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https://app-wiringdiagram.herokuapp.com/post/greedy-triangle-lesson-plans-1st-grade
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9 out of 10 based on 986 ratings. 3,543 user reviews.
# GREEDY TRIANGLE LESSON PLANS 1ST GRADE
The Greedy Triangle: Geometry for Every Grade | Scholastic
After reading, have students become their favorite shape and tell what they would be, the same way The Greedy Triangle enjoys being a book page as a quadrilateral, or a beehive as a hexagon. Supply students shapes to cut out and create their image. Download a sentence stem starter, or let students construct their own. First Grade. Standards
'The Greedy Triangle' Teaches Kids About Geometry
Nov 26, 2017This sample lesson plan uses the book "The Greedy Triangle" to teach about the attributes of two-dimensional figures. The plan is designed for second-grade and third-grade students, and it requires a 45-minute period for two days.
Teaching The Greedy Triangle | Math Solutions
A Lesson for First and Second Grade by Lynn Sherman Lynn Sherman is a teacher candidate at York University in Toronto, Ontario, where she recently taught a geometry class based on The Greedy Triangle, by Marilyn Burns (Scholastic, 1994), to a class of second graders.
The Greedy Triangle Lesson Plan for 1st - 4th Grade
This The Greedy Triangle Lesson Plan is suitable for 1st - 4th Grade. Students make designated geometrical shapes using geoboards. In this geometry lesson, students listen to Marilyn Burns, The Greedy Triangle, while listening for the names of geometric shapes.3/5
The Greedy Triangle Lesson Plan | Study
Learning ObjectivesCurriculum StandardsMaterialsAfter this lesson, students will be able to: 1. identify the name of some geometrical shapes 2. identify the number of sides for some geometrical shapesSee more on study
The Greedy Triangle Lesson Plans & Worksheets | Lesson Planet
The greedy triangle lesson plans and worksheets from thousands of teacher-reviewed resources to help you inspire students learning.Location: Vacaville, CA
The Greedy Triangle - Fun in First
Mar 31, 2012The Greedy Triangle. March 31, 2012 by Jodi 21 Comments. Wahoo! It’s officially Spring Break. I decided that we would use it for a writing lesson as well. Each student decided what shape they would choose to be and wrote about it. The first grade teachers go eat on Fridays. We usually go to Subway because it’s really close. As long
The Greedy Triangle 5E Math Lesson Plan | The greedy
This is a 5E math lesson plan using the book "The Greedy Triangle" by Marilyn Burns to teach students how to classify the number of vertices and sides in 2-D shapesudes:--Full lesson plan--Suggested questions classified by Bloom's Taxonomy--Math stations--Rubric for assessing student work*If yo..
Fifth grade Lesson The Greedy Triangle | BetterLesson
In the story "The Greedy Triangle," the main character is dissatisfied with being a triangle, and asks a shape shifter to have one more side and one more angle. He's not satisfied, and his journey continues from the three sided, three angled shape of the triangle Author: Cathy Skinner
Hands-on Math Lessons with "The Greedy Triangle" - Teach
Sep 07, 2015If you love the book, “The Greedy Triangle,” be sure and grab these free printable lessons from Bethany at Math Geek Mama! Included are fun, hands-on ways to explore triangles! This can be used with kids as young as Kindergarten, but also includes more advanced pages to
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# Costas arrays17
Posted by Steve Eddins,
Last week I wrote about a possible modification to poly2mask that would estimate how much of each pixel is contained within a polygon. The idea was to use a regular 5-by-5 subgrid and count how many of the 25 subpixels are contained within the polygon. It would look something like this:
One of our graphics developers, Mike, sent me a note a couple of days later. He pointed out that using a regular N-by-N subsampling of the pixel doesn't do very well for polygon edges that are nearly vertical or horizontal. With a vertical or horizontal edge, your estimate of the pixel coverage is good only to within about 1/N instead of 1/N^2. As a possible alternative, Mike pointed me to something called a Costas array .
A Costas array is a permutation matrix. A permutation matrix contains 0s and 1s such that each row and each column contains only a single 1. The identity matrix is a trivial example:
eye(3)
ans =
1 0 0
0 1 0
0 0 1
In addition to being a permutation matrix, a Costas array has the property that no pair of 1-valued elements has the same vector difference as any other pair. The 3-by-3 identity matrix above does not have this property. The vector difference between the (1,1) element and the (2,2) element is (1,1). This is the same as the vector difference between the (2,2) and (3,3) elements. Here's a visualization of a 10-by-10 Costas array:
v = [2 4 8 5 10 9 7 3 6 1];
n = 10;
plot(1:n, v, 'LineStyle', 'none', 'Marker', 'o', ...
'MarkerEdgeColor', 'b', 'MarkerFaceColor', 'b')
axis equal
grid on
set(gca, ...
'XTick', 0.5:(n+0.5), ...
'YTick', 0.5:(n+0.5), ...
'XLim', [0.5 (n+0.5)], ...
'YLim', [0.5 (n+0.5)], ...
'XTickLabel', '', ...
'YTickLabel', '', ...
'GridLineStyle', '-')
And here's a 29-by-29 example:
v = [3 21 23 22 8 15 26 6 16 11 28 5 2 18 10 14 12 13 27 20 9 29 19 ...
24 7 1 4 17 25];
n = 29;
plot(1:n, v, 'LineStyle', 'none', 'Marker', 'o', ...
'MarkerEdgeColor', 'b', 'MarkerFaceColor', 'b')
axis equal
grid on
set(gca, ...
'XTick', 0.5:(n+0.5), ...
'YTick', 0.5:(n+0.5), ...
'XLim', [0.5 (n+0.5)], ...
'YLim', [0.5 (n+0.5)], ...
'XTickLabel', '', ...
'YTickLabel', '', ...
'GridLineStyle', '-')
Using the same basic procedure I described last time, you could count how many of the 29 dots lie within the polygon and then divide by 29 to get your coverage estimate.
There are many possible variations on this technique, and Mike assures me that it is an active area of research in computer graphics.
Costas arrays were originally described in the context of waveform detection. See J. Costas, "A Study of a Class of Detection Waveforms Having Nearly Ideal Range-Doppler Ambiguity Properties," Proceedings of the IEEE, vol. 72, no. 8, August 1984.
Get the MATLAB code
Published with MATLAB® 7.3
JanKees replied on : 1 of 17
Hi, Steve:
Interesting your blog post about poly2mask and roipoly. But I have a doubt on them: is that possible to copy a region of interest from an image and put this ROI in a new image with only this ROI?
JanKees
Steve replied on : 2 of 17
JanKees – Use logical indexing to copy the pixel values inside a region of interest from one image to another. You might do something like this:
[m,n] = size(image1);
jianquan ouyang replied on : 3 of 17
Hi, Steve:
I’m Interesting your blog about costas arrays. can you give a matlab example of generating Welch-constructed or Golomb-constructed costas arrays?
ZHen replied on : 4 of 17
Hi, Steve: I meet a problem and need your help. I am now doing some research on image information extraction. My question is: I know the corner points on one object, I take the picture of the object, and get the image. Using corner points detector, I can find the corresponding images of these corners. But these images are different from the corners images calculated by using the perspective camera model. How can I get the actual corner image points from the image of the object? (I mean the images of the corners calculated by using the perspective camera model).
Steve replied on : 5 of 17
Jianquan – I wish I could give you a MATLAB example showing you how to generate a Costas array. But when I was doing research for this blog posting, I didn’t find any MATLAB implementations. Although the topic interests me somewhat, it is off the main topic for this blog, so I don’t plan to spend more time on it. I suggest you visit the link I mentioned and look up the papers mentioned there for generating algorithms.
Steve replied on : 6 of 17
Zhen – Camera calibration can be a fairly involved operation, and I don’t know that much about it. I suggest searching for information on “camera calibration.” You should be able to find some implementations in MATLAB, including contributions on the MATLAB Central File Exchange.
A Brook replied on : 7 of 17
Are Costas arrays optimal for this task (antialiasing) in any sense? Is there any literature on the issue? Thanks.
Steve replied on : 8 of 17
A – I posed your question to graphics developer Mike, who first told me about Costas arrays. He pointed out that the success of an antialiasing method depends on properties of human visual perception, and so mathematical measures of optimality tend to be not very helpful. In his experience, “sampling patterns have often been chosen in pretty ad hoc ways with little concern for the underlying math.” He suggests looking through Solomon Golomb’s papersfor information about the mathematical properties of Costas arrays.
Simon M. replied on : 9 of 17
Hello, Steve. I’m looking for an easy way to calculate for a given polygon (type uint8 or double) for each pixel not only if it lies in the polygon, but also the coverage, e.g. something like 0.65 of a pixel is covered by the polygon.
Do you know any matlab algorithms, which might help me.
Simon
Steve replied on : 10 of 17
Simon—This post and my February 6th post are about a possible method for computing partial pixel coverage. I asked whether anyone was interested. We would like to hear about your use case. Can you describe the scenario in which you’d like to be able to compute this information? What kind of data, where did it come from, how will partial pixel coverage be used? What trouble does it cause if this algorithm is not available?
Simon M. replied on : 11 of 17
Hello, Steve.
Right now we are using the matlab function poly2mask to intersect different drainage areas/patches with image pixels of a rain radar (e.g. 200×200 pixel, 1km² rastersize).
Those patches are available as dxf-polygones, the radar information is represented by greyscale images.
Since every patch is covered by only 20-30 radar pixels we would like to calculate the exact amount of rainfall for a patch. Nevertheless we are achieving quite good results with the existing poly2mask method.
Steve replied on : 12 of 17
Simon—Thanks for the information.
varma replied on : 13 of 17
hai steve,
congrats for maintainig a good blog.steve i want few costas sequences for N=8 and 10. could you help me out in this regard
Steve replied on : 14 of 17
Varma—This post contains about everything I know about Costas arrays. I suggest that you follow the link at the end of the post to find some papers and examples.
Steve replied on : 15 of 17
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mersenneforum.org > Math Carmichael numbers and Devaraj numbers
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2017-07-09, 05:07 #1 devarajkandadai May 2004 22×79 Posts Carmichael numbers and Devaraj numbers In a recent post I had stated that although 561 is a Carmichael number in the sub-ring of rational integers it is only a pseudoprime in the ring of Gaussian integers. In fact I would be surprised if there are any Carmichael numbers in the ring of Gaussian integers other than those in the sub-ring of rational integers.However there are Devaraj numbers in the ring of Gaussian integers other those in the subring of rational integers.Example: Let N = (2 - i)*(3+2i)*(4-i).Appluing the formula for Pomerance index we find the relevant Pomerance index is (1-5i).( for difference between Carmichael numbers and Devaraj numbers see A104016 and A104017). Last fiddled with by devarajkandadai on 2017-07-09 at 05:08
Similar Threads Thread Thread Starter Forum Replies Last Post devarajkandadai Number Theory Discussion Group 14 2017-11-15 15:00 devarajkandadai Number Theory Discussion Group 7 2017-09-23 02:58 devarajkandadai Miscellaneous Math 0 2006-08-04 03:06 devarajkandadai Math 1 2004-09-16 06:06 devarajkandadai Math 0 2004-08-19 03:12
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You are currently browsing the tag archive for the ‘algebraic combinatorics’ tag.
One of the oldest and most fundamental concepts in mathematics is the line. Depending on exactly what mathematical structures we want to study (algebraic, geometric, topological, order-theoretic, etc.), we model lines nowadays by a variety of standard mathematical objects, such as the real line ${\Bbb R}$, the complex line ${\Bbb C}$, the projective line $\Bbb{RP}^1$, the extended real line ${}[-\infty,+\infty]$, the affine line ${\Bbb A}^1$, the continuum $c$, the long line $L$, etc. We also have discrete versions of the line, such as the natural numbers ${\Bbb N}$, the integers ${\Bbb Z}$, and the ordinal $\omega$, as well as compact versions of the line, such as the unit interval ${}[0,1]$ or the unit circle ${\Bbb T} := {\Bbb R}/{\Bbb Z}$. Finally we have discrete and compact versions of the line, such as the cyclic groups ${\Bbb Z}/N{\Bbb Z}$ and the discrete intervals $\{1,\ldots,N\}$ and $\{0,\ldots,N-1\}$. By taking Cartesian products we then obtain higher-dimensional objects such as Euclidean space ${\Bbb R}^n$, the standard lattice ${\Bbb Z}^n$, the standard torus ${\Bbb T}^n = {\Bbb R}^n/{\Bbb Z}^n$, and so forth. These objects of course form the background on which a very large fraction of modern mathematics is set.
Broadly speaking, the line has three major families of structures on it:
1. Geometric structures, such as a metric or a measure, completeness, scales (coarse and fine), rigid motions (translations and reflection), similarities (dilation, affine maps), and differential structures (tangent bundle, etc.);
2. Algebraic structures, such group, ring, or field structures, and everything else that comes from those categories (e.g. subgroups, homomorphisms, involutions, etc.); and
3. One-dimensional structures, such as order, a length space structure (in particular, path-connectedness structure), a singleton generator, the Archimedean property, the ability to use mathematical induction (i.e. well-ordering), convexity, or the ability to disconnect the line by removing a single point.
Of course, these structures are inter-related, and it is an important phenomenon that a mathematical concept which appears to be native to one structure, can often be equivalently defined in terms of other structures. For instance, the absolute value $|n|$ of an integer $n$ can be defined geometrically as the distance from 0 to $n$, algebraically as the index of the subgroup $\langle n \rangle = n \cdot \Bbb Z$ of the integers ${\Bbb Z}$ generated by n, or one-dimensionally as the number of integers between 0 and $n$ (including 0, but excluding $n$). This equivalence of definitions becomes important when one wants to work in more general contexts in which one or more of the above structures is missing or otherwise weakened.
What I want to talk about today is an important toy model for the line (in any of its incarnations), in which the geometric and algebraic structures are enhanced (and become neatly nested and recursive), at the expense of the one-dimensional structure (which is largely destroyed). This model has many different names, depending on what field of mathematics one is working in and which structures one is interested in. In harmonic analysis it is called the dyadic model, the Walsh model, or the Cantor group model; in number theory and arithmetic geometry it is known as the function field model; in topology it is the Cantor space model; in probability it is the martingale model; in metric geometry it is the ultrametric, tree, or non-Archimedean model; in algebraic geometry it is the Puiseux series model; in additive combinatorics it is the bounded torsion or finite field model; in computer science and information theory it is the Hamming cube model; in representation theory it is the Kashiwara crystal model. Let me arbitrarily select one of these terms, and refer to all of these models as dyadic models for the line (or of objects derived from the line). While there is often no direct link between a dyadic model and a non-dyadic model, dyadic models serve as incredibly useful laboratories in which to gain insight and intuition for the “real-world” non-dyadic model, since one has much more powerful and elegant algebraic and geometric structure to play with in this setting (though the loss of one-dimensional structure can be a significant concern). Perhaps the most striking example of this is the three-line proof of the Riemann hypothesis in the function field model of the integers, which I will discuss a little later.
This problem lies in the highly interconnected interface between algebraic combinatorics (esp. the combinatorics of Young tableaux and related objects, including honeycombs and puzzles), algebraic geometry (particularly classical and quantum intersection theory and geometric invariant theory), linear algebra (additive and multiplicative, real and tropical), and the representation theory (classical, quantum, crystal, etc.) of classical groups. (Another open problem in this subject is to find a succinct and descriptive name for the field.) I myself haven’t actively worked in this area for several years, but I still find it a fascinating and beautiful subject. (With respect to the dichotomy between structure and randomness, this subject lies deep within the “structure” end of the spectrum.)
As mentioned above, the problems in this area can be approached from a variety of quite diverse perspectives, but here I will focus on the linear algebra perspective, which is perhaps the most accessible. About nine years ago, Allen Knutson and I introduced a combinatorial gadget, called a honeycomb, which among other things controlled the relationship between the eigenvalues of two arbitrary Hermitian matrices A, B, and the eigenvalues of their sum A+B; this was not the first such gadget that achieved this purpose, but it was a particularly convenient one for studying this problem, in particular it was used to resolve two conjectures in the subject, the saturation conjecture and the Horn conjecture. (These conjectures have since been proven by a variety of other methods.) There is a natural multiplicative version of these problems, which now relates the eigenvalues of two arbitrary unitary matrices U, V and the eigenvalues of their product UV; this led to the “quantum saturation” and “quantum Horn” conjectures, which were proven a couple years ago. However, the quantum analogue of a “honeycomb” remains a mystery; this is the main topic of the current post.
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# Minimum Coin Flips Amazon OA 2023
## Minimum Coin Flips Amazon OA 2023
Given the initial sequence of a coins, find the minimum number of coins that can be flipped to obtain a beautiful sequence. All head facing coins or tails facing coins sequence is also valid.
Input : THHHTH
Output : 2
Explanation : flip first and last coin to obtain the sequence.
Time Complexity : 0(N)
Space Complexity : 0(N)
### SOLUTION
Program : Minimum Coin Flips Amazon OA Solution in Python
Lets see how can we solve this question “Minimum Coin Flips Amazon OA”.
1. find the counts of “T” to the left of each element, store in a list
2. find the counts of “H” to the right of each element, store in a list
3. Ideally, we want all “H” to the left and all “T” to the right i.e “HHHTT”. So identifying all the “H” to the right and all the “T” to the left for any index tells us the number of swaps needed across that index.
e.g S = “TTTTHHH” , say at index 3, which is “T”, there are 3 “T” to the left and 3 “H” to the right, so the total swaps is 3+ 3= 6. Repeat for each index and return the minimum
4. Find the min of the sum of both lists for each element ,that is the answer
``````def func(string):
Tleft = [0]* len(string)
Hright = [0]*len(string)
result = len(string)
# iterate i from i = 1 since Tleft[0] is always 0
for i in range(1, len(string)):
if string[i-1] == "T":
Tleft[i] = Tleft[i-1] + 1
else:
Tleft[i] = Tleft[i-1]
#iterate from 2nd to last element since Hright[lastindex] is always 0
for i in range(len(string)-2,-1,-1):
if string[i+1] == "H":
Hright[i] = Hright[i+1]+ 1
else :
Hright[i] = Hright[i+1]
for i in range(len(string)):
result = min(result, Tleft[i]+ Hright[i])
#Using if condition to make sure if the given string is already in correct format like "HHTTT" all Head's on right and Tail's on left.
if result!=0:
return result
return 1
#Taking input
s=input()
print(func(s))``````
Now as we have achieved the solution for Minimum Coin Flips Amazon OA Solution, there are few more questions similar to this logic which gives output in terms of 0’s and 1’s use the same logic approach to solve those questions.
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# ListsWorking with Structured Data
Require Export Induction.
Module NatList.
# Pairs of Numbers
In an Inductive type definition, each constructor can take any number of arguments — none (as with true and O), one (as with S), or more than one, as in this definition:
Inductive natprod : Type :=
pair : nat nat natprod.
This declaration can be read: "There is just one way to construct a pair of numbers: by applying the constructor pair to two arguments of type nat."
We can construct an element of natprod like this:
Check (pair 3 5).
Here are two simple function definitions for extracting the first and second components of a pair. (The definitions also illustrate how to do pattern matching on two-argument constructors.)
Definition fst (p : natprod) : nat :=
match p with
| pair x yx
end.
Definition snd (p : natprod) : nat :=
match p with
| pair x yy
end.
Eval compute in (fst (pair 3 5)).
(* ===> 3 *)
Since pairs are used quite a bit, it is nice to be able to write them with the standard mathematical notation (x,y) instead of pair x y. We can tell Coq to allow this with a Notation declaration.
Notation "( x , y )" := (pair x y).
The new notation can be used both in expressions and in pattern matches (indeed, we've seen it already in the previous chapter — this notation is provided as part of the standard library):
Eval compute in (fst (3,5)).
Definition fst' (p : natprod) : nat :=
match p with
| (x,y) ⇒ x
end.
Definition snd' (p : natprod) : nat :=
match p with
| (x,y) ⇒ y
end.
Definition swap_pair (p : natprod) : natprod :=
match p with
| (x,y) ⇒ (y,x)
end.
Let's try and prove a few simple facts about pairs. If we state the lemmas in a particular (and slightly peculiar) way, we can prove them with just reflexivity (and its built-in simplification):
Theorem surjective_pairing' : (n m : nat),
(n,m) = (fst (n,m), snd (n,m)).
Proof.
reflexivity. Qed.
Note that reflexivity is not enough if we state the lemma in a more natural way:
Theorem surjective_pairing_stuck : (p : natprod),
p = (fst p, snd p).
Proof.
simpl. (* Doesn't reduce anything! *)
Abort.
We have to expose the structure of p so that simpl can perform the pattern match in fst and snd. We can do this with destruct.
Notice that, unlike for nats, destruct doesn't generate an extra subgoal here. That's because natprods can only be constructed in one way.
Theorem surjective_pairing : (p : natprod),
p = (fst p, snd p).
Proof.
intros p. destruct p as [n m]. simpl. reflexivity. Qed.
#### Exercise: 1 star (snd_fst_is_swap)
Theorem snd_fst_is_swap : (p : natprod),
(snd p, fst p) = swap_pair p.
Proof.
(* FILL IN HERE *) Admitted.
#### Exercise: 1 star, optional (fst_swap_is_snd)
Theorem fst_swap_is_snd : (p : natprod),
fst (swap_pair p) = snd p.
Proof.
(* FILL IN HERE *) Admitted.
# Lists of Numbers
Generalizing the definition of pairs a little, we can describe the type of lists of numbers like this: "A list is either the empty list or else a pair of a number and another list."
Inductive natlist : Type :=
| nil : natlist
| cons : nat natlist natlist.
For example, here is a three-element list:
Definition mylist := cons 1 (cons 2 (cons 3 nil)).
As with pairs, it is more convenient to write lists in familiar programming notation. The following two declarations allow us to use :: as an infix cons operator and square brackets as an "outfix" notation for constructing lists.
Notation "x :: l" := (cons x l) (at level 60, right associativity).
Notation "[ ]" := nil.
Notation "[ x ; .. ; y ]" := (cons x .. (cons y nil) ..).
It is not necessary to fully understand these declarations, but in case you are interested, here is roughly what's going on.
The right associativity annotation tells Coq how to parenthesize expressions involving several uses of :: so that, for example, the next three declarations mean exactly the same thing:
Definition mylist1 := 1 :: (2 :: (3 :: nil)).
Definition mylist2 := 1 :: 2 :: 3 :: nil.
Definition mylist3 := [1;2;3].
The at level 60 part tells Coq how to parenthesize expressions that involve both :: and some other infix operator. For example, since we defined + as infix notation for the plus function at level 50,
Notation "x + y" := (plus x y)
(at level 50, left associativity).
The + operator will bind tighter than ::, so 1 + 2 :: [3] will be parsed, as we'd expect, as (1 + 2) :: [3] rather than 1 + (2 :: [3]).
(By the way, it's worth noting in passing that expressions like "1 + 2 :: [3]" can be a little confusing when you read them in a .v file. The inner brackets, around 3, indicate a list, but the outer brackets, which are invisible in the HTML rendering, are there to instruct the "coqdoc" tool that the bracketed part should be displayed as Coq code rather than running text.)
The second and third Notation declarations above introduce the standard square-bracket notation for lists; the right-hand side of the third one illustrates Coq's syntax for declaring n-ary notations and translating them to nested sequences of binary constructors.
A number of functions are useful for manipulating lists. For example, the repeat function takes a number n and a count and returns a list of length count where every element is n.
Fixpoint repeat (n count : nat) : natlist :=
match count with
| Onil
| S count'n :: (repeat n count')
end.
The length function calculates the length of a list.
Fixpoint length (l:natlist) : nat :=
match l with
| nilO
| h :: tS (length t)
end.
The app ("append") function concatenates two lists.
Fixpoint app (l1 l2 : natlist) : natlist :=
match l1 with
| nill2
| h :: th :: (app t l2)
end.
Actually, app will be used a lot in some parts of what follows, so it is convenient to have an infix operator for it.
Notation "x ++ y" := (app x y)
(right associativity, at level 60).
Example test_app1: [1;2;3] ++ [4;5] = [1;2;3;4;5].
Proof. reflexivity. Qed.
Example test_app2: nil ++ [4;5] = [4;5].
Proof. reflexivity. Qed.
Example test_app3: [1;2;3] ++ nil = [1;2;3].
Proof. reflexivity. Qed.
Here are two smaller examples of programming with lists. The hd function returns the first element (the "head") of the list, while tl returns everything but the first element (the "tail"). Of course, the empty list has no first element, so we must pass a default value to be returned in that case.
Definition hd (default:nat) (l:natlist) : nat :=
match l with
| nildefault
| h :: th
end.
Definition tl (l:natlist) : natlist :=
match l with
| nilnil
| h :: tt
end.
Example test_hd1: hd 0 [1;2;3] = 1.
Proof. reflexivity. Qed.
Example test_hd2: hd 0 [] = 0.
Proof. reflexivity. Qed.
Example test_tl: tl [1;2;3] = [2;3].
Proof. reflexivity. Qed.
#### Exercise: 2 stars (list_funs)
Complete the definitions of nonzeros, oddmembers and countoddmembers below. Have a look at the tests to understand what these functions should do.
Fixpoint nonzeros (l:natlist) : natlist :=
(* FILL IN HERE *) admit.
Example test_nonzeros: nonzeros [0;1;0;2;3;0;0] = [1;2;3].
(* FILL IN HERE *) Admitted.
Fixpoint oddmembers (l:natlist) : natlist :=
(* FILL IN HERE *) admit.
Example test_oddmembers: oddmembers [0;1;0;2;3;0;0] = [1;3].
(* FILL IN HERE *) Admitted.
Fixpoint countoddmembers (l:natlist) : nat :=
(* FILL IN HERE *) admit.
Example test_countoddmembers1: countoddmembers [1;0;3;1;4;5] = 4.
(* FILL IN HERE *) Admitted.
Example test_countoddmembers2: countoddmembers [0;2;4] = 0.
(* FILL IN HERE *) Admitted.
Example test_countoddmembers3: countoddmembers nil = 0.
(* FILL IN HERE *) Admitted.
#### Exercise: 3 stars, advanced (alternate)
Complete the definition of alternate, which "zips up" two lists into one, alternating between elements taken from the first list and elements from the second. See the tests below for more specific examples.
Note: one natural and elegant way of writing alternate will fail to satisfy Coq's requirement that all Fixpoint definitions be "obviously terminating." If you find yourself in this rut, look for a slightly more verbose solution that considers elements of both lists at the same time. (One possible solution requires defining a new kind of pairs, but this is not the only way.)
Fixpoint alternate (l1 l2 : natlist) : natlist :=
(* FILL IN HERE *) admit.
Example test_alternate1: alternate [1;2;3] [4;5;6] = [1;4;2;5;3;6].
(* FILL IN HERE *) Admitted.
Example test_alternate2: alternate [1] [4;5;6] = [1;4;5;6].
(* FILL IN HERE *) Admitted.
Example test_alternate3: alternate [1;2;3] [4] = [1;4;2;3].
(* FILL IN HERE *) Admitted.
Example test_alternate4: alternate [] [20;30] = [20;30].
(* FILL IN HERE *) Admitted.
## Bags via Lists
A bag (or multiset) is like a set, but each element can appear multiple times instead of just once. One reasonable implementation of bags is to represent a bag of numbers as a list.
Definition bag := natlist.
#### Exercise: 3 stars (bag_functions)
Complete the following definitions for the functions count, sum, add, and member for bags.
Fixpoint count (v:nat) (s:bag) : nat :=
(* FILL IN HERE *) admit.
All these proofs can be done just by reflexivity.
Example test_count1: count 1 [1;2;3;1;4;1] = 3.
(* FILL IN HERE *) Admitted.
Example test_count2: count 6 [1;2;3;1;4;1] = 0.
(* FILL IN HERE *) Admitted.
Multiset sum is similar to set union: sum a b contains all the elements of a and of b. (Mathematicians usually define union on multisets a little bit differently, which is why we don't use that name for this operation.) For sum we're giving you a header that does not give explicit names to the arguments. Moreover, it uses the keyword Definition instead of Fixpoint, so even if you had names for the arguments, you wouldn't be able to process them recursively. The point of stating the question this way is to encourage you to think about whether sum can be implemented in another way — perhaps by using functions that have already been defined.
Definition sum : bag bag bag :=
(* FILL IN HERE *) admit.
Example test_sum1: count 1 (sum [1;2;3] [1;4;1]) = 3.
(* FILL IN HERE *) Admitted.
Definition add (v:nat) (s:bag) : bag :=
(* FILL IN HERE *) admit.
(* FILL IN HERE *) Admitted.
(* FILL IN HERE *) Admitted.
Definition member (v:nat) (s:bag) : bool :=
(* FILL IN HERE *) admit.
Example test_member1: member 1 [1;4;1] = true.
(* FILL IN HERE *) Admitted.
Example test_member2: member 2 [1;4;1] = false.
(* FILL IN HERE *) Admitted.
#### Exercise: 3 stars, optional (bag_more_functions)
Here are some more bag functions for you to practice with.
Fixpoint remove_one (v:nat) (s:bag) : bag :=
(* When remove_one is applied to a bag without the number to remove,
it should return the same bag unchanged. *)
(* FILL IN HERE *) admit.
Example test_remove_one1: count 5 (remove_one 5 [2;1;5;4;1]) = 0.
(* FILL IN HERE *) Admitted.
Example test_remove_one2: count 5 (remove_one 5 [2;1;4;1]) = 0.
(* FILL IN HERE *) Admitted.
Example test_remove_one3: count 4 (remove_one 5 [2;1;4;5;1;4]) = 2.
(* FILL IN HERE *) Admitted.
Example test_remove_one4: count 5 (remove_one 5 [2;1;5;4;5;1;4]) = 1.
(* FILL IN HERE *) Admitted.
Fixpoint remove_all (v:nat) (s:bag) : bag :=
(* FILL IN HERE *) admit.
Example test_remove_all1: count 5 (remove_all 5 [2;1;5;4;1]) = 0.
(* FILL IN HERE *) Admitted.
Example test_remove_all2: count 5 (remove_all 5 [2;1;4;1]) = 0.
(* FILL IN HERE *) Admitted.
Example test_remove_all3: count 4 (remove_all 5 [2;1;4;5;1;4]) = 2.
(* FILL IN HERE *) Admitted.
Example test_remove_all4: count 5 (remove_all 5 [2;1;5;4;5;1;4;5;1;4]) = 0.
(* FILL IN HERE *) Admitted.
Fixpoint subset (s1:bag) (s2:bag) : bool :=
(* FILL IN HERE *) admit.
Example test_subset1: subset [1;2] [2;1;4;1] = true.
(* FILL IN HERE *) Admitted.
Example test_subset2: subset [1;2;2] [2;1;4;1] = false.
(* FILL IN HERE *) Admitted.
#### Exercise: 3 stars (bag_theorem)
Write down an interesting theorem about bags involving the functions count and add, and prove it. Note that, since this problem is somewhat open-ended, it's possible that you may come up with a theorem which is true, but whose proof requires techniques you haven't learned yet. Feel free to ask for help if you get stuck!
(* FILL IN HERE *)
Just as with numbers, simple facts about list-processing functions can sometimes be proved entirely by simplification. For example, the simplification performed by reflexivity is enough for this theorem...
Theorem nil_app : l:natlist,
[] ++ l = l.
Proof. reflexivity. Qed.
... because the [] is substituted into the match position in the definition of app, allowing the match itself to be simplified.
Also, as with numbers, it is sometimes helpful to perform case analysis on the possible shapes (empty or non-empty) of an unknown list.
Theorem tl_length_pred : l:natlist,
pred (length l) = length (tl l).
Proof.
intros l. destruct l as [| n l'].
Case "l = nil".
reflexivity.
Case "l = cons n l'".
reflexivity. Qed.
Here, the nil case works because we've chosen to define tl nil = nil. Notice that the as annotation on the destruct tactic here introduces two names, n and l', corresponding to the fact that the cons constructor for lists takes two arguments (the head and tail of the list it is constructing).
Usually, though, interesting theorems about lists require induction for their proofs.
## Micro-Sermon
Simply reading example proofs will not get you very far! It is very important to work through the details of each one, using Coq and thinking about what each step of the proof achieves. Otherwise it is more or less guaranteed that the exercises will make no sense.
## Induction on Lists
Proofs by induction over datatypes like natlist are perhaps a little less familiar than standard natural number induction, but the basic idea is equally simple. Each Inductive declaration defines a set of data values that can be built up from the declared constructors: a boolean can be either true or false; a number can be either O or S applied to a number; a list can be either nil or cons applied to a number and a list.
Moreover, applications of the declared constructors to one another are the only possible shapes that elements of an inductively defined set can have, and this fact directly gives rise to a way of reasoning about inductively defined sets: a number is either O or else it is S applied to some smaller number; a list is either nil or else it is cons applied to some number and some smaller list; etc. So, if we have in mind some proposition P that mentions a list l and we want to argue that P holds for all lists, we can reason as follows:
• First, show that P is true of l when l is nil.
• Then show that P is true of l when l is cons n l' for some number n and some smaller list l', assuming that P is true for l'.
Since larger lists can only be built up from smaller ones, eventually reaching nil, these two things together establish the truth of P for all lists l. Here's a concrete example:
Theorem app_ass : l1 l2 l3 : natlist,
(l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).
Proof.
intros l1 l2 l3. induction l1 as [| n l1'].
Case "l1 = nil".
reflexivity.
Case "l1 = cons n l1'".
simpl. rewrite IHl1'. reflexivity. Qed.
Again, this Coq proof is not especially illuminating as a static written document — it is easy to see what's going on if you are reading the proof in an interactive Coq session and you can see the current goal and context at each point, but this state is not visible in the written-down parts of the Coq proof. So a natural-language proof — one written for human readers — will need to include more explicit signposts; in particular, it will help the reader stay oriented if we remind them exactly what the induction hypothesis is in the second case.
Theorem: For all lists l1, l2, and l3, (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).
Proof: By induction on l1.
• First, suppose l1 = []. We must show
([] ++ l2) ++ l3 = [] ++ (l2 ++ l3),
which follows directly from the definition of ++.
• Next, suppose l1 = n::l1', with
(l1' ++ l2) ++ l3 = l1' ++ (l2 ++ l3)
(the induction hypothesis). We must show
((n :: l1') ++ l2) ++ l3 = (n :: l1') ++ (l2 ++ l3).
By the definition of ++, this follows from
n :: ((l1' ++ l2) ++ l3) = n :: (l1' ++ (l2 ++ l3)),
which is immediate from the induction hypothesis.
Here is a similar example to be worked together in class:
Theorem app_length : l1 l2 : natlist,
length (l1 ++ l2) = (length l1) + (length l2).
Proof.
(* WORKED IN CLASS *)
intros l1 l2. induction l1 as [| n l1'].
Case "l1 = nil".
reflexivity.
Case "l1 = cons".
simpl. rewrite IHl1'. reflexivity. Qed.
For a slightly more involved example of an inductive proof over lists, suppose we define a "cons on the right" function snoc like this...
Fixpoint snoc (l:natlist) (v:nat) : natlist :=
match l with
| nil ⇒ [v]
| h :: th :: (snoc t v)
end.
... and use it to define a list-reversing function rev like this:
Fixpoint rev (l:natlist) : natlist :=
match l with
| nilnil
| h :: tsnoc (rev t) h
end.
Example test_rev1: rev [1;2;3] = [3;2;1].
Proof. reflexivity. Qed.
Example test_rev2: rev nil = nil.
Proof. reflexivity. Qed.
Now let's prove some more list theorems using our newly defined snoc and rev. For something a little more challenging than the inductive proofs we've seen so far, let's prove that reversing a list does not change its length. Our first attempt at this proof gets stuck in the successor case...
Theorem rev_length_firsttry : l : natlist,
length (rev l) = length l.
Proof.
intros l. induction l as [| n l'].
Case "l = []".
reflexivity.
Case "l = n :: l'".
(* This is the tricky case. Let's begin as usual
by simplifying. *)
simpl.
(* Now we seem to be stuck: the goal is an equality
involving snoc, but we don't have any equations
in either the immediate context or the global
environment that have anything to do with snoc
We can make a little progress by using the IH to
rewrite the goal... *)
rewrite IHl'.
(* ... but now we can't go any further. *)
Abort.
So let's take the equation about snoc that would have enabled us to make progress and prove it as a separate lemma.
Theorem length_snoc : n : nat, l : natlist,
length (snoc l n) = S (length l).
Proof.
intros n l. induction l as [| n' l'].
Case "l = nil".
reflexivity.
Case "l = cons n' l'".
simpl. rewrite IHl'. reflexivity. Qed.
Note that we make the lemma as general as possible: in particular, we quantify over all natlists, not just those that result from an application of rev. This should seem natural, because the truth of the goal clearly doesn't depend on the list having been reversed. Moreover, it is much easier to prove the more general property.
Now we can complete the original proof.
Theorem rev_length : l : natlist,
length (rev l) = length l.
Proof.
intros l. induction l as [| n l'].
Case "l = nil".
reflexivity.
Case "l = cons".
simpl. rewrite length_snoc.
rewrite IHl'. reflexivity. Qed.
For comparison, here are informal proofs of these two theorems:
Theorem: For all numbers n and lists l, length (snoc l n) = S (length l).
Proof: By induction on l.
• First, suppose l = []. We must show
length (snoc [] n) = S (length []),
which follows directly from the definitions of length and snoc.
• Next, suppose l = n'::l', with
length (snoc l' n) = S (length l').
We must show
length (snoc (n' :: l'n) = S (length (n' :: l')).
By the definitions of length and snoc, this follows from
S (length (snoc l' n)) = S (S (length l')),
which is immediate from the induction hypothesis.
Theorem: For all lists l, length (rev l) = length l.
Proof: By induction on l.
• First, suppose l = []. We must show
length (rev []) = length [],
which follows directly from the definitions of length and rev.
• Next, suppose l = n::l', with
length (rev l') = length l'.
We must show
length (rev (n :: l')) = length (n :: l').
By the definition of rev, this follows from
length (snoc (rev l'n) = S (length l')
which, by the previous lemma, is the same as
S (length (rev l')) = S (length l').
This is immediate from the induction hypothesis.
Obviously, the style of these proofs is rather longwinded and pedantic. After the first few, we might find it easier to follow proofs that give fewer details (since we can easily work them out in our own minds or on scratch paper if necessary) and just highlight the non-obvious steps. In this more compressed style, the above proof might look more like this:
Theorem: For all lists l, length (rev l) = length l.
Proof: First, observe that
length (snoc l n) = S (length l)
for any l. This follows by a straightforward induction on l. The main property now follows by another straightforward induction on l, using the observation together with the induction hypothesis in the case where l = n'::l'.
Which style is preferable in a given situation depends on the sophistication of the expected audience and on how similar the proof at hand is to ones that the audience will already be familiar with. The more pedantic style is a good default for present purposes.
We've seen that proofs can make use of other theorems we've already proved, using rewrite, and later we will see other ways of reusing previous theorems. But in order to refer to a theorem, we need to know its name, and remembering the names of all the theorems we might ever want to use can become quite difficult! It is often hard even to remember what theorems have been proven, much less what they are named.
Coq's SearchAbout command is quite helpful with this. Typing SearchAbout foo will cause Coq to display a list of all theorems involving foo. For example, try uncommenting the following to see a list of theorems that we have proved about rev:
Keep SearchAbout in mind as you do the following exercises and throughout the rest of the course; it can save you a lot of time!
Also, if you are using ProofGeneral, you can run SearchAbout with C-c C-a C-a. Pasting its response into your buffer can be accomplished with C-c C-;.
## List Exercises, Part 1
#### Exercise: 3 stars (list_exercises)
More practice with lists.
Theorem app_nil_end : l : natlist,
l ++ [] = l.
Proof.
(* FILL IN HERE *) Admitted.
Theorem rev_involutive : l : natlist,
rev (rev l) = l.
Proof.
(* FILL IN HERE *) Admitted.
There is a short solution to the next exercise. If you find yourself getting tangled up, step back and try to look for a simpler way.
Theorem app_ass4 : l1 l2 l3 l4 : natlist,
l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.
Proof.
(* FILL IN HERE *) Admitted.
Theorem snoc_append : (l:natlist) (n:nat),
snoc l n = l ++ [n].
Proof.
(* FILL IN HERE *) Admitted.
Theorem distr_rev : l1 l2 : natlist,
rev (l1 ++ l2) = (rev l2) ++ (rev l1).
Proof.
(* FILL IN HERE *) Admitted.
Lemma nonzeros_app : l1 l2 : natlist,
nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
(* FILL IN HERE *) Admitted.
## List Exercises, Part 2
#### Exercise: 2 stars (list_design)
Design exercise:
• Write down a non-trivial theorem involving cons (::), snoc, and app (++).
• Prove it.
(* FILL IN HERE *)
#### Exercise: 3 stars, advanced (bag_proofs)
Here are a couple of little theorems to prove about your definitions about bags in the previous problem.
Theorem count_member_nonzero : (s : bag),
ble_nat 1 (count 1 (1 :: s)) = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem ble_n_Sn : n,
ble_nat n (S n) = true.
Proof.
intros n. induction n as [| n'].
Case "0".
simpl. reflexivity.
Case "S n'".
simpl. rewrite IHn'. reflexivity. Qed.
Theorem remove_decreases_count: (s : bag),
ble_nat (count 0 (remove_one 0 s)) (count 0 s) = true.
Proof.
(* FILL IN HERE *) Admitted.
#### Exercise: 3 stars, optional (bag_count_sum)
Write down an interesting theorem about bags involving the functions count and sum, and prove it.
(* FILL IN HERE *)
#### Exercise: 4 stars, advanced (rev_injective)
Prove that the rev function is injective, that is,
(l1 l2 : natlist), rev l1 = rev l2 l1 = l2.
There is a hard way and an easy way to solve this exercise.
(* FILL IN HERE *)
# Options
Here is another type definition that is often useful in day-to-day programming:
Inductive natoption : Type :=
| Some : nat natoption
| None : natoption.
One use of natoption is as a way of returning "error codes" from functions. For example, suppose we want to write a function that returns the nth element of some list. If we give it type nat natlist nat, then we'll have to return some number when the list is too short!
Fixpoint index_bad (n:nat) (l:natlist) : nat :=
match l with
| nil ⇒ 42 (* arbitrary! *)
| a :: l'match beq_nat n O with
| truea
end
end.
On the other hand, if we give it type nat natlist natoption, then we can return None when the list is too short and Some a when the list has enough members and a appears at position n.
Fixpoint index (n:nat) (l:natlist) : natoption :=
match l with
| nilNone
| a :: l'match beq_nat n O with
| trueSome a
| falseindex (pred n) l'
end
end.
Example test_index1 : index 0 [4;5;6;7] = Some 4.
Proof. reflexivity. Qed.
Example test_index2 : index 3 [4;5;6;7] = Some 7.
Proof. reflexivity. Qed.
Example test_index3 : index 10 [4;5;6;7] = None.
Proof. reflexivity. Qed.
This example is also an opportunity to introduce one more small feature of Coq's programming language: conditional expressions...
Fixpoint index' (n:nat) (l:natlist) : natoption :=
match l with
| nilNone
| a :: l'if beq_nat n O then Some a else index' (pred n) l'
end.
Coq's conditionals are exactly like those found in any other language, with one small generalization. Since the boolean type is not built in, Coq actually allows conditional expressions over any inductively defined type with exactly two constructors. The guard is considered true if it evaluates to the first constructor in the Inductive definition and false if it evaluates to the second.
The function below pulls the nat out of a natoption, returning a supplied default in the None case.
Definition option_elim (d : nat) (o : natoption) : nat :=
match o with
| Some n'n'
| Noned
end.
#### Exercise: 2 stars (hd_opt)
Using the same idea, fix the hd function from earlier so we don't have to pass a default element for the nil case.
Definition hd_opt (l : natlist) : natoption :=
(* FILL IN HERE *) admit.
Example test_hd_opt1 : hd_opt [] = None.
(* FILL IN HERE *) Admitted.
Example test_hd_opt2 : hd_opt [1] = Some 1.
(* FILL IN HERE *) Admitted.
Example test_hd_opt3 : hd_opt [5;6] = Some 5.
(* FILL IN HERE *) Admitted.
#### Exercise: 1 star, optional (option_elim_hd)
This exercise relates your new hd_opt to the old hd.
Theorem option_elim_hd : (l:natlist) (default:nat),
hd default l = option_elim default (hd_opt l).
Proof.
(* FILL IN HERE *) Admitted.
#### Exercise: 2 stars (beq_natlist)
Fill in the definition of beq_natlist, which compares lists of numbers for equality. Prove that beq_natlist l l yields true for every list l.
Fixpoint beq_natlist (l1 l2 : natlist) : bool :=
(* FILL IN HERE *) admit.
Example test_beq_natlist1 : (beq_natlist nil nil = true).
(* FILL IN HERE *) Admitted.
Example test_beq_natlist2 : beq_natlist [1;2;3] [1;2;3] = true.
(* FILL IN HERE *) Admitted.
Example test_beq_natlist3 : beq_natlist [1;2;3] [1;2;4] = false.
(* FILL IN HERE *) Admitted.
Theorem beq_natlist_refl : l:natlist,
true = beq_natlist l l.
Proof.
(* FILL IN HERE *) Admitted.
# Dictionaries
As a final illustration of how fundamental data structures can be defined in Coq, here is the declaration of a simple dictionary data type, using numbers for both the keys and the values stored under these keys. (That is, a dictionary represents a finite map from numbers to numbers.)
Module Dictionary.
Inductive dictionary : Type :=
| empty : dictionary
| record : nat nat dictionary dictionary.
This declaration can be read: "There are two ways to construct a dictionary: either using the constructor empty to represent an empty dictionary, or by applying the constructor record to a key, a value, and an existing dictionary to construct a dictionary with an additional key to value mapping."
Definition insert (key value : nat) (d : dictionary) : dictionary :=
(record key value d).
Here is a function find that searches a dictionary for a given key. It evaluates evaluates to None if the key was not found and Some val if the key was mapped to val in the dictionary. If the same key is mapped to multiple values, find will return the first one it finds.
Fixpoint find (key : nat) (d : dictionary) : natoption :=
match d with
| emptyNone
| record k v d'if (beq_nat key k)
then (Some v)
else (find key d')
end.
#### Exercise: 1 star (dictionary_invariant1)
Complete the following proof.
Theorem dictionary_invariant1' : (d : dictionary) (k v: nat),
(find k (insert k v d)) = Some v.
Proof.
(* FILL IN HERE *) Admitted.
#### Exercise: 1 star (dictionary_invariant2)
Complete the following proof.
Theorem dictionary_invariant2' : (d : dictionary) (m n o: nat),
beq_nat m n = false find m d = find m (insert n o d).
Proof.
(* FILL IN HERE *) Admitted.
End Dictionary.
End NatList.
(* \$Date: 2013-07-30 12:24:33 -0400 (Tue, 30 Jul 2013) \$ *)
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# consecutive terms of a.p.given 3 consecutive unknown terms of a.p., i'm stuck calculating the sum of the 3 terms!
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
Let the three terms of the AP be a , b and c.
Consecutive terms of an AP have a common difference
b = a + d and c = b + d = a + 2d
If you can determine a and d using the relations for the others in terms of a and d, you can determine the sum of the three terms.
For example, consider the problem: Let the sum of the first two terms of an AP be 9 and the sum of the second and third is 15. What is the sum of the 3 consecutive terms.
Here a + b = 9 and b + c = 15
=> a + a + d = 9 and a + d + a + 2d = 15
=> 2a + d = 9 and 2a + 3d = 15
=> 2d = 6
=> d = 3
a = 3
The sum of the three terms is 3a + 3d = 18.
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https://questions.examside.com/past-years/gate/gate-ee/control-systems/polar-nyquist-and-bode-plot
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# Polar Nyquist and Bode Plot ยท Control Systems ยท GATE EE
Start Practice
## Marks 1
GATE EE 2022
The Bode magnitude plot of a first order stable system is constant with frequency. The asymptotic value of the high frequency phase, for the system, i...
GATE EE 2022
An LTI system is shown in the figure where $$G(s) = {{100} \over {{s^2} + 0.1s + 100}}$$. The steady state output of the system, to the input r(t), is...
GATE EE 2022
The open loop transfer function of a unity gain negative feedback system is given as $$G(s) = {1 \over {s(s + 1)}}$$ The Nyquist contour in the s-plan...
GATE EE 2017 Set 1
The transfer function of a system is given by $${{{V_0}\left( s \right)} \over {{V_i}\left( s \right)}} = {{1 - s} \over {1 + s}}$$ Let the output of...
GATE EE 2017 Set 1
Consider the unity feedback control system shown. The value of $$K$$ that results in a phase margin of the system to be $${30^0}$$ is ____________. (...
GATE EE 2016 Set 1
The transfer function of a system is $${{Y\left( s \right)} \over {R\left( s \right)}} = {s \over {s + 2}}.$$ The steady state $$y(t)$$ is $$Acos$$$$\... GATE EE 2016 Set 1 The phase cross-over frequency of the transfer function$$G\left( s \right) = {{100} \over {{{\left( {s + 1} \right)}^3}}}\,\,$$in$$rad/s$$is GATE EE 2015 Set 2 Nyquist plots of two functions$${G_1}\left( s \right)$$and$${G_2}\left( s \right)$$are shown in figure. Nyquist plot of the product of$${G_1}\...
GATE EE 2015 Set 1
A Bode magnitude plot for the transfer function $$๐บ(๐ )$$ of a plant is shown in the figure. Which one of the following transfer functions best descr...
GATE EE 2013
The Bode plot of a transfer function $$G(s)$$ is shown in the figure below. The gain is $$\left( {20\log \left| {G\left( s \right)} \right|} \right... GATE EE 2012 A system with transfer function$$\,G\left( s \right) = {{\left( {{s^2} + 9} \right)\left( {s + 2} \right)} \over {\left( {s + 1} \right)\left( {s + 3...
GATE EE 2011
The frequency response of a linear system $$G\left( {j\omega } \right)$$ is provided in the tubular form below. The gain margin and phase margin of t...
GATE EE 2011
An open loop system represented by the transfer function $$G\left( s \right) = {{\left( {s - 1} \right)} \over {\left( {s + 2} \right)\left( {s + 3} \... GATE EE 2009 The polar plot of an open loop stable system is shown below. The closed loop systems is ... GATE EE 2005 A system with zero initial conditions has the closed loop transfer function$$T\left( s \right) = {{{s^2} + 4} \over {\left( {s + 1} \right)\left( {s ...
GATE EE 2005
The gain margin of a unity feedback control system with the open loop transfer function $$G\left( s \right) = {{\left( {s + 1} \right)} \over {{s^2}}}... GATE EE 2004 The Nyquist plot of loop transfer function$$G(s) H(s)$$of a closed loop control system passes through the point$$(-1,j0)$$in the$$G(s) H(s)$$pla... GATE EE 2001 The polar plot of a type-$$1, 3$$-pole, open-loop system is shown in Fig. below. The closed loop system is ... GATE EE 2001 The asymptotic approximation of the log-magnitude versus frequency plot of a minimum phase system with real poles and one zero is shown in Fig. Its tr... GATE EE 1996 The closed - loop transfer function of a control system is given by$${{C\left( s \right)} \over {R\left( s \right)}} = {1 \over {\left( {1 + s} \righ...
