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https://miprofe.com/en/graphing-logarithmic-functions/ | 1,596,742,555,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00057.warc.gz | 365,918,807 | 23,166 | # Graphing logarithmic functions
When we talk about the logarithm function, we are also referring to the inverse function of the exponential, since the domain of the exponential function becomes the range of the logarithm function. The general expression of the logarithm function is as follows:
y (x) = logax
Where, a is the base and x is the variable. It is important to know that there is no logarithm of zero or a negative number. Its domain is (0, + ∞) and its range is all reals of (-∞, + ∞).
Analyzing the logarithm function, we must know that it can be increasing or decreasing depending on the value of its base. Also, x will always be greater than zero (x> 0), since its domain does not contain negative numbers.
y (x) = logax
If a> 1 your graph will be increasing:
If a is greater than zero but less than one (0 <a <1) your graph will be decreasing:
Example: Graph the following logarithmic function y (x) = log5 (2x-3).
The first step is to find the vertical asymptote, in order to know its domain and know where the graph starts from. To do this, we will solve what is inside the parentheses, knowing that always x> 0:
2x - 3> 0
2x> 3
x> 3/2
x> 1.5
That is, we have a vertical asymptote at the point (3/2, 0):
The second step will be to find the cut-off point on the x axis, for this we must set the entire logarithm to zero:
log5(2x-3) = 0
We apply the exponential to both sides of the equation:
and log5 (2x-3) = e0
2x - 3 = 1
2x = 1 + 3
X = 4/2
x = 2
This tells us that our graph cuts at x = 2:
The function has domain (2/3, + ∞) and range (-∞, + ∞). | 458 | 1,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2020-34 | latest | en | 0.918416 |
https://everything2.com/user/hobyrne/writeups/statistical+clustering | 1,544,984,013,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827963.70/warc/CC-MAIN-20181216165437-20181216191437-00083.warc.gz | 570,629,601 | 7,872 | (This writeup inspired by the Prime Spiral node.)
It is human nature to try and discern patterns in everything we see. Pattern recognition is widely regarded in the field of artificial intelligence as being one of the very fundamentals of what makes intelligence, intelligence. Sometimes this works against us, and gives us the perception of a pattern where there really is none. This kind of occurrence is called statistical clustering. A very good test of whether a significant pattern is really present is to examine the real extent of the field in which a coincidence occurs, and to examine which knowledge is a priori and which is a posteriori.
If a conjecture is made a priori, and is borne out to be supported by the data, it is likely a true pattern with a true cause. If an observation is made a posteriori, the best test is to conjecture "things will be more normal from now on" (without dwelling on patterns of the past), and seeing how well that conjecture holds up in the future. If the conjecture passes, you have a case of statistical clustering - if the conjecture fails, there is a valid statistical link.
The field of interest in which a pattern is found is often difficult to recognise at first. For example, there may be an article in the newspaper that someone got dealt the perfect Bridge hand today - all 13 spades cards. The chances of that happening are tiny. Or that two drivers in a race got exactly the same time on the qualifying lap, to within a thousandth of a second. This seems newsworthy too (at least for the Sports section). But think about what else there might be, that would be just as unusual. This is somewhat difficult, because we think about what we expect ( - evaluating relevance is another core issue of intelligence). A perfect 147-point break in snooker. The finalists in a karate championship who have the same initials and the same birthday. Two ballet dancers who both trip in their hotel rooms before a big show. Two horses, whose names are anagrams of each other, in the same horse race. Very outlandish, yes - but there are so many outlandish things that it's really no wonder one or two crop up occasionally. And numerology, don't get me started, there are so many combinations of mathematical operators that some sequence of tricks will link several unrelated items together from a large pool of information.
See The Psychology of Randomness. We are taught that random is the opposite of order, the imposed absence of order, where a better conceptual model might be to think of random as being the absence of imposed order. 'Order', in the psychology of randomness node above, is perceived as being a long run of heads or tails. Imposed order would cause 5 heads in a row, but given enough tosses of the coin without any impositions (of perceived order or of perceived randomness), 5 heads in a row will occur naturally.
- However, just as statistical arguments can be abused, the statistical clustering argument can be abused too. Caution must be exercised in believing either claim. The only way to see through either type of abuse is to use the scientific method impartially (which is impossible for most of us who don't have access to the amounts of data required to do so, especially for events which are big enough to be newsworthy, like elections, so the next best thing we can do is put trust in someone else to use the scientific method and then report honestly to us). | 710 | 3,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-51 | latest | en | 0.963644 |
http://mathhelpforum.com/advanced-algebra/122487-lagranges-theorem.html | 1,480,899,944,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541517.94/warc/CC-MAIN-20161202170901-00400-ip-10-31-129-80.ec2.internal.warc.gz | 173,460,043 | 13,672 | 1. ## Lagranges theorem
I have tried to understand this using various websites and m,y notes but i still dont understand whats going on.
Could someone please explain what im meant to do to prove a a set in lagranges theorem thanks.
I have tried to understand this using various websites and m,y notes but i still dont understand whats going on.
Could someone please explain what im meant to do to prove a a set in lagranges theorem thanks.
What is it your are wanting us to explain how to prove?
Lagranges Theorem states that the order of every subgroup $H$ of a finite group $G$ divides the order of $G$, $H \leq G \Rightarrow |H| \mid |G|$.
This is, perhaps, a result which is easier to understand once you have played around with it a bit. So, pick a few of your favourite groups and verify the result.
For instance, the Klein 4-group has order 4 and subgroups of order 1, 2, 2, 2 and 4, and cyclic groups of prime order (for instance, $\mathbb{Z}/7\mathbb{Z}$) have no proper subgroups.
3. Originally Posted by Swlabr
What is it your are wanting us to explain how to prove?
Lagranges Theorem states that the order of every subgroup $H$ of a finite group $G$ divides the order of $G$, $H \leq G \Rightarrow |H| \mid |G|$.
This is, perhaps, a result which is easier to understand once you have played around with it a bit. So, pick a few of your favourite groups and verify the result.
For instance, the Klein 4-group has order 4 and subgroups of order 1, 2, 2, 2 and 4, and cyclic groups of prime order (for instance, $\mathbb{Z}/7\mathbb{Z}$) have no proper subgroups.
What i dont understand is how we calculate g and h
What i dont understand is how we calculate g and h
Do you mean how do you calculate the orders of the group and the subgroup? This is just the number of elements in your group.
For instance, the Klein 4-group has order 4 as it has precisely 4 elements, and the cyclic group of order 7 has, well, order 7. It is just the elements $\{0, 1, 2, 3, 4, 5, 6\}$ under addition modulo 7.
What i dont understand is how we calculate g and h
We don't!
G is a given group and H is a given subgroup of G. If by "g" and "h" you mean the number of elements in G and H, respectively, when we are given G and H, we are given g and h.
I think you should look closely at the concept of "left cosets" of H. They are crucial in Lagrange's theorem and important for other things as well. For any x in G, its left coset is the set {xy| y in H}. (A "right" coset would be of the form {gx| y in H}.) You should look at the proofs that
1) Every left coset contains the same number of elements.
(And since {1y |y in H} is just H itself, that is just h.)
2) Every member of G is in exactly one such left coset.
If there are, say, n left cosets, each containing h members, and each member of G is in exactly one, we must have g= nh so h divides g.
6. Originally Posted by Swlabr
Do you mean how do you calculate the orders of the group and the subgroup? This is just the number of elements in your group.
For instance, the Klein 4-group has order 4 as it has precisely 4 elements, and the cyclic group of order 7 has, well, order 7. It is just the elements $\{0, 1, 2, 3, 4, 5, 6\}$ under addition modulo 7.
In my notes it says If G is a finite group of order g = ¦G¦ and H is a subgroup of order h = ¦H¦, then h must be a factor of G.
So if the set was G = {1, -1, i, -i}, i = ((-1)^0.5). I know how to show its a group by the axioms.
But then the question says obtain a non-trivial solution subgoup H which i ahve no idea to do. And use it to illustrate lagranges theorem for a finite group.
Would G and H be 4 in this case?
In my notes it says If G is a finite group of order g = ¦G¦ and H is a subgroup of order h = ¦H¦, then h must be a factor of G.
So if the set was G = {1, -1, i, -i}, i = ((-1)^0.5). I know how to show its a group by the axioms.
But then the question says obtain a non-trivial solution subgoup H which i ahve no idea to do. And use it to illustrate lagranges theorem for a finite group.
Would G and H be 4 in this case?
A non-trivial subgroup is a subgroup which is not equal to the group itself nor the trivial group. Thus, it must have order strictly greater to one and not equal to that of the group. The group you have been given is the Klein 4-group which I havev mentioned above. It has 3 non-trivial subgroups each of order 2. Can you find them? (Hint: Pick an element. Any element...other than 1...)
8. Originally Posted by Swlabr
A non-trivial subgroup is a subgroup which is not equal to the group itself nor the trivial group. Thus, it must have order strictly greater to one and not equal to that of the group. The group you have been given is the Klein 4-group which I havev mentioned above. It has 3 non-trivial subgroups each of order 2. Can you find them? (Hint: Pick an element. Any element...other than 1...)
why cant it be one?
thanks for your help by the way it is much appreciated.
why cant it be one?
thanks for your help by the way it is much appreciated.
By definition, if $H$ is a non-trivial subgroup of $G$, then $H \neq G ~ \text{and} ~ H \neq \{1_G\}$ where $1_{G}$ is the identity element of G and the group $\{1_G\}$ is called the trivial subgroup of G.
10. Originally Posted by Defunkt
By definition, if $H$ is a non-trivial subgroup of $G$, then $H \neq G ~ \text{and} ~ H \neq \{1_G\}$ where $1_{G}$ is the identity element of G and the group $\{1_G\}$ is called the trivial subgroup of G.
well -1, i and -i cant be in it as well then?
well -1, i and -i cant be in it as well then?
Why not?
13. Originally Posted by Swlabr
Why not?
because they are in G
because they are in G
A subgroup is a set of elements from G which form a group under the operation of G.
So, a subgroup will always contain the identity, in this case denoted 1. If it is non-trivial it will contain other elements too, for instance -1.
So, take the set $\{-1, 1\}$. Firstly, you should note that you have associativity as you inherit this from the group itself. You also have the identity element, this is just 1.
Can you find an inverse for -1? Is this in the set?
What about closure? If you multiply two elements from this set are we still in the set? The only non-trivial product is $(-1)^2$, but this is still easy...
15. Originally Posted by Swlabr
A subgroup is a set of elements from G which form a group under the operation of G.
So, a subgroup will always contain the identity, in this case denoted 1. If it is non-trivial it will contain other elements too, for instance -1.
So, take the set $\{-1, 1\}$. Firstly, you should note that you have associativity as you inherit this from the group itself. You also have the identity element, this is just 1.
Can you find an inverse for -1? Is this in the set?
What about closure? If you multiply two elements from this set are we still in the set? The only non-trivial product is $(-1)^2$, but this is still easy...
the inverse of -1 is -1 and is in the set
1 times -1 = -1 which is in the set which shows closure
but i thought you said 1 couldnt be in the subset?
Page 1 of 2 12 Last | 1,945 | 7,112 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-50 | longest | en | 0.941659 |
https://algebra-answer.com/tutorials-2/greatest-common-factor/math-1314-prerequisites.html | 1,631,842,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053918.46/warc/CC-MAIN-20210916234514-20210917024514-00547.warc.gz | 149,138,841 | 18,884 | # Math 1314 Prerequisites
This is a review of the topics you will need from previous math classes. Note that we expect these
to be skills with which you are already proficient.
This review should help you prepare for the prerequisite test. You can also refer to this review
throughout the semester. The numbers in parentheses after the topic indicate the lesson numbers
when you will need this skill as you progress through the course. For example, you’ll use the skills
reviewed in topic 1 in Lessons 3 and 8.
You will be allowed to use the online calculator on tests. You should become familiar with it as
soon as possible. It is a bit different from most calculators. If you have problems, see your
instructor in the Math Lab during his/her office hours. Note: you will not be allowed to use your
own calculator, only the online calculator. Do not become dependent on a graphing
calculator!!!
Topic 1: Writing equations of lines (L3, L8)
Given a slope and a point, write an equation of the line.
Example 1: Suppose the slope of a line is -6 and the line passes through the point (3, -5). Write the
equation of the line.
Use the point-slope form of the equation of the line to solve.
Substitute in the given values:
y - (-5) = -6(x - 3)
Simplify:
y + 5 = -6x +18
Solve for y:
y = -6x +13
Sometimes, you will need to find the slope of the line.
Example 2: Write an equation of the line that passes through the points (-3, 7) and (4, 12).
Use the slope formula to find the slope of the line:
so
Now proceed as for example 1:
Feel free to use a calculator to help compute with fractions. Note that the answer is given in
fraction form, so if you use your calculator, you’ll need to be able to convert back to fractions.
Topic 2: Solving an equation for a given variable (L14, L24)
You should be able to solve an equation for a stated variable, usually y.
Example 3: Solve for y: 2x - 3y = 5
Start by isolating the term containing y by subtracting 2x from both sides of the equation.
- 3y = -2x + 5
Now divide both sides of the equation by -3 and simplify.
Example 4: Solve for y: xy = -16
To isolate y, you need to divide both sides of the equation by x and simplify.
Note that x cannot be 0, since division by 0 is not defined.
Example 5: Solve for y:
To isolate y, subtract 4x2 from both sides of the equation, then divide by 2 and simplify.
Topic 3: Using interval notation (L10, 11, 12, 13, 14)
You will need to use interval notation extensively throughout the course. The interval (2, 5) means
all real numbers between 2 and 5, not including either 2 or 5. If you want to include an endpoint,
you use a bracket instead of a parenthese. So [2, 5) means all real numbers between 2 and 5,
including 2 but not including 5. The interval [2, 5] includes both endpoints.
You can use interval notation to indicate, say, all real numbers less than 3. You will write this as
(- ∞, 3). This can also be written as x < 3. You will most likely be asked to state answers in
interval notation, so you should be able to rewrite the inequality using an interval.
Example 6: Write using interval notation: x ≥ 6 .
This inequality includes 6 and all real numbers bigger than 6. In interval notation:
[6, ∞).
Example 7: Write using interval notation: - 3 < x ≤ 12 .
This inequality includes 12, does not include -3 and includes all real numbers between -3 and 12. In
interval notation:
(- 3, 12].
Topic 4: Solving quadratic inequalities (L10, 11, 12, 13, 14)
Example 8: Solve the inequality:
Start by factoring the left hand side of the inequality: (x - 3)(x + 2) ≥ 0
Next, set each of the binomials on the left hand side equal to zero and solve each equation for x.
Now draw a number line and use these two values to subdivide your number line into intervals:
Next, we’ll test each of the three intervals to see if (x - 3)(x + 2) is positive or negative. We’ll
choose test values in each of the three intervals. Any number in each interval will do. Convenient
choices for this problem would be -3 (it’s less than -2), 0 (it’s between -2 and 3) and 4 (it’s bigger
than 3).
Now try each of the test points in (x - 3)(x + 2).
If x = -3: (- 3 - 3)(- 3 + 2) = (- 6)(-1) = 6
If x = 0: (0 - 3)(0 + 2) = (- 3)(2) = -6
If x = 4: (4 - 3)(4 + 2) = (1)(6) = 6
We’ll record the sign (+/-) of each of these answers in its interval:
We see from the original problem that we want (x - 3)(x + 2) to be greater than or equal to zero.
From our sign chart, we can see that (x - 3)(x + 2) is positive when x is less than -2 or greater than
3. Our expression (x - 3)(x + 2) = 0 when x = 3 or -2. Putting this information together, we have
that (x - 3)(x + 2) ≥ 0 when x ≤ -2 or when x ≥3 . Now write the answer using interval notation:
(- ∞, - 2]∪[3, ∞).
Topic 5: Finding the domain of a function (L 10, 11, 12, 13, 14)
The domain of a function is the set of all x values that “work” in the function. We start by assuming
that all real numbers will work, then check to see if we have to exclude any numbers from the
domain. For example, given the function f (x) = x2 , there are no trouble spots. Any real number
works in this function. The domain is “all real numbers” which we will write as (−∞,∞).
The main trouble spots are radical functions , rational functions and logarithmic functions.
Example 9: Find the domain:
This is a rational function. We have the binomial 3x – 6 in the denominator. Since we cannot let
the denominator be equal to zero, 3x – 6 cannot be equal to zero. So we have:
3x - 6 ≠ 0 which we solve for x to find x≠ 2 .
This tells us that all real numbers except 2 will work. We will write our answers using interval
(-∞, 2)∪(2, ∞) .
Example 10: Find the domain:
This is a radical function. The quantity under the square root sign must be non-negative (i.e., either
0 or positive), so we must have
x +1 ≥ 0.
Then x ≥ -1.
In interval notation, [-1, ∞).
Example 11: Find the domain:
Since we have a square root in the problem, we know that what is under the square root sign cannot
be negative. So to find the domain, we need to solve the problem, .
Factor the left side: (x - 3)(x - 2) ≥ 0 .
The two values at which this inequality equals zero are 3 and 2.
Next, we’ll draw a number line, and split it into three regions, the region from - ∞ to 2, the region
from 2 to 3 and the region from 3 to ∞:
We’ll choose a test value in each of these regions.
We’ll test these points to determine if the region gives positive y values or negative y values. To do
this, substitute each test value into (x – 3)(x – 2) and determine if you get a positive result or a
negative result.
We’ll choose as our answer the intervals which give positive y values, as well as 2 and 3 which
make the inequality equal to zero:
(-∞, 2]∪[3, ∞) .
Example 12: Find the domain: k(x) = ln(x - 5)
The domain of the function f (x) = ln(x) is (0, ∞). From this, we see that x - 5must be positive.
This is,
In interval notation, (5, ∞).
Topic 6: Finding information from a graph (L10, 11, 12, 13, 21)
Sometimes you’ll be given the graph of a function. You can read the domain from the graph. You
can also find the range of the function from the graph, that is, the set of all y values that “work” in
the function. You’ll also be able to read other information from the graph, such as the y value
associated with a given x value.
Example 13: Given the graph of f (x), find the domain, range, f (2), and f (5).
From the graph, we see that there are no x values less than -4 and there are no x values greater than
5. So the domain is [- 4, 5]. There are no y values below -5 and there are no y values above 4. So
the range is [- 5, 4]. We can locate f (2) on the graph. Look along the x axis until you find the
number 2. Then move up or down until your finger is on the graph of the function. Read off the
corresponding y value. In this case, it’s -2. So f (2) = -2 . Similarly, to find f (5), move your
finger along the x axis until you find the number 5. Then move up or down until your finger is on
the graph of the function. Read off the y value. In this case, it’s -5. So f (5) = -5 .
Topic 7: Multiplying polynomials (Throughout)
Example 14: Simplify: (2x - 5)(3x +1).
Multiply using FOIL (First, Outer, Inner, Last), then combine any like terms. You are actually
distributing first the 2x and then the -5 to both terms in the second binomial.
Example 15: Simplify:
Distribute x to all terms in the second factor, then distribute -1 to all terms in the second factor.
Then combine any like terms.
Topic 8: Function notation (Throughout, especially L1, 2, 3, 8, 9, 15)
You should be able to use function notation.
Example 16: Suppose
Find and
The next two parts of this example will require the use of a calculator.
You should also be able to evaluate a variety of problems.
Example 17: SupposeFind f (-3) and f (5) .
This is a piecewise defined function. The problem divides the domain of the function into two
parts, those values less than 1 and those values greater than or equal to 1. Each part of the domain
is assigned a function. You have to figure out which line applies to the given value of x, then use
that line to evaluate the function.
First, find f (-3) . Since -3 < 1, we’ll look at the top line.
Next, find f (5) . Since 5 > 1, we’ll look at the bottom line.
You will not be able to escape word problems in this class!
Example 18: A study of driving costs based on 1992 model compact cars found that the average
cost measured in cents per mile is approximated by the function 32.082095 , where x (in
thousands) denotes the number of miles the car is driven in one year. (The average cost includes car
payments, gas, insurance, maintenance and depreciation.) Find the average cost of driving a
compact car 25000 in one year.
We are just being asked to evaluate this function at a number, but we must be careful! The problem
states that x is given in thousands of miles. Since we want to find the cost of driving 25,000 miles,
we should substitute 25 for x.
or approximately 34 cents per mile driven.
Topic 9: Factoring (Throughout, especially L1, 3, 5, 6, 10, 11, 12, 13)
You must (repeat, must!!) be proficient at factoring.
Example 19: Factor:
This is the difference of two squares , so it has the factoring pattern
So we have (x - 7)(x + 7) .
Example 20: Factor:
You should always look for a common factor before you try factoring. In this case, we have a
common factor, 5ab . So we start by factoring that out.
Next we look at the expression inside the parentheses, but it cannot be factored further.
Example 21: Factor
We can factor this by grouping. Group the first two terms together and the last two terms together.
Then we can factor x2 out of the first two terms and –4 out of the last two terms:
Now our two terms have the binomial (x - 4) in common, so we factor it out:
The second binomial can factored some more:
(x - 4)(x - 2)(x + 2)
Example 22: Factor:
Method 1: Trial and error. 6x2 can have 6x and x as factors or it can have 3x and 2x as factors.
We’ll start with the factors that are closer together, 3x and 2x. If those don’t work, we’ll try 6x and
x. Next, the constant -20 has lots of factors, -1 and 20, 1 and -20, -2 and 10, 2 and -10, -4 and 5, 4
and -5. So we’ll try pairs of these factors until we find the one that works. We check each trial by
multiplying the factors together to see if our factoring gives us what we started with.
no!
no!
close but no!
yes!
There are many other possibilities, but we came across the right one, so we stopped.
Method 2: Use factoring by grouping.
The trinomial is in the form , so find ac . In our case, it’s (6)(-20) = -120. Next, find
factors of -120 that add up to b, which is -7 in this case. Factors of 120 are 1 and 120, 2 and 60, 3
and 40, 4 and 30, 5 and 24, 6 and 20, 8 and 15 and 10 and 12. Since we have -120, one or the other
of the factors must be negative (e.g., -10 and 12 or 10 and -12). We need factors that add up to -7,
so 8 and -15 will do the trick.
Rewrite -7x as 8x – 15x. Then factor by grouping.
So we have
2x(3x + 4) - 5(3x + 4)
which is
(3x + 4)(2x - 5) .
Method 3: Ms. Leigh’s never-fail method: If you had Leigh Hollyer for Math 1300 or Math 1310,
she teaches another method for factoring which also works. For this problem, we’ll start by
writing down two factors – we know these won’t work, but this is just our starting point. We’ll
choose the 6 because it’s the leading coefficient in the problem:
Now find ac. In our case, that’s (6)(-20) = -120. We need factors of – 120 that add up to our value
for b which is -7. Factors of 120 are 1 and 120, 2 and 60, 3 and 40, 4 and 30, 5 and 24, 6 and 20, 8
and 15 and 10 and 12. Since we have -120, one or the other of the factors must be negative (e.g., -
10 and 12 or 10 and -12). We need factors that add up to -7, so 8 and -15 will do the trick.
We can now rewrite our factors:
(6x + 8)(6x -15)
Now get rid of any common factors in the two factors. We can get rid of 2 in the first set of
parentheses and 3 in the second.
(3x + 4)(2x - 5)
Example 23: Factor:
We can use any of the methods shown in example 20, but trial and error is easier here because the
x2 factors into x and x only. We need factors of -10 that add up to -3. Those are -5 and 2. So we
(x - 5)(x + 2)
Check by multiplying.
Example 24: Factor completely:
There is a common factor of x, so begin by factoring it out.
We can factor x2 - 9 , so we continue factoring.
Example 25: Factor completely:
Look for a common factor. This time it includes the exponential function. We can take out .
We cannot factor 3 + x any further, so we are done.
Example 26: Factor completely: .
This is the difference of cubes. There is a factoring pattern: . In our
case, a = x and b = 3. We have
which cannot be factored any further.
Topic 10: Synthetic Division
You should be able to find a quotient and a remainder using synthetic division. You should also be
able to tell if a binomial is a factor of the original polynomial.
Synthetic division is an alternative to using long division of polynomials when you are dividing by
x ± a . You cannot use synthetic division if the power of x is your divisor is bigger (e.g., x2-5 ).
It easiest to explain synthetic division by example:
Example 27: Use synthetic division to find the quotient and the remainder:
We’ll need to divide by the number that will make x - 2 = 0 , that is 2. To perform the division,
we’ll use only the coefficients of the polynomial.
Now, bring down the one. Then multiply by 2 and add what you get to -5. Write that down. Then
multiply that number by 2 and add what you get to 2. Write that down. Repeat the process until
you run out of numbers. The last number you get is the remainder. The rest of the numbers are
coefficients of the quotient, which will be one degree lower than the original polynomial. So
continuing with our example:
So the quotient is and the remainder is -27.
Example 28: Use synthetic division to find the quotient and the remainder:
Quotient: ; Remainder 25
Example 29: Use synthetic division to find the quotient and the remainder:
All powers must be accounted for when you set up your work. In this problem, we have no x3 or x
term in the problem. We’ll put in placeholders (zeros) when we set up the synthetic division.
Quotient: ; Remainder 0. Since the remainder is zero, x – 1 is a factor of
.
Topic 11: Using the rules of exponents (Throughout, especially L4, 10, 11, 12, 13, 16)
The definitions and laws of exponents are summarized in Section 9.1 of the online text. Assume all
variables in these examples are positive.
Following are some examples and frequent student errors:
Example 30: Simplify: - 52
The correct answer is – 25. Note, only the 5 is raised to the second power. The negative sign is not.
Do not confuse this with (-5)2, which raises -5 to the second power.
Example 31: Simplify: 7x0
The correct answer is 7. Only the x is raised to the zero power, so x0 = 1. Then multiply by 7.
Example 32: Simplify: (7x )0
The correct answer is 1. Anything raised to the zero power is 1. In this case, 7x is raised to the zero
power, so it simplifies to 1. Do not confuse Examples 28 and 29.
Example 33: Write using rational exponents:
In a rational exponent, the numerator is the power and the denominator is the root, so in this case,
Example 34: Write using a radical:
Same as above, so
Example 35: Rewrite this expression so that it is not a fraction: .
We can bring the x2 to the numerator by rewriting it as x-2 , so our answer is 3x-2
Example 36: Rewrite this expression so that it does not contain any negative exponents: x-4 .
We can think of this expression as. To eliminate the negative exponent, we’ll move the x-4 to
the denominator and change its sign. So our answer is .
Example 37: Rewrite this expression so that it is not a fraction:.
We can begin by rewriting the denominator using a rational exponent:.
Now we’ll bring the to the numerator by changing its sign: .
Example 38: Simplify: .
To simplify this, we will split 32 into two factors, one of which is a perfect square (i.e., 1, 4, 9, 16,
25, 36, …) In this case, 16 is a factor of 32, so we rewrite 32 as 16 * 2.
Example 39: Assume x > 0. Simplify:
Split the radicand into two factors. Both factors are perfect squares. Take the square root of each.
Note, if we had not assumed that x > 0, we’d write the answer as .
Topic 13: Exponential functions, logarithmic functions and the number e (L 4, 5, 6, 7, 10, 11,
12, 15)
The number e is the value that the expression approaches as you let n get bigger and
bigger. (This idea is called a limit and we will study them this semester.) e ≈ 2.718281828... You
will find an ex key on your calculator. (It is exp( ) on the online calculator.) It is sometimes
convenient to remember that e ≈ 3 .
You should be able to recognize and draw the graph of an exponential function.
Example 40: Sketch .
To sketch this, you can use a table of values, then plot the points . However, you should be able to
draw a rough sketch from memory.
x -3 -2 -1 0 1 2 3 f(x) 1/8 1/4 1/2 1 2 4 8
Recall that all basic exponential functions have this same shape. So any function of the form
f (x) = ax , a > 1, including y = ex , will basically look like this:
Any function of the form f (x) = ax , 0 < a < 1will be a reflection of the above graph over the y axis:
Some features of exponential functions of the form y = ax :
Domain is (-∞,∞)
Range is (0,∞)
Graph passes through the point (0, 1)
If a > 1, the function is increasing on (-∞,∞) . If 0 < a < 1, the function is
decreasing on (-∞,∞)
Now we’ll turn our attention to logarithms.
A logarithmic function is the inverse of an exponential function. So if we have y = 2x , the inverse
of this function is x = 2y . This equation is inconvenient to deal with. Logarithms help us with this
and allow us to rewrite this equation as . Now we can choose values for x, and evaluate
the function to find y. If we choose carefully, we’ll get integers.
The first thing you should be able to do is evaluate a logarithm. You’ll use the definition of a
logarithm to do this. The definition of a logarithm says that if and only if by = x.
Example 41: Evaluate .
We can rewrite this as We want to find y.
Use the definition of a logarithm: 3y = 81
Rewrite the right hand side as 34 . We have 3y = 34 , so y = 4.
Example 42: Graph
x 1/8 1/4 1/2 1 2 4 8 f(x) -3 -2 -1 0 1 2 3
All functions of the form will have the same basic shape.
All functions of the form will be the reflection of the above graph over the x
axis.
features of basic logarithmic graphs:
Domain is (0,∞)
Range is (−∞,∞)
Graph passes through the point (1, 0)
The function is increasing on (0,∞) if b > 1, and it’s decreasing on (0,∞) if 0 < b < 1.
You will work with the natural logarithmic function most of the time in this course. This is the
function, . This function is so important in mathematics that it has its own notation:
y = ln x.
Log properties are very useful in calculus.
Here are the log properties you need to know:
Suppose M, N and b are positive real numbers, b ≠ 1, and r is any real number. Then:
Example 43: Expand using log properties:
Using property 1, . Note that in this case, b = 10.
Example 44: Expand using log properties:
We’ll use properties 2 and 3.
Example 45: Expand using log properties:
We’ll use all 3 properties.
Example 46: Express as a single logarithm:
We reverse the process from the previous examples.
Other properties of logarithms that you need to remember:
Example 47: Expand and simplify using log properties: .
Using property 1,
Using property 3,
Using property 6, since ln e = 1, we have .
Topic 14: Solving equations (Throughout, especially L8, 10, 11, 12, 13, 14, 15)
You must be able to solve equations of all types.
Example 48: Solve for x:
To solve, you must isolate x, so start by subtracting 4 on both sides. Then you will multiply both
sides by 2.
so,
Then
So x = 2 .
Example 49: Solve for x: x2 - 25 = 0
Factor the left side of the equation: (x - 5)(x + 5) = 0 .
Now if ab = 0, then either a = 0 or b = 0, so we have
x - 5 = 0 or x + 5 = 0 , so
x = 5 or x = -5 .
Example 50: Solve for x:
Factor the left side of the equation: which factors further into
Set each factor equal to zero and solve for x.
x3 = 0 or x - 4 = 0 or x + 4 = 0 , so x = 0, x = 4, or x = -4 .
Example 51: Solve for x: x2 - x -12 = 0 .
Factor the left side of the equation: (x - 4)(x + 3) = 0 , then set each factor equal to zero and solve
each equation for x to find your answers, x = 4,or x = -3 .
Example 52: Solve for x: 3x2 - 4x - 7 = 3
To begin, we must have one side of the equation as 0. So start by subtracting 3 from both sides.
This gives us 3x2 - 4x -10 = 0 .
We can try to factor this, but it doesn’t factor. In this case, we will use the quadratic formula,
You must know the quadratic formula!!
We have a = 3, b = -4, and c = -10. Substituting these values, we have
We can simplify the radical to get
More likely, we will want to have decimal answers for x, so we will
computeand
Example 53: Solve for x:
Write both sides of the equation with a base of 3:
Then 2(2x + 3) = -3.
So 4x + 6 = -3.
Then 4x = -9 , and x = -9 / 4 .
Example 54: Solve for x: 5x= 11
Since we can’t rewrite this using common bases, we must use logarithms. Rewrite the problem in
logarithmic form:
You might need to rewrite this using the change of bases formula:
You might see the answer in this form, or it could be computed using a calculator. In that case,
you’d see
x = 1.4899 .
Example 55: Solve for x:
Rewrite in exponential form:
34 = 2x + 4
Solve for x.
81 = 2x + 4
77 = 2x
Example 56: Solve for x: 4ln(x -1) = 2
Start by dividing both sides by 4.
Rewrite in exponential form. Recall that the base of the natural log is e.
so
Topic 15: Graphing (Throughout, especially L12, 19, 19, 20, 21)
There are several remaining topics from graphing.
Example 57: Suppose . Find the x intercept(s), y intercept(s), vertical
asymptote(s) and horizontal asymptote(s).
To start, factor everything that is factorable:
Recall that if you have a common factor in the numerator and the denominator, you can reduce by
that factor. This will generate a hole in the graph. We don’t have that here.
To find the x intercepts, set f (x) equal to 0 and solve for x.
Multiply both sides of the equation by the denominator to clear the problem of fractions. We have:
(x - 2)(x -1) = 0
So, x - 2 = 0 or x -1 = 0
So x = 2 or x = 1.
We have two x intercepts, 2 and 1.
To find the y intercept, compute f (0) . In this case,
We could also have easily found the y intercept by looking at the original problem. Sub in 0 for x in
that function, and you can probably find the y intercept in your head.
By the way, we will often ask you to find the y intercept(s), implying that there can be more than
one. Don’t be fooled. If there are two or more y intercepts for your graph, it’s not a function.
We’ll only deal with functions in this class.
To find the vertical asymptote(s), set the denominator equal to 0 and solve for x. Note, if you have
a common factor in the numerator and the denominator (which we don’t have here), you will reduce
the fraction before finding the vertical asymptote(s).
(x - 3)(x + 3) = 0
So, x - 3 = 0 or x + 3 = 0
And x = 3 or x = -3 . These are the two vertical asymptotes.
Finally, to find the horizontal asymptote, you will need to compare the highest power of x in the
numerator with the highest power of x in the denominator. This number is called the degree of the
function. There are three possibilities:
Degree of numerator > Degree of denominator. In this case, there is no horizontal asymptote.
Degree of numerator < Degree of denominator. In this case, the horizontal asymptote is 0. We
write this as y = 0.
Degree of numerator = Degree of denominator. In this case, we make a fraction out of the
coefficients of the terms with the highest power and reduce it to lowest terms. This is the horizontal
asymptote.
Our problem falls into the third category, since we have x2 in both the numerator and the
denominator. Both coefficients are 1, so our horizontal asymptote is y = 1.
You should be able to tell by looking at a graph of an equation whether or not it is a function.
Recall the vertical line test: if a vertical line can cross a graph in more than one place, the graph is
not the graph of a function.
Example 58: Determine which of these are graphs of functions:
A is not a function. At x = 1, for example, we have two y values.
B is a function. It passes the vertical line test.
Given a function, you should be able to pick out its graph from a selection of graphs.
Example 59: What do you know about the graph of the function, f (x) = x3 - 9x ?
1. We know that it is a positive cubic function. Its basic shape is like this:
This shows that as x gets bigger, the y values will get bigger and the graph will be in the first
quadrant. As x moves towards - ∞, (really big, but negative), the y values will do the same, and the
graph will be in the third quadrant. Recall that cubic functions often have “humps” in the graph.
You should know the basic shapes of positive quadratic functions, negative quadratic functions,
positive cubic functions, negative cubic functions, positive quartic (4th degree) functions and
negative quartic (4th degree) functions.
2. We can find the zeros (x intercepts) of the function. To do this, we set the function equal to 0
and solve for x.
0 = x3 - 9x
0 = x(x - 3)(x + 3) (See Example 24 for details on the factoring)
So x = 0, x - 3 = 0, or x + 3 = 0 .
So x = 0, x = 3, or x = -3 are the x intercepts.
3. We can find the y intercept of the function. f (0) = 03 - 9(0) = 0 , so the graph passes through
the origin.
With no more information than this, we can graph the function. Plot the x and y intercepts. Use the
shape information to fill in the rest.
You will not know how high or low the “humps” need to be, but you can draw a fairly accurate
graph just from this information. Here, we generated a graph, but with the information we found
you should be able to pick out the correct graph from a selection of graphs.
You should be able to graph certain functions quickly, without a table of values or looking at a
calculator. Here are some examples.
Example 60: Graph y = 2x + 3.
From this equation, you know that the slope of the line is 2 and the y intercept is 3. Graph the point
(0, 3) and use the slopeo get another point or two. Connect the dots.
Example 61: Graph f (x) = x2 .
You should just know this one.
Example 62: Graph
You can use what you have learned about transformations to graph each of these. The first one is a
bit “narrower”, or closer to the y axis than the problem in Example 60. The second one is “wider”,
or closer to the x axis than the problem in Example 60. The third one is a reflection of the function
in Example 60 over the x axis. The fourth one is the graph from Example 60 shifted up one unit.
The last one is the graph from Example 60 shifted two units to the right. (Recall f (x) = (x + 2)2
would shift to the left.) Here are the graphs:
In the next example, we’ll need to look at the graph of a function and see what we can read from it.
Example 63: What do you know about the equation whose graph is given?
We start by observing that there are two vertical asymptotes. These must come from the
denominator of the function. The vertical asymptotes are at x = 1 and x = -1, so we have factors of
(x -1) and (x +1) in the denominator of the rational function.
We have two x intercepts, -2 and 2. This tells us that the function has factors of (x - 2) and
(x + 2) in the numerator.
We can also see a horizontal asymptote at y = 1. This tells us that the function has the same highest
power of x in the numerator as it has in the denominator, and that the ratio of their coefficients is 1,
so they are the same.
We have a y intercept at 4.
So a good guess of the function isThis function meets all the
criteria we noted above.
Topic 16: Solving systems of equations (L24)
Example 64: Solve the system of equations:
We can solve this by substitution or by multiplying/adding:
By substitution:
x - 5y = 9 so x = 5y + 9 . Substitute this into the other equation for x and solve for y:
Now that we know what y is, we can find x by substituting into either one of the original equations.
So the solution to the system is the ordered pair (4, -1).
Now we work the same problem, this time by multiplying and adding:
Multiply the bottom equation by -2 to eliminate the x column,
Now that we know what y is, we can find x by substituting into either one of the original equations.
So the solution to the system is the ordered pair (4, -1).
Example 65: Solve the system:
Each equation involves only one variable. Solve each independent of the other.
We’ll pair 0 with each of the y values we found. So our answers are (0, 3) and (0, -2).
Example 66: Solve the system:
If we just add the two equations together, the y column drops out. We have
Look at the bottom equation: 4x - 4y = 0
Add 4y to both sides of the equation and divide by 4:
So for our solutions, x = y. Our answers are (0, 0) and
Topic 17: Simplifying expressions (L4, 5, 6, 16)
You’ll need to be able to simplify some algebraic expressions.
Example 67: Simplify:
Split up the problem so that each term in the numerator is divided by the term in the denominator:
Simplify each term.
Note: this only works when the single term is in the denominator. You can’t simplify, for example,
using this method.
Example 68: Simplify:
Distribute in the numerator:
Combine like terms in the numerator
Do not multiply out the denominator.
Example 69: Simplify:
There is a common factor in this problem. Common to the two terms is . So we can factor
this out:
Now simplify inside the brackets:
Topic 18: Formulas (L14, 24)
You are expected to remember some simple formulas:
Perimeter of a square: P = 4s , s is the length of a side Perimeter of a rectangle: P = 2x + 2y , x and y are length and width, respectively Circumference of a circle : C = 2πr , r is the radius of the circle Area of a square: Area of a rectangle: A = xy , x is length, y is width Area of a circle: , r is the radius of the circle Volume of a box: V = xyh , x is length of base, y is width of the base, h is height Volume of a box with a square base: Pythagorean Theorem n right triangle ABC, where
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https://playworksheet.com/sheet/addition-table-charts-12x12 | 1,680,221,739,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00430.warc.gz | 498,206,138 | 4,614 | # Addition Table 12 by 12
The 12 by 12 addition table is a list of charts that shows the results of adding numbers from 1 to 12 to each other. It is a very useful tool for learning addition and can help kids quickly and easily memorize their addition facts.
Here's how you can introduce the 12 by 12 addition table to little kids:
Start with the easier sums: Start by practicing the easier sums, such as 1+1, 1+2, 2+1, and 2+2. You can have the kids point to the boxes on the chart to show the results of each sum.
Use visual aids: To help the kids remember the addition facts, use visual aids such as flashcards or manipulatives like counting bears or blocks. You can also use rhymes or songs to make it more fun and memorable.
Practice regularly: Make sure to practice regularly with the kids to help them memorize the addition facts. You can use worksheets, games, or quizzes to make it more engaging and interactive.
Learning the 12 by 12 addition chart is useful for several reasons:
Develops basic math skills: Addition is one of the basic math skills that kids need to master in order to progress in their math education. By learning the 12 by 12 addition chart, kids can build a solid foundation in addition, which will help them in more advanced math concepts like multiplication and division.
Speeds up mental calculations: Memorizing the 12 by 12 addition chart can help kids perform mental calculations quickly and accurately. This is a useful skill for everyday life, such as when shopping, cooking, or budgeting.
Saves time in school: Teachers often expect students to know their addition facts by heart, especially in the lower grades. By learning the 12 by 12 addition chart, kids can save time on homework and classwork, and have more time to focus on other subjects.
Boosts confidence: Knowing the addition facts can boost kids' confidence and make them feel more comfortable with math. This can help them perform better on tests and improve their overall academic performance.
Provides a solid foundation for more advanced math: As mentioned earlier, the 12 by 12 addition chart is a building block for more advanced math concepts like multiplication and division. By mastering the addition facts, kids can build a strong foundation for future math learning.
Remember, learning the 12 by 12 addition table takes time and practice, but with patience and perseverance, kids can master it!
Number Blocks: Additions 5 by 5
Number Blocks: Odd and Even numbers
Number Blocks: Twelve
Number Blocks: One Hundred
Number Blocks: Triangular Numbers
Number Blocks: First Ten Triangle Numbers
Multiplication Table 10x10
Multiplication Table 12 by 12
Number Blocks: Number Line to 20
Number Blocks: Multiplications 5 by 5
Number Blocks: Counting to Ten | 602 | 2,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-14 | latest | en | 0.959361 |
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How fewer math problems can help kids become math masters While children who are numerate can apply the math and problem solving skills they have to problems that are similar to the ones they have already solved, mathematical mastery prepares youngsters to combine the tools they already have to solve problems unlike ones they have seen before, Rusczyk said. He's the head of school for Proof School, a San Francisco based private liberal arts secondary school that caters to "students are internally driven to spend more than two hours on math in school every day," according to its website. Since solving unfamiliar problems requires contemplation, less is more when it comes to the number of problems kids tackle at once.
(For instance, 14 is one such number, since we can write 14 = 2 3 4 5.) For most, this qualifies as a hard problem, in that we’ve never encountered a question like this before, so can’t base our approach on any known method, and consequently may have no idea where to even begin.
In this situation, the teacher adopts the role of a coach, providing problem-solving strategies (try specific cases, collect data, look for patterns), supplying encouragement not to give up, celebrating partial results, pushing for a thorough explanation of conjectures, and guiding the student to make an effective presentation of their findings," Vandervelde explained, adding "I would rather a student occasionally spend several days wrestling with a single problem in this manner than forever plowing through more routine exercises." If more than one student is ready for an extra challenge?
Vandervelde has a classroom tip: "Assign each student, or pair of students, unrelated problems so that they can own their progress and not compete with or be scooped by classmates." Are math contests the answer? Some kids simply aren't interested in competitive math.
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A website dedicated to the fascinating world of mathematics and programming. That is, by solving one problem it will expose you to a new concept that allows.…
• ###### Problem Solving NZ Maths
This section of the nzmaths website has problem-solving lessons that you can use in your maths programme. The lessons provide coverage of Levels 1 to 6 of.… | 853 | 4,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-17 | longest | en | 0.953576 |
https://www.inchcalculator.com/convert/millinewton-to-meganewton/ | 1,719,178,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00810.warc.gz | 722,894,500 | 15,577 | # Millinewtons to Meganewtons Converter
Enter the force in millinewtons below to get the value converted to meganewtons.
## Result in Meganewtons:
1 mN = 1.0E-9 MN
Hint: use a scientific notation calculator to convert E notation to decimal
Do you want to convert meganewtons to millinewtons?
## How to Convert Millinewtons to Meganewtons
To convert a measurement in millinewtons to a measurement in meganewtons, divide the force by the following conversion ratio: 1,000,000,000 millinewtons/meganewton.
Since one meganewton is equal to 1,000,000,000 millinewtons, you can use this simple formula to convert:
meganewtons = millinewtons ÷ 1,000,000,000
The force in meganewtons is equal to the force in millinewtons divided by 1,000,000,000.
For example, here's how to convert 5,000,000,000 millinewtons to meganewtons using the formula above.
meganewtons = (5,000,000,000 mN ÷ 1,000,000,000) = 5 MN
## What Is a Millinewton?
One millinewton is equal to 1/1,000 of a newton, which is equal to the force needed to move one kilogram of mass at a rate of one meter per second squared.
The millinewton is a multiple of the newton, which is the SI derived unit for force. In the metric system, "milli" is the prefix for thousandths, or 10-3. Millinewtons can be abbreviated as mN; for example, 1 millinewton can be written as 1 mN.
## What Is a Meganewton?
One meganewton is equal to 1,000,000 newtons, which are equal to the force needed to move one kilogram of mass at a rate of one meter per second squared.
The meganewton is a multiple of the newton, which is the SI derived unit for force. In the metric system, "mega" is the prefix for millions, or 106. Meganewtons can be abbreviated as MN; for example, 1 meganewton can be written as 1 MN.
## Millinewton to Meganewton Conversion Table
Table showing various millinewton measurements converted to meganewtons.
Millinewtons Meganewtons
1 mN 0.000000001 MN
2 mN 0.000000002 MN
3 mN 0.000000003 MN
4 mN 0.000000004 MN
5 mN 0.000000005 MN
6 mN 0.000000006 MN
7 mN 0.000000007 MN
8 mN 0.000000008 MN
9 mN 0.000000009 MN
10 mN 0.00000001 MN
100 mN 0.0000001 MN
1,000 mN 0.000001 MN
10,000 mN 0.00001 MN
100,000 mN 0.0001 MN
1,000,000 mN 0.001 MN
10,000,000 mN 0.01 MN
100,000,000 mN 0.1 MN
1,000,000,000 mN 1 MN | 761 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.78239 |
https://jonathan-hui.medium.com/rl-value-learning-24f52b49c36d | 1,725,980,158,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00269.warc.gz | 301,306,857 | 45,813 | # RL — Value Learning
--
Value learning is a fundamental concept in reinforcement learning RL. It is the entry point to learn RL and as basic as the fully connected network in Deep Learning. It estimates how good to reach certain states or to take certain actions. While it may not be sufficient to use value-learning alone to solve complex problems, it is a key building block for many RL methods. In this article, we will use examples to demonstrate its concept.
Let’s plan a trip from San Francisco to San Jose. Say you are a devoted data scientist and you include many factors in your decision. These factors may include the remaining distance, the traffic, the road condition and even the chance of getting a ticket. After the analysis, you score every city and you always pick the next route with the highest score.
For example, when you are in San Bruno (SB), you have two possible routes. Based on their scores, SM has a higher score and therefore, we will travel to SM instead of WSM.
In reinforcement learning RL, the value-learning methods are based on a similar principle. We estimate how good to be in a state. We take actions for the next state that will collect the highest total rewards.
# Value function
Intuitively, value function V(s) measures how good to be in a specific state. By definition, it is the expected discounted rewards that collect totally following the specific policy:
where γ is the discount factor. If γ is smaller than one, we value future rewards with a lower current value. In most of the examples here, we set γ to one for simplicity in illustration. Our objective is finding a policy that maximizes the expected rewards.
There are many methods to find the optimal policy through value learning. We will discuss them in the next few sections.
# Value iteration
First, we can use dynamic programming to calculate the optimal value of V iteratively.
Then, we can use the value function to derive the optimal policy.
When we are in SB, we have two choices.
The SB to SM route receives a -10 rewards because SB is further away. We get an additional -5 rewards for the SB to WSM route because we can get a speeding ticket easily in that stretch of the highway.
The optimal V*(SB) = max( -10 + V*(SM), -15 + V*(WSM)) = 60.
Value Iteration Example
Let’s get into a full example. Below is a maze with the exit on the top left. At every location, there are four possible actions: up, down, left or right. If we hit the boundary, we bounce back to the original position. Every single-step move receives a negative one reward. Starting from the terminal state, we propagate the value of V* outwards using:
The following is the graphical illustration from iteration one to seven.
Once it is done, for every location, we locate the neighbor with the highest V-value as our best move.
# Policy evaluation
The second method is the policy evaluation. A policy tells us what to do from a particular state.
We can evaluate a random policy continually to calculate its value functions. A random policy is simply a policy that take any possible action randomly. Let’s consider another maze with exits on the top left and the bottom right. The value function is calculated as:
For a random policy, each action has the same probability. For four possible actions, π(a|s) equals 0.25 for any state.
For iteration 3 below, V[2, 2] = -2.9: we subtract one from each neighbor (negative one reward for every move), and take their average.
As we continue the iteration, V will converge and we can use the value function to determine the optimal policy again.
# Policy Iteration
The third method is the policy iteration. Policy iteration performs policy evaluation and policy improvement alternatively:
We continuously evaluate the current policy but we also refine the policy in each step.
As we keep improving the policy, it will converge to the optimal policy.
Here is the example which we can find the optimal policy in four iterations, much faster than the policy evaluation.
Algorithm
Let’s formulate the equations. The value-function at time step i+1 equals
Where P is the model (system dynamics) determining the next state after taking an action. The refined policy will be
For a deterministic model, the equation can be simplified to:
Here is the general flow of the algorithm:
# Bellman Optimality Equation
In the previous section, we use the dynamic programming to learn the value iteratively. The equation below is often mentioned in RL and is called the Bellman equation constraint.
# Value-Function Learning with Stochastic Model
In the previous value iteration example, we spread out the optimal value V* calculation to its neighbors in each iteration
using the equation:
In those examples, the model P is deterministic and is known. P is all zero except one state (s’) that is one. Therefore, it is simplified to:
But for the stochastic model, we need to consider all possible future states.
Let’s demonstrate it with another maze example using a stochastic model. This model has a noise of 0.2. i.e., if we try to move right, there is 0.8 chance that we do move right. But there is a 0.1 chance that we move up and 0.1 chance that we move down instead. If we hit a wall or boundary, we bounce back to the original position.
We assume the discount factor γ will be 0.9 and we receive a zero reward for every move unless we hit the terminate state which is +1 for the green spot and -1 for the red spot above.
Let’s fast forward to iteration 5, and see how to compute V*[2, 3] (underlined in white below) from the result of iteration 4.
The state above [2, 3] has the highest V value. So the optimal action for V*[2, 3] is going up. The new value function is
In each iteration, we will re-calculate V* for every location except the terminal state. As we keep iterating, V* will converge. For example, V*[2, 3] eventually converges to 0.57.
Algorithm
Here is the pseudocode for the value iteration:
# Model-Free
Regardless whether it is a deterministic or a stochastic model, we need the model P to compute the value function or to derive the optimal policy. (even though in a deterministic model, P is all zero except one state which is one.)
Monte-Carlo method
Whenever we don’t know the model, we fall back to sampling and observation to estimate the total rewards. Starting from the initial state, we run a policy and observe the total rewards (G) collected.
G is equal to
If the policy or the model is stochastic, the sampled total rewards can be different in each run. We can run and reset the system multiple times to find the average of V(S). Or we can simply keep a running average like the one below so we don’t need to keep all the previous sampled results.
Monte-Carlo method samples actions until the end of an episode to approximate total rewards.
Monte-Carlo control
Even we can estimate V(S) by sampling, how can we determine the action from one state to another?
Without knowing the model, we don’t know what action can lead us to the next optimal state s’. For example, without the road signs (the model), we don’t know whether the left lanes or the right lanes of the highway lead us to SM or WSM?
In the pong game below, we know what state we want to reach. But without the model, we don’t know how far (or how hard) should we push the joystick.
# Action-value Learning
This comes to the action-value function, the cousin of value function but without the need of a model. Instead of measuring how good a state V(s) is, we measure how good to take an action at a state Q(s, a). For example, when we are at SB, we ask how good to take the right lanes or the left lanes on Highway 101 even though we don’t know where it leads us to. So at any state, we can just take the action with the highest Q-value. This allows us to work without a model at the cost of more bookkeeping for each state. For a state with k possible actions, we have now k Q-values.
The Q-value (action-value function) is defined as the expected rewards for an action under a policy.
Similarly to the previous discussion, we can use the Monte-Carlo method to find Q.
In our example, we will keep on the left lanes when we are in SB.
This is the Monte-Carlo method for Q-value function.
# Policy Iteration with Q-value (model-free)
We can apply the Q-value function in the policy iteration. Here is the flow:
# Issues
The solution presented here has some practical problems. First, it cannot scale well for a large state space. The memory to keep track of V or Q for each state is impractical. Can we build a function estimator for value functions to save memory, just like the deep network for a classifier? Second, the Monte-Carlo method has very high variance. A stochastic policy may compute very different reward results in different runs. This high variance hurts the training. How can train the model with better convergence?
# Recap
Before looking into the solution. Let’s recap what we learn. We want to find an optimal policy that can maximize the expected discounted rewards.
We can solve it by computing the value function or the Q-value function:
And this can be solved using dynamic programming:
or one-step lookahead which is also called Temporal Difference TD.
# Thoughts
Stay tuned for the next part where we will solve some of the problems mentioned before and apply them in practice. | 2,037 | 9,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.942327 |
ps3ita.com | 1,601,428,003,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00359.warc.gz | 588,552,931 | 10,582 | # A Card Counter’s Guide to Basic Strategy
– The single most important thing you will ever learn in playing Blackjack, or becoming a card counter, is a system called Basic Strategy. So what it Basic Strategy, and why do you need to know it? I’m gonna answer that in this video.
(funky music) The absolute foundation to beating Blackjack is Basic Strategy. If you don’t master this, you have no chance of beating the game. Okay, so what is Basic Strategy?
Basic Strategy tells you exactly how to play every possible hand in Blackjack; when to Hit, when to Stand, when to Split, when to Double Down, or even when to Surrender. So, where did Basic Strategy come from and how do we know it we can trust it? Well, Basic Strategy was figured out like, a long time ago, using very early computers, but it’s been revised over the years based on how Blackjack is normally dealt today, and it’s completely run off math and computer simulations. And we’ll tell you the mathematically correct decision for every hand you can possibly be dealt at the Blackjack tables. Because Basic Strategy is the statistically correct answer for every hand you can be dealt, it’s really important that you don’t just learn 80 or 90% of it. These aren’t suggestions that you can then build your own strategy around.
These are the correct decisions to make every time. When I say they’re the mathematically correct decision, that doesn’t mean you’re gonna win every hand when you play this way. But it means it’s gonna help you in one of two ways. The first way Basic Strategy helps you is it helps you, at times, to win more money.
So, as an example of that, it’ll tell you the correct times to Double Down. Doubling Down when you have the advantage is a way to have more money on the table when you actually have an advantageous hand. It doesn’t work out 100% of the time, but it does work out over the long run if you follow it properly. The second way Basic Strategy will help you is it’ll help you lose less money. So as an example, consider a 16 against an ace.
Most players don’t like hitting a 16 against an ace because the odds are they’re going to lose. And the reality is, the odds are you are going to lose if you hit a 16 against an ace. But you’re gonna lose less money by following Basic Strategy and hitting a 16 against an ace, than by ignoring Basic Strategy and Standing a 16 against an ace. So, if you have a hand that is a negative expectation, maybe you’re gonna only lose 50 cents on average by following Basic Strategy, rather than losing 75 cents by ignoring Basic Strategy. So by following the math, the expectation over enough time is that you’re going to lose less money.
Okay, as far as learning Basic Strategy, it’s put together in a simple chart. There’s a link below where you can download it and start memorizing it. And if you commit yourself to memorizing it, you can actually learn it fairly quickly. And, for our members in our video course section, we actually have a series of videos that will help you learn it even faster by teaching you how to memorize the patterns, and different ways of reciting it.
We also provide a Basic Strategy drill that you can use for free on BlackjackApprenticeship.com to start practicing playing Basic Strategy. So, some people point out that our chart might look a little bit different from other charts that you could find on the internet. And, you can trust our chart. The reason ours might look different is we have it based off of the most common rules that are found today at the Blackjack tables, and some of the other charts might be based off of rules that were more common 20 or 30 years ago.
But, don’t worry about it. Commit this chart to memory, and then if you wanna actually beat the game, you can move on to the next steps that’ll take you to the next level. Now, I wanna be honest with you. Even with perfect Basic Strategy, you won’t have a winning game. Basic Strategy cuts the casino’s advantage to roughly half of one percent, but the casino still has a very slim advantage with perfect Basic Strategy. To actually gain the advantage, you have to learn how to count cards.
That will tell you, with perfect Basic Strategy, when you have the edge, and when the casino has the edge, and how do you use that to your advantage. If you’re interested in learning how to legally beat Blackjack with card counting, we have a free card counting mini course that you can sign up for below, where I will teach you what card counting is and isn’t, how to do it, and the biggest mistakes Blackjack players make, and how to avoid them. So check out the card counting mini course, and learn how to play Blackjack like a pro. | 1,021 | 4,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-40 | latest | en | 0.93971 |
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# STA641 - Statistical Packages and its Applications Online Quiz No 01 Fall 2020 Solution / Discussion Last Date: 30-11-2020
STA641 - Statistical Packages and its Applications Online Quiz No 01 Fall 2020 Solution / Discussion Last Date: 30-11-2020
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STA641-Solved_Online_Quiz_01_Fall_2020
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# STA-641 Quiz No-1 Solution November 2020 | STA-641 Quiz Solution | STA-641 Quiz last date 30 Nov
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# help
0
270
1
ABCD is a square. M is the midpoint of BC and N is the midpoint of CD. A point is selected at random in the square. Calculate the probability that it will liein the triangle MCN.
Feb 14, 2020
#1
+117546
+1
See the diagram :
Let the side of the square = S
Then the area of the square = S^2
Note that MC, NC = (1/2)S
And triangle MCN is a right triangle with MN, NC the legs
The area of this triangle = (1/2) (Product of the leg lengths) = (1/2) [ (1/2S * (1/2) S [ = (1/8) S^2
So....the probability that the point lies within triangle MCN =
Area of triangle MCN
_________________ =
Area of square
(1/8) S^2
________ =
S^2
1
_______
8
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# CHAPTER 9 - PowerPoint PPT Presentation
CHAPTER 9. APT AND MULTIFACTOR MODELS OF RISK AND RETURN. Arbitrage Exploitation of security mispricing, risk-free profits can be earned No arbitrage condition, equilibrium market prices are rational in that they rule out arbitrage opportunities. 9.1 MULTIFACTOR MODELS. Single Factor Model.
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### CHAPTER 9
APT AND MULTIFACTOR MODELS OF RISK AND RETURN
Arbitrage
• Exploitation of security mispricing, risk-free profits can be earned
• No arbitrage condition, equilibrium market prices are rational in that they rule out arbitrage opportunities
Single Factor Model
Returns on a security come from two sources
Common macro-economic factor
Firm specific events
Focus directly on the ultimate sources of risk, such as risk assessment when measuring one’s exposures to particular sources of uncertainty
Factors models are tools that allow us to describe and quantify the different factors that affect the rate of return on a security
Single Factor Model
ri= Return for security I
F = Surprise in macro-economic factor
(F could be positive, negative or zero)
ei = Firm specific events
F and ei have zero expected value, uncorrelated
Single Factor Model
• Example
• Suppose F is taken to be news about the state of the business cycle, measured by the unexpected percentage change in GDP, the consensus is that GDP will increase by 4% this year.
• Suppose that a stock’s beta value is 1.2, if GDP increases by only 3%, then the value of F=?
• F=-1%, representing a 1% disappointment in actual growth versus expected growth, resulting in the stock’s return 1.2% lower than previously expected
Multifactor Models
Macro factor summarized by the market return arises from a number of sources, a more explicit representation of systematic risk allowing for the possibility that different stocks exhibit different sensitivities to its various components
Use more than one factor in addition to market return
Examples include gross domestic product, expected inflation, interest rates etc.
Estimate a beta or factor loading for each factor using multiple regression.
Multifactor models, useful in risk management applications, to measure exposure to various macroeconomic risks, and to construct portfolios to hedge those risks
Two factor models
GDP, Unanticipated growth in GDP, zero expectation
IR, Unanticipated decline in interest rate, zero expectation
Multifactor model: Description of the factors that affect the security returns
Multifactor Models
Factor betas
Example
• One regulated electric-power utility (U), one airline (A), compare their betas on GDP and IR
• Beta on GDP: U low, A high, positive
• Beta on IR: U high, A low, negative
• When a good news suggesting the economy will expand, GDP and IR will both increase, is the news good or bad ?
• For U, dominant sensitivity is to rates, bad
• For A, dominant sensitivity is to GDP, good
• One-factor model cannot capture differential responses to varying sources of macroeconomic uncertainty
Multifactor Models
Multifactor Models
• Expected rate of return=13.3%
• 1% increase in GDP beyond current expectations, the stock’s return will increase by 1%*1.2
Multifactor model, a description of the factors that affect security returns, what determines E(r) in multifactor model
Expected return on a security (CAPM)
Multifactor Security Market Line
Compensation for bearing the macroeconomic risk
Compensation for time value of money
Multifactor Security Market Line for multifactor index model, risk premium is determined by exposure to each systematic risk factor and its risk premiumMultifactor Security Market Line
Arbitrage Pricing Theory
• Stephen Ross, 1976, APT, link expected returns to risk
• Three key propositions
• Security returns can be described by a factor model
• Sufficient securities to diversify away idiosyncratic risk
• Well-functioning security markets do not allow for the persistence of arbitrage opportunities
Arbitrage Pricing Theory
Arbitrage - arises if an investor can construct a zero investment portfolio with a sure profit
Since no investment is required, an investor can create large positions to secure large levels of profit
In efficient markets, profitable arbitrage opportunities will quickly disappear
Arbitrage
• Law of One Price
• If two assets are equivalent in all economically relevant respects, then they should have the same market price
• Arbitrage activity
• If two portfolios are mispriced, the investor could buy the low-priced portfolio and sell the high-priced portfolio
• Market price will move up to rule out arbitrage opportunities
• Security prices should satisfy a no-arbitrage condition
Well-diversified portfolio, the firm-specific risk negligible, only systematic risk remain
n-stock portfolio
Well-diversified portfolios
The portfolio variance
If equally-weighted portfolio , the nonsystematic variance
N lager, the nonsystematic variance approaches zero, the effect of diversification
Well-diversified portfolios
This is true for other than equally weighted one
Well-diversified portfolio is one that is diversified over a large enough number of securities with eachweight small enough that the nonsystematic variance is negligible, eP approaches zero
For a well-diversified portfolio
Well-diversified portfolios
Betas and Expected Returns
• Only systematic risk should command a risk premium in market equilibrium
• Well-diversified portfolios with equal betas must have equal expected returns in market equilibrium, or arbitrage opportunities exist
• Expected return on all well-diversified portfolio must lie on the straight line from the risk-free asset
Betas and Expected Returns
Expected rate=10%,completely determined by Rm
Subject to nonsystematic risk
Only systematic risk should command a risk premium in market equilibrium
Solid line: plot the return of A with beta=1 for various realization of the systematic factor (Rm)
B: E(r)=8%. beta=1; A:E(r)=10%. beta=1
• Arbitrage opportunity exist, so A and B can’t coexist
• Long in A, Short in B
• Factor risk cancels out across the long and short positions, zero net investment get risk-free profit
• infinitely large scale until return discrepancy disappears
• well-diversified portfolios with equal betas must have equal expected return in market equilibrium, or arbitrage opportunities exist
• A: beta=1,E(r)=10%;
• C: beta=0.5,E(r)=6%;
• D: 50% A and 50% risk-free (4%) asset,
• beta=0.5*1+0.5*0=0.5, E(r)=7%
• C and D have same beta (0.5)
• different expected return
• arbitrage opportunity
Expected Return %
A
10
D
7
6
C
Risk-free rate=4
0
0.5
beta
1
An arbitrage opportunity
A/C/D, well-diversified portfolio,
D : 50% A and 50% risk-free asset,
C and D have same beta (0.5),
different expected return,
arbitrage opportunity
M, market index portfolio, on the line and beta=1
• no-arbitrage condition to obtain an expected return-beta relationship identical to that of CAPM
EXAMPLE
• Market index, expected return=10%;Risk-free rate=4%
• Suppose any deviation from market index return can serve as the systematic factor
• E, beta=2/3, expected return=4%+2/3(10%-4%)=8%
• If E’s expected return=9%, arbitrage opportunity
• Construct a portfolio F with same beta as E,
• 2/3 in M, 1/3 in T-bill
• Long E, short F
One-Factor SML
• M, market index portfolio, as a well-diversified portfolio, no-arbitrage condition to obtain an expected return-beta relationship identical to that of CAPM
• three assumptions: a factor model, sufficient number of securities to form a well-diversified portfolios, absence of arbitrage opportunities
• APT does not require that the benchmark portfolio in SML be the true market portfolio
Multifactor APT
Use of more than a single factor
Several factors driven by the business cycle that might affect stock returns
Exposure to any of these factors will affect a stock’s risk and its expected return
Two-factor model
Each factor has zero expected value, surprise
Factor 1, departure of GDP growth from expectations
Factor 2, unanticipated change in IR
e, zero expected ,firm-specific component of unexpected return
A MULTIFACTOR APT
• Requires formation of factor portfolios
• Factor portfolio:
• Well-diversified
• Beta of 1 for one factor
• Beta of 0 for any other
• Or Tracking portfolio: the return on such portfolio track the evolution of particular sources of macroeconomic risk, but are uncorrelated with other sources of risk
• Factor portfolios will serve as the benchmark portfolios for a multifactor SML
A MULTIFACTOR APT
Example: Suppose two factor Portfolio 1, 2,
Risk-free rate=4%
Consider a well-diversified portfolio A ,with beta on the two factors
Multifactor APT states that the overall risk premium on portfolio A must equal the sum of the risk premiums required as compensation for each source of systematic risk
Total risk premium on the portfolio A:
Total return on the portfolio A: 9%+4%=13%
Factor Portfolio 1 and 2, factor exposures of any portfolio P are given by its and
Consider a portfolio Q formed by investing in factor portfolios with weights
in portfolio 1
in portfolio 2
in T-bills
Return of portfolio Q
A MULTIFACTOR APT
Form a portfolio Q from the factor portfolios with same betas as A, with weights:
0.5 in factor 1 portfolio
0.75 in factor 2 portfolio
-0.25 in T-bill
Invest \$1 in Q, and sell % in A, net investment is 0, but with positive riskless profit
Q has same exposure as A to the two sources of risk, their expected return also ought to be equal
A MULTIFACTOR APT
Two principles when specify a reasonable list of factors
Limit ourselves to systematic factors with considerable ability to explain security returns
Choose factors that seem likely to be important risk factors, demand meaningful risk premiums to bear exposure to those sources of risk
Multifactor APT
Chen, Roll, Ross 1986
Chose a set of factors based on the ability of the factors to paint a broad picture of the macro-economy
IP: % change in industrial production
EI: % change in expected inflation
UI: % change in unexpected inflation
CG: excess return of long-term corporate bonds over long-term government bonds
GB: excess return of long-term government bonds over T-bill
Multidimensional SCL, multiple regression, residual variance of the regression estimates the firm-specific risk
Multifactor APT
Fama, French, three-factor model
Use firm characteristics that seem on empirical grounds to proxy for exposure to systematic risk
SMB: return of a portfolio of small stocks in excess of the return on a portfolio of large stocks
HML: return of a portfolio of stocks with high book-to-market ratio in excess of the return on a portfolio of stocks with low ratio
Market index is expected to capture systematic risk
Multifactor APT
Fama, French, three-factor model
Long-standing observations that firm size and book-to-market ratio predict deviations of average stock returns from levels with the CAPM
High ratios of book-to-market value are more likely to be in financial distress, small stocks may be more sensitive to changes in business conditions
The variables may capture sensitivity to risk-factors in macroeconomy
APT
highlight the crucial distinction between factor risk and diversifiable risk
APT assumption: rational equilibrium in capital markets precludes arbitrage opportunities (not necessarily to individual stocks)
APT yields expected return-beta relationship using a well-diversified portfolio (not a market portfolio)
APT and CAPM Compared
APT and CAPM Compared
• APT applies to well diversified portfolios and not necessarily to individual stocks
• APT is more general in that it gets to an expected return and beta relationship without the assumption of the market portfolio
• APT can be extended to multifactor models
The Multifactor CAPM and the APM
• A multi-index CAPM
• Derived from a multi-period consideration of a stream of consumption
• will inherit its risk factors from sources of risk that a broad group of investors deem important enough to hedge, from a particular hedging motive
• The APT is largely silent on where to look for priced sources of risk | 2,874 | 12,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-47 | latest | en | 0.860563 |
https://www.in2013dollars.com/us/inflation/1840?amount=1&endYear=1837 | 1,586,497,288,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371886991.92/warc/CC-MAIN-20200410043735-20200410074235-00243.warc.gz | 962,179,948 | 24,514 | # \$1 in 1840 → \$1.10 in 1837
## Inflation Calculator
\$
### U.S. Inflation Rate, \$1 from 1840 to 1837
According to the Bureau of Labor Statistics consumer price index, prices in 1837 are 10.34% higher than average prices since 1840. The U.S. dollar experienced an average deflation rate of -3.23% per year during this period, meaning the real value of a dollar increased.
In other words, \$1 in 1840 is equivalent in purchasing power to about \$1.10 in 1837, a difference of \$0.10 over 3 years.
The 1837 inflation rate was 3.23%. The inflation rate in 1840 was -6.45%. The 1840 inflation rate is lower compared to the average inflation rate of 1.90% per year between 1840 and 2020.
Cumulative price change 10.34% Average inflation rate -3.23% Converted amount (\$1 base) \$1.10 Price difference (\$1 base) \$0.10 CPI in 1840 8.700 CPI in 1837 9.600 Inflation in 1837 3.23% Inflation in 1840 -6.45%
USD Inflation since 1635
Annual Rate, the Bureau of Labor Statistics CPI
### Buying power of \$1 in 1837
This chart shows a calculation of buying power equivalence for \$1 in 1837 (price index tracking began in 1635).
