMath: regular language A regular language (also known as a regular set or a regular event) is the set of strings generated by a regular grammar. Note that since the set of regular languages is a subset of context-free languages, any deterministic or non-deterministic finite automaton can be simulated by a pushdown automaton. This is version 11 of regular language, born on 2002-02-23, modified 2007-08-19. www.planetmath.org /encyclopedia/RegularLanguage.html (309 words)
Syntax and Semantics of Regular Expressions - Xerox XRCE Enclosing a regular expression in round parentheses (as opposed to square brackets) represents a union with the empty-string language. The opposite of the null language is the universal language. " in the sense of "the language denoted by the regular expression www.xrce.xerox.com /competencies/content-analysis/fsCompiler/fssyntax.html (2518 words)
Regular expression - Wikipedia, the free encyclopedia A regular expression (abbreviated as regexp, regex, or regxp, with plural forms regexps, regexes, or regexen) is a string that describes or matches a set of strings, according to certain syntax rules. Regular expressions are used by many text editors and utilities to search and manipulate bodies of text based on certain patterns. Regular expressions correspond to the type 3 grammars of the Chomsky hierarchy and may be used to describe a regular language. en.wikipedia.org /wiki/Regular_expression (3269 words)
Regular language - Wikipedia, the free encyclopedia If a language is not regular, it requires a machine with at least O(log log n) space to recognize (where n is the input size). To prove that a language such as this is not regular, one uses the Myhill-Nerode theorem or the pumping lemma. The language L is regular if and only if the number of equivalence classes of ~ is finite; if this is the case, this number is equal to the number of states of the minimal deterministic finite automaton accepting L. en.wikipedia.org /wiki/Regular_language (583 words)
CMSC 451 Lecture 9, Intersection of languages, closure Regular languages are closed under operations: concatenation, union, intersection, complementation, difference, reversal, Kleene star, substitution, homomorphism and any finite combination of these operations. We have seen that regular languages are closed under union because the "+" is regular expressions yields a union operator for regular languages. Regular set properties: One way to show that an operation on two regular languages produces a regular language is to construct a machine that performs the operation. www.csee.umbc.edu /~squire/cs451_l9.html (782 words)
Encyclopedia: Regular language The regular grammars describe exactly all regular languages and are in that sense equivalent with finite state automata and regular expressions. Moreover, the right regular grammars by themselves are also equivalent to the regular languages, as are the left regular grammers. Regular grammars, which use either left-regular or right-regular rules but not both, can only express a smaller set of languages, called the regular languages. www.nationmaster.com /encyclopedia/Regular-language (595 words)
Formal Language Definitions L(G) is the notation for a language defined by a grammar G. The grammar G recognizes a certain set of strings, thus a language. The building blocks of regular languages are symbols, concatenation of symbols to make strings (words), set union of strings and Kleene closure (denoted as *, also called the Kleene star, it should be typed as a superscript but this is plain text.) Informally, we use a syntax for regular expressions. The language class P is the set of languages for which there exists a deterministic Turing machine that accepts each language in a number of transitions bounded by a fixed polynomial in the length of the input string. www.cs.umbc.edu /help/theory/lang_def.shtml (1263 words)
Regular languages * (negation) of a regular language is a regular language. A regular expression is a compact notation for representing a particular regular language, using concatenation, union and unbounded repetition. The language generated by RG is the set L consisting of strings that have terminated derivations. dingo.sbs.arizona.edu /~langendoen/LING501/LING501regular.htm (757 words)
[No title] phi (the empty-set symbol) is a regular language expression and the corresponding regular expression is phi. {lambda} is a regular language, and the corresponding regular expression is lambda. For each a in E, {a} is a regular language, and a is the corresponding regular expression. www.ecst.csuchico.edu /~kent/Cs256/student256/chap3notes (1141 words)
[No title] phi is a regular expression denoting language {} 2. For all a in Sigma, a is a regular expression denoting {a} 4. Regular expressions are "language generators", they provide a mechanism to list elements of a language. ranger.uta.edu /~cook/tcs/l5.html (1095 words)
Syntax of Regular Expressions (Finite-State Calculus) In general, the concatenation of two regular languages consists of strings that extend each string of the first language with all the strings of the second language. Enclosing a regular expression in round parentheses (as opposed to square brackets) represents a union with the empty-string language. " in the sense of "the language denoted by the regular expression www.cis.upenn.edu /~cis639/docs/fssyntax.html (2451 words)
[No title] (Site not responding. Last check: 2007-10-30) These formal languages are characterized by grammars which are essentially a set of rewrite rules for generating strings belonging to a language as we see later. In asddition two of the formal languages, regular and context-free languages, are quite useful for modeling systems used in practice such as co9mputer network communication protocols, lexical analyzers and parser for compilers for programming languages. Then we study regular languages, the simplest of the four formal languages, together with regular expressions which are a method of representing regular languages. www.cs.odu.edu /~toida/nerzic/390teched/language/intro.html (525 words)
Cover Pages: Regular Language Description for XML (RELAX) RELAX (REgular LAnguage description for XML) "is a specification for describing XML-based languages. RELAX (REgular LAnguage description for XML) is a namespace-aware specification for describing XML-based languages which borrows rich datatypes from XML Schema Part 2; the RELAX grammar can be written as an XML document. Although the XMLNews-Story markup language has been superseded by the News Industry Text Format, I've chosen it because it's simple, quite widely used, looks a great deal like HTML, and its RELAX specification will use most of the features we want to focus on. xml.coverpages.org /relax.html (2827 words)
Regular Language Regular expressions and regular grammars can be thought of as “generators” of strings in a language and automata can be thought of as “acceptors” of strings in a language. The set of all languages that these constructs are capable of working with is known as the “regular” languages. Invoke this option to create a regular expression in memory which represents the same regular language as the automaton currently in memory. The algorithm RLM uses to accomplish this is taken from [1], but a slight optimization has been made. Invoke this option to create a regular grammar in memory which represents the same regular language as the automaton currently in memory. The algorithm RLM uses to accomplish this is taken from [1]. www.theese.com /rlm/rlm.html (4175 words)
AWK Language Programming - Regular Expressions A regular expression, or regexp, is a way of describing a set of strings. A regular expression can be used as a pattern by enclosing it in slashes. Regular expressions can also be used in matching expressions. www.math.utah.edu /docs/info/gawk_5.html (2909 words)
Formal Language Definitions (Site not responding. Last check: 2007-10-30) L(M) is the notation for a language defined by a machine M. The machine M accepts a certain set of strings, thus defines a language. L(G) is the notation for a language defined by a grammar G. The grammar G recognizes a certain set of strings, thus defines a language. L = L1 â© L2 The complement of a language is a language. cs.wwc.edu /~aabyan/CC2001/AL/lang_def.html (1520 words)
Regular language (Site not responding. Last check: 2007-10-30) The results of the union, intersection and set-difference operations when applied to regular languages is itself a regularlanguage; the complement of every regular language is a regular language as well. Concatenating two regular languages (in the sense of concatenating every string from the firstlanguage with every string from the second one) also yields a regular language. The rightquotient and the leftquotient of a regular language by an arbitrary language is also regular. www.therfcc.org /regular-language-34498.html (503 words)
[No title] This true because every description of a regular language is of finite length, so there is a countably infinite number of such descriptions. Showing That a Language is Regular Techniques for showing that a language L is regular: Show that L has a finite number of elements. Using Closure Properties Once we have some languages that we can prove are not regular, such as anbn, we can use the closure properties of regular languages to show that other languages are also not regular. www.cs.utexas.edu /users/cline/ear/Slides/Regular/RegularSlides5.doc (1994 words)
[No title] (Site not responding. Last check: 2007-10-30) Regular Expressions, Languages, and Grammars A class of languages called “regular languages” is of interest to us. There are 3 ways to describe a regular language: regular expression the set containing the elements in the language regular grammar Regular Expressions We can construct regular expressions from primitive constituents by repeatedly applying recursive rules: Let A be a given alphabet. These are primitive regular expressions if w1 and w2 are regular expressions, then w1(w2, w1 (w2, w1*, and (w1) are regular expressions a string is a regular expression if and only if it can be derived from the primitive regular expressions by a finite number of applications of the rules in (2). www.utdallas.edu /~kcooper/teaching/2305/FSA/FSA.doc (1170 words)
Cellular Automata as Models of Complexity (1984) The words in a regular language correspond to the possible paths through a finite graph representing a finite state machine. in the smallest graph corresponding to a particular set of configurations may be defined as the `regular language complexity' of the set. Formal languages are recognized or generated by idealized computers with a `central processing unit' containing a fixed finite number of internal states, together with a `memory'. www.stephenwolfram.com /publications/articles/ca/84-cellular/9/text.html (597 words)
Closure Proof Template The language A' is the concatenation of two languages: A and B, where language B = {"1"}. B is a regular language [this is pretty obvious but if want to be very rigorous you could draw a little DFA that recognizes B]. Since A' is the concatenation of two regular languages, and we know regular languages are closed under concatenation, then A' is also regular. www-cse.ucsd.edu /users/clbailey/ClosureProofTemplate.htm (1413 words)
Regular Expression HOWTO Since regular expressions are used to operate on strings, we'll begin with the most common task: matching characters. Regular expressions are often used to dissect strings by writing a RE divided into several subgroups which match different components of interest. Regular expressions are a powerful tool for some applications, but in some ways their behaviour isn't intuitive and at times they don't behave the way you may expect them to. www.amk.ca /python/howto/regex (6489 words)
Cover Pages: Regular Language Description for XML (RELAX) RELAX (REgular LAnguage description for XML) "is a specification for describing XML-based languages. RELAX (REgular LAnguage description for XML) is a namespace-aware specification for describing XML-based languages which borrows rich datatypes from XML Schema Part 2; the RELAX grammar can be written as an XML document. Although the XMLNews-Story markup language has been superseded by the News Industry Text Format, I've chosen it because it's simple, quite widely used, looks a great deal like HTML, and its RELAX specification will use most of the features we want to focus on. www.oasis-open.org /cover/relax.html (2827 words)
English Language School-Regular Programs (Site not responding. Last check: 2007-10-30) The Primary Mission of the intensive English program of the English Language Institute is to prepare international students for study at the graduate or undergraduate level in institutions of higher learning in the USA. The ELI also provides intensive language instruction to persons already trained in their professions and others who wish to improve their level of English proficiency and cultural awareness. For these reasons, it is the policy of the English Language Institute that English be used in all communications among students as well as students and employees, and we assume that, by applying for admission, you accept our rule on the use of English. www.eli.ufl.edu /firsttime/regular.htm (776 words)
Talk:Regular language The word regular is used because these expressions are built up using very distinct limited and regulated syntax rules. I think one could argue that context-free languages are also built according to limited syntax rules. I'm pretty sure the word "regular" was chosen without much thinking; it's one of those meaningless terms, like "normal", that are used all over the place. www.termsdefined.net /ta/talk:regular-language.html (357 words)
[No title] Theorem 1.13 47 The class of regular languages is closed under the concatenation operation. Theorem 1.22 59 The class of regular languages is closed under the union operation. Theorem 1.23 60 The class of regular languages is closed under the concatenation operation. www.geneffects.com /briarskin/860/thms.txt (1936 words)
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About us | Why use us? | Reviews | Press | Contact us | 3,195 | 14,576 | {"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-2018-26 | latest | en | 0.815346 |
http://mathoverflow.net/questions/13116/c-sieve-of-atkin-overlooks-a-few-prime-numbers | 1,469,350,173,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823989.0/warc/CC-MAIN-20160723071023-00062-ip-10-185-27-174.ec2.internal.warc.gz | 169,058,885 | 14,957 | # C++ Sieve of Atkin overlooks a few prime numbers [closed]
Hello All!
Recently I've been working on a C++ prime generator that uses the Sieve of Atkin ( http://en.wikipedia.org/wiki/Sieve_of_atkin ) to generate its primes. My objective is to be able to generate any 32-bit number. I'll use it mostly for project euler problems. mostly it's just a summer project.
The program uses a bitboard to store primality: that is, a series of ones and zeros where for example the 11th bit would be a 1, the 12th a 0, and the 13th a 1, etc. For efficient memory usage, this is actually and array of chars, each char containing 8 bits. I use flags and bitwise-operators to set and retrieve bits. The gyst of the algorithm is simple: do a first pass using some equations I don't pretend to understand to define if a number is considered "prime" or not. This will for the most part get the correct answers, but a couple nonprime numbers will be marked as prime. Therefore, when iterating through the list, you set all multiples of the prime you just found to "not prime". This has the handy advantage of requiring less processor time the larger a prime gets.
I've got it 90% complete, with one catch: some of the primes are missing.
Through inspecting the bitboard, I have ascertained that these primes are omitted during the first pass, which basically toggles a number for every solution it has for a number of equations (see wikipedia entry). I've gone over this chunk of code time and time again. I even tried increasing the bounds to what is shown in the wikipedia articles, which is less efficient but I figured might hit a few numbers that I have somehow omitted. Nothing has worked. These numbers simply evaluate to not prime. Most of my test has been on all primes under 128. Of this range, these are the primes that are omitted:
23 and 59.
I have no doubt that on a higher range, more would be missing (just don't want to count through all of them). I don't know why these are missing, but they are. Is there anything special about these two primes? I've double and triple checked, finding and fixing mistakes, but it is still probably something stupid that I am missing.
anyways, here is my code:
#include <iostream>
#include <limits.h>
#include <math.h>
using namespace std;
const unsigned short DWORD_BITS = 8;
unsigned char flag(const unsigned char);
void printBinary(unsigned char);
class PrimeGen
{
public:
unsigned char* sieve;
unsigned sievelen;
unsigned limit;
unsigned bookmark;
PrimeGen(const unsigned);
void firstPass();
unsigned next();
bool getBit(const unsigned);
void onBit(const unsigned);
void offBit(const unsigned);
void switchBit(const unsigned);
void printBoard();
};
PrimeGen::PrimeGen(const unsigned max_num)
{
limit = max_num;
sievelen = limit / DWORD_BITS + 1;
bookmark = 0;
sieve = (unsigned char*) malloc(sievelen);
for (unsigned i = 0; i < sievelen; i++) {sieve[i] = 0;}
firstPass();
}
inline bool PrimeGen::getBit(const unsigned index)
{
return sieve[index/DWORD_BITS] & flag(index%DWORD_BITS);
}
inline void PrimeGen::onBit(const unsigned index)
{
sieve[index/DWORD_BITS] |= flag(index%DWORD_BITS);
}
inline void PrimeGen::offBit(const unsigned index)
{
sieve[index/DWORD_BITS] &= ~flag(index%DWORD_BITS);
}
inline void PrimeGen::switchBit(const unsigned index)
{
sieve[index/DWORD_BITS] ^= flag(index%DWORD_BITS);
}
void PrimeGen::firstPass()
{
unsigned nmod,n,x,y,xroof, yroof;
//n = 4x^2 + y^2
xroof = (unsigned) sqrt(((double)(limit - 1)) / 4);
for(x = 1; x <= xroof; x++){
yroof = (unsigned) sqrt((double)(limit - 4 * x * x));
for(y = 1; y <= yroof; y++){
n = (4 * x * x) + (y * y);
nmod = n % 12;
if (nmod == 1 || nmod == 5){
switchBit(n);
}
}
}
xroof = (unsigned) sqrt(((double)(limit - 1)) / 3);
for(x = 1; x <= xroof; x++){
yroof = (unsigned) sqrt((double)(limit - 3 * x * x));
for(y = 1; y <= yroof; y++){
n = (3 * x * x) + (y * y);
nmod = n % 12;
if (nmod == 7){
switchBit(n);
}
}
}
xroof = (unsigned) sqrt(((double)(limit + 1)) / 3);
for(x = 1; x <= xroof; x++){
yroof = (unsigned) sqrt((double)(3 * x * x - 1));
for(y = 1; y <= yroof; y++){
n = (3 * x * x) - (y * y);
nmod = n % 12;
if (nmod == 11){
switchBit(n);
}
}
}
}
unsigned PrimeGen::next()
{
while (bookmark <= limit)
{
bookmark++;
if (getBit(bookmark))
{
unsigned out = bookmark;
for(unsigned num = bookmark * 2; num <= limit; num += bookmark)
{
offBit(num);
}
return out;
}
}
return 0;
}
inline void PrimeGen::printBoard()
{
for(unsigned i = 0; i < sievelen; i++)
{
if (i % 4 == 0)
cout << endl;
printBinary(sieve[i]);
cout << " ";
}
}
inline unsigned char flag(const unsigned char bit_index)
{
return ((unsigned char) 128) >> bit_index;
}
inline void printBinary(unsigned char byte)
{
unsigned int i = 1 << (sizeof(byte) * 8 - 1);
while (i > 0) {
if (byte & i)
cout << "1";
else
cout << "0";
i >>= 1;
}
}
I did my best to clean it up and make it readable. I'm not a professional programmer, so please be merciful.
Here is the output I get, when I initialize a PrimeGen object named pgen, print its initial bitboard with pgen.printBoard() (please note that 23 and 59 are missing before next() iteration), and then iterate through next() and print all of the returned primes:
00000101 00010100 01010000 01000101
00000100 01010001 00000100 00000100
00010001 01000001 00010000 01000000
01000101 00010100 01000000 00000001
5
7
11
13
17
19
29
31
37
41
43
47
53
61
67
71
73
79
83
89
97
101
103
107
109
113
127
DONE
Process returned 0 (0x0) execution time : 0.064 s
Press any key to continue.
-
## closed as no longer relevant by Pete L. Clark, Loop Space, Michael Lugo, Scott Morrison♦Jan 27 '10 at 17:26
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
Closed. It seems SpaceMunkee has found the problem himself, but I'm closing it anyway on the principle that the question wasn't appropriate here -- if it were about the underlying math of the implementation, fine, a long code posting might have been okay, but "please debug this" is not a valid question on mathoverflow, regardless of the underlying mathematics. – Scott Morrison Jan 27 '10 at 17:28
## 1 Answer
Eureka!!!
As expected, it was a stupid error on my part.
The 3x^2 - y^2 equation has a small caveat that I overlooked: x > y. With this taken into account, I was switching 23 and 59 too many times, leading to them failing.
Sorry for posting, hitting my head against the wall for bad math practices.
- | 1,850 | 6,786 | {"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} | 2.765625 | 3 | CC-MAIN-2016-30 | latest | en | 0.954697 |
http://www.downloadmela.com/tcs-nbspplacement-papernbspnbspnbspgeneral-othernbspnbspnbsp-1-jan-2004-11%3Cfd%3E2xAvG%2FZ5XtGIwNbXse9Q0w%3D%3D | 1,555,629,520,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00196.warc.gz | 227,700,318 | 9,558 | TCS Placement Paper General - Other -1 Jan 2004
#### TCS Placement Paper General - Other -1 Jan 2004
• Posted by FreshersWorld
7 Jan, 2012
Test Pattern
1. VOCABULARY.(SYNONYMS) 40 BITS. 20 MARKS. 20MIN.
2. QUANTITATIVE APTITUDE 15 BITS 30 MARKS 15
MIN.
3. CRITICAL REASIONING.(COMPREHENSION) 50 BITS 50 MARKS 25
MIN.
4. PSYHOMETRIC TEST. 150 BITS 150 MARKS 30 MIN
PART II QUANTITATIVE APTITUDE .
-----------------------------------------------------------------
1. Two pencils costs 8 cents, then 5 pencils cost how much
(Ans:20 cents).
2. A work is done by the people in 24 min. one of them can do
this work a lonely in 40 min. how much time required to do the
same
work for the second person.
(ans:60 min.)
3. A car is filled with four and half gallons of oil for full
round
trip. fuel is taken 1/4 gallons mor3 in going than coming. what is
the fiel consumed in coming up? (2 gallons)
4. low temperature at the night in a city is 1/3 more than 1/2
hinge as higher temperature in a day. sum of the low temp and
higherst temp is 100C. then what is the low temperature (40 C)
5. A person who decided to go weekend trip should not exceed 8
hours
driving in a day Average speed of forward journy is 40 mph. due
to
traffic in sundays, the return journey average speed is 30 mph.
how far he can select a picnic spot (120 miles).
6. A sales person multiplied a number and get the answer is 3,
instead of that number divided by 3. what is th answer he actually
has to get ? (1/3).
7. A ship started from port and moving with I mph and another ship
started from L and moving with H mph. At which place these two
ships
meet ? ( Ans is between I and J and close to J)
!_____!_____!_____!_____!_____!_____!
port G H I J K L
8. A building with hight D ft shadow upto G A neighbour building
with
what height shadow C ft is (B ft.)
!_____!_____!_____!_____!_____!_____!_____!
A B C D E F G H
9. A person was fined for exceeding the speed limit by 10
mph.Another
person was also fined for exceeding the same speed limit by twice
the same. If the second person was travelling at a speed of 35
mph.
find the speed limit (15 mph)
10. A bus started from bustand at 8.00a m and after 30 min staying
at
destination, it returned back to the bustand. the destination is
27
miles from the bustand. the speed of the bus 50 percent fast
speed.
at what time it retur4ns to the bustand (11.00)
11.in a mixture, R is 2 parts, S is 1 part. in order to make S to
25%
of the mixture, howmuch R is to be added ( one part).
12. wind flows 160 miles in 330 min, for 80 miles how much time
required.
13. with 4/5 full tank vehicle travels 12 miles, with 1/3 full
tank
how much distance travels ( 5 miles).
14. two trees are there. one grows at 3/5 of the other. in 4
years,
total growth of trees is 8 ft. what growth will smaller tree will
have in 2 years. (<2ft)
15. A storm will move with a velocity of towords the center in
hours. At the same rate how much far will it move in hrs.
(but Ans is 8/3 or 2 2/3).
PART III:
-----------------------------------------------
1. My father has no brothers. he has three sisters who has two childs
each.
1 my grandfather has two sons (f)
2 three of my aunts have two sons(can't say)
3 my father is only child to his father(f)
4 i have six cousins from my mother side(f)
5 i have one uncle(f)
2. Ether injected into gallablader to dissolve galstones. this type
oneday
treatment is enough for gallstones not for calcium stones. this method is
alternative to surgery for millions of people who are suffering from this
disease.
1 calcium stones can be cured in oneday (f)
2 hundreds of people contains calcium stones(can't say)
3 surgery is the only treatment to calcium stones(t)
4 Eather will be injected into the gallbleder to cure the
cholestrol
based gall stones(t).
3. Hacking is illigal entry into other computer. this is done mostly
because of lack of knowledge of computer networking with networks one
that each network is accredited to use network facility.
1 Hacking people never break the code of the company which they
work for (can't say).
2 Hacking is the only vulnerability of the computers for the
usage
of the data.(f)
3 Hacking is done mostly due to the lack of computer knowledge
(f).
(there will be some more questions in this one )
4. alphine tunnels are closed tunnels. in the past 30 yrs not even a
single
accident has been recorded for there is one accident in the rail road
system. even in case of a fire accident it is possible to shift the
passengers
into adjacent wagons and even the living fire can be detected and
extinguished
with in the duration of 30 min.
1 no accident can occur in the closed tunnels (True)
2 fire is allowed to live for 30 min. (False)
3 All the care that travel in the tunnels will be carried by rail
shutters.(t)
4
5. In the past helicopters are forced to ground or crash because of the
formation of the ice on the rotors and engines. a new electronic device
has
been developed which can detect the watercontent in the atmosphere and
warns
the pilot if the temp.is below freezing temp. about the formation of the
ice
on the rotors and wings.
1 the electronic device can avoid formation of the ice on the
wigs (False).
2 There will be the malfunction of rotor & engine because of
fomation of ice (t)
3 The helicopters are to be crashed or down (t)
4 There is only one device that warn about the formation of
ice(t).
6.In the survey conducted in mumbai out of 63 newly married house wives
not a single house
wife felt that the husbands should take equal part in the household work
as
they felt they loose their power over their husbands. inspite of their
careers
they opt to do the kitchen work themselves after coming back to home. the
wives get half as much leisure time as the husbands get at the week ends.
1 housewives want the husbands to take part equally in the
household(f)
2 wives have half as much leisure time as the husbands have(f)
3 39% of the men will work equally in the house in cleaning and
washing
3
7. copernicus is the intelligent. In the days of copernicus the transport
and technology
development was less & it took place weeks to comunicate a message at
that
time.wherein we can send it through satellite with in no time ----------.
even with this fast developments it has become difficult to understand
each other.
1 people were not intelligent during Copernicus days (f).
2 Transport facilities are very much improved in noe a days (can'
say)
3 Even with the fast developments of the techonology we can't
live happily.(can't say)
4 We can understand the people very much with the development of
communication(f).
Q8) senior managers warned the workers that because of the
intfoductors
of japanese industry in the car market. There is the threat to the
workers.
They also said that there will be the reduction in the purchase of
the sales of
car in public.the interest rates of the car will be increased with
the loss in demand.
1 japanese workers are taking over the jobs of indian
industry.(false)
2 managers said car interests will go down after seeing the raise
in
interest rates.(true)
3 japanese investments are ceasing to end in the car
industry.(false)
4 people are very much interested to buy the cars.(false)
Q9) In the totalitariturican days,the words have very much
devalued.In
the present day,they are becoming domestic that is the words will be
much more devalued.
In that days, the words will be very much effected in political
area.but
at present,the words came very cheap .we can say they come free at
cost.
1 totalitarian society words are devalued.(false)
2 totalitarian will have to come much about words(t)
3 The art totalitatian society the words are used for the
political
speeches.
4
Q10) There should be copyright for all arts. The reele has came
that
all the
arts has come under one copy right society,they were use the money
that come from
the arts for the developments . There may be a lot of money will
come from
the Tagore works. We have to ask the benifiters from Tagore work
to
help for
the development of his works.
1 Tagore works are came under this copy right rule.(f)
2 People are free to go to the because of the copy right
rule.(can't say)
3 People gives to theater and collect the money for
development.(can't say)
4 We have ask the Tagore resedents to help for the developments
of
art.(can't say) | 2,161 | 8,354 | {"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} | 3.078125 | 3 | CC-MAIN-2019-18 | latest | en | 0.893011 |
https://ccssmathanswers.com/mcgraw-hill-my-math-grade-4-chapter-7-lesson-3-answer-key/ | 1,723,676,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00350.warc.gz | 130,792,044 | 57,975 | # McGraw Hill My Math Grade 4 Chapter 7 Lesson 3 Answer Key Sequences
All the solutions provided in McGraw Hill Math Grade 4 Answer Key PDF Chapter 7 Lesson 3 Sequences will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 4 Answer Key Chapter 7 Lesson 3 Sequences
Math in My World
Example 1
Crystal starts reading her book on Monday. She reads 25 pages on the first day. Each day, she reads 25 pages. How many total pages will she have read by Tuesday, Wednesday, Thursday, and Friday?
The first term of the sequence is 25.
Extend the pattern.
So, Crystal will have read ____ pages by Tuesday, ___ pages by Wednesday, ___ pages by Thursday, and ___ pages by Friday.
So, Crystal will have read 50 pages by Tuesday, 75 pages by Wednesday, 100 pages by Thursday, and 125 pages by Friday.
Explanation:
Number of pages on the first day she reads = 25.
Number of pages each day she reads = 25.
Number of pages each day she reads on Monday = Number of pages on the first day she reads = 25.
Number of pages each day she reads on Tuesday = Number of pages each day she reads on Monday + Number of pages each day she reads
= 25 + 25
= 50.
Number of pages each day she reads on Wednesday = Number of pages each day she reads on Tuesday + Number of pages each day she reads
= 50 + 25
= 75.
Number of pages each day she reads on Thursday = Number of pages each day she reads on Wednesday + Number of pages each day she reads
= 75 + 25
= 100.
Number of pages each day she reads on Friday = Number of pages each day she reads on Thursday + Number of pages each day she reads
= 100 + 25
= 125.
Example 2
The first term of a sequence is 65. The rule of the sequence is subtract 4. Find the next four terms in the sequence. Then make observations about the pattern.
1. Find the next four terms.
The next four terms in the sequence are ___, ____, ___, and ____
2. Make observations about the pattern.
Circle whether the terms are all odd or even. odd even
Circle whether the terms increase or decrease. increase decrease
Extend the pattern to a total of 10 terms.
65, ____, ___, ___, ___, ___, ___, ___, ___, ___
Make another observation about the pattern.
The ones digits repeat the pattern 5, 1, ___, ____, and ____
Explanation:
The first term of a sequence is 65.
The rule of the sequence is subtract 4.
Next four terms in the sequence:
61 – 4 = 57.
57 – 4 = 53.
53 – 4 = 49.
49 – 4 = 45.
Next Six terms in the sequence:
45 – 4 = 41.
41 – 4 = 37.
37 – 4 = 33.
33 – 4 = 29.
29 – 4 = 25.
25 – 4 = 21.
The pattern to a total of 10 terms: 65, 57, 53, 49, 45, 41, 37, 33, 29, 25, 21.
The ones digits repeat the pattern 5, 1, 7, 3, and 9.
Talk Math
How does the operation of a rule affect the terms of a sequence?
The operation of a rule affect the terms of a sequence as it helps to find any term in the problem.
Explanation:
An operation is a mathematical action. Addition, subtraction, multiplication, division, and calculating the root are all examples of a mathematical operation
Guided Practice
Extend each pattern by four terms. Write an observation about the pattern.
Question 1.
Pattern: 8, ___, ___, ___, ___
Observation: ________
Pattern: 8, 15, 22, 29, 36.
Observation: It is observed that four numbers are prime and 1 is composite number.
Explanation:
8 + 7 = 15.
15 + 7 = 22.
22 + 7 = 29.
29 + 7 = 36.
Pattern: 8, 15, 22, 29, 36.
Question 2.
Rule: subtract 10
Pattern: 90, ___, ___, ___, ____
Observation: __________
Pattern: 90, 80, 70, 60, 50.
Observation: It is observed the numbers are Multiplies of 10.
Explanation:
Rule: subtract 10
90 – 10 = 80.
80 – 10 = 70.
70 – 10 = 60.
60 – 10 = 50.
Pattern: 90, 80, 70, 60, 50.
### McGraw Hill My Math Grade 4 Chapter 7 Lesson 3 My Homework Answer KeyPractice
Extend each pattern by four terms. Write an observation about the pattern.
Question 1.
Pattern: 5, ___, ___, ___, ____
Observation: ________
Pattern: 5, 13, 21, 29, 37.
Observation: It is observed all numbers are prime numbers.
Explanation:
5 + 8 = 13.
13 + 8 = 21.
21 + 8 = 29.
29 + 8 = 37.
Pattern: 5, 13, 21, 29, 37.
Question 2.
Rule: multiply by 2
Pattern: 3, ___, ___, ___, ____
Observation: __________
Pattern: 3, 6, 12, 24, 48.
Observation: It is observed all numbers are even numbers.
Explanation:
Rule: multiply by 2
3 × 2 = 6.
6 × 2 = 12.
12 × 2 = 24.
24 × 2 = 48.
Question 3.
Rule: subtract 20
Pattern: 175, ___, ___, ____
Observation: ________
Pattern: 175, 155, 135, 115, 95.
Observation: It is observed all numbers are multiplies of 5.
Explanation:
Rule: subtract 20
175 – 20 = 155.
155 – 20 = 135.
135 – 20 = 115.
115 – 20 = 95.
Question 4.
Extend the pattern below by four terms. Write an observation about the pattern.
Rule: multiply by 10
Pattern: 26, ___, ___, ____, ____
Observation: ___________
Pattern: 26, 260, 2600, 26000, 260000.
Observation: It is observed all numbers are even and are increasing.
Explanation:
Rule: multiply by 10
26 × 10 = 260.
260 × 10 = 2,600.
2,600 × 10 = 26,000.
26,000 × 10 = 2,60,000.
Problem Solving
Question 5.
Mathematical PRACTICE 8 Look for a Pattern Brad puts an equal amount of money in his savings account once a month. He started with $25. The next month, he had$35 in his account. Two months after that, he had $55 in his account. How much money will Brad have in his account after 6 months? Describe a rule. Then solve. Answer: Amount of savings the next six months he had in his account =$175.
Explanation:
Amount of savings he started = $25. Amount of savings the next month he had in his account =$35 .
Amount of savings on three months he had in his account = $55. Amount of savings in a month he did = Amount of savings the next month he had in his account – Amount of savings he started =$35 – $25 =$10.
Amount of savings the next three months he had in his account = Amount of savings in two month he did + (Number of months + Number of months)
= $35 + (2 × Amount of savings in a month he did) =$35 + (2 × $10) =$35 + $20 =$55.
Amount of savings the next four months he had in his account = Amount of savings in three months he did + (Number of months + Amount of savings in a month he did)
= $55 + (3 ×$10)
= $55 +$30
= $85. Amount of savings the next five months he had in his account = Amount of savings in four months he did + (Number of months + Amount of savings in a month he did) =$85 + (4 × $10) =$85 + $40 =$125.
Amount of savings the next six months he had in his account = Amount of savings in five months he did + (Number of months + Amount of savings in a month he did)
= $125 + (5 ×$10)
= $125 +$50
= \$175.
Question 6.
On Monday, a toy store sold 4 race cars. On Tuesday, it sold 8 race cars. On Wednesday, it sold 16 race cars. . Suppose this pattern continues. How many race cars will be sold on Friday? Describe a rule. Then solve.
Number of race cars a toy store sold on Friday = 64.
Explanation:
Number of race cars a toy store sold on Monday = 4.
Number of race cars a toy store sold on Tuesday = 8.
Number of race cars a toy store sold on Wednesday = 16
Number of race cars a toy store sold on Thursday = Number of race cars a toy store sold on Wednesday × 2
= 16 × 2
= 32.
Number of race cars a toy store sold on Friday = Number of race cars a toy store sold on Thursday × 2
= 32 × 2
= 64.
Vocabulary Check
Write a vocabulary word to complete each sentence.
sequence term
Question 7.
Each number in a numeric pattern is a ____
Each number in a numeric pattern is a term.
Explanation:
A term is a single mathematical expression. It may be a single number (positive or negative), a single variable ( a letter ), several variables multiplied but never added or subtracted. Some terms contain variables with a number in front of them.
Question 8.
A ____ is the ordered arrangement of terms that make up a pattern.
A sequence is the ordered arrangement of terms that make up a pattern.
Explanation:
A Sequence is a list of things (usually numbers) that are in order.
Test Practice
Question 9.