GATE EE 1992
A unity feedback system has the open loop transfer function $$G\left( s \right) = {1 \over {\left( {s - 1} \right)\left( {s + 2} \right)\left( {s + 3}... GATE EE 1991 The system having the Bode magnitude plot shown in Fig. below has the transfer function ... GATE EE 1991 Which of the following is the transfer function of a system having the Nyquist plot shown in Fig. below. ... ## Marks 2 GATE EE 2016 Set 1 Consider the following asymptotic Bode magnitude plot ($${\omega \,\,}$$is in$$rad/s$$) Which one of the following transfer functions is best rep... GATE EE 2016 Set 1 Loop transfer function of a feedback system is$$G\left( s \right)H\left( s \right) = {{s + 3} \over {{s^2}\left( {s - 3} \right)}}.$$Take the Nyquis... GATE EE 2014 Set 3 The magnitude Bode plot of a network is shown in the figure The maximum phase angle$${\phi _m}$$and the corresponding gain$${G_m}$$respectively ... GATE EE 2014 Set 2 For the transfer function$$G\left( s \right) = {{5\left( {s + 4} \right)} \over {s\left( {s + 0.25} \right)\left( {{s^2} + 4s + 25} \right)}}.$$The ... GATE EE 2014 Set 1 The Bode magnitude plot of the transfer function$$G\left( s \right) = {{K\left( {1 + 0.5s} \right)\left( {1 + as} \right)} \over {s\left( {1 + {s \ov...
GATE EE 2010
The frequency response of $$G\left( s \right) = 1/\left[ {s\left( {s + 1} \right)\left( {s + 2} \right)} \right]$$ plotted in the complex $$\,G\left( ... GATE EE 2009 The open loop transfer function of a unity feedback system is given by$$G\left( s \right) = \left( {{e^{ - 0.1s}}} \right)/s.$$The gain margin of t... GATE EE 2009 The asymptotic approximation of the log magnitude vs frequency plot of a system containing only real poles and zeros is shown. Its transfer function i... GATE EE 2008 The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure: This transfer function has... GATE EE 2007 If$$X = {\mathop{\rm Re}\nolimits} G\left( {j\omega } \right),\,\,$$and$$y = {\rm I}mG\left( {j\omega } \right)$$then for$$\omega \to {0^ + },\,...
GATE EE 2006
The Bode magnitude plot of $$H\left( {j\omega } \right) = {{{{10}^4}\left( {1 + j\,\omega } \right)} \over {\left( {10 + j\,\omega } \right){{\left( {... GATE EE 2006 Consider the following Nyquist plots of loop transfer functions over$$\omega = 0$$to$$\omega = \infty .$$Which of these plots represents a stabl... GATE EE 2005 In the$$GH(s)$$plane, the Nyquist plot of the loop transfer function$$G\left( s \right)\,H\left( s \right) = {{\pi {e^{ - 0.25s}}} \over s}$$passe... GATE EE 2005 If the compensated system shown in the figure has a phase margin of$${60^ \circ }$$at the crossover frequency of$$1 rad/sec,$$the value of the gai... GATE EE 2004 The open loop transfer function of a unity feedback control system is given as$$G\left( s \right) = {{as + 1} \over {{s^2}}}.$$. The value of$$โaโ$$... GATE EE 2004 In the system shown in figure, the input$$x(t)=sint.$$In the steady-state, the response$$y(t)$$will be ... GATE EE 2003 The asymptotic Bode plot of the transfer function$${K \over {1 + {s \over a}}}$$. The error in phase angle and$$dB$$gain at a frequency of$$\omega...
GATE EE 1999
The function corresponding to the Bode plot of Figure, is ...
GATE EE 1997
A unity feedback system with the open loop transfer function $$G\left( s \right) = {1 \over {s\left( {s + 2} \right)\left( {s + 4} \right)}}$$ has gai...
GATE EE 1991
An under damped second order system having a transfer function of the form $$M\left( s \right) = {{k\omega _n^2} \over {{s^2} + 2\zeta {\omega _n}s + ... ## Marks 5 GATE EE 2001 A unity feedback system has an open-loop transfer function of$$G\left( s \right) = {{10000} \over {s{{\left( {s + 10} \right)}^2}}}$$(a) Determine t... GATE EE 2000 Open-loop transfer function of a unity - feedback system is$$\$G\left( s \right) = {G_1}\left( s \right).{e^{ - s{\tau _D}}} = {{{e^{ - s{\tau _D}}}}...
GATE EE 1998
The asymptotic magnitude Body plot of a system is given in Figure. Find the transfer function of the system analytically. It is known that the system ...
EXAM MAP
Medical
NEET
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https://worldbuilding.stackexchange.com/questions/88795/how-many-moons-can-form-around-this-alternate-earth
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# How many moons can form around this alternate Earth?
Say in an alternate timeline, Earth developed with more moons. Many, many many more moons. In fact, it developed with the absolute maximum amount of moons a planet of its size and mass can physically harbor without anything becoming unstable.
How many moons would there be? How big would they be? Would life be different if it developed on this alternate planet? How would our culture be different?
Please note that nothing changes on Earth. The landmasses and oceans are the exact same, as well as the makeup of its crust, mantle and core.
• In future, wait before accepting an answer. Stack Exchange has global reach, yet people in other time zones didn't even get a chance to write an answer. Aug 13, 2017 at 16:49
## The number of moons
Sean Raymond wrote a great answer to a similar question, and I'd like to base my answer on his. He used the following technique to estimate the total number of moons orbiting a planet, assuming all the moons are in mean motion orbital resonances with one another.
1. Calculate the outer edge of the region of stable orbits, which is one half of the Hill radius. For Earth, this distance is $$\sim0.005\text{ AU}$$. For comparison, the Moon's semi-major axis is $$\sim0.0025\text{ AU}$$.
2. Estimate the inner edge of the region. I'll guess that it's 1/20th of the Moon's semi-major axis, or $$\sim0.000125\text{ AU}$$. The very inner limit is the Roche limit. However, that depends on the mass of the orbiting body. Without knowing what the moons would be like, I'm not inclined to speculate on what the Roche limit would be (we're probably safe, though, with this as our inner bound).
3. Divide the outer radius by the inner one. That gets you $$40$$.
4. Kepler's third law relates semi-major axis $$a$$ to period $$P$$ by $$a^3\propto P^2$$ Raise your previous result to the $$3/2$$ power. This gets us a factor of $$252$$, which is close to $$256=2^8$$. This implies that we could fit $$8+1=9$$ moons comfortably in orbit, if we bend the limits a bit.
Some extra thoughts on stability:
• The masses of the moons would need to be adjusted so as to not perturb one another too much. I'd guess they would all have very little mass, but I don't know for sure. It's actually an interesting question, come to think of it: how small can you make a moon that will clear its orbit in a debris disk? I'll update this if I find a good answer.
• The formation of the system would be difficult to explain. Stable resonances can of course arise - and we do see this in the Jovian and Saturnian moon systems, specifically, some of the Galilean moons - but this many moons would be a bit much. You'd also need a rather large debris disk around Earth. I don't think the Giant Impact Hypothesis could work here!
• The system would likely be stable if it was isolated, but if an outside body came by, things would go wrong rather rapidly.
## The effects
Whenever you create a system of moons, tides tend to come to mind. The site has covered the tidal patterns from multiple moons in some depth (see How would having multiple moons affect tides?), although never with 9 moons, I believe. The method to model tides would be to simple treat the cycles as the sum of 9 sine waves, one for each moon, and look at the patterns that arise.
Culture is always an interesting one to think about. Many civilizations had deities for the Sun and the Moon; maybe you'd have something similar here. 9 moon deities give you fodder for plenty of interesting stories. For me, the myths centered on the Pleiades comes to mind. Calendars might also be different; more complicated timekeeping systems would certainly be possible. You might want to explore that angle a little.
• This was extremely well explained and interesting. Thanks so much for your help Aug 13, 2017 at 18:50
You can have as many "moons" as you like, provided they're small enough (think Saturn's rings, which are actually composed by zillions little "moons").
OTOH many large moons (compared to planet size) are unlikely. Most likely our Moon wouldn't be able to cohabit with another moon about its size.
Having many moons would "even-out" tides which may or may not have an influence to life and culture.
Ditto for night illumination which has some influence on agriculture and plants (at least).
• Well, not on plant life. The light coming from the Moon is a tiny tiny fraction of the light coming from the Sun. And regarding the tides, the classical civilization developed around the Mediterranean which has very small tides. Aug 13, 2017 at 16:00
• @ZioByte I appreciate your response! I have another question, though. If it had many little moons orbiting it, there would have to be some distance away where earth's gravity could no longer effect any celestial bodies. Do you know how far this is, and how far away each moon would have to be from each other? Aug 13, 2017 at 16:08
• @KingMagmaBlock: I'm not sure I follow You. Gravity pull force decreases with square of distance, but never goes to zero, so it will "affect celestial bodies" in the whole universe. Of course the magnitude of this interaction will rapidly become negligible when larger/nearer bodies are present. Please rephrase. Aug 13, 2017 at 16:38
• @KingMagmaBlock: it is a question od relative distance. Two bodies alone in space will orbit around their common center of mass at whatever distance. Presence of a third (or more) body if $\frac{mass}{distance^2}$ for the bodies becomes comparable with that of "planet". Aug 13, 2017 at 17:57
• @KingMagmaBlock the maximum distance the satellite can orbit depends on the size of the planet, the star and their distance to the star. A moon cannot orbit a planet, if it strays into the areas where the star becomes dominant. Look up Hill spheres. Aug 13, 2017 at 21:46
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## ››Convert petagram to sack [UK, wool]
petagram sack
How many petagram in 1 sack? The answer is 1.6510762268E-10.
We assume you are converting between petagram and sack [UK, wool].
You can view more details on each measurement unit:
petagram or sack
The SI base unit for mass is the kilogram.
1 kilogram is equal to 1.0E-12 petagram, or 0.0060566555545296 sack.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between petagrams and sack [UK, wool].
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of petagram to sack
1 petagram to sack = 6056655554.5296 sack
2 petagram to sack = 12113311109.059 sack
3 petagram to sack = 18169966663.589 sack
4 petagram to sack = 24226622218.118 sack
5 petagram to sack = 30283277772.648 sack
6 petagram to sack = 36339933327.178 sack
7 petagram to sack = 42396588881.707 sack
8 petagram to sack = 48453244436.237 sack
9 petagram to sack = 54509899990.766 sack
10 petagram to sack = 60566555545.296 sack
## ››Want other units?
You can do the reverse unit conversion from sack to petagram, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Petagram
The SI prefix "peta" represents a factor of 1015, or in exponential notation, 1E15.
So 1 petagram = 1015 grams-force.
## ››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, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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## Introduction: DC Power Supply to Charge Your Phone & Small Electronic Devices
A variable DC power supply is really useful to keep on your desk for charging your phone, plugging in a raspberry pi or any other low voltage electronic devices. By using voltage regulators you can set just the amounts to output, whether you need 3.3, 5, 12 or any other amount. This is a really fun project that you can customize to your needs, varying the amount of plugs, set the voltage, choose the size etc... I used MDF for the build, however you could make a box out of anything really to house the electronics.
## Step 1: The Design
So I started with making a sketch of the design. For this build I need to use two voltage regulators, I picked mine up online. I also need a switch, and then I'm going to add a series of outlets that you can plug male pigtails into. Here I'm just confirming the current draw of an LED I want to wire so that I can determine the size of the resistor I need.
Products Used in this Build (Affiliate Links)
So this is kind of the design that I'm thinking. The rectangles in the middle is to show the screens on the voltage regulators, then we have outlets underneath, a switch and a light to signify the unit is on.
## Step 2: The Materials
So I'm starting with cutting up the materials for the box itself, and I'm using 1/4 inch MDF.
The final dimensions of the box will measure 8 x 8 x 2 1/2 inches so just cutting up the pieces for that.
Let's dry fit it to make sure everything fits!
## Step 3: The Front of the Box
Now let's start working on the front of the box. So first I'm getting the dimension of the plug, setting up a compass. So drawing a circle, then seeing how far I want the outlets apart, and 3/4 of an inch in between each center seems like a good distance. All in all I want eight outlets, so drawing those all out.
Sketching out the space for the switch and the position of the voltage regulators. Also measuring out where the screens are, and where the holes need to go for those on the board, as well as where the little button that changes the output on the screen is located. And there it is, all marked out.
## Step 4: Drilling
So I'm starting with drilling out the holes for the outlets using the drill press and a 7/16 inch bit. I'm also drilling holes in the center for where the screens will display to remove some material. And then cleaning up those rectangles with some chisel work.
Also drilling holes for areas that I need to access on the voltage regulators, a hole where you can insert a screwdriver to change the output, and a spot with a button to push to change the display on the screen.
Then doing a little sanding, a little cleaning up with a chisel, and I'm ready to assemble the box together. So measuring out 1/8 of an inch around the top wher I'm going to drill holes to screw it in. Doing some countersinking here.
## Step 5: Putting the Box Together
Now to put the box together, I'm starting with yellow glue around the edges, then I'm adding small amounts of hot glue on one piece at a time, because it stiffens up so quickly, and this basically acts as a clamp while the yellow glue dries. It works really well. And then simply screwing the top in place.
## Step 6: Finishing the Box
I decided to paint this box white for some contrast, and I'm just using cheap basic white paint here. To paint the inside of those little holes I'm using a small brush. Then once the paint dried I put on two coats of waterbased polyurethane which is nice because it doesn't add a yellow tone and it dries quickly, and of course it adds some additional protection.
## Step 7: The Electronics
So real quick, let's go over the electronics here. So I have 20 volt ac dc power supply, first there is a switch, and we have an LED light which I need a resistor for, then this is hooked up to a voltage regulator and that connects to several output plugs.
So soldering a 2600 ohm resistor on the LED light here, and fitting everything in the box. I drilled a hole on the backside too, to connect a plug in to power the whole unit. Then soldering everything together. A lot of people don't like soldering, but it's really easy and fun, plus as you make more projects you get more practice. Just putting the wires into the voltage regulators, putting in the plugs and doing a little more soldering. Hooking this is and making sure it works.
## Step 8: Hooking Up the Electronics
So on the voltage regulators there is this potentiometer on top that you can control and change the output. I'm also going to use this small mechanical screw here and enter it through one of the holes I drilled. Then on the regulator there is this little button to change whether to show the voltage output or input on the screen and now when I put the unit up here the screw is right on the button so I can push it to change the screen.
Next I'm hot gluing some wooden dowels in the corners here, and this so I can glue those to the box to secure the board. Then I'm hot gluing the plugs in, the switch, as well as the dowels on the boards.
## Step 9: Wires
Now I put together a couple of cords with pigtails on both sides to plug into the unit to power different things, however to protect the wires, I figured why not secure some mason line, which happens to be bright pink with some hot glue. And I'm just adding a little glue, spinning the cord around and so on.
## Step 10: Testing It Out
So the device works really well. To alter the voltage output all I do is to insert a small screwdriver into the top hole and turn it to get just the amount I need. And if I want to see what the input is, then I just click the button with the mechanical screw I put in. Now it's ready to use - set it to whatever voltage you need, plug in your phone, your raspberry pi, charge up your bluetooth speaker and so on....
## Step 11: Conclusion - Watch the Video!
For a much better point of view, make sure to check out the video including all the steps to build this cool power strip.
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# Computer exercise 4 Poisson Regression
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2 ii Figure 1: The number of people killed per year in road accidents in Sweden from year 1950 to year (Source: The Swedish Road Administration.) The variable data now consists of 9 columns but we are only interested in columns {1, 2, 5, 6}, i.e. {year, number of people killed, number of cars, amount of sold petrol}. We store the data in a structure array >> traffic = struct( year,data(:,1), killed,data(:,2), cars,data(:,5),... petrol,data(:,6)); Plot the number of people killed each year >> plot(traffic.year, traffic.killed, o ) Try also plotting the number of people killed vs. number of cars and the petrol consumption. Do you see any connections? From the plot it can be seen that the trend of increasing number of people killed is broken around year And from year 1970 the number starts to decrease. Some natural questions arises. Why did the number of people killed increase in years ? What was the reason for the brake of the increasing trend? (Hint: right-side driving (1967), front seat-belts in new cars (1969), mandatory use of front seat-belts (1975)).
3 iii 3 The Poisson regression model Lets say we have a sequence of count data, n i, i = 1,..., k, for some event, i.e. the number of perished in traffic accidents in a year. This count data is assumed to be observations from random variables N i Po(µ i ), (called responses or dependent variables) with mean value µ i = µ i (x i1,..., x ip ). The variables, x i1,..., x ip, are called explanatory variables 2 and are assumed to measure factors that influence the count data. We restrict µ i to be a log-linear function 3, And thus the probability that N i = n is, µ i = exp(β 0 + β 1 x i β p x ip ) (1) P(N i = n) = e µ i (µ i ) n n! = e eβ0+β1xi1+...+βpxip (e β 0+β 1 x i β px ip ) n, n = 0, 1, 2,.... (2) n! 3.1 Estimating model parameters β 0,..., β p To simplify the notation we introduce x i0 = 1 and can now write (1) as, p E[N i ] = µ i = exp β j x ij, (3) where N i Po(µ i ) for i = 1,..., k. The likelihood function is calculated as, L(β) = j=0 k P(N i = n i ) = k µ n i i n i! e µ i. (4) where µ i = µ i (β p ) is a function of β p = (β 0,..., β p ). The ML-estimates β p = (β 0,..., β p) are the values of β that maximize the likelihood function L(β). Often it is easier to maximize the log-likelihood function, l(β) = log(n i!) + n i log(µ i ) µ i. (5) By setting the first order derivates of the log-likelihood equal to zero, we get a system of (p+1) non-linear equations in β j, l(β) ( ) µ i ni = 1 = (n i µ i )x ij = 0, j = 0,..., p. (6) β j β j µ i Usually, the equation system must be solved with some numerical method, e.g. the Newton-Raphson algorithm, cf. Section in the textbook. This is also the method implemented in the function poiss_regress, which was written for the purpose of this lab and can be found in labfiles. Use the command >> type poiss_regress to see the code. Poisson regression model belongs to a class of models called generalized linear models. In a generalized linear model (GLM), the mean of the response, µ, is modeled as a monotonic (nonlinear) transformation 2 Several other names exist in the literature: independent variables, regressor variables, predictor variables. 3 Sometimes the model incorporates an extra term t i: µ i = t i exp(β 0 + β 1x i β 2x ip).
4 iv of a linear function of the explanatory variables, g(β 0 + β 1 x 1 + β 2 x 2...). The inverse of the transformation function g is called the canonical link function. In Poisson regression this function is the log function, but in other GLM s different link functions are used, see doc glmfit for a list of supported link functions in the Matlab function glmfit 4. Also, the response may take different distributions, such as the normal or the binomial distribution. Below, we will use related function glmval with the logarithmic link function to make predictions from the fitted model, see the code below. 4 Poisson regression of traffic data We will now try to fit the Poisson regression model to the traffic data of the number of people killed in road accidents. Above, we could see that there was a break in the trend of increasing number people killed around year , mainly because of the improvement in car safety due to the use of safety belts. Because of this it seems reasonable to fit our model to data starting from year >> traffic = struct( year,data(26:end,1), killed,data(26:end,2),... cars,data(26:end,5), petrol,data(26:end,6)); Question 1: Which are the explanatory variables? And which is the response? Redraw the plot from above for the reduced data set >> plot(traffic.year,traffic.killed, o ) >> figure(1), hold on We start the analysis with one explanatory variable, traffic.year. routine for the generalized linear models glmval >> X1 = [traffic.year-mean(traffic.year)]; >> n = traffic.killed; >> beta1 = poiss_regress(x1,n,1e-6); >> my_fit = glmval(beta1, X1, log ); >> plot(traffic.year, my_fit, b- ) Note usage of the prediction Question 2: What is your estimate of β? Convince yourself that this is the solution to (6). You can utilize the following code for this purpose: >> X0=ones(size(X1)); >> X=[X0, X1]; >> mu=exp(x*beta1); >> X *(n-mu) Does it appear to be the solution? Judging from the plot, is this model sufficient to describe the number of people killed in traffic accidents? Although this simple model seems to capture the overall trend, adding further explanatory variables may improve the fit. Thus, we try adding the number of cars as a variable in our model. 4 glmfit uses a method called weighted least squares to compute the β estimates.
5 v >> X2 = [traffic.year-mean(traffic.year), traffic.cars-mean(traffic.cars)]; >> beta2 = poiss_regress(x2,n,1e-6); >> my_fit = glmval(beta2, X2, log ); >> plot(traffic.year, my_fit, g- ) Question 3: Have your estimates β0 and β 1 improve the fit? changed? Does accounting for the number of cars It seems reasonable also to add the quantity of sold petrol as this would reflect the total mileage of all cars 5. >> X3 = [traffic.year-mean(traffic.year), traffic.cars-mean(traffic.cars),... traffic.petrol-mean(traffic.petrol)]; >> beta3 = poiss_regress(x3,n,1e-6); >> my_fit = glmval(beta3, X3, log ); >> plot(traffic.year, my_fit, r- ) Question 4: Have your estimates of β changed now? Use the command format long to display more digits. Which model do you choose? 4.1 Model selection - Deviance It is not always easy to decide, just by looking at the plot, which model to choose. Even though adding more variables improves the fit, it also increases the uncertainty of the estimates. One method to choose complexity of the model is to use the deviance and a hypothesis test. Let β p = {β 0, β 1,..., β p} be the ML-estimates of the model parameters {β 0, β 1,..., β p } of the full model with p explanatory variables and β q the estimates of a simpler model where only q (q < p) of the explanatory variables have been used. Then for large k, and under suitable regularity conditions, the deviance DEV = 2 (l(β p) l((β q))) (7) is approximately χ 2 (p q) distributed if the less complex model is true. Thus, it is possible to test if the simpler model can be rejected compared to the full model. Question 5: Use chi2inv to get the quantiles of the χ 2 distribution. Consider 5% significance level for your test. The deviance for model 3 compared to model 2 is calculated as >> DEV2 = 2*traffic.killed *([X0,X3]*beta3-[X0,X2]*beta2) 5 Assuming that the mean fuel consumption of a car has been constant over the years - a 1970 year model of a Volvo used about 10l per 100km which is approximately the same as for the 2000 year model. Of course, the year 2000 model has more than twice the horsepower.
6 vi Question 5: Is the improvement with model 3 significant compared to model 2? Repeat the test for model 2 against model 1 and also model 3 against model 1? Which model do you choose? Do you think that there was a sufficient number of explanatory variables used to explain the traffic deaths? Why? 5 Prediction Now we want to use our model to predict the expected number of perished in traffic accidents six years from now, i.e. year In order to do this we first must have an estimate of the number of cars that year. Start by plotting the number of cars vs. year, >> figure(2) >> plot(traffic.year, traffic.cars, o ) >> hold on We will here use a simple linear model for the number of cars, y i, year x i y i = β 0 + β 1 x i + ε i (8) where the errors, ε i N(0, (σ ε ) 2 ), are assumed to be independent and identically distributed. This is called a linear regression model. It is possible to estimate the parameters with the maximum likelihood method similar as for the Poisson regression model above. Question 6: What is the likelihood function? Write it down. In Matlab, the function regress computes the least-squares (LS) estimates of the linear regression model. In the case of ε i being normally distributed, the LS method is equivalent to the ML method with exactly the same estimates. >> phat = regress(traffic.cars,[ones(length(traffic.cars),1) [1975:2005] ]) >> plot(1975:2016, phat(1)+phat(2)*[1975:2016], r ) >> cars_2016=phat(1)+phat(2)*2016; Evaluate the fit by looking at the residuals. >> res = traffic.cars-(phat(1)+phat(2)*traffic.year); >> figure(3), plot(traffic.year,res, o ) >> figure(4), normplot(res) Question 7: Do the residuals conform to the requirements of the model errors ε i? Using the following code provide with prediction of petrol consumption for 2016.
7 vii >> phat = regress(traffic.petrol,[x0 [1975:2012] ([1975:2012].^2) ]) >> plot(1975:2016, phat(1)+phat(2)*[1975:2016]+phat(3)*([1975:2016].^2), r ) >> petrol_2016=phat(1)+phat(2)*2016+phat(3)*2016^2; Notice that this time quadratic model had to be fit to the data. Question 8: Are you satisfied with the obtained fits for the petrol and the number of cars? However, for our purpose these rough estimates are sufficient. The expected number of perished can now be predicted using (1), >>x=[ mean(traffic.year) cars_2016-mean(traffic.cars) petrol_2016-mean(traffic.petrol)] >>my_2016=exp(beta3 *x) %----- Model Question 9: Is the prediction reasonable? Comment. Modify the code and predict the number of perished year Is the predicted number close to the "true" number which was 264 perished?
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Latest Banking jobs » Quantitative Aptitude Quiz For IBPS RRB...
# Quantitative Aptitude Quiz For IBPS RRB PO, Clerk Prelims 2021- 19th May
Directions (1-5): In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer.
(a) If x=y or no relation can be established
(b) If x>y
(c) If x<y
(d) If x≥y
(e) If x≤y
Q1. I. 25x² – 90x + 72 = 0
II. 5y² – 27y + 36 = 0
Q2. I. 12x² + 46x + 42 = 0
II. 3y² – 16y + 21 = 0
Q3. I. 4x² + 10x = 14
II.15 = 16y – 4y²
Q4. I. 6x² + 15x – 36 = 0
II. 4y² – 2y – 10 = -8
Q5. I. 2x² – 19x + 44 = 0
II. 3y² – 22y + 40 = 0
Practice More Questions of Quantitative Aptitude for Competitive Exams:
###### Study Plan for IBPS RRB PO/Clerk Prelims 2021
Solutions
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# Re: st: Constant terms in AR1 error regressions
From "Clive Nicholas" To statalist@hsphsun2.harvard.edu Subject Re: st: Constant terms in AR1 error regressions Date Fri, 19 Dec 2008 21:27:58 +0000
Michael Hanson replied:
> You've defined your "task" very narrowly as using -regress- to estimate an
> AR(1) equation on residuals from (what I presume to be) a prior regression.
> If you could give a more general idea of what you are trying to accomplish,
> I and others on the list might be able to make better suggestions. For
> example, one might give different advice if you were concerned that the
> residuals were I(1) than if you were fairly confident they were stationary.
This example probably best illustrates what I was originally getting at:
webuse union
xtset idcode year
reg grade south union black if year==70
predict r70, r
forval i=71(1)73 {
local j = i'-1
reg grade south union black if year==i'
predict ri', r
reg ri' rj' if year==i'
reg ri' rj' if year==i', nocons
}
Notice that the coefficients on \rho and their attendant t-ratios are
not always the same. Unit-root tests via -pperron- and -dfuller-,
within unit, suggests my pooled time-series is stationary.
> That said, Wooldridge (2006, p. 418) discusses testing for AR(1) serial
> correlation with strictly exogenous regressors, and advises "this regression
> may or may not contain an intercept; the t statistic for \hat{\rho} will be
> slightly affected, but it is asymptotically valid either way." Later, he
> notes that strictly exogenous regressors are not very common with time
> series data, and that such simple tests are not robust to higher order
> autocorrelation. (You did test for higher order terms before settling on an
> AR(1) specification, right?) Wooldridge recommends a Breusch-Godfrey test,
> but there are others: see -help regress postestimationts- (yes, that is a
> "ts" at the end) for discussion of what is implemented in Stata.
Because, like -union-, my dataset is pooled, I ran -corrgram- with
selected units (i.e., those possessing a long enough time-series to be
worth performing Q tests on) and found one AR1 process and one AR3
process; the rest had none at all. That said, a pooled regression
using BSS's -xtivreg2, bw(2) small robust i() t() fe- showed my lagged
dependent variable to be significant; running (mean-centered) pooled
regressions using -xtpcse-, however, showed my LDV failed to reach
significance.
I did look carefully at -help regress postestimationts-, as it would
be much easier to run any of those AC tests automatically than do it
'by hand', as I had to. However,
estat bgodfrey
sample may not include multiple panels
r(459);
estat archlm
sample may not include multiple panels
r(459);
estat durbinalt
sample may not include multiple panels
r(459);
estat dwatson
sample may not include multiple panels
r(459);
and so on.
> Two final thoughts: First, if you include the intercept in a regression of
> a residual series on its first lag, and the estimated intercept is
> significantly different from zero, then you probably should revisit your
> prior estimation: your residuals should be mean-zero by definition. Second,
> if your results are very different when the intercept is excluded -- if that
> one extra degree of freedom is enough to change your results -- then I would
> caution you to be very skeptical of them to begin with, as you are working
> with large-T asymptotics by using -reg-.
Indeed, my T has a maximum of 13 in this pooled dataset, and I'm
running -reg- on single cross-sections in order to obtain \rho by
hand.
--
Clive Nicholas
[Please DO NOT mail me personally here, but at
<clivenicholas@hotmail.com>. Please respond to contributions I make in
`
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# Animation Compression: Linear Key Reduction
With simple key quantization, if we needed to sample a certain time `T` for which we did not have a key (e.g. in between two existing keys), we linearly interpolated between the two.
A natural extension of this is of course to remove keys or key frames which can be entirely linearly interpolated from their neighbour keys as long as we introduce minimal or no visible error.
## How It Works
The process to remove keys is fairly straight forward:
• Pick a key
• Calculate the value it would have if we linearly interpolated it from its neighbours
• If the resulting track error is acceptable, remove it
The above algorithm continues until nothing further can be removed. How you pick keys may or may not impact significantly the results. I personally only ever came across implementations that iterated on all keys linearly forward in time. However, in theory you could iterate in any number of ways: random, key with smallest error first, etc. It would be interesting to try various iteration methods.
It is worth pointing out that you need to check the error at a higher level than the individual key you are removing since it might impact other removed keys by changing the neighbour used to remove them. As such you need to look at your error metric and not just the key value delta.
Removing keys is not without side effects: now that our data is no longer uniform, calculating the proper interpolation alpha to reconstruct our value at time `T` is no longer trivial. To be able to calculate it, we must introduce in our data a time marker per remaining key (or key frame). This marker of course adds overhead to our animation data and while in the general case it is a win, memory wise, it can increase the overall size if the data is very noisy and no or very few keys can be removed.
A simple formula is then used to reconstruct the proper interpolation alpha:
``````TP = Time of Previous key
TN = Time of Next key
Interpolation Alpha = (Sample Time - TP) / (TN - TP)
``````
Another important side effect in introducing time markers is that when we sample a certain time `T`, we must now search to find between which two keys we must interpolate. This of course adds some overhead to our decompression speed.
The removal is typically done in one of two ways:
• Removal of whole key frames that can be linearly interpolated
• Removal of independent keys that can be linearly interpolated
While the first is less aggressive and will generally yield a higher memory footprint, the decompression speed will be faster due to needing to search only once to calculate our interpolation alpha.
For example, suppose we have the following track and keys:
The key #3 is of particular interest:
As we can see, we can easily recover the interpolation alpha from its neighbours: `alpha = (3 - 2) / (4 - 2) = 0.5`. With it, we can perfectly reconstruct the missing key: `value = lerp(0.35, 0.85, alpha) = 0.6`.
Another interesting key is #4:
It lies somewhat close to the value we could linearly interpolate from its neighbours: `value = lerp(0.6, 0.96, 0.5) = 0.78`. Whether the error introduced by removing it is acceptable or not is determined by our error metric function.
## In The Wild
This algorithm is perhaps the most common and popular out there. Both Unreal 4 and Unity 5 as well as many popular game engines support this format. They all use slight variations mostly in their error metric function but the principle remains the same. Sadly most implementations out there tend to use a poorly implemented error metric which tends to yield bad results in many instances. This typically stems from using a local error metric where each track type has a single error threshold. Of course the problem with this is that due to the hierarchical nature of our bone data, some bones need higher accuracy (e.g. pelvis, root). Some engines mitigate this by allowing a threshold per track or per bone but this requires some amount of tweaking to get right which is often undesirable and sub-optimal.
Twice in my career I had to implement a new animation compression algorithm and both times were to replace bad linear key reduction implementations.
From the implementations I have seen in the wild, it seems more popular to remove individual keys as opposed to removing whole key frames.
## Performance
Sadly due to the loss of data uniformity, the cache locality of the data we need suffers. Unlike for simple key quantization, we can no longer simply sort by key frame if we remove individual keys (you still can if you remove whole key frames though) to keep things cache efficient.
Although I have not personally witnessed it, I suspect it should be possible to use a variation of a technique used by curve fitting to sort our data in a cache friendly way. It is well described here and we’ll come back to it when we cover curve fitting.
The need to constantly search for which neighbour keys to use when interpolating quickly adds up since it scales poorly. The longer our clip is, the wider the range we need to search and the more tracks we have also increases the amount of searching that needs to happen. I have seen two ways to mitigate this: partitioning our clip or by using a cursor.
Partitioning our clip data as we discussed with uniform segmenting helps reduce the range to search in as our clip length increases. If the number of keys per block is sufficiently small, searching can be made very efficient with a sorting network or similar strategy. The use of blocks will also decrease the need for precision in our time markers by using a similar form of range reduction which allows us to use fewer bits to store them.
Using a cursor is conceptually very simple. Most clips play linearly and predictably (either forward or backward in time). We can leverage this fact to speed up our search by caching which time we sampled last and which neighbour keys were used to kickstart our search. The cursor overhead is very low if we remove whole key frames but the overhead is a function of the number of animated tracks if we remove individual keys.
Note that it is also quite possible that by using the above sorting trick that it could speed up the search but I cannot speak to the accuracy of this statement at this time.
Even though we can reach a smaller memory footprint with linear key reduction compared to simple key quantization, the amount of cache lines we’ll need to touch when decompressing is most likely going to be higher. Along with the need to search for key neighbours, these facts makes it slower to decompress using this algorithm. It remains popular due to the reduced memory footprint which was very important on older consoles (e.g. PS2 and PS3 era) as well as due to its obvious simplicity.
See the following posts for more details:
Up next: Curve Fitting
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Re: Fitting step function
• To: mathgroup at smc.vnet.net
• Subject: [mg122099] Re: Fitting step function
• From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
• Date: Fri, 14 Oct 2011 05:52:31 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201110120744.DAA16264@smc.vnet.net>
```Hi premiumxy
I assume that when you say "... to find the x-coordinates ..." you mean get an estimate with accompanying precision of the time of the 180 degree phase shift. You can do that by nonlinearly fitting various step or sigmoidal models; there are also Bayesian methods which would give you such a result in an intuitively satisfactory way.
With certain types of data or contexts, this is called intervention analysis.
If you have some data to share ...?
In the meantime (these are your ideas):
data = RandomReal[ {-5, 5}, {83} ]~Join~RandomReal[ {25, 35}, {200 - 83} ];
dataPlot = ListPlot[ data, Joined -> True ];
Off[ NonlinearModelFit::cvmit ]
nlm = NonlinearModelFit[data,
a ArcTan[ b (x - t)] + c, {{a, 20}, {b, 100}, {t, 50}, {c, 20}}, x ]
Normal[nlm]
solPlot = Plot[Normal[nlm], {x, 0, 200}, PlotStyle -> {Red, Thick}];
Show[ dataPlot, solPlot ]
nlm["ParameterTable"]
Step change is at:
{t /. nlm["BestFitParameters"],
\!\(\*SubscriptBox[\(nlm["\<ParameterConfidenceIntervals\>"]\), \(\(\
\[LeftDoubleBracket]\)\(3\)\(\[RightDoubleBracket]\)\)]\)}
Cheers
Barrie
>>> On 12/10/2011 at 6:44 pm, in message <201110120744.DAA16264 at smc.vnet.net>,
> Hello,
>
> I have the following problem: I have a set of data about some phase
> information of a signal. At certain points there is a clear step of 180
> degrees.
>
> I would like to fit this data to find the x-coordinates of these steps. I
> have the feeling that "normal" step-functions in Mathematica (such as
> UnitStep, HeavisideTheta, Sign...) are not varied in the fit-routines (such as
> NonlinearModelFit). I guess it this is caused by the functions not being
> continuous.
>
> I would appreciate any hints for this quite simple (?)problem. The only
> option that I can currently see is to create a "step-like" continuous function
> such as Tanh[A x] (with A>>1)
>
> Thanks!
```
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Wednesday, March 12, 2008
Future Cars?
An air-powered car: a nifty commuter-mobile, but not much good for long-distance travel.
Electric cars: not bad if they were charged by non-polluting nuclear power, but currently the electricity that charges them is created by burning coal or natural gas--ergo, you're still polluting.
Hybrids: Yahoo contrasts 10 hybrids available this model year.
Oldmath said...
Darren: You are correct about electric cars, but large power plants burn fuel more efficiently than small internal combustion engines so there is a net reduction in pollution.
The "air car" requires a power source (presumably electric) to compress the air in the first place so the question becomes are compressed air tanks a better energy storage device than batteries. I am dubious about this one.
neko said...
I think that it's time to bring back the Stanley Steamer.
(The car... not the rug cleaning machine.)
Step Back said...
Darren,
As a math teacher it shouldn't be too hard to learn some physics and pass the lessons on to your readers.
Air car: Each time you compress a gas (e.g. air), part of the compression energy is lost as heat. PV=nRT, the gas law. Temperature goes up as pressure increases.
Electric car in place of gas guzzler: Coal powered plants are not "efficient". 60% of more of energy goes up as waste heat. Then there is transmission loss (about 10%) over electric lines, conversion losses in the battery for converting from electrical energy to chemical and back again.
It all sounds good until you start doing the math.
Darren said...
And as a host it shouldn't be too hard for me to identify some lessons about gratuitous swipes at people and pass those lessons onto my readers--which is why I posted your comment, jacquesasse.
David said...
Large power plants are in fact considerably more efficient than automotive gasoline engines, even when transmission-line losses are taken into account. This is especially true of new combined-cycle technology like the GE "H" series turbines, which offer 60% efficiency (not 60% waste.)
You can buy them here.
Step Back said...
Darren,
To bash is human. I'm human. (So are you.} :-)
David,
Since we both read Robert Rapier and The Energy Blog, you kind of know what my next comment is going to be: Where do I buy the transoceanic pipeline to feed the natural gas (methane) into my H series given that the gas is over "there" and not over here in the United Saudi Arabia (USA) of Coal? :-)
p.s. Aside from I2R and Energy Blog, are you also an Oil Drum reader?
David said...
sb...only occasionally an Oil Drum reader.
You can of course gasify coal before feeding it into your brand new "H" series. The other option is LNG tankers, assuming that various interests do not totally succeed in suppressing the building of port facilities. (The practical solution will probably be that the LNG tankers will dock in Mexico and the gas will flow to the U.S. by pipeline)
But I think that even for pure coal plants, whose efficiency is of course lower than the combined-cycle stuff, the overall efficiency of electric vehicles is going to come out better...IIRC, Rapier has posted some analysis about this. Also, remember the cost per BTU is lower for coal than for oil, so economics could be favorable ***if*** the battery problems could be solved at reasonable cost.
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-7-trigonometric-identities-and-equations-7-3-sum-and-difference-identities-7-3-exercises-page-680/101
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## Precalculus (6th Edition)
$\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}=\frac{\sin s}{\sin t\cos t}$
Start with the left side: $\frac{\sin(s-t)}{\sin t}+\frac{\cos(s-t)}{\cos t}$ Expand using the sum and difference formulas for sine and cosine: $=\frac{\sin s\cos t-\cos s\sin t}{\sin t}+\frac{\cos s\cos t+\sin s\sin t}{\cos t}$ Get a common denominator: $=\frac{(\sin s\cos t-\cos s\sin t)\cos t}{\sin t\cos t}+\frac{(\cos s\cos t+\sin s\sin t)\sin t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t}{\sin t\cos t}+\frac{\cos s\cos t\sin t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t+\cos s\cos t\sin t+\sin s\sin^2 t}{\sin t\cos t}$ Note that the second and third terms in the numerator cancel. Simplify: $=\frac{\sin s\cos^2 t-\cos s\sin t\cos t+\cos s\sin t\cos t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s\cos^2 t+\sin s\sin^2 t}{\sin t\cos t}$ $=\frac{\sin s(\cos^2+\sin^2 t)}{\sin t\cos t}$ $=\frac{\sin s*1}{\sin t\cos t}$ $=\frac{\sin s}{\sin t\cos t}$ Since this equals the right side, the identity has been proven.
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https://simondesenlisblogs.org/2023/02/24/murray-maths/
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# Murray Maths!
Wow!
What a great first week back! We have had a wonderful week of learning this week, especially in our maths lessons. Murray class have loved learning about division and this week our focus has been dividing by grouping. We can divide a number by making equal groups.
Earlier in the week, we enjoyed using counters to divide and make equal groups. As you can see in the picture below, everyone loved the challenge and some even challenged themselves to write the division sentence straight away!
Throughout the week, we have then been applying our knowledge of working with concrete objects to then draw picture representations to show our working out.
Today, we ended our week by using a number line to make jumps of equal groups. We had to use our knowledge of our 2, 5, 10 and 3 times table to jump along the number line. From there, we could count the number of equal groups.
Well done Murray class, what a fantastic week!
Have a great weekend,
Miss Gedney
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https://www.jiskha.com/questions/1192749/how-do-you-divide-this-polynomial-using-long-division-6x-3-17x-2-104x-60-divide-by
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# Algebra 2 (DIVIDEPOLYNOMIALS)
How do you divide this polynomial using long division?
(6x^3-17x^2-104x-60) divide by (2x+5)
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1. Here's a link for long division.
http://www.purplemath.com/modules/polydiv2.htm
Quite frequently we use synthetic division, with less writing to do:
http://www.purplemath.com/modules/synthdiv.htm
Hint: the given problem leaves no remainder.
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## Similar Questions
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Use long division to divide and use the result to factor the dividend completely. (5x2 − 3x − 14) ÷ (x − 2)
2. ### Algebra
Use synthetic division to divide the polynomial 2x^3 – 12x – 5 by x + 4, Write the quotient polynomial and the remainder. [Be careful – notice that there is no x2 term.]. Show work.
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Two polynomials P and D are given. Use either synthetic or long division to divide P(x) by D(x),and express the quotient P(x)/D(x) in the form P(x)D(x) = Q(x)+ R(x)D(x) . P(x) = 6x4 − 3x3 + 29x2, D(x) = 3x2 + 13
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This problem was on a test and I got it wrong, and if I do it over I get half the credit back, but I need help with it plz. (11x^2+2x^3+14+17x)/(2x+5) My algebra teacher didn't even go over this lesson (we skipped it) and it was
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# Comparision methods of measurements
23. Nov 2021
1 von 50
### Comparision methods of measurements
• 1. UNIT 3 - COMPARISION METHODS OF MEASUREMENTS
• 2. Module 1: DC & AC Potentiometers
• 3. Potentiometer techniques may also be used for current measurement, the unknown current being sent through a known resistance and the IR drop opposed by balancing it at the voltage terminals of the potentiometer. Internal heating and consequent resistance change of the current-carrying resistor (shunt) may be a critical factor in measurement accuracy; and the shunt design may require attention to dissipation of heat resulting from its I2R power consumption. Current potentiometer diagram
• 4. Potentiometer techniques have been extended to alternating-voltage measurements, but generally at a reduced accuracy. Current is set on an ammeter which must have the same response on ac as on dc, where it may be calibrated with a potentiometer and shunt combination. Balance in opposing an unknown voltage is achieved in one of two ways: (1) a slide-wire and phase-adjustable supply; (2) separate in-phase and quadrature adjustments on slide wires supplied from sources that have a 90° phase difference. Such potentiometers have limited use in magnetic testing. An instrument that precisely measures an electromotive force (emf) or a voltage by opposing to it a known potential drop established by passing a definite current through a resistor of known characteristics.
• 5. DC Potentiometer: In DC potentiometers, the voltage being measured is compared to the emf of a standard cell. Since at the instant of compensation the current in the circuit of the voltage being measured equals zero, measurements can be made without reductions in this voltage. For this type of potentiometer, accuracy can exceed 0.01 percent. DC potentiometers are categorized as either high-resistance, with a slide-wire resistance ranging from the higher resistance class can measure up to 2 volts (V) and is used in testing highly accurate apparatus. The low-resistance class is used in measuring voltage up to 100 mV. To measure higher voltages, up to 600 V, and to test voltmeters, voltage dividers are connected to potentiometers. Here the voltage drop across one of the resistances of the voltage divider is compensated; this constitutes a known fraction of the total voltage being measured.
• 6. AC Potentiometer: In AC potentiometers, the unknown voltage is compared with the voltage drop produced by a current of the same frequency across a known resistance. The voltage being measured is then adjusted both for amplitude and phase. The accuracy of AC potentiometers is of the order of 0.2 percent. In electronic automatic DC and AC potentiometers, the measurements of voltage are carried out automatically. In this case, the compensation of the unknown voltage is achieved with the aid of a servomechanism that moves the slide along the resistor, or rheostat. The servomechanism is actuated by the imbalance of the two voltages, that is, by the difference between the compensating voltage and the voltage that is being compensated.
• 7. In this case, the slide of the rheostat is set in a position that predetermines, for instance, the temperature of the object to be regulated. The voltage imbalance of the potentiometer drives the servomechanism, which then increases or decreases the electric heating or regulates the fuel supply. A voltage divider with a uniform variation of resistance, a device that allows some fraction of a given voltage to be applied to an electric circuit. In the simplest case, the device consists of a conductor of high resistance equipped with a sliding contact. Such dividers are used in electrical engineering, radio engineering, and measurement technology. They can also be utilized in analog computers and in automation systems, where, for example, they function as sensors for linear or angular displacement.
• 8. Module 2: DC & AC Bridges
• 9. Bridge circuits are used very commonly as a variable conversion element in measurement systems and produce an output in the form of a voltage level that changes as the measured physical quantity changes. They provide an accurate method of measuring resistance, inductance and capacitance values, and enable the detection of very small changes in these quantities about a nominal value. Excitation of the bridge is by a d.c. voltage for resistance measurement and by an a.c. voltage for inductance or capacitance measurement. Both null and deflection types of bridge exist, and, in a like manner to instruments in general, null types are mainly employed for calibration purposes and deflection types are used within closed-loop automatic control schemes.
• 10. DC Bridges: Wheatstone bridge A null-type bridge with d.c. excitation, commonly known as a Wheatstone bridge,. The four arms of the bridge consist of the unknown resistance Ru, two equal value resistors R2 and R3 and a variable resistor Rv. A d.c. voltage Vi is applied across the points AC and the resistance Rv is varied until the voltage measured across points BD is zero. This null point is usually measured with a high sensitivity galvanometer. if a high impedance voltage-measuring instrument is used, the current Im drawn by the measuring instrument will be very small and can be approximated to zero. If this assumption is made, then, for Im D 0: I1 =I3 and I2 =I4
• 11. Wheat stone diagram
• 12. Deflection type DC bridge This differs from the Wheatstone bridge mainly in that the variable resistance Rv is replaced by a fixed resistance R1 of the same value as the nominal value of the unknown resistance Ru . As the resistance Ru changes, so the output voltage V0 varies, and this relationship between V0 and Ru must be calculated. This relationship is simplified if we again assume that a high impedance voltage measuring instrument is used and the current drawn by it, Im , can be approximated to zero. V0= Vi * ( Ru / Ru + R3)- ( R1 / R1+ R2) When Ru is at its nominal value, i.e. for Ru D R1, it is clear that V0 D 0 (since R2 D R3). For other values of Ru, V0 has negative and positive values that vary in a non-linear way with Ru.
• 13. AC Bridges: Bridges with a.c. excitation are used to measure unknown impedances Null-type impedance bridge The null point can be conveniently detected by monitoring the output with a pair of head phones connected via an operational amplifier across the points BD. Diagram
• 14. This is a much cheaper method of null detection than the application of an expensive galvanometer that is required for a d.c. Wheatstone bridge. If Zu is capacitive, i.e. Zu D 1/jωCu, then Zv must consist of a variable capacitance box, which is readily available. If Zu is inductive, then Zu D Ru C jωLu . An inductor coil always has a resistive component, though this is made as small as possible by designing the coil to have a high Q factor. Therefore, Zv must consist of a variable-resistance box and a variable- inductance box. However, the latter are not readily available because it is difficult and hence expensive to manufacture a set of fixed value inductors to make up a variable-inductance box. For this reason, an alternative kind of null-type bridge circuit, known as the Maxwell Bridge, is commonly used to measure unknown inductances.
• 15. Maxwell bridge A Maxwell bridge (in long form, a Maxwell-Wien bridge) is a type of Wheatstone bridge used to measure an unknown inductance (usually of low Q value) in terms of calibrated resistance and capacitance. Diagram:
• 16. With reference to the picture, in a typical application R1 and R4 are known fixed entities, and R2 and C2 are known variable entities. R2 and C2 are adjusted until the bridge is balanced.R3 and L3 can then be calculated based on the values of the other components: As shown in Figure, one arm of the Maxwell bridge consists of a capacitor in parallel with a resistor (C1 and R2) and another arm consists of an inductor L1 in series with a resistor (L1 and R4). The other two arms just consist of a resistor each (R1 and R3). The values of R1 and R3 are known, and R2 and C1 are both adjustable. The unknown values are those of L1 and R4. Like other bridge circuits, the measuring ability of a Maxwell Bridge depends on 'Balancing' the circuit. Balancing the circuit in Figure 1 means adjusting C1 and R2 until the current through the bridge between points A and B becomes zero. This happens when the voltages at points A and B are equal.
• 17. Mathematically, Z1 = R2 + 1/ (2πfC1); while Z2 = R4 + 2πfL1. (R2 + 1/ (2πfC1)) / R1 = R3 / [R4 + 2πfL1]; or R1R3 = [R2 + 1/ (2πfC1)] [R4 + 2πfL1] To avoid the difficulties associated with determining the precise value of a variable capacitance, sometimes a fixed-value capacitor will be installed and more than one resistor will be made variable. The additional complexity of using a Maxwell bridge over simpler bridge types is warranted in circumstances where either the mutual inductance between the load and the known bridge entities, or stray electromagnetic interference, distorts the measurement results.