For example, if you started with \$1, you would need to end with \$1.10 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
According to the Bureau of Labor Statistics, each of these USD amounts below is equal in terms of what it could buy at the time:
Year Dollar Value Inflation Rate
1837 \$1.00 3.23%
1838 \$0.97 -3.12%
1839 \$0.97 0.00%
1840 \$0.91 -6.45%
1841 \$0.91 0.00%
1842 \$0.85 -5.75%
1843 \$0.77 -9.76%
1844 \$0.78 1.35%
1845 \$0.79 1.33%
1846 \$0.80 1.32%
1847 \$0.85 6.49%
1848 \$0.82 -3.66%
1849 \$0.80 -2.53%
1850 \$0.81 1.30%
1851 \$0.80 -1.28%
1852 \$0.80 0.00%
1853 \$0.80 0.00%
1854 \$0.88 9.09%
1855 \$0.91 3.57%
1856 \$0.89 -2.30%
1857 \$0.91 2.35%
1858 \$0.85 -5.75%
1859 \$0.86 1.22%
1860 \$0.86 0.00%
1861 \$0.92 6.02%
1862 \$1.05 14.77%
1863 \$1.31 24.75%
1864 \$1.64 24.60%
1865 \$1.70 3.82%
1866 \$1.66 -2.45%
1867 \$1.54 -6.92%
1868 \$1.48 -4.05%
1869 \$1.42 -4.23%
1870 \$1.36 -3.68%
1871 \$1.27 -6.87%
1872 \$1.27 0.00%
1873 \$1.25 -1.64%
1874 \$1.19 -5.00%
1875 \$1.15 -3.51%
1876 \$1.11 -2.73%
1877 \$1.09 -1.87%
1878 \$1.04 -4.76%
1879 \$1.04 0.00%
1880 \$1.06 2.00%
1881 \$1.06 0.00%
1882 \$1.06 0.00%
1883 \$1.05 -0.98%
1884 \$1.02 -2.97%
1885 \$1.01 -1.02%
1886 \$0.98 -3.09%
1887 \$0.99 1.06%
1888 \$0.99 0.00%
1889 \$0.96 -3.16%
1890 \$0.95 -1.09%
1891 \$0.95 0.00%
1892 \$0.95 0.00%
1893 \$0.94 -1.10%
1894 \$0.90 -4.44%
1895 \$0.88 -2.33%
1896 \$0.88 0.00%
1897 \$0.86 -1.19%
1898 \$0.86 0.00%
1899 \$0.86 0.00%
1900 \$0.88 1.20%
1901 \$0.89 1.19%
1902 \$0.90 1.18%
1903 \$0.92 2.33%
1904 \$0.93 1.14%
1905 \$0.92 -1.12%
1906 \$0.94 2.27%
1907 \$0.98 4.44%
1908 \$0.96 -2.13%
1909 \$0.95 -1.09%
1910 \$0.99 4.40%
1911 \$0.99 0.00%
1912 \$1.01 2.11%
1913 \$1.03 2.06%
1914 \$1.04 1.01%
1915 \$1.05 1.00%
1916 \$1.14 7.92%
1917 \$1.33 17.43%
1918 \$1.57 17.97%
1919 \$1.80 14.57%
1920 \$2.08 15.61%
1921 \$1.86 -10.50%
1922 \$1.75 -6.15%
1923 \$1.78 1.79%
1924 \$1.78 0.00%
1925 \$1.82 2.34%
1926 \$1.84 1.14%
1927 \$1.81 -1.69%
1928 \$1.78 -1.72%
1929 \$1.78 0.00%
1930 \$1.74 -2.34%
1931 \$1.58 -8.98%
1932 \$1.43 -9.87%
1933 \$1.35 -5.11%
1934 \$1.40 3.08%
1935 \$1.43 2.24%
1936 \$1.45 1.46%
1937 \$1.50 3.60%
1938 \$1.47 -2.08%
1939 \$1.45 -1.42%
1940 \$1.46 0.72%
1941 \$1.53 5.00%
1942 \$1.70 10.88%
1943 \$1.80 6.13%
1944 \$1.83 1.73%
1945 \$1.88 2.27%
1946 \$2.03 8.33%
1947 \$2.32 14.36%
1948 \$2.51 8.07%
1949 \$2.48 -1.24%
1950 \$2.51 1.26%
1951 \$2.71 7.88%
1952 \$2.76 1.92%
1953 \$2.78 0.75%
1954 \$2.80 0.75%
1955 \$2.79 -0.37%
1956 \$2.83 1.49%
1957 \$2.93 3.31%
1958 \$3.01 2.85%
1959 \$3.03 0.69%
1960 \$3.08 1.72%
1961 \$3.11 1.01%
1962 \$3.15 1.00%
1963 \$3.19 1.32%
1964 \$3.23 1.31%
1965 \$3.28 1.61%
1966 \$3.38 2.86%
1967 \$3.48 3.09%
1968 \$3.63 4.19%
1969 \$3.82 5.46%
1970 \$4.04 5.72%
1971 \$4.22 4.38%
1972 \$4.35 3.21%
1973 \$4.63 6.22%
1974 \$5.14 11.04%
1975 \$5.60 9.13%
1976 \$5.93 5.76%
1977 \$6.31 6.50%
1978 \$6.79 7.59%
1979 \$7.56 11.35%
1980 \$8.58 13.50%
1981 \$9.47 10.32%
1982 \$10.05 6.16%
1983 \$10.38 3.21%
1984 \$10.82 4.32%
1985 \$11.21 3.56%
1986 \$11.42 1.86%
1987 \$11.83 3.65%
1988 \$12.32 4.14%
1989 \$12.92 4.82%
1990 \$13.61 5.40%
1991 \$14.19 4.21%
1992 \$14.61 3.01%
1993 \$15.05 2.99%
1994 \$15.44 2.56%
1995 \$15.88 2.83%
1996 \$16.34 2.95%
1997 \$16.72 2.29%
1998 \$16.98 1.56%
1999 \$17.35 2.21%
2000 \$17.94 3.36%
2001 \$18.45 2.85%
2002 \$18.74 1.58%
2003 \$19.17 2.28%
2004 \$19.68 2.66%
2005 \$20.34 3.39%
2006 \$21.00 3.23%
2007 \$21.60 2.85%
2008 \$22.43 3.84%
2009 \$22.35 -0.36%
2010 \$22.71 1.64%
2011 \$23.43 3.16%
2012 \$23.92 2.07%
2013 \$24.27 1.46%
2014 \$24.66 1.62%
2015 \$24.69 0.12%
2016 \$25.00 1.26%
2017 \$25.53 2.13%
2018 \$26.17 2.49%
2019 \$26.63 1.76%
2020 \$26.95 1.18%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
### Inflation by Country
Inflation can also vary widely by country. For comparison, in the UK £1.00 in 1840 would be equivalent to £0.91 in 1837, an absolute change of £-0.09 and a cumulative change of -9.01%.
Compare these numbers to the US's overall absolute change of \$0.10 and total percent change of 10.34%.
### Inflation by Spending Category
CPI is the weighted combination of many categories of spending that are tracked by the government. This chart shows the average rate of inflation for select CPI categories between 1840 and 1837.
Compare these values to the overall average of -3.23% per year:
Category Avg Inflation (%) Total Inflation (%) \$1 in 1837 → 1840
Food and beverages 0.00 0.00 1.00
Housing 0.00 0.00 1.00
Apparel 0.00 0.00 1.00
Transportation 0.00 0.00 1.00
Medical care 0.00 0.00 1.00
Recreation 0.00 0.00 1.00
Education and communication 0.00 0.00 1.00
Other goods and services 0.00 0.00 1.00
The graph below compares inflation in categories of goods over time. Click on a category such as "Food" to toggle it on or off:
For all these visualizations, it's important to note that not all categories may have been tracked since 1840. This table and charts use the earliest available data for each category.
### How to Calculate Inflation Rate for \$1, 1837 to 1840
This inflation calculator uses the following inflation rate formula:
CPI in 1837CPI in 1840
×
1840 USD value
=
1837 USD value
Then plug in historical CPI values. The U.S. CPI was 8.7 in the year 1840 and 9.6 in 1837:
9.68.7
×
\$1
=
\$1.10
\$1 in 1840 has the same "purchasing power" or "buying power" as \$1.10 in 1837.
To get the total inflation rate for the 3 years between 1837 and 1840, we use the following formula:
CPI in 1837 - CPI in 1840CPI in 1840
×
100
=
Cumulative inflation rate (3 years)
Plugging in the values to this equation, we get:
9.6 - 8.78.7
×
100
=
10%
Politics and news often influence economic performance. Here's what was happening at the time:
• Identification of Antarctica as a new continent by American naval expedition lead by Charles Wilkes.
• Charles Wilkes discovers Antarctica in an American naval expedition.
• Charles Wilkes discovers the Shackleton Ice Shelf in Antarctica.
• The world’s first postage stamp, the Penny Black, is released.
• Samuel Morse patents the telegraph.
### Data Source & Citation
Raw data for these calculations comes from the Bureau of Labor Statistics' (CPI), established in 1913. Inflation data from 1665 to 1912 is sourced from a historical study conducted by political science professor Robert Sahr at Oregon State University.
You may use the following MLA citation for this page: “\$1 in 1840 → 1837 | Inflation Calculator.” Official Inflation Data, Alioth Finance, 10 Apr. 2020, https://www.officialdata.org/us/inflation/1840?amount=1&endYear=1837. | 3,245 | 7,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.839469 |
https://community.fabric.microsoft.com/t5/DAX-Commands-and-Tips/Need-help-with-DAX-measure-to-show-user-defined-value-in-row/m-p/3263827/highlight/true | 1,721,261,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00736.warc.gz | 151,730,479 | 50,195 | cancel
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Anonymous
Not applicable
## Need help with DAX measure to show user defined value in row subtotals in matrix
Hi All ,
I have matrix below where I am showing contact data mismtatch count based on firstname and lastname column .
Count of firstname and lastname is correct but in parent row - Contact I want to calcualte distinct of contactid where
(fistname <> "matched" or lastname <> "matched") so its basiscally not sum of firstname and lastname count 7252 (so count will be less in Contacts row ) . Similarly, Totals also would not be sum of each row (7252) . Apart from Contacts , I have some other factor as well for data mismatch which will be getting added to matrix as row .
Contacts FirstName Data Mismatch = CALCULATE(DISTINCTCOUNT(DataMismatch[CRMContactId]),KEEPFILTERS (DataMismatch[FirstNameCheck]<>"MATCHED"))
Data Mismatch =
SWITCH(SELECTEDVALUE('List of Tables'[Level1]),
"Contacts",
SWITCH (SELECTEDVALUE('List of Tables'[Level2]),
"Firstname",[Contacts FirstName Data Mismatch]+0,
"Lastname",[Contacts LastName Data Mismatch Count]+0))
Final Data Mismatch = SUMX('List of Tables',[Data Mismatch])
1 ACCEPTED SOLUTION
Super User
Hi @Anonymous
It is clear why this is hapenning but the question is what should be the correct value at the total and based on what? I would suggest tto create a third measure
``````Contacts Data Mismatch =
CALCULATE (
DISTINCTCOUNT ( DataMismatch[CRMContactId] ),
KEEPFILTERS ( DataMismatch[FirstNameCheck] <> "MATCHED"
|| DataMismatch[LastNameCheck] <> "MATCHED" )
)``````
Then the final measure to use in the visual would be
``````Data Mismatch =
IF (
SELECTEDVALUE ( 'List of Tables'[Level1] ) = "Contacts",
IF (
HASONEVALUE ( 'List of Tables'[Level2] ),
IF (
VALUES ( 'List of Tables'[Level2] ) = "Firstname",
[Contacts FirstName Data Mismatch] + 0,
[Contacts LastName Data Mismatch Count] + 0
),
[Contacts Data Mismatch]
)
)``````
2 REPLIES 2
Anonymous
Not applicable
Thanks a lot @tamerj1 , seems its working . I will apply the login in my actual dataset and check. thanks again!
Super User
Hi @Anonymous
It is clear why this is hapenning but the question is what should be the correct value at the total and based on what? I would suggest tto create a third measure
``````Contacts Data Mismatch =
CALCULATE (
DISTINCTCOUNT ( DataMismatch[CRMContactId] ),
KEEPFILTERS ( DataMismatch[FirstNameCheck] <> "MATCHED"
|| DataMismatch[LastNameCheck] <> "MATCHED" )
)``````
Then the final measure to use in the visual would be
``````Data Mismatch =
IF (
SELECTEDVALUE ( 'List of Tables'[Level1] ) = "Contacts",
IF (
HASONEVALUE ( 'List of Tables'[Level2] ),
IF (
VALUES ( 'List of Tables'[Level2] ) = "Firstname",
[Contacts FirstName Data Mismatch] + 0,
[Contacts LastName Data Mismatch Count] + 0
),
[Contacts Data Mismatch]
)
)`````` | 760 | 2,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.853989 |
http://mathhelpforum.com/algebra/34671-mixed-questions-help-needed-print.html | 1,529,635,106,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864337.41/warc/CC-MAIN-20180622010629-20180622030629-00420.warc.gz | 202,781,104 | 4,238 | # Mixed Questions: Help Needed
• Apr 15th 2008, 06:32 PM
Quester
Mixed Questions: Help Needed
3 5/8 ½ =
1Ύ divided by 2 =
Write as a ratio in simplest form
a) 24:40 (would it be 3:5)
b) 220 cm to 1 m
Leah and Rachael both drive the same car, and use it basically at the same rate.
If Leah drives for 8 hours at 60 km/h to visit her best friend, the trip costs $63 in fuel. Rachael however drives to see her father but has to drive a further 256 km. How much will the trip cost in fuel? Three People contribute money to a raffle -$15, $20 and$25. If the winnings are distributed in same ratio as the contributions how much would each person get, if they won $270? • Apr 15th 2008, 06:42 PM topsquark Quote: Originally Posted by Quester 3 5/8 ½ = 1Ύ divided by 2 = Write as a ratio in simplest form a) 24:40 (would it be 3:5) b) 220 cm to 1 m For the first one convert the mixed numbers into improper fractions:$\displaystyle 3~\frac{5}{8} = \frac{29}{8}$so$\displaystyle 3~\frac{5}{8} - \frac{1}{2} = \frac{29}{8} - \frac{1}{2}\displaystyle = \frac{29}{8} - \frac{4}{8} = \frac{25}{8}$If you really must have the answer as a mixed number, then$\displaystyle \frac{25}{8} = 3 + \frac{1}{8} = 3~\frac{1}{8}$The same thing goes for the other one. For the ratios: a) 24:40$\displaystyle \frac{24}{40} = \frac{3 \cdot 8}{5 \cdot 8} = \frac{3}{5}$and this is 3:5. Again, the other one works the same way. -Dan • Apr 15th 2008, 07:01 PM Quester Simplify (expand and collect like terms) a)4q + 12q 3 b)4 (3a + 1) + 2a c)Subtract 3x 5 from 8x 18 Factorise Following a)3a + 15 ( is it: 3a + 15 = 3 x a + 3 x 5 = 3(a+5) b)4 x xy + 3x2 Solve Following a)2n + 3 = 17 b)3x 4/2 6 = 4 c)5(x 2) = 4 (x + 9) But this particually, I cant get: a) The adjacent sides of a rectangle are (3x 8) and 6 cm. Given that the area of the rectangle is 96 cm^2, find the length of the rectangle and the value of x. c)The Three sides of a triangle are 1/1 y + 2 and y = 4. The perimeter is 87 cm. What are the lengths of the three sides? • Apr 15th 2008, 07:02 PM Quester oh sorry when i wrote 1/2 i mean the fraction 1/2 • Apr 15th 2008, 10:26 PM earboth Quote: Originally Posted by Quester Simplify (expand and collect like terms) a)4q + 12q – 3 b)4 (3a + 1) + 2a c)Subtract 3x – 5 from 8x – 18 Factorise Following a)3a + 15 ( is it: 3a + 15 = 3 x a + 3 x 5 = 3(a+5) b)4 x – xy + 3x2 Solve Following a)2n + 3 = 17 b)3x – 4/2 – 6 = 4 c)5(x – 2) = 4 (x + 9) But this particually, I can’t get: a) The adjacent sides of a rectangle are (3x – 8) and 6 cm. Given that the area of the rectangle is 96 cm^2, find the length of the rectangle and the value of x. c)The Three sides of a triangle are 1/1 y + 2 and y = 4. The perimeter is 87 cm. What are the lengths of the three sides? 1. Do yourself and do us a favor and start a new thread if you have new questions. 2. to #1.c.): Translate the sentence into a mathematical operation:$\displaystyle (8x-18)-(3x-5)~\buildrel {expand} \over \longrightarrow~8x-18-3x+5 = 5x-13$to #2.a.): Yes to 2.b.):$\displaystyle 4x - xy + 3x^2= x(4-y+3x)$to #3.c.):$\displaystyle 5(x-2) = 4 (x + 9)~\iff~ 5x-10=4x+36~\iff~x=46$....... The last step is: add (-4x+10) on both sides of the equation - you only have to know why this is necessary! to #4.a.): You are suposed to know that the area of a rectangle is calculated by:$\displaystyle area = length\ \cdot \ width$....... with$\displaystyle length = 3x-8$....... and .......$\displaystyle width = 6$So you have to solve for x:$\displaystyle (3x-8) \cdot 6 = 96 ~\iff~ 18x - 48 = 96~\iff~ 18x = 144 ~\iff~ x = \frac{144}{18} = 8$Therefore the rectangle has the dimensions:$\displaystyle l = 16$and$\displaystyle w = 6$• Apr 16th 2008, 03:52 PM Quester Quote: Originally Posted by earboth So you have to solve for x:$\displaystyle (3x-8) \cdot 6 = 96 ~\iff~ 18x - 48 = 96~\iff~ 18x = 144 ~\iff~ x = \frac{144}{18} = 8$Therefore the rectangle has the dimensions:$\displaystyle l = 16$and$\displaystyle w = 6\$
how did you get the answer for (3x-8).6 = 96 what does the dot stand for?
• Apr 16th 2008, 08:24 PM
Isomorphism
Quote:
Originally Posted by Quester
how did you get the answer for (3x-8).6 = 96 what does the dot stand for?
Multiplication | 1,576 | 4,218 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-26 | latest | en | 0.870703 |
https://www.coursehero.com/file/6785417/s4/ | 1,481,444,910,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544358.59/warc/CC-MAIN-20161202170904-00264-ip-10-31-129-80.ec2.internal.warc.gz | 930,471,353 | 21,605 | # s4 - Math 5615H: Introduction to Analysis I. Fall 2011...
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Unformatted text preview: Math 5615H: Introduction to Analysis I. Fall 2011 Homework #4. Problems and Solutions. #1. Let f be a mapping of A to B . Show that for each B 1 B and B 2 B , their inverse images satisfy the properties ( i ) f- 1 ( B 1 B 2 ) = f- 1 ( B 1 ) f- 1 ( B 2 ) , ( ii ) f- 1 ( B 1 B 2 ) = f- 1 ( B 1 ) f- 1 ( B 2 ) . Proof. (i) We have x f- 1 ( B 1 B 2 ) f ( x ) B 1 B 2 ( f ( x ) B 1 ) or ( f ( x ) B 2 ) ( x f- 1 ( B 1 ) ) or ( x f- 1 ( B 2 ) ) x f- 1 ( B 1 ) f- 1 ( B 2 ) . Therefore, both sides in (i) coincide. The proof of (ii) is quite similar, with being replaced by , and or by and. #2. Let f be a mapping of A to B . Verify whether of not the images of subsets A 1 A and A 2 A in general satisfy the properties ( i ) f ( A 1 A 2 ) = f ( A 1 ) f ( A 2 ) , ( ii ) f ( A 1 A 2 ) = f ( A 1 ) f ( A 2 ) ....
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## s4 - Math 5615H: Introduction to Analysis I. Fall 2011...
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Ask a homework question - tutors are online | 473 | 1,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2016-50 | longest | en | 0.741934 |
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For questions regarding circulant matrices, where each row vector is rotated one element to the right relative to the preceding row vector.
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### How to compute the determinant of a block circulant matrix?
I am curious if there are any general formulas for problems like this or special cases. I want to compute the determinant of $2n \times 2n$ complex matrices made of identical $2 \times 2$ matrices. If ...
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### Determinant of the circulant matrix corresponding to the $r$-tuple $(1, 1, 0, 0, \ldots , 0, 0)$
For any integer $r \geq 3$, consider the $r$-tuple $(1, 1, 0, 0, \ldots , 0, 0)$ (involving $r - 2$ zeros) which represents the first row of the corresponding $r \times r$ circulant matrix. Show that ...
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### Determinant of circulant $(0,1)$ matrices of certain form
I am interested in computing the determinant of the following circulant matrices: let $n=p^k$ for $p$ a prime and $k\in \mathbb{N}$, take $a\in \mathbb{N}$ to be such that $a<p$ and $(a,p)=1$. ...
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### Diagonalize matrix of linear operator
Let $f : \mathbb{C} \to \mathbb{C}$ linear map such $$f(e_{i}) = \begin{cases} e_{i+1} & 1 \leq i<n \\ e_{1} & i=n\end{cases}$$ Diagonalize $f$. Thoughts I know the characteristic ...
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### Operations with Circulant Matrix using GAP
I am newbie using GAP software. I need to know how to use GAP software for algebraic computations with circulant matrix. Some examples would suffice. Just for clarity Circulant Matrix: In linear ...
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### results of calculating eigenvalues and eigenvectors of permutation matrix
My question is actually about the derivation of eigenvalues and eigenvectors of the circulant matrix. I am not good at doing that in a straight way, i.e. calculating them with a general form of ...
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### Diagonalizing a matrix with 4 circulant blocks
I have the following matrix: $$\mathbf{M} = \begin{pmatrix} G_{1}^{(N)} & G_{2}^{(N)} \\ G_{2}^{(N)} & G_{3}^{(N)} \end{pmatrix}$$, where $G^{(N)}_{j}$ are symmetric circulant matrices of size ...
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### The square root of symmetric and circulant matrix is symmetric
Let $A$ a circulant symmetric matrix. From the definition of $\sqrt{A}$ it follows that also $\sqrt{A}$ is symmetric ( and circulant ). How can I show the symmetry without using the concept of ...
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• 3,847 | 775 | 2,624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-30 | latest | en | 0.726104 |
https://www.hackmath.net/en/example/1067 | 1,484,913,280,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280834.29/warc/CC-MAIN-20170116095120-00119-ip-10-171-10-70.ec2.internal.warc.gz | 915,869,289 | 14,639 | # Equation
Equation
has one root x1 = 10. Determine the coefficient b and the second root x2.
Result
b = 1.6
x2 = -8.4
#### Solution:
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Calculate: ? | 473 | 1,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-04 | longest | en | 0.884923 |
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# 26 Fresh What is A Series Parallel Circuit
series and parallel circuits ponents of an electrical circuit or electronic circuit can be connected in many different ways the two simplest of these are called series and parallel and occur frequently circuit construction kit dc series circuit experiment with an electronics kit build circuits with batteries resistors light bulbs and switches determine if everyday objects are conductors or insulators and take measurements with an ammeter and voltmeter view the circuit as a schematic diagram or switch to a lifelike view electrical dc series and parallel circuit an example of series dc circuit suppose three resistors r 1 r 2 and r 3 are connected in series across a voltage source of v quantified as volts as shown in the figure parallel and series circuit instructables how to make in a parallel circuit each ponent has its own direct path to both the negative and positive sides of the circuit a simple schematic of a parallel circuit is shown below series and parallel circuit worksheet furrey s physics those fifty 15 ohm series connected christmas tree lights calculate the total current in the circuit if they are connected to a 115 vac source those fifty 15 ohm parallel connected christmas tree lights calculate the total current in the circuit if they are connected to a 115 vac source series or parallel circuit pedagogy series or parallel circuit led series parallel array wizard linear1 the linear1 network where all my interests e to her discussion forum ask your questions in the forums linear1 case mods many illustrated case mod walkthroughs series and parallel circuits furrey s physics classroom 1 series and parallel circuits direct current series circuits a series circuit is a circuit in which the ponents are connected in a line one after the differences between short circuit in a series and a as is the case nowadays a circuit breaker or a fuse is used in wiring the fuse can blow or the circuit breaker can trip if there is a short circuit and that will cut off the current supply to all the ponents irrespective of a series or a parallel arrangement series & parallel circuits free mobile website templates series & parallel circuits introduction so far we have discussed circuits with only two ponents a source of current such as a battery and a single resistance such as a lightbulb or resistor
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You can download all 26 of 26 Fresh What is A Series Parallel Circuit image to your device by right clicking image and then save image as. Do not forget to click share if you love with this wallpaper. | 881 | 4,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-47 | latest | en | 0.921273 |
https://sites.google.com/site/e90e50fx/home/create-list-of-sheet-names-based-on-prefix | 1,563,850,153,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528687.63/warc/CC-MAIN-20190723022935-20190723044935-00089.warc.gz | 532,723,207 | 19,015 | ### Formula challenge: create a list of sheet names based on a prefix
posted Jan 10, 2014, 8:25 AM by Krisztina Szabó [ updated Jun 12, 2014, 2:48 AM ]
Freddy: When I give the word, throw the first switch.
Igor: You've got it, master.
Freddy: Get set.
Freddy: Go.
Freddy: Throw the second switch.
Freddy: Throw the third switch.
Igor: Not the third switch?
by The FrankensTeam
A formula which lists sheet names according to a rule could be useful for example to summarize data from particular sheets. But could it be done without VBA, using Excel formulas?
The challenge:
In the file sheets are named according to a rule: prefix and number, so for example:
prefix = “Company”
sheets:
Company1
Company2
Company3
Company5
The numbering is not necessary to be continuous.
The task is to create a formula which will produce the list of existing sheet names in array: {“Company1”;“Company2”;“Company3”;“Company5”}
And it will automatically capture the newly added sheets too.
So, if you would like to think… do not scroll down!
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Two formula solutions
Solution with ERROR.TYPE
The idea is to test the existence of the possible sheet names ( prefix&ROW(\$1:\$100) ) and acquire the numbers which results existing sheet. Then concatenate these numbers with the prefix will give the list we wants to have.
`=prefix & SMALL( IF( ``ERROR.TYPE(INDIRECT(prefix&ROW(\$1:\$100)&"!A1"))=3`` , ``ROW(\$1:\$100)`` ), ``ROW( INDIRECT("1:"&SUM( --(ERROR.TYPE(INDIRECT(prefix&ROW(\$1:\$100)&"!A1"))=3)) ) )`` )`
to be array-entered
How does it work?
The test is based on INDIRECT formula, which returns the reference specified by a text string. If the reference is invalid (so the sheet does not exists) the result is #REF! error. Using ERROR.TYPE we can check the result: it will give 3 if the reference exists and 4 if not. The test statement is nested to an IF formula, where you can see the array of the same numbers we concatenated to the prefix. The false argument of IF is omitted, so the result of the formula will contain numbers and FALSE values only, for example:
{1;2;FALSE;FALSE;5;6;7;FALSE;FALSE;FALSE;FALSE;.....}
The reason of leaving out the false argument (no comma separator is used!) is that we will use SMALL function in the next step, and in SMALL logical values are not counted.
SMALL function will help to choose the numbers from this array. We need a correctly dimensioned array of numbers from 1 to the number of existing sheets - this is the second argument of SMALL.
Solution using IFERROR and AREAS:
`=prefix & LARGE( IFERROR( AREAS(INDIRECT(prefix&ROW(\$1:\$100)&"!A1"))*ROW(\$1:\$100) ,0), ROW(\$A\$1:INDEX(\$A:\$A,SUM(IFERROR(AREAS(INDIRECT(prefix&ROW(\$1:\$100)&"!A1")),0)))) )`
to be array-entered
How does it work?
The base logic is the same, only the test part is different. We used a relatively rarely used formula: AREAS. The argument of the formula is a reference (or in our case: array of references, generated by the INDIRECT function). The result of the formula is the number of areas in the reference. Area is a range of continuous cells or a single cell. In our case, the references contain only one cell (A1) so the result will always be 1 or #REF! error if the sheet does not exist:
{1;1;#REF!;#REF!;1;1;1;#REF!;#REF!;#REF!;#REF!;.....}
After multiplying it by an array of the same numbers concatenated to the prefix, then put it into an IFERROR to replace the #REF!s with 0, we have this array:
{1;2;0;0;5;6;7;0;0;0;0;.....}
Now we can use LARGE to retrieve the non-0 numbers. The second argument of LARGE is built with the same logic as in the first solution for SMALL.
Based on these solutions you can create your own formula for different rules, for example year numbers.
Please remeber, adding a new worksheet does not trigger calculation, but renaming a sheet triggers it, so the formula will be re-calculated after renaming the sheet.
Example
Maybe you have already read the previous post about how to sumif from more sheets. This sheetname-list challenge originates from that post, so here we share a file contains the full model: sumif from more sheets with the possibility of adding new sheets with prefix-based names. Please download the file - you can find the first solution as name: MySheets.
UDF solution
Finally, here is a user defined function with the possibility of excluding sheets from the sheet list. You can add the sheet names to be excluded as a range or a text array. If the argument is empty, the result will contain all the sheets of the workbook.
`Function ListSheets(Optional vExcludeSheets As Variant)`
` Application.Volatile`
` `
` Dim asSheetNames() As String`
` Dim vExcludedName, vWorkSheet As Variant`
` Dim i As Long, bNeed As Boolean`
` `
` For Each vWorkSheet In ThisWorkbook.Worksheets`
` bNeed = True`
` If Not IsMissing(vExcludeSheets) Then`
` If IsArray(vExcludeSheets) Or TypeOf vExcludeSheets Is Range Then`
` For Each vExcludedName In vExcludeSheets`
` If vExcludedName = vWorkSheet.Name Then`
` bNeed = False`
` Exit For`
` End If`
` Next`
` Else`
` If vExcludeSheets = vWorkSheet.Name Then`
` bNeed = False`
` End If`
` End If`
` End If`
` If bNeed Then`
` ReDim Preserve asSheetNames(i)`
` asSheetNames(i) = vWorkSheet.Name`
` i = i + 1`
` End If`
` Next`
` ListSheets = Application.Transpose(asSheetNames)`
`End Function`
Formulas: Igo-r, IngaKris
UDF: Igo-r | 1,469 | 5,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-30 | latest | en | 0.772166 |
https://www.jiskha.com/archives/2012/10/22 | 1,590,682,680,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00577.warc.gz | 787,960,575 | 44,313 | # Questions Asked onOctober 22, 2012
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A car moving 30m/s slows uniformly to a speed of 10 m/s in a time of 5 sec determine 1) the acceleration of the car 2) the distance it moved in 3 sec
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An elevator accelerates upward at 1.2 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the upward force exerted by the floor of the elevator on a(n) 92 kg passenger? Answer in units of N
9. ## Physics
An elevator accelerates upward at 1.2 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the upward force exerted by the floor of the elevator on a(n) 92 kg passenger? Answer in units of N
10. ## physics
A human hair is approximately 55 µm in diameter. Express this diameter in meters. Answer in units of m
11. ## geometry
The volume of a sphere is 4,000 m3. What is the surface area of the sphere to the nearest square meter?
12. ## Physics
Compute the angular momentum of the earth arising from the following motions. (a) Earth's orbital motion around the sun. (b) Earth's rotation on its axis.
13. ## Physics
A current of 60 mA passes through a loudspeaker of resistance 6 Ω. What is the potential difference across the resistance of the loudspeaker? 36.0 V 10.0 V 0.36 V 0.10 V Answer B seems like it would be the obvious answer..
14. ## Physics
An electric kettle operates at 110V. What is the resistance of the kettle if the power is 1.5kW? 8.1 Ω 1.4 Ω 81 Ω 14 Ω I'm stuck between A and B
15. ## Physics
Which of the following is correct for a step-down transformer? Number of turns in the primary coil is less than the number of turns in the secondary coil. Step-down transformer increases the voltage from low value to high value. Number of turns in the
21. A candle is 17 in. tall after burning for 3 hours. After 5 hours, it is 15 in. tall. Write a linear equation to model the relationship between height h of the candle and time t. Predict how tall the candle will be after burning 8 hours. (2 points)
17. ## Physics
the input and output of hydraulic jack are respectively 1 cm and 4 cm in diameter. A lever with a mechanical advantage of 6 is used to apply force to the input piston. How much mass can the jack lift if a force of 180 N is applied to the lever and
18. ## Physics
A girl is skipping stones across a lake. One of the stones accidentally ricochets off a toy boat that is initially at rest in the water (see the drawing below). The 0.094-kg stone strikes the boat at a velocity of 14 m/s, 15° below due east, and ricochets
19. ## Physics
In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer space. From stationary positions they push against each other. Bonzo flies off with a velocity of +1.7 m/s, while Ender recoils with a velocity of −2.7 m/s. (b) Determine the
20. ## Physics
A proton and an electron enter perpendicular to the direction of the magnetic field. The speed of the proton is twice the speed of the electron. What is the ratio of the force experienced by the proton to the electron? 1:2 1:3 2:1 3:1 I would think A, but
21. ## Chemistry I
In determining the heat capacity of a calorimeter, 50ml of water at 56*C is added to 50ml of water at 23*C. After about five minutes of mixing, the final temperature of the solution inside the calorimeter reached 37*C. Calculate the heat capacity of the
22. ## collegemathwordproblem
When a number is decreased by 30%of itself, the result is 28. What is the number?
23. ## soicial studys help plz
1. Who was the French absolute monarch who called himself the "Sun King" to show that he was the center of the French nation? (1 point)Cardinal Richelieu Louis XIII Louis XIV Versailles 2. The Elizabethan Age was considered a golden age due to all of the
24. ## Geometry
The angle of elevation of a ladder leaning against a wall is 45°. If the distance from the base of the ladder to the wall is 15 feet, find the length of the ladder.
25. ## physics
A solid sphere of mass 0.604 kg rolls without slipping along a horizontal surface with a translational speed of 5.25 m/s. It comes to an incline that makes an angle of 38 with the horizontal surface. Neglecting energy losses due to friction, (a) what is
26. ## Math
On a standardized test, Phyllis scored 84, exactly one standard deviation about the mean. If the standard deviation for the test is 6, what is the mean score for the test?
27. ## Physics
By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor, with v1 = 2.75 m/s and v2 = 1.65 m/s. As the plate falls, its momentum has only a vertical component, and no component parallel to the floor.
28. ## physics
John runs at 4.6 m/s for 7.53 s along the x axis. How far does John run? Answer in units of m AND What relationship should be used to solve this situation? 1. x = v − t 2. x = v t 3. x = v + t 4. None of these 5. x = v
29. ## algebra
can anyone please help me set this problem up, I can do the math but I have tried for hours... Air fare club offers membership at $300.00, at least 50 people must join.For every member over 50 their fare will be reduced by$2 for every member. Due to space
30. ## Gen Chem
epsom salts, household ammonia, and rubbing alcohol are also consumer chemicals that could be used in this experiment. what are the chemical formulas for these substances?
31. ## Precalculus
cotx-cos^(3)xcscx
32. ## Physics
Two resistors 10 Ω and 5 Ω are connected in parallel across a 15 V battery. What is the potential difference across the 10 Ω resistor? 5 V 10 V 15 V 20 V I would think the answer would be B, or 15V
33. ## physics
A chair of weight 100N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 43.0N directed at an angle of 35.0^\circ below the horizontal and the chair slides along the floor. Using Newton's laws, calculate
34. ## Physics
A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.20 104 rad/s to an angular speed of 3.14 104 rad/s. In the process, the bit turns through 1.92 104 rad. Assuming a constant angular acceleration, how long would it take
35. ## physics
If a rod is moving at a velocity equal to 1/2 the speed of light parallel to its length, what will a stationary observer observe about its length? The length of the rod will become exactly half of its original value. The length of the rod remains the same.
36. ## trig
Show that the height, h, of the A-frame is equal to the expression 5sinc(1+2cosc). Im not sure what to do.
37. ## Physics
Why is it necessary to collimate the light source before using the prism to disperse the light? A. If white light enters the prism with different angles, some of the blue light will exit the prism with the same angle as the orange light. B. If white light
38. ## math
Write a number sentence for the model. Let one white tile equal +1 and one black tile equal –1. A. –14 – (–6) = –8 B. 14 – 6 = 8 C. –14 – 6 = –8 D. –14 – (–6) = 8 i put d but it was rong
39. ## Organic chem
What gases escaped during the sodium bicarbonate washing for the preparation of Synthetic Banana Oil lab. The isopentyl acetate, acetic acid, isopentyl alcohol, sulfuric acid, water , by-products goes through the sodium bicarbonate wash and the organic
40. ## math
Name a number that can be evenly divided by 6 and 7
41. ## Chemistry
Why is a Gas Chromatography separation more efficient than a fractional distillation? What characteristics must the liquid stationary phase have? describe a method for identifying a compound using GC analysis. If you could answer anyone of these that would
42. ## Physics
A volleyball is spiked so that its incoming velocity of +3.6 m/s is changed to an outgoing velocity of -16 m/s. The mass of the volleyball is 0.27 kg. What impulse does the player apply to the ball? Answer will be in kg*m/s
43. ## Statistics
A set of X and Y scores has SSx=10, SSY=20, and SP=8, what is the slope for the regression equation? a. 8/10 b. 8/20 c. 10/8 d. 20/8 For the regression equation, Y=bX+a, which of the following X,Y points will be on the regression line? a. a,b b. b,a c. 0,
44. ## CHEMISTRY
Not for homework, just practice problems I'm having trouble with. Knowing that one mole of KHP, C8H5O4K, reacts with one mole of NaOH, what mass of KHP is required to neutralize 30.0 mL of the 0.10 M NaOH solution? if 24.5 mL of the 0.10 M NaOH solution is
45. ## Math
Cutting a circle into equal sections of a small central angle to find the area of a circle by using the formula A=pi*r*r
88. ## calculus
compute the maximum profit for the profit function P(x)= x^3/3-9/2x^2+8x
89. ## Chemistry
A student starts with 25 grams of sodium chloride as the limiting reactant and obtained 17.9 grams of dry pure sodium carbonate. What is the theoretical yield of sodium carbonate? What is the percent yield obtained by the student?
90. ## Algebra
2 cyclists travel opposite on circular trail that is 5 miles long. one cyclist travels 12 miles per hour, the other travels 18 miles per hour. How long before they meet?
91. ## physics
Your projectile launching system is partially jammed. It can only launch objects with an initial vertical velocity of 42.0 m/s, though the horizontal component of the velocity can vary. You need your projectile to land 214 m from its launch point. What
92. ## Physics
007 (part 1 of 2) An elevator accelerates upward at 1.2 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the upward force exerted by the floor of the elevator on a(n) 92 kg passenger? Answer in units of N 008 (part 2 of 2) If the same elevator
93. ## geometry
The volume of 2 similar solids are 27 ft3 and 216 ft3. The surface area of the larger solid is 300 ft2. What is the surface area of the smaller solid? Round your answer to the nearest tenths place
94. ## Chemistry 9
a) A 2.56 balloon contains 3.66 g of gas. Find the density b) A 45.3 g sample placed in a graduated cylinder causes the water level to rise from 25.0 mL to 41.8 mL . Find the density
95. ## chemistry
Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium? I got Qsp 1.38 * 10^-8 and Ksp = 3.45*10^-11 does that seem right? and does this mean fluride concentration will be soluble in water containing calcium?
96. ## Physics
007 (part 1 of 2) An elevator accelerates upward at 1.2 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the upward force exerted by the floor of the elevator on a(n) 92 kg passenger? Answer in units of N 008 (part 2 of 2) If the same elevator
97. ## Math
A shaded area reprsents approximately 95% of the scores on a standardized test. If these scores ranged from 78 to 92, whic could be the standard deviation?
98. ## CHEMISTRY
Mass (before reaction): test tube + HCl(aq) + stir bar + capsule 26.600 g Mass (after reaction): test tube + HCl(aq) + stir bar + capsule 25.300 g Volume of water displaced from the squirt bottle 148 mL Temperature of the CO2(g) 287.4 K Pressure (atm)
99. ## Science Help!
How much heat is needed to cause the following irreversible reaction? 200 g of liquid H2O at 50°C is heated to form 200 g of steam at 120°C
100. ## PHYSICAL SCIENCE
How do the pressure variation with depth affect our ability to explore the ocean?
101. ## history 103
The voyages of Zheng He reveal a powerful, accomplished, and restored Confucian state and civilization in China that was at least equal to, but very different from, emerging early modern Western civilization. List and describe the major factors that made
102. ## Chemistry 20 titration
1) oi50 . tinypic . com / ejh6vm . jpg The most suitable indicator for the titration is: A)methyl violet B)chlorophenol red C)methyl orange D)phenolphthalein 2-A base is titrated with an acid solution until the indicator changes colour. This change in the
103. ## Chemistry
A student starts with 25 grams of sodium chloride as the limiting reactant and obtained 17.9 grams of dry pure sodium carbonate. What is the theoretical yield of sodium carbonate? What is the percent yield obtained by the student?
104. ## Punctuation and period
Commas are used to set off _____ phrases. A) essential B) nonessential I answered nonessentials
105. ## chemistry 20 MC questions
1- The equipment needed to carry out a titration is: A) hot plate, pipet, and filter paper B) heat source, test tube, and filter paper C) heat source, pipet, and burette D) erlenmeyer flask, pipet, and burette 2- A student correctly predicted that 10.0 g
106. ## Physics
:'( Please help me....? (7 points) 2. Contrast the metric units of distance, time, and speed. Describe the relationship between them. Answer: (11 points) 3. A microbiologist measures the speed of a swimming bacterium. The bacterium covers a distance of
107. ## physics
two kids push on a 10kg box. one pushes 100N to the west and the other pushes north at 200N. what is the acceleration of the pbject?
108. ## Physics help PLZ!!
Estimate the energy required from fuel to launch a 1465 kg satellite into orbit 1375 km above the Earth’s surface. Consider two cases: (a) the satellite is launched into an equatorial orbit from a point on the Earth’s equator, and (b) it is launched
109. ## Science
How much heat must be removed from 500 grams of water at 0°C to form ice at 0°C?
123. ## English
essy in mockingbird how Caitlin over Family struggle abput 700 word see example of different essay
124. ## physics
When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm
125. ## Chemistry
25.0mL of an acidic 0.100 mol/L tin (II) chloride solution required an average volume of 12.7 mL of potassium dichromate solution for a complete reaction. The amount of concentration of the potassium dichromate solution is _______ mmol/L How do you go
126. ## organic chemistry
a careless organic chemistry student was performing "competing nucleophiles in nucleophilic substitution reactions" experiment , and left his container of tert-butyl chloride and tert-butyl bromide open to the air for several hours. what happened to the
127. ## math
23 divided by 6=3.83333333333 the 23 is my points and i divide it to find my gpa how do i round the answer
128. ## Chemistry
In the theoretical reaction 2 B(aq) + 3 D(aq) ---> E(aq) + 4 F(aq) the concentration of D(aq) is 0.40 mol/L. If 25.0 mL of aqueous B is needed to react completely with 30.0 mL of aqueous D, the concentration of B(aq) is A-0.48 mol/L B) B-0.32 mol/L C)
129. ## geometry
A square pyramid has base edges 10 in. long and a height of 4 in. Sketch the pyramid and find its surface area. Round your answer to the nearest tenth
130. ## math
sharon is diving her green and blue rock collection into bags. Each bag will contain the same number of each color of rock. How many rocks of each color will be in each bag? there are 16 green marbles and 24 blue
131. ## chemistry
What is the chemical formula of the compound that forms between lithium (Li) and phosphorus (P)?
132. ## college percentages
One number exceeds another by 24. The sum of the numbers are 58. What are the numbers? please help!
133. ## Math
Which Statements of congruence are true and which are false and why? 177 _= 17 (mod 8) 871 _= 713 (mod 29) 1322 _= 5294 (mod 12) 5141 _= 8353 (mod 11) 13944 _= 8919 (mod 13) 67 x 73 _= 1 x 3 (mod 5) 17 x 18 x 19 x 20 _= 4! (mod 8) 83 (144 power) _= 15 (144
134. ## cis210
Assignment 1: Inventory Management Systems Due Week 2 and worth 100 points Your sister owns a small clothing store. During a conversation at a family dinner, she mentions her frustration with having to manually track and reorder high demand items. She
135. ## Calculus
Air pressure at sea level is 30 inches of mercury. At an altitude of h feet above the, air pressure, P, in inches of mercury, is given by, P =30e^(-3.23x10-5h) a)Find th equation of the tangent line at h=0. b)A rule of thumb is given by travelers is that
136. ## Algebra
Quick question. The sum of two numbers is 48. One number is 3 times as large as the other. What are the numbers?
137. ## PHYSICS
When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm.
138. ## Chemistry
What volume of a 0.332 M hydrobromic acid solution is required to neutralize 29.6 mL of a 0.155 M calcium hydroxide solution ?
139. ## physics
A bullet is fired straight up from a gun with a muzzle velocity of 253 m/s. Neglecting air resistance, what will be its displacement after 2.9 s? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
141. ## physics help i still dont get it
A picture of width 43.0 cm, weighing 52.0 N, hangs from a nail by means of flexible wire attached to the sides of the picture frame. The midpoint of the wire passes over the nail, which is 2.50 cm higher than the points where the wire is attached to the
142. ## Chemistry
lime, Ca(OH)2 can be used to neutralize an acid spill.A 5.06-g sample of Ca(OH)reacts with excess of hydrochloric acid:6.74 g of calcium chloride is collected what is the percent yield of this experiment Ca(OH)2 + 2HCI ====> CaCl2 + 2H2O
143. ## Physics
A car traveling in a straight line has a velocity of 6.64 m/s at some instant. After 5.55s, its velocity is 10.9 m/s. What is its average acceleration in this time interval? Answer in units of m/s
144. ## Punctuation
Select the correctly punctuated sentence. A) No, I do not agree with that diagnosis. B) No I do not agree with that diagnosis. C) No I do not agree, with that diagnosis. My answer is A
145. ## chemistry
How many grams of solid potassium hydroxide are needed to exactly neutralize 10.8 mL of a 1.25 M hydrobromic acid solution
146. ## science
A balloon with an initial volume of 2.8L at a temperature of 296K is warmed to 375K . What is its volume at the final temperature? (Assume constant pressure.)
147. ## Math
What is the volume in cubic feet, of a box that is 3ft long, 8th wide, and 16ft high?
148. ## lens surface
- a spherical generator is used to produce a -15.00D surface on glass of refractive index 1.80. The diameter of the cutting tool is 80mm and the radius of the cutting surface is 4 mm. What is the angle between the axis of the tool and the axis of the lens?
149. ## Physics
How much heat must be removed from 500 grams of water at 0 degrees C to form ice at 0 degrees c?
A biologist finds that 32% of fruit flies have short wings and 27% have eye defects. Assuming these two events are independent, what is the probability that a randomly selected fruit fly has neither short wings nor eye defects?
151. ## english2
Bee balm, coneflowers, and black-eyed Susans, brightly colored flowers with darker bulbous centers, contrast beautifully when planted together. is it correct
152. ## math
23 divided by 6=3.83333333333 the 23 is my points and i divide it to find my gpa how do i round the answer
153. ## Physics
A simple pendulum consists of a light rod of length L = 0.9 m anchored at a fixed point (like the ceiling), and a small mass m = 0.5 kg attached to the free end. The configuration is moved to an angle θ = 39 degrees from the vertical and released from
16. Which of the following would be the best revision of these sentences? "Our present receivables are in line with last year's. However, they exceed the budget. The reason they exceed the budget is that our goal for receivable investment was very
155. ## Math
Complete the analogy. Monday : Tuesday :: Saturday : _
156. ## Chemistry
25.0mL of an acidic 0.100 mol/L tin (II) chloride solution required an average volume of 12.7 mL of potassium dichromate solution for a complete reaction. The amount of concentration of the potassium dichromate solution is _______ mmol/L How do you go
158. ## Abnormal Psychology
In comparison to children with autism, children with schizophrenia A. tend to be younger at diagnosis. B. show a more chronic and declining course. C. show similar social and language deficits. D. show less intellectual impairment. I think its C or D but
159. ## physics
When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm
160. ## science
you are sledding on a snow-covered hill at a speed of 6 m/s there is a drop that increases your speed to 12 m/s in 4 s. find your acceleration? i need help answer?
161. ## college math word problem
A new car worth 45K is depreciating in value by 5K per year. How can I write a formula that models the cars' value,y, in dollars, after x years?Then using the formula determine after how many years the cars' value will be 9K.
162. ## Spanish Beg. 1 - help!
spanish words to talk about names so far I got... comom te llamas and me llamo is there a website that list all spanish words to talk about names I just need two more please help!
163. ## Accounting
I am given several hypothetical amounts and interest rates and need to find the amount of the rental payments. (Lessee-Lessor Entries; Sales-Type Lease) On January 1, 2011, Palmer Company leased equipment to Woods Corporation. The following information
164. ## math
1,405ft 1 and 3/4in.cut into 1/2 in.thick slices.how many slices are there?
165. ## Chemistry
What is the ionic equation for hydrolysis reaction of NH4C2H2O2 ? Thanks.
166. ## physics
A 81.0 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 28.0 m/s. If the player was moving upward with a speed of 4.40 m/s just before impact. (a) What will be the speed of the ball
167. ## physics
When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm
168. ## Can comeone check my answer for me. Many Thanks!
A set of data is normally distributed with a mean of 1000 and standard deviation of 100. · What would be the standard score for a score of 1100? · What percentage of scores is between 1000 and 1100? · What would be the percentile rank for a score of
169. ## physics
wile e coyote pushes a 1500n rock toward the edge of a cliff with the road runner below. he gives a rock acceleration of .8m/s. what is the net force of the rcok?
170. ## physics
wile e coyote pushes a 1500n rock toward the edge of a cliff with the road runner below. he gives a rock acceleration of .8m/s^2. what is the net force of the rcok?
171. ## collegemathwordproblem
One number exceeds another by 24. The sum of the numbers are 58. What are the numbers?
172. ## physics
When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm
173. ## math word fu
A new car worth 45K is depreciating in value by 5K per year. How can I write a formula that models the cars' value,y, in dollars, after x years?Then using the formula determine after how many years the cars' value will be 9K.
174. ## algebra
a barrel contains 163 gallons of water and is being drained at a constant rate of gallons per hour write an equation that models the number of gallons g after t hours
175. ## Physics
The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is 0.015 kg, and its velocity is +726 m/s. Her
176. ## physics need help bad
A picture of width 43.0 cm, weighing 52.0 N, hangs from a nail by means of flexible wire attached to the sides of the picture frame. The midpoint of the wire passes over the nail, which is 2.50 cm higher than the points where the wire is attached to the
177. ## chemistry
Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 70. cm3. If the combustion of this mixture releases 885 J of energy, to what volume will the gases expand against a constant pressure of 645 torr if all the
178. ## Physics
A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 92.5 kg. The mass of the rock is 0.310 kg. Initially the wagon is rolling forward at a
179. ## Math Calculus 2
At a certain instant an aircraft flying due east at 240 miles per hour passes directly over a car traveling due southeast at 60 miles per hour on a straight, level road. If the aircraft is flying at an altitude of .5mile, how fast is the distance between
180. ## physics
Just after opening a parachute of negligible mass, a parachutist of mass 85.5 kg experiences an instantaneous upward acceleration of 1.12 m/s2. Find the force of the air on the parachute.
181. ## input/output tables
I have a blank table to fill in.It says divide m by 3 m÷3
182. ## Punctuation
Select the correctly punctuated sentence. A) Depression, often a genetic disorder, can be safely treated with medication. B) Depression, often a genetic disorder can be safely treated with medication. C) Depression often a genetic disorder can be safely
183. ## Discrete Math
A bag contains 4 red, 6 white and 9 green marbles. How many ways can 3 red, 2 white and 1 green marbles be selected? Do you used permutation or combination, give your reasoning.
184. ## physics
A 81.0 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 28.0 m/s. If the player was moving upward with a speed of 4.40 m/s just before impact. (a) What will be the speed of the ball
3.4t+0.08=11
186. ## Poltics
Who is the President of The Department of Human Services for Wayne County in Michigan?
187. ## Word problem
A new car worth 45K is depreciating in value by 5K per year. How can I write a formula that models the cars' value,y, in dollars, after x years?Then using the formula determine after how many years the cars' value will be 9K.
188. ## xeco/212
Wall Street Journal in which economists are quoted using positive and normative statements. o Provide one quote and its context for both a positive and normative statement. Explain your choices. o Evaluate the statement choices of your fellow students.
189. ## prealgebra
0.85km/s=m/min
190. ## Punctuation
Select the correctly punctuated sentence. A) Dr. Christiaan Barnard, who performed the first heart transplant, is a well-known cardiologist. B) Dr. Christiaan Barnard, who performed the first heart transplant. Is a well-known cardiologist. C) Dr.
191. ## Punctuation and period
The most commonly used mark of punctuation is the _____ A) comma B) period I answered B
192. ## Physics
A 42.4-kg boy, riding a 2.29-kg skateboard at a velocity of +5.37 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 6.30 m/s, 9.09° above the
193. ## chemistry
how many moles of NaBr are produced from reacting 20.05 moles of AuBr3 with excess NaCN? AuBr3 +NaCN -> Au(CN)3 + NaBr
194. ## physics
A 16.0 g bullet is fired horizontally into a 87 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 168 N/m. The bullet becomes embedded in the block. The bullet block system
195. ## Physics
The length of the rod will become exactly half of its original value. The length of the rod remains the same. The length of the rod will decrease. The length of the rod will increase. Wouldn't the length double, thus being answer D? No one has answered
196. ## phys!!! how do you do this
Just after opening a parachute of negligible mass, a parachutist of mass 85.5 kg experiences an instantaneous upward acceleration of 1.12 m/s2. Find the force of the air on the parachute.
197. ## Math
Sare estimated 32x45 by using 30x40. How could Sara make a more accurate estimate?
198. ## Math
Use mathematical induction to prove that 2^(3n) - 3^n is divisible by 5 for all positive integers. ThankS!
199. ## physics
How many significant figures are in 24.010
200. ## chemistry
onic compounds have names that... (choose all that apply) might contain Roman numerals. might start with ammonium. start with the name of an anion. have prefixes, like di or penta. start with the name of a nonmetal.
201. ## physics
John runs at 4.6 m/s for 7.53 s along the x axis. How far does John run? Answer in units of m
202. ## Grammar
Select the correctly punctuated sentence. A) The implant was ruptured, it was removed, the capsule was irrigated. B) The implant was ruptured it was removed, the capsule was irrigated. C) The implant was ruptured, it was removed the capsule was irrigated.
203. ## chemistry
Is this equation correctly balanced? If not why. Al2S3+H2O----> Al2S3+3H2O---->Al2(SO)3+3H2
204. ## Word problem
A new car worth 45K is depreciating in value by 5K per year. How can I write a formula that models the cars' value,y, in dollars, after x years?Then using the formula determine after how many years the cars' value will be 9K.
205. ## MATHS
how much money would you have after 20 weeks if you were given 5c the first week, 10c the second, 20c the third, 40c the fourth, and so on?
206. ## physics
a wagon is pulled at a constant speed at an angle of 46 dregress with a force of 100N. ehat is the force of friction between the wagon and the ground ?
207. ## language
what does scv, s, f, and scs mean when identifying sentences?
208. ## Physics
Which doesn't describe motion? mass or position eg. The position of a golf ball standing still on a tee at 7:30 A.M. Sunday morning. or The mass of Voyager as it plunges into a black hole.
Just after opening a parachute of negligible mass, a parachutist of mass 85.5 kg experiences an instantaneous upward acceleration of 1.12 m/s2. Find the force of the air on the parachute.
210. ## Physics
A railroad car with a mass of 1.86 104 kg moving at 2.80 m/s joins with two railroad cars already joined together, each with the same mass as the single car and initially moving in the same direction at 1.40 m/s. (a) What is the speed of the three joined
211. ## collegemathwordproblem
When 80% of a number is added to the number, the result is 252. How do I find the number?
212. ## Word problem
One number exceeds another by 24. The sum of the numbers are 58. What are the numbers?
216. ## chemistry
each element Na, Mg, Al, Si, P, S has a very big jump in ionization energies. Please explain what causes these jumps in energy?
217. ## Sci 207 wk 2 lab 2
Which species appear to benefit from increasing fragmentation? Which species are negatively impacted? Based on your knowledge of environmental science, what could explain these observations? Support your answer with specific results from your data and
218. ## math
if you have a 6 foot long board and it's cut into 4/5 pieces how many pieces can you have a what portion of the board is left over
219. ## physics
A 59.0-kg skater is traveling due east at a speed of 1.80 m/s. A 61.0-kg skater is moving due south at a speed of 6.20 m/s. They collide and hold on to each other after the collision, managing to move off at an angle south of east, with a speed of vf. Find
220. ## Math
Find all solutions of the equation (sec(x))^2−2=0 The answer is A+Bk where k is any integer and 0
221. ## Calculus
The population (in thousands) of the Tzitzit bird is well described by a function of the form P(t) = ae^kt, where t is the time in years and a and k are constants. If the population was 10 thousand when t-0 and 300 thousand when t=3, determine the
222. ## accounting
Okay so I need help Identifying the reporting issues in this case with references to the conceptual framework pyramid, such as revenue recognition, full disclosure, historical cost, etc. So this gym offers one year memberships. Members can use any of the
223. ## chemistry
What is the final temperature of water when 2.70 grams of calcium chloride is dissolved in 25 mls of water at 25.5 Celcius.
224. ## Math
Monday, Mr. C stacked a display case of 80 heads of lettuce. By end of day, some were sold. Tuesday Mr. C counted lettuce left and decided to add an equal number of lettuce (**hint - he doubled the leftovers). At end of Tuesday, he sold the same number as
225. ## Punctuation
Select the two (2) correctly punctuated sentences. A) I also explained antepartum fetal surveillance, i.e., fetal heart rate monitoring and assessment of amniotic fluid volume. B) I also explained antepartum fetal surveillance, i.e. fetal heart rate
226. ## chemistry
how many moles of NaBr are produced from reacting 20.05 moles of AuBr3 with excess NaCN? AuBr3 +NaCN -> Au(CN)3 + NaBr
227. ## calculus
What is the degree 5 coefficient of the polynomial p(x)= x^2 (x+1) (x+3x^3 + 4x^4)?
228. ## Punctuation
Select the correctly punctuated letter closing. A) Very sincerely yours, Joseph Blow, President B) Very sincerely yours, Joseph Blow President C) Very sincerely yours Joseph Blow, President I answered A
229. ## Physics
How many calories will take to convert 5 grams of 90C degree water to 0 C degree?
230. ## Punctuation
Select the correctly punctuated sentence. A) He arrived at the hospital with a heavily, bleeding gunshot wound. B) He arrived at the hospital with a heavily bleeding, gunshot wound. C) He arrived at the hospital with a heavily bleeding gunshot wound. I
231. ## CALC 251
Suppose that f(x)= (e^x)/(x^2+16. Find f'(0).
232. ## math
I need a set of 12 numbers that have maximum 8, range 6, mode 6, and median 5. Any answers?
233. ## Chemistry
What is the molar concentration of nitrate ions in a 0.161M magnesium nitrate(aq) solution?
234. ## Pre-Cal
State the transformations. Identify holes, vertical asymptotes, and horizontal asymptotes of each. (I'm fine with the asymptotes, just need help with stating the trandformations) f(x)= (3x+9)/(x+2)
235. ## Writing
I have to do a research paper & I want my topic to be on Pixar's influences on the world. My thesis: Pixar is a positive influence on the world. My supporting evidences - 1) Has helped the animation industry 2) Has helped the film industry 3) Has inspired
236. ## physics
A driver of a car traveling at 14.5 m/s applies the brakes, causing a uniform deceleration of 1.7 m/s 2 . How long does it take the car to accelerate to a final speed of 9.50 m/s? Answer in units of s
237. ## Social Studies
List the unique circumstances of the interwar period that may have led to ultranationalism.
238. ## math
Is this the right answer? 42+x-3(x-2)+11 42+8-3(8-2)+11 47×6+11 282+11 293
a particle moves in a straight line such that its position x from a fixed point 0 at time 't' is given by x= 5 + 8sin2t + 6cos2t 1. Find the period and amplitude of the particle. 2. Find the greatest speed of the particle. Could you please explain the
240. ## Finite Math
A fair coin is flipped. If the flip results in a head, then a marble is selected from an urn containing 1 red, 4 white, and 9 blue marbles. If the flip results in a tail then a marble is selected from an urn containing 3 red and 3 white marbles. If the
241. ## science
These are the 3 questions that will help me study for my test. i need help understanding each of these thankyou :) Distinguish between microevolution and macroevolution. Explain how forest organisms (producers and consumers) separated by a highway could
242. ## Calculus
Im finding local Min and max of x^2/x-2 I know I take the derivative of the first one but how do i derive this one. And the second one especially don't know how to do.
243. ## Punctuation
Select the correctly punctuated sentence. A) It was an exhausting complicated surgery. B) It was an exhausting, complicated surgery. C) It was an exhausting, complicated surgery My answer is B
244. ## physical science
how much heat is absorbed by contacting 10 grams of 100 degree C steam by a students hand.
245. ## Punctuation
_____ are primarily used to separate sentences. A) commas B) periods My answer is B
246. ## Physics
A massless string connects three blocks as shown below. A force of 24 N acts on the system. What is the acceleration of the blocks and the tension in the rope attached to the 2 kg block?
247. ## Chemistry
how many grams of hydrogen occupy 40 ml at 300.15 K and 710 torr?
254. ## math
Find the second derivative. f(x)=(2x^2+4)^7/2
255. ## Social Studies
What national interests lead to ultranationalism?
256. ## english-dr.faustus
the warning Dr.Faustus teaches is? A-the devil is evil B-do not wish for more than heavenly power permits C-the end will justify the means D-both A and C
257. ## Calc
Find all solutions of the equation (sec(x))^2−2=0 The answer is A+Bk where k is any integer and 0
258. ## Precalc
#1. (3x^3-4x^2-3x+4)/(x^3-5x) my answers: y-int: NONE x-int: 1, -1, and 4/3 x asymtope: x= 0 and +- square root of 5 y asym: y=3 If it crosses horiz asym: idk i need help on this #2. (x^4-7x^2+12)/(x^2-5x+4) x-int: -2, 2, and +- square root of 3 y-int: 3
259. ## Calculus 1
Find the derivative of y with respect to x. y=(x^6/6)(lnx)-(x^6/36) So far this is what I've gotten: y=(x^6/6)(lnx)-(x^6/36) y=(1/6)x^6(lnx)-(1/36)x^6 y'=(1/6)x^5(1/x)+lnx(x^5)-(1/6)x^5 What do I do now?
260. ## MATH
HELP? Exam in 20 mins -dont' have any idea how to do this question; please helpp.. anyone! PLEAZE A.S.A.P. minimize p= 15x +18y subject to x+2y= 36 x>=0 y>=0
261. ## math
Can these functions be defined for all real numbers? y= e^cos(x) y= ln(sin(x)) y= sqr(cos(x)+2) I'm trying to find some systematic way to test these, but I'm not sure how I can proe it for any possible input.
262. ## physics help
Just after opening a parachute of negligible mass, a parachutist of mass 85.5 kg experiences an instantaneous upward acceleration of 1.12 m/s2. Find the force of the air on the parachute.
263. ## Chemistry Lab
Does the pressure exerted by a gas depend on its molar mass?
264. ## Physics
Three objects with masses m1 = 7.4 kg, m2 = 11 kg, and m3 = 18 kg, respectively, are attached by strings over frictionless pulleys (M1 hangs off the left side of the table and M3 hangs off the right side of the table with M2 between them on the table). The
265. ## Calculus
simplify log2(4x^2 2^x)
266. ## Calculus (math)
Write a formula for f^-1 (x) if f(x) = (x+5)/(3x-4) and state the domain of f-1
267. ## Math
Eliminate the parameter in the pair of parametric equations to find the corresponding rectangular question. x=h+ asecQ y=k+ tanQ
268. ## math
For the functions f(x)=3x-12 and g(x)=(x/3)+4, find each of the following a) f o g (x) b) g o f (x)
269. ## English
Hi i need help with an example of a Nominative Absolute sentences using an example from Emily Bronte's wuthering heights
270. ## Physics
A 59.1 N/m spring is unstretched next to the 90 degree angle of an incline, and the system is released from rest. The mass of the block on the incline is m1 = 21.4 kg. (Neglect the mass of the pulley). If the coefficient of kinetic friction between m1 (the
271. ## College math
What do these mean? They are supposed to represent things but I don't understand them. C=2╥r for r, S=P+Prt for t, and A=2lw+2lh+2wh for h
272. ## English
I have to come up a reason for my claim help please? Thanks! My claim: technology is harmful to children and teens.
273. ## calculus
Find the equation of the tangent line to the graph of the function f(x)=x-3/3x-5 at x=1
33x650x-43
275. ## math
which measures is closest to 59mm? A 50mm B 65mm C 6mm D 6cm
276. ## Chemistry
I titrated 0.25 grams of soda ash with 43.7 mL of HCl (0.115M) and 25 mL of H2O. I am getting over 100% sodium carbonate in my answer here are my steps that I took to get the answer can somebody help me see what I did wrong? mols HCl=(0.115M)x
277. ## mathe
aufgabe 1.rechne für die folgenden massstäbe aus,welche Länge fünf zentimeter auf der wirklichkeit haben. massstab 1--1000000-- massstab 1--40000000,massstab1--50000--massstab--10000-- massstab 1--500000
278. ## math
which length is between 7m and 9m A 80 cm
279. ## Health 8
what happen to a person tolerance as they drink regularly?
280. ## CALC 251
Let f(x)= [5x^(1/4)]*(x^3-8) Evaluate at f'(x).
281. ## Calculus
Find the limits if they exist lim_x->0 x/|x| lim_x->0+ sqr(x+4)
282. ## calculus
Find the limits lim_x->2 x^3+x lim_x->2 (x^2-4)/(x-2)
283. ## science
Is a mineral a crystal?
284. ## chem
a new compound composed of just Hex-Nuts and bolts is formed.A 13.7 Kg sample of the new compound is separated into its component elements with the following result 3.84 kg. of hex-nuts and 9.90 kg of bolts. 1. How many hex-nuts are present in the 13.7 kg
285. ## need help
A picture of width 43.0 cm, weighing 52.0 N, hangs from a nail by means of flexible wire attached to the sides of the picture frame. The midpoint of the wire passes over the nail, which is 2.50 cm higher than the points where the wire is attached to the
286. ## Math
How do you find the percentage of a number? I have no clue how you do that. An example I have is 850 students are attending a camp, 38% of them play musical instruments. What number of people play instruments? I don't get how to solve that and here's
287. ## maths
BEGGING FOR HELP!!! EXAM IN 20 MINS LOST ON HOW TO DO THIS Q. Company has \$ 20 million dollars for two types insurance. Insurance A-homeowners have a 10% annual return. Insurance B-Auto has a 12% annual return. The total amount of Homeowners Loans should
288. ## Finite Math
A polygraph machine correctly identifies a lie 81% of the time, and incorrectly identifies a true statement as a lie 8% of the time. If a person being examined with the machine lies 10% of the time, what is the probability that a statement identified by
289. ## Physics
How many significant figures are in 1.30470
290. ## Grammar
The plan was to quickly attack England and afterwards Turkey. Is the sentence correct?
291. ## math
a disk has an area of 250 square inches.what is theradius of the disk?
292. ## MATH
Given Pr(a)= 0.9, Pr(C|A)= 0.78 and Pr(D|B) = 0.38. find a. (A|C) = b. (A|D) = c. (B|C) = d. (B|D) =
293. ## Computers
How can computers affect our health?
294. ## calculus
Left f(x) = 1/x+2 and g(x) = x-1/x. Simplify as much as possible and sstate the domain. f 0 g(x) g 0 f(x)
295. ## math
if there are 7 boxes of items and the least amount of items in any one of the boxes is 2 and the greatest amount in any one of the boxes is 4. What are all the possible total amounts of items in the 7 boxes combined?
296. ## math/algebra
How do I find the quotient of 5/22 & 6/11?
What do these mean? They are supposed to represent things but I don't understand them. C=2╥r for r, S=P+Prt for t, and A=2lw+2lh+2wh for h
298. ## math
height of a stack of 30 sheets of paper A 2mm B 2m
299. ## math
which length is between 7m and 9m A 80 cm B 8km C 74cm D 740cm
300. ## Algebra 2
To rationalize the denominator of 4/(2+√8), Brittany multiplied by (2-√8)/(2-√8) and Justin multiplied by (1-√2)/(1-√2). Explain why both are correct.
301. ## Physics
How many calories it take to elevate the temperature of 10 gr of water-from 26C to 50C?
302. ## Math
Solve using Augmented Matrix Methods -4(x1) + 6(x2) = -8 6(x1) - 9(x2) = 12
303. ## calculus
Find the derivative of (x^2)(e^-3x)
304. ## English
I need a paragraph of 50sentences?
305. ## Social Studies
Which underlying circumstances occurring during the interwar period of 1919 to 1939 may have influenced whether or not nationalist interest was nationalist or ultranationalist?
306. ## Social Studies 8 - help!
What kind of business(es) did Vanderbilt control (location, dates number of employees, etc.)?
307. ## math
5n+2 divided by 6 n is -4 5×-4+2÷6 -20+2÷6 -19÷6
308. ## Maths
Question: I graphed -2x+y=-6 Y=-(1/3)x+1 & w/ the options No Solution, Infinante Solutions, or ( , ) I found the answer is (3,1) am I correct? What would be a better way to double check my answers?
309. ## algabra
why are prime numbers beter when writing codes?
310. ## math
(3x+1)^3/(3x+1)^3 +1 use the chain rule to complete the derivative
311. ## Calc
Set the first derivative and second derivative of x^2/x-2 equal to 0. Solve for 0.
312. ## Biology (cell-molec)
You wish to perform a restriction digest in a 20uL volume. Your DNA is 0.20 ug/uL and you wish to digest 1.0 ug. Your restriction enzyme is provided as 0.2 U/uL. Fill in the appropriate volumes: H2O, 10Xbuffer, DNA, and restriction enzyme. Please show
313. ## math
Angelique has five more than 3times the number of movies passes that she had last week. Let x represent the number of movie passes that she had last week. X(5x3)=15x
314. ## Punctuation
Select the correctly punctuated sentence. A) His greatest fears were a stroke, a heart attack, and blindness. B) His greatest fears were a stroke a heart attack and blindness. C) His greatest fears were a stroke, a heart attack, and blindness I answered A
315. ## math
if f(x)=3/x+x^3 and g(x)=1-x^2 find d/dx f(g(x)).
316. ## social studies
what is venezuela's human development index
317. ## social studies
what challenges face venezuela today
318. ## math
distance for a 1-hour bike ride A 800m B 8km
319. ## history
what are some of the natural resources found at the grand canyon
320. ## math
How can i get on xtramath
321. ## College chemistry
When your body metabolizes amino acids, one of the final end products is urea, a water so liable compound that is removed from the body as urine. Why is urea soluble in water, when hexamide, a related compound is not?
322. ## math
write two equations that prove the associative property is not true for division
323. ## south piedmont
convert to a mixed number 89/6
324. ## Math
Estimate the sum by rounding: 7.68+3.52
325. ## College chemistry
When your body metabolizes amino acids, one of the final end products is urea, a water so liable compound that is removed from the body as urine. Why is urea soluble in water, when hexamide, a related compound is not?
326. ## Health 8 (2nd question)
Describe the effects of alcohol Answer - There were 283 deaths recorded as alcohol-related and could cause blackouts, memory loss and anxiety. Am I correct and is my answer is good????
327. ## math
what is the prime factorization of 1 bilion using exponents?
328. ## math
if there are 7 boxes of items and the least amount of items in any one of the boxes is 2 and the greatest amount in any one of the boxes is 4. What are all the possible total amounts of items in the 7 boxes combined?
329. ## Social Studies 8 - help!!!!! (2nd question)
What organizations or buildings named after cornelius vanderbilt? So far I only got Vanderbily University is there anything else????? I google it and that's all it gave me please help me!!!!!
330. ## Science 207
As we reach a possible peak in oil production, what alternatives are forming? Consider the role of OPEC, perhaps in reorganizing and monitoring oil production. What do you think is the most viable way to address after effects of this possible peak?
331. ## math
3a+2-5a=-14 combine like terms right?
332. ## voltage drop
12v r1=10 r2=15 r3=2 r3=18 how would one solve this
333. ## mathURGENT!!!!
the midpoints of the sides of a triangle are (1,1),(4,3),and (3,5). find the area of the triangle. what is the answer and can u explain how u got it step by step? please...
334. ## english
they spent a year in france is an example of how fitzgerald
335. ## Math
2x-1/3 +3 or=6
336. ## math
4(3) exponent is 2
337. ## Math
2(2+7)+(6x-5)=37
338. ## math
what is 37% of 100
339. ## science
at what age does brain stops developing
340. ## math
Suppose f is a linear function such that f(1) = -1 and f(3)=2. Then f(4)=?
341. ## math
find the first derivative y= 2/5square root2x+1
7(s-5)=42
343. ## math
42+×-3(×-2)+11 × IS 8
-4b-5+2b=10
345. ## math
(-7)(-4)(-2)=-56
346. ## vocabulary
very few starting pitchers have the to pich well for nine innings.
23×17
348. ## Math
2(2+7)+(6x-5)=37
349. ## Math
2(2x+7)+(3x-5)=37
350. ## Grammar
She is fascinated to monkeys. Or She is fascinated by monkeys? Which do I use? To or By?
351. ## math
5n+2÷6 N is -4 5×-4=20+2=22÷6 3 4/6
1.48=w+x/10 | 13,661 | 50,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-24 | latest | en | 0.892468 |
https://yokolet.com/algo/strings/2022-11-02-word-ladder | 1,695,332,837,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506045.12/warc/CC-MAIN-20230921210007-20230922000007-00006.warc.gz | 1,208,579,287 | 10,562 | Published: Nov 2, 2022
Hard Breadth-First Search String
## Introduction
When words are given, and each character in the word should be replaced and checked one by one, the breadth-first search is a good algorithm to find the answer. However, the problem is not just the breadth-first search. It needs some more tweak to run faster. The solution here uses two sets to save word path from start and end. While checking one character changed words one by one, it switches the begin and end word sets. By doing this, the process will quickly comes to the end.
## Problem Description
A transformation sequence from word `beginWord` to word `endWord` using a dictionary `wordList` is a sequence of words `beginWord -> s1 -> s2 -> ... -> sk` such that:
• Every adjacent pair of words differs by a single letter.
• Every `s[i]` for `1 <= i <= k` is in wordList. Note that beginWord does not need to be in `wordList`.
• `sk == endWord`
Given two words, `beginWord` and `endWord`, and a dictionary `wordList`, return the number of words in the shortest transformation sequence from `beginWord` to `endWord`, or 0 if no such sequence exists.
Constraints:
• `1 <= beginWord.length <= 10`
• `endWord.length == beginWord.length`
• `1 <= wordList.length <= 5000`
• `wordList[i].length == beginWord.length`
• `beginWord`, `endWord`, and `wordList[i]` consist of lowercase English letters.
• `beginWord != endWord`
• All the words in `wordList` are unique.
## Examples
``````Example 1
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.