Identify the next term in the sequence. 171, 141, 111, 81, ____
A. 61
B. 51
C. 41
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https://codedump.io/share/pnluaPhIrvs5/1/check-if-data-follows-planned-order | 1,529,762,690,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00014.warc.gz | 571,984,815 | 11,119 | Felipe Deguchi - 1 year ago 77
SQL Question
Check if data follows planned order
So I'm having a hard time trying solve this.
I currently have a Material production plan like this (each row is a batch):
``````SELECT
Material,
Quantity,
Range,
OrderBy
FROM
ProductionPlan
ORDER BY
OrderBy
+----------+----------+-------+---------+
| Material | Quantity | Range | OrderBy |
+----------+----------+-------+---------+
| A | 120 | 5 | 1 |
| B | 120 | 5 | 2 |
| A | 120 | 5 | 3 |
| C | 120 | 10 | 4 |
| A | 120 | 5 | 5 |
| A | 120 | 5 | 6 |
+----------+----------+-------+---------+
``````
And our actual production data looks like this:
``````SELECT
Material,
Quantity,
BatchNm
FROM
ProducedMaterials
ORDER BY
BatchNm
+----------+----------+---------+
| Material | Quantity | BatchNm |
+----------+----------+---------+
| A | 120 | 101 |
| B | 113 | 102 |
| C | 111 | 103 |
| A | 353 | 104 |
+----------+----------+---------+
``````
What I need to know is if each planned Material was achieved by checking a couple of things, if it is it should show GOOD, otherwise show BAD:
• If the produced quantity is equal the planned quantity (+-Range)
• Was produced the same order as planned.
When the order is broken: It keeps looking for the produced material.
In the example he was looking for C, but the correct order was A. It shows a BAD for that row and looks at the next row if the material matches. Until the produced material matches it would keep showing BAD for the planned quantities;
So I should end with something like this:
``````+----------+----------+-------+---------+--------+
| Material | Quantity | Range | OrderBy | Status |
+----------+----------+-------+---------+--------+
| A | 120 | 5 | 1 | GOOD | <-- Was produced first and quantity is within range
| B | 120 | 5 | 2 | BAD | <-- The Produced quantity(113) is not withing planned range
| A | 120 | 5 | 4 | BAD | <-- Bad because it didn't follow the plan (Suposed to be Material C)
| C | 120 | 10 | 3 | GOOD | <-- Good because it IS the next produced material AND it's quantity matched the planned quantity
| A | 120 | 5 | 5 | GOOD | <-- Good because it matches the next planned material AND the quantity is withing range(because the next row is the same material)
| A | 120 | 5 | 6 | BAD | <-- Bad because even thought the planned order is ok (Same material as the above row), the remaining quantity is above the quantity range (353 - 120(from the above row) = 233 "remaining" material)
+----------+----------+-------+---------+--------+
``````
P.S. I'm not on my dev machine right now. Once I get to that, I will post the function I was working on;
The following solution is a bit long winded but hopefully it will do the trick. I have not tested this approach on all possible scenarios so chances of mistakes in code is still good. The query will not properly handle the production of material that was never on the plan in the first place. Example if actual production included ‘D’ but ‘D’ is not in the plan.
``````DECLARE @ProductionPlan TABLE
(
[Order] BIGINT NOT NULL IDENTITY(1,1) PRIMARY KEY CLUSTERED -- Use identity column to define order of plan
,Material CHAR(1) not null
,Quantity INT not null
,TolerenceRange INT not null
)
DECLARE @ProducedMaterials TABLE
(
[Order] BIGINT NOT NULL IDENTITY(1,1) PRIMARY KEY CLUSTERED -- Use identity column to define order of actual production
,Material CHAR(1) not null
,QuantityProduced INT not null
)
INSERT @ProductionPlan (Material, Quantity, TolerenceRange)
VALUES ('A',120,5)
,('B',120,5)
,('A',120,5)
,('C',120,10)
,('A',120,5)
,('A',120,5)
,('A',120,5);
INSERT @ProducedMaterials (Material,QuantityProduced)
VALUES ('A',120)
,('B',113)
,('C',111)
,('A',353);
WITH CTE_PlanAndActual
AS
(
-- Join actual production with scheduled production on Material type only. For each possible
-- combination the query will calculate the “Distance” in order of execution between the actual
-- production step and the planned production steps.
SELECT [Plan].[Order] AS [PlanedOrder]
,[Actual].[Order] AS [ActualOrder]
,[Plan].[Material]
,ABS([Plan].[Order] - [Actual].[Order]) AS [Distance] -- Distance between actual production order and planned production order.
,[Plan].[Quantity] AS [PlannedQuantity]
,[Actual].[QuantityProduced] AS [ActualQuantity]
,[Plan].[TolerenceRange]
FROM @ProductionPlan [Plan]
LEFT OUTER JOIN @ProducedMaterials [Actual] ON [Actual].[Material] = [Plan].[Material]
),
CTE_PlanAndActualBestMatch
AS
(
--Next step we will use windowing function to determine the minimum distance between planned production step and actual production step.
--This will help us determine the best match with the information we have thus far.
SELECT [PlanedOrder]
,[ActualOrder]
,[Material]
,[Distance]
,MIN([Distance]) OVER (PARTITION BY [PlanedOrder]) AS [MinDistance]
,[PlannedQuantity]
,[ActualQuantity]
,[TolerenceRange]
FROM CTE_PlanAndActual
),
CTE_PlanAndActualOrderValidated
AS
(
-- Next eliminate records which does not meet the minimum distance criteria for each planned production step.
-- Now that we have only the records that matches the minimum distance criteria we need to determine if any
-- of the actual production execution steps was out of order. We will use the LEAD windowing function to determine this.
SELECT [PlanedOrder]
,(
-- If one or more step is out of order, then it means the production plan was not followed. Simply set the Actual order value for the record to null.
CASE
WHEN [ActualOrder] > LEAD([ActualOrder], 1, [ActualOrder]) OVER (ORDER BY [PlanedOrder]) THEN NULL
WHEN ( ([PlanedOrder] = 1) AND ([ActualOrder] <> 1) ) THEN NULL
ELSE [ActualOrder]
END
) [ActualOrder]
,[Material]
,[Distance]
,[PlannedQuantity]
,[ActualQuantity]
,[TolerenceRange]
FROM CTE_PlanAndActualBestMatch
WHERE [MinDistance] = [Distance] -- Eliminate records that is not the minimum distance between plan and actual.
),
CTE_PlanAndActualWithRepeats
AS
(
-- Next determine repeated planned orders this will be needed to correctly determine if the
-- production quantiles were within in planned tolerance range.
-- Also calculate the Cumulative Planed Quantity for planned entries that repeat, this will
-- be needed to determine if repeated production entries are within tolerance range.
SELECT [PlanedOrder]
,[ActualOrder]
,[Material]
,[Distance]
,[PlannedQuantity]
,IIF([ActualOrder] IS NULL, NULL, [ActualQuantity]) AS [ActualQuantity]
,[TolerenceRange]
,IIF([ActualOrder] IS NULL, 0, 1) AS PlanFollowed
,COUNT([PlanedOrder]) OVER (PARTITION BY [ActualOrder]) AS RepeatCount
,ROW_NUMBER() OVER (PARTITION BY [ActualOrder] ORDER BY [PlanedOrder]) AS RepeatIndex
,SUM([PlannedQuantity]) OVER (PARTITION BY [ActualOrder] ORDER BY [PlanedOrder]) AS [CumulativePlanedQuantity]
FROM CTE_PlanAndActualOrderValidated
)
,
CTE_PlanAndEffectiveProduction
AS
(
-- Calculate the effective production. In the event that production plan entry repeats the
-- effective production will use, the final effective production value will be calculated
-- from total actual production and cumulative planned production.
SELECT [PlanedOrder]
,[ActualOrder]
,[Material]
,[Distance]
,[PlannedQuantity]
,[ActualQuantity]
,[TolerenceRange]
,[PlanFollowed]
,[RepeatCount]
,[RepeatIndex]
,[CumulativePlanedQuantity]
,(
CASE
WHEN ([RepeatIndex] < [RepeatCount]) AND ([CumulativePlanedQuantity] < [ActualQuantity]) THEN [PlannedQuantity]
WHEN ([RepeatIndex] > 1) AND ([RepeatIndex] = [RepeatCount]) THEN [ActualQuantity] - ([CumulativePlanedQuantity] - [PlannedQuantity])
ELSE [ActualQuantity]
END
) AS EffectiveQuantity
FROM CTE_PlanAndActualWithRepeats
)
-- Finally determine status
SELECT [PlanedOrder]
,[ActualOrder]
,[Material]
,[Distance]
,[PlannedQuantity]
,[ActualQuantity]
,[TolerenceRange]
,[PlanFollowed]
,[RepeatCount]
,[RepeatIndex]
,[CumulativePlanedQuantity]
,[EffectiveQuantity]
,(
CASE
WHEN ([PlanFollowed] = 1) AND (ABS([EffectiveQuantity] - [PlannedQuantity]) <= [TolerenceRange]) THEN 'Good'
WHEN ([PlanFollowed] = 1) AND (ABS([EffectiveQuantity] - [PlannedQuantity]) > [TolerenceRange]) THEN 'Bad - Out of Range'
WHEN ([PlanFollowed] = 0) THEN 'Bad - Plan Not Followed' | 2,206 | 8,493 | {"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-2018-26 | latest | en | 0.831269 |
http://www.chicagotribune.com/suburbs/daily-southtown/lifestyles/ct-stn-ahern-column-st-0419-20150412-story.html | 1,495,537,268,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607620.78/warc/CC-MAIN-20170523103136-20170523123136-00322.warc.gz | 445,619,505 | 47,446 | # St. Barnabas students learn quick ways to solve math problems
Daily Southtown
'Brainetics' creator shows St. Barnabas students quick ways to compute math problems
Here's a math test: Without a calculator and without writing down any numbers, multiply 105 by 105, and by the way, have the answer in mind by the end of this sentence.
Solve 208 times 208, and have that answer by the end of this sentence.
Did the multiplying intimidate? Did the speed make it difficult? It is likely that most people would find it somewhat daunting to multiply two sets of three digit numbers, but children in fourth through eighth grades at St. Barnabas School in the Beverly community of Chicago recently witnessed these calculations, and more, done at lightning fast speeds.
The speedy math was performed by Mike Byster, who refers to himself as one of the world's fastest mathematical minds. Byster said that growing up, math was not his best subject.
"Math was my worst subject," Byster told the St. Barnabas children. "I was not born this way. I got one or two As in math, but I also got Cs, Ds, and an F in geometry."
Byster explained, however, that he wanted very much to improve his math skills, so he worked very hard, and began to recognize patterns in numbers. The three digit multiplications with a zero in the middle could be solved by following a special pattern that he taught the children, and in no time, he was throwing out challenges to the children, who responded quickly, giving correct answers in a matter of seconds.
Patterns, according to Byster, worked for other learning efforts as well. During his grade school years, he was given the assignment to memorize all the planets of the solar system.
"My very educated mother just served us nine pizzas" was the sentence that helped him remember the planets, because the first letter of each word to that sentence is the first letter for a planet. (Byster quickly pointed out to the children that in his childhood, Pluto was still considered a planet).
When Byster had to learn state capitals to all 50 states, he said he stared at the states and capitals over and over again. He was not able to remember everything until his mother suggested that the two of them make up a song to remember the capitals.
Similarly, he made up a song to remember all the presidents of the United States, and easily remembered those facts as well. Patterns became Byster's method of doing math, and memory tools became his way to remember facts.
As an example, Byster asked the children to remember a list of these 10 items: a silver tray, goblets, soap, a banana peel, dental floss, a loaf of bread, a red rose, a tomato, chocolate pudding and laundry detergent. After reading the list to the students, he asked them to write as many items down as they could remember. On average, the children remembered about five items.
When Byster told them a silly story, though, that included silly details such as falling into a tub full of chocolate pudding, or using dental floss to pull himself up a hill, the children's average went up significantly, and many children remembered between eight and all 10 items.
Mary Wolf, a volunteer with the St. Barnabas Family and School Association, said the decision to have Byster speak came as the result of recommendations from parents, as well as feedback from school administrators describing goals for the students. She and her fellow volunteer, Megan Curran-Hurless, worked hard to get Byster to come.
"He was long ago referred to me, but he was hard to get," Wolf said. "Megan finally got him, and it was good to have him come and maybe make math less daunting. The students might think, 'If this guy can do this kind of work, then why can't I?' I think he opened their eyes to what they can do.
"I think he gave the children a 'Hey, this is cool. I can do this' attitude."
Byster agreed with Wolf, and encouraged the children to try his pattern method of learning, as well as his memory tools.
"Any brain can do this," Byster said. He added that his methods do not come without effort, but for the most part, he insisted that the effort to learn patterns, or create stories to remember things, were methods for learning that could change the students' lives.
After showing his method to the students, Byster gave the children many math problems. Hands raised up quickly in an effort to be the first with the answer.
Eighth-grade student David Wolf said he enjoyed the program and was eager to try some of the methods that Byster taught.
"Mike Byster showed me a new, fun way to find the answer to math problems," David said. "With these new skills, I will help others and myself in furthering education."
Byster, who always gives presentations at school for free, calls his system "Brainetics" and sells his method over the Internet.
"I think this is a great thing for the kids to be exposed to," Mary Wolf said. "It made math less scary." | 1,067 | 4,942 | {"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} | 3.546875 | 4 | CC-MAIN-2017-22 | latest | en | 0.987946 |
https://gitlab.kwant-project.org/solidstate/lectures/commit/513bb80af067fe15910d0cd9da9c3d23209df455 | 1,601,229,753,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00790.warc.gz | 425,909,976 | 15,210 | Commit 513bb80a by Iacopo Bertelli
### Update src/7_tight_binding.md
parent c13f346d
Pipeline #15937 passed with stages
in 1 minute and 45 seconds
... ... @@ -261,14 +261,19 @@ Also a sanity check: when the energy is close to the bottom of the band, \$E = E_ 1. What is a normal mode? What is a phonon? Why do phonons obey Bose statistics? 2. From the dispersion relation of a 1D monatomic chain given in the lecture notes, calculate the group velocity \$v_g\$ and density of states \$g(\omega)\$. ??? hint To deal easily with the absolute value, you can split the dispersion for positive and negative \$k\$. 3. Sketch them. 4. From the 1D dispersion relation \$\omega(k)\$ in the picture below, sketch the group velocity \$v_g(k)\$ and the density of states \$g(\omega)\$. 4. From the 1D dispersion relation \$\omega(k)\$ in the picture below, sketch the group velocity \$v_g(k)\$ and the density of states \$g(\omega)\$ as a function of \$k\$.  ### Exercise 2: Vibrational heat capacity of a 1D monatomic chain 1. Give an integral expression for the total energy \$U\$ of a 1D monatomic chain (similarly to what was done within the Debye theory). To do so, first derive the density of states from the appropriate dispersion relation given in the lecture notes. 1. Give an integral expression for the total energy \$U\$ of a 1D monatomic chain (similarly to what was done within the Debye theory). Use the density of state calculated in exercise 1.2. 2. Give an integral expression for the heat capacity \$C\$. 3. Compute the heat capacity numerically, using e.g. MatLab or Python. 4. Do the same for \$C\$ in the Debye model and compare the two. What differences do you see? ... ...
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Finish editing this message first! | 488 | 1,851 | {"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} | 3.53125 | 4 | CC-MAIN-2020-40 | latest | en | 0.792373 |
https://oeis.org/A129506 | 1,576,381,763,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541301014.74/warc/CC-MAIN-20191215015215-20191215043215-00315.warc.gz | 486,900,902 | 5,572 | This site is supported by donations to The OEIS Foundation.
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A129506 Number of partitions of a {2n-1}-set into n nonempty subsets. 7
1, 3, 25, 350, 6951, 179487, 5715424, 216627840, 9528822303, 477297033785, 26826851689001, 1672162773483930, 114485073343744260, 8541149231801585700, 689692892575539953400, 59932861644880019603520, 5576731051262006158950735, 553234633385550257808059085 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS B^{-1}(x) = Sum_{n>0} a(n)/(2*n-1)!*(n-1)! x^n is inverse function for x*B(x), where B(x) is g.f. for Bernoulli number (see A027641). - Vladimir Kruchinin, Jan 19 2012 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..200 D. Kruchinin and V. Kruchinin, A Method for Obtaining Generating Function for Central Coefficients of Triangles, Journal of Integer Sequences, Vol. 15 (2012), article 12.9.3. FORMULA Central Stirling numbers of the second kind: a(n) = A008277(2n-1,n) for n >= 1. G.f.: Sum_{n>=1} n^(2*n-1) * exp(-n^2*x) * x^n / n!, an integer series. - Paul D. Hanna, Oct 15 2012 a(n) = 1/n! * Sum_{k=1..n} (-1)^(n-k) * binomial(n,k) * k^(2*n-1). - Paul D. Hanna, Oct 15 2012 a(n) = ((2*n-1)*((sum(i=1..n-2, (stirling2(2*i-1,i)*C(2*n-2,2*i-1)*stirling2(2*(n-i)-1,n-i-1))/((n-i-1)*C(n-1,i))))+(n-1)*stirling2(2*n-3,n-1) +stirling2(2*n-2,n-1)))/n. - Vladimir Kruchinin, Feb 28 2013 a(n-1) = sum(j=0..n, binomial(2*n,j)*stirling2(2*n-j,n)). - Vladimir Kruchinin, Jun 14 2013 a(n) ~ 2^(2*n-3/2) * n^(n-3/2) * (2-c)^(1-n) / (sqrt(Pi*(1-c)) * exp(n) * c^n), where c = -LambertW(-2*exp(-2)) = 0.4063757399599599... = 2*A106533. - Vaclav Kotesovec, Dec 15 2013 a(n) = A258170(2*n-1,n). - Alois P. Heinz, Mar 16 2018 EXAMPLE G.f.: A(x) = x + 3*x^2 + 25*x^3 + 350*x^4 + 6951*x^5 + 179487*x^6 + ... where A(x) = 1^1*x*exp(-1^2*x) + 2^3*exp(-2^2*x)*x^2/2! + 3^5*exp(-3^2*x)*x^3/3! + 4^7*exp(-4^2*x)*x^4/4! + 5^9*exp(-5^2*x)*x^5/5! + ... forms a power series in x with integer coefficients. - Paul D. Hanna, Oct 15 2012 MAPLE a:= n-> Stirling2(2*n-1, n): seq(a(n), n=1..25); # Alois P. Heinz, Dec 15 2013 MATHEMATICA a[n_] := Sum[ Binomial[2*n - 2, j]*StirlingS2[2*n - j - 2, n-1], {j, 0, n-1}]; Table[a[n], {n, 1, 18}] (* Jean-François Alcover, Jun 14 2013, after Vladimir Kruchinin *) Table[StirlingS2[2*n-1, n], {n, 1, 20}] (* Vaclav Kotesovec, Dec 15 2013 *) PROG (PARI) a(n)=polcoeff(1/prod(k=1, n, 1-k*x +x*O(x^n)), n-1) (PARI) vector(66, n, abs( stirling(2*n-1, n, 2) ) ) /* Joerg Arndt, Jun 09 2012 */ (PARI) {a(n)=1/n!*sum(k=0, n, (-1)^(n-k)*binomial(n, k)*k^(2*n-1))} \\ Paul D. Hanna, Oct 15 2012 (PARI) {a(n)=polcoeff(sum(m=1, n, m^(2*m-1)*x^m*exp(-m^2*x+x*O(x^n))/m!), n)} for(n=1, 20, print1(a(n), ", ")) (Maxima) a(n):=((2*n-1)*((sum((stirling2(2*i-1, i)*binomial(2*n-2, 2*i-1)*stirling2(2*(n-i)-1, n-i-1))/((n-i-1)*binomial(n-1, i)), i, 1, n-2))+(n-1)* stirling2(2*n-3, n-1)+stirling2(2*n-2, n-1)))/(n); makelist(a(n), n, 1, 10); \\ Vladimir Kruchinin, Feb 28 2013 CROSSREFS Cf. A008277, A129505, A217900, A217910, A217913, A258170. Sequence in context: A085527 A093360 A161629 * A143139 A231637 A295765 Adjacent sequences: A129503 A129504 A129505 * A129507 A129508 A129509 KEYWORD nonn AUTHOR Paul D. Hanna, Apr 18 2007 STATUS approved
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Last modified December 14 22:42 EST 2019. Contains 329987 sequences. (Running on oeis4.) | 1,656 | 3,976 | {"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} | 3.28125 | 3 | CC-MAIN-2019-51 | longest | en | 0.58614 |
http://oeis.org/A005217 | 1,618,979,039,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039508673.81/warc/CC-MAIN-20210421035139-20210421065139-00220.warc.gz | 75,116,282 | 4,123 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A005217 Number of unlabeled unit interval graphs with n nodes. (Formerly M1186) 2
1, 2, 4, 9, 21, 55, 151, 447, 1389, 4502, 15046, 51505, 179463, 634086, 2265014, 8163125, 29637903, 108282989, 397761507, 1468063369, 5441174511, 20242989728, 75566702558, 282959337159, 1062523000005, 4000108867555, 15095081362907, 57088782570433 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 REFERENCES S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 5.6.7. R. W. Robinson, personal communication. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). R. W. Robinson, Numerical implementation of graph counting algorithms, AGRC Grant, Math. Dept., Univ. Newcastle, Australia, 1980. LINKS R. W. Robinson, Table of n, a(n) for n = 1..190 Phil Hanlon, Counting interval graphs, Trans. Amer. Math. Soc. 272 (1982), no. 2, 383-426. FORMULA G.f. A(x) = x + 2x^2 + 4x^3 + 9x^4 + 21x^5 + ... satisfies 1 + A(x) = exp( Sum_{k >= 1} psi(x^k)/k ), where psi(x) = (1+2*x-sqrt(1-4*x)*sqrt(1-4*x^2))/(4*sqrt(1-4*x^2)) is the g.f. for A007123. For asymptotics, see for example Finch. MATHEMATICA m = 30; A[x_] = (-1 + Exp[Sum[psi[x^k]/k, {k, 1, m}]] /. psi[x_] -> (1 + 2 x - Sqrt[1 - 4 x] Sqrt[1 - 4 x^2])/(4 Sqrt[1 - 4 x^2])) + O[x]^m; CoefficientList[A[x], x] // Rest (* Jean-François Alcover, Oct 24 2019 *) CROSSREFS Sequence in context: A198304 A032129 A304914 * A148072 A001430 A148073 Adjacent sequences: A005214 A005215 A005216 * A005218 A005219 A005220 KEYWORD nonn AUTHOR STATUS approved
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Last modified April 20 23:46 EDT 2021. Contains 343143 sequences. (Running on oeis4.) | 742 | 2,090 | {"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} | 3.28125 | 3 | CC-MAIN-2021-17 | latest | en | 0.56788 |
https://medium.com/conectric-networks/playing-with-raspberry-pi-traffic-lights-with-a-finite-state-machine-c6dc7035b8a1?source=post_internal_links---------1---------------------------- | 1,620,757,717,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.10/warc/CC-MAIN-20210511153555-20210511183555-00286.warc.gz | 421,497,095 | 35,098 | # Playing with Raspberry Pi: Traffic Lights with a Finite State Machine
Jun 5, 2017 · 4 min read
This is the third article in our series where we’re playing with the Low Voltage Labs LED Traffic Lights using Python on the Raspberry Pi. In the first article, we set up a basic traffic light pattern, then enhanced this in the second article to run the lights in either the UK or USA patterns according to the value of an environment variable.
This time, we’ll look at restructuring the code to use a Finite State Machine, so let’s go…
A Finite State Machine (sometimes known as a Finite State Automaton) is a computing model of a theoretical machine which can be in exactly one state at a given time. The machine will have a known (finite) number of states, with each state having one or more transitions to a new state. Modeling traffic lights is a simple example, but one that is often used in teaching materials to introduce the concept to students.
In order to figure out what our states and transitions are, we can draw out a basic state diagram to visualize them:
From the diagram, we can see that the traffic lights have five distinct states, each with a single transition to the next state, and that the state machine enters an infinite loop:
• INITIALIZING: set up and turn off all the lights
• RED: turn on the red light only, wait, then transition to REDAMBER if using the UK pattern, or GREEN if using the USA pattern. This is the only state in which we have to make a decision about the next state to transition to
• REDAMBER: turn on the red and amber lights only, wait, then transition to GREEN
• GREEN: turn on the green light only, wait, then transition to AMBER
• AMBER: turn on the amber light only, wait, then transition to RED
To make our code read better, we’ll introduce a couple of convenience classes… the first allows us to name all the states:
`class TrafficLightStates: INITIALIZING = 1 RED = 2 REDAMBER = 3 GREEN = 4 AMBER = 5`
And another allows us to give the GPIO pin numbers for each light on the traffic lights more meaningful names:
`class TrafficLightLEDs: RED = 9 AMBER = 10 GREEN = 11`
There are many ways we could implement a Finite State Machine, and there are even libraries for Python and many other languages specifically designed to help with this. For example, Transitions is a Finite State Machine framework for Python.
In this case, we have a very simple set of states and transitions, so we can just use an `if` / `elif` / `else`statement to build out the Finite State Machine. This wouldn’t scale well for more complex models, but is fine for traffic lights.
First up, we need to keep track of which state the machine is in at any given time… we’ll do that with a variable `currentState`. At the beginning, this is set to the first state that the machine enters when “booted”:
`currentState = TrafficLightStates.INITIALIZING`
Then the bulk of our code becomes an infinite loop, in which we:
• Check the value of `currentState`
• Perform actions appropriate for the current state (turn LEDs on and off etc)
• Set the value of `currentState`to whatever the next state in our state diagram is
• Go around the loop again
If we apply the above rules to each of our states, we end up with:
`while True: if (currentState == TrafficLightStates.INITIALIZING): if ('TRAFFIC_LIGHT_COUNTRY' in os.environ) and (os.environ['TRAFFIC_LIGHT_COUNTRY'] in ['UK', 'USA']): pattern = os.environ['TRAFFIC_LIGHT_COUNTRY'].lower() else: print('TRAFFIC_LIGHT_COUNTRY should be set to UK or USA') sys.exit(1) # Setup Hardware GPIO.setmode(GPIO.BCM) GPIO.setup(TrafficLightLEDs.RED, GPIO.OUT) GPIO.setup(TrafficLightLEDs.AMBER, GPIO.OUT) GPIO.setup(TrafficLightLEDs.GREEN, GPIO.OUT) currentState = TrafficLightStates.RED elif (currentState == TrafficLightStates.RED): GPIO.output(TrafficLightLEDs.RED, True) GPIO.output(TrafficLightLEDs.AMBER, False) GPIO.output(TrafficLightLEDs.GREEN, False) time.sleep(3) if pattern == 'uk': currentState = TrafficLightStates.REDAMBER else: currentState = TrafficLightStates.GREEN elif (currentState == TrafficLightStates.REDAMBER): GPIO.output(TrafficLightLEDs.RED, True) GPIO.output(TrafficLightLEDs.AMBER, True) GPIO.output(TrafficLightLEDs.GREEN, False) time.sleep(1) currentState = TrafficLightStates.GREEN elif (currentState == TrafficLightStates.GREEN): GPIO.output(TrafficLightLEDs.RED, False) GPIO.output(TrafficLightLEDs.AMBER, False) GPIO.output(TrafficLightLEDs.GREEN, True) time.sleep(5) currentState = TrafficLightStates.AMBER elif (currentState == TrafficLightStates.AMBER): GPIO.output(TrafficLightLEDs.RED, False) GPIO.output(TrafficLightLEDs.AMBER, True) GPIO.output(TrafficLightLEDs.GREEN, False) if pattern == 'uk': time.sleep(2) else: time.sleep(3) currentState = TrafficLightStates.RED else: print 'Invalid state!'`
To make each state self-contained and not dependent on a prior state, we’ve added code to explicitly turn each LED to the required on or off status for each state. This means that should we choose to run the lights in a different order, we just change the code that sets `currentState`, without changing the actual implementation of a state.
When set to operate using the UK pattern, the lights will cycle as follows:
And with the USA pattern, expect:
# Next Steps
In this article, we changed the way that the traffic light code works to incorporate a Finite State Machine. Whilst this is a very simple example, Finite State Machines have a great many uses in programming — you’ll find them used in compiler implementations, game development, and in some of the C language firmware for connected sensors that we’re building at Conectric.
I’d love to hear what you’re up to with the Raspberry Pi — find me on Twitter or via the responses here. If you enjoyed this article, please hit the ❤ and share it far and wide!
## Conectric Networks
Conectric Networks eliminates wasted energy consumption…
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## Simon Prickett
Software Professional, builder of things with Arduino and Raspberry Pi.
## Conectric Networks
Conectric Networks eliminates wasted energy consumption from heating and air conditioning in hotel buildings using our patented occupancy monitoring and HVAC control system.
Written by
## Simon Prickett
Software Professional, builder of things with Arduino and Raspberry Pi.
## Conectric Networks
Conectric Networks eliminates wasted energy consumption from heating and air conditioning in hotel buildings using our patented occupancy monitoring and HVAC control system.
## Configure web server on different type os using ansible and dynamically take software name…
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Get the Medium app | 1,651 | 7,299 | {"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} | 3.265625 | 3 | CC-MAIN-2021-21 | latest | en | 0.882693 |
https://bedtimemath.org/fun-math-iowa/ | 1,720,827,090,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00719.warc.gz | 107,459,089 | 20,798 | # I Owe a Lot to Iowa
Today’s the perfect time to travel to Iowa, because it’s holding its big state fair! The Iowa State Fair is one of our country’s biggest and best-known. More than 1 million people show up to check out crafts, animals, and of course, food. Lots of the treats from the 200+ food stands are fried on a stick. But one food doesn’t need a stick, because it comes on a handy cob. Yes, Iowa grows more corn than any other state. Corn has some cool math. Ears grow an even number of rows, usually 16. Those rows add up to around 800 kernels in an average ear. An Iowan farmer holds the world record for most ears grown on a single plant, with 16. It’s only fair that Iowa holds that record!
Wee ones: If the Iowa State Fair lasts for 11 days, does it run longer than 1 week?
Little kids: If you eat 4 sticks of deep-fried butter – they really do serve that – what numbers do you say to count down from 4 as you eat them? Bonus: If you’ve nibbled off 9 rows from your ear of corn, what numbers do you say to count up the next 5 rows?
Big kids: Corn is sold in big containers called “bushels.” Does a 56-pound bushel of corn weigh more than you? What’s the difference between the 2 weights? Bonus: The Iowa State Fair was first held in 1854. What’s the biggest number you can make from those digits?
The sky’s the limit: If an ear of corn has 800 kernels in 16 equal rows, how many kernels are in each row? (Hint if needed: 16 is 2 x 2 x 2 x 2, so to divide a number by 16, just cut it in half 4 times!)
Wee ones: Yes, because 11 is more than 7.
Little kids: 4, 3, 2, 1. Bonus: 10, 11, 12, 13, 14.
Big kids: Different for everyone…find out your weight, then subtract it from 56 if it’s a smaller number, or subtract 56 from it if your weight is greater than 56. Bonus: 8,541.
The sky’s the limit: 50 kernels per row. | 519 | 1,837 | {"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} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.919657 |
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Web Encyclopedia4u.com
# Logical operator
In logical calculus, logical operators or logical connectors serve to connect statements into more complicated compound statements. For example, considering the assertionss "It's raining", and "I'm inside", we can form the compound assertions "it's raining, and I'm inside" or "it's not raining" or "if it's raining, then I'm inside."
A new statement or proposition combing two statements are called compound statement or compound proposition.
The basic operators are "not" (¬), "and" (∧, or ), "or" (∨), "conditional" (→), and "biconditional" (↔). "Not" is a unary operator--it takes a single term ( ¬ P ). The rest are binary operators, taking two terms to make a compound statement ( P ∧ Q, P ∨ Q, P → Q, P ↔ Q ).
Note the similarity between the symbols for "and" () and "set theoretic intersection" ( ∩ ); likewise for "or" ( ∨ ) and "union ( ∪ ). This is not a coincidence: the definition of the intersection uses "and" and the definition of union uses "or".
Truth tables for these connectives:
P Q ¬P P ∧ Q P ∨ Q P → Q P ↔ Q
T T F T T T T
T F F F T F F
F T T F T T F
F F T F F T T
In order to reduce the number of necessary parentheses, one introduces precendence rules: ¬ has higher precedence than ∧, ∧ higher than ∨, and ∨ higher than →. So for example, P ∨ Q ∧ ¬ R → S is short for (P ∨ (Q ∧ (¬ R)) → S.
Note that the logical equivalence of certain compound statements entails that not all of these operators are necessary for a full-blooded logical calculus. For example, ¬ P ∨ Q is logically equivalent to P → Q; since logical equivalence means that equivalent terms may be subsituted for each other in an expression, it's not necessary to have a conditional operator. The five operators listed above are the basic set for the sake of convenience (and brevity).
One can also consider other connectives, such as NAND, XOR and NOR. It can be shown that all connectives can be expressed with NAND alone, and they can also all be expressed with NOR alone. | 512 | 2,038 | {"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-2013-20 | latest | en | 0.912491 |
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Invert Binary Tree
Inverting a Binary Tree is a problem asked in a lot of interviews and we will be tackling it in a simple but efficient manner. Any tree-related problem should be attempted to be solved in a recursive approach since Tree is a localized data structure i.e small trees make up large trees.
Isomorphic Strings
Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.
Min Stack
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Set Matrix Zeroes
Given a matrix, A of size M x N of 0’s and 1’s. If an element is 0, set its entire row and column to 0.
Maximum Subarray Sum
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Count Complete Tree Nodes
Given a Complete Binary Tree, count the number of nodes. Definition of a Complete Binary Tree: In a Complete Binary Tree every level, except possibly the last, is completely filled, and all […]
K Closest Points to Origin
Given a list of points on the plane. Find the K closest points to the origin (0, 0). The distance between two points on a plane is the Euclidean Distance.
Next Closest Time
Given a time represented in the format “HH:MM”, Form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused. You may assume the given input string is always valid. For example, “01:34”, “12:09” are all valid. “1:34”, “12:9” are all invalid.
Maximum Number of Events That Can Be Attended
Let’s say we had a calendar for a four day week. The calendar is full of events, which each can span multiple days. Each color block represents an event:
Search for a target value in a 2D Matrix
Write an efficient algorithm that searches for a value in an m x n matrix | 488 | 2,117 | {"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-2021-39 | latest | en | 0.898958 |
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#16
November 27th, 2012, 07:19 PM
NormR
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Looks like you're making progress.