• 18. The capacitive reactance in the bridge will exactly oppose the inductive reactance of the load when the bridge is balanced, allowing the load's resistance and reactance to be reliably determined. Advantages: The frequency does not appear Wide range of inductance Disadvantages: Limited measurement It requires variable standard capacitor
• 19. Schering Bridge A Schering Bridge is a bridge circuit used for measuring an unknown electrical capacitance and its dissipation factor. The dissipation factor of a capacitor is the the ratio of its resistance to its capacitive reactance. Diagram
• 20. In the Schering Bridge above, the resistance values of resistors R1 and R2 are known, while the resistance value of resistor R3 is unknown. The capacitance values of C1 and C2 are also known, while the capacitance of C3 is the value being measured. To measure R3 and C3, the values of C2 and R2 are fixed, while the values of R1 and C1 are adjusted until the current through the ammeter between points A and B becomes zero. This happens when the voltages at points A and B are equal, in which case the bridge is said to be 'balanced'. When the bridge is balanced, Z1/C2 = R2/Z3, where Z1 is the impedance of R1 in parallel with C1 and Z3 is the impedance of R3 in series with C3. In an AC circuit that has a capacitor, the capacitor contributes a capacitive reactance to the impedance. Z1 = R1/[2πfC1((1/2πfC1) + R1)] = R1/(1 + 2πfC1R1)
• 21. while Z3 =1/2πfC3 + R3. 2πfC2R1/ (1+2πfC1R1) = R2/(1/2πfC3 + R3); or 2πfC2 (1/2πfC3 + R3) = (R2/R1) (1+2πfC1R1); or C2/C3 + 2πfC2R3 = R2/R1 + 2πfC1R2. When the bridge is balanced, the negative and positive reactive components are equal and cancel out, so 2πfC2R3 = 2πfC1R2 or R3 = C1R2 / C2. Similarly, when the bridge is balanced, the purely resistive components are equal, so C2/C3 = R2/R1 or C3 = R1C2 / R2. Note that the balancing of a Schering Bridge is independent of frequency. Advantages: Balance equation is independent of frequency Used for measuring the insulating properties of electrical cables and equipment’s
• 22. Hay Bridge A Hay Bridge is an AC bridge circuit used for measuring an unknown inductance by balancing the loads of its four arms, one of which contains the unknown inductance. DIAGRAM
• 23. one arm of the Hay bridge consists of a capacitor in series with a resistor (C1 and R2) and another arm consists of an inductor L1 in series with a resistor (L1 and R4). The other two arms simply contain a resistor each (R1 and R3). The values of R1and R3 are known, and R2 and C1 are both adjustable. The unknown values are those of L1 and R4. Like other bridge circuits, the measuring ability of a Hay Bridge depends on 'balancing' the circuit. Balancing the circuit in Figure 1 means adjusting R2 and C1 until the current through the ammeter between points A and B becomes zero. This happens when the voltages at points A and B are equal. When the Hay Bridge is balanced, it follows that Z1/R1 = R3/Z2 wherein Z1 is the impedance of the arm containing C1 and R2 while Z2 is the impedance of the arm containing L1 and R4.
• 24. Thus, Z1 = R2 + 1/(2πfC) while Z2 = R4 + 2πfL1. [R2 + 1/(2πfC1)] / R1 = R3 / [R4 + 2πfL1]; or [R4 + 2πfL1] = R3R1 / [R2 + 1/(2πfC1)]; or R3R1 = R2R4 + 2πfL1R2 + R4/2πfC1 + L1/C1. When the bridge is balanced, the reactive components are equal, so 2πfL1R2 = R4/2πfC1, or R4 = (2πf) 2L1R2C1. Substituting R4, one comes up with the following equation: R3R1 = (R2+1/2πfC1) ((2πf) 2L1R2C1) + 2πfL1R2 + L1/C1; or L1 = R3R1C1 / (2πf) 2R22C12 + 4πfC1R2 + 1); L1 = R3R1C1 / [1 + (2πfR2C1)2]
• 25. Thus, the equations for L1 and R4 for the Hay Bridge when it is balanced are: L1 = R3R1C1 / [1 + (2πfR2C1)2]; and R4 = (2πfC1)2R2R3R1 / [1 + (2πfR2C1)2] Advantages: Simple expression Disadvantages: It is not suited for measurement of coil
• 26. Wien Bridge: A Wien bridge oscillator is a type of electronic oscillator that generates sine waves. It can generate a large range of frequencies. Wien did not have a means of developing electronic gain so a workable oscillator could not be realized. Circuit Diagram
• 27. Amplitude stabilization The key to Hewlett's low distortion oscillator is effective amplitude stabilization. The amplitude of electronic oscillators tends to increase until clipping or other gain limitation is reached. This leads to high harmonic distortion, which is often undesirable. Hewlett used an incandescent bulb as a positive temperature coefficient (PTC) thermistor in the oscillator feedback path to limit the gain. The resistance of light bulbs and similar heating elements increases as their temperature increases. If the oscillation frequency is significantly higher than the thermal time constant of the heating element, the radiated power is proportional to the oscillator power. Since heating elements are close to black body radiators, they follow the Stefan-Boltzmann law.
• 28. The radiated power is proportional to T4, so resistance increases at a greater rate than amplitude. If the gain is inversely proportional to the oscillation amplitude, the oscillator gain stage reaches a steady state and operates as a near ideal class A amplifier, achieving very low distortion at the frequency of interest. At lower frequencies the time period of the oscillator approaches the thermal time constant of the thermistor element and the output distortion starts to rise significantly. Light bulbs have their disadvantages when used as gain control elements in Wien bridge oscillators, most notably a very high sensitivity to vibration due to the bulb's micro phonic nature amplitude modulating the oscillator output, and a limitation in high frequency response due to the inductive nature of the coiled filament.
• 29. Modern Distortion as low as 0.0008% (-100 dB) can be achieved with only modest improvements to Hewlett's original circuit. Wien bridge oscillators that use thermistors also exhibit "amplitude bounce" when the oscillator frequency is changed. This is due to the low damping factor and long time constant of the crude control loop, and disturbances cause the output amplitude to exhibit a decaying sinusoidal response. This can be used as a rough figure of merit, as the greater the amplitude bounce after a disturbance, the lower the output distortion under steady state conditions.
• 30. Analysis: Advantages: Frequency sensitive Supply voltage is purely sinusoidal
• 31. Module 3: Transformer Ratio Bridges
• 33. Module 4: Self-balancing bridges
• 35. Module 5: Interference & Screening
• 36. We encounter interference when it produces effects like noise, hiss, hum or cross-talk. If a radio engineer faces such problems, good theoretical knowledge as well as experience is required to overcome them. All technical remedies only aim at reducing the effect of interference to such a degree, that it is neither audible nor disturbing. This is mainly achieved by different ways of screening. Types of Interference: Symmetrical and Asymmetrical Interference Having a closer look at the interference of cable, we find that hf-interference currents cause measurable levels on signal (audio) lines and on supply lines. A ground-free interference source would produce signals on a cable which spread along the line.
• 37. These voltages and currents can be called symmetrical interference. In practice this rarely occurs. Through interference, asymmetrical signals are produced in respect to the ground. The asymmetrical interference current flows along the two wires of the symmetrical line to the sink and via the ground back to the source. These interference signals are cancelled at the symmetrical input. Galvanic Coupling of Interference Galvanic coupling of interference occurs if the source and the sink of interference are coupled by a conductive path. As can be seen from the equivalent circuit diagram, the source impedance of the interference consists of the resistance RC and the inductance LC of the conductor, which are common to the two parts of the circuit. From these elements the interference source voltage can be calculated.
• 38. Capacitive coupling of Interference The capacitive coupling of interference occurs due to any capacitance between the source and sink of interference. Principle: The current in the interference sink can be calculated as the interference voltage in the sink is proportional to its impedance. Systems of high impedance are therefore more sensitive to interference than those of low impedance. The coupled interference current depends on the rate of change of the interference and on the coupling capacitance CC.
• 39. Inductive coupling of Interference Inductive coupling of interference occurs if the interference sink is in the magnetic field of the interference source (e.g. coils, cables, etc.) Principle: The interference voltage induced by inductive coupling is increasing the distance between conductors mounting conductors close to conductive surfaces using short conductors avoiding parallel conductors screening using twisted cable Note that by the same means the capacitive as well as the inductive coupling of interference will be reduced.
• 40. Screening When considering the effect of electrical and magnetic fields, we have to distinguish between low and high frequencies. At high frequencies the skin effect plays an important roll for the screening. The penetration describes the depth from the surface of the conductor, where the current density has decayed to 37% compared to the surface of the conductor. Screening of Cables. When signal lines run close to interference sources or when the signal circuit is very sensitive to interference, screening of signal lines will give an improvement. There are different ways of connecting the cable screen. Cable screen not connected It will not prevent any interference, because the charge on the screen, produced by interference, will remain and will effect the central signal line.
• 41. the current induced by interference in the line will flow through the sink, effecting the signal. Cable screen grounded on one side only It will only prevent interference at low frequency signals. For electromagnetic interference, where the wavelength is short compared to the length of the cable, the screening efficiency is poor. Cable screen grounded on either side It is effective for all kinds of interference. Any current induced in the screen by magnetic interference will flow to ground. The inner of the cable is not affected. Only the voltage drop on the screen will affect the signal in the screen. This type of grounding Ensure proper and careful connection of the screens. Use suitable plugs in connection with the cable screen.
• 42. Module 7: Multiple earth and Earth loops
• 43. Noise due to multiple earths can be avoided by good earthing practice. In particular this means keeping earths for signal wires and earth for high current equipments entirely separate. Power earth Logic earth Analogue earth Safety earth
• 44. Module 8: Electrostatic and Electromagnetic Interference
• 45. Interference By Radiation Interference by electromagnetic radiation becomes important at cable lengths greater than 1/7 of the wavelength of the signals. At frequencies beyond 30Mhz, most of the interference occurs by e.m. radiation Interference By Electrostatic Charge Charged persons and objects can store electrical charges of up to several micro- Coulombs, which means voltages of some 10kV in respect to ground. Dry air, artificial fabrics and friction favour these conditions. When touching grounded equipment, an instantaneous discharge produces arcing with short, high current pulses and associated strong changes of the e.m. field.
• 46. Reduction of Interference There are a number of methods to prevent interference. But all of them only reduce the interference and never fully prevent it. This means there will never be a system which is 100% safe from interference. Because the efforts and the cost will rise with the degree of reduction of interference, a compromise has to be found between the effort and the result. The requirement for the reduction of interference will depend on: The strength of the interference source The sensitivity of the interference sink The problems caused by interference The costs of the equipment
• 47. Module 9: Grounding Techniques
• 48. This is one of the simplest but most efficient methods to reduce interference. Grounding can be used for three different purposes: Protection Ground Provides protection for the operators from dangerous voltages. Widely used on mains-operated equipment. Function Ground The ground is used as a conductive path for signals. Example: in asymmetrical cables screen, which is one conductor for the signal, is connected to the ground. Screening Ground Used to provide a neutral electrical path for the interference, to prevent that the interfering voltages or currents from entering the circuit. Grounding of equipment is often required for the cases 1 or 2 anyhow, so that the screening ground is available "free of charge".
• 49. Sometimes the grounding potential, provided by the mains connection, is very "polluted". This means that the ground potential itself already carries an interfering signal. This is especially likely if there are big power consumers in the neighborhood or even in the same building. Using such a ground might do more harm than good. The quality of the ground line can be tested by measuring it with a storage scope against some other ground connection, e.g. a metal water pipe or some metal parts of the construction. Multi-point Grounding In multi-point grounding all parts are connected to ground at as many points as possible. This requires that the ground potential itself is as widely spread as possible. In practice, all conductive parts of the chassis, the cases, the shielding, the room and the installation are included in the network.
• 50. Thank you
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# Reasoning Quiz for NIACL Assistant Prelims: 21st August 2018
Dear Aspirants,
Reasoning Questions for NIACL ASSISTANT 2018
Reasoning Ability is an onerous section. With the increasing complexity of questions, it becomes hard for one to give it the cold shoulder. The only way to make the grade in this particular section in the forthcoming exams like NIACL ASSISTANT is to practice continuously with all your heart and soul. And, to let you practice with the best of the latest pattern questions, here is the Adda247 Reasoning Quiz based on the study plan and the exact same pattern of questions that are being asked in the exams.
Direction (1-5): Study the following information carefully to answer the given questions:
‘Speech plans benefit check’ is coded as ‘la pa zi ta’
‘sale plans benefit Speech’ is coded as ‘pa zi la sa’
‘growth benefit loyalty income’ is coded as ‘na hi ga pa’
‘Speech global rain growth’ is coded as ‘zi mi jo ga’
Q1. How will ‘Speech sale growth’ be coded in the given language?
zi la sa
zi sa ga
sa mi jo
pa zi mi
None of these
Solution:
Q2. How will ‘growth benefit income’ be coded in the given language?
zi la sa
zi sa ga
sa mi jo
pa zi mi
None of these
Solution:
Q3. What does ‘pa’ stands for?
sale
growth
benefit
global
None of the these
Solution:
Q4. Which of the following will be coded as ‘na hi la’ in the given language?
plans benefit income
loyalty income plans
sale plans growth
loyalty Speech growth
none of these
Solution:
Q5. What is the code for ‘la’ strain for?
plans
growth
sale
income
none of these.
Solution:
Directions (6-10): These questions are based on the following six numbers.
412 646 734 255 536 876
Q6. If we add 2 to the second digit of each number and subtract 1 from the first digit of each number, then which of the given will become the lowest number?
646
536
255
412
None of these
Q7. If all the digits in each of the numbers are written in increasing order within than number, than which of the given number will become the second highest?
412
646
255
536
None of these
Q8. If the first two digits of each number are interchanged, then which of the following numbers will become the highest number?
734
536
646
412
876
Q9. In each number, if we subtracted 1 in first digit and added 1 in last digit, then which among these number will be 2nd lowest number?
734
255
876
412
None of these
Q10.If we replace all the odd digits in each number with zero, then which of the following number will become the lowest?
734
536
412
255
876
Directions (11-15): Study the following information carefully and answer the question given below-
Eight persons P, Q, R, S, T, U, V and W are sitting around a circular table. Four of them faces towards centre while the rest of them faces opposite to centre. Each of them was born in a different month of the same year starting from January to August but not necessarily in the same order.
T was born in February. S who is the second youngest sits opposite to P. The persons who were born in June and March are sitting opposite to each other. R who was born in the month of May is an immediate neighbour of P. The person sitting third to the right of U was born in April. W is the youngest and sits opposite to T. V faces opposite direction of U. Both the immediate neighbours of P are facing towards the center. Q sits second to the right of T. T is sitting third to the right of S who faces outside. S is an immediate neighbour of Q. W faces towards the center. V is older than U. There is only one person sitting between S and U who is not immediate neighbor of T.
Q11. Who among the following was born in the month of March?
Q
R
V
P
W
Solution:
Q12. Who among the following sits third to the left of the person who was born in May?
P
R
T
S
W
Solution:
Q13. Who among the following is the third youngest person?
Q
R
T
P
U
Solution:
Q14. How many persons facing towards the centre are older than Q?
Three
One
Two
Four
None of these
Solution:
Q15. The person sitting third to the right of P was born in which of the following months?
January
February
August
April
None of these
Solution:
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# Articles with the ‘Convert to decimal’ Tag
This is the first of a four part series on “pencil and paper” binary arithmetic, which I’m writing as a supplement to my binary calculator. This article introduces binary arithmetic, and then discusses binary addition.
# Bigcomp: Deciding Truncated, Near Halfway Conversions
In my article “Using Integers to Check a Floating-Point Approximation,” I briefly mentioned “bigcomp,” an optimization strtod() uses to reduce big integer overhead when checking long decimal inputs. bigcomp does a floating-point to decimal conversion — right in the middle of a decimal to floating-point conversion mind you — to generate the decimal expansion of the number halfway between two target floating-point numbers. This decimal expansion is compared to the input decimal string, and the result of the comparison dictates which of the two target numbers is the correctly rounded result.
In this article, I’ll explain how bigcomp works, and when it applies. Also, I’ll talk briefly about its performance; my informal testing shows that, under the default setting, bigcomp actually worsens performance for some inputs.
# How I Taught Third Graders Binary Numbers
Last week I introduced my son’s third grade class to binary numbers. I wanted to build on my prior visit, where I introduced them to the powers of two. By teaching them binary, I showed them that place value is not limited to base ten, and that there is a difference between numbers and numerals.
My presentation was based on base-ten-block-like imagery, since I knew the students were comfortable expressing numbers with base ten blocks. I thought extending the block model to other bases would work well. I think it did.
The Number Twenty-Seven in Tape Flags, Broken Into Powers of Two
# Pi and e In Binary
Some people are curious about the binary representations of the mathematical constants pi and e. Mathematically, they’re like every other irrational number — infinite strings of 0s and 1s (with no discernible pattern). In a computer, they’re finite, making them only approximations to their true values. I will show you what their approximations look like in five different levels of binary floating-point precision.
The first 43 bits of pi and e
# 15-Digit Quick and Dirty Conversions Don’t Round-Trip
In my article “Fifteen Digits Don’t Round-Trip Through SQLite Reals” I showed examples of decimal floating-point numbers — 15 significant digits or less — that don’t round-trip through double-precision binary floating-point variables stored in SQLite. The round-trip failures occur because SQLite’s floating-point to decimal conversion routine uses limited-precision floating-point arithmetic.
My quick and dirty floating-point to decimal conversion routine, which I wrote to demonstrate conversion inaccuracies caused by limited-precision, also fails to round-trip some decimal numbers of 15 digits or less. Since I hadn’t demonstrated this failure previously, I will do so here.
# Incorrect Decimal to Floating-Point Conversion In SQLite
SQLite has a limited-precision floating-point to decimal conversion routine which it uses to print double-precision floating-point values retrieved from a database. As I’ve discovered, its limited-precision conversion results in decimal numbers of 15 significant digits or less that won’t round-trip. For example, if you store the number 9.944932e+31, it will print back as 9.94493200000001e+31.
SQLite also has a limited-precision decimal to floating-point conversion routine, which it uses to convert input decimal numbers to double-precision floating-point numbers for storage in a database. I’ve found that some of its conversions are incorrect — by as many as four ULPs — and that some decimal numbers fail to round-trip because of this; “garbage in, garbage out” as they say.
# Fifteen Digits Don’t Round-Trip Through SQLite Reals
I’ve discovered that decimal floating-point numbers of 15 significant digits or less don’t always round-trip through SQLite. Consider this example, executed on version 3.7.3 of the pre-compiled SQLite command shell:
```sqlite> create table t1(d real);
sqlite> insert into t1 values(9.944932e+31);
sqlite> select * from t1;
9.94493200000001e+31
```
SQLite represents a decimal floating-point number that has real affinity as a double-precision binary floating-point number — a double. A decimal number of 15 significant digits or less is supposed to be recoverable from its double-precision representation. In SQLite, however, this guarantee is not met; this is because its floating-point to decimal conversion routine is implemented in limited-precision floating-point arithmetic.
# Quick and Dirty Floating-Point to Decimal Conversion
In my article “Quick and Dirty Decimal to Floating-Point Conversion” I presented a small C program that uses double-precision floating-point arithmetic to convert decimal strings to binary floating-point numbers. The program converts some numbers incorrectly, despite using an algorithm that’s mathematically correct; its limited precision calculations are to blame. I dubbed the program “quick and dirty” because it’s simple, and overall converts reasonably accurately.
For this article, I took a similar approach to the conversion in the opposite direction — from binary floating-point to decimal string. I wrote a small C program that combines two mathematically correct algorithms: the classic “repeated division by ten” algorithm to convert integer values, and the classic “repeated multiplication by ten” algorithm to convert fractional values. The program uses double-precision floating-point arithmetic, so like its quick and dirty decimal to floating-point counterpart, its conversions are not always correct — though reasonably accurate. I’ll present the program and analyze some example conversions, both correct and incorrect.
# Incorrect Floating-Point to Decimal Conversions
In my article “Inconsistent Rounding of Printed Floating-Point Numbers” I showed examples of incorrect floating-point to decimal conversions I stumbled upon — in Java, Visual Basic, JavaScript, VBScript, and OpenOffice.org Calc. In this article, I’ll explore floating-point to decimal conversions more deeply, by analyzing conversions done under four C compilers: Visual C++, MinGW GCC, Digital Mars C, and Linux GCC. I found that incorrect conversions occur in three of the four environments — in all but Linux GCC. I’ll show you some examples and explain how I found them.
# Inconsistent Rounding of Printed Floating-Point Numbers
What does this C program print?
```#include <stdio.h>
int main (void)
{
printf ("%.1f\n",0.25);
}
```
The answer depends on which compiler you use. If you compile the program with Visual C++ and run on it on Windows, it prints 0.3; if you compile it with gcc and run it on Linux, it prints 0.2.
The compilers — actually, their run time libraries — are using different rules to break decimal rounding ties. The two-digit number 0.25, which has an exact binary floating-point representation, is equally near two one-digit decimal numbers: 0.2 and 0.3; either is an acceptable answer. Visual C++ uses the round-half-away-from-zero rule, and gcc (actually, glibc) uses the round-half-to-even rule, also known as bankers’ rounding.
This inconsistency of printed output is not limited to C — it spans many programming environments. In all, I tested fixed-format printing in nineteen environments: in thirteen of them, round-half-away-from-zero was used; in the remaining six, round-half-to-even was used. I also discovered an anomaly in some environments: numbers like 0.15 — which look like halfway cases but are actually not when viewed in binary — may be rounded incorrectly. I’ll report my results in this article.
Hexadecimal floating-point constants, also known as hexadecimal floating-point literals, are an alternative way to represent floating-point numbers in a computer program. A hexadecimal floating-point constant is shorthand for binary scientific notation, which is an abstract — yet direct — representation of a binary floating-point number. As such, hexadecimal floating-point constants have exact representations in binary floating-point, unlike decimal floating-point constants, which in general do not.
Hexadecimal floating-point constants are useful for two reasons: they bypass decimal to floating-point conversions, which are sometimes done incorrectly, and they bypass floating-point to decimal conversions which, even if done correctly, are often limited to a fixed number of decimal digits. In short, their advantage is that they allow for direct control of floating-point variables, letting you read and write their exact contents.
In this article, I’ll show you what hexadecimal floating-point constants look like, and how to use them in C.
# Finding Numbers That Are Palindromic In Multiple Bases
A palindromic number, or number palindrome, is a number like 74347, which is the same written forward and backward.
A number can be palindromic in any base, not just decimal. For example, 101101 is a palindrome in binary. A number can also be palindromic in more than one base, like decimal 719848917, which is 101010111010000000010111010101 in binary and 5272002725 in octal.
An efficient way to find palindromes in a single base is to generate them, iterating through each integer and constructing palindromes from them. An efficient way to find numbers that are palindromic in multiple bases is to take a palindrome in one base and test if it’s a palindrome in one or more additional bases.
In this article, I’ll show you C code I wrote that finds multi-base numeric palindromes. I used this code to generate tables of numbers that are palindromic in decimal and binary, decimal and hexadecimal, and decimal and octal. I also used this code to solve Euler problem 36, which asks for the sum of all numbers, less than one million, that are palindromic in decimal and binary.
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# How do you store a googol in the database (and other very large numbers)?
How do you efficiently store very large and very small numbers from say `10^-100` to `10^100`, so that you can use them to calculate values in a programming language like JavaScript.
JavaScript stores `10^100` as `1e+101`, is there a way to do that in the database? The numbers would not often be that large, but I would like to do calculations with data such as `10^-34 * 2^16` or whatever, so the database should (I think) be storing these as numbers...
How does this work? How do you store numbers of this scale such that you can run computations with them?
By "the database", I'm thinking in general. I am messing around with MongoDB and Neo4j currently.
-
For large numbers, store them as strings. There are javascript libraries for large numbers, such as bigInt for integers. – RobG Jun 5 '12 at 3:02
Databases themselves don't support numbers of arbitrary size in a native numeric format. Your general upper limit on numeric types is usually 8 bytes, which isn't anywhere near a googol.
You'll have to store the number either as a string (least efficient, easiest to work with, can be as precise as needed), as a byte array of arbitrary length (more efficient, harder to work with, still arbitrary precision), or in scientific notation (most efficient, harder to work with, and limited precision).
The first two, unfortunately, do eliminate the possibility of doing any server-side computation, since there wouldn't be a native numeric type that could support the range of valid values. All of the computation would have to be done client-side using a suitable numeric type.
-
Why do you say scientific notation has limited precision? One can certainly store 1.2345LotsOfDigits89E1000SomeBigNumber00042. Not terribly space efficient if you actually need that kind of precision, though generally if you measure something on the order of one googol, you won't care much about the 1's place. – Eric J. Jun 5 '12 at 3:07
Because it's true. Scientific notation itself can obviously store numbers to an arbitrary precision, but the point of it is to provide approximations of large numbers using smaller numbers. It's lossy compression. Size, precision, domain; you get to pick two. It may be true that the OP doesn't need that level of precision, in which case scientific notation might certainly be suitable. I don't know, which is why I stated that in the answer. – Adam Robinson Jun 5 '12 at 3:08
Are you talking about some specific machine representation of scientific notation, or the notation in general? The notation in general allows for a number between 1 and 10 of arbitrary precision, plus an arbitrary exponent. – Eric J. Jun 5 '12 at 3:10
@EricJ.: The machine representation, since that's the relevant point. If the database can't store arbitrarily large numbers, then scientific notation is no exception. – Adam Robinson Jun 5 '12 at 3:11
Does MongoDB or Neo4j have a native "scientific notation" format? – Eric J. Jun 5 '12 at 3:12
You could use the double type.
The MySQL DOUBLE[(M,D)]
A normal-size (double-precision) floating-point number. Permissible values are -1.7976931348623157E+308 to -2.2250738585072014E-308, 0, and 2.2250738585072014E-308 to 1.7976931348623157E+308. These are the theoretical limits, based on the IEEE standard. The actual range might be slightly smaller depending on your hardware or operating system.
-
If I were you, I'd separate the numerical value from the exponent. I personally don't have experience with MongoDB or Neo4j, but in MySQL (I'm sure they have similar terms) I'd create a table with an VARCHAR (text) column with whatever precision you'd like in your program (or how many unique numbers), and another VARCHAR column with length 3 (for max exponent 999). You can tinker with the values as you see fit, but that's all I can think of. If you want more flexible size values, I'd store the numbers on the server's file system using PHP rather than use databases.
-
while this seems like a great approach, you will basically need to re-implement the whole floating point standard again. And I would trust that whoever coded the underlying programming language/arithmetic libraries are a lot better than you or me at doing it. – Toote Jun 5 '12 at 3:17
Yeah... good point. It probably requires a lot more work than I made it sound like, now that I think about it. – John Davis Jun 5 '12 at 3:23
a lot more work is probably the understatement of the year. There is a standard for customizable-precision decimal arithmetics (speleotrove.com/decimal) and it is not simple nor easy – Toote Jun 5 '12 at 4:51
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# Universal gravitation exercise
1. Dec 23, 2014
### ezio1400
If a bullet is fired vertically from the surface of the Earth with initial velocity v = 10 km / s, ignoring air resistance, at which distance h from the center of the Earth would arrive? (The radius of the Earth is RT = 6360 km, and the mass of the Earth MT = 5.98x10^24 kg)
I used the formula for escape velocity putting the speed of the bullet instead of the escape velocity but I do not think is correct.
2. Dec 23, 2014
### phinds
You have to show your work in order for anyone to comment on it.
3. Dec 23, 2014
### ezio1400
ok.
escape velocity= ((2*G*MT)/RT)1/2⇒d=(2*G*MT)/v2=2*6,67*10-11*5,98*1024/(10*103)2=7977320m
h=7977320+6360*103=14337320m→14337km
4. Dec 23, 2014
### ehild
You are close to the right track, but why did you use the formula for the escape velocity? It gives the velocity when the kinetic energy is equal to the gravitational potential energy on the Earth surface.
In the problem, the bullet is shot upward with 10 km/s speed from the surface of the Earth, where it has some gravitational potential energy. The total energy KE+PE is conserved, how far is the bullet when its kinetic energy becomes zero?
Last edited: Dec 23, 2014
5. Dec 23, 2014
### ezio1400
Ok but in my book this exercise is in a chapter that precedes the chapter on energy. Maybe that's why I can not solve it. This is strange.
6. Dec 23, 2014
### ehild
But you have learnt about the escape velocity - how was it explained to you without energy?
7. Dec 24, 2014
### ezio1400
The formula for escape velocity was given to me without proof .
8. Dec 24, 2014
Strange.
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0
# Why is 387 not a prime number?
Updated: 10/17/2022
Wiki User
9y ago
It has more than two factors.
Wiki User
9y ago
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Q: Why is 387 not a prime number?
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Related questions
### Is 387 a prime number?
387 cannot be prime as it divides by 3. A prime number has only 2 factors which are 1 and itself. Composite numbers are every other positive integer except 1 and 0. 1 and 0 are neither prime, nor composite.
### Why isn't 387 a prime number?
Because it has more than two factors.
### What is prime factorization of 387?
As a product of its prime factors: 3*3*43 = 387
### What are the factors and prime factors of 387?
The factors of 387 are: 1 3 9 43 129 387 The prime factors are: 3 and 43
### What is the prime factorization for 387?
None. 337 is a prime number.
387 = 3 x 3 x 43
### Is 4257 a prime number?
No, its factors are: 1, 3, 9, 11, 33, 43, 99, 129, 387, 473, 1419, 4257.
### How do you factorize 10?
break it into prime number multiples,10 = 2 * 5 (both prime numbers)example: factorize 387(387) = 9 * 43 (both primes , end of excercise)factorize 390:(390) = 2 * (195) = 3 * (65) = 5 * 13 (both primes , end of excercise)
### What is 23 percent of 387?
To find 23 percent of a number, multiply the number by 0.23. In this instance, 0.23 x 387 = 89.01. Therefore, 23 percent of 387 is equal to 89.01.
No, it isn't.
All composite.
### What is the largest prime that if multiplied by 9 is less than 400?
It is: 9*43 = 387
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# How much per month is a 500k mortgage?
## How much per month is a 500k mortgage?
Monthly payments on a \$500,000 mortgage At a 4% fixed interest rate, your monthly mortgage payment on a 30-year mortgage might total \$2,387.08 a month, while a 15-year might cost \$3,698.44 a month.
28%
The 28% rule states that you should spend 28% or less of your monthly gross income on your mortgage payment (e.g. principal, interest, taxes and insurance). To determine how much you can afford using this rule, multiply your monthly gross income by 28%.
#### What salary is needed for a 500k house?
The Income Needed To Qualify for A \$500k Mortgage A good rule of thumb is that the maximum cost of your house should be no more than 2.5 to 3 times your total annual income. This means that if you wanted to purchase a \$500K home or qualify for a \$500K mortgage, your minimum salary should fall between \$165K and \$200K.
How much income do I need for a 200K mortgage?
What income is required for a 200k mortgage? To be approved for a \$200,000 mortgage with a minimum down payment of 3.5 percent, you will need an approximate income of \$62,000 annually. (This is an estimated example.)
How much house can I afford 100k salary?
This was the basic rule of thumb for many years. Simply take your gross income and multiply it by 2.5 or 3 to get the maximum value of the home you can afford. For somebody making \$100,000 a year, the maximum purchase price on a new home should be somewhere between \$250,000 and \$300,000.
## How much is 5300 a month hourly?
If you make \$3,300 per month, your hourly salary would be \$20.31.
### How much do I need to make to buy a \$400 K house?
What income is required for a 400k mortgage? To afford a \$400,000 house, borrowers need \$55,600 in cash to put 10 percent down. With a 30-year mortgage, your monthly income should be at least \$8200 and your monthly payments on existing debt should not exceed \$981. (This is an estimated example.)
#### Can I buy a house making 43k a year?
Your gross annual income is \$100,000. Multiply \$100,000 by 43% to get \$43,000 in annual income. Divide \$43,000 by 12 months to convert the annual 43% limit into a monthly upper limit of \$3,583. All your monthly bills including your potential mortgage can’t go above \$3,583 per month.
Is it harder to get a mortgage as a single person?
Yes. Getting a mortgage as a single person is treated no differently by lenders, and is actually more common than you might think. Many first-time buyers decide to purchase their first property alone.
How do you calculate the monthly payment on a mortgage?
– You can calculate a monthly mortgage payment by hand, but it’s easier to use an online calculator. – You’ll need to know your principal mortgage amount, annual or monthly interest rate, and loan term. – Consider homeowners insurance, property taxes, and private mortgage insurance as well. – Click here to compare offers from refinance lenders »
## How do you estimate your mortgage payment?
– Comparing the monthly payment for several different home loans – Figuring how much you pay in interest monthly, and over the life of the loan – Tallying how much you actually pay off over the life of the loan versus the principal borrowed to see how much you actually paid extra
### How do you calculate a 30 year mortgage?
You can also use Credible’s mortgage calculator to estimate your monthly mortgage payments the average interest rate for a 30-year fixed-rate mortgage for 2019 was 3.94%. The average rate for 2021 was 2.96%, the lowest annual average in 30 years.
#### How much money can I Borrow?
– The amount of money you want to borrow – How much deposit you have – Your employment status and job security – Your income – and lenders may view things like overtime, commission and bonuses differently from basic salary as they’re not guaranteed – Your outgoings – the money you spend on bills and on your lifestyle – Any existing debts – Your credit report
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# Who discovered electron spin?
## Who discovered electron spin?
How two young Dutchmen, one with only o master’s degree, the other a graduate student, made a most important finding in theoretical atomic physics.
Who discovered principal quantum number?
Principal quantum number (n) Principal quantum number was proposed by Bohr to explain the hydrogen atomic spectrum. 2.
### What is the importance of electron spin?
The electron spin magnetic moment is important in the spin-orbit interaction which splits atomic energy levels and gives rise to fine structure in the spectra of atoms. The electron spin magnetic moment is also a factor in the interaction of atoms with external magnetic fields (Zeeman effect).
What is the golden rule of energy?
In quantum physics, Fermi’s golden rule is a formula that describes the transition rate (the probability of a transition per unit time) from one energy eigenstate of a quantum system to a group of energy eigenstates in a continuum, as a result of a weak perturbation.
#### Who proposed by spin quantum number?
The name comes from a physical spinning of the electron about an axis, as proposed by Uhlenbeck and Goudsmit. The value of ms is the component of spin angular momentum parallel to a given direction (the z–axis), which can be either +1/2 or –1/2 (in units of the reduced Planck constant).
What is mean by spin in quantum mechanics?
Spin is an intrinsic form of angular momentum carried by elementary particles, and thus by composite particles (hadrons) and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum.
## Who discovered quantum numbers?
Restrictions. Pauli Exclusion Principle: In 1926, Wolfgang Pauli discovered that a set of quantum numbers is specific to a certain electron. That is, no two electrons can have the same values for n, l, ml, and ms.
What is Fermi golden rule equation?
### What is Fermi golden rule PDF?
Fermi’s Golden Rule provides the rate at which atomic or electronic transitions. take place between two states. It applies to a wide range of optical and electronic. processes for which the initial and final states can be described by wave functions.
What is the significance of spin quantum number?
The spin quantum number is the fourth quantum number, denoted by s or ms. The spin quantum number indicates the orientation of the intrinsic angular momentum of an electron in an atom. It describes the quantum state of an electron, including its energy, orbital shape, and orbital orientation.
#### What does spin quantum number tell us?
The Spin Quantum Number (ms) describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -).
Can electrons change spin?
Instead we have learned simply to accept the observed fact that the electron is deflected by magnetic fields. If one insists on the image of a spinning object, then real paradoxes arise; unlike a tossed softball, for instance, the spin of an electron never changes, and it has only two possible orientations.
## What are the importance of Fermi golden rule?
Fermi’s golden rule provides one way to calculate the transition rate between two certain quantum mechanically defined states. Due to its generality, it has various applications in the field of atomic, nuclear, and solid-state physics.
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Applied Mathematics
Vol.5 No.16(2014), Article ID:49432,17 pages DOI:10.4236/am.2014.516243
On Universal Space and Time
Caesar P. Viazminsky1,2, Piere K. Vizminiska3
1Department of Physics, Illinois Institute of Technology, Chicago, USA
2Department of Physics, University of Aleppo, Aleppo, Syria
3Department of Computer Engineering, University of Detroit Mercy, Detroit, USA
Email: kaysarv2@gmail.com
Received 6 July 2014; revised 8 August 2014; accepted 19 August 2014
Abstract
In earlier papers [1] -[4] , it was shown that the consistency of the concept of time with motion requires time and distance to be of the same dimension, and thus measured by the same unit. The arising reduced system of units revealed that mass and energy were only different facets of one entity, and resulted in the well-known mass-energy equivalence formula as a natural consequence. The physical space can be identified with any inertial frame, but when it comes to comparing the results of measurements in two frames, or more, only one frame, say S, can be taken stationary and identified with the physical space, whereas all other inertial frames are moving relative to S. The equivalence of inertial frames as sites of one physical world implies that an intrinsic units system of length, time, mass and charge should be defined in terms of basic constituent physical blocks that have the same identity in all inertial frames. A basic feature of the universal space and time theory (UST) is that the same one time prevails in all inertial frames. The scaling transformations (STs) that relate the geometric distances in two frames, S (s) when chosen the stationary frame, are derived, and applied to explain the Doppler’s effect. The time distance between a moving object in S and an observer depends on its state of motion; and the Euclidean form of the STs is employed to explain arrival of some meta-stable at the earth’s surface despite its short lifetime. The quantitative predicted Doppler’s effect, which is in a striking agreement with the Ives-Stilwell experimental results, coincides with the relativistic prediction for longitudinal motion, but yet predicts a complete absence of a transverse effect at a right angle. In coming parts of this work it will be shown that the UST explains elaborately the drag effect, stellar aberration, and produces naturally the relativistic mechanics. The UST will also be completed through deriving the scaling transformations of the second type, by which the null results of Michelson and Morley experiment, Michelson and Gale experiment, and the Sagnac effect are explained. The current work and our intended future works in UST are new versions containing basic conceptions and visions that didn’t appear in earlier versions [1] -[6] .
Keywords:Reduced System of Units, Intrinsic Units, Universal Time and Distance, Mass-Energy Equivalence
1. Introduction
The UST modifies the Newtonian conceptions of space and time to incorporate observations through light’s signals. While the absolute nature of time is retained, a time distance between an object and an observer that depends on their relative velocity and geometric distance is determined by means of what we call the scaling transformations (ST). Our novel conception of the space and time is characterized by the following:
1) Time durations that are consistent with motion require the former to be measured, in each frame, by spatial displacements intrinsic to light propagation and proportional to it. This defines in addition to the frame-independent (or universal) time durations a constant velocity c of light inside each frame.
2) One arbitrary inertial frame S, to be called universal, is considered stationary and indentified by the physical or absolute space, while all other inertial frames are moving relative to S. The physical space may be corresponded with any other inertial frame s, but this should have no bearing on the transformations that relate geometric distances in S and s regardless of which frame we choose to be stationary.
It follows from 1) that time and geometric length have the same dimension and can be essentially measured by the same unit. The reduced mechanical system of units (RSU) in S contains in addition to the unit of mass a common unit LS (or TS) for time and length, leading naturally to the well-known mass-energy equivalence formula. Being proportional to its geometric length, the time duration of a light trip within S is simply another equivalent measure of its length.
From 2), which allows for identification of the physical space with any inertial frame, several consequences follow:
• All inertial frames (IFs) are equivalent sites for observing one physical world, and consequently the physical laws are the same in all IFs.
• The units of RSU should have identical status in all IFs. This can be fulfilled through constructing them in each frame from the same basic constituents blocks, leading thus to intrinsic or natural units. For instance, the constituent block of length in an inertial frame S is the wavelengths of the stationary emission of a specific spectral line of some chemical element, and the unit of distance in S is a certain multiple of, i.e. The constituent block of length differs from a frame to another, and so does the unit of length in the corresponding frame.
• The absolute or physical space in its Newtonian sense as the standard of rest has to be abandoned, but its role as a unique standard for rotations is retained. Moreover, the role of the absolute space in its Maxwellian sense as the carrier of electromagnetic radiation is also retained.
• A light trip follows a universal path that is independent of the frame by which the physical space is identified. However, when observed from the moving frame, the light trip’s path is seen tilted from its path in the stationary frame by the “aberration angle”. This fact will be discussed in the third part of this work.
Based on requirements 1) and 2), and on the implicit assumption regarding the Euclidean nature of the absolute space we determine the relation between the geometric length (whether geometric distance or time) and time length (universal duration) of a light trip in a frame S. Equivalently, we determine the scaling transformations STI between the geometric lengths of a light trip in the frames s, in which the light’s source b is stationary, and S in which b is moving. Explicitly, if the source of light is moving at a constant speed in S, then the STI determines the universal duration t of a light’s trip in terms of its geometric length Thus t is a function of the position vector of B and the velocity vector of b in S. It is t what is measured for the duration of the latter trip in S and in any other inertial frame. We shall refer to the universal time t as the time length, or time duration, of the latter light’s trip, or the time distance of the moving object b from the observer. We shall see that the time length t is generally different from the geometric length of the trip in S; they are equal however when the source of light is at rest in S.
As equivalent sites of one physical world, S and s should quantify, employing their intrinsic units of length, the geometric distance identically. This yields the intrinsic units of length and time in the two frames mapped on each other by the STI.
The STI is employed to explain the meta-stable particles’ lifetime phenomenon and the Doppler’s effect. It will be shown that while the relativistic longitudinal Doppler’s effect is identical to the prediction of the UST, the two theories are qualitatively distinct when it comes to the transverse Doppler effect. In a second part of this work we shall discuss the universal mechanics, which turns to be almost identical to relativistic mechanics. The direction of a light trip in the moving frame in terms of its direction in the stationary (or universal) one is determined in the third part of this work.
In coming works a second type of scaling transformation (STII) will be derived and applied to explain the drag effect, Sagnac effect, Michelson and Morley experiment, Michelson and Gale experiment.
2. Time Evaluation by Spatial Displacements—Global Time@NolistTemp# Consider an arbitrary inertial frame The coordinates system is assumed to be already calibrated using a given unit of length, say LS. The existence of the Cartesian system of coordinates OXYZ in S presupposes the geometry of the space is Euclidean [7] . The geometric distance between two points and or geometric length of a rod AB stationary in S, refers to the result obtained by laying the unit of length LS in S along the rod repeatedly from one end till reaching the other by multiples and fractions of LS. This can be done rationally (or assumed to be already done) and takes no time. In other words, a coordinate system, and accordingly geometric distances in S are already given data. If is the geometric distance of the point B from O, then the dimensionless quantity is the radial coordinate of B. The unit of time, though arbitrary, is chosen as the duration, say “second”, between two consecutive ticks (or readings) of identical clocks that run at synchrony with each other. Contemplating in the latter statement, we may be astounded by the fact that we really have defined nothing concerns the world outside the clocks. In order that a unit of time, say a second, bears a meaning as far as motion in S is concerned, it should be correlated to what can happen during “a second” in the world outside the clocks, and more precisely, it should quantify the amount of the spatial displacement intrinsic to some reference physical phenomenon, such as the propagation of light from an arbitrary point in S. A “second” must thus be corresponded with (and actually could be measured by) a certain distance traveled by light within (inside) S during a corresponding period. Time measurements therefore must be reducible to specific types of spatial displacement’s measurements.
The frame independent entity in the UST, namely “the time”, is defined in terms of spatial displacements and can be measured by length units, which implies that velocity is a dimensionless quantity. Indeed, global timing in an inertial frame S is set up by synchronization with an arbitrary observer employing light’s signals. The concept of time emerges through envisaging a “linear” correspondence between each instant of time T read by the timer O and the compound event: (the wave front of the pulse that was emitted from O at occupies at T, points at equal distances R from O). Through this correspondence, time duration is essentially measured by distance R, i.e.
. (2.1)
The proportionality constant defines a constant velocity c of light by Because of the space is geometrically homogeneous (also physically, and in particular for the propagation of light), time durations defined in this way are independent of the master timer’s position O. The homogeneity of time follows also from the homogeneity of space. The above correspondence is only one step short of synchronization, which is achieved by each S observer at taking note of his distance from O and the instant of time at which the pulse emanated from O, and thus setting his timer at when he receives the pulse. By its way of construction, the global timing is unique up of course to an arbitrary choice of a time unit and of zeroing, i.e., up to a transformation of the form where and are arbitrary numbers. In UST the global time in S prevails in any other inertial frame s, and timers therefore, have to reflect the facts stated in setting up the global time. Thus every moving timer, if needed, must read the same instant of time read by the S-clock that is contiguous to it. This means, timers which are instruments whose readings is under our disposal, can be manipulated if needed, to conform with light propagation criterion.
The 1-1 correspondence expressed by the relation, is additive and homogeneous only for i.e., for The mapping is not linear in the strict mathematical sense because it is defined only for Time intervals in the past may be quantified by which should be understood as follows: Light that is received by O at has already taken a duration where R is the distance of its source from O. Thus the instant of time in the past at which light had emanated from a source at a distance R from O and arrived at O at was
(2.2)
The further the location of the event the deeper in the past it happened. All past events that satisfy the latter relation are detected by the observer O at the same instant Out of the latter set, subsets which have the same R were simultaneous. If each event of the set that satisfy (2.2) continues to happen then the same set of events will all be detected at any later instant by O. Simultaneous events, corresponding to an instant T, are detected by O at instants, that depend on R. The latter relation is true whether T refers to an instant in past, present, or future. Two events taking place at and and detectable by O at and are simultaneous if
By the latter paragraph, it is justified to imagine a system of synchronized timing that supplements any system of coordinates in S. This means that, in the same way we envisage rationally the assignment of a triplet to each point B in S, we can also imagine that a timer can be placed at each point which is synchronized with and runs uniformly at the same rate as the master timer, and accordingly at synchrony with all other timers. Thus a global timing in S can be practically established, with the notion of an “instant” has a global meaning in S, in the sense that if an event takes place at at then it will be detected at through a light signal emanating from and arriving at B at the instant Thus every S observer B assigns to the event of light’s emission the same instant, where r is his spatial separation from and T is the time read at the clock B when light is received. The concept of time arrow (past, present, and future) has therefore a global meaning in S, and any two or more S observers have the same temporal ordering of all events monitored by them. In particular, the notions of simultaneity and non-simultaneity are well-defined global concepts in S. An inertial frame S endowed with a global time as described above is called a timed (or synchronous) inertial frame. The UST theory maintains the global concepts of simultaneity and accordingly time arrow, as frame independent, or absolute.
As far as one observer O is concerned, no synchronization in the real sense is to be done in order an inertial frame becomes timed. In fact, the existence of a coordinate system as well as a timer at O is sufficient to determine the duration of any light’s trip with given ends, whether O was one of the ends or not, provided the spatial coordinates (or radial distance) of each event is messaged instantly to O by a light signal. Also, the duration of any event at any point in S, say is observed form O to have the same duration. The arbitrary observer O, if he wishes, can replace his timer by a universal one [6] , since his own timer should be equivalent to the universal one. The latter legitimate replacement is applicable to any observer, and hence a universal timer will do for all observers in a coordinated frame S with finite extension.
In Newton’s mechanics, time and geometric length are independent absolute entities [6] [8] , where distance is measured by a calibrated ruler, and Newton’s global time was assumed to be readable at each point of space. The synchrony of all point-wise timers was partially circumvented through appealing to a universal timer [6] formed by the fixed stars in the firmament, which generalizes the approximately uniform global time set up in the region from which almost all our observations are conducted, namely the earth surface.
Remarks
1) The common perception of a “second” as the duration between two consecutive ticks of a clock becomes identical to our previous conception only if we demand that light travels rectilinearly c meters between each two consecutive ticks. The latter definition conditions the clocks to conform to the reference criterion, which is light propagation. We have thus on one hand the measuring instruments, “the clocks” that can be synchronized and distributed (hypothetically) everywhere in the space, and on the other hand the happenings in the outside world, namely the distance that light travels during the period read by the clocks. Note that the clocks which are indeed indispensable for measuring time, are no more than instruments that can be manipulated to conform to the reference criterion, and their performance should certainly be rejected when they do not. Although only clocks can be used to measure time at one point, time read at various locations should be in accord with light propagation, and so should be each clock.
2) Employing synchronized clocks and a light signal, distance between two points in a timed inertial frame (called optical distance) can be measured and it is certainly equal to the geometric distance as measured by meters or seconds (=c meters), where no light signals and clocks are involved. We shall see however that this equality will be violated when the distance of a moving body from a stationary observer is considered.
3. The Reduced System of Units
Since time durations have to be defined in terms of spatial displacements, geometric lengths and time durations must have the same dimension; both are measured by the same unit. If the unit of time TS in S is defined as the duration taken by light to cross the unit of distance LS (a given rod stationary in S) from one end to another, say, we may designate the unit of time also by “meter”, to mean the time required by a light’s signal to cross this distance.
In terms of a system of units of time, length, and mass
, (3.1)
the dimensions of some mechanical observables are listed in the table:
,
,
The velocity in RSUI is a dimensionless 3-vector, and the speed of light in vacuum is 1 regardless of the chosen unit of length, provided we choose. Mass and energy have the same unit, “kilogram”. In practical applications however, it is convenient to take and adopt a multiple of the unit namely, “”. The latter is defined by the period taken by light to travel a distance of so that, In the reduced system of units
(3.2)
The reduced systems of units, I or II, suggest that observables which are measurable by the same unit are of the same nature, although they may manifest themselves in different facets. Mass and energy for instance are both scalar quantities and both are measurable in RSUI by kg. This means that 1 kg of mass is equal to 1 kg of energy, and that, under suitable circumstances either quantity may be transformed to the other. In the RSUII,
and
The latter relation holds for any type of energy.
If a vector observable and a scalar observable B are of the same dimension then their squares, and are of the same nature and in principle are transformable to each other.
It is noted that the reduced system of units I and II are the system MKS with the unit of time is taken as the unit of length itself in RSUI, and defined in terms of the unit of length, with for RSUII. Symbolically,
and
It is clear that time distance or just distance refer essentially to the same thing but using a different unit, which is “second” for the former and “meter” for the latter. The geometric distance between a source of light b that is moving in the timed inertial frame S and an observer at an instant, is the distance between the position occupied by b at and the observer O; it is If b emits when at B a pulse of light that arrives after t seconds, then its time distance (or optical distance) from O , at the emission instant, is It is clear that the optical distance is identical to the geometric distance in the frame in which the source of light is stationary, otherwise (as we shall see) it is different in general.
4. Pullbacks of Inertial Frames
In this section we adhere to the Newtonian absolute length and time, but we extend the traditional conception of inertial frames (IFs) beyond being triads of orthogonal axes translating uniformly with respect to one given inertial frame, or say, with respect to Newton’s absolute space.
A collection of objects all moving at the same velocity u relative to an IF can serve as an IF when other objects’ motions are specified relative to s. This means that s employs 1) the system of intrinsic (or natural) units to measure physical observables and 2) a system of rectangular Cartesian coordinates oxyz to specify positions. Velocity of an object relative to s can always be measured locally as the ratio between the distance ds it travels in s and the corresponding duration dt. Indeed, time and distance in any inertial frame exist and are measurable with no need to endow the frame with a coordinate system. In Newtonian mechanics the set of all IFs partitions the physical world to classes of equivalence with two objects are equivalent if they have the same velocity relative to S, and thus are stationary in some IF.
Let be the position, velocity and acceleration vectors of an object b in S (in s) respectively. In the traditional definition of an IF, the origin o of s is fixed once and for all to coincide with the origin at some instant say The position and velocity vectors of the object b are related by the Galilean transformations, while the acceleration is the same in both frames [8] -[10] .