``````
``````Example 2
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
``````
## Analysis
The solution starts from checking the end word is in the given word list. If not, we don’t do any, just return 0. Instead of a queue, it uses a set to save one character changed words for the next layer. Create a new word set which is one character changed from a to z in each index. If the end word is found while creating the new word set, the answer is found. Otherwise, if the created word is in the given word list, add it to the next layer set. If all words in the next layer set are checked, switch end and start word set based on the lengths. The shorter one will be used for the next iteration. This way, the computation burden can stay lesser.
## Solution
``````class WordLadder:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
if endWord not in wordList: return 0
n, visited, wordSet = len(beginWord), set(), set(wordList)
begin_set, end_set = {beginWord}, {endWord}
level = 1 # begin word
while begin_set and end_set:
level += 1
nextLayer = set()
for word in begin_set:
for i in range(n):
prefix, suffix = word[:i], word[i+1:]
for c in "abcdefghijklmnopqrstuvwxyz":
nextWord = prefix + c + suffix
if nextWord in end_set: return level
if nextWord in wordSet and nextWord not in visited:
• Time: `O(m^2 * n)` – m: length of each word, n: total number of words in the wordList
• Space: `O(m^2 * n)` | 871 | 3,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-40 | latest | en | 0.837692 |
http://gradestack.com/CBSE-Class-8th-Complete/Playing-with-Numbers/Numbers-in-General-Form/14827-2854-3068-study-wtw | 1,484,951,111,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280888.62/warc/CC-MAIN-20170116095120-00089-ip-10-171-10-70.ec2.internal.warc.gz | 123,456,350 | 9,589 | # Numbers in General Form
Let us take the number 32 and write it as
Similarly,
In general, any two digit number ab made of two digits a and b can be written as
Let us take 423. This is a three digit number. It can be written as
In general, a 3 digit number abc made up of digits, can be written as
In the same way, | 84 | 319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-04 | longest | en | 0.956993 |
https://codereview.stackexchange.com/questions/196204/numbers-as-strings-multiplication-function-in-swift | 1,701,264,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100081.47/warc/CC-MAIN-20231129105306-20231129135306-00349.warc.gz | 219,523,121 | 45,994 | # Numbers as Strings Multiplication Function in Swift
I was solving a problem 20, Factorial Digit Sum, on Project Euler.
## Factorial Digits Sum
$n!$ means $n \cdot (n − 1) \cdot \ldots \cdot 3 \cdot 2 \cdot 1$
For example, $10! = 10 \cdot 9 \cdot \ldots \cdot 3 \cdot 2 \cdot 1 = 3628800$, and the sum of the digits in the number $10!$ is $3 + 6 + 2 + 8 + 8 + 0 + 0 = 27$.
Find the sum of the digits in the number $100!$.
While doing this problem, I was working with really big numbers ($100!$ or $9.33 \cdot 10^{157}$). So at first, I decided to try to use Float80 and format it properly.
Since I have no I idea and could not find anything to properly explain to me how to do this with Float80 and String(format:_:) or NumberFormatter, I decided to make a function takes in two String parameters and returns their values multiplied as a String. With my multiplyLong (_:_:) function, I can multiply two numbers of any size. I wrote this function using concepts and principals from long multiplication from elementary school.
I wanted to find out if there is a better way to make a function that can multiply numbers of any size. Also, although my multiplyLong(_:_:) function works perfectly, it is quite long and I would like dramatically refine it, shorten it, and make it more efficient.
Note: I am using Swift 4.1 and Xcode 9.4
### My multiplyLong(_:_:) Function
func multiplyLong(_ x: String, _ y: String) -> String {
precondition(!x.isEmpty && !y.isEmpty, "Error, valid numbers are required for multiplication")
var prefix = ""
var containsInvalidCharacters : Bool {
switch (x, y) {
case let (s1, s2) where s1.first! == "-" && s2.first! == "-":
return !((s1.dropFirst() + s2.dropFirst()).contains { ["0","1","2","3","4","5","6","7","8","9"].contains($0) ? false : true }) case let (s1, s2) where s2.first! == "-": prefix = "-" return !((s1 + s2.dropFirst()).contains { ["0","1","2","3","4","5","6","7","8","9"].contains($0) ? false : true })
case let (s1, s2) where s1.first! == "-":
prefix = "-"
return !((s1.dropFirst() + s2).contains { ["0","1","2","3","4","5","6","7","8","9"].contains($0) ? false : true }) case let (s1, s2): return !((s1 + s2).contains { ["0","1","2","3","4","5","6","7","8","9"].contains($0) ? false : true })
}
}
precondition(containsInvalidCharacters, "Error, multiplicand contains invalid non-numeric characters")
let s1 = x.replacingOccurrences(of: "-", with: "").map { Int(String($0))! } let s2 = y.replacingOccurrences(of: "-", with: "").map { Int(String($0))! }
var a = [Int]()
var b = [Int]()
switch (s1, s2) {
case let (arr1, arr2) where arr1.count > arr2.count || arr1.first! >= arr2.first!:
a = Array(arr1.reversed())
b = Array(arr2.reversed())
case let (arr1, arr2):
a = Array(arr2.reversed())
b = Array(arr1.reversed())
}
var lines = [[Int]]()
for (bi, bn) in b.enumerated() {
var line = Array(repeating: 0, count: bi)
var carriedNumber = 0
for an in a {
let v = (bn * an) + carriedNumber
carriedNumber = (v - (v % 10)) / 10
line.insert(v % 10, at: 0)
}
if carriedNumber != 0 {
if carriedNumber >= 10 {
line.insert(carriedNumber % 10, at: 0)
line.insert((carriedNumber - (carriedNumber % 10)) / 10, at: 0)
} else {
line.insert(carriedNumber, at: 0)
}
}
lines.append(line)
}
let maxIndex = lines.max { $0.count <$1.count }!.count
var result = [Int]()
for i in 0...maxIndex {
for line in lines {
v += line.indices.contains(i) ? line.reversed()[i] : 0
}
if v >= 10 {
result.insert(Int("\("\(v)".last!)")!, at: 0)
} else {
result.insert(v, at: 0)
}
}
result.insert(Int("\(i)")!, at: 0)
}
}
return f.map { Int("\($0)") ?? 0 }.reduce(0, +) } print(factorialDigitSum()) //Prints "648" Note: 648 is indeed the right answer. ## 1 Answer The main performance bottleneck of your solution is the representation of big integers as a string. In each step of the factorial computation: • The intermediate result is converted from a string to an integer array, • the current factor is converted from Int to a string to an integer array, • a “long multiplication” is done on the two integer arrays, and finally, • the resulting integer array is converted back to a string. Also in every step, both input strings are validated to contain only decimal digits (plus an optional minus sign). This is not very efficient, and I'll address this point later. Instead of validating each possible combination of input strings x and y, it is more simple to define both strings separately, using a common utility function: func multiplyLong(_ x: String, _ y: String) -> String { func isValidNumber(_ s: String) -> Bool { guard let first = s.first else { return false // String is empty } let rest = first == "-" ? s.dropFirst() : s[...] return !rest.isEmpty && !rest.contains { !"0123456789".contains($0) }
}
precondition(isValidNumber(x) && isValidNumber(y), "Valid numbers are required for multiplication")
// ...
}
Note also how "0123456789" is used as a collection instead of ["0","1","2","3","4","5","6","7","8","9"], and that a conditional expression
<someCondition> ? false : true
can always be simplified to !<someCondition>.
The multiplication algorithm can also be improved. Instead of computing separate results for the product of one big integer with a single digit of the other big integer (your lines array), and “adding” those intermediate results later, it is more efficient to accumulate all products into a single array of result digits directly.
You also insert elements at the start each line array repeatedly, which requires moving all existing elements. This could be avoided by reversing the order of elements in line.
Finally note that the expression
(v - (v % 10)) / 10
which occurs at two places, can be simplified to v / 10.
### A better representation
All the conversion from strings to integer arrays and back to strings can be avoided if we represent a “big integer” not as a string, but as an integer array directly. Let's start by defining a suitable type:
struct BigInt {
typealias Digit = UInt8
let digits: [Digit]
let negative: Bool
}
digits holds the decimal digits of the big integer, starting with the least-significant one. UInt8 is large enough to hold decimal digits (and all intermediate results during the long multiplication). For reasons that become apparent later, we use a type alias instead of hard-coding the UInt8 type.
Now we define some initializers:
• The first one takes an array of digits, which is truncated for a compact representation. The lastIndex(where:) method has been added in Swift 4.2, see In Swift Array, is there a function that returns the last index based in where clause? on Stack Overflow for a possible approach in earlier Swift versions.
• The second initializer takes an Int and is based on the first one.
• We also add a digitSum computed property for obvious reasons.
Then we have:
struct BigInt {
typealias Digit = UInt8
let digits: [Digit]
let negative: Bool
init(digits: [Digit], negative: Bool = false) {
if let idx = digits.lastIndex(where: { $0 != 0 }) { self.digits = Array(digits[...idx]) } else { self.digits = [] } self.negative = negative } init(_ value: Int) { var digits: [Digit] = [] var n = value.magnitude while n > 0 { digits.append(Digit(n % 10)) n /= 10 } self.init(digits: digits, negative: value < 0) } var digitSum: Int { return digits.reduce(0) {$0 + Int($1) } } } In order to print a big integer, we adopt the CustomStringConvertible protocol: extension BigInt: CustomStringConvertible { var description: String { if digits.isEmpty { return "0" } return (negative ? "-" : "") + digits.reversed().map { String($0) }.joined()
}
}
Now we can implement the multiplication. This is essentially your method of long multiplication, but simplified to accumulate all digits into a single result buffer (which is allocated only once):
extension BigInt {
func multiplied(with other: BigInt) -> BigInt {
var result = Array(repeating: Digit(0), count: digits.count + other.digits.count)
for (i, a) in digits.enumerated() {
var carry: Digit = 0
for (j, b) in other.digits.enumerated() {
let r = result[i + j] + a * b + carry
result[i + j] = r % 10
carry = r / 10
}
while carry > 0 {
let r = result[i + other.digits.count] + carry
result[i + other.digits.count] = r % 10
carry = r / 10
}
}
return BigInt(digits: result, negative: self.negative != other.negative)
}
static func *(lhs: BigInt, rhs: BigInt) -> BigInt {
return lhs.multiplied(with: rhs)
}
}
Note how the init(digits:negative:) is used to compact the result array, i.e. to remove unnecessary zero digits.
As a bonus, we implement the * method so that big integers can be multiplied more naturally.
With all that, the digit sum of 100! is computed as
let sum = (1...100).reduce(BigInt(1), { $0 * BigInt($1) }).digitSum
print(sum) // 648
In my tests on a 1.2 GHz Intel Core m5 MacBook, with the code compiled in the Release configuration (i.e. with optimization on), the above code runs in approximately 0.5 milliseconds, compared to 30 milliseconds with your original code.
### Further suggestions
• Currently, the digits array contains the base-10 digits of the big integer. If we choose a larger Digit type and store more decimal digits in each array element, then less multiplications are required, making the multiplication more efficient.
With UInt64 as the Digit type, one can perform multiplication of 9-digit decimal numbers without overflow. This requires changing the Digit type alias, and modifying all places in the code where base-10 arithmetic is done.
• Add a init?(string: String) initializer so that a big integer can be created with
let bn = BigInt(string: "713847891471894789471894789478917417439274")
• Adopt the ExpressibleByIntegerLiteral protocol, so that a big integer can be created as
let bn: BigInt = 123
• Add addition and subtraction to the BigInt type.
• If you feel courageous, add division and remainder calculation.
• I don't understand what you mean, I've since changed the Digit type to UInt64, and it seems to work perfectly, so then what do I need to do in regards to: "modifying all places in the code where base-10 arithmetic is done"? Also, when conforming to the ExpressibleByIntegerLiteral protocol, it means that I can only declare small instances of BigInt with IntegerLiterals, but not large, yes? In my conformance extension I have: typealias IntegerLiteralType = Int and init(integerLiteral value: IntegerLiteralType) { self.init(value) }, is that correct? Thanks @MartinR for your help! Jun 12, 2018 at 5:21
• @NoahWilder: Changing UInt8 to UInt64 alone does not improve the efficiency. The idea is that every array element holds not values 0...9, but 0...999999999 (for example). – You are right with respect to ExpressibleByIntegerLiteral. Jun 12, 2018 at 5:25 | 2,800 | 10,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-50 | longest | en | 0.789473 |
https://ja.scribd.com/document/312944904/Counting-in-Two-Ways | 1,560,711,881,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998291.9/warc/CC-MAIN-20190616182800-20190616204800-00338.warc.gz | 485,636,965 | 61,131 | You are on page 1of 4
# Counting In Two Ways
January 25, 2016
Introduction
It is often very useful and rewarding (especially if you consider a 7/7 as a reward)
to count a quantity in two ways. This method provides us an efficient method
to bound the size and sometimes even find the values of unknown quantities. It
also has a few remarkable applications, which we shall see shortly. This is what
you basically do in double counting: Pick a quantity, count it up in two ways,
and hence compare the expressions so obtained. This is sometimes useful even
in proving the existence of some objects. There doesnt seem to be anything
more that I would like to put in this section, so lean back, have a nice warm
cup of tea, and .... enjoy.
## Double Counting - Illustrations
Okay, so in this section you shall find problems which shall neither be trivial
nor hard. If you are already well-versed in this topic, then go ahead and try
1. (Iran 2010) There are n points in the plane such that no three of them are
collinear. Prove that the number of triangles, whose vertices are chosen from
these n points and whose area is 1, is not greater than 23 (n2 n).
Solution. ([1]) Let the number of such triangles be k. For each edge between
two points in the set we count the number of triangles it is part of. Let the
total number over all edges be T . On the one hand, for any edge AB, there are
at most 4 points such that the triangles they form with A and B have the same
area. This is because those points have to be the same
distance from line AB,
and no three of them are collinear. Thus, T 43 n2 . On the other hand, each
triangle has 3 edges so T 3k. These two together imply the required result.
Just a few things to keep in mind while applying this technique:
i. Try counting ordered pairs or triples or n-tuples satisfying certain conditions. They will often help you kill the problem
ii. If you wish to bound an unknown quantity, say, x, then set up the doublecount in such a way that one of the ways of coutning involves x, while the other
doesnt.
2. (IMO Shortlist 2004) There are 10001 students at a university. Some students
1
join together to form several clubs (a student may belong to different clubs).
Some clubs join together to form several societies (a club may belong to different
societies). There are a total of k societies. Suppose that the following conditions
hold:
(i) Each pair of students is in exactly one club.
(ii) For each student and each society, the student is in exactly one club of
the society.
(iii) Each club has an odd number of students. In addition, a club with
2m + 1 students (m is a positive integer) is in exactly m societies.
Find all possible values of k.
Solution. Let us double-count the number of triples of the form (s, C, S) where
s is a student, C is a community, S is a society, and s C, C S. Let the
number of such triples be k. Suppose we fix s, then S then C. Thus we find
k = 10001 n, where n is the number of
Now,
suppose we fix C, then
Psocieties.
a
=
S then s. We get that k is equal to
a a1
2 , where the sum is over
2
all clubs, and a denotes the number of members of the particular club. On the
other hand, due to condition (i), another simple double-count shows that this
sum is equal to the number of pairs of students, which is equal to 5000 10001.
Hence k = 5000 is the only possible value of k.
Now, to show that this value works, take only one club having all students
and 5000 societies having this club.
We now leave a few problems for the readers.
1. (USA TST 2005) Let n be an integer greater than 1. For a positive integer
m, let Sm = {1, 2, ..., mn}. Suppose that there exists a 2n-element set T such
that
(a) each element of T is an m-element subset of Sm ,
(b) each pair of elements of T shares at most one common element; and
(c) each element of Sm is contained in exactly two elements of T .
Determine the maximum possible value of m in terms of n.
2. (IMO 1987) Let pn (k) be the number of permutations of the set {1, 2, ..., n}
which have exactly k fixed points. Prove that
n
P
## kpk (n) = n!.
k=0
3. (Hong Kong 2007) In a school there are 2007 girls and 2007 boys. Each
student joins no more than 100 clubs in the school. It is known that any two
students of opposite genders have joined at least one common club. Show that
there is a club with at least 11 boys and 11 girls.
4. (IMO 1998) In a competition, there are a contestants and b judges, where
b 3 is an odd integer. Each judge rates a contestant as either pass or fail.
Suppose k is a number such that for any two judges, their ratings coincide for
at most k contestants. Prove that ka b1
2b .
5. (APMO 1989) Show that a graph on n vertices and k edges has at least
k(4kn2 )
triangles.
3n
6. (USAMO 1995) Suppose that in a certain society, each pair of persons can
## be classified as either amicable or hostile. We shall say that each member of an
amicable pair is a friend of the other, and each member of a hostile pair is a
foe of the other. Suppose that the society has n people and q amicable pairs,
and that for every set of three persons, at least one pair is hostile. Prove that
there is at least one member of the society whose foes include q(1 n4q2 ) or fewer
amicable pairs.
## Some Cool Proofs Using Double Counting
We start with a problem that is simply amazing. We borrow its solution and
motivation from [2].
1. (IMO Shortlist 2003/C4) Let x1 , x2 , ..., xn and y1 , y2 , ..., yn be real numbers.
Let A = {aij } (with 1 i, j n) be an n n matrix with entries aij = 1 if
xi + yj 0, and aij = 0 otherwise.
Let B be an n n matrix with entries 0 or 1 such that the sum of elements
of each row and each column of B equals the corresponding sum for matrix A.
Show that A = B.
Solution. ([2]) Let bij denote the entry in the i-th row and j-th column of B.
Define
P
S=
(aij bij )(xi + yi ).
1i,jn
On one hand, S =
n
P
i=1
xi (
n
P
aij
j=1
n
P
bij ) +
j=1
n
P
j=1
yi (
n
P
i=1
aij
n
P
bij ) = 0 by the
i=1
## conditions of the problem.
On the other hand, note that if xi + yj 0 then aij = 1 so aij bij 0. If
xi + yj < 0 then aij = 0 so aij bij 0. In both cases, (xi + yj )(aij bij ) 0.
But the total sum S is 0, so whenever xi + yj 6= 0 we have aij = bij . Whenever
xi + yj = 0, aij = 1. In these cases we must have bij = 1 as well, as the sums
of entries in both matrices is the same. This completes the solution.
Motivation. ([2]) Where on earth does the expression
P
S=
(aij bij )(xi + yi ).
1i,jn
come from?!?! Note that one way of proving that several different real numbers
are 0 is to show that their squares sum to 0, since no square is negative. Thus,
a first approach to the problem may be to show that the sum
P
S0 =
(aij bij )2
1i,jn
is zero. This doesnt work as it doesnt utilize information about the xs and
ys. Instead we try the following modification: we seek to weight each term
by some other quantity that still ensures that each term in the summation is
nonnegative, and additionally enables us to use the information about the xs
and ys to show that the entire sum is 0.
3
## The rest are left as problems for the reader:
1. (Cayleys Formula) The number of different unrooted trees that can be
formed from a set of n distinct vertices is Tn = nn2 .
2. (Fermats Little Theorem!!!!) If a is an integer and p is a prime, then
ap a(mod p).
3. (IMO 2001) Let n be an odd positive integer greater than 1 and let c1 , c2 , ..., cn
be integers. For each permutation a = {a1 , ..., an } of {1, 2, ..., n}, define S(a) =
n
P
ai ci . Prove that there exist permutations a 6= b such that n! is a divisor of
i=1
S(a) S(b).
References | 2,134 | 7,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2019-26 | latest | en | 0.952152 |
https://excelnotes.com/hectometer-to-dekameter/ | 1,632,814,297,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00588.warc.gz | 282,523,525 | 10,783 | # Convert Hectometer to Dekameter
Hectometer (hm):
Dekameter (dam):
The unit of hectometer is one of the length units in the International System of Units. One hectometer is equal to 10 dekameters.
Formula:
1 hectometer = 10 dekameters
Symbol:
• Hectometer: hm
• Dekameter: dam
Convert to Meter:
• 1 hectometer = 100 meters
• 1 dekameter = 10 meters
Examples:
• 10 hectometers = 10 x 10 dekameters = 100 dekameters
• 25 hectometers = 25 x 10 dekameters = 250 dekameters
The following is the conversion from hectometers to dekameters: | 164 | 545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.58003 |
https://www.numbersaplenty.com/35346143 | 1,695,706,461,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510149.21/warc/CC-MAIN-20230926043538-20230926073538-00497.warc.gz | 1,016,964,733 | 3,049 | Search a number
35346143 = 77171119
BaseRepresentation
bin1000011011010…
…1011011011111
32110111202210102
42012311123133
533022034033
63301331315
7606302660
oct206653337
973452712
1035346143
1118a52098
12ba06b3b
1374274a1
1449a1367
153182de8
hex21b56df
35346143 has 8 divisors (see below), whose sum is σ = 40965120. Its totient is φ = 29869560.
The previous prime is 35346139. The next prime is 35346163. The reversal of 35346143 is 34164353.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 35346143 - 22 = 35346139 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (35346163) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 35063 + ... + 36056.
It is an arithmetic number, because the mean of its divisors is an integer number (5120640).
Almost surely, 235346143 is an apocalyptic number.
35346143 is a deficient number, since it is larger than the sum of its proper divisors (5618977).
35346143 is an equidigital number, since it uses as much as digits as its factorization.
35346143 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 71197.
The product of its digits is 12960, while the sum is 29.
The square root of 35346143 is about 5945.2622313906. The cubic root of 35346143 is about 328.1814352877.
The spelling of 35346143 in words is "thirty-five million, three hundred forty-six thousand, one hundred forty-three".
Divisors: 1 7 71 497 71119 497833 5049449 35346143 | 506 | 1,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-40 | latest | en | 0.851926 |
https://docsbay.net/20-pre-requisites-for-calculus | 1,642,467,206,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00697.warc.gz | 284,396,410 | 8,991 | # 20 Pre-Requisites for Calculus
AP Calculus Summer Project Solutions
1. Show all work on separate paper.
2. Write your name in pen at the top of each page.
3. Use graph paper for all graphs. Do all examples.
4. Prepare to be tested on all topics on the first day of class.
5. If you have any questions during the summer, email the instructor at w at least 48 hours for a response.
Topic: Slope, Lines, and Linear Equations
1. Find slope from two given points. Example: (4 , –7) and (–5 , 8)
Slope (m) of the line through (x1,y1) and (x2,y2) is m =
Slope of the line through (4 , –7) and (–5 , 8) is m = = =
2. Write the slope and an equation for a horizontal line through (4, 7). m = 0
y = 7
3. Write the slope and an equation for a vertical line through (-3 , 5).
no slope or undefined slope ; equation: x = –3
Point-Slope Form: y – y1 = m(x – x1)
4. Use point-slope form to write an equation for the line through (6 , –5) with
slope 2.
y – (–5) = 2(x – 6)
y + 5 = 2x – 12
y = 2x – 17
5. Use point-slope form to write an equation for the line through (4 , –7) and (–5 , 8).
First find the slope: m =
Then choose either point and use point-slope form: y – y1 = m(x – x1)
y – 8 = ( x – (–5) )
y – 8 = ( x + 5)
y – 8 =
y = or 5x + 3y = –1
6. Convert from point slope form to slope-intercept form: y – 7 =
Distribute and simplify: y – 7 =
y =
7. Define and illustrate a tangent line.
A tangent line is a line that touches a curve at exactly one point.
8. Define and illustrate a secant line.
A secant line is a line that crosses a curve at two points.
9. Define “average rate of change” algebraically and graphically.
The ARC on [ a , b ] is the slope of the secant line through (a , f (a)) and (b , f (b)).
Algebraically: ARC = or
Geometrically: Average rate of change is slope of a secant line.
10. Define “instantaneous rate of change” algebraically and graphically.
Algebraically: IRC is the derivative or the limit as h approaches zero of the DQ.
Geometrically: IRC is the instantaneous slope of a tangent line
Do the following AP released free response questions found online at ap central.
1. 2010 FRQ #2(a)ARC = hundred entries per hour
2. 2011 FRQ #2(a)ARC@ 3.5 = ° Celsius/minute
13. 2008 FRQ #2(a)ARC@(5.5) =
14. 2005 FRQ #3(a)ARC@(7) = ° Celsius/minute
Topic: Functions
Graph the 15 parent functions listed below. Use graph paper. State the domain and
range for each.
DomainRange
15. f (x) = x2all Ry ≥ 0
16. f (x) = x3all Rall R
17. f (x) =x ≥ 0y ≥ 0
18. f (x) = all Rall R
19. f (x) = |x|all Ry ≥ 0
20. f (x) = [x]all Rall integers
21. f (x) = sin(x)all R[ –1 , 1 ]
22. f (x) = cos(x)all R[ –1 , 1 ]
23.f (x) = tan(x)all R except odd multiples of all R
24. f (x) =all R except x = 0all R except y = 0
Topic: Semi-Circles
25. Write the general function for a semi-circle: f (x) = , where r is the radius.
26. Give examples of three semi-circle functions and graphs. (Answers may vary.)
y = ; y = ; y = ; y =
Be able to instantly recognize the equation and graph of a semi-circle!
Topic: Special Right Triangles
1. Describe and illustrate the ratios for the sides of a 30º–60º–90ºtriangle.
In a 30º-60º-90º triangle, the hypotenuse is double the length of the short side
and the longer leg is times the length of the short side.
The ratio of the sides is 1 : 2 :(short leg : hypotenuse : long leg)
1. Describe and illustrate the ratios for the sides of a 45º–45º–90º right triangle.
In a 45º-45º-90º triangle, the hypotenuse is times the length of either leg.
The ratio of the sides is 1 : 1 :(leg : leg : hypotenuse)
1. Label all three sides for 30º–60º–90º triangles with hypotenuse of length
1 , 2 , 5 , 8 , 10, and x.
Short Leg / Long Leg / Hypotenuse
/ / 1
1 / / 2
/ / 5
4 / 4 / 8
5 / 5 / 10
/ / x
30. Label all three sides for 45º–45º–90º triangles with hypotenuse of length
1 , 2 , 5 , 8 , 10, and x.
Leg / Hypotenuse
or / 1
/ 2
or / 5
4 / 8
10 / 10
or / x
Topic: Trigonometric Ratios
31. Know sine, cosine, and tangent of the following angles to automaticity.
θ =
If necessary, make 48 flashcards. Identify all 48 trigonometric ratios from randomly-ordered flashcards to 99% accuracy in four minutes or less.
Prove these seven identities and their corollaries and memorize the results.
32. Pythagorean Identity #1: sin2θ + cos2θ = 1
Proof: From a right triangle diagram with acute angle θ, adjacent leg labeled x,
opposite leg labeled y, and hypotenuse h:
x2 + y2 = h2 by the Pythagorean Theorem
Divide through by h2:
cos2θ + sin2θ = 1 or sin2θ+ cos2θ = 1
Remember: “This sign costs one dollar.”
33. Pythagorean Identity #2: tan2θ + 1 = sec2θ
sin2θ + cos2θ= 1
Divide through by sin2θ:
Remember: “One vampire got caught in a casket.” or
You cannot buy just “one cotton ball at Costco”.
34. Pythagorean Identity #3: 1 + cot2θ = csc2θ
sin2θ +cos2θ = 1
Divide through by cos2θ:
Remember: You should only “tan for one second”.
Note: You need to prove the addition and subtraction identities before the double angle identities. I use a geometric proof starting with a rectangle ABCD. Right triangle ∆AED is inscribed in the rectangle with right angle AEF and hypotenuse AF = 1.
35. Sine Addition and Sine Subtraction: sin(α ± β) = sin α cos β ± cos α sin β
36. Cosine Addition and Cosine Subtraction: cos(α ± β) = cos α cos β ± sin α sin β
Next we can prove these:
37. Sine Double Angle Identity: sin 2θ = 2sinθ cosθ
= sin Ө cos Ө + cos Ө sin Ө
= 2 sin Ө cos Ө
38. Cosine Double Angle Identities
(a) cos 2θ = cos2θ – sin2θ
= cosӨ cos Ө – sin Ө sin Ө
= cos2Ө – sin2Ө
(b) cos 2θ = 2cos2θ – 1
Use the first Pythagorean Identity: sin2Ө = 1 – cos2Ө
Substitute: cos 2Ө = cos2Ө – (1 – cos2Ө)
cos 2Ө = cos2Ө – 1 + cos2Ө
cos 2Ө = 2cos2Ө – 1
(c): cos 2θ = 1 – 2sin2θ
Use the second Pythagorean Identity: cos2Ө = 1 – sin2Ө
Substitute: cos 2Ө = (1 – sin2Ө) – sin2Ө
cos 2Ө = 1 – 2sin2Ө
Topic: Sinusoidal Waves
Know the amplitude, period, horizontal shift, and vertical shift for trigonometric functions in the following forms. f (x) = a sin b(x – c) + d and f (x) = a cos b(x – c) + d
Graph and label these sinusoidal waves. Use a separate graph for each.
39. f (x) = 3 sin 2 + 1 40.f (x) = –2 cos 3 – 3
Topic: Absolute Value Functions
41. Define the absolute value function, f (x) = | x | as a piecewise function.
Algebraically: | x | = x when x ≥ 0 and | x | = –x when x < 0.
Graphically: | x | is the distance from 0 to x on the number line.
Graph and label the following absolute value functions on four separate graphs.
42.f (x) = | x + 3 | – 5 V-shape with vertex at ( –3 , –5 )
43. f (x) = 2| x | – 4 Skinny V-shape with vertex at ( 0 , –4 )
44. f (x) = –| x – 2 | + 1 A-shape with vertex at ( 2 , 1 )
45. f (x) = | 3 – x |V-shape with vertex at ( 3 , 0 )
Topic: Greatest Integer Function
46. Define the greatest integer function.
[[ x ]] is the greatest integer that is less than or equal to x. It is a rounding DOWN function.
Graph and label the following on four separate graphs. These are all “stair step” functions with an open circle on the left and a closed circle on the right.
47. f (x) = 2[x]49. f (x) = [x + 3]
48. f (x) = –[x]50. f (x) = [0.5x]
Topic: Piecewise Functions Graph the following piecewise functions.
51. f (x) =52. f (x) =
53. f (x) =
Topic: Exponential Functions
Graph. Label the y-intercept, the horizontal asymptote, and one anchor point.
54. f (x) = 2xincreasing for all Ry = 0all Ry > 0(0 , 1)
55. f (x) = exincreasing for all Ry = 0all Ry > 0(0 , 1)
56. f (x) = 10xincreasing for all R y = 0all Ry > 0(0 , 1)
57. f (x) = ex+2– 4 same as ex shifted y = –4all Ry > –4(0, –3)
left 2 and down 4
58. f (x) = –5xdecreasing for all Ry = 0all Ry < 0(0 , –1)
59. f (x) = 4x + 3increasing for all Ry = 3all Ry > 3(0 , 4)
60. f (x) = 10x+2 + 3same as 10x shifted
left 2 and up 3y = 3all Ry > 3(0 , 4)
31. f (x) = log(x – 3) same as log(x) shifted right 3x > 3all R
32. f (x) = log(x + 5) – 1 same as ln(x) shifted left 5 and down 1x > –5all
Topic: Logarithmic Functions
61. Graph f (x) = ln(x) and g(x) = ex on one graph.
Label the x- and y-intercepts and one anchor point on each graph.
( 1 , 0 ) and (e , 1) are on f (x) = ln(x)
( 0 , 1 ) and (1 , e) are on g(x) = ex
The y-intercept for g(x) = exis y = 1.
The x-intercept for f (x) = ln(x) is ( 1 , 0 ).
62. Graph f (x) = log(x) and g(x) = 10x on one graph.
Label the x- and y-intercepts and one anchor point on each graph.
The y-intercept for g(x) = 10xis y = 1.
( 0 , 1 ) and (1 , 10) are on g(x) = 10x
The x-intercept for f (x) = log(x) is ( 1 , 0 ).
( 1 , 0 ) and (10 , 1) are on f (x) = log(x)
Topic: Transformations of Logarithmic Functions
Graph. Label the x-intercept, the vertical asymptote (VA) and one anchor point. Use four separate graphs.
1. f (x) = ln(x – 1) + 2VA: x = 1Anchor Point: ( 2 , –3)
1. f (x) = ln(x – 3) + 5VA: x = 3Anchor Point: ( 4 , 5 )
2. f (x) = ln(x + 2) + 4VA: x = –2Anchor Point: (–1 , 4)
1. f (x) = ln(x + 4) – 2VA: x = –4Anchor Point: ( –3 , –2)
Topic: Solving Logarithmic Equations
67. ln(x – 2) + 4 = ln x
4 = ln x – ln(x – 2)
4 = ln
e4 =
e4 (x – 2) = x
xe4 – 2e4 = x
xe4 – x = 2e4
x( e4 – 1 ) = 2e4
x =
68. 3(x+1) = 7(x–2)
ln 3(x+1) = ln 7(x–2)
( x + 1) ln 3 = ( x – 2 ) ln 7
x ln 3 + ln 3 = x ln 7 – 2 ln 7
ln 3 + 2 ln 7 = x ln 7 – x ln 3
ln 3 + 2 ln 7 = x ( ln 7 – ln 3 )
= x
= x
= x
69. log5(x–2) = log5x + log57
log5(x–2) – log5x = log57
log5 = log57
= 7
x – 2 = 7x
–2 = 6x
= x
70. ex(ex–4 ) = 1
ex–4 =
ex–4= e-x
x – 4 = –x
–4 = –2x
2 = x
Topic: Inverse Functions
1. What is inverse function notation for f (x)? f -1(x)
2. How do you find the inverse of f (x) algebraically? Switch x and y, then solve for x.
73. How do you find the inverse of f (x) graphically?
Reflect the graph across the diagonal line y = x.
74. Give 3 examples of points on a function and points on the inverse.
( 5 , –2 ) (–2 , 5 )
( 3 , 4 ) ( 4 , 3 )
(–7 , –10 ) (–10 , –7 )
75. If f (x) and g(x) are inverses, what is f (g(x))? What is g(f (x))?
f (g(x)) = x
g(f (x)) = x
76. Give 3 examples of slopes of functions and their inverses at corresponding points.
Slope of Function at ( x , y ) / Slope of Inverse at ( y , x )
5 / 1/5
–3 / –⅓
½ / 2
77. Inverse functions are reflections across what line? y = x
78. The slopes of inverse functions are reciprocals of each other at corresponding points reflected across the line y = x.
Topic: Polynomials and Rational Expressions
79. X-Intercepts occur when the numerator of a rational expression equals zero.
80. Vertical Asymptotes occur when the denominator of a rational expression equals zero.
81. Horizontal Asymptotes occur when:
the degree of the numerator ≤ the degree of the denominator.
82. Holes occur when there is a:
zero of both the numerator and denominator with multiplicity ≥ in the numerator.
Non-Vertical Asymptotes:In a rational expression, let n = the degree of the numerator and
d = the degree of the denominator. Describe the end behavior.
83. If n < d, then the horizontal asymptote is y = 0 (the x-axis).
84. If n = d, then the horizontal asymptote is y =
where LCON is the leading coefficient of the numerator and
LCOD is the leading coefficient of the denominator.
85. If n > d, and n = d + 1, then the non-vertical asymptote is a line.
86. If n > d, and n = d + 2, then the non-vertical asymptote is a parabola.
87. If n > d, and n = d + 3, then the non-vertical asymptote is a cubic.
Find the x-intercepts, vertical asymptotes, horizontal asymptotes, and holes.
Graph each function separately on graph paper. Describe the end behavior.
Optional: Find and .
88.
HA: y = 0
VA: none
Holes: none
x-intercepts: x = 0 and x = –7
= 0 and = 0
89. =
HA: none, but there is a parabolic asymptote at approx. y = –2
VA: none
Holes: x = 0
x-intercepts: none
= -2and = -2
90. =
HA: none, but there is a parabolic asymptote at approx. y = x2
VA: none
Holes: x = 0
x-intercepts: x =
= ∞and = ∞
91. =
HA: y = 1
VA: none
Holes: none
x-intercepts: x = 4 and x = –3
= 1 and = 1
Topic: Area Formulas for Basic Geometric Shapes
92. Area of a Square with side s.A = s2
93. Area of a Semi-Circle with diameter DA = or A =
94. Area of an Isosceles Right Triangle with leg x.A =
95. Area of an Isosceles Right Triangle with Hypotenuse h. A =
96. Area of a TrapezoidA =
97. Area of a Rectangle with width x and length equal to three times the width. A = 3x2
98. Area of an Equilateral Triangle with side s. A =
Do the following AP released free response questions at
99. 2011B #6: Find the area of the triangle on [–2π , 4π].A = 6π2
100. 2011 #4: Find the area between the graph and the x-axis on the intervals
(a) [–4 , –3]:A = units2
(b) [–3 , 0]:A = units2
(c) [0 , 1.5]:A = units2
101. 2010 #5: Find the area between the graph and the x-axis on the intervals
[–7, –2]:A = units2
[–2 , 2]: A = units2
[2 , 4.5]:A = units2
[4.5, 5]:A = units2
102. 2010B #4: Find the area of the three trapezoids on [0 , 18].
1st Trapezoid on [0 , 9]:A = = 140 units2
2nd Trapezoid on [9 , 15]:A = = 50 units2
3rd Trapezoid on [15 , 18]:A = = 25 units2
Topic: Difference Quotient and Derivative
1. Find the difference quotient.
2. Find the derivative, f '(x). Use the limit of the difference quotient as h approaches 0.
103. f (x) = –x – 4
DQ =
DQ =
DQ =
DQ = = –1f '(x) =
104. f (x) = x2
DQ =
DQ =
DQ =
DQ =
DQ = 2x + hf '(x) =
105. f (x) = x3 + x
DQ =
DQ =
DQ =
DQ =
DQ = f '(x) =
DQ = 3x2 + 3xh + h2 + 1f '(x) = 3x2 + 1
106. f (x) = 4x2 – 3x – 2
DQ =
DQ =
DQ =
DQ =
DQ = f '(x) =
DQ = 8x + 4h – 3 f '(x) = 8x – 3
107. f (x) =
DQ =
DQ =
DQ = .
DQ = f '(x) =
DQ = f '(x) =
DQ = f '(x) =
108. f (x) =
DQ =
DQ =
DQ =
DQ =
DQ = f '(x) =
DQ = f '(x) =
109. f (x) = sin(x)
DQ =
DQ =
DQ = You can STOP here.
We will find the derivative, f '(x) = cos(x) soon.
110. f (x) = cos(x)
DQ =
DQ =
DQ = You can STOP here.
We will find the derivative, f '(x) = –sin(x) soon. | 5,029 | 14,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-05 | latest | en | 0.891284 |
http://www.physicsforums.com/showthread.php?t=581750 | 1,386,604,832,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163988740/warc/CC-MAIN-20131204133308-00068-ip-10-33-133-15.ec2.internal.warc.gz | 487,167,078 | 11,734 | # What kind of op-amp is that?
by Femme_physics
Tags: kind, opamp
PF Patron P: 2,544 1. The problem statement, all variables and given/known data At first I thought a comparator, then I noticed a line connecting the Vout and the lines that connect to V- and V+, so it can't be a comparator, can it? I don't see any other option.... As far as the question, I need to find out the unknown resistors, but I'll start working on it once I figure that basic fact.
PF Patron HW Helper P: 3,786 As a start, you can either ignore Rx or the circuit is in saturated mode ... this is a consideration since the output current is defined to be 3 mA. The rest is just the usual KVL equations of an active op amp circuit (input voltages equal etc.).
Mentor
P: 10,764
Quote by Femme_physics 1. The problem statement, all variables and given/known data At first I thought a comparator, then I noticed a line connecting the Vout and the lines that connect to V- and V+, so it can't be a comparator, can it? I don't see any other option.... As far as the question, I need to find out the unknown resistors, but I'll start working on it once I figure that basic fact.
Looks like an amplifier with feedback, although the feedback mechanism is made a bit trickier due to the presence of Rx. What effect do you suppose the zener diode is going to have?
Rb and R1 form a voltage divider. What voltage are they dividing? What's the op-amp going to try to do with that voltage?
HW Helper
P: 6,164
## What kind of op-amp is that?
I'd say it looks a bit like an inverting amplifier, but it's not quite it.
KVL and KCL should tell what it does.
PF Patron HW Helper P: 3,786 I hadn't noticed the output voltage is given. Since R1 is given, all components are defined implicitly. You can solve this problem without writing any KVL equations. Hints: what must Ra be to satisfy Iz_min? What must Rb be to give a 12V output? And finally, what must Rx be to give 3 mA of current flowing into the op amp?
P: 1,761 There are a few opamps with open collector outputs. This could be one of them.
P: 1,508 I would say it is a non-inverting amplifier with voltage of 5V on the + input and voltage gain of R1/(R1+Rb). Vout =+12V so Rb can be calculated.... and so on
PF Patron
HW Helper
P: 3,786
Quote by technician I would say it is a non-inverting amplifier with voltage of 5V on the + input and voltage gain of R1/(R1+Rb). Vout =+12V so Rb can be calculated.... and so on
Nope on your gain expression ... but you're warm ...
P: 1,508 gain of (R1+Rb)/R1 = 12/5 = 2.4
PF Patron P: 2,544 Well, this is the solution my classmate offered I agree on Ra. It's just KVL. But on Rb-- I'm confused as far as how to use a voltage divider. I CAN indeed apply KVL Since I know that the 3 mA split at this point marked in red: I know that there's only 1 mA going through Rb and R1, and I know they have a 12V potential difference to the ground. So, Sum of all V = 0 ; 12 - 1ma x Rb -1ma x R1 = 0 I get that Rb 2000 ohms. Makes sense? As far as Rx -- well, I don't really understand something fundemental about the circuit. Is Vs some type of another Vout? Or does it just define the limits of the Op-Amp like we see in those Vcc+ Vcc- sort of thing? I really don't know how to approach Vs. How can 25 Volt comes out of an op-amp who only produces a Vout of 12v?!? And how does any of that helps me with Rx? Sorry-- A lot of questions, I know. Just a confusing circuit!
HW Helper P: 4,192 +Vs is the power supply (you've previously seen it as +Vcc) to power the OP-AMP. There is no -Vcc here, the negative power supply here is ground (you've previously seen OP-AMP's negative supply as -Vcc, but not with this arrangement here).
HW Helper
P: 4,192
I know that there's only 1 mA going through Rb and R1
HW Helper
P: 4,192
Quote by Femme_physics Well, this is the solution my classmate offered
Are you referring to the words he's written to the right of the resistor string?
PF Patron
P: 2,544
+Vs is the power supply (you've previously seen it as +Vcc) to power the OP-AMP. There is no -Vcc here, the negative power supply here is ground (you've previously seen OP-AMP's negative supply as -Vcc, but not with this arrangement here).
Shouldn't the power supply ENTER the op-amp as opposed to EXIT from it?
Well, sum of all I entering that crossection I marked in red
3 -2 -I1 = 0
I1 = 1 mA
There you go...
Are you referring to the words he's written to the right of the resistor string?
LOL I didn't c that..sorry..ignore that part.
HW Helper
P: 6,164
Quote by Femme_physics Shouldn't the power supply ENTER the op-amp as opposed to EXIT from it?
It does.
The arrow indicates that the wire goes to a power supply.
It does not indicate the direction of the current.
The actual current flows away from the 25V power supply.
Well, sum of all I entering that crossection I marked in red 3 -2 -I1 = 0 I1 = 1 mA There you go...
What happened to the current coming from the 25V power supply that is coming in through Rx?
LOL I didn't c that..sorry..ignore that part.
How can we now that it's out there! ;)
PF Patron
P: 2,544
It does. The arrow indicates that the wire goes to a power supply. It does not indicate the direction of the current. The actual current flows away from the 25V power supply.
OOOOOhhhhhhhhh!!!! AHHHHHHA!!!
OH! OH!
Now I get it :)
What happened to the current coming from the 25V power supply that is coming in through Rx?
It's kinda late now but i'll sit with it tomorrow trying t finalize my results based on this new evidence!
How can we now that it's out there! ;)
Well, it just says "Or take it"
Or being my name. Telling me that I should take this exercise and try to solve it. He just put the "e" in front of the "k" by accident :)
HW Helper P: 6,164 OOOOOhhhhhhhhh!!!! AHHHHHHA!!! Or take it! Yes, I think I understand now what he meant. ;)
HW Helper
P: 4,192
Quote by Femme_physics Well, sum of all I entering that crossection I marked in red 3 -2 -I1 = 0 I1 = 1 mA There you go.../
I believe you realise this is so not right that you need to make a fresh start.
Hint: you know V+ so determine V_.
Related Discussions Introductory Physics Homework 63 Differential Equations 1 Academic Guidance 6 Special & General Relativity 9 Introductory Physics Homework 1 | 1,720 | 6,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-48 | longest | en | 0.952014 |
http://protestira.me/algoritma-euclid-39/ | 1,566,247,153,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314959.58/warc/CC-MAIN-20190819201207-20190819223207-00370.warc.gz | 157,260,751 | 10,819 | # ALGORITMA EUCLID PDF
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This failure of unique algorotma in some cyclotomic fields led Ernst Kummer to the concept of ideal numbers and, later, Richard Dedekind to ideals. In other projects Wikimedia Commons. A Euclidean domain is always a principal ideal domain PIDan integral domain in which every ideal is a principal ideal.
These quasilinear methods generally scale as O h log h 2 log log h. The winner is the first player to reduce one pile to zero stones. If two numbers have no prime factors in algoirtma, their greatest common algortima is 1 obtained here as an instance of the empty productin other words they are coprime.
Thus the iteration of eucllid Euclidean algorithm becomes simply. In other words, the set of all possible sums of integer multiples of two numbers a and b is equivalent to the set of multiples of gcd ab. Instead of representing an integer by its digits, it may be represented by its remainders x i modulo a set of N alforitma numbers m i: A finite field is a set of numbers with four generalized operations.
The Euclidean algorithm has many theoretical and practical applications. For example, it can be used to solve linear Diophantine equations and Chinese remainder problems for Gaussian integers; [] continued fractions of Gaussian integers can also be defined.
The Elements of Algebra in Ten Books.
For comparison, Euclid’s original subtraction-based algorithm can be much slower. Substituting the approximate formula for T a into this equation yields an estimate for Y n [].
Thus, the solutions may be expressed as. It is named after the ancient Greek mathematician Euclidwho first described it in his Elements c.
Since the determinant of M is never zero, the vector of the algoirtma remainders can be solved using the inverse of M. High primes and misdemeanours: Since the degree is a nonnegative integer, and since it decreases with every step, the Euclidean algorithm concludes in a finite number of steps.
The process of substituting remainders by formulae involving their predecessors can be continued until the original numbers a and b are reached:. Euclidean division reduces all the steps between two exchanges into a single step, which is thus more efficient. Finally, the coefficients of the polynomials need not be drawn from integers, real numbers or even the complex numbers. Excursions in number theory. When that occurs, they are the GCD of the original two numbers.
The binary GCD algorithm is an efficient alternative that substitutes division with faster operations by exploiting the binary representation used by computers.
The solution is to combine the multiple equations into a single linear Diophantine equation with a much larger modulus M that is the product of all the individual moduli m i euclie, and define M i as. Another inefficient approach is to find the prime factors of one or both numbers.
### Euclidean algorithm – Wikipedia
The Euclidean algorithm proceeds in a series of steps such that the output of each step is used as an input for the next one. The sides of the rectangle can be divided into segments of length cwhich alggoritma the rectangle into a grid of squares of side length c.
You may also like - BOTTESINI GRAN DUO CONCERTANTE PDF
If r k is replaced by e k. In this field, the results of any mathematical operation addition, subtraction, multiplication, or division is reduced modulo 13; that is, multiples of 13 are added or subtracted until the result is brought within the range euclld The Euclidean algorithm has almost the same relationship to another binary tree on the rational numbers called the Calkin—Wilf tree.
## Euclidean algorithm
The players take turns removing m multiples of the smaller pile from the larger. InCole and Davie developed a two-player game based on the Euclidean algorithm, called The Game of Euclid[49] which has an optimal strategy.
The integers s and t can be calculated from the quotients q 0q 1etc. The Universal Book of Mathematics: It is used for reducing eucclid to their simplest form and for performing division in modular arithmetic.
Binary Euclidean Extended Euclidean Lehmer’s. Let g be the greatest common divisor of a and b. Since the remainders decrease with every step but can never be negative, a remainder r N must eventually equal zero, at which point the algorithm stops.
Polynomials in a single variable x can be added, multiplied and factored into irreducible polynomialswhich are the analogs of the prime numbers for integers. Theory of Numbers 2nd ed.
Begin typing your search term above and press enter to search. Press ESC to cancel. | 1,047 | 4,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-35 | latest | en | 0.897714 |
http://clickspay.ru/difference-between-stress-and-strain/ | 1,529,859,487,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00534.warc.gz | 71,519,937 | 14,496 | # Difference Between Stress and Strain
## Main Difference – Stress vs. Strain
When deforming forces act on an object, they can change the object’s shape. The main difference between stress and strain is that stress measures the deforming force per unit area of the object, whereas strain measures the relative change in length caused by a deforming force.
## What is Stress
Whenever a force attempts to deform an object, we say that the object is under stress. Stress is defined as the deforming force per unit area of the object. Since we can resolve any force on an object into directions parallel and perpendicular to the surface, we define normal stress to be equal to the force perpendicular to the surface per unit area. Similarly, we define shear stress as the force parallel to the surface per unit area. If the force acting on a surface is $F$ and the area of the surface is $A$, then the stress $\sigma$ is given by:
$\sigma=\frac{F}{A}$
Stress has the same dimensions as pressure, so the units used for measuring stress are also N m-2 or Pa (1 Pa=1 N m-2). When forces act to lengthen the material, the stress is referred to as tensile stress. When the forces try to compress a material, the stress is referred to as compressive stress.
## What is Strain
Strain measures the amount of relative deformation caused by a force acting on an object. For simplicity here, we will only consider the normal strain, created by normal stress. Suppose the original length of the object is $l_0$ and due to stress, the length changes to $l_1$. The change in length is $\Delta l=l_1-l_0$. The strain $\epsilon$ is then given by,
$\epsilon =\frac{\Delta l}{l_0}$
Since strain is given by a fraction where the numerator and denominator both have units of length, the strain itself has no units. i.e. it is a “dimensionless quantity”. It is common to see strain expressed in terms of percentages.
## Stress vs. Strain Curve
We can draw a graph of how the strain in a body changes as we vary the stress acting on the object (this can be done, for example, by adding weights). These graphs, called stress vs. strain curves, reveal lots of information about the nature of the material that the object is made of. The figure below shows the typical stress-strain curve for a ductile material (“ductile” means that the material can be stretched out well):
Stress-strain curve for a ductile material
The gradient of the elastic region of the curve is called the Young Modulus. This is a very important number for materials engineers, as it gives how much strain would be caused by a given stress in a material.
## Difference Between Stress and Strain
### What it Measures
Stress gives the force acting per unit area of an object.
Strain gives the relative change in length due to deforming forces.
### Units
Stress is measured in pascals (Pa).
Strain has no units; it is simply a ratio.
Image Courtesy
“Typical Stress vs. Strain diagram for a ductile material (e.g. Steel).” by Breakdown (Own work) [], via (Modified)
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https://flatearthsoc.com/flat-earth-proofs-flat-earth-news-by-nick-davies.html | 1,568,900,957,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573519.72/warc/CC-MAIN-20190919122032-20190919144032-00295.warc.gz | 474,769,544 | 14,096 | 139) Not only is the disappearance of ship’s hulls explained by the Law of Perspective on flat surfaces, it is proven undeniably true with the aid of a good telescope. If you watch a ship sailing away into the horizon with the naked eye until its hull has completely disappeared from view under the supposed “curvature of the Earth,” then look through a telescope, you will notice the entire ship quickly zooms back into view, hull and all, proving that the disappearance was caused by the Law of Perspective, not by a wall of curved water! This also proves that the horizon is simply the vanishing line of perspective from your point of view, NOT the alleged “curvature” of Earth.
Another false law of Newton is that gravity increases with the increase of the mass of the object. There is no such thing as mass (no one in the world can define it) - there is only density of the object (total density volume of the object, including it's electric field that surrounds it), and it is enough to understand how the laws work. Rubber ball pumped with helium goes up irrespective of the "gravity law" which supposed to bring everything down. Ball goes up because the density of the helium is smaller than the density of air above it. There is also no resistance of the environment above the ball.
55.) The Newtonian theory of astronomy requires that the Moon "borrow" her light from the Sun. Now, since the Sun's rays are hot and the Moon's light sends with it no heat at all, it follows that the Sun and Moon are "two great lights," as we somewhere read; that the Newtonian theory is a mistake; and that, therefore, we have a proof that the Earth is not a globe.
###### 114) Quoting, “On the False Wisdom of the Philosophers” by Lacantius, “A sphere where people on the other side live with their feet above their heads, where rain, snow and hail fall upwards, where trees and crops grow upside-down and the sky is lower than the ground? The ancient wonder of the hanging gardens of Babylon dwindle into nothing in comparison to the fields, seas, towns and mountains that pagan philosophers believe to be hanging from the earth without support!”
64) Quoting “Earth Not a Globe!” by Samuel Rowbotham, “It is known that the horizon at sea, whatever distance it may extend to the right and left of the observer on land, always appears as a straight line. The following experiment has been tried in various parts of the country. At Brighton, on a rising ground near the race course, two poles were fixed in the earth six yards apart, and directly opposite the sea. Between these poles a line was tightly stretched parallel to the horizon. From the center of the line the view embraced not less than 20 miles on each side making a distance of 40 miles. A vessel was observed sailing directly westwards; the line cut the rigging a little above the bulwarks, which it did for several hours or until the vessel had sailed the whole distance of 40 miles. The ship coming into view from the east would have to ascend an inclined plane for 20 miles until it arrived at the center of the arc, whence it would have to descend for the same distance. The square of 20 miles multiplied by 8 inches gives 266 feet as the amount the vessel would be below the line at the beginning and at the end of the 40 miles.”
The ancient Hebrew view of the earth was of a flat earth with a glass firmament that separated the “waters below” from the “waters above” (Gen. 1:6-7; Job 37:28), and in which the sun and moon were placed and had their daily circuit around and above the Flat earth (Gen. 1:14-18; Joshua 10:13; Job 22:14). The earth did not move and was set on a firm foundation, on pillars in the waters of the “deep”, a great ocean underneath the flat earth (Gen. 1:2; 1 Samuel 2:8; 1 Chronicles 16:30; Psalm 104:5; Proverbs 8:28-29). They did not believe in a spinning globe. If the false globe ideas were taught in ancient times, they were taught by the Babylonian Pagans, not God’s people.
17) “Olber’s Paradox” states that if there were billions of stars which are suns the night sky would be filled completely with light. As Edgar Allen Poe said, “Were the succession of stars endless, then the background of the sky would present us a uniform luminosity, since there could exist absolutely no point, in all that background, at which would not exist a star.” In fact Olber’s “Paradox” is no more a paradox than George Airy’s experiment was a “failure.” Both are actually excellent refutations of the heliocentric spinning ball model.
94) From the highland near Portsmouth Harbor in Hampshire, England looking across Spithead to the Isle of Wight, the entire base of the island, where water and land come together composes a perfectly straight line 22 statute miles long. According to the ball-Earth theory, the Isle of Wight should decline 80 feet from the center on each side to account for the necessary curvature. The cross-hairs of a good theodolite directed there, however, have repeatedly shown the land and water line to be perfectly level.
32.) It is often said that, if the Earth were flat, we could see all over it! This is the result of ignorance. If we stand on the level surface a plain or a prairie, and take notice, we shall find that the horizon is formed at about three miles all around us: that is, the ground appears to rise up until, at that distance, it seems on a level with the eye-line or line of sight. Consequently, objects no higher than we stand – say, six feet – and which are at that distance (three miles), have reached the "vanishing point," and are beyond the sphere of our unaided vision. This is the reason why the hull of a ship disappears (in going away from us) before the sails; and, instead of there being about it the faintest shadow of evidence of the, Earth's rotundity, it is a clear proof that Earth is not a globe.
It is a fact not so well known as it ought to be that when a ship, in sailing away from us, has reached the point at which her hull is lost to our unaided vision, a good telescope will restore to our view this portion of the vessel. Now, since telescopes are not made to enable people to see through a "hill of water," it is clear that the hulls of ships are not behind a hill of water when they can be seen through a telescope though lost to our unaided vision. This is a proof that Earth is not a globe.
179) If the Earth were constantly spinning Eastwards 1000mph then airplane flight durations going Eastwards vs. Westwards should be significantly different. If the average commercial airliner travels 500mph, it follows that Westbound equatorial flights should reach their destination at approximately thrice the speed as their Eastbound return flights. In reality, however, the differences in East/Westbound flight durations usually amount to a matter of minutes, and nothing near what would occur on a 1000mph spinning ball Earth.
43.) The circumstances which attend bodies which are caused merely to fall from a great height prove nothing as to the motion or stability of the Earth, since the object, if it be on a thing that is in motion, will participate in that motion; but, if an object be thrown, upwards from a body at rest, and, again, from a body in motion, the circumstances attending its descent will be very different. In the former case, it will fall, if thrown vertically upwards, at the place from whence it was projected; in the latter case, it will fall behind the moving body from which it is thrown will leave it in the rear. Now, fix a gun, muzzle upwards, accurately, in the ground; fire off a projectile; and it will fall by the gun. If the Earth traveled eleven hundred miles a minute, the projectile would fall behind the gun, in the opposite direction to that of the supposed motion. Since, then, this is NOT the case, in fact, the Earth's fancied motion is negatived and we have a proof that the Earth is not a, globe.
11) A surveyor and engineer of thirty years published in the Birmingham Weekly Mercury stated, “I am thoroughly acquainted with the theory and practice of civil engineering. However bigoted some of our professors may be in the theory of surveying according to the prescribed rules, yet it is well known amongst us that such theoretical measurements are INCAPABLE OF ANY PRACTICAL ILLUSTRATION. All our locomotives are designed to run on what may be regarded as TRUE LEVELS or FLATS. There are, of course, partial inclines or gradients here and there, but they are always accurately defined and must be carefully traversed. But anything approaching to eight inches in the mile, increasing as the square of the distance, COULD NOT BE WORKED BY ANY ENGINE THAT WAS EVER YET CONSTRUCTED. Taking one station with another all over England and Scotland, it may be stated that all the platforms are ON THE SAME RELATIVE LEVEL. The distance between Eastern and Western coasts of England may be set down as 300 miles. If the prescribed curvature was indeed as represented, the central stations at Rugby or Warwick ought to be close upon three miles higher than a chord drawn from the two extremities. If such was the case there is not a driver or stoker within the Kingdom that would be found to take charge of the train. We can only laugh at those of your readers who seriously give us credit for such venturesome exploits, as running trains round spherical curves. Horizontal curves on levels are dangerous enough, vertical curves would be a thousand times worse, and with our rolling stock constructed as at present physically impossible.”
126) The Sun’s annual journey from tropic to tropic, solstice to solstice, is what determines the length and character of days, nights and seasons. This is why equatorial regions experience almost year-round summer and heat while higher latitudes North and especially South experience more distinct seasons with harsh winters. The heliocentric model claims seasons change based on the ball-Earth’s alleged “axial tilt” and “elliptical orbit” around the Sun, yet their flawed current model places us closest to the Sun (91,400,000 miles) in January when its actually winter, and farthest from the Sun (94,500,000 miles) in July when its actually summer throughout most of the Earth.
so if the earth is not a sphere in space revolveing around the sphereical sun, then what is it. Its one thing to say that "its not that way" but its different to say "its actually this way not that way". So what way is it? what way are you proposing is the correct way? do you beleive this is the only planet in the universe? do you believe that the stars are only decorations on a flat backdrop? I'm not certain what idea you are proposing is the correct way of looking at this...
Starting with Indelibly Stamped in 1971, Davies shared lead vocals with Supertramp songwriting partner, Roger Hodgson until the latter's departure in 1983,[4] at which point he became the sole lead vocalist of the group. Davies's voice is deeper than Hodgson's, and he usually employs a raspy baritone which stands in stark contrast to his bandmate's tenor. However, he occasionally sings in a falsetto which superficially resembles Hodgson's vocals, such as on "Goodbye Stranger" and "My Kind of Lady". He also plays harmonica for the group.
Both Davies and Hodgson talked of a reunion a couple of times, however, this would never come to pass. The first hint of a reunion came in 1993 when Davies and Hodgson reunited for an A & M dinner honoring Jerry Moss, co-founder of A & M Records. This dinner resulted in writing and demoing new songs, but it never went anywhere due to disagreements over management. Another hint of a reunion came in 2010 when Roger Hodgson approached Rick Davies about a fortieth anniversary of their very first album Supertramp (rogerhodgson.com). Rick Davies declined the invitation and any chance of Supertramp reuniting was squashed.
```106) The so-called “South Pole” is simply an arbitrary point along the Antarctic ice marked with a red and white barbershop pole topped with a metal ball-Earth. This ceremonial South Pole is admittedly and provably NOT the actual South Pole, however, because the actual South Pole could be demonstrably confirmed with the aid of a compass showing North to be 360 degrees around the observer. Since this feat has never been achieved, the model remains pure theory, along with the establishment’s excuse that the geomagnetic poles supposedly constantly move around making verification of their claims impossible.
```
If the Moon is a disc which faces us and whose orientation can be explained, for instance, as the head of a tennis racket as someone holds it out and turns, thus the same side of the head always faces to the inside of the circle path it travels, and if this disc faces down all the time on our FE, what purpose the craters and how did they get there?
God's Truth never - no, never - requires a falsehood to help it along. Mr. Proctor, in his " Lessons," says: Men " have been able to go round and round the Earth in several directions." Now, in this case, the word " several will imply more than two, unquestionably: whereas, it is utterly impossible to circumnavigate the Earth in any other than an easterly or a westerly direction; and the fact is perfectly consistent and clear in its relation to Earth as a Plane.. Now, since astronomers would not be so foolish as to damage a good cause by misrepresentation, it is presumptive evidence that their cause is a bad one, and - a proof that Earth is not a globe.
24.) When a man speaks of a "most complete" thing amongst several other things which claim to be what that thing is, it is evident that they must fall short of something which the "most complete" thing possesses. And when it is known that the "most complete" thing is an entire failure, it is plain that the others, all and sundry, are worthless. Proctor's "most complete proof that the Earth is a globe" lies in what he calls "the fact" that distances from place to place agree with calculation. But, since the distance round the Earth at 45 " degrees" south of the equator is twice the distance it would be on a globe, it follows that what the greatest astronomer of the age calls "a fact" is NOT a fact; that his "most complete proof' is a most complete failure; and that be might as well have told us, at once, that he has NO PROOF to give us at all. Now, since, if the Earth be a globe, there would, necessarily, be piles of proofs of it all round us, it follows that when astronomers, with all their ingenuity, are utterly unable to point one out – to say nothing about picking one up – that they give us a proof that Earth is not a globe.
## 172) If you pick any cloud in the sky and watch for several minutes, two things will happen: the clouds will move and they will morph gradually changing shape. In official NASA footage of the spinning ball Earth, such as the “Galileo” time-lapse video however, clouds are constantly shown for 24+ hours at a time and not moving or morphing whatsoever! This is completely impossible, further proof that NASA produces fake CGI videos, and further evidence that Earth is not a spinning ball.
The only explanation which has been given of this phenomenon is the refraction caused by the earth’s atmosphere. This, at first sight, is a plausible and fairly satisfactory solution; but on carefully examining the subject, it is found to be utterly inadequate; and those who have recourse to it cannot be aware that the refraction of an object and that of a shadow are in opposite directions. An object by refraction is bent upwards; but the shadow of any object is bent downwards, as will be seen by the following very simple experiment. Take a plain white shallow basin, and place it ten or twelve inches from a light in such a position that the shadow of the edge of the basin touches the centre of the bottom. Hold a rod vertically over and on the edge of the shadow, to denote its true position. Now let water be gradually poured into the basin, and the shadow will be seen to recede or shorten inwards and downwards; but if a rod or a spoon is allowed to rest, with its upper end towards the light, and the lower end in the bottom of the vessel, it will be seen, as the water is poured in, to bend upwards–thus proving that if refraction operated at all, it would do so by elevating the moon above its true position, and throwing the earth’s shadow downwards, or directly away from the moon’s surface. Hence it is clear that a lunar eclipse by a shadow of the earth is an utter impossibility.
# It is plain that a theory of measurements without a measuring-rod is like a ship without a rudder; that a measure that is not fixed, not likely to be fixed, and never has been fixed, forms no measuring-rod at all; and that as modern theoretical astronomy depends upon the Sun's distance from the Earth as its measuring-rod, and the distance is not known, it is a system of measurements without a measuring-rod - a ship without a rudder. Now, since it is not difficult to foresee the dashing of this thing upon the rock on which Zetetic astronomy is founded, it is a proof that Earth is not a globe.
```The amazing thing is that we can truly find God when we discover His true creation. There is no longer a false foundation for Atheism and Evolution. God’s Flat Earth can only exist by design, and knowing the reality of His creation leads us to the Bible and salvation in Jesus! The sun is IN THE FIRMAMENT as described in Genesis 1! As you can see in the picture below, the sun is small and close, in the firmament, just as the Bible describes. You can see the sun’s hot spot directly underneath the sun on the clouds, proving it is very close.
```
69) The New York City skyline is clearly visible from Harriman State Park’s Bear Mountain 60 miles away. If Earth were a ball 25,000 miles in circumference, viewing from Bear Mountain’s 1,283 foot summit, the Pythagorean Theorem determining distance to the horizon being 1.23 times the square root of the height in feet, the NYC skyline should be invisible behind 170 feet of curved Earth.
```93) The St. George’s Channel between Holyhead and Kingstown Harbor near Dublin is 60 miles across. When half-way across a ferry passenger will notice behind them the light on Holyhead pier as well as in front of them the Poolbeg light in Dublin Bay. The Holyhead Pier light is 44 feet high, while the Poolbeg lighthouse 68 feet, therefore a vessel in the middle of the channel, 30 miles from either side standing on a deck 24 feet above the water, can clearly see both lights. On a ball Earth 25,000 miles in circumference, however, both lights should be hidden well below both horizons by over 300 feet!
```
130) From “Earth Not a Globe!” by Samuel Rowbotham, “Take two carefully-bored metallic tubes, not less than six feet in length, and place them one yard asunder, on the opposite sides of a wooden frame, or a solid block of wood or stone: so adjust them that their centres or axes of vision shall be perfectly parallel to each other. Now, direct them to the plane of some notable fixed star, a few seconds previous to its meridian time. Let an observer be stationed at each tube and the moment the star appears in the first tube let a loud knock or other signal be given, to be repeated by the observer at the second tube when he first sees the same star. A distinct period of time will elapse between the signals given. The signals will follow each other in very rapid succession, but still, the time between is sufficient to show that the same star is not visible at the same moment by two parallel lines of sight when only one yard asunder. A slight inclination of the second tube towards the first tube would be required for the star to be seen through both tubes at the same instant. Let the tubes remain in their position for six months; at the end of which time the same observation or experiment will produce the same results--the star will be visible at the same meridian time, without the slightest alteration being required in the direction of the tubes: from which it is concluded that if the earth had moved one single yard in an orbit through space, there would at least be observed the slight inclination of the tube which the difference in position of one yard had previously required. But as no such difference in the direction of the tube is required, the conclusion is unavoidable, that in six months a given meridian upon the earth's surface does not move a single yard, and therefore, that the earth has not the slightest degree of orbital motion."
55.) The Newtonian theory of astronomy requires that the Moon "borrow" her light from the Sun. Now, since the Sun's rays are hot and the Moon's light sends with it no heat at all, it follows that the Sun and Moon are "two great lights," as we somewhere read; that the Newtonian theory is a mistake; and that, therefore, we have a proof that the Earth is not a globe.
# There is much real evidence to prove the Flat stationary earth and it’s not just photos and hearsay. Many professional pilots and military personnel have come out as Flat Earthers, and have confirmed that there is no curvature on the earth, flights don’t make sense on a ball, and GPS is ground-based and more. Here are some of these very interesting interviews by Mark Sargent, a Flat Earther. They bring up some very compelling evidence and that will make even the most staunch Ball Earther think twice, or at least we can hope so. If anything, it should make you think about where we really live. These interviews are not made up or staged. They are real! You can prove their evidence correct with your own research.
```The way our vision works makes everything converge to a single vanishing point on the flat horizon, including airplanes and the sun. Artists understand this. Airplanes appear to drop below the horizon when in reality they are flying level to the flat earth and never dip their noses down to account for any supposed curve. It's the same with the sun. It is moving across the sky on a flat circular path but it appears to rise and fall due to perspective.
```
146) The ball-Earth model claims the Moon orbits around the Earth once every 28 days, yet it is plain for anyone to see that the Moon orbits around the Earth every single day! The Moon’s orbit is slightly slower than the Sun’s, but follows the Sun’s same path from Tropic to Tropic, solstice to solstice, making a full circle over the Earth in just under 25 hours.
When astronomers assert that it is "necessary" to make "allowance for curvature" in canal construction, it is, of course, in order that, in their idea, a level cutting may be had, for the water. How flagrantly, then, do they contradict themselves when the curved surface of the Earth is a "true level!" What more can they want for a canal than a true level? Since they contradict themselves in such an elementary point as this, it is an evidence that the whole thing is a delusion, and we have a proof that the Earth is not a globe.
The group had relocated to the United States by 1977, and it was there that they recorded their best-selling album, Breakfast in America. With more hit singles than their first five albums combined, it reached number three in the UK,[10] and top of the charts in America. The album is reckoned to have sold over 20 million copies since its release on 29 March 1979.[12]
80.) It is supposed,"in the regular course of the Newtonian theory, that the Earth is, in June, about 190 millions of miles (190,000,000) away from its position in December. Now, since we can, (in middle north latitudes), see the North Star, on looking out of a window that faces it – and out of the very same corner of the very same pane of glass in the very same window – all the year round, it is proof enough for any man in his senses that we have made no motion at all. It is a proof that the Earth is not a globe.
There is much real evidence to prove the Flat stationary earth and it’s not just photos and hearsay. Many professional pilots and military personnel have come out as Flat Earthers, and have confirmed that there is no curvature on the earth, flights don’t make sense on a ball, and GPS is ground-based and more. Here are some of these very interesting interviews by Mark Sargent, a Flat Earther. They bring up some very compelling evidence and that will make even the most staunch Ball Earther think twice, or at least we can hope so. If anything, it should make you think about where we really live. These interviews are not made up or staged. They are real! You can prove their evidence correct with your own research.
136) Many people think that modern astronomy’s ability to accurately predict lunar and solar eclipses is a result and proof positive of the heliocentric theory of the universe. The fact of the matter however is that eclipses have been accurately predicted by cultures worldwide for thousands of years before the “heliocentric ball-Earth” was even a glimmer in Copernicus’ imagination. Ptolemy in the 1st century A.D. accurately predicted eclipses for six hundred years on the basis of a flat, stationary Earth with equal precision as anyone living today. All the way back in 600 B.C. Thales accurately predicted an eclipse which ended the war between the Medes and Lydians. Eclipses happen regularly with precision in 18 year cycles, so regardless of geocentric or heliocentric, flat or globe Earth cosmologies, eclipses can be accurately calculated independent of such factors.
It is a well-known fact that clouds are continually seen moving in all manner of directions - yes, and frequently, in different directions at the same time - from west to east being as frequent a direction as any other. . Now, if the Earth were a globe, revolving through space from west to east at the rate of nineteen miles in a second, the clouds appearing to us to move towards the east would have to move quicker than nineteen miles in a second to be thus seen; whilst those which appear to be moving in the opposite direction would have no necessity to be moving at all, since the motion of the Earth would be more than sufficient to cause the appearance. But it only takes a little common sense to show us that it is the clouds that move just as they appear to do, and that, therefore, the Earth is motionless. We have, then a proof that the Earth is not a globe.
## 32.) It is often said that, if the Earth were flat, we could see all over it! This is the result of ignorance. If we stand on the level surface a plain or a prairie, and take notice, we shall find that the horizon is formed at about three miles all around us: that is, the ground appears to rise up until, at that distance, it seems on a level with the eye-line or line of sight. Consequently, objects no higher than we stand – say, six feet – and which are at that distance (three miles), have reached the "vanishing point," and are beyond the sphere of our unaided vision. This is the reason why the hull of a ship disappears (in going away from us) before the sails; and, instead of there being about it the faintest shadow of evidence of the, Earth's rotundity, it is a clear proof that Earth is not a globe.
```5) The sun is much closer than we have been told. It is, in fact, in our atmosphere. You can clearly see that it is not 93 million miles away. Many times you can see the sun’s rays shooting out of a cloud forming a triangle. If you follow the rays to their source it will always lead to a place above the clouds. If the sun was truly millions of miles away, all the rays would come in at a straight angle. Also the sun can be seen directly above clouds in some balloon photos, creating a hot spot on the clouds below it and in other photos you can clearly see the clouds dispersing directly underneath the close small sun.