#17
November 27th, 2012, 08:36 PM
Seahawk9892
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I have really hit a wall. I really have no idea how to create this for loop; I struggle mightily with loops and arrays. I looked at those links you posted NormR and I'm sure that the answer lies in there, however I just cannot figure it out.
Anybody want to give me a pointer? I am pretty sure I could figure it out if I could just get started.
#18
November 27th, 2012, 09:13 PM
NormR
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Quote:
have no idea how to create this for loop
Here is a sample for loop from the tutorial:
Code:
``` for(int i=1; i<11; i++){
System.out.println("Count is: " + i);
}```
What logic is supposed to go in the for loop? Why have you chosen a for loop? A for loop is used when you know how any iterations the loop should make.
Are talking about what steps should be taken inside the loop?
Start by making a list of the simple steps the code needs to do inside the loop.
#19
November 27th, 2012, 09:29 PM
Seahawk9892
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Oh oops didn't even notice that I typed for loop, I meant to type just a loop. As for the steps to be taken inside the loop, good point about listing them, so here it goes.
I need to create a loop that shuffles the puzzle randomly a certain number of times. So I need to design a loop that does the following:
1. Chooses two of the 16 tiles at random.
2. Swaps the values of the two integers enclosed in brackets (i.e. board [0][0] and board [2][3]).
I guess I could use a for loop and set the number of times the loop is executed to something like 20 as to ensure that the tiles are sufficiently scattered. Am I on the right track here?
#20
November 27th, 2012, 09:50 PM
NormR
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So this loop is to shuffle the numbers. I wouldn't think that was important now because you can manually set the numbers to give a starting position for the play of the game.
See the Random class for a method to return a random number between 0 and 15.
Call it twice for two array indexes and swap those two numbers.
Here's a way to get a row and column index from a single number:
Code:
``` // 4 rows, 6 columns
int NbrCols = 6;
for(int i=0; i < NbrCols*4; i++) {
int row = i / NbrCols;
int col = i % NbrCols;
System.out.println("i=" + i + ", row="+row + ", col="+col);
}
```
That's it for tonight. Back tomorrow.
Last edited by NormR : November 27th, 2012 at 09:58 PM.
#21
November 28th, 2012, 02:23 AM
Seahawk9892
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Been working on my code extensively, yet still struggling. Updates posted below:
Code:
``` public class PuzzleMethods
{
int [][] board = new int [4][4];
final static int NUMBER_OF_ROWS = 4;
final static int NUMBER_OF_COLUMNS = 4;
int rowNumber = 0;
int columnNumber = 0;
public PuzzleMethods()
{
for(int i = 0; i < NUMBER_OF_ROWS * 4; i++)
{
rowNumber = i / NUMBER_OF_ROWS;
columnNumber = i % NUMBER_OF_ROWS;
board[rowNumber][columnNumber] = i;
}
}
public static int randomNumberGenerator()
{
int minimum = 1;
int maximum = 15;
return minimum + (int)(Math.random() * ((maximum - minimum) + 1));
}
public <board> void swap(int n1, int n2)
{
}
public static void scrambleTiles()
{
}
public void printBoard()
{
String lineSeparator = "+---+---+---+---+";
String row1 = "| " + board[0][0] + " | " + board[1][0] + " | " + board[2][0] + " | " + board[3][0] + " |";
String row2 = "| " + board[0][1] + " | " + board[1][1] + " | " + board[2][1] + " | " + board[3][1] + " |";
String row3 = "| " + board[0][2] + " |" + board[1][2] + " |" + board[2][2] + " |" + board[3][2] + " |";
String row4 = "|" + board[0][3] + " |" + board[1][3] + " |" + board[2][3] + " |" + board[3][3] + " |";
System.out.println(lineSeparator);
System.out.println(row1);
System.out.println(lineSeparator);
System.out.println(row2);
System.out.println(lineSeparator);
System.out.println(row3);
System.out.println(lineSeparator);
System.out.println(row4);
System.out.println(lineSeparator);
}
} ```
I am absolutely clueless as what to do now.
#22
November 28th, 2012, 08:09 AM
NormR
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Quote:
what to do now.
Make a list of the steps the program needs to do.
Refine and add to that list until there is good and complete logic
Write code to implement the steps.
Test the code.
Fix problems.
Have you compiled and tested the code you posted?
Does it work?
You need to test what you've coded before writing any more code.
Last edited by NormR : November 28th, 2012 at 08:12 AM.
#23
November 28th, 2012, 08:28 AM
Seahawk9892
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Quote:
Originally Posted by NormR Make a list of the steps the program needs to do. Refine and add to that list until there is good and complete logic Write code to implement the steps. Test the code. Fix problems. Have you compiled and tested the code you posted? Does it work? You need to test what you've coded before writing any more code.
Yes, I have compiled the code and the parts I know that I will keep in tact work. I just don't know how to create this method to scramble the initial tiles of the board. Maybe using the Collections.shuffle method? I am very new to java however and a lot of this information goes above my head...it is very frustrating.
#24
November 28th, 2012, 08:30 AM
NormR
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The code you posted won't execute. It needs a main() method.
How do you test the code?
Quote:
method to scramble the initial tiles
See posts 19 and 20
Last edited by NormR : November 28th, 2012 at 08:33 AM.
#25
November 28th, 2012, 08:35 AM
Seahawk9892
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Quote:
Originally Posted by NormR The code you posted won't execute. It needs a main() method. How do you test the code?
I made another class with a main method for testing purposes, however I didn't post it. Here it is:
Code:
``` public class Puzzle
{
public static void main(String[] args)
{
int [][] board = new int [4][4];
int i = 1;
for (int x = 0; x < 4; x++) {
for (int y = 0; y < 4; y++) {
if (i < 16) {
board[x][y] = i++;
System.out.println(board[x][y]);
} else {
board[x][y] = -1;
}
}
}
}
} ```
Here is an example of me testing the constructor to receive the desired output.
I have edit my code further, and it is as follows:
Code:
``` public class PuzzleMethods
{
int [][] board = new int [4][4];
final static int NUMBER_OF_ROWS = 4;
final static int NUMBER_OF_COLUMNS = 4;
public PuzzleMethods()
{
int i = 1;
for (int x = 0; x < NUMBER_OF_ROWS; x++) {
for (int y = 0; y < NUMBER_OF_COLUMNS; y++) {
if (i < 16)
{
board[x][y] = i++;
} else {
board[x][y] = -1;
}
}
}
}
public static int randomNumberGenerator()
{
int minimum = 1;
int maximum = 15;
return minimum + (int)(Math.random() * ((maximum - minimum) + 1));
}
public void scrambleTiles()
{
}
public void printBoard()
{
String lineSeparator = "+---+---+---+---+";
String row1 = "| " + board[0][0] + " | " + board[1][0] + " | " + board[2][0] + " | " + board[3][0] + " |";
String row2 = "| " + board[0][1] + " | " + board[1][1] + " | " + board[2][1] + " | " + board[3][1] + " |";
String row3 = "| " + board[0][2] + " |" + board[1][2] + " |" + board[2][2] + " |" + board[3][2] + " |";
String row4 = "|" + board[0][3] + " |" + board[1][3] + " |" + board[2][3] + " |" + board[3][3] + " |";
System.out.println(lineSeparator);
System.out.println(row1);
System.out.println(lineSeparator);
System.out.println(row2);
System.out.println(lineSeparator);
System.out.println(row3);
System.out.println(lineSeparator);
System.out.println(row4);
System.out.println(lineSeparator);
}
} ```
I deleted the swap method as I don't think it is necessary. Not sure though
#26
November 28th, 2012, 08:44 AM
NormR
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Where are any of the methods in the PuzzleMethods class called for testing?
This should be coded this way:
Code:
``` final static int NUMBER_OF_ROWS = 4;
final static int NUMBER_OF_COLUMNS = 4;
int [][] board = new int [NUMBER_OF_ROWS][NUMBER_OF_COLUMNS];```
There should NOT be any 4's in the code. All references to rows and columns should either use those variables or the .length attribute of the array: board.length and board[i].length
Last edited by NormR : November 28th, 2012 at 08:49 AM.
#27
November 28th, 2012, 10:27 AM
Seahawk9892
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After countless hours of hair-pulling madness, I am convinced I have lost my mind. I fear I may go on a murderous rampage if I continue to work on this program, so in the best interests of those around me, I have decided to retire this program, chalk it up as a loss, and take the 0.
I really do appreciate the help though, NormR. I know you were trying to steer me in the right direction but java coding is simply NOT my forte.
#28
November 28th, 2012, 10:43 AM
NormR
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Java coding is one thing
Program design is another.
You need to be able to "think" like a computer to be able to design a program. Many people try to code their programs before they have a design and often end up with a mess.
Given a design, the program can be coded in many different languages.
Good luck.
Last edited by NormR : November 28th, 2012 at 10:45 AM.
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The complex solution to a quadratic equation is x equals start fraction negative 10 plus or minus square root of negative 700 end square root over two end fraction full stop Write this solution in standard form, x = a ± bi, where a and b are real numbers. Justify your answer by showing your work.
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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# 22. Unlike most warbler species, the male and female
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22. Unlike most warbler species, the male and female [#permalink]
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22. Unlike most warbler species, the male and female
blue-winged warbler are very difficult to tell apart
.
(A) Unlike most warbler species, the male and
female blue-winged warbler are very difficult
to tell apart.
(B) Unlike most warbler species, the gender of the
blue-winged warbler is very difficult to
distinguish.
(C) Unlike those in most warbler species, the male
and female blue-winged warblers are very
difficult to distinguish.
(D) It is very difficult, unlike in most warbler species,
to tell the male and female blue-winged warbler
apart.
(E) Blue-winged warblers are unlike most species of
warbler in that it is very difficult to tell the
male and female apart.
If you have any questions
New!
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15 Jul 2004, 02:47
1 min, and again E.
I also found D close but the construction of E seemed just a tad better
OA?[/img]
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15 Jul 2004, 07:05
had to see a prev post
AB misplaced modifier
C "those in" should be "those of"
D seems to be wrong because "difficult" modifies the phrase "unlike in most warbler species'
E seems to be the right answer by POE.
Now my question is Aren't we comparing warblers and species in choice E ?
Blue-winged warblers are unlike most species of warbler in that it is very difficult to tell the male and female apart. [/b]
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15 Jul 2004, 08:03
any alternative/additional explanation to why (C) is wrong?
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15 Jul 2004, 08:08
usage of distinguish looks weird.
"it is very difficult to distinguish between x and y"
"it is very difficult to distinguish x from y"
are examples of correct usage.
C has it wrong.
Unlike X, Y and Z are very difficult to distinguish.
Doesnt that strike you as wrong ?
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15 Jul 2004, 11:17
ashkg wrote:
usage of distinguish looks weird.
"it is very difficult to distinguish between x and y"
"it is very difficult to distinguish x from y"
are examples of correct usage.
C has it wrong.
Unlike X, Y and Z are very difficult to distinguish.
Doesnt that strike you as wrong ?
yes, you are right!
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15 Jul 2004, 11:50
ashkg wrote:
had to see a prev post
AB misplaced modifier
C "those in" should be "those of"
D seems to be wrong because "difficult" modifies the phrase "unlike in most warbler species'
E seems to be the right answer by POE.
Now my question is aren't we comparing warblers and species in choice E ?
Blue-winged warblers are unlike most species of warbler in that it is very difficult to tell the male and female apart. [/b]
Does anyone agree with what I feel about E ? Please see my quote above.
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15 Jul 2004, 12:46
ashkg wrote:
ashkg wrote:
had to see a prev post
AB misplaced modifier
C "those in" should be "those of"
D seems to be wrong because "difficult" modifies the phrase "unlike in most warbler species'
E seems to be the right answer by POE.
Now my question is aren't we comparing warblers and species in choice E ?
Blue-winged warblers are unlike most species of warbler in that it is very difficult to tell the male and female apart. [/b]
Does anyone agree with what I feel about E ? Please see my quote above.
My understanding is we are comparing "blue-winged warblers" with "most species of warblers" ..Why are you ignoring "of warbler" part of "most species?
Btw, Now my question is aren't we comparing warblers and species in choice E ?
Idioms is "compare with" in such cases right? You are using "compare ...and .." .. You've been caught!
15 Jul 2004, 12:46
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17 Unlike most warbler species, the male and female 5 02 Jul 2015, 07:59
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2 Unlike most warbler species, the male and female blue-winged 18 04 Oct 2007, 00:31
13 Unlike most warbler species, the male and female blue-winged 17 06 Jul 2007, 07:53
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What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?
A. 1.4
B. $$\sqrt{2}$$
C. 1.7
D. $$\sqrt{3}$$
E. 2.0
[Reveal] Spoiler: OA
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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Bunuel wrote:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?
A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0
Good problem again. Takes into account a lot of co ordinate geometry fundas.
eqn of circle = x^2 + y^2 = 1, center = (0,0) radius = 1
min dist of line from circle = dist of line from the center - radius
Make the distance of the line from the circle to be 0 and we see that it becomes a tangent to the circle.
Now we know if we draw a line from the center to the point where the tangent touches the circle the line and the tangent are perpendicular to each other.
So we need to find the equation of this line first.
We can take the line back where it was now
Since the lines are perpendicular m1 x m2 = -1
m of line = 3/4
so slope of the new line = -4/3
Since the line passes through the origin (center of circle) its eqn => y=-4/3x
now we need to get the point of intersection of our two lines, which comes out to be (36/25,-48/25)
now get the distance of this point from the origin and subtract the radius from it.
Comes to 1.4 (may have made calculation errors )
So A.
Comes under 2 mins.
Bunuel, great work with the Questions. I suggest you make a single thread and keep updating it. People can subscribe to that and also it will help new guys read all the probs and solutions in one thread.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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12 Oct 2009, 05:45
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Lets do in 1 min.
Use the formula
D = | Am+Bn+C|/ SQRT(A^2 + B^2) where Ax+By+C = 0
put (m,n) =0,0 = center of circle
we get D = 12/5 thus required distance is D-1 = 12/5 -1 = 7/5 = 1.4
we don't require points
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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12 Oct 2009, 08:28
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Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html
You can continue discussions and see the solutions there.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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I have learnt this in my school time that distance from one point to a line is what stated above.
Actually its same like finding the line which originates from the distant point and then intersect at the required line,
and then we find the point of intersection and then using the formula we calculate distance between the two points.
But even if u dont rem the formula what u can do is....
suppose line is ax+by-c=0 now when y=0 , x=c/a and when x=0 , y = c/b
area of triangle formed by these 2 points and center (0,0) is 1/2 * c/a * c/b = c^2/2ab
now this is equal to 1/2 * D1 * D2 , where D1 is distance between points on x and y coordinates of the line which is sqrt [ (c/a)^2 + (b/a) ^2 ] and D2 is the required perpendicular distance on the line.
Equate them and u will get the ans. Its very easy concept and I m not that good in explaining here I think sry for that.
If you still dont get this please letme know I will explain again.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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14 Dec 2009, 19:55
gurpreetsingh wrote:
Lets do in 1 min.
Use the formula
D = | Am+Bn+C|/ SQRT(A^2 + B^2) where Ax+By+C = 0
put (m,n) =0,0 = center of circle
we get D = 12/5 thus required distance is D-1 = 12/5 -1 = 7/5 = 1.4
we don't require points
I am too weak in this section, could you please explain...what is this formula and why did you do D-1?
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ISBtarget wrote:
I am too weak in this section, could you please explain...what is this formula and why did you do D-1?
First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).
Now we can do this by finding the equation of a line perpendicular to given line $$y=\frac{3}{4}*x-3$$ (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.
There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.
We know the formula to calculate the distance between two points $$(x1,y1)$$ and $$(x2,y2)$$: $$d=\sqrt{(x1-x2)^2+(y1-y2)^2}$$ BUT there is a formula to calculate the distance between the point (in our case origin) and the line:
DISTANCE BETWEEN THE LINE AND POINT:
Line: $$ay+bx+c=0$$, point $$(x1,y1)$$
$$d=\frac{|ay1+bx1+c|}{\sqrt{a^2+b^2}}$$
DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is $$(0,0)$$ -->
$$d=\frac{|c|}{\sqrt{a^2+b^2}}$$
So in our case it would be: $$d=\frac{|-3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4$$
So the shortest distance would be: $$2.4-1(radius)=1.4$$
P.S. Also note that when we have $$x^2+y^2=k$$, we have circle (as we have $$x^2$$ and $$y^2$$), it's centered at the origin (as coefficients of $$x$$ and $$y$$ are $$1$$) and the radius of that circle $$r=\sqrt{k}$$.
You can check the link of Coordinate Geometry below for more.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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15 Dec 2009, 12:50
Bunuel wrote:
ISBtarget wrote:
I am too weak in this section, could you please explain...what is this formula and why did you do D-1?
First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).
Now we can do this by finding the equation of a line perpendicular to given line $$y=\frac{3}{4}*x-3$$ (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.
There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.
We know the formula to calculate the distance between two points $$(x1,y1)$$ and $$(x2,y2)$$: $$d=\sqrt{(x1-x2)^2+(y1-y2)^2}$$ BUT there is a formula to calculate the distance between the point (in our case origin) and the line:
DISTANCE BETWEEN THE LINE AND POINT:
Line: $$ay+bx+c=0$$, point $$(x1,y1)$$
$$d=\frac{|ay1+bx1+c|}{\sqrt{a^2+b^2}}$$
DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is $$(0,0)$$ -->
$$d=\frac{|c|}{\sqrt{a^2+b^2}}$$
So in our case it would be: $$d=\frac{|-3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4$$
So the shortest distance would be: $$2.4-1(radius)=1.4$$
P.S. Also note that when we have $$x^2+y^2=k$$, we have circle (as we have $$x^2$$ and $$y^2$$), it's centered at the origin (as coefficients of $$x$$ and $$y$$ are $$1$$) and the radius of that circle $$r=\sqrt{k}$$.
You can check the link of Coordinate Geometry below for more.
Awesome man, why wouldnt you start a quant training program....excellent
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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17 Dec 2009, 12:55
Hi
One more question...sounds silly but can you help
why are you doing D-1 , you are calculating the distance between a point on the circle and the line , 0,0 is a point on the circle, why cant 2.4 be the answer
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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17 Dec 2009, 13:11
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Minimum distance from the circle to the line would be:
Length of perpendicular from the origin to the line (as the circle is centered at the origin) - The radius of a circle (which is 1).
(0,0) is not the point on the circle, it's the center of the circle with radius 1 (circle is centered at the origin).
2.4 is the distance from the line to the origin (the center of the circle), so we should subtract the length of the radius to get the distance from the line to the circle.
ANOTHER SOLUTION:
In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
$$(x-a)^2+(y-b)^2=r^2$$
If the circle is centered at the origin (0, 0), then the equation simplifies to:
$$x^2+y^2=r^2$$
So, the circle represented by the equation $$x^2+y^2 = 1$$ is centered at the origin and has the radius of $$r=\sqrt{1}=1$$.
Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).
So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line $$y = \frac{{3}}{{4}}x-3$$.
The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> $$leg_1=4$$.
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> $$leg_2=3$$.
So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: $$\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}$$ --> $$\frac{height}{3}=\frac{4}{5}$$ --> $$height=2.4$$.
$$Distance=height-radius=2.4-1=1.4$$
Hope it's helps.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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31 Dec 2010, 07:45
Have questions based on this concept (i.e. Circle on a plane) ever been tested by the GMAT?
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rahul321 wrote:
Have questions based on this concept (i.e. Circle on a plane) ever been tested by the GMAT?
Yes. Check the following GMAT prep questions:
point-on-a-circle-106016.html
does-line-k-touch-circle-or-not-101471.html
plane-geometry-semicircle-from-gmatprep-85154.html
does-the-point-p-touch-the-circle-101485.html
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A rather nice question and I think Bunuel has already explained sufficient alternative solutions. I just want to add a method of finding the altitude of a right triangle here that I think is particularly neat (discussed in Veritas Geometry book)
Attachment:
Ques2.jpg [ 8.1 KiB | Viewed 19016 times ]
I want to find x here since (x - 1) will be the minimum distance from the circle to the line (as explained above)
Area of the given right triangle = (1/2)*3*4 (3 is base and 4 is altitude)= (1/2)*5*x (5 is base and x is altitude)
So x = 2.4
Finding the area of the original triangle in two different ways and equating it will help you find the altitude.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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VeritasPrepKarishma wrote:
A rather nice question and I think Bunuel has already explained sufficient alternative solutions. I just want to add a method of finding the altitude of a right triangle here that I think is particularly neat (discussed in Veritas Geometry book)
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I want to find x here since (x - 1) will be the minimum distance from the circle to the line (as explained above)
Area of the given right triangle = (1/2)*3*4 (3 is base and 4 is altitude)= (1/2)*5*x (5 is base and x is altitude)
So x = 2.4
Finding the area of the original triangle in two different ways and equating it will help you find the altitude.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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01 Jan 2011, 02:08
Bunuel wrote:
rahul321 wrote:
Have questions based on this concept (i.e. Circle on a plane) ever been tested by the GMAT?
Yes. Check the following GMAT prep questions:
point-on-a-circle-106016.html
does-line-k-touch-circle-or-not-101471.html
plane-geometry-semicircle-from-gmatprep-85154.html
does-the-point-p-touch-the-circle-101485.html
Thanks. I'm just reading the GMATclub Coordinate Geometry post now...Great stuff!!
It's surprising that the OG Quant book doesn't remotely touch upon the 'Circle on a plane' topic even though it is tested!
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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22 Oct 2011, 10:35
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Bunuel wrote:
ISBtarget wrote:
I am too weak in this section, could you please explain...what is this formula and why did you do D-1?
First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).
Now we can do this by finding the equation of a line perpendicular to given line $$y=\frac{3}{4}*x-3$$ (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.
There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.
We know the formula to calculate the distance between two points $$(x1,y1)$$ and $$(x2,y2)$$: $$d=\sqrt{(x1-x2)^2+(y1-y2)^2}$$ BUT there is a formula to calculate the distance between the point (in our case origin) and the line:
DISTANCE BETWEEN THE LINE AND POINT:
Line: $$ay+bx+c=0$$, point $$(x1,y1)$$
$$d=\frac{|ay1+bx1+c|}{\sqrt{a^2+b^2}}$$
DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is $$(0,0)$$ -->
$$d=\frac{|c|}{\sqrt{a^2+b^2}}$$
So in our case it would be: $$d=\frac{|-3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4$$
So the shortest distance would be: $$2.4-1(radius)=1.4$$
P.S. Also note that when we have $$x^2+y^2=k$$, we have circle (as we have $$x^2$$ and $$y^2$$), it's centered at the origin (as coefficients of $$x$$ and $$y$$ are $$1$$) and the radius of that circle $$r=\sqrt{k}$$.
You can check the link of Coordinate Geometry below for more.
Brilliant explanation.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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06 Dec 2012, 11:45
Bunuel wrote:
Minimum distance from the circle to the line would be:
Length of perpendicular from the origin to the line (as the circle is centered at the origin) - The radius of a circle (which is 1).
(0,0) is not the point on the circle, it's the center of the circle with radius 1 (circle is centered at the origin).
2.4 is the distance from the line to the origin (the center of the circle), so we should subtract the length of the radius to get the distance from the line to the circle.
ANOTHER SOLUTION:
In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
$$(x-a)^2+(y-b)^2=r^2$$
If the circle is centered at the origin (0, 0), then the equation simplifies to:
$$x^2+y^2=r^2$$
So, the circle represented by the equation $$x^2+y^2 = 1$$ is centered at the origin and has the radius of $$r=\sqrt{1}=1$$.
Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).
So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line $$y = \frac{{3}}{{4}}x-3$$.
The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> $$leg_1=4$$.
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> $$leg_2=3$$.
So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: $$\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}$$ --> $$\frac{height}{3}=\frac{4}{5}$$ --> $$height=2.4$$.
$$Distance=height-radius=2.4-1=1.4$$
Hope it's helps.
Hi Bunuel,
I could not understand the last but one step.
how did you take the ratios?
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09 Feb 2014, 06:50
Took quite some time(6+ mins) to solve . But I took the same approach as explained by VeritasPrepKarishma. I just hope these ideas click on time. Nice question.
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For the people still troubled with this questions -
First of all this questions tests relatively advanced skills in mathematics and hence IMO can not be a part of the real GMAT.
However, to find the answer of this Q we need to follow below mentioned steps:
1) find the distance of the line from origin : this distance should be the shortest possible distance
2) as all points on circle are equi distance from the origin, we need to find the shortest distance of line from origin and subtract radius from it to get our answer
3) to get shortest distance we need to actually find length of perpendicular line which starts from origin and ends at our given line
3-a) One of the method to solve for (3) is using the equation - |ax1 + by1 + c| /sqrt (a^2+b^2) (this formula you need to remember) - read posts by bunuel or gurpreet for more details.
3-b) Another method is to find the equation of perpendicular line and then find an intersection point of this perpendicular line with our given line. Now find the distance between this point to origin (PHEW) - I surely can't do all this in less 2 minutes and be accurate to the second decimal point [ remember our options are 1.4 and sqrt (2) = 1.41 ]
Finally, for the purpose of GMAT only, i would advise you should not be worried if you can't remember this formula or find this question too difficult.
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STAT 509 – Practice Final – Spring 2010 Name: __________________________________ Show work unless specified otherwise. Answers alone will not receive credit. Good Luck. (1) Below is a stem & leaf diagram for final exam scores from a previous semester. The data has been ordered and there are 50 scores in all. 4 8 9 5 0 4 5 6 8 9 6 0 0 0 3 4 4 6 8 7 0 0 2 2 2 3 4 6 7 9 9 8 1 2 3 4 4 4 5 6 6 6 7 8 8 9 9 0 0 1 2 2 4 6 8 9 Draw and label the corresponding boxplot on the scale below. (5 points) (2) If time to failure of an electrical component follows an exponential distribution with a mean of 1000 hours: (a) What is the probability that the time to failure of the next component will be more than 750 hours? (5 points) (b) What is the probability that the time to failure of the next 36 components will be more than 750 hours? (5 points)
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(03) An arsenal contains several identical boxes of ammunition. The number of defective bullets per box has the following distribution. Y 0 1 2 3 P(y) 0.90 .07 0.027 0.003 What is the expected number of defective bullets per box in the arsenal? (5 points) (04) The expected number of radioactive particles passing by a counter every 2 milliseconds is 4.9. What is the expected number of radioactive particles passing a counter over a 6 millisecond time frame? (5 points) (05) Below is summary output for a test of hypothesis that tests to see if two variances from independent populations can be considered equal. State your conclusions at .05 level of significance. Assume samples came from normal populations. (5 points) Two Sample Variance results: σ 2 1 - variance of Pop'n 1 σ 2 2 - variance of Pop'n 2 H 0 : σ 2 1 / σ 2 2 = 1 H A : σ 2 1
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## Presentation on theme: "Two protons, each of charge 1. 6 x C are 2 x 10-5m apart"— Presentation transcript:
Two protons, each of charge 1. 6 x 10-19 C are 2 x 10-5m apart
Two protons, each of charge 1.6 x C are 2 x 10-5m apart. What is the change in potential energy if they are brought 10-5 m closer together? 1.15 x J 3.20 x J 3.20 x J 1.60 x10-14 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
There is a hollow, conducting, uncharged sphere with a negative charge inside the sphere. Consider the electrical potential at the inner and outer surfaces of the sphere The potential on the inner surface is greater. The potential on the outer surface is greater. The potentials on both surfaces are zero. The potentials on both surfaces are equal but not zero. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
There are two capacitors with CA bigger than CB and they are connected in series with a battery.
There is more charge stored on CA. There is more charge stored on CB There is the same charge stored on each capacitor. There is the same potential difference across both capacitors. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
The potential at P1 is 7 V and the electric field there is 3 V/m
The potential at P1 is 7 V and the electric field there is 3 V/m. When I move to P2 , the electric field decreases. However, if I now triple the size of the charge +Q, the electric field at point P2 becomes 3 V/m. What is the potential at P2 now? 7/3 V 7 V 12 V 21 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
The potential at P1 is 7 V and the electric field there is 3 V/m
The potential at P1 is 7 V and the electric field there is 3 V/m. When I move to point P2 the potential decreases. However, if I now triple the size of charge +Q the potential at point P2 now becomes 7 V. What is electric field at P2 now? 1 V/m 3 V/m 5.2 V/m 9 V/m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
An electron is released from rest at the negative plate of a parallel plate capacitor. If the distance across the plate is 5 mm and the potential difference across the plate is 5 V, with what velocity does the electron hit the positive plate? (melectron = 9.1 x kg, qe = 1.6 x C) 2.65 x 105 m/s 5.30 x 106 m/s 1.06 x 106 m/s 1.33 x 106 m/s 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If two parallel, conducting plates have equal positive charge, the electric field lines will
leave one plate and go straight to the other plate leave both plates and go to infinity enter both plates from infinity none of the choices 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If the distance between two isolated parallel plates that are oppositely charged is doubled, the electric field between the plates is essentially unchanged. However, the potential difference between the plates will double. the charge on each plate will double. the force on a charged particle that is half way between the plates will get twice as small. the force on a charged particle that is half way between the plates will get four times as small. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
the potential energy of the proton
Consider two charged spheres, one with charge +2 C and the other with -2 C. A proton (a positively charged particle) is located at the point halfway between the spheres. What is not zero? the potential energy of the proton the work to move the proton from infinity to that point the force on the proton all of these are zero 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
An electronics technician wishes to construct a parallel plate capacitor using Rutile (k = 100) as the dielectric. If the cross-sectional area of the plates is 1.0 cm2 , what is the capacitance if the Rutile thickness is 1.0 mm? ( e0 = 8.85 x MKS units) 88.5 pF 177.0 nF 8.85 µ F 100.0 µ F 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
The dielectric strength of Rutile is 6 x 106 V/m, which corresponds to the maximum electric field that the dielectric can sustain before breakdown. What is the maximum charge that a F capacitor with a 1-mm thickness of Rutile can hold? 1.67 nC 0.60 µ C 0.30 mC 6.0 C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
In which case does an electric field do positive work on a charged particle?