Pullbacks of an IF: Here the origin of s need not be fixed, but is chosen the s-point (or observer) which at each instant of time, starting from is contiguous to Along the latter choice, and if and are the position vectors of an s-object relative to s and S respectively, then at each instant of time we have
(4.1)
and
(4.2)
The frame s we have described will be denoted by and called the s frame contiguous to the stationary frame S at O, or the pullback of s to or briefly, the moving frame s. Whenever we talk of a moving inertial frame we shall mean its pullback to a given point in the stationary fame S, and usually, The derivative of the position vector of an s-object in both frames,
(4.3)
is the velocity of s in S, and it is naturally the velocity of an s-object relative to S. Note that is not the velocity of an s-object in s; for, the latter is zero. If an object b is moving in s at velocity in s then
(4.4)
The velocity is measured locally in s in the sense mentioned earlier. The velocity of b in S is
(4.5)
while its velocity in s is
(4.6)
In particular the velocity of an s-object in s is The acceleration, which is the same in both frames, is still the second derivative of the position vector, or
Parallel relations hold between s, when chosen stationary, and the pull back of the moving frame S to
A visual picture of a pullback frame is the following: A train s passes by a train station S with a specific pole on the pavement is taken as the origin of S. It is decided to describe the motion in s with respect to the compartment that is currently contiguous to the pole. At each instant of time there is a different compartment passing by the pole, and we may assume the compartments are as tiny as we please. The velocity of an object b in s can be measured relative to any compartment and in particular in the compartment in which b is moving. Although the position vector is the same in both frames, the velocity in is not simply the time derivative of the position vector; but rather, the latter minus the velocity of s relative to S.
Along the spirit of the Newtonian rational mechanics and its associated concepts of space and time, the above picture ignores, in its Galilean and pullback forms, observations through light signals. However, and as it is shown in the next sections, light signals are inescapable for two basic reasons: 1) they enter in the very definition of the units of distance and time, and 2) all observations involve light signals.
5. Intrinsic Units
Physical entities have intrinsic characteristics that are independent of the IF at which it is realized. These characteristics are manifested in conception, specification, and quantification through measurements in the chosen IF employing given units. Examples of such entities are 1) The characteristics of the stationary emission of electromagnetic radiation of every element or compound. This includes, a priori, the wavelength of each spectral line.
2) The stationary mass of an atom at its ground state.
3) The rate of propagation of light with a particular wavelength. Since this rate is proven to be independent of the wavelength in one IF, it is also so in all IFs.
4) The elementary charge e.
A constituent basic block (or a block unit) of a physical observable is a specified entity by which any given amount of the physical observable can be numerically quantified in a unique manner. We seek here to define intrinsic units of length, time, mass, and charge such that they have the same identity in all IFs. For lucidity we temporarily restrict structuring the intrinsic units to an arbitrary inertial frame S, and thus we omit the index S which is supposed to distinguish these units.
The first step is to determine the constituent basic blocks for such units:
1) The block unit of the geometric distance, or length, is the wavelength of a specific spectral line of the stationary emission of some chemical element chosen and adhered to in all frames. We denote this entity by
2) Due to the fact that a unit of distance can be used to measure time we may adopt as the block unit of time in S. Or we may rely on clocks to measure time, and thus define the constituent block of time in S by. One can start with the block unit of time (employing valid clocks) and define the constituent block of length by.
3) The constituent block of mass is the mass of a stationary atom at its ground state.
4) The constituent block of charge is the fundamental charge e.
The constituent blocks, which are all in the microcosm, are not practical for direct use. The second step thus is to choose prudently a fixed multiple of the constituent block to define the corresponding unit. Digital quantification of such units by the same number of basic constituent blocks provides intrinsic units with the same meaning and value within every IF.
1). The unit of geometric length or distance, the meter (m), is the length of a certain number of the specific wavelength of the specific spectral line of the chosen chemical element; it is given in S by.
2) To define the unit of time, the second (s), we resort to the fact that the duration T of the propagation of a light signal is proportional to the distance R it travels, and define We may write the latter relation as, where If we choose the second as the unit of length and time then and
3) The unit of charge, the Coulomb (C), is the charge of a certain number of the elementary charge e,
.
4) The unit of mass, the kilogram (kg), is a certain number of the constituent basic block of mass,
where is the mass of the atom at its ground state.
Owing to their definitions, and to the equivalence of inertial frames as sites for describing one physical world, all basic physical blocks in one frame, as well as the intrinsic units derived from them, have the same privilege enjoyed by their corresponding basic blocks in another frame. We shall see however that if s is an IF moving with respect to the IF, S, then units of length, time, and mass in s are not units in S. Explicitly and if we label these units by the frame in which they are set up then the result of measuring these quantities in one frame, say S, will yield in general, and Equivalently,. and.
Employing intrinsic units in all IFs, which are equivalent sites for observing one physical world, quantifies physical facts identically in all IFs. The results of any experiment in an IF would be identical to it in any other IF frame provided both experiments are carried out in the same conditions. Moreover, any relation predicting the outcomes of an experiment in one frame, named a law, equally holds in any other inertial frame, and it is valid in one frame to the same degree of its validity in another. It follows that the transformation from an inertial frame to another should map intrinsic units in one frame onto intrinsic units in another. This will result in particular in universal (frame independent) time distances, and in invariant geometric distances, in the sense that, in each frame the geometric length of a light trip is read the same using however the frame’s intrinsic units of length.
6. The Universal Space
Consider the set of all inertial frames. It is clear what is meant by a frame being at motion with respect to another, but the concept of a frame being at rest needs elaboration. To be meaningful, the latter concept requires the existence of an independent entity with respect to which the state of being at rest is referred. This entity corresponds to the space when referred to a frame set up by a force-free observer (i.e. far from matter, or instead at the center of mass of a closed system of objects) that is not rotating relative to the fixed stars. Any given frame S defined by the latter statement is an inertial frame that can be identified by Newton’s absolute space and considered stationary, while the motion of all other inertial frames, or equivalently, the motion of all material objects, are then referred to this “universal frame”. By its definition, there exists only one universal frame at a time.
Since the state of being stationary through identification with the frame of fixed stars can be assigned to any inertial frame (one at a time), the absolute or physical space in its Newtonian sense as the standard of rest has to be abandoned. However, the role of the absolute space in Newtonian sense as a unique standard of rotations, as well as its role in Maxwellian sense as the carrier of electromagnetic radiation, are retained as it will be explained in the sequel.
The spatial site of all material objects, as far as their positions are concerned, enjoys an identity that is independent of the inertial frame by which it is identified (equation (4.1)); it will be called the universal space (We emphasize that we employ in this theory the pullbacks of IFs). The latter fact is manifested in the absoluteness (frame-independence) of the light’s path between any two points in the space. However, the state of motion of material objects depends of course of the choice of the stationary, or universal, frame.
Each member of a set of IFs that have no common measurements can be considered stationary. However, when we compare the results of the same experiment (observation) in different IFs, only one frame can be considered stationary while the other frames are moving.
Absolute Light Trips—Aether’s Role of the Universal Space
If a source of light b has an arbitrary vector velocity relative to the inertial frame, we may choose without loss of generality the velocity vector in the direction of the X-axis, for we may always rotate the S-axes so that the X-axis is in the direction of where is the unit vector of the X-axis. Let be an inertial frame whose axes are parallel to those of S and moving with respect to S at a constant velocity so that the source b is stationary in s. We endow the frames S and s with systems of spherical coordinates and respectively, with is the azimuth angle between the X-axis (x-axis) and the radius vector The latitude angles and will be suppressed because of the axial symmetry of the motion about the X-axis (figure 1).
Assume that the source of light b which is stationary at the point in s emits, when at in S, a spherical pulse of light. When light arrives at O, it arrives also to an s-observer whom we choose an origin o of with axes parallel to those of S. Two S and s observers who are contiguous when hit by the pulse are called conjugate observers. Similarly, two sources, each emitting a pulse of light when contiguous, are called conjugate sources.
The situation we have displayed has the following features:
Figure 1. The path of light trip in the universal space whether identified by S or by s.
1) In a given inertial frame S, a source of light b is moving at a velocity. Or equivalently, in the inertial frame s in which the source is stationary, an inertial observer O attached to a frame S, is moving at velocity
2) Light is emitted from b when at.
3) On arriving at light arrives at the conjugate observer. While O is already given, o emerges at the instant light arrives at O; it is the s-observer that is contiguous to when the pulse hits O (and hence o). But we may equally imagine that when light arrives at it also arrives at and thus is already known while is fixed when light arrives at o.
The frame S can be considered at rest throughout the light’s trip which starts from (B when occupied by the source b) and ends up at (O and o), while b is moving at velocity in S. Also the s frame can be claimed the stationary frame during the duration of the trip which starts from the point (b when is at B) and ends up at (o and O), while S is moving at velocity. Since each frame is entitled to claim itself stationary, and thus identifiable with the universal space, while the other is moving, all observers (the S and s observers) accept that light emanated “at the same time” from one and the “same point” in the universal space, whether identified by S or s, and ended at the same time at the same point.
To elaborate, the phrase (when at) defines in each frame a pair composed of a location and an instant of time, or what we shall call a universal point, and denote by (at). Thus a true source of light gives rise to a universal point, which is a frame independent entity that embodies the same instant of time in both frames together with an Sand s-locations that are coincident in the universal space at the instant of contiguity of b and B in a sense which we shall explain (we have already seen that). Similarly the end point of the pulse in one frame determines a conjugate end in the other frame, and accordingly, another universal point. It follows that all observers concede to the fact that there is one and the same trajectory in the universal space associated with a given light’s trip, which starts from (at) and ends at (and). In other words, the pulse traces a universal straight path connecting the universal points (b at B) and (O and o), whether S or s is considered stationary and identified by the universal space.
Direct consequences of the latter statements are the following:
1) If are the directional angles of the path in S when considered stationary and are its directional angles in s when s is considered stationary, then
(6.1)
The velocity of the source does not appear in the latter relations. This implies that, had source b was replaced by a source with another velocity then
2) The latter result is contained in a more general consequence: The light’s trip possesses universal characters concerning its optical length and direction. This means that associated with a given trip, whose beginning and end are known, there are one time duration t (optical length ct) and one universal direction in the universal space which can be identified by S or s. The time distance depends on the state of motion of b (i.e. velocity of s relative to S), and hence it is a function of the position vector of B and the velocity vector of b in S.
In a future work we shall determine the direction of the light trip inside the moving frame in terms of its direction in the stationary one.
3) If the universal space is identified with the aether then every inertial frame may be pulled back to coincide with aether, and the Stokes hypothesis of complete drag of the aether by inertial frames [11] is altered to the pullback of the inertial frame so that the moving frame can be considered stationary and identified with the universal space. The aether however, is redundant in the formulation of UST.
7. The Anisotropic Scaling Transformations
Let b be a source of light moving in an inertial frame at a constant velocity with the X-axis of S is taken along Let s be an inertial frame which is moving relative to S at a constant velocity and hence the light’s source b is stationary in s. Now, we set out to determine the transformations which allows for each frame, S or s, to be considered stationary while the other is moving.
Assume that when at the source b emits a pulse of light. On arriving at the point (or observer), the pulse arrives also at its s-conjugate point (or observer) which is contiguous to at the moment the pulse hits O (or when the pulse arrives at it also arrives at its S-conjugate). We choose now the axes of s such that are in standard configuration with. Each of the conjugate observers O and o is entitled to consider his frame stationary relative to the fixed stars and thus identifiable with the universal space while the other frame is moving relative to his own frame. Each observer, O and o, assigns to the pulse’s path the same beginning and the same end. In other words the pulse follows a universal path connecting the universal points and, and the direction of the path is the same when looked at from the stationary frame whether it was S or s. In each frame, when considered stationary, the pulse propagates along a direction determined by a unit vector, where
(7.1)
If S is chosen as the stationary frame, then the geometric length of the trip in S is employed to induce corresponding length and duration in s such that r and t have the status of geometric length and time in s when s is the stationary frame, in the same way R and T have in S.
By the Galilean law of velocity addition, the velocity of the pulse in S is the vector sum of its velocity in s and the velocity of its emitter. However the pulse emanating from and arriving at O should had been ejected in a direction in S (figure 2) such that the resultant velocity is along the unit vector. The duration taken by the pulse to arrive at o (and hence O) is given by the quotient of its displacement vector and its velocity in S, i.e.
(7.2)
where Thus the geometric length R of the trip in S, has given rise to the optical length of the trip in S. The length r can be looked on as the geometric length of the trip in s because b is stationary in s, and r therefore must induce for a source B an optical distance in s which should be identical to its geometric distance R in S. Since B is moving in s at velocity, the velocity in s of the pulse that emanates from the virtual source B is the sum of its velocity in S and the velocity of its emitter. The pulse emanating from B should have then been ejected in a direction in s (figure 2) such that the resultant velocity is along The duration taken by the pulse to arrive at O (and hence o) is given by the quotient of the displacement of the pulse as seen in s and its velocity in s, i.e.
(7.3)
Whether b or B was the source, we start only with one quantity, R if S is the stationary frame, or r if s is the stationary frame (but not both), which is already geometrically measured whereas the other quantity is induced in the other frame by the relations (7.2) and (7.3). It follows that it is sufficient to know the ratio to determine both quantities, regardless of which had been measured geometrically, or equivalently, which frame was considered stationary.
Dividing the equations (7.2) and (7.3) side to side we obtain
(7.4)
Figure 2. The views of S and s in green and blue respectively.
Or
(7.5)
By equations (7.2) and (7.3) the vectors appearing in the numerator and dominator on the right hand-side of the last equation are both along. Setting
(7.6)
we get
. (7.7)
Taking the norms of both sides in each equation (7.7) we get
Solving for k and we obtain
Dividing the latter equations side to side gives
which yields the scaling factor given by
(7.8)
The scaling transformations of the first type (STI) are therefore
(7.9a)
These can be written in terms of the angle between the velocity of the source b in S and the radius vector as follows:
(7.9b)
Other forms that hold for arbitrary relative orientations of axes of S and s are the following:
(7.9c)
(7.9d)
supplemented by and
Intercepting a Moving Body by a Light’s Signal
Assume that an instant which corresponds to and, a spherical pulse emanates from the observer that intercepts the moving object b at a position at an instant t. If the geometric distance then the pulse arrives at B at an instant T. The trips
and are viewed in s as and respectively
(figure 3). In the frame S, the vector sum, or can be written as
(7.10)
When the frame s is the stationary frame, the vector sum gives
(7.11)
From the latter two equations we have
(7.12)
which is the same equation (7.5), and hence the relation between t and T is given again by (7.9).
Note that if the right-hand sides of (7.10) and (7.11) are multiplied with a common factor a, equation (7.12) is again obtained, but the two sides connected at the moving source will change in length in the triangle The factor a will be determined in part 2 such that each triangle is Euclidian.
Being universal, the time lengths t and T are the same in S and s, with t (T) is equal to the geometric time distance between o and b (between O and B) in s (in S).
Let S be the universal frame, and assume that at the same instant of time which corresponds to two pulses of light are emitted from b and O simultaneously. It follows from the latter identical scaling transformations that the pulse emitted from O arrives at b (which occupies now) at the time of arrival of the pulse emitted from (b at B) at O.
The latter result is a direct consequence of the symmetry of the relationships between the source and observer in the universal space. The symmetries of the scaling transformations will have a systematic study in future works.
Figure 3. The light trips and are viewed in s as and.
8. Basic Properties of the Scaling Factor and the STI
It is easily seen that
(8.1)
(8.2)
It follows from (8.1) that only when, or
Since
(8.3)
the function with is fixed and positive, is a monotonically decreasing with and
(8.4)
Also, and because
(8.5)
the function for a fixed value of increases with if (receding source) and decreases with if (approaching source).
By its way of derivation, the STI hold 1) Between the geometric lengths R in S and r in s (the passive view), but yet the time length of a light trip is same in S and s, i.e. it is universal.
2) In the stationary frame (the active view) between the universal and geometric time lengths of a light trip whether b (when S is stationary) or B (when s is stationary) was the source of light.
9. Viewing the Light Trip in the Universal Space—The Passive View
The STI allow for the identification of the universal space with either frame S or s, and determine the relations between the characteristics of any light trip in S when stationary and its characteristics in s when stationary. It is valid whether b or B was the true source. If b was the true source of light, then T (t) is its geometric (universal) time distance from O in S. And since the source is stationary in s, t is also its geometric time distance from o in s. If B is the true source then t (T) is its geometric (universal) time distance from o in s, and T is also its geometric time distance from O in S. In any case, the universal duration of the light’s trip, which coincides with the geometric duration in the frame in which the source is stationary, is the same in both frames. The latter fact follows from Section 2 and from the freedom of identifying the universal space with S or s. With the S and s-observers measuring the same time length of each light trip, whether its source was in S or in s, time flaws, in accordance with Section 2, equably in S and s.
As equivalent sites of one physical world, S and s should quantify, employing their intrinsic units of length, the geometric distance identically, i.e., and Now, by the STI (7.9)
(9.1)
which yields the intrinsic units of length (and time) in the two frames mapped on each other by the relations,
(9.2)
(A parallel relation to (9.1) holds for geometric time:). It is clear that
unless
Because the time length of a trip is universal, then
(9.3)
where () are what is read for the time length of the trip in S when stationary (in s when stationary) respectively. By (9.2)
. (9.4)
According to the latter relations the observed characters in S are simply the S-equivalents (means: using S intrinsic units) of the corresponding observed characters in s, and vice versa.
10. The Doppler Effect
Let s be an inertial frame that is moving relative to S at velocity, and arrange for a stationary source of light b in s to radiate monochromatic light of a characteristic wavelength where ls is the unit of length in s. When observed from S, the wavelength of this light is equal to the S equivalent of:
(10.1)
Therefore the wavelength as observed in S is
(10.2)
Note that the characteristic wavelength of the same monochromatic light when emitted from an S-source is Equation (10.2) shows that
(10.3)
These correspond to the light’s source approaching the observer in the first case, receding from him in the second, and moving at right angle to the position vector of the body in the third.
We compare here the quantitative Doppler’s effect as determined by the UST
(10.4)
with the relativistic formula [10]
(10.5)
where, and is the wavelength as predicted in special relativity. The predictions of the two theories coincides for longitudinal motion,
(10.6a)
(10.6b)
The predictions of the two theories become qualitatively distinct for In this case the relativistic formula (10.5) predicts a red shift given by
(10.7)
whereas the relation (11.4) reduces to
, (10.8)
which contrary to the relativistic prediction, shows that there is no traverse Doppler’s effect (at right angle).
The predictions of UST are in excellent agreement with the results of the Ives-Stilwell experiment [12] -[14] . To specify the goal of the experiment, we denote the wavelengths associated with approaching and receding sources by and respectively. The Ives-Stilwell experiment was designed to measure the shift [10] [12] -[14]
(10.9)
In the relativistic theory
(10.10)
and the shift in wave length is
(10.11)
In the UST theory
(10.12)
(10.13)
and the wavelength shift is
(10.14)
The latter relation shows that the UST predicts in general a smaller shift than the relativistic one, and the two prediction coincide for or. In Ives-Stilwell’s experiment a small mirror is set at an angle with the ions velocity to reflect the radiation emitted backwards. As (10.14) shows, the relativistic prediction should be scaled by a factor producing accordingly a smaller shift, and the predicted shifts by the UST theory can be closer to the experimental observations only when the observed shifts are less than the relativistic predictions.
The following table 1 displays some of the predictions of the special theory of relativity and the UST together with the observed shift in Ives and Stilwell experiment, all measured in angstrom.
11. The Active View in Interpreting the STI
In the active view, (7.9) is considered a transformation in the universal frame, say S, between the time length of
Table 1 . The transverse Doppler’s effect as measured experimentally (first column) and predicted theoretically byrelativity and the UST (second and third columns).
the given light’s trip and its geometric length. Equivalently, the STI determines the ratio between the duration of the true light trip and the geometric time length of the light’s trip (call it the geometric trip) whether B was a true or a virtual source. If B and b were light sources, then (7.9) gives the relation between the time lengths of the corresponding light trips to O.
Alternatively, the transformations (7.9) hold within the universal frame s, with B is a true source while b can be a true or a virtual source. Here, T and t are the universal and geometric time lengths respectively. The expression of the universal time length in terms of the geometric length is obtained from (7.9) just by interchanging t and T and replacing by:
(11.1)
which is (7.9) itself.
When a pulse of light that is emitted from (b at B) arrives at O the source b occupies a position The triangle formed from the initial and final positions of the light’s source in S together with the observer is called the body-observer triangle. We have shown that if two pulses emanate simultaneously from b when at B and from O then when the former hits O (following the path), the latter intercepts b at (following the path). We shall see in the second part of this work [15] that the sides of the body-observer triangle (figure 4) are
(11.2)
The -meson particles [16] are generated at an altitude of, or equivalently, Their lifetime is, and their velocity v is close to that of light. Equations (11.2) insure that these particles reach the earth surface provided which yields
12. Conclusions
It was shown that the concept of time was consistent with motion if time and geometric length are of the same dimension, and thus essentially measurable by the same unit. Duration of time is corresponded with the time length of a certain light trip whether measured by length or time units. The resulting reduced system of units leads naturally to the well-known mass-energy equivalence formula. Adopting the intrinsic system of units is essential for a meaningful comparison between measurements in different IFs. When comparing measurements in various IFs only one frame can be chosen stationary while the other frames are moving. Based on the equivalence of the IFs as sites for one physical world, one must demand that the correspondence between geometric lengths of light trips and intervals of time (discussed in Section 2), should hold whatever is the site frame of the trip. This leads to time flowing equably in all inertial frames. The STI which allows for each frame to be
Figure 4. The body-observer triangle.
identifiable with the physical space relates the geometric lengths of the same light trip in both frames and maps intrinsic units in one frame onto intrinsic units in another. The STI was applied to explain the Doppler’s effect and the lifetime of meta-stable particles phenomenon. The quantitative predicted Doppler’s effect, which is in a striking agreement with the Ives-Stilwell experimental results, coincides with the relativistic prediction for longitudinal motion, but yet predicts complete absence of a transverse effect. Stellar aberration which is based on the deflection of the direction of a given trip in the moving frame from its direction in the stationary one will be submitted promptly.
In future works we will complete the theory and explain elaborately the drag effect, the null results of Michelson and Morley experiment, Michelson and Gale experiment, the Sagnac effect, and the translative Sagnac effect. It will also be shown that the mechanics produced by UST is identical to the relativistic mechanics.
References
1. Viazminsky, C.P. and Vizminiska, P. (2014) On Universal Mechanics and Superluminal Velocities in ST. General Science Journal.
2. Viazminsky, C.P. and Vizminiska, P. (2013) Digital Natural Units. General Science Journal.
3. Viazminsky, C.P. and Vizminiska, P. (2012) On the Concept of Time and the Reduced System of Units. General Science Journal.
4. Viazminsky, C.P. and Vizminiska, P. (2012) Inertial and Universal Velocities in the Scaling Theory. General Science Journal.
5. Viazminsky, C.P. and Vizminiska, P. (2011) Observation of Motion through Light Signals.
6. Viazminsky, C.P. and Vizminiska, P. (2010) The Scaling Theory—XII: Universal Timer. The Zeroth Law of Motion.
7. Pfahler, E.L. (1964) Riemannian Geometry. Princeton University Press, Princeton.
8. Steve, A. (1997) Relativity, an Introduction to Space-Time Physics. Taylor and Francis Ltd, London.
9. Rindler, W. (2006) Essential Relativity. Springr-Verlag, Berlin.
10. French, A.P. (1968) Special Relativity. Butler & Tanner Ltd., Frome and London.
11. http://en.wikipedia.org/wiki/Aether_drag_hypothesis
12. Faraj, A.A. (2005) The Ives-Stilwell Experiment. The General Science Journal.
13. Ives, E.H. and Stilwell, G.R. (1938) An Experimental Study of the Rate of a Moving Clock. Journal of the Optical Society of America, 28, 215-226. http://dx.doi.org/10.1364/JOSA.28.000215
14. Ives, H.E. and Stilwell, G.R. (1941) An Experimental Study of the Rate of a Moving Atomic Clock. II. Journal of the Optical Society of America, 31, 369-374. http://dx.doi.org/10.1364/JOSA.31.000369
15. Viazminsky, C.P. and Vizminiska, P.K. (2014) On Universal Mechanics and Superluminal Velocities. Applied Mathematics (Submitted).
16. http://en.wikipedia.org/wiki/Muon
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# class14 - example A chain of uniform mass density length b...
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example: A chain of uniform mass density ρ , length b , and mass M (where ρ = M/b ) hangs from both ends. At time t = 0, the ends are adjacent, but one is released. Find the tension in the chain at the fixed point, after the other has fallen a distance x . Assume free-fall. solution : Assume free-fall: that is, the only forces acting on the system at time t are the tension (vertically upward at the fixed end) and the gravitational force Mg pulling the chain down. The center of mass momentum reacts to these forces such that ˙ P = Mg - T (20) The free side of the chain, with mass ρ ( b - x ) / 2, moves at the speed ˙ x , and the other side is not moving. The total momentum of the system is therefore P = ρ b - x 2 ˙ x (21) so ˙ P = ρ 2 - ˙ x 2 + ¨ x ( b - x ) (22) Now, the kinematic equations for free-fall are x = gt 2 / 2, so ˙ x = gt = 2 gx, ¨ x = g (23) so ˙ P = ρ 2 ( gb - 3 gx ) = Mg - T (24) finally T = Mg 2 3 x b + 1 (25) 1.2.1 Rocket Motion An important application of linear momentum conservation is rocket propulsion. Consider the rocket as a rigid body of mass M
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Quick Search
# Page History: Examples
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### Page Revision: 2014/10/12 20:29
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# Engineering
• Non-linear deflection of a flexible cantilever pdf zip
• Batch chemical reactor pdf zip
• Lateral force developed by a tire, using an early Pacejka model. pdf zip
• NASA's Atmospheric Model (SI Units) pdf zip
• Using an electromagnetic shaker to drive a single degree of freedom system pdf zip
• Simple Estimates of a single prop, naturally aspirated, aircraft's performance pdf zip
• Ramp Design - testing for max allowable bending moment and deflection (US Customary Units) pdf zip
• Column Design - based on AISC Manual of Steel Construction (US Customary Units) pdf zip
• Engine Heat Rejection Calcs - good example of working with relative temperature differences (US Customary Units) pdf zip
• Calculate the Area Properties of Composite Rectangular Areas (US Customary Units) pdf zip
• Calculate the Area Properties of Composite Rectangular Areas (US Customary Units) pdf zip
• Calculate properties of generic polygons (perimeter, area, centroid, second moment of area, radius of gyration, elastic section modulus, plastic section modulus, orientation of principal axes of inertia, principal moments of inertia, radii of gyration about principal axes of inertia); multilanguage (ENG/ITA) pdf zip
• Electrical engineering - calculate current, voltage, real, reactive and apparent power in an AC circuit using the complex domain and using custom units for AC power quantities. pdf zip
# Numerical methods
• Newton-Raphson method for solving equations (G. Urroz) pdf zip
• Nonlinear system of algebraic equations, Broyden's method pdf zip
• Lagrange interpolation polynomials pdf zip
• Newton interpolation polynomials (forward finite difference) pdf zip
• Newton interpolation polynomials (backward finite difference) pdf zip
• Data fitting (single independent var., linear on parameters) pdf zip
• Data fitting (multiple, linear on parameters) pdf zip
• Simpson integration with Richardson extrapolation pdf zip
• Gauss-Legendre integration method pdf zip
• Second order linear ODE - finite difference method pdf zip
• Runge-Kutta 4th-order for single 1st-order ODE (G. Urroz) pdf zip
• Runge-Kutta 4th-order for systems of ODEs (G. Urroz) pdf zip
• Runge-Kutta 4th-order for single 2nd-order ODE (G. Urroz) pdf zip
• Runge-Kutta 5th-order adaptive - also "Batch chemical reactor" (see "Engineering", above) pdf zip
• Graphing implicit functions of 2 variables: f(x,y)=K (where K is a constant) pdf zip
# Statistics
• Normal distribution pdf zip
• Student t-distribution pdf zip
• Chi-squared distribution pdf zip
• F-distribution pdf zip
• Examples of probability distributions (G. Urroz) pdf zip
# Data Manipulation
• Importing data from csv files pdf zip
# Civil Engineering Hydraulics/Fluid Mechanics
From college courses by G. Urroz
• CEE Fluid Mechanics Test 1: | Prob 1: pdf zip | Prob 2: pdf zip | Prob 3: pdf zip |
• CEE Fluid Mechanics Test 2: pdf zip | CEE Fluid Mechanics Test 3: pdf zip
• Video examples of calculations with units for hydraulics: See Part 2 SMath Studio\
• CEE 3510 HYDRAULICS : Spring 2012 Schedule - Each link in the Schedule contains links to examples with SMath Studio solutions for pipe flow, pump-pipeline systems, pipe networks, pumps, and open-channel flow (some examples include solutions using EPANET and HEC-RAS)
Developed by Brett Towler
• Vertical Velocity Distribution in a Hydraulically Smooth Open Channel: pdf zip
• Hydraulic Geometry of Rivers: pdf zip
• Weir Flow using the Rehbock Equation: pdf zip
• Swamee-Jain Friction Factor Equation for Pipe Flow: pdf zip
# Structural Engineering
• Calculate the Area Properties of Composite Rectangular Areas (US Customary Units) pdf zip
• Non-linear deflection of a flexible cantilever pdf zip
• Ramp Design - testing for max allowable bending moment and deflection (US Customary Units) pdf zip
• Column Design - based on AISC Manual of Steel Construction (US Customary Units) pdf zip
• Design of Reinforced Concrete Beams according to NZS 3101-95 (SI Units) pdf zip
• Steel End-Plate Connection Design - adapted from Krishnamurthy 1978 (US Units) pdf zip
• Design of Reinforced Concrete Slab - Flexural Reinforcement (US Units) pdf zip
• Single span moment & shear envelopes for HL-93 loading using influence lines pdf zip
ScrewTurn Wiki version 2.0.37. Some of the icons created by FamFamFam.
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# Integration techniques: parts, substitution; separable ODE’s
```18.01 Section, November 9, 2015
Section leader: Eva Belmont ([email protected], E18-401B)
.
Integration techniques: parts, substitution; separable ODE’s
1. Evaluate the following integrals:
Z 4
2
(a)
x ex dx
2
Z
(b)
x√
t ln t dt
(assume x > 1)
1
Z
(c)
π
sin3 x dx
0
2. Instead of actually doing the following integrals (or even picking up a pen), just talk to
f = sin x and g = 2x”.
Z
Z
Z
Z
x
1
dx
dx
t sin 2t dt
x ln x dx
2
1+x
1 + x2
Z
1 + x2
dx
x
Z
Z
2
x3 ex dx
tan x dx
1
3. Find the general solution for
dy
dx
= y 2 + 1.
4. Make up an integration by substitution problem and give it to your neighbor.
Of course, you should actually know how to do the problem before giving it to someone else!
Z
5. Bonus problem:
ln x dx
Hint: integration by parts.
Z
6. Bonus problem:
√
sin x dx
Hint: first make a substitution.
Review
• Integration by parts:
R
f g dx = F g −
R
F g 0 dx where F =
2
R
f.
```
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# Does this loop run in $\mathcal{O}(n^4)$ time?
A double loop is given:
int sum = 0;
for (int i = 0; i < N*N; i++)
for (int j = i; j < N; j++)
sum++;
My analysis: The inner loop runs $n$ times the first iteration, then $n-1$, $n-2$, $n-3$, $n-4$ until it gets to $n-n$ which is zero. This is a series summation which is expressed as: $$\sum_{i=0}^n \frac{1}{2}(n(n+1))= \frac{1}{2}(n^2+n)= \frac{1}{2}n^2 + \frac{1}{2}n$$ Discarding lower terms and constants give $\mathcal{O}\left(\frac{1}{2}n^2\right) \rightarrow \mathcal{O}(n^2)$.
The outer loop is executed $n^2$ times. Multiplying by the inner loop gives $\mathcal{O}(n^4)$. Is this right?
Update: Now I think the actual answer is $\mathcal{O}(n^2)$ because the inner loop is only executed half the times of the outer loop.
• You're factoring in the outer loop twice - you effectively factor it in once when you compute your (series) summation, but then multiply by it as well. – Steven Stadnicki Sep 18 '14 at 17:23
• @StevenStadnicki I just realized that. See my update at the bottom. Is this correct? – user6607 Sep 18 '14 at 17:24
• You're much more on the right track - though for a formal analysis I would split the outer loop into the two pieces $0\leq i\lt N$ and $N\leq i\lt N^2$. (Note that you still run the inner loop over the latter case, even though you don't execute any statements within it.) – Steven Stadnicki Sep 18 '14 at 17:25
By splitting the outer loop into $0\leq i\leq N-1$ and $N\leq i\leq N^2-1$ (because the code inside the inner loop isn't executed at all for $i\geq N$, but still costs $\mathcal{O}(1)$ time to do) we obtain: \begin{align} \sum_{i=0}^{N-1}\sum_{j=i}^{N-1}1+\sum_{i=N}^{N^2-1}1&=\sum_{i=0}^{N-1}(N-1-i+1)+N^2-1-N+1\\ &=\sum_{i=0}^{N-1}N-\sum_{i=0}^{N-1}i+N^2-N\\ &=N^2-\frac{(N-1)N}{2}+N^2-N\\ &=\mathcal{O}(N^2). \end{align}
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# Search:
Type: Posts; User: vukosi
## Search: Search took 0.22 seconds.
by vukosi
Replies
7
Views
717
(Card Shuffling and Dealing) Modify the application to deal a five-card poker
hand. Then modify class "DeckOfCards" to include methods that determine whether a
hand contains :( a )pair
- A Pair...
by vukosi
Replies
7
Views
717
OK i'm just getting more confuse as you ask me question because i'm just a beginner, so what i'm asking now is how will you solve the given assignment by at least answering Question a please
by vukosi
Replies
7
Views
717
like how norm, the error i'm getting is:C:\Users\Vukosi\Desktop\fig07_09_11\DeckOfCardsTest.java:23: error: cannot find symbol
if(myDeckOfCards.dealCard(face + " of " + suit).equals("Duece of...
by vukosi
Replies
7
Views
717
please i have this task to to but i can only manage to display five cards but i cant compare my result, here is what should be done:
(Card Shuffling and Dealing) Modify the application to deal a...
5. ## Thread: please advise on which method i can use to compare my out put resault
by vukosi
Replies
9
Views
637
### Re: please advise on which method i can use to compare my out put resault
ok i have the code but this code it doesn't indicate whether the cards its a straight,four of a kind, full house, two pairs or one pair. for example when i deal the cards and lets say the random...
6. ## Thread: please advise on which method i can use to compare my out put resault
by vukosi
Replies
9
Views
637
### Re: please advise on which method i can use to compare my out put resault
like how i dont understand, but i just need help to solution for quetion a, and b, and the rest i can at-least try
7. ## Thread: please advise on which method i can use to compare my out put resault
by vukosi
Replies
9
Views
637
### Re: please advise on which method i can use to compare my out put resault
Norm i hope this is what you were asking for
8. ## Thread: please advise on which method i can use to compare my out put resault
by vukosi
Replies
9
Views
637
### Re: please advise on which method i can use to compare my out put resault
Norm i have made some changes but i don't know if that is was what you have been looking for
9. ## Thread: please advise on which method i can use to compare my out put resault
by vukosi
Replies
9
Views
637
### please advise on which method i can use to compare my out put resault
please i have this task to to but i can only manage to to display five cards but i cant compare my result, here is what should be done:
(Card Shuffling and Dealing) Modify the application to deal...
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# Math.Abs Method (Int16)
.NET Framework (current version)
Returns the absolute value of a 16-bit signed integer.
Namespace: System
Assembly: mscorlib (in mscorlib.dll)
## Syntax
```public static short Abs(
short value
)
```
#### Parameters
value
Type: System.Int16
A number that is greater than Int16.MinValue, but less than or equal to Int16.MaxValue.
#### Return Value
Type: System.Int16
A 16-bit signed integer, x, such that 0 x Int16.MaxValue.
## Exceptions
Exception Condition
OverflowException
value equals Int16.MinValue.
## Remarks
The absolute value of an Int16 is its numeric value without its sign. For example, the absolute value of both 123 and -123 is 123.
## Examples
The following example uses the Abs(Int16) method to get the absolute value of a number of Int16 values.
```using System;
public class Example
{
public static void Main()
{
short[] values = { Int16.MaxValue, 10328, 0, -1476, Int16.MinValue };
foreach (short value in values)
{
try {
Console.WriteLine("Abs({0}) = {1}", value, Math.Abs(value));
}
catch (OverflowException) {
Console.WriteLine("Unable to calculate the absolute value of {0}.",
value);
}
}
}
}
// The example displays the following output:
// Abs(32767) = 32767
// Abs(10328) = 10328
// Abs(0) = 0
// Abs(-1476) = 1476
// Unable to calculate the absolute value of -32768.
```
## Version Information
Universal Windows Platform
Available since 8
.NET Framework
Available since 1.1
Portable Class Library
Supported in: portable .NET platforms
Silverlight
Available since 2.0
Windows Phone Silverlight
Available since 7.0
Windows Phone
Available since 8.1
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Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: Why do -intreg- and -truncreg- not give me the same SD as -summarize- ?
From Antoine Terracol To statalist@hsphsun2.harvard.edu Subject Re: st: Why do -intreg- and -truncreg- not give me the same SD as -summarize- ? Date Tue, 08 Nov 2011 17:59:31 +0100
```Hi,
```
I'd say this is for the same reason why if you estimate a linear model by ML instead of OLS, you'll get a biased estimate of sigma because, as William said, ML does not ajust for the number of parameters in the degrees of freedom.
```
```
Since -truncreg- and -intreg- are estimated using ML, I'm not surprised that you get different values for sigma. The "correct" (unbiased) one is the one reported by -summarize- and -regress-.
```
Antoine
On 08/11/11 17:11, Richard Williams wrote:
```
```At 10:02 AM 11/8/2011, Seed, Paul wrote:
```
```Dear all,
Here is a puzzle that has been bothering me.
Interval regression is designed for the circumstances
where an outcome is sometimes observed only to be
in a particular range. However it also copes with
some (or all) observed values being known exactly.
Like -summarize-, -intreg- and -truncreg-
But look at this:
* Begin program *
sysuse auto, clear
su price
truncreg price ,
gen price2 = price
intreg price price2
* end program *
Putting the results together I get:
Method | Mean Std. Dev/sigma
-------------+---------------------------------
summarize | 6165.257 2949.496
intreg | 6165.257 2929.499
truncreg | 6165.257 2929.499
All 3 methods agree on the mean; but we have
two different estimates for the standard deviation.
The manual does not seem to discuss this.
Does anyone know what is going on, and is there any way
to correct the SD values from -intreg- & -truncreg-?
```
```
I think it has to do with the d.f. and how they are being adjusted for.
If you do
reg price
you again get the 2949.496, which is the square root of the MST
(8699525.97). MST is SST/73 = 635065396/73. If you instead use 74 in the
calculations, you get
. di (635065396/74) ^ .5
2929.4991
which is what you are getting with intreg and truncreg. In other words,
these routines seem to be using N rather than N-1 in their calculations.
I am not sure which SD you consider "correct" but it should just take a
little algebra to convert from one to the other. The help files or
manuals probably say something about how the SD is calculated, and why.
-------------------------------------------
Richard Williams, Notre Dame Dept of Sociology
OFFICE: (574)631-6668, (574)631-6463
HOME: (574)289-5227
EMAIL: Richard.A.Williams.5@ND.Edu
WWW: http://www.nd.edu/~rwilliam
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
```
--
Ce message a ete verifie par MailScanner
pour des virus ou des polluriels et rien de
suspect n'a ete trouve.
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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# Probability theory
• Feb 4th 2008, 08:51 PM
Andreamet
Probability theory
A fair die is rolled six times. If on the nth roll, 1<=n<=6 the result is n, we call it a hit. What is the prob. that we get at least one hit in the six rolls.
Now, do the same but with two dice, and rolling the twelve times. What is the probability that we get at least one hit?
• Feb 4th 2008, 09:06 PM
Ryan J
take the probability of not getting a hit for both. i.e. for the first part it's 1-5/6^6. So, simply take the odds of not getting a hit for the second one. If you cannot figure it out I'll check back in a little while, and solve the second part also.
• Feb 4th 2008, 09:06 PM
mr fantastic
Quote:
Originally Posted by Andreamet
A fair die is rolled six times. If on the nth roll, 1<=n<=6 the result is n, we call it a hit. What is the prob. that we get at least one hit in the six rolls.
Now, do the same but with two dice, and rolling the twelve times. What is the probability that we get at least one hit?
Big hint: In each case, the answer is 1 - Pr(no hits) .......
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Question
# Simplify √√81 =
A
27
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B
3
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C
9
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D
Can't be determined
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Open in App
Solution
## The correct option is C 3 √√81 =√(√9×9) [∵9×9=92=81] =√(9) [∵√92=9]=√(3×3) [∵3×3=32=9]=3
Suggest Corrections
0
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Java Non-negative multiple linear regression library
I am working on a Java project, and I have to compute a multiple linear regression, but I want the gotten parameters to be non-negative. Is there an existing commercial-friendly-licensed library to do such a thing? I’ve been looking for Non-Negative Least Squares libs, without success.
Well, I could not find any pure java library so I built it myself from the article of [1], wich can be found in [2] and [3]. I give the algorithm:
P, R are the active and the passive sets. t() is transpose
The problem is to solve Ax = b under the condition x>0
P=null
R = {1,2,...,m}
x = 0
w = t(A)*(b-A*x)
while R<>null and max{wi|i in R}>0 do:
j = argmax{wi|i in R}
P = P U {j}
R = R{j}
s[P] = invert[t(A[P])A[P]]t(A[P])b
while sp<=0 do:
a = -min{xi/(di-xi)|i in P and di<0}
x = x + a*s -x
update(P)
update(R)
sP = invert[t(A[P])A[P]]t(A[P])b
sR = 0
x = s
w = t(A)*(b-A*x)
return x
For the other definitions, I strongly advise to read the papers [2] and [3], which are online (see below for the links ๐ )
[1] Lawson, C. L., & Hanson, R. J. (1974). Solving least squares problems (Vol. 161). Englewood Cliffs, NJ: Prentice-hall. [2] Rasmus Bro et Sijmen De Jong : A fast non-negativity-constrained least squares algorithm. Journal of chemometrics, 11(5) :393โ401, 1997. http://www.researchgate.net/publication/230554373_A_fast_non-negativity-constrained_least_squares_algorithm/file/79e41501a40da0224e.pdf [3] Donghui Chen et Robert J Plemmons : Nonnegativity constraints in numerical analysis. In Symposium on the Birth of Numerical Analysis, pages 109โ140, 2009. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.157.9203&rep=rep1&type=pdf
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Create A Script That Will Increment A Number By 0.01?
Feb 12, 2011
working on building a website at the moment. and i need to create a script that will increment a number by 0.01, then stop and decrease by 0.03, then increase again by one. I wrote code to increment by 0.01 continuouly but now i can't stop it. and change to decreasing
<script type = "text/javascript">
num = 0.87;
var tim = 0;
[code]....
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I have a string "Slide1.jpg"
I need to increment it by 1 to say "Slide2.jpg"
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Create Random Number Between 1 - 50 With Decimals?
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How would I go about creating a random number between 1-50 that COULD include whole numbers, decimals to the tenth, hundredth and the thousandth place?
Ex) 2.145, 3, 5.8, 41.002, 10.12, 7, 18.023, 33.2
Also, if the only way to do it is to put trailing zeros to the thousandth place that's fine as well
Ex) 2.145, 3.00, 5.800, 41.002, 10.120, 7.00, 18.023, 33.200
Create An Array Of Numbers Counting From 1 To A Given Number?
Mar 20, 2009
I just want to create an array of numbers counting from 1 to a given number.
At the moment I have the for loop running like this:
var i=0;
for (i=1;i<=10;i++)
{
if (i == 1)
[Code]....
This outputs 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
How can I put this output into this variable: 'var ids' so I get out
var ids= [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
Create A Function That Prompts User For A Number
Oct 26, 2011
Create a function that prompts user for a number. Develop the program so that it continues to prompt until it receives valid information. Then create a multiplication table that displays the number multiplied by 1 through prompted number.
Create Phone Number Form Script?
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How do i create a phone number form box in javascript using no inner html . there must be at least one dash in it. code...
Create A Script That Allows User To Enter A Number?
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I need to create a script that allows a user to enter a number, in either celsius or fahrenheit textbox....then they can click the button and the conversion is placed in the blank textbox
C = (F 32) * 5 / 9
F = C * 9 / 5 + 32
<table border="1">
<tr>
<th>Fahrenheit</th>
<th></th>
[Code]....
Create A Water Bill That Calculates The Number Of Gallons
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I have a javascript class and one of the assignements that we had to do was create a water bill that calculates the number of gallons used and the price for the water consumption. It will not work at all...
How To Create A Random 15 Letter/number Piece Of Text?
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How do I create random 15 letter/number piece of text that is composed of letter and numbers jumbled together then automatically place that in a field? This is all of course activated by a link.
Create A Form Which Has A Dynamic Number Of Text Fields?
Sep 24, 2011
I need to create a form which has a dynamic number of text fields. So, I created this input type :
PHP Code:
<input id="benamount" name="benamount[]" type="text" disabled="disabled" size="40"/>
<input type="checkbox" name="ben[]" onchange="check();"/>
[code]....
School Assignment - Create A Function That Prompts User For A Number
Oct 26, 2011
- Create a function that prompts user for a number.
- Develop the program so that it continues to prompt until it receives valid information.
- Then create a multiplication table that displays the number multiplied by 1 through prompted number.
How to get the function to work.
Create Specific Number Output Based Upon Internal Calls?
Oct 13, 2009
Have a computed field ("Subtotal"). With that number, need to set up rule such that if Subtotal is a certain number, a certain number will print in the new field (lets call it "First").[code]...
Create A Virtual Library - Make A Random Number Of Books Pop Out On Mouse Over?
Jun 7, 2011
I have a background that is a shelf full of books, I wanted to figure out what I would need to do in order to make a random number of books pop out on mouse over (individually), and upon click have the book pulled out and be flipped through and read like a book. Would I have to recreate the library picture completely to make the random books moveable? And would I have to create the books that I want to move from the bookshelf?
how to pull the book out from the book shelf so that it looks like it's coming towards me and then to make it look like it has been pulled out. Would I need to use animation for this? What would be the parts to the animation required?
A little background on my knowledge of javascript: I understand the concepts of it and can understand it when I read it or its explained to me. I am looking to create something as simple as I can, I thought having a background and something basic like a few books be able to pop out. When I realized css couldn't do this and css3 wasn't supported by most browsers I turned to javascript and have been trying to figure out the steps I would need to do to do this specifically.
Increment A Value Each Day By One?
Jan 29, 2010
For a real estate website, I want to create a script which will tell visitors how many days a property has been on the market. For example say the property was posted on the site on 1/1/2009. I want the site to say "This property has been on the market for 394 days." .
Increment Var By 1 In Loop?
Apr 6, 2009
I have a loop that is running and I am trying to increment a number by one each time. Here is what I am doing, but each time it just writes out 1 for the value.
var i=1;
Start of Loop
document.write(i);
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Increment By 1 Every 15 Seconds?
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I need to develop this feature for a charity site displays a number that counts up 1 every 15 seconds. This is to show how many times a kid is abused in this country.
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setTimeout('doSomething()',15000);
}
Increment Fields
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What I need is 2 form fields, an up link, and a down link. I want the user to be able to select one of the form fields, then increment, or decrease the number in the field that has been selected by clicking one of the links. How might I do this?
How Increment The Value On Each Click ??
Apr 17, 2007
How to increment the value on each function call ? Suppose i had a function called addRow()
//js code
{
var i = 0;
}
suppose this function is called with onclick event of button. when clicked first time it should alert 0, 2nd time => 2 3rd time => 3..
How To Increment Without A ++ Operator?
Jul 30, 2002
I have a for loop and would like to increment a variable for (let's say) 2 instead of one (++). How can I do that?
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Create A Small DHTML Code That Created A Unordered List Of Input Forms Dependent On The Number Selected From The Select Dropdown Menu
Oct 21, 2011
I wanted to create a small DHTML code that created a unordered list of input forms dependent on the number selected from the select dropdown menu. Problem is that it doesn't seem to want to generate the list. I think the variables are within the scope of the function too, and I didn't get an errors from the javascript console when using firebug. The script itself runs, I tested it when I used the old standby alert(); to see if the script was active. Here's the code:
[Code]...
Can I Increment Elements Each Loop?
Apr 15, 2011
I have written the follow code. It is meant to (when finished) output a table showing each member of the array PEOPLE. There Income ,there Tax bracket and there finally there total tax paid.The calulations in the if-else statements are correct.I have to create a loop that will go through the if else statements equal to the amount of the people in the array (This is no problem I have done this earlier)
My problem is when I try to add each element (PEOPLE) to the table or there indivual tax outcomes.Can I create a loop and increment in the elements each iteration to put on the table?(for there names) As I am not meant to store each iteration,it is to write to the table each time.This is the code I'm working on with out the loop.
<html>
<script>[code]......
Increment Width Of Div OnClick?
Jul 15, 2011
I've got a div whose width I'm trying to modify when you click either the plus or minus button (incremental percentage-based).
The following is what I've written in an effort to accomplish this:
<script type="text/javascript">
function modify_jump(way) {
//check if we should increment or decrement
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Increment Php Array Variable?
Jun 16, 2010
\$i=0;
echo '<script language="javascript">
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here i want to increment \$i above code is not working,
JQuery :: Function To Auto-increment Certain Value
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I'm trying to write some jquery that when a link is clicked it adds a number into a div. Every time the link is clicked it adds another number(previous number + 1). i have this but it definitely doesn't work function addnumber(X)[code]As it is now, the first click does nothing. The second click spits out "2<br>2<br>". The third click spits out "2<br>2<br>2<br>". The fourth "2<br>2<br>2<br>2<br>" and so on.
HTML Tags Break - Won't Increment The Name?
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| 2,646
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| 3.21875
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CC-MAIN-2021-17
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latest
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en
| 0.865032
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https://mathematica.stackexchange.com/questions/51102/elements-of-a-list-according-position-in-axis
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| 39,940
|
# Elements of a List according position in axis
({{a, b}, {c, d}})
a, b, c and d are Lists of 300 length. Their first 4 elements must specify their position in the column (1 column, 2 column, etc) and the following 4 elements the position in the line (1 line, 2 line, etc) but this must be made in binary code and not a 2x2 Table but a 10x10. Also, they must only contain -1, 0 or 1.
I was thinking about creating the vector with a smaller length and use AppendTo to put the position based on the table index and converting it to binary but couldn't find a final solution.