```
95) On a clear day from the highland near Douglas Harbor on the Isle of Man, the whole length of the coast of North Wales is often plainly visible to the naked eye. From the Point of Ayr at the mouth of the River Dee to Holyhead comprises a 50 mile stretch which has also been repeatedly found to be perfectly horizontal. If the Earth actually had curvature of 8 inches per mile squared, as NASA and modern astronomy claim, the 50 mile length of Welsh coast seen along the horizon in Liverpool Bay would have to decline from the center-point an easily detectable 416 feet on each side!
55) If the Sun circles over and around the Earth every 24 hours, steadily travelling from Tropic to Tropic every 6 months, it follows that the Northern, central region would annually receive far more heat and sunlight than the Southern circumferential region. Since the Sun must sweep over the larger Southern region in the same 24 hours it has to pass over the smaller Northern region, its passage must necessarily be proportionally faster as well. This perfectly explains the differences in Arctic/Antarctic temperatures, seasons, length of daylight, plant and animal life; this is why the Antarctic morning dawn and evening twilight are very abrupt compared with the North; and this explains why many midsummer Arctic nights the Sun does not set at all!
Both Davies and Hodgson talked of a reunion a couple of times, however, this would never come to pass. The first hint of a reunion came in 1993 when Davies and Hodgson reunited for an A & M dinner honoring Jerry Moss, co-founder of A & M Records. This dinner resulted in writing and demoing new songs, but it never went anywhere due to disagreements over management. Another hint of a reunion came in 2010 when Roger Hodgson approached Rick Davies about a fortieth anniversary of their very first album Supertramp (rogerhodgson.com). Rick Davies declined the invitation and any chance of Supertramp reuniting was squashed. | 6,520 | 29,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-39 | latest | en | 0.962528 |
https://ninetynine.haskell.chungyc.org/src/Problems.P50.html | 1,723,346,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00256.warc.gz | 325,283,680 | 5,585 | ```{- |
Description: Huffman codes
Copyright: Copyright (C) 2021 Yoo Chung
License: GPL-3.0-or-later
Maintainer: dev@chungyc.org
Part of Ninety-Nine Haskell "Problems". Some solutions are in "Solutions.P50".
-}
module Problems.P50
( huffman
-- * Supporting functions
-- | The functions below are not part of the problem.
-- Instead, they are used to illustrate the use of Huffman coding.
, countCharacters
, encodeHuffman
, decodeHuffman
, loweralpha
, ascii
, text
) where
import Data.List (group, sort)
import Data.Map.Lazy (Map)
import qualified Data.Map.Lazy as Map
import Data.Maybe (mapMaybe)
import qualified Solutions.P50 as Solution
{- |
Given a list of characters and their number of occurrences,
construct a list of the characters and their [Huffman encoding](https://brilliant.org/wiki/huffman-encoding/).
In the encoding, @'0'@ and @'1'@ will denote the 0 and 1 bits.
=== Examples
>>> huffman [('a',45),('b',13),('c',12),('d',16),('e',9),('f',5)]
[('a',"0"),('b',"101"),('c',"100"),('d',"111"),('e',"1101"),('f',"1100")]
The encoding table computed by 'huffman' can be used to compress data:
>>> length \$ encodeHuffman (huffman \$ countCharacters text) text
3552
Compare this to the length of a fixed-length 5-bit encoding:
>>> length \$ encodeHuffman loweralpha text
4375
or the length of the more standard ASCII encoding with 8 bits:
>>> length \$ encodeHuffman ascii text
7000
Huffman coding is unambiguous, so we can get back the original text:
>>> let table = huffman \$ countCharacters text
>>> let encodedText = encodeHuffman table text
>>> let decodedText = decodeHuffman table encodedText
>>> decodedText == text
True
-}
huffman :: [(Char,Int)] -> [(Char,String)]
huffman :: [(Char, Int)] -> [(Char, String)]
huffman = [(Char, Int)] -> [(Char, String)]
Solution.huffman
-- | Count the number of occurrences of a character in a string.
countCharacters :: String -> [(Char,Int)]
countCharacters :: String -> [(Char, Int)]
countCharacters String
s = (String -> Maybe (Char, Int)) -> [String] -> [(Char, Int)]
forall a b. (a -> Maybe b) -> [a] -> [b]
mapMaybe String -> Maybe (Char, Int)
forall {a}. [a] -> Maybe (a, Int)
count ([String] -> [(Char, Int)]) -> [String] -> [(Char, Int)]
forall a b. (a -> b) -> a -> b
\$ String -> [String]
forall a. Eq a => [a] -> [[a]]
group (String -> [String]) -> String -> [String]
forall a b. (a -> b) -> a -> b
\$ String -> String
forall a. Ord a => [a] -> [a]
sort String
s
where count :: [a] -> Maybe (a, Int)
count [] = Maybe (a, Int)
forall a. Maybe a
Nothing
count xs :: [a]
xs@(a
x:[a]
_) = (a, Int) -> Maybe (a, Int)
forall a. a -> Maybe a
Just (a
x, [a] -> Int
forall a. [a] -> Int
forall (t :: * -> *) a. Foldable t => t a -> Int
length [a]
xs)
-- | Given an encoding table and a string, encode the string.
--
-- While this is intended for use in illustrating Huffman coding, it is not limited to such.
-- In particular, it can encode with a fixed-length encoding table.
encodeHuffman :: [(Char,String)] -> String -> String
encodeHuffman :: [(Char, String)] -> String -> String
encodeHuffman [(Char, String)]
table String
string = Map Char String -> String -> String -> String
encodeHuffman' ([(Char, String)] -> Map Char String
forall k a. Ord k => [(k, a)] -> Map k a
Map.fromList [(Char, String)]
table) String
string String
""
encodeHuffman' :: Map Char String -> String -> String -> String
encodeHuffman' :: Map Char String -> String -> String -> String
encodeHuffman' Map Char String
_ String
"" String
encoded = String -> String
forall a. [a] -> [a]
reverse String
encoded
encodeHuffman' Map Char String
table (Char
c:String
string) String
encoded =
case Char -> Map Char String -> Maybe String
forall k a. Ord k => k -> Map k a -> Maybe a
Map.lookup Char
c Map Char String
table of
Maybe String
Nothing -> String
forall a. HasCallStack => a
undefined
Just String
e -> Map Char String -> String -> String -> String
encodeHuffman' Map Char String
table String
string (String -> String) -> String -> String
forall a b. (a -> b) -> a -> b
\$ String -> String
forall a. [a] -> [a]
reverse String
e String -> String -> String
forall a. [a] -> [a] -> [a]
++ String
encoded
-- | Given an encoding table and a string, decode the string.
--
-- While this is intended for use in illustrating Huffman coding, it is not limited to such.
-- In particular, it can decode with a fixed-length encoding table.
decodeHuffman :: [(Char,String)] -> String -> String
decodeHuffman :: [(Char, String)] -> String -> String
decodeHuffman [(Char, String)]
table String
string = Map String Char -> String -> String -> String -> String
decodeHuffman' ([(String, Char)] -> Map String Char
forall k a. Ord k => [(k, a)] -> Map k a
Map.fromList ([(String, Char)] -> Map String Char)
-> [(String, Char)] -> Map String Char
forall a b. (a -> b) -> a -> b
\$ ((Char, String) -> (String, Char))
-> [(Char, String)] -> [(String, Char)]
forall a b. (a -> b) -> [a] -> [b]
map (\(Char
x,String
y) -> (String -> String
forall a. [a] -> [a]
reverse String
y, Char
x)) [(Char, String)]
table) String
"" String
string String
""
decodeHuffman' :: Map String Char -> String -> String -> String -> String
decodeHuffman' :: Map String Char -> String -> String -> String -> String
decodeHuffman' Map String Char
_ String
"" String
"" String
decoded = String -> String
forall a. [a] -> [a]
reverse String
decoded
decodeHuffman' Map String Char
table String
code String
encoded String
decoded =
case String -> Map String Char -> Maybe Char
forall k a. Ord k => k -> Map k a -> Maybe a
Map.lookup String
code Map String Char
table of
Maybe Char
Nothing -> case String
encoded of
(Char
c:String
encoded') -> Map String Char -> String -> String -> String -> String
decodeHuffman' Map String Char
table (Char
cChar -> String -> String
forall a. a -> [a] -> [a]
:String
code) String
encoded' String
decoded
String
"" -> String
forall a. HasCallStack => a
undefined
Just Char
c -> Map String Char -> String -> String -> String -> String
decodeHuffman' Map String Char
table String
"" String
encoded (Char
cChar -> String -> String
forall a. a -> [a] -> [a]
:String
decoded)
-- | Fixed-length encoding of lower case letters and a space using 5 bits.
loweralpha :: [(Char,String)]
loweralpha :: [(Char, String)]
loweralpha = [Char -> (Char, String)
forall {a}. Enum a => a -> (a, String)
encode Char
c | Char
c <- Char
' ' Char -> String -> String
forall a. a -> [a] -> [a]
: [Char
'a'..Char
'z']]
where encode :: a -> (a, String)
encode a
c = (a
c, Int -> String
toBits (Int -> String) -> Int -> String
forall a b. (a -> b) -> a -> b
\$ a -> Int
forall a. Enum a => a -> Int
fromEnum a
c Int -> Int -> Int
forall a. Num a => a -> a -> a
- Char -> Int
forall a. Enum a => a -> Int
fromEnum Char
'a')
toBits :: Int -> String
toBits Int
n = Int -> Int -> String
getBits Int
n Int
5
-- | Fixed-length encoding of ASCII characters using 8 bits.
ascii :: [(Char,String)]
ascii :: [(Char, String)]
ascii = [Char -> (Char, String)
forall {a}. Enum a => a -> (a, String)
encode Char
c | Char
c <- [Char
'\0'..Char
'\127']]
where encode :: a -> (a, String)
encode a
c = (a
c, Int -> String
toBits (Int -> String) -> Int -> String
forall a b. (a -> b) -> a -> b
\$ a -> Int
forall a. Enum a => a -> Int
fromEnum a
c Int -> Int -> Int
forall a. Num a => a -> a -> a
- Char -> Int
forall a. Enum a => a -> Int
fromEnum Char
'\0')
toBits :: Int -> String
toBits Int
n = Int -> Int -> String
getBits Int
n Int
8
getBits :: Int -> Int -> String
getBits :: Int -> Int -> String
getBits Int
_ Int
0 = []
getBits Int
n Int
b = Char
bit Char -> String -> String
forall a. a -> [a] -> [a]
: Int -> Int -> String
getBits (Int
n Int -> Int -> Int
forall a. Integral a => a -> a -> a
`div` Int
2) (Int
bInt -> Int -> Int
forall a. Num a => a -> a -> a
-Int
1)
where bit :: Char
bit | Int
n Int -> Int -> Int
forall a. Integral a => a -> a -> a
`mod` Int
2 Int -> Int -> Bool
forall a. Eq a => a -> a -> Bool
== Int
1 = Char
'1'
| Bool
otherwise = Char
'0'
-- | Long text against which various encoding schemes can be tried.
text :: String
text :: String
text =
String
"this is going to be a very long string of text which tries to use all letters " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"of the alphabet there is no punctuation and no upper case letters because i " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"did not want to write more code to create the encoding table and it also means " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"it can use fewer bits to encode each letter with a fixed size encoding " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"using most of the letters means that the five bit encoding is the smallest " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"fixed size encoding for this text and i actually find it easier to write out " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"this text randomly instead of carefully gathering the letters used by a certain " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"text and getting a smaller fixed size encoding from just those letters " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"it is trying to be zany for the sake of being zany being quite long it is also " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"more convincing that the extra space for the huffman coding table can be worth " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"the extra cost because it is less than the savings we get from encoding a long text " String -> String -> String
forall a. [a] -> [a] -> [a]
++
String
"with huffman coding"
``` | 2,870 | 9,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-33 | latest | en | 0.736541 |
https://infocenter-archive.sybase.com/help/topic/com.sybase.dc33621_33620_33619_1250/html/ptallbk/X17690.htm | 1,701,924,370,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00172.warc.gz | 358,790,926 | 2,227 | # subq
### Description
Identifies a subquery.
### Syntax
```( subq subquery_id
)
```
### Parameters
subquery_id
is an integer identifying the subquery. In abstract plans, subquery numbering is based on the order of the leading parenthesis for the subqueries in a query.
### Examples
#### Example 1
```select c11 from t1
where c12 =
(select c21 from t2 where c22 = t1.c11)
```
```( nested
( t_scan t1 )
( subq 1
( t_scan ( table t2 ( in ( subq 1 ) ) ) )
)
)
```
A single nested subquery.
#### Example 2
```select c11 from t1
where c12 =
(select c21 from t2 where c22 = t1.c11)
and c12 =
(select c31 from t3 where c32 = t1.c11)
```
```( nested
( nested
( t_scan t1 )
( subq 1
( t_scan ( table t2 ( in ( subq 1 ) ) ) )
)
)
( subq 2
( t_scan ( table t3 ( in ( subq 2 ) ) ) )
)
)
```
The two subqueries are both nested in the main query.
#### Example 3
```select c11 from t1
where c12 =
(select c21 from t2 where c22 =
(select c31 from t3 where c32 = t1.c11))
```
```( nested
( t_scan t1 )
( subq 1
( nested
( t_scan ( table t2 ( in ( subq 1 ) ) ) )
( subq 2
( t_scan ( table t3 ( in ( subq 2 ) ) ) )
)
)
)
)
```
A level 2 subquery nested into a level 1 subquery nested in the main query.
### Usage
• The subq operator has two meanings in an abstract plan expression:
• Under a nested operator, it describes the attachment of a nested subquery to a table
• Under an in operator, it describes the nesting of the base tables and views that the subquery contains
• To specify the attachment of a subquery without providing a plan specification, use an empty hint:
```( nested
( t_scan t1)
( subq 1
()
)
```
```)
```
• To provide a description of the abstract plan for a subquery, without specifying its attachment, specify an empty hint as the derived table in the nested operator:
```( nested
()
( subq 1
(t_scan ( table t1 ( in ( subq 1 ) ) ) )
)
)
```
• When subqueries are flattened to a join, the only reference to the subquery in the abstract plan is the identification of the table specified in the subquery:
```select *
from t2
where c21 in (select c12 from t1)
```
```( nl_g_join
( t_scan t1 )
( t_scan ( table t2 ( in ( subq 1 ) ) ) )
```
• When a subquery is materialized, the subquery appears in the store operation, identifying the table to be scanned during the materialization step:
```select *
from t1
where c11 in (select max(c22) from t2 group by c21)
```
```( plan
( store Worktab1
( t_scan ( table t2 ( in ( subq 1 ) ) ) )
)
( nl_g_join
( t_scan t1 )
( t_scan ( work_t Worktab1 ) )
)
)
``` | 774 | 2,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | latest | en | 0.815219 |
https://www.formulas.today/formulas/ModulationTransferFunction/ | 1,726,619,369,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00554.warc.gz | 716,438,513 | 4,671 | ## Unlocking the Mystery of MTF: How Modulation Transfer Function Shapes Optical Clarity
Contrast In Image: Contrast In Object:
Output: `Press calculate`
# Understanding the Modulation Transfer Function (MTF) in Optics
When you look at the clarity and sharpness of an image, whether in photography, microscopy, or astronomy, what you often scrutinize is the Modulation Transfer Function (MTF). The MTF is pivotal in optics, as it measures the ability of an optical system to transfer various levels of detail from the object to the image. This metric gauges the performance of lenses and imaging systems and answers the ultimate question: How well does this lens capture fine details?
Formula:`MTF = (contrastInImage / contrastInObject) × 100%`
## Decoding the MTF Formula
The MTF is expressed as a percentage and is calculated using the formula:
``MTF = (contrastInImage / contrastInObject) × 100%``
Where:
• `contrastInImage` represents the contrast as observed in the produced image.
• `contrastInObject` denotes the contrast of the original object being imaged.
Let’s break down the inputs:
• Contrast in Image: This is measured by the difference in luminance or color that makes an object distinguishable in the produced image. It ranges from 0 to 1, where 0 means no contrast and 1 means maximum contrast.
• Contrast in Object: Similar to the image contrast, this measures the original object’s intrinsic contrast before it has passed through the optical system.
## What's the Output?
The output, `MTF`, is a percentage. A higher MTF percentage indicates better optical performance, meaning the optical system can transfer higher detail levels from the object to the image.
## Real Life Example: The Photographer's Lens
Consider a photographer capturing breathtaking landscapes. Their passion lies in ensuring each leaf, mountain ridge, and wave crest is crisp and detailed. Here's how MTF comes into play:
``MTF = (contrast in image / contrast in object) × 100%``
Suppose the photographer's lens displays an `MTF` value of `70%` when capturing a high contrast object (e.g., a checkerboard pattern with black and white squares). This means 70% of the original contrast detail is preserved in the final photograph.
## Detailed Example Calculation
Imagine we have a high contrast checkerboard pattern with black and white squares. This pattern passes through an optical system, and we wish to calculate the MTF. If the measure of contrast in the original pattern (Contrast in Object) is 1 and the measured contrast in the resulting image (Contrast in Image) is 0.85, we can use the MTF formula:
``MTF = (0.85 / 1) × 100% = 85%``
An MTF of 85% means the optical system retains a significant portion of the original contrast, resulting in a sharp, detailed image.
## Applications in Various Fields
The importance of MTF extends beyond photography and is vital in:
• Microscopy: Researchers rely on high MTF values to observe cellular structures accurately.
• Astrophotography: High MTF values help capture the fine details of celestial bodies.
• Surveillance Systems: Ensuring security cameras deliver sharp images with high detail fidelity.
### Why is MTF crucial in optics?
MTF quantifies the clarity and sharpness of images, directly reflecting an optical system's ability to transfer detail.
### How is MTF typically measured?
MTF is often measured using test patterns, like bar targets or edge charts, to evaluate an optical system's performance.
### What factors affect MTF?
Several factors, including lens quality, aperture settings, and diffraction, can influence MTF.
## Conclusion
The Modulation Transfer Function is an invaluable tool in optics, enabling precise evaluation of image clarity and detail preservation. Whether you're a photographer, scientist, or engineer, understanding and utilizing MTF ensures optimal performance of your optical systems.
Tags: Optics, Imaging, Photography | 834 | 3,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.865231 |
https://stats.stackexchange.com/questions/347523/is-it-possible-to-fit-mixed-models-via-gls | 1,713,246,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00268.warc.gz | 490,057,820 | 40,400 | # Is it possible to fit mixed-models via gls?
Is it possible to fit multivariate Gaussian models implied by mixed-models through generalised least squares in R, by using, for instance, the gls function?
For instance, the random intercept model via lme is
mod.lme <- lme(score~Machine, random = ~1|Worker, data=Machines)
s2.lme <- as.numeric(VarCorr(mod.lme)[2,1]) #residual variance
s2.ranef.lme <- as.numeric(VarCorr(mod.lme)[1,1]) # ran. eff. variance
(tot.var.lme <- s2.lme + s2.ranef.lme) # sum of variance components
The corresponding Gaussian model can be fitted by gls as follows:
mod.gls <- gls(score~Machine, correlation = corCompSymm(form = ~1| Worker),
data=Machines)
(tot.var.gls <- as.numeric(exp(attributes(mod.gls$apVar)$Pars[2]))^2)
mSt <- mod.gls$modelStruct cSt <- mSt$corStruct
(rho <- coef(cSt, unconstrained = FALSE))
all.equal(tot.var.gls, tot.var.lme) # total variances equal?
s2.ranef.gls <- rho*tot.var.gls # get variance of the ran. eff from gls
all.equal(as.numeric(s2.ranef.gls), s2.ranef.lme) # is equal to lme ?
mod2.lme <- lme(score~Machine, random = ~1|Worker/Machine, data = Machines)
How would you fit it by gls ? Is it possible ?
• All LMMs correspond to a multivariate normal model (while the converse is not true) with a structured variance covariance matrix, so "all" you have to do is to work out the marginal variance covariance matrix for the nested random-effect model and fit that - whether gls is then able to parameterize that model is then the next question. I haven't worked out the math, so I don't know, but my guess is that you may have to write your own corStruct class if you need to use gls. May 23, 2018 at 6:22
• Also beware of the difference in parameter spaces: the parameter space for the compound symmetry model is bigger than it is for the random intercept model. The random-effect variance is necessarily non-negative which leads to a non-negative corr but the corr in the compound symmetry model can also be negative (though not too much). So while two model fits can be equivalent (if $\rho \ge 0$) they need not be (if $\rho < 0$) and strictly speaking the underlying models are not the same. May 23, 2018 at 6:35
• Last comment :-) Your last model indicates that you are interested in a model with nested random effects but in the Machines data it is actually not the case that Machine is nested in Worker - rather, these variables are crossed. In fact the last model "tricks" lme to fit a model with a random main effect of Worker and the random interaction for Worker:Machine. Was that intentional? May 23, 2018 at 6:47
• Maybe I can ask for more clarification at this point: (1) Are you primarily asking how to derive the Gaussian model that corresponds to a mixed-effect model or are you asking how to write a corStruct class in order to fit it with gls? (2) If you have a particular and not any mixed-model in mind, please specify it. For a general (any) mixed model of the form $Y = X\beta + Z b + e$ with $e \sim MVN(0, R)$ and $b \sim MVN(0, G)$ we have $Cov(Y) = Z G Z' + R$ which does not simplify and therefore infeasible to fit with gls. Some structure is needed. May 23, 2018 at 11:16 | 865 | 3,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.78897 |
https://en.wikipedia.org/wiki/Huzita%27s_axioms | 1,469,674,987,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827782.42/warc/CC-MAIN-20160723071027-00322-ip-10-185-27-174.ec2.internal.warc.gz | 853,506,806 | 18,656 | # Huzita–Hatori axioms
(Redirected from Huzita's axioms)
The Huzita–Hatori axioms or Huzita–Justin axioms are a set of rules related to the mathematical principles of paper folding, describing the operations that can be made when folding a piece of paper. The axioms assume that the operations are completed on a plane (i.e. a perfect piece of paper), and that all folds are linear. These are not a minimal set of axioms but rather the complete set of possible single folds.
The axioms were first discovered by Jacques Justin in 1989.[1] Axioms 1 through 6 were rediscovered by Japanese-Italian mathematician Humiaki Huzita and reported at the First International Conference on Origami in Education and Therapy in 1991. Axioms 1 though 5 were rediscovered by Auckly and Cleveland in 1995. Axiom 7 was rediscovered by Koshiro Hatori in 2001; Robert J. Lang also found axiom 7.
## The seven axioms
The first 6 axioms are known as Huzita's axioms. Axiom 7 was discovered by Koshiro Hatori. Jacques Justin and Robert J. Lang also found axiom 7. The axioms are as follows:
1. Given two points p1 and p2, there is a unique fold that passes through both of them.
2. Given two points p1 and p2, there is a unique fold that places p1 onto p2.
3. Given two lines l1 and l2, there is a fold that places l1 onto l2.
4. Given a point p1 and a line l1, there is a unique fold perpendicular to l1 that passes through point p1.
5. Given two points p1 and p2 and a line l1, there is a fold that places p1 onto l1 and passes through p2.
6. Given two points p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2.
7. Given one point p and two lines l1 and l2, there is a fold that places p onto l1 and is perpendicular to l2.
Axiom 5 may have 0, 1, or 2 solutions, while Axiom 6 may have 0, 1, 2, or 3 solutions. In this way, the resulting geometries of origami are stronger than the geometries of compass and straightedge, where the maximum number of solutions an axiom has is 2. Thus compass and straightedge geometry solves second-degree equations, while origami geometry, or origametry, can solve third-degree equations, and solve problems such as angle trisection and doubling of the cube. The construction of the fold guaranteed by Axiom 6 requires "sliding" the paper, or neusis, which is not allowed in classical compass and straightedge constructions. Use of neusis together with a compass and straightedge does allow trisection of an arbitrary angle.
## Details
### Axiom 1
Given two points p1 and p2, there is a unique fold that passes through both of them.
In parametric form, the equation for the line that passes through the two points is :
${\displaystyle F(s)=p_{1}+s(p_{2}-p_{1}).}$
### Axiom 2
Given two points p1 and p2, there is a unique fold that places p1 onto p2.
This is equivalent to finding the perpendicular bisector of the line segment p1p2. This can be done in four steps:
• Use Axiom 1 to find the line through p1 and p2, given by ${\displaystyle P(s)=p_{1}+s(p_{2}-p_{1})}$
• Find the midpoint of pmid of P(s)
• Find the vector vperp perpendicular to P(s)
• The parametric equation of the fold is then:
${\displaystyle F(s)=p_{\mathrm {mid} }+s\cdot \mathbf {v} ^{\mathrm {perp} }.}$
### Axiom 3
Given two lines l1 and l2, there is a fold that places l1 onto l2.
This is equivalent to finding a bisector of the angle between l1 and l2. Let p1 and p2 be any two points on l1, and let q1 and q2 be any two points on l2. Also, let u and v be the unit direction vectors of l1 and l2, respectively; that is:
${\displaystyle \mathbf {u} =(p_{2}-p_{1})/\left|(p_{2}-p_{1})\right|}$
${\displaystyle \mathbf {v} =(q_{2}-q_{1})/\left|(q_{2}-q_{1})\right|.}$
If the two lines are not parallel, their point of intersection is:
${\displaystyle p_{\mathrm {int} }=p_{1}+s_{\mathrm {int} }\cdot \mathbf {u} }$
where
${\displaystyle s_{int}=-{\frac {\mathbf {v} ^{\perp }\cdot (p_{1}-q_{1})}{\mathbf {v} ^{\perp }\cdot \mathbf {u} }}.}$
The direction of one of the bisectors is then:
${\displaystyle \mathbf {w} ={\frac {\left|\mathbf {u} \right|\mathbf {v} +\left|\mathbf {v} \right|\mathbf {u} }{\left|\mathbf {u} \right|+\left|\mathbf {v} \right|}}.}$
And the parametric equation of the fold is:
${\displaystyle F(s)=p_{\mathrm {int} }+s\cdot \mathbf {w} .}$
A second bisector also exists, perpendicular to the first and passing through pint. Folding along this second bisector will also achieve the desired result of placing l1 onto l2. It may not be possible to perform one or the other of these folds, depending on the location of the intersection point.
If the two lines are parallel, they have no point of intersection. The fold must be the line midway between l1 and l2 and parallel to them.
### Axiom 4
Given a point p1 and a line l1, there is a unique fold perpendicular to l1 that passes through point p1.
This is equivalent to finding a perpendicular to l1 that passes through p1. If we find some vector v that is perpendicular to the line l1, then the parametric equation of the fold is:
${\displaystyle F(s)=p_{1}+s\cdot \mathbf {v} .}$
### Axiom 5
Given two points p1 and p2 and a line l1, there is a fold that places p1 onto l1 and passes through p2.
This axiom is equivalent to finding the intersection of a line with a circle, so it may have 0, 1, or 2 solutions. The line is defined by l1, and the circle has its center at p2, and a radius equal to the distance from p2 to p1. If the line does not intersect the circle, there are no solutions. If the line is tangent to the circle, there is one solution, and if the line intersects the circle in two places, there are two solutions.
If we know two points on the line, (x1, y1) and (x2, y2), then the line can be expressed parametrically as:
${\displaystyle x=x_{1}+s(x_{2}-x_{1})\,}$
${\displaystyle y=y_{1}+s(y_{2}-y_{1}).\,}$
Let the circle be defined by its center at p2=(xc, yc), with radius ${\displaystyle r=\left|p_{1}-p_{2}\right|}$. Then the circle can be expressed as:
${\displaystyle (x-x_{c})^{2}+(y-y_{c})^{2}=r^{2}.\,}$
In order to determine the points of intersection of the line with the circle, we substitute the x and y components of the equations for the line into the equation for the circle, giving:
${\displaystyle (x_{1}+s(x_{2}-x_{1})-x_{c})^{2}+(y_{1}+s(y_{2}-y_{1})-y_{c})^{2}=r^{2}.\,}$
Or, simplified:
${\displaystyle as^{2}+bs+c=0\,}$
where:
${\displaystyle a=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}\,}$
${\displaystyle b=2(x_{2}-x_{1})(x_{1}-x_{c})+2(y_{2}-y_{1})(y_{1}-y_{c})\,}$
${\displaystyle c=x_{c}^{2}+y_{c}^{2}+x_{1}^{2}+y_{1}^{2}-2(x_{c}x_{1}+y_{c}y_{1})-r^{2}.\,}$
Then we simply solve the quadratic equation:
${\displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}$
If the discriminant b2 − 4ac < 0, there are no solutions. The circle does not intersect or touch the line. If the discriminant is equal to 0, then there is a single solution, where the line is tangent to the circle. And if the discriminant is greater than 0, there are two solutions, representing the two points of intersection. Let us call the solutions d1 and d2, if they exist. We have 0, 1, or 2 line segments:
${\displaystyle m_{1}={\overline {p_{1}d_{1}}}\,}$
${\displaystyle m_{2}={\overline {p_{1}d_{2}}}.\,}$
A fold F1(s) perpendicular to m1 through its midpoint will place p1 on the line at location d1. Similarly, a fold F2(s) perpendicular to m2 through its midpoint will place p1 on the line at location d2. The application of Axiom 2 easily accomplishes this. The parametric equations of the folds are thus:
{\displaystyle {\begin{aligned}F_{1}(s)&=p_{1}+{\frac {1}{2}}(d_{1}-p_{1})+s(d_{1}-p_{1})^{\perp }\\[8pt]F_{2}(s)&=p_{1}+{\frac {1}{2}}(d_{2}-p_{1})+s(d_{2}-p_{1})^{\perp }.\end{aligned}}}
### Axiom 6
Given two points p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2.
This axiom is equivalent to finding a line simultaneously tangent to two parabolas, and can be considered equivalent to solving a third-degree equation as there are in general three solutions. The two parabolas have foci at p1 and p2, respectively, with directrices defined by l1 and l2, respectively.
This fold is called the Beloch fold after Margharita P. Beloch, who in 1936 showed using it that origami can be used to solve general cubic equations.[2]
### Axiom 7
Given one point p and two lines l1 and l2, there is a fold that places p onto l1 and is perpendicular to l2.
This axiom was originally discovered by Jacques Justin in 1989 but was overlooked and was rediscovered by Koshiro Hatori in 2002.[3] Robert J. Lang has proven that this list of axioms completes the axioms of origami.
## Constructibility
Subsets of the axioms can be used to construct different sets of numbers. The first three can be used with three given points not on a line to do what Alperin calls Thalian constructions.[4]
The first four axioms with two given points define a system weaker than compass and straightedge constructions: every shape that can be folded with those axioms can be constructed with compass and straightedge, but some things can be constructed by compass and straightedge that cannot be folded with those axioms.[5] The numbers that can be constructed are called the origami or pythagorean numbers, if the distance between the two given points is 1 then the constructible points are all of the form ${\displaystyle (\alpha ,\beta )}$ where ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are Pythagorean numbers. The Pythagorean numbers are given by the smallest field containing the rational numbers and ${\displaystyle {\sqrt {1+\alpha ^{2}}}}$ whenever ${\displaystyle \alpha }$ is such a number.
Adding the fifth axiom gives the Euclidean numbers, that is the points constructible by compass and straightedge construction.
Adding the neusis axiom 6, all compass-straightedge constructions, and more, can be made. In particular, the constructible regular polygons with these axioms are those with ${\displaystyle 2^{a}3^{b}\rho \geq 3}$ sides, where ${\displaystyle \rho }$ is a product of distinct Pierpont primes. Compass-straightedge constructions allow only those with ${\displaystyle 2^{a}\phi \geq 3}$ sides, where ${\displaystyle \phi }$ is a product of distinct Fermat primes. (Fermat primes are a subset of Pierpont primes.)
The seventh axiom does not allow construction of further axioms. The seven axioms give all the single-fold constructions that can be done rather than being a minimal set of axioms.
## References
1. ^ Justin, Jacques, "Resolution par le pliage de l'equation du troisieme degre et applications geometriques", reprinted in Proceedings of the First International Meeting of Origami Science and Technology, H. Huzita ed. (1989), pp. 251–261.
2. ^ Thomas C. Hull (April 2011). "Solving Cubics With Creases: The Work of Beloch and Lill" (PDF). American Mathematical Monthly: 307–315. doi:10.4169/amer.math.monthly.118.04.307.
3. ^ Roger C. Alperin; Robert J. Lang (2009). "One-, Two-, and Multi-Fold Origami Axioms" (PDF). 4OSME (A K Peters).
4. ^ Alperin, Roger C (2000). "A Mathematical Theory of Origami Constructions and Numbers" (PDF). New York Journal of Mathematics 6: 119–133.
5. ^ D. Auckly and J. Cleveland. "Totally real origami and impossible paperfolding". American Mathematical Monthly 102: 215–226. arXiv:math/0407174. doi:10.2307/2975008. | 3,346 | 11,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 32, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-30 | latest | en | 0.95058 |
https://chandoo.org/wp/2015/03/10/check-if-2-ranges-have-same-values-set-equality-problem/ | 1,524,253,944,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944682.35/warc/CC-MAIN-20180420194306-20180420214306-00498.warc.gz | 611,734,671 | 28,835 | # Check if 2 ranges have same values (set equality problem)
Posted on March 10th, 2015 in Excel Challenges - 69 comments
Hello folks,
Time for another homework problem. Assuming you have 2 ranges of values like below, how do you check if both of them have same set of values?
You may assume that these ranges are named range1 range2.
### A bonus question…
How do you find the first equal set of source in the range target?
For example, for below scenario, source = 2nd column of target.
So the expected answer is 2 for this question.
Please assume the source & target ranges are named as source and target.
Go ahead and share your comments. I am very keen to see what kind of creative & elegant solutions we can come up to solve this problem.
Want more homework or Excel challenges?
Check out our Excel challenges page & homework tag for more problems. Be warned though, serious workouts for your brain ahead. Your mind will have a six pack at the end of it all.
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Written by Chandoo Tags: advanced excel, duplicates, homework, Microsoft Excel Formulas, sets, unique items Home: Chandoo.org Main Page ? Doubt: Ask an Excel Question
### 69 Responses to “Check if 2 ranges have same values (set equality problem)”
1. Somendra Misra says:
Hi Chandoo,
May be something like:
For first problem:
=IFERROR(SUMPRODUCT(--(MATCH(Range1,Range2,0)>0))=COUNTA(Range1),FALSE)
For Bonus Question, WITH CSE:
=MATCH(TRUE,MMULT(COUNTIF(Source,TRANSPOSE(Target)),{1;1;1;1;1})=COUNTA(Source),0)
2. Desk Lamp says:
Problem1: =AND(range1=range2)
Ctrl+Shift+Enter
• Rob says:
This is great but only works if the items in each range are in the same order. Ranges 1 & 2 above are identical but in different orders
3. Satish says:
For First Problem It May be Just Compare 2 Columns
=Range1=Range2 --- Result will be "True" if column Data match
---- Result will be "False" if Column Data Don't Match
Second Question i not completely understand. what to do in That
4. First of all, I would put an additional column using VLOOKUP function to find whether a value in Range 1 exists in Range 2. And if so I will bring that value using the same VLOOKUP function.
Then using IF function will show me whether value in Range 1 corresponds to the value generated by VLOOKUP function.
5. Dick Byrne says:
Main question:
=AND(NOT(ISNA(MATCH(Range1, Range2, 0))))
Enter as an array formula (CSE)
6. Dick Byrne says:
Ah but what if the ranges are different sizes? Then you need a two way check ...
=AND(NOT(ISNA(MATCH(Range2, Range1, 0))), NOT(ISNA(MATCH(Range1, Range2, 0))))
Entered as an array function (CSE)
7. Jason Morin says:
Assuming ranges are same dimensions as in the example:
=SUMPRODUCT(1*ISNA(MATCH(range1,range2,0)))=0
8. Michael says:
First problem (Array formula):
=AND(COUNTIF(range1,range2)=IF(range1=range1,1,0),COUNTIF(range2,range1)=IF(range2=range2,1,0))
9. Ericson says:
Problem #2
=(Geomean(source)^var.s(source))-(geomean(target2)^var.s(target2))
A zero indicates same. A variety of math equations can be used this is just one. I could have chosen logs. The point is to make the range a unique value and compare the next range using same methodology. Only works for numbers and the probability of this failing falls well below an "I trust excel enough for the array to work"
10. XOR LX says:
Hi Chandoo.
In all of your examples each of the values from a given range is unique within that range. Is it safe to assume that this will always be the case?
Or could you have a case such as:
range1: "DEF", "DEF", "GHI", "JKL", "MNO"
range2: "JKL", "GHI", "MNO", "DEF", "DEF"
?
Regards
• Chandoo says:
Hi XOR LX... you can assume that all values will be unique and ranges will have same size (number of cells).
11. Miguel says:
I know is a easier way to do it and some one is going to clean this bit of code
Thanks a lot Chandoo
========================================
Public Sub sameRange()
Dim setOne As Range
Dim setTwo As Range
Set setOne = Sheets("Sheet1").Range("range1")
Set setTwo = Sheets("Sheet1").Range("range2")
'REMOVE THE COLOR FILL
setOne.Interior.ColorIndex = xlNone
For Each cellitem In setOne
For Each cellItem2 In setTwo
If cellitem = cellItem2 Then
MsgBox "Found match at cell.." & cellitem.Address
Range("E2").Select
ActiveCell.Value = cellitem
cellitem.Interior.ColorIndex = 50
End If
Next cellItem2
Next cellitem
End Sub
======================
🙂
12. Maribeth EM says:
Use function EXACT. It compares two text strings and returns True if exactly the same and False if not. This function however is case-sensitive so if one is capitalized and the other is not, then it will return False.
=EXACT(A1,A2)
assumption is that the text are in cells A1 and A2
13. Problem 1:
=SUM((COUNTIFS(D3:D7,C3:C7))*1)=COUNTA(C3:C7)
CTL ALT DLT
Returns true if exact match, false otherwise
14. Steve S. says:
I would use conditional formatting for each range with the following formula in the conditional formatting :
For Range 1:
=NOT(IFERROR(VLOOKUP(A2,Range_2,1,0)=A2,0))
For Range 2:
=NOT(IFERROR(VLOOKUP(C2,Range_1,1,0)=C2,0))
This gives me true/false for each cell and highlights any cells that are different.
15. Ysahme says:
Hi Guys...
Good day!!!
Problem #1
I use IF Statement to another column and it return OK if same and blank if it is not...
I also try match but it return me a Number meaning that range2 is in the number base on range1...
16. Harry S says:
' got a bit if VBA instead
'use the data as at least 1 space around to use CurrentRegion
TRUE Range 1 Range 2 Source Target
ABC ABC 123 128 127 123 134 127 140
DEF GHI 124 129 123 126 135 123 141
GHI MNO 125 130 124 125 136 124 142
Not In Col 1 JKL JKL 126 131 126 124 137 125 143
In Col 2 MNO DEF 127 132 125 127 138 132 144
In Col 3 126
Not In Col 4
In Col 5
Not In Col 6
'
Then code as
[code]
Option Explicit
Function AllRaInRB(Ra As Range, Rb As Range) As Boolean
Dim CE As Range
For Each CE In Ra
If Rb.Find(CE) Is Nothing Then GoTo NotThere
' see the Find ( Options .. ) like below if needed
' If Rb.Find(CE, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False) Is Nothing Then GoTo NotThere
Next CE
AllRaInRB = True ' found them all
NotThere:
End Function
Private Sub CommandButton21_Click()
Dim Ra As Range, Rb As Range, Rc%, Rot%
'
' assuming Range header one space above the data and the range data is spaced from other data
' the headers for the ranges can be whatever and anywhere in the active sheet
'
With ActiveSheet.UsedRange
Set Ra = .Find("Range 1")(3, 1).CurrentRegion
Set Rb = .Find("Range 2")(3, 1).CurrentRegion
[b2] = AllRaInRB(Ra, Rb) ' part A
Set Ra = .Find("Source")(3, 1).CurrentRegion
Set Rb = .Find("Target")(3, 1).CurrentRegion
Rot = 6 ' row Out start
End With
For Rc = 1 To Rb.Columns.Count ' finding all columns in Target
If AllRaInRB(Ra, Rb.Columns(Rc)) Then
Cells(Rot + Rc, 1) = " In Col " & Rc ' results down column A starting at Rot +1
Else
Cells(Rot + Rc, 1) = "Not In Col " & Rc
End If
Next Rc
End Sub
[/code]
17. Krishna says:
Hi all,
For first question:
{=(ISNUMBER(MATCH(Range1,Range2,0)))}
or as mentioned by Maribeth, we can use Exact function
{=EXACT(Range1, range2)}
Use of CSE as we are dealing with ranges
18. VISHAL says:
For First Problem
=IF(A2=B2,TRUE,FALSE)
19. XOR LX says:
Since Chandoo confirmed that the values in each range are unique within that range, CSE:
=AND(COUNTIF(range1,range2))
Bonus Question, non-array:
=MATCH(5,MMULT({1,1,1,1,1},COUNTIF(source,target)),0)
if the number of rows is fixed at 5, or, CSE:
=MATCH(COUNTA(source),MMULT(TRANSPOSE(ROW(source))^0,COUNTIF(source,target)),0)
if the number of rows is dynamic.
Regards
• Oscar says:
XOR LX
Congratulations, your two first formulas are exactly the same formulas I was about to post here but you were faster.
• Somendra Misra says:
XOR,
Clever way of changing array1 with array2. 🙂
See my formula in comment #1.
Regards,
• XOR LX says:
@Somendra
Yes - yours was of course the first post I saw, and a very good one as well.
I just saw the opportunity to tweak your MMULT construction so that we donn't require the TRANSPOSE. 🙂
Cheers
20. Gabriel says:
=AND(SUM(SUM(--(vector_2=vector_1)))=COUNTA(vector_1),COUNTA(vector_1)=COUNTA(vector_2))
Control+Shift + Enter. ( Array Formula )
example:( for ture)
vector 2 vector 1
abc abc
dei dei
copy copy
try try
bug bug
loop loop
example for False:
vector 2 vector 1
abc abc
bei dei
sopi copy
try try
bug bug
loop loop
21. XOR LX says:
@Oscar
Cheers! And sorry! 🙂
• Oscar says:
@XOR LX
I wouldn´t have made the bonus question formula without learning the MMULT function from your blog.
22. Ilyas says:
For problem 1 i tried the following formula:
{=ISNUMBER(SUM(MATCH(Range2,Range1,0)))}
For problem 2 i used the following formula in conditional formatting so that the matching column gets highlighted:
=ISNUMBER(SUM(MATCH(source,OFFSET(target,1,C\$9-1,5,1),0)))
23. Bhavesh Soni says:
Sum(match(range1)=(match(range2) will show true if all items are there. Note enter this as Array formula.
24. Bhavesh Soni says:
Sum(match(range1)=(match(range2) will show true if all items are there. Note enter this as Array formula
25. cllach says:
For first one:
=NOT(ISERROR(PRODUCT(MATCH(IF(Range1="""",""""&COUNTBLANK(Range1),Range1 & COUNTIF(Range1,Range1)),IF(Range2="""",""""&COUNTBLANK(Range2),Range2& COUNTIF(Range2,Range2)),0))))
I think that works returning True for vertical ranges even if values are not unique (must appear same number of times), or empty or numeric, if ranges are diferent size it returns False.
Easy to adapt to second one.
• cllach says:
Sorry, pasting doubled quotes...
=NOT(ISERROR(PRODUCT(MATCH(IF(Range1="",""&COUNTBLANK(Range1),Range1 & COUNTIF(Range1,Range1)),IF(Range2="",""&COUNTBLANK(Range2),Range2& COUNTIF(Range2,Range2)),0))))
============================================
• cllach says:
Think that works for unique or not, values, returns True if found same number of times, numeric or text and return False if ranges are diferent size. Easy to adapt to second problem.
26. Lokesh says:
For the main question.
=VLOOKUP([@[Range 2]],Table1[Range 1],1,0)=[@[Range 2]]
If all values are true than the values in both range match
=IFERROR(SUM(MATCH(C3:C7,E3:E7,0)),"No Match")
Array Formula
I am expecting it work irrespective of number of items in both the lists,,
28. Nelson says:
"=IF(NOT(ISERROR(VLOOKUP(C1,Range2,1,0))),"True","False")"
29. Fern says:
Highlight the 2 ranges
Under Conditional Formatting, Select "Highlight Cell Rules", select "Duplicate Values"
The values that match in each range will be highlighted.
You can also choose to highlight only Unique value in each range
30. CraigM says:
use {=OR(Exact(A1,range2))}
where:
the cells in range1 are A1 to A5
the cells in range2 are B1 to B5
Enter the formula in cell C1. This will check to see if the value in A1 exists in range2. Copy down for the rest of the values.
To check if each of the values in range1 exist in range2, use:
{=OR(Exact(B1,range1))} in D1 & then copy down for the rest.
This is useful for lists that aren't too large. If they are, then resources/performance becomes an issue.
31. Dev says:
Use Conditional Formatting and use formula:
where A1 is the first cell in Range 1 i.e. ABC
So, if ABC is not listed in Range 2, ABC will be highlighted in Blue
32. Bhavesh Soni says:
=SUM(MATCH(RANGE1,RANGE1,0))=(SUM(MATCH(RANGE2,RANGE2,0)))
Enter this as array formula.
33. shanmugam says:
Assumtion Range1 A1 to A6, Range 2 B1 to B6
=Vlookup(A2,\$A\$2:\$A\$6,2,0)
34. Shaji says:
first one is so simple...
{=if(COUNTIF(Range1,Range2)=1,"True","false")}
2nd one to find out...
• Shaji says:
Sorry,
this one is correct for the first question...
{=SUM(COUNTIF(Range2,Range1))=ROWS(Range1)}
35. Aravind says:
Hai,Good Morning
In the first one even though the values are same but the cells are different know so,I will make the criteria with cell numbers and find the answer
36. Ian says:
=IF(IFERROR(MATCH(A1,B:B,0),0)=0,"No Match","Matched")
37. Anand says:
I simply use conditional formatting feature.
Select both range and use conditional formatting to see duplicate/unique values.
38. Belle-Isle says:
For first problem :
CSE : =AND(COUNT(Range1)=COUNT(Range2), COUNTIF(Range2, Range1)=1, COUNTIF(Range1, Range2)=1)
This returns true in the wanted situation (identical arrays of unique values, whatever their order), but returns false if the arrays are of different sizes or if there are any duplicates in any one array. All in a single formula.
39. Abhijeet says:
My source data: A3:A7
Target Ranges: C3:F7
Select Target Range and use conditional formatting formula "=COUNTIF(\$A\$3:\$A\$7,C3)" with some colour to highlight cells matching criteria. If all cells in a column are colored, it's a match target range.
40. DJP says:
Good morning / Good afternoon,
May I suggest this formula:
SUM(--(Range1=TRANSPOSE(Range2)))=COUNTA(Range1)
As an array formula, Ctrl+Shift+Enter is required
For Frenchies, it gives:
SOMME(--(Range1=TRANSPOSE(Range2)))=NBVAL(Range1)
Ctrl+Shift+Enter needed too 😉
Cheers
41. sam says:
Problem 1
=AND(COUNTIF(Rng1,Rng2))
Array Entered
True if both are the same, False if they are different
42. Jorge Eduardo (brazil) says:
=contREP(O3:S7;H3:L17)
vai informar a quantidade de valores repetidos entre 2 range
funciona para mim e uso muito dentro de macros
Dim reg1() As Variant, reg2() As Variant, ttl As Long
Dim lc1 As Long, cc1 As Long, lc2 As Long, cc2 As Long
Function ContREP(ByVal rag1 As Range, ByVal rag2 As Range) As Long
ttl = 0
reg1 = rag1.Value2
reg2 = rag2.Value2
lc1 = UBound(reg1, 1)
cc1 = UBound(reg1, 2)
lc2 = UBound(reg2, 1)
cc2 = UBound(reg2, 2)
For l1 = 1 To lc1
For c1 = 1 To cc1
If reg1(l1, c1) "" Then GoSub lk
Next
Next
ContREP = ttl
Exit Function
lk:
For l2 = 1 To lc2
For c2 = 1 To cc2
If reg1(l1, c1) = reg2(l2, c2) Then ttl = ttl + 1: Return
Next
Next
Return
End Function
43. Abbott Katz says:
Assuming the two ranges are in B8:B12 and C8:C12 respectively:
{=IF(ISNA(SUM(MATCH(B8:B12,C8:C12,0))),"Incomplete match","Complete match")}
44. QL says:
My approach to the first question:
Assume range_1 and range_2 are arranged as vertical column range with same size, and there might be duplicate values in the range.
The array formula
=SUM(--(MMULT(--(TRANSPOSE(Range_1)=Range_1),--(ROW(Range_1)=ROW(Range_1)))=MMULT(--(TRANSPOSE(Range_1)=Range_2),--(ROW(Range_1)=ROW(Range_1))))) = ROWS(Range_1)
45. Katie Grimes says:
Assuming the two ranges are in A2:A6 and B2:B6, I created a row beneath the two ranges to tell me how many values are MISSING from Range 1 in Range 2 and vice versa. If they are a perfect match, there would be 0 missing values in each column. (I decided to do missing values instead of matching values so that I wouldn't have to know HOW MANY values were in the range to begin with.)
My array formula for A7 = SUM(IF(ISNA(MATCH(A2:A6, B2:B6, 0)), 1, 0)), Ctrl + Shift + Enter. This would give 0 if every value in Range 1 could be found in Range 2.
Similarly, for B7 = SUM(IF(ISNA(MATCH(B2:B6, A2:A6, 0)), 1, 0)). If both A7 and B7 = 0, then you have a perfect match.
In addition, I wanted to be able to tell which values WERE missing if they did not match perfectly. So I created a conditional formatting in cells A2:A6 using the formula = ISNA(MATCH(A2, \$B\$2:\$B\$6,0)). This would highlight any cells in Range 1 that could not be found in Range 2. I then repeated the formatting for B2:B6 using the formula condition = ISNA(MATCH(B2, \$A\$2:\$A\$6, 0)).
For question 2:
Assume the Source Range is in K3:K7, and the Target Table has headers in M2:R2, and the data is in M3:R7.
Beneath each of the ranges in the Target table, I inserted a row with the same array formula used above to compare it to the Source Range:
M8 = SUM(IF(ISNA(MATCH(M3:M7, \$K\$3:\$K\$7, 0)), 1, 0)), Ctrl + Shift + Enter.
N8 = SUM(IF(ISNA(MATCH(N3:N7, \$K\$3:\$K\$7, 0)), 1, 0)), Ctrl + Shift + Enter.
O8 = SUM(IF(ISNA(MATCH(O3:O7, \$K\$3:\$K\$7, 0)), 1, 0)), Ctrl + Shift + Enter.
P8 = SUM(IF(ISNA(MATCH(P3:P7, \$K\$3:\$K\$7, 0)), 1, 0)), Ctrl + Shift + Enter.
Q8 = SUM(IF(ISNA(MATCH(Q3:Q7, \$K\$3:\$K\$7, 0)), 1, 0)), Ctrl + Shift + Enter.
R8 = SUM(IF(ISNA(MATCH(R3:R7, \$K\$3:\$K\$7, 0)), 1, 0)), Ctrl + Shift + Enter.
Then, in cell K10 (where I wanted to pull the Range that matched, i.e. where I wanted my answer to appear), I used a Match formula to find the range with 0 missing values.
K10 = MATCH(0, M8:R8, 0). This returned 2, our given range.
I also decided that I wanted to be able to return the Range Name, even if it wasn't 1-6. Changing Range 2 to be called "Apple" (no reason, just the word I picked), in cell K11, I used an Index-Match formula.
K11 = INDEX(M2:R2, ,MATCH(0,M8:R8,0)) This then returned "Apple" instead of "2".
The only thing I realize now as I am typing this is that I didn't do a back check: I didn't check that the Source range had all the values in the Target Range 2. Given they had the same number of values, this isn't a big deal, but ideally, I would have added that as well.
46. David says:
Problem 1:
{=AND(COUNTIF(RANGE1,RANGE2)*COUNTIF(RANGE2,RANGE1))}
47. Alex Groberman says:
If there can be duplicate values and/or a different number of rows in each set:
{=IFERROR(AND(SMALL(MATCH(range1,range2,0),ROW(INDIRECT("1:"&ROWS(range1))))=ROW(INDIRECT("1:"&ROWS(range2)))),FALSE)}
48. Hi sir,
for Q1
two option I tried correct me if I am wrong
a) conditional formatting- cell value(range 1) (giving colour) stop if true
b) match function - by which range1 order is matched with range 2
showing a result like
Rang1 Range2 Order
ABC ABC 1
DEF GHI 3
GHI MNO 5
JKL JKL 4
MNO DEF 2
49. Neil says:
For question two here is my starter for ten that has not been tidied up but should provide some food for thought:
CSE enter:
{=MATCH(MAX(MMULT(COUNTIF(CHK_RNG,TRANSPOSE(CHOOSE(COLUMN(A1:F1),TEST1,TEST2,TEST3,TEST4,TEST5,TEST6))),ROW(A1:A5)^0)),
MMULT(COUNTIF(CHK_RNG,TRANSPOSE(CHOOSE(COLUMN(A1:F1),TEST1,TEST2,TEST3,TEST4,TEST5,TEST6))),ROW(A1:A5)^0),0)}
I set up a range called CHK_RNG in cells B4:B8 for the data to be checked against. And six ranges for the guesses called TEST1 (D4:D8), TEST2 (E4:E8), ...TEST6 (I4:I8)
50. Micah Dail says:
First problem is pretty straighforward.
{=AND(ISNUMBER(MATCH(range1, range2, 0)))}
Second one is rather tricky. Haven't figured it out yet.
51. parveen says:
Hello Chandoo,
As per me exact function should be used.
52. Ale says:
My solution to the first question:
=AND(MATCH(INDEX(Range1,1),Range2,0),MATCH(INDEX(Range1,2),Range2,0),MATCH(INDEX(Range1,3),Range2,0),MATCH(INDEX(Range1,4),Range2,0),MATCH(INDEX(Range1,5),Range2,0))
53. Ale says:
And this is my solution for the bonus question:
IF(NOT(ISERROR(AND(MATCH(INDEX(Source,1),Target1,0),MATCH(INDEX(Source,2),Target1,0),MATCH(INDEX(Source,3),Target1,0),MATCH(INDEX(Source,4),Target1,0),MATCH(INDEX(Source,5),Target1,0)))),D15,IF(NOT(ISERROR(AND(MATCH(INDEX(Source,1),Target2,0),MATCH(INDEX(Source,2),Target2,0),MATCH(INDEX(Source,3),Target2,0),MATCH(INDEX(Source,4),Target2,0),MATCH(INDEX(Source,5),Target2,0)))),E15,IF(NOT(ISERROR(AND(MATCH(INDEX(Source,1),Target3,0),MATCH(INDEX(Source,2),Target3,0),MATCH(INDEX(Source,3),Target3,0),MATCH(INDEX(Source,4),Target3,0),MATCH(INDEX(Source,5),Target3,0)))),F15,IF(NOT(ISERROR(AND(MATCH(INDEX(Source,1),Target4,0),MATCH(INDEX(Source,2),Target4,0),MATCH(INDEX(Source,3),Target4,0),MATCH(INDEX(Source,4),Target4,0),MATCH(INDEX(Source,5),Target4,0)))),G15,IF(NOT(ISERROR(AND(MATCH(INDEX(Source,1),Target5,0),MATCH(INDEX(Source,2),Target5,0),MATCH(INDEX(Source,3),Target5,0),MATCH(INDEX(Source,4),Target5,0),MATCH(INDEX(Source,5),Target5,0)))),H15,IF(NOT(ISERROR(AND(MATCH(INDEX(Source,1),Target6,0),MATCH(INDEX(Source,2),Target6,0),MATCH(INDEX(Source,3),Target6,0),MATCH(INDEX(Source,4),Target6,0),MATCH(INDEX(Source,5),Target6,0)))),I15,"no match"))))))
Tip: any idea on how to make it shorter?
54. Eric Lind says:
Count if makes the most sense to me.
The basic formula would be:
=COUNTIF(Range2,A2)
which is nice because you can count how many instances of a value in the range.
55. Ryan Wells says:
I know you can do this with formulae, but I'm partial to VBA. Since I have to compare strings regularly for nuclear engineering design verification, I created a VBA code that doesn't stop at simply comparing the ranges. It will tell you where in Range 2 the value in Range 1 appears and it will highlight the cells not found in both ranges. If you're interested: http://wellsr.com/vba/2015/excel/examples/excel-compare-two-columns-for-differences/
56. Ramesh Deo says:
(=OR(range1=range2)}
57. Ramesh Deo says:
{=OR(range1=range2)}
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## Bandwidth consecutive multicolorings of graphs
Theoretical Computer Science 532 (2014) 64–72 Contents lists available at SciVerse ScienceDirect Theoretical Computer Science journal homepage: www...
Theoretical Computer Science 532 (2014) 64–72
Contents lists available at SciVerse ScienceDirect
Theoretical Computer Science journal homepage: www.elsevier.com/locate/tcs
Bandwidth consecutive multicolorings of graphs✩ Kazuhide Nishikawa a , Takao Nishizeki a , Xiao Zhou b,∗ a
School of Science and Technology, Kwansei Gakuin University, 2-1 Gakuen, Sanda, 669-1337, Japan
b
Graduate School of Information Sciences, Tohoku University, Sendai, 980-8579, Japan
article Keywords: Bandwidth coloring Channel assignment Multicoloring Series-parallel graph Partial k-tree Algorithm Acyclic orientation Approximation FPTAS
info
abstract Let G be a simple graph in which each vertex v has a positive integer weight b(v) and each edge (v, w) has a nonnegative integer weight b(v, w). A bandwidth consecutive multicoloring of G assigns each vertex v a specified number b(v) of consecutive positive integers so that, for each edge (v, w), all integers assigned to vertex v differ from all integers assigned to vertex w by more than b(v, w). The maximum integer assigned to a vertex is called the span of the coloring. In the paper, we first investigate fundamental properties of such a coloring. We then obtain a pseudo polynomial-time exact algorithm and a fully polynomial-time approximation scheme for the problem of finding such a coloring of a given series-parallel graph with the minimum span. We finally extend the results to the case where a given graph G is a partial k-tree, that is, G has a bounded tree-width. © 2013 Elsevier B.V. All rights reserved.
1. Introduction An ordinary coloring of a graph G assigns each vertex a color so that, for each edge (v, w), the color assigned to v differs from the color assigned to w [7]. The problem of finding a coloring of a graph G with the minimum number χ(G) of colors often appears in the scheduling, task-allocation, etc. [7]. However, it is NP-hard, and difficult to find a good approximate solution. More precisely, for all ε > 0, approximating χ(G) within n1−ε is NP-hard [16], where n is the number of vertices in G. The ordinary coloring has been extended in various ways [3–7,9,14,15]. A multicoloring assigns each vertex a specified number of colors so that, for each edge (v, w), the set of colors assigned to v is disjoint with the set of colors assigned to w [3–5,15]. A bandwidth coloring assigns each vertex a positive integer as a color so that the two integers assigned to the ends of each edge (v, w) differ by at least the specified weight ω(v, w) of (v, w) [9]. In this paper we deal with another generalized coloring, called a ‘‘bandwidth consecutive multicoloring’’. Let G = (V , E ) be a simple graph with vertex set V and edge set E. Each vertex v ∈ V has a positive integer weight b(v), while each edge (v, w) ∈ E has a non-negative integer weight b(v, w). A bandwidth consecutive multicoloring F of G is an assignment of positive integers to vertices such that (a) each vertex v ∈ V is assigned a set F (v) of b(v) consecutive positive integers; and (b) for each edge (v, w) ∈ E, all integers assigned to v differ from all integers assigned to vertex w by more than b(v, w). We call such a bandwidth consecutive multicoloring F simply a b-coloring of G for a weight function b. The maximum integer assigned to a vertex is called the span of a b-coloring F , and is denoted by span(F ). We define the b-chromatic number χb (G) of a graph G to be the minimum span over all b-colorings F of G. A b-coloring F is called optimal if span(F ) = χb (G). A b-coloring problem is to compute χb (G) for a given graph G. ✩ Work partly supported by MEXT-supported Program for the Strategic Research Foundation at Private Universities.
Corresponding author. Tel.: +81 22 795 7166. E-mail addresses: [email protected] (K. Nishikawa), [email protected] (T. Nishizeki), [email protected] (X. Zhou).
K. Nishikawa et al. / Theoretical Computer Science 532 (2014) 64–72
(a) G.
⃗ (b) G.
(c) Gσ .
(d) G⃗σ .
65
⃗ and the longest path P, (c) graph Gσ with weights scaled Fig. 1. (a) series-parallel weighted graph G and its optimal b-coloring F , (b) acyclic orientation G by σ = 2 and its optimal bσ -coloring Fσ , and (d) acyclic orientation G⃗σ and the longest path Pσ .
Fig. 1(a) depicts a weighted graph G together with an optimal b-coloring F of G, where a weight b(e) is attached to an edge e, a weight b(v) is written in a circle representing a vertex v , and a set F (v) is attached to a vertex v . Since span(F ) = 11, χb (G) = 11. The ordinary vertex-coloring is merely a b-coloring for the case b(v) = 1 for every vertex v and b(v, w) = 0 for every edge (v, w). The ‘‘bandwidth coloring’’ or ‘‘channel assignment’’ [9] is a b-coloring for the case b(v) = 1 for every vertex v and b(v, w) = ω(v, w) − 1 for every edge (v, w). It should be noted that our edge weight b(v, w) is one less than the ordinary edge weight ω(v, w) of a bandwidth coloring. (This convention will make the arguments and algorithms simple and transparent.) A b-coloring arises in the assignment of radio channels in cellular communication systems [9] and in the non-preemptive task scheduling [10]. The b(v) consecutive integers assigned to a vertex v correspond to the contiguous bandwidth of a channel v or the consecutive time periods of a task v . The weight b(v, w) assigned to edge (v, w) represents the requirement that the frequency band or time period of v must differ from that of w by more than b(v, w). The span of a b-coloring corresponds to the minimum total bandwidth or the minimum makespan. One can find a multicoloring of a graph G with the minimum number of colors in time polynomial in the output size if G is a series-parallel graph or a partial k-tree, that is, a graph of bounded tree-width [5,15]. The problem of finding a bandwidth coloring with the minimum number of colors is NP-hard even for partial 3-trees [9], and there is a fully polynomial-time approximation scheme (FPTAS) for the problem on partial k-trees [9]. Since our b-coloring problem is also NP-hard for partial 3-trees, it is desirable to obtain a good approximation algorithm. However, there are only heuristics for the b-coloring problem so far [8]. In this paper, we first investigate fundamental properties of a b-coloring. In particular, we characterize the b-chromatic number χb (G) of a graph G in terms of the longest path in acyclic orientations of G. We then obtain a pseudo polynomialtime exact algorithm for the b-coloring problem on series-parallel graphs, which often appear in the task scheduling and electrical circuits [10,12]. The algorithm takes time O(B3 n), where B is the maximum weight of G: B = maxx∈V ∪E b(x). Using the algorithm, we then give a fully polynomial-time approximation scheme (FPTAS) for the problem. We finally extend these results to the case where G is a partial k-tree. It should be noted that a series-parallel graph is a partial 2-tree. An early version of the paper was presented at a conference [11]. 2. Preliminaries In this section, we first give some definitions and then present three lemmas on a b-coloring.
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K. Nishikawa et al. / Theoretical Computer Science 532 (2014) 64–72
Let G = (V , E ) be a simple graph without selfloops and multiple edges. We denote by n and m the number of vertices and edge in G, respectively. The chromatic number χ (G) of G is the minimum number of colors required by an ordinary coloring of G. Let N be the set of all positive integers, that are regarded as colors. A b-coloring F : V → 2N of G must satisfy the following (a) and (b): (a) for every vertex v ∈ V , the set F (v) consists of b(v) consecutive positive integers, and hence min F (v) = max F (v) − b(v) + 1;
and
(b) for every edge (v, w) ∈ E, all integers in F (v) differ from those in F (w) by more than b(v, w). A b-coloring F can be represented by a function f : V → N such that f (v) = max F (v) for every vertex v ∈ V . Clearly, for every vertex v ∈ V , b(v) ≤ f (v).
(1)
For every edge (v, w) ∈ E, f (v) ̸= f (w)
(2)
since b(v, w) ≥ 0. For every edge (v, w) ∈ E with f (v) < f (w), b(v, w) < (f (w) − b(w) + 1) − f (v) and hence f (v) + b(v, w) + b(w) ≤ f (w).
(3)
Conversely, every function f : V → N satisfying Eqs. (1)–(3) represents a b-coloring F such that F (v) = {f (v) − b(v) + 1, f (v) − b(v) + 2, . . . , f (v)}. Thus, such a function f is also called a b-coloring of G. Obviously, span(F ) = maxv∈V f (v). We often denote span(F ) by span(f ). A b-coloring f is called optimal if span(f ) = χb (G). The b-coloring problem is to compute χb (G) for a given graph G with weight b(x), x ∈ V ∪ E. The graph in Fig. 1(a) has the maximum weight B = 7. One can easily observe the following lemmas. Lemma 2.1. For every weighted graph G = (V , E ), B ≤ χb (G) ≤ B(2χ(G) − 1). Proof. Obviously B ≤ χb (G). There is an ordinary coloring of G which uses a number χ(G) of colors ci , 1 ≤ i ≤ χ(G). Let f : V → N be a function such that f (v) = 2(i − 1)B + b(v) if v is colored by ci . Then f satisfies Eqs. (1)–(3), and hence f is a b-coloring of G. Therefore, χb (G) ≤ span(f ) ≤ B(2χ (G) − 1). Lemma 2.2. Let G = (V , E ) be a bipartite graph in which every vertex has degree one or more, and let B¯ = max{b(v) + b(v, w) + b(w) | (v, w) ∈ E }.
¯ Then χb (G) = B. ¯ Since G is a bipartite graph, G has an ordinary coloring Proof. Obviously B¯ ≤ χb (G). Thus, it suffices to prove that χb (G) ≤ B. with two colors c1 and c2 . Define f : V → N as follows: f (v) = b(v) if v is colored by c1 ; and f (w) = B¯
if w is colored by c2 .
¯ Then f satisfies Eqs. (1)–(3), and hence f is a b-coloring of G. Hence χb (G) ≤ span(f ) = B. Lemma 2.2 implies that the b-coloring problem can be solved in linear time for bipartite graphs and hence for trees. We then characterize χb (G) in terms of the longest path in acyclic orientations of G. Orient all edges of G so that the
− →
− →
resulting directed graph G is acyclic. The directed graph G is called an acyclic orientation of G. Fig. 1(b) depicts an acyclic orientation of the graph G in Fig. 1(a). The length ℓ(P , D) of a directed path P in an acyclic graph D is the sum of the weights of all vertices and edges in P. We
− →
− →
denote by ℓmax (D) the length of the longest directed path in D. For the acyclic graph G in Fig. 1(b) ℓmax ( G ) = 11, and the
− →
longest directed path P in G is drawn by thick lines. Extending the Gallai–Roy theorem on the ordinary coloring (see for example [13]), we have the following lemma on the b-coloring.
K. Nishikawa et al. / Theoretical Computer Science 532 (2014) 64–72
67
Lemma 2.3. For every graph G = (V , E ) with a weight function b
− → χb (G) = min ℓ ( G ), − → max G
− →
where the minimum is taken over all acyclic orientations G of G.
− →
→ ℓmax ( G ). Let f be an optimal b-coloring of G. Then span(f ) = χb (G). Orient each Proof. We first prove that χb (G) ≥ min− G edge (v, w) ∈ E from v to w if and only if f (v) < f (w). Then clearly the resulting directed graph D is acyclic. Let P = v1 , e1 , v2 , e2 , . . . , vp−1 , ep−1 , vp be the longest directed path in D, where edge ei , 1 ≤ i ≤ p − 1, goes from vertex vi to vi+1 . Then
ℓmax (D) =
p
b(vi ) +
i=1
p−1
b(vi , vi+1 ).
i=1
Since f is a b-coloring of G, by Eqs. (1) and (3) we have b(v1 ) ≤ f (v1 )
(4)
b(vi , vi+1 ) + b(vi+1 ) ≤ f (vi+1 ) − f (vi )
(5)
and
for every i, 1 ≤ i ≤ p − 1. Taking the sum of Eqs. (4) and (5) for all i, 1 ≤ i ≤ p − 1, we have
ℓmax (D) ≤ f (vp ) ≤ span(f ) = χb (G). − → − → → ℓmax ( G ) ≤ ℓmax (D), we have χb (G) ≥ min− → ℓmax ( G ). Since min− G G − → → ℓmax ( G ). Let D be an acyclic orientation of G such that We then prove that χb (G) ≤ min− G − → ℓmax (D) = min ℓ ( G ). − → max G
Let f : V → N be a mapping such that f (v) is the length of the longest directed path in D ending at v for each vertex v of D. Then, for every directed edge (v, w) of D, f (v) + b(v, w) + b(w) ≤ f (w) and hence f (v) ̸= f (w). The definition of f implies that b(v) ≤ f (v) for every vertex v ∈ V . Thus f is a b-coloring of G and span(f ) = ℓmax (D). Hence χb (G) ≤
− →
→ ℓmax ( G ). span(f ) = min− G
− →
− →
There are at most 2m acyclic orientations of G, and one can compute ℓmax ( G ) in time O(m + n) for each acyclic orientation
G of G, where m and n are the numbers of edges and vertices in G, respectively. Thus, Lemma 2.3 implies that χb (G) can be computed in time O((m + n)2m ), regardless of how large the weights are.
3. Exact algorithm for series-parallel graphs Many problems can be solved for series-parallel graphs in polynomial time or even in linear time [12]. In this section we show that the b-coloring problem can be solved for series-parallel graphs in pseudo polynomial-time O(B3 n). It should be noted that B3 n is polynomial in n and B. A series-parallel graph is recursively defined as follows [12]: 1. A graph G of a single edge is a series-parallel graph, and has the ends of the edge as terminals s and t of G. (See Fig. 2(a).) 2. Let G1 be a series-parallel graph with terminals s1 and t1 , and let G2 be a series-parallel graph with terminals s2 and t2 . (See Fig. 2(b).) (a) A graph G obtained from G1 and G2 by identifying t1 with s2 is a series-parallel graph, whose terminals are s1 and t2 . Such a connection is called a series connection. (See Fig. 2(c).) (b) A graph obtained from G1 and G2 by identifying s1 with s2 and identifying t1 with t2 is a series-parallel graph, whose terminals are s1 = s2 and t1 = t2 . Such a connection is called a parallel connection. (See Fig. 2(d).) Every series-parallel graph G can be represented by a ‘‘binary decomposition tree’’. Fig. 3 illustrates a decomposition tree T of the series-parallel graph G in Fig. 1(a). Labels s and p attached to internal nodes in T indicate series and parallel connections, respectively. Every leaf of T represents a subgraph of G induced by a single edge. A node u of T corresponds to a subgraph Gu of G induced by all edges represented by the leaves that are descendants of u in T . Thus G = Gr for the root r of T . One can find a decomposition tree of a given series-parallel graph in linear time [12]. The definition immediately implies that every series-parallel graph G has an ordinary coloring with at most three colors, that is, χ (G) ≤ 3. Therefore, by Lemma 2.1, we have χb (G) ≤ 5B. For a series-parallel graph G with terminals s and t and
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Fig. 2. Definition of series-parallel graphs.
Fig. 3. Decomposition tree T of the series-parallel graph in Fig. 1(a).
integers i and j, 1 ≤ i, j ≤ 5B, we define
χij (G) = min span(f ) f
where the minimum is taken over all b-colorings f of G such that f (s) = i and f (t ) = j. Let χij (G) = ∞ if there is no such b-coloring. One can recursively compute χij (G), 1 ≤ i, j ≤ 5B, as follows. Consider first the case where G consists of a single edge e = (s, t ) as illustrated in Fig. 2(a). Then χij (G) = max{i, j} if the following (a)–(c) hold: (a) i ̸= j, b(s) ≤ i, and b(t ) ≤ j; (b) i < j implies i + b(s, t ) + b(t ) ≤ j; and (c) j < i implies j + b(s, t ) + b(s) ≤ i. Otherwise, χij (G) = ∞. Consider next the case where G is obtained from G1 and G2 by a series connection as illustrated in Fig. 2(c). Then
χij (G) = min max{χik (G1 ), χkj (G2 )}. 1≤k≤5B
(6)
Consider finally the case where G is obtained from G1 and G2 by a parallel connection as illustrated in Fig. 2(d). Then
χij (G) = max{χij (G1 ), χij (G2 )}.