When a negative charge moves opposite to the direction of the electric field. When a positive charge is moved to a point of higher potential energy. When a positive charge completes one circular path around a stationary positive charge. When a positive charge completes one elliptical path around a stationary positive charge. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Two capacitors with capacitances of 1. 0 and 0
Two capacitors with capacitances of 1.0 and 0.5 microfarads, respectively, are connected in series. The system is connected to a 100 V battery. What electrical potential energy is stored in the 1.0 microfarad capacitor? 0.065 x 10-3 J 4.30 x 10-3 J 0.80 x 10-3 J 5.45 x 10-4 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
At what distance from a point charge of 8
At what distance from a point charge of 8.0 microcoul would the electrical potential be 4.2 x 104 V? (k = 9 x 109 N-m2 /C2) 0.58 m 0.76 m 1.71 m 2.94 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A point charge of +3 microcoul is located at the origin of a coordinate system and a second point charge of -6 microcoul is at x = 1.0 m. What is the electric potential at the x = 0.5 m point? (k = 9 x 109 N-m2 /C2 ) 16.2 x 104 V 10.8 x 104 V -10.8 x 104 V -5.4 x 104 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A point charge of +3 microcoul is located at the origin of a coordinate system and a second point charge of -6 microcoul is at x = 1.0 m. At what point on the x-axis is the electrical potential zero? -0.25 m +0.25 m +0.33 m +0.75 m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A 0. 25 microfarad capacitor is connected to a 400 V battery
A 0.25 microfarad capacitor is connected to a 400 V battery. What potential energy is stored in the capacitor? 1.2 x J 1.0 x 10-4 J 0.040 J 0.020 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
150 microcoul 100 microcoul 50 microcoul 33 microcoul
Two capacitors with capacitances of 1.0 and 0.5 microfarads, respectively, are connected in parallel. The system is connected to a 100 V battery. What charge accumulates on the 1.0 microfarad capacitor? 150 microcoul 100 microcoul 50 microcoul 33 microcoul 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Two capacitors with capacitances of 1. 0 and 0
Two capacitors with capacitances of 1.0 and 0.5 microfarads, respectively, are connected in parallel. The system is connected to a 100 V battery. What electrical potential energy is stored in the 1.0 microfarad capacitor? 1.7 x 10-3 J 7.5 x 10-3 J 5.0 x 10-3 J 10.0 x 10-3 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
150 microcoul 100 microcoul 50 microcoul 33 micorcoul
Two capacitors with capacitances of 1.0 and 0.5 microfarads, respectively, are connected in series. The system is connected to a 100 V battery. What charge accumulates on the 1.0 microfarad capacitor? 150 microcoul 100 microcoul 50 microcoul 33 micorcoul 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Two point charges of values +3. 4 and +6
Two point charges of values +3.4 and +6.6 microcoul, respectively, are separated by 0.10 m. What is the electrical potential at the point midway between the two point charges? (k = 9 x 109 N-m2 /C2 ) +1.8 x 106 V -0.9 x 106 V +0.9 x 106 V +3.6 x 106 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
An electron in a TV picture tube is accelerated through a potential difference of 10 kV before it hits the screen. What kinetic energy does the electron gain in the process? (qe = 1.6 x C) 1.0 x 104 J 1.6 x J 1.6 x J 6.25 x 1022 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
An electron in a TV picture tube is accelerated through a potential difference of 10 kV before it hits the screen. What is the kinetic energy of the electron in electron volts? (1 ev = 1.6 x J) 1.0 x 104 eV 1.6 x eV 1.6 x eV 6.25 x 1022 eV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A 0. 25 microfarad capacitor is connected to a 400 V battery
A 0.25 microfarad capacitor is connected to a 400 V battery. What is the charge on the capacitor? 1.2 x C 1.0 x 10-4 C 0.040 C 0.020 C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Very large capacitors have been considered as a means for storing electrical energy. If we constructed a very large parallel plate capacitor of plate area 1 m2 using Pyrex (k = 5.6) of thickness 2 mm as a dielectric, how much electrical energy would it store at a plate voltage of 6000 V? ( e0 = 8.85 x C/N-m2 ) 0.45 J 90 J 9,000 J 45,000 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
How much charge can be placed on a capacitor of plate area 10 cm2 with air between the plates before it reaches "atmospheric breakdown" where E = 3 x 106 V/m? ( e0 = 8.85 x C/N-m2 ) 2.66 x 10-8 C 3.99 x 10-7 C 5.32 x 10-6 C 6.65 x 10-5 C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A uniform electric field, with a magnitude of 5 x 102 N/C, is directed parallel to the positive x-axis. If the potential at x = 5 m is 2500 V, what is the potential at x = 2 m? 1000 V 2000 V 4000 V 4500 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A uniform electric field, with a magnitude of 5 x 102 N/C, is directed parallel to the positive x-axis. If the potential at x = 5 m is 2500 V, what is the change in potential energy of a proton as it moves from x = 5 m to x = 2 m? (qp = 1.6 x C) 8.0 x J 2.4 x J 1.9 x 1021 J 500 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
What will be the electrical potential at a distance of 0
What will be the electrical potential at a distance of 0.15 m from a point charge of 6.0 microcoul? (k = 9 x 109 N-m2 /C2 ) 5.4 x 104 V 3.6 x 105 V 2.4 x 106 V 1.2 x 107 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Two point charges of values +3. 4 and +6
Two point charges of values +3.4 and +6.6 microcoul, respectively, are separated by 0.20 m. What is the potential energy of this 2-charge system? (k = 9 x 109 N-m2 /C2 ) +0.34 J -0.75 J +1.0 J -3.4 J 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
The unit of capacitance, the farad, is dimensionally equivalent to which of the following?
volt/coulomb volt x coulomb joule/volt coulomb/volt 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
increase charge increase voltage increase capacitance
Inserting a dielectric material between two charged parallel conducting plates, originally separated by air and disconnected from a battery, will produce what effect on the capacitor? increase charge increase voltage increase capacitance decrease capacitance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
increase charge decrease charge increase capacitance
Increasing the separation of the two charged parallel plates of a capacitor which are disconnected from a battery will produce what effect on the capacitor? increase charge decrease charge increase capacitance decrease capacitance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Increasing the voltage across the two plates of a capacitor will produce what effect on the capacitor? increase charge decrease charge increase capacitance decrease capacitance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
inverse square distance law applies forces are conservative
Which of the following characteristics are held in common by both gravitational and electrostatic forces when dealing with either point masses or charges? inverse square distance law applies forces are conservative potential energy is a function of distance of separation all of these choices are valid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If three capacitors of values 1.0, 1.5, and 2.0 microfarads each are connected in parallel, what is the combined capacitance? 4.5 microfarads 4.0 microfarads 2.17 microfarads 0.46 microfarads 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If a 10.0 microfarad capacitor is charged so that it stores 2 x 10-3 J of electrical potential energy, what is its electrical potential? 20 V 15 V 10 V 5 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
The unit of electrical potential, the volt, is dimensionally equivalent to which of the following?
joule x coulomb joule/coulomb coulomb/joule farad x coulomb 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
electric field x distance electric field/distance
The quantity of electrical potential, the volt, is dimensionally equivalent to which of the following? force/charge force x charge electric field x distance electric field/distance 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
along a constant potential line
A free electron in an electric field will experience a force acting in what direction with respect to the field? parallel anti-parallel perpendicular along a constant potential line 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
near the positive plate near the negative plate
At which location will the electric field between the two parallel plates of a charged capacitor be the strongest in magnitude? near the positive plate near the negative plate midway between the two plates electric field is constant throughout space between plates 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Four point charges are positioned on the rim of a circle of radius 10 cm. The charge on each of the four (in microcoul) is +0.5, +1.5, -1.0, If we are told that the electrical potential at the center of the circle due to the +0.5 charge alone is 4.5 x 104 V, what is the total potential at the center due to the four charges combined? 18.0 x 104 V 4.5 x 104 V zero -4.5 x 104 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If an electron is accelerated from rest through a potential difference of 1200 V, what is its approximate velocity at the end of this process? (qe = 1.6 x C; me = 9.1 x kg) 1.0 x 107 m/s 1.4 x 107 m/s 2.1 x 107 m/s 2.5 x 107 m/s 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If three capacitors of values 1.0, 1.5, and 2.0 microfarads each are connected in series, what is the combined capacitance? 4.5 microfarads 4.0 microfarads 2.17 microfarads 0.46 microfarads 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If a 12 V battery is connected between two parallel plates separated by 0.6 cm, what is the magnitude of the electric field between the plates? 2.0 x 103 N/C 7.2 x 10-2 N/C 0.5 x 10-3 N/C 0.75 x 106 N/C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
An electron (charge = -1.6 x C) moves 10 cm on a path perpendicular to the direction of a uniform electric field of strength 3.0 N/C. How much work is done on the electron in this process? 4.8 x J -4.8 x J 1.6 x J zero 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Two positive point charges are initially separated by a distance of 2 cm. If their separation is increased to 6 cm, the resultant potential energy is what factor times the initial potential energy? 3.0 9.0 1/3 1/9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A proton (+1.6 x C) moves 10 cm on a path parallel to the direction of a uniform electric field of strength 3.0 N/C. How much work is done on the proton by the electrical field in this process? 4.8 x J -4.8 x J 1.6 x J zero 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
When a proton (+1. 6 x 10-19 C) moves 0
When a proton (+1.6 x C) moves 0.10 m along the direction of an electric field of strength 3.0 N/C, what is the magnitude of the electrical potential difference between the proton's initial and ending points? 4.8 x V 0.30 V 0.033 V 30.0 V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
An electron with velocity v = 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. If the distance the electron travels through the field is 1 cm, how far is it deviated (Y) in its path when it emerges from the electric field? (melectron = 9.1 x kg, qelectron = 1.6 x C) 2.1 mm 4.2 mm 2.1 cm 4.2 cm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Electrons in an X-ray machine are accelerated from rest through a potential difference of 50,000 V. What is the kinetic energy of each of these electrons in eV? (1 eV = 1.6 x J) 50 eV 80 eV 330 eV 50 keV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A parallel-plate capacitor has dimensions 2 cm x 3 cm
A parallel-plate capacitor has dimensions 2 cm x 3 cm. The plates are separated by a 1 mm thickness of paper (dielectric constant k = 3.7). What is the charge that can be stored on this capacitor, when connected to a 9-volt battery? ( e0 =8.85 x C2 /N-m2 ) 19.6 x C 4.75 x C 4.75 x C 1.76 x10-10 C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
What is the equivalent capacitance of the combination shown?
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If C1 = 15 µ F, C2 = 10 µ F, C3 = 20 µ F, and V0 = 18 V, determine the energy stored by C2 .
0.72 mJ 0.32 mJ 0.50 mJ 0.18 mJ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A parallel-plate capacitor has a capacitance of 20 µ F
A parallel-plate capacitor has a capacitance of 20 µ F. What charge on each plate will produce a potential difference of 36 V between the plates of the capacitor? 7.2 x 10-4 C 3.6 x 10-4 C 1.8 x 10-4 C 0.9 x 10-4 C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
What is the equivalent capacitance of the combination shown?
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
What is the equivalent capacitance between points a and b
What is the equivalent capacitance between points a and b ? All capacitors are 1 microfarad capacitors. 4.00 µ F 1.67 µ F 0.60 µ F 0.25 µ F 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If C = 45 µ F, determine the equivalent capacitance for the combination shown.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
If C = 10 µ F, what is the equivalent capacitance for the combination shown?
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
There are two capacitors with CA bigger than CB and they are connected in parallel with a battery.
There is more potential difference across CA There is more potential difference across CB There is the same charge stored on each capacitor. There is the same potential difference across both capacitors. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
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THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
Chapter 1
Nonlinear and Chaotic Maps
1.1 One-Dimensional Maps
In this section we consider nonlinear and chaotic one-dimensional maps (Devaney
[47],Arrowsmith and Place [7],Holmgren [98],Collet and Eckmann [37],Gumowski
and Mira [77],Ruelle [177],Baker and Gollub [10],van Wyk and Steeb [209])
f:S →S,S ⊂ R.
In most cases the set S will be S = [0,1] or S = [−1,1].The one-dimensional map
can also be written as a difference equation
x
t+1
= f(x
t
),t = 0,1,2,...x
0
∈ S.
Starting from an initial value x
0
∈ S we obtain,by iterating the map,the sequence
x
0
,x
1
,x
2
,...
or
x
0
,f(x
0
),f(f(x
0
)),f(f(f(x
0
))),....
For any x
0
∈ S the sequence of points x
0
,x
1
,x
2
,...is called the forward orbit (or
forward trajectory) generated by x
0
.The goal of a dynamical system is to under-
stand the nature of all orbits,and to identify the set of orbits which are periodic,
eventually periodic,asymptotic,etc.Thus we want to understand what happens if
t →∞.In some cases the long-time behaviour is quite simple.
Example.Consider the map f:R
+
→R
+
with f(x) =
x.For all x
0
∈ R
+
and
x
0
> 0 the forward trajectory tends to 1.The fixed points of f are 0 and 1.♣
Example.Consider the map f:[0,1] → [0,1],f(x) = x
2
.If x
0
= 0,then
f(x
0
) = 0.Analogously,if x
0
= 1,then f(x
0
) = 1.The points 0 and 1 we call fixed
points for this map.For x ∈ (0,1),the forward trajectory tends to 0.♣
1
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
2 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
In most cases the behaviour of a map is much more complex.
Next we introduce some basic definitions for dynamical systems.
Denition.A point x
∈ S is called a fixed point of the map f if f(x
) = x
.
Example.Consider the map f:[0,1] → [0,1] with f(x) = 4x(1 − x).Then
x
= 3/4 and x
= 0 are fixed points.♣
Denition.A point x
∈ S is called a periodic point of period n if
f
(n)
(x
) = x
where f
(n)
denotes the n-th iterate of f.The least positive integer n for which
f
(n)
(x
) = x
is called the prime period of x
.The set of all iterates of a periodic
point form a periodic orbit.
Example.Consider the map f:R →R and f(x) = x
2
−1.Then the points 0 and
−1 lie on a periodic orbit of period 2 since f(0) = −1 and f(−1) = 0.♣
Denition.A point x
is eventually periodic of period n if x
is not periodic but
there exists m> 0 such that
f
(n+i)
(x
) = f
(i)
(x
)
for all i ≥ m.That is,f
(i)
(x
) is periodic for i ≥ m.
Example.Consider the map f:R →R with f(x) = x
2
−1.Then with x
0
=
2 we
have the orbit x
1
= 1,x
2
= 0,x
3
= −1,x
4
= 0,i.e.,the orbit is eventually periodic.♣
Denition.Let x
be a periodic point of prime period n.The point x
is hyperbolic
if
|(f
(n)
)
(x
)| = 1
where
denotes the derivative of f
(n)
with respect to x.
Example.Consider the map f
c
:R → R with f
c
(x) = x
2
+ c and c ∈ R.Then
the fixed points are x
±
= 1/2 ±
p
1/4 −c.We have f
c
≡ df
c
/dx = 2x and
df
c
(x
+
)/dx = 1 +
1 −4c.With c = 1/4 we have |f
c
(x
+
)| = 1 and the fixed
point is non-hyperbolic.♣
Theorem.Let x
be a hyperbolic fixed point with |f
(x
)| < 1.Then there is an
such that if x ∈ U,then
lim
n→∞
f
(n)
(x) = x
.
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
1.1.ONE-DIMENSIONAL MAPS 3
Denition.Let M be a differentiable manifold.A mapping f:M →M is called
a diffeomorphism if f is a bijection with f and f
−1
continuously differentiable (of
class C
1
).
Example.The map f:R →R,f(x) = sinh(x) is a diffeomorphism.♣
Example.The map f:R →R,f(x) = x
3
is not a diffeomorphism since its deriva-
tive vanishes at 0.♣
In the following sections we introduce the following concepts important in the study
of nonlinear and chaotic maps.The concepts are
1) Fixed points
2) Liapunov exponent
3) Invariant density
4) Autocorrelation functions
5) Moments
6) Fourier transform
7) Bifurcation diagrams
8) Feigenbaum number
9) Symbolic dynamics
10) Chaotic repeller
The one-dimensionial maps we study in the examples are the logistic map,the tent
map,the Bernoulli map,the Gauss map,a bungalow-tent map and the circle map.
A necessary condition for a one-dimensional map to show chaotic behaviour is that
the map is non-invertible.
1.1.1 Exact and Numerical Trajectories
In this section we calculate trajectories for one-dimensional maps.In the first ex-
ample we consider the map f:N →N defined by
f(x):=
x/2 if xis even
3x +1 if x is odd
where N denotes the natural numbers.For this map it is conjectured that for all ini-
tial values the trajectory finally tends to the periodic orbit...4 2 1 4 2 1....
The data type unsigned long (4 bytes) in C++ is restricted to the range
0...4294967295
and the data type long (4 bytes) in C++ is restricted to the range
-2147483648...+2147483647
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
4 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
To check the conjecture for larger initial values we use the abstract data type
Verylong in SymbolicC++.In this class all arithmetic operators are overloaded.
+,-,*,/,%,+=,-=,*=,/=.
For example for the initial value 28 we find the sequence
14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,...
Thus the orbit is eventually periodic.Two different initial values are considered in
the program trajectory1.cpp,namely 28 and 998123456789.
//trajectory1.cpp
#include <iostream>//for cout
#include"verylong.h"//for data type Verylong of SymbolicC++
using namespace std;
int main(void)
{
unsigned long y = 28;//initial value
unsigned long T = 20;//number of iterations
unsigned long t;
for(t=0;t<T;t++)
{ if((y%2)==0) y = y/2;else y = 3*y+1;cout << y << endl;}
Verylong x("998123456789");//initial value
Verylong zero("0"),one("1"),two("2"),three("3");
T = 350;
for(t=0;t<T;t++)
{ if((x%two)==zero) x = x/two;else x = three*x + one;cout << x << endl;}
return 0;
}
Java provides a class BigInteger.Since operators such as +,-,*,/,% cannot
be overloaded in Java,Java uses methods to do the arithmetic operations.The
methods are
where divide() provides integer division.The constructor BigInteger(String val)
translates the decimal String representation of a BigInteger into a BigInteger.
The class BigInteger also provides the data fields
BigInteger.ONE BigInteger.ZERO
//Trajectory1.java
import java.math.*;
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
1.1.ONE-DIMENSIONAL MAPS 5
public class Trajectory1
{
public static void main(String[] args)
{
BigInteger X = new BigInteger("998123456789");
BigInteger TWO = new BigInteger("2");
BigInteger THREE = new BigInteger("3");
int T = 350;
for(int t=0;t<T;t++)
{
if((X.mod(TWO)).equals(BigInteger.ZERO)) X = X.divide(TWO);
else { X = X.multiply(THREE);X = X.add(BigInteger.ONE);}
System.out.println("X ="+ X);
}
}
}
In the second example we consider the trajectories for the logistic map.The logistic
map f:[0,1] → [0,1] is given by f(x) = 4x(1 − x).The logistic map can also be
written as the difference equation
x
t+1
= 4x
t
(1 −x
t
)
where t = 0,1,2,...and x
0
∈ [0,1].It follows that x
t
∈ [0,1] for all t ∈ N.Let
x
0
= 1/3 be the initial value.Then we find that
x
1
=
8
9
,x
2
=
32
81
,x
3
=
6272
6561
,x
4
=
7250432
43046721
,...
The exact solution of the logistic map is given by
x
t
=
1
2
1
2
cos(2
t
arccos(1 −2x
0
)).
In the C++ program trajectory2.cpp we evaluate the exact trajectory up to
t = 10 using the abstract data type Verylong of SymbolicC++.For t = 7 we find
x
7
=
3383826162019367796397224108032
3433683820292512484657849089281
.
//trajectory2a.cpp
#include <iostream>
#include"verylong.h"//for data type Verylong
#include"rational.h"//for data type Rational
using namespace std;
inline void map(Rational<Verylong>& x)
{
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6 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
Rational<Verylong> one("1");//number 1
Rational<Verylong> four("4");//number 4
Rational<Verylong> x1 = four*x*(one-x);
x = x1;
}
int main(void)
{
Rational<Verylong> x0("1/3");//initial value 1/3
unsigned long T = 10;//number of iterations
Rational<Verylong> x = x0;
cout <<"x[0] ="<< x << endl;
for(unsigned long t=0;t<T;t++)
{ map(x);cout <<"x["<< t+1 <<"] ="<< x << endl;}
return 0;
}
In the C++ program trajectory3.cpp we evaluate the numerical trajectory using
the basic data type double.We find that the difference between the exact value
and the numerical value for t = 40 is
x
40exact
−x
40approx
= 0.055008 −0.055015 = −0.000007.
//trajectory2b.cpp
#include <iostream>
using namespace std;
inline void map(double& x) { double x1 = 4.0*x*(1.0-x);x = x1;}
int main(void)
{
double x0 = 1.0/3.0;//initial value
unsigned long T = 10;//number of iterations
double x = x0;
cout <<"x[0] ="<< x << endl;
for(unsigned long t=0;t<T;t++)
{ map(x);cout <<"x["<< t+1 <<"] ="<< x << endl;}
return 0;
}
As a third example we consider the Bernoulli map.Let f:[0,1) → [0,1).It is
defined by
f(x):= 2x mod 1 ≡ frac(2x).
The map can be written as the difference equation
x
t+1
=
2x
t
for 0 ≤ x
t
< 1/2
(2x
t
−1) for 1/2 ≤ x
t
< 1
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1.1.ONE-DIMENSIONAL MAPS 7
where t = 0,1,2,...and x
0
∈ [0,1).The map admits only one fixed point x
= 0.
The fixed point is unstable.Let x
0
= 1/17.Then we find the periodic orbit
2
17
,
4
17
,
8
17
,
16
17
,
15
17
,
13
17
,
9
17
,
1
17
,
2
17
,...
If x
0
is a rational number in the interval [0,1),then x
t
is either periodic or tends to
the fixed point x
= 0.The solution of the Bernoulli map is given by
x
t
= 2
t
x
0
mod1
where x
0
is the initial value.For almost all initial values the Liapunov exponent is
given by ln2.Every x
0
∈ [0,1) can be written (uniquely) in the binary representation
x
0
=
X
k=1
a
k
2
−k
,a
k
∈ {0,1}.
One defines
(a
1
,a
2
,a
3
,...):=
X
k=1
a
k
2
−k
and considers the infinite sequence (a
1
,a
2
,a
3
,...).For example,x
0
= 3/8 can be
represented by the sequence (0,1,1,0,0,...).Instead of investigating the Bernoulli
map we can use the map τ defined by
τ(a
1
,a
2
,a
3
,...):= (a
2
,a
3
,a
4
,...).
This map is called the Bernoulli shift.In the C++ program trajectory3.cpp we
find the trajectory of the Bernoulli map using the data type Rational and Verylong
of SymbolicC++.The initial value is 1/17.The orbit is periodic.
//trajectory3.cpp
#include <iostream>
#include"verylong.h"//for data type Verylong
#include"rational.h"//for data type Rational
using namespace std;
inline void map(Rational<Verylong>& x)
{
Rational<Verylong> one("1");//number 1
Rational<Verylong> two("2");//number 2
Rational<Verylong> half("1/2");//number 1/2
Rational<Verylong> x1;
if(x < half) x1 = two*x;else x1 = two*x-one;
x = x1;
}
int main(void)
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8 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
{
Rational<Verylong> x0("1/17");//initial value 1/17
unsigned long T = 10;//number of iterations
Rational<Verylong> x = x0;
cout <<"x[0] ="<< x << endl;
for(unsigned long t=0;t<T;t++)
{ map(x);cout <<"x["<< t+1 <<"] ="<< x << endl;}
return 0;
}
As a fourth example we consider the tent map.The tent map f:[0,1] → [0,1] is
defined as
f(x):=
2x if x ∈ [0,1/2)
2(1 −x) if x ∈ [1/2,1]
.
The map can also be written as the difference equation
x
t+1
=
2x
t
if x
t
∈ [0,1/2)
2(1 −x
t
) if x
t
∈ [1/2,1]
where t = 0,1,2,...and x
0
∈ [0,1].Let x
0
= 1/17 be the initial value.Then the
exact orbit is given by
x
0
=
1
17
,x
1
=
2
17
,x
2
=
4
17
,x
3
=
8
17
,x
4
=
16
17
,x
5
=
2
17
,...
This is an example of an eventually periodic orbit.If the initial value is a rational
number then the orbit is eventually periodic,periodic or tends to a fixed point.For
example the initial value 1/16 tends to the fixed point 1.To find chaotic orbits the
initial value must be an irrational number,for example x
0
= 1/π.The fixed points
of the map are given by x
= 0,x
= 2/3.These fixed points are unstable.The map
shows fully developed chaotic behaviour.The invariant density is given by ρ(y) = 1.
For almost all initial values the Liapunov exponent is given by λ = ln2.For the
autocorrelation function we find
C
xx
(τ) =
1
12
for τ = 0
0 for τ ≥ 1
.
The tent map f:[0,1] →[0,1] given above and the logistic map g:[0,1] →[0,1],
g(x) = 4x(1 −x) are topologically conjugate,i.e.
f = h ◦ g ◦ h
−1
where the homeomorphism h:[0,1] →[0,1] is given by
h(x) =
2
π
arcsin(
x),h
−1
(x) =
1 −cos(πx)
2
.
In the C++ program trajectory4.cpp we find the trajectory of the tent map with
the inital value 1/17.
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1.1.ONE-DIMENSIONAL MAPS 9
//trajectory4.cpp
#include <iostream>
#include"verylong.h"
#include"rational.h"
using namespace std;
inline void map(Rational<Verylong>& x)
{
Rational<Verylong> one("1"),two("2");
Rational<Verylong> half("1/2");//number 1/2
Rational<Verylong> x1;
if(x < half) x1 = two*x;else x1 = two*(one-x);
x = x1;
}
int main(void)
{
Rational<Verylong> x0("1/17");//initial value 1/17
unsigned long T = 10;//number of iterations
Rational<Verylong> x = x0;
cout <<"x[0] ="<< x << endl;
for(unsigned long t=0;t<T;t++)
{ map(x);cout <<"x["<< t+1 <<"] ="<< x << endl;}
return 0;
}
As a fifth example we consider a bungalow-tent map.Our bungalow-tent map f
r
:
[0,1] →[0,1] is defined by
f
r
(x):=
8
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
:
1 −r
r
x for x ∈ [0,r)
2r
1 −2r
x +
1 −3r
1 −2r
for x ∈ [r,1/2)
2r
1 −2r
(1 −x) +
1 −3r
1 −2r
for x ∈ [1/2,1 −r)
1 −r
r
(1 −x) for x ∈ [1 −r,1]
where r ∈ (0,1/2) is the control parameter.The map is continuous,but not differ-
entiable at the points r,1 −r (r = 1/3) and x = 1/2.The map is piecewise linear.
The fixed points are 0 and 1 −r.For r = 1/3 we obtain the tent map.The map f
r
is a special bungalow-tent map.The intersection point P of the line in the interval
[1/2,1 − r) and the line in the interval [1 − r,1] lies on the diagonal y = x.The
invariant density is given by
ρ
r
(x) =
1
2 −3r
χ
[0,1−r]
(x) +
1 −2r
r(2 −3r)
χ
(1−r,1]
(x)
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10 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
where χ is the indicator function,i.e.χ
A
(x) = 1 if x ∈ A and χ
A
(x) = 0 if x/∈ A.
Thus the invariant density is constant in the interval [0,1−r).At 1−r the invariant
density jumps to another constant value.The Liapunov exponent is given by
λ(r) =
1 −r
2 −3r
ln
1 −r
r
+
1 −2r
2 −3r
ln
2r
1 −2r
.
For r = 1/3 we obviously obtain λ(1/3) = ln2.This is the Liapunov exponent for
the tent map.For r → 0 we obtain λ(r → 0) =
1
2
ln2.For r → 1/2 we obtain
λ(r →1/2) = 0.λ(r) has a maximum for r = 1/3 (tent map).Furthermore λ(r) is
a convex function in the interval (0,1/2).Thus we have
λ(r) ≤ ln2.
The C++ programtrajectory5.cpp finds the trajectory of the bungalow-tent map
for the control parameter r = 1/7 and the initial value x
0
= 1/17.We find x
1
= 6/17,
x
2
= 16/17,x
3
= 6/17.Thus the orbit is eventually periodic.
//trajectory5.cpp
#include <iostream>
#include"verylong.h"
#include"rational.h"
using namespace std;
inline void map(Rational<Verylong>& x,Rational<Verylong>& r)
{
Rational<Verylong> one("1"),two("2"),three("3");
Rational<Verylong> half("1/2");//number 1/2
Rational<Verylong> x1;
if(x < r) x1 = (one-r)*x/r;
else if((x >= r) && (x < half))
x1 = two*r*x/(one-two*r) + (one-three*r)/(one-two*r);
else if((x >= half) && (x < one-r))
x1 = two*r*(one-x)/(one-two*r)+(one-three*r)/(one-two*r);
else if((x <= one) && (x > one-r))
x1 = (one-r)*(one-x)/r;
x = x1;
}
int main(void)
{
Rational<Verylong> x0("1/17");//initial value 1/17
Rational<Verylong> r("1/7");//control parameter 1/7
unsigned long T = 10;//number of iterations
Rational<Verylong> x = x0;
cout <<"x[0] ="<< x << endl;
for(unsigned long t=0;t<T;t++)
{ map(x,r);cout <<"x["<< t+1 <<"] ="<< x << endl;}
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1.1.ONE-DIMENSIONAL MAPS 11
return 0;
}
In the sixth example we consider the Gauss map.The Gauss map f:[0,1] →[0,1]
is defined as
f(x):=
0 if x = 0
1/x if x = 0
where y denotes the fractional part of y.For example
3.2 = 0.2,17/3 =
2
3
.
Owing to the definition x
= 0 is a fixed point.Let x
0
= 23/101 be the initial value.
Then the orbit is given by
x
1
=
9
23
,x
2
=
5
9
,x
3
=
4
5
,x
4
=
1
4
,x
5
= 0
where x
5
= 0 is a fixed point.The Gauss map possesses an infinite number of
discontinuities and is not injective since each x
0
∈ [0,1] has countable infinite images.
The map admits an infinite number of unstable fixed points and shows chaotic
behaviour.For example x
= (
5 − 1)/2 (golden mean number) is a fixed point,
since x
= f(x
).The Gauss map preserves the Gauss measure on [0,1] which is
given by
m(A):=
1
ln2
Z
A
1
1 +x
dx.
The periodic points of the Gauss map are the reciprocal of the reduced quadratic
irrationals.These numbers are dense in [0,1).
//trajectory6.cpp
#include <iostream>
#include"verylong.h"
#include"rational.h"
using namespace std;
inline void map(Rational<Verylong>& x)
{
Rational<Verylong> zero("0"),one("1");
Rational<Verylong> x1;
if(x==zero) return;
x1 = one/x;
while(x1 >= one) x1 = x1-one;
x = x1;
}
int main(void)
{
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12 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
Rational<Verylong> x0("23/101");//initial value
unsigned long T = 10;//number of iterations
Rational<Verylong> x = x0;
cout <<"x[0] ="<< x << endl;
for(unsigned long t=0;t<T;t++)
{ map(x);cout <<"x["<< t+1 <<"] ="<< x << endl;}
return 0;
}
In the last example we define a fractal signal as the discrete time series
x
2t
= b(1 +x
t
),x
2t+1
= a(1 +x
t
),t = 0,1,2,...
where a < 1 and b < 1.Owing to the first equation the initial value x
0
is given
by x
0
= b/(1 − b).This sequence is generated when the elements of the two-scale
Cantor set are taken in a definite order.A C++ implementation using templates is
//trajectory7.cpp
#include <iostream>
using namespace std;
template <class T> void sequence(T* y,int N,T a,T b)
{
for(int t=1;t<N;t++)
{
if(t%2==0) y[t] = b*(T(1)+y[t/2]);
else y[t] = a*(T(1)+y[(t-1)/2]);
}
}
int main(void)
{
int N = 20;
double* y = new double[N];
double a = 1.0/3.0;double b = 1.0/2.0;
y[0] = b/(1.0-b);
sequence(y,N,a,b);
for(int j=0;j<N;j++) cout <<"y["<< j <<"] ="<< y[j] << endl;
delete[] y;
return 0;
}
The output from a time series should be stored in a file and then plotted.The next
program datagnu.cpp shows how to write the output data from an iteration to a
file.We consider the logistic map as an example.The output is stored in a file
called timeev.dat.We use the C++ style for the file manipulation.
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1.1.ONE-DIMENSIONAL MAPS 13
//datagnu.cpp
#include <fstream>//for ofstream,close
using namespace std;
int main(void)
{
ofstream data("timeev.dat");//filename timeev.dat
unsigned long T = 100;//number of iterations
double x0 = 0.618;//initial value
double x1;
for(unsigned long t=0;t<T;t++)
{ x1 = 4.0*x0*(1.0-x0);data << t <<""<< x0 <<"\n";x0 = x1;}
data.close();
return 0;
}
The data files can now be used to draw a graph of the time evolution using GNU-
plot.After we entered GNU-plot using the command gnuplot the plot command is
as follows
plot [0:10] ’timeev.dat’
This command plots the first eleven points of the time evolution.Furthermore we
can create a postscript file using the commands:
set term postscript default
set output"timeev.ps"
plot ’timeev.dat’
The next program shows how to write data to a file and read data from a file
using JAVA.We consider the logistic map.The data are stored in a file called
"timeev.dat".In the second part of the program we read the data back.In JAVA
the filename and the class name which includes the
public static void main(String args[])
method must coincide (case sensitive).We output data using a DataOutputStream
that is connected to a FileOutputStream via a technique called chaining of stream
objects.When the DataOutputStream object output is created,its constructor
is supplied a FileOutputStream object as an argument.The statement creates
a DataOutputStream object named output associated with the file timeev.dat.
The argument"timeev.dat"is passed to the FileOutputStream constructor which
opens the file.
Class DataOutputStream.A data output stream lets an application write primi-
tive (basic) Java data types to an output stream in a portable way.
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14 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
Class FileOutputStream.A file output stream is an output stream for writing
data to File or a FileDescriptor.
The method void writeDouble(double v) converts the double argument to a long
using the doubleToLongBits() method in class Double,and then writes that long
value to the underlying output stream as an 8-byte quantity,high byte first.The
method double readDouble() reads eight input bytes and returns a double value.
//FileManipulation.java
import java.io.*;
import java.lang.Exception;
public class FileManipulation
{
public static void main(String args[])
{ DataOutputStream output;
try
{ output = new DataOutputStream(new FileOutputStream("timeev.dat"));
int T = 10;
double x0 = 0.618;double x1;
output.writeDouble(x0);
System.out.println("The output is"+ x0);
for(int t=0;t<T;t++) { x1 = 4.0*x0*(1.0-x0);//logistic map
System.out.println("The output is"+ x1);
output.writeDouble(x1);
x0 = x1;
}
try { output.flush();output.close();}
catch(IOException e)
{
System.err.println("File not closed properly\n"+ e.toString());
System.exit(1);
}
}
catch(IOException e)
{
System.err.println("File not opened properly\n"+ e.toString());
System.exit(1);
}
try
{
FileInputStream fin = new FileInputStream("timeev.dat");
DataInputStream in = new DataInputStream(fin);
}
catch(Exception e) { }
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1.1.ONE-DIMENSIONAL MAPS 15
}
}
1.1.2 Fixed Points and Stability
Consider a map f:[0,1] → [0,1].The fixed points are defined as the solutions of
the equation
f(x
) = x
.
Assume that the map f is differentiable.Then the variational equation of x
t+1
=
f(x
t
) is defined as
y
t+1
=
d
d
f(x +y)
=0,x=x
t
,y=y
t
=
df
dx
(x = x
t
)y
t
.
A fixed point is called stable if
df
dx
(x = x
)
< 1.
Example.Consider the logistic map f:[0,1] →[0,1],f(x) = 4x(1 −x).We have
4x
(1 −x
) = x
to find the fixed points.The fixed points are given by x
1
= 0,x
2
= 3/4.Since
df
dx
= 4 −8x
we find that the fixed points x
1
= 0 and x
2
= 3/4 are unstable.♣
In the C++ program fixpointlog.cpp we consider the stability of the fixed points
for the logistic map x
t+1
= 4x
t
(1 −x
t
).We evaluate the variational equation of the
logistic equation and determine the stability of the fixed points.We test whether
the fixed points of the logistic map f(x) = 4x(1 − x) are unstable.We use the
header file derive.h from SymbolicC++ to do the differentiation.
//fixpointlog.cpp
#include <iostream>
#include <cmath>//for fabs
#include"verylong.h"
#include"rational.h"
#include"derive.h"
using namespace std;
int main(void)
{
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16 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
double x1 = 0.0;double x2 = 3.0/4.0;
Derive<double> C1(1.0);//constant 1.0
Derive<double> C4(4.0);//constant 4.0
Derive<double> X1,X2;
X1.set(x1);
Derive<double> R1 = C4*X1*(C1-X1);
double result1 = df(R1);
cout <<"result1 ="<< result1 << endl;
if(fabs(result1) > 1) cout <<"fixpoint x1 unstable"<< endl;
X2.set(x2);
Derive<double> R2 = C4*X2*(C1-X2);
double result2 = df(R2);
cout <<"result2 ="<< result2 << endl;
if(fabs(result2) > 1) cout <<"fixpoint x2 unstable";
return 0;
}
1.1.3 Invariant Density
Consider a one-hump fully developed chaotic map f:[0,1] →[0,1]
x
t+1
= f(x
t
)
where t = 0,1,2,....We define the invariant density ρ (also called probability
density) of the iterates,starting from an initial point x
0
,by
ρ(x):= lim
T→∞
1
T
T−1
X
t=0
δ(x −f
(t)
(x
0
))
where f
(0)
(x
0
) = x
0
and
f
(1)
(x
0
) = f(x
0
) = x
1
,...,f
(t)
(x
0
) = f
(t−1)
(f(x
0
)) = f(f
(t−1)
(x
0
)) = x
t
with t > 1.Here δ denotes the delta function.Not all starting points x
0
∈ [0,1] are
allowed in the definition.Those belonging to an unstable cycle must be excluded
since we are only interested in the stable chaotic trajectory.For any arbitrary (but
integrable in the Lebesgue sense) function g on the unit interval [0,1] the mean value
of that function along the chaotic trajectory is defined by
g(x):= lim
T→∞
1
T
T−1
X
t=0
g(x
t
) =
Z
1
0
ρ(x)g(x)dx.