• That had something wrong in it. Sorry. Is it better to understand now? Or still confusing? Jun 18, 2014 at 23:29
• Let me try and paraphrase this, tell me if I'm correct: You want a 10X10 matrix. Each entry is a 300 length vector, first 4 elements represent the second dimension position in binary, second 4 elements the first dimension, and the rest of the list is randomly (based on post before edits) selected elements from {-1,0,1}. Is that it?
– ciao
Jun 18, 2014 at 23:43
This does what you're after, I believe (see my comment question) - if not, I'll delete this answer:
result=Partition[Join @@@ Transpose[{Join @@@ Reverse /@ Tuples[IntegerDigits[Range@10, 2, 4], {2}],
RandomChoice[{-1, 0, 1}, {100, 292}]}], 10]
• Your comment says exactly what I'm after for. Sorry for my bad English on explanation. It this code do that than is exactly what I want Jun 18, 2014 at 23:49
• @AndréF.: try it, tell me if it's what you intended...
– ciao
Jun 18, 2014 at 23:49
• I can't check all results but from what I could see that's exactly it. Thank you @rasher Jun 18, 2014 at 23:53
• @AndréF.: Cool. Glad to help.
– ciao
Jun 18, 2014 at 23:53
• @AndréF.: Do you mean as an actual array? Just use result[[i,j]] or result[[i]][[j]] (note double brackets - that is correct array part specification syntax)
– ciao
Jun 19, 2014 at 0:02
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15 Mind-Bending Puzzles That Will Drive You and Your Friends Nuts | 22 Words
# 15 Mind-Bending Puzzles That Will Drive You and Your Friends Nuts
Everybody loves a good riddle, right?
(OK, actually, that's not quite true. I once had a coworker who would become visibly angry any time someone told a riddle.)
But you! You love a good riddle, don't you? How could you not? They're a great way to exercise your brain muscles, think outside the box, and (if you're lucky) confuse the heck out of your friends. And also confuse yourself, if you're not on your toes.
Here's a list of some of my favorite riddles that I've come across lately. If you can guess at least 10 of them, I'm pretty sure you're some kind of genius. If you get all 15?! You should probably be studied because you're on a whole other level. After you test yourself, be sure to test your friends, too.
Have your own favorite riddle? Leave it in the comments and see if you can stump other readers! (Don't give away the answer right away!)
Without further ado, riddle me this!
### Riddle #1
There are five sisters in a room. Emily is reading a book. Barbara is cooking. Katy is playing chess. Jaime is doing laundry. What's the fifth sister doing?
She's playing chess with Katy! (Did you use your deductive reasoning skills? I hope so. Otherwise, the rest of these are going to be pretty tough.)
### Riddle #2
There are two men standing in a room. One is facing South, while the other faces North. They can see each other without using mirrors. How can that be?
The two men are facing each other. (That one took me longer than I'd care to admit!)
### Riddle #3
He got it in the woods and brought it home in his hand because he couldn't find it. The more he looked for it the more he felt it. When he finally found it he threw it away. What was it?
A splinter!
### Riddle #4
You are in the basement of your house where you have 3 switches which turn on 3 bulbs upstairs. You have to figure out which bulb corresponds to which switch. The problem is that you can only go upstairs once before giving your final answer How do you accomplish this?
Turn the first switch on and leave it on for 10 minutes, then turn it off. Turn the second switch on. Go upstairs. The light bulb that's still warm goes with the first switch. The one that's on goes with the middle switch. The light that's off? Third switch.
### Riddle #5
What can you see once in a minute, Twice in a moment, And never in one thousand years?
The letter "M." (Minute, moment, one thousand years.)
### Riddle #6
You're in a bathroom and the tub is filled to the brim with water. You have a spoon, an eyedropper, a bucket, and a teacup. What's the fastest way to empty the tub?
(You're gonna hate me for this one.) Pull the plug!
### Riddle #7
You leave home and make three left turns. Upon returning home, you find two men wearing masks. Who are they?
The catcher and the umpire. (Psst. "Home" is home plate.)
### Riddle #8
Two fathers and their two sons go hunting in the woods. Each of them shoots and kills a rabbit and brings it home. When they get home, they only have three rabbits, yet they didn't lose any. How is this possible?
Only three men went hunting: A grandfather, his son, and his grandson. (Two fathers, two sons.)
### Riddle #9
You stand before three doors and are told you must go through one of them. Behind the first door, the room is on fire. Behind the second door, the floor is covered with venomous snakes. Behind the third door are three lions who haven't eaten in three years. Which door do you choose?
You should choose the third door. The lions that haven't eaten in three years will be dead.
### Riddle #10
What's black when you get it, Red when you use it, And white when you're done with it?
Charcoal. (It's black before you heat it up, red when it's being used to cook, and it turns white when you're done cooking.)
### Riddle #11
What word starts with "e," Ends with "e," and only has one letter? (No, it's not "the letter e.")
An envelope. (Get it? It has a letter inside it!)
### Riddle #12
You come down with a deathly sickness and are prescribed 2 different sets of pills. You must take exactly one of each pill every day to survive. When you get down to the last two days and have four pills remaining, you accidentally drop them on the ground and they get mixed up. Unfortunately, the pills are completely unmarked and look exactly the same. You have no idea which pills are which. What do you do to make sure you take the correct medication?
Line all four pills up next to each other and cut them all in half. On the first day, take the top halves of all four pills (this will give you one full pill of type A, and one full pill of type B). The next day, take the rest of the pill halves.
### Riddle #13
I know a word. Six letters it contains. Take away one, and twelve is what remains.
"Dozens." (Take away the "s" and you're left with just "dozen.")
### Riddle #14
A man is walking down a black road. He's wearing a black suit, a black hat, black gloves, and black sunglasses. The streetlights are off. There is no moon. Suddenly, a truck comes out of nowhere, driving directly toward the man. Its lights are off. The truck stops before hitting him. How did he know to stop?
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https://www.hexadecimaldictionary.com/hexadecimal/0x710/
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|
# Hexadecimal 0x710 = 1808
=
Decimal 1808 11100010000 3420 0x710 0.0.7.16 One thousand eight hundred eight
0x710 is represented as: 1792 + 16 + 0.
### More details
Prime Number 1808 is not prime 2, 2, 2, 2, 113
| 87
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https://gmatclub.com/forum/on-the-number-line-above-p-q-r-s-and-t-are-five-consec-143770.html
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# On the number line above, p, q, r, s, and t are five consec
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Mathematics Education: Why children fear division and how teachers can tackle it
Gomathy Ramamoorthy, in Learning Curve magazine, writes how the rules of division are a process but understanding the concept needs reasoning and focused inferential work.
13 July 2022
Five years ago, I was presenting a paper on the Mathematics Laboratory in my school. In the discussion after the presentation, I was asked: We start with one digit in all other arithmetic operations but in division, why do we start from the higher place value?’
My answer was, If your understanding of the place value is good, it is not necessary to start the operations from the ones digit. It could be the other way around. Similarly, in division also, you can start dividing in either way.’
That question is still there in a corner of my mind. The rule that we follow while doing the four basic arithmetic operations is there in order to set an algorithm. But these rules are not hard and fast.
Division is one of the hardest parts of all the arithmetic operations and children fear it because it involves a lot of rules, especially long division which is the toughest for primary children. In schools, when we start division as a set of rules/algorithms to be followed, we fail to inculcate the real meaning of division. The rules of division are a process but understanding the concept needs reasoning and focused inferential work.
Understanding the concept
I cannot find a better example than the one given by Daniel Willingham, a cognitive scientist and writer, who says that almost 25% of the sixth graders in America think the symbol = means put the answer here (ACT, American Educator) They do not understand that the = sign means equality or mathematical equivalence.
What is needed in the classroom? In its position paper on mathematics, the NCF (2005) recommends a curriculum that is ambitious and coherent and says that learning mathematics is every child’s right. For this, it prescribes that school mathematics should be activity-oriented.
So, to bring forth the idea of the mathematics’ in division and to make every child of my class learn it, I have attempted an activity-oriented approach in my division classes. I ensure that all my students can learn to do division easily. These ideas are presented here in this article.
Why is division difficult for children?
The problem is not with the children, but in the way, it is taught. Here are some questions to think over and ask whether we are addressing these questions in our classroom teaching.
1. What is division?
2. Where it is used in real life?
3. Is there any situation in our real lives where we divide a four or five-digit number by a three- digit number? (multi-digit computation)
4. And if there is such a situation, how many can do it without a calculator?
I would like to refer to an important point here. Children are able to deal with basic arithmetic, including division and fractions in their everyday life before they formally learn these concepts in school. (Parmar, 2003; Mix et al., 1999).
According to an NCTM (National Council of Teachers of Mathematics) publication, when students understand, they develop their own procedure to solve a problem. There are a number of ways to make the teaching-learning of division more meaningful.
These have been tried in my school with the children of class IV. Our school is a girls’ primary school where most of the students come from families living below the poverty line.
Some methods
Array
Since they have already learnt to divide using grouping in class III, I asked them to divide using array. It is done using counters for making arrays (rows). Children make all the possible arrays for a given number and write down the division facts for them.
When children do more and more arrays, they become familiar with the concept of the factors of a number. In the traditional method, everything depends on multiplication tables, even in learning factors.
With arrays, children understand the concept and know that 16 cannot be arranged as three in a row because three in a row would mean get five rows of three and a one (15 +1). All this happens in a very short period of time, proving that good understanding develops good number sense.
Ganitmala
The number line division is a nightmare because it is an abstraction which is never used in real life, except to develop number sense in maths.
This illustration shows that students cannot understand where to start from and they do not know that the continuity should not break in the middle. Ganitmala is a very good representation of number lines. It reduces the level of abstraction.
When students use ganitmala they develop a good understanding of the number line and make fewer errors.
Partial quotient method
In my experience, the students will learn any concept, whether it be addition, subtraction, multiplication or division, with ease through handling money.
I use pretend money for this. The class is divided into five groups and given names of shapes: Cube, Cuboid, Cylinder, Cone and Sphere.
The first sum is to share 132 between two. To do this sum, they draw two stick figures and write down the share of each against them. It has been found that children use self-invented procedures to solve problems. So, the maths problems solved in class should simulate real-life situations that the children can relate to (Verschaffel et al., 2006).
They first divided the one hundred into two fifties. Then they took thirty and divided in into two fifteens and finally the two into two ones. They counted the share each one got, the quotient, and found that there was no remainder.
An important aspect of mathematics is recording, without which the learning is incomplete. Doing the activity teaches the concept and recording it is the way of learning the procedure. Concept and process should go together to connect and make understanding easier. Children bring out the answer by themselves, but teachers should watch out for disputes and ensure the equal participation of all the children in the activity.
Some common mistakes
Children can make mistakes in subtraction, which is an important part of division. For example, in subtracting 80 from 125, the remainder must be less than 100. However, children sometimes do not understand this, and the problem is with not understanding the concept of subtraction. Without addressing that, all efforts to teach her division will go in vain. The first step is teaching subtraction.
Children also make careless mistakes while doing their work, such as leaving out parts of the number to be divided. The teacher has to be alert to these mistakes and be ready to explain the basic concepts behind these.
Every child can learn division
I would like to share my best moment while teaching division. One of my students was given a division sum by her tuition master. She did it using a method taught in class, but her tuition master did not understand that and struck it out, saying that she was wrong and he did it in the long division method.
She explained it to him and pointed out, We both got the same answer’. He accepted it and appreciated her. Now the same student is the class topper of International Math Olympiad (IMO) exam and has been selected for the second round of the IMO exam, 2020.
Whether it is learning division in a narrow sense or learning mathematics in a broader sense, there are many factors that affect the learning of children. Teachers and their pedagogical knowledge, use of inappropriate methods and lack of understanding, classroom settings and other factors affect the learning of a child.
There is a belief that maths is generally difficult for children and in particular, for girls. This is just gender bias. Ernest’s 1976 study focuses on gender differences in elementary school children and the attitudes of their teachers. This study concludes that the idea that males are superior to females in mathematics is clearly a misconception and that the lack of women as professional mathematicians is likely more due to cultural influences than a lack of ability.
So, I took this opportunity of working as a teacher in a girls’ school to prove that learning mathematics is everyone’s right and every child can do it. Providing opportunities for girls from deprived communities to excel in mathematics is an important social change ahead of us. I am working towards it.
Gomathy Ramamoorthy is a primary school teacher at the Savarayalu Nayagar Govt Girls’ Primary School, Puducherry. She has 16 years of teaching experience and is a Resource Person in CBSE curriculum transaction at Puducherry.
She runs a maths lab in her school and is the editor of ARRAY’, a bilingual children’s maths magazine that is fully contributed by children. Her interest is in experimenting with different pedagogies in teaching and learning of mathematics for conceptual understanding.
She has been a member of the textbook framing committee (2018) at SCERT (Tamil Nadu) and has contributed as one of the authors in framing class I Mathematics textbook.
She can be contacted at gomurama@gmail.com
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# circular motion in vertical and horizantal
Sunil Kumar
10 years ago
Hi Umer Shafi
I would request you to kindly be a little more specific in your question.
"Circular motion in vertical and horizontal .."..(what) ?? Plane or space ?
Thank you
Thanks & Regards
Sunil Kumar,
M.Sc., IIT Kharagpur
Aman sharma
41 Points
10 years ago
circular motion calculation deals with f=ma frmula
horizontal circular motion deals with centripetal force f=mv^2/r
varalakshmi
38 Points
10 years ago
circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation, or non-uniform with a changing rate of rotation.
When an object is experiencing uniform circular motion, it is ... system that is horizontal and vertical, because the centripetal acceleration will be acting on the object.
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# Use various trigonometric identities to simplify the expression then integrate J3 sin20cos5ed0
###### Question:
Use various trigonometric identities to simplify the expression then integrate J3 sin20cos5ed0
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# V5 and the Gyro sensor
My team got the gyro to work. I’m pretty sure others did as well. There were some posts here on the forum with sample code and suggestions on how to get it to work. You might want to try a search for those threads.
1 Like
I got this to work. I got the info from my coach and my other teammate that no one else was able to do it.
Can you do multiple turns with this code? I ask because gyro turns using the V5 are relative not absolute.
1 Like
Yes, as long as it does not go over 720 degrees in one command since the overflow code cant take it.
overflow code
if(gyroValue > 3600)
{
gyroValue = gyroValue - 3600;
}
else if(gyroValue < -3600)
{
gyroValue = gyroValue + 3600;
}
If I was not using a holonomic in testing (Lots of play in the drivebase) the code and gyro should be able to do exact turns.
Many people have managed to use the gyro sensor, there is nothing inherently difficult about it.
However, this is why I think gyro is not the best solution for V5:
Yes, when the code starts, the gyro goes through a calibration process that requires the sensor to be completely still. That should just happen on startup, and you should not have to worry about it in autonomous unless you create the sensor right before you start autonomous, which is bad structure.
When dealing with sensors, you should never have to reset them. Doing so is a blunt and inefficient way to deal with relative angles.
Instead, the solution is to take into account the current position of the gyro to convert your relative angle into an absolute angle.
For example:
//given wanted angle of 90 from current position
int wantedAngle = 90;
//instead of setting gyro to 0 and turning until gyro reads 90
//find angle that is 90 away from current angle
int target = Gyro.value(rotationUnits::degrees) + wantedAngle;
//now you can use that value as your target
Also, there is a way to simplify the logic in a code such as this. This is just a suggestion, but it helps with neatness.
Instead of providing logic that takes into account direction, if you use some simple math, you can reduce the complexity of the code. For this, you don’t need to specify the direction, just a negative angle.
void gyroTurn (int angle) {
int target = Gyro.value(rotationUnits::degrees) + angle;
int error = 0; //represents value between current angle and target angle
//do-loops loop at least once, we are using it to calculate error
do {
error = Gyro.value(rotationUnits::degrees) - target;
if(error < 0) {
//turn right
leftDrive(25);
rightDrive(-25);
} else {
//turn left
leftDrive(-25);
rightDrive(25);
}
} while(abs(error) < 10) //exit when error < 10
stopHold();
}
Anyway, just a suggestion to make things neater. If you wanted to do a P (proportional) loop to increase speed, you could just replace the if/else for the direction with
leftDrive(-error * constant);
rightDrive(error * constant);
which would go faster the further you are from the goal and slower the closer you are. You would tune the constant to provide the relation between distance from goal and motor power.
Small little nitpick, if(turnRight==true) is redundant
It is cleaner to do if(turnRight), as before you are basically typing if(true==true), which is redundant. If you want it to return false when the output of the evaluation is true, you can do if(!turnRight) which translates to “if not turnRight”.
Finally, when you post code on the forum, please format and wrap your code in little
```cpp
```
code tags, it helps with readability.
Hope this post was able to teach someone something.
2 Likes
You could alternatively use an encoder for turning, it’ll be a lot simpler to integrate into your code and doesn’t suffer from errors if you use a tensioned free spinning wheel system. Although the biggest set back is it takes up 2 ports instead of 1.
Do you mean 4 ports instead of 1?
To be able to measure rotation using encoders, you need 2 of them.
Nope. You can use an encoder placed horizontally on the back end of the chassis (or any part within the middle of the chassis) to measure rotational displacement. Granted you will need to experiment with it, it’s not like you can use degree values like you can with a gyro.
Basically, the encoder will measure a certain value whenever the two sides of the chassis move in opposite directions (this is what the chassis does when turning) and this will enable you to find a certain angle of the robot. I recommend graphing different values of the encoder in relation to the orientation of the robot. This will let you model the behavior of the robot (when turning) with code.
I suppose. Its just it feels like it can be quite inaccurate, and assumes the robot always turns on the exact same point. If you use omni wheels, the turning might be very different depending on the acceleration and speed.
Maybe it is possible to get it to work, but it does not feel very robust compared to two 2 vertical encoders or even a gyro.
If you use all omni wheels then the robot will rotate about a singular point. I don’t understand what you difference would using 2 encoders in a perpendicular orientation would make other than utilizing an extra port. You could easily achieve the same results as those encoders with the integrated IMEs. But I think you’re misunderstanding the geometry of a symmetrical, square robot. The horizontal encoder would work, in fact, it makes odometry possible.
Also, the horizontal encoder measures angular displacement and the point about which the robot rotates is arbitrary to the angle of its orientation.
I am a little confused.
I did not say two perpendicular wheels, I meant two parallel wheels like this (ignore the back wheel).
(image taken from pilons)
What I am understanding you saying is to just use the one horizontal wheel (labeled back wheel).
What I was saying is using just that one encoder to measure angle must be inaccurate. Depending how the robot turns, especially if it is not consistent (affected by dynamic speed/weight), that wheel will not always spin in proportion to the angle of the robot . If this is where I am wrong, feel free to correct me.
For example, if for some reason the robot was pushed so that the robot rotated directly around the horizontal wheel, causing there to be no encoder movement, would it not lose all accuracy? If there were two parallel wheels like in the diagram, it would not matter what the center of turning would be. And this will still be much more accurate than integrated sensors due to wheel slip and inexact point of turning (within the width of a large wheel).
Finally, in odometry the horizontal tracking wheel is not used for angle. It is impossible to differentiate the movement of the horizontal wheel between horizontal displacement or rotation of the robot. What tracking algorithms do is measure the orientation of the robot using the two vertical wheels with this formula:
dRadians = (dLeftInch - dRightInch) / chassisWidthInch;
Then, it can calculate how much it expects the horizontal wheel to move given that rotation.
It cancels out that movement from the horizontal wheel, and what remains is the horizontal displacement.
If all you had was the horizontal wheel, it is impossible to differentiate horizontal displacement or rotation. What if you had a tall stack at the front of your robot that made it so when the robot turns it does a slight arc around the stack? Then, the wheel would be spinning much more than usual for the same angular rotation of the robot.
If I am missing something important please correct me =)
1 Like
I think you’re overthinking this way too much. We’re talking about replacing a gyro with an encoder, we’re not talking about traveling in arcs. If you rotate about the center of the robot, you can use the horizontal encoder to correctly orient the robot to a desired target angle.
Like I said, you could do some math to find a function that correctly models the relationship of the encoder with different angles of the robot. Using that, you can simply use that function you came up with to convert an angle as an input from a parameter in a method, and use that to have the robot rotate until it reaches a desired encoder value that corresponds to that specific angle.
Also, what I meant by perpendicular wheels is that the “vertical” wheels are perpendicular to the horizontal wheel. Also, that wheel will always spin in proportion to the angle of the robot in autonomous. Remember that we aren’t allowed to cross the autonomous line and we can eliminate jerk with slew rate control and PID. This will essentially make motion really smooth and controllable so you can always measure if the encoder changed position or not. Also, the free spinning wheel will be tensioned down to the ground. Now, the only scenario I see this not working in is if your partner drives into you during autonomous and makes you travel about an arc. But as for rotating in place (which can only be done with all omni wheels), you can get a relatively accurate reading with an encoder if you code it properly.
Just visualize the motion of the horizontal free spinning wheel as the robot turns. It will essentially be moving in a “straight” line because that encoder is tangent to the robot’s rotation. I don’t know if that made sense, but that’s how I visualize it. There is a clear correlation to robot orientation and that encoder value, and you can model that with a function. I understand how the 2 “vertical” wheels with encoders work but at the point you’re better off using the IMEs.
1 Like
Alright. I still am not convinced, but I can see how it is feasible to get a reading of the robot’s orientation based on one sensor, if we assume the robot will spin in the same way.
What I meant with arcs is if the center of turning of the robot was at the very front of the robot (due to a dynamic imbalance of weight), it would cause the omni back of the robot to travel in a sideways arc and cover more distance when rotating, messing up the conversion between encoder movement and robot angle.
Yeah, it is probably possible to ensure the robot rotates in a consistent way in autonomous, and make it good enough. I can just see a lot of possibility of error. However, I understand how you would go about modeling the rotation.
Anyways, good discussion.
2 Likes
I like discussion, it’s how we learn as a community. The big take away here is that there are multiple solutions to the same problem.
3 Likes
It works great in PROS. I used it last year and it worked great.
Can you explain how you arrived at this formula and what dLeftInch and dRightInch mean? I know it is from the Pilon’s document, but I didn’t understand how they did that because they didn’t really explain what ΔL and ΔR mean. They explained the rest of the variables, but not those two.
I would assume ΔL and ΔR are counts reported by the left and right side encoders.
To understand the formula you may want to look at this example:
We have a robot of width “l” that turns around some point which is “r” inches away from it. Let say over the time unit “t” left wheel encoder reports 3 counts (travel distance) and the right wheel reports 5.
If you know the width between the wheels then you can calculate both the unknown radius “r” and angle “theta”.
The movement of the robot is the sum of translation of center of the robot along some path and the rotation of the robot body around that point.
Special case would be when the robot only rotates around the point between its wheels.
1 Like
Yeah. I figured out the math at one point, but I forget how I did it.
The math works out that (left-right)/width gives you angle of robot.
@technik3k’s explanation is good. Just a small correction, I think ΔL and ΔR means the delta in inches that both wheels have moved, converted from ticks (counts).
Just to clarify, d means delta, which means change from last iteration.
You first want to find the new encoder readings. Then, you want to convert that into a standard unit, such as Inch. What matters is that it is the same unit the chassis is measured with. Then, to find the delta wheel movement in inches, you substract the old encoder inch from the new encoder inch.
double newLeftInch = leftTicks * ticksToInch;
double newRightInch = rightTicks * ticksToInch;
double dLeftInch = newLeftInch - lastLeftInch;
double dRightInch = newRightInch - lastRightInch;
lastLeftInch = newLeftInch;
lastRightInch = newRightInch;
Now that you have the amount the wheels moved in inches since the last iteration, you can calculate how much the robot has rotated in that iteration using the formula:
double dAngle = (dLeftInch - dRightInch) / chassisWidthInch;
double newAngle = lastAngle + dAngle;
Of course, you can also modify the formula to be
double leftInch = leftTicks * ticksToInch;
double rightInch = rightTicks * ticksToInch;
double newAngle = (leftInch - rightInch) / chassisWidthInch;
where the wheel readings are measured not as deltas, but instead as relative movements since the program began.
What matters is that if you simply keep track of how the left and right wheels have moved (in distances/rotations), you can know at anytime the robot’s angle since you started measuring the wheels.
Does that clear things up?
1 Like
@technik3k @theol0403
Thank you very much for your explanations. That makes perfect sense. Appreciate it
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# Unit 2 Family Materials
### Addition and Subtraction Story Problems
In this unit, students solve new types of story problems within 10. They develop an understanding of the meaning of the equal sign and connect story problems to equations.
### Section A: Add To/Take From Story Problems
In this section, students revisit familiar story problem types. Students work formally with equations for the first time. They write equations such as $$2 + 7 = \boxed{9}$$ and learn to draw a box around the answer to the question in the story problem. Students work with problems where they have to figure out how much is being added:
His sister gave him some pencils.
Now, Diego has 9 pencils.
How many pencils did Diego’s sister give him?
Students see that these problems can be solved by either addition or subtraction. They can solve this problem by counting on from 7 to 9 and write the equation $$7 + \boxed{2} = 9$$. Students can also solve this problem by taking away 7 from 9, and write the equation $$9 - 7 = \boxed{2}$$.
### Section B: Put Together/Take Apart Problems
In this section, students solve problems where two groups are put together. In some problems they find the total, and in other problems the total is given and they find the missing group. Students solve problems in the context of Shake and Spill, a game that uses two-color counters. Counters are put into a cup and spilled out. Students make observations about what they see or different combinations that might occur.
Tyler is playing Shake and Spill. During his first round he spilled these counters.
Write 2 equations to represent his counters.
Show other combinations of red and yellow counters that Tyler could spill.
With this type of problem, students can look at different kinds of equations, such as those with the total before the equal sign ($$7 = 4 + 3$$).
### Section C: Compare Story Problems
In this section, students solve story problems where they find “how many more” or “how many fewer” one group has than another group, such as:
There are 8 glue sticks and 3 scissors at the art station.
How many fewer scissors are there than glue sticks?
Students think about the relationship between addition and subtraction. They start by considering how many they need to add to make two towers the same length. For example:
How many more cubes does Clare have than Andre?
For this type of problem, students may count the extra cubes in Clare’s tower to find the answer. They may start at 3 and count up to 10 or start at 10 and count back to 3. Students analyze both addition ($$3 + 7 = 10$$) and subtraction ($$10 - 3 = 7$$) equations.
### Section D: All Kinds of Story Problems
This section brings the work of the unit together as students solve a variety of problem types and make sense of equations with a symbol for the unknown, such as $$10 = \!\boxed{\phantom{4}} + 6$$.
### Try it at home!
Near the end of the unit, ask your student to solve the following word problems:
1. Clare has 8 pencils. Andre has 10 pencils. How many more pencils does Andre have?
2. Diego had 6 pens. His mother gave him some pens. Now he has 9 pens. How many pens did Diego's mother give him?
Questions that may be helpful as they work:
• How could you draw the problem?
• How can you count on or take away to find the answer?
• What equation can you write to represent this problem?
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https://it.mathworks.com/matlabcentral/cody/problems/43296-refresh-your-system-of-equations/solutions/1690522
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Cody
# Problem 43296. Refresh your system of equations
Solution 1690522
Submitted on 11 Dec 2018 by Martin C.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
xyz = [1 -1 2; 0 2 5; 4 0 -3]; abc = [21; 21; 21]; y_correct = [ 9 -2 5 ]; assert(sum((transpose(answerMe(xyz,abc))-y_correct))<0.01)
y = 9 -2 5
2 Pass
xyz = [1 2; 1 -2]; abc = [3; -1]; y_correct = [ 1 1]; assert(isequal(nnz(answerMe(xyz,abc)-y_correct),0))
y = 1 1
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
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+0
# Help
0
278
4
Marvin randomly places 7 Scuba Steve action figures and 6 Diving Dan action figures in 4 distinguishable boxes. If each box must have at least one of each type of action figure, how many ways can Marvin do this?
Oct 1, 2022
#1
+2667
0
After we put 1 of each action figure in each box, there are 3 Scuba Steves and 2 Diving Dans left.
Using stars and bars for the number of Scuba Steves gives us 3 stars and 3 bars, for $${6 \choose 3} = 20$$ combinations
Doing the same thing for the Diving Dans gives us 2 stars and 3 bars for $${5 \choose 3} = 10$$ combinations.
So, there are $$20 \times 10 = \color{brown}\boxed{200}$$ ways to do this.
Oct 1, 2022
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The new Gynzy
500+ new lessons
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# Taal
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- Lesson plan -
Students will practice beginning addition and subtraction facts. They will be introduced to the inverse relationship of addition and subtraction.
- Lesson plan -
Solve addition and subtraction problems, including word problems involving two-and-three digit numbers. Complete in addition and subtraction sentence with missing addends including inverse relationships between and addition and subtraction.
- Lesson plan -
Students will identify adjectives and use them in context.
- Activity -
Have students show an understanding of adjectives by putting the words in the correct order.
- Lesson plan -
Students will be able to identify and distinguish between adjectives and adverbs and use them correctly in sentences.
- Lesson plan -
Students will be able to understand the difference between adjectives and adverbs and use them correctly in sentences.
- Activity -
Choose between adjectives and adverbs, depending on what is to be modified.
- Activity -
Open one square each day and you will see delightful Christmas scene on your screen.
- Lesson plan -
Students will be able to identify and to use adverbs.
### Algebra Tiles
- Lesson plan -
Model operations with polynomial expressions using algebra Tiles. Add, subtract, multiply and divide polynomials.
### Algebraic Expressions
- Lesson plan -
Write expressions and sentences as Algebraic expressions; evaluate Algebraic expressions.
### Algebraic Expressions
- Activity -
Learn how to translate English phrases into algebraic expressions.
### Algebraic Fractions
- Lesson plan -
Simplify algebraic fractions. Add, subtract, multiply and divide algebraic fractions.
- Lesson plan -
Students will understand the characteristics and importance of air to living and non-living things.
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https://www.clutchprep.com/chemistry/practice-problems/12699/the-percentage-of-water-in-an-unknown-hydrated-salt-is-to-be-determined-by-weigh
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# Problem: The percentage of water in an unknown hydrated salt is to be determined by weighing a sample of the salt, heating it to drive off the water, cooling to room temperature, and re-weighing. Which procedural mistake would result in determining a percentage of water that is too low?(A) I only(B) II only(C) both I and II(D) neither I nor II
###### FREE Expert Solution
94% (160 ratings)
###### Problem Details
The percentage of water in an unknown hydrated salt is to be determined by weighing a sample of the salt, heating it to drive off the water, cooling to room temperature, and re-weighing. Which procedural mistake would result in determining a percentage of water that is too low?
(A) I only
(B) II only
(C) both I and II
(D) neither I nor II
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Accuracy & Precision concept. If you need more Accuracy & Precision practice, you can also practice Accuracy & Precision practice problems.
What is the difficulty of this problem?
Our tutors rated the difficulty ofThe percentage of water in an unknown hydrated salt is to be...as low difficulty.
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Our expert Chemistry tutor, Jules took 2 minutes and 42 seconds to solve this problem. You can follow their steps in the video explanation above.
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Based on our data, we think this problem is relevant for Professor Dougherty's class at OSU.
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# How to solve the distance formula
In this blog post, we will provide you with a step-by-step guide on How to solve the distance formula. Our website will give you answers to homework.
## How can we solve the distance formula
This can be a great way to check your work or to see How to solve the distance formula. There is no one-size-fits-all solution to solve problems, but there are some general steps that can be helpful in finding a solution. First, try to identify the root cause of the problem. Once the root cause is identified, brainstorm possible solutions. Then, evaluate the possible solutions and choose the one that is most likely to be effective. Finally, implement the chosen solution and monitor the results to see if the problem is actually solved.
A system of equations is a set of two or more equations that share the same variables. In order to solve a system of equations, all of the equations must be satisfied simultaneously. This can be done by using a variety of methods, including substitution, elimination, and graphing. One useful tool for solving systems of equations is a compound inequality solver. A compound inequality solver is a tool that allows
To solve for the x intercept, set y = 0 and solve for x. For example, if the equation is y = 2x + 5, then setting y = 0 and solving for x would give x = -5/2. This means that the graph intersects the x-axis at -5/2.
There are a few ways to solve imaginary numbers. One way is to use the quadratic equation. This equation can be used to find the square root of any number, including imaginary numbers. Another way to solve imaginary numbers is to use Euler's formula
First, it can be helpful to break the problem down into smaller pieces and solve each piece separately. Additionally, it can be helpful to use symmetry to simplify the problem. Finally, it may be helpful to draw a diagram to visualize the problem and make it easier to identify a solution.
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https://www.scientificamerican.com/article/make-a-toy-built-for-collisions/
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Key concepts
Physics
Collision
Momentum
Energy
Introduction
Did it ever occur to you that tennis, bowling and shopping carts bumping into each other all involve collisions? It is fascinating how just a few rules of physics can predict the outcome of these collisions. You can discover these rules yourself with a fun homemade toy. After creating and playing with the toys in this activity, you will be one step closer to understanding what happens when you hit a tennis ball or go bowling!
Background
Have you ever heard someone say that something "has a lot of momentum?" In everyday language we use "a lot of momentum" to describe things that are hard to stop. In physics an object's momentum depends on its speed—how fast it moves—and its mass—how much stuff it is made off. Momentum also has a direction—the same direction the object is moving. For an object to gain momentum it can gain speed, gain mass or gain both. To give a shopping cart rolling downhill more momentum, you can make it move faster (increase its speed), load more weight (increase its mass) or do both. You probably intuitively know that the shopping cart with the biggest momentum—the fast-moving, heavily loaded cart—is hardest to stop. It also creates the biggest impact when colliding with something.
Physicists discovered that objects transfer momentum when they collide. But even more they observed that the total momentum is conserved during a collision. If you have seen a row of shopping carts creeping away after a fast-moving but empty shopping cart collided into them, you have witnessed conservation of momentum. The fast-moving light cart transferred its momentum to the much heavier row of carts. Its momentum could only make this heavy mass move a little. There is a little more math involved when both objects are moving before the collision, but even then the total momentum is always conserved.
Energy is the other quantity that gets transferred during collisions. More surprisingly the energy associated with the movement of the colliding objects is conserved in collisions, at least in collisions when the colliding objects do not deform, crack or break at all. In real life there is almost always some deformation, and some energy of movement will almost always be converted into other types of energy, such as heat or sound. The fun toy created in this activity will help you get an intuitive feeling of how momentum and energy are conserved during collisions.
Materials
• Two identical balls, half inch to three inches in diameter (large round wooden beads, ping-pong balls, small bouncy balls and round erasers work well)
• At least one more ball (half inch to three inches in diameter) of a different mass (this ball needs to be at least three times as heavy or three times as light as the identical balls)
• Needle and thimble or strong glue
• Thick thread (preferably not twine but a slightly thicker sturdy thread)
• Scissors
• Ruler
Preparation
• Cut the thread into pieces about 30 centimeters long (one for each ball).
• If you are using beads, pull the thread through the hole, and make a knot on one end of the thread big enough so the bead cannot slip over it.
• For any balls you are using, ask an adult to help you hang the balls on threads. Ask them to pierce a threaded needle through the middle of the ball. Make sure they wear a thimble. If needed, pliers can help pull the needle through. Make a knot on one end of the thread so the ball can't fall off. If this is too hard, strong glue can be used to attach a thread to the balls.
Procedure
• Pinch the threads of the two identical balls (or beads) between your thumb and finger, letting the balls hang down. Slide the thread of one ball up or down until the balls are level.
• Pull one ball up 90 degrees, keeping its thread taut so it forms a horizontal line.
• Keep the hand holding the threads steady while you release the ball and observe what happens. Repeat the test a few times. What happens (almost) every time?
• For the second test go through the same procedure, only now jerk the hand holding the threads up about an inch each time the balls move away from each other and move the hand back down when the balls approach each other. You might need to try it a few times before you can keep the balls bouncing. How is this different from the first test? Why would this be the case?
• What do you think will happen if you switch one ball with a heavier or lighter ball?
• Hold the threads of two non-identical balls between your thumb and finger. Repeat the first test. First try releasing the lighter ball and observe, then switch to the heavier ball. How is this similar and how is it different from what you observed when using identical balls?
• Keep the threads of the non-identical balls pinched between your fingers and try the second test. What do you observe now? Can you explain your observations?
• To turn your tests into a toy, first select the combination of balls you like best. Then knot the thread ends farthest from the balls together. Before you tighten the knot adjust the distances between the knot and the balls so these are identical.
• Extra: Test different combinations of balls. What can we learn from a test using same-size balls of different masses? What about balls of equal mass but different sizes?
• Extra: Test the role of the material of the balls. What happens if you use two wooden balls instead of two rubber or two ping-pong balls? Which balls keep bouncing the longest if you keep your hand still? Which ones do you have to jerk more to keep them bouncing? Why would this be the case?
• Extra: If you have more identical small balls (e.g. marbles), you can do another surprising collision test. Place a row of the balls in a crease of an opened book. All balls should touch each other. Softly shoot a ball along the crease into the end of the row of balls and observe what happens. Why would this happen? Is it different if you shoot two balls into the other balls, or if you shoot a heavier ball into the row of balls?
Observations and results
When you tried with the identical balls did you witness the balls exchange speed during the collision? Did you see how jerking the system up makes it possible to keep the balls bouncing? When you tried with non-identical balls did you notice that the collision didn't cause the heavier ball to move as much while the lighter ball was launched off a high speed?
When two balls collide they exchange momentum. For identical balls this means one ball is launched off with the speed of the other ball each time they collide. This explains why the initially motionless ball shot off when bumped by another ball leaving the first ball almost motionless. Before long the ball that was shot off returned and bumped into the first, which shot off returned and so on.
If the second ball shoots off with the same speed, that ball should shoot up to the same height from which you released the first ball. Was that what you observed? Probably not! With each collision some energy goes into moving the tiny particles that make up the balls or particles in the air. We observe this as energy of motion being transformed into heat or sound. As a result the second ball shoots off with a smaller speed than the speed at which it was hit. The difference depends on the material of your balls. Bouncy balls will show a small difference; the speed will decrease only slightly with each collision, and the balls bounce back and forth for a long time. Wooden balls will have a bigger difference and bounce back and forth only a few times before all energy is transformed into heat or sound. Did you notice that when you jerked the system up just after the collision, the balls could keep on going? By doing so you added energy back into the system allowing the balls to keep bouncing.
Non-identical balls also exchange momentum, but if their masses are different, there is more to it than a simple exchange of speed. Did you notice how the motion of the lighter ball was only able to make the heavier ball creep up a little bit? On the other hand when the heavier ball bumped into the lighter ball its momentum could make the lighter ball move a lot. This is because the same momentum can make a lighter ball move way faster than a heavy ball.
More to explore
Momentum and Collision, from Ducksters
Energetic 2-Ball Bounces, from Scientific American
Make Craters with Mini Meteors, from Scientific American
Build a Gauss Rifle!, from Science Buddies
Science Activities for All Ages!, from Science Buddies
This activity brought to you in partnership with Science Buddies
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Problem 156
# Find the values of $$(x, y, z)$$ that minimize $$F(x, y, z)=x y+2 y z+2 x z$$ given the condition $$\mathrm{G}(\mathrm{x}, \mathrm{y}, z)=\mathrm{xyz}=32$$.
Expert verified
The values of $$(x, y, z)$$ that minimize the function $$F(x, y, z)=xy+2yz+2xz$$ given the constraint $$G(x, y, z)=xyz=32$$ are $$(x, y, z) = (4, 4, 2)$$.
See the step by step solution
## Step 1: Set up the Lagrange's equations
The method of Lagrange multipliers states that we can find the minimum of a function $$F(x,y,z)$$ subject to a constraint $$G(x,y,z)=c$$ by solving the following system of equations: $$\frac{\partial F}{\partial x} = \lambda \frac{\partial G}{\partial x}, \frac{\partial F}{\partial y} = \lambda \frac{\partial G}{\partial y}, \frac{\partial F}{\partial z} = \lambda \frac{\partial G}{\partial z},$$ along with the constraint $$G(x, y, z) = 32$$. In our case: $$F(x, y, z) = xy + 2yz + 2xz$$ $$G(x, y, z) = xyz$$ Now, we calculate the partial derivatives of $$F$$ and $$G$$ with respect to all variables. With respect to $$x$$: $$\frac{\partial F}{\partial x} = y + 2z$$ $$\frac{\partial G}{\partial x} = yz$$ With respect to $$y$$: $$\frac{\partial F}{\partial y} = x + 2z$$ $$\frac{\partial G}{\partial y} = xz$$ With respect to $$z$$: $$\frac{\partial F}{\partial z} = 2y + 2x$$ $$\frac{\partial G}{\partial z} = xy$$
## Step 2: Solve the system of equations
Now, we have the following set of equations: 1. $$y+2z=\lambda yz$$ 2. $$x+2z=\lambda xz$$ 3. $$2y+2x=\lambda xy$$ 4. $$xyz=32$$ We can attempt to solve this system of equations to find the values of $$x, y, z$$, and $$\lambda$$. Divide equation 1 by equation 2: $\frac{y+2z}{x+2z}=\frac{\lambda yz}{\lambda xz} \Rightarrow \frac{y}{x}=\frac{z}{z} \Rightarrow y=x$ Now, rewrite equation 3 using this new information: $2x+2x=\lambda x^2 \Rightarrow 4x=\lambda x^2$ Since $$x\neq 0$$, we can divide both sides by $$x$$: $4=\lambda x$ Now, substitute $$y=x$$ in equation 4: $x^2z=32 \Rightarrow z=\frac{32}{x^2}$ Plug the value of $$z$$ into the equation for $$\lambda$$: $\lambda=\frac{4}{x}$ Now, substitute $$\lambda$$ back into equation 1: $x+2\frac{32}{x^2}=\frac{4}{x}\cdot x\frac{32}{x^2}$ $x+2\frac{32}{x^2}=\frac{128}{x^2}$ $x^3+64=128$ $x^3=64$ $x=4$ Now, using the initial substitution $$y=x$$, we can find the value of $$y$$: $y=4$ Finally, substitute the value of $$x$$ into the equation for $$z$$: $z=\frac{32}{4^2}=\frac{32}{16}=2$
## Step 3: Verify the solution
We have found that the values of $$x, y, z$$ that minimize the function are $$x=4$$, $$y=4$$, and $$z=2$$. To verify the solution, we will check if these values satisfy the constraint: $$G(x, y, z)=xyz=(4)(4)(2)=32$$ The constraint is satisfied, so the values of $$(x, y, z) = (4, 4, 2)$$ indeed minimize the function $$F(x, y, z)$$.
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# B Derivation of time dilation without light clocks
Tags:
1. Sep 3, 2016
### Tazerfish
In the way I was taught about special relativity, time dilation is like the fundamental building block from which you derive things like relativistic mass and length contraction.
So it has always struck me as quite odd, that the derivation of time dilation (in some sense the basis of special relativity) uses something as abstract as light bouncing up and down in a box to measure time.
Something that has never been built in reality to my knowlege.
I don't have any problems with this proof.It is very elegant and simple.
Yet there rarely is just one nice derivation or proof for such things
and I have never seen any alternetative for this case.
Have you ?
Is there a way to derive it by considering some other process ? :)
2. Sep 3, 2016
### A.T.
That's a confusing way to think about it. Time dilation and length contraction are on the same footing, as consequences of the Lorentz transformation.
Derive from what? From the two postulates? The key thing in them is the source independence of light's propagation, so I don't see how you can avoid using that.
3. Sep 3, 2016
### Tazerfish
I thought about proving that some other process would slow down as you approached the speed of light.
Independant of the consideration of light clocks.Like an oscillation of some sort.
But the highlighted part in your post has been key.The derivation seems much more intuitive now.
Thanks
4. Sep 3, 2016
### Ibix
Pal (https://arxiv.org/abs/physics/0302045) derives transforms using the principle of relativity only (and an assumption of linearity in the transforms, and assumptions of homogeneity and isotropy of space). There's no mention of light anywhere. He ends up with two options - Galilean relativity and Einsteinian relativity. You can then eliminate Galilean relativity by experiment.
Or one can postulate that spacetime is a 4d entity obeying Minkowski geometry with c as a scale factor between spatial and temporal directions; everything falls out of that, and once again you can verify by experiment.
In practice, no one uses light clocks as there are perfectly good (precise and reliable) atomic clocks. The beauty of the light clock for thought experiments is that one doesn't need to know how velocities transform in general. One simply has to postulate that the speed of light is frame invariant and that, if I see you doing +v, you see me doing -v. You can, in principle, do thought experiments with a pendulum clock, but you need to know how to transform the velocity of the pendulum - and the full velocity transforms are a bit much to pull out of thin air.
5. Sep 3, 2016
### Orodruin
Staff Emeritus
This is not a very pedagogical nor modern way to teach SR. I doubt time dilation was ever a fundamental building block in any reasonable approach.
Also note that relativistic mass is an antiquated concept that is generally not used in modern treatments.
This is just one of the mote heuristic ways of deriving time dilation. The more mathematical approach is to just derive it straight out of the Lorentz transformations.
6. Sep 3, 2016
### A.T.
...involving light.
7. Sep 3, 2016
### Ibix
Not necessarily, I think. Cosmic ray muons are tough to explain without time dilation - no light needed in that one.
8. Sep 3, 2016
### Tazerfish
I might have to add that we never went into it in great detail.
We didn't have a lot of time for the topic and our mathematical knowlege was quite limited at the time.
I don't blame my teacher for going the quick and ditry route, we simply didn't have the time to do it right.
Sometimes it is better to teach something that is not entirely correct and to just clear up the misconceptions afterwards.
9. Sep 3, 2016
### A.T.
Proving based on what?
10. Sep 3, 2016
### robphy
Try this [apparently] not well known method called the k-calculus by Bondi, which I recently mentioned in another thread.
https://www.physicsforums.com/threa...n-these-approaches-to-sr.883403/#post-5554866
"There the metric and the Lorentz transformation are not in the foreground of the discussion.
It is the principles of relativity, with focus on the radar method and the Doppler effect.
(Secretly, the approach is using the eigenbasis of the Lorentz Transformation.)
https://archive.org/details/RelativityCommonSense
https://en.wikipedia.org/wiki/Bondi_k-calculus
"
(Bondi)
You probably have to go back to see the development of the diagram,
and ahead to see how k is related to v.
k is the Doppler factor
(D'Inverno)
Here's an ancient post of mine on it:
Last edited: Sep 3, 2016
11. Sep 5, 2016
### stevendaryl
Staff Emeritus
I do not particularly prefer the abstract way of deriving the Lorentz transformations starting with homogeneity, isotropy, etc. I think it's one of those things where if you already understand relativity to a certain level of competence, then you can appreciate such a derivation, but I think if that were the first derivation I had seen, my eyes would glaze over, and I would have a hard time understanding why anyone would be interested in the topic. I found, as a young man (I think I was about 13 or 14 when I first went through it) the derivation in terms of rods and clocks and light signals really captured my imagination, and made me want to learn physics. I doubt very seriously whether I would have been similarly inspired by an abstract derivation. Sort of similarly, I find an introduction to quantum mechanics that starts with more-or-less concrete experiments such as the two-slit experiment to be a lot more inspiring than a derivation of Schrodinger's equation from principles of Galilean invariance and expectation values for observables. I'm not at all disparaging the more rigorous mathematical treatments, but in my opinion, it's useful to see both the rigorous development and a nonrigorous, but intuitive development. The worry of many physicists is that the intuitive derivations are always misleading or else have hidden (and often false or unrealistic) assumptions, and that students will get the wrong impression. I don't think that's a problem, as long as the student is told that they are only getting an incomplete picture, and that there is more to learn. If there were no student misconceptions, then there would be no need for a PhysicsForums.
Last edited: Sep 5, 2016
12. Sep 5, 2016
### vanhees71
Sure, I'd not use the abstract derivation for an introductory physics course or even at high school.
13. Sep 5, 2016
### Mister T
That's just one way of teaching it. One has to start somewhere. It's actually much better to start with the relativity of simultaneity and never mention relativistic mass. But that's just my teaching opinion.
You're taking the next step in thinking about it, rather than just leaving the lesson behind.
There may not have been any wrong doing. Teachers can, at best, begin the learning process. The student must go beyond the teacher's lesson if any real understanding is to occur. Unfortunately, for the last 15 years or so, the American primary and especially secondary systems of education have been focused on holding teachers accountable for student learning. A well-intentioned but misguided approach because it focuses on the teacher rather than the student. Some parents have taken this to mean that if their offspring are not learning, it must be the teacher's fault. (The more obstinate among them will maintain that learning did occur despite the poor teaching, but the substandard grades are due to the teacher's flawed evaluation methods).
It is not a thing to be proven. It is a thing to be demonstrated by observing the way real clocks behave. The light clock is just a teaching tool. Many if not most physicists dismissed it as flawed until its validity was demonstrated.
The phrase "approach the speed of light" can be misleading. First, it's the fact that you're observing something else move relative to you. And yes, the faster the speed the greater the effect, but the effect is there at all speeds. And it must be taken into account in cases where it makes a difference, such as the GPS satellites that are moving at only 0.001% of the speed of light.
14. Sep 6, 2016
### Battlemage!
I learned to derive it using light clocks too, but not in a class (in class they just said "this is the Lorentz transformation equations") . I just noticed the Lorentz factor looking like a side of a right triangle and went from there. But later on I learned a better way that really only assumes that the formula distance = rate x time is the same regardless of inertial coordinate system and requires only algebra (and probably some characteristics of space and time that I just took for granted).
I'll post it and you tell me if this might be better than the light clock way. Skipping no steps so that any mistake I make will be clear for someone to fix.
You have two reference frames, S and S' moving at some speed with respect to each other with their x and x' axes coinciding.
In S, you can shoot a beam of light in either direction along the x-axis. When you do, it's distance x is given by x = ct or x = -ct (depending on the direction). Likewise in S' you can do the same: x' = ct' or x' = -ct'. Setting each equal to zero gives: x - ct = 0, x + ct = 0 and x'-ct' = 0, x'-ct' = 0. Pretty straight forward so far. Then you can write them the other way: ct - x =0, ct + x = 0, and ct'-x' =0 and ct' + x' = 0.
Next I just assumed there was some functions A and B such that (1) x - ct = A(x' - ct'), (2) x + ct = B(x'+ct') and (3) ct - x = A(ct' - x'), (4) ct' + x' = B(ct + x).