(7)
One may assume that a series-parallel graph G has no multiple edges. Then one can easily prove by induction that m ≤ 2n − 3. Since the binary decomposition tree T of G has m leaves, T has exactly m − 1(≤ 2n − 4) internal nodes. We compute χij (Gu ), 1 ≤ i, j ≤ 5B, for all nodes u of T from leaves to the root r. It takes time O(B3 n). Since G = Gr , we compute χb (G) from χij (Gr ) in time O(B2 ) as follows:
χb (G) = min χij (Gr ). 1≤i,j≤5B
Thus we have the following theorem. Theorem 3.1. The b-coloring problem can be solved in time O(B3 n) for a series-parallel graph G, where n is the number of vertices in G and B is the maximum weight of G. Clearly, B3 n is polynomial in n if B is bounded above by a polynomial in n.
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4. FPTAS In this section we give a fully polynomial-time approximation scheme (FPTAS) for the b-coloring problem on seriesparallel graphs. Let G be a graph with a weight function b, and let σ be a scaling factor which is a positive integer. Then we denote by Gσ a graph which is isomorphic with G but has a weight function bσ such that bσ (x) = ⌈b(x)/σ ⌉
(8)
for every element x ∈ V ∪ E. Fig. 1(c) depicts Gσ with σ = 2 for the graph G in Fig. 1(a). An optimal bσ -coloring Fσ of Gσ is also depicted in Fig. 1(c). We now have the following lemma. Lemma 4.1. Let G = (V , E ) be a graph with a weight function b, let σ be a positive integer, and let fσ be an optimal bσ -coloring of Gσ . Then, a function f such that f (v) = σ fσ (v) for every vertex v is a b-coloring of G, and hence χb (G) ≤ σ χbσ (Gσ ). Proof. Since fσ is an optimal bσ -coloring of Gσ , we have span(fσ ) = χbσ (Gσ ), bσ (v) ≤ fσ (v) for every vertex v ∈ V , fσ (v) ̸= fσ (w) for every edge (v, w), and fσ (v) + bσ (v, w) + bσ (w) ≤ fσ (w) for every edge (v, w) with fσ (v) < fσ (w). Therefore, we have b(v) ≤ σ bσ (v) ≤ σ fσ (v) = f (v) for every vertex v . Similarly, we have f (v) ̸= f (w) for every edge (v, w), and f (v) + b(v, w) + b(w) ≤ f (w) for every edge (v, w) with f (v) < f (w). Thus f is a b-coloring of G, and hence χb (G) ≤ span(f ) = σ · span(fσ ) = σ χbσ (Gσ ). Consider the following approximation scheme. Approximation scheme 1. Choose a scaling factor σ appropriately. (We will later choose σ = ⌊ε B/4n⌋ for a desired approximation error rate ε .) 2. Find an optimal bσ -coloring fσ of Gσ (by a pseudo polynomial-time exact algorithm, say the algorithm in Section 3). 3. Output, as an approximate solution, a b-coloring f of G such that f (v) = σ fσ (v) for every vertex v .
− →
− →
We now have the following lemma on the longest paths in acyclic orientations G and Gσ .
− → − → − → directed path in G . Let σ be a positive integer, let Gσ be the acyclic graph obtained from Gσ by orienting each edge in the same − → − →
Lemma 4.2. Let G = (V , E ) be a weighted graph of n vertices, let G be an acyclic orientation of G, and let P be the longest direction as in G , and let Pσ be the longest directed path in Gσ . (See Fig. 1.) Then
− → − → ℓ(Pσ , Gσ ) < ℓ(P , Gσ ) + 2n.
(9)
Proof. By Eq. (8) we have bσ (x) < b(x)/σ + 1 for every element x ∈ V ∪ E. Clearly there are at most 2n − 1 elements (vertices and edges) in Pσ . Therefore, we have
− → σ ℓ(Pσ , Gσ ) = σ b σ ( x) x∈Pσ
b(x) x∈Pσ
σ
+1
b(x) + σ (2n − 1)
x∈Pσ
− → < ℓ(Pσ , G ) + 2σ n
(10)
− →
where the summation is taken over all elements x in Pσ . Since P is the longest path in G ,
− → − → ℓ(Pσ , G ) ≤ ℓ(P , G ).
(11)
Since b(x) ≤ σ bσ (x) for every element x ∈ V ∪ E, we have
− → ℓ(P , G ) = b(x) x∈ P
≤σ
b σ ( x)
x∈ P
− → = σ ℓ(P , Gσ ). − → − → From Eqs. (10)–(12) we have σ ℓ(Pσ , Gσ ) < σ ℓ(P , Gσ ) + 2σ n. We have thus proved Eq. (9). Using Lemmas 2.3 and 4.2, we then have the following lemma on the error of the approximation scheme above.
(12)
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Lemma 4.3. For a positive integer σ and a graph G = (V , E ) with a weight function b
σ χbσ (Gσ ) < χb (G) + 4σ n. − →
− →
Proof. Lemma 2.3 implies that there is an acyclic orientation G of G and the longest path P in G such that
− → − → χb (G) = ℓmax ( G ) = ℓ(P , G ).
(13)
− →
− →
Let Gσ be the acyclic orientation obtained from Gσ by orienting each edge in the same direction as in G . The path P contains at most 2n − 1 elements (vertices and edges). Therefore, we have
χb (G) + 2σ n =
b(x) + 2σ n
x∈P
b(x) x∈P
σ
+1
b σ ( x)
x∈P
− → = σ ℓ(P , Gσ ). − →
(14)
− → − → ℓ(P , Gσ ) + 2n > ℓ(Pσ , Gσ ).
(15)
Let Pσ be the longest path in Gσ , then by Lemma 4.2 we have
By Lemma 2.3 we have
− → ℓ(Pσ , Gσ ) ≥ χbσ (Gσ ).
(16)
By Eqs. (14)–(16) we have
− → χb (G) + 4σ n > σ ℓ(P , Gσ ) + 2σ n − → > σ ℓ(Pσ , Gσ ) ≥ σ χbσ (Gσ ). Let ε (>0) be a desired approximation error rate. If ε B/4n ≤ 1, then we compute χb (G) by a pseudo polynomial-time exact algorithm, say the algorithm in Section 3; the computation time is bounded by a polynomial in n and 1/ε since B ≤ 4n/ε . One may thus assume that ε B/4n > 1. We then choose σ = ⌊ε B/4n⌋(≥1), and find an approximately optimal b-coloring f (=σ fσ ) of G by the approximation scheme above. By Lemmas 2.1 and 4.3 one can bound the error as follows: span(f ) − χb (G) = σ χbσ (Gσ ) − χb (G)
< 4σ n ≤ εB ≤ εχb (G).
(17)
We thus have the following theorem. Theorem 4.1. If there is an exact algorithm to solve the b-coloring problem for a class of graphs in time polynomial in n and B, then there is a fully polynomial-time approximation scheme for the class. Proof. Suppose that the algorithm finds an optimal b-coloring of a graph G in the class in time p(n, B), where p(n, B) is a polynomial in n and B. Find an optimal bσ -coloring fσ of Gσ in time p(n, Bσ ) by the algorithm, and output a b-coloring f = σ fσ of G by the approximation scheme above, where σ = ⌊εB/4n⌋ and Bσ = ⌈B/σ ⌉ is the maximum weight of Gσ . By Eq. (17) the error is less than εχb (G). Since Bσ = O(n/ε), the computation time p(n, Bσ ) of the scheme is bounded by a polynomial in n and 1/ε . From Theorems 3.1 and 4.1 we thus have the following corollary. Corollary 4.1. There is a fully polynomial-time approximation scheme for the b-coloring problem on series-parallel graphs, and the computation time is O(Bσ 3 n) = O(n4 /ε 3 ).
K. Nishikawa et al. / Theoretical Computer Science 532 (2014) 64–72
71
5. Partial k-Trees The class of partial k-trees, that is, graphs with bounded tree-width, contains trees, outerplanar graphs, series-parallel graphs, etc. A series-parallel graph is indeed a partial 2-tree. In this section we show that the results in Sections 3 and 4 can be extended to partial k-trees. For a bounded positive integer k, a k-tree is recursively defined as follows [1,2]: (1) A complete graph with k vertices is a k-tree. (2) If G = (V , E ) is a k-tree and k vertices v1 , v2 , . . . , vk induce a complete subgraph of G, then G′ = (V ∪ {w}, E ∪ {(vi , w) : 1 ≤ i ≤ k}) is a k-tree where w is a new vertex not contained in G. Any subgraph of a k-tree is called a partial k-tree. A binary tree T = (VT , ET ) is called a tree decomposition of a partial k-tree G = (V , E ) if T satisfies the following conditions (a)–(e): (a) (b) (c) (d) (e)
every node X ∈ VT is a subset of V and |X | = k + 1; X ∈V T X = V ; for each edge e = (u, v) ∈ E, T has a leaf X ∈ VT such that u, v ∈ X ; if node Xp lies on the path in T from node Xq to node Xr , then Xq ∩ Xr ⊆ Xp ; and each internal node Xi of T has exactly two children, say Xℓ and Xr , such that |Xℓ − Xr | = 1 and either Xi = Xℓ or Xi = Xr .
Each node X of T corresponds to a subgraph GX of G. If X is a leaf of T , then GX is a subgraph of G induced by the vertices in X . If Xℓ and Xr are the two children of an internal node Xi of T , then GXi is a union of GXℓ and GXr , whose common vertices are all contained in Xi . Thus G = GXroot for the root Xroot of T . One can easily observe from the definitions above that χ(G) ≤ k + 1 for every partial k-tree G. Therefore, by Lemma 2.1 we have χb (G) ≤ (2k + 1)B. Similarly as in Section 3, we compute the counterparts of χij from leaves to the root of a tree decomposition T of G. Since χb (G) ≤ (2k + 1)B and |X | = k + 1 for every node X of T , there are a number ((2k + 1)B)k+1 of counterparts of χij . Since T has O(n) leaves, the counterparts of χij can be computed in time
(k + 1)2 × ((2k + 1)B)k+1 × n = O(Bk+1 n) for all leaves of T . Since T has O(n) nodes and every internal node Xi of T has two children Xℓ and Xr such that |Xℓ − Xr | = 1 and either Xi = Xℓ or Xi = Xr , the counterparts of χij can be computed in time
((2k + 1)B)k+1 × (2k + 1)B × n = O(Bk+2 n) for all internal nodes Xi of T . From the counterparts of χij for the root of T , χb (G) can be computed in time O(Bk+1 ). Thus χb (G) can be computed in time O(Bk+2 n). Since the time is bounded by a polynomial in n and B, by Theorem 4.1 the scheme in Section 4 is an FPTAS and takes time Bσ
k+2
n=O
k+2 n ε
n .
We thus have the following corollary. Corollary 5.1. There is a fully polynomial-time approximation scheme for the b-coloring problem on partial k-trees. 6. Conclusions We first investigated the fundamental properties of a b-coloring. We then gave a pseudo polynomial-time exact algorithm and a fully polynomial-time approximation scheme for the b-coloring problem on series-parallel graphs and partial k-trees. It is desired to improve the time complexities. It is open whether the b-coloring problem can be solved in polynomial time or is NP-hard for series-parallel graphs or partial 2-trees. Acknowledgments We thank anonymous referees, whose comments and suggestions helped us to improve the presentation of the paper. References [1] S. Arnborg, A. Proskurowski, Linear time algorithms for NP-hard problems restricted to partial k-trees, Discrete Appl. Math. 23 (1989) 11–24. [2] H.L. Bodlaender, Treewidth: algorithmic techniques and results, in: Proc. MFCS 1997, in: Springer Lect. Notes in Computer Science, vol. 1295, 1997, pp. 19–36.
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[3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
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M.M. Halldórsson, G. Kortsarz, Tools for multicoloring with applications to planar graphs and partial k-trees, J. Algorithms 42 (2) (2002) 334–366. M.M. Halldórsson, G. Kortsarz, A. Proskurowski, Multicoloring trees, Inform. and Comput. 180 (2) (2003) 113–129. T. Ito, T. Nishizeki, X. Zhou, Algorithms for multicolorings of partial k-trees, IEICE Trans. Inf. Syst. E86-D (2) (2003) 191–200. K. Jansen, P. Scheffler, Generalized coloring for tree-like graphs, Discrete Appl. Math. 75 (2) (1997) 135–155. T.R. Jensen, B. Toft, Graph Coloring Problems, Wiley, New York, 1994. E. Malaguti, P. Toth, An evolutionary approach for bandwidth multicoloring problems, European J. Oper. Res. 189 (2008) 638–651. C. McDiamid, B. Reed, Channel assignment on graphs of bounded treewidth, Discrete Math. 273 (2003) 183–192. M.L. Pinedo, Scheduling: Theory, Algorithms and Systems, Springer Science, New York, 2008. K. Nishikawa, T. Nishizeki, X. Zhou, Algorithms for bandwidth consecutive multicolorings of graphs (Extended Abstract), in: Proc. FAW-AIIM 2012, in: Lecture Notes in Computer Science, vol. 7285, 2012, pp. 117–128. K. Takamizawa, T. Nishizeki, N. Saito, Linear-time computability of combinatorial problems on series-parallel graphs, J. Assoc. Comput. Mach. 29 (1982) 623–641. D.B. West, Introduction to Graph Theory, Prentice-Hall, Englewood Cliffs, NJ, 1996. X. Zhou, Y. Kanari, T. Nishizeki, Generalized vertex-colorings of partial k-trees, IEICE Trans. Fundam. E83-A (4) (2000) 671–678. X. Zhou, T. Nishizeki, Multicolorings of series-parallel graphs, Algorithmica 38 (2004) 271–297. D. Zuckerman, Linear degree extractors and the inapproximability of max clique and chromatic number, Theory Comput. 3 (2007) 103–128. | 8,693 | 27,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-34 | latest | en | 0.812791 |
https://mykonos-greece.biz/5v-battery-for-arduino-5v-battery-for-arduino/ | 1,716,521,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00340.warc.gz | 345,841,726 | 24,130 | # 5V battery for arduino. 5v battery for arduino
## Follow me down the optimization rabbit hole
I optimize software for a living. This blog is to share some personal projects which others may find interesting.
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### The Premise
So, you’ve created an Arduino project and you want to power it with batteries to take it on the road. Your board’s components are designed to run on 5 Volts and you know you can’t feed 9V directly into the Vcc of a 5V board because it will damage it. Arduino has you covered. the pin marked RAW is for that purpose and according to the documentation, you can feed it between 6 and 12V and it will regulate that voltage down to the 5V needed by the board. Perfect, right? Well, not quite.
### Voltage Regulation
There are 2 main ways to regulate (aka control) the voltage. A linear regulator allows you to supply a higher voltage than desired (in our case 9V) and get a stable, lower voltage as output. It essentially does this by generating heat from the excess energy. Let’s say your board uses 100mA @5V while executing your code. If you’re powering it from a 9V battery through a linear regulator, then you’re creating 4V x 100mA = 400mW of waste heat (9V. 5V = 4V). If you measure the current coming out of your 9V battery, it will be very close to 100mA, so 9V @ 100mA is going into the regulator and 5V @ 100mA is coming out. The means that the linear regulator effectively has an energy efficiency of 56% in this case. For many many years, the 78xx series of linear regulators is what you would use for cases like this:
Aside from the simplicity, there isn’t much to like about linear regulators. When running on battery power, you’re throwing away a large percentage of your battery’s energy when you regulate the output this way. In circuits that use a lot of current, heat becomes a major concern too.
The older Arduino boards have a linear regulator built in to make it easier to power them from various energy sources. The assumption is usually that you’ll be running from a wall wart (A/C power brick) so wasted energy isn’t much of a concern and that you’ll be running a small current through your circuit, so waste heat isn’t a concern either.
If you’re using a 9V alkaline battery that has a typical energy capacity of 500mAh and you connect it to the Arduino’s linear regulator.
In effect, instead of having a 4.5Wh battery, you have a 2.5Wh battery because the rest of the energy is given off as heat.
Buck (or step down) converters are another way to convert a high voltage source to a lower voltage. Buck converters are a more complex circuit that relies on an oscillator and inductor (coil) to change the voltage. The advantage of the buck converter is that it doesn’t waste nearly as much energy as a linear regulator. Here are the efficiency curves for a typical buck converter:
With 5V output at low current, it approaches 97% efficiency. That means that your 9V battery could potentially provide closer to 4.5Wh of energy to your circuit instead of the 2.5Wh you get with the linear regulator. Electronut Labs sells a convenient buck converter specifically designed for easy use with 9V batteries:
What about going in the other direction? This is the option that’s usually not mentioned in Arduino project articles. It’s also possible to boost the voltage from a lower voltage to a higher voltage. DC-DC boost converters work similarly to buck converters and use a high frequency oscillator and a coil to generate a higher voltage. They also have typical efficiencies greater than 85%. This frees you to use other power sources such as a single AA battery. An Alkaline AA battery typically has a capacity of around 2500mAh. At 1.5V, this translates to about 3.75Wh of energy. Less total energy than a 9V battery, but used efficiently, it can save space and cost compared to a 9V. If we boost 1.5V to 5V and assume that the boost converter has an efficiency of 90%, we should be able to squeeze about 3.375Wh of energy out of it. Here’s a typical DC-DC boost converter sold by various vendors in China for 0.45-1.00 each:
I like to use these in my projects because they’re tiny and inexpensive. They operate down to about 0.8V as input and the output is clean enough (low noise) to use in most microcontroller projects.
### Power Down
The original set of Arduino boards were all based on AVR microcontrollers and all set to run at 5V. This made sense at the time because the AVR MCU can operate on any voltage between 1.8V and 5.5V, but at the higher voltages, the clock can run at up to 20Mhz (see chart below):
The reality is that your project probably doesn’t need the MCU to run at 20Mhz. If you’re reading a few sensors and updating a display, you could accomplish the same work at a lower clock rate. Another reason is that the amount of energy used by the MCU and peripherals does not perfectly follow a linear scale. Even so, you can potentially accomplish the same amount of work running at 8Mhz and 3.3V as you could at 16Mhz and 5V. Running the CPU slower or at a lower voltage uses less energy.
A lot of newer MCUs in the Arduino lineup operate at 3.3v (e.g. ARM Cortex-M MCUs) and so do many add-on boards, so it makes a lot of sense to run your project at 3.3v. Without having to know too much about AVR fuses and hardware, it’s possible to run a board designed for 5V and 16Mhz at 3.3V and 8Mhz with a simple trick in software. The main CPU clock divider can be set in software. We can use this to cause a 16Mhz part to run at 8Mhz so that it can run reliably at 3.3V. The clock prescaler is normally set to 1 on Arduino boards. By setting it to 2, the CPU will run at a more stable 8Mhz:
Now with this new information, let’s look at cost and battery life of our original 5V project running on a 9V battery versus our new idea of running it at 3.3V from a single AA battery. For the cost, I’m making the assumption of buying a 4-pack of each battery type (Duracell) from Amazon.com.
## v battery for arduino
• Pressure Sensor Module
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• Touch Sensor Module
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• DM Strong Series
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• For Arduino
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• Wireless Camera
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• 1-Channel Delay Relay
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### 18650 Battery Shield V8 Mobile Power Bank 3V/5V for Arduino ESP32 ESP8266
• guarantee Quality checked
• Free return Within 60 days
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This module is a portable mobile power supply that supports 3V / 1A and 5V / 2.2A two voltage outputs!
5V voltage output rated current is 2.2A, maximum support 3A current (not recommended for overload, easy to damage the module, speed up the service life), the output of current depends on the quality of the 18650 battery!
MICRO USB charging current 600mA-800mA,
Supports up to two 18650 batteries. It is recommended to use two 18650 batteries, which can be longer!
Special attention: the installation of the battery must be determined positive and negative, the board has been clearly marked positive and negative! Installation errors will burn out the module!
Battery protection(Over charge or Over discharge)Micro USB port InputType-A USB Output0.5A current charging1 switch control USB output5~8V Input Voltage3V 1A Output5V 2A OutputLED indicate(Green means full,Red means charging)3V output port x35V output port x3Be careful of and., You should follow the direction of. on the PCB.If you put wrong direction, charging chip will be destroyed.This product does not include 18650 battery
Size: 9.8×2.9cm/3.86×1.14inchQuantity: 1 Set
1.Please allow 1-2cm error due to manual measurement. pls make sure you do not mind before you bid.
2.Due to the difference between different monitors, the picture may not reflect the actual color of the item. Thank you!
100% brand new and high quality High Speed USB 2.0 Cable Compatible with Cellphones, GPS systems, PDAs, OTG devices and digital cameras Cable lenght: 30CM Complies with fully rated cable specification using braid-and-foil shield protection
This 30CM USB A to Micro USB B 5pin Cable provides one USB A Male connector, and one Micro USB B connector to create a simple way to connect mobile devices to a USB capable computer for every day tasks such as data synchronization and file transfers.
The micro-USB connector is smaller and thinner than the previous USB Mini-b standard while offering better performance.
Before you make your purchase, it’s helpful to know the measurements of the area you plan to place the furniture. You should also measure any doorways and hallways through which the furniture will pass to get to its final destination.
Picking up at the store
Shopify Shop requires that all products are properly inspected BEFORE you take it home to insure there are no surprises. Our team is happy to open all packages and will assist in the inspection process. We will then reseal packages for safe transport. We encourage all customers to bring furniture pads or blankets to protect the items during transport as well as rope or tie downs. Shopify Shop will not be responsible for damage that occurs after leaving the store or during transit. It is the purchaser’s responsibility to make sure the correct items are picked up and in good condition.
## rlogiacco/BatterySense
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This is a simple Arduino library to monitor battery consumption of your battery powered projects, being LiPo, LiIon, NiCd or any other battery type, single or multiple cells: if it can power your Arduino you can monitor it!
The principle is simple: we are going to measure our battery capacity by measuring the voltage across the battery terminals.
The big assumption here is that battery capacity is linearly correlated to its voltage: the assumption itself is wrong, but in most cases it’s close enough to reality, especially when it comes to the battery higher capacity side.
In reality, the relation between battery capacity and its voltage is better represented by a curve and there are many factors affecting it: current drawn, temperature, age, etc.
The library requires at least 1 analog pin (we will call this the sense pin ) and no less than 2 pieces of info on your battery: the voltage you will consider the minimum acceptable level, below which your project/product becomes unreliable and should be shut down, and the maximum voltage you can expect when the battery is fully charged.
Additionally, you can provide a second pin (either analog or digital) to activate the battery measurement circuit (we call it the activation pin ), useful in all those situations where you can sacrifice a pin to further increase your battery duration.
If you want your readings to be more accurate we strongly suggest to calibrate the library by providing your board reference voltage: most of the times you assume your board has exactly 5V between Vcc and GND. but this is rarely the case. To improve this we suggest using the VoltageReference library to obtain a better calibration value for all analog readings.
The sense pin wiring can vary depending on your battery configuration, but here are a few examples based on the assumption you are using a 5V board: in case of a 3.3V board you should be performing the necessary adjustments.
Lesser than 5V, with voltage booster
Voltage sources made of single cell LiPo or LiIon, along with some single or multi-cell NiCd configurations (like up to 3 AA or AAA), are not able to provide the suggested 5.0 volts input to your board and a voltage booster can solve your problem. What does that mean when it comes to measuring your battery level? We need to measure the battery voltage before it gets boosted, which means your sense pin must be connected between the battery positive terminal and the booster positive input and we don’t need any additional components as the voltage is already in the acceptable range:
SENSE | | |. | | | | | | | BAT.- IN | 5V | OUT. 5V | Arduino | | BOOSTER | | | BAT-. IN- | | OUT-.- GND | |. | |.
Higher than 5V, with internal voltage regulator
Voltage sources made of multiple cells LiPo or LiIon, along with some single or multi-cell NiCd configurations (like up the classic 9V battery or 4 AA or AAA), provide voltages above the 5.0 volts input : most of the Arduino boards are equipped with voltage regulators able to dissipate into heat all the excess. To measure such batteries we need to hook our sense pin before it gets regulated, between the battery positive terminal and the Arduino unregulated input VIN or RAW. but we require two resistors to reduce the voltage to acceptable values:
. BAT.- VIN | | | | | R1 | | | | |. SENSE | Arduino | | | | R2 | | | | | BAT-. GND | |.
The values of R1 and R2 determine the voltage ratio parameter for this library: for information about this value refer to the section below.
Because the resistors in this configuration will constantly draw power out of your battery, you shouldn’t pick values under 1k Ohm. or you’ll deplete your batteries much faster than normal. On the other end, going too high on the resistor values will impede the library from getting accurate readings.
Higher than 5V, with external voltage regulator
Whenever your battery maximum voltage exceeds the onboard regulator (if there is any) an external voltage regulator is required. Once again, to measure such batteries we need to hook our sense pin before it gets regulated, between the battery positive terminal and the voltage regulator positive input VIN or RAW and, as before, we require two resistors to reduce the voltage to acceptable values:
– |. |. | BAT.- IN | | SENSE | | | | | | | | | R1 | | | | | | | | | |. | REG | OUT.- 5V | Arduino | | | | | | R2 | | | | | | | | | BAT-.- IN- | | OUT-. GND | |
The values of R1 and R2 determine the voltage ratio parameter for this library: for information about this value refer to the section below.
Higher than 5V, activated on demand
Batteries are a precious resource and you want to prolong their life as much as you can so, deplete your battery to determine its capacity is not desirable.
As a consequence of connecting the battery terminals through two resistors we are drawing some energy out of the battery: for a 9V battery and 1k Ohm for R1 and R2, you will be adding a constant 4.5mA current consumption to your circuit. Not a huge amount, but definitely not desirable.
If you have an unused pin on your Arduino it will be easy to limit this additional current consumption to be drawn only when needed: during battery measurement. We will be turning the activation pin HIGH during battery measurement so that the voltage divider will be disconnected most of the time:
. BAT.- VIN | | | | | SW- ACT | | | | | R1 | | | | Arduino |. SENSE | | | | | R2 | | | | | BAT-. GND | |.
In the above schematics SW is a circuit which can connect or disconnect the sensing circuit depending on the voltage on ACT : the most common and cheap circuit is made of a NPN transistor Q1, a p-channel MOSFET Q2, a 1k-4.7k Ohm resistor R3 and a 5k-20k Ohm resistor R4:
BAT |. | | R4 | |\ | ACT. R3.Q1 \ Q2 | | | | GND VDIV to R1/R2/SENSE
Feel free to refer to this circuit simulation to better understand how the circuit works and how much current draws when in operation.
Whenever your battery voltage is above your board voltage you need a voltage divider to constraint your readings within the 0-5V range allowed by your Arduino and you will have to provide this library with its ratio.
BAT. | R1 |. SENSE | R2 | BAT-.-
The voltage divider ratio is determined by the formula (R1 R2) / R2 : if you use two resistors of the same value the ratio will be 2, which can be interpreted as whatever value we read it will be half of the actual value. This allows us to sense batteries up to 10V. If you use a 22k Ohm resistor for R1 and a 10k Ohm for R2 than your voltage ratio will be 3.2 and you will be able to safely monitor a 12-15V battery.
You must select the resistors in order to get a ratio which will produce values between the 0-5V range (or 0-3.3V for 3.3V devices) at all the times and to obtain that the process is quite simple: divide your battery maximum voltage by 5V and you’ll get the absolute minimum value for the voltage ratio. then pick any two resistors values whose combination produce a ratio equal or higher than the absolute minimum. For a 12V battery the absolute minimum voltage ratio is 12/5=2.4, meaning you can’t use a split supply divider made of two equal resistors: you need R1 to be a higher value than R2! Get this wrong and you will probably burn your sense pin.
You can use this nice website to find some appropriate values for the resistors setting your battery maximum voltage as Voltage source and aiming at obtaining a Output voltage value lesser than your board voltage ( 5V or 3.3V ) but as close as possible.
The voltage divider total resistance, made of R1 R2. will determine the current drawn from your battery by the sensing circuit: lower is the total resistance and more accurate are your readings, higher the resistance and less current is drawn from your battery (Ohm’s law rulez!). My suggestion is to keep this value within 20k-22k Ohm when using an always-connected circuit and under 10k Ohm if you use an on-demand configuration.
When determining the ratio don’t stick with the resistors nominal values, instead, if possible, use a multimeter to actually measure their resistance so to improve your results: a 4.7kΩ resistor could easily be a 4.75kΩ in reality!
Remaining capacity approximation
The level available functions aim at providing an approximation of the remaining battery capacity in percentage. This is not an easy task when you want to achieve reliable values and it is something the industry of mobile devices invests a decent amount of resources. When an accurate estimate is desireable the battery voltage is not the sole parameter you want to take into consideration:
• cell chemistry has a very high influence, obviously
• cells based on the same chemistry might produce pretty different results depending on the production process
• each chemistry has a different ideal operating cell temperature
• the rate you draw current from the battery influences the remaining capacity
• batteries are not everlasting: as the cell ages, the battery capacity gets reduced
• and more
The library itself doesn’t aim at providing accurate estimates, but what I consider an improvable but good enough estimate.
The library can be configured to use a mapping function of your choice, given the function complies with the mapFn_t interface:
uint8_t mapFunction(uint16_t voltage, uint16_t minVoltage, uint16_t maxVoltage)
To configure your personalized function you only have to provide a pointer to it during initialization:
Battery batt = Battery(3000, 4200, SENSE_PIN); uint8_t myMapFunction(uint16_t voltage, uint16_t minVoltage, uint16_t maxVoltage) // your code here void setup batt.begin(3300, 1.47, myMapFunction);
You are not limited in considering only the parameters listed in the function interface, meaning you can take into consideration the cell(s) temperature, current consumption or age: that’s open to your requirements and circuitry.
After collecting a few data points on battery voltage vs. battery capacity, I’ve used the https://mycurvefit.com/ and https://www.desmos.com online tools to calculate the math functions best representing the data I’ve collected.
In the above plot I represent the battery percentage (Y axis) as a function of the difference between the current battery voltage and the minimum value (X axis): the graph represents a battery with a voltage swing of 1200mV from full to empty, but the functions scale accordingly to the minVoltage and maxVoltage parameters.
The library ships with three different implementations of mapping function:
• linear is the default one (dashed red), probably the least accurate but the easiest to understand. It’s main drawback is, for most chemistries, it will very quickly go from 25-20% to 0%, meaning you have to select the minVoltage parameter for your battery accordingly. As an example, a typical Li-Ion battery having a 3V to 4.2V range, you want to specify a 3.3V configuration value as minimum voltage.
• sigmoidal (in blue) is a good compromise between computational effort and approximation, modeled after the tipical discharge curve of Li-Ion and Li-Poly chemistries. It’s more representative of the remaining charge on the lower end of the spectrum, meaning you can set the minimum voltage accordingly to the battery safe discharge limit (typically 3V for a Li-Ion or Li-Poly).
• asymmetric sigmoidal (in green) is probably the best approximation when you only look at battery voltage, but it’s more computational expensive compared to sigmoidal function and, in most cases, it doesn’t provide a great advantage over it’s simmetric counterpart.
I strongly encourage you to determine the function that best matches your particular battery chemistry/producer when you want to use this library in your product.
Here follow a few real case scenarios which can guide you in using this library.
Single-cell Li-Ion on 3.3V MCU
As an example, for a single cell Li-Ion battery (4.2V. 3.7V) powering a 3.3V MCU. you’ll need to use a voltage divider with a ratio no less than 1.3. Considering only E6 resistors, you can use a 4.7kΩ (R1) and a 10kΩ (R2) to set a ratio of 1.47 : this allows to measure batteries with a maximum voltage of 4.85V. well within the swing of a Li-Ion. It’s a little too current hungry for my tastes in an always-connected configuration, but still ok. Considering the chemistry maps pretty well to our sigmoidal approximation function I’m going to set it accordingly along with the minimum voltage which lowest safe value clearly is 3.0V (if a Li-Ion is drained below 3.0V the risk of permanent damage is high), so your code should look like:
Battery batt = Battery(3000, 4200, SENSE_PIN); void setup // specify an activationPin activationMode for on-demand configurations //batt.onDemand(3, HIGH); batt.begin(3300, 1.47, sigmoidal);
Double cell Li-Ion (2S) on 5V MCU
For a double cell Li-Ion battery (8.4V. 7.4V) powering a 5V MCU. you’ll need to use a voltage divider with a ratio no less than 1.68 : you can use a 6.8kΩ (R1) and a 10kΩ (R2) to set the ratio precisely at 1.68. perfect for our 8.4V battery pack. The circuit will continuously draw 0.5mA in an always-connected configuration, if you can live with that. As we don’t want to ruin our battery pack and we don’t want to rush from 20% to empty in afew seconds, we’ll have to set the minimum voltage to 6.8V (with a linear mapping) to avoid the risk of permanent damage, meaning your code should look like:
Battery batt = Battery(6800, 8400, SENSE_PIN); void setup // specify an activationPin activationMode for on-demand configurations //batt.onDemand(3, HIGH); batt.begin(5000, 1.68);
NOTE: I could have used the sigmoidal approximation, as the chemistry fits pretty well on the curve, in which case a 6V minimum voltage would have been a better configuration value.
Another classic example might be a single 9V Alkaline battery (9V. 6V) powering a 5V MCU. In this case, you’ll need to use a voltage divider with a ratio no less than 1.8 and, for sake of simplicity, we’ll go for a nice round 2 ratio. Using a nice 10kΩ both for R1 and R2 we’ll be able to measure batteries with a maximum voltage of 10V consuming only 0.45mA. The trick here is to determine when our battery should be considered empty: a 9V Alkaline, being a non-rechargeable one, can potentially go down to 0V, but it’s hard our board can still be alive when this occurs. Assuming we are using a linear regulator to step down the battery voltage to power our board we’ll have to account for the regulator voltage drop: assuming it’s a 1.2V drop, we might safely consider our battery empty when it reaches 6.2V (5V 1.2V), leading to the following code:
Battery batt = Battery(6200, 9000, SENSE_PIN); void setup // specify an activationPin activationMode for on-demand configurations //batt.onDemand(3, HIGH); batt.begin(5000, 2.0);
NOTE: Most 5V MCU can actually continue to operate when receiving 4.8V or even less: if you want to squeeze out as much energy as you can you can fine tune the low end, but also consider there is not much juice left when a battery voltage drops that much.
## Introduction
This tutorial demonstrates how to power your Arduino Uno with a solar cell. Solar cells can be a useful solution for powering projects that require portability or remote monitoring. This tutorial uses concepts drawn from the following resources:
## Parts
This project requires the following components:
## Wiring
he following steps describe how to set up your Arduino Uno with solar power. As a note, components should be soldered together for stability.
Step 1: Solder M-M jumper wires to the positive and negative (-) terminals of the solar cell.
Step 2: Solder the other end of the M-M jumper wires to the input terminals of the TP4056 battery charge controller.
Step 3: Solder the output wires from the battery holder to the TP4056 battery charge controller B and B- terminals.
Step 4: Solder a second set of M-M jumper wires to the output terminals of the TP4056 battery charge controller.
Step 5: Solder the other end of the M-M jumper wires to the input terminal of the XL6009 – Voltage Adjustable DC-DC (5v-35v) Boost Converter. Use a voltmeter connected to the output terminals to determine the output voltage. Powering the Arduino Uno through the Vin port requires an input between 7 and 12 Volts, so the desired output from the Boost Converter is 9V. The voltage output can be adjusted by turning the knob located on the blue rectangle.
Step 6: Solder another set of M-M jumper wires to the output terminals of the Boost Converter. Insert the other end of the M-M jumper wires to the Arduino Uno with the positive terminal connected to the Vin pin and the negative terminal connected to the GND pin (-).
If working properly, the green light of your Arduino Uno should light up and it should now be ready to use!
## Can I connect the solar cell directly to the Arduino Uno?
This is not a good idea for several reasons. Due to variability in sun This is not a good idea for several reasons. Due to variability in sun exposure, the solar cell may not provide a steady stream of power. The Arduino Uno may not be able to draw the maximum power at any given instant from the solar cell. Additionally, the power demands from the Arduino Uno may overload the solar cell. Using a rechargeable battery provides a constant, reliable energy source.
## Are lithium-ion batteries safe to work with?
Lithium-ion batteries are extremely sensitive to charging characteristics and can easily catch fire or explode. It is necessary to take precautions when working with these batteries, considering they contain a high amount of energy and volatile chemical content.
The TP4056 battery charge controller works to mitigate the risks of working with lithium-ion batteries. The controller regulates the current produced by the solar cell to protect the batteries from overcharging. The controller detects when the battery is fully charged and can stop or limit the current received by the battery. Additionally, the controller also protects the solar cell by stopping reverse current flowing back from the batteries when there is no sunlight.
## How do I choose a solar cell and battery?
The TP4056 battery charge controller has a maximum input of 6V, thus, the solar cell should be at maximum 6V. The voltage of the solar cell should be at least 1.5 times the voltage of the battery. So a 3.7V lithium-ion battery needs a solar cell of at least 5.55V. The current of the solar cell should have 1/10th of the capacity of the battery divided by 1 hour. So a lithium-ion battery of 2000 mAh, should be supported by a solar cell with around 200 mAh.
## Why do I need a boost converter?
The power source that connects to the Vin pin on the Arduino Uno has to be 7 to 12 volts for the regulator to work reliably. The Vin pin converts unregulated input voltage to a stable 5V. The output voltage from the lithium-ion battery is 3.7V. A boost converter converter can step up the voltage from its input to its output to meet the desired input range of the Vin pin of between 7 and 12 volts.
## Solar Power Manager 5V
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### Introduction
The DFRobot Solar Power Manager series are designed for IoT projects and renewable energy projects, providing safe and high-efficiency embedded solar power management modules for makers and application engineers.
Solar Power Manager 5V is a small power and high-efficiency solar power management module designed for 5V solar panel. It features as MPPT (Maximum Power Point Tracking) function, maximizing the efficiency of the solar panel. The module can provide up to 900mA charging current to 3.7V Li battery with USB charger or solar panel. The ON/OFF controllable DC-DC converters with 5V 1A output satisfies the needs of various solar power projects and low-power applications. The module also employs various protection functions for battery, solar panel and output, which greatly improves the stability and safety of your solar projects.
Version History V1.0: It is highly recommended to use a 3.7V lipo battery with protection circuits to improve battery safety. V1.1(Newest): Battery (BAT IN) overcurrent and overdischarge protection is added to improve battery safety. It can be use for 3.7 lipo battery no matter it is packed with protection circuits or not.
Solar Power Manager 5V
A complete multifunction solar power management module.
Applications: Small Solar Street Lamp, Solar Powered Robots
For 9V/12V/18V Solar Panels within20w
A small and easy-to-use 5V solar power management module.
Applications: Solar Power Bank, Solar Environment Monitors
For 5V Solar Panels within 10W
A micro power solar power management module for low-power sensors and controllers.
Applications: Wireless Sensor Network, BLE iBecon
For 1V/2V/3V Solar Panels within 0.5W
A medium power solar management module for 12V lead-acid batteries.
Applications: Street lighting, intelligent agriculture, environmental monitoring station, UPS
For 18V Solar Panels within 100W
### Features
• Constant voltage MPPT algorithm, maximizing solar panel efficiency
• Designed for 5V solar panel
• Double charging mode: solar/USB charger (900mA max charge current)
• 5V ON/OFF controllable regulated power supply for low-power applications
• All-round protection functions
• USB connector with ESD shell
### Applications
• Smart solar environment monitor system
• Solar powered robot
• Small solar street lamp
• Solar power bank
### Specification
• Solar Power Management IC: CN3165
• Solar Input Voltage (SOLAR IN): 4.5V~6V
• Battery Input (BAT IN): 3.7V Single cell Li-polymer/Li-ion Battery
• Charge Current(USB/SOLAR IN): 900mA Max trickle charging, constant current, constant voltage three phases charging
• Charging Cutoff Voltage (USB/SOLAR IN): 4.2V±1%
• Regulated Power Supply: 5V 1A
• Regulated Power Supply Efficiency (3.7V BAT IN): 86%@50%Load
• USB/Solar Charge Efficiency: 73%@3.7V 900mA BAT IN
• Quiescent Current: 2023-01-11 00:17:07 | 7,601 | 32,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-22 | latest | en | 0.941005 |
https://www.physicsforums.com/threads/calculating-symmetric-components-in-a-three-phase-4-wire-system.774274/ | 1,723,608,112,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00174.warc.gz | 733,562,382 | 25,553 | # Calculating Symmetric Components in a Three Phase 4 Wire System
• CasterSkux
In summary, this student is trying to find out the symmetrical components of the current in a three-phase system with a phase-to-neutral voltage of 230 V when one phase has a blown fuse. The student first calculates the balanced current and then uses Kirchoff's law to find the neutral current.
CasterSkux
Hello fellow engineers!
I am a student doing a simple course in Electrical Engineering.
I've got an enquiry regarding this question
"
1. In a three phase 4 wire system with phase to neutral voltage of 230V, a balanced set of resistive loads of 8 ohms are connected between each phase and neutral.
If the fuse cartridge of phase “C” blows, what will be the currents in the corresponding phases and the symmetrical components of the load currents?
""
We use a program to calculate the symmetric components. I believe I need to construct the resultant phasor of this system and derive the symmetric components
This is my approach:
1) Assume this is a star connected load, therefore the phase voltage is Vp = Vl / sqrt(3) = 230/sqrt(3) = 132.7906 V
2) Calculate the balanced current that flows through each resistor I = V/R, I = 132.7906V/8 = 16.5988.
3) If Phase C is eliminated from the fuse, I think that the line/phase voltages stay the same (as there is a neutral) and hence the current is the same.
4) Therefore Phase A is 16.5988 amps and so is Phase B. Phase B is offsetted by 120 degrees taking A as the relative point.
5) Phase C is 0 and it's on the 240 degree mark on the phasor diagram.I was wondering if this is all right in approach?
CasterSkux said:
This is my approach:
1) Assume this is a star connected load, therefore the phase voltage is Vp = Vl / sqrt(3) = 230/sqrt(3) = 132.7906 V
Didn't you say phase to neutral voltage was 230 V?
CasterSkux
Yes, oh have I overlooked something?
You should show by using symmetrical components what is new voltage phasor between phase B and star neutral after phase C is gone. Looks like a typical homework problem .
CasterSkux
zoki85 said:
Didn't you say phase to neutral voltage was 230 V?
Ah I ,maybe somehow I've interpreted "phase" as the "line", so the current is simply 230/8 = 28.75 amps.
Indeed It's a typical homework problem, the problem is my course is an online one: they didn't go through the basics as through as I would've liked and we hardly do math problems, so I'm not as well practiced unfortunately.
Zoki85, do you mean the new Phase A?
I was under the impression I work backwards to find the symmetrical components, not the other way around?
CasterSkux said:
Ah I ,maybe somehow I've interpreted "phase" as the "line", so the current is simply 230/8 = 28.75 amps.
Indeed It's a typical homework problem, the problem is my course is an online one: they didn't go through the basics as through as I would've liked and we hardly do math problems, so I'm not as well practiced unfortunately.
230 V is a common phase voltage (phase to neutral) in LV networks in Europe, so I guess it isn't mistake. This is simple situation and can be quickly solved even without method of symmetrical components. Phase to phase voltage is 230×√3 ≈ 400 V. Since phase C is open, current flowing through phases A and B is the same current that flows through series resistance 8+8=16 Ω. Current has magnitude 400/16=25 A. Note that in a phase diagram currents in phase A and phase B have same magnitude but are in phase oposition (angle difference between phasors is 1800 ).
Cheers
Last edited:
CasterSkux
Hello Zoki85,
Thanks for the reply. I was wondering if we should've considered current flowing to the neutral? would the phase C being broken, cause the load to be unbalanced and because it's (I assume a star connection): would cause current to flow into the neutral?
Cheers,
Indeed, I forgot you were referring to the case with connection between neutral points of two stars. In that case you were right! Current through the phase A: IA= VA/R=230/8=28.75 A, and current through the phase B: IB=VB/R= 28.75 ∠1200 A.
By Kirchoff's law, current through the neutral is I0=IA + IB. That would give I0= 28.75 ∠60° A (if I didn't make mistake without pencil&paper). For excersise, find from that, for both cases, direct, inverse, and zero sequence components of currents.
CasterSkux
I started my calculation when nobody was on the web. Now some of the correspondents already answered [partially in my opinion].
230 V it is the voltage phase-to-neutral [your declaration].So phase-to-phase will be sqrt(3)*230=400 V [as usual].
The current in phase A [let's take it as origin of the angles] will be IA= 230/8=28.75<0;
IB=28.75<-120=28.75<240 and IC=0.Neutral current will be IN = - (IA+IB+IC) = -28.7*{[cos(0)+cos(240)]+j*sin(240)}
cos(0)=1; cos(240)=cos(pi()/180*240)=-0.5; sin(240)=-0.866
IN=-28.75*(0.5-0.866j)=-14.38+24.9j
CasterSkux
Hello Zoki and Babadag!, Thanks for the help! This was vaguely what I thought I was supposed to do.
I got Zoki's answer of 28.75 ∠60° by converting everything to the complex number and adding, but that's only because I used the convention that all the phase angles are referenced from Phase A and rotate counterclockwise. Babadag you're also right and I know that convention of everything anticlockwise of A is negative and clockwise is positive.
You're both legends!
The symmetric components are (if I take 28.75 arg(60)) for this question are:
Zero =19.2 arg(60)
Positive = 19.2 arg(-60)
Negative = 9.58 arg(0)
The symmetric components are (if I take 28.75 arg(240)) for this question are:
Zero = 0
Postive = 0
Negative = 28.8
Interesting differences.
Cheers,
Just been told that there shouldn't be any zero sequences through the neutral, don't know why?
CasterSkux said:
The symmetric components are (if I take 28.75 arg(60)) for this question are:
Zero =19.2 arg(60)
Positive = 19.2 arg(-60)
Negative = 9.58 arg(0)
The symmetric components are (if I take 28.75 arg(240)) for this question are:
Zero = 0
Postive = 0
Negative = 28.8
Interesting differences.
Cheers,
?
Symmetrical components here you get from known phase currents IA,IB,IC .
Zero sequence phasor definitelly isn't 0 due to fact sum of the currents in phases A and B isn't 0.
I see that in my calcs I took oposite way of system rotation, and not the convential, counterclockwise one (like Baba did) .
Better stick to the convention to avoid confusion with arguments
CasterSkux
Sorry I made a mistake in the inputting to calculate the symmetrical values.
Start again: the resultant phasor values are (using Badabung's convention):
Ia = 28.75 arg(0)
Ib = 28.75 arg(240)
Ic from Badabung =-14.38+24.9j = 28.75 arg(-60) = 28.75 arg(300) = this is the resultant neutral phasorThe calculator for Symmetric components gave me
I0 = 19.2 arg(-60)
I1 = 9.60 arg(0)
I2 =19.2 arg(60)
CasterSkux said:
Start again: the resultant phasor values are (using Badabung's convention):
Ia = 28.75 arg(0)
Ib = 28.75 arg(240)
Ic from Badabung =-14.38+24.9j = 28.75 arg(-60) = 28.75 arg(300) = this is the resultant neutral phasor
Oh dear, Ic = 0. Phase C is dead broken (remember?) and no current can be flowing through it. You should really learn basics of this stuff...
CasterSkux
Sorry I've attached Ic as the resultant vector of A + B, it's a terminology error. I'd like to learn more but we don't get enough practice excersizes unfortunately:(
I really do appreciate your help though Zoki!
Hello again, I was wondering if anyone can confirm my answer?
IA= 28.75 A
IB= 28.75 ∠240° A
IC= 0 A
Use formulae:)
CasterSkux
Ah the take home lesson I guess is to "always allocate A B and C to their appropriate phases and you MUST do so" I made Ic = In subconsciously because I thought it would be needed in a Phasor diagram, but these Phasor Diagrams are strictly for Phase A, B and C - something that I wasn't so brought on about.
Thank you heaps Zoki85, I've learn't more from you about these diagrams than reading them bit by bit in a year!
Hello all:)
Have to revive this thread just to clarify a new question:
Now we must answer the same question, but with no Neutral, this was answered without intention in the beginning of the post
230 V is a common phase voltage (phase to neutral) in LV networks in Europe, so I guess it isn't mistake. This is simple situation and can be quickly solved even without method of symmetrical components. Phase to phase voltage is 230×√3 ≈ 400 V. Since phase C is open, current flowing through phases A and B is the same current that flows through series resistance 8+8=16 Ω. Current has magnitude 400/16=25 A. Note that in a phase diagram currents in phase A and phase B have same magnitude but are in phase oposition (angle difference between phasors is 1800 ).
I can do the maths involved, but I don't know how the phasors are 180 degrees apart and I don't know how to mathematically derive it. I'm under the impression that Phase A and B are still 120 degrees separated.
I was wondering how A and B can be 180?
With no neutral, and phase C open, the line currents add at the "centre junction" to equal zero. ( Σ current in = Σ current out )
So ##\mathbf{I_A = -I_B}## (equal and opposite means 180° apart)
The voltage across one load is ##\mathbf{V_A - V_B}## and across the other load it's ##\mathbf{V_B - V_A}##
Construct the phasor diagrams and you'll see these phase relationships.
## 1. What is the purpose of calculating symmetric components in a three phase 4 wire system?
The purpose of calculating symmetric components is to analyze and understand the behavior of a three phase 4 wire system. It helps to simplify complex systems into more manageable components, making it easier to identify and troubleshoot issues.
## 2. How do you calculate the positive, negative, and zero sequence components in a three phase 4 wire system?
To calculate the positive, negative, and zero sequence components, you will need to use mathematical formulas and equations. These formulas involve taking the phase voltages and currents and applying them to the appropriate equations. Alternatively, you can use software or online calculators specifically designed for this purpose.
## 3. What is the significance of symmetric components in power system analysis?
Symmetric components are important in power system analysis because they help to understand the behavior of a system under different fault conditions. By analyzing the positive, negative, and zero sequence components, engineers can determine the impact of different types of faults on the system and plan for appropriate protection measures.
## 4. How does unbalanced load affect symmetric components in a three phase 4 wire system?
Unbalanced loads can create asymmetrical conditions in a three phase 4 wire system, which can affect the symmetric components. This can result in increased voltage and current imbalances, leading to potential equipment damage and power quality issues. Calculating symmetric components can help identify and mitigate these effects.
## 5. Are there any limitations to calculating symmetric components in a three phase 4 wire system?
Yes, there are limitations to calculating symmetric components in a three phase 4 wire system. The methodology assumes a balanced system and does not account for non-linear loads or harmonics. Additionally, it may not accurately represent the system behavior under certain fault conditions, such as open or short circuits. Therefore, it is important to validate the results with actual measurements and consider other factors in power system analysis.
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1K | 3,088 | 12,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-33 | latest | en | 0.934749 |
https://www.physicsforums.com/threads/net-force-problem.535624/ | 1,526,813,719,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00605.warc.gz | 818,951,852 | 15,518 | Homework Help: Net force problem
1. Oct 1, 2011
1. The problem statement, all variables and given/known data
An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 850 kg traveling initially at a speed of 51.0 km/h in a distance equal to the diameter of a dime, which is 1.8 cm? Answer must be in two sig. figs.
2. Relevant equations
V² = u² + 2 a*s
F(net) = m*a
3. The attempt at a solution
51 km/h = 14.16 m/s
0= (14.16)^2 + 2a(0.018)
a= 200.5 / -0.036
a= -5,569.4 m/s^2
850 kg * -5569.4 = Fnet
Fnet = 4,734,027.78 N
(this is 2 sig figs right)?
Did I do this right? If I did, why exactly do I receive -0.036?(I just multiply 0.018 but I don't really understand why.)
Thanks.
2. Oct 1, 2011
Your answer has 2 decimal places. Two sig figs would be $$4.7 * 10^6 N$$. That's probably what they're looking for. You dont' get the negative value from multiplying, that came from solving your equation for "a."
$$V_o^2=-2a_x\Delta x$$
3. Oct 1, 2011
Oh ok thanks. I forgot about that notation.
Ok, even without the negative value, the final answer should still be the same right?(positive)
Last edited: Oct 1, 2011
4. Oct 1, 2011
No, it should be negative because if you're taking the direction of the initial velocity to be in the +x direction (which you are of course) the acceleration is in the opposite (negative) direction because it is bringing the car to a stop. For the force, you wrote:
850 kg * -5569.4 = Fnet
This is correct, but you have to keep the negative when you multiply. You're multiplying a negative and a postive, the result is $$-4.7*10^6 N$$
5. Oct 1, 2011 | 518 | 1,689 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.895998 |
http://mathhelpforum.com/pre-calculus/120204-find-functions-fog.html | 1,527,288,875,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00545.warc.gz | 188,784,693 | 9,686 | 1. ## Find Functions (FoG)
Find functions f and g so that (fog)(x)=2/(6x+1)
(6x+1) is all square rooted. I tried to enter alt251 for the square root symbol, but it didn't work.
I am not sure what the question is asking and how to start the problem...
2. Originally Posted by RenSully
Find functions f and g so that (fog)(x)=2/(6x+1)
(6x+1) is all square rooted. I tried to enter alt251 for the square root symbol, but it didn't work.
I am not sure what the question is asking and how to start the problem...
You can learn to use LaTex
The code $$f \circ g(x) = \frac{2}{{\sqrt {6x + 1} }}$$ gives nice output: $\displaystyle f \circ g(x) = \frac{2}{{\sqrt {6x + 1} }}$.
Let $\displaystyle f (x) = \frac{2}{{\sqrt {x} }}$ and $\displaystyle g(x) = {6x + 1}$.
3. what are the steps to get there?
I understand (fog)(x) is f(g(x)). I can't find a similar example.
4. Originally Posted by RenSully
what are the steps to get there?
I understand (fog)(x) is f(g(x)). I can't find a similar example.
Actually there are no steps.
It is a matter of understanding how functions work.
It is a matter of practice. | 346 | 1,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-22 | latest | en | 0.938304 |
http://blogs.harvard.edu/hoanga/2004/01/04/80211-wireless-channel-plan/ | 1,675,029,050,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00696.warc.gz | 5,015,034 | 11,333 | 802.11 Wireless Channel Plan
I got curious about what the channel allocation plan for 802.11 (Wi-Fi) is. After alway choosing a random channel that is not the same as any other nearby WiFi point I beacme curious to see what the channel plan is. After some googling around I dug up a channel plan on the net. If you know what channel plans are you can just skip the next paragraph and read on. Otherwise…
WHAT’S A CHANNEL PLAN?
If channel plan doesn’t make much sense. What happens is when you’re given a range of frequences you can broadcast on. You end up segmenting the range of frequences into a set. So for example 802.11 for the U.S. covers 2.412Ghz – 2.462Ghz frequency range. Now in theory you could just things wherever you want but it makes sense to put them in intervals that gives the best multiple for the range you have. Once you have those set of intervals, most people seem to find it a pain in the butt to refer to them as the interval 2.412 – 2.417. Hence the creation of a channel map. For the interval I just mentioned, in 802.11 parlor that is Channel 1. (Actually the above could be slightly wrong if 2.412 is the center frequency for Channel 1 but the general idea is the same). So now we can just say channel 1 or 4 or 8 and people can reference the channel map and figure out what’s what. This works for TV as well. Of course the tricky thing is most folk don’t know where to find this channel map most of the times as this data might be floating around SOMEWHERE but it’s usually not under an easy name to find unless you’re a RF junkie.
I ALREADY KNOW WHAT A CHANNEL PLAN IS, SHOW ME THE MONEY
Now that I’ve given a brief overview of what a Wireless Channel Plan here’s the channel plan for 802.11 and the countries which can use which set of channels. Thanks to this link for the channel map.
Channel Frequency 1 2.412GHz 2 2.417GHz 3 2.422GHz 4 2.427GHz 5 2.432GHz 6 2.437GHz 7 2.442GHz 8 2.447GHz 9 2.452GHz 10 2.457GHz 11 2.462GHz 12 2.467GHz 13 2.472GHz 14 2.484GHz
Country Channels Europe (ETSI) 01 – 13 USA (FCC) 01 – 11 France 10 – 13 Japan 01 – 14
Be Sociable, Share! | 578 | 2,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-06 | latest | en | 0.894303 |
http://mathhelpforum.com/number-theory/186396-factorial-problem.html | 1,527,076,169,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00083.warc.gz | 175,397,947 | 10,615 | 1. ## Factorial Problem
Hi All,
I want to know the value of below:
Sum of [(n-2)! / (n-k)!], where n=59 and k begins with 2 and ends with 59.
2. ## Re: Factorial Problem
You mean $\displaystyle \displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)}$ ?
3. ## Re: Factorial Problem
Originally Posted by pickslides
You mean $\displaystyle \displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)}$ ?
yes absolutely !!
(59-k) !
5. ## Re: Factorial Problem
Originally Posted by pickslides
You mean $\displaystyle \displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)!}$ ?
I want to know the value for above
6. ## Re: Factorial Problem
Originally Posted by pickslides
You mean $\displaystyle \displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)}$ ?
Should the be $\displaystyle !$ in the denominator?
$\displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)\color{red}!}$
yes .
8. ## Re: Factorial Problem
Here is the value.
$\displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)!}=110163588853530184980727482893521304479399855 114493057608241578732074256106650$
9. ## Re: Factorial Problem
Originally Posted by Plato
Here is the value.
$\displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)!}=110163588853530184980727482893521304479399855 114493057608241578732074256106650$
Thank U man
10. ## Re: Factorial Problem
Hi How did u calculated this ??? can u tell me the procedure ?? | 455 | 1,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-22 | latest | en | 0.583075 |
https://www.physicsforums.com/threads/lorentz-transformation.400699/ | 1,521,604,586,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647567.36/warc/CC-MAIN-20180321023951-20180321043951-00071.warc.gz | 847,752,174 | 14,591 | # Lorentz transformation
1. May 3, 2010
### kathykoo9
Using the Lorentz transformation, at what speed relative to speed of light c must you travel so that your length along the direction of motion is seen to decrease by a factor of 2? For this speed, hwo much would your mass be increased?
Please show steps, I'm confused with this question!
2. May 4, 2010
### ansgar
first of all my dear friend, this is the wrong forum...
second, show us what you have tried so far - otherwise it is pointless and there is always a suspicion that this is homework.
3. May 4, 2010
### Andrew Mason
Can you write out the Lorentz transformation for distance co-ordinates in two frames whose relative speed is v?
You are moving at speed v relative to an earth observer, say.
Let one end of your length be at x=0 at time t=0 and let the other end be at x = L at time t=0 in your frame. Use the Lorentz transformation to translate those co-ordinates to the earth observer's frame.
It is not considered correct to say that the mass increases. It certainly does not increase in your frame. Relativistic (apparent) mass may increase for an earth observer and this is probably what you are being asked to find. Do you know the relationship between rest mass and relativistic (apparent) mass?
AM | 311 | 1,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-13 | longest | en | 0.939012 |
https://www.cadtutor.net/forum/topic/31293-help-what-wrong-in-fix-defun/ | 1,618,992,431,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00446.warc.gz | 730,495,997 | 22,795 | # [Help] What wrong in Fix defun ?
## Recommended Posts
I've a problem with this function (in my case it isn't have answer)
```(defun test (a)
(setq b (atof a))
(setq c (fix B))
(setq d (rtos C 2 0 ))
(setq e (- b (fix B)))
(setq e1 (* e 100))
(setq e2 (fix e1))
(setq e3(rtos e2 2 0))
(setq g (- e1 (fix e1)))
(setq g1 (* g 100))
(setq g2 (fix g1))
(setq g3(rtos g2 2 0))
(princ (strcat d " d " e3 " m " g3 " s "))(princ)
)```
Everything seem to be Ok, but, when i test with some string with 00 in last, it return wrong.
Ex
Command: (test "45.2300") ;want it return 45 23 00
45 d 22 m 99 s ;but it return this
;So I check value in progress
Command: !b
45.23
Command: !c
45
Command: !e
0.23
Command: !e1
23.0
Command: !e2
22
Command: !e3
"22"
Command: !g
1.0
Command: !g1
100.0
Command: !g2
99
Command: !g3
"99"
Why e1 return 22 and g2 return 99 ?
Many thanks all ^^
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Because of the rounding of Doubles, perhaps look into the angtof/angtos functions.
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I just guess it cause of Rounding, but don't know what happend in progress calculate ^^ So i worry about making the same problem in other defun, not only in Degree convert...
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More fun in Fix ^^
Command: (- 3.2 (fix 3.2) 0.2)
1.66533e-016
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Yep, try to stear clear of using the fix function - there's some issues when mixing integers & floating point numbers. I'm not exactly sure why you want to do this - it's as if you simply type in the dms, but with the wrong notation. If you perform the next: (angtos (angtof "45.2300") 1) ... you get a value of 45d13'48" since your entry is in decimal 45 + 23/100 (instead of 45 + 23/60).
Anyway, try this one:
```(defun testang (str / d)
(if (setq d (vl-string-search "." str))
(setq str (strcat (substr str 1 d) "d" (substr str (+ d 2) 2) "'" (substr str (+ d 4)) "\""))
)
(angtos (angtof str) 1)
)```
You could actually remove the last line. I.e. you simply manipulate the string. In which case you could simply have done this to get exactly "45 23 00":
```(defun testang (str / d)
(if (setq d (vl-string-search "." str))
(setq str (strcat (substr str 1 d) " " (substr str (+ d 2) 2) " " (substr str (+ d 4))))
(strcat str " 00 00")
)
)```
Or one which works a bit better if they "forget" to enter the minutes / seconds:
```(defun testang (str / d emtystring->00)
(defun emtystring->00 (str)
(if (or (not str) (eq str ""))
"00"
str
)
)
(if (setq d (vl-string-search "." str))
(setq str (strcat (substr str 1 d)
" "
(emtystring->00 (substr str (+ d 2) 2))
" "
(emtystring->00 (substr str (+ d 4)))
)
)
(strcat str " 00 00")
)
)```
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Many Thanks irneb I can solve it by string functions, but I still problem with "what wrong with FIX" in #4 post. I don't know what exactly problem in here and i'm so vague
ANW, should i forgot it to do something else ^^
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but I still problem with "what wrong with FIX" in #4 post.
There's nothing 'wrong' with the fix function. Consider that numbers cannot be expressed to an inifinite number of decimal places as they are in mathematics (since this would obviously require an infinite amount of memory), hence some calculations can lose some degree of accuracy.
Note that your result of 1.66533e-016 is extremely small and is indeed on the limit of accuracy for the IEEE 754 double-precision floating-point format (double) wherein a number can be approximated to about 16 decimal places.
When equating values of doubles (in conditional statements, for example) always use the equal function with a small tolerance to compensate for this rounding effect.
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As a method to show what's actually happening, consider showing each variable using as much precision as possible:
```(rtos b 1 16) ;--> "4.522000000000000E+01"
(rtos e 1 16) ;--> "2.199999999999988E-01"```
Already you can see an error occurring. Though extremely small, this starts accumulating the more you use the value in more calculations. And since you keep multiplying and then fix the result to get the next 2 digits, this gets worse and worse.
In computer terms these errors "shouldn't" occur as long as all operations and values are in powers of 2. Since then it would simply be a shift in position. But seeing as many of your calculations work in powers of 10 a simple x10 is not a "simple" binary calculation. And since there's only a finite amount of bits to store this number, once it gets to the point where all the bits are needed the "so-called" least-significant-bits are dropped.
The fix function doesn't account for this, since a fix of 21.99999999999999988 would still only give 21. It doesn't round off to the nearest integer, it simply cuts off the floating point piece. You could use a round-off function of your own instead of fix (there's no standard one in ALisp though), or you could convert to and from strings all the time - though that has other problems as well, and simply passes the error along while possible causing worse.
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Thanks all Lee and irneb, it' cleared my mind . Have a good day ( i don't know how to say alot of "THank you " because of limiting in my English )
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sample-mt-sol
# sample-mt-sol - ECE-C490(Programming for Engineers...
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ECE-C490 (Programming for Engineers) Solutions to Sample Mid-Term Examination Winter 2005-06 Department of ECE, Drexel University (100 points, 1 hour and 20 minutes) Problem 1. (a) o o v = (b) What is k now: 5 or 7 (c) The answer is either 8 or 1 (d) That it is lunch time is false? true? (e) Thursday Monday Tuesday Wednesday Thursday Tuesday Thursday Tuesday Thursday Thursday (f) D E (g) The total is 14 1
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Problem 2. public class MaxTwo { public static void main ( String[] args ) { int a, b; Scanner input = new Scanner( System.in ); System.out.print( "Enter first integer: " ); a = input.nextInt(); System.out.print( "Enter second integer: " ); b = input.nextInt(); MaxTwo m = new MaxTwo(); if ( m.mystery( a, b ) ) System.out.println( "The maximum of the two is " + a ); else System.out.println( "The maximum of the two is " + b ); } public boolean mystery ( int x, int y ) { if ( x >= y ) return true; else return false; } } 2
Problem 3. import java.util.*;
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Ask a homework question - tutors are online | 445 | 1,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-09 | latest | en | 0.704233 |
https://www.bankrate.com/calculators/mortgages/mortgage-calculator.aspx?ic_id=home_smart%20spending_global-nav_mortgage_mortgage-calculator | 1,585,486,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00189.warc.gz | 833,181,102 | 40,907 | # Mortgage Calculator
The listings that appear on this page are from companies from which this website receives compensation, which may impact how, where and in what order products appear. This table does not include all companies or all available products. Bankrate does not endorse or recommend any companies.
Use our mortgage calculator to estimate your monthly mortgage payment. You can input a different home price, down payment, loan term and interest rate to see how your monthly payment changes.
Our monthly payment estimates are broken down by principal and interest, property taxes and homeowners insurance. We take our calculator a step further by factoring in your credit score range, zip code and HOA fees to give you a more precise payment estimate. You’ll also go into the home-buying process with a more accurate picture of how to calculate mortgage payments and purchase with confidence. After you run some estimates, read on for more education and home-buying tips.
## How to calculate mortgage payments
Want to figure out how much your monthly mortgage payment will be? For the mathematically inclined, here’s a formula to help you calculate mortgage payments manually:
### Equation for mortgage payments
M = P[r(1+r)^n/((1+r)^n)-1)]
• M = the total monthly mortgage payment.
• P = the principal loan amount.
• r = your monthly interest rate. Lenders provide you an annual rate so you’ll need to divide that figure by 12 (the number of months in a year) to get the monthly rate. If your interest rate is 5%, your monthly rate would be 0.004167 (0.05/12=0.004167)
• n = number of payments over the loan’s lifetime. Multiply the number of years in your loan term by 12 (the number of months in a year) to get the number of payments for your loan. For example, a 30-year fixed mortgage would have 360 payments (30x12=360)
This formula can help you crunch the numbers to see how much house you can afford. Using Bankrate.com’s tool to calculate your mortgage payments can take the work out of it for you and help you decide whether you’re putting enough money down or if you need to adjust your loan term. It’s always a good idea to rate-shop with several lenders to ensure you’re getting the best deal available.
## How a mortgage calculator can help
Buying a home is often life’s largest financial transaction, and how you finance it shouldn’t be a snap decision. Setting a budget upfront -- long before you look at homes -- can help you avoid falling in love with a home you can’t afford. That’s where a simple mortgage calculator can help.
A mortgage payment includes four components called PITI: principal, interest, taxes and insurance. Many homebuyers know about these costs but what they’re not prepared for are the hidden costs of homeownership. These include homeowners association fees, private mortgage insurance, routine maintenance, larger utility bills and major repairs.
Bankrate.com’s mortgage loan calculator can help you factor in PITI and HOA fees. You also can adjust your loan and down payment amounts, interest rate and loan term to see how much your payments might change. It’s important to know that your specific interest rate will depend on your overall credit profile and debt-to-income, or DTI, ratio (the sum of all of your debts and new mortgage payment divided by your gross monthly income). The riskier the borrower, the higher the interest rate in many cases.
## Deciding how much house you can afford
If you’re not sure how much of your income should go toward housing, follow the tried-and-true 28/36 percent rule. Most financial advisers agree that people should spend no more than 28 percent of their gross income on housing (i.e. mortgage payment), and no more than 36 percent of their gross income on total debt, including mortgage payments, credit cards, student loans, medical bills and the like.
Here’s an example of what this looks like:
Joe makes \$60,000 a year. That’s a gross monthly income of \$5,000 a month.
\$5,000 x 0.28 = \$1,400 total monthly mortgage payment (PITI)
Joe’s total monthly mortgage payments -- including principal, interest, taxes and insurance -- shouldn’t exceed \$1,400 per month. That’s a maximum loan amount of roughly \$253,379.
You can qualify for a mortgage with a DTI ratio of up to 50 percent for some loans, but you might not have enough wiggle room in your budget for other living expenses, retirement and emergency savings, and discretionary spending. Lenders don’t take those budget items into account when they preapprove you for a loan; it’s up to you to factor those expenses into your housing affordability picture.
Depending on where you live, your annual income could be more than enough to cover a mortgage -- or it could fall short. Knowing what you can afford can help you take financially sound next steps. The last thing you want to do is jump into a 30-year home loan that’s too expensive for your budget, even if a lender willing to loan you the money.
## Next steps
A mortgage calculator is a springboard to helping you estimate your monthly mortgage payment and understand what it includes. Your next step after playing with the numbers: getting preapproved by a mortgage lender.
Applying for a mortgage will give you a more definitive idea of how much house you can afford after a lender has vetted your employment, income, credit and finances. You’ll also have a clearer idea of how much money you’ll need to bring to the closing table.
Learn more about specific loan type rates
Loan Type Purchase Rates Refinance Rates
The table above links out to loan-specific content to help you learn more about rates by loan type.
30-Year Loan 30-Year Mortgage Rates 30-Year Refinance Rates
20-Year Loan 20-Year Mortgage Rates 20-Year Refinance Rates
15-Year Loan 15-Year Mortgage Rates 15-Year Refinance Rates
10-Year Loan 10-Year Mortgage Rates 10-Year Refinance Rates
FHA Loan FHA Mortgage Rates FHA Refinance Rates
VA Loan VA Mortgage Rates VA Refinance Rates
ARM Loan ARM Mortgage Rates ARM Refinance Rates
Jumbo Loan Jumbo Mortgage Rates Jumbo Refinance Rates
## About our Mortgage Rate Tables
• The above mortgage loan information is provided to, or obtained by, Bankrate. Some lenders provide their mortgage loan terms to Bankrate for advertising purposes and Bankrate receives compensation from those advertisers (our "Advertisers"). Other lenders' terms are gathered by Bankrate through its own research of available mortgage loan terms and that information is displayed in our rate table for applicable criteria. In the above table, an Advertiser listing can be identified and distinguished from other listings because it includes a "Next" button that can be used to click-through to the Advertiser's own website or a phone number for the Advertiser.
Each Advertiser is responsible for the accuracy and availability of its own advertised terms. Bankrate cannot guaranty the accuracy or availability of any loan term shown above. However, Bankrate attempts to verify the accuracy and availability of the advertised terms through its quality assurance process and requires Advertisers to agree to our Terms and Conditions and to adhere to our Quality Control Program. Click here for rate criteria by loan product.
Advertisers may have different loan terms on their own website from those advertised through Bankrate.com. To receive the Bankrate.com rate, you must identify yourself to the Advertiser as a Bankrate.com customer. This will typically be done by phone so you should look for the Advertiser's phone number when you click-through to their website. In addition, credit unions may require membership.
If you are seeking a loan for more than \$424,100, lenders in certain locations may be able to provide terms that are different from those shown in the table above. You should confirm your terms with the lender for your requested loan amount.
The loan terms (APR and Payment examples) shown above do not include amounts for taxes or insurance premiums. Your monthly payment amount will be greater if taxes and insurance premiums are included.
If you have used Bankrate.com and have not received the advertised loan terms or otherwise been dissatisfied with your experience with any Advertiser, we want to hear from you. Please click here to provide your comments to Bankrate Quality Control.
#### Quality Assurance
Compare rates with confidence. Rates are accurate and available as of the date seen for Bankrate customers. Identify yourself as a Bankrate consumer to get the Bankrate.com rate.
## Mortgage Calculators: Alternative Use
Most people use a mortgage calculator to estimate the payment on a new mortgage, but it can be used for other purposes, too.
Here are some other uses:
1. Planning to pay off your mortgage early.
Use the "Extra payments" functionality of Bankrate's mortgage calculator to find out how you can shorten your term and net big savings by paying extra money toward your loan's principal each month, every year or even just one time.
To calculate the savings, click "Amortization / Payment Schedule" link and enter a hypothetical amount into one of the payment categories (monthly, yearly or one-time) and then click "Apply Extra Payments" to see how much interest you"ll end up paying and your new payoff date.
2. Decide if an ARM is worth the risk.
The lower initial interest rate of an adjustable-rate mortgage, or ARM, can be tempting. But while an ARM may be appropriate for some borrowers, others may find that the lower initial interest rate won't cut their monthly payments as much as they think.
To get an idea of how much you'll really save initially, try entering the ARM interest rate into the mortgage calculator, leaving the term as 30 years. Then, compare those payments to the payments you get when you enter the rate for a conventional 30-year fixed mortgage. Doing so may confirm your initial hopes about the benefits of an ARM -- or give you a reality check about whether the potential plusses of an ARM really outweigh the risks.