Choosing g(x) = 1 we obtain the normalization condition
Z
1
0
ρ(x)dx = 1.
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1.1.ONE-DIMENSIONAL MAPS 17
Since the probability density is independent of the starting point x
0
,the expression
for ρ can also be written as
ρ(x) = lim
T→∞
1
T
T−1
X
t=0
δ(x −x
t+k
),k = 0,1,2,....
An integral equation for ρ can be derived as follows:Let σ be defined as
σ(y):=
Z
1
0
δ(y −f
(k)
(x))ρ(x)dx.
Let g be an arbitrary (but integrable in the sense of Lebesgue) function on [0,1].
Then
Z
1
0
σ(y)g(y)dy =
Z
1
0
Z
1
0
δ(y −f
(k)
(x))ρ(x)g(y)dydx.
Therefore we obtain
Z
1
0
σ(y)g(y)dy = lim
T→∞
1
T
T−1
X
t=0
Z
1
0
δ(x −f
(t)
(x
0
))g(f
(k)
(x))dx.
Using the properties of the delta function we arrive at
Z
1
0
σ(y)g(y)dy = lim
T→∞
1
T
T−1
X
t=0
g(f
(t+k)
(x
0
)).
Hence
Z
1
0
σ(y)g(y)dy = lim
T→∞
1
T
T−1
X
t=0
Z
1
0
δ(y −f
(t+k)
(x
0
))g(y)dy =
Z
1
0
ρ(y)g(y)dy.
Since the function g is arbitrarily chosen,we have to set σ(y) = ρ(y).Thus the
probability density ρ obeys the integral equation
ρ(y) =
Z
1
0
dxδ(y −f(x))ρ(x).
This equation is called the Frobenius-Perron integral equation.This equation has
many solutions (Kluiving et al [115]).Among these are the solutions associated
with the unstable periodic orbits.If these unstable solutions are left out of con-
sideration and the map is one-hump fully developed chaotic,then there is only one
stable chaotic trajectory exploring the unit interval [0,1] and the Frobenius-Perron
equation has a unique solution associated with the chaotic orbit.
Example.Consider the logistic map f:[0,1] → [0,1],f(x) = 4x(1 − x).For
the stable chaotic trajectory exploring the unit interval [0,1] the Frobenius-Perron
integral equation has the unique solution
ρ(x) =
1
p
x(1 −x)
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18 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
where
Z
1
0
ρ(x)dx = 1
and ρ(x) > 0 for x ∈ [0,1].We see that ρ(x) →∞for x →0 and x →1,respectively.
The solution can be found by iteration of
ρ
t+1
(y) =
Z
1
0
dxδ(y −f(x))ρ
t
(x)
with the initial density ρ
0
(x) = 1.♣
In the C++ program invdensity.cpp we determine numerically the invariant den-
sity for the logistic map x
t+1
= 4x
t
(1 − x
t
).We find the histogram for the logis-
tic map.We divide the unit interval [0,1] into 20 bins with bin size 0.05 each.
We calculate how many points exist in the intervals [0.05 ∙ i,0.05 ∙ (i + 1)),where
i = 0,1,2,...,19.This gives an approximation for the invariant density defined
above.For example,the number of points in the intervals [0,0.05) and [0.95,1.0] is
much higher than in the other intervals (bins).
//invdensity.cpp
#include <iostream>
#include <cmath>//for floor,sqrt
using namespace std;
void histogram(double* x,int* hist,double T,double xmax,
double xmin,int n_bins)
{
}
int main(void)
{
int T = 10000;//number of iterations
double xmax = 1.0;//length of interval xmax-xmin
double xmin = 0.0;
double bin_width = 0.05;
double* x = new double[T];//memory allocation
int n_bins = (int)(xmax-xmin)/bin_width;
cout <<"number of bins ="<< n_bins << endl;
//generating the data for the histogram
x[0] = (sqrt(5.0)-1.0)/2.0;//initial value
for(int t=0;t<(T-1);t++) x[t+1] = 4.0*x[t]*(1.0-x[t]);
int* hist = new int[n_bins];//memory allocation
//setting hist[i] to zero
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1.1.ONE-DIMENSIONAL MAPS 19
for(int i=0;i<n_bins;i++) hist[i] = 0;
histogram(x,hist,T,xmax,xmin,n_bins);
for(int i=0;i<n_bins;i++)
cout <<"hist["<< i <<"] ="<< hist[i] << endl;
delete[] x;delete[] hist;
return 0;
}
Consider the sine map.The sine map f:[0,1] →[0,1] is defined by
f(x):= sin(πx).
The map can also be written as a difference equation
x
t+1
= sin(πx
t
),t = 0,1,...
where x
0
∈ [0,1].The fixed points are determined by the solution of the equation
x
= sin(πx
).
The map admits two fixed points.One fixed point is given by x
1
= 0.The other
fixed point is determined from x
2
= sin(πx
2
) and x
2
> 0.We find x
2
= 0.73648....
The variational equation of the sine map takes the form
y
t+1
= π cos(πx
t
)y
t
,t = 0,1,....
Both fixed points are unstable.This can be seen by inserting x
1
and x
2
into the
variational equation.
To find the invariant density for the sine-map we replace in programinvdensity.cpp
the line
x[t+1] = 4.0*x[t]*(1.0-x[t]);
with
x[t+1] = sin(pi*x[t]);
and add const double pi=3.1415927;in front of this statement.The numerical
result suggests that the density for the sine map is quite similar to that of the lo-
gistic map.
Example.We find the invariant density for the bungalow-tent map f
r
:[0,1] →[0,1]
f
r
(x):=
8
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
:
1 −r
r
x for x ∈ [0,r)
2r
1 −2r
x +
1 −3r
1 −2r
for x ∈ [r,1/2)
2r
1 −2r
(1 −x) +
1 −3r
1 −2r
for x ∈ [1/2,1 −r)
1 −r
r
(1 −x) for x ∈ [1 −r,1]
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20 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
where r ∈ (0,1/2).To find the invariant density exactly we solve the Frobenius-
Perron integral equation.The Frobenius-Perron integral equation is given by
ρ
r
(x) =
Z
1
0
ρ
r
(y)δ(x −f
r
(y))dy.
We apply the identities for the delta function
δ(cy) ≡
1
|c|
δ(y),δ(g(y)) ≡
X
n
1
|g
(y
n
)|
δ(y −y
n
)
where the sum over n runs over all zeros with multiplicity 1 and g
(y
n
) denotes the
derivative of g taken at y
n
.Taking these identities into account and differentiating
in the sense of generalized functions we obtain the invariant density
ρ
r
(x) =
1
2 −3r
χ
[0,1−r]
(x) +
1 −2r
r(2 −3r)
χ
(1−r,1]
(x)
where χ is the indicator function,i.e.,
χ
A
(x):=
1 if x ∈ A
0 if x/∈ A
.
Thus the invariant density is constant in the interval [0,1−r).At 1−r the invariant
density jumps to another constant value.In the calculations we have to consider
two domains for x,[0,1 − r) and [1 − r,1].The Liapunov exponent is calculated
using
λ(r) =
Z
1
0
ρ
r
(x) ln
df
r
dx
dx
where we differentiated in the sense of generalized functions.Thus we find that the
Liapunov exponent as a smooth function of the control parameter r is given by
λ(r) =
1 −r
2 −3r
ln
1 −r
r
+
1 −2r
2 −3r
ln
2r
1 −2r
.
For r = 1/3 we obviously obtain λ(1/3) = ln2.This is the Liapunov exponent for
the tent map.For r →0 we obtain
λ(r →0) =
1
2
ln2.
For r → 1/2 we obtain λ(r → 1/2) = 0.The Liapunov exponent λ(r) has a
maximum for r = 1/3 (tent map).Furthermore λ(r) is a convex function in the
interval (0,1/2).We have λ(r) ≤ ln2.The numerical simulation confirms the result
for the invariant density,i.e.constant in the interval [0,1−r) and another constant
in the interval [1 −r,1].In the numerical simulation we have to set one of the bins
boundary points to 1 −r.♣
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1.1.ONE-DIMENSIONAL MAPS 21
1.1.4 Liapunov Exponent
Here we calculate the Liapunov exponent λ for one-dimensional chaotic maps.Con-
sider the one-dimensional map
x
t+1
= f(x
t
)
where t = 0,1,2,...and x
0
∈ [0,1].The variational equation (also called the
linearized equation) of this map takes the form
y
t+1
=
df
dx
(x
t
)y
t
with y
0
= 0.We assumed that f is differentiable.The Liapunov exponent λ is
defined as
λ(x
0
,y
0
):= lim
T→∞
1
T
ln
y
T
y
0
.
Example.We calculate the Liapunov exponent for the logistic map.Thus f(x) =
4x(1 −x) and
df
dx
= 4 −8x.
Consequently we obtain the variational equation
y
t+1
= (4 −8x
t
)y
t
,t = 0,1,...
with y
0
= 0.The exact solution of the logistic map is given by
x
t
=
1
2
1
2
cos(2
t
arccos(1 −2x
0
)).
For almost all initial values the Liapunov exponent is given by λ = ln2.♣
In the C++ program Liapunov1.cpp we evaluate the Liapunov exponent by using
the variational equation.Overflow occurs if T is made too large.In an alternative
method we use nearby trajectories and reset the distance between the two trajecto-
ries after each time step.Thus we avoid overflow for large T.
//Liapunov1.cpp
#include <iostream>
#include <cmath>//for fabs,log
using namespace std;
int main(void)
{
unsigned long T = 200;//number of iterations
double x = 0.3;//initial value for logistic map
double y = 1.0;//initial value for variational map
double x1,y1;
for(unsigned long t=0;t<T;t++)
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22 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
{ x1 = x;y1 = y;
x = 4.0*x1*(1.0-x1);//logistic map
y = (4.0-8.0*x1)*y1;//variational map
}
//notice that y becomes large very quickly
double lambda = log(fabs(y))/((double) T);//Liapunov exponent
cout <<"lambda ="<< lambda << endl;
//alternative method
double eps = 0.001;double xeps,xeps1;
x = 0.3;xeps = x-eps;
//x and xeps are nearby points
double sum = 0.0;
T = 1000;
double distance;
for(unsigned long t=0;t<T;t++)
{
x1 = x;xeps1 = xeps;
x = 4.0*x1*(1.0-x1);xeps = 4.0*xeps1*(1.0-xeps1);
double distance = fabs(x-xeps);
sum += log(distance/eps);
xeps = x-eps;
}
lambda = sum/((double) T);
cout <<"lambda ="<< lambda << endl;
return 0;
}
In the following C++ program we use the Rational,Verylong and Derive class
of SymbolicC++ to find an approximation of the Liapunov exponent.The Derive
class provides the derivative.Thus the variational equation is obtained via exact
differentiation
//Liapunov2.cpp
//iteration of logistic equation and variational equation
#include <iostream>
#include <cmath>//for fabs,log
#include"verylong.h"
#include"rational.h"
#include"derive.h"
using namespace std;
int main(void)
{
int T = 100;//number of iterations
double x = 1.0/3.0;//initial value
double x1;
double y = 1.0;
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1.1.ONE-DIMENSIONAL MAPS 23
Derive<double> C1(1.0);//constant 1.0
Derive<double> C4(4.0);//constant 4.0
Derive<double> X;
cout <<"t = 0 x ="<< x <<""<<"y ="<< y << endl;
for(int t=1;t<=T;t++)
{ x1 = x;x = 4.0*x1*(1.0-x1);
X.set(x1);
Derive<double> Y = C4*X*(C1-X);
y = df(Y)*y;
cout <<"t ="<< t <<""<<"x ="<< x <<""
<<"y ="<< y << endl;
}
double lambda = log(fabs(y))/((double) T);
cout <<"approximate value for lambda ="<< lambda << endl;
int M = 9;
Rational<Verylong> u1;
Rational<Verylong> u("1/3"),v("1");
Rational<Verylong> K1("1"),K2("4");
Derive<Rational<Verylong> > D1(K1);//constant 1
Derive<Rational<Verylong> > D4(K2);//constant 4
Derive<Rational<Verylong> > U;
cout <<"j = 0 u ="<< u <<""<<"v ="<< v << endl;
for(int j=1;j<=M;j++)
{
u1 = u;u = K2*u1*(K1-u1);
U.set(Rational<Verylong>(u1));
Derive<Rational<Verylong> > V = D4*U*(D1-U);
v = df(V)*v;
cout <<"j ="<< j <<""
<<"u ="<< u <<""<<"v ="<< v << endl;
}
lambda = log(fabs(double(v)))/((double) M);
cout <<"approximate value for lambda ="<< lambda << endl;
return 0;
}
Consider the sine map.The sine map f:[0,1] →[0,1] is defined by
f(x):= sin(πx).
The map can be written as the difference equation
x
t+1
= sin(πx
t
)
where t = 0,1,2,...and x
0
∈ [0,1].The variational equation of the sine equation is
given by
y
t+1
=
df
dx
(x = x
t
)y
t
= π cos(πx
t
)y
t
.
To find the Liapunov exponent for the sine-map we replace in the C++ program
Liapunov1.cpp the line
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24 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
x = 4.0*x1*(1.0-x1);xeps = 4.0*xeps1*(1.0-xeps1);
by
x = sin(pi*x1);xeps = sin(pi*xeps1);
and add const double pi = 3.14159;in front of this statement.For T = 5000 we
find λ = 0.689.Thus there is numerical evidence that the sine-map shows chaotic
behaviour.
1.1.5 Autocorrelation Function
Consider a one-dimensional difference equation f:[0,1] →[0,1]
x
t+1
= f(x
t
)
where t = 0,1,2,....The time average is defined as
x
t
:= lim
T→∞
1
T
T−1
X
t=0
x
t
.
Obviously,x
t
depends on the initial value x
0
.The autocorrelation function is
defined as
C
xx
(τ):= lim
T→∞
1
T
T−1
X
t=0
(x
t
−x
t
)(x
t+τ
−x
t
)
where τ = 0,1,2,....The autocorrelation function depends on the initial value x
0
.
Example.For the logistic map f:[0,1] →[0,1],f(x) = 4x(1 −x) we find that the
time average for almost all initial conditions is given by
x
t
=
1
2
.
The autocorrelation function is given by
C
xx
(τ) =
1
8
for τ = 0
0 otherwise
for almost all initial conditions.♣
The C++ program autocorrelation.cpp calculates the time average and autocor-
relation function for the logistic map.
//autocorrelation.cpp
#include <iostream>
using namespace std;
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1.1.ONE-DIMENSIONAL MAPS 25
double average(double* x,int T)
{
double sum = 0.0;
for(int t=0;t<T;t++) { sum += x[t];}
double av = sum/((double) T);
return av;
}
void autocorr(double* x,double* CXX,int T,int length,double av)
{
for(int tau=0;tau<length;tau++)
{
double C = 0.0;
double diff = (double) (T-length);
for(int t=0;t<diff;t++) { C += (x[t]-av)*(x[t+tau]-av);}
CXX[tau] = C/(diff+1.0);
}//end for loop tau
}
int main(void)
{
const int T = 4096;
double* x = new double[T];
x[0] = 1.0/3.0;//initial value
for(int t=0;t<(T-1);t++) { x[t+1] = 4.0*x[t]*(1.0-x[t]);}
double av = average(x,T);
cout <<"average value ="<< av << endl;
int length = 11;
double* CXX = new double[length];
autocorr(x,CXX,T,length,av);
delete[] x;
for(int tau=0;tau<length;tau++)
cout <<"CXX["<< tau <<"] ="<< CXX[tau] << endl;
delete[] CXX;
return 0;
}
The output is (exact solution is 0.5,CXX[0]=1/8,CXX[1]=0,CXX[2]=0 etc.)
average value = 0.497383
CXX[0] = 0.125707
CXX[1] = 0.00134996
CXX[2] = -0.000105384
CXX[3] = -0.000289099
CXX[4] = 0.00477107
CXX[5] = -0.00186259
CXX[6] = 0.00383531
CXX[7] = -0.00425356
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26 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
CXX[8] = -0.00288615
CXX[9] = -0.00110183
CXX[10] = -0.00148765
1.1.6 Discrete One-Dimensional Fourier Transform
The discrete Fourier transform is an approximation of the continuous Fourier trans-
form.The discrete transform is used when a set of function sample values,x(t),are
available at equally spaced time intervals numbered t = 0,1,...,T − 1.The dis-
crete Fourier transform maps the given set of function values into a set of uniformly
spaced sine waves whose frequencies are numbered k = 0,1,...,T −1,and whose
amplitudes are given by
ˆx(k) =
1
T
T−1
X
t=0
x(t) exp
−i2πk
t
T
.
This equation can be written as
ˆx(k) =
1
T
T−1
X
t=0
x(t) cos
2πk
t
T
i
T
T−1
X
t=0
x(t) sin
2πk
t
T
.
The inverse discrete Fourier transformation is given by
x(t) =
T−1
X
k=0
ˆx(k) exp
i2πt
k
T
.
To find the inverse Fourier transformation we use the fact that
T−1
X
k=0
exp
i2πk
(n −m)
T
= Tδ
nm
where δ
nm
denotes the Kronecker symbol.
In the first C++ program (Fourier.cpp) we consider the time series
x(t) = cos(2πt/T)
where T = 8 and t = 0,1,2,...,T −1.We find the discrete Fourier transform ˆx(k)
(k = 0,1,...,T −1).We have
ˆx(k) =
1
8
7
X
t=0
cos
2πt
8
e
−i2πkt/8
.
Using the identity
cos(2πt/8) ≡
e
i2πt/8
+e
−i2πt/8
2
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1.1.ONE-DIMENSIONAL MAPS 27
we find
ˆx(k) =
1
16
7
X
t=0
(e
i2πt(1−k)/8
+e
−i2πt(1+k)/8
).
Consequently,
ˆx(k) =
8
>
<
>
:
1
2
for k = 1
1
2
for k = 7
0 otherwise
.
//fourier.cpp
#include <iostream>
#include <cmath>//for cos,sin
using namespace std;
int main(void)
{
const double pi = 3.14159;
int T = 8;
double* x = new double[T];
for(int t=0;t<T;t++) x[t] = cos(2.0*pi*t/((double) T));
double* rex = new double[T];double* imx = new double[T];
for(int k=0;k<T;k++)
{ double cossum = 0.0,sinsum = 0.0;
for(int t=0;t<T;t++)
{
cossum += x[t]*cos(2.0*pi*k*t/((double) T));
sinsum += x[t]*sin(2.0*pi*k*t/((double) T));
}
rex[k] = cossum/((double) T);imx[k] = -sinsum/((double) T);
}
//display the output
for(int k=0;k<T;k++)
{
cout <<"rex["<< k <<"] ="<< rex[k] <<"";
cout <<"imx["<< k <<"] ="<< imx[k] << endl;
}
delete[] x;delete[] rex;delete[] imx;
return 0;
}
In the next C++ program (fourierlog.cpp) we consider the logistic map x
t+1
=
4x
t
(1 −x
t
),where t = 0,1,2,...and x
0
∈ [0,1].We assume that we have a set of T
samples from the logistic map,i.e.,x
0
,x
1
,x
2
,...,x
T−1
.
//fourierlog.cpp
#include <iostream>
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28 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
#include <cmath>//for cos,sin
using namespace std;
int main(void)
{
const double pi = 3.1415927;
int T = 256;
double* x = new double[T];
x[0] = 0.5;
for(int t=0;t<(T-1);t++) x[t+1] = 4.0*x[t]*(1.0-x[t]);
double* rex = new double[T];double* imx = new double[T];
for(int k=0;k<T;k++)
{ double cossum = 0.0,sinsum = 0.0;
for(int t=0;t<T;t++)
{
cossum += x[t]*cos(2.0*pi*k*t/((double) T));
sinsum += x[t]*sin(2.0*pi*k*t/((double) T));
}
rex[k] = cossum/((double) T);imx[k] = -sinsum/((double) T);
}
//display the output
for(int k=0;k<T;k++)
{
cout <<"rex["<< k <<"] ="<< rex[k] <<"";
cout <<"imx["<< k <<"] ="<< imx[k] << endl;
}
delete[] x;delete[] rex;delete[] imx;
return 0;
}
1.1.7 Fast Fourier Transform
Let n ≥ 1.The discrete Fourier transform transforms an n-vector with real compo-
nents into a complex n-vector.Methods that compute the discrete Fourier trans-
form in O(N log N) complex floating-point operations are referred to as fast Fourier
transforms,FFT for short.Based on the odd-even decomposition of a trigonomet-
ric polynomial,a problem of size n = 2
k
is reduced to two problems of size 2
k−1
.
Subsequently,two problems of size 2
k−1
are reduced to two problems of size 2
k−2
.
Finally,n = 2
k
problems of size 1 are obtained,each of which is solved trivially.Let
ω be a primitive nth root of 1,i.e.
ω:= exp(2πi/n).
The matrix F
n
denotes the n ×n matrix with entries
f
jk
:= ω
jk
≡ e
2πijk/n
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1.1.ONE-DIMENSIONAL MAPS 29
where 0 ≤ j,k ≤ n −1.The discrete Fourier transform of the n-vector
P
T
= (p
0
,p
1
,...,p
n−1
)
is the product F
n
P.The components of F
n
P are
(F
n
P)
0
0
p
0
0
p
1
+∙ ∙ ∙ +ω
0
p
n−2
0
p
n−1
(F
n
P)
1
0
p
0
+ωp
1
+∙ ∙ ∙ +ω
n−2
p
n−2
n−1
p
n−1
.
.
.
(F
n
P)
i
0
p
0
i
p
1
+∙ ∙ ∙ +ω
i(n−2)
p
n−2
i(n−1)
p
n−1
.
.
.
(F
n
P)
n−1
0
p
0
n−1
p
1
+∙ ∙ ∙ +ω
(n−1)(n−2)
p
n−2
(n−1)(n−1)
p
n−1
.
Rewritten in a slightly different form,the ith component is
p
n−1
i
)
n−1
+p
n−2
i
)
n−2
+∙ ∙ ∙ +p
1
ω
i
+p
0
.
If we interpret the components of P as coefficients of the polynomial
p(x) = p
n−1
x
n−1
+p
n−2
x
n−2
+∙ ∙ ∙ +p
1
x +p
0
then the ith component is p(ω
i
) and computing the discrete Fourier transform of P
means evaluating the polynomial p(x) at ω
0
,ω,ω
2
,...,ω
n−1
,i.e.,at each of the nth
roots of 1.A Divide and Conquer algorithm is as follows.Assume that n = 2
k
for
some k ≥ 0.The problem is divided into smaller instances,solve those,and use the
solutions to get the solution for the current instance.To evaluate p at n points,we
evaluate two smaller polynomials at a subset of the points and then combine the
results appropriately.Since ω
n/2
= −1 we have for 0 ≤ j ≤ n/2 −1,
ω
(n/2)+j
= −ω
j
.
We order the terms of p(x) with even powers and the terms with odd powers as
follows
p(x) =
n−1
X
i=0
p
i
x
i
n/2−1
X
i=0
p
2i
x
2i
+x
n/2−1
X
i=0
p
2i+1
x
2i
.
We define
p
even
(x):=
n/2−1
X
i=0
p
2i
x
i
,p
odd
(x):=
n/2−1
X
i=0
p
2i+1
x
i
.
It follows that
p(x) = p
even
(x
2
) +x ∙ p
odd
(x
2
),p(−x) = p
even
(x
2
) −x ∙ p
odd
(x
2
).
To evaluate p at
1,ω,...,ω
(n/2)−1
,−1,−ω,...,−ω
(n/2)−1
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30 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
it suffices to evaluate p
even
and p
odd
at
1,ω
2
,...,(ω
(n/2)−1
)
2
and then do n/2 multiplications (for x ∙ p
odd
(x
2
)) and n additions and subtractions.
The polynomials p
even
and p
odd
can be evaluated recursively by the same scheme.
They are polynomials of degree n/2 −1 and will be evaluated at the n/2th roots of
unity
1,ω
2
,...,(ω
(n/2)−1
)
2
.
//fft1.cpp
#include <iostream>
#include <cmath>//for cos,sin
using namespace std;
void p(double wre,double wim,double *re,double *im,
double &fftre,double &fftim,const int M,int step,int init)
{
double pre,pim,w2re,w2im;
if(step==(1 << M))//<< shift operator
{
fftre = re[init]*wre-im[init]*wim;
fftim = im[init]*wre+re[init]*wim;
return;
}
w2re = wre*wre-wim*wim;w2im = 2.0*wre*wim;
p(w2re,w2im,re,im,pre,pim,M,step<<1,init);//peven
fftre = pre;fftim = pim;
p(w2re,w2im,re,im,pre,pim,M,step<<1,init+step);//podd
fftre += wre*pre-wim*pim;fftim += wre*pim+wim*pre;
}
void fft(double *re,double *im,double *ftre,double *ftim,const int M)
{
const double pi = 3.1415927;
int N = 1 << M;
double fftre,fftim,wre,wim,w2re,w2im;
for(int i=0;i<(N>>1);i++)
{
wre = cos(i*2.0*pi/N);wim = sin(i*2.0*pi/N);
w2re = wre*wre-wim*wim;w2im = 2.0*wre*wim;
p(w2re,w2im,re,im,fftre,fftim,M,2,0);//peven
ftre[i] = ftre[i+(N>>1)] = fftre;
ftim[i] = ftim[i+(N>>1)] = fftim;
p(w2re,w2im,re,im,fftre,fftim,M,2,1);//podd
ftre[i] += wre*fftre-wim*fftim;
ftre[i+(N>>1)] -= wre*fftre-wim*fftim;
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1.1.ONE-DIMENSIONAL MAPS 31
ftim[i] += wre*fftim+wim*fftre;
ftim[i+(N>>1)] -= wre*fftim+wim*fftre;
}
}
int main(void)
{
const double pi = 3.1415927;
const int M = 3;
int T = 1 << M;
double* re = new double[T];double* im = new double[T];
double* fftre = new double[T];double* fftim = new double[T];
for(int i=0;i<T;i++) { re[i] = cos(2.0*i*pi/T);}
for(int k=0;k<T;k++) { im[k] = 0.0;}
fft(re,im,fftre,fftim,M);
for(int k=0;k<T;k++)
cout <<"fftre["<< k <<"]="<< fftre[k]/T << endl;
cout << endl;
for(int k=0;k<T;k++)
cout <<"fftim["<< k <<"]="<< fftim[k]/T << endl;
delete[] re;delete[] im;delete[] fftre;delete[] fftim;
return 0;
}
A nonrecursive version is given below.We use in place substitution.
//FFT2.cpp
#include <iostream>
#include <cmath>//for sqrt,cos
using namespace std;
//dir = 1 gives the FFT tranform
//dir = -1 gives the inverse FFT transform
//n = 2^m is the length of the time series
//x[] is the real part of the signal
//y[] is the imaginary part of the signal
void FFT(int dir,unsigned long m,double* x,double* y)
{
unsigned long n,i,i1,j,k,i2,l,l1,l2;
double c1,c2,tx,ty,t1,t2,u1,u2,z;
//number of points n = 2^m
n = 1;
for(i=0;i<m;i++) n *= 2;
//bit reversal
i2 = n >> 1;
j = 0;
for(i=0;i<n-1;i++)
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32 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
{
if(i < j)
{
tx = x[i];ty = y[i];
x[i] = x[j];y[i] = y[j];x[j] = tx;y[j] = ty;
}
k = i2;
while(k <= j) { j -= k;k >>= 1;}
j += k;
}//end for loop
//compute the FFT
c1 = -1.0;c2 = 0.0;l2 = 1;
for(l=0;l<m;l++)
{
l1 = l2;l2 <<= 1;u1 = 1.0;u2 = 0.0;
for(j=0;j<l1;j++)
{
for(i=j;i<n;i+=l2)
{
i1 = i + l1;
t1 = u1*x[i1]-u2*y[i1];t2 = u1*y[i1]+u2*x[i1];
x[i1] = x[i]-t1;y[i1] = y[i]-t2;
x[i] += t1;y[i] += t2;
}
z = u1*c1-u2*c2;
u2 = u1*c2+u2*c1;u1 = z;
}
c2 = sqrt((1.0-c1)/2.0);
if(dir==1) c2 = -c2;
c1 = sqrt((1.0+c1)/2.0);
}
if(dir==1)
{ for(i=0;i<n;i++) { x[i]/= n;y[i]/= n;} }
}//end function FFT
unsigned long power(unsigned long m)
{
unsigned long r = 1;
for(unsigned long i=0;i<m;i++) r *= 2;
return r;
}
int main(void)
{
unsigned long m = 3;
const double pi = 3.1415927;
unsigned long n = power(m);
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1.1.ONE-DIMENSIONAL MAPS 33
double* x = new double[n];double* y = new double[n];
unsigned long k;
for(k=0;k<n;k++) { x[k] = cos(2.0*pi*k/n);y[k] = 0.0;}
//call FFT
FFT(1,m,x,y);
for(k=0;k<n;k++) { cout << x[k] <<""<< y[k] << endl;}
//call inverse FFT
cout <<"calling inverse FFT"<< endl;
FFT(-1,m,x,y);
for(k=0;k<n;k++) { cout << x[k] <<""<< y[k] << endl;}
return 0;
}
1.1.8 Logistic Map and Liapunov Exponent for r 2 [3;4]
We consider the logistic map
x
t+1
= rx
t
(1 −x
t
)
where t = 0,1,2,...,x
0
∈ [0,1] and r ∈ [3,4].Here r is the bifurcation parameter.
Thus the Liapunov exponent depends on r.We evaluate the Liapunov exponent for
r ∈ [3,4].The variational equation is given by
y
t+1
= r(1 −2x
t
)y
t
.
The Liapunov exponent is defined as
λ(x
0
,y
0
):= lim
T→∞
1
T
ln
y
T
y
0
.
The point r = 3 is a bifurcation point.The Liapunov exponent is given by λ = 0
for r = 3.In the range 3 < r < 3.5699.....we find periodic solutions.The Liapunov
exponent is negative.We also find period doubling.In the region
3.5699...< r < 4
we find chaotic behaviour (positive Liapunov exponent) but also periodic windows.
For example in the region
3.828...< r < 3.842...
we have a trajectory with period 3.The Liapunov exponent can be evaluated exactly
only for r = 4.One finds λ(r = 4) = ln2 for almost all initial values.In the program
lambdaf.cpp the Liapunov exponent is evaluated for the interval r ∈ [3.0,4.0] with
step size 0.001.
//lambdaf.cpp
#include <fstream>
#include <cmath>//for fabs,log
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34 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
using namespace std;
int main(void)
{
ofstream data("lambda.dat");
int T = 10000;//number of iterations
double x = 0.618;//initial value
double x1;
double eps = 0.0005;
double xeps = x-eps;
double xeps1;
double r = 3.0;double sum = 0.0;
while(r <= 4.0)
{
for(int t=0;t<T;t++)
{ x1 = x;x = r*x1*(1.0-x1);
xeps1 = xeps;xeps = r*xeps1*(1.0-xeps1);
double distance = fabs(x-xeps);
sum += log(distance/eps);
xeps = x-eps;
}
double lambda = sum/((double) T);
data << r <<""<< lambda <<"\n";
sum = 0.0;
r += 0.001;
}//end while
data.close();
return 0;
}
1.1.9 Logistic Map and Bifurcation Diagram
We consider the logistic map
x
t+1
= rx
t
(1 −x
t
)
where r ∈ [2,4] and x
0
∈ [0,1].Here r is a bifurcation parameter.We now study
the bifurcation diagram.For r ∈ [2,3) the fixed point x
= 1 −1/r is stable.The
fixed point x
= 0 is unstable in the range (2,4].For r = 3 (bifurcation point) the
stable fixed point x
= 1−1/r becomes unstable.We find a stable orbit of period 2.
With increasing r we find a period doubling process with repeated bifurcation from
2,4,8,...,2
n
,....
There is a threshold value
r
= 3.5699...
for the parameter r where the limit 2
n
,n → ∞ of the periodicity is reached.For
r = 4 the logistic map and all its iterates are ergodic and mixing.Within the in-
terval (r
,4) period triplings p3
n
n
etc.also occur (so-called
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1.1.ONE-DIMENSIONAL MAPS 35
periodic windows)
In the Java program Bifurcationlo.java we display the bifurcation diagram for
the interval r ∈ [2.0,4.0].
//Bifurcationlo.java
import java.awt.*;
import java.awt.Frame;
import java.awt.event.*;
import java.awt.Graphics;
public class Bifurcationlo extends Frame
{
public Bifurcationlo()
{
setSize(600,500);
{ public void windowClosing(WindowEvent event) { System.exit(0);}});}
public void paint(Graphics g)
{
int xmax = 600;int ymax = 400;
int j,k,m,n;
double x,xplot,yplot;
double r = 2.0;//bifurcation parameter
while(r <= 4.0)
{ xplot = xmax*(r-2.0)/2.0;x = 0.5;
for(j=0;j<400;j++) { x = r*x*(1.0-x);}
for(k=0;k<400;k++)
{ x = r*x*(1.0-x);
yplot = ymax*(1.0-x);
m = (int) Math.round(xplot);n = 50 + (int) Math.round(yplot);
g.drawLine(m,n,m,n);
}
r += 0.0005;
}//end while
}
public static void main(String[] args)
{ Frame f = new Bifurcationlo();f.setVisible(true);}
}
1.1.10 Random Number Map and Invariant Density
We consider methods for generating a sequence of random fractions,i.e.,random
real numbers u
t
,uniformly distributed between zero and one.Since a computer can
represent a real number with only finite accuracy,we shall actually be generating
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36 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
integers x
t
between zero and some number m.The fraction
u
t
= x
t
/m,t = 0,1,2,...
will then lie between zero and one.Usually m is the word size of the computer,so
x
t
may be regarded as the integer contents of a computer word with the radix point
assumed at the extreme right,and u
t
may be regarded as the contents of the same
word with the radix point assumed at the extreme left.The most popular random
number generators are special cases of the following scheme.We select four numbers
m,the modulus;m> 0
a,the multiplier;0 ≤ a < m
c,the increment;0 ≤ c < m
x
0
,the initial value;0 ≤ x
0
< m.
The desired sequence of pseudo-random numbers x
0
,x
1
,x
2
,...is then obtained by
the one-dimensional difference equation
x
t+1
= (ax
t
+c) mod m,t = 0,1,2,....
This is also called a linear congruential sequence.Taking the remainder mod m is
somewhat like determining where a ball will land in a spinning roulette wheel.