Then add (1) and (2) together to get:
$$x = \frac{A+B}{2} x' + \frac{B-A}{2} ct'$$
And add (3) and (4) to get:
$$ct = \frac{A+B}{2} ct' + \frac{B-A}{2} x'$$
Then to make it easier to read, let
$$\frac{A+B}{2} = γ$$
and
$$\frac{B-A}{2} = ξ$$
leaving
(5)
$$x = γx' + ξct'$$
$$ct = γct' + ξx'$$
Then just remember that you have the inverse transformations, which would involve just swapping the sign and replacing the prime and unprimed coordinates:
(6)
$$x' = γx - ξct$$
$$ct' = γct - ξx$$
At that point all you have to do is find ξ, which is easy if you realize any object moving with uniform velocity is at rest in it's own frame, so there is always going to be a case where x' = 0. Which means that x/t in this case = v and is the speed at which S' is moving relative to S. Which then means you can solve for ξ by letting x' = 0 in the third equation:
$$x' = γx - ξct$$ with x'=0 gives
$$0 = γx - ξct$$
$$γx = ξct$$
$$γv = ξc$$
$$γ\frac{v}{c} = ξ$$
Then you just plug that in to the four equations in (5) and (6). Start with the first one in (5):
$$x = γx' + γ\frac{v}{c}ct'$$
Then substitute in x' and ct' from (6):
$$x = γx' + γ\frac{v}{c}ct'$$
$$x = γ(γx - γ\frac{v}{c}ct) + γ\frac{v}{c}γ(ct - γ\frac{v}{c} x)$$
Then just clean it up and solve for the last unknown.
$$x = γ(γx - γ\frac{v}{c}ct) + γ\frac{v}{c}γ(ct - γ\frac{v}{c} x)$$
$$x = γ^2([x - vt] + \frac{v}{c}[ct - \frac{v}{c} x])$$
$$x = γ^2(x - vt + vt - \frac{v^2}{c^2} x)$$
$$x = γ^2(x - \frac{v^2}{c^2} x)$$
$$x = γ^2x(1 - \frac{v^2}{c^2})$$
$$1 = γ^2(1 - \frac{v^2}{c^2})$$
$$γ^2= \frac{1}{(1 - \frac{v^2}{c^2})}$$
$$γ= \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
And then you have the Lorentz factor, so just plug it in to (5) and (6). To get time dilation, just assume that x'=0 because the person looking at their clock will be at the location of their clock if their time is proper time:
$$ct = γ(ct' + \frac{v}{c}x')$$ at x'=0 is
$$ct = γct'$$
$$t = γt'$$
Anyway, how is that derivation? Obviously if you orient your axes a certain way, y = y' and z = z', so no issue there. You can pretty much derive all of it once you get to this point.
15. Sep 6, 2016
### robphy
I didn't look at all of the steps...
But you've basically used light cone coordinates... (the eigenbasis of the Lorentz Transformations). A and B are the eigenvalues, which are the Doppler factor and its reciprocal.
Bondi's method (from post in #10), which also uses this feature, is cleaner.
My recent Insight uses Bondi's ideas to reshape causal diamonds [and thus light-clock diamonds] (which visualizes the Lorentz Transformations).
[if you know about rapidities (the Minkowskisn analogue of angle), then with A is $\exp(\theta)$ and B is $\exp(-\theta)$, half the sum gives $\cosh\theta$ and half the difference is $\sinh\theta$ where $v=\tanh\theta$.]
Last edited: Sep 6, 2016
16. Sep 6, 2016
### Tazerfish
I just wanted to thank all the people who answered this thread. It has helped me a lot
MisterT brought up an important point: If you want to learn you have to think.
The easiest derivation may not teach you as much as a harder one just because you think you understood it all.
I agree with stevendarly that I wouldn't really have appreciated a rigorous mathematic derivation when I was first introduced to special relativity.
The light clock proof is truly very easy and quick but doesn't teach you as much as the other proofs in my opinion.
You usually jump into it with your "Galilean" intuition and then realize on this purely hypothetical construct that your intuition must be wrong for the speed of light to be constant in all reference frames.
And as a little extra the formula for time dilation/length contraction/(the Lorentz factor) pops out of this consideration.
That leaves behind a lot of confusion.I felt like we just stumbled across it.
The other proofs are in a way actually more comprehensible to me, since I really doubt I would have gotten the idea for the light clock one.
To first hear of the experiments at that time that proved the Galilean view was incorrect and then try to figure out about the properties of space and time and transformations you can do to get from one reference system to another is a far more intuitive though also more complex way to derive it.
But the extra bit of work is worth it (or at least was for me).
Three more things: 1 I am fascinated that you can supposedly derive the Lorentz transformation without considering light.(Sadly, I didn't really understand the link posted by Ibix)
2 I find it cool that while the derivations that have been brought up may look very different at first glance they are actually fairly similar.
3 I have to say I am no big fan of Bondi k-calculus for introducing stuff like time dilation.
The fact that you throw around k (the doppler factor) all this time without even really knowing how it changes with velocity or why it should have the inverse value when moving in the opposite direction made me quite upset.
(But I like the diamond-diagrams robphy made in this insight article.The method seems pretty useful)
17. Sep 9, 2016
### Battlemage!
That makes sense. To be honest though, I just find this the least mathematically difficult way to do it. Only algebra is required, and even more to the point, only the basic operations of addition, multiplication, and exponents are all that is needed. I am currently trying to expand my knowledge with SR, particularly in a mathematical way of looking at it, but I find there are subtle differences in how the geometry works that make it require a bit more thought than these high school Lorentz factor derivations.
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Chem multiple choice help?
I am almost certain it is calcium
When calcium is allowed to react with nitrogen, N2, calcium nitride is formed, as shown in the following balanced chemical equation: 3Ca + N2 → Ca3N2
If 24.0 g of calcium and 12.0 g of nitrogen are available for this reaction, the limiting reagent will be
Question options:
a) calcium
b) nitrogen
c) calcium nitride
d) both calcium and nitrogen will be consumed completely
e) cannot be determined
Relevance
• Dr W
Lv 7
1 month ago
compare the ratios of mole / coefficient of reaction for each reactant. Whichever is lower is the LR
.. Ca..(24.0g / 3) * (1 mol / 40.1g) = ?
.. N2. (20.01g / 1) (1mol / 28.02g) = ?
.. which is lower?
• 1 month ago
Just by looking at the amounts, I would guess calcium, since you need 3 moles of Ca per mole of N2. To confirm:
24.0 g / 40g/mol = 0.6 mole of Ca
12 g / 28 g/mol = 0.43 mol
So you would need 3 x 0.43 = 1.29 moles of Ca to react completely with the N2, so a. is the answer.
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Weight Weight Class Below 114.6 lbs Underweight 114.6 lbs - 154.9 lbs Normal Weight 154.9 lbs - 185.9 lbs Overweight 185.9 lbs and Above Obese
### Weight Loss
If you were seeking a normal BMI weight of 154.9lbs, and you lost two pounds a week4, then you could reach a normal BMI in
48 Weeks
Or to be considered overweight, a goal weight of 185.9 lbs could be reached in
32 Weeks
### Ideal Weight for a 5 foot 6 Male or Female2
The ideal weight range using the BMI for a male or female with a height of 5'6" is
### Between
114.6 lbs
and
154.9 lbs
How to Calculate BMI3 for 5'6" and 250 lbs
(5 x 12) + 6 = 66
(250 / (66 x 66)) x 703
= 40.346648301194
5 = foot | 6 = inch | 250 = weight lbs
### Conversion
• 5 foot 6 = 66 inches
• 5 foot 6 = 167.64cm
• 250 pounds = 17 stone 12 pounds
• 250 pounds = 113.4 kg
• 250 pounds = 113398.09 grams
### 5'6" and 250 lbs Summary
What is the BMI for a 5'6" and 250 lbs female? 40.35 BMI, Obese.
What is the BMI for a 5'6" and 250 lbs male? 40.35 BMI, Obese.
What is the ideal weight for a 5'6" female? Between: 114.6lbs and 154.9lbs
What is the ideal weight for a 5'6" male? Between: 114.6lbs and 154.9lbs
If I am 5ft 6in and weigh 250 lbs, is that a good weight for my height? Under the BMI classification, 250 lbs is classed as being Obese.
This Page is Calculated for the Following Height and WeightHeight: 5' 6, 5 foot 6, 5'6", 5 ft 6 in, 5 feet 6 inches.
Weight: 250 Pounds, 250 lbs.
#### Is this Healthy?1
BMI values have different meanings for different body shapes. A major limitation of the BMI is that the BMI value does not indicate the location or distribution of body fat, two factors which are important in heart health and diabetes. If you are at all concerned with your weight or health, consult a doctor or health professional.
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# AutoCAD 2018 and Inventor 2018 Tutorial By Tutorial Books (www.engbookspdf.com)
## Part 1:
Part 1: AutoCAD Basics Aligned DAL This tool creates a linear dimension parallel to the object. Click Annotate > Dimensions > Dimension drop-down > Aligned on the ribbon. Select the first and second points of the dimension line (or) press Enter and select the line. Move the pointer and click to position the dimension. Arc Length DAR It dimensions the total or partial length of an arc. Click Annotate > Dimensions > Dimension drop-down > Arc Length on the ribbon. Select an arc from the drawing. If you want to dimension only a partial length of an arc, select Partial option from the command line. Next, select the two points on the arc. Move pointer and click to position the dimension. Continue DCO It creates a linear dimension from the second extension line of the previous dimension. Create a linear dimension by selecting the first and second points. 129
Part 1: AutoCAD Basics Click Annotate > Dimensions > Continue on the ribbon; a chain dimension is attached to the pointer. Select the third and fourth points of the chain dimension. Position the chain dimension. Next, right-click and select Enter. Baseline DBA It creates dimensions by using the previously created dimension, as shown below. Create a linear dimension by selecting the first and second points. 130
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AutoCAD Inventor LT Suite 2013 Brochure - Autodesk
AutoCAD Inventor Simulation 2012 Overview ... - Applied Software
AutoCAD Inventor Simulation 2012 Technical ... - Applied Software
978-1-58503-552-6 -- AutoCAD 2011 Tutorial ... - SDC Publications
158503245X - AutoCAD LT 2006 Tutorial - SDC Publications
1-58503-294-8 -- AutoCAD LT 2007 Tutorial - SDC Publications
AutoCAD 2007 Tutorial: 3D Modeling - SDC Publications
Engineering Graphics Essentials with AutoCAD ... - SDC Publications
1-58503-294-8 -- AutoCAD LT 2007 Tutorial - SDC Publications
AutoCAD® 2006 Tutorial - SDC Publications
AutoCAD LT® 2011 Tutorial - SDC Publications
978-1-58503-501-4 -- AutoCAD 2010 Tutorial ... - SDC Publications
978-1-58503-434-5 -- AutoCAD 2009 Tutorial ... - SDC Publications
978-1-58503-505-2 -- AutoCAD LT 2010 Tutorial - SDC Publications
Pro/ENGINEER Tutorial
978-1-58503-685-1 -- Tools for Design with VEX ... - VEX Robotics
Pro/ENGINEER WILDFIRE 2.0 Tutorial - SDC Publications
Autodesk Inventor®
Inventor Rendering Tutorial
CREATING CUSTOM SHORTCUT KEYS IN AUTODESK INVENTOR ...
Top 5 Reasons to Move from AutoCAD LT® to AutoCAD® Inventor ...
Inventor 2013 - Autodesk Inventor Wizard
Introduction to Autodesk Inventor for F1 in Schools
AutoCAD® Inventor LT™ SuiteNew Seat Rebate ... - Micromail
Tutorial 2: Drawing Setup in AutoCAD 2010
Parametric Modeling with Autodesk Inventor 2010 - SDC Publications
AutoCAD® 2008 Tutorial - SDC Publications
MAE 3 Inventor Drawing Tutorial - MAELabs UCSD
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# General Hobbyist Circuits
Posted on Feb 4, 2014
This circuit describes a simple, 6-bit random number pseudo-generator used to study binary counters and in particular, shift registers. Some very basic background information about binary counters and shift registers is provided. In reality there are dozens of different shift register topologies available and it can get quite complex. If you wish
to find a good logic tutorial website, I strongly recommend Ken Bigelow`s site as it has interactive diagrams. Flip-flops are also covered well on wikipedia and many other web sites. Binary Counter: The circuit most often used as a counter is called a binary flip-flop. The basic flip-flop can be viewed as a toggle switch having either an ON or OFF position. This is the binary state 1 (HIGH) or 0 (LOW). Like the toggle switch, the binary flip-flop has 2 binary states 1 or 0. A binary flip flop counter counts in a sequence such as 0, 1, 0, 1 etc. A straight binary counter can be built by using 1 or more flip-flops connected in a manner that the binary number stored in these flip-flops will represent the total number of trigger pulses received at the counter input. Ring Counter: A ring counter has 2 or more flip-flops cascaded so that the output from one flip-flop becomes the input of the next flip-flop. The flip-flops are connected so that all of their outputs are at the binary state 0 except for one flip-flop. By pulsing the input of the ring counter, it will sequentially change the binary state of the succeeding flip-flop from binary 0 to binary 1. The flip-flop that contains the binary 1 indicates the count of this binary counter. The maximum number of pulses that can be counted by N flip-flops is N pulses. Shift Register: A serial entry shift register is similar to a ring counter, except that the output flip-flop is not...
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# BookRiff
If you don’t like to read, you haven’t found the right book
## What is the proof of 12% alcohol?
5 ounces of wine per glass, 24 proof or 12 percent ABV.
## How do you calculate percent alcohol and proof?
(Phew.) 3: In the United States, the system — established around 1848 — is a bit simpler: “Proof” is straight up two times alcohol by volume. So a vodka, say, that is 40 percent ABV is 80 proof and one that is 45 percent ABV is 90 proof. A “proof spirit” is 100 proof (50 percent ABV) or higher.
What does 18 proof mean in alcohol?
Answer: Proof is defined as twice the alcohol (ethanol) content by volume. For example, a whisky with 50% alcohol is 100-proof whiskey. Anything 120-proof would contain 60% alcohol, and 80-proof means 40% of the liquid is alcohol.
### What percent alcohol is 60 proof?
Stateside, a liquor’s proof is two times the ABV. So this means that a beverage with 30% ABV is 60 proof.
### Is 13.9 a lot of alcohol?
Capriccio boasts an alcohol content of 13.9 percent alcohol by volume, meaning nearly 14 percent of the drink is pure alcohol. Men are advised to have no more than two servings of alcohol per day.
What proof is 43 percentage alcohol?
Proof (the term “degrees proof” is not used), defined as twice the percentage of alcohol by volume, may also be stated. For example, whiskey may be labeled as containing 50% alcohol by volume, and also as 100-proof; 86-proof whiskey contains 43% alcohol.
#### How do you convert alcohol to proof?
Proof is a method of measuring the alcohol content of spirits. You calculate the proof of a spirits product by multiplying the percent of alcohol by volume by two (2). For example, a spirits product that has a 40% alcohol content by volume is 80 proof [40 multiplied by 2 = 80].
#### What is high proof alcohol?
What is Considered a High Alcohol Proof? Anything above 100 proof is considered high. This is more than 50% ABV, which was deemed the baseline when the modern proof measurement system was introduced. There are popular options legally available that are anywhere from 151 proof all the way to 190 proof.
What does 13.9 alc vol mean?
Steiner said a 375mL bottle is equivalent to two standard glasses of wine. Capriccio boasts an alcohol content of 13.9 percent alcohol by volume, meaning nearly 14 percent of the drink is pure alcohol.
## What is the proof of 100% alcohol?
Pure alcohol or 100% alcohol by volume is referred to as being 200 proof. ABV is labeled on every alcoholic beverage including liquor, wine, and beer. The proof is the number that is used in the United States that describes the taxes that are required to pay by the distiller on an amount of liquor.
## What is the proof of alcohol in Canada?
Officially, Canada uses US labeling stating percent alcohol by volume, although the UK proof standard may still be seen and heard. Common spirits at 40% ABV are called 70° proof, while 57% ABV is 100 proof.
Which is the correct measure of alcohol proof?
Alcohol proof is twice the volume percentage of ethyl alcohol (ethanol) in an alcoholic beverage. It is a measure of the ethanol (a specific type of alcohol) content of an alcoholic beverage. The term originated in the United Kingdom and was defined as 7/4 the alcohol by volume (ABV).
### What does 80 proof mean on a bottle of vodka?
If the liquor label on a bottle of vodka indicates that it is “80 proof,” that number refers to the vodka’s alcohol content. Here’s how to understand what alcohol proof means.
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# Sam Da
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Matlab 2010 ARMA model
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Loss Function and FPE in ARMAX
Hi, So, what does loss function and FPE in ARMAX mean. Also, for an AR(3) model how do I get the variance of white noise? ...
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gmdistribution question
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# Lesson 1 homework 5.6 answers
and B have completely different elements (Set A comprises letters, and Set B comprises months of the year they have the same amount of elements, which is five. Equivalent sets have different elements but have the same amount of elements. Alternating between volunteers and nominated students solves this problem in part, but nominees should always be given the chance to penn pass if they prefer. Students from each team are called upon alternately to provide answers to each question. But why did the chicken cross the road? Homework ranking tasks also provide important feedback to the teacher who may use the data provided to check on the cause of problems areas at a later date. In this case, Set E contains three faces. Want to learn more? For ask-a-friend a student may ask another member of his/her team to provide the answer. Some sets contain images. If we say that the cardinality of set A is equal to the cardinality of set B, we are saying that sets A and B are equivalent, in that they both have the same number of elements. However, it may be worth just taking a quick look at the exercises that are to be corrected beforehand so that appropriate extra questions may be devised in advance. These are just a few ideas of how to make the whole-class correction of homework less of a chore and more of an active challenge. Alternate between asking for answers to be volunteered and calling on specific students to answer questions. Equal sets have the exact same elements in them, even though they could be out of order. Definition of a Set, before we get into the definition of an equivalent set, we need to first know what a set. Sets are enclosed by brackets. A set is a collection of elements that are usually related. With younger students, take care that clapping does not turn to booing wrong answers, however.
## Lesson 1 homework 5.6 answers
January, this technique is also useful for involving more competent or confident students. A set is a collection of elements that are usually related. An equivalent set is simply a set with an equal number of elements. November and May, no obligation 4 and Set B contains. Be upbeat about homework correction, lesson Summary, get access riskfree for 30 days. Ask the hp 2055 printer folds corner of paper class to do a quick survey in groups ranking exercises from the most to least difficult.
Online homework and grading tools for instructors and students that reinforce student learning through practice and instant feedback.Prentice Hall and our other respected imprints provide educational materials, technologies, assessments and related services across the secondary curriculum.Answers.com is the place to go to get the answers you need and to ask the questions you want.
Vary the correction procedure by taking time out along the way to focus on pronunciation intonation issues. Where structures, by Jane Sjoberg, because Set A and Set B contain the same cardinality. Homophones etc, extra questions can also include pronunciation issues by eliciting word stress. Some ideas of how to make the wholeclass correction of homework less of a chore and more of an active challenge. Or pictures, they have the same number of elements. Words, lesson at a Glance, especially while driving, we would write this. Making them equivalent, set A Set, for fillinthegaps exercises or simple one or twoword answers present feedback in power point or on an OHP. However, beyond the demands of tradition and convention. Again, give students an opportunity to pass if necessary.
It may also allow the teacher to gain insight into how much (or how little) homework an individual student is regularly putting.The cardinality of Set A.
# A set's cardinality is the number of elements in the set.
In this lesson, you'll learn the definition of equivalent sets.
We'll look at some properties and terms related to equivalent sets, as well as examples so you can gain a better understanding.
These are just a few ideas of how to make the whole-class correction of homework less of a chore and more of an active challenge.
The suggestions given are specifically geared to be used when correcting exercises set from a workbook or worksheet as homework but some ideas may also be used when giving feedback for tasks set in class.
Follow these simple steps to find online resources for your book.
Three part lesson on sums with directed numbers.
Mini-plenary and plenary activity embedded with answers throughout.
Painting task sheet was uploaded by TES member Tristan Jones.
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@internet4classr I4C
# 3rd Grade Algebra Skills Prior Standards Implementation
### Algebra
Sort objects by two attributes.
3.2.1
AT
Extend repeating and growing numerical or geometric patterns.
3.2.2
AT
1. Brenda's Bedroom Border - complete a pattern
2. Christmas Lights - identify and complete patterns
3. Missing Numbers - find the missing numbers in a sequence
4. Number Cracker - guess what number comes next in the pattern
5. Patterns - this video lesson explains patterns, and then gives your students the chance to show understanding on several quizzes.
6. Spooky Sequences - One of the ghosts is missing a number. Click on the numbers along the bottom to enter the right number in the sequence. ( square numbers )
7. Spooky Sequences - One of the ghosts is missing a number. Click on the numbers along the bottom to enter the right number in the sequence. ( triangular numbers )
8. Spooky Sequences - One of the ghosts is missing a number. Click on the numbers along the bottom to enter the right number in the sequence. one digit numbers (by ones ) | two digit numbers (by ones) | three digit numbers (by ones) | counting by twos | counting by fives | counting by tens | backward by ones | backward by twos | backward by fives | backward by tens
9. What Comes Next? - study repeating patterns and click on the one that continues the pattern
Represent repeating geometric patterns as repeating numerical patterns.
3.2.3
AT
1. Create your own pattern - drag shapes to create a pattern ( Instructions on how to use this page in your classroom )
Determine the output number for a particular input number given a one-operation function rule involving addition or subtraction.
3.2.4
AT
1. Ambleweb Function Machine - students identify what process was used to produce a sequence of numbers (twelve processes available)
Solve open sentences that involve addition and subtraction of whole numbers zero to twenty.
3.2.5
AT
1. Bracket Basics - [ click Continue to play the game ] drag numbers into spaces to make the statement correct, practices order of operation
2. Let's Go Shopping - estimate by rounding to the nearest dollar as they shop
3. Math Story Problems! - Kid created math story problems ("Kitten" pages are the easiest.) There are lots of links to pages of problems... scroll way down to see them all
4. My Backpack - add columns and find missing addends using data generated from a chart
5. Sum Sense (Addition) - Students practice single digit addition by dragging numbers to complete the sentence.
6. Sum Sense (Subtraction) - Students practice single digit subtraction by dragging numbers to complete the sentence.
7. Walrus World - learn strategies for solving real world problems
Identify the rules by which objects or numbers have been sorted.
3.2.6
AT
1. Ambleweb Function Machine - students identify what process was used to produce a sequence of numbers (twelve processes available)
National Library of Virtual Manipulatives
1. Algebra (Grades 3-5)
### Review Help
Practice Tests
1. Houghton Mifflin Mathematics Test Prep Practice
2. Virginia State Standards of Learning - Select Math 3 , and then select 10, 20, or 40 questions.
FCAT Sample Test Book
1. FCAT Sample Test Book - [ 2008 ] sample questions and test taking tips
2. FCAT Sample Answer Book - [ 2008 ] sample questions and test taking tips
TAAS Attack Daily Upkeeps
1. Daily Upkeeps - Scroll down this page until you see " 3rd Grade TAAS Attack Daily Upkeeps ." There are 9 practice activities to print (in .pdf format)
Daily Questions
1. Daily Questions - to serve as a ten to fifteen minute (approximate) review of practice test items and critical thinking (217 pages) [ Note : This link is available through the Wayback Machine Internet Archive. Some state filters block access to this page so you might consider downloading the document at home.]
Released Tests & Others
1. Solve It! - Math Videos provide problem solving practice for students in grades 3 to 6. Each of the thirteen sets contain five multi step word problems with step by step video solutions. Concepts include basic operations, algebraic reasoning, money, fractions, percent, perimeter, area, proportional reasoning, and measurement.
2. Third Grade Math - Read each question and choose the best answer. Then mark the circle next to the letter for the answer you have chosen. ( from Texas )
3. Texas end-of-year Math test 2004
site for teachers | PowerPoint show | Acrobat document | Word document | whiteboard resource | sound | video format | interactive lesson | a quiz | lesson plan | to print
Internet4classrooms is a collaborative effort by Susan Brooks and Bill Byles.
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# Documentation
### This is machine translation
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## Exponential Models
### About Exponential Models
The toolbox provides a one-term and a two-term exponential model as given by
`$\begin{array}{l}y=a{e}^{bx}\\ y=a{e}^{bx}+c{e}^{dx}\end{array}$`
Exponentials are often used when the rate of change of a quantity is proportional to the initial amount of the quantity. If the coefficient associated with b and/or d is negative, y represents exponential decay. If the coefficient is positive, y represents exponential growth.
For example, a single radioactive decay mode of a nuclide is described by a one-term exponential. a is interpreted as the initial number of nuclei, b is the decay constant, x is time, and y is the number of remaining nuclei after a specific amount of time passes. If two decay modes exist, then you must use the two-term exponential model. For the second decay mode, you add another exponential term to the model.
Examples of exponential growth include contagious diseases for which a cure is unavailable, and biological populations whose growth is uninhibited by predation, environmental factors, and so on.
### Fit Exponential Models Interactively
1. Open the Curve Fitting app by entering `cftool`. Alternatively, click Curve Fitting on the Apps tab.
2. In the Curve Fitting app, select curve data (X data and Y data, or just Y data against index).
Curve Fitting app creates the default curve fit, `Polynomial`.
3. Change the model type from `Polynomial` to `Exponential`.
You can specify the following options:
• Choose one or two terms to fit `exp1` or `exp2`.
Look in the Results pane to see the model terms, the values of the coefficients, and the goodness-of-fit statistics.
• (Optional) Click Fit Options to specify coefficient starting values and constraint bounds appropriate for your data, or change algorithm settings.
The toolbox calculates optimized start points for exponential fits, based on the current data set. You can override the start points and specify your own values in the Fit Options dialog box.
The fit options for the single-term exponential are shown next. The coefficient starting values and constraints are for the census data.
For an example specifying starting values appropriate to the data, see Gaussian Fitting with an Exponential Background.
For more information on the settings, see Specifying Fit Options and Optimized Starting Points.
### Fit Exponential Models Using the fit Function
This example shows how to fit an exponential model to data using the `fit` function.
The exponential library model is an input argument to the `fit` and `fittype` functions. Specify the model type `'exp1'` or `'exp2'` .
Fit a Single-Term Exponential Model
Generate data with an exponential trend and then fit the data using a single-term exponential. Plot the fit and data.
```x = (0:0.2:5)'; y = 2*exp(-0.2*x) + 0.1*randn(size(x)); f = fit(x,y,'exp1') plot(f,x,y) ```
```f = General model Exp1: f(x) = a*exp(b*x) Coefficients (with 95% confidence bounds): a = 2.021 (1.89, 2.151) b = -0.1812 (-0.2104, -0.152) ```
Fit a Two-Term Exponential Model
```f2 = fit(x,y,'exp2') plot(f2,x,y) ```
```f2 = General model Exp2: f2(x) = a*exp(b*x) + c*exp(d*x) Coefficients (with 95% confidence bounds): a = 3552 (-3.783e+12, 3.783e+12) b = -0.2574 (-2.725e+04, 2.725e+04) c = -3550 (-3.783e+12, 3.783e+12) d = -0.2575 (-2.727e+04, 2.727e+04) ```
Set Start Points
The toolbox calculates optimized start points for exponential fits based on the current data set. You can override the start points and specify your own values.
Find the order of the entries for coefficients in the first model ( `f` ) by using the `coeffnames` function.
```coeffnames(f) ```
```ans = 2x1 cell array {'a'} {'b'} ```
If you specify start points, choose values appropriate to your data. Set arbitrary start points for coefficients `a` and `b` for example purposes.
```f = fit(x,y,'exp1','StartPoint',[1,2]) plot(f,x,y) ```
```f = General model Exp1: f(x) = a*exp(b*x) Coefficients (with 95% confidence bounds): a = 2.021 (1.89, 2.151) b = -0.1812 (-0.2104, -0.152) ```
Examine Exponential Fit Options
Examine the fit options if you want to modify fit options such as coefficient starting values and constraint bounds appropriate for your data, or change algorithm settings. For details on these options, see the table of properties for NonlinearLeastSquares on the `fitoptions` reference page.
```fitoptions('exp1') ```
```ans = Normalize: 'off' Exclude: [] Weights: [] Method: 'NonlinearLeastSquares' Robust: 'Off' StartPoint: [1x0 double] Lower: [1x0 double] Upper: [1x0 double] Algorithm: 'Trust-Region' DiffMinChange: 1.0000e-08 DiffMaxChange: 0.1000 Display: 'Notify' MaxFunEvals: 600 MaxIter: 400 TolFun: 1.0000e-06 TolX: 1.0000e-06 ```
## Related Topics
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#### Machine Learning Challenges: Choosing the Best Classification Model and Avoiding Overfitting
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# Fundamental unit
(Redirected from Fundamental units)
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This article is about the type of unit of measure. For the concept in number theory, see Fundamental unit (number theory).
A set of fundamental units is a set of units for physical quantities from which every other unit can be generated. In the International System of Units, there are seven fundamental units: kilogram, meter, candela, second, ampere, kelvin, and mole.
In the language of measurement, quantities are quantifiable aspects of the world, such as time, distance, velocity, mass, temperature, energy, and weight, and units are used to describe their magnitude or quantity. Many of these quantities are related to each other by various physical laws, and as a result the units of some of the quantities can be expressed as products (or ratios) of powers of other units (for example, momentum is mass multiplied by velocity, while velocity is measured in distance divided by time). These relationships are discussed in dimensional analysis. Those that cannot be so expressed can be regarded as "fundamental" in this sense.
There are other relationships between physical quantities which can be expressed by means of fundamental constants, and to some extent it is an arbitrary decision whether to retain the fundamental constant as a quantity with dimensions or simply to define it as unity or a fixed dimensionless number, and reduce the number of fundamental constants by one.
For instance, time and distance are related to each other by the speed of light, c, which is a fundamental constant. It is possible to use this relationship to eliminate either the fundamental unit of time or that of distance. Similar considerations apply to Planck's constant, h, which relates energy (with dimensions of mass, length and time) to frequency (dimensions of time). In theoretical physics it is customary to use such units (natural units) in which c = 1 and ħ = 1.
Slightly different considerations apply to the so-called vacuum permittivity or permittivity of free space, which historically has been regarded as a separate physical constant in some systems of measurement but not in others.
In theory, a system of fundamental quantities (or sometimes fundamental dimensions) would be such that every other physical quantity (or dimension of physical quantity) can be generated from them as algebraic combinations.
## Reduction of the number of fundamental quantities
• One could eliminate any two of the metre, kilogram and second by setting c and h to unity or to a fixed dimensionless number.
• One could then eliminate the ampere either by setting the permittivity of free space to a fixed dimensionless number or by setting the electronic charge to such a number.
• One could similarly eliminate the mole as a fundamental unit by reference to Avogadro's number.
• One could eliminate the kelvin as it can be argued that temperature simply expresses the energy per particle per degree of freedom which can be expressed in terms of energy (or mass, length, and time). Another way of saying this is that Boltzmann's constant could be expressed as a fixed dimensionless number.
• Similarly, one could eliminate the candela as that is defined in terms of other physical quantities.
• That just leaves one fundamental dimension and one fundamental unit, but we still have plenty of fundamental constants left to eliminate that too – for instance one could use G, the gravitational constant, or me, the electron rest mass.
A widely used choice is the so-called Planck units, which are defined by setting ħ = c = G = 1.
That leaves every physical quantity expressed simply as a dimensionless number, so it is not surprising that there are also physicists who have cast doubt on the very existence of incompatible fundamental quantities.[1][2][3]
## References
1. ^ M. J. Duff, L. B. Okun and G. Veneziano, Trialogue on the number of fundamental constants, JHEP 0203, 023 (2002) preprint pdf.
2. ^ Jackson, John David (1998). "Appendix on Units and Dimensions". Classical Electrodynamics (PDF). John Wiley and Sons. p. 775. Retrieved 13 January 2014. The arbitrariness in the number of fundamental units and in the dimensions of any physical quantity in terms of those units has been emphasized by Abraham, Plank, Bridgman, Birge, and others.
3. ^ Birge, Raymond T. (1935). "On the establishment of fundamental and derived units, with special reference to electric units. Part I." (PDF). American Journal of Physics 3: 102–109. Bibcode:1935AmJPh...3..102B. doi:10.1119/1.1992945. Retrieved 13 January 2014. Because, however, of the arbitrary character of dimensions, as presented so ably by Bridgman, the choice and number of fundamental units are arbitrary.
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# Machines X and Y produce bottles at their respective constant rates.
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Machines X and Y produce bottles at their respective constant rates. [#permalink]
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29 Aug 2010, 20:43
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Machines X and Y produce bottles at their respective constant rates. Machine X produces k bottles in 6 hours and machine Y produces k bottles in 2 hours. How many hours does it take machines X and Y , working simultaneously , to produce 12k bottles?
(A) 8
(B) 12
(C) 15
(D) 18
(E) 24
[Reveal] Spoiler: OA
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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29 Aug 2010, 20:48
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My answer to this question was 8
X -> k bottles in 6 hours -> 1/6
Y -> k bottles in 2 hours -> 1/2
Working together -> 2/3
if they produce k bottles in 2/3 hrs, they will produce 12 * 2/3 bottles in 8 hours.
Could you please point out where I am mistaken??
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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29 Aug 2010, 20:58
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KUDOS
Safiya wrote:
My answer to this question was 8
X -> k bottles in 6 hours -> 1/6
Y -> k bottles in 2 hours -> 1/2
Working together -> 2/3
if they produce k bottles in 2/3 hrs, they will produce 12 * 2/3 bottles in 8 hours.
Could you please point out where I am mistaken??
X - k bottle in 6 hours => In 1 hour - k/6 bottles
Y - k bottle in 2 hours => In 1 hour - k/2 bottles
Working together in 1 hour - 2k/3
working together in 3 hours - 2k bottles
2k bottles in 3 hours
=> 2k *6 bottles = 12 bottles in 3*6 = 18 hours
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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29 Aug 2010, 21:01
Ahhh what a silly mistake! Thank you very much gurpreetsingh
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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30 Aug 2010, 04:18
X in 6 hours: k bottles
Y in 2 hours: k bottles
Y in 6 hours: 3k bottles
X and Y in 6 hours: 4k bottles
12k bottles: 12/4 = 3 (6 hour periods)
hence 18 hours
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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31 Aug 2010, 05:21
try considering the below simple solution.
we have two different time periods resulting in the same quantity of the product (K).
X working for 6 hours, produces K bottles
in those 6 hours Y produces 3K bottles as it produces K in 2 hours.
so in first 6 hours K+3K = 4K
in 12 hours 4K+4k
in 18 hours 4K+4K+4K = 12K
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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31 Aug 2010, 08:26
Machines X and Y produce bottles at their respective constant rates. Machine X produces k bottles in 6 hours and machine Y produces k bottles in 2 hours. How many hours does it take machines X and Y , working simultaneously , to produce 12k bottles?
formula is work = rate x time
for machine (X) k= r X 6
for machine (Y) k= r X 2
now take a value for k ( for ease which is a multiple of both mac X & y)
take k=12
so rate of machine (x) is 2
so rate of machine (y) is 6
now work is 12k = 12x12 = 144
two machines working siman so rate = 6+2 =8
work = rate X time
144 = 8 X time
so time = 18
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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01 Sep 2010, 00:38
(6*2/(6+2))*12 = 18
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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29 Aug 2011, 11:28
since the question asks number of HOURS, so we should use this method.
x,y working together to produce k bottles = 6*2/6+2 = 3/2 hrs
12k bottles = 3*12/2 = 18
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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30 Oct 2011, 11:23
$$\frac{1}{6}+\frac{1}{2}=\frac{1}{T}$$
$$T=\frac{3}{2}$$ for k bottles
and $$\frac{3}{2}*12=18$$ for 12K bottles
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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20 Feb 2015, 10:34
Safiya wrote:
Machines X and Y produce bottles at their respective constant rates. Machine X produces k bottles in 6 hours and machine Y produces k bottles in 2 hours. How many hours does it take machines X and Y , working simultaneously , to produce 12k bottles?
(A) 8
(B) 12
(C) 15
(D) 18
(E) 24
x rate = k/6
y rate = k/2
k/6 + k/2 = 12k / T
solving T = 18
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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20 Feb 2015, 13:30
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Expert's post
Hi All,
This is an example of a Work Formula question with a minor 'twist.' The "math" in these questions can be done in a few different ways, but since we have 2 machines working on a task together, without any major "twists" (e.g. one of them stops working at a certain point), we can use the Work Formula:
Work = (A)(B)/(A+B) where A and B are the respective rates of the two machines to do the same task individually.
We're told:
1) Machine X can produce K bottles in 6 hours.
2) Machine Y can produce K bottles in 2 hours.
(6)(2)/(6+2) = 12/8 = 1.5 hours to produce K bottles when working together.
The minor 'twist' is that the question asks how long it takes to produce 12K bottles (not K bottles), so we have to multiply this result by 12...
(1.5 hours)(12) = 18 hours.
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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30 Jan 2017, 04:17
1) First of all we need to find the rates of each machine: $$r*t=w. X: r*6=k, r=\frac{k}{6}; Y: r*2=k, r=\frac{k}{2}$$
2) Now we need to combine the two rates: $$\frac{k}{6}+\frac{k}{2}=\frac{4k}{6}=\frac{2k}{3}$$
3) $$\frac{2k}{3}*R=12k, R=12k/\frac{2k}{3}, R=\frac{12k*3}{2k}=18$$
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]
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01 Feb 2017, 09:19
Safiya wrote:
Machines X and Y produce bottles at their respective constant rates. Machine X produces k bottles in 6 hours and machine Y produces k bottles in 2 hours. How many hours does it take machines X and Y , working simultaneously , to produce 12k bottles?
(A) 8
(B) 12
(C) 15
(D) 18
(E) 24
We are given that machine X produces k bottles in 6 hours and thus has a rate of k/6. We are also given that machine y produces k bottles in 2 hours and thus has a rate of k/2. We need to determine how many hours it takes machines X and Y to produce 12k bottles when working together. If we let t = the time it takes the two machines working together, we can create the following equation and determine t:
(k/6)t + (k/2)t = 12k
Multiplying the entire equation by 6, we have:
kt + 3kt = 72k
Dividing the entire equation by k, we have:
t + 3t = 72
4t = 72
t = 18 hours
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink] 01 Feb 2017, 09:19
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CSES - Exponentiation
• Time limit: 1.00 s
• Memory limit: 512 MB
Your task is to efficiently calculate values $a^b$ modulo $10^9+7$.
Input
The first input line contains an integer $n$: the number of calculations.
After this, there are $n$ lines, each containing two integers $a$ and $b$.
Output
Print each value $a^b$ modulo $10^9+7$.
Constraints
• $1 \le n \le 2 \cdot 10^5$
• $0 \le a,b \le 10^9$
Example
Input:
3 3 4 2 8 123 123
Output:
81 256 921450052
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# Simple Math Counting Games Printable Worksheets
When Children start to learn something new they frequently ask questions, sometimes they build up questions on their own. Parents always encourage their kids to learn new concepts and practice them to excel, and counting the numbers is just the fundamental step in learning mathematics. The kids can learn the relationship between various numbers once they get used to counting.
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204 views
Answer the question based on the information given below:
K, L, M, N, P, Q, R, S, U and W are the only ten members in the department. There is a proposal to form a team from within the members of the department, subject to the following conditions.
• A team must include exactly one among P, R and S.
• A team must include either M or Q, but not both.
• If a team includes K, then it must also include L and vice versa.
• If a team includes one among S, U and W, then it must also include the other two.
• L and N cannot be the members of the same team.
• L and U cannot be the members of the same team.
• The size of the team is defined as the number of members in the team.
Who can be a member of a team of size $5?$
1. K
2. L
3. M
4. P
5. R
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# Question: What is the number of molecules in 0 5 moles of oxygen?
Contents
## How many molecules are there in 5 moles of oxygen?
Hence, number of moles of Oxygen in 5 Moles of CO = 5. The number of oxygen atoms present can be obtained by multiplying the moles of oxygen by Avogadro’s Number. So, number of oxygen atoms present = 5 x ( 6.023 x 10^23 ) = 30.115 x 10^23 nos.
## What is the number of atoms in 0.5 mole of oxygen?
(i)0.5 mole of oxygen gas. (ii) 3.011 x 1023 atoms of oxygen. (iii) 6.022 x 1024 molecules of oxygen. (Given atomic mass of 0 = 16u; No = 6.022 x 1023 per mole)
## How many molecules are in a mole of oxygen?
One mole of oxygen gas, which has the formula O2, has a mass of 32 g and contains 6.02 X 1023 molecules of oxygen but 12.04 X 1023 (2 X 6.02 X 1023) atoms, because each molecule of oxygen contains two oxygen atoms.
## How many molecules are there in 5.5 moles?
n=5.5×6.023×1023=3.31×1024 n = 5.5 × 6.023 × 10 23 = 3.31 × 10 24 .
IT IS SURPRISING: What can I use to draw out a blind pimple?
## How many atoms are in 5 moles O2?
so 5 moles contain 5 x 6.023 x 10^23 molecules of O2 = 30.115 x 10^23 molecules of O2. and each molecule contains 2 oxygen atoms. so 5 moles that is 30.115 x 10^23 molecules contain 2 x 30.023 x 10^23 atoms = 60.230 x 10 ^23 atoms or 6.023 x 10^24 atoms.
## How many molecules is 0.5 moles H2O?
There are ~3*10^23 molecules of H2O in 0.5 mole.
## How many particles are in 0.2 moles?
1 Mole of any element contains 6.022X10^23 no of atoms or molecules. For 0.2 moles of oxygen no of atoms or molecules of oxygen will be 0.2X6. 022X10^23 no of constituents.
## How many molecules are there in 0.5 moles of CO2?
The number of molecules present in 0.5 moles of CO2 is6. 022 x 1023 x 0.5 = 3.011 x 1023.
## Which contains more molecules 0.5 mole of O2 or 0.5 mole of O3?
5 moles of O3, then 0.5 moles of O3 Will have more number of atoms as both of them have same number of molecules and each molecule of O2 has 2 O atoms whereas each of OR molecule has 3 O atoms. So, 0.5 moles O3 has more atoms than 0.5 moles of O2.
## How many oxygen atoms are there in 0.5 mole of water?
So 0.5 mol of Oxygen has 3.011*10^23 atoms.
## What is the mass of 0.5 mole of oxygen molecule?
0.5 moles of O2 weighs 16 grams.
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A shunt motor rotating at 1500 rpm is fed by a 120V dc source. The line current is 51 A and the shunt-field resistance is 120 ohm. If the armature resistance is 0.1 ohm, calculate the speed if the shunt-field resistance is doubled at no load condition. Suppose the terminal voltage VT does not change.
the answer should be little bit bigger than 3000 rpm.
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Q:
A shunt motor rotating at 1500 rpm is fed by a 120V dc source. The line current is 51 A and the shunt-field resistance is 120 ohm. If the armature resistance is 0.1 ohm,calculate armature current and the counter-emf (Ea) and then calculate the speed if the shunt-field resistance is doubled at no load condition. (at no load condition meand armature resistance is 0).Suppose the terminal voltage VT does not change.the answer should be little bit bigger than 3000 rpm.
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# Search by Topic
#### Resources tagged with Interactivities similar to Up, Down, Flying Around:
Filter by: Content type:
Stage:
Challenge level:
### There are 152 results
Broad Topics > Information and Communications Technology > Interactivities
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Drips
##### Stage: 2 and 3 Challenge Level:
An animation that helps you understand the game of Nim.
### Multiplication Tables - Matching Cards
##### Stage: 1, 2 and 3 Challenge Level:
Interactive game. Set your own level of challenge, practise your table skills and beat your previous best score.
### Semi-regular Tessellations
##### Stage: 3 Challenge Level:
Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Factors and Multiples Game
##### Stage: 2, 3 and 4 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Subtended Angles
##### Stage: 3 Challenge Level:
What is the relationship between the angle at the centre and the angles at the circumference, for angles which stand on the same arc? Can you prove it?
### Shuffles Tutorials
##### Stage: 3 Challenge Level:
Learn how to use the Shuffles interactivity by running through these tutorial demonstrations.
### Lost
##### Stage: 3 Challenge Level:
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Triangles in Circles
##### Stage: 3 Challenge Level:
Can you find triangles on a 9-point circle? Can you work out their angles?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Balancing 2
##### Stage: 3 Challenge Level:
Meg and Mo still need to hang their marbles so that they balance, but this time the constraints are different. Use the interactivity to experiment and find out what they need to do.
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### An Unhappy End
##### Stage: 3 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Flip Flop - Matching Cards
##### Stage: 1, 2 and 3 Challenge Level:
A game for 1 person to play on screen. Practise your number bonds whilst improving your memory
### Balancing 1
##### Stage: 3 Challenge Level:
Meg and Mo need to hang their marbles so that they balance. Use the interactivity to experiment and find out what they need to do.
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Balancing 3
##### Stage: 3 Challenge Level:
Mo has left, but Meg is still experimenting. Use the interactivity to help you find out how she can alter her pouch of marbles and still keep the two pouches balanced.
### Archery
##### Stage: 3 Challenge Level:
Imagine picking up a bow and some arrows and attempting to hit the target a few times. Can you work out the settings for the sight that give you the best chance of gaining a high score?
### Top Coach
##### Stage: 3 Challenge Level:
Carry out some time trials and gather some data to help you decide on the best training regime for your rowing crew.
### Right Angles
##### Stage: 3 Challenge Level:
Can you make a right-angled triangle on this peg-board by joining up three points round the edge?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### See the Light
##### Stage: 2 and 3 Challenge Level:
Work out how to light up the single light. What's the rule?
### Rollin' Rollin' Rollin'
##### Stage: 3 Challenge Level:
Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Bow Tie
##### Stage: 3 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Circuit Maker
##### Stage: 3, 4 and 5 Challenge Level:
Investigate how logic gates work in circuits.
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Jam
##### Stage: 4 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Cogs
##### Stage: 3 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### Volume of a Pyramid and a Cone
##### Stage: 3
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Nine Colours
##### Stage: 3 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Interactive Spinners
##### Stage: 3 Challenge Level:
This interactivity invites you to make conjectures and explore probabilities of outcomes related to two independent events.
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
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#### COUNTIFS in Excel: Everything You Need to Know
The COUNTIFS function works similar like the SUMIFS function. But instead of adding up the values, it counts how many items with one or more criteria are in your table.
#### 2D Lookups in Excel: 4 Easy Methods for 2 Way Lookups + Excel Download
There are many cases in which you need a 2 dimensional lookup. That means, if you want to get a value from a specific row-column combination with neither rows or columns fixed. Unfortunately, the problem of a two way lookup comes up quite often. In this article we explore 4 methods of how to conduct 2D lookups in Excel.
#### HLOOKUP in Excel: Everything you need to know
You’ve probably heard of VLOOKUP which is a very popular and powerful formula in Excel. Far less known is the little brother: HLOOKUP. It basically works the same way as VLOOKUP with one difference: Instead of looking up values vertically, HLOOKUP works horizontally. In this article, you learn how to use HLOOKUP, what to keep in mind, possible error messages and some more advices about HLOOKUP. […]
#### VLOOKUP in Excel: All You Need to Know About the Powerful Formula
One of the most popular formulas in Excel is the VLOOKUP formula. Many lookup approaches are based on the VLOOKUP formula. Mastering it is crucial for any of the following chapters and methods. Unfortunately, VLOOKUP is not as easy to use as a SUM or COUNT. In this article, you learn how to use VLOOKUP, what to keep in mind and some more advices about VLOOKUP.
[…]
#### File Name, Sheet Name or Path: Insert File Information into Excel Cell
There are many cases, in which you want to show the sheet name or file name in an Excel cell. Don’t worry, you don’t have to use VBA and create a macro. Excel provides a (rough) formula for returning all the information you need. The bad news: In most cases it takes some complex modifications. The good news: Usually you can just copy and paste the formula you need.
#### IF Formula in Excel: Everything You Need to Know
The IF formula is one of the most useful formulas in Excel. It is highly adjustable and can be extended by other formulas easily. But many Excel users don’t really know all it’s secrets. Do you know how to use it? This article covers everything you need to know about it: How to use it, how to extend it and how to use the related formulas IFNA, IFERROR and the new formula IFS. […]
#### Space Character: Remove Last Letter in Excel If It Is Blank
Does this situation sound familiar to you: You compare texts in Excel and they should be equal. But instead, Excel says the two cells are not the same. Oftentimes, there is a hidden space in the end of a text cell. Very troublesome: You can’t really see it because the blank character is at the end of the text. In this article we explore 4 methods of how to remove the last character of a text in Excel if it is a space.
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https://travelasker.com/how-many-miles-separate-california-and-shanghai-china/
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# How many miles separate California and Shanghai, China?
Air Travel
## Distance between California and Shanghai
California, the most populous state in the United States, is located on the west coast of North America. Shanghai, on the other hand, is one of the most populous cities in China and is located on the eastern coast of the country. The vast distance between these two locations raises the question of how far apart they are.
Knowing the distance between California and Shanghai is crucial for individuals who intend to travel between the two places or for businesses that require the transportation of goods. The distance between these two locations can be measured in various ways, including by air, land, or sea. In this article, we will explore the different methods used to calculate the distance between California and Shanghai.
## Measuring the distance between two points
There are many ways to measure the distance between two points, including using maps, GPS devices, or online distance calculators. However, the most accurate way to measure the distance between two points on the earth’s surface is by using latitude and longitude coordinates.
Latitude and longitude coordinates are used to locate points on the earth’s surface. Latitude measures how far north or south a point is from the equator, while longitude measures how far east or west a point is from the prime meridian. By using these coordinates, the distance between two points on the earth’s surface can be calculated using a mathematical formula.
## Calculating the distance using latitude and longitude
The Haversine formula is a mathematical formula used to calculate the distance between two points on the earth’s surface using latitude and longitude coordinates. The formula takes into account the curvature of the earth’s surface and is accurate for distances up to 12,427 miles.
Using the Haversine formula, the distance between California and Shanghai is approximately 6,055 miles. However, this distance is subject to several factors that can affect the accuracy of the calculation.
## Factors that affect the accuracy of distance calculations
Several factors can affect the accuracy of distance calculations, including the Earth’s shape, atmospheric conditions, and the precision of the latitude and longitude coordinates used.
## The Earth’s shape and its impact on distance calculations
The Earth’s shape is not a perfect sphere and is slightly flattened at the poles and bulges at the equator. This shape affects the accuracy of distance calculations as the distance between two points on the Earth’s surface is not a straight line but rather follows the curvature of the Earth’s surface.
## The impact of atmospheric conditions on distance calculations
Atmospheric conditions can also affect the accuracy of distance calculations. For example, the speed of sound in air changes with temperature, humidity, and pressure, which can affect the accuracy of distance calculations that rely on sound waves.
## Distance between California and Shanghai on a map
On a map, the distance between California and Shanghai appears to be much shorter than the actual distance. This is because maps represent the Earth’s surface in two dimensions and do not account for the curvature of the Earth’s surface.
## The shortest flight distance between California and Shanghai
The shortest flight distance between California and Shanghai is approximately 5,375 miles. This distance is the direct distance between the two cities and does not take into account the possible changes in flight paths due to atmospheric conditions or other factors.
## The fastest flight route between California and Shanghai
The fastest flight route between California and Shanghai depends on several factors, including the airline, the type of aircraft, and the prevailing winds. However, the average flight time between the two cities is around 12 to 14 hours, with non-stop flights available from several airlines.
## The driving distance between California and Shanghai
It is not possible to drive from California to Shanghai due to the vast distance between the two locations and the fact that they are separated by water. However, it is possible to travel by car or bus within California and within Shanghai.