3. Find out when to get rid of private mortgage insurance.
You can use the mortgage calculator to determine when you"ll have 20 percent equity in your home. This percentage is the magic number for requesting that a lender wave private mortgage insurance requirement.
Simply enter in the original amount of your mortgage and the date you closed, and click "Show Amortization Schedule." Then, multiply your original mortgage amount by 0.8 and match the result to the closest number on the far-right column of the amortization table to find out when you'll reach 20 percent equity.
## Mortgage Calculator Help
Using an online mortgage calculator can help you quickly and accurately predict your monthly mortgage payment with just a few pieces of information. It can also show you the total amount of interest you"ll pay over the life of your mortgage. To use this calculator, you"ll need the following information:
Home price - The dollar amount you expect to pay for a home.
Down payment - The down payment is money you give to the home's seller. At least 20% down typically lets you avoid mortgage insurance.
Mortgage Amount - If you're getting a mortgage to buy a new home, you can find this number by subtracting your down payment from the home's price. If you're refinancing, this number will be the outstanding balance on your mortgage.
Mortgage Term (Years) - This is the length of the mortgage you're considering. For example, if you're buying new, you may choose a mortgage loan that lasts 30 years. On the other hand, a homeowner who is refinancing may opt of a loan that lasts 15 years.
Interest Rate - Estimate the interest rate on a new mortgage by checking Bankrate's mortgage rate tables for your area. Once you have a projected rate (your real-life rate may be different depending on your overall credit picture) you can plug it into the calculator.
Mortgage Start Date - Select the month, day and year when your mortgage payments will start. | 2,473 | 12,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-16 | longest | en | 0.946064 |
https://www.coursehero.com/file/6381027/2-Questions-6-4-nested-loops-1/ | 1,527,290,775,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00388.warc.gz | 732,939,914 | 124,760 | {[ promptMessage ]}
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# 2_Questions_6_4_nested_loops-1 - printf""...
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CIS 15AG Questions Chapter 6: Repetitions (nested loops) 25. Write a fragment of code that prints 3 rows of 5 asterisks each. ***** ***** ***** 26. Write a function that prints n rows of asterisks. The first row has 2 asterisks, the second 4, the third 6, and so on. Below is the pattern for n = 5 ** **** ****** ******** ********** 1 int i; for ( i = 1; i <= 5; i++ ) printf( "*" ); printf( "\n" ); for ( i = 1; i <= 5; i++ ) printf( "*" ); printf( "\n" ); for ( i = 1; i <= 5; i++ ) printf( "*" ); printf( "\n" ); int i; void printTriangle(int height) { int row; int col; return; }
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CIS 15AG Questions Chapter 6: Repetitions (nested loops) 27. Predict the output. int i; int j; for ( i = 1; i <= 3; i++ ) { for ( j = 1; j <= 3; j++ ) printf("%d %d\n", i, j); printf("\n"); } 28. How many asterisks and '\n's are displayed? (A). for ( k = 1; k <= 5; k++ ) printf("*"); printf("\n"); (B). for ( j = 1; j <= 10; j++ ) { for ( k = 1; k <= 5; k++ ) printf("*"); printf("\n"); } printf("\n"); (C). for ( i = 1; i <= 3; i++ ) { for ( j = 1; j <= 10; j++ ) {
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Unformatted text preview: printf("*"); printf("\n"); } printf("\n"); } 29. Predict the output. s = 0; x = 5; y = 2; while (x > y) { for ( i = 1; i < x; i++ ) s += i; x -= y; } printf("%d\n", s); 2 i i < 3 j j < 3 Output x x > y i i < x s Output 5 CIS 15AG Questions Chapter 6: Repetitions (nested loops) 30. Write a program fragment that uses nested loops to produce the output 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 int row; int col; 31. Write a program fragment that uses nested loops to produce the output 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 int row; int col; 3 CIS 15AG Questions Chapter 6: Repetitions (nested loops) 32. Write a program fragment that uses nested loops to produce the output 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5 int row; int col; 33 * . Write a program fragment that uses nested loops to produce the output 1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9 6 7 8 9 7 8 9 0 1 8 9 0 1 2 9 0 1 2 3 int row; int col; 4...
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Jill Tulane University ‘16, Course Hero Intern | 1,140 | 3,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-22 | latest | en | 0.675152 |
https://breldigital.com/for-the-reaction-given-in-part-a-how-much-heat-is-absorbed-when-3-80-mol-of-a-reacts/ | 1,685,349,188,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00227.warc.gz | 174,942,219 | 19,030 | # For the reaction given in part a, how much heat is absorbed when 3.80 mol of a reacts?
By John
(A) Calculate the standard enthalpy change for the reaction:
2A+B <===> 2C+2D
Use the following data:
Substance:
A (?H^f [kJ/mol] = -231)
B (-409)
C (217)
D (-475)
(B) For the reaction given in Part A, how much heat is absorbed when 3.80 mol of A reacts?
Concepts and reason
The concept used to solve this problem is based on enthalpy change of a chemical reaction.
The enthalpy change is the difference of enthalpy of formation of products and reactants multiplied by their stoichiometric coefficient.
Fundamentals
The formula for enthalpy change is written as follows.
image
image.png
700×81 11.3 KB
Part A
The reaction is as follows.
image
image.png
756×192 30.7 KB
Part A
The enthalpy change of the reaction is 355 kJ.
The enthalpy change is calculated by subtracting the enthalpy of formation of reactants from enthalpy of formation of products.
Part B
In the reaction,
image
Two moles of A reacted with one mole of B to give the products. Thus, 3.80 mol of A reacted is calculated by dividing 3.80 mol and 2.
image
Part B
The heat absorbed in the reaction is 674.5 kJ.
The heat absorbed in the reaction is calculated by multiplying the change in enthalpy and number of moles of substance reacted. | 352 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-23 | latest | en | 0.91884 |
https://www.calculatorbit.com/en/length/33-femtometer-to-feet | 1,695,652,071,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508977.50/warc/CC-MAIN-20230925115505-20230925145505-00525.warc.gz | 765,133,739 | 7,949 | # 33 Femtometer to Feet Calculator
Result:
33 Femtometer = 1.0826771653543308e-13 Feet (ft)
Rounded: ( Nearest 4 digits)
33 Femtometer is 1.0826771653543308e-13 Feet (ft)
33 Femtometer is 3.3e-11mm
## How to Convert Femtometer to Feet (Explanation)
• 1 femtometer = 3.2808398950131234e-15 ft (Nearest 4 digits)
• 1 feet = 304800000000000.06 fm (Nearest 4 digits)
There are 3.2808398950131234e-15 Feet in 1 Femtometer. To convert Femtometer to Feet all you need to do is multiple the Femtometer with 3.2808398950131234e-15.
In formula distance is denoted with d
The distance d in Feet (ft) is equal to 3.2808398950131234e-15 times the distance in femtometer (fm):
### Equation
d (ft) = d (fm) × 3.2808398950131234e-15
Formula for 33 Femtometer (fm) to Feet (ft) conversion:
d (ft) = 33 fm × 3.2808398950131234e-15 => 1.0826771653543308e-13 ft
## How many Feet in a Femtometer
One Femtometer is equal to 3.2808398950131234e-15 Feet
1 fm = 1 fm × 3.2808398950131234e-15 => 3.2808398950131234e-15 ft
## How many Femtometer in a Feet
One Feet is equal to 304800000000000.06 Femtometer
1 ft = 1 ft / 3.2808398950131234e-15 => 304800000000000.06 fm
## femtometer:
The femtometre (symbol: fm) is unit of length in the International System of Units (SI), equal to .000000000001 meteres or 10^-15 meteres or 1/1000000000000000 meteres. The distance is sometimes called fermi and was so named in honour of Italian-American physicist Enrico Fermi, as it is a typical length-scale of nuclear physics.
## feet:
The foot (symbol: ft) also called feet(plural) is a unit of length in International System of Units (SI), equal to 0.3048 meters exactly. One foot is made of 12 inches. Historically the 'foot' was part of the local systesm of units in Greek, Chinese, Roaman, English and French systems although length of a 'foot' varied from one city to another.
## Femtometer to Feet Calculations Table
Now by following above explained formulas we can prepare a Femtometer to Feet Chart.
Femtometer (fm) Feet (ft)
29 9.514435695538058e-14
30 9.842519685039371e-14
31 1.0170603674540682e-13
32 1.0498687664041995e-13
33 1.0826771653543308e-13
34 1.1154855643044619e-13
35 1.1482939632545933e-13
36 1.1811023622047243e-13
37 1.2139107611548556e-13
38 1.2467191601049869e-13
Nearest 4 digits
## Convert from Femtometer to other units
Here are some quick links to convert 33 Femtometer to other length units.
## Convert to Femtometer from other units
Here are some quick links to convert other length units to Femtometer.
## FAQs About Femtometer and Feet
Converting from one Femtometer to Feet or Feet to Femtometer sometimes gets confusing.
### Is 3.2808398950131234e-15 Feet in 1 Femtometer?
Yes, 1 Femtometer have 3.2808398950131234e-15 (Nearest 4 digits) Feet.
### What is the symbol for Femtometer and Feet?
Symbol for Femtometer is fm and symbol for Feet is ft.
### How many Femtometer makes 1 Feet?
304800000000000.06 Femtometer is euqal to 1 Feet.
### How many Feet in 33 Femtometer?
Femtometer have 1.0826771653543308e-13 Feet.
### How many Feet in a Femtometer?
Femtometer have 3.2808398950131234e-15 (Nearest 4 digits) Feet. | 1,051 | 3,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-40 | longest | en | 0.793368 |
http://en.wikipedia.org/wiki/Software_Reliability_Testing | 1,409,647,544,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535921869.7/warc/CC-MAIN-20140901014521-00054-ip-10-180-136-8.ec2.internal.warc.gz | 189,024,455 | 15,634 | # Software reliability testing
(Redirected from Software Reliability Testing)
Software reliability testing is a field of software testing that relates to testing a software's ability to function, given environmental conditions, for a particular amount of time. Software reliability testing helps discover many problems in the software design and functionality.
## Overview
Software reliability is the probability that software will work properly in a specified environment and for a given amount of time. Using the following formula, the probability of failure is calculated by testing a sample of all available input states.
Probability = Number of failing cases / Total number of cases under consideration
The set of all possible input states is called the input space. To find reliability of software, we need to find output space from given input space and software.[1]
For reliability testing, data is gathered from various stages of development, such as the design and operating stages. The tests are limited due to restrictions such as cost and time restrictions. Statistical samples are obtained from the software products to test for the reliability of the software. Once sufficient data or information is gathered, statistical studies are done. Time constraints are handled by applying fixed dates or deadlines for the tests to be performed. After this phase, design of the software is stopped and the actual implementation phase starts. As there are restrictions on costs and time, the data is gathered carefully so that each data has some purpose and gets its expected precision.[2] To achieve the satisfactory results from reliability testing one must take care of some reliability characteristics. For example Mean Time to Failure (MTTF)[3] is measured in terms of three factors:
1. operating time,
2. number of on off cycles,
3. and calendar time.
If the restrictions are on operation time or if the focus is on first point for improvement, then one can apply compressed time accelerations to reduce the testing time. If the focus is on calendar time (i.e. if there are predefined deadlines), then intensified stress testing is used.[2]
## Measurement
Software reliability is measured in terms of mean time between failures(MTBF).[4]
MTBF consists of mean time to failure (MTTF) and mean time to repair(MTTR). MTTF is the difference of time between two consecutive failures and MTTR is the time required to fix the failure.[5] Reliability for good software is a number between 0 and 1. Reliability increases when errors or bugs from the program are removed.[6]
For example, if MTBF = 1000 hours for average software, then the software should work for 1000 hours for continuous operations.
## Objectives of reliability testing
The main objective of the reliability testing is to test software performance under given conditions without any type of corrective measure using known fixed procedures considering its specifications.
### Secondary objectives
The secondary objectives of reliability testing is:
1. To find perceptual structure of repeating failures.
2. To find the number of failures occurring in a specified amount of time.
3. To find the mean life of the software.
4. To discover the main cause of failure.
5. Checking the performance of different units of software after taking preventive actions.
### Points for defining objectives
Some restrictions on creating objectives include:
1. Behaviour of the software should be defined in given conditions.
2. The objective should be feasible.
3. Time constraints should be provided.[7]
## Importance of reliability testing
The application of computer software has crossed into many different fields, with software being an essential part of industrial, commercial and military systems. Because of its many applications in safety critical systems, software reliability is now an important research area. Although software engineering is becoming the fastest developing technology of the last century, there is no complete, scientific, quantitative measure to assess them. Software reliability testing is being used as a tool to help assess these software engineering technologies.[8]
To improve the performance of software product and software development process, a thorough assessment of reliability is required. Testing software reliability is important because it is of great use for software managers and practitioners.[9]
## Types of reliability testing
Software reliability testing includes feature testing, load testing, and regression testing.[10]
### Feature test
Feature testing checks the features provided by the software and is conducted in the following steps:
• Each operation in the software is executed once.
• Interaction between the two operations is reduced and
• Each operation is checked for its proper execution.
The feature test is followed by the load test.[10]
This test is conducted to check the performance of the software under maximum work load. Any software performs better up to some amount of workload, after which the response time of the software starts degrading. For example, a web site can be tested to see how many simultaneous users it can support without performance degradation. This testing mainly helps for Databases and Application servers. Load testing also requires software performance testing, which checks how well some software performs under workload.[10]
### Regression test
Regression testing is used to check if any new bugs have been introduced through previous bug fixes. Regression testing is conducted after every change or update in the software features. This testing is periodic, depending on the length and features of the software.[10]
## Test planning
Reliability testing is more costly compared to other types of testing. Thus while doing reliability testing, proper management and planning is required. This plan includes testing process to be implemented, data about its environment, test schedule, test points etc.
### Problems in designing test cases
Some common problems that occur when designing test cases include:
• Test cases can be designed simply by selecting only valid input values for each field in the software. When changes are made in a particular module, the previous values may not actually test the new features introduced after the older version of software.
• There may be some critical runs in the software which are not handled by any existing test case. Therefore, it is necessary to ensure that all possible types of test cases are considered through careful test case selection.[10]
## Reliability enhancement through testing
Studies during development and design of software help for improving the reliability of a product. Reliability testing is essentially performed to eliminate the failure mode of the software. Life testing of the product should always be done after the design part is finished or at least the complete design is finalized.[11] Failure analysis and design improvement is achieved through testings.
### Reliability growth testing
[11] This testing is used to check new prototypes of the software which are initially supposed to fail frequently. The causes of failure are detected and actions are taken to reduce defects. Suppose T is total accumulated time for prototype. n(T) is number of failure from start to time T. The graph drawn for n(T)/T is a straight line. This graph is called Duane Plot. One can get how much reliability can be gained after all other cycles of test and fix it.
\begin{alignat}{5} ln\left[ \frac {n\left( T\right)} {T}\right] = -\alpha ln\left( T\right) + b ; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..... Eq:1 \end{alignat}
solving eq.1 for n(T),
\begin{alignat}{5} n \left( T\right) = KT^{1-\alpha} ; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...... Eq:2 \end{alignat}
where K is e^b. If the value of alpha in the equation is zero the reliability can not be improved as expected for given number of failure. For alpha greater than zero, cumulative time T increases. This explains that number of the failures doesn't depends on test lengths.
### Designing test cases for current release
If we are adding new features to the current version of software,then writing a test case for that operation is done differently.
• First plan how many new test cases are to be written for current version.
• If the new feature is part of any existing feature, then share the test cases of new and existing features among them.
• Finally combine all test cases from current version and previous one and record all the results.[10]
There is a predefined rule to calculate count of new test cases for the software. if N is the probability of occurrence of new operations for new release of the software, R is the probability of occurrence of used operations in the current release and T is the number of all previously used test cases then
\begin{alignat}{5} New Test cases_{(current release)} = \left( \frac {N} {R}\right) * T \end{alignat}
## Reliability evaluation based on operational testing
The method of operational testing is used to test the reliability of software. Here one checks how the software works in its relevant operational environment. The main problem with this type of evaluation is constructing such an operational environment. Such type of simulation is observed in some industries like nuclear industries, in aircraft etc. Predicting future reliability is a part of reliability evaluation.
There are two techniques used for this:
Steady state reliability estimation
In this case, we use feedback from delivered software products. Depending on those results, we can predict the future reliability for the next version of product. This is similar to sample testing for physical products.
Reliability growth based prediction
This method uses documentation of the testing procedure. For example, consider a developed software and that we are creating different new versions of that software. We consider data on the testing of each version and based on the observed trend, we predict the reliability of the new version of software.[12]
### Reliability growth assessment and prediction
In the assessment and prediction of software reliability, we use the reliability growth model. During operation of the software, any data about its failure is stored in statistical form and is given as input to the reliability growth model. Using this data, the reliability growth model can evaluate the reliability of software. Lots of data about reliability growth model is available with probability models claiming to represent failure process. But there is no model which is best suited for all conditions. Therefore we must choose a model based on the appropriate conditions.
### Reliability estimation based on failure-free working
In this case, the reliability of the software is estimated with assumptions like the following:
• If a bug is found, then it is sure that it is going to be fixed by someone.
• Fixing the bug will not have any effect on the reliability of the software.
• Each fix in the software is accurate.[12]
## References
1. ^ Software Reliability. Hoang Pham.
2. ^ a b E.E.Lewis. Introduction to Reliability Engineering.
3. ^
4. ^ Roger Pressman. Software Engineering A Practitioner's Approach. McGrawHill.
5. ^
6. ^ Aditya P. Mathur. Foundations of Software Testing. Pearson publications.
7. ^ Reliability and life testing handbook. Dimitri kececioglu.
8. ^ A Statistical Basis for Software Reliability Assessment. M. xie.
9. ^ Software Reliability modelling. M. Xie.
10. John D. Musa. Software reliability engineering: more reliable software, faster and cheaper. McGraw-Hill. ISBN 0-07-060319-7.
11. ^ a b E.E.Liwis. Introduction to Reliability Engineering. ISBN 0-471-01833-3.
12. ^ a b "Problem of Assessing reliability". CiteSeerX: 10.1.1.104.9831. | 2,359 | 11,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-35 | latest | en | 0.936027 |
https://www.justintools.com/unit-conversion/force.php?k1=pound-yard-per-square-second&k2=joules-per-meter | 1,718,360,007,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00736.warc.gz | 755,278,333 | 26,716 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# FORCE Units Conversionpound-yard-per-square-second to joules-per-meter
1 Pound Yard Per Square Second
= 0.414765 Joules Per Meter
Category: force
Conversion: Pound Yard Per Square Second to Joules Per Meter
The base unit for force is newtons (Derived SI Unit)
[Pound Yard Per Square Second] symbol/abbrevation: (lb.yd/s2)
[Joules Per Meter] symbol/abbrevation: (J/m)
How to convert Pound Yard Per Square Second to Joules Per Meter (lb.yd/s2 to J/m)?
1 lb.yd/s2 = 0.414765 J/m.
1 x 0.414765 J/m = 0.414765 Joules Per Meter.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [force] => (newtons), 1 Pound Yard Per Square Second (lb.yd/s2) is equal to 0.414765 newtons, while 1 Joules Per Meter (J/m) = 1 newtons.
1 Pound Yard Per Square Second to common force units
1 lb.yd/s2 = 0.414765 newtons (N)
1 lb.yd/s2 = 0.093242873219907 pounds force (lbf)
1 lb.yd/s2 = 5034336.5995284 atomic units of force (auf)
1 lb.yd/s2 = 4.14765E+17 attonewtons (aN)
1 lb.yd/s2 = 4229.4259507579 centigrams force (cgf)
1 lb.yd/s2 = 41.4765 centinewtons (cN)
1 lb.yd/s2 = 0.0414765 decanewtons (daN)
1 lb.yd/s2 = 4.14765 decinewtons (dN)
1 lb.yd/s2 = 41476.5 dynes (dyn)
1 lb.yd/s2 = 4.14765E-19 exanewtons (EN)
Pound Yard Per Square Secondto Joules Per Meter (table conversion)
1 lb.yd/s2 = 0.414765 J/m
2 lb.yd/s2 = 0.82953 J/m
3 lb.yd/s2 = 1.244295 J/m
4 lb.yd/s2 = 1.65906 J/m
5 lb.yd/s2 = 2.073825 J/m
6 lb.yd/s2 = 2.48859 J/m
7 lb.yd/s2 = 2.903355 J/m
8 lb.yd/s2 = 3.31812 J/m
9 lb.yd/s2 = 3.732885 J/m
10 lb.yd/s2 = 4.14765 J/m
20 lb.yd/s2 = 8.2953 J/m
30 lb.yd/s2 = 12.44295 J/m
40 lb.yd/s2 = 16.5906 J/m
50 lb.yd/s2 = 20.73825 J/m
60 lb.yd/s2 = 24.8859 J/m
70 lb.yd/s2 = 29.03355 J/m
80 lb.yd/s2 = 33.1812 J/m
90 lb.yd/s2 = 37.32885 J/m
100 lb.yd/s2 = 41.4765 J/m
200 lb.yd/s2 = 82.953 J/m
300 lb.yd/s2 = 124.4295 J/m
400 lb.yd/s2 = 165.906 J/m
500 lb.yd/s2 = 207.3825 J/m
600 lb.yd/s2 = 248.859 J/m
700 lb.yd/s2 = 290.3355 J/m
800 lb.yd/s2 = 331.812 J/m
900 lb.yd/s2 = 373.2885 J/m
1000 lb.yd/s2 = 414.765 J/m
2000 lb.yd/s2 = 829.53 J/m
4000 lb.yd/s2 = 1659.06 J/m
5000 lb.yd/s2 = 2073.825 J/m
7500 lb.yd/s2 = 3110.7375 J/m
10000 lb.yd/s2 = 4147.65 J/m
25000 lb.yd/s2 = 10369.125 J/m
50000 lb.yd/s2 = 20738.25 J/m
100000 lb.yd/s2 = 41476.5 J/m
1000000 lb.yd/s2 = 414765 J/m
1000000000 lb.yd/s2 = 414765000 J/m
(Pound Yard Per Square Second) to (Joules Per Meter) conversions | 1,130 | 2,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.643386 |
https://learn.careers360.com/school/question-please-solve-rd-sharma-class-12-chapter-inverse-trigonometric-function-exercise-3-10-question-1-subquestion-ii-maths-textbook-solution/?question_number=1.2 | 1,719,109,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00001.warc.gz | 318,707,874 | 39,639 | #### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 1 Subquestion (ii) Maths Textbook Solution.
Answer:$-1$
Hint: Check if there is any relation between$\sec ^{-1}x$ and $\cos^{-1}x$ . Use inverse trigonometric functions properties to solve.$( \therefore \tan^{-1}\frac{1}{x} = - \pi + \cot^{-1}x )$
Given: $\sin\left (\tan^{-1}x+ \tan^{-1}\frac{1}{x} \right ) f\! or\: x < 0$
Solution: Let replace $\tan^{-1}\frac{1}{x}$ by $(- \pi + \cot^{-1}x) , x < 0$
Now,$\sin\left (\tan^{-1}x+ \tan^{-1}\frac{1}{x} \right )$
$\Rightarrow \sin\left (\tan^{-1}x- \pi + \cot^{-1}x \right )$
Again by property, $\tan^{-1}x+ \cot^{-1}x = \frac{\pi }{2}$
$\! \! \! \! \! \! \! \! \! \! = \sin\frac{\pi }{2}- \pi \\ = \sin-\frac{\pi }{2}\\ = - 1$
Concept: Properties and relations between inverse trigonometric functions.
Note: Inverse trigonometric functions remember relation between all trigonometric functions. Also, try to remember value of trigonometric functions. | 354 | 1,005 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.439255 |
https://groups.yahoo.com/neo/groups/primenumbers/conversations/topics/22221 | 1,503,263,437,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106990.33/warc/CC-MAIN-20170820204359-20170820224359-00251.warc.gz | 786,697,475 | 21,492 | ## Primes between consecutive squares
Expand Messages
• By prime number theorem, the approximate number of primes less than N**2 is N**2/(2 ln(N)). I considered using this relationship to estimate whether or not
Message 1 of 3 , Dec 7, 2010
By prime number theorem, the approximate number of primes less
than N**2 is N**2/(2 ln(N)).
I considered using this relationship to estimate whether or not
there always exist a prime between a square and the next larger
square.
I derive the conjecture that there exist infinitely many
integers, N, such that there are no primes between N**2 and
(N+1)**2.
I "reasoned" as follows:
Suppose we had a graph of the function,
f(x) = number of primes < x**2.
For large x, this function is approximated by
f(x) = x**2/ln(x**2) = x**2/(2 ln(x))
I calculate the first and second derivatives of this
approximation.
y = x**2/(2 ln(x))
2 ln(x) y = x**2
2 ln(x) y' + 2y/x = 2x
ln(x) y' + y/x = x
x ln(x) y' + y = x**2
(2 ln(x)) x ln(x) y' + (2 ln(x)) y = (2 ln(x)) x**2
2 ln(x) y = x**2
(2 ln(x)) x ln(x) y' + x**2 = (2 ln(x)) x**2
2x [ln(x)]**2 y' = (2 ln(x) - 1) x**2
2 [ln(x)]**2 y' = (2 ln(x) - 1) x
2 [ln(x)]**2 y'' + 4 ([ln(x)] /x) y' = (2/x)x + (2 ln(x) - 1)
2x [ln(x)]**2 y'' + 4 [ln(x)] y' = x(2 ln(x) + 1)
2x [ln(x)]**2 y' = (2 ln(x) - 1) x**2
4x [ln(x)]**2 y' = 2 (2 ln(x) - 1) x**2
2x [ln(x)]**2 y'' + 4 [ln(x)] y' = x(2 ln(x) + 1)
2x [ln(x)]**3 y'' + 4 [ln(x)]**2 y' = x(2 ln(x) + 1)[ln(x)]
4x [ln(x)]**2 y' = 2 (2 ln(x) - 1) x**2
2x [ln(x)]**3 y'' + 2 (2 ln(x) - 1) x**2 = x(2 ln(x) + 1)[ln(x)]
2x [ln(x)]**3 y'' =x(2 ln(x) + 1)[ln(x)] -2(2 ln(x) - 1) x**2
2[ln(x)]**3 y'' =(2 ln(x) + 1)[ln(x)] -2(2 ln(x) - 1) x
showing that the second derivative is negative for all x.
This suggests that the rate of increase of the number of primes
between consecutive squares will slow, and eventually, on the
average, decrease.
As a test, I calculated the actual number of primes between
consecutive squares, for the first thousand squares.
Here is the result for the last 100 of those calculated:
First Column: N
Second column: Number of primes < N**2
Third column: Number of primes between N**2 and (N+1)**2
Fourth Column: Number of primes between (N+1)**2 and (N+2)**2
minus the number of primes between N**2 and (N+1)
squared.
These results indicate that the second difference of the actual
prime count is often negative.
Perhaps someone can calculate, by finding when the first
derivative of x**2/(2 ln(x)), becomes less than 1, the order of
magnitude of the smallest value of N, such that there is no prime
between N**2 and (N+1)**2.
900 64683 137 -7
901 64820 130 -3
902 64950 127 11
903 65077 138 -18
904 65215 120 24
905 65335 144 -7
906 65479 137 2
907 65616 139 1
908 65755 140 -3
909 65895 137 -13
910 66032 124 16
911 66156 140 0
912 66296 140 -19
913 66436 121 14
914 66557 135 -3
915 66692 132 -6
916 66824 126 14
917 66950 140 0
918 67090 140 -17
919 67230 123 12
920 67353 135 3
921 67488 138 -3
922 67626 135 -10
923 67761 125 7
924 67886 132 13
925 68018 145 -14
926 68163 131 11
927 68294 142 0
928 68436 142 -13
929 68578 129 12
930 68707 141 2
931 68848 143 0
932 68991 143 -13
933 69134 130 -5
934 69264 125 23
935 69389 148 -16
936 69537 132 12
937 69669 144 0
938 69813 144 -12
939 69957 132 0
940 70089 132 8
941 70221 140 2
942 70361 142 -12
943 70503 130 11
944 70633 141 -2
945 70774 139 0
946 70913 139 3
947 71052 142 -17
948 71194 125 10
949 71319 135 -6
950 71454 129 15
951 71583 144 8
952 71727 152 4
953 71879 156 -21
954 72035 135 -14
955 72170 121 14
956 72291 135 7
957 72426 142 -1
958 72568 141 -2
959 72709 139 5
960 72848 144 10
961 72992 154 -16
962 73146 138 -1
963 73284 137 0
964 73421 137 1
965 73558 138 9
966 73696 147 -11
967 73843 136 -1
968 73979 135 4
969 74114 139 7
970 74253 146 2
971 74399 148 -19
972 74547 129 7
973 74676 136 5
974 74812 141 2
975 74953 143 -3
976 75096 140 -2
977 75236 138 -6
978 75374 132 9
979 75506 141 7
980 75647 148 0
981 75795 148 -9
982 75943 139 2
983 76082 141 2
984 76223 143 11
985 76366 154 -18
986 76520 136 14
987 76656 150 -18
988 76806 132 2
989 76938 134 8
990 77072 142 -2
991 77214 140 3
992 77354 143 -3
993 77497 140 3
994 77637 143 5
995 77780 148 -16
996 77928 132 16
997 78060 148 3
• Hi Kermit, ... Unless I ve made a mistake, the first and second derivative look like this: f (x) = x/log(x) - 1/(2*log^2(x)) f (x) = 1/log(x) -
Message 2 of 3 , Dec 7, 2010
Hi Kermit,
> f(x) = x**2/ln(x**2) = x**2/(2 ln(x))
>
> I calculate the first and second derivatives of this
> approximation.
>
> [ long calculation snipped ]
>
> showing that the second derivative is negative for all x.
Unless I've made a mistake, the first and second derivative
look like this:
f'(x) = x/log(x) - 1/(2*log^2(x))
f''(x) = 1/log(x) - (3/2)*(1/log^2(x)) + 1/log^3(x)
Unfortunately, they're both positive for x > sqrt(e) :-)
Peter
• ... Thank you. I trust your calculations more than my own. Kermit.
Message 3 of 3 , Dec 7, 2010
On 12/7/2010 11:34 PM, Peter Kosinar wrote:
>
> Hi Kermit,
>
>> f(x) = x**2/ln(x**2) = x**2/(2 ln(x))
>>
>> I calculate the first and second derivatives of this
>> approximation.
>>
>> [ long calculation snipped ]
>>
>> showing that the second derivative is negative for all x.
>
> Unless I've made a mistake, the first and second derivative look like this:
>
> f'(x) = x/log(x) - 1/(2*log^2(x))
> f''(x) = 1/log(x) - (3/2)*(1/log^2(x)) + 1/log^3(x)
>
> Unfortunately, they're both positive for x > sqrt(e) :-)
Thank you.
I trust your calculations more than my own.
Kermit.
Your message has been successfully submitted and would be delivered to recipients shortly. | 2,266 | 5,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-34 | latest | en | 0.776949 |
https://stackoverflow.com/questions/77623041/square-root-of-big-real-in-ada | 1,721,602,988,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00506.warc.gz | 466,660,415 | 41,523 | # Square root of Big_Real in Ada
I am trying to calculate the square root of big numbers (around 16 digits) using the `Big_Reals` package. I have the following square root function which uses the Newton-Raphson method
``````pragma Ada_2022;
function Big_Sqrt(X: Big_Real) return Big_Real is
package Converter is new Float_Conversions(Float);
use Converter;
Z: Big_Real := X;
Big_Half: Big_Real := To_Big_Real(0.5);
begin
for I in 1..32 loop
Z := Big_Half * (Z+X/Z);
Put_Line(Z'Image);
end loop;
return Z;
end Big_Sqrt;
``````
The output with input 1813789079679324 is
``````906894539839662.500
453447269919832.249
226723634959918.124
113361817479963.062
56680908739989.531
28340454370010.765
14170227185037.382
raised STORAGE_ERROR : Ada.Numerics.Big_Numbers.Big_Integers.Bignums.Normalize: big integer limit exceeded
``````
I'm assuming this happens because although the whole part of the number is getting smaller there is too much space being used for the decimal part but I can't find a way to reduce the precision of the decimal part.
As you have observed here, Ada support for big numbers offers arbitrary precision but not arbitrary size; some details on the limits are discussed here. Truncating intermediate, undesired results is a reasonable approach. In addition,
Iteration: Note that the approach converges. Instead of iterating for a fixed number of loops, consider exiting the loop when the difference falls below a specified threshold, `Epsilon` in the first example below. A related example is shown here.
Output Precision: Note that the `Big_Reals` function `To_String` provides output control for the number of digits after the decimal point. The first example below compares 64 digit results with known values. A related example is seen here.
Available Precision: Also examine the suitability the precision available on the target platform. The implementation-defined value of `Max_Digits` may be found in the `System` package; the second example below illustrates your result.
`Big_Sqrt`:
``````--https://stackoverflow.com/q/77623041/230513
procedure Big_Sqrt is
N : constant Natural := 64;
function Sqrt (X : Big_Real) return Big_Real is
Epsilon : constant Big_Real := 1.0 / 10.0**N;
One_Half : constant Big_Real := 0.5;
Z0 : Big_Real := X;
Z1 : Big_Real;
begin
loop
Z1 := One_Half * (Z0 + X / Z0);
exit when Z0 - Z1 < Epsilon;
Z0 := Z1;
end loop;
return Z1;
end Sqrt;
procedure Compare_Square_Root (S1, S2 : String) is
V1 : constant Big_Real := Sqrt (From_String (S1));
V2 : constant Big_Real := From_String (S2);
begin
Put_Line (To_String (V1, 0, N));
Put_Line (To_String (V2, 0, N));
end Compare_Square_Root;
begin
Compare_Square_Root
("2.0",
"1.4142135623730950488016887242096980785696718753769480731766797379");
Compare_Square_Root
("5.0",
"2.2360679774997896964091736687312762354406183596115257242708972454");
end Big_Sqrt;
``````
Console:
``````\$ ./obj/big_sqrt
1.4142135623730950488016887242096980785696718753769480731766797379
1.4142135623730950488016887242096980785696718753769480731766797379
2.2360679774997896964091736687312762354406183596115257242708972454
2.2360679774997896964091736687312762354406183596115257242708972454
``````
`Stock_Sqrt`:
``````with Ada.Text_IO; use Ada.Text_IO;
with System;
procedure Stock_Sqrt is
type Real is digits System.Max_Digits;
package Real_IO is new Float_IO (Real);
package Functions is new Ada.Numerics.Generic_Elementary_Functions (Real);
begin
Put_Line ("Max_Digits:" & System.Max_Digits'Image);
Real_IO.Put (Functions.Sqrt (1_813_789_079_679_324.0), 0, 4, 0);
New_Line;
end Stock_Sqrt;
``````
Console:
``````\$ ./obj/stock_sqrt
Max_Digits: 18
42588602.6970
``````
• That seems like a much cleaner way to do it, thank you Commented Dec 10, 2023 at 16:36
A fixed precision decimal part can be achieved by multiplying by 10 for each digit after the decimal point, converting it to a `Big_Integer`, converting it back to a `Big_Real` and then dividing by the original multiplier. The `Big_Sqrt` function becomes
``````function Big_Sqrt(X: Big_Real) return Big_Real is
package Converter is new Float_Conversions(Float);
use Converter;
Z: Big_Real := X;
Make_4dp: Big_Real := To_Big_Real(10000);
Big_Half: Big_Real := To_Big_Real(0.5);
begin
for I in 1..32 loop
Z := Big_Half * (Z+X/Z) * Make_4dp;
Z := To_Big_Real(Numerator(Z)/Denominator(Z))/Make_4dp;
end loop;
return Z;
end Big_Sqrt;
``````
This will truncate to 4 decimal points by multiplying by 10^4 = 10000, converting to a `Big_Integer` by dividing the numerator by the denominator (both of which are `Big_Integer`, making the result `Big_Integer`), converting it back to a `Big_Real` and then dividing by 10000. I could not find a better way to convert a `Big_Real` to a `Big_Integer` but this seems to work. | 1,374 | 4,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.761586 |
http://www.riddlesandanswers.com/puzzles-brain-teasers/what-seven-letter-woord-has-hundreds-of-letters-in-it-riddles/ | 1,575,692,261,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540495263.57/warc/CC-MAIN-20191207032404-20191207060404-00023.warc.gz | 226,020,003 | 28,387 | # WHAT SEVEN LETTER WOORD HAS HUNDREDS OF LETTERS IN IT RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS
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## Two Letters
Hint:
Eye ('e' and 'y' are the only letters).
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YES NO
Solved: 45%
## Remove My Letters Riddle
Hint:
Queue, Q
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Solved: 53%
## Consecutive Double Letters Riddle
What English word has three consecutive double letters?
Body parts remaining: 6
?
?
?
?
?
?
?
?
?
?
Hint: Someone who keeps record of financial transactions.
Bookkeeper
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## How Many Letters In The Alphabet Riddle
Hint:
There are 11 letters in "THE ALPHABET."
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YES NO
Solved: 59%
## Letters In A Series
Hint:
N and D. These are initials of the months in a year.
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Solved: 50%
## What Is A Word Made Up Of 4 Letters Riddle
Hint:
The word 'what' has 4 letters in it, 'yet' has three, 'sometimes' has 9, 'then' has 4, 'rarely' has 6, and 'never' has 5.
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## Sentence With All Letters
Hint:
The quick brown fox jumps over the lazy dog.
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## What Has Four Letters, Sometimes Nine, But Never Has Five Letters
Hint:
What has 4 letters.
Sometimes has 9 letters.
Never has 5 letters.
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Solved: 19%
## Thousands Of Letters Riddle
Hint:
Post Office
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Solved: 64%
## Common Letters Riddle
Hint: Hint: These combinations dont have it in common. AK IN WY
They are the only three US state abbreviations whose two letters are next to each other in the alphabet.
DE Delaware
HI Hawaii
MN Minnesota
Did you answer this riddle correctly?
YES NO | 692 | 2,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-51 | latest | en | 0.865157 |
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# Which of the following best completes the passage below? In
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Which of the following best completes the passage below? In [#permalink]
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02 Sep 2005, 08:14
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Which of the following best completes the passage below?
In opposing government regulation of business, conservatives often appeal to the Jeffersonian ideal of limited government, expressing the wish that government would “get off the backs of the American people.â€
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Director
Joined: 03 Nov 2004
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02 Sep 2005, 09:06
I will go with E
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02 Sep 2005, 09:17
E'
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Senior Manager
Joined: 13 Jan 2005
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02 Sep 2005, 09:22
E.
GA
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SVP
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02 Sep 2005, 11:20
E ends correctly with the right contrast.
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02 Sep 2005, 12:31
IMO E it is..
_________________
Fear Mediocrity, Respect Ignorance
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02 Sep 2005, 17:02
Shuld be E
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03 Sep 2005, 01:36
OA is E.
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03 Sep 2005, 01:36
Display posts from previous: Sort by | 826 | 2,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-43 | latest | en | 0.82703 |
http://openstudy.com/updates/563a4534e4b01656df95849c | 1,519,291,269,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814079.59/warc/CC-MAIN-20180222081525-20180222101525-00363.warc.gz | 260,658,842 | 11,582 | • anonymous
Please help??? Identify the vertex and the axis of the symmetry of the graph of the function y=3(x+2)^2-3. a. vertex: (2,-3); axis: x=2 b. vertex: (-2,-3); axis: x=-2 c. vertex: (2,3); axis: x=2 d. vertex: (-2,3); axis: x=-2 I sort of thinking that it is C but I need someone to check and explain. Please :D
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
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Not the answer you are looking for? Search for more explanations. | 353 | 1,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-09 | latest | en | 0.307548 |
https://www.kingexcel.info/2020/01/sorting-data-in-excel.html | 1,606,467,602,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141191511.46/warc/CC-MAIN-20201127073750-20201127103750-00167.warc.gz | 715,093,788 | 75,317 | Sorting data in Excel - KING OF EXCEL
KINGEXCEL.INFO ( KING OF EXCEL ) Welcome KINGEXCEL.INFO - Nothing Is Unable ... About Excel Tricks, Learning VBA Programming, Dedicated Software, Accounting, Living Skills ...
## Friday, January 24, 2020
Sorting data in Excel is easy. Unless, you are building a dashboard for your manager's manager. You can't ask that person to select C3, go to the Data tab and click the AZ button every time they want an updated report. The new SORT and SORTBY functions allow you to easily sort with a formula.
You can pass three arguments to the SORT function. The first is the range to be sorted. Leave the headings out of this argument. Next, which column do you want to sort by. If your data is in B:D and you want to sort by column D, you would specify column 3 as the sort column. The third argument is a 1 for ascending or -1 for descending.
In this figure, the data is sorted by Amount descending:
What if you want to do a two-level sort? You can specify an array constant for both the second and third argument. In this case, the data is sorted by Team ascending and Amount descending. For the sort column, specify {2;3}. For the sort order, specify {1,-1}.
The Excel Calc team also gave you the SORTBY function. Say you want to return a list of products but not the associated amounts. You want the products to be sorted by the amount. The formula below says to return the products from B3:B9 sorted descending by the amounts in D3:D9.
#evba #etipfree #kingexcel
📤You download App EVBA.info installed directly on the latest phone here : https://www.evba.info/p/app-evbainfo-setting-for-your-phone.html?m=1 | 391 | 1,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-50 | longest | en | 0.857835 |
http://proofindex.com/inverse-relation.html | 1,566,408,094,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316075.15/warc/CC-MAIN-20190821152344-20190821174344-00051.warc.gz | 157,097,961 | 1,386 | # Inverse relation
## Definition
THEOREM $$\left(\mathrel{R}^{-1}\right)^{-1} = \mathrel{R}$$
Proof available
THEOREM
Let $$\mathrel{R}$$ be a relation on a set $$S$$. If $$\mathrel{R}$$ is reflexive, then so is (\mathrel{R^{-1}}\).
Proof available
THEOREM
Let $$\mathrel{R}$$ be a relation on a set $$S$$. If $$\mathrel{R}$$ is antisymmetric, then so is $$\mathrel{R^{-1}}$$.
Proof available | 146 | 404 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-35 | latest | en | 0.482927 |
https://raw.githubusercontent.com/AlgebraicJulia/Catlab.jl/gh-pages/dev/generated/graphics/layouts_vs_drawings.ipynb | 1,653,394,565,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00639.warc.gz | 519,730,317 | 72,985 | { "cells": [ { "cell_type": "markdown", "source": [ "# Layouts versus drawings of wiring diagrams\n", "\n", "\n", "In Catlab, layout and drawing (rendering) of wiring diagrams are mostly\n", "decoupled. This notebook shows how to lay out diagrams using Graphviz's\n", "rank-based layout or Catlab's series-parallel layout and then render them\n", "using Compose.jl or TikZ." ], "metadata": {} }, { "cell_type": "markdown", "source": [ "The morphism we will visualize is:" ], "metadata": {} }, { "outputs": [ { "output_type": "execute_result", "data": { "text/plain": "f⊗(f⋅g)⊗(f⋅g⋅h): X⊗X⊗X → X⊗X⊗X", "text/latex": "$f \\otimes \\left(f \\cdot g\\right) \\otimes \\left(f \\cdot g \\cdot h\\right) : X \\otimes X \\otimes X \\to X \\otimes X \\otimes X$" }, "metadata": {}, "execution_count": 1 } ], "cell_type": "code", "source": [ "using Catlab.Theories\n", "\n", "X = Ob(FreeSymmetricMonoidalCategory, :X)\n", "f, g, h = (Hom(sym, X, X) for sym in (:f, :g, :h))\n", "\n", "expr = otimes(f, compose(f,g), compose(f,g,h))" ], "metadata": {}, "execution_count": 1 }, { "cell_type": "markdown", "source": [ "Let's convert this expression into a wiring diagram. This yields a purely\n", "combinatorial object, as evidenced by its underlying graph." ], "metadata": {} }, { "outputs": [ { "output_type": "execute_result", "data": { "text/plain": "Catlab.WiringDiagrams.DirectedWiringDiagrams.WiringDiagramGraphACSet{Int64} with elements V = 1:8, E = 1:9\n┌───┬─────┐\n│ V │ box │\n├───┼─────┤\n│ 1 │ 1 │\n│ 2 │ 2 │\n│ 3 │ 3 │\n│ 4 │ 4 │\n│ 5 │ 5 │\n│ 6 │ 6 │\n│ 7 │ -2 │\n│ 8 │ -1 │\n└───┴─────┘\n┌───┬─────┬─────┬──────┐\n│ E │ src │ tgt │ wire │\n├───┼─────┼─────┼──────┤\n│ 1 │ 2 │ 3 │ 1 │\n│ 2 │ 5 │ 6 │ 2 │\n│ 3 │ 4 │ 5 │ 3 │\n│ 4 │ 7 │ 1 │ 1 │\n│ 5 │ 7 │ 2 │ 2 │\n│ 6 │ 7 │ 4 │ 3 │\n│ 7 │ 1 │ 8 │ 1 │\n│ 8 │ 3 │ 8 │ 2 │\n│ 9 │ 6 │ 8 │ 3 │\n└───┴─────┴─────┴──────┘\n", "text/html": [ "
\n", "Catlab.WiringDiagrams.DirectedWiringDiagrams.WiringDiagramGraphACSet{Int64} with elements V = 1:8, E = 1:9\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
Vbox
11
22
33
44
55
66
7-2
8-1
\n", "\n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", " \n", "
Esrctgtwire
1231
2562
3453
4711
5722
6743
7181
8382
9683
\n", "
\n" ] }, "metadata": {}, "execution_count": 2 } ], "cell_type": "code", "source": [ "using Catlab.WiringDiagrams, Catlab.Graphics\n", "\n", "diagram = to_wiring_diagram(expr)\n", "WiringDiagrams.graph(diagram)" ], "metadata": {}, "execution_count": 2 }, { "cell_type": "markdown", "source": [ "## Graphviz layout" ], "metadata": {} }, { "cell_type": "markdown", "source": [ "Calling to_graphviz both lays out and draws the diagram, entirely within\n", "Graphviz." ], "metadata": {} }, { "outputs": [ { "output_type": "execute_result", "data": { "text/plain": "Catlab.Graphics.Graphviz.Graph(\"G\", true, \"dot\", Catlab.Graphics.Graphviz.Statement[Catlab.Graphics.Graphviz.Subgraph(\"\", Catlab.Graphics.Graphviz.Statement[Catlab.Graphics.Graphviz.Node(\"n0in1\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:id => \"in1\")), Catlab.Graphics.Graphviz.Node(\"n0in2\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:id => \"in2\")), Catlab.Graphics.Graphviz.Node(\"n0in3\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:id => \"in3\")), Catlab.Graphics.Graphviz.Edge(Catlab.Graphics.Graphviz.NodeID[Catlab.Graphics.Graphviz.NodeID(\"n0in1\", \"\", \"\"), Catlab.Graphics.Graphviz.NodeID(\"n0in2\", \"\", \"\"), Catlab.Graphics.Graphviz.NodeID(\"n0in3\", \"\", \"\")], OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}())], OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:rank => \"source\", :rankdir => \"TB\"), OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:style => \"invis\", :shape => \"none\", :label => \"\", :width => \"0\", :height => \"0.333\"), OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:style => \"invis\")), Catlab.Graphics.Graphviz.Subgraph(\"\", Catlab.Graphics.Graphviz.Statement[Catlab.Graphics.Graphviz.Node(\"n0out1\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:id => \"out1\")), Catlab.Graphics.Graphviz.Node(\"n0out2\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:id => \"out2\")), Catlab.Graphics.Graphviz.Node(\"n0out3\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:id => \"out3\")), Catlab.Graphics.Graphviz.Edge(Catlab.Graphics.Graphviz.NodeID[Catlab.Graphics.Graphviz.NodeID(\"n0out1\", \"\", \"\"), Catlab.Graphics.Graphviz.NodeID(\"n0out2\", \"\", \"\"), Catlab.Graphics.Graphviz.NodeID(\"n0out3\", \"\", \"\")], OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}())], OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:rank => \"sink\", :rankdir => \"TB\"), OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:style => \"invis\", :shape => \"none\", :label => \"\", :width => \"0\", :height => \"0.333\"), OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:style => \"invis\")), Catlab.Graphics.Graphviz.Node(\"n1\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:color => \"black\", :comment => \"f\", :fillcolor => \"white\", :id => \"n1\", :label => Catlab.Graphics.Graphviz.Html(\"\\n\\n\\n\\n\\n\\n
f
\"), :style => \"solid\")), Catlab.Graphics.Graphviz.Node(\"n2\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:color => \"black\", :comment => \"f\", :fillcolor => \"white\", :id => \"n2\", :label => Catlab.Graphics.Graphviz.Html(\"\\n\\n\\n\\n\\n\\n
f
\"), :style => \"solid\")), Catlab.Graphics.Graphviz.Node(\"n3\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:color => \"black\", :comment => \"g\", :fillcolor => \"white\", :id => \"n3\", :label => Catlab.Graphics.Graphviz.Html(\"\\n\\n\\n\\n\\n\\n
g
\"), :style => \"solid\")), Catlab.Graphics.Graphviz.Node(\"n4\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:color => \"black\", :comment => \"f\", :fillcolor => \"white\", :id => \"n4\", :label => Catlab.Graphics.Graphviz.Html(\"\\n\\n\\n\\n\\n\\n
f
\"), :style => \"solid\")), Catlab.Graphics.Graphviz.Node(\"n5\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:color => \"black\", :comment => \"g\", :fillcolor => \"white\", :id => \"n5\", :label => Catlab.Graphics.Graphviz.Html(\"\\n\\n\\n\\n\\n\\n
g
\"), :style => \"solid\")), Catlab.Graphics.Graphviz.Node(\"n6\", OrderedCollections.OrderedDict{Symbol, Union{String, Catlab.Graphics.Graphviz.Html}}(:color => \"black\", :comment => \"h\", :fillcolor => \"white\", :id => \"n6\", :label => Catlab.Graphics.Graphviz.Html(\"\\n\\n\\n\\n\\n\\n
h
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We\n", "can then render this layout using Compose.jl. 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nothing, 0.0, Symbol(\"\"))]), List([]), List([]), 0, false, false, false, false, nothing, nothing, 0.0, Symbol(\"\"))]), List([]), List([]), 0, false, false, false, false, nothing, nothing, 0.0, :box)]), List([]), List([]), 0, false, false, false, false, nothing, nothing, 0.0, :boxes)]), List([]), List([]), 0, false, false, false, false, nothing, nothing, 0.0, :diagram), 71.2611111111111mm, 38.80555555555555mm)", "text/html": [ "\n", "\n" ], "image/svg+xml": [ "\n", "\n" ] }, "metadata": {}, "execution_count": 4 } ], "cell_type": "code", "source": [ "import Compose\n", "\n", "layout = graphviz_layout(diagram, orientation=LeftToRight)\n", "layout_to_composejl(layout, base_unit=Compose.pt)" ], "metadata": {}, "execution_count": 4 }, { "cell_type": "markdown", "source": [ "The same layout can be rendered in TikZ:" ], "metadata": {} }, { "outputs": [ { "output_type": "execute_result", "data": { "text/plain": "Catlab.Graphics.TikZ.Document(Catlab.Graphics.TikZ.Picture(Catlab.Graphics.TikZ.Statement[Catlab.Graphics.TikZ.Node(\"root\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"outer box\", nothing), Catlab.Graphics.TikZ.Property(\"minimum width\", \"202\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"minimum height\", \"110\\\\tikzunit\")], Catlab.Graphics.TikZ.Coordinate(\"0\", \"0\"), \"\"), Catlab.Graphics.TikZ.Node(\"n1\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"box\", nothing), Catlab.Graphics.TikZ.Property(\"minimum width\", \"16\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"minimum height\", \"25\\\\tikzunit\")], Catlab.Graphics.TikZ.Coordinate(\"54\", \"42.5\"), \"\\$f\\$\"), Catlab.Graphics.TikZ.Node(\"n2\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"box\", nothing), Catlab.Graphics.TikZ.Property(\"minimum width\", \"16\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"minimum height\", \"25\\\\tikzunit\")], Catlab.Graphics.TikZ.Coordinate(\"-56\", \"0.5\"), \"\\$f\\$\"), Catlab.Graphics.TikZ.Node(\"n3\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"box\", nothing), Catlab.Graphics.TikZ.Property(\"minimum width\", \"20\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"minimum height\", \"25\\\\tikzunit\")], Catlab.Graphics.TikZ.Coordinate(\"-2\", \"0.5\"), \"\\$g\\$\"), Catlab.Graphics.TikZ.Node(\"n4\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"box\", nothing), Catlab.Graphics.TikZ.Property(\"minimum width\", \"16\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"minimum height\", \"25\\\\tikzunit\")], Catlab.Graphics.TikZ.Coordinate(\"-56\", \"-42.5\"), \"\\$f\\$\"), Catlab.Graphics.TikZ.Node(\"n5\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"box\", nothing), Catlab.Graphics.TikZ.Property(\"minimum width\", \"20\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"minimum height\", \"25\\\\tikzunit\")], Catlab.Graphics.TikZ.Coordinate(\"-2\", \"-42.5\"), \"\\$g\\$\"), Catlab.Graphics.TikZ.Node(\"n6\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"box\", nothing), Catlab.Graphics.TikZ.Property(\"minimum width\", \"20\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"minimum height\", \"25\\\\tikzunit\")], Catlab.Graphics.TikZ.Coordinate(\"54\", \"-42.5\"), \"\\$h\\$\"), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"\\$(root.west)+(-0.0,41.5)\\$\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"-180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"n1.west\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"\\$(root.west)+(-0.0,-0.5)\\$\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"-180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"n2.west\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"\\$(root.west)+(-0.0,-42.5)\\$\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"-180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"n4.west\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"n2.east\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"-180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"n3.west\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"n4.east\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"-180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"n5.west\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"n5.east\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"-180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"n6.west\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"n1.east\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"\\$(root.east)+(0.0,41.5)\\$\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"n3.east\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"\\$(root.east)+(0.0,-0.5)\\$\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)]), Catlab.Graphics.TikZ.Edge(Catlab.Graphics.TikZ.PathExpression[Catlab.Graphics.TikZ.NodeCoordinate(\"n6.east\"), Catlab.Graphics.TikZ.PathOperation(\"to\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"out\", \"0\"), Catlab.Graphics.TikZ.Property(\"in\", \"180\")]), Catlab.Graphics.TikZ.NodeCoordinate(\"\\$(root.east)+(0.0,-42.5)\\$\")], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"wire\", nothing)])], Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"unit length/.code\", \"{{\\\\newdimen\\\\tikzunit}\\\\setlength{\\\\tikzunit}{#1}}\"), Catlab.Graphics.TikZ.Property(\"unit length\", \"1pt\"), Catlab.Graphics.TikZ.Property(\"x\", \"\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"y\", \"\\\\tikzunit\"), Catlab.Graphics.TikZ.Property(\"semithick\", nothing), Catlab.Graphics.TikZ.Property(\"box/.style\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"rectangle\", nothing), Catlab.Graphics.TikZ.Property(\"draw\", nothing), Catlab.Graphics.TikZ.Property(\"solid\", nothing), Catlab.Graphics.TikZ.Property(\"rounded corners\", nothing)]), Catlab.Graphics.TikZ.Property(\"outer box/.style\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"draw\", \"none\")]), Catlab.Graphics.TikZ.Property(\"wire/.style\", Catlab.Graphics.TikZ.Property[Catlab.Graphics.TikZ.Property(\"draw\", nothing)])]), [\"calc\", \"shapes.geometric\"], [\"amssymb\"])", "image/svg+xml": [ "\n", "\n", "\n" ] }, "metadata": {}, "execution_count": 5 } ], "cell_type": "code", "source": [ "import TikzPictures\n", "\n", "layout_to_tikz(layout, base_unit=\"1pt\")" ], "metadata": {}, "execution_count": 5 }, { "cell_type": "markdown", "source": [ "## Series-parallel layout" ], "metadata": {} }, { "cell_type": "markdown", "source": [ "Catlab has its own layout system based on series-parallel decomposition. In\n", "this case, the layout exactly recovers the structure of the morphism\n", "expression created at the beginning." ], "metadata": {} }, { "outputs": [ { "output_type": "execute_result", "data": { "text/plain": "ComposePicture(Compose.Context(Measures.BoundingBox{Tuple{Measures.Length{:w, Float64}, Measures.Length{:h, Float64}}, Tuple{Measures.Length{:w, Float64}, Measures.Length{:h, Float64}}}((0.0w, 0.0h), (1.0w, 1.0h)), Compose.UnitBox{Float64, Float64, Float64, Float64}(-7.0, -5.0, 14.0, 10.0, 0.0mm, 0.0mm, 0.0mm, 0.0mm), nothing, nothing, nothing, List([Compose.Context(Measures.BoundingBox{Tuple{Measures.Length{:w, Float64}, Measures.Length{:h, Float64}}, Tuple{Measures.Length{:w, Float64}, Measures.Length{:h, Float64}}}((0.0w, 0.0h), (1.0w, 1.0h)), nothing, nothing, nothing, nothing, List([Compose.Context(Measures.BoundingBox{Tuple{Measures.Length{:w, Float64}, Measures.Length{:h, Float64}}, Tuple{Measures.Length{:w, Float64}, Measures.Length{:h, Float64}}}((0.0w, 0.0h), 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"nbformat": 4 } | 56,451 | 186,543 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | latest | en | 0.383025 |
https://statsidea.com/set-axis-limits-of-ggplot2-facet-plot-in-r-4-examples-using-facet_wrap-scales/ | 1,670,569,188,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711390.55/warc/CC-MAIN-20221209043931-20221209073931-00714.warc.gz | 567,759,955 | 16,306 | # Set Axis Limits of ggplot2 Facet Plot in R (4 Examples) | Using facet_wrap & scales
This page shows how to manipulate the axis limits of a ggplot2 facet plot in R programming.
Sound good? Here’s how to do it!
## Creation of Exemplifying Data
First, we need to create some example data that we can draw as facet plots later on:
``` set.seed(374658) # Create example data x <- rnorm(1000) y <- rnorm(1000) + x^3 group <- rbinom(1000, 1, 0.1) x[group == 1] <- x[group == 1] * 2 y[group == 1] <- y[group == 1] * 2 data <- data.frame(x, y, group) head(data) # Print first lines of data # x y group # 1 0.27298625 1.0002430 0 # 2 -2.15253919 -9.7464363 0 # 3 0.01161435 -1.8907440 1 # 4 -0.65622032 0.7712863 0 # 5 -0.44922753 0.6075897 0 # 6 0.09518892 0.3644900 0 ```
As you can see based on the output of the RStudio console, our example data consists of three columns. It contains two numeric columns x and y as well as a group indicator.
If we want to draw data with the ggplot2 package, we need to install and load the package to R:
``` install.packages("ggplot2") # Install & load ggplot2 library("ggplot2") ```
Now, we can create a facet plot with default specifications as shown in the R code below:
``` ggp <- ggplot(data, aes(x, y)) + # Default plot with facet_wrap geom_point() + facet_wrap(~ group) ggp # Draw plot in RStudio ```
Figure 1: Facet Plot with Default Scales.
Figure 1 shows the output of the previous R syntax: A facet plot consisting of two ggplot2 scatterplots.
In the following examples, I’ll explain how to manipulate the axis scales of the panels of our plot. Keep on reading!
## Example 1: Create Facet Plot with Free Scales
As you have seen in the previous plot, by default the x-axis and the y-axis of our panels are set to be the same. Example 1 illustrates how to disentangle the scales of both plots, so that each scale fits the values shown in each panel:
``` ggp + # Draw plot with free scales facet_wrap(~ group, scales = "free") ```
Figure 2: Facet Plot with Free Scales.
In Figure 2 you can see that our new facet graph shows panels with different scales on the x-axis as well as on the y-axis.
## Example 2: Create Facet Plot with Free X-Axis
In Example 2, you’ll learn how to keep the y-axis the same for both panels, while the x-axis is free:
``` ggp + # Draw plot with free x-axis facet_wrap(~ group, scales = "free_x") ```
Figure 3: Facet Plot with Free X-Axis.
## Example 3: Create Facet Plot with Free Y-Axis
In Example 3, we keep the x-axis at a fixed size and disentangle the y-axes (i.e. the opposite as in Example 2):
``` ggp + # Draw plot with free y-axis facet_wrap(~ group, scales = "free_y") ```
Figure 4: Facet Plot with Free Y-Axis.
## Example 4: Create Facet Plot with Individual Axes
In the previous examples, we used the predefined scales that are showing all values of each panel with the least possible empty space. However, you may also choose the axis limits manually by using the coord_cartesian function and the xlim (or ylim) argument.
The following R code creates a facet graphic with free y-axis and an individually specified x-axis:
``` ggp + # Free y-axis & manual x-axis facet_wrap(~ group, scales = "free_y") + coord_cartesian(xlim = c(- 10, 10)) ```
Figure 5: Facet Plot with Manual X-Axis.
## Video, Further Resources & Summary
If you need further info on the R programming syntax of this article, you could watch the following video of my YouTube channel. I’m explaining the R syntax of this article in the video.
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Furthermore, you may want to have a look at the related tutorials on this website:
To summarize: You learned in this article how to modify and change the axis limits of different panels in a ggplot2 facet plot in the R programming language. Please tell me about it in the comments, in case you have additional comments or questions. | 1,084 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-49 | latest | en | 0.760245 |
https://en.ppt-online.org/180450 | 1,701,806,929,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00694.warc.gz | 289,517,115 | 8,561 | 9.12M
Category: astronomy
Similar presentations:
# Inflation and string cosmology
Andrei Linde
## 2. Contents:
From the Big Bang theory to Inflationary
Cosmology
Eternal inflation and string theory landscape
## 3. Two major cosmological discoveries:
Inflation
Starobinsky, 1980 – modified gravity, R + R2
a complicated but almost working model
Guth, 1981 - old inflation (inflation in a false
vacuum)
A.L., 1982 - new inflation
1983 - chaotic inflation
1991
- hybrid inflation
## 4. Closed, open or flat universe
What was before the Big Bang?
Why is our universe so homogeneous (better than 1
part in 10000) ?
Why is it isotropic (the same in all directions)?
Why all of its parts started expanding
simultaneously?
Why is it flat? Why parallel lines do not intersect?
Why it contains so many particles?
## 5. Big Bang Theory
Energy of matter in the Big Bang theory
According to the Big Bang theory, the total mass of matter
soon after the Big Bang was greater than 1080 ton
Mass = Energy: E = mc2
Before the Big Bang there was NOTHING, and then
suddenly we got A HUGE AMOUNT OF ENERGY
Where did it come from?
To create our universe we would need more than 1080 tons
of high tech explosive compressed to a size of 1cm, and
exploded simultaneously, with accuracy 10-43 s.
Who could do it?…
## 6. Inflationary Universe
Inflationary theory
solves many problems of the old Big
Bang theory, and explains how the
whole universe could be created from
less than a milligram of matter
## 7. Inflation
as a theory of a harmonic oscillator
Eternal Inflation
## 8. Inflation was invented in an attempt to answer almost metaphysical questions:
Equations of motion:
Einstein equation:
Klein-Gordon equation:
Compare with equation for the harmonic oscillator with
friction:
## 9. Where did the energy come from?
Logic of Inflation:
Large φ
large H
large friction
field φ moves very slowly, so that its potential
energy for a long time remains nearly constant
This is the stage of inflation
## 10. Energy of matter in the Big Bang theory
Inflation makes the universe flat,
homogeneous and isotropic
In this simple model the
universe typically grows
101000000000000 times
during inflation.
Now we can see just a
tiny part of the universe
of size ct = 1010 light yrs.
That is why the universe
looks homogeneous,
isotropic, and flat.
## 11.
Quantum fluctuations produced during inflation
x
Small quantum fluctuations of all physical fields exist everywhere. They are
similar to waves, which appear and then rapidly oscillate, move and
disappear. Inflation stretched them, together with stretching the universe.
When the wavelength of the fluctuations becomes sufficiently large, they stop
moving and oscillating, and do not disappear. They look like frozen waves.
## 12. Inflation as a theory of a harmonic oscillator
x
When expansion of the universe continues, new quantum fluctuations
become stretched, stop oscillating, and freeze on top of the previously
frozen fluctuations.
## 13.
x
This process continues, and eventually the universe becomes
populated by inhomogeneous scalar field. Its energy takes different
values in different parts of the universe. These inhomogeneities are
responsible for the formation of galaxies.
Sometimes these fluctuations are so large that they can increase the
value of the scalar field in some parts of the universe. Then inflation in
these parts of the universe occurs again and again. In other words,
the process of inflation becomes eternal.
We will illustrate it now by computer simulation of this process.
## 14. Logic of Inflation:
WMAP5 + Acbar + Boomerang + CBI
Observations
## 16.
WMAP and the temperature of the sky
## 17.
This is a photographic image
of quantum fluctuations
blown up
to the size of
the universe
## 18.
On a much, much larger scale…
Inflationary
## 19.
Predictions of Inflation:
1) The universe should be homogeneous, isotropic
and flat, = 1 + O(10-4)
[
Observations: the universe is homogeneous, isotropic
and flat, = 1 + O(10-2)
2) Inflationary perturbations should be gaussian
and adiabatic, with flat spectrum, ns = 1+ O(10-1)
Observations: perturbations are gaussian and adiabatic,
with flat spectrum, ns = 1 + O(10-2)
## 20.
Big Bang
Earth
Astronomers use our universe as a “time machine”. By looking at the stars
close to us, we see them as they were several hundreds years ago.
## 21. WMAP and the temperature of the sky
Big Bang
Earth
The light from distant galaxies travel to us for billions of years, so we see
them in the form they had billions of years ago.
## 22.
Big Bang
Earth
Looking even further, we can detect photons emitted 400000 years after
the Big Bang. But 30 years ago everyone believed that there is nothing
beyond the cosmic fire created in the Big Bang at the time t = 0.
## 23. On a much, much larger scale…
Big Bang
Earth
Inflationary theory tells us that this cosmic fire was created not at the time
t = 0, but after inflation. If we look beyond the circle of fire surrounding us,
we will see enormously large empty space filled only by a scalar field.
## 24. Predictions of Inflation:
Big Bang
Inflation
If we look there very carefully, we will see small perturbations of space, which
are responsible for galaxy formation. And if we look even further, we will see
how new parts of inflationary universe are created by quantum fluctuations.
## 25.
From the Universe to the Multiverse
In realistic theories of elementary particles there are many scalar fields,
and their potential energy has many different minima. Each minimum
corresponds to different masses of particles and different laws of their
interactions.
V
SU(5)
SU(4)xU(1)
SU(3)xSU(2)xU(1)
Quantum fluctuations during eternal inflation can bring the scalar fields
to different minima in different exponentially large parts of the universe.
The universe becomes divided into many exponentially large parts with
different laws of physics operating in each of them. (In our computer
simulations we will show them by using different colors.)
## 26.
Genetic code of the Universe
There may be one fundamental law of physics, like a single
genetic code for the whole Universe. However, this law may
have different realizations. For example, water can be liquid,
solid or gas. In elementary particle physics, the effective laws
of physics depend on the values of the scalar fields.
Quantum fluctuations during inflation can take the scalar
fields from one minimum of their potential energy to another,
altering its genetic code. Once it happens in a small part of
the universe, inflation makes this part exponentially big.
This is the cosmological
mutation mechanism
## 27.
In string theory, genetic code is written in the
properties of compactification of extra dimensions
Up to 10500 different combinations
## 28.
String Theory Landscape
100
1000
Perhaps 10 - 10
different minima in string
theory
## 29.
Kandinsky Universe
## 30.
Landscape of eternal inflation
## 31.
Self-reproducing Inflationary Universe
We live here
Big Bang ?
## 32.
"It is said that there is no such thing as a free
lunch. But the universe is the ultimate free
lunch".
Alan Guth
Now we know that the universe
is not just a free lunch: It is an
eternal feast were ALL possible
types of dishes are served.
## 33.
All vacuum states in string theory are METASTABLE.
After a very long time, vacuum will decay. At that time,
our part of the universe will become ten-dimensional, or
it will collapse and disappear.
But because of eternal inflation,
the universe as a whole is
immortal | 1,811 | 7,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.912409 |
http://openstudy.com/updates/50654f2be4b08d185211e888 | 1,449,026,949,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398538144.64/warc/CC-MAIN-20151124205538-00205-ip-10-71-132-137.ec2.internal.warc.gz | 164,134,082 | 13,504 | ## Sup_open~study 3 years ago Use the formula for slope to verify that a horizontal line has a slope of zero and that a vertical has an undefined slope. I have no idea what to do can some one help explain it to me thanks. :)
1. hartnn
The slope of the line through points (x1,y1) and (x2,y2) is given by : $$\huge m=\frac{y_1-y_2}{x_1-x_2}$$ now horizontal line has equation of the form y=c, here xterm has '0' co-efficient take any two points on this, say (x1,c),(x2,c) so put y1 = c and y2 = c in m and tell me what u get.
2. Sup_open~study
i'm sorry i don't get what the y = c part ??
3. hartnn
i'll draw a horizontal line|dw:1348817053391:dw| can u see that y-co ordinate of every point on that line is same??
4. Sup_open~study
yes
5. hartnn
that means for any value of x-coordinate , y co-ordinate will be some constant 'c' only.....so we can write its equation as y=c got this?
6. Sup_open~study
i think so the y will stay the same ( constant) meaning y = c ??
7. hartnn
yup. similarly for vertical line , the equation is x=c
8. Sup_open~study
i don't get the x=c part ??
9. hartnn
|dw:1348817443207:dw| can u see that x-co ordinate of every point on that vertical line is same
10. Sup_open~study
one is smaller than the other ??
11. Sup_open~study
so they are not really the same ??
12. hartnn
for a vertical line, x co-ordinate always remains constant. hence its equation can be written as x=c did u understand this?
13. Sup_open~study
yes i think i kind of get what what you are saying
14. hartnn
now reply to my first comment
Find more explanations on OpenStudy | 466 | 1,600 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2015-48 | longest | en | 0.90743 |
davidbeede.com | 1,501,253,078,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550969387.94/warc/CC-MAIN-20170728143709-20170728163709-00514.warc.gz | 74,804,652 | 3,257 | Fret positions found graphically using a refined “18 rule”
With all the free fret scale calculator floating around the internet, it might seem silly to bother learning a manual method.
But what if you were shipwrecked on a deserted island and wanted to build an instrument?
Then you’d wish you took a minute to glimpse this page… or maybe that you had stayed awake in geometry…;-)
Don’t worry this actually requires almost no math, just the ability to use a ruler, compass and a square.
You'll need a piece of poster board, a straight edge, a tape measure, a 90 degree triangle, a pencile and a compass.
To make the math really easy start with a 27” string length.
That will be the long leg, or base of a pointy isosceles triangle we'll draw. It will look like a long wedge or ramp.
The short leg comes from the “18” part of our rule, found by dividing 18 into the string length of 27 producing 1.5”
That's all the math we'll need to do for this system. 27/18= 1.5
More mathamatical methods like the 12th root of 2 tells us a more accurate figure would be 17.817 instead of 18. If you use that figure, 17.817, instead of 18 you get this length for our vertical leg..
1.5154066341134871190436100353595 of an inch.
Now we see why folks might round it off to 18, but we can do a little better than that.
If you take a close look you’ll see it is .0154 thousandths larger than our 18 rule.
1/64 th of an inch = 0.015625< o:p > < /O:P >
So add a 1/64 of an inch to our vertical leg and you have it pretty darn close.
Connect the ends to close the triangle. Let's call that top line the "ramp line."
Now get out your compass with the sharpest point you have, and scribe an arc from the top of the short leg to intersect the long leg.
That’s our first fret position.
Draw a right angle line from that point up to intersect our "ramp line." Strike an arc from there to the long leg and you get fret 2… and so on.
That will give you as many frets as you want to draw.
For clarity, I’ll just show twelve frets.
You can check how well you did by measuring from your nut point to your 12th fret and see how close it is to 13.5”.
Now this would be fine if you’re making an instrument with a 27” string length, but what if we want a 25" string length pattern?
Here’s what you do. Draw lines from all your critical points, nut, frets and bridge, to an arbitrary point some distance away from your 12th fret. For clarity I’m using 13” [it stays within my screen capture] but you could make it farther away.
There now you have a handy dandy "Fret Fan."
Here's a TIP:
You could skip all the geometry above, and borrow a known accurate fret scale, say from a decent guitar, and make a "Fret Fan" from that.
With this method you can make an instrument “by the seat of your pants” without even knowing what the string length is, make a paper template from nut to bridge…
Lay that template or any desired string length on this “fan” and if you keep it parallel to the original line you can transfer the fret points to your template. If you want to do a longer string length, for a bass perhaps, just continue your fan lines out thru your points.