Example.The sequence obtained when m= 10 and x
0
= a = c = 7 is
7,6,9,0,7,6,9,0,....
This example shows that the sequence is not always very “random” for all choices
of m,a,c,and x
0
.♣
The example also illustrates the fact that congruential sequences always “get into a
loop”;i.e.,there is ultimately a cycle of numbers which is repeated endlessly.The
repeating cycle is called the period.The sequence given above has a period of length
4.A useful sequence will of course have a relatively long period.
In the C++ program we implement a linear congruential sequence.
//modulus.cpp
#include <iostream>
using namespace std;
int main(void)
{
unsigned long a = 7,c = 7;
unsigned long m = 10;//modulus
unsigned long x0 = 7;//initial value
int T = 10;//number of iterations
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1.1.ONE-DIMENSIONAL MAPS 37
unsigned long x1;
for(int t=0;t<T;t++) { x1 = a*x0 + c;
while(x1 >= m) x1 = x1-m;
x0 = x1;
cout <<"x["<< t <<"] ="<< x0 << endl;
}
a = 3125;c = 47;m = 2048;//m modulus
x0 = 3;//initial value
T = 12;//number of iterations
for(int t=0;t<T;t++)
{ x1 = a*x0 + c;
while(x1 >= m) x1 = x1-m;
x0 = x1;
cout <<"x["<< t <<"] ="<< x0 << endl;
}
return 0;
}
In the following C++ program we consider the chaotic sequence
x
t+1
= (π +x
t
)
5
mod 1 ≡ frac(π +x
t
)
5
and ask whether the sequence is uniformly distributed.
//random1.cpp
#include <iostream>
#include <cmath>//for sqrt,fmod
using namespace std;
int main(void)
{
const double pi = 3.14159;
int T = 6000;//number of iterations
double* x = new double[T];
x[0] = (sqrt(5.0)-1.0)/2.0;//initial value
for(int t=0;t<(T-1);t++)
{ double r = x[t]+pi;x[t+1] = fmod(r*r*r*r*r,1);}
const int N = 10;
double hist[N];
for(int j=0;j<N;j++) hist[j] = 0.0;
for(int k=0;k<T;k++)
hist[(int) floor(N*x[k])] = hist[(int) floor(N*x[k])]+1;
for(int l=0;l<N;l++)
cout <<"hist["<< l <<"] ="<< hist[l] << endl;
return 0;
}
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38 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
1.1.11 Random Number Map and Random Integration
We describe the Monte Carlo method for the calculation of integrals.We demon-
strate the technique on one-dimensional integrals.Let f:[0,1] → [0,1] be a con-
tinuous function.Consider the integral
I =
Z
1
0
f(x)dx.
We choose N number pairs (x
j
,y
j
) with uniform distribution and define z
j
by
z
j
:=
0 if y
j
> f(x
j
)
1 if y
j
≤ f(x
j
).
Putting
n =
N
X
j=1
z
j
we have n/N I.More precisely,we find
I = n/N +O(N
−1/2
).
The accuracy is not very good.The traditional formulas,such as Simpson’s for-
mula,are much better.However,in higher dimensions the Monte Carlo technique
is favourable,at least if the number of dimensions is ≥ 6.We consider the integral
as the mean value of f(ξ) where ξ is uniform.An estimate of the mean value is
I
1
N
N
X
j=1
f(ξ
j
).
This formula can easily be generalized to higher dimensions.
In the C++ program we use the map f:[0,1) →[0,1)
f(x) = (x +π)
5
mod 1
as random number generator and evaluate
Z
1
0
sin(x)dx = 0.459697694132.
//randint.cpp
#include <iostream>
#include <cmath>
using namespace std;
void randval(double* x,double pi)
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
1.1.ONE-DIMENSIONAL MAPS 39
{ *x = fmod((*x+pi)*(*x+pi)*(*x+pi)*(*x+pi)*(*x+pi),1);}
int main(void)
{
const double pi = 3.1415927;
unsigned long T = 20000;//number of iterations
double x = 0.5;//initial value
double sum = 0.0;
for(int t=0;t<T;t++) { randval(&x,pi);sum += sin(x);}
cout <<"The integral is ="<< sum/((double) T);
return 0;
}
The output is 0.461403.
In the Java program we use the same map as in the C++ progam for the random
number generator.
//Random1.java
class WrappedDouble
{
WrappedDouble(final double value) { this.value = value;}
public double value() { return value;}
public void value(final double newValue) { value = newValue;}
private double value;
}
class MathUtils
{
public static void randval(WrappedDouble x)
{ double y = Math.pow(x.value()+Math.PI,5);x.value(y-Math.floor(y));}
}
class Random1
{
public static void main(String[] args)
{
int n = 20000;
double sum = 0.0;
WrappedDouble x = new WrappedDouble(0.5);
for(int i=0;i<n;++i)
{ MathUtils.randval(x);sum += Math.sin(x.value());}
System.out.println("The integral is"+ sum/n);
}
}
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
40 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
1.1.12 Circle Map and Rotation Number
The circle map is given by
x
t+1
= f(x
t
) ≡ x
t
+Ω−
r
sin(2πx
t
),t = 0,1,...
which may be regarded as a transformation of the phase of one oscillator through
a period of the second one.The map depends on two bifurcation parameters:Ω
describes the ratio of undisturbed frequencies while the bifurcation parameter r
governs the strength of the nonlinear interaction.The subcritical (r < 1) mappings
are diffeomorphisms (and thus invertible) whereas the supercritical ones (r > 1) are
non-invertible and may exhibit chaotic behaviour.The borderline between these two
cases consists of the critical circle mappings - homeomorphisms with one (usually
cubic) inflection point.This corresponds to r = 1 in the family of this map.The
dynamics of the map may be characterized by the rotation number (also called
winding number)
ρ:= lim
T→∞
1
T
(f
(T)
(x) −x).
When f is invertible,the rotation number is well defined and independent of x.The
inverse function f
−1
does not exist for r > 1.For subcritical and critical maps this
number does not depend on the initial point x.The dependence ρ(Ω) is the so-called
devil’s staircase,in which each rational ρ = p/q is represented by an interval of Ω
values (which is named the p/q-locking interval).The set of all these intervals has
a full measure in the critical case.The locked motion in subcritical and critical
cases is represented by a stable periodic orbit of period q.The rotation number
is the mean number of rotations per iteration,i.e.,the frequency of the underlying
dynamical system.If r = 0 we obviously find ρ = Ω.Under iteration the variable
x
i
may converge to a series which is either periodic,
x
i+Q
= x
i
+P
with rational rotation number ρ = P/Q;quasiperiodic,with irrational rotation
number ρ = q;or chaotic where the sequence behaves irregularly.
//circle.cpp
#include <fstream>
#include <cmath>//for sin
using namespace std;
int main(void)
{
ofstream data("circle.dat");
const double pi = 3.1415927;
int T = 8000;//number of iterations
double r = 1.0;//parameter of map
double Omega = 0.0;
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
1.1.ONE-DIMENSIONAL MAPS 41
while(Omega <= 1.0)
{ double x = 0.3;//inital value
double x0 = x;
double x1;
for(int t=0;t<T;t++)
{ x1 = x;x = x1+Omega-r*sin(2.0*pi*x1)/(2.0*pi);}
double rho = (x-x0)/((double) T);
data << Omega <<""<< rho <<"\n";
Omega += 0.005;
}//end while
data.close();
return 0;
}
1.1.13 One-Dimensional Newton Method
Consider the equation f(x) = 0 where it is assumed that f:R → R is at least
twice differentiable.Let I be some interval containing a root of f.A root is a point
ex such that f(ex) = 0.We assume that the root is simple (also called multiplicity
one).The Newton method (Fr¨oberg [64]) can be derived by taking the tangent line
to the curve y = f(x) at the point (x
t
,f(x
t
)) corresponding to the current estimate,
x
t
of the root.The intersection of this line with the x-axis gives the next estimate
to the root,x
t+1
.The gradient of the curve y = f(x) at the point (x
t
,f(x
t
)) is
f
(x
t
),where
denotes differentiation.The tangent line at this point has the form
y = f
(x)x+b.Since this passes through (x
t
,f(x
t
)) we see that b = f(x
t
)−x
t
f
(x
t
).
Therefore the tangent line is
y = f
(x
t
)x +f(x
t
) −x
t
f
(x
t
).
To determine where this line cuts the x-axis we set y = 0.Taking this point of
intersection as the next estimate,x
t+1
,to the root we have
0 = f
(x
t
)x
t+1
+f(x
t
) −x
t
f
(x
t
).
We obtain the first order difference equation
x
t+1
= x
t
f(x
t
)
f
(x
t
)
,t = 0,1,2,....
This is the Newton method.This scheme has the form ’next estimate = current
estimate + correction term’.The correction term is −f(x
t
)/f
(x
t
) and this must
be small when x
t
is close to the root if convergence is to be achieved.This will
depend on the behaviour of f
(x) near the root and,in particular,difficulty will be
encountered when f
(x) and f(x) have roots close together.The Newton method is
of the form x
t+1
= g(x
t
) with
g(x):= x −
f(x)
f
(x)
.
THE NONLINEAR WORKBOOK (5th Edition) - Chaos, Fractals, Cellular Automata, Genetic Algorithms, Gene Expression
Programming, Support Vector Machine, Wavelets, Hidden Markov Models, Fuzzy Logic with C++, Java and SymbolicC++ Programs
© World Scientific Publishing Co. Pte. Ltd.
http://www.worldscibooks.com/chaos/8050.html
42 CHAPTER 1.NONLINEAR AND CHAOTIC MAPS
The order of the method can be examined.Differentiating this equation leads to
g
(x) =
f(x)f
(x)
(f
(x))
2
.
For convergence we require that
f(x)f
(x)
(f
(x))
2
< 1
for all x in some interval I containing the root.Since f(ex) = 0,the above condition
is satisfied at the root x = ex provided that f
(ex) = 0.Then provided that the
function g is continuous,an interval I must exist in the neighbourhood of the root | 25,009 | 72,220 | {"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} | 3.515625 | 4 | CC-MAIN-2021-39 | latest | en | 0.797683 |
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# Energy Methods | Strength of Materials (SOM) - Mechanical Engineering PDF Download
### Strain Energy
The work done by the load in straining the body is stored within the strained material in the form of strain energy.
Strain energy,
U = 1 / 2 P(Al)
U = P2L / 2AE
Put
P = AEAℓ / L
Or
U = σ/ 2E x V
Strain Energy Diagram
### Proof Resilience
The maximum strain energy that can be stored in a material is known as proof resilience.
U = σ/ 2E1
Where,
σ = p / A, u = 1 / 2 Ee2
Strain energy of prismatic bar with varying sections
Prismatic bar
Strain energy of non-prismatic bar with varying axial force
Ax = Cross-section of differential section.
Non-Prismatic Bar
Stresses due to
Work stored in the bar
.....(b)
By equating, stress will be
and If h is very small then
Strain Energy in Torsion
For solid shaft,
U = τ/ 4G x Volume of Shaft
For hollow shaft,
x Volume of Shaft
### Castigliano’s First Theorem
It the strain energy of an elastic structure can be expressed as a function of generalized displacement, then the partial derivative of the strain energy with respect to generalized displacement gives the generalized force
[Where M is function of W (load)]
Deflection:
Slope:
Theories of Failure
Theories of failure are defined as following groups:
1. Maximum Principal Stress Theory (Rankine theory)
• According to this theory, permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.
• For design, critical maximum principal stress (σ1) must not exceed the working stress (s1) for the material.
σ< σy
Note: For bittle material, it gives satisfactory result. Yield criteria for 3D stress system,
σ1 = σy or |σ3| = σry
Where, σy = Yield stress point in simple tension, and σy = Yield stress point in simple compression.
Stresses on rectangular Section
2. Maximum Principal Strain Theory (St. Venant’s theory)
According to this theory, a ductile material begins to yield when the maximum principal strain at which yielding occurs in simple tension.
For 3D stress system,
If ey = Yield point strain tensile σ/ E
ery = Yield point strain compressive σr/ E
According to theory, e1 = ey
Yield criteria:
And
For 2D system,
Rhombus
Note: This theory can estimate the elastic strength of ductile material.
3. Maximum Shear Stress Theory (Guest & Tresca’s theory)
According to this theory, failure of specimen subjected to any combination of loads when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.
For 3D system:
Yielding criteria,
τmax = 1 / 2 (σ1 - σ3) = σ/ 2
In case of 2D: σ1 – σ3 = σy
Yielding criteria, σ1 - σ2 = σy
This theory gives well estimation for ductile material.
4. Maximum Strain Energy Theory (Haigh’s theory)
• According to this theory, a body under complex stress fails when the total strain energy on the body is equal to the strain energy at elastic limit in simple tension. For 3D stress system yield criteria,
For 2D stress system,
EllipseThis theory does not apply to brittle material for which elastic limit stress in tension and in compression are different.
5. Maximum shear strain energy/Distortion energy theory/Mises-Henky theory
• It states that inelastic action at any point in a body, under any combination of stress begins, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point an a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension/compression test.
1/2[(σ1 - σ2)2 + (σ1 - σ3)2 + (σ3 - σ1)2] ≤ σ2For no failure
1/2[(σ1 - σ2)2 + (σ2 - σ3)2 + (σ3 - σ1)2] ≤ (σ/ FOS)2 For no failure
The document Energy Methods | Strength of Materials (SOM) - Mechanical Engineering is a part of the Mechanical Engineering Course Strength of Materials (SOM).
All you need of Mechanical Engineering at this link: Mechanical Engineering
## Strength of Materials (SOM)
37 videos|39 docs|45 tests
## FAQs on Energy Methods - Strength of Materials (SOM) - Mechanical Engineering
1. What are some energy methods used in the field of engineering?
Ans. Energy methods commonly used in engineering include the Finite Element Method (FEM), the Boundary Element Method (BEM), the Finite Difference Method (FDM), the Discrete Element Method (DEM), and the Rayleigh-Ritz Method. These methods are used to analyze and solve complex problems in various engineering disciplines.
2. How does the Finite Element Method (FEM) work in energy analysis?
Ans. The Finite Element Method (FEM) is a numerical technique used to solve problems in engineering and physics. In energy analysis, FEM divides a complex system into smaller, simpler elements or subdomains. These elements are interconnected at discrete points called nodes. By applying mathematical equations and boundary conditions to these elements, FEM calculates the energy distribution and behavior of the system.
3. What is the role of the Boundary Element Method (BEM) in energy analysis?
Ans. The Boundary Element Method (BEM) is a numerical technique used to solve problems in engineering and physics, particularly in energy analysis. BEM focuses on solving problems by discretizing only the boundary of a system instead of the entire volume. By applying mathematical equations and boundary conditions on the boundary elements, BEM determines the energy distribution and behavior of the system.
4. How does the Finite Difference Method (FDM) contribute to energy analysis?
Ans. The Finite Difference Method (FDM) is a numerical technique used to solve partial differential equations in energy analysis. FDM discretizes the domain of the problem into a grid of points and approximates the derivatives of the energy variables using finite difference formulas. By solving these equations iteratively, FDM calculates the energy distribution and behavior of the system.
5. What is the significance of the Rayleigh-Ritz Method in energy analysis?
Ans. The Rayleigh-Ritz Method is a numerical technique used to approximate the solutions of complex energy problems. It is particularly useful when analytical solutions are not feasible. In energy analysis, the Rayleigh-Ritz Method approximates the energy distribution and behavior of a system by representing the variables as a linear combination of known basis functions. By minimizing the energy functional, the Rayleigh-Ritz Method determines the most accurate approximation for the system's behavior.
## Strength of Materials (SOM)
37 videos|39 docs|45 tests
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# A man running along a straight road with uniform velocity →u=uˆi feels that the rain is falling vertically down along −ˆj. If he doubles his speed,he finds that the rain is coming at an angle θ with the vertical. The velocity of the rain with respect to the ground is :
A
uiutanθ^j
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B
uiutanθˆj
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C
utanθuj
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D
utanθiuj
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Solution
## The correct option is B ui−utanθˆjGiven,Velocity of man =→u=u^iLet the velocity of rain =→v=x^i+y^jNow, in first case velocity of rain with respectto man =→VR=→v−→u =(x−u)^i+y^jGiven that It x− component is zero asrain is falling vertically down, and y component is along −ve directionSo, x−u=0⇒x=u∴ velocity of rain =u^i+y^jIn second case when be will doobles his speed velocity of rain with respect to man =→VR⇒→VR=→v−→u=u^i+y^j−2u^i=−u^i+y^jNow this case rain is coming at an angle θ to the vertical∴tanθ=−uy⇒y=−utanθ∴ velocity of rain with respect to ground=→V=x^i+y^j=u^i−ytanθ^j∴ Option B is correct
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Thread starter 6 years ago
#1
I have spent over 2 hours of attempting this question, reviewing online guides, forums, solutions etc, and the only conclusion I've come to is that I really have no idea what I'm doing.
The question is this: http://imgur.com/lF7JA6g
I have absolutely no idea where to start or what to do, but after pages and pages of failed attempts I'm tearing my hair out.
Help, please?
0
reply
6 years ago
#2
(Original post by Kalia_)
I have spent over 2 hours of attempting this question, reviewing online guides, forums, solutions etc, and the only conclusion I've come to is that I really have no idea what I'm doing.
The question is this: http://imgur.com/lF7JA6g
I have absolutely no idea where to start or what to do, but after pages and pages of failed attempts I'm tearing my hair out.
Help, please?
Get the position vectors of M and Q in terms of a and c, then get the vector equations of MQ and AC and solve simultaneously.
0
reply
Thread starter 6 years ago
#3
(Original post by brianeverit)
Get the position vectors of M and Q in terms of a and c, then get the vector equations of MQ and AC and solve simultaneously.
I have no idea what a position vector or a vector equation is, or how you would go about finding them, sorry.
As I said before, I have no idea what I'm doing, this question is in the very first exercise of the topic and I've never encountered vectors before.
0
reply
6 years ago
#4
(Original post by Kalia_)
I have no idea what a position vector or a vector equation is, or how you would go about finding them, sorry.
As I said before, I have no idea what I'm doing, this question is in the very first exercise of the topic and I've never encountered vectors before.
Ok, did you do GCSE maths?
Vectors will come up in that course.
Section 19 below will help:
http://www.cimt.plymouth.ac.uk/proje...se/allgcse.htm
From memory I used similar triangles when teaching Edexcel M1 last year.
If I am going to be 100% honest the question serves no great purpose in M1 but its a lovely piece of maths to be enjoyed regardless.
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Thread starter 6 years ago
#5
(Original post by m4ths/maths247)
Ok, did you do GCSE maths?
Vectors will come up in that course.
Section 19 below will help:
http://www.cimt.plymouth.ac.uk/proje...se/allgcse.htm
Thanks for the link, gonna have a read through and take notes - I honestly have no memory of ever having seen vectors before but I suppose I did them!
(Original post by m4ths/maths247)
From memory I used similar triangles when teaching Edexcel M1 last year.
If I am going to be 100% honest the question serves no great purpose in M1 but its a lovely piece of maths to be enjoyed regardless
If you say it's not important as a key topic or anything, I think I'll leave it - I was panicking because so far I've been fine with all the questions and then suddenly this one totally throws me... so that puts me at ease a bit, thanks.
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6 years ago
#6
You can build knowledge or follow rules and algorithms when studying maths.
I always think its better to get and understanding of what is going on here. It will help if you decide to do C4 too.
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6 years ago
#7
(Original post by Kalia_)
If you say it's not important as a key topic or anything, I think I'll leave it
Vectors are very important in Mechanics!
The point that the previous poster was making was that this question isn't really an M1 question - it's testing your basic knowledge of vectors, not any underlying mechanical principles.
So you should still try to solve this question - you will find the technique useful in other work.
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# Kysymys
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Milja talletti 64145 € ajalle 31.3. – 28.8. ja sai talletukselleen nettokorkoa 292,4 €, kun lähdevero oli 30 %. Mikä oli talletuksen nimellinen vuotuinen korkokanta (= korkokanta, joka sisältää sekä tallettajan että verottajan osuuden)?
Apr 25, 2023
#1
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This is in Finnish:
Milja deposited €64,145 for 31.3. – 28.8. and received a net interest of €292.4 on his deposit, when the withholding tax was 30%. What was the nominal annual interest rate of the deposit (= interest rate that includes both the depositor's and the taxpayer's share)?
March 31 to August 28 ==150 days
292.40 ==net interest before 30% tax.
292.40 / [100% - 30%] ==292.40 / 0.70 ==417.71 - gross amount of interest before 30% tax.
64,145 x [150 / 365] x R ==417.71
26,360.96 x R ==417.71
R==417.71 / 26,360.96 ==1.584578 % - gross annual simple interest rate.
R==292.40 / 26,360.96==1.109216% - net annual interest rate after 30% tax.
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Unformatted text preview: e that the variance of the uniform distribution will only depend on the length of the interval. To see that it will be a multiple of (b a)2 note that Z = X/(b a) is uniform on [ 1/2, 1/2] and then use part (d) of Theorem A.3 to conclude var (X ) = (b a)2 var (Z ). Of course one needs calculus to conclude that var (Z ) = 1/12. Generating functions will be used at several points in the text. If pk = P (X = k ) is the distribution of X then the generating function is (x) = P1 P1 k (1) = k=0 pk = 1. Di↵erentiating (and not worrying about the k=0 pk x . detail of interchanging the sum and th...
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Ask a homework question - tutors are online | 277 | 986 | {"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-2017-13 | longest | en | 0.86022 |
https://www.kylesconverter.com/length/metric-miles-to-links-(gunter's-surveyor's) | 1,675,171,941,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00694.warc.gz | 902,891,801 | 5,584 | # Convert Metric Miles to Links (gunter's Surveyor's)
### Kyle's Converter > Length > Metric Miles > Metric Miles to Links (gunter's Surveyor's)
Metric Miles Links (gunter's Surveyor's) (lnk) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
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Unit Descriptions
1 Metric Mile:
1609.344 Meters (SI base unit). An approximation of a statute mile in a round figure of meters. 1 Metric mile = 1609.344 m.
1 Link (Gunter's Surveyor's):
One hundredth (1/100) of a chain (Gunter's, Surveyor's). A chain having 66 ft or 4 rods. A link being approximately 0.201168 meters (SI base unit). 1 lnk ≈ 0.201168 m.
Conversions Table
1 Metric Miles to Links (gunter's Surveyor's) = 800070 Metric Miles to Links (gunter's Surveyor's) = 560000
2 Metric Miles to Links (gunter's Surveyor's) = 1600080 Metric Miles to Links (gunter's Surveyor's) = 640000
3 Metric Miles to Links (gunter's Surveyor's) = 2400090 Metric Miles to Links (gunter's Surveyor's) = 720000
4 Metric Miles to Links (gunter's Surveyor's) = 32000100 Metric Miles to Links (gunter's Surveyor's) = 800000
5 Metric Miles to Links (gunter's Surveyor's) = 40000200 Metric Miles to Links (gunter's Surveyor's) = 1600000
6 Metric Miles to Links (gunter's Surveyor's) = 48000300 Metric Miles to Links (gunter's Surveyor's) = 2400000
7 Metric Miles to Links (gunter's Surveyor's) = 56000400 Metric Miles to Links (gunter's Surveyor's) = 3200000
8 Metric Miles to Links (gunter's Surveyor's) = 64000500 Metric Miles to Links (gunter's Surveyor's) = 4000000
9 Metric Miles to Links (gunter's Surveyor's) = 72000600 Metric Miles to Links (gunter's Surveyor's) = 4800000
10 Metric Miles to Links (gunter's Surveyor's) = 80000800 Metric Miles to Links (gunter's Surveyor's) = 6400000
20 Metric Miles to Links (gunter's Surveyor's) = 160000900 Metric Miles to Links (gunter's Surveyor's) = 7200000
30 Metric Miles to Links (gunter's Surveyor's) = 2400001,000 Metric Miles to Links (gunter's Surveyor's) = 8000000
40 Metric Miles to Links (gunter's Surveyor's) = 32000010,000 Metric Miles to Links (gunter's Surveyor's) = 80000000
50 Metric Miles to Links (gunter's Surveyor's) = 400000100,000 Metric Miles to Links (gunter's Surveyor's) = 800000000
60 Metric Miles to Links (gunter's Surveyor's) = 4800001,000,000 Metric Miles to Links (gunter's Surveyor's) = 8000000000 | 780 | 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} | 2.671875 | 3 | CC-MAIN-2023-06 | longest | en | 0.77236 |
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# For a trade embargo against a particular country to succeed,
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19 Oct 2005, 02:56
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16. For a trade embargo against a particular country to succeed, a high degree of both international accord and ability to prevent goods from entering or leaving that country must be sustained. A total blockade of Patria’s ports is necessary to an embargo, but such an action would be likely to cause international discord over the embargo.
The claims above, if true, most strongly support which of the following conclusions?
(A) The balance of opinion is likely to favor Patria in the event of a blockade.
(B) As long as international opinion is unanimously against Patria, a trade embargo is likely to succeed.
(C) A naval blockade of Patria’s ports would ensure that no goods enter or leave Patria.
(D) Any trade embargo against Patria would be likely to fail at some time.
(E) For a blockade of Patria’s ports to be successful, international opinion must be unanimous.
11 An experiment was done in which human subjects recognize a pattern within a matrix of abstract designs and then select another design that completes that pattern. The results of the experiment were surprising. The lowest expenditure of energy in neurons in the brain was found in those subjects who performed most successfully in the experiments.
Which of the following hypotheses best accounts for the findings of the experiment?
(A) The neurons of the brain react less when a subject is trying to recognize patterns than when the subject is doing other kinds of reasoning.
(B) Those who performed best in the experiment experienced more satisfaction when working with abstract patterns than did those who performed less well.
(C) People who are better at abstract pattern recognition have more energy-efficient neural connections.
(D) The energy expenditure of the subjects brains increases when a design that completes the initially recognized pattern is determined.
(E) The task of completing a given design is more capably performed by athletes, whose energy expenditure is lower when they are at rest than is that of the general population.
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19 Oct 2005, 06:16
D
C
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19 Oct 2005, 09:01
I still wait for some more ans on Q 1
but for II ...why cant the ans be A...
neurons in the brain react less = lowest expenditure of energy in neurons
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19 Oct 2005, 09:13
(1)
(a) international accord AND
from the premises(for Patria):
ports blockade -> ~ international accord
D is correct.
so (ports blockade) & (international accord) = (~ international accord
) & (international accord)
(E) is saying:
we cannot infer it from the argument.
(2)
(A) is qrong as we donot know neurons energy efficiency behavior for "other kind of reasoning". and therefore cannot compare.
(C) is correct. all we know from the experiment is that subjects' neurons are more energy efficient while performing "abstract pattern recognition"
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19 Oct 2005, 10:03
1) I am stuck btw A and D...
while accord is needed, no where does it mention that accord needs to unanimous...so D it
2) C
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19 Oct 2005, 10:30
I would choose A for the first one. "such an action would be likely to cause international discord over the embargo" gives the idea that international community would not likely support the embargo.
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19 Oct 2005, 11:11
I would go with D and C
For the first, all the other choices weaken/don't apply to the conclusion.
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19 Oct 2005, 12:10
I choose E and C.
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19 Oct 2005, 12:16
I agree with D for the first question too. If A and B has to happen for C to happen, and if A would lead to non B, then A and B can never happen in the same time and C would not happen, in other words, the embago would fail.
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19 Oct 2005, 16:02
sushom101 wrote:
I still wait for some more ans on Q 1
but for II ...why cant the ans be A...
neurons in the brain react less = lowest expenditure of energy in neurons
Q1)
(A) The balance doesn't have to be in favor of Patria. It only has to be "not a high degree of accord", which doesn't mean higher than 50%(=the balance).
(B) Sounds true, but not relevant with the conclusion.
(C) Highly likely, but not necessarily true.
(D) Any trade embargo against Patria would be likely to fail at some time. Though I'm a little bit suspicious about the term "any trade embargo", I guess this is the best answer.
(E) It doesn't have to be "unanimous(100%)". It only has to be "high degree of accord(80~90%)". High degree doesn't mean unanimousness.
Q2)
Neurons in the brain of a subject react less =/= lowest expenditure of energy in neurons
Neurons in the brain of a successful performer = lowest expenditure of energy in neurons
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19 Oct 2005, 17:09
Question 1
Premise
1) For trade embargo to succeed, both intl' accord and ability to prevent good from entering or leaving that country must be sustained
2) Total blockade of Patria's ports is nescessary to an embargo, although such an action will cause international discord
If there is going to be international discord, then the conclusion that is most likely to happen is given in D. The trade embargo might not succeed.
In A), we can't support if the international community will favor Patria. We are only told the community will not be happy with a blockade.
In B), it changes the requirement of a successful trade embargo
In C), the word 'naval embargo' isn't even mentioned in the passage
In E), the conclusion is twisted. For the trade embargo to succed, international opinion must be unaminous, but for the blockade to succeed, opinion is not important.
D is the best choice.
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19 Oct 2005, 17:14
Question 2
I'll take C for the second question.
A) - if true, then it would apply to all subjects and not just a certain group
B) - can't reason from the passage
D) - same as B
E) - to far-fetched. We're not sure if athletes will perform better as they may be other requirements apart from energy expenditure
C is the best choice.
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20 Oct 2005, 12:28
In 1 one can discard all the answer except D. In 2 one can discard all the questions except C. SO I choose D and C.
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21 Oct 2005, 02:18
OA is D,C
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21 Oct 2005, 02:18
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# For a trade embargo against a particular country to succeed,
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,493 | 9,192 | {"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} | 3.390625 | 3 | CC-MAIN-2017-51 | latest | en | 0.926944 |
http://mathoverflow.net/revisions/11242/list | 1,369,161,297,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700438490/warc/CC-MAIN-20130516103358-00077-ip-10-60-113-184.ec2.internal.warc.gz | 161,303,558 | 7,180 | For a totally real field F, the Deligne-Ribet p-adic L-function provides a generalization of the p-adic Riemann zeta function exists, namely the p-adic Dedekind zeta function $\zeta_{F,p}$. \zeta_{F,p}$(as proved independently by Deligne–Ribet (Inv Math 59), Cassou-Noguès (Inv Math 51), and Barsky (1978)). One link between these and the Leopoldt conjecture is through the p-adic analytic class number formula which is the main theorem of Colmez's "Résidue en s = 1 des fonctions zêta p-adiques":adiques" (Inv Math 91): Added (2010/04/09): So here's how you can use von Staudt–Clausen to see that the$p$-adic zeta function (of Q) has a pole at s = 1. It is clear from your statment of vS–C that it is saying that for$k\equiv0\text{ (mod }p-1)$,$B_k\equiv -1/p\text{ (mod }\mathbf{Z}_p)$(i.e. it is not$p$-integral). Let$k_i=(p-1)p^i$, the$k_i$is$p$-adically converging to 0, so$\zeta_p(1-k_i)$is approaching$\zeta_p(1)$(since$\zeta_p(s)$is$p$-adically continuous, at least for$s\neq1$). By the aforementioned interpolation property of$\zeta_p(1-k)$, we have$$v_p(\zeta_p(1-k_i))=v_p(B_{k_i}/k_i)=-1-i\rightarrow -\infty$$hence$1/\zeta_p(1-k_i)$is approaching 0. 3 clean up I also can't answer the question, but I'll say some things that could help. One thing von Staudt-Clausen tells you is the denominator of the Bernoulli number$B_k$when p-1|k: B_k$: it is precisely, the product of primes p for which p-1|k $p-1\mid k$ (when $p-1\nmid k$, a result of Kummer says that $B_k/k$ is p-integral). As Buzzard commented, the Bernoulli numbers should be thought of (at least in this situation) as appearing in special values of p-adic L-functions, p-adic L-functions, specifically, for k a positive integer $$\zeta_p(1-k)=(1-p^{k-1})(-B_k/k),$$ where $\zeta_p$ is the p-adic Riemann zeta function (see chapter II of Koblitz's "p-adic numbers, p-adic analysis, and zeta-functions", for example). For a totally real field F, the Deligne-Ribet p-adic L-function p-adic L-function provides a generalization of the p-adic p-adic Riemann zeta function, namely the p-adic p-adic Dedekind zeta function $\zeta_{F,p}$. One link between these and the Leopoldt conjecture is through the p-adic p-adic analytic class number formula which is the main theorem of Colmez's "Résidue en s=1 s = 1 des fonctions zêta p-adiques": p-adiques": $$\lim_{s\rightarrow1}(s-1)\zeta_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R_phE_p}{w\sqrt{D}}$$ where h is the class number, $E_p$ $E_p=\prod_{\mathfrak{p}\mid p}\left(1-\mathcal{N}(\mathfrak{p})^{-1}\right)$$is a product of Euler Euler-like factors, w = 2 is the number of roots of unity, D is the discriminant and R_p is the interesting part here: the p-adic p-adic regulator (as Colmez notes, \sqrt{D} and R_p both depend on a choice of sign, but their ratio does not). Theorem: The Leopoldt conjecture is equivalent to the non-vanishing of the p-adic p-adic regulator. (For this, see, for example, chapter X of Neukirch-Schmidt-Wingberg's "Cohomology of number fields"). A clear consequence of this is that if \zeta_{F,p} does not have a pole at s=1s = 1, then the Leopoldt conjecture is false for (F,p). F, p). Perhaps an understanding of the denominators of values of \zeta_{F,p} could lead to an understanding of the pole at s=1 s = 1 of \zeta_{F,p}. 2 Fixed typo in first displayed equation ("B_k/b") I also can't answer the question, but I'll say some things that could help. One thing von Staudt-Clausen tells you is the denominator of the Bernoulli number B_k when p-1|k: it is precisely, the product of primes p for which p-1|k (when p-1\nmid k, a result of Kummer says that B_k/k is p-integral). As Buzzard commented, the Bernoulli numbers should be thought of (at least in this situation) as appearing in special values of p-adic L-functions, specifically, for k a positive integer$$\zeta_p(1-k)=(1-p^{k-1})(-B_k/b),$$\zeta_p(1-k)=(1-p^{k-1})(-B_k/k),$$ where$\zeta_p$is the p-adic Riemann zeta function (see chapter II of Koblitz's "p-adic numbers, p-adic analysis, and zeta-functions", for example). For a totally real field F, the Deligne-Ribet p-adic L-function provides a generalization of the p-adic Riemann zeta function, namely the p-adic Dedekind zeta function$\zeta_{F,p}$. One link between these and the Leopoldt conjecture is through the p-adic analytic class number formula which is the main theorem of Colmez's "Résidue en s=1 des fonctions zêta p-adiques": $$\lim_{s\rightarrow1}(s-1)\zeta_{F,p}(s)=\frac{2^{[F:\mathbf{Q}]}R_phE_p}{w\sqrt{D}}$$ where h is the class number,$E_p$is a product of Euler factors, w is the number of roots of unity, D is the discriminant and$R_p$is the interesting part here: the p-adic regulator. Theorem: The Leopoldt conjecture is equivalent to the non-vanishing of the p-adic regulator. A clear consequence of this is that if$\zeta_{F,p}$does not have a pole at s=1, then the Leopoldt conjecture is false for (F,p). Perhaps an understanding of the denominators of values of$\zeta_{F,p}$could lead to an understanding of the pole at s=1 of$\zeta_{F,p}\$. | 1,590 | 5,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2013-20 | latest | en | 0.776171 |
http://isabelle.in.tum.de/repos/isabelle/file/85ed00c1fe7c/src/HOL/Fields.thy | 1,571,169,227,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00349.warc.gz | 95,259,906 | 23,917 | src/HOL/Fields.thy
author hoelzl Fri Feb 19 13:40:50 2016 +0100 (2016-02-19) changeset 62378 85ed00c1fe7c parent 62347 2230b7047376 child 62481 b5d8e57826df permissions -rw-r--r--
generalize more theorems to support enat and ennreal
1 (* Title: HOL/Fields.thy
2 Author: Gertrud Bauer
3 Author: Steven Obua
4 Author: Tobias Nipkow
5 Author: Lawrence C Paulson
6 Author: Markus Wenzel
8 *)
10 section \<open>Fields\<close>
12 theory Fields
13 imports Rings
14 begin
16 subsection \<open>Division rings\<close>
18 text \<open>
19 A division ring is like a field, but without the commutativity requirement.
20 \<close>
22 class inverse = divide +
23 fixes inverse :: "'a \<Rightarrow> 'a"
24 begin
26 abbreviation inverse_divide :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "'/" 70)
27 where
28 "inverse_divide \<equiv> divide"
30 end
32 text\<open>Lemmas \<open>divide_simps\<close> move division to the outside and eliminates them on (in)equalities.\<close>
34 named_theorems divide_simps "rewrite rules to eliminate divisions"
36 class division_ring = ring_1 + inverse +
37 assumes left_inverse [simp]: "a \<noteq> 0 \<Longrightarrow> inverse a * a = 1"
38 assumes right_inverse [simp]: "a \<noteq> 0 \<Longrightarrow> a * inverse a = 1"
39 assumes divide_inverse: "a / b = a * inverse b"
40 assumes inverse_zero [simp]: "inverse 0 = 0"
41 begin
43 subclass ring_1_no_zero_divisors
44 proof
45 fix a b :: 'a
46 assume a: "a \<noteq> 0" and b: "b \<noteq> 0"
47 show "a * b \<noteq> 0"
48 proof
49 assume ab: "a * b = 0"
50 hence "0 = inverse a * (a * b) * inverse b" by simp
51 also have "\<dots> = (inverse a * a) * (b * inverse b)"
52 by (simp only: mult.assoc)
53 also have "\<dots> = 1" using a b by simp
54 finally show False by simp
55 qed
56 qed
58 lemma nonzero_imp_inverse_nonzero:
59 "a \<noteq> 0 \<Longrightarrow> inverse a \<noteq> 0"
60 proof
61 assume ianz: "inverse a = 0"
62 assume "a \<noteq> 0"
63 hence "1 = a * inverse a" by simp
64 also have "... = 0" by (simp add: ianz)
65 finally have "1 = 0" .