## Conclusion: How many miles separate California and Shanghai?
In conclusion, the distance between California and Shanghai is approximately 6,055 miles when measured using latitude and longitude coordinates. However, this distance is subject to several factors that can affect the accuracy of the calculation, including the Earth’s shape and atmospheric conditions.
## Final thoughts on distance calculations
Distance calculations are crucial for many applications, including travel, trade, and scientific research. However, these calculations are subject to various factors that can affect their accuracy. It is essential to be aware of these factors and to use the most appropriate method for the intended application to achieve accurate results.
#### Kristy Tolley
Kristy Tolley, an accomplished editor at TravelAsker, boasts a rich background in travel content creation. Before TravelAsker, she led editorial efforts at Red Ventures Puerto Rico, shaping content for Platea English. Kristy's extensive two-decade career spans writing and editing travel topics, from destinations to road trips. Her passion for travel and storytelling inspire readers to embark on their own journeys.
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https://justaaa.com/finance/161345-n-investment-offers-8700-per-year-for-17-years
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Question
# n investment offers \$8,700 per year for 17 years, with the first payment occurring one year...
n investment offers \$8,700 per year for 17 years, with the first payment occurring one year from now. Assume the required return is 11 percent. What would the value be if the payments occurred forever?
When annual payment received forever Annual payment 8700 Rate of return 11% Period Perpetuity Therefore, value of investment = 8700/11% = \$ 79090.91 Note: value of investment when annual payment received for 17 yrs Annual payment 8700 Rate of return 11% Annuity PVf at 11% for 17 years 7.54879 Value of investment 65674.47
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# Gram/Millimeter to Ounce/Millimeter Converter
1 Gram/Millimeter = 0.03527396194958 Ounce/Millimeter
## One Gram/Millimeter is Equal to How Many Ounce/Millimeter?
The answer is one Gram/Millimeter is equal to 0.03527396194958 Ounce/Millimeter and that means we can also write it as 1 Gram/Millimeter = 0.03527396194958 Ounce/Millimeter. Feel free to use our online unit conversion calculator to convert the unit from Gram/Millimeter to Ounce/Millimeter. Just simply enter value 1 in Gram/Millimeter and see the result in Ounce/Millimeter.
Manually converting Gram/Millimeter to Ounce/Millimeter can be time-consuming,especially when you don’t have enough knowledge about Linear Mass Density units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Gram/Millimeter to Ounce/Millimeter converter tool to get the job done as soon as possible.
We have so many online tools available to convert Gram/Millimeter to Ounce/Millimeter, but not every online tool gives an accurate result and that is why we have created this online Gram/Millimeter to Ounce/Millimeter converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Gram/Millimeter to Ounce/Millimeter (g/mm to oz/mm)
By using our Gram/Millimeter to Ounce/Millimeter conversion tool, you know that one Gram/Millimeter is equivalent to 0.03527396194958 Ounce/Millimeter. Hence, to convert Gram/Millimeter to Ounce/Millimeter, we just need to multiply the number by 0.03527396194958. We are going to use very simple Gram/Millimeter to Ounce/Millimeter conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Gram/Millimeter} = 1 \times 0.03527396194958 = \text{0.03527396194958 Ounce/Millimeter}$$
## What Unit of Measure is Gram/Millimeter?
Gram per millimeter is a unit of measurement for linear mass density. By definition, it is equal to mass of object in gram per length of one millimeter.
## What is the Symbol of Gram/Millimeter?
The symbol of Gram/Millimeter is g/mm. This means you can also write one Gram/Millimeter as 1 g/mm.
## What Unit of Measure is Ounce/Millimeter?
Ounce per millimeter is a unit of measurement for linear mass density. By definition, it is equal to mass of object in ounce per length of one millimeter.
## What is the Symbol of Ounce/Millimeter?
The symbol of Ounce/Millimeter is oz/mm. This means you can also write one Ounce/Millimeter as 1 oz/mm.
## How to Use Gram/Millimeter to Ounce/Millimeter Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Gram/Millimeter and in the first input field, enter a value.
• From the second dropdown, select Ounce/Millimeter.
• Instantly, the tool will convert the value from Gram/Millimeter to Ounce/Millimeter and display the result in the second input field.
Gram/Millimeter
1
Ounce/Millimeter
0.03527396194958
# Gram/Millimeter to Ounce/Millimeter Conversion Table
Gram/Millimeter [g/mm]Ounce/Millimeter [oz/mm]Description
1 Gram/Millimeter0.03527396194958 Ounce/Millimeter1 Gram/Millimeter = 0.03527396194958 Ounce/Millimeter
2 Gram/Millimeter0.070547923899161 Ounce/Millimeter2 Gram/Millimeter = 0.070547923899161 Ounce/Millimeter
3 Gram/Millimeter0.10582188584874 Ounce/Millimeter3 Gram/Millimeter = 0.10582188584874 Ounce/Millimeter
4 Gram/Millimeter0.14109584779832 Ounce/Millimeter4 Gram/Millimeter = 0.14109584779832 Ounce/Millimeter
5 Gram/Millimeter0.1763698097479 Ounce/Millimeter5 Gram/Millimeter = 0.1763698097479 Ounce/Millimeter
6 Gram/Millimeter0.21164377169748 Ounce/Millimeter6 Gram/Millimeter = 0.21164377169748 Ounce/Millimeter
7 Gram/Millimeter0.24691773364706 Ounce/Millimeter7 Gram/Millimeter = 0.24691773364706 Ounce/Millimeter
8 Gram/Millimeter0.28219169559664 Ounce/Millimeter8 Gram/Millimeter = 0.28219169559664 Ounce/Millimeter
9 Gram/Millimeter0.31746565754622 Ounce/Millimeter9 Gram/Millimeter = 0.31746565754622 Ounce/Millimeter
10 Gram/Millimeter0.3527396194958 Ounce/Millimeter10 Gram/Millimeter = 0.3527396194958 Ounce/Millimeter
100 Gram/Millimeter3.53 Ounce/Millimeter100 Gram/Millimeter = 3.53 Ounce/Millimeter
1000 Gram/Millimeter35.27 Ounce/Millimeter1000 Gram/Millimeter = 35.27 Ounce/Millimeter
# Gram/Millimeter to Other Units Conversion Table
ConversionDescription
1 Gram/Millimeter = 1 Kilogram/Meter1 Gram/Millimeter in Kilogram/Meter is equal to 1
1 Gram/Millimeter = 0.01 Kilogram/Centimeter1 Gram/Millimeter in Kilogram/Centimeter is equal to 0.01
1 Gram/Millimeter = 0.001 Kilogram/Millimeter1 Gram/Millimeter in Kilogram/Millimeter is equal to 0.001
1 Gram/Millimeter = 0.3048 Kilogram/Foot1 Gram/Millimeter in Kilogram/Foot is equal to 0.3048
1 Gram/Millimeter = 0.0254 Kilogram/Inch1 Gram/Millimeter in Kilogram/Inch is equal to 0.0254
1 Gram/Millimeter = 0.9144 Kilogram/Yard1 Gram/Millimeter in Kilogram/Yard is equal to 0.9144
1 Gram/Millimeter = 1000 Gram/Meter1 Gram/Millimeter in Gram/Meter is equal to 1000
1 Gram/Millimeter = 10 Gram/Centimeter1 Gram/Millimeter in Gram/Centimeter is equal to 10
1 Gram/Millimeter = 304.8 Gram/Foot1 Gram/Millimeter in Gram/Foot is equal to 304.8
1 Gram/Millimeter = 25.4 Gram/Inch1 Gram/Millimeter in Gram/Inch is equal to 25.4
1 Gram/Millimeter = 914.4 Gram/Yard1 Gram/Millimeter in Gram/Yard is equal to 914.4
1 Gram/Millimeter = 35.27 Ounce/Meter1 Gram/Millimeter in Ounce/Meter is equal to 35.27
1 Gram/Millimeter = 0.3527396194958 Ounce/Centimeter1 Gram/Millimeter in Ounce/Centimeter is equal to 0.3527396194958
1 Gram/Millimeter = 0.03527396194958 Ounce/Millimeter1 Gram/Millimeter in Ounce/Millimeter is equal to 0.03527396194958
1 Gram/Millimeter = 10.75 Ounce/Foot1 Gram/Millimeter in Ounce/Foot is equal to 10.75
1 Gram/Millimeter = 0.89595863351934 Ounce/Inch1 Gram/Millimeter in Ounce/Inch is equal to 0.89595863351934
1 Gram/Millimeter = 32.25 Ounce/Yard1 Gram/Millimeter in Ounce/Yard is equal to 32.25
1 Gram/Millimeter = 2.2 Pound/Meter1 Gram/Millimeter in Pound/Meter is equal to 2.2
1 Gram/Millimeter = 0.022046226218488 Pound/Centimeter1 Gram/Millimeter in Pound/Centimeter is equal to 0.022046226218488
1 Gram/Millimeter = 0.0022046226218488 Pound/Millimeter1 Gram/Millimeter in Pound/Millimeter is equal to 0.0022046226218488
1 Gram/Millimeter = 0.67196897513951 Pound/Foot1 Gram/Millimeter in Pound/Foot is equal to 0.67196897513951
1 Gram/Millimeter = 0.055997414594959 Pound/Inch1 Gram/Millimeter in Pound/Inch is equal to 0.055997414594959
1 Gram/Millimeter = 2.02 Pound/Yard1 Gram/Millimeter in Pound/Yard is equal to 2.02
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# What does $\#P\subseteq FP^{PPAD}$ imply?
We know $$\#P\subseteq {PPAD}\implies PH\subseteq P^{{PPAD}}\subseteq P^{{NP}}$$ and the polynomial hierarchy collapses ($$FP^{PPAD}=PPAD$$ following Emil Jerabek's comment).
1. Can $$\#P\subseteq {PPAD}$$ give a stronger collapse such as $$\oplus P=UP=NP=coNP=PP$$ or $$NP=coNP=P^{PPAD}$$?
2. What other consequences follow in $$\#P\subseteq {PPAD}$$ holds?
• Would you mind adding a reference for your opening claim? – Daniel Apon Jul 30 '17 at 2:23
• @DanielApon $PPAD\subseteq NP$ is known. – 1.. Jul 30 '17 at 2:42
• @DanielApon I thought it follows from Toda's theorem. – 1.. Jul 30 '17 at 4:55
• PPAD is a class of search problems, not decision problems. Thus "NP = PPAD" is a meaningless string of symbols. – Emil Jeřábek Jul 30 '17 at 18:13
• @EmilJeřábek taken your suggestion comment. – 1.. Jul 31 '17 at 11:24
First, $\mathrm{PPAD\subseteq FP^{NP}}$, hence $\mathrm{\#P^{PPAD}\subseteq\#P^{NP}\subseteq FP^{\#P}}$. Moreover, $\mathrm{PPAD}$ is closed under Turing reductions, i.e., $\mathrm{FP^{PPAD}\subseteq PPAD}$. Thus, if we assume $$\mathrm{\#P\subseteq PPAD},$$ then $$\mathrm{\#P^{PPAD}\subseteq PPAD},$$ which by induction implies $$\mathrm{FCH=PPAD}.$$ Passing to decision problems, since $\mathrm{P^{PPAD}\subseteq P^{TFNP}\subseteq NP\cap coNP}$, this shows $$\mathrm{\#P\subseteq PPAD}\implies\mathrm{CH=P^{PPAD}=NP=coNP}.$$ (Note that using the closure of $\mathrm{PPAD}$ under Turing reductions, $\mathrm{P^{PPAD}}$ consists of predicates whose characteristic functions can be computed as projections of $\mathrm{PPAD}$ problems.)
As for $\oplus\mathrm P$, I believe $\mathrm{PPA\supseteq PPAD}$ can be solved by binary search on the predicate “the sum of degrees of vertices whose labels start with a given string is odd”, which means $$\mathrm{PPAD\subseteq FP^{\oplus P}},$$ thus (using $\mathrm{P^{\oplus P}=\oplus P}$) $$\mathrm{\#P\subseteq PPAD}\implies\mathrm{CH=P^{PPAD}=NP=coNP=\oplus P}.$$ A similar argument applies with $\mathrm{Mod}_p\mathrm P$ in place of $\oplus\mathrm P$ for any prime $p$.
• interesting that this gives $\oplus P=PP$. – 1.. Jul 31 '17 at 18:42
• is it believed $Mod_pP\subseteq P/poly$ possible at $p\neq 2$? – 1.. Aug 1 '17 at 10:54
• It’s possible in the sense that we don’t know how to disprove it at the moment. However, it is quite unlikely, just like for $\oplus P$. To begin with, it would imply $\mathrm{PH\subseteq P/poly}$ by Toda’s theorem. – Emil Jeřábek Aug 1 '17 at 13:28
• The inclusion of PP in any subclass $C$ of ModPH collapses CH to $P^C$. However, you won’t get $PP=\oplus P$ that way, this relied on the connection of PPAD to parity. – Emil Jeřábek Nov 29 '17 at 9:50
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Audio harmonics.
Can someone explain to me precisely what harmonics are, and the technical differences between even and odd numbered harmonics?
Are you familiar with normal modes in oscillating systems? Harmonics are simply different normal modes. The lowest possible mode is the "fundamental" or "first harmonic." The next excited mode is the "second harmonic" and so on. In stringed instruments these are different transverse waves on a string. In wind instruments these are different longitudinal compression waves in air. The numbers simply label the different modes, there is not any other technical difference.
It so happens, that by their construction some instruments do not support exciting certain modes. For instance, flutes and saxophones can only sustain fundamental tones and even harmonics. This why the first register change (from the 1st to 2nd harmonic) in these instruments skips an octave. By contrast, the clarinet only allows the fundamental and odd harmonics, which is why its first register change (from the 1st to 3rd harmonic) skips a major twelfth instead of an octave. But that has to do with the physical differences (specifically, the boundary conditions) of the instruments. Odd or even, modes (harmonics) are just modes (harmonics).
I'm a total layman when it comes to practically all branches of physics, so no I don't know about normal modes or oscilating systems.
I'm not sure if you're familliar with different types of amplification or not, but largely vaccumm tube aplifiers are considered worlds better thans solid state amplifiers, the only scientifical reason i can find for this is that tubes create even harmonics and Solid state creates odd number harmonics (didn't know what it meant, so I asked here). Does the harmonic being even/odd nessicarily make something sound better (or remarkably different?) or are there other things in play with tube amps vs. solid state amps?
At very simplified level you could think of harmonics as a byproduct of main sine frequency, that is multiple of the main sine. It appears always along with distortions in amps, and harmonics goes as 1,2,3,4,5,6,7... where odd and even is obvious.
Issue with odd harmonics is that they are perceived as least pleasant by humans, while even harmonics "seem" more natural. (sorta unadjusted instrument vs adjusted instrument playing a chord) Human ear is less sensitive to even harmonics (as they sound more like "chords"), thats why vacuum tube amps are perceived as sounding softer. Technically they might be much worse (like 1-2% THD).
There are some more specifics of tubes that makes use of them much simpler, but that would be quite difficult to explain.
ps. it could fit better into engineering forums
OK, I am not too familiar with audio amplification equipment. I have also heard that tube amps are preferred, but I have never heard that it is because analog amps produce even harmonics and electronic devices produce odd ones (I am a little dubious of this claim).
What I will say though, is that the harmonics that are present give tones their character, or timbre. You may have heard pure tones, with no harmonics (overtones) present. They sound like sterile beeps (think of a Moog synth tone). What is the difference between, say, a saxophone and a clarinet's sound when they are playing "the same pitch?" When a sax and a clarinet play the same note, the sound is different because, in addition to the fundamental note played, each instrument also adds other pitches (harmonics) in different amounts. These additional pitches give the overall sound its individual character, and make them sound different to your ear, even though they are both "middle C."
Edit To put it succinctly: Sounds produced by musical instruments are not beep-like pure pitches, they are really complicated chords. The upper notes in these "chords" are the harmonics, and the particular mix of these are what give instruments their characteristic sounds.
Again, I am skeptical that the reason you give (a simple even/odd dichotomy) is the reason some peope prefer tube amps. If that were really the case then analog amplification of a clarinet sound would leave only a pure tone, since all the clarinet overtones are odd harmonics. But this is not what happens. Faithful amplification means that you accurately amplify a sound -- it's fundamental and all of the harmonic overtones present (both even and odd). More likely in my mind is something along the lines of: electronic amps don't have the bandwidth to accurately amplify all of the harmonic overtones present in a complex sound. Maybe for a given price you can only get a digital amp to reproduce the first 7 or 8 harmonics, but subtle and nuanced parts of, say, a piano sound, might come from higher harmonics. In which case the digital amp would only be able to "throw away" these important higher harmonics, or amplify them by a disproportionate amount, and we would hear this as distortion.
But that's just my speculation; I will leave the rest to those who know.
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For those who aren't familiar with instrument amplification:
All amps are "electronic", Solid state amps are not digital, they are analouge along with tube amps. I really have no idea as to the real reason tube amps are so vastly preferred over solid state amps, I've merely heard that it has to do with the fact that tube amps produce even harmonics which are more pleasing to humans, just trying to verify and quantify that statement.
Perhaps tube amps simply emphasise even harmonics instead of only producing them?
Perhaps tube amps simply emphasise even harmonics instead of only producing them?
I just don't buy it. Original sounds to be amplified may have both odd and even harmonics. This is not up to the amp, it has to deal with whatever it gets. If an amp only emphasized even harmonics, then it would not be faithfully amplifying. In particular, any sound that consists mostly of odd harmonics would be distorted.
I don't think there is anything to the idea that odd or even harmonics are more or less pleasing to the ear, either. The second harmonic is at the octave, sure, but the third harmonic is at an octave plus a fifth, which is also considered "sonorous." Also, after more than a few harmonics, both odd and even harmonics are spaced extremely close together (ultimately, less than a semitone apart) and produce "dischordant" intervals.
Originally posted by wasteofo2
Can someone explain to me precisely what harmonics are, and the technical differences between even and odd numbered harmonics?
Somebody can, but I'm going to have a go at it.
Ah, well, unfortunately I'm still confused about a few things; but... A frequency that is faster than another frequency by a whole number multiple, 2 times as fast, 3 times as fast..., is called the harmonic of the original frequency. Take one frequency, say, 60 hertz. 60 times 2 is 120, so 120 hertz is the second harmonic of 60, and so on through 3, 4, 5, 6...
Are you familiar with destructive and constructive interference? If not, I first learned about it reading about holograms. Well, take one frequency and then combine it with one or more of its harmonics and they will add to or subtract from that original frequency. All the signals combine to create one big final signal that is shaped differently from the original. I hope that made sense. At any rate (har), it turns out that adding odd numbered harmonics to a signal tends to push it towards a saw tooth shape, so you get that harsh sawtooth sound.
That's all I know so far.
Originally posted by Vosh
I'm sorry, but it's just idiotic to come back with an abstruse answer like that to a question asked by a lay person. I've worked out a list of reasons why some ppl. behave this way: They don't really understand
I'm sorry, but its just stupid to jump on someone when you have no idea of what you are talking about.
I gave short and succint answer to the question. If you can't understand it, even after reading twice, ask.
Person coming from listening music and asking about tube amps is assumed to have rudimentary background to understand what is said without starting with a lecture.
BigRedDot,
you are talking about harmonics already present in the signal, which is what needs to be amplified. Distortion in amps means _addition_ of harmonics that were NOT there. It happens because electronic is not perfect. Tubes and Solid amps are not perfect in different ways, and that is reason why they produce different kinds of distortion harmonics. Tubes are more or less uncapable of adding odd harmonics, while Solids are more prone to produce exactly them.
This is true that some instruments have more odd harmonics, and that amps that add them where unappropriate, are bound to sound more sharp and unpleasant, because they are basically overamplifying what was there and also adding what wasn't there.
And that odd harmonics from distortions sound less pleasant than even harmonics has been established eons ago (heck, ask your grandma). One reason is exactly because there are less even harmonics in the original signal, thus what amps adds there simply doesn't stick out.
Perhaps tube amps simply emphasise even harmonics instead of only producing them?
wasteofo2, all amps are meant to reproduce input signal precisely. Distortion of signal happens due to imperfections. As Vosh has found out, distortions that make sinewave look more like sawtooth is adding odd harmonics (in reality its different from sawtooth, but idea would do).
Why this happens is question about electronics engineering, which is way beyond layman explanations.
Nevertheless, don't rush into buying a tube amp, its mostly attribute of hopeless audiophiles who refuse to consider their issues.
I probably over reacted. Have a hair trigger when I think someone is being unnecessarily abstruse (and this happens all too often and is practically an institution and it irritates me). Beg 'pardon.
I had to start over several times with my own answer because it is after all much much easier to explain by just drawing the sine wave as it appears on an oscilliscope and then adding the harmonics and then draw the final wave that results -- trying to describe all that verbally is tricky even though the actual basic concept is simple.
I am now wondering now if my modem kicks off periodically because of harmonics in the phone lines. Perhaps at certain times there are more ppl. in this apartment complex using their phones (or using appliances; would the mains effect the phones?) than at other times, building up the harmonic distortions that ruin the signal (I'm told that a little gap in the data stream makes the modem think it's supposed to hang up -- and *man* it's annoying!). Or old rickety wires perhaps could create these harmonics... I feel like I'm on Star Trek...
Originally posted by BigRedDot
OK, I am not too familiar with audio amplification equipment. I have also heard that tube amps are preferred, but I have never heard that it is because analog amps produce even harmonics and electronic devices produce odd ones (I am a little dubious of this claim).
Solid state devices have distortion characteristics that result in even harmonics being produced while tube amplifiers have distortion characteristics that result in odd harmonics. This is true for every single pure tone that is input into the amp. If you put in a flute sound or a clarinet sound then each harmonic of the sound while suffer some distortion.
Again, I am skeptical that the reason you give (a simple even/odd dichotomy) is the reason some peope prefer tube amps. If that were really the case then analog amplification of a clarinet sound would leave only a pure tone, since all the clarinet overtones are odd harmonics. But this is not what happens. Faithful amplification means that you accurately amplify a sound -- it's fundamental and all of the harmonic overtones present (both even and odd).
You are right. Any decent amplifier will produce all the harmonics of the sound input to it. The distortion caused by the amplifier is additional harmonics caused by its imperfections. All the overtones are reproduced but each tone suffers some distortion. Since the distortion increases with signal amplitude, the strongest tones suffer the most distortion.
The difference in tube vs. transistor distortion is often held up for the reason many people prefer tube amps. I don't think that's the whole story although I do prefer tube amps for my guitar.
If you put in a flute sound or a clarinet sound then each harmonic of the sound while suffer some distortion.
Ok that makes perfect sense. Thank you for the clarification.
Originally posted by BigRedDot
I just don't buy it. Original sounds to be amplified may have both odd and even harmonics. This is not up to the amp, it has to deal with whatever it gets. If an amp only emphasized even harmonics, then it would not be faithfully amplifying. In particular, any sound that consists mostly of odd harmonics would be distorted.
When I'm referring to amplifiers, I mean instrument amplifiers, not stereo speakers or anything. Instrument amplifiers goal is not to faithfully reproduce the exact sound coming from the instrument, but to be a sort of extention of the instrument and add it's own tone to it. If you play the same instrument through different kinds of amps it will sound very different, which is why there is such a large variety of different amplifiers, becuase people like different tones for their instruments.
Originally posted by wimms
wasteofo2, all amps are meant to reproduce input signal precisely. Distortion of signal happens due to imperfections. As Vosh has found out, distortions that make sinewave look more like sawtooth is adding odd harmonics (in reality its different from sawtooth, but idea would do).
Nevertheless, don't rush into buying a tube amp, its mostly attribute of hopeless audiophiles who refuse to consider their issues.
Either I'm misunderstanding your point, or you're just plain wrong in saying "all amps are meant to reproduce input signal precisely". Any given amplifier company will have many different models which are supposed to have distinct tonal qualities.
I was always under the impression that distortion of a signal (at least in the sense of distortion used in heavy metal recordings) was caused due to the fact that the signal being fed into the amp was too great and the amp needed to compress the signal, kind of like if you have an image and make it smaller using MS paint the quality will go down.
As a confessed audiophile, tube amps are the only amps for me. They're louder than solid state amps with the same size speaker and wattage, they're much more easy to customize and, at least to me, they sound much better as a whole.
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Originally posted by wasteofo2
Either I'm misunderstanding your point, or you're just plain wrong in saying "all amps are meant to reproduce input signal precisely". Any given amplifier company will have many different models which are supposed to have distinct tonal qualities.
You are confused. Function of amp is to have precise amplification. You are talking about something else. Of amps that sound very differently one is pile of crap. Perhaps musicians call about anything that draws power an amp.
As a confessed audiophile, tube amps are the only amps for me. They're louder than solid state amps with the same size speaker and wattage, they're much more easy to customize and, at least to me, they sound much better as a whole.
Oh. Though, "sound much better" is typically achieved by pulling bass and treble to the sky in these rings. If its more important that it sounds good instead of right, then sure.
Though, with really good amps you'd starve before you could note any difference.
Originally posted by wimms
You are confused. Function of amp is to have precise amplification. You are talking about something else. Of amps that sound very differently one is pile of crap. Perhaps musicians call about anything that draws power an amp.
The following are clips of the same person, playing the same song, with the same guitar and amp, the only difference being the speakers used in the amp. I don't know if maybe my ear is more delicate than yours, but I can hear a very noticable difference between the seperate recordings, though none sound like crap.
http://members.cox.net/savage/tmp/ShivaV30.MP3
http://members.cox.net/savage/tmp/ShivaGreenback.MP3
http://members.cox.net/savage/tmp/ShivaCL80.MP3
http://members.cox.net/savage/tmp/ShivaV30.MP3
If you have any guitarist friends, ask to go to a guitar shop with them and have them try out a fender amp and then a mesa boogie amp of simmilar wattage with the same guitar and have the amps set simmilarly, Perhaps you won't hear a difference, but the guitarist probabally will.
If you are a guitarist and still maintain that all good amps sound the same, then either your perception of what a "good amp" is is very very particular and you only like one particular sound, you don't have a refined ear for tone, or you haven't heard many different amps in your life.
Originally posted by wimms
Oh. Though, "sound much better" is typically achieved by pulling bass and treble to the sky in these rings. If its more important that it sounds good instead of right, then sure.
Though, with really good amps you'd starve before you could note any difference.
I actually hate that scooped (prominant lows and highs) sound and prefer something more mid oriented.
I just don't see how you can say that good amps will all sound the same, especially considering any "good" amp company will have many different models all designed to sound unique. Again, go try out a fender then a mesa boogie, both are highly regarded and both sound ridiculously different
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You are talking about different thing. What you call amp is really "Guitar Amp" thats far from being simply amplifier. I don't know what they do with signal to achieve different sounds, but I can assure you that its not general state of all audio amplifiers to sound differently, and I'm pretty convinced that they have something else besides amp inside the units you test that actually are responsible for differing sound.
What you call "good amp" and what I mean by "good amp" are different things, apparently. Amplification is a function of increasing input signal by amount of some gain factor. There is only ONE way to do it right, so that output signal is related to input signal exactly by factor of N. Any deviation, and its called distortion. What you seem to call "good amp" is really "tasteful distortion", ala that same "overdrive" thingy used in heavy metal.
What I meant by "crap amp" was not "crap sound", these things are separate. I meant that the amp deviated from precise amplification.
So, in my dictionary, "good amp" means that which represents input signal precisely. And two good amps have to be damn similar in that. Thus if you have really good amps to listen, it must be damn hard to make a difference. And thats reality in audio amps.
Hope that clarifies confusion.
Integral
Staff Emeritus
Gold Member
This has really turned into a discussion best dealt with by the Engineers, so off to engineering we go.
As to why odd harmonics are "less pleasing" then even. In Fourier Analysis the sum of all odd harmonics produces a square wave. It is generally true that the human ear and psychic prefer "smooth" variations rather then abrupt ones. So if for some reason the electronics emphasize odd harmonics over even the result will be a final wave form which is approaching a square wave, thus less pleasing to the ear.
The connection between odd harmonics and square waves goes both ways, in the electronics if there is anything that attempts to cause a step function change (that is a square wave form) it will instead add odd harmonics to the signal. The more abrupt the transition the more odd harmonics that will be created.
Keep in mind that electronic devices maintain a continuous signal, by continuous here I am referring to a mathematical definition of continuity. This means a picture of the signal can be drawn without picking up you pencil and that it is always moving forward with time. In other word perfect square waves are NOT allowed. Any attempt at a square wave results in many (as in high n) odd harmonics.
We must separate Heavy Metal from main stream when it comes to sound amplification. In Heavy Metal distortion is a desirable feature. I would bet that the signal is over driven at every possible step of the way. Over driving an amp tends to induce clipping, which results in square waves, thus odd harmonics.
Now with this in consideration, I am surprised that you prefer the the tube amp and its tendency to favor even harmonics (as you claim) as this would tend to damp out the odd harmonics you are attempting to force. Hummm....
What you seem to be saying is that the solid state amp reproduces the effects I am trying cause TO FAITHFULLY.
Perhaps some of the engineering types can add to this.
Originally posted by Integral
We must separate Heavy Metal from main stream when it comes to sound amplification. In Heavy Metal distortion is a desirable feature. I would bet that the signal is over driven at every possible step of the way. Over driving an amp tends to induce clipping, which results in square waves, thus odd harmonics.
Now with this in consideration, I am surprised that you prefer the the tube amp and its tendency to favor even harmonics (as you claim) as this would tend to damp out the odd harmonics you are attempting to force. Hummm....
What you seem to be saying is that the solid state amp reproduces the effects I am trying cause TO FAITHFULLY.
Perhaps some of the engineering types can add to this.
This is just a guess, but perhaps why I find tube amps to give a better distorted sound is that it's more evenly balanced between even and odd harmonics, while a solid state amp would heavily favor odd harmonics.
Integral
Staff Emeritus
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# Science
posted by .
There Are Three Resistance In A Parallel Circuit. Two Of Them Are Rated At 20 Ohms, The Other At 10 Ohms. What Is The Circuit's Total Resistance?
I believe the answer D 5ohms
A. 20 ohms
b. 12 ohms
c. 8 ohms
d. 5 ohms
• Science -
1/R = 1/20 + 1/20 + 1/10 = 4/20 = 1/5
R = 5
## Similar Questions
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there are three sources of resistance in a parallel ciruit. Two of them are rated at 20ohms, the other at ohms. What is the curuit's total resistance?
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More Similar Questions
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# Solving Equations Maths Worksheet Algebra Worksheets Linear Gcse Simultaneous Genie Edexcel Formula
Slainie Héloïse October 2, 2020 Math Worksheet
How Do You Find Points In A Graph? This set of numbers (2, 3) is an example of an ordered pair. The first number refers to the value of x while the second number stands for the value of y. When ordered pairs are used to find points on the grid, they are called the coordinates of the point. In above example, the x coordinate is 2 while the y coordinate is 3. Together, they enable you to locate the point (2, 3) on the grid. What’s the point of all this? Well, ever wondered how ships describe exactly where they are in the vastness of the ocean? To be able to locate places, people have to draw a grid over the map and describe points with the help of x and y coordinates. Why don’t you give it a try? Imagine left side wall of your room to be y axis and the wall at your back to be the x axis. The corner that connects them both will be your origin. Measure both in feet. If I say stand on coordinates (3, 2), would you know where to go? That means from the corner (origin) you should move 3 feet to the right and 2 feet forward.
Granted, that will always be a tough uphill battle for math to win out over most video games, but the point is, students today are much more immersed in technology than ever before. So even if you need to pass out a math worksheet to review concepts and formulas, it will greatly benefit your cause if you design the worksheet to be as stimulating as possible.
At the grassroots level, teachers in schools are given a packed curriculum for the year. Schools try to teach the students a number of procedures without delving much into its finer details. Hence, the student is left in a confounding position as to when a particular procedure must be used. The key ingredient to understanding math is constant practice and math assignment help. Unfortunately, this is not a common scenario among the popular math classes. It is widely understood that math has a global use and acceptance. People are also aware of the rate at which math is advancing today at various fields of research and study. Many mathematicians will talk about the pattern and structure of math worksheets which are helpful for people in working fields. Math has helped science and technology reach a higher level of advancement.
It’s easy to see how free worksheets can save you money. If you want, you can skip buying math books and just use worksheets that you get for free on the internet. All you need to do is use a ”scope and sequence” book that tells you what your child needs to be doing in math by age and grade. This book is essential when you homeschool. I recommend getting one of these books when you first begin homeschooling and use it as a reference throughout your homeschool journey. Regardless of how long you homeschool, you’ll always have doubts and questions about how your child is performing.A scope and sequence book can put your mind at ease.
### Find The Message Math Worksheet Answers94
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#### Math And Reading Printable Worksheets
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##### Distance Time Graphs Maths Worksheets
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###### Math Course 1 Printable Worksheets
Nov 07, 2020
Let us discuss some tangible advantages of Mathematics in today’s world. One should also be aware of the wide importance of Mathematics, and the way in which it is advancing at a spectacular rate. Mathematics is about pattern and structure; it is about logical analysis, deduction, calculation within these patterns and structures. When patterns are found, often in widely different areas of science and technology, the mathematics of these patterns can be used to explain and control natural happenings and situations. Mathematics has a pervasive influence on our everyday lives, and contributes to the wealth of the individual. Now that we know the benefits of mathematics what is needed to conquer this hard to tackle and arduous subject. Answer is quite simple actually, practice, a lot of practice from early childhood. But in order to perform any task or practice any task, one requires resources. Obtaining a resource is not difficult in today’s modern world but affordable resource is definitely a rare commodity. Today education is a sector which gets very little funding from the federal government.
1st grade math worksheets and my Mom’s math teaching style. Math won’t be as terrible as it seems if parents take interest in preparing their little ones for math before school age. I grew up not understanding how it is that people talk about math as difficult as they do, it was my best subject at school. It was easy because of my upbringing that ensured that math and I got acquainted long before school. My mother who was a primary grade teacher told me how she began teaching me math in different guises at home before I got to school age.
### Photos of Gcse Maths Linear Equations Worksheet
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# 23600 in Words
The number 23600 is written in words as “twenty-three thousand six hundred”. It’s six hundred more than twenty-three thousand. For instance, if you have twenty-three thousand six hundred flowers, it means you have twenty-three thousand flowers and six hundred additional flowers.
Thousands Hundreds Tens Ones 23 6 0 0
## How to Write 23600 in Words?
The number 23600 is written as ‘Twenty-Three Thousand Six Hundred’ in words. It has a ‘2’ in the ten-thousands place, a ‘3’ in the thousands place, a ‘6’ in the hundreds place, and ‘0’ in both the tens and ones places. Like counting twenty-three thousand six hundred items, you say, “I have twenty-three thousand six hundred.” Therefore, 23600 is written in words as ‘Twenty-Three Thousand Six Hundred’.
Example:
1. Place Value Chart:
Thousands: 23, Hundreds: 6, Tens: 0, Ones: 0
2. Writing it down: 23600 = Twenty-Three Thousand Six Hundred
This method is an essential part of early mathematics education for understanding how larger numbers are formed and written in words.
## FAQ on 23600 in Words
### How do you express the number 23600 in words?
The number 23600 is expressed as ‘Twenty-three thousand six hundred’.
### Write the number 23600 using letters.
Twenty-three thousand six hundred’ is how you write the number 23600.
### If you see 23600 birds, how would you spell ‘twenty-three thousand six hundred’?
Seeing 23600 birds, you spell it as ‘Twenty-three thousand six hundred’.
## Other Numbers in the Words:
After-School Math Program
• Boost Math Skills After School!
After-School Math Program
Boost Your Child's Math Abilities! Ideal for 1st-8th Graders, Perfectly Synced with School Curriculum!
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ソースコード 拡げる
Copy
```#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long LL;
typedef double db;
int get(){
char ch;
while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
if (ch=='-'){
int s=0;
while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
return -s;
}
int s=ch-'0';
while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
return s;
}
const int L = 10395+10;
const int N = 13;
const int M = (1<<12)+1;
bitset<L>cp[N][N],f[M];
int n,m;
char s[N][N];
bool bz[N];
int tp[L][N];
int k;
int ct[N];
void dfs(int x,bool tg){
if (x>m){
k++;
fo(i,1,m)tp[k][i]=ct[i];
return;
}
if (bz[x]){dfs(x+1,tg);return;}
fo(i,x+1,m)
if (!bz[i]){
bz[x]=bz[i]=1;
ct[x]=i,ct[i]=x;
dfs(x+1,tg);
ct[x]=ct[i]=0;
bz[x]=bz[i]=0;
}
if (!tg&&(m&1)>0){
bz[x]=1;
ct[x]=x;
dfs(x+1,1);
ct[x]=x;
bz[x]=0;
}
}
int cnt[M];
int main(){
//freopen("data.in","r",stdin);
n=get();m=get();
fo(i,1,n)scanf("%s",s[i]+1);
dfs(1,0);
fo(i,1,n){
fo(j,i,n){
fo(w,1,k){
bool pd=1;
fo(x,1,m)pd&=(s[j][tp[w][x]]==s[i][x]);
if (pd)cp[i][j][w-1]=1;
}
}
}
if (n&1){
fo(st,1,n)
f[(1<<(st-1))]=cp[st][st];
}
else
fo(i,0,k-1)f[0][i]=1;
fo(i,1,(1<<n)-1)cnt[i]=cnt[i-(i&-i)]+1;
fo(i,0,(1<<n)-1)
if (cnt[i]+2<=n){
int x;
fo(w,1,n)
if (((1<<(w-1))&i)==0){x=w;break;}
fo(y,x+1,n)
if (((1<<(y-1))&i)==0)f[i|(1<<(x-1))|(1<<(y-1))]=f[i|(1<<(x-1))|(1<<(y-1))]|(f[i]&cp[x][y]);
}
bool ans=0;
fo(i,0,k-1)ans=ans|f[(1<<n)-1][i];
if (ans)printf("YES\n");
else printf("NO\n");
return 0;
}```
#### 提出情報
提出日時 2018-04-14 22:05:15+0900 C - Many Medians samjia2000 C++14 (GCC 5.4.1) 0 1801 Byte RE 1437 ms 29568 KB
#### コンパイルエラー
```./Main.cpp: In function ‘int main()’:
./Main.cpp:72:29: warning: ignoring return value of ‘int scanf(const char*, ...)’, declared with attribute warn_unused_result [-Wunused-result]
fo(i,1,n)scanf("%s",s[i]+1);
^
```
#### ジャッジ結果
セット名 Sample All
WA × 3
WA × 7 RE × 11
セット名 テストケース
Sample sample1.txt, sample2.txt, sample3.txt
All sample1.txt, sample2.txt, sample3.txt, 1.txt, 10.txt, 11.txt, 12.txt, 2.txt, 3.txt, 4.txt, 5.txt, 6.txt, 7.txt, 8.txt, 9.txt, sample1.txt, sample2.txt, sample3.txt
ケース名 結果 実行時間 メモリ
1.txt WA 1312 ms 6144 KB
10.txt RE 1380 ms 24960 KB
11.txt RE 1362 ms 27392 KB
12.txt RE 1428 ms 24448 KB
2.txt RE 1269 ms 29568 KB
3.txt RE 1409 ms 24960 KB
4.txt RE 1292 ms 24960 KB
5.txt RE 1309 ms 24960 KB
6.txt RE 1386 ms 24960 KB
7.txt RE 1422 ms 25088 KB
8.txt RE 1437 ms 24960 KB
9.txt RE 1235 ms 24960 KB
sample1.txt WA 1 ms 256 KB
sample2.txt WA 1 ms 256 KB
sample3.txt WA 1 ms 384 KB
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https://marocstreetfood.com/northern-territory/set-of-ordered-pairs-example.php
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SOLUTION Provide an example of at least five ordered. 3.1 functions a relation is a set of ordered pairs (x, y). example: the set {(1,a), (1, b), (2,b), (3,c), (3, a), (4,a)} is a relation a function is a relation (so, printable ordered pairs and coordinate plane worksheets. math. addition. algebra tell what ordered pair is represented by the picture icons on the coordinate.
## Graphs Basic Terms Victoria University Australia
Pre ctivity Ordered Pairs Intercepts and Slopes PrePArAtion. A motivating example for equivalence relations is the problem of con-structing the rational numbers. happened to be a set of ordered pairs. 3., for example the ordered pair <1, 2> is not equal to the ordered pair <2, 1>. definition (cartesian product): the set of all ordered pairs ,.
Discrete mathematics/set theory/page 2. we have examples of ordered pairs. as the name says, an ordered pair is simply a suppose that the menu in example 7 is compound sets and indexing the set of all ordered pairs from two given sets appears frequently in for example, to represent the pairs in the first row of the
14/08/2012в в· for more free math tutorials visit mathgotserved.com in this clip we go over the definition of relation as an ordered pairs. we then considered four 3.1 linear systems with two variables and their solutions. all of the examples have been of consistent systems which reads вђњthe set of all ordered pairs
Definition of ordered pair explained with real life illustrated examples. also learn the facts to easily understand math glossary with fun math worksheet online at ordered pairs, intercepts, and slopes section 5.2 new terms to learn slope slope-intercept form example 1: find an ordered pair solution for the equation 2x + 3y = 10
Relations deffining relations as sets of ordered if x and y are sets then any set of ordered pairs example. let us consider the following set of ordered pairs you can put this solution on your website! a "function" can provide only a "single" solution for any given input.. the sets provided are in (x,y) pairs:
10/05/2014в в· example of a function: y=2x+1 where x is a member of the set of real numbers. i'm assuming that the set of ordered pairs would be the set/domain where the a pair of numbers used to locate a point on....complete information about ordered pair, definition of an ordered pair, examples of an ordered pair, step by step
An ordinal number is the order type of a well ordered set. algebra. is an example of a set that is not well ordered. ordered pairs. you can put this solution on your website! a "function" can provide only a "single" solution for any given input.. the sets provided are in (x,y) pairs:
## 5.3 Ordered Sets Whitman College
Set Operations cs.odu.edu. Question 393212: provide an example of at least five ordered pairs that do not model a function. the domain will be any five integers between 0 and 20., 7 relations and functions a relation rfrom a set ato a set bis a set of ordered pairs represented by ordered pairs using the set-builder notation. for example.
## Compound Sets and Indexing AMPL
Graphs Basic Terms Victoria University Australia. 3.1 linear systems with two variables and their solutions. all of the examples have been of consistent systems which reads вђњthe set of all ordered pairs https://simple.m.wikipedia.org/wiki/Ordered_pair For example, if the domain is a set subset of ordered pairs drawn from the set of all possible ordered pairs very important in discrete mathematics,.
• SparkNotes Graphing Equations Ordered Pairs
• https://sco.m.wikipedia.org/wiki/Ordered_pair
• Order Pair peter-lo.com
• How to make unordered set of pairs of integers in C++
• Watch videoв в· we're plotting an ordered pair on the x (horizontal) points on the coordinate plane examples. plotting a point (ordered pairs) 319 chapter viii ordered sets, ordinals and transfinite methods 1. introduction in this chapter, we will look at certain kinds of ordered sets. if a set is ordered in
The ordered pair is called a poset example 9: the poset of the set of natural numbers with the less-than-or-equal-to relation is a well order, printable ordered pairs and coordinate plane worksheets. math. addition. algebra tell what ordered pair is represented by the picture icons on the coordinate
For example, the graph of this functions as sets of ordered pairs . you know what that meansвђ”this set of ordered pairs is not a function. ordered pairs, intercepts, and slopes section 5.2 new terms to learn slope slope-intercept form example 1: find an ordered pair solution for the equation 2x + 3y = 10
25/11/2012в в· how to determine if a function represented by a set of ordered pairs is a one-to-one function exercise : give some examples of ordered pairs (a;b ) 2 n 2 that ordered pairs (a;a ) appear in a relation on a set a for every a 2 a then it is called re exive.
The first set of ordered pairs is a function, because no two ordered pairs have the same first coordinates with different second coordinates. the second example is for example, definition of an ordered pair. the problem with your proposal is that it does not have the defining property we want for ordered pairs: for example
Cartesian product: the cartesian or cartesian coordinate plane is an example of the cartesian product. we consider the set of all the ordered pairs describing a relation describes certain properties of things and is a set of ordered pairs. example: in calculus, universal set is real topics in sets and relations.
In this lesson you will learn to tell if a set of ordered pairs represents a function by matching the x-values to the y-values. order pair nan ordered pair consists of two elements, nexample: uthe relation on the set nthe set of all ordered pairs of the function f
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https://www.hackmath.net/en/example/2683?tag_id=104
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crawl-data/CC-MAIN-2018-43/segments/1539583511872.19/warc/CC-MAIN-20181018130914-20181018152414-00413.warc.gz
| 936,313,424
| 6,571
|
# AP - simple
Determine the first nine elements of sequence if a10 = -1 and d = 4
Result
a9 = -5
a8 = -9
a7 = -13
a6 = -17
a5 = -21
a4 = -25
a3 = -29
a2 = -33
a1 = -37
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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In what ratio must two sorts of sugar, costing #390 and #315 per kg respectively, be mixed in order to produce a mixture worth #369 per kg?
12. Overtaking
On the direct road, the passenger car overtakes the slower bus by starting to overtake 20 meters from the bus and after passing it ahead of it again 20 meters away. The car overtakes at a steady speed of 72 km/h, the bus goes at a steady speed of 54 km/h..
13. Paratrooper
After the parachute is opened, the paratrooper drops to the ground at a constant speed of 2 m/s, with the sidewinding at a steady speed of 1.5 m/s. Find: a) the magnitude of its resulting velocity with respect to the ground, b) the distance of his land fr
14. Mixed with percentages
Calculate 33 1/3% of 570.
15. Square metal sheet
Four squares of 300 mm side were cut out from a square sheet metal plate with a side of 0,7 m. Express the fraction and the percentage of waste from the square metal sheet.
16. Exchange rates
If the Canadian dollar appreciated by C\$0.005 relative to the US dollar, what would be the new value of the Canadian dollar per US dollar? Assume the current exchange rate was US\$1 = C\$0.907.
17. The swimmer
The swimmer swims at a constant speed of 0.85 m/s relative to water flow. The current speed in the river is 0.40 m/s, the river width is 90 m. a) What is the resulting speed of the swimmer with respect to the tree on the riverbank when the swimmer motion
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# Difference between revisions of "Southern Celestial Rotation"
The Southern Celestial Rotation refers to the anti-rotation of the stars observed in the southern regions of the earth.
## Monopole Model
P-Brane explains the counter-rotation of the stars as a consequence of perspective. The author uses the sun's crepuscular rays as an example for how motion and perspective can seem to cause anti-rotations in two different directions, much like how spinning in the chair and looking up and down can cause the appearance of two rotations in different directions.
According to P-Brane's explanation the observer is seeing a wide range of stars squished into a small area by perspective. Rotation opposite from the Northern rotation appears as a mirrored anti-rotation due to opposite converging perspective elements.
Q. How can two people on opposite sides of the earth both see the South Pole Stars simultaneously?
A. Since those areas are many hours apart from each other, when it is night or dusk for one area it is likely day or dawn for the other. It is questioned whether it is the case that those observers see the same stars simultaneously.
## Bi-Polar Model
Under the Bi-Polar Model the rotation in the south is explained with the existence of a South Pole and a rotating southern celestial system above it.
See the Bi-Polar Model
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# Which of the following most logically completes the
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Which of the following most logically completes the argument?
The irradiation of food kills bacteria and thus retards spoilage. However, it also lowers the nutritional value of many foods. For example, irradiation destroys a significant percentage of whatever vitamin B1 a food may contain. Proponents of irradiation point out that irradiation is no worse in this respect than cooking. However, this fact is either beside the point, since much irradiated food is eaten raw, or else misleading, since _______.
A. many of the proponents of irradiation are food distributors who gain from food’s having a longer shelf life
B. it is clear that killing bacteria that may be present on food is not the only effect that irradiation has
C. cooking is usually the final step in preparing food for consumption, whereas irradiation serves to ensure a longer shelf life for perishable foods
D. certain kinds of cooking are, in fact, even more destructive of vitamin B1 than carefully controlled irradiation is
E. for food that is both irradiated and cooked, the reduction of vitamin B1 associated with either process individually is compounded
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Betweeb A and C .I would go for A
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14 Jun 2007, 20:04
I would go for E. It is miselading if each effect is compounded resulting in even less vitamin B1.
14 Jun 2007, 20:04
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# Weighted Poincare Inequality
I'm trying to prove a result I found in a paper, and I think I'm being a bit silly.
The paper claims the following: By the Poincare inequality on the unit square $\Omega \subset \mathbb{R}^2$ we have that $$\int_{\Omega} f(x)^2 dx \leq C \int_{\Omega}|\nabla f|^2 dx + \left(\int_{\Omega}f(x)dx\right)^2.$$ Thus, if $w(x)$ is a weight satisfying $\int_\Omega w(x)\,dx =1$ and $0 < W_{-} \leq w(x)$ one can prove the Poincare inequality with respect to the measure $w(x)dx$.
I'm assuming that the author means the following result $$\int_{\Omega} f(x)^2 w(x)dx \leq C' \int_{\Omega}|\nabla f|^2 w(x)dx + \left(\int_{\Omega}f(x)w(x)dx\right)^2.$$
Does anybody have any idea how to prove this? I'm particularly interested in the relationship between the Poincare constant $C$ and $C'$.
• Which paper? (In case you are missing something there) – 40 votes Jul 16 '13 at 18:50
• I am curious about the paper as well. – Shuhao Cao Jul 17 '13 at 3:04
• The paper in question is "Rate of Convergence for Ergodic Continuous Markov processes: Lyapunov vs. Poincare", by Bakry et al. They only use this result in passing (Section 4.1) without the details. However, I have seen a similar result in several other paper, always stated without justification, which led me to believe it's obvious given the right assumptions. – Il-Bhima Jul 17 '13 at 9:02
The natural form of Poincaré inequality is $$\int_\Omega |f-f_\Omega|^2 \le C\int_\Omega |\nabla f|^2\tag1$$ where $f_\Omega=\int_\Omega f$ is the mean of $f$. This is exactly your first inequality, but I think (1) captures the meaning better. The weighted Poincaré inequality would be $$\int_\Omega |f-f_{\Omega, w}|^2w \le C'\int_\Omega |\nabla f|^2w\tag2$$ where $f_{\Omega,w}=\int_\Omega fw$ is the weighted mean of $f$. Again, this is what you have but written in a more natural way.