As you can see, the accuracy of this system depends on your drawing tools and your skills. Any errors that get made become cumulative. Add to this the problem of having had to make your compass and square and paper and pencil from sea shells and palm leaves on your island, and you may wish you’d brought your laptop and inkjet printer…;-) | 868 | 3,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2017-30 | longest | en | 0.911146 |
http://www.cfd-online.com/Forums/main/95216-drag-coefficient-over-flat-plate.html | 1,438,520,666,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042989043.35/warc/CC-MAIN-20150728002309-00339-ip-10-236-191-2.ec2.internal.warc.gz | 364,026,375 | 16,155 | # Drag Coefficient over a Flat Plate
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December 9, 2011, 11:07 Drag Coefficient over a Flat Plate #1 New Member Russell Join Date: Sep 2011 Posts: 13 Rep Power: 5 How can you find the drag coefficient over a 2D flat plate? Thanks
December 9, 2011, 13:16
#2
Senior Member
Join Date: Jan 2011
Posts: 236
Blog Entries: 5
Rep Power: 8
Quote:
Originally Posted by rfar0028 How can you find the drag coefficient over a 2D flat plate? Thanks
You can find it by using google or a standard fluid dynamics textbook
December 9, 2011, 14:08 #3 New Member AeroSuresh Join Date: Nov 2011 Posts: 11 Rep Power: 5 Drag coefficient cd examples In general, is not an absolute constant for a given body shape. It varies with the speed of airflow (or more generally with Reynolds number). A smooth sphere, for example, has a that varies from high values for laminar flow to 0.47 for turbulent flow. cd Shapes 0.001 laminar flat plate parallel to the flow (Re<10^5) 0.005 turbulent flat plate parallel to the flow (Re>10^5) 0.1 smooth sphere (Re=10^6) 0.195 General Motors EV1 1996 0.24 Mercedes-Benz E-Class Coupé 2010 0.25 3rd Generation Toyota Prius 0.295 bullet (not ogive, at subsonic velocity) 0.48 rough sphere (Re = 10^6) 0.7 a typical bicycle plus cyclist 1.0–1.1 skier 1.0–1.3 wires and cables 1.0–1.3 man (upright position) 1.1-1.3 ski jumper 1.28 flat plate perpendicular to flow (3D) 1.3–1.5 Empire State Building 1.8–2.0 Eiffel Tower 1.98–2.05 flat plate perpendicular to flow (2D) 2.1 a smooth brick Reference: http://en.wikipedia.org/wiki/Drag_coefficient __________________ ------------ AeroSuresh ------------
December 10, 2011, 04:17 #4 New Member Russell Join Date: Sep 2011 Posts: 13 Rep Power: 5 I'm sorry I didn't explain very well but I requested a method of finding the drag coefficient over a 2D flat plate using CFD (Ansys Fluent)
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Contact Us - CFD Online - Top | 780 | 2,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2015-32 | longest | en | 0.807933 |
https://rewildtv.com/what-are-the-factors-of-243/ | 1,653,282,962,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00250.warc.gz | 543,941,875 | 4,930 | Did you know that 243 is a herbal number i beg your pardon is the 5th power of 3? In this lesson, we will certainly calculate the factors of 243, prime factors of 243, and also factors of 243 in pairs in addition to solved instances for a much better understanding.
You are watching: What are the factors of 243
Factors that 243: 1, 3, 9, 27, 81 and also 243Prime administer of 243: 243 = 35
1 What are the components of 243? 2 How to Calculate components of 243? 3 Important Notes 4 Factors the 243 by element Factorization 5 Factors the 243 in Pairs 6 Challenging Questions
## What are the components of 243?
The determinants of 243 are every the number that give the value 243 when multiplied. Together 243 is odd, it will certainly not have actually 2 or any type of multiples the 2 together its factor. To know why it is composite, let"s remind the meaning of a composite number.
A number is said to be composite if the has more than 2 factors.
## How to calculation the components of 243?
Let"s learn exactly how to calculate the factors of 243. Let"s shot calculating the components of 243, starting with the smallest totality number, i.e., 1.
Did it result in the remainder 0?
Yes! So, we will certainly get,
243 ÷1= 243243 ×1= 243
Now shot to find out because that other entirety numbers. As already discussed in the earlier section, also numbers cannot divide 249.
Hence, us only require to check odd numbers.
Thus, the factors that 243 are 1, 3, 9, 27, 81 and 243.
Explore factors using illustrations and interactive examples.
Important Notes:
As 243 is one odd number, every its factors will be odd.243 is a non-perfect square number. Thus, it will have actually an even number of factors. This residential property holds true for every non-perfect square number.
## Factors of 243 by element Factorization
Prime factorization way to express a composite number together the product of its prime factors. To acquire the element factorization that 243, we division it by its the smallest prime variable which is 3.
243/3= 81
Now divide 81 by its the smallest prime factor. This procedure goes ~ above till we acquire the quotient together 1
The prime factorization the 243 is shown below:
Have friend noticed any type of pattern?
243 = 3 × 81 = 3 × 3 × 3 × 3 × 3 = 243 = 35
Hence, the determinants of 243 are 1, 3, 9, 27, 81 and also 243.
## Factors of 243 in Pairs
Let"s create the components of 243 in pairs. The pair determinants of 243 can be given as below:
FactorsPair factors
1 × 243 1, 243
3 × 81 3, 81
9 × 279, 27
The above-given components are hopeful pair factors.
See more: How Far Is Brunswick From Savannah From Brunswick, Savannah To Brunswick Distance (Sav To Bqk)
The product of two an unfavorable numbers also gives a confident number. Hence, it is feasible to have negative pair determinants as well. | 763 | 2,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-21 | latest | en | 0.884326 |
https://docs.scipy.org/doc/numpy-1.17.0/reference/generated/numpy.hypot.html | 1,679,433,074,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00115.warc.gz | 262,462,624 | 3,461 | numpy.hypot¶
`numpy.``hypot`(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'hypot'>
Given the “legs” of a right triangle, return its hypotenuse.
Equivalent to `sqrt(x1**2 + x2**2)`, element-wise. If x1 or x2 is scalar_like (i.e., unambiguously cast-able to a scalar type), it is broadcast for use with each element of the other argument. (See Examples)
Parameters: x1, x2 : array_like Leg of the triangle(s). If `x1.shape != x2.shape`, they must be broadcastable to a common shape (which becomes the shape of the output). out : ndarray, None, or tuple of ndarray and None, optional A location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated array is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs. where : array_like, optional This condition is broadcast over the input. At locations where the condition is True, the out array will be set to the ufunc result. Elsewhere, the out array will retain its original value. Note that if an uninitialized out array is created via the default `out=None`, locations within it where the condition is False will remain uninitialized. **kwargs For other keyword-only arguments, see the ufunc docs. z : ndarray The hypotenuse of the triangle(s). This is a scalar if both x1 and x2 are scalars.
Examples
```>>> np.hypot(3*np.ones((3, 3)), 4*np.ones((3, 3)))
array([[ 5., 5., 5.],
[ 5., 5., 5.],
[ 5., 5., 5.]])
```
Example showing broadcast of scalar_like argument:
```>>> np.hypot(3*np.ones((3, 3)), [4])
array([[ 5., 5., 5.],
[ 5., 5., 5.],
[ 5., 5., 5.]])
```
numpy.arctan
numpy.arctan2 | 513 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.742226 |
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# Differential Equations for Applications MATH 3410
Marketplace > University of Connecticut > Mathematics (M) > MATH 3410 > Differential Equations for Applications
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Math 3410 Fall 2009 Second semester differential equations 1 Some review 11 First order equations We will need to know how to do separable equations and linear equations A separable equation is one like dy 2 7 t dt 9 We can rewrite this as d y 1 7 7 tdt Integrating both sides7 ii it C y 2 so 1 y 7 1 39 it c A linear equation is one like 2 it 9 It i One multiplies by an integrating factor pgf 521ntt2 to get tzy Qty R 01 tsz t3 which leads to 2 714 tyilt 0 andthen ylt2 4 13239 When the linear equation has constant coef cients and is homogeneous ie7 the right hand side is 07 things are much easier To solve y 7 4y 0 we guess a solutions of the form y 6 so y requot Then re 7 4e 07 or r 4 and therefore the solution is y 06 To identify 0 one needs an initial condition7 eg7 y0 2 Then so we then have For non hornogeneous equations7 such as 17 4y 63 one way to solve it is to solve the homogeneous equation y 7 4y 07 and then use y 664 yp where yp is a particular solution One way to nd a particular solution is to make an educated guess If we guess yp A65 then we have y 7 4 3A63t 7 4A 7 and this will equal 63 if A 71 We conclude the solution to the non hornogeneous equation is y 664 7 63 12 Series From calculus we have the Taylor series 2 3 em1x m 2 4 cosz17 gi and 3 5 s1nx7 7 lf 239 V71 substituting and doing some algebra shows that e cos z39 sinx 2 Second order linear 21 Applications First consider a spring hung from the ceiling with a weight hanging from it Let u be the distance the weight is below equilibrium There is a restoring force upwards of amount ku by Hooke7s law There is darnping resistance against the motion which is iRu And the net force is related to accelera tion by Newton7s laws so ku 7 RM F mu This leads to mu Ru 7 ku 0 If there is an external force acting on the spring then the right hand side is replaced by The second example is that of a circuit with a resistor inductance coil and capacitor hooked up in series Let I be the current Q the charge R the resistance L the inductance and C the capacitance We know that I dQdt The voltage drop across the resistor is IR across the capacitor QC397 and across the inductance coil L So if Et is the potential put into the current7 l Et LQ RQ 5Q Sometimes this is differentiated to give 1 E t LI RI 51 22 Linear constant coef cients homogeneous Let7s look at an example y 7 5y 4y 0 From Math 2117 we know a way of solving this Let 1 y and this one equation becomes a system 711 5117419 We then set up matrices7 where and the equation is X 7 AX We assume W w1 w2 and that our solution is of the form X We for some 7 wl and LU2 We will review this method later when we want to generalize it7 but lets look at an easier method 4
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Non-trivial extension of max(n,1)-1 to the reals and its iteration. MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/16/2014, 01:37 PM (This post was last modified: 05/16/2014, 02:40 PM by MphLee.) I was very interested in the problem of extending $max(n,1)-1$ to the reals (that is actual the "incomplete" predecessor function $S^{\circ (-1)}$ over the naturals) I've asked the same question on MSE but it was a bit ignored...I hope because it is trivial! http://math.stackexchange.com/questions/...its-fracti The question is about extensions of $A(n):=max(n,1)-1$ to the reals with some conditions Quote:A-$A(x)=max(x,1)-1$ only if $x \in \mathbb{N}$ B-$A(x)$ is not discontinuous I just noticed that I've made a lot of errors in my MathSE question, I'll fix it in this post (and later on mathSE) From successor and inverse successor we can define the subtraction in this way $x-0:=x$ and $x-(y+1):=S^{\circ(-1)}(x-y)$ with $A$ , that is a modified predecessor function we could define its iteration using an "esotic subtraction" that is "incomplete" for naturals and is "complete" for reals (like we are cutting all the negative integers) Quote:$x -^* 0:=x$ $x-^* (n+1)=A(x-^* n)$ In this way we have Quote:$x-^*1=A(x)$ and $x-^*n=A^{\circ n}(x)$ How we can go in order to extend $x-^*y$ to real $y$? For example what can we know about $x-0,5$ ? for example $(x-^*0.5)-^*1=(x-^*1)-^*0.5=x-^*1.5$ if we put $x=0$ $(0-^*0.5)-^*1=(0-^*1)-^*0.5=0-^*1.5$ then if $0_^*0.5=\alpha$ $\alpha-^*1=0-^*0.5=0-^*1.5$ if $\alpha$ is not a natural number $\alpha-^*1=\alpha=0-^*1.5$ !?? what is going on here? If $\alpha$ is natural $max(\alpha,1)-1=\alpha=0-^*1.5$ in this case we should have that $\alpha=0$. What do you think about this? MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ tommy1729 Ultimate Fellow Posts: 1,640 Threads: 366 Joined: Feb 2009 05/16/2014, 09:21 PM Your function is equal to (x + abs(x))/2. abs(x) can be written as sgn(x) x. sgn(x) is well approximated by tanh(100x). This gives that your function is very close to (x+tanh(100x) x)/2 The problem with your function is that it has all positive integers as fixpoints. Too many fixpoints to have half-iterates valid everywhere. Im not sure if you want an interpolation or an approximation like I just gave. Also the reason you get little response is probably because your mainly asking " what makes this question more intresting " ? If you know what I mean. Asking what properties to look for or asking what questions to ask is similar. You have to decide what you want to do , want to see solved or what properties you desire. Otherwise it sounds weird. Kinda like asking for " a special integer ". Math is like driving a car without a map. You dont know where you will end up. But if you want to end up somewhere you have too start , stop and drive. I hope my metaphor is understood. I assume you are still young. You dont have to tell me about your age but I suspect it. Hope you dont mind me saying. regards tommy1729 MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/16/2014, 10:20 PM (This post was last modified: 05/17/2014, 11:48 AM by MphLee.) (05/16/2014, 09:21 PM)tommy1729 Wrote: Your function is equal to (x + abs(x))/2. abs(x) can be written as sgn(x) x. sgn(x) is well approximated by tanh(100x). This gives that your function is very close to (x+tanh(100x) x)/2 I've plotted this and is the same as $max(0,x)$ or the same as $lim_{h \rightarrow 0^+}h(ln(e^{0/h}+e^{x/h}))=max(0,x)$ ... I know this but I was loking for a function that coincides with $max(1,x)-1=max(0,x-1)$ only for the naturals... If I did not understand something of your formula of approximation tell me. Quote:The problem with your function is that it has all positive integers as fixpoints. Too many fixpoints to have half-iterates valid everywhere. Ok..I don't get this (I'm not good with analysis and the iteration theory)... but help to to understand pls. A fixpoint is a x such that $A(x)=x$ in the case of $A(0)=0$... is it the only fixpoint of A? Maybe you talk about the fact that $A^{n}(0)=0$ and $A^{n}(m)=max(0,m-n)$ for every $n$. This is really a big problem for the real iteration problem? Quote:Im not sure if you want an interpolation or an approximation like I just gave. Also the reason you get little response is probably because your mainly asking " what makes this question more intresting " ? If you know what I mean. Asking what properties to look for or asking what questions to ask is similar. You have to decide what you want to do , want to see solved or what properties you desire. Otherwise it sounds weird. Kinda like asking for " a special integer ". Math is like driving a car without a map. You dont know where you will end up. But if you want to end up somewhere you have too start , stop and drive. I hope my metaphor is understood. Yea I know, the question is a bit unclear and is because I don't exatly know what to ask..or how to ask it. I'm a bit confused about this problem but it is very interesting for me and I think it can be important for oher things I'm doing. Quote:I assume you are still young. You dont have to tell me about your age but I suspect it. Hope you dont mind me saying. regards tommy1729 Don't worry I'm enough young (but not very very young) Maybe you feel more my lack of knowledge on some really basic topics...I don't study math at school probably thats why. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/17/2014, 07:10 PM Update: I voted for the closure of the question on MSE because it is not clear at all. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Catullus Fellow Posts: 201 Threads: 45 Joined: Jun 2022 06/09/2022, 11:06 PM (This post was last modified: 07/29/2022, 10:28 PM by Catullus.) (05/16/2014, 01:37 PM)MphLee Wrote: Quote:A-$A(x)=max(x,1)-1$ only if $x \in \mathbb{N}$ B-$A(x)$ is not discontinuous$\dpi{110} max(n,1)-1=n-1\forall n\in\mathbb{N}$. Defining $\dpi{110} A(x)$ as $\dpi{110} x-\sin(x*\pi)-1$ might work. It has a lot of fixed points though. Can someone please tell me a special integer? Like maybe one related to hyper-operations. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 06/10/2022, 09:07 AM This will work, but is not really what I was looking for in my question. Back in the days my mind was really foggy on this problem because I wasn't able to phrase the real problem I had in mind. Here indeed your solution $$(x-1)-\sin(x\pi)$$ is correct. But probably I was looking for somethign like a smooth approximation of $$T(x)\sim \max(x,1)-1$$ and then the family $$T_\theta=T(x)+\theta(x)$$ for $$\theta(x+1)=\theta(x)$$ and $$\theta(n)=0$$ for each $$n\in\mathbb Z$$. So Tommy's solution fits better since $$\max(x,1)-1=\lim_{k\to\infty} \frac{(x-1)+(x-1){\rm tanh}(k(x-1))}{2}$$ Here Tommy's approximation for $$k=6$$ Here then $$T_\theta(x)=\frac{(x-1)+(x-1){\rm tanh}(k(x-1))}{2} + \lambda \sin(x\pi)$$ for $$\lambda=0.2$$ After all this years I have to think more about this and see if those approximations can be useful for my endgame... One day I'll post something about it. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Catullus Fellow Posts: 201 Threads: 45 Joined: Jun 2022 06/10/2022, 10:17 AM (06/10/2022, 09:07 AM)MphLee Wrote: After all this years I have to think more about this and see if those approximations can be useful for my endgame...Your endgame? Endgame of what? ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 06/10/2022, 11:04 AM Excuse me, I mean that I'm not sure anymore that those approximations are useful for the purpose of my research. I'm, and I was, researching about a far reaching generalization of hyperoperations. I don't know your background so I'll make it self-contained. I'm studying special functions of the kind $${\bf g}:J\to X$$ and maps of kind $$\rho:X\to \mathbb N$$ for $$J$$ a dynamical system. That means that fixing a $$\rho$$ of that kind we can send each $$\bf g$$ to a map $$\rho{\bf g}:J\to \mathbb N$$. Here in this post, 8 years ago, I was initiating the study, with scarce success, of functions $$f:J\to \mathbb N$$ that are of the form $$f=\rho{\bf g}$$. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Catullus Fellow Posts: 201 Threads: 45 Joined: Jun 2022 06/11/2022, 06:22 AM (This post was last modified: 06/19/2022, 09:31 AM by Catullus.) (06/10/2022, 09:07 AM)MphLee Wrote: But probably I was looking for somethign like a smooth approximation of $$T(x)\sim \max(x,1)-1$$ and then the family $$T_\theta=T(x)+\theta(x)$$ for $$\theta(x+1)=\theta(x)$$ and $$\theta(n)=0$$ for each $$n\in\mathbb Z$$.log(sqrt(2),sqrt(2)^x+sqrt(2))-1 is a smooth approximation of max(x,1)-1. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 06/15/2022, 10:59 PM Catullus, this option was already pointed back then. It is the common representation of the tropical operation. max, as as the "Litvinov-Maslov dequantization" of the bennet preaddition. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ « Next Oldest | Next Newest »
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# Ackerman through bump and steer
## Ackerman through bump and steer
(OP)
I've come to wanting to check how steady my ackerman is through bump and steer. My static ackerman is around 30%. This is about as good as I can get as my steering rack is quite far forward of my outer tie rods. But for off road race use low ackerman seems to be favoured. I've been using Lotus Suspension Analysis.
The ackerman appears steady throughout steering:
However, through bump it goes nuts:
I spoke to Lotus and they said to me that as ackerman definitions rely on the difference between angles of the wheels they can become unstable when the wheels are both facing forwards. Their exact words:
#### Quote:
Ackermann as it is one of those parameters that seems to attract a number of differing definitions. In addition as most of the algorithms tend to rely on differing steer angles between the left and right wheel, they all have the potential to be unstable as you approach the straight ahead case, (i.e when both steer angles are the same).
Unfortunately my tech support with Lotus has expired so I can't show them my graph to get any comments on the graph. To me it looks like a problem with infinities around the zero point, i.e. a problem with the formula. When I animate my design the wheels appear quite parallel with each other:
Any ideas how to check my ackerman further?
### RE: Ackerman through bump and steer
I suggest you plot the difference in steer angle (toe) between each wheel vs bump/rebound. This should allay your fears and confirm Lotus' diagnosis. By the by, who were you talking to at Lotus?
Ackermann at straight ahead is dominated by the static toe setting and is unhelpful.
Parallel steer (ie zero Ackermann) has been used successfully by circuit racers, as has 100% Ackermann. An experienced and successful race engineer told me it is just about the least used tool in his tuning box. If he had the time he would investigate the ideal Ackermann by running different toe settings and comparing slow and fast corners. Toe is fake Ackermann.
For production cars the main use of Ackermann is to minimise the turning circle at full lock, essentially by preventing the outer wheel from fighting the inner wheel. That is not the same as 100% Ackermann incidentally. That is not really a consideration for a circuit car.
For a circuit car there is theory that excess A may help to rotate the car into a corner, because the inner wheel will drag the car around. As theories go, plausible but not confirmed. If I was less lazy I would run that in ADAMS. I am lazy. Given that a race car should be operating with each tire on the limit of its friction circle at all times, (for a hot lap), I don't think you can really be prescriptive like that for a kinematic effect.
Cheers
Greg Locock
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### RE: Ackerman through bump and steer
Greg - I think I've even read where anti-Ackermann (negative A?) has some supporters. Something about the peak "mu" of the lightly loaded inboard tire occurring at a smaller slip angle than is the situation on the outboard tire.
Norm
### RE: Ackerman through bump and steer
That's another plausible hypothesis. Hmm, that could be a nice little project for a vehicle dynamics student. I'd run a swept steer to establish max latacc at a few speeds, and some sort of step steer and look at the details of the build up of yaw rate. Do that for a few different cars and tires and see if there is a general trend. You might get a different answer with trail braking as well. Here's a nice paper on the latacc part https://www.google.com.au/url?sa=t&source=web&...
Cheers
Greg Locock
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### RE: Ackerman through bump and steer
The use of anti-Ackermann is very easy to document and analyze. Simple road test procedures reveal the effects of using pro vs. parallel vs. anti Ackermann and the reasons for its specification:
Yes, the inside tire can be seen dragging down the front grip levels. However, this characteristic is highly dependent on the tire characteristics themselves as well as the load transfer distribution. In other words, its not always the case for a fixed steering geometry setting. Even tire pressure gets into the recipe.
One point of confusion is usually that 'correcting' the Ackermann either with toe or geometry will loosen the car up because it reduces understeer if you gain front grip (reduce the front axle sideslip gradient). This effect will usually be misconstrued as a 'bad' thing because the potential for higher max lat is there, just not the reduced control sensitivity necessary to attain it. Toe (out) changes of 1 to 2 degrees are commonly needed to get this extra sidebite. This can be a little 'tireing' on the tread on long straight runs. Change tire brand or construction on the same brand and it can be a whole new ball game.
Needless to say, the effect is the same on the rear, with the same tendencies: extra rear grip adds understeer and this will be viewed as 'worse' unless the front is also rebalanced to bring the control sensitivity back up. This is done (in the rear) by toe and roll steer alterations.
If I put my 'Student' hat on, I'll produce some tire carpet plots with 4 wheel tire force & moment traces on them. The effect(s) are pretty obvious, graphically. It makes a wonderfull excercise for simulation when you have tire data worthy of engineering the car instead of wrenching it.
Happy Holidays to all !
### RE: Ackerman through bump and steer
Here's one typical response example taken from analysis of a small student competition car subject to rules and engineering constraints. Tires are race slicks run on a belt surface tire test machine and fitted to an appropriate, non-Pacejka tire model. Tire surface asymmetries have been removed to make the 4 tires have no conicity or plysteer effects (Not that this makes a big difference for the example purpose)..
The car is still able to achieve steady state conditions in a constant speed increasing steer angle test, but does better with some toe-out added. Is not driveable with same amount of toe-in because of an oversteering level beyond its negative Ackermann gradient.
Put a different brand of tire on the car, different results in the limit. One key aspect of this analysis is the high amount of tire load reserve in play (The wheel weights are maybe only 1/2 of the tire's rated Fz load at specified pressure). This analysis is not limited to small rear weight biased race cars. Its a useful analysis for all high performance (as in race prepped track) cars.
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https://blogs.sap.com/2016/03/13/dynamic-time-period-axis-date-month-quarter-year-based-on-the-date-range-selection/ | 1,695,538,160,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00177.warc.gz | 163,617,525 | 20,642 | # Dynamic Time Period Axis (Date, Month, Quarter, Year) based on the date range selection
Problem Statement:
We often get a requirement to generate a report based on a given date range, a typical requirement would be to display charts based on the date range selected. if user selects date range as too big then the chart will clutter with so many points and impacts chart readability.
Do we have any option to change the chart axis dynamically based on the date range.
For example –
If user selects data range less than a month then show week wise data
If selection is more than a week and less than a month then show month wise data
If selection is more than a month and less than a quarter then show month wise data
If selection is more than a quarter and less than a year then show quarter wise data
If selection is more than a year and less then show year wise data
Solution:
This is how we could approach the solution, create a variable which gives the output as Date, Week, Month, Quarter and Year based on the Number of Days between the date range selected by the user, use this variable while creating the chart.
Below are the detailed steps for creating variables and chart.
Step1 : Create Report:
Create a sample report with Island Resorts marketing Universe with objects like Country, Invoice Date and Revenue
Create Range Filter on Invoice date as with prompt text as “Start Date” and “End Date” respectively.
Make these prompts as optional so that user can run the report fir entire data range
Query will look like below
SELECT
Resort_Country.country, Sales.invoice_date, sum(Invoice_Line.days * Invoice_Line.nb_guests * Service.price)
FROM
Country Resort_Country,
Sales, Invoice_Line, Service, Service_Line, Resort
WHERE
( Resort_Country.country_id=Resort.country_id )
AND ( Sales.inv_id=Invoice_Line.inv_id )
AND ( Invoice_Line.service_id=Service.service_id )
AND ( Resort.resort_id=Service_Line.resort_id )
AND ( Service.sl_id=Service_Line.sl_id )
GROUP BY
Resort_Country.country,
Sales.invoice_date
Step 2: Create these variables in the report level
1. User Start Date
2. User End Date
3. No of Days
=DaysBetween([User Start Date];[User End Date])
4. Year
=FormatNumber(Year([Invoice Date]);”####”)
5. Year_Number(YYYY)
=Year([Invoice Date])
6. MonthYear(Mon-YYYY)
=Left(Month([Invoice Date]);3)+” – “+[Year]
7. Month_Number(YYYYMM)
=[Year_Number]*100+MonthNumberOfYear(([Invoice Date]))
8. Quarter(YYYY-QQ)
=[Year]+”-Q”+Quarter([Invoice Date])
9. Quarter_Number(YYYYQ)
=[Year_Number]*10+Quarter([Invoice Date])
10. Date_Number(DDMMYYYY)
=(DayNumberOfMonth([Invoice Date])*100+MonthNumberOfYear([Invoice Date]))*10000+[Year_Number]
11.Week
=”w”+Week([Invoice Date])+”-“+[MonthYear(Mon-YYYY)]
12.WeekNumber
=Week([Invoice Date])*1000000+[Month_Number(YYYYMM)]
13. Dynamic_Period
=If([NoofDays]<=7;[Invoice Date];If([NoofDays]<=31;[Week];If([NoofDays]<=90;[MonthYear(Mon-YYYY)];If([NoofDays]>90 And [NoofDays] <=365;[Quarter(YYYY-QQ)];[Year]))))
14. Dynamic_Period_Number
=If([NoofDays]<=7;[Date_Number(DDMMYYYY)];If([NoofDays]<=31;[WeekNumber];If([NoofDays]<=90;[Month_Number(YYYYMM)];If([NoofDays]>90 And [NoofDays] <=360 ;[Quarter_Number(YYYYQ)]; [Year_Number]))))
Output of above variables will be like below
Dynamic _Period User Start Date User End Date Noof Days Invoice Date Date_Nu mber (DDMMY YYY) Week WeekNu mber MonthYear (Mon- YYYY) Month_N umber (YYYYM M) Quarter (YYYY- QQ) Quarter_Nu mber (YYYYQ) Year Year_Number 1,199,801 1/1/98 1/31/98 30 1/1/98 1,011,998 w1-Jan – 1998 1,199,801 Jan – 1998 199,801 1998-Q1 19,981 1998 1,998 1,199,801 1/1/98 1/31/98 30 1/2/98 2,011,998 w1-Jan – 1998 1,199,801 Jan – 1998 199,801 1998-Q1 19,981 1998 1,998 2,199,801 1/1/98 1/31/98 30 1/6/98 6,011,998 w2-Jan – 1998 2,199,801 Jan – 1998 199,801 1998-Q1 19,981 1998 1,998 2,199,801 1/1/98 1/31/98 30 1/7/98 7,011,998 w2-Jan – 1998 2,199,801 Jan – 1998 199,801 1998-Q1 19,981 1998 1,998 2,199,801 1/1/98 1/31/98 30 1/11/98 11,011,998 w2-Jan – 1998 2,199,801 Jan – 1998 199,801 1998-Q1 19,981 1998 1,998 3,199,801 1/1/98 1/31/98 30 1/12/98 12,011,998 w3-Jan – 1998 3,199,801 Jan – 1998 199,801 1998-Q1 19,981 1998 1,998
Step 3: Create Chart using Country, Dynamic_Period,Dynamic_period_Number,Revenue objects
Step 4: Sort the chart on Dynamic_Period_Number variable
note:
Here i have hidden the Dynamic period Number variable in the chart and sorted on it, but display variable will be Dynamic Period. This to avoid the sorting issue which you get while using the Dynamic Period Object and the explanation of the issue is at the end of this solution.
Step 5: Create chart using Country, Invoice_Date and Revenue object
Step 6: Report output with different set of date parameters
1) Date Range: Entire Date Range
No value selected for Start and End date (left them blank) as they are optional
Report output with normal date as the x-axis value
Report output with Dynamic Period as the x-axis value shows year wise data, as the entire data is for 3 Years in the data base
2) Date Range: One week
Start Date: 1/1/1998
End date: 7/1/1998
Report output with normal date as the x-axis value
Dynamic Period Chart gives day wise data
3) Date Range: One Month of Data
Start Date: 1/1/1998
End date: 31/1/1998
Report output with normal date as the x-axis value
Report output with Dynamic Period as the x-axis value shows week wise data
4) Date Range: Three Months of Data
Start Date: 1/1/1998
End date: 31/3/1998
Report output with default date axis
Dynamic axis chart output shows month wise data
5) Date Range: One Year
Start Date: 1/1/1998
End date: 12/1/1998
Report output with normal date as the x-axis value
Report output with Dynamic Period as the x-axis value shows quarter wise data
6) Date Range: Two Years
Start Date: 1/1/1998
End date: 12/1/1999
Report output with normal date as the x-axis value
Report output with Dynamic Period as the x-axis value gives year wise data
Note:
Reason for having Number and string for every formula is to avoid sorting issue of the Date field. This is because the output of the Dynamic Period is a String. Since the if statement contains combination of Date and String data types even the date is considered as String.
=If([NoofDays]<=31;[Invoice Date];If([NoofDays]<=90;[MonthYear(Mon-YYYY)];If([NoofDays]>90 And [NoofDays] <=365;[Quarter(YYYY-QQ)];[Year])))
The sorting issue in chart created only Dynamic Period which is of String Data type will be like below (if you see the order of dates as 11,12 & 7)
ASCII of 12 less than 7 so it came first)
Date Range Selected is : 7/1/1998 to 14/1/1998
Chart Created using both Dynamic Period and Dynamic Period Number will be like this | 1,979 | 6,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.858778 |
http://www.dhs.state.il.us/page.aspx/page.aspx?item=17645 | 1,500,554,973,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423183.57/warc/CC-MAIN-20170720121902-20170720141902-00445.warc.gz | 404,692,125 | 5,244 | (FCRC) Authorize an Initial Prorated Entitlement (IPE), if eligible for an increased Payment Level, to meet the needs of the eligible person before receipt of the first regular check.
Complete the current Form 552 for both the IPE and the regular roll actions (see WAG 25-07-10).
To meet the needs of the eligible person before the receipt of the first regular check and to add the person's needs to the regular benefit check, authorize the IPE action, if eligible, and the regular roll action using the current Form 552 (see WAG 25-07-10).
## (1) Central Proration
(System) Computes the number of days in the IPE period and authorizes the correct amount of the IPE for additions to the benefit unit when a 6-digit date is entered in Item 2 and TA 34 or TA 42 is entered in Item 3 of Form 552, unless an exception to central prorating is required or Item 78 is coded to suppress the IPE.
Suppresses the IPE check when a 4-digit date is entered in Item 2. Issues the IPE in the designated amount if an exception to central prorating is authorized.
## (2) Exception to Central Proration
Compute the IPE period and authorize the correct amount of the IPE as an exception to central prorate by entering code 100 in Item 80 when the person being added has budgetable income. Always budget prospectively the income of a person being added.
Round down the IPE issued as an exception to central prorate, code 100. If the Family Community Resource Center enters cents in code 100, Data Processing will drop them.
Calculate an exception to central proration as follows:
1. Determine the number of days in the IPE period. This is the date entered in Item 2, through the day before the first day of the fiscal month for the first regular roll action.
2. Subtract the Payment Level amount for the smaller benefit unit size from the Payment Level amount for the larger benefit unit size, including the added person. The difference is the monthly amount for the added person.
3. Divide the monthly amount in (b) above, by 30 to arrive at a daily amount. Round to the nearest cent. If the 3rd position to the right of the decimal is a number 5 or greater, round up to the nearest cent. If the number is less than 5, round down to the nearest cent.
For example, 2.877 = 2.88 or 1.833 = 1.83. Multiply the daily amount by the number of days in the IPE period. Do not drop cents. This is the prorated Payment Level amount for the IPE period.
4. Deduct the income received or anticipated to be received by the added person in the IPE period from the prorated Payment Level amount for the IPE period.
5. Round down to the nearest dollar. This is the amount of the IPE check.
6. Enter code 100 and the amount from (e) above, in Item 80.
The amount of the centrally computed IPE check or locally computed IPE check appears on the:
• Mercury Payroll, and
• Client Payroll Inquiry (PF7/F7), and
• Recipient Ledger Inquiry Screen (PF6/F6).
Example: Locally Computed IPE (TANF)
A request is made on 02/15 to add a child to the case who is not required to be included. The benefit unit is eligible for an increased Payment Level because the child was born to a minor mother receiving cash in her parent's case.
The fiscal month is the first through the last day of calendar month. On the 3rd of each month the child receives monthly social security benefit of \$50. IPE runs from 02/15 - 03/31.
\$623 6-person Payment Level amount for the new benefit unit size
-555 5-person Payment Level amount for the old benefit unit size
\$ 68 Monthly prorated Payment Level amount for the added person
To figure the daily amount, take \$68.00 and divide by 30 days:
\$68 ÷ 30 days = \$2.22 = daily amount. Multiply \$2.22 x 45 days of IPE = \$99.90 = prorated payment.
\$99.90 Prorated Payment Level amount for 45-day IPE period
-50.00 SSA income anticipated to be received in 45 day period (02/15 through 03/31)
\$49.00 Code 100 amount (rounded down to nearest dollar)
Example: Locally Computed IPE (GA-FCA)
Request is made on 02/15 to add child to the GA-FCA case. The fiscal month is first through the last day of the calendar month. On 03/03 the child is to receive monthly Social Security benefit of \$50. IPE runs from 02/15 - 03/31.
\$555 5-person Payment Level Amount for the new benefit unit size
-474 4-person Payment Level Amount for the old benefit unit size
\$ 81 Monthly prorated Payment Level amount for the additional person
To figure the daily amount, take \$81.00 and divide by 30 days (81 ÷ 30 days = \$2.70 = daily amount). Multiply \$2.70 x 45 days of IPE = \$121.50 = prorated Payment Level amount.
\$121.50 Prorated Payment Level amount for 45-day IPE period
- 50.00 SSA income anticipated to be received in 45-day period (02/15 through 03/31)
\$71.00 Code 100 amount (rounded down to nearest dollar) | 1,212 | 4,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-30 | latest | en | 0.900887 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=PAL | 1,526,969,809,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864626.37/warc/CC-MAIN-20180522053839-20180522073839-00199.warc.gz | 406,973,337 | 98,367 | Back to list of Stocks See Also: Seasonal Analysis of PALGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
# Fourier Analysis of PAL (North American Palladium Ltd)
PAL (North American Palladium Ltd) appears to have interesting cyclic behaviour every 69 weeks (5.6719*cosine), 76 weeks (5.6372*cosine), and 59 weeks (5.4445*cosine).
PAL (North American Palladium Ltd) has an average price of 7.82 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
## Fourier Analysis
Using data from 10/16/2000 to 4/30/2018 for PAL (North American Palladium Ltd), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
07.82282 0
14.73643 2.32649 (1*2π)/764764 weeks
25.83857 -.52217 (2*2π)/764382 weeks
36.94325 .83709 (3*2π)/764255 weeks
46.88712 -.94946 (4*2π)/764191 weeks
55.93431 .30863 (5*2π)/764153 weeks
65.66148 -.69101 (6*2π)/764127 weeks
75.82248 -1.67877 (7*2π)/764109 weeks
86.44365 -1.74903 (8*2π)/76496 weeks
95.74113 -1.15734 (9*2π)/76485 weeks
105.63717 -1.7742 (10*2π)/76476 weeks
115.67188 -2.20163 (11*2π)/76469 weeks
124.81135 -2.09232 (12*2π)/76464 weeks
135.44455 -2.89089 (13*2π)/76459 weeks
145.28262 -2.48916 (14*2π)/76455 weeks
154.57457 -3.22451 (15*2π)/76451 weeks
164.74602 -3.21853 (16*2π)/76448 weeks
175.01351 -3.54636 (17*2π)/76445 weeks
184.36839 -3.46553 (18*2π)/76442 weeks
194.3506 -3.7818 (19*2π)/76440 weeks
204.26153 -3.83081 (20*2π)/76438 weeks
213.99488 -4.2012 (21*2π)/76436 weeks
223.88099 -3.94279 (22*2π)/76435 weeks
233.53595 -4.26217 (23*2π)/76433 weeks
243.48044 -4.47883 (24*2π)/76432 weeks
253.26467 -4.38174 (25*2π)/76431 weeks
263.01071 -4.57557 (26*2π)/76429 weeks
272.84696 -4.52718 (27*2π)/76428 weeks
282.48828 -4.76926 (28*2π)/76427 weeks
292.38694 -4.83228 (29*2π)/76426 weeks
302.12068 -4.73425 (30*2π)/76425 weeks
311.94277 -4.93611 (31*2π)/76425 weeks
321.70496 -4.87165 (32*2π)/76424 weeks
331.53461 -5.06477 (33*2π)/76423 weeks
341.38938 -4.82322 (34*2π)/76422 weeks
351.00292 -5.06067 (35*2π)/76422 weeks
36.86804 -4.86414 (36*2π)/76421 weeks
37.63366 -5.0076 (37*2π)/76421 weeks
38.52569 -4.8126 (38*2π)/76420 weeks
39.19179 -5.00671 (39*2π)/76420 weeks
40.10382 -4.70845 (40*2π)/76419 weeks
41-.30344 -4.76548 (41*2π)/76419 weeks
42-.31328 -4.80178 (42*2π)/76418 weeks
43-.48261 -4.732 (43*2π)/76418 weeks
44-.64078 -4.49037 (44*2π)/76417 weeks
45-.97613 -4.48639 (45*2π)/76417 weeks
46-.89116 -4.45579 (46*2π)/76417 weeks
47-1.26125 -4.22398 (47*2π)/76416 weeks
48-1.26908 -4.05096 (48*2π)/76416 weeks
49-1.56665 -4.01108 (49*2π)/76416 weeks
50-1.60146 -3.92329 (50*2π)/76415 weeks
51-1.79585 -3.72935 (51*2π)/76415 weeks
52-1.89529 -3.49015 (52*2π)/76415 weeks
53-2.10081 -3.41888 (53*2π)/76414 weeks
54-2.13772 -3.28468 (54*2π)/76414 weeks
55-2.3376 -3.14769 (55*2π)/76414 weeks
56-2.32785 -2.86767 (56*2π)/76414 weeks
57-2.54743 -2.78609 (57*2π)/76413 weeks
58-2.58235 -2.60918 (58*2π)/76413 weeks
59-2.74443 -2.43168 (59*2π)/76413 weeks
60-2.6542 -2.28626 (60*2π)/76413 weeks
61-2.81764 -2.07499 (61*2π)/76413 weeks
62-2.86352 -1.93984 (62*2π)/76412 weeks
63-2.83266 -1.75444 (63*2π)/76412 weeks
64-2.95027 -1.52806 (64*2π)/76412 weeks
65-2.98036 -1.40143 (65*2π)/76412 weeks
66-3.01369 -1.24956 (66*2π)/76412 weeks
67-2.96292 -1.05075 (67*2π)/76411 weeks
68-3.01148 -.85298 (68*2π)/76411 weeks
69-2.96691 -.78615 (69*2π)/76411 weeks
70-2.94736 -.48871 (70*2π)/76411 weeks
71-2.89489 -.50203 (71*2π)/76411 weeks
72-2.7531 -.20759 (72*2π)/76411 weeks
73-2.85502 .02756 (73*2π)/76410 weeks
74-2.72747 .12924 (74*2π)/76410 weeks
75-2.72377 .23813 (75*2π)/76410 weeks
76-2.66022 .40331 (76*2π)/76410 weeks
77-2.57147 .56982 (77*2π)/76410 weeks
78-2.46838 .73527 (78*2π)/76410 weeks
79-2.46872 .79234 (79*2π)/76410 weeks
80-2.27023 .95902 (80*2π)/76410 weeks
81-2.19856 1.09756 (81*2π)/7649 weeks
82-2.08196 1.19584 (82*2π)/7649 weeks
83-2.03255 1.27575 (83*2π)/7649 weeks
84-1.89563 1.41024 (84*2π)/7649 weeks
85-1.69246 1.48288 (85*2π)/7649 weeks
86-1.59615 1.63074 (86*2π)/7649 weeks
87-1.53204 1.70905 (87*2π)/7649 weeks
88-1.41606 1.79905 (88*2π)/7649 weeks
89-1.21589 1.82423 (89*2π)/7649 weeks
90-1.1483 1.91688 (90*2π)/7648 weeks
91-.99462 1.93799 (91*2π)/7648 weeks
92-.86167 2.02445 (92*2π)/7648 weeks
93-.69194 2.02317 (93*2π)/7648 weeks
94-.56658 2.08873 (94*2π)/7648 weeks
95-.42799 2.07853 (95*2π)/7648 weeks
96-.32187 2.12737 (96*2π)/7648 weeks
97-.13875 2.10933 (97*2π)/7648 weeks
98-.00036 2.14181 (98*2π)/7648 weeks
99.09953 2.11767 (99*2π)/7648 weeks
100.30657 2.09861 (100*2π)/7648 weeks
101.39715 2.10619 (101*2π)/7648 weeks
102.56608 2.13064 (102*2π)/7647 weeks
103.64437 2.03072 (103*2π)/7647 weeks
104.7785 2.01574 (104*2π)/7647 weeks
105.91815 1.98578 (105*2π)/7647 weeks
1061.06085 1.91345 (106*2π)/7647 weeks
1071.12847 1.86185 (107*2π)/7647 weeks
1081.27703 1.87397 (108*2π)/7647 weeks
1091.36955 1.70532 (109*2π)/7647 weeks
1101.52139 1.68049 (110*2π)/7647 weeks
1111.57924 1.66136 (111*2π)/7647 weeks
1121.64376 1.53864 (112*2π)/7647 weeks
1131.78095 1.43419 (113*2π)/7647 weeks
1141.86745 1.3709 (114*2π)/7647 weeks
1151.89499 1.27567 (115*2π)/7647 weeks
1162.0086 1.22163 (116*2π)/7647 weeks
1172.07468 1.03229 (117*2π)/7647 weeks
1182.17822 .9907 (118*2π)/7646 weeks
1192.19396 .91898 (119*2π)/7646 weeks
1202.23779 .82922 (120*2π)/7646 weeks
1212.27414 .63395 (121*2π)/7646 weeks
1222.38721 .59774 (122*2π)/7646 weeks
1232.34852 .47438 (123*2π)/7646 weeks
1242.37897 .3877 (124*2π)/7646 weeks
1252.44706 .18967 (125*2π)/7646 weeks
1262.46497 .17129 (126*2π)/7646 weeks
1272.48093 .053 (127*2π)/7646 weeks
1282.47395 -.04362 (128*2π)/7646 weeks
1292.4713 -.18952 (129*2π)/7646 weeks
1302.5017 -.2371 (130*2π)/7646 weeks
1312.44501 -.3602 (131*2π)/7646 weeks
1322.45528 -.45937 (132*2π)/7646 weeks
1332.43971 -.58713 (133*2π)/7646 weeks
1342.39337 -.65222 (134*2π)/7646 weeks
1352.38749 -.75198 (135*2π)/7646 weeks
1362.33868 -.86655 (136*2π)/7646 weeks
1372.3172 -.93135 (137*2π)/7646 weeks
1382.25609 -1.02341 (138*2π)/7646 weeks
1392.22168 -1.13281 (139*2π)/7645 weeks
1402.18889 -1.21809 (140*2π)/7645 weeks
1412.13426 -1.26143 (141*2π)/7645 weeks
1422.06775 -1.34507 (142*2π)/7645 weeks
1432.05073 -1.44761 (143*2π)/7645 weeks
1441.95741 -1.46356 (144*2π)/7645 weeks
1451.89763 -1.54367 (145*2π)/7645 weeks
1461.85951 -1.62383 (146*2π)/7645 weeks
1471.80749 -1.67783 (147*2π)/7645 weeks
1481.70695 -1.68445 (148*2π)/7645 weeks
1491.64719 -1.72719 (149*2π)/7645 weeks
1501.57025 -1.79891 (150*2π)/7645 weeks
1511.53057 -1.876 (151*2π)/7645 weeks
1521.44731 -1.83875 (152*2π)/7645 weeks
1531.38481 -1.88328 (153*2π)/7645 weeks
1541.30855 -1.92795 (154*2π)/7645 weeks
1551.23285 -1.91339 (155*2π)/7645 weeks
1561.15973 -1.94485 (156*2π)/7645 weeks
1571.09762 -1.94832 (157*2π)/7645 weeks
1581.00427 -1.97983 (158*2π)/7645 weeks
159.95589 -1.95517 (159*2π)/7645 weeks
160.87122 -1.98431 (160*2π)/7645 weeks
161.79853 -1.97905 (161*2π)/7645 weeks
162.75152 -1.98429 (162*2π)/7645 weeks
163.69857 -1.97007 (163*2π)/7645 weeks
164.63434 -1.9787 (164*2π)/7645 weeks
165.56381 -1.93443 (165*2π)/7645 weeks
166.51098 -1.96058 (166*2π)/7645 weeks
167.4915 -1.91001 (167*2π)/7645 weeks
168.42217 -1.8894 (168*2π)/7645 weeks
169.37008 -1.86648 (169*2π)/7645 weeks
170.31758 -1.80781 (170*2π)/7644 weeks
171.26654 -1.86327 (171*2π)/7644 weeks
172.25252 -1.78342 (172*2π)/7644 weeks
173.17696 -1.75173 (173*2π)/7644 weeks
174.17349 -1.74907 (174*2π)/7644 weeks
175.14956 -1.69993 (175*2π)/7644 weeks
176.09433 -1.68463 (176*2π)/7644 weeks
177.08983 -1.6654 (177*2π)/7644 weeks
178.07629 -1.56581 (178*2π)/7644 weeks
179.03016 -1.59342 (179*2π)/7644 weeks
180.0232 -1.54467 (180*2π)/7644 weeks
181.02848 -1.50182 (181*2π)/7644 weeks
182.00701 -1.48988 (182*2π)/7644 weeks
183.01199 -1.42881 (183*2π)/7644 weeks
184-.03628 -1.40421 (184*2π)/7644 weeks
185.0087 -1.41238 (185*2π)/7644 weeks
186-.02835 -1.33011 (186*2π)/7644 weeks
187-.0166 -1.31882 (187*2π)/7644 weeks
188-.01133 -1.29136 (188*2π)/7644 weeks
189-.00009 -1.30469 (189*2π)/7644 weeks
190-.00498 -1.25913 (190*2π)/7644 weeks
191.01047 -1.21813 (191*2π)/7644 weeks
192.04353 -1.2075 (192*2π)/7644 weeks
193.05013 -1.22757 (193*2π)/7644 weeks
194.07295 -1.15585 (194*2π)/7644 weeks
195.05602 -1.14929 (195*2π)/7644 weeks
196.10302 -1.15256 (196*2π)/7644 weeks
197.10837 -1.1474 (197*2π)/7644 weeks
198.10039 -1.11375 (198*2π)/7644 weeks
199.12789 -1.13513 (199*2π)/7644 weeks
200.16838 -1.13194 (200*2π)/7644 weeks
201.16298 -1.14119 (201*2π)/7644 weeks
202.19589 -1.12808 (202*2π)/7644 weeks
203.23238 -1.14116 (203*2π)/7644 weeks
204.22321 -1.13117 (204*2π)/7644 weeks
205.23409 -1.17292 (205*2π)/7644 weeks
206.26979 -1.14906 (206*2π)/7644 weeks
207.28847 -1.18745 (207*2π)/7644 weeks
208.28561 -1.17484 (208*2π)/7644 weeks
209.26992 -1.22457 (209*2π)/7644 weeks
210.34521 -1.282 (210*2π)/7644 weeks
211.32758 -1.24222 (211*2π)/7644 weeks
212.31731 -1.29174 (212*2π)/7644 weeks
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6292.38749 .75198 (629*2π)/7641 weeks
6302.39337 .65222 (630*2π)/7641 weeks
6312.43971 .58713 (631*2π)/7641 weeks
6322.45528 .45937 (632*2π)/7641 weeks
6332.44501 .3602 (633*2π)/7641 weeks
6342.5017 .2371 (634*2π)/7641 weeks
6352.4713 .18952 (635*2π)/7641 weeks
6362.47395 .04362 (636*2π)/7641 weeks
6372.48093 -.053 (637*2π)/7641 weeks
6382.46497 -.17129 (638*2π)/7641 weeks
6392.44706 -.18967 (639*2π)/7641 weeks
6402.37897 -.3877 (640*2π)/7641 weeks
6412.34852 -.47438 (641*2π)/7641 weeks
6422.38721 -.59774 (642*2π)/7641 weeks
6432.27414 -.63395 (643*2π)/7641 weeks
6442.23779 -.82922 (644*2π)/7641 weeks
6452.19396 -.91898 (645*2π)/7641 weeks
6462.17822 -.9907 (646*2π)/7641 weeks
6472.07468 -1.03229 (647*2π)/7641 weeks
6482.0086 -1.22163 (648*2π)/7641 weeks
6491.89499 -1.27567 (649*2π)/7641 weeks
6501.86745 -1.3709 (650*2π)/7641 weeks
6511.78095 -1.43419 (651*2π)/7641 weeks
6521.64376 -1.53864 (652*2π)/7641 weeks
6531.57924 -1.66136 (653*2π)/7641 weeks
6541.52139 -1.68049 (654*2π)/7641 weeks
6551.36955 -1.70532 (655*2π)/7641 weeks
6561.27703 -1.87397 (656*2π)/7641 weeks
6571.12847 -1.86185 (657*2π)/7641 weeks
6581.06085 -1.91345 (658*2π)/7641 weeks
659.91815 -1.98578 (659*2π)/7641 weeks
660.7785 -2.01574 (660*2π)/7641 weeks
661.64437 -2.03072 (661*2π)/7641 weeks
662.56608 -2.13064 (662*2π)/7641 weeks
663.39715 -2.10619 (663*2π)/7641 weeks
664.30657 -2.09861 (664*2π)/7641 weeks
665.09953 -2.11767 (665*2π)/7641 weeks
666-.00036 -2.14181 (666*2π)/7641 weeks
667-.13875 -2.10933 (667*2π)/7641 weeks
668-.32187 -2.12737 (668*2π)/7641 weeks
669-.42799 -2.07853 (669*2π)/7641 weeks
670-.56658 -2.08873 (670*2π)/7641 weeks
671-.69194 -2.02317 (671*2π)/7641 weeks
672-.86167 -2.02445 (672*2π)/7641 weeks
673-.99462 -1.93799 (673*2π)/7641 weeks
674-1.1483 -1.91688 (674*2π)/7641 weeks
675-1.21589 -1.82423 (675*2π)/7641 weeks
676-1.41606 -1.79905 (676*2π)/7641 weeks
677-1.53204 -1.70905 (677*2π)/7641 weeks
678-1.59615 -1.63074 (678*2π)/7641 weeks
679-1.69246 -1.48288 (679*2π)/7641 weeks
680-1.89563 -1.41024 (680*2π)/7641 weeks
681-2.03255 -1.27575 (681*2π)/7641 weeks
682-2.08196 -1.19584 (682*2π)/7641 weeks
683-2.19856 -1.09756 (683*2π)/7641 weeks
684-2.27023 -.95902 (684*2π)/7641 weeks
685-2.46872 -.79234 (685*2π)/7641 weeks
686-2.46838 -.73527 (686*2π)/7641 weeks
687-2.57147 -.56982 (687*2π)/7641 weeks
688-2.66022 -.40331 (688*2π)/7641 weeks
689-2.72377 -.23813 (689*2π)/7641 weeks
690-2.72747 -.12924 (690*2π)/7641 weeks
691-2.85502 -.02756 (691*2π)/7641 weeks
692-2.7531 .20759 (692*2π)/7641 weeks
693-2.89489 .50203 (693*2π)/7641 weeks
694-2.94736 .48871 (694*2π)/7641 weeks
695-2.96691 .78615 (695*2π)/7641 weeks
696-3.01148 .85298 (696*2π)/7641 weeks
697-2.96292 1.05075 (697*2π)/7641 weeks
698-3.01369 1.24956 (698*2π)/7641 weeks
699-2.98036 1.40143 (699*2π)/7641 weeks
700-2.95027 1.52806 (700*2π)/7641 weeks
701-2.83266 1.75444 (701*2π)/7641 weeks
702-2.86352 1.93984 (702*2π)/7641 weeks
703-2.81764 2.07499 (703*2π)/7641 weeks
704-2.6542 2.28626 (704*2π)/7641 weeks
705-2.74443 2.43168 (705*2π)/7641 weeks
706-2.58235 2.60918 (706*2π)/7641 weeks
707-2.54743 2.78609 (707*2π)/7641 weeks
708-2.32785 2.86767 (708*2π)/7641 weeks
709-2.3376 3.14769 (709*2π)/7641 weeks
710-2.13772 3.28468 (710*2π)/7641 weeks
711-2.10081 3.41888 (711*2π)/7641 weeks
712-1.89529 3.49015 (712*2π)/7641 weeks
713-1.79585 3.72935 (713*2π)/7641 weeks
714-1.60146 3.92329 (714*2π)/7641 weeks
715-1.56665 4.01108 (715*2π)/7641 weeks
716-1.26908 4.05096 (716*2π)/7641 weeks
717-1.26125 4.22398 (717*2π)/7641 weeks
718-.89116 4.45579 (718*2π)/7641 weeks
719-.97613 4.48639 (719*2π)/7641 weeks
720-.64078 4.49037 (720*2π)/7641 weeks
721-.48261 4.732 (721*2π)/7641 weeks
722-.31328 4.80178 (722*2π)/7641 weeks
723-.30344 4.76548 (723*2π)/7641 weeks
724.10382 4.70845 (724*2π)/7641 weeks
725.19179 5.00671 (725*2π)/7641 weeks
726.52569 4.8126 (726*2π)/7641 weeks
727.63366 5.0076 (727*2π)/7641 weeks
728.86804 4.86414 (728*2π)/7641 weeks
7291.00292 5.06067 (729*2π)/7641 weeks
7301.38938 4.82322 (730*2π)/7641 weeks
7311.53461 5.06477 (731*2π)/7641 weeks
7321.70496 4.87165 (732*2π)/7641 weeks
7331.94277 4.93611 (733*2π)/7641 weeks
7342.12068 4.73425 (734*2π)/7641 weeks
7352.38694 4.83228 (735*2π)/7641 weeks
7362.48828 4.76926 (736*2π)/7641 weeks
7372.84696 4.52718 (737*2π)/7641 weeks
7383.01071 4.57557 (738*2π)/7641 weeks
7393.26467 4.38174 (739*2π)/7641 weeks
7403.48044 4.47883 (740*2π)/7641 weeks
7413.53595 4.26217 (741*2π)/7641 weeks
7423.88099 3.94279 (742*2π)/7641 weeks
7433.99488 4.2012 (743*2π)/7641 weeks
7444.26153 3.83081 (744*2π)/7641 weeks
7454.3506 3.7818 (745*2π)/7641 weeks
7464.36839 3.46553 (746*2π)/7641 weeks
7475.01351 3.54636 (747*2π)/7641 weeks
7484.74602 3.21853 (748*2π)/7641 weeks
7494.57457 3.22451 (749*2π)/7641 weeks
7505.28262 2.48916 (750*2π)/7641 weeks
7515.44455 2.89089 (751*2π)/7641 weeks
7524.81135 2.09232 (752*2π)/7641 weeks
7535.67188 2.20163 (753*2π)/7641 weeks
7545.63717 1.7742 (754*2π)/7641 weeks
7555.74113 1.15734 (755*2π)/7641 weeks
7566.44365 1.74903 (756*2π)/7641 weeks
7575.82248 1.67877 (757*2π)/7641 weeks
7585.66148 .69101 (758*2π)/7641 weeks
7595.93431 -.30863 (759*2π)/7641 weeks
7606.88712 .94946 (760*2π)/7641 weeks
7616.94325 -.83709 (761*2π)/7641 weeks
7625.83857 .52217 (762*2π)/7641 weeks | 14,789 | 30,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-22 | latest | en | 0.753329 |
http://quant.stackexchange.com/questions/tagged/options?page=14&sort=active&pagesize=15 | 1,467,405,555,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403825.35/warc/CC-MAIN-20160624155003-00018-ip-10-164-35-72.ec2.internal.warc.gz | 254,975,815 | 19,270 | # Tagged Questions
A contract that gives the owner the right, but not the obligation, to buy or sell a security at a fixed price in the future.
232 views
### What is more appropriate: the EMA of the option price or the EMA of the underlying?
I'm progressing, all too slowly, on a site that aims to show real-time numbers for options that are listed on the CBOE. Most of the instantaneous numbers are all set. Now I'm going to pay attention to ...
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### How should FX options be priced when a currency is artificially capped?
The question is inspired by yesterday's (06/09/11) historic announcement by the Swiss National Bank that it would impose a ceiling on the franc of 1.20 against the euro. I would like to know if there ...
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We know that $$C-P = PV(F_{0,T}-K)$$ When we create a synthetic forward, we buy call and sell a put at the same strike price $K$. When we buy the call why do we assume the premium is positive? When ...
1k views
### How should I estimate the implied volatility skew term when calculating the skew-adjusted delta?
I'm trying to come up with the implied volatility skew adjusted delta for SPY options. I'm working with the following formula: Skew Adjusted Delta = Black Scholes Delta + Vega * Vol Skew Slope. I ...
381 views
### How do you calculate the implied liquidity of an option?
How does one calculate the implied liquidity of a specific option contract given a set of vanilla puts and calls with various strikes and maturities on a single underlying?
5k views
### Why hold options when you can dynamically replicate their payoff?
When holding vanilla options, you can cancel out, theoretically, all risk with dynamic (delta) hedging. Then you earn the "risk free rate of return". Why would you make such a portfolio when you can ...
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### Can one use options on Treasury futures to hedge a portfolio?
Can one use options on Treasury bond futures to hedge a typical fixed income portfolio? If so, how can one estimate the duration for an option on a Treasury futures contract, and taking this a step ...
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### Is QuantLib more trouble than it's worth?
I'm just starting to work with QuantLib and wonder if I'm going down a very wrong path. I'm working on a site that presents the visitor with a table of streamed real-time options data, including ...
405 views | 519 | 2,342 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-26 | latest | en | 0.904617 |
https://www.mathlearnit.com/area-of-shapes.html | 1,563,608,168,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00549.warc.gz | 777,240,728 | 5,056 | # Area of Shapes 3D
Working out the area of shapes in solid 3D form, isn’t entirely different to working out the area of flat 2D shapes.
There’s just usually a bit more work to do, to get the whole surface area of a solid shape, as opposed to a flat shape. Although the notation is still the same, for example cm² or m², for squared centimetres and metres.
When dealing with the surface area of 3D shapes. It’s important not to confuse area with volume. As surface area is concerned with measuring over the outside of a shape, where as volume is measuring what can go inside a shape.
It would be like if you had a box with a gift inside, to give to a friend or relative.
The surface area is how much wrapping paper you would need to cover the outside surface of the box.
Where as the volume is how much you can fit inside the box.
So for the area of 3D shapes, the key fact to remember is that area is a measure of the outside, volume is a measure of the inside.
Below are a list of pages featuring examples and explanations on how to establish the surface area of some of the most common shapes seen not just in Math, but in everyday life in general.
## Area of Shapes 3D Pages
- Area of a Cuboid
A cuboid is one of the best shapes to provide a good introduction to 3D area.
- Area of a Cylinder
From food packaging to tall buildings, many objects that we see everyday are cylinder shaped.
- Area of a Sphere
A sphere is the 3D version of a circle, and like a circle, has its own surface area.
- Area of a Pyramid
Examples of how to establish the area of a right pyramid with a square or rectangular base.
- Surface Area, Right Regular Pyramid
How to work out the area of a right regular pyramid with a polygon base.
- Area of a Tetrahedron
A specific type of pyramid, where each side/face is a triangle.
› Area of 3D Shapes | 433 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-30 | latest | en | 0.952701 |
https://cboard.cprogramming.com/c-programming/18314-prolog-inquiry.html | 1,495,567,804,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607649.26/warc/CC-MAIN-20170523183134-20170523203134-00514.warc.gz | 744,304,978 | 12,024 | Hi all,
Hopefully you all didn't mind if i ask some prolog question at here. Cos i finds that this board is the place that i can find the experience and best programmer from others site.
Ya, important part.
I have a difficulties in converting the the decimals number to hexadecimals and then convert it back. I know that we need to convert it first to binary rite and the nchange it to hex. But ont thing is i don have any clue on what command i need to use to convert it to binary and then convert it to hex. I have crack my head and still does not found any solutions.
So, does anyone willing to help me pls. I am using the SWI-prolog 3.2.8
Thanks first.
2. And the Engrish of the day award goes to:
*snickers*
Seriously, if you're having trouble with it, post some code and im sure people will be able to point you in the right direction.
3. Hi there,
Thanks for giving me the green light to ask for help.
Currently i have think out an algorithm for the hexadecimals problem.
Which is:
-First i make a database like this for the helping to convert back the
hex(0,0).
hex(1,1).
hex(2,2).
hex(3,3).
hex(4,4).
hex(5,5).
hex(6,6).
hex(7,7).
hex(8,8).
hex(9,9).
hex(a,10).
hex(b,11).
hex(c,12).
hex(d,13).
hex(e,14).
hex(f,15).
-Then i put a function to read the input from the keyboard the
hexadecimals values thats need to change to decimals.
And then, i change the input to a list as below:
[2,0].
-Using the mathematicals calculation, suppose the '2' should multiply
with the '16 power of 1' and adding with the '0' multiply with the
'16 power of 0'. And then adding together will comes out the value of
'32' which is the decimals value.
-What i have currently now are like this. I want to check with you
and see if this algorithm works first before i start the coding part.
Please advice. BTW, i have check some information from the web that
there is a predicate for the swi-prolog '\x' which use for the | 516 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-22 | longest | en | 0.89455 |
http://raborak.com/index.php/lib/fermats-last-theorem-for-regular-primes | 1,632,530,044,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00048.warc.gz | 59,316,229 | 8,618 | # Fermat's Last Theorem for Regular Primes by Jao D.
By Jao D.
By Jao D.
Similar nonfiction_1 books
FUDGE Role-Playing Game
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"
Additional info for Fermat's Last Theorem for Regular Primes
Sample text
2 is just one formulation of Rouch e 's Theorem C. j j j on C implies that neither f (z ) nor f (z )+ g(z ) Note that the condition may have a zero on but it is sucient for our purposes. The next theorem is a simple application of Rouch e's Theorem. It is however most useful since it applies to polynomials. 6) and consider a circle k , of radius rk , centered at sk which is a root of P (s) of multiplicity tk . Let rk be xed in such a way that, 0 < rk < min sk sj ; for j = 1; 2; ; k 1; k + 1; ; m: (1:7) Then, there exists a positive number , such that i ; for i = 0; 1; ; n, implies that Q(s) has precisely tk zeros inside the circle k .
It turns out that this result on exposed edges also follows from a more powerful result, namely the Edge Theorem, which is established in Chapter 6. Here we show that this stability testing property of the exposed edges carries over to complex polynomials as well as to quasipolynomials which arise in control systems containing time-delay. A computationally ecient solution to testing the stability of general polytopic families is given by the Bounded Phase Lemma, which reduces the problem to checking the maximal phase di erence over the vertex set, evaluated along the boundary of the stability region.
In [52] Brasch and Pearson showed that arbitrary pole placement could be achieved in the closed loop system by a controller of order no higher than the controllability index or the observability index. In the late 1960's and early 1970's the interest of control theorists turned to the servomechanism problem. Tracking and disturbance rejection problems with persistent signals such as steps, ramps and sinusoids could not be solved in an obvious way by the existing methods of optimal control. The reason is that unless the proper signals are included in the performance index the cost function usually turns out to be unbounded. | 539 | 2,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.918129 |
https://en.wikipedia.org/wiki/Friis_formulas_for_noise | 1,726,882,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00490.warc.gz | 198,589,197 | 22,502 | # Friis formulas for noise
Friis formula or Friis's formula (sometimes Friis' formula), named after Danish-American electrical engineer Harald T. Friis, is either of two formulas used in telecommunications engineering to calculate the signal-to-noise ratio of a multistage amplifier. One relates to noise factor while the other relates to noise temperature.
## The Friis formula for noise factor
Friis's formula is used to calculate the total noise factor of a cascade of stages, each with its own noise factor and power gain (assuming that the impedances are matched at each stage). The total noise factor can then be used to calculate the total noise figure. The total noise factor is given as
${\displaystyle F_{\text{total}}=F_{1}+{\frac {F_{2}-1}{G_{1}}}+{\frac {F_{3}-1}{G_{1}G_{2}}}+{\frac {F_{4}-1}{G_{1}G_{2}G_{3}}}+\cdots +{\frac {F_{n}-1}{G_{1}G_{2}\cdots G_{n-1}}}}$
where ${\displaystyle F_{i}}$ and ${\displaystyle G_{i}}$ are the noise factor and available power gain, respectively, of the i-th stage, and n is the number of stages. Both magnitudes are expressed as ratios, not in decibels.
### Consequences
An important consequence of this formula is that the overall noise figure of a radio receiver is primarily established by the noise figure of its first amplifying stage. Subsequent stages have a diminishing effect on signal-to-noise ratio. For this reason, the first stage amplifier in a receiver is often called the low-noise amplifier (LNA). The overall receiver noise "factor" is then
${\displaystyle F_{\mathrm {receiver} }=F_{\mathrm {LNA} }+{\frac {F_{\mathrm {rest} }-1}{G_{\mathrm {LNA} }}}}$
where ${\displaystyle F_{\mathrm {rest} }}$ is the overall noise factor of the subsequent stages. According to the equation, the overall noise factor, ${\displaystyle F_{\mathrm {receiver} }}$, is dominated by the noise factor of the LNA, ${\displaystyle F_{\mathrm {LNA} }}$, if the gain is sufficiently high. The resultant Noise Figure expressed in dB is:
${\displaystyle \mathrm {NF} _{\mathrm {receiver} }=10\log(F_{\mathrm {receiver} })}$
### Derivation
For a derivation of Friis' formula for the case of three cascaded amplifiers (${\displaystyle n=3}$) consider the image below.
A source outputs a signal of power ${\displaystyle S_{i}}$ and noise of power ${\displaystyle N_{i}}$. Therefore the SNR at the input of the receiver chain is ${\displaystyle {\text{SNR}}_{i}=S_{i}/N_{i}}$. The signal of power ${\displaystyle S_{i}}$ gets amplified by all three amplifiers. Thus the signal power at the output of the third amplifier is ${\displaystyle S_{o}=S_{i}\cdot G_{1}G_{2}G_{3}}$. The noise power at the output of the amplifier chain consists of four parts:
• The amplified noise of the source (${\displaystyle N_{i}\cdot G_{1}G_{2}G_{3}}$)
• The output referred noise of the first amplifier ${\displaystyle N_{a1}}$ amplified by the second and third amplifier (${\displaystyle N_{a1}\cdot G_{2}G_{3}}$)
• The output referred noise of the second amplifier ${\displaystyle N_{a2}}$ amplified by the third amplifier (${\displaystyle N_{a2}\cdot G_{3}}$)
• The output referred noise of the third amplifier ${\displaystyle N_{a3}}$
Therefore the total noise power at the output of the amplifier chain equals
${\displaystyle N_{o}=N_{i}G_{1}G_{2}G_{3}+N_{a1}G_{2}G_{3}+N_{a2}G_{3}+N_{a3}}$
and the SNR at the output of the amplifier chain equals
${\displaystyle {\text{SNR}}_{o}={\frac {S_{i}G_{1}G_{2}G_{3}}{N_{i}G_{1}G_{2}G_{3}+N_{a1}G_{2}G_{3}+N_{a2}G_{3}+N_{a3}}}}$.
The total noise factor may now be calculated as quotient of the input and output SNR:
${\displaystyle F_{\text{total}}={\frac {{\text{SNR}}_{i}}{{\text{SNR}}_{o}}}={\frac {\frac {S_{i}}{N_{i}}}{\frac {S_{i}G_{1}G_{2}G_{3}}{N_{i}G_{1}G_{2}G_{3}+N_{a1}G_{2}G_{3}+N_{a2}G_{3}+N_{a3}}}}=1+{\frac {N_{a1}}{N_{i}G_{1}}}+{\frac {N_{a2}}{N_{i}G_{1}G_{2}}}+{\frac {N_{a3}}{N_{i}G_{1}G_{2}G_{3}}}}$
Using the definitions of the noise factors of the amplifiers we get the final result:
${\displaystyle F_{\text{total}}=\underbrace {1+{\frac {N_{a1}}{N_{i}G_{1}}}} _{=F_{1}}+\underbrace {\frac {N_{a2}}{N_{i}G_{1}G_{2}}} _{={\frac {F_{2}-1}{G_{1}}}}+\underbrace {\frac {N_{a3}}{N_{i}G_{1}G_{2}G_{3}}} _{={\frac {F_{3}-1}{G_{1}G_{2}}}}=F_{1}+{\frac {F_{2}-1}{G_{1}}}+{\frac {F_{3}-1}{G_{1}G_{2}}}}$.
General derivation for a cascade of ${\displaystyle n}$ amplifiers:
The total noise figure is given as the relation of the signal-to-noise ratio at the cascade input ${\displaystyle \mathrm {SNR_{i}} ={\frac {S_{\mathrm {i} }}{N_{\mathrm {i} }}}}$ to the signal-to-noise ratio at the cascade output ${\displaystyle \mathrm {SNR_{o}} ={\frac {S_{\mathrm {o} }}{N_{\mathrm {o} }}}}$ as
${\displaystyle F_{\mathrm {total} }={\frac {\mathrm {SNR_{i}} }{\mathrm {SNR_{o}} }}={\frac {S_{\mathrm {i} }}{S_{\mathrm {o} }}}{\frac {N_{\mathrm {o} }}{N_{\mathrm {i} }}}}$.