66 thus False by (simp add: eq_commute)
67 qed
69 lemma inverse_zero_imp_zero:
70 "inverse a = 0 \<Longrightarrow> a = 0"
71 apply (rule classical)
72 apply (drule nonzero_imp_inverse_nonzero)
73 apply auto
74 done
76 lemma inverse_unique:
77 assumes ab: "a * b = 1"
78 shows "inverse a = b"
79 proof -
80 have "a \<noteq> 0" using ab by (cases "a = 0") simp_all
81 moreover have "inverse a * (a * b) = inverse a" by (simp add: ab)
82 ultimately show ?thesis by (simp add: mult.assoc [symmetric])
83 qed
85 lemma nonzero_inverse_minus_eq:
86 "a \<noteq> 0 \<Longrightarrow> inverse (- a) = - inverse a"
87 by (rule inverse_unique) simp
89 lemma nonzero_inverse_inverse_eq:
90 "a \<noteq> 0 \<Longrightarrow> inverse (inverse a) = a"
91 by (rule inverse_unique) simp
93 lemma nonzero_inverse_eq_imp_eq:
94 assumes "inverse a = inverse b" and "a \<noteq> 0" and "b \<noteq> 0"
95 shows "a = b"
96 proof -
97 from \<open>inverse a = inverse b\<close>
98 have "inverse (inverse a) = inverse (inverse b)" by (rule arg_cong)
99 with \<open>a \<noteq> 0\<close> and \<open>b \<noteq> 0\<close> show "a = b"
101 qed
103 lemma inverse_1 [simp]: "inverse 1 = 1"
104 by (rule inverse_unique) simp
106 lemma nonzero_inverse_mult_distrib:
107 assumes "a \<noteq> 0" and "b \<noteq> 0"
108 shows "inverse (a * b) = inverse b * inverse a"
109 proof -
110 have "a * (b * inverse b) * inverse a = 1" using assms by simp
111 hence "a * b * (inverse b * inverse a) = 1" by (simp only: mult.assoc)
112 thus ?thesis by (rule inverse_unique)
113 qed
116 "a \<noteq> 0 \<Longrightarrow> b \<noteq> 0 \<Longrightarrow> inverse a + inverse b = inverse a * (a + b) * inverse b"
119 lemma division_ring_inverse_diff:
120 "a \<noteq> 0 \<Longrightarrow> b \<noteq> 0 \<Longrightarrow> inverse a - inverse b = inverse a * (b - a) * inverse b"
123 lemma right_inverse_eq: "b \<noteq> 0 \<Longrightarrow> a / b = 1 \<longleftrightarrow> a = b"
124 proof
125 assume neq: "b \<noteq> 0"
126 {
127 hence "a = (a / b) * b" by (simp add: divide_inverse mult.assoc)
128 also assume "a / b = 1"
129 finally show "a = b" by simp
130 next
131 assume "a = b"
132 with neq show "a / b = 1" by (simp add: divide_inverse)
133 }
134 qed
136 lemma nonzero_inverse_eq_divide: "a \<noteq> 0 \<Longrightarrow> inverse a = 1 / a"
139 lemma divide_self [simp]: "a \<noteq> 0 \<Longrightarrow> a / a = 1"
142 lemma inverse_eq_divide [field_simps, divide_simps]: "inverse a = 1 / a"
145 lemma add_divide_distrib: "(a+b) / c = a/c + b/c"
146 by (simp add: divide_inverse algebra_simps)
148 lemma times_divide_eq_right [simp]: "a * (b / c) = (a * b) / c"
149 by (simp add: divide_inverse mult.assoc)
151 lemma minus_divide_left: "- (a / b) = (-a) / b"
154 lemma nonzero_minus_divide_right: "b \<noteq> 0 ==> - (a / b) = a / (- b)"
155 by (simp add: divide_inverse nonzero_inverse_minus_eq)
157 lemma nonzero_minus_divide_divide: "b \<noteq> 0 ==> (-a) / (-b) = a / b"
158 by (simp add: divide_inverse nonzero_inverse_minus_eq)
160 lemma divide_minus_left [simp]: "(-a) / b = - (a / b)"
163 lemma diff_divide_distrib: "(a - b) / c = a / c - b / c"
164 using add_divide_distrib [of a "- b" c] by simp
166 lemma nonzero_eq_divide_eq [field_simps]: "c \<noteq> 0 \<Longrightarrow> a = b / c \<longleftrightarrow> a * c = b"
167 proof -
168 assume [simp]: "c \<noteq> 0"
169 have "a = b / c \<longleftrightarrow> a * c = (b / c) * c" by simp
170 also have "... \<longleftrightarrow> a * c = b" by (simp add: divide_inverse mult.assoc)
171 finally show ?thesis .
172 qed
174 lemma nonzero_divide_eq_eq [field_simps]: "c \<noteq> 0 \<Longrightarrow> b / c = a \<longleftrightarrow> b = a * c"
175 proof -
176 assume [simp]: "c \<noteq> 0"
177 have "b / c = a \<longleftrightarrow> (b / c) * c = a * c" by simp
178 also have "... \<longleftrightarrow> b = a * c" by (simp add: divide_inverse mult.assoc)
179 finally show ?thesis .
180 qed
182 lemma nonzero_neg_divide_eq_eq [field_simps]: "b \<noteq> 0 \<Longrightarrow> - (a / b) = c \<longleftrightarrow> - a = c * b"
183 using nonzero_divide_eq_eq[of b "-a" c] by simp
185 lemma nonzero_neg_divide_eq_eq2 [field_simps]: "b \<noteq> 0 \<Longrightarrow> c = - (a / b) \<longleftrightarrow> c * b = - a"
186 using nonzero_neg_divide_eq_eq[of b a c] by auto
188 lemma divide_eq_imp: "c \<noteq> 0 \<Longrightarrow> b = a * c \<Longrightarrow> b / c = a"
189 by (simp add: divide_inverse mult.assoc)
191 lemma eq_divide_imp: "c \<noteq> 0 \<Longrightarrow> a * c = b \<Longrightarrow> a = b / c"
192 by (drule sym) (simp add: divide_inverse mult.assoc)
195 "z \<noteq> 0 \<Longrightarrow> x + y / z = (x * z + y) / z"
199 "z \<noteq> 0 \<Longrightarrow> x / z + y = (x + y * z) / z"
202 lemma diff_divide_eq_iff [field_simps]:
203 "z \<noteq> 0 \<Longrightarrow> x - y / z = (x * z - y) / z"
204 by (simp add: diff_divide_distrib nonzero_eq_divide_eq eq_diff_eq)
207 "z \<noteq> 0 \<Longrightarrow> - (x / z) + y = (- x + y * z) / z"
210 lemma divide_diff_eq_iff [field_simps]:
211 "z \<noteq> 0 \<Longrightarrow> x / z - y = (x - y * z) / z"
214 lemma minus_divide_diff_eq_iff [field_simps]:
215 "z \<noteq> 0 \<Longrightarrow> - (x / z) - y = (- x - y * z) / z"
218 lemma division_ring_divide_zero [simp]:
219 "a / 0 = 0"
222 lemma divide_self_if [simp]:
223 "a / a = (if a = 0 then 0 else 1)"
224 by simp
226 lemma inverse_nonzero_iff_nonzero [simp]:
227 "inverse a = 0 \<longleftrightarrow> a = 0"
228 by rule (fact inverse_zero_imp_zero, simp)
230 lemma inverse_minus_eq [simp]:
231 "inverse (- a) = - inverse a"
232 proof cases
233 assume "a=0" thus ?thesis by simp
234 next
235 assume "a\<noteq>0"
236 thus ?thesis by (simp add: nonzero_inverse_minus_eq)
237 qed
239 lemma inverse_inverse_eq [simp]:
240 "inverse (inverse a) = a"
241 proof cases
242 assume "a=0" thus ?thesis by simp
243 next
244 assume "a\<noteq>0"
245 thus ?thesis by (simp add: nonzero_inverse_inverse_eq)
246 qed
248 lemma inverse_eq_imp_eq:
249 "inverse a = inverse b \<Longrightarrow> a = b"
250 by (drule arg_cong [where f="inverse"], simp)
252 lemma inverse_eq_iff_eq [simp]:
253 "inverse a = inverse b \<longleftrightarrow> a = b"
254 by (force dest!: inverse_eq_imp_eq)
257 "a + b / z = (if z = 0 then a else (a * z + b) / z)"
258 "a / z + b = (if z = 0 then b else (a + b * z) / z)"
259 "- (a / z) + b = (if z = 0 then b else (-a + b * z) / z)"
260 "a - b / z = (if z = 0 then a else (a * z - b) / z)"
261 "a / z - b = (if z = 0 then -b else (a - b * z) / z)"
262 "- (a / z) - b = (if z = 0 then -b else (- a - b * z) / z)"
264 minus_divide_diff_eq_iff)
266 lemma [divide_simps]:
267 shows divide_eq_eq: "b / c = a \<longleftrightarrow> (if c \<noteq> 0 then b = a * c else a = 0)"
268 and eq_divide_eq: "a = b / c \<longleftrightarrow> (if c \<noteq> 0 then a * c = b else a = 0)"
269 and minus_divide_eq_eq: "- (b / c) = a \<longleftrightarrow> (if c \<noteq> 0 then - b = a * c else a = 0)"
270 and eq_minus_divide_eq: "a = - (b / c) \<longleftrightarrow> (if c \<noteq> 0 then a * c = - b else a = 0)"
271 by (auto simp add: field_simps)
273 end
275 subsection \<open>Fields\<close>
277 class field = comm_ring_1 + inverse +
278 assumes field_inverse: "a \<noteq> 0 \<Longrightarrow> inverse a * a = 1"
279 assumes field_divide_inverse: "a / b = a * inverse b"
280 assumes field_inverse_zero: "inverse 0 = 0"
281 begin
283 subclass division_ring
284 proof
285 fix a :: 'a
286 assume "a \<noteq> 0"
287 thus "inverse a * a = 1" by (rule field_inverse)
288 thus "a * inverse a = 1" by (simp only: mult.commute)
289 next
290 fix a b :: 'a
291 show "a / b = a * inverse b" by (rule field_divide_inverse)
292 next
293 show "inverse 0 = 0"
294 by (fact field_inverse_zero)
295 qed
297 subclass idom_divide
298 proof
299 fix b a
300 assume "b \<noteq> 0"
301 then show "a * b / b = a"
302 by (simp add: divide_inverse ac_simps)
303 next
304 fix a
305 show "a / 0 = 0"
307 qed
309 text\<open>There is no slick version using division by zero.\<close>
311 "a \<noteq> 0 \<Longrightarrow> b \<noteq> 0 \<Longrightarrow> inverse a + inverse b = (a + b) * inverse a * inverse b"
314 lemma nonzero_mult_divide_mult_cancel_left [simp]:
315 assumes [simp]: "c \<noteq> 0"
316 shows "(c * a) / (c * b) = a / b"
317 proof (cases "b = 0")
318 case True then show ?thesis by simp
319 next
320 case False
321 then have "(c*a)/(c*b) = c * a * (inverse b * inverse c)"
322 by (simp add: divide_inverse nonzero_inverse_mult_distrib)
323 also have "... = a * inverse b * (inverse c * c)"
324 by (simp only: ac_simps)
325 also have "... = a * inverse b" by simp
326 finally show ?thesis by (simp add: divide_inverse)
327 qed
329 lemma nonzero_mult_divide_mult_cancel_right [simp]:
330 "c \<noteq> 0 \<Longrightarrow> (a * c) / (b * c) = a / b"
331 using nonzero_mult_divide_mult_cancel_left [of c a b] by (simp add: ac_simps)
333 lemma times_divide_eq_left [simp]: "(b / c) * a = (b * a) / c"
334 by (simp add: divide_inverse ac_simps)
336 lemma divide_inverse_commute: "a / b = inverse b * a"
337 by (simp add: divide_inverse mult.commute)
340 assumes "y \<noteq> 0" and "z \<noteq> 0"
341 shows "x / y + w / z = (x * z + w * y) / (y * z)"
342 proof -
343 have "x / y + w / z = (x * z) / (y * z) + (y * w) / (y * z)"
344 using assms by simp
345 also have "\<dots> = (x * z + y * w) / (y * z)"
347 finally show ?thesis
348 by (simp only: mult.commute)
349 qed
351 text\<open>Special Cancellation Simprules for Division\<close>
353 lemma nonzero_divide_mult_cancel_right [simp]:
354 "b \<noteq> 0 \<Longrightarrow> b / (a * b) = 1 / a"
355 using nonzero_mult_divide_mult_cancel_right [of b 1 a] by simp
357 lemma nonzero_divide_mult_cancel_left [simp]:
358 "a \<noteq> 0 \<Longrightarrow> a / (a * b) = 1 / b"
359 using nonzero_mult_divide_mult_cancel_left [of a 1 b] by simp
361 lemma nonzero_mult_divide_mult_cancel_left2 [simp]:
362 "c \<noteq> 0 \<Longrightarrow> (c * a) / (b * c) = a / b"
363 using nonzero_mult_divide_mult_cancel_left [of c a b] by (simp add: ac_simps)
365 lemma nonzero_mult_divide_mult_cancel_right2 [simp]:
366 "c \<noteq> 0 \<Longrightarrow> (a * c) / (c * b) = a / b"
367 using nonzero_mult_divide_mult_cancel_right [of b c a] by (simp add: ac_simps)
369 lemma diff_frac_eq:
370 "y \<noteq> 0 \<Longrightarrow> z \<noteq> 0 \<Longrightarrow> x / y - w / z = (x * z - w * y) / (y * z)"
373 lemma frac_eq_eq:
374 "y \<noteq> 0 \<Longrightarrow> z \<noteq> 0 \<Longrightarrow> (x / y = w / z) = (x * z = w * y)"
377 lemma divide_minus1 [simp]: "x / - 1 = - x"
378 using nonzero_minus_divide_right [of "1" x] by simp
380 text\<open>This version builds in division by zero while also re-orienting
381 the right-hand side.\<close>
382 lemma inverse_mult_distrib [simp]:
383 "inverse (a * b) = inverse a * inverse b"
384 proof cases
385 assume "a \<noteq> 0 & b \<noteq> 0"
386 thus ?thesis by (simp add: nonzero_inverse_mult_distrib ac_simps)
387 next
388 assume "~ (a \<noteq> 0 & b \<noteq> 0)"
389 thus ?thesis by force
390 qed
392 lemma inverse_divide [simp]:
393 "inverse (a / b) = b / a"
394 by (simp add: divide_inverse mult.commute)
397 text \<open>Calculations with fractions\<close>
399 text\<open>There is a whole bunch of simp-rules just for class \<open>field\<close> but none for class \<open>field\<close> and \<open>nonzero_divides\<close>
400 because the latter are covered by a simproc.\<close>
402 lemma mult_divide_mult_cancel_left:
403 "c \<noteq> 0 \<Longrightarrow> (c * a) / (c * b) = a / b"
404 apply (cases "b = 0")
405 apply simp_all
406 done
408 lemma mult_divide_mult_cancel_right:
409 "c \<noteq> 0 \<Longrightarrow> (a * c) / (b * c) = a / b"
410 apply (cases "b = 0")
411 apply simp_all
412 done
414 lemma divide_divide_eq_right [simp]:
415 "a / (b / c) = (a * c) / b"
416 by (simp add: divide_inverse ac_simps)
418 lemma divide_divide_eq_left [simp]:
419 "(a / b) / c = a / (b * c)"
420 by (simp add: divide_inverse mult.assoc)
422 lemma divide_divide_times_eq:
423 "(x / y) / (z / w) = (x * w) / (y * z)"
424 by simp
426 text \<open>Special Cancellation Simprules for Division\<close>
428 lemma mult_divide_mult_cancel_left_if [simp]:
429 shows "(c * a) / (c * b) = (if c = 0 then 0 else a / b)"
430 by simp
433 text \<open>Division and Unary Minus\<close>
435 lemma minus_divide_right:
436 "- (a / b) = a / - b"
439 lemma divide_minus_right [simp]:
440 "a / - b = - (a / b)"
443 lemma minus_divide_divide:
444 "(- a) / (- b) = a / b"
445 apply (cases "b=0", simp)
447 done
449 lemma inverse_eq_1_iff [simp]:
450 "inverse x = 1 \<longleftrightarrow> x = 1"
451 by (insert inverse_eq_iff_eq [of x 1], simp)
453 lemma divide_eq_0_iff [simp]:
454 "a / b = 0 \<longleftrightarrow> a = 0 \<or> b = 0"
457 lemma divide_cancel_right [simp]:
458 "a / c = b / c \<longleftrightarrow> c = 0 \<or> a = b"
459 apply (cases "c=0", simp)
461 done
463 lemma divide_cancel_left [simp]:
464 "c / a = c / b \<longleftrightarrow> c = 0 \<or> a = b"
465 apply (cases "c=0", simp)
467 done
469 lemma divide_eq_1_iff [simp]:
470 "a / b = 1 \<longleftrightarrow> b \<noteq> 0 \<and> a = b"
471 apply (cases "b=0", simp)
473 done
475 lemma one_eq_divide_iff [simp]:
476 "1 = a / b \<longleftrightarrow> b \<noteq> 0 \<and> a = b"
477 by (simp add: eq_commute [of 1])
479 lemma times_divide_times_eq:
480 "(x / y) * (z / w) = (x * z) / (y * w)"
481 by simp
484 "y \<noteq> 0 \<Longrightarrow> x / y + z = (x + z * y) / y"
488 "y \<noteq> 0 \<Longrightarrow> z + x / y = (x + z * y) / y"
491 end
494 subsection \<open>Ordered fields\<close>
496 class linordered_field = field + linordered_idom
497 begin
499 lemma positive_imp_inverse_positive:
500 assumes a_gt_0: "0 < a"
501 shows "0 < inverse a"
502 proof -
503 have "0 < a * inverse a"
504 by (simp add: a_gt_0 [THEN less_imp_not_eq2])
505 thus "0 < inverse a"
506 by (simp add: a_gt_0 [THEN less_not_sym] zero_less_mult_iff)
507 qed
509 lemma negative_imp_inverse_negative:
510 "a < 0 \<Longrightarrow> inverse a < 0"
511 by (insert positive_imp_inverse_positive [of "-a"],
514 lemma inverse_le_imp_le:
515 assumes invle: "inverse a \<le> inverse b" and apos: "0 < a"
516 shows "b \<le> a"
517 proof (rule classical)
518 assume "~ b \<le> a"
519 hence "a < b" by (simp add: linorder_not_le)
520 hence bpos: "0 < b" by (blast intro: apos less_trans)
521 hence "a * inverse a \<le> a * inverse b"
522 by (simp add: apos invle less_imp_le mult_left_mono)
523 hence "(a * inverse a) * b \<le> (a * inverse b) * b"
524 by (simp add: bpos less_imp_le mult_right_mono)
525 thus "b \<le> a" by (simp add: mult.assoc apos bpos less_imp_not_eq2)
526 qed
528 lemma inverse_positive_imp_positive:
529 assumes inv_gt_0: "0 < inverse a" and nz: "a \<noteq> 0"
530 shows "0 < a"
531 proof -
532 have "0 < inverse (inverse a)"
533 using inv_gt_0 by (rule positive_imp_inverse_positive)
534 thus "0 < a"
535 using nz by (simp add: nonzero_inverse_inverse_eq)
536 qed
538 lemma inverse_negative_imp_negative:
539 assumes inv_less_0: "inverse a < 0" and nz: "a \<noteq> 0"
540 shows "a < 0"
541 proof -
542 have "inverse (inverse a) < 0"
543 using inv_less_0 by (rule negative_imp_inverse_negative)
544 thus "a < 0" using nz by (simp add: nonzero_inverse_inverse_eq)
545 qed
547 lemma linordered_field_no_lb:
548 "\<forall>x. \<exists>y. y < x"
549 proof
550 fix x::'a
551 have m1: "- (1::'a) < 0" by simp
552 from add_strict_right_mono[OF m1, where c=x]
553 have "(- 1) + x < x" by simp
554 thus "\<exists>y. y < x" by blast
555 qed
557 lemma linordered_field_no_ub:
558 "\<forall> x. \<exists>y. y > x"
559 proof
560 fix x::'a
561 have m1: " (1::'a) > 0" by simp
562 from add_strict_right_mono[OF m1, where c=x]
563 have "1 + x > x" by simp
564 thus "\<exists>y. y > x" by blast
565 qed
567 lemma less_imp_inverse_less:
568 assumes less: "a < b" and apos: "0 < a"
569 shows "inverse b < inverse a"
570 proof (rule ccontr)
571 assume "~ inverse b < inverse a"
572 hence "inverse a \<le> inverse b" by simp
573 hence "~ (a < b)"
574 by (simp add: not_less inverse_le_imp_le [OF _ apos])
575 thus False by (rule notE [OF _ less])
576 qed
578 lemma inverse_less_imp_less:
579 "inverse a < inverse b \<Longrightarrow> 0 < a \<Longrightarrow> b < a"
580 apply (simp add: less_le [of "inverse a"] less_le [of "b"])
581 apply (force dest!: inverse_le_imp_le nonzero_inverse_eq_imp_eq)
582 done
584 text\<open>Both premises are essential. Consider -1 and 1.\<close>
585 lemma inverse_less_iff_less [simp]:
586 "0 < a \<Longrightarrow> 0 < b \<Longrightarrow> inverse a < inverse b \<longleftrightarrow> b < a"
587 by (blast intro: less_imp_inverse_less dest: inverse_less_imp_less)
589 lemma le_imp_inverse_le:
590 "a \<le> b \<Longrightarrow> 0 < a \<Longrightarrow> inverse b \<le> inverse a"
591 by (force simp add: le_less less_imp_inverse_less)
593 lemma inverse_le_iff_le [simp]:
594 "0 < a \<Longrightarrow> 0 < b \<Longrightarrow> inverse a \<le> inverse b \<longleftrightarrow> b \<le> a"
595 by (blast intro: le_imp_inverse_le dest: inverse_le_imp_le)
598 text\<open>These results refer to both operands being negative. The opposite-sign
599 case is trivial, since inverse preserves signs.\<close>
600 lemma inverse_le_imp_le_neg:
601 "inverse a \<le> inverse b \<Longrightarrow> b < 0 \<Longrightarrow> b \<le> a"
602 apply (rule classical)
603 apply (subgoal_tac "a < 0")
604 prefer 2 apply force
605 apply (insert inverse_le_imp_le [of "-b" "-a"])
607 done
609 lemma less_imp_inverse_less_neg:
610 "a < b \<Longrightarrow> b < 0 \<Longrightarrow> inverse b < inverse a"
611 apply (subgoal_tac "a < 0")
612 prefer 2 apply (blast intro: less_trans)
613 apply (insert less_imp_inverse_less [of "-b" "-a"])
615 done
617 lemma inverse_less_imp_less_neg:
618 "inverse a < inverse b \<Longrightarrow> b < 0 \<Longrightarrow> b < a"
619 apply (rule classical)
620 apply (subgoal_tac "a < 0")
621 prefer 2
622 apply force
623 apply (insert inverse_less_imp_less [of "-b" "-a"])
625 done
627 lemma inverse_less_iff_less_neg [simp]:
628 "a < 0 \<Longrightarrow> b < 0 \<Longrightarrow> inverse a < inverse b \<longleftrightarrow> b < a"
629 apply (insert inverse_less_iff_less [of "-b" "-a"])
630 apply (simp del: inverse_less_iff_less
632 done
634 lemma le_imp_inverse_le_neg:
635 "a \<le> b \<Longrightarrow> b < 0 ==> inverse b \<le> inverse a"
636 by (force simp add: le_less less_imp_inverse_less_neg)
638 lemma inverse_le_iff_le_neg [simp]:
639 "a < 0 \<Longrightarrow> b < 0 \<Longrightarrow> inverse a \<le> inverse b \<longleftrightarrow> b \<le> a"
640 by (blast intro: le_imp_inverse_le_neg dest: inverse_le_imp_le_neg)
642 lemma one_less_inverse:
643 "0 < a \<Longrightarrow> a < 1 \<Longrightarrow> 1 < inverse a"
644 using less_imp_inverse_less [of a 1, unfolded inverse_1] .
646 lemma one_le_inverse:
647 "0 < a \<Longrightarrow> a \<le> 1 \<Longrightarrow> 1 \<le> inverse a"
648 using le_imp_inverse_le [of a 1, unfolded inverse_1] .
650 lemma pos_le_divide_eq [field_simps]:
651 assumes "0 < c"
652 shows "a \<le> b / c \<longleftrightarrow> a * c \<le> b"
653 proof -
654 from assms have "a \<le> b / c \<longleftrightarrow> a * c \<le> (b / c) * c"
655 using mult_le_cancel_right [of a c "b * inverse c"] by (auto simp add: field_simps)
656 also have "... \<longleftrightarrow> a * c \<le> b"
657 by (simp add: less_imp_not_eq2 [OF assms] divide_inverse mult.assoc)
658 finally show ?thesis .
659 qed
661 lemma pos_less_divide_eq [field_simps]:
662 assumes "0 < c"
663 shows "a < b / c \<longleftrightarrow> a * c < b"
664 proof -
665 from assms have "a < b / c \<longleftrightarrow> a * c < (b / c) * c"
666 using mult_less_cancel_right [of a c "b / c"] by auto
667 also have "... = (a*c < b)"
668 by (simp add: less_imp_not_eq2 [OF assms] divide_inverse mult.assoc)
669 finally show ?thesis .
670 qed
672 lemma neg_less_divide_eq [field_simps]:
673 assumes "c < 0"
674 shows "a < b / c \<longleftrightarrow> b < a * c"
675 proof -
676 from assms have "a < b / c \<longleftrightarrow> (b / c) * c < a * c"
677 using mult_less_cancel_right [of "b / c" c a] by auto
678 also have "... \<longleftrightarrow> b < a * c"
679 by (simp add: less_imp_not_eq [OF assms] divide_inverse mult.assoc)
680 finally show ?thesis .
681 qed
683 lemma neg_le_divide_eq [field_simps]:
684 assumes "c < 0"
685 shows "a \<le> b / c \<longleftrightarrow> b \<le> a * c"
686 proof -
687 from assms have "a \<le> b / c \<longleftrightarrow> (b / c) * c \<le> a * c"
688 using mult_le_cancel_right [of "b * inverse c" c a] by (auto simp add: field_simps)
689 also have "... \<longleftrightarrow> b \<le> a * c"
690 by (simp add: less_imp_not_eq [OF assms] divide_inverse mult.assoc)
691 finally show ?thesis .
692 qed
694 lemma pos_divide_le_eq [field_simps]:
695 assumes "0 < c"
696 shows "b / c \<le> a \<longleftrightarrow> b \<le> a * c"
697 proof -
698 from assms have "b / c \<le> a \<longleftrightarrow> (b / c) * c \<le> a * c"
699 using mult_le_cancel_right [of "b / c" c a] by auto
700 also have "... \<longleftrightarrow> b \<le> a * c"
701 by (simp add: less_imp_not_eq2 [OF assms] divide_inverse mult.assoc)
702 finally show ?thesis .
703 qed
705 lemma pos_divide_less_eq [field_simps]:
706 assumes "0 < c"
707 shows "b / c < a \<longleftrightarrow> b < a * c"
708 proof -
709 from assms have "b / c < a \<longleftrightarrow> (b / c) * c < a * c"
710 using mult_less_cancel_right [of "b / c" c a] by auto
711 also have "... \<longleftrightarrow> b < a * c"
712 by (simp add: less_imp_not_eq2 [OF assms] divide_inverse mult.assoc)
713 finally show ?thesis .
714 qed
716 lemma neg_divide_le_eq [field_simps]:
717 assumes "c < 0"
718 shows "b / c \<le> a \<longleftrightarrow> a * c \<le> b"
719 proof -
720 from assms have "b / c \<le> a \<longleftrightarrow> a * c \<le> (b / c) * c"
721 using mult_le_cancel_right [of a c "b / c"] by auto
722 also have "... \<longleftrightarrow> a * c \<le> b"
723 by (simp add: less_imp_not_eq [OF assms] divide_inverse mult.assoc)
724 finally show ?thesis .
725 qed
727 lemma neg_divide_less_eq [field_simps]:
728 assumes "c < 0"
729 shows "b / c < a \<longleftrightarrow> a * c < b"
730 proof -
731 from assms have "b / c < a \<longleftrightarrow> a * c < b / c * c"
732 using mult_less_cancel_right [of a c "b / c"] by auto
733 also have "... \<longleftrightarrow> a * c < b"
734 by (simp add: less_imp_not_eq [OF assms] divide_inverse mult.assoc)
735 finally show ?thesis .