The industry of weighted Poincaré inequalities is huge, but the most fundamental result is that the Muckenhoupt condition $w\in A_2$ is sufficient for (2). This is proved in detail, e.g., in Chapter 15 of Nonlinear Potential Theory of Degenerate Elliptic Equations by Heinonen, Kilpeläinen, and Martio (now published by Dover, $10). The constant$C'$of course depends on the$A_2$norm of$w$, but not in any explicit way. There has been some recent interest in estimates of the form$C'\le c\|w\|_{A_2}$with weight-independent constant$c$, but I do not know if this particular one has been proved. Anyway, the assumptions you stated are not enough for (2) to hold. Let$f(x)=\min(M, \log\log (e+|x|^{-1}))$where$M$is a large number to be chosen later. Spread one half of available weight uniformly on the square, and put the other half onto the set where$f=M$. Since on most of the square$f$is much smaller than$M$, the weighted mean of$f$is about$M/2$. Hence, the left hand side of (2) is of order$M$. But the right hand side of (2) is bounded independently of$M$. • Hi, thanks for this answer. I managed to find a proof of the result for when$w$is a smooth density. The approach is quite similar to that of Shuhao Cao below, though mine requires more assumptions. I also managed to obtain a bound for the weighted Poincare' constant$C'$. The counterexample you provided is extremely interesting as well as the Heinonen reference, I was not aware of this, nor of the relation to the Muckenhoupt condition. I came to this problem from a stochastic point of view, for me the constant$C'$is a lower bound for the$L^2$-spectral gap of a diffusion process. – Il-Bhima Jul 17 '13 at 8:58 Like 40 votes said, there is an industry of this. I happened to learn some of the results of the weighted Poincaré type inequality for some of my previous researches. There is a famous book by A. Kufner: Weighted Sobolev spaces. A more recent treatise on this is Nonlinear Potential Theory and Weighted Sobolev Spaces, please refer to a dedicated section 4.3 for Poincaré type inequality in weighted Sobolev space. I myself learned the proof from this paper: An explicit right inverse of the divergence operator which is continuous in weighted norms. The section 3 of this paper proved the weighted Friedrichs' inequality for the weight$w = (|x-x_0|^2 + \theta^2)^{1/2}$. Following is roughly the adaptation of the proof for any positive smooth weight$w>0$,$\int_{\Omega} w = 1$(the weight can vanish on the boundary), under the same assumption "$\Omega$is star-shaped with respect to a ball$B$". Also what you wrote in your question is actually okay with a slight modification, which is the weighted Friedrichs' inequality (notice the following is much stronger version of 40 votes wrote): For smooth function$f$such that$\displaystyle \int_{\Omega} f w = 0$with smooth weight$w>0$in$\Omega$, and$\displaystyle \int_{\Omega} w = 1$: $$\int_{\Omega}f^2 w \leq \int_{\Omega}f^2 \leq C\int_{\Omega} |\nabla f|^2w + (\text{Optional } L^2\text{-term}).\tag{\star}$$ A sketch of the proof: For simplicity we just assume$\Omega$is a ball$B$with radius 2 centered at the origin, and we can find some ball$\hat{B}\subset \Omega$such that inside$B$, with positive radius$\hat{r} \geq c r_B$for some positive$c$. For$x\in \hat{B}$, and$y\in B$: $$|y-x|\leq C w. \tag{1}$$ Why? Think a mollifier centered at origin, and this is a sufficient condition for the weight to satisfy for there to be a Poincaré type inequality. Let$\phi$be a smooth weight vanishing at boundary of$\hat{B}$, and$\bar{f} = \int_{B} f\phi$the weighted mean of$f$vanishing outside this smaller ball then: for any$y\in B$$$f(y) - \bar{f} = \int_{B} \big(f(y) - f(z)\big)\phi(z)dx \\ = \int_{B} \int^1_0 (y-z)\cdot \nabla f(y + t(z-y)) \phi(z) dt \,dz,$$ which follows from the multi-dimensional Taylor formula. Now the trick is to rescale the variables: let$x = y + t(z-y)$$$f(y) - \bar{f} = \int_{B} \int^1_0 \frac{y-x}{t}\cdot \nabla v(x) \phi\left(y + \frac{x-y}{t}\right) \frac{1}{t^2} dx\,ds.$$ Now$\phi(y + \frac{x-y}{t})$vanishes when$0< t < \gamma|x-y|$for$\gamma$relying on$\hat{r}$, hence for the kernel $$K := \int^1_0 \frac{y-x}{t^3} \phi\left(y + \frac{x-y}{t}\right) dt,$$ we have $$|K| \leq \sup|\phi| \int^1_{\gamma|x-y|} \frac{|y-x|}{t^3} dt \leq C |y-x|^{-1}.$$ Now use condition (1),$w^{1/2} \leq C |x-y|^{-1/2}$inside the$\hat{B}$, and$K$vanishes outside$\hat{B}$we have $$|f(y) - \bar{f}| \leq \int_B |K| |\nabla f| \\ = \int_B |K|w^{-1/2} w^{1/2}|\nabla f| \\ \leq C\int_B |y-x|^{-3/2} w^{1/2}|\nabla f|.$$ Now Young's inequality for convolution reads: $$\|f - \bar{f}\|_{L^2(B)} \leq \Big\| |x|^{-3/2}\Big\|_{L^1(B)} \Big\|w^{1/2}|\nabla f|\Big\|_{L^2(B)}.\tag{2}$$ For$|x|^{-3/2}$is$L^1$when$B\subset \mathbb{R}^2$. Now if$\int_B fw = 0$, we have $$\bar{f}_{B} := \frac{1}{|B|}\int_B f \phi = \frac{1}{|B|}\int_B f(\phi-w) \leq \frac{1}{|B|} \|f\|_{L^2(B)} \left(\int_B (\phi-w)^2\right)^{1/2},$$ and the optional$L^2$-term in$(\star)$is determined by how we can manipulate the weight so that$\|\bar{f}_{B} \|_{L^2(B)}$can be absorbed by$ \|f\|_{L^2(B)}$, then we have: $$\|f\|_{L^2(B)}^2\leq \|f-\bar{f}_{B}\|_{L^2(B)}^2 + \|\bar{f}_{B} \|^2_{L^2(B)}.$$ By (2) we have the result, for domain with radius not$O(1)$, a scaling argument will be in place. I took these from some old notes I wrote to myself, there are maybe many loose ends. The$C$in$(\star)$depends on how this smooth weight$w$"concentrates" itself with this domain, if it is very concentrated around some point, then$C$in$(\star)$has to be really large to control$\|f\|_{L^2(B)}$. • I was just about to answer my own question with a proof of this result for when$w$is smooth, but the sketch proof you provided is quite similar, using the Taylor to express the$L^2$norm of$f$in terms of its derivative. I need some time to process your proof, however, I noticed you managed to prove this without requiring a positive lower bound for$w\$. In my approach this is an essential assumption in the final step of the proof to obtain the correct weighted Dirichlet form on the RHS of the inequality. Thanks again! I'll comment further after I read it carefully. – Il-Bhima Jul 17 '13 at 8:47
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# Density of smooth functions in Sobolev space, respecting nonnegative traces
I am looking for a reference for the following result (if it is true, which I would expect):
Let $\Omega\subset\mathbb{R}^n$ be a bounded Lipschitz domain. Let $\Gamma_0\subset\partial\Omega$ be sufficiently regular.
Let $V_1:=\left\{\phi\in C^\infty\left(\overline{\Omega}\right):\phi\geq 0 ~\text{on}~ \Gamma_0\right\}$.
Let $V_2$ be the set of functions $u\in W^{1,p}(\Omega)$ with non-negative trace on $\Gamma_0$.
Then $V_1$ is dense in $V_2$.
If we consider the case of "$\phi = 0$" instead of "$\phi \geq 0$", of course an analogous result for $\Gamma_0=\partial\Omega$ is well known, and there is a result that it stays true if $\Gamma_0\subset\partial\Omega$ is relatively open and has relative Lipschitz boundary.
It would already be very helpful to have a reference for the result with $\phi \geq 0$ and $\Gamma_0=\partial\Omega$.
I don't know a reference, but perhaps you could prove it as follows. For $u \in W^{1,p}(\Omega)$ with nonnegative trace, write $u = u^+ - u^-$ which are both in $W^{1,p}(\Omega)$. Since $u^+$ is nonnegative everywhere, you should be able to approximate it by nonnegative smooth functions $\phi_n$. And since $u^-$ has zero trace, by the result you quote, you can approximate it by smooth functions $\psi_n$ vanishing on the boundary (maybe even compactly supported). Then $\phi_n - \psi_n$ are nonnegative on the boundary and converge to $u$.
• Actually, a Theorem due to Stampacchia says that if $\phi$ is a Lipschit functin, then $u\mapsto\phi\circ u$ is a Lipschitz function from $W^{1,p}$ into itself. Apply this to $\phi(s)=s^+$. Mar 10, 2015 at 15:29
• @DenisSerre: That will certainly work. You can also do it by hand, by approximating $\phi(s) = s^+$ pointwise by $C^\infty$ functions with controlled derivatives. Mar 10, 2015 at 15:31
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## Ooooc
ever, it would be practically impossible to find every case of Huntington's disease, because it would require knowing the medical condition of every person in a population. Instead, genetic researchers sample a small subset of the population that they believe is representative of the whole. (The same technique is used in political polling.)
Whenever a sample is used, the possibility exists that it is unrepresentative, generating misleading data. Statisticians have a variety of methods to minimize sampling error, including sampling at random and using large samples. But sampling errors cannot be eliminated entirely, so data from the sample must be reported not just as a single number but with a range that conveys the precision and possible error of the data. Instead of saying the prevalence of Huntington's disease in a population is 10 per 100,000 people, a researcher would say the prevalence is 7.8-12.1 per 100,000 people.
The potential for errors in sampling also means that statistical tests must be conducted to determine if two numbers are close enough to be considered the same. When we take two samples, even if they are both from exactly the same population, there will always be slight differences in the samples that will make the results differ.
A researcher might want to determine if the prevalence of Huntington's disease is the same in the United States as it is in Japan, for example. The population samples might indicate that the prevalences, ignoring ranges, are 10 per 100,000 in the United States and 11 per 100,000 in Japan. Are these numbers close enough to be considered the same? This is where the Chi-square test is useful.
First we state the "null hypothesis," which is that the two prevalences are the same and that the difference in the numbers is due to sampling error alone. Then we use the Chi-square test, which is a mathematical formula, to test the hypothesis.
The test generates a measure of probability, called a p value, that can range from 0 percent to 100 percent. If the p value is close to 100 percent, the difference in the two numbers is almost certainly due to sampling error alone. The lower the p value, the less likely the difference is due solely to chance.
Scientists have agreed to use a cutoff value of 5 percent for most purposes. If the p value is less than 5 percent, the two numbers are said to be significantly different, the null hypothesis is rejected, and some other cause for the difference must be sought besides sampling error. There are many statistical tests and measures of significance in addition to the Chi-square test. Each is adapted for special circumstances.
Another application of the Chi-square test in genetics is to test whether a particular genotype is more or less common in a population than would be expected. The expected frequencies can be calculated from population data and the Hardy-Weinberg Equilibrium formula. These expected frequencies can then be compared to observed frequencies, and a p value can be calculated. A significant difference between observed and expected frequencies would indicate that some factor, such as natural selection or migration, is at work in the population, acting on allele frequencies. Population geneticists use this information to plan further studies to find these factors.
oooc ,oc
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Estimation of Partial Linear Error-in-Variables Models under Martingale Difference Sequence
Received:April 01, 2014 Revised:June 18, 2014
Key Words: partial linear error-in-variables models martingale difference sequence validation data strong consistency
Fund Project:Supported by National Natural Science Foundation of China (Grant Nos.11271155; 11371168; 11001105; 11071126; 11071269), Specialized Research Fund for the Doctoral Program of Higher Education (Grant No.20110061110003), the Natural Science Foundation of Jilin Province (Grant Nos.20130101066JC; 20130522102JH; 20101596), Twelfth Five-Year Plan' Science and Technology Research Project of the Education Department of Jilin Province (Grant No.2012186).
Author Name Affiliation Zhuoxi YU School of Management Science and Information Engineering, Jilin University of Finance and Economics, Jilin 130117, P. R. China Dehui WANG Department of Statistic, College of Mathematics, Jilin University, Jilin 130021, P. R. China Na HUANG School of Information Management and Engineering, Shanghai University of Finance and Economics, Shanghai 200433, P. R. China School of Management Science and Information Engineering, Jilin University of Finance and Economics, Jilin 130117, P. R. China
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Consider the partly linear model $Y=x\beta+g(t)+e$ where the explanatory $x$ is erroneously measured, and both $t$ and the response $Y$ are measured exactly, the random error $e$ is a martingale difference sequence. Let $\widetilde{x}$ be a surrogate variable observed instead of the true $x$ in the primary survey data. Assume that in addition to the primary data set containing $N$ observations of $\{(Y_{j},\widetilde{x}_{j},t_{j})_{j=n+1}^{n+N}\}$, the independent validation data containing $n$ observations of $\{(\widetilde{x}_{j},x_{j},t_{j})_{j=1}^{n}\}$ is available. In this paper, a semiparametric method with the primary data is employed to obtain the estimator of $\beta$ and $g(\cdot)$ based on the least squares criterion with the help of validation data. The proposed estimators are proved to be strongly consistent. Finite sample behavior of the estimators is investigated via simulations too.
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# Physics Lab Report Guidelines
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1 Physics Lab Report Guidelines Summary The following is an outline of the requirements for a physics lab report. A. Experimental Description 1. Provide a statement of the physical theory or principle observed during the experiment. (page 3) 2. Briefly describe the experiment you performed. (page 3) 3. Discuss the relevance of the experiment to the theory. Consider answering the following question: This experiment was designed to prove theory x. Did it do so? (page 3) B. Discussion of Graphs 1. For each graph, discuss the proximity of the data points to the line of best fit. (pages 4-6) 2. For each graph, state the theoretical values for slope and y-intercept. (pages 4-6) 3. For each graph, perform a percent difference calculation between the theoretical slope and y-intercept, and the measured values of slope and y-intercept. (page 6) C. Uncertainty Analysis Discussion 1. Discuss the method used to determine the measured uncertainty value for each measured quantity. (page 7) 2. Consider each sub-experiment and determine the dominant source of uncertainty. (page 7) 3. Determine the percent difference, percent uncertainty and percent variation. (pages 6-8) Then, (a) Determine if each percent difference value is larger than or smaller than the dominant percent uncertainty for the experiment. (page 8)
2 2 Guidelines (b) Draw conclusions based on your findings; for example determine the factors contributing to percent difference values greater than the percent uncertainty value. (page 8) (c) Determine if the percent variation value corresponds to constant measured values, i.e. determine if each percent variation is smaller than the percent uncertainty in the experiment. (page 8) (d) Determine experimental factors which could have contributed to variation of measurements, i.e. how would you account for measured values that are not constant. (page 8) 4. What did you think of the experiment? 5. What would you do to improve the experiment? Inquiry Based Lab Conclusions (page 9)
3 Guidelines 3 Section A: Experimental Description Most experiments performed during the semester will require a typewritten lab report. The guidelines attempt to illustrate the essential topics necessary in lab reports. The lab instructor reserves the right to augment any portion of the guidelines in any way suitable to his/her intent. Therefore you, the lab student, should always refer to your lab instructor with any disputes concerning the guidelines. This text often makes reference to a sample lab report titled The Measurement of the Gravitation Acceleration g. The sample lab report can be found on pages Though the Measurement of g report provides an accurate description of how to write a lab report, the successful lab student will benefit from learning the procedures necessary to completely satisfy all requirements of a lab report. The experimental description section of your lab report provides a brief explanation of the purpose of the experiment. Each lab in the lab manual contains the following subsections: Objective, Equipment List, Theoretical Background, Procedure, Data Analysis, and Selected Questions. The Objective and Theoretical Background subsections are presented to give you some idea of the physics supporting the theory, while the experimental process or the means of testing the theory is provided in the Procedure subsection. Generally the experimental description provides, (1) a summary of the theoretical background, and (2) a summary of the method you followed to test the theory. For example, if you wished to perform an experiment to determine the free fall rate of a ball dropped from the roof of your dormitory, you would first need to know the theory of physics governing the motion of the ball, and you would want to propose a method for testing the theory. Additionally, the experimental description may include a list of the equipment used to perform the experiment. For example, if you performed the Measurement of g experiment you would state that you used a stopwatch to measure the drop time of a ball, a meter stick to measure the distance of the ball s fall and a mass scale to measure the mass of the ball. Several lab experiments have multiple components. The separate components of a lab allow you to make broad conclusions about a theory. Often times this involves an examination of the relationship between variables of an equation. For example, the first component for the Measurement of g experiment may instruct you to position a ball at a fixed height while for each drop you vary the mass of the ball, whereas the second component instructs you to use only one ball while varying the height of the ball s fall. In general the components of a lab are printed in boldface in the procedure section of the lab. Labs containing multiple components must have conclusions for all parts of the experiment within the lab report, i.e. you will provide 5 experimental descriptions, 5 graphs/graphical conclusions (where applicable) and 5 uncertainty conclusions for an experiment that has 5 components.
4 4 Guidelines Section B: Graphs/Graphical Analysis General Graph Properties In most experiments you will be required to construct a graph. It is the lab instructor s prerogative to allow you to use Excel or other software packages to construct your graphs. All graphs (manual and electronic) have the following properties: 1. A good graph fills the whole page. The data points should extend across as much of the page as possible. This implies that there should be only one graph per page. 2. A good graph has a title that describes what is being plotted. 3. A good graph has axis labels that describe what is being plotted on each axis. 4. A good graph has tick marks both major and minor to indicate the divisions of both the x and y axis. 5. A good graph has axis units describing the units represented by the tick marks. Acceptable axis units are those that clearly label each axis with correct units corresponding to the measured quantity plotted along the axis. For example, for the x-axis 2d [cm] denotes that the quantity 2d is measured in cm and each major tick mark represents a multiple of centimeters scaled according to the graph page. 6. A good graph has evenly spaced tick marks, unless otherwise indicated by the lab instructor. 7. A good graph must clearly show the y-intercept plotted on the graph, that is, graph the point (x = 0, y = b), for b is the value of y when x = 0. Consequently, the continuity of the x and y axes must not be broken. Showing the y- intercept may be easier if the x and y axes are placed inside the graph-paper lines. See the graph examples on pages 12 and On a good graph, the numerical values of the slope and y-intercept are clearly written. 9. A good graph has the points used to measure the slope clearly indicated. These should be points on the line and not data points. The points used for the slope calculation should be marked differently from the data points. Choose points on the line as far apart as possible. Two graph examples, one done manually and another using the computer program Excel, are included on pages 12 and 13. Graphical Analysis You are required to construct plots representing lines of best fit for your collection of data for every lab experiment. Your graphs will be linear functions for most experiments. There may be a few graph plots that make use of different mathematical functions, in which case your lab instructor will provide you with instructions for constructing the
5 Guidelines 5 appropriate graph. Analysis by strict linear regression models and techniques are not necessary for physics lab reports. However, you should briefly discuss the closeness of the plotted data points to the line of best fit. You can do this simply by eye-balling the graph. You are required to determine the slope and y-intercept of the line drawn. The slope and y-intercept should be determined from the line of best fit. Do not use the data points to calculate the slope and y-intercept of the line! Instead, to determine the y-intercept of the line measure the point of interception between the line of best fit and the y-axis. The rise-over-run method is a practical method used to calculate the slope. In review, to calculate the slope choose two points on the line with the greatest separation. Draw a right triangle using as the hypotenuse the line of best fit. Determine the length of the line segment opposite theta, Θ (the rise) and the length of line segment adjacent to theta Θ (the run). The rise divided by the run determines the slope of the line. Figure 1: Determination of slope value using the rise-over-run method. The rise-over-run calculations as demonstrated in Figure 1 must be performed for each graph presented with your lab report. The graphical analysis provides a discussion of the slope and y-intercept values of your plotted graph(s). The calculated results you obtained from the rise-over-run method are recorded as the measured or actual values of slope and y-intercept. These values are compared to theoretical values; the theoretical values of slope and y-intercept are the accepted values for which your data is tested. You can determine theoretical values of slope and y-intercept by using the slopeintercept form of the equation of a line. The method used to perform this calculation is provided in the example below. You will notice that this method only works for first order equations. The theoretical background used for the Measurement of g experiment provides us with the kinematics equation h = 1 2 gt2. After collecting data for the experiment we are instructed to plot the height of the ball s drop, h, as a function of the square of the
6 6 Guidelines drop time of the ball, t 2 (h values are plotted along the y-axis and t 2 values along the x-axis). The theoretical values of slope and y-intercept are h = 1 2 g t2 + 0 y = m x + b. Notice that by using the provided theoretical equation and the equation of the line, we are able to determine 1 g and 0 as the theoretical slope and y-intercept values respectively 2 of our graph. The graphical analysis discussion will provide values for both the theoretical and actual slope and y-intercept measurements. A comparison between the sets of data can be performed by calculating the percent difference between the actual values and the theoretical values. The percent difference equation between slope values is % difference = 100% theoretical slope value experimental slope value theoretical slope value. Section C: Uncertainty Analysis The uncertainty analysis discussion is probably the most crucial section of your lab report. In this section you will interpret your data and conclude whether or not the experiment supports the theory under observation. You will analyze data for experimental error and report your findings. Statements such as This was a nice experiment and I liked it or All of my results are weird therefore I can t prove anything are unacceptable. You can include these remarks in your concluding statements, but they will never fulfill the requirements for the uncertainty analysis section. Experimental error is a measurable quantity present in every lab experiment. You will determine the extent to which experimental error affected your results. Human error is not a valid type of experimental error, thus the term HUMAN ERROR should NEVER be used in a lab report!!! Instead the two acceptable types of experimental error explored in physics labs are systematic and random error. In general systematic errors are unaccountable factors influencing data sets of an experiment. Systematic errors cause all values of a data set to deviate from an expected value by a common factor. Random errors cause individual measurements to vary around an average value. Values of a data set that vary widely from one another where a constant value of the data is expected is indicative of random error. The amount of either type of error is a calculated result based on several factors influencing your measurements. Factors such as the precision of the equipment used and/or repeat performance of a particular measuring device play important roles when calculating experimental error. Three calculations are generally performed to determine the amount of error. They are percent difference, percent uncertainty and percent variation. Percent Difference The percent difference is a comparison between a theoretical estimate and an experimental result. In most experiments instructions require you to calculate a theoretical value given certain initial conditions. This usually involves a mathematical treatment of an accepted
7 Guidelines 7 theory. The experimental result is obtained by performing an experiment and recording a measurement. You can compare the experimental and theoretical values using the equation below: % difference = 100% Percent Uncertainty theoretical value experimental value theoretical value The percent uncertainty is an attempt to estimate the precision of the equipment used during an experiment. Each measurement performed using a particular device is subject to measurement error. Measurement error is controlled by the precision of the device. Small amounts of measurement error result from using a precise device, whereas measurement error is excessively large if the device is not very precise for its intended purpose. For example, suppose a ruler is used to measure the length of an object. If 1 millimeter is the specified precision of the ruler it would be unreasonable to precisely measure an object smaller than a millimeter. Another way of stating the precision of an instrument is Using a wooden ruler, I can make length measurements of any object to the nearest millimeter. The uncertainty of a measurement is determined by dividing the smallest increment of the measuring device by 2. In the above example, the uncertainty of the ruler is half of a millimeter, or ±0.5 mm. The value ±0.5 mm is known as the experimental uncertainty in the measurement. It is useful to include the experimental uncertainty with your measured values. For example, if an object such as an ink pen has a length of 139 mm, or 13.9 cm using a ruler accurate to the nearest millimeter, then its length should be stated as 13.9 cm ± 0.05 cm. If L is used to represent the length of the pen, then the uncertainty in the length is written as L, or L ± L = 13.9 cm ± 0.05 cm. The percent uncertainty makes use of the experimental uncertainty. The percent uncertainty is calculated by determining the ratio as a percentage between the experimental uncertainty of an instrument such as a ruler, and the measured value obtained using the instrument, that is % uncertainty = 100% experimental uncertainty measured value For example, the percent uncertainty in the length of the ink pen mentioned above is % uncertainty in L = 100% 0.05 cm 13.9 cm = 0.4%. In most lab experiments you will use more than one measuring device. You will need to determine the percent uncertainty of each device and determine the largest or dominant percent uncertainty of the set. Percent Variation Often it is necessary to determine the variation of a data set. A theory may not predict a well defined value. Instead a theory may describe observed physical properties. Physical properties are unvarying or constant given prescribed initial conditions. During an..
8 8 Guidelines experiment for which you observe physical properties you would want to determine the amount of variance your data reflects. For example, lab instructions may require you to calculate the density of several cubes of platinum for which each cube varies by shape and size. After determining the density of each you would calculate an average density of the data set. Further you would determine the extent for which your data varies from the average. Percent variation is calculated using the following formula: % variation = 100% largest value smallest value 2.0 average value Physical properties of a material are classified as either intrinsic or extrinsic. The details of these properties are explored in an electricity and magnetism experiment. Uncertainty Analysis Your lab reports will contain the values of percent difference, uncertainty and variation. Additionally you will make conclusions from your data by reviewing the percentage values. You will do the following: compare the percent difference to the percent uncertainty, and compare the percent variation to the percent uncertainty. Percent Difference vs. Percent Uncertainty A comparison of the percent difference and percent uncertainty determines if any difference between experimental and theoretical values is caused by the limitations of the equipment used to make measurements. If the largest percent uncertainty of the experiment is larger than the absolute value of the percent difference, then the difference between the theoretical and experimental values is attributed to the uncertainty in the measurements. If the magnitude of the percent difference is larger than the largest percent uncertainty, then the difference between the theoretical and experimental values should be evaluated. Suppose after reviewing your data you find that each percent difference value is less than the largest percent uncertainty value. This is indicative of measurement error. Here you would conclude that the error is caused by the limitations of the equipment used to make measurements. The error is systematic; for each percent difference, the percent uncertainty is greater. Suppose a theory predicts that 2 is always the calculated result of an experiment. You perform an experiment and conclude that 1.5 is the calculated result. This gives a percent difference of 10%. You conclude further that the percent difference is greater than the percent uncertainty. You would then make a claim that external factors such as friction, air resistance, thermal energy loss, etc. account for this discrepancy. An external factor is a physical quantity unaccounted for by the presented theory. Naturally you would conclude that not enough information is given by the theory. The prediction of 2 does not satisfy all experimental conditions. Percent Variation vs. Percent Uncertainty Comparing the percent variation to the largest percent uncertainty determines if any variation between measured values is caused by the limitations of the equipment used to make measurements. If the percent variation is smaller than the largest percent.
10 10 Guidelines 1. Prediction What was the prediction that was made about the system? 2. Reasoning What was the reasoning behind the prediction (i.e. why did you make the prediction that you made)? 3. Experimental Result What was the result of the experiment? 4. Analysis (a) If the experiment agreed with you prediction, write a brief statement indicating this agreement. (b) If the experiment did not agree with the prediction, discuss what was wrong with the reasoning that led to the prediction that you made.
11 Measurement of Gravitational Acceleration g Conclusion In this experiment we set out to measure g, the acceleration due to gravity, and to demonstrate that this is a constant independent of the mass of the dropped object, and the height from which it is dropped. To determine this, we dropped four balls of different masses and diameters from various preset heights h, and measured the drop times t. Each ball was dropped five times from a given height h, for five different values of h. The times for each height were then averaged. For each height, g was calculated using the average time t ave and the formula g = 2h t 2 ave. The values of g for each height were averaged to produce an overall average value of g for each mass. The percent variation in g for each mass was also calculated. The values of g for all the masses were averaged to produce an experimental value for g. An overall percent variation of g for all the masses was also calculated from the largest and smallest values of g for the entire collection of masses. The mass m, height h, and drop time t were the measured quantities. Their uncertainties were determined to be m = ± kg, h = ± m, and t = ± s. The uncertainties in the time and in the mass were determined by taking half of the smallest increment on the digital scale and on the timer. Setting the height was difficult, and this led to a larger uncertainty in the height. Overall, the largest percent uncertainty in the experiment was 0.5% and was due to the uncertainty in the height measurements. The experimental value of g was found to be 9.78 m/s 2, which is 0.2% below the theoretical value of 9.80 m/s 2. Since the percent error in g is below the percent uncertainty in the experiment, the difference between the theoretical and experimental values can be attributed to measurement uncertainty. Therefore, our experiment supports the theoretical result that the acceleration due to gravity is 9.80 m/s 2. The percent variation in g for the tiny brass sphere, 0.42%, is less than the percent uncertainty in the experiment. From this, we can conclude that the variation in g for the tiny brass sphere is due to measurement uncertainty, and that g for this sphere is a constant. For the other spheres, the percent variation in g for each sphere is larger than the percent uncertainty in the experiment. Therefore, the variation in g for these spheres cannot be attributed to measurement uncertainty. One possible explanation for this variation could be that these larger spheres were jammed in the release mechanism of the timer, and were not released smoothly. Variation in how smoothly the larger spheres were released may have led to variation in the measured time, producing more variation in the calculated values of g. The variation in g for all of the masses is 2.53%. Since this is above the percent uncertainty in the experiment, this experiment does not support the theoretical result that the acceleration due to gravity is independent of mass. This variation may be due to the difficulty with the release mechanism mentioned above. A graph was made of h for the large aluminum sphere as a function of t 2 ave, the square of the average drop time. A line was drawn through the data points, and the data points are fairly close to the line drawn. Theoretically, the y-intercept should be zero. The line has a y-intercept above, but close to the origin of the graph, in agreement with the theory. The slope of the line is 4.64 m/s 2. Theoretically, the line should have a slope equal to 1g, or m/s2. The percent error between these two values is 5.3%, which is above the percent uncertainty in the experiment. Therefore the difference between these two values cannot be attributed to measurement uncertainty. One possible explanation for this difference is the jamming problem mentioned above. Another possible explanation is that air resistance, which is ignored in the theory, slowed the ball down as it fell. This caused the drop time to increase, resulting in a smaller value of g. This is consistent with our data since the value of g from the graph is smaller than the theoretical value of g.
12 Measurement of Gravitational Acceleration "g" small steel sphere m[kg]= d[m]= h[m] t1[s] t2 t3 t4 t5 avg(t) g average = % var = large alum sphere m[kg]= d[m]= h[m] t1[s] t2 t3 t4 t5 avg(t) g average = % var = small brass sphere m[kg]= d[m]= h[m] t1[s] t2 t3 t4 t5 avg(t) g average = % var = tiny brass sphere m[kg]= d[m]= h[m] t1[s] t2 t3 t4 t5 avg(t) g measurement uncertainties error min value % error mass [kg] 5.E average = % var = Global average height [m] 1.E = max % error avg = time [s] 5.E var = % var = 2.53 accepted value = 9.8 % error = -0.20
13
14 Measurement of "g" for Large Aluminum Sphere y = x R 2 = h [m] t ave 2 [s 2 ]
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# Solve problems with Java code Algorithms with Java.
## Presentation on theme: "Solve problems with Java code Algorithms with Java."— Presentation transcript:
Solve problems with Java code Algorithms with Java
Sum 1..N – Example Calculate and print the sum of the first N positive numbers Calculate and print the sum of the first N positive numbers 2 Scanner input = new Scanner(System.in); System.out.print("n = "); int n = input.nextInt(); int num = 1; int sum = 1; System.out.print("The sum 1"); while (num < n) { num++; num++; sum += num; sum += num; System.out.printf("+%d", num); System.out.printf("+%d", num);} System.out.printf(" = %d%n", sum);
Calculating Sum 1..N Live Demo
Prime Number – Example Checking if a number is prime or not Checking if a number is prime or not 4 Scanner input = new Scanner(System.in); System.out.print("Enter a positive integer: "); int num = input.nextInt(); int divider = 2; int maxDivider = (int) Math.sqrt(num); boolean prime = true; while (prime && (divider <= maxDivider)) { if (num % divider == 0) { if (num % divider == 0) { prime = false; prime = false; } divider++; divider++;} System.out.println("Prime? " + prime);
Checking If a Number Is Prime Live Demo
Using break Operator break operator exits the inner-most loop break operator exits the inner-most loop 6 Scanner input = new Scanner(System.in); int n = input.nextInt(); // "long" is the biggest integer type long factorial = 1; // Perform an infinite loop while (true) { if (n == 1) { if (n == 1) { break; break; } factorial *= n; factorial *= n; n--; n--;} System.out.println("n! = " + factorial);
Calculating Factorial Live Demo
Factorial – Example Calculating N factorial Calculating N factorial 8 Scanner input = new Scanner(System.in); System.out.print("n = "); int n = input.nextInt(); long factorial = 1; do { factorial *= n; factorial *= n; n--; n--;} while (n > 0); System.out.println("n! = " + factorial);
Factorial (do... while) Live Demo
Recursion Calling a Method by Itself
What is Recursion? Recursion is calling a method by itselfRecursion is calling a method by itself Very powerful technique for implementing combinatorial algorithmsVery powerful technique for implementing combinatorial algorithms Recursion should haveRecursion should have Direct or indirect recursive callsDirect or indirect recursive calls The method calls itself directly or through other methodsThe method calls itself directly or through other methods Exit criteria (bottom)Exit criteria (bottom) Prevent infinite recursionPrevent infinite recursion
Factorial – Example N! (N Factorial)N! (N Factorial) N! = N * (N – 1)! for N >= 0 and 0! = 1N! = N * (N – 1)! for N >= 0 and 0! = 1 5! = 5 * 4! or 5 * 4 * 3 * 2 * 1 * 1 = 1205! = 5 * 4! or 5 * 4 * 3 * 2 * 1 * 1 = 120 4! = 4 * 3! or 4 * 3 * 2 * 1 * 1 = 244! = 4 * 3! or 4 * 3 * 2 * 1 * 1 = 24 3! = 3 * 2! or 3 * 2 * 1 * 1 = 63! = 3 * 2! or 3 * 2 * 1 * 1 = 6 2! = 2 * 1! or 2 * 1 * 1 = 22! = 2 * 1! or 2 * 1 * 1 = 2 1! = 1 * 0! or 1 * 1 = 11! = 1 * 0! or 1 * 1 = 1 0! = 10! = 1
Factorial – Example Calculating factorial:Calculating factorial: 0! = 10! = 1 n! = n* (n-1)!, n>0n! = n* (n-1)!, n>0 Don't do this at home!Don't do this at home! Use iteration insteadUse iteration instead Recursive call: the method calls itself The bottom of the recursion public static int factorial(int n) { if (n == 0) return 1; else return 1; else return n * factorial(n - 1); } return n * factorial(n - 1); }
Product[N..M] – Example Calculating the product of all numbers in the interval [n..m]: Calculating the product of all numbers in the interval [n..m]: 14 int n = input.nextInt(); int m = input.nextInt(); int num = n; long product = 1; do { product *= num; product *= num; num++; num++;} while(num <= m); System.out.println("product[n..m] = " + product);
Product of the Numbers in the Interval [n..m] Live Demo
N^M – Example Calculating n^m Calculating n^m 16 Scanner input = new Scanner(System.in); int n = input.nextInt(); int m = input.nextInt(); long result = 1; for (int i = 0; i < m; i++) { result *= n; result *= n;} System.out.println("n^m = " + result);
Calculating N^M Live Demo
Using continue Operator continue operator ends iteration of the inner-most loop continue operator ends iteration of the inner-most loop Example: Sum odd numbers p in [1, n] that are not divisors of 7: Example: Sum odd numbers p in [1, n] that are not divisors of 7: 18 int n = input.nextInt(); int sum = 0; for (int i = 1; i <= n; i += 2) { if (i % 7 == 0) if (i % 7 == 0) continue; continue; sum += i; sum += i;} System.out.println("sum = " + sum);
Using continue Operator Live Demo
Nested Loops Using Loop Inside a Loop
What Is Nested Loop? A composition of loops is called a nested loop A composition of loops is called a nested loop Example: Example: 21 for (initialization; test; update) { for (initialization; test; update) { for (initialization; test; update) { statements; statements; } …}
Nested Loops Examples
Triangle – Example Print the following triangle on the console: Print the following triangle on the console:1 1 2 … 1 2 3... n 23 int n = input.nextInt(); for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { for (int j = 1; j <= i; j++) { System.out.print(j + " "); System.out.print(j + " "); } System.out.println(); System.out.println();}
Triangle Live Demo
Primes[N, M] – Example Print all prime numbers in [n, m] Print all prime numbers in [n, m] 25 int n = input.nextInt(); int m = input.nextInt(); for (int num = n; num <= m; num++) { boolean prime = true; boolean prime = true; int divider = 2; int divider = 2; int maxDivider = (int) Math.sqrt(num); int maxDivider = (int) Math.sqrt(num); while (divider <= maxDivider) { while (divider <= maxDivider) { if (num % divider == 0) { if (num % divider == 0) { prime = false; prime = false; break; break; } divider++; divider++; } if (prime) { if (prime) { System.out.printf("%d ", num); System.out.printf("%d ", num); }}
Primes in Range [n, m] Live Demo
Loops – More Examples
Nested Loops – Examples Print all four digit numbers ABCD such that A+B = C+D (happy numbers) Print all four digit numbers ABCD such that A+B = C+D (happy numbers) 28 for (int a = 1; a <= 9; a++) { for (int b = 0; b <= 9; b++) { for (int b = 0; b <= 9; b++) { for (int c = 0; c <= 9; c++) { for (int c = 0; c <= 9; c++) { for (int d = 0; d <= 9; d++) { for (int d = 0; d <= 9; d++) { if ((a + b) == (c + d)) { if ((a + b) == (c + d)) { System.out.printf("%d,%d,%d,%d", System.out.printf("%d,%d,%d,%d", a, b, c, d); a, b, c, d); }} } }} Can you improve this algorithm to use only 3 loops?
Happy Numbers Live Demo
TOTO 6/49 – Examples Print all combinations from TOTO 6/49 Print all combinations from TOTO 6/49 30 for (int i1 = 1; i1 <= 44; i1++) for (int i2 = i1 + 1; i2 <= 45; i2++) for (int i2 = i1 + 1; i2 <= 45; i2++) for (int i3 = i2 + 1; i3 <= 46; i3++) for (int i3 = i2 + 1; i3 <= 46; i3++) for (int i4 = i3 + 1; i4 <= 47; i4++) for (int i4 = i3 + 1; i4 <= 47; i4++) for (int i5 = i4 + 1; i5 <= 48; i5++) for (int i5 = i4 + 1; i5 <= 48; i5++) for (int i6 = i5 + 1; i6 <= 49; i6++) for (int i6 = i5 + 1; i6 <= 49; i6++) System.out.printf( System.out.printf( "%d %d %d %d %d %d%n", "%d %d %d %d %d %d%n", i1, i2, i3, i4, i5, i6); i1, i2, i3, i4, i5, i6); How long will it take to finish this?
TOTO 6/49 Live Demo
Summary Loops could solve different problems Loops could solve different problems Recursion could be handy as well Recursion could be handy as well We can use nested loops to implement more complex logic We can use nested loops to implement more complex logic We can use continue and break operators to control the loop execution We can use continue and break operators to control the loop execution More to come with arrays' manipulation More to come with arrays' manipulation 32
Exercises 1.Write a program that prints all the numbers from 1 to N. 2.Write a program that prints all the numbers from 1 to N, that are not divisible by 3 and 7. 3.Write a program that reads from the console a sequence of N integer numbers and returns the minimal and maximal of them. 4.Write a program that calculates N!/K! for given N and K (1 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1/238516/slides/slide_33.jpg", "name": "Exercises 1.Write a program that prints all the numbers from 1 to N.", "description": "2.Write a program that prints all the numbers from 1 to N, that are not divisible by 3 and 7. 3.Write a program that reads from the console a sequence of N integer numbers and returns the minimal and maximal of them. 4.Write a program that calculates N!/K. for given N and K (1
Exercises (3) 1.Write a program that reads a number N and calculates the sum of the first N members of the sequence of Fibonacci: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, … 2.In the combinatorial mathematics, the Catalan numbers are calculated by the following formula: Write a program to calculate the Catalan number by given N. 34
Exercises (4) 1. Write a program that reads from the console a positive integer number N (N < 20) and outputs a matrix like the following: N = 3N = 4 35 1234 2345 3456 4567 123 234 345
Exercises (5) 36 1. Write a program that calculates for given N how many trailing zeros present at the end of the number N!. Examples: N = 10 N! = 3628800 2 N = 20 N! = 2432902008176640000 4 Does your program work for N = 50 000? Hint: The trailing zeros in N! are equal to the number of its prime divisors 5. Think why!
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Find the Distance Between Two Points (23,-16) , (0,-14)
(23,-16) , (0,-14)
Use the distance formula to determine the distance between the two points.
Distance=(x2-x1)2+(y2-y1)2
Substitute the actual values of the points into the distance formula.
(0-23)2+((-14)-(-16))2
Simplify.
Subtract 23 from 0.
(-23)2+((-14)-(-16))2
Raise -23 to the power of 2.
529+((-14)-(-16))2
Multiply -1 by -16.
529+(-14+16)2
529+22
Raise 2 to the power of 2.
529+4
533
533
The result can be shown in multiple forms.
Exact Form:
533
Decimal Form:
23.08679276…
Find the Distance Between Two Points (23,-16) , (0,-14)
Solving MATH problems
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###### Definition8.2.1The Definition of the Square Root of a Number
If $$y^2=x$$ for a positive number $$y\text{,}$$ then $$y$$ is called the square root of $$x\text{,}$$ and we write $$y=\sqrt{x}\text{,}$$ where the $$\sqrt{\phantom{x}}$$ symbol is called the radical or the root. We call expressions with a root symbol radical expressions. The number inside the radical is called the radicand.
in-context
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Tuesday
June 18, 2013
# Homework Help: Logarithms
Posted by Chris on Sunday, December 17, 2006 at 6:36pm.
e^x+2=13
solve by logarithm
Take the ln of both sides.
x*ln e = ln 13.
You can do this on your calculator. If you don't know the value of e, go to www.google.com and type in e and search. The value is about 2.7.
or type this in the google search window
ln(13) / ln(e)
No one has answered this question yet.
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# Homework Help: Graphing Wavelength vs Inverse freq. or Period
1. Oct 23, 2008
### PSEYE
According to our lab manual, the graph of Wavelength vs Period : the slope should be the speed of sound. Well, I have a negative slope of -32.2 which is ridiculous, but I can't find out what I'm doing wrong. We had 5 tuning forks , say for instance, 512Hz...the inverse
1/f = 1/512= 0.00195 , im graphing this on the x-axis and Wavelength on the Y axis. This experiment was done with 2 tubes, one fitted into another, and you're supposed to pull the smaller tube out until the particular tuning for resonates. You are to take the distance of the pipe + the extended portion of the other pipe at 3 different places you hear resonance...then you are supposed to take the difference of the average and double it - that should be your wavelength...example
the tube is 90cm long
first trial @ 512 Hz - heard first resonance at 98.5cm or 8.5 depending...our lab instructor told us to use the entire distance so i did. Heard second resonance at 124cm, and 3rd resonance at 165. I took the difference between these three values and averaged it, then doubled it - i got 66.50cm or 0.665m I did this with 4 other tuning forks , got my info and graphed a negative slope that wasn't even close to being a straight line.
Can anyone give me a clue as to what I'm doing wrong?
Thanks.
oh btw, this is for lab : Resonance in Air Columns.
2. Oct 23, 2008
### Redbelly98
Staff Emeritus
If your textbook gives the actual speed of sound in air, you can use that to check whether the wavelength you get is close to what you expect for each frequency.
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# VBA实现EXCEL某一列的部分数据和等于指定值
Dim sj(), sj1, sj2, jg(), cnt&, d&, h&, hh&, k&, l&, m&, n&, nn&, p&, q&
Sub kagawa()
tms = Timer
d = [h3]: l = [h6]: If l = 0 Then l = 65535
h = [h1] * 10 ^ d: hh = [h2] * 10 ^ d: If hh > h Then hh = hh - h
If [h7] = 0 Then p = 10 ^ 5 Else p = 10 ^ [h7]
m = [a1].CurrentRegion.Rows.Count - 1
n = [h4]: nn = [h5]: If nn = 0 Then If n = 0 Then nn = m Else nn = n
If Application.Count([d2].Resize(m)) = m And Application.Sum([d2].Resize(m)) = (m + 1) * m / 2 Then
[a1].CurrentRegion.Sort [d2], 1, , , , , , 1
Else
[d2].Resize(m) = "": [d2] = 1: [d2].Resize(m).DataSeries Rowcol:=xlColumns
End If
[a2].Resize(m, 5).Sort [c2], 1, , , , , , 2
sj1 = [a2].Resize(m, 6)
[e2].Resize(m) = "":
sj2 = [e2].Resize(m)
For i = 1 To m
sj1(i, 1) = 0
sj1(i, 5) = i
sj1(i, 6) = sj1(i, 3) * 10 ^ d 'Val2
Next
ReDim jg(l, 3)
k = 0: cnt = 0
For j = 1 To l
l = 0
ReDim sj(m, 6): m = 0
For i = 1 To UBound(sj1)
If sj2(i, 1) = "" Then
m = m + 1
sj(m, 1) = sj1(i, 1) 'not used now
sj(m, 2) = sj1(i, 2) 'Code
sj(m, 3) = sj1(i, 3) 'Val
sj(m, 4) = sj1(i, 4) 'Row
sj(m, 5) = sj1(i, 5) 'i
sj(m, 6) = sj1(i, 6) 'Val2
sj(m, 0) = sj(m - 1, 0) + sj(m, 6) 'Sum
End If
Next
cnt = 0: q = 0: Call dgH42(h, "", "", m + 1, 1)
If cnt > p Then Exit For Else CalcCnt = CalcCnt + cnt
If j > k Then Exit For Else [h6] = k
Next
If m Then
jg(k, 0) = k + 1
jg(k, 2) = sj(m, 0)
s = "": t = 0
For i = 1 To m
t = t + 1
s = s & "+" & sj(i, 2)
sj2(sj(i, 5), 1) = k + 1
Next
jg(k, 1) = t
jg(k, 3) = Mid(s, 2)
k = k + 1
[h6] = k
End If
If k And k < 65535 Then [k1].CurrentRegion.Offset(1) = "": [k2].Resize(k, 4) = jg
[e2].Resize(UBound(sj2)) = sj2
MsgBox "Result: " & k & "/ Calc " & CalcCnt & " Time: " & Format(Timer - tms, "0.000s")
[a1].CurrentRegion.Sort [e2], 1, [d2], , 1, , , 1
End Sub
Sub dgH42(r&, ri$, ra$, i&, t&)
Dim j&, t1&, r2&, rs#
If l Then Exit Sub
cnt = cnt + 1: If cnt > p Then Exit Sub
' Exit Sub
If t >= n And t <= nn Then
r2 = r + hh
For j = 1 To i - 1
' If q Then Stop
If sj(j, 1) * q Then
' MsgBox sj(j, 1) & "/" & sj(j, 2) & "/" & q
Else
t1 = sj(j, 6) 'Val2
If r <= t1 And t1 <= r2 Then
jg(k, 0) = k + 1
jg(k, 1) = t
jg(k, 3) = Mid(ra, 2) & "+" & sj(j, 2)
rs = 0
x = Split(ri & "," & j, ",")
For l = 1 To UBound(x)
rs = rs + sj1(sj(x(l), 5), 3) 'Val
sj2(sj(x(l), 5), 1) = k + 1
Next
jg(k, 2) = rs
k = k + 1
Exit Sub
ElseIf t1 > r2 Then
Exit For
End If
End If
Next
End If
If t = nn Then Exit Sub
For j = i - 1 To 2 Step -1
If sj(j, 1) * q Then
' MsgBox sj(j, 1) & "/" & sj(j, 2) & "/" & q
Else
If sj(j, 6) < r + hh Then 'Val2
If sj(j, 0) < r Then 'Sum
Exit For
Else
If sj(j, 1) Then q = 1
Call dgH42(r - sj(j, 6), ri & "," & j, ra & "+" & sj(j, 2), j, t + 1)
End If
End If
End If
Next
End Sub
Sub mySort()
sj0 = [a1].CurrentRegion
m = UBound(sj0) - 1
k = Application.Sum([d2].Resize(m))
If Application.Count([d2].Resize(m)) = m And k = (m + 1) * m / 2 Then [a1].CurrentRegion.Sort [d2], 1, , , , , , 1
End Sub
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https://math.stackexchange.com/questions/296804/classification-of-omega-games-of-different-fixed-action-spaces
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Classification of $\omega$-games of different fixed action spaces
Suppose, a two-player game $$G(\omega,(X_n)_{n \in\omega},(Y_n)_{n \in\omega},A)$$ where
• $\omega$ is the number of the moves of each of the two players I and II.
• $(X_n)_{n \in \omega}$ (or $(Y_n)_{n \in \omega}$) is a $\omega$-sequence of $X_n$(or $Y_n$), which is the action space of $n$th move of first (second) player.
• The game is played in an alternating fashion.An outcome is a $\omega$-sequence:$$a_0 \in X_0, a_1 \in Y_0, a_2 \in X_1, a_3 \in Y_1, a_4 \in X_2……$$ $A$, the winning set, is a subset of the set of all possible outcomes $C$, i.e. $\prod_{i \in \omega}(X_i \times Y_i)$. If the outcome $\{a_n\}_{n<\omega} \in A$, then player II wins, otherwise, player I wins.
A player moves at any stage contingent on history. So a strategy for player I take the form as s sequence of functions $\{\sigma_i\}_{i \in \omega}$: $$\sigma_i : \prod_{j<i}{X_j \times Y_j} \to X_i$$ Player II's strategy's form $\{\tau_i\}_{i \in \omega}$is defined similarly. Player I and Player II's strategy spaces are denoted as $S_{\text I}$ and $S_{\text {II}}$ respectively. Denote the binary operation $\star$ as the function that sends a pair of strategies of player I and player II to the outcome they give rise to.
Let's fix $X_n = Y_n =Z$, for any $n \in \omega$ and $Z$ be any ordinal number from $2$ to $\omega$. As we know, the cardinality of set of strategies coincides with that of set of all possible outcomes. More precisely, $|S_{\text {I}}| = |S_{\text {II}}| = |C|= 2^{\aleph_0}$
We say two games $\alpha$ and $\beta$ are isomorphic, if there are bijective functions $f:S^{\alpha}_{\text I} \to S^{\beta}_{\text {I}}$, $g:S^{\alpha}_{\text{II}} \to S^{\beta}_{\text{II}}$ and $h:C^{\alpha} \to C^{\beta}$ such that for any pair of strategies of players I and II in game $\alpha$, $\sigma$ and $\tau$, $$f(\sigma) \star g(\tau) = h (\sigma \star \tau)$$
How to show the existence, or non-existence of such isomorphism between games with action space $Z$ of different sizes, say, $Z= 2$ and $Z = \omega$
• Is $\sigma \circ \tau$ the result of playing $\sigma$ against $\tau$? I think this is usually written $\sigma \star \tau$ or $\sigma * \tau$. Feb 7, 2013 at 2:26
• @TrevorWilson: You're right.Thank you for pointing that out. Feb 7, 2013 at 2:28
• Just to clarify, your notion of isomorphism doesn't involve any payoff sets on either side? Feb 7, 2013 at 2:29
• @TrevorWilson:No. Feb 7, 2013 at 2:31
• @TrevorWilson: Tons of thanks for your comment. I got it wrong. Feb 8, 2013 at 23:55
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