736 qed
738 text\<open>The following \<open>field_simps\<close> rules are necessary, as minus is always moved atop of
739 division but we want to get rid of division.\<close>
741 lemma pos_le_minus_divide_eq [field_simps]: "0 < c \<Longrightarrow> a \<le> - (b / c) \<longleftrightarrow> a * c \<le> - b"
742 unfolding minus_divide_left by (rule pos_le_divide_eq)
744 lemma neg_le_minus_divide_eq [field_simps]: "c < 0 \<Longrightarrow> a \<le> - (b / c) \<longleftrightarrow> - b \<le> a * c"
745 unfolding minus_divide_left by (rule neg_le_divide_eq)
747 lemma pos_less_minus_divide_eq [field_simps]: "0 < c \<Longrightarrow> a < - (b / c) \<longleftrightarrow> a * c < - b"
748 unfolding minus_divide_left by (rule pos_less_divide_eq)
750 lemma neg_less_minus_divide_eq [field_simps]: "c < 0 \<Longrightarrow> a < - (b / c) \<longleftrightarrow> - b < a * c"
751 unfolding minus_divide_left by (rule neg_less_divide_eq)
753 lemma pos_minus_divide_less_eq [field_simps]: "0 < c \<Longrightarrow> - (b / c) < a \<longleftrightarrow> - b < a * c"
754 unfolding minus_divide_left by (rule pos_divide_less_eq)
756 lemma neg_minus_divide_less_eq [field_simps]: "c < 0 \<Longrightarrow> - (b / c) < a \<longleftrightarrow> a * c < - b"
757 unfolding minus_divide_left by (rule neg_divide_less_eq)
759 lemma pos_minus_divide_le_eq [field_simps]: "0 < c \<Longrightarrow> - (b / c) \<le> a \<longleftrightarrow> - b \<le> a * c"
760 unfolding minus_divide_left by (rule pos_divide_le_eq)
762 lemma neg_minus_divide_le_eq [field_simps]: "c < 0 \<Longrightarrow> - (b / c) \<le> a \<longleftrightarrow> a * c \<le> - b"
763 unfolding minus_divide_left by (rule neg_divide_le_eq)
765 lemma frac_less_eq:
766 "y \<noteq> 0 \<Longrightarrow> z \<noteq> 0 \<Longrightarrow> x / y < w / z \<longleftrightarrow> (x * z - w * y) / (y * z) < 0"
767 by (subst less_iff_diff_less_0) (simp add: diff_frac_eq )
769 lemma frac_le_eq:
770 "y \<noteq> 0 \<Longrightarrow> z \<noteq> 0 \<Longrightarrow> x / y \<le> w / z \<longleftrightarrow> (x * z - w * y) / (y * z) \<le> 0"
771 by (subst le_iff_diff_le_0) (simp add: diff_frac_eq )
773 text\<open>Lemmas \<open>sign_simps\<close> is a first attempt to automate proofs
774 of positivity/negativity needed for \<open>field_simps\<close>. Have not added \<open>sign_simps\<close> to \<open>field_simps\<close> because the former can lead to case
775 explosions.\<close>
777 lemmas sign_simps = algebra_simps zero_less_mult_iff mult_less_0_iff
779 lemmas (in -) sign_simps = algebra_simps zero_less_mult_iff mult_less_0_iff
781 (* Only works once linear arithmetic is installed:
782 text{*An example:*}
783 lemma fixes a b c d e f :: "'a::linordered_field"
784 shows "\<lbrakk>a>b; c<d; e<f; 0 < u \<rbrakk> \<Longrightarrow>
785 ((a-b)*(c-d)*(e-f))/((c-d)*(e-f)*(a-b)) <
786 ((e-f)*(a-b)*(c-d))/((e-f)*(a-b)*(c-d)) + u"
787 apply(subgoal_tac "(c-d)*(e-f)*(a-b) > 0")
789 apply(subgoal_tac "(c-d)*(e-f)*(a-b)*u > 0")
792 done
793 *)
795 lemma divide_pos_pos[simp]:
796 "0 < x ==> 0 < y ==> 0 < x / y"
799 lemma divide_nonneg_pos:
800 "0 <= x ==> 0 < y ==> 0 <= x / y"
803 lemma divide_neg_pos:
804 "x < 0 ==> 0 < y ==> x / y < 0"
807 lemma divide_nonpos_pos:
808 "x <= 0 ==> 0 < y ==> x / y <= 0"
811 lemma divide_pos_neg:
812 "0 < x ==> y < 0 ==> x / y < 0"
815 lemma divide_nonneg_neg:
816 "0 <= x ==> y < 0 ==> x / y <= 0"
819 lemma divide_neg_neg:
820 "x < 0 ==> y < 0 ==> 0 < x / y"
823 lemma divide_nonpos_neg:
824 "x <= 0 ==> y < 0 ==> 0 <= x / y"
827 lemma divide_strict_right_mono:
828 "[|a < b; 0 < c|] ==> a / c < b / c"
829 by (simp add: less_imp_not_eq2 divide_inverse mult_strict_right_mono
830 positive_imp_inverse_positive)
833 lemma divide_strict_right_mono_neg:
834 "[|b < a; c < 0|] ==> a / c < b / c"
835 apply (drule divide_strict_right_mono [of _ _ "-c"], simp)
836 apply (simp add: less_imp_not_eq nonzero_minus_divide_right [symmetric])
837 done
839 text\<open>The last premise ensures that @{term a} and @{term b}
840 have the same sign\<close>
841 lemma divide_strict_left_mono:
842 "[|b < a; 0 < c; 0 < a*b|] ==> c / a < c / b"
843 by (auto simp: field_simps zero_less_mult_iff mult_strict_right_mono)
845 lemma divide_left_mono:
846 "[|b \<le> a; 0 \<le> c; 0 < a*b|] ==> c / a \<le> c / b"
847 by (auto simp: field_simps zero_less_mult_iff mult_right_mono)
849 lemma divide_strict_left_mono_neg:
850 "[|a < b; c < 0; 0 < a*b|] ==> c / a < c / b"
851 by (auto simp: field_simps zero_less_mult_iff mult_strict_right_mono_neg)
853 lemma mult_imp_div_pos_le: "0 < y ==> x <= z * y ==>
854 x / y <= z"
855 by (subst pos_divide_le_eq, assumption+)
857 lemma mult_imp_le_div_pos: "0 < y ==> z * y <= x ==>
858 z <= x / y"
861 lemma mult_imp_div_pos_less: "0 < y ==> x < z * y ==>
862 x / y < z"
865 lemma mult_imp_less_div_pos: "0 < y ==> z * y < x ==>
866 z < x / y"
869 lemma frac_le: "0 <= x ==>
870 x <= y ==> 0 < w ==> w <= z ==> x / z <= y / w"
871 apply (rule mult_imp_div_pos_le)
872 apply simp
873 apply (subst times_divide_eq_left)
874 apply (rule mult_imp_le_div_pos, assumption)
875 apply (rule mult_mono)
876 apply simp_all
877 done
879 lemma frac_less: "0 <= x ==>
880 x < y ==> 0 < w ==> w <= z ==> x / z < y / w"
881 apply (rule mult_imp_div_pos_less)
882 apply simp
883 apply (subst times_divide_eq_left)
884 apply (rule mult_imp_less_div_pos, assumption)
885 apply (erule mult_less_le_imp_less)
886 apply simp_all
887 done
889 lemma frac_less2: "0 < x ==>
890 x <= y ==> 0 < w ==> w < z ==> x / z < y / w"
891 apply (rule mult_imp_div_pos_less)
892 apply simp_all
893 apply (rule mult_imp_less_div_pos, assumption)
894 apply (erule mult_le_less_imp_less)
895 apply simp_all
896 done
898 lemma less_half_sum: "a < b ==> a < (a+b) / (1+1)"
899 by (simp add: field_simps zero_less_two)
901 lemma gt_half_sum: "a < b ==> (a+b)/(1+1) < b"
902 by (simp add: field_simps zero_less_two)
904 subclass unbounded_dense_linorder
905 proof
906 fix x y :: 'a
907 from less_add_one show "\<exists>y. x < y" ..
908 from less_add_one have "x + (- 1) < (x + 1) + (- 1)" by (rule add_strict_right_mono)
909 then have "x - 1 < x + 1 - 1" by simp
910 then have "x - 1 < x" by (simp add: algebra_simps)
911 then show "\<exists>y. y < x" ..
912 show "x < y \<Longrightarrow> \<exists>z>x. z < y" by (blast intro!: less_half_sum gt_half_sum)
913 qed
915 lemma nonzero_abs_inverse:
916 "a \<noteq> 0 ==> \<bar>inverse a\<bar> = inverse \<bar>a\<bar>"
917 apply (auto simp add: neq_iff abs_if nonzero_inverse_minus_eq
918 negative_imp_inverse_negative)
919 apply (blast intro: positive_imp_inverse_positive elim: less_asym)
920 done
922 lemma nonzero_abs_divide:
923 "b \<noteq> 0 ==> \<bar>a / b\<bar> = \<bar>a\<bar> / \<bar>b\<bar>"
924 by (simp add: divide_inverse abs_mult nonzero_abs_inverse)
926 lemma field_le_epsilon:
927 assumes e: "\<And>e. 0 < e \<Longrightarrow> x \<le> y + e"
928 shows "x \<le> y"
929 proof (rule dense_le)
930 fix t assume "t < x"
931 hence "0 < x - t" by (simp add: less_diff_eq)
932 from e [OF this] have "x + 0 \<le> x + (y - t)" by (simp add: algebra_simps)
933 then have "0 \<le> y - t" by (simp only: add_le_cancel_left)
934 then show "t \<le> y" by (simp add: algebra_simps)
935 qed
937 lemma inverse_positive_iff_positive [simp]:
938 "(0 < inverse a) = (0 < a)"
939 apply (cases "a = 0", simp)
940 apply (blast intro: inverse_positive_imp_positive positive_imp_inverse_positive)
941 done
943 lemma inverse_negative_iff_negative [simp]:
944 "(inverse a < 0) = (a < 0)"
945 apply (cases "a = 0", simp)
946 apply (blast intro: inverse_negative_imp_negative negative_imp_inverse_negative)
947 done
949 lemma inverse_nonnegative_iff_nonnegative [simp]:
950 "0 \<le> inverse a \<longleftrightarrow> 0 \<le> a"
951 by (simp add: not_less [symmetric])
953 lemma inverse_nonpositive_iff_nonpositive [simp]:
954 "inverse a \<le> 0 \<longleftrightarrow> a \<le> 0"
955 by (simp add: not_less [symmetric])
957 lemma one_less_inverse_iff: "1 < inverse x \<longleftrightarrow> 0 < x \<and> x < 1"
958 using less_trans[of 1 x 0 for x]
959 by (cases x 0 rule: linorder_cases) (auto simp add: field_simps)
961 lemma one_le_inverse_iff: "1 \<le> inverse x \<longleftrightarrow> 0 < x \<and> x \<le> 1"
962 proof (cases "x = 1")
963 case True then show ?thesis by simp
964 next
965 case False then have "inverse x \<noteq> 1" by simp
966 then have "1 \<noteq> inverse x" by blast
967 then have "1 \<le> inverse x \<longleftrightarrow> 1 < inverse x" by (simp add: le_less)
968 with False show ?thesis by (auto simp add: one_less_inverse_iff)
969 qed
971 lemma inverse_less_1_iff: "inverse x < 1 \<longleftrightarrow> x \<le> 0 \<or> 1 < x"
972 by (simp add: not_le [symmetric] one_le_inverse_iff)
974 lemma inverse_le_1_iff: "inverse x \<le> 1 \<longleftrightarrow> x \<le> 0 \<or> 1 \<le> x"
975 by (simp add: not_less [symmetric] one_less_inverse_iff)
977 lemma [divide_simps]:
978 shows le_divide_eq: "a \<le> b / c \<longleftrightarrow> (if 0 < c then a * c \<le> b else if c < 0 then b \<le> a * c else a \<le> 0)"
979 and divide_le_eq: "b / c \<le> a \<longleftrightarrow> (if 0 < c then b \<le> a * c else if c < 0 then a * c \<le> b else 0 \<le> a)"
980 and less_divide_eq: "a < b / c \<longleftrightarrow> (if 0 < c then a * c < b else if c < 0 then b < a * c else a < 0)"
981 and divide_less_eq: "b / c < a \<longleftrightarrow> (if 0 < c then b < a * c else if c < 0 then a * c < b else 0 < a)"
982 and le_minus_divide_eq: "a \<le> - (b / c) \<longleftrightarrow> (if 0 < c then a * c \<le> - b else if c < 0 then - b \<le> a * c else a \<le> 0)"
983 and minus_divide_le_eq: "- (b / c) \<le> a \<longleftrightarrow> (if 0 < c then - b \<le> a * c else if c < 0 then a * c \<le> - b else 0 \<le> a)"
984 and less_minus_divide_eq: "a < - (b / c) \<longleftrightarrow> (if 0 < c then a * c < - b else if c < 0 then - b < a * c else a < 0)"
985 and minus_divide_less_eq: "- (b / c) < a \<longleftrightarrow> (if 0 < c then - b < a * c else if c < 0 then a * c < - b else 0 < a)"
986 by (auto simp: field_simps not_less dest: antisym)
988 text \<open>Division and Signs\<close>
990 lemma
991 shows zero_less_divide_iff: "0 < a / b \<longleftrightarrow> 0 < a \<and> 0 < b \<or> a < 0 \<and> b < 0"
992 and divide_less_0_iff: "a / b < 0 \<longleftrightarrow> 0 < a \<and> b < 0 \<or> a < 0 \<and> 0 < b"
993 and zero_le_divide_iff: "0 \<le> a / b \<longleftrightarrow> 0 \<le> a \<and> 0 \<le> b \<or> a \<le> 0 \<and> b \<le> 0"
994 and divide_le_0_iff: "a / b \<le> 0 \<longleftrightarrow> 0 \<le> a \<and> b \<le> 0 \<or> a \<le> 0 \<and> 0 \<le> b"
995 by (auto simp add: divide_simps)
997 text \<open>Division and the Number One\<close>
999 text\<open>Simplify expressions equated with 1\<close>
1001 lemma zero_eq_1_divide_iff [simp]: "0 = 1 / a \<longleftrightarrow> a = 0"
1002 by (cases "a = 0") (auto simp: field_simps)
1004 lemma one_divide_eq_0_iff [simp]: "1 / a = 0 \<longleftrightarrow> a = 0"
1005 using zero_eq_1_divide_iff[of a] by simp
1007 text\<open>Simplify expressions such as \<open>0 < 1/x\<close> to \<open>0 < x\<close>\<close>
1009 lemma zero_le_divide_1_iff [simp]:
1010 "0 \<le> 1 / a \<longleftrightarrow> 0 \<le> a"
1013 lemma zero_less_divide_1_iff [simp]:
1014 "0 < 1 / a \<longleftrightarrow> 0 < a"
1017 lemma divide_le_0_1_iff [simp]:
1018 "1 / a \<le> 0 \<longleftrightarrow> a \<le> 0"
1021 lemma divide_less_0_1_iff [simp]:
1022 "1 / a < 0 \<longleftrightarrow> a < 0"
1025 lemma divide_right_mono:
1026 "[|a \<le> b; 0 \<le> c|] ==> a/c \<le> b/c"
1027 by (force simp add: divide_strict_right_mono le_less)
1029 lemma divide_right_mono_neg: "a <= b
1030 ==> c <= 0 ==> b / c <= a / c"
1031 apply (drule divide_right_mono [of _ _ "- c"])
1032 apply auto
1033 done
1035 lemma divide_left_mono_neg: "a <= b
1036 ==> c <= 0 ==> 0 < a * b ==> c / a <= c / b"
1037 apply (drule divide_left_mono [of _ _ "- c"])
1038 apply (auto simp add: mult.commute)
1039 done
1041 lemma inverse_le_iff: "inverse a \<le> inverse b \<longleftrightarrow> (0 < a * b \<longrightarrow> b \<le> a) \<and> (a * b \<le> 0 \<longrightarrow> a \<le> b)"
1042 by (cases a 0 b 0 rule: linorder_cases[case_product linorder_cases])
1043 (auto simp add: field_simps zero_less_mult_iff mult_le_0_iff)
1045 lemma inverse_less_iff: "inverse a < inverse b \<longleftrightarrow> (0 < a * b \<longrightarrow> b < a) \<and> (a * b \<le> 0 \<longrightarrow> a < b)"
1046 by (subst less_le) (auto simp: inverse_le_iff)
1048 lemma divide_le_cancel: "a / c \<le> b / c \<longleftrightarrow> (0 < c \<longrightarrow> a \<le> b) \<and> (c < 0 \<longrightarrow> b \<le> a)"
1049 by (simp add: divide_inverse mult_le_cancel_right)
1051 lemma divide_less_cancel: "a / c < b / c \<longleftrightarrow> (0 < c \<longrightarrow> a < b) \<and> (c < 0 \<longrightarrow> b < a) \<and> c \<noteq> 0"
1052 by (auto simp add: divide_inverse mult_less_cancel_right)
1054 text\<open>Simplify quotients that are compared with the value 1.\<close>
1056 lemma le_divide_eq_1:
1057 "(1 \<le> b / a) = ((0 < a & a \<le> b) | (a < 0 & b \<le> a))"
1058 by (auto simp add: le_divide_eq)
1060 lemma divide_le_eq_1:
1061 "(b / a \<le> 1) = ((0 < a & b \<le> a) | (a < 0 & a \<le> b) | a=0)"
1062 by (auto simp add: divide_le_eq)
1064 lemma less_divide_eq_1:
1065 "(1 < b / a) = ((0 < a & a < b) | (a < 0 & b < a))"
1066 by (auto simp add: less_divide_eq)
1068 lemma divide_less_eq_1:
1069 "(b / a < 1) = ((0 < a & b < a) | (a < 0 & a < b) | a=0)"
1070 by (auto simp add: divide_less_eq)
1072 lemma divide_nonneg_nonneg [simp]:
1073 "0 \<le> x \<Longrightarrow> 0 \<le> y \<Longrightarrow> 0 \<le> x / y"
1074 by (auto simp add: divide_simps)
1076 lemma divide_nonpos_nonpos:
1077 "x \<le> 0 \<Longrightarrow> y \<le> 0 \<Longrightarrow> 0 \<le> x / y"
1078 by (auto simp add: divide_simps)
1080 lemma divide_nonneg_nonpos:
1081 "0 \<le> x \<Longrightarrow> y \<le> 0 \<Longrightarrow> x / y \<le> 0"
1082 by (auto simp add: divide_simps)
1084 lemma divide_nonpos_nonneg:
1085 "x \<le> 0 \<Longrightarrow> 0 \<le> y \<Longrightarrow> x / y \<le> 0"
1086 by (auto simp add: divide_simps)
1088 text \<open>Conditional Simplification Rules: No Case Splits\<close>
1090 lemma le_divide_eq_1_pos [simp]:
1091 "0 < a \<Longrightarrow> (1 \<le> b/a) = (a \<le> b)"
1092 by (auto simp add: le_divide_eq)
1094 lemma le_divide_eq_1_neg [simp]:
1095 "a < 0 \<Longrightarrow> (1 \<le> b/a) = (b \<le> a)"
1096 by (auto simp add: le_divide_eq)
1098 lemma divide_le_eq_1_pos [simp]:
1099 "0 < a \<Longrightarrow> (b/a \<le> 1) = (b \<le> a)"
1100 by (auto simp add: divide_le_eq)
1102 lemma divide_le_eq_1_neg [simp]:
1103 "a < 0 \<Longrightarrow> (b/a \<le> 1) = (a \<le> b)"
1104 by (auto simp add: divide_le_eq)
1106 lemma less_divide_eq_1_pos [simp]:
1107 "0 < a \<Longrightarrow> (1 < b/a) = (a < b)"
1108 by (auto simp add: less_divide_eq)
1110 lemma less_divide_eq_1_neg [simp]:
1111 "a < 0 \<Longrightarrow> (1 < b/a) = (b < a)"
1112 by (auto simp add: less_divide_eq)
1114 lemma divide_less_eq_1_pos [simp]:
1115 "0 < a \<Longrightarrow> (b/a < 1) = (b < a)"
1116 by (auto simp add: divide_less_eq)
1118 lemma divide_less_eq_1_neg [simp]:
1119 "a < 0 \<Longrightarrow> b/a < 1 \<longleftrightarrow> a < b"
1120 by (auto simp add: divide_less_eq)
1122 lemma eq_divide_eq_1 [simp]:
1123 "(1 = b/a) = ((a \<noteq> 0 & a = b))"
1124 by (auto simp add: eq_divide_eq)
1126 lemma divide_eq_eq_1 [simp]:
1127 "(b/a = 1) = ((a \<noteq> 0 & a = b))"
1128 by (auto simp add: divide_eq_eq)
1130 lemma abs_inverse [simp]:
1131 "\<bar>inverse a\<bar> =
1132 inverse \<bar>a\<bar>"
1133 apply (cases "a=0", simp)
1135 done
1137 lemma abs_divide [simp]:
1138 "\<bar>a / b\<bar> = \<bar>a\<bar> / \<bar>b\<bar>"
1139 apply (cases "b=0", simp)
1141 done
1143 lemma abs_div_pos: "0 < y ==>
1144 \<bar>x\<bar> / y = \<bar>x / y\<bar>"
1145 apply (subst abs_divide)
1147 done
1149 lemma zero_le_divide_abs_iff [simp]: "(0 \<le> a / \<bar>b\<bar>) = (0 \<le> a | b = 0)"
1150 by (auto simp: zero_le_divide_iff)
1152 lemma divide_le_0_abs_iff [simp]: "(a / \<bar>b\<bar> \<le> 0) = (a \<le> 0 | b = 0)"
1153 by (auto simp: divide_le_0_iff)
1155 lemma inverse_sgn:
1156 "sgn (inverse a) = inverse (sgn a)"
1159 lemma field_le_mult_one_interval:
1160 assumes *: "\<And>z. \<lbrakk> 0 < z ; z < 1 \<rbrakk> \<Longrightarrow> z * x \<le> y"
1161 shows "x \<le> y"
1162 proof (cases "0 < x")
1163 assume "0 < x"
1164 thus ?thesis
1165 using dense_le_bounded[of 0 1 "y/x"] *
1166 unfolding le_divide_eq if_P[OF \<open>0 < x\<close>] by simp
1167 next
1168 assume "\<not>0 < x" hence "x \<le> 0" by simp
1169 obtain s::'a where s: "0 < s" "s < 1" using dense[of 0 "1::'a"] by auto
1170 hence "x \<le> s * x" using mult_le_cancel_right[of 1 x s] \<open>x \<le> 0\<close> by auto
1171 also note *[OF s]
1172 finally show ?thesis .
1173 qed
1175 end
1177 text \<open>Min/max Simplification Rules\<close>
1179 lemma min_mult_distrib_left:
1180 fixes x::"'a::linordered_idom"
1181 shows "p * min x y = (if 0 \<le> p then min (p*x) (p*y) else max (p*x) (p*y))"
1182 by (auto simp add: min_def max_def mult_le_cancel_left)
1184 lemma min_mult_distrib_right:
1185 fixes x::"'a::linordered_idom"
1186 shows "min x y * p = (if 0 \<le> p then min (x*p) (y*p) else max (x*p) (y*p))"
1187 by (auto simp add: min_def max_def mult_le_cancel_right)
1189 lemma min_divide_distrib_right:
1190 fixes x::"'a::linordered_field"
1191 shows "min x y / p = (if 0 \<le> p then min (x/p) (y/p) else max (x/p) (y/p))"
1192 by (simp add: min_mult_distrib_right divide_inverse)
1194 lemma max_mult_distrib_left:
1195 fixes x::"'a::linordered_idom"
1196 shows "p * max x y = (if 0 \<le> p then max (p*x) (p*y) else min (p*x) (p*y))"
1197 by (auto simp add: min_def max_def mult_le_cancel_left)
1199 lemma max_mult_distrib_right:
1200 fixes x::"'a::linordered_idom"
1201 shows "max x y * p = (if 0 \<le> p then max (x*p) (y*p) else min (x*p) (y*p))"
1202 by (auto simp add: min_def max_def mult_le_cancel_right)
1204 lemma max_divide_distrib_right:
1205 fixes x::"'a::linordered_field"
1206 shows "max x y / p = (if 0 \<le> p then max (x/p) (y/p) else min (x/p) (y/p))"
1207 by (simp add: max_mult_distrib_right divide_inverse)
1209 hide_fact (open) field_inverse field_divide_inverse field_inverse_zero
1211 code_identifier
1212 code_module Fields \<rightharpoonup> (SML) Arith and (OCaml) Arith and (Haskell) Arith
1214 end | 14,285 | 43,812 | {"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-2019-43 | latest | en | 0.527753 |
https://teamtreehouse.com/videos/55772/captions | 1,653,553,615,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00410.warc.gz | 622,045,544 | 2,775 | 1 00:00:00,400 --> 00:00:03,050 Imagine we're still working on an e-commerce site. 2 00:00:03,050 --> 00:00:06,590 And the designers come up to us to ask a question. 3 00:00:06,590 --> 00:00:10,040 They're updating the design on a customer profile page. 4 00:00:10,040 --> 00:00:13,910 They want to know the length of the largest user name in the database 5 00:00:13,910 --> 00:00:18,350 to test out if their design works at the extremes of what's in our database. 6 00:00:19,470 --> 00:00:23,670 We can use the SQL function called length to help us answer this question. 7 00:00:25,190 --> 00:00:28,130 We have a select statement that only brings back the username 8 00:00:28,130 --> 00:00:30,180 from our customers table. 9 00:00:30,180 --> 00:00:34,620 Let's show the username and their length of the username in the report so 10 00:00:34,620 --> 00:00:37,980 we can verify that it's actually working. 11 00:00:37,980 --> 00:00:42,818 Remember, a function starts with the name of the function, length, 12 00:00:42,818 --> 00:00:45,688 a pair of parenthesis and then the value or 13 00:00:45,688 --> 00:00:49,060 the column name that you want to be transformed. 14 00:00:50,070 --> 00:00:55,799 Watch, when we run this, we get the username with their lengths. 15 00:00:56,850 --> 00:01:00,470 Notice here, in the column names that length username is shown. 16 00:01:01,490 --> 00:01:04,600 That's what happens when you don't alias a result of a function. 17 00:01:05,610 --> 00:01:08,990 Let's alias this to something that's easier to understand. 18 00:01:10,740 --> 00:01:14,840 That's better, but what about the longest username? 19 00:01:14,840 --> 00:01:16,170 What is it? 20 00:01:16,170 --> 00:01:20,380 What new keywords have we learned that we could use to figure this out? 21 00:01:20,380 --> 00:01:23,620 We can order by their length in descending order. 22 00:01:27,619 --> 00:01:30,420 And then limit it to just the longest one. 23 00:01:31,540 --> 00:01:36,640 When we run it only the record with the longest username is returned. 24 00:01:36,640 --> 00:01:40,850 There may be more with the same length, but those results are irrelevant. 25 00:01:40,850 --> 00:01:42,570 All we want to know is the length. 26 00:01:43,800 --> 00:01:46,940 You're also not limited to using the results of the function 27 00:01:46,940 --> 00:01:49,300 in the order by criteria. 28 00:01:49,300 --> 00:01:54,710 Let's say we wanted to find all usernames under the length of seven characters. 29 00:01:54,710 --> 00:01:57,490 We could modify our query to look like this. 30 00:01:58,800 --> 00:02:02,910 Where length is less than seven. 31 00:02:06,699 --> 00:02:10,210 In fact, you don't even need to select it. 32 00:02:10,210 --> 00:02:13,970 You can simply move it over to the condition like this. 33 00:02:15,970 --> 00:02:19,981 You may want to do this, because seeing the length doesn't matter, but 34 00:02:19,981 --> 00:02:22,430 the actual username is what's important. 35 00:02:23,490 --> 00:02:28,040 You've seen the length function used in several different ways. 36 00:02:28,040 --> 00:02:31,620 You've seen it used in this select portion of a query. 37 00:02:31,620 --> 00:02:35,360 We even saw it being used as part of a work condition. 38 00:02:35,360 --> 00:02:39,050 You can include it anywhere you'd put a value or column name. 39 00:02:41,070 --> 00:02:45,380 You should start to see how using functions can help answer more questions 40 00:02:45,380 --> 00:02:46,920 or form an output. 41 00:02:46,920 --> 00:02:49,880 We're going to explore many more functions throughout this course. | 1,256 | 3,601 | {"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-2022-21 | latest | en | 0.917804 |
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• solving equations game | 1,230 | 5,268 | {"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-2017-22 | longest | en | 0.932522 |
https://ru.scribd.com/document/395630070/Physics-Project-Balloon-Racers | 1,607,017,643,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00272.warc.gz | 407,093,787 | 91,090 | Вы находитесь на странице: 1из 3
# Name
## And Newton said, “Let it rip!”
Newton’s laws of motion will always seem to work in class, but in real life, we find that surfaces are seldom
frictionless and that conditions are never truly ideal. This is because outside non-constant forces interfere with
Newton’s laws as we study them in introductory physics. Balloon racers heavily rely on Newton’s laws of motion
and will provide an excellent opportunity to investigate them as we continue this study on forces. While building
balloon racers, we will discover how difficult it can be to make something work consistently as it is intended.
RULES…
• Must work individually.
• You may build the car out of anything you have laying around!
• It must have at least THREE wheels. Wheels are defined as anything that is round and spins
• The wheels CANNOT be wheels from a toy car. They must be made out of something that was not
originally meant to be used as wheels.
• The car MUST NOT leave the ground.
• The car MUST travel in a straight line.
• Balloons, straws, tape, pens, pencils, paper clips, balsa wood, CDs/DVDs, Styrofoam, PVC
_______________________________________________________________________
_______________________________________________________________________
Total points possible = 50 points
• 10 pts - Class participation; Materials brought to class on time and use of class time!
• 15 pts - Completed planning guide: Must be completed in your LAB NOTEBOOK!
• 25 pts - Car moved 5 meters or more
23 pts - Car moved 4 – 5 m
21 pts - Car moved 3 – 4 m
19 pts - Car moved 2 – 3 m
17 pts - Car moved 1 – 2 m
15 pts - Car moved 0 – 1 m
13 pts - Completed car – Does not move! =-(
Extra Credit
1 pt for each meter beyond 5 meters
DUE DATES…
• Monday, December 3rd Planning Guide
• Friday, December 7th Materials brought to class; Construction Days
• Monday, December 10th Race Day!
PLANNING GUIDE
Balloon Racer
Incorporate Newton’s laws of motion into the design and construction of a balloon race car.
Define Newton’s 3 laws of motion in your LAB NOTEBOOK!
Answer the following questions in you LAB NOTEBOOK!
1. What type of force causes an object to move?
2. What must be overcome in order to get the car to move?
3. What is inertia?
4. Do resting objects have inertia?
5. Do moving objects have inertia?
6. What property of matter determines how much inertia an object has?
7. What will cause your car to stop?
8. Explain how will you minimize friction in your car?
9. Will all cars have roughly the same amount of force available? Explain
10. If force = mass x acceleration what will happen to the acceleration of the car if mass
is increased? Decreased?
11. What caused your car to move forward?
12. Explain why the open end of the balloon must face backwards if you want the car to
move forward.
13. List 3 different types of objects you could use for wheels.
14. How will you connect the wheel to the axle?
15. How will you connect the axle to the body of the car?
16. What could you do to reduce friction between the wheels, axle and body of the car?
17. List 3 different objects you could use as a car body.
18. How will you attach the balloon to the car body?
19. Describe how you could control the amount of force released from the balloon.
20. What are the advantages/disadvantages of controlling how the air is released from
the balloon? | 800 | 3,371 | {"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-2020-50 | latest | en | 0.918238 |
http://www.finderchem.com/math-binomial-theorem-help.html | 1,500,664,112,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423808.34/warc/CC-MAIN-20170721182450-20170721202450-00445.warc.gz | 447,182,439 | 7,935 | # Math Binomial Theorem Help?
## We found this answers
Binomial Theorem Examples . Algebra > Binomial Theorem Examples 13. Let us understand the Binomial Theorem concepts discussed above with the following numerous solved ... - Read more
Math video on defining and solving combinations (choosing), used in determining coefficients of the binomial theorem. Problem 1. - Read more
## Math Binomial Theorem Help? resources
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### Binomial Theorem -- from Wolfram MathWorld
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### Mathguru – Help – Example:Finding Middle Term using ...
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### Binomial Theorem - IB Math Stuff - Wikidot
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### The Binomial Theorem and Pascal’s Triangle | Math Tricks
Using Pascal's triangle to find the binomial coefficients when performing a binomial expansion. A great example of math tricks!
### what is the binomial theorem for - Math Homework Answers
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Can anyone help me? I've just started doing the Binomial theorem, and I'm stuck on the question. There are two parts to the question, part one is:
### Binomial Theorem Calculator | Online Tutoring, Math Online ...
Online Tutoring, Math Online Help, Algebra Help, Test Prep, Homework Help for Math, Science, English, Essay Writing Online Tutoring is no more Monotonous with the use ...
### Binomial Theorem algebra topics - Math . info
Binomial Theorem A general formula for expanding (a + b) n for any positive integer n is provided by the binomial theorem:
### Maths Tutor - Mathtutor
Pascal's triangle and the binomial theorem
### Binomial theorem - Math Planet - Math help online, math ...
Expanding a binomial expression that has been raised to some large power could be troublesome; one way to solve it is to use the binomial theorem:
### Help with the Binomial Theorem - Math Help Forum
I am somewhat confused. Some text books refer to the following as the "General Form" math] (a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k [/tex
### Binomial Theorem | MathCaptain.com - Online Math Help ...
Binomial theorem is a mathematical theorem that gives the expansion of any binomial raised to a positive integral power n and contains n + 1 terms where n is a ...
### Binomial Theorem Assignment Help | Binomial Theorem ...
Binomial theorem is described as the algebraic expansion of powers of a binomial in the form (x + y) n. According to the theorem we can expand (x + y) n into a sum ...
### Sequences & Series - Cool math Algebra Help Lessons - The ...
This algebra lesson introduces the binomial theorem and shows how it's related to Pascal's Triangle.
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Depdendent Variable
Number of equations to solve: 23456789
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Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
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Ineq. #5:
Ineq. #6:
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Solve for:
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Author Message
interTisd
Registered: 11.04.2002
From: esperanto.org.il
Posted: Sunday 31st of Dec 10:41 Hi Fellows ! I have a serious issue regarding algebra and I was wondering if someone might have the ability to help me out a little . I have a algebra test pretty soon and even though I have been studying math seriously, there are still a some parts that cause a lot of headache , such as lowest common denominator calculator and system of equations especially. Last week I had a meeting with a math teacher, but many things still remain vague to me. Can you suggest a good way of studying or a good private teacher that you know already?
kfir
Registered: 07.05.2006
From: egypt
Posted: Tuesday 02nd of Jan 09:18 Hey. I believe I can lend you a hand . Can you explain some more on what your troubles are? What specifically are your problems with lowest common denominator calculator? Getting a excellent coach would have been the best thing. But do not lose sleep . I think there is a way out . I have come across a number of math software programs. I have tried them out myself. They are pretty smart and excellent. These might just be what you need. They also do not cost a lot. I think what you need is Algebrator. Why not try this out? It could be just be the answer for your problems .
alhatec16
Registered: 10.03.2002
From: Notts, UK.
Posted: Thursday 04th of Jan 09:04 Hey, even I made use of Algebrator to learn more about lowest common denominator calculator. This was just a remarkable tool that assisted me with all the basic principles. I would suggest you to use this before resorting to the assistance from coaches, which is often very costly .
Svizes
Registered: 10.03.2003
From: Slovenia
Posted: Thursday 04th of Jan 16:45 I advise using Algebrator. It not only helps you with your math problems, but also displays all the necessary steps in detail so that you can enhance the understanding of the subject.
dxe_cin
Registered: 20.04.2003
From: Minnesota
Posted: Saturday 06th of Jan 11:23 Is it really true that a program can perform like that? I don’t really know much anything about this Algebrator but I am really seeking for some help so would you mind sharing me where could I find that software ? Is it downloadable over the internet ? I’m hoping for your fast response because I really need assistance desperately.
Registered: 03.07.2001
From:
Posted: Monday 08th of Jan 10:12 This one is actually quite unique . I am recommending it only after trying it myself. You can find the information about the software at https://algebrahomework.org/flash/noflash/35.html. | 875 | 3,535 | {"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-2019-39 | latest | en | 0.932138 |
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