The total input power of the ${\displaystyle k}$-th amplifier in the cascade (noise and signal) is ${\displaystyle S_{k-1}+N_{k-1}}$. It is amplified according to the amplifier's power gain ${\displaystyle G_{k}}$. Additionally, the amplifier adds noise with power ${\displaystyle N_{\mathrm {a} ,k}}$. Thus the output power of the ${\displaystyle k}$-th amplifier is ${\displaystyle G_{k}\left(S_{k-1}+N_{k-1}\right)+N_{\mathrm {a} ,k}}$. For the entire cascade, one obtains the total output power
${\displaystyle S_{\mathrm {o} }+N_{\mathrm {o} }=\left(\left(\left(\left(S_{\mathrm {i} }+N_{\mathrm {i} }\right)G_{1}+N_{\mathrm {a} ,1}\right)G_{2}+N_{\mathrm {a} ,2}\right)G_{3}+N_{\mathrm {a} ,3}\right)G_{4}+\dots }$
The output signal power thus rewrites as
${\displaystyle S_{\mathrm {o} }=S_{\mathrm {i} }\prod _{k=1}^{n}G_{k}}$
whereas the output noise power can be written as
${\displaystyle N_{\mathrm {o} }=N_{\mathrm {i} }\prod _{k=1}^{n}G_{k}+\sum _{k=1}^{n-1}N_{\mathrm {a} ,k}\prod _{l=k+1}^{n}{G_{l}}+N_{\mathrm {a} ,n}}$
Substituting these results into the total noise figure leads to
${\displaystyle F_{\mathrm {total} }={\frac {N_{\mathrm {i} }\prod _{k=1}^{n}G_{k}+\sum _{k=1}^{n-1}N_{\mathrm {a} ,k}\prod _{l=k+1}^{n}{G_{l}}+N_{\mathrm {a} ,n}}{N_{\mathrm {i} }\prod _{k=1}^{n}G_{k}}}=1+\sum _{k=1}^{n-1}{\frac {N_{\mathrm {a} ,k}\prod _{l=k+1}^{n}{G_{l}}}{N_{\mathrm {i} }\prod _{m=1}^{n}G_{m}}}+{\frac {N_{\mathrm {a} ,n}}{N_{\mathrm {i} }\prod _{k=1}^{n}G_{k}}}=1+\sum _{k=1}^{n-1}{\frac {N_{\mathrm {a} ,k}}{N_{\mathrm {i} }\prod _{m=1}^{k}G_{m}}}+{\frac {N_{\mathrm {a} ,n}}{N_{\mathrm {i} }\prod _{k=1}^{n}G_{k}}}}$
${\displaystyle =1+{\frac {N_{\mathrm {a} ,1}}{N_{\mathrm {i} }G_{1}}}+\sum _{k=2}^{n-1}{\frac {N_{\mathrm {a} ,k}}{N_{\mathrm {i} }\prod _{m=1}^{k}G_{m}}}+{\frac {N_{\mathrm {a} ,n}}{N_{\mathrm {i} }\prod _{k=1}^{n}G_{k}}}}$
Now, using ${\displaystyle F_{k}=1+{\frac {N_{\mathrm {a} ,k}}{N_{\mathrm {i} }G_{k}}}}$ as the noise figure of the individual ${\displaystyle k}$-th amplifier, one obtains
${\displaystyle F_{\mathrm {total} }=F_{1}+\sum _{k=2}^{n}{\frac {F_{k}-1}{\prod _{l=1}^{k-1}G_{l}}}}$
${\displaystyle =F_{1}+{\frac {F_{2}-1}{G_{1}}}+{\frac {F_{3}-1}{G_{1}G_{2}}}+{\frac {F_{4}-1}{G_{1}G_{2}G_{3}}}+\dots +{\frac {F_{n}-1}{G_{1}G_{2}\dots G_{n-1}}}}$
## The Friis formula for noise temperature
Friis's formula can be equivalently expressed in terms of noise temperature:
${\displaystyle T_{\text{eq}}=T_{1}+{\frac {T_{2}}{G_{1}}}+{\frac {T_{3}}{G_{1}G_{2}}}+\cdots }$
## Published references
• J.D. Kraus, Radio Astronomy, McGraw-Hill, 1966. | 2,724 | 7,490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 44, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-38 | latest | en | 0.86311 |
https://sorumatik.co/t/test-2-soru8/85416 | 1,726,687,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00495.warc.gz | 489,841,922 | 5,665 | # Test 2 soru8
x, y ve z pozitif tam sayılardır
X×Y=22
Y×Z=26
Olduğna göre x+y+z toplamı kaçtır
@sorumatikbot
## karacakader said x, y ve z pozitif tam sayılardır, X×Y=22 ve Y×Z=26 olduğuna göre x+y+z toplamı kaçtır?
To solve this problem, we need to determine the values of ( x, y, ) and ( z ) such that ( x \times y = 22 ) and ( y \times z = 26 ) with ( x, y, ) and ( z ) being positive integers. Let’s follow these steps:
1. Analyze the Equations:
• Equation 1: ( x \times y = 22 )
• Equation 2: ( y \times z = 26 )
2. Prime Factorization:
• Factorize 22: ( 22 = 2 \times 11 )
• Factorize 26: ( 26 = 2 \times 13 )
3. Identify Possible Values for ( y ):
• ( y ) must be a common factor of both 22 and 26.
• The common factors are 1 and 2, considering ( y ) must be a common divisor of their prime factors.
4. Determine ( (x, y, z) ) for Each Possible ( y ):
• Case 1: ( y = 1 )
• If ( y = 1 ), then from ( x \times 1 = 22 ), we get ( x = 22 ).
• From ( 1 \times z = 26 ), we get ( z = 26 ).
• This solution yields ( x + y + z = 22 + 1 + 26 = 49 ).
• Case 2: ( y = 2 )
• If ( y = 2 ), then from ( x \times 2 = 22 ), we get ( x = 11 ).
• From ( 2 \times z = 26 ), we get ( z = 13 ).
• This solution yields ( x + y + z = 11 + 2 + 13 = 26 ).
Examining our cases, we see that ( y = 2 ) provides a consistent set of positive integers for ( x, y, ) and ( z ), whereas ( y = 1 ) does not satisfy the integer constraint for ( z ). | 510 | 1,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.618794 |
https://www.scribd.com/document/305542556/CH03 | 1,560,885,080,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998813.71/warc/CC-MAIN-20190618183446-20190618205446-00324.warc.gz | 917,338,128 | 94,115 | You are on page 1of 32
# CHAPTER
CHARACTERISTICS
OF AERIAL LINES
Original Authors:
Sherwin
H. Wright
D. F. Shankle
and C. F. Hall
and R. L. Tremaine
## N the design, operation,
and expansion
of electrical
power systems it is necessary to know electrical and
physical characteristics
of conductors used in the construction
of aerial distribution
and transmission
lines.
This chapter presents a description of the common types
of conductors
along with tabulations
of their important
electrical and physical characteristics.
General formulas
are presented
with their derivation
to show the basis of the
tabulated
values and as a guide in calculating
data for
other conductors
of similar shapes, dimensions,
composition and operating conditions.
Also included are the more commonly used symmetricalcomponent-sequence
impedance equations that are applicable to the solution of power system problems involving
voltage regulation,
system currents,
and voltages under fault conditions,
or other system problems where the electrical characteristics
of aerial lines are
involved.
formulas are given to permit calculation
of
approximate
current-carrying
capacity
of conductors
taking into account such factors as convection
by ambient, temperature,
wind
velocity, and permissible temperature
rise.
I.
TYPES
## Fig. 1A typical stranded
conductor,
(bare copper).
OF CONDUCTORS
Fig. 2A typical ACSR conductor.
In the electric-power
field the following types of conductors are generally used for high-voltage
power transmission lines: stranded copper conductors,
hollow copper conductors, and ACSR (aluminum
cable, steel reinforced).
Other types of conductors such as Copperweld and Copperweld-Copper
conductors
are also used for transmission
and distribution
lines. Use is made of Copperweld, bronze,
copper bronze, and steel for current-carrying
conductors
on rural lines, as overhead ground wires for transmission
lines, as buried counterpoises
at the base of transmission
towers, and also for long river crossings.
A stranded conductor,
typical of both copper and stee1
conductors in the larger sizes, is shown in Fig. 1. A stranded conductor is easier to handle and is more flexible than a,
solid conductor,
particularly
in the larger sizes.
A typical ACSR conductor
is illustrated
in Fig. 2. In
this type of conductor, aluminum strands are wound about,
a core of stranded
steel. Varying relationships
between
tensile strength
and current-carrying
capacity as well as
overall size of conductor can be obtained by varying the
proportions
of steel and aluminum.
By the use of a filler,
such as paper, between the outer aluminum
strands and
the inner steel strands, a conductor of large diameter can
be obtained for use in high voltage lines. This type of con32
ACSR conductor.
## ductor is known as expanded
ACSR and is shown in
Fig. 3.
In Fig. 4 is shown a representative
Anaconda
Hollow
Copper Conductor.
It consists of a twisted copper I
Chapter
33
## ors as shown in Fig. 7. Different relationships
between
current-carrying
capacity,
outside diameter,
and tensile
strength can be obtained by varying the number and size
of the Copperweld and copper strands.
II. ELECTRICAL
AERIAL
Fig. S-A
Fig. 6-A
typical General
## Cable Type HH.
typical Copperweld
conductor.
CHARACTERISTICS
CONDUCTORS
OF
## The following discussion is primarily concerned with the
development,
of electrical characteristics
and constants
of
aerial conductors,
particularly
those required for analysis
of power-system
problems.
The constants
developed are
particulary
useful in the application
of the principles of
symmetrical
components
to the solution of power-system
problems involving positive-, negative-, and zero-sequence
impedances
of transmission
and distribution
lines. The
basic quantities
needed are the positive-, negative-,
and
zero-sequence
resistances,
inductive
reactances and shunt
capacitive
reactances
of the various types of conductors
and some general equations showing how these quantities
are used.
1. Positive-
and Negative-Sequence
Resistance
## The resistance of an aerial conductor is affected by the
three factors:
temperature,
frequency,
current
density.
Practical formulas and methods will now be given to take
into account these factors.
Temperature
Effect on ResistanceThe
resistance
of copper and aluminum
conductors
varies almost directly with temperature.
While this variation is not strictly
linear for an extremely
wide range of temperatures,
for
practical
purposes it can be considered
linear over the
range of tempertures
normally encountered.
When the d-c resistance of a conductor at a given temperature is known and it is desired to find the d-c resistance
at some other temperature,
the following general formula
may be used.
Fig. 7Typical
Copperweld-Copper
(a) Upper
(b) Lower
photographType
photographType
conductors
V
F
## beam as a core about which strands of copper wire are
wound. The I beam is twisted in a direction opposite to
that of the inner layer of strands.
Another form of hollow copper conductor
is shown in
Fig. 5. Known as the General Cable Type HH hollow copper conductor, it is made up of segmental section
of copper mortised
into each other to form a self-supporting
hollow cylinder.
Hollow copper conductors
result in conductors of large diameter for a given cross section of copper.
Corona losses are therefore smaller. This construction
also
produces a reduction in skin effect as well as inductance
as
compared with stranded conductors. A discussion
of large
diameter conductors
and their characteristics
is given in
reference 1.
Copperweld
conductors
consist of different numbers of
copper-coated
steel strands, a typical conductor being illustrated in Fig. 6. Strength is provided
by thecore of steel
and protection by the outer coating of copper.
When high current-carrying
capacities
are desired as
well as high tensile strength, copper stands
are used with
Copperweld strands to form Copperweld
Coppcr conduct-
## Rt2 =d-c resistance at any temperature
t2 degree C.
Rtl =d-c resistance at any other temperature
t1 degree C.
M =a constant for any one type of conductor material.
=inferred
absolute zero temperature.
=234.5 for annealed 100 percent conductivity
copper.
=241.5 for hard drawn 97.3 percent conductivity
copper.
=228.1 for aluminum.
The above formula is useful for evaluating
changes in
d-c resistance only, and cannot be used to give a-c resistance variations
unless skin effect can be neglected.
For
small conductor sizes the frequency has a negligible effect
on resistance in the d-c to GO-cycle range. This is generally
true for conductor sizes up to 2/0.
The variations of resistance with temperature
are usually
unimportant
because the actual ambient
temperature
is
indefinite as well as variable along a transmission
line. An
illustration
of percentage change in resistance is when temperature varies from winter to summer over a range of
0 degree C to 40 degrees C (32 degrees F to 104 degrees F)
in which case copper resistance increases 17 percent.
34
Chapter
## Skin Effect in Straight Round Wires- The resistance of non-magnetic
conductors varies not only with temperature but also with frequency.
This is due to skin effect.
Skin effect is due to the current flowing nearer the outer
surface of the conductor
as a result of non-uniform
flux
distribution
in the conductor.
This increases the resistance
of the conductor by reducing the effective cross section of
the conductor through which the current flows.
The conductor tables give the resistance at commercial
frequencies of 25, 50, and GO cycles. For other frequencies
the following formula should be used.
## Table 5 (skin effect table) is carried in the Bureau of
Standards
Bulletin No. 169 on pages 226-8, to values of
X = 100. To facilitate interpolation
over a small range of
the table, it is accurate as well as convenient
to plot a curve
of the values of K vs. values of X.
Combined
on Resistance
## Skin Effect and Temperature
Effect
of Straight Round WiresWhen
both
temperature
and skin effect are considered in determining
conductor resistance,
the following procedure is followed.
First calculate the d-c resistance at the new temperature
using Eq. (1). Then substitute
this new value of d-c resistance and the desired frequency in the equation defining
X. Having calculated X, determine K from Table 5. Then
using Eq. (2), calculate the new a-c resistance rf, using the
new d-c resistance for rdc and the value of K obtained from
Table 5.
Effect of Current on ResistanceThe
resistance of
magnetic conductors varies with current magnitude
as well
as with the factors that affect non-magnetic
conductors
(temperature
and frequency).
Current magnitude
determines the flux and therefore the
iron or magnetic losses inside magnetic conductors.
The
factor complicates the determination of resistance of magnetic conductors as well as any
tabulation
of such data.
For these reasons the effect of
current magnitude
will not be analyzed in detail.
However, Fig. 8 gives the resistance
of steel conductors
as a
function of current, and the tables on magnetic conductors
such as Copperweld-copper,
Copperweld,
and ACISR conductors include resistance tabulations
at two current carrying levels to show this effect. These tabulated
resistances
are generally values obtained by tests.
Zero-Sequence
ResistanceThe
zero-sequence
resistance of aerial conductors
is discussed in detail in the
section on zero-sequence
resistance
and inductive
reactance given later in the chapter since the resistance and in-
Fig. 8Electrical
Characteristics
ductive reactance
presented
influenced by the distribution
in the earth return path.
2.
Positiveactance
of Steel Ground
Wires*
to zero-sequence
currents is
of the zero-sequence
current
and Negative-Sequence
Inductive
Re-
## To develop the positive- and negative-sequence
inductive reactance of three-phase aerial lines it is first necessary
to develop
a few concepts
that greatly
simplify
the
problem.
First, the total inductive reactance of a conductor carrying current will be considered
as the sum of two components:
*This figure has been taken from Symmetrical
Components
(a book)
by C. F. Wagner and R. D. Evans, McGraw-Hill
Book Company,
1933.
Chapter
35
## The inductive reactance due to the flux within a radius of
one foot from the conductor
center, including the flux
inside the conductor.
(2) The inductive
reactance
due to the flux external to a
radius of one foot and out to some finite distance.
(1)
## This concept was first given in Wagner and Evans book on
Symmetrical
Components2
and was suggested by W. A.
Lewis.48
It can be shown most easily by considering
a two-conductor single-phase
circuit with the current flowing out in
one conductor and returning in the other. In Fig. 9 such a circuit is shown with only the flux produced by conductor 1 for
simplicity.
Conductor 2 also produces similar lines of flux.
The classic inductance formula for a single round straight
wire in the two-conductor
single-phase
circuit is:
## Fig. 10Inductance due to flux between radius a and radius b
(2
lnabhenries
b/a
per cm.)
D12=distance
between conductor
1 and conductor 2.
D12 and r must be expressed in the same units for the above
For practical purposes one foot is
equation to be valid.
used as the unit of length since most distances between
aerial conductors
are in feet. In cable circuits, however,
the distance between conductors
is less than one foot and
the inch is a more common unit (see Chap. 4).
From
derivation
formulas
a general
## term such as 21n
represents
the flux and associated
inductance
between
a conductor
carrying current.
(See Fig. 10).
Rewriting Eq. (4) keeping in mind the significance of the
general
= inductance
=
## inductance due to the flux outside the conductor
to a radius of one foot.
= inductance
due to the flux external to a one foot
radius out to D12 feet where D12 is the distance
between conductor
1 and conductor 2.
## From Fig. 9 it can be seen that it is unnecessary
to include the flux beyond the return conductor 2 because this
flux does not link any net current and therefore does not
affect the inductance
of conductor
1.
Grouping the terms in Eq. (5) we have:
term 21n t,
Examining
the terms in the first bracket, it is evident
that this expression is the sum of the flux both inside the
single
phase
circuit
(inductance)
## contains terms that are strictly a function of the conductor
characteristics
of permeability
The term in the second bracket of Eq. (6) is an expression for inductance
due to flux external to a radius of one
foot and out to a distance of D12, which, in the two-conductor case, is the distance between conductor 1 and conductor 2. This term is not dependent
upon the conductor
characteristics
and is dependent
only upon conductor
spacing.
Equation
(6) can be written again as follows:
36
Chapter
## GMR in the first term is the conductor
geometric mean
It can be defined as the radius of a tubular
conductor with an infinitesimally
thin wall that has the same
external flux out to a radius of one foot as the internal and
external flux of solid conductor
1, out to a radius of one
foot.
In other words, GMR is a mathematical
assigned to a solid conductor
(or other configuration
such
as stranded conductors),
which describes in one term the
inductance
of the conductor
due to both its internal flux
pendent
upon the
expressed in feet.
Converting
Eq.
reactance,
condu&or
(7)
to
characteristics.
practical
## ohms per conductor
units
GMR
is
of inductive
per mile
(8)
where j--frequency
in cps.
GMR = conductor geometric mean radius in feet.
D12=distance
between conductors
1 and 2 in feet.
If we let the first term be called xa and the second
xd, then
z=&+xd
ohms per conductor per mile
term
(9)
where
za= inductive
reactance
due to both the internal
flux and that external to conductor 1 to a radius
of one foot.
xd = inductive reactance due to the flux surrounding
conductor
1 from a radius of one foot out to a
For the two-conductor,
total inductive
reactance
x = 2(x,+xd)
single-phase
circuit,
then,
the
Fig. 11Geometric
is
## ohms per mile of circuit
(10)
since the circuit has two conductors,
or both a (go and
return conductor.
Sometimes a tabulated
or experimental
reactance with
1 foot spacing is known, and from this it is desired to calculate the conductor
GMR.
By derivation
from, Eq. (8)
## When reactance is known not to a one-foot radius but out
to the conductor
surface, it is called the internal
reactance.
The formula for calculating
the GMR from the
internal reactance is:
physical
GMR=
Antiloglo
Internal
Reactance
0.2794
(60 cycles)
feet
(12)
## The values of GMR at GO cycles and xB at 25, 50, and
60 cycles for each type of conductor are given in the tables
of electrical characteristics
of conductors.
They are given
## Fig. 12A Three-conductor
three-phase
spacing).
circuit (symmetrical
Chapter
## in these tables because they are a function
of conductor
characteristics
Values of xa
for various spacings are given in separate tables in this
Chapter
for 25, 50, and 60 cycles.
This factor is dependent on distance between conductors
only, and is not
associated with the conductor characteristics
in any way.
In addition to the GMR given in the conductor
characteristics tables, it is sometimes necessary to determine
this quantity
for other conductor
configurations.
Figure
11 is given for convenience
in determining
such values of
GMR.
This table is taken from the Wagner and Evans
book Symmetrical Components, page 138.
Having developed xa and xd in terms of a two-conductor,
single-phase
circuit, these quantities
can be used to deand negative-sequence
inductive
termine
the positivereactance of a three-conductor,
three-phase
circuit.
Figure 12 shows a three-conductor,
three-phase
circuit
by line to
carrying
phase currents
Is, Ib, I, produced
ground voltages Ea, Eb, and Ec. First, consider the case
where the three conductors are symmetrically
spaced in a
triangular
configuration
so that no transpositions
are required to maintain equal voltage drops in each phase along
the line. Assume that the three-phase
voltages Ea, Eb, E,
are balanced (equal in magnitude
and 120 apart) so that
they may be either positive- or negative-sequence
voltages. Also assume the currents Ia, Ib, I, are also balanced
so that I,+Ib+l,=O.
Therefore no return current flows
in the earth, which practically
eliminates
mutual effects
between the conductors
and earth, and the currents I,,
Ib, I, can be considered as positive- or negative-sequence
currents.
In the following solution, positive- or negativesequence voltages E,, Eb, E,, are applied to the conductors
and corresponding
positive- or negative-sequence
currents
are assumed to flow producing voltage drops in each conductor.
The voltage drop per phase, divided by the current per phase results in the positive- or negative-sequence
inductive
reactance per phase for the three-phase
circuit.
To simplify the problem further, consider only one current
flowing at a time. With all three currents flowing simultaneously, the resultant effect is the sum of the effects produced by each current flowing alone.
Taking phase a, the voltage drop is:
Ea - Ea = Iaxaa+ Ibxab + Icxao
(13)
where
xaa = self inductive
reactance of conductor a.
xab = mutual
inductive
reactance
between conductor a, and conductor b.
xac = mutual inductive
reactance
between conductor a and conductor c.
In terms
reactance
Xaa = xa+Xd(ak)
spacing
factor,
(14)
## where only Ia is flowing and returning by a remote path e
feet away, assumed to be the point k.
Considering only Ib flowing in conductor b and returning
by the same remote path f feet away,
Xab =
xd(bk)
-xd(ba)
(15)
## where xab is the inductive
reactance associated with the
flux produced by rb that links conductor a out to the return
path f feet away.
Finally,
returning
37
considering
only I, flowing in conductor
by the same remote path g feet away.
X ac =
Xd(ck)
c and
(16)
-xd(ca)
## where xac is the inductive
reactance associated with the
flux produced by I, that links conductor a out to the return
path 9 feet away.
With all three currents I,, 1h, I, flowing simultaneously,
we have in terms of xa and xd factors:
E,-E,=ja(xa+~d(nk))+Ib(xd(bk)-xd(ba))
+Ic(xd(ck)
Expanding
-xd(ca)).
and regrouping
(17)
## the terms we have:
Ea-E,=/.x,-Itxd(ba)-Icxd(ca)
+v
Since
written
I,=
-I,-Ib,
I
Using
axd(:&k)+Ibxd(bk)
the terms
a(Xd(ak) -xd(ck)
the definition
+hd(ck)].
(18)
in the bracket
> +Ib(xd(bk)
-xd(ck)
.f
a2
of x d, 0.2794 -log--,
60
1
may
be
>*
this expression
can be written
f
I;, 0.2794% log :Assuming
path
the distances
approach
d(W .
0.2794@f log d
(ck)
dc3kj, dcckj, and d(bk) to the remote
(ck)
infinity,
then
the
ratios
d (ck)
and
y
(ck)
approach unity.
Since the log of unity is zero, the two
terms in the bracket are zero, and Eq. (18) reduces to
Ea-Ea
=IaXa-Ibxd(ba)
-IcXd(ca)
(19)
since
xd(ba)
=Zd(ca)
=xd(h~)
=xd,
Ea-Ea=Ia(~a+~d).
and
Ia=
-Ib-lo,
(20)
Dividing
## the equation by Ia,
E,- E,
x1=x2 = -1= xa+xd ohms per phase per mile
(21)
a
where
xa= inductive
reactance for conductor
a due to the flux
out to one foot.
Xd=inductive
reactance
corresponding
to the flux external to a one-foot radius from conductor
a out to
the center of conductor
b or conductor
c since the
spacing between conductors is symmetrical.
Therefore, the positive- or negative-sequence
inductive
reactance
per phase for a three-phase
circuit with equilateral spacing is the same as for one conductor of a singlephase circuit as previously derived.
Values of xa for various conductors
are given in the tables of electrical characteristics
of conductors
later in the chapter,
and the
values of xd are given in the tables of inductive reactance
spacing factors for various conductor spacings.
When the conductors
are unsymmetrically
spaced, the
voltage drop for each conductor is different, assuming the
currents to be equal and balanced.
Also, due to the unsymmetrical conductor spacing, the magnetic field external to
the conductors
is not zero, thereby causing induced voltages in adjacent electrical circuits, particularly
telephone
circuits, that may result in telephone interference.
To reduce this effect to a minimum,
the conductors are
transposed so that each conductor occupies successively the
## Characteristics of Aerial Lines
38
Expressed
Chapter
in general
terms,
(log d(12)+log d(s) +log d(u))
2d = + 0.2794G
xd=o.2794%
xd=o.2794%
Fig.
13A Three-conductor
three-phase
rical spacing).
circuit (unsymmet-
## same positions as the other two conductors in two successive
line sections. For three such transposed line sections, called
the total voltage drop for each
a barrel of transposition,
conductor is the same, and any electrical circuit parallel to
the three transposed sections has a net voltage of very low
magnitude
induced in it due to normal line currents.
In the following derivation
use is made of the general
equations developed for the case of symmetrically
spaced
conductors.
First, the inductive
reactance voltage drop of
phase a in each of the three line sections is obtained.
by three gives the
average inductive reactance voltage drop for a line section.
Referring to Fig. 13 and using Eq. (19) for the first line
section where I, is flowing in conductor
1,
E,-E,,
Iah-
Ibxd(l2)
Taking
log +d,,d,,d,l
log GMD
## where GMD (geometrical mean distance) = qd12d23d31, and
is mathematically
defined as the nth root of an n-fold
product.
For a three-phase
circuit where the conductors
are not
symmetrically
spaced, we therefore have an expression for
or negative-sequence
inductive
reactance,
the positivewhich is similar to the symmetrically
spaced case except xd
is the inductive-reactance
spacing factor for the GMD
(geometric mean distance) of the three conductor separations. For xd, then, in the case of unsymmetrical
conductor spacing, we can take the average of the three inductivereactance spacing factors
xd
+(xd(12)
or we can calculate
+xd(23)+xd(31))
the GMD
ohms
per
phase
per
mile
(23)
## and use the inductive-reactance
spacing factor for this
This latter procedure is perhaps the easier of the
distance.
two methods.
x8 is taken from the tables of electrical characteristics
of
conductors
presented later in the chapter, and xd is taken
2,
=IaXa-Ibxd(23)-Icxd(21).
E, -
.f
log dndmd31
Icxd(l3).
## In the second line section where I, is flowing in conductor
EL-E,
3,
E,=Iax,-Ibxd(3l)-Icxd(32).
the average
E
voltage
## drop per line section,
we have
_(Ea-E,I)+(Ea-E,)+(E,-EBI)
avg
C-
3
31aXa
_ 1
3
Ic(xd(12)
xd (12) +
xd (23) +
b(
xd (31))
+xd(23)
+xd(31))
3
E aw
(xd(12)
=Iaxa-
+xd(23)
(Ib+Ic)
+xd(31))
Since
Ia=
Xd(12)
Eavg=Ia(xa+
Dividing
inductive
(Ib+Ic)
+xd(23)
+xd(31)
-------I*
## by Ia, we have the positivereactance per phase
x1 = 52 = (xa+xd)
or negative-sequence
## ohms per phase per mile
where
xd=\$( xd(12) +xd(23)
_
per mile.
+xd(31)
## ) ohms per phase
(22)
Fig. 14Quick reference curves for 60-cycle inductive reactance of three-phase lines (per phase) using hard drawn copper
conductors. For total reactance of single-phase lines multiply
these values by two. See Eqs.
and (21).
## Fig. 15Quick reference
ance of three-phase lines
For total reactance of
values by two.
## curves for 60-cycle inductive react(per phase) using ACSR conductors.
single-phase
lines, multiply
these
See Eqs. (10) and (21).
Fig. 17Quick reference curves for 60-cycle inductive reactance of three-phase lines (per phase) using Copperweld conductors. For total reactance of single-phase
lines multiply
these values by two. See Eqs. (10) and (21).
## Fig. 16-Quick reference curves for 60-cycle inductive reactance of three-phase
lines (per phase) using CopperweldCopper conductors. For total reactance of single-phase lines
multiply these values by two. See Eqs. (10) and (21).
## from the tables of inductive-reactance
spacing factors.
Geometric mean distance (GMD) is sometimes referred to
as equivalent
conductor
spacing.
For quick reference
the curves of Figs. (14), (15), (16), and (17) have been
plotted giving the reactance
(z,+z)
for different conductor sizes and equivalent
conductor
spacings.
Since most three-phase lines or circuits do not have conductors symmetrically
spaced, the above formula for positive- or negative-sequence
inductive reactance is generally
used. This formula, however, assumes that the circuit is
transposed.
When a single-circuit
line or double-circuit
line is not
transposed,
either the dissymmetry
is to be ignored in the
calculations,
in which case the general symmetrical
components methods can be used, or dissymmetry
is to be considered, thus preventing
the use of general symmetricalIn considering
this dissymmetry,
components
methods.
unequal currents and voltages are calculated for the three
phases even when terminal
conditions
are balanced.
In
most cases of dissymmetry
it is most practical to treat the
circuit as transposed
and use the equations for x1 and x1
derived for an unsymmetrically-spaced
transposed
circuit.
Some error results from this method but in general it is
small as compared with the laborious
calculations
that
must be made when the method of symmetrical
components cannot be used.
Characteristics
40
Positiveand Negative-Sequence
Parallel Circuits-When
two parallel
Reactance
of Aerial Lines
Chapter
of
three-phase
circuits are close together, particularly
on the same tower, the
effect of mutual inductance
between the two circuits is
not entirely eliminated
by transpositions.
By referring to
Fig. 18 showing two transposed
circuits on a single tower,
the positive- or negative-sequence
reactance of the paralleled circuit is:
Fig. 19Arrangement
of conductors on a single tower which
materially increases the inductance per phase.
(24)
## in which the distances are those between conductors in the
first section of transposition.
The first term in the above equation is the positive- or
negative-sequence
reactance for the combined circuits. The
second term represents
the correction
factor due to the
Fig. 8
## ductors results in five to seven percent greater inductive
reactance than the usual arrangement
of conductors.
This
has been demonstrated
in several references.3
3.
Zero-Sequence
actance
Resistance
and
Re-
The development
of zero-sequence
resistance
and inductive reactance of aerial lines will be considered simultaneously
as they are related quantities.
Since zero-sequence currents for three-phase
systems are in phase and
equal in magnitude,
they flow out through the phase conductors and return by a neutral path consisting
of the
earth alone, neutral
conductor
ground
wires, or any combination
of these. Since the return path
often consists of the earth alone, or the earth in parallel
with some other path such as overhead ground wires, it is
necessary to use a method that takes into account the resistivity of the earth as well as the current distribution
in
the earth.
Since both the zero-sequence
resistance
and
inductive-reactance
of three-phase
circuits are affected by
these two factors, their development
is considered jointly.
As with the positive- and negative-sequence
inductive
reactance, first consider a single-phase
circuit consisting of
a single conductor
grounded at its far end with the earth
acting as a return conductor to complete the circuit. This
permits the development
of some useful concepts for calculating
the zero-sequence
resistance
and inductive
reactance of three-phase
circuits.
Figure 20 shows a single-phase
circuit consisting
of a
single outgoing conductor a, grounded at its far end with
the return path for the current consisting of the earth. A
second conductor,
b, is shown to illustrate
the mutual
effects produced by current flowing in the single-phase
circuit. The zero-sequence resistance and inductive reactance
of this circuit are dependent
upon the resistivity
of the
earth and the distribution
of the current returning
in the
earth.
This problem has been analyzed by Rudenberg,
Mayr,
## parallel three-phase circuits on a single tower
showing transpositions.
## mutual reactance between the two circuits and may reduce
the reactance three to five percent.
The formula assumes
transposition
of the conductor as shown in Fig. 18.
The formula also assumes symmetry
axis but not necessarily about the horizontal
axis.
As contrasted
with the usual conductor arrangement
as
shown in Fig. 18, the arrangement
of conductors shown in
Fig. 19 might be used. However, this arrangement
of con-
Inductive
return.
Chapter
## and Pollaczek in Europe, and Carson and Campbell in this
The more commonly
used method is that of
country.
Carson, who, like Pollaczek, considered the return current
to return through the earth, which was assumed to have
uniform resistivity
and to be of infinite extent.
The solution of the problem is in two parts: (I) the determination
of the self impedance z, of conductor a with
earth return (the voltage between a and earth for unit current in conductor a), and (2), the mutual impedance
zgm
between conductors
a and b with common earth return
(the voltage between b and earth for unit current in a and
earth return).
As a result of Carsons formulas,
and using average
heights of conductors
above ground, the following fundamental simplified equations may be written:
l-P21601( j
loglo GMR
z,=r0+0.00159j+j0.004657j
ohms per mile
(25)
!?
2160 J j
loglo
d
z, = 0.00159j+j0.004657j
ab
## ohms per mile
Rewriting
Carsons
depth of return, D,,
equations
in terms
## ohms per mile.
w9
z,, = 0.00l59~+~0.004657jlog~~~
## A useful physical concept for analyzing
earth-return
circuits is that of concentrating
the current
returning
through the earth in a fictitious conductor
at some considerable depth below the outgoing
conductor
a. This
equivalent
depth of the fictitious return conductor is represented as De,.
For the single-conductor,
single-phase circuit with earth
return now considered as a single-phase,
two-wire circuit,
the self-inductive
reactance is given by the previously de
I
rived j0.279460J loglo sR
(See Eq. (8)) for a single-phase,
or jO.OO4657j loglo &
where
D, is
substituted
for D12, the distance between conductor a and
the fictitious return conductor in the earth.
This expression is similar to the inductive-reactance
as given in
Carsons simplified equation for self impedance.
Equating
the logarithmic
expressions of the two equations,
tDO
(29)
These equations
can be applied to multiple-conductor
circuits if rc, the GMR and d&brefer to the conductors as a
group. Subsequently
the GMR of a group of conductors
are derived for use in the above equations.
To convert the above equations to zero-sequence
quantities the following considerations
Considering three conductors
for a three-phase
system, unit
zero-sequence
current consists of one ampere in each phase
conductor
and three amperes in the earth return circuit.
To use Eqs. (28) and (29), replace the three conductors by
a single equivalent
conductor in which three amperes flow
for every ampere of zero-sequence
current.
Therefore the
corresponding
zero-sequence
self and mutual impedances
per phase are three times the values given in Carsons
Calling the zero sequence impedsimplified equations.
ances zo and zOm,we have:
(26)
rc =resistance
of conductor a per mile.
f=frequency
in cps.
p =earth
resistivity
in ohms per meter cube.
GMR = geometric mean radius of conductor a in feet.
d ab = distance between conductors
a and b in feet.
j0.004657jlog~o~R
## ohms per mile.
ab
20 = 3r,+o.oo477j+jo.o1397j
circuit,
of equivalent
DO
loglo GTMR
z,= r0+0.00159j+j0.004657j
where
two-wire
41
## Characterastics of Aerial Lines
=jO.O04657jlogl,,-
or De=2160
% feet.
J
2160 J ;
## ohms per phase per mile.
20(m)= o.oo477j+jo.o1397j
(30)
DO
log10 d
ab
## ohms per phase per mile
(31)
where j=frequency
in cps.
rc = resistance of a conductor equivalent
to the three
conductors
in parallel.
3r, therefore equals the
resistance
of one conductor
for a three-phase
circuit.
GMR= geometric mean radius for the group of phase
conductors.
This is different than the GMR for
a single conductor
and is derived subsequently
as GMR
d ab=distance
from the equivalent
conductor
to a
parallel
conductor,
or some other equivalent
conductor if the mutual impedance between two
parallel three-phase
circuits is being considered.
For the case of a single overhead ground wire, Eq. (30)
gives the zero-sequence
self impedance.
Equation
(31)
gives the zero-sequence
mutual impedance
between two
Zero-sequence self impedance of two ground wires with
earth return
Using Eq. (30) the zero-sequence
self impedance of two
ground wires with earth return can be derived.
DO
z. = 3r,+O.OO477j+jO.Ol397j
log10 mR
(27)
## This defines De, equivalent
depth of return, and shows
that it is a function of earth resistivity,
p, and frequency, j.
Also an inspection of Carsons simplified equations show
that the self and mutual impedances
contain a resistance
component 0.00159f
which is a function of frequency.
1ogll-J&R
where
## ohms per phase per mile
of a single conductor
TO =resistance
the two ground wires in parallel.
becomes
(30)
equivalent
to
(r, therefore
wires).
of one of
## Characteristics of Aerial Lines
42
GMR=
geometric
wires. (GMR therefore becomes
q(GMR)2
where
conductor
x and y.)
Substituting
or
## (30), the zero-sequence
wires with earth return
20= 2+000477f
3ra
G2
between
Q(GMR)
the
(4,)
ground
(A,)
two
conductors
for GMR
self impedance
becomes
+jo.o1397j
two
of two
Chapter
## The expression for self impedance
is then
transpositions.
converted
to zero-sequence
self impedance
in a manner
analogous
to the case of single conductors
with earth
return.
Consider three phase conductors a, b, and c as shown in
Fig. 21. With the conductors
transposed
the current
in Eq.
ground
DO
log10 q(GMR)
(&,)
(32)
## Zero-sequence self impedance of n ground wires with earth
return
Again using Eq. (30), the zero-sequence
self impedance
of n ground wires with earth return can be developed.
z. = 3ro+0.00477j+j0.01397j
DO
log,0 GMR
(30)
n ground
## of a single conductor equivalent
to
wires in parallel,
then r. =- ra where ra is the
n
of one of the n ground wires, in ohms per phase
resistance
per mile.
GMR is the geometric mean radius of the n ground wires
as a group, which may be written as follows in terms of all
possible distances,
## This expression can also be written
pairs of distances as follows.
Fig. 21-Self
impedance
with
earth
For conductor
b:
y+y+7
## The equation for zero-sequence
self impedance of n ground
wires with earth return can therefore be obtained by sub
z for r. and Eq. (33) for GMR
conductors
## divides equally between the conductors so that for a total
current of unity, the current in each conductor is one third.
The voltage drop in conductor
a for the position indicated in Fig. 21 is
stituting
of parallel
return.
in Eq. (30).
## Self impedance of parallel conductors with earth return
In the preceeding
discussion
the self and mutual impedances between single cylindrical
conductors with earth
return were derived from which the zero-sequence
self and
mutual reactances were obtained.
These expressions were
expanded to include the case of multiple overhead ground
wires, which are not transposed.
The more common case
is that of three-phase
conductors
in a three-phase
circuit
which can be considered
to be in parallel when zero-sequence currents are considered.
Also the three conductors
in a three-phase
circuit are generally
transposed.
This
factor was not considered in the preceeding cases for multiple overhead ground wires.
In order to derive the zero-sequence
self impedance
of
three-phase
circuits it is first necessary to derive the self
impedance
of three-phase
circuits taking
into account
c:
f+Y+%
of the
in which Zaa, zbb, and zco are the self impedances
three conductors with ground return and .&b, &or and 2.0
are the mutual impedances
between the conductors.
Since conductor
a takes each of the three conductor
positions successively
for a transposed
line, the average
drop per conductor is
1
g(Zaa+zbb+z,of2zabf2Zbof22..).
Substituting
the values of self and mutual
impedances
given by Eqs. (28) and (29) in this expression,
## The ninth root in the denominator
of the logarithmic
term
is the GMR of the circuit and is equal to an infinitely thin
tube which would have the same inductance
as the threeconductor system with
earth return shown in Fig. 21.
GMRclrcult= q(GMR)3conductor dsb2dbc2dca2 feet.
GMRcl,cuft= ~(GMR)3,,,d,,t,,
(&t&A~)2 feet.
______~GM.Lcult= ~GMR,,,ductor(~dahdbodca ) 2 feet.
By previous
43
## Characteristicsof Aerial Lines
Chapter 3
derivation
=ma
Zero-sequence self impedance of two identical parallel circuits with earth return
For the special case where the two parallel
circuits
are identical,
following
the same
derivation
20 = ~+0.00477j+j0.01397j
log10
three-phase
method
of
D,
~(GMR)
(GMD)
## (See Eq. (23)), GMDseparatlon
in which GMR
is the geometric
(39)
of one set of
feet.
conductors,
( (GMR;,,,,,,,,,(GMD)Zaeparatlon
), and GMD
is
the
geometric
mean
distance
between
the
two sets of
Therefore GMR
= ~(GMR),,,,,,,,,(GMD)
2seDarst on
conductors
or
the
ninth
root
of
the
product
of the nine
feet.
(35)
possible distances between conductors
in one circuit and
(35) in equation
Substituting
GMRClrotit from equation
conductors
in the other circuit.
(34),
This equation
is the same as \$(zo+zom)
where zo is
the
zero-sequence
self
impedance
of
one
circuit
by equazfz=~+o.o0159j
mutual impedance
tion (37) and zom) is the zero-sequence
between
two circuits as given by Eq. (38). For nonD,
+jO.O04657j loglo J
identical circuits it is better to compute the mutual and
~(GMR)conduetor(GMD)2,,Daratlon
self impedance
for the individual
circuits,
and using
ohms per mile.
(36)
+(~o+z0~~,) compute the zero-sequence
self impedance.
In equations (34) and (36), r0 is the resistance per mile of
one phase conductor.
Zero-sequence mutual impedance between one circuit (with
earth return) and n ground wires (with earth return)
Zero-sequence self impedance of three parallel conductors
Figure 22 shows a three-phase
circuit with n ground
with earth return
Equation
(36) gives the self impedance of three parallel
conductors
with earth return and was derived for a total
current of unity divided equally among the three conductors.
Since zero-sequence
current consists of unit current in each conductor or a total of three times unit current
for the group of three conductors,
the voltage drop for
zero-sequence
currents is three times as great. Therefore
Eq. (36) must be multiplied
by three to obtain the zerosequence self impedance
of three parallel conductors
with
earth return.
Therefore,
z. = r,+O.O0477f
.
+jo.o13g?f
log10 ,JGMR
DC3
conmor (GMD)
9,
0
## Fig. 22A three-conductor
return) and n ground
2twmmn
wires. Equation
pedance between
## Using a similar method of derivation
the zero-sequence
mutual
impedance
between 2 three-phase
circuits with
common earth return is found to be
## where GMD is the geometric
2 three-phase
circuits or the
the nine possible distances
group and conductors in the
larity between Eq. (38) and
earth return)
## (31) gives the zero sequence
two conductors:
ZO(@= o.oo477j+jo.o1397j
Zero-sequence mutual impedance between two circuits with
earth return
log10 \$D
## ohms per phase per mile
three-phase
wires (with
(37)
where ~GMRcondUctorGMD~
is the GMLuit
derived in equation
(35) or \$(GMR)3Conductor dab2dbc2dCs2
ZO(l?l)
=o.oo477j+jo.o1397j
0.2
(38)
## mean distance between the
ninth root of the product of
between
conductors
in one
other group. Note the simiEq. (31)
mutual
im-
log10 %
aab
(31)
## where dab is the distance between the two conductors.
This
equation
can be applied to two groups of conductors
if
dab is replaced by the GMD or geometric mean distance
between the two groups. In Fig. 22, if the ground wires are
considered as one group of conductors,
and the phase conductors a, b, c, are considered as the second group of conductors
then
, the GMD between the two groups is
GMD = 3i/d,gldbgldcgl-dagndbgndcgn
feet
Substituting
this quantity
for dab in Eq. (31) results in an
equation for the zero-sequence mutual impedance between
one circuit and n ground wires. This zoCrn)is z,,(,).
Chapter 3
44
## General Method for Zero-Sequence
Calculations
The preceding sections have derived the zero-sequence
self and mutual impedances
for the more common circuit
arrangements
both with and without ground wires. For
more complex circuit and ground wire arrangements
a
## ohms per phase per mile.
Zero-sequence impedance of one circuit with n ground wires
(and earth) return.
Referring to Fig. 20 the zero-sequence
self impedance of
a single conductor,
and the zero-sequence
mutual
impedance between a single conductor
and another single
conductor
with the same earth return path was derived.
These values are given in Eqs. (30) and (31). As stated before, these equations
can be applied to multi-conductor
circuits by substituting
the circuit GMR for the conductor
GMR in Eq. (30) and the GMD between the two circuits
for dab in Eq. (31).
First, consider the single-conductor,
single-phase
circuit
with earth return and one ground wire with earth return.
Referring to Fig. 20 conductor a is considered as the single
conductor of the single-phase
circuit and conductor b will
be used as the ground wire.
Writing-the
equations for Ea and Eb, we have:
Ea
IaZaa
Eb
l&m
+
+
Ibzrn
(41)
IbZbb-
(42)
## If we assume conductor b as a ground wire, then & =0
since both ends of this conductor are connected to ground.
Therefore
solving Eq. (42) for Ib and substituting
this
Value Of Ib in Eq. (41),
.
To obtain
z8, divide
za =
zaa
--
2m2
Zbb
(43)
## The zero-sequence impedance of a single-conductor,
singlephase circuit with one ground wire (and earth) return is
therefore defined by Eq. (43) when zero-sequence
self impedances of single-conductor,
single-phase circuits are substituted for zaa and zbb and the zero-sequence mutual
impedance between the two conductors is substituted
for zm.
Equation
(43) can be expanded to give the zero-sequence
impedance
of a three-phase
circuit with n ground wires
(and earth) return.
zo=
20(a)
xO2Ca9)
-
20w
(44)
## impedance of one circuit with n
z o= zero-sequence
ground wires (and earth) return.
zoa) = zero-sequence
self impedance
of the threephase circuit.
self impedance
of n ground
20(lx)= zero-sequence
wires.
zo(,) = zero-sequence
mutual impedance
between the
phase conductors
as one group of conductors
and the ground wire(s) as the other conductor
group.
Where
Equation
(44) results in the equivalent
circuit of Fig. 23
for determining
the zero-sequence impedance of one circuit
with n ground wires (and earth) return.
Fig. 23Equivalent
circuit for zero-sequence
impedance of
one circuit (with earth return) and n ground wires (with earth
return).
## general method must be used to obtain the zero-sequence
impedance
of a particular
circuit in such arrangements.
The general method consists of writing the voltage drop
for each conductor or each group of conductors in terms of
zero-sequence
self and mutual impedances
with all conductors or groups of conductors
present.
Ground wire
conductors
or groups of conductors
have their voltage
drops equal to zero. Solving these simultaneous
equations
for F
of the desired
circuit
## gives the zero-sequence
im-
pedance of that circuit in the presence of all the other zerosequence circuits.
This general method is shown in detail in Chap. 2,
Part X, Zero-Sequence
Reactances.
Two circuits, one
with two overhead ground wires and one with a single overhead ground wire are used to show the details of this more
general method.
Practical Calculation
of Zero-Sequence
Impedance of Aerial Lines-In
the preceding
discussion
a
number of equations have been derived for zero-sequence
self and mutual impedances
of transmission
lines taking
into account overhead ground wires. These equations can
be further simplified to make use of the already familiar
quanquantities
ra, x8, and x,J. To do this two additional
tities, re and x, are necessary that result from the use of the
earth as a return path for zero-sequence
currents.
They
are derived from Carsons formulas and can be defined as
follows:
r,=O.O0477j
per mile.
X 10 6p-ohms
per
phase
per
mile.
(46)
## It is now possible to write the previously derived equations
for zero-sequence
self and mutual impedances
in terms of
and xe. The quantities
r,, xa, xd are given
rap xa,
xd,
?,,
in the tables of Electrical
Characteristics
of Conductors
and Inductive
Reactance Spacing Factors.
The quantities
re and xe are given in Table 7 as functions
of earth
resistivity,
p, in meter ohms for 25, 50, and 60 cycles per
second.
The following derived equations
are those most
commonly
used in the analysis
of power system problems.
Chapter
Characteristics
(38)
## phase per mile.
z~(~)= r,+jO.O06985j
where xd
is
)(xd(as*)
+xd(sbt)
+xd(ac)
+Xd(bc))+Xd(cs\$)
+Xd(cb#)
## Zero-sequence self impedance-one
return)
+xd(ba\$)
&I@,&
= 0.00477j
DC3
log10 3n
X &gdbgdcgr- - dsgndbgndcgn
ohms per phase per mile.
(40)
+jo.o1397j
conductor
(30)
106p
f
ZO(~)
= r,+jO.O06985j
## log,, 4.6656 X lo6 e
f
-jO.O06985j loglo ( zdsgldbgldegl- - -dagndbgndcgn)
2
zOcag)
= r,+j(x, - 3xd) ohms per phase per mile
(52)
2conductor
## Zero sequence self impedance-two
return)
20(g)= 3~+o.oo477j+~o.01397j
log --GMR
zow 3\$+re+j(xe+zxa-5xd)
conductordw
(32)
zo2w
Zo=Zo(a) - all*\
- \o,
0.8382
&,
1% 1
2
3
6 for spacing
where
(50)
between
+xd(agn)
ground
+xd(bgn)
Zero-sequence impedance-One
(and earth return)
## ohms per phase per mile
zd = - (Xd(ngl)+xd(bgl)
3n
---
WIhere
xd= xd from Table
wires, c&.
where
D,
log10
i? (GMR)
QE) = \$+r,+jo.O06985j
(49)
+0.8382
2
## Zero-sequence mutual impedance between one circuit (with
earth return) and n ground wires (with earth return)
1
t GMR)
n(n--
+xd(bbr)
## ground wire (with earth
zocgI=3r,+r,+j0.006985jlog1o4.6656X
zocg)=3r,+r,+,j(x.+3x,)
## t sum of xde for all possible distances
1)
between all possible pairs of ground wires).
or xd=---
(48)
## ohms per phase per mile.
loglo
distances
+xd(ccq)
Dt!
log10 ___
GMR
zo(g)= 3r,+O.OO477j+jO.O1397j
+O.O06985j
x,-J= -- l
(sum of xds for all possible
n(n - 1)
between all ground wires.)
where
## -jO.O06985j log,, GMD2
zO(m)=re-i--j(x,--3xd)
ohms per phase per mile
45
of Aerial Lines
+xd(cgl)
+xd(cgn)
>*
## circuit with n ground wires
(44)
self impedance
of the three~0~~)= zero-sequence
phase circuit.
zocg)= zero-sequence
self impedance
of n ground
wires.
zo(aR)= zero-sequence mutual impedance between the
three-phase circuit as one group of conductors
and the ground wire(s) as the other condue tor group.
Characteristics
46
Shunt
Positive-, Negative-,
and Zero-sequence
Capacitive Reactance
The capacitance
of transmission
lines is generally
a
negligible factor at the lower voltages under normal operHowever,
it becomes an appreciable
ating conditions.
effect for higher voltage lines and must be taken into consideration
when determining
efficiency,
power factor,
regulation,
and voltage distribution
under normal operUse of capacitance
in determining
the
ating conditions.
performance
of long high voltage lines is covered in detail
and Losses of Transmission
in Chap. 9, Regulation
Lines.
Capacitance
effects of transmission
lines are also useful
in studying such problems as inductive interference,
lightning performance of lines, corona, and transients
on power
systems such as those that occur during faults.
For these reasons formulas are given for the positive-,
and zero-sequence
shunt capacitive
reactance
negative-,
for the more common transmission
line configurations.
The
case of a two-conductor,
single-phase
circuit is considered
to show some of the fundamentals
used to obtain these
formulas.
For a more detailed analysis of the capacitance
problem a number of references are available.
2,4*5.
In deriving capacitance
formulas the distribution
of a
charge, q, on the conductor surface is assumed to be uniform. This is true because the spacing between conductors
in the usual transmission
circuit is large and therefore the
charges on surrounding
conductors produce negligible distortion in the charge distribution
on a particular
conductor.
Also, in the case of a single isolated charged conductor, the voltage between any two points of distances x
and y meters radially from the conductor can be defined as
the work done in moving a unit charge of one coulomb from
point P2 to point Pr through the electric field produced by
the charge on the conductor.
(See Fig. 24.) This is given
4.
Chapter 3
of Aerial Lines
## This equation shows the work done in moving a unit
charge from conductor 2 a distance D12 meters to the surface of conductor
1 through the electric field produced by
ql. Now assuming
only conductor 2, having a charge 42,
the voltage between conductors
1 and 2 is
VI2 = 18 X log q2 In uz- volts.
## This equation shows the work done in moving a unit
charge from the outer radius of conductor 2 to conductor 1
a distance D12 meters away through the electric field produced by qz.
With both charges q1 and q2 present, by the principle of
superposition
the voltage VI2 is the sum of t!he voltages
resulting from q1 and q2 existing one at a time. Therefore
VIZ is the sum of Eqs. (54) and (55) when both charges q1
and q2 are present.
V12= 18X log q1 In G-/-q,
r
Also if the charges
their sum is zero,
or
ql+q2=0
v12
V 12= 18 X log q1 In \$
volts.
(54)
per meter.
(58)
per meter.
reactance
60
012
loglo -
= O.,,,,+)
megohms
## (or per con-
to neutral
or in more practical
__
27rjc
is x,,=
(59)
mile.
This can be written
(53)
## where q is the conductor charge in coulombs per meter.
By use of this equation and the principle of superposition, the capacitances
of systems of parallel conductors can
be determined.
Applying Eq. (53) and the principle of superposition
to
the two-conductor,
single-phase circuit of Fig. 24 assuming
conductor 1 alone to have a charge ql, the voltage between
conductors
1 and 2 is
In 12
r
shunt-capacitive
X cn
## V XY = 18X log q In x volts
(57)
The capacitance
to neutral is twice that given in Eq. (58)
because the voltage to neutral is half of Vu.
xcn= 0.0683-
by
volts.
12 =
36X10g
ductor)
(capacitance).
and
(56)
The capacitance
between conductors
the charge to the voltage or
-=
q1
are equal
-ql
q2=
V12=36X 10gql In 2
The
volts.
12
## on the two conductors
two conductor
ln \$
for q2 in equation
Substitutingql
C,=
Fig. 24-A
(55)
12
units
(60)
as
log10 ;+0.0683?
per conductor
log,,1012
per mile
(61)
## where D12 and r are in feet and j is cycles
Eq. (61) may be written
2
cn = x, +s:
megohms
per conductor
per second.
per mile.
(62)
The derivation
of shunt-capacitive
reactance formulas
brings about terms quite analogous to those derived for
inductive reactance, and as in the case of inductive reactance, these terms can be resolved into components
as
shown in Eq. (62). The term xa accounts for the electrostatic flux within a one foot radius and is the term
It is a function
of the con-
where
x(+
## ductor outside Radius only. The term xd accounts for the
electric flux between a one foot radius and the distance
D12 to the other conductor
D12
in Eq.
(61).
Note
## and is the term 0.0683 \$ log,,
that
unlike
inductive-reactance
## where the conductor
used, in capacitance calculations
used is the actual physical radius of the conductor in feet.
Zero-sequence
capacitive
reactance
is, like inductivereactance, divided into components x, taking into account
the electrostatic
flux within a one-foot radius, xd taking
into account the electrostatic
flux external to a radius of
one foot out to a radius D feet, and x, taking into account
the flux external to a radius of one foot and is a function
of the spacing to the image conductor.
I
where
## Shunt-Capacitive Reactance, xc, of Three-Phase
(Conductors a, b, c)
(a) Positive (and negative) sequence xc.
x~=x~=x,+x~megohms~erconductorpermile.
\$sum
## of all three xds for distances
Circuits
See Table
between
all
(65)
(8)
(b) Zero-Sequence
xc of one circuit (and earth).
xl&, =x:+x,-22: megohms per conductor per
mile.
xd = value given in Eq. (65).
(c)
4,)
Zero-Sequence
=3x,(,)+&)
mile.
Table
(66)
## xc of one ground wire (and earth).
megohms per conductor
per
(67)
(d) Zero-Sequence
xc of two ground wires (and earth).
3
3
x dcp)= -x,/(,) +x,(,, - -xd megohms per conductor per
2
2
mile.
(6%)
\--,
xd = xd(glg2)= xd for distance
between
+d
distances
(bgn) +d(cgn)
1.
(g) Zero-Sequence
x0 of one circuit with n ground wires
x0(wJ2
x0 = x&q - ____ megohms per conductor per mile. (71)
d (69
ground
## (h) xc of single-phase circuit of two identical
5 = 2(x: \$-xi) megohms per mile of circuit.
x~ = xd for spacing between conductors.
wires.
## (e) Zero Sequence xc of n ground wires (and earth).
3
3(n-1)
-xi
megohms per conductor
x0(9)=x,+-x:n
n
mile
per
(69)
Circuits
conductors
(72)
(i) xc of single-phase
circuit of two non-identical
conductors a and b.
x = x,(a) + d (b)+ 2s: megohms per mile of circuit. (73)
(j) xc of one conductor
X = x,1++x,l megohms
and earth.
per mile.
(74)
## In using the equations it should be remembered that the
shunt capacitive reactance in megohms for more than one
mile decreases because the capacitance
increases. For more
than one mile of line, therefore, the shunt-capacitive
reactance as given by the above equations should be divided
(64)
possible. pairs).
= ~(x~&-txd,,+x~t,~).
l
(sum of all xds for all possible
n(n-1)
between all ground wires).
(f) Zero-Sequence
xc between one circuit (and earth)
and n ground wires (and earth)
xd (a&\$)
= 2, - 3~: megohms per conductor per mile. (70)
(63)
## xl is given in the tables of Electrical Characteristics
of
conductors,
xl is given in Table
8, Shunt-Capacitive
Reactance
Spacing Factor, and xQ is given in Table 9,
Zero-Sequence
Shunt-Capacitive
Reactance
Factor.
The following equations have been derived in a manner
similar to those for the two-conductor,
single-phase
case,
making use of the terms x,, x~ and xl. They are summarized in the following tabulation.
xi=
orxd=--
____2
(sum of all xds for all possible distances
n(n-1)
between all possible pairs of ground wires)
## Shunt Capacitive Reactance, xc, of Single-Phase
(Conductors a and 6)
12.30
## log,, 2 2hmegohms per mile per
f
conductor
h = conductor height above ground.
j=frequency
in cps.
x *=-
47
Chapter 3
## by the number of miles of line.
5.
Conductor
Temperature
Carrying Capacity
Rise
and
Current-
In distributionand transmission-line
design the temperature rise of conductors
above ambient while carrying
While power loss, voltage regulacurrent is important.
tion, stability and other factors may determine the choice
of a conductor for a given line, it is sometimes necessary to
consider the maximum continuous
current carrying capacity of a conductor.
The maximum
continuous
current
rating is necessary because it is determined
by the maximum operating temperature
of the conductor.
This temperature affects the sag between towers or poles and determines
the loss of conductor
tensile strength
due to
annealing.
For short tie lines or lines that must carry
excessive loads under emergency conditions, the maximum
continuous
current-carrying
capacity may be important
in selecting the proper conductor.
The following discussion presents the Schurig and Fricks
formulas for calculating
the approximate
current-carrying
capacity of conductors under known conditions of ambient
temperature,
wind velocity, and limiting temperature
rise.
The basis of this method is that the heat developed in
the conductor by 12R loss is dissipated
(1) by convection
Characteristics
48
of Aerial
Lines
Chapter 3
in the surrounding
to surrounding
objects.
This can expressed
as follows:
12R = (IV,+ W,)A watts.
where I
R
W,
W,
05)
## = conductor current in amperes.
= conductor resistance per foot.
= watts per square
inch dissipated by
= watts per square inch dissipated by
A = conductor
of length.
## The watts per square inch dissipated
can be determined
from the following
0.01284
W, = -- - -- ---At
T;?1232/d
watts
convection.
## inches per foot
by convection,
equation:
per square
inch
Wc,
(76)
where
p =pressure
in atmospheres
(p = 1.0 for atmospheric pressure).
v= velocity
in feet per second.
T,= (degrees Kelvin) average of absolute temperatures of conductor and
air.
d = outside diameter of conductor in inches.
At = (degrees C) temperature
rise.
This formula is an approximation
applicable
to conductor diameters ranging from 0.3 inch to 5 inches or more
when the velocity of air is higher than free convection air
currents (0.2O.5 ft/sec).
The watts per square inch dissipated by radiation,
Wr,
can be determined
from thc following equation:
watts
where
per square
Fig. 25Copper conductor current carrying capacity in Amperes VS. Ambient Temperature in C. (Copper Conductors at
75 C, wind velocity at 2 fps.).
inch
## E = relative emissivity of conductor surface
(E= 1.0 for black body, or 0.5 for average
oxidized copper).
T= (degrees Kelvin) absolute temperature
of conductor.
To = (degrees Kelvin) absolute temperature
of surroundings.
By calculating
(W,+ W,), A, and R, it is then possible
to determine I from El. (75). The value of R to use is the
a-c resistance
at the conductor
temperature
(ambient
temperature
plus temperature
rise) taking into account
skin effect as discussed previously
in the section on positive- and negative-sequence
resist mccs.
This method is, in general, applicable
to both copper
and aluminum
conductors.
Tests have shown that aluminum conductors dissipate heat at, about the same rate as
copper conductors of the same outside diameter when the
temperature
rise is the same. Where test data is available
on conductors,
it should be used.
The above general
method can be used when test data is not available, or to
check test results.
The effect of the sun upon conductor temperature
rise is
generally
neglected,
being some 3 to 8 C. This small
effect is less important
under conditions of high temperature rise above ambient.6
The tables of Electrical Characteristics
of Conductors
include tabulations
of the approximate
maximum current-
Fig. 26Aluminum
conductor current carrying capacity in
Amperes VS. Ambient Temperature
in C. (Aluminum
Conductors at 75C, wind velocity at 2 fps).
Chapter
Characteristics
of Aerial
Lines
49
## carrying capacity based on 50C rise above an ambient of
25C, (75C total conductor temperature),
tarnished
surface (E = 0.5), and an air velocity of 2 feet per second. These
conditions were used after discussion and agreement with
the conductor manufacturers.
These thermal limitations
are based on continuous
The technical literature shows little variation from these
conditions
as line design limits.
The ambient
air temperature is generally assumed to be 25C to 40C whereas
the temperature
rise is assumed to be 10C to 60C. This
gives a conductor
total temperature
range of 35C to
100C. For design purposes copper or ACSR conductor
total temperature
is usually assumed to be 75C as use of
this value has given good conductor performance
from an
annealing
standpoint,
the limit being about 100C where
annealing of copper and aluminum
begins.
Using Schurig and Fricks formulas, Fig. 25 and Fig. 26
have been calculated to show how current-carrying
capncity of copper and aluminum
conductors
varies with ambient temperature
assuming a conductor
temperature
of
75.C and wind velocity of 2 feet per second. These values
are conservative
and can be used as a guide in normal line
design.
For those lines where a higher conductor
tem-
## perature may be obtained that approaches l00C, the conductor manufacturer
should be consulted for test data or
other more accurate information
as to conductor temperature limitations.
Such data on copper conductors has been
presented rather thoroughly
in the technical literature.
## III TABLES OF CONDUCTOR
CHARACTERISTICS
The following tables contain data on copper, ACSR,
hollow copper, Copperweld-copper,
and Copperweld conductors, which along with the previously derived equations,
permit the determination
of positive-, negative-, and zerosequence impedances of conductors for use in the solution
of power-system
problems.
Also tabulated
are such conductor characteristics
as size, weight, and current-carrying
capacity as limited by heating.
The conductor data (rn, x,, x,1) along with inductive and
shunt-capacitive
reactance
spacing factors (xd, zd) and
zero-sequence
resistance,
inductive
and shunt-capacitive
reactance factors (re, x,, x,) permit easy substitution
in
the previously derived equations for determining
the symmetrical component sequence impedances of aerial circuits.
The cross-sectional
inserts in the tables are for ease in
Characteristics
TABLE 2-ACHARACTERISTICS
OF ALUMINUM
(Aluminum
TABLE ~-B-CHARACTERISTICS
of Aerial
Company
OF EXPANDED
(Aluminum
Lines
Chapter 3
of America)
ALUMINUM
Company
ofYAmerica)
## CABLE STEEL REINFORCED
Chapter
Characteristics
TABLE3-ACHARACTERISTICS
TABLE 3-B-CHARACTERISTICSOF
OF ANACONDA
GENERAL
Notes:
HOLLOW
CABLE TYPE
(General
51
of Aerial Lines
Cable
COPPER
HH HOLLOW
CONDUCTORS
COPPER
CONDUCTORS
Corporation)
*Thickness
at edges of interlocked
segments.
Thickness
uniform
throughout.
1) Conductors
of smaller diameter
for given cross-sectional
area also available;
in the naught sizes, some
2) For conductor.at,
75C., air at 25C., wind 1.4 miles per hour (2 ft/sec),
frequency=60
cycles.
diameter
expansion
is possible.
Characteristics
of Aerial
Lines
Chapter
TABLE 4-ACHARACTERISTICSOFCOPPERWELD-COPPER
(Copperweld
CONDUCTORS
Steel Company)
*Based on a conductor temperature of i5C. and an ambient of 25C., wind 1.4 miles per hour 12 ft/scc.), frequency=60
**Resistances at 50C. total temperature, based on an ambrent of 25C. plus 25Y. rrse due to heating effect of current.
25 C. rrse IS 75: of the Approxrmate (urrent Carrymg Capacity at 60 cycles.
## finding the appropriate
table for a particular
conductor.
For these figures open circles, solid circles, and crosshatched circles represent copper, steel, and aluminum
conductors respectively.
The double cross hatched area in the
insert for Table 2-B, Characteristics
of EXPANDED
## cycles, average tarnished surface.
The approximate magmtude of current necessary
to produce
the
Aluminum
Cable Steel Reinforced,
represents
stranded
paper.
The authors wish to acknowledge the cooperation of the
conductor manufacturers
in supplying the information
for
compiling these tables.
Chapter
Characteristics
of Aerial
Lines
TABLE 4-B-CHARACTERISTICS
OF COPPERWELD
(Copperweld
Steel Company)
CONDUCTORS
Characteristics
54
TABLE
6INDUCTIVE
of Aerial
Lines
Chapter
Chapter
Characteristics
Table 8
SHUNT CAPACITIVE
of Aerial
Lines
55
Characteristics
56
of Aerial
## With the increased use of high-voltage
transmission
lines
and the probability
of going to still higher operating voltages, the common aspects of corona (radio influence and
corona loss) have become more important
in the design of
transmission
lines.
In the early days of high-voltage
transmission,
corona
was something which had to be avoided, largely because of
the energy loss associated with it. In recent years the RI
aspect of corona has become more important.
In areas where RI must be considered, this factor
might establish the limit of acceptable corona performance.
Under conditions
where abnormally
high voltages are
present, corona can affect system behavior.
It can reduce
the overvoltage
on long open-circuited
lines. It will attenuate lightning
voltage surges (see Sec. 29 Chap. 15) and
switching
surges. 177 By increasing
the electrostatic
coupling between the shield wire and phase conductors,
corona at times of lightning strokes to towers or shield wires
reduces the voltage across the supporting
string of insulators and thus, in turn, reduces the probability
of flashover and improves system performance.
On high-voltage
lines grounded through a ground-fault
neutralizer,
the inphase current due to corona loss can prevent extinction
of
the arc during a line to ground fault.28
## 6. Factors Affecting Corona
At a given voltage, corona is determined
by conductor
diameter, line configuration,
type of conductor,
condition
of its surface, and weather.
Rain is by far the most important aspect of weather in increasing corona. Hoarfrost
and fog have resulted in high values of corona loss on experimental
test lines. However, it is believed that these
high losses were caused by sublimation
or condensation
of
water vapor, which are conditions not likely to occur on an
operating line because the conductor
temperature
would
normally
be above ambient.
For this reason, measurements of loss made under conditions
of fog and hoarfrost
might be unreliable unless the conductors
were at operating temperatures.
Falling snow generally causes only a
moderate increase in corona. Also, relative humidity,
temperature,
atmospheric
pressure, and the earths electric
field can affect corona, but their effect is minor compared
to that of rain. There are apparently
other unknown factors
found under desert conditions which can increase corona.19
The effect of atmospheric
pressure and temperature
is
generally considered to modify the critical disruptive voltage of a conductor directly, or as the 2/3 power of the air
density factor, 6, which is given by:
where
b = barometric
F = temperature
The temperature
erally considered
17.9b
459+OF
(7%
## pressure in inches of mercury
in degrees Fahrenheit.
to be used in the above equation
to be the conductor temperature.
Chapter
## TABLE 10STANDARD BAROMETRIC
AS A FUNCTION OF ALTITUDE
IV CORONA
6=
Lines
is genUnder
standard
conditions
(29.92 in. of Hg. and 77F) the air
density factor equals 1.00. The air density factor should
be considered in the design of transmission
lines to be built
in areas of high altitude or extreme temperatures.
Table
10 gives barometric
pressures as a function
of altitude.
Corona in fair weather is negligible or moderate up to
a voltage near the disruptive
voltage for a particular
conductor. Above this voltage corona effects increase very
rapidly.
The calculated disruptive
voltage is an indicator
A high value of critical disruptive
of corona performance.
voltage is not the only criterion
of satisfactory
corona
Consideration
should also be given to the
performance.
sensitivity
of the conductor
to foul weather.
Corona increases somewhat more rapidly on smooth conductors than
it does on stranded conductors.
Thus the relative corona
characteristics
of these two types of conductors
might
interchange
between fair and foul weather.
The equation
for critical disruptive
voltage is:
E,=g,, 6% T m log, D/r
(79a)
where :
E, = critical disruptive voltage in kv to neutral
g,=critical
(Ref. 10 and 16 use
g,=21.1 Kv/cm rms. Recent work indicates value given in
Sec. 10 is more
accurate.)
## r =radius of conductor in centimeters
D = the distance in centimeters between conductors,
## for singlephase, or the equivalent phase spacing, for three-phase
vo1 tages.
m= surface factor (common values, 0.84 for stranded, 0.92 for
segmental conductors)
6 = air density factor
## The more closely the surface of a conductor approaches
a smooth cylinder, the higher the critical disruptive
voltage assuming constant diameter.
For equal diameters,
a
stranded conductor is usually satisfactory
for 80 to 85 percent of the voltage of a smooth conductor.
Any distortion
of the surface of a conductor
such as raised strands, die
burrs, and scratches will increase corona. Care in handling
conductors
should be exercised, and imperfections
in the
surface should be corrected, if it is desired to obtain the
best corona performance
from a conductor.
Die burrs and
die grease on a new conductor, particularly
the segmental
type, can appreciably
increase corona effects when it is
first placed in service. This condition improves with time,
taking some six months to become stable.
Strigel44 concluded that the material from which a conductor is made has no effect on its corona performance.
In
Chapter
Characteristics
of Aerial
57
Lines
## in. HH copper. 6=0.88. Ref. 19. Corona loss test
made in desert at a location where abnormally high corona loss is
observed on the Hoover-Los Angeles 287.5-kv line, which is strung
with this conductor.
Measurement made in three-phase test line.
This particular curve is plotted for 6 =0.88 to show operating condition in desert. All other curves are for 6 = 1.00.
Curve aSame as curve 1, except converted to 6 = 1.00.
Curve 3-1.4
in. HH copper. Ref. 12. Corona loss test made in
Curve l-l.4
## California. Comparison with curve 2 shows effect of desert conditions. Measurements
made on three-phase test line, 30-foot flat
spacing, 16-foot sag, 30-foot ground clearance, 700 feet long.
Curve 41.1 in. HH. Ref. 13. Measurements made on three-phase
test line, 22-foot flat spacing, 16-foot sag, 30-foot clearance to
ground, 700 feet long.
Curve 5-1.65
in. smooth. Ref. 12. This conductor had a poor surface. Measurements made on three-phase test line, 30-foot spacing, 16-foot sag, 30-foot ground clearance, 700 feet long.
Curve 6-1.65
in. smooth aluminum. Ref. 27. Reference curve obtained by converting per-phase measurement to loss on three-phase
line. Dimensions of line not given.
Curve 7-1.04
in. smooth cylinder. Ref. 23. In reference this conductor is referred to as having an infinite number of strands.
Plotted curve obtained by conversion of per-phase measurements
to three-phase values, using an estimated value for charging kva,
to give loss on a line having 45-foot flat configuration.
Curve 8l .96 in. smooth aluminum. Ref. 28. Reference curve gives
three-phase loss, but line dimensions are not given.
Curve 9-1.57 in. smooth. Ref. 23. This conductor was smooth and
clean. Reference curve gives per-phase values. Plotted curve is
for 45-foot flat spacing.
Fig. 27Fair-Weather
ductors;
industrial
areas, foreign material
deposited
on the conductor can, in some cases, seriously reduce the corona performance.
(Reference 28 gives some measurements
in an industrial
area.)
Corona is an extremely variable phenomenon.
On a conductor energized at a voltage slightly above its fair weather
corona-starting
voltage, variations
up to 10 to 1 in corona
factor have been recorded during
fair weather.
The presence of rain produces corona loss on
a conductor at voltages as low as 65 percent of the voltage
at which the same loss is observed during fair-weather.
Thus it is not practical to design a high-voltage
line such
that it will never be in corona.
This also precludes expressing a ratio between fair- and foul-weather
corona,
since the former might be negligibly small.
If a conductor is de-energized for more than about a day,
corona is temporarily
increased.
This effect is moderate
compared to that of rain. It can be mitigated
by re-energizing a line during fair weather where such a choice is possible.
7. Corona Loss
Extensive work by a large number of investigators
has
been done in determining
corona loss on conductors
operated at various voltages.
This work has lead to the devel-
## Corona-Loss Curves for Smooth
Air Density Factor, 6 = 1.
Con-
## opment of three formu1as(10~14~16)generally
used in this
country (Reference 18 gives a large number of formulas).
The Carroll-Rockwell
and the Peterson formulas are considered the most accurate especially in the important
low
loss region (below 5 kw per three-phase mile). The Peterson formula, when judiciously used, has proved to be a reliable indicator
of corona performance
(see Sec. 9) for
transmission
voltages in use up to this time. Recent work
on corona loss has been directed toward the extra-highvoltage range and indicates that more recent information
should be used for these voltages.
Fair-weather
corona-loss measurements
made by a number of different investigators
are shown in Figs. 27, 28, and
29. All curves are plotted in terms of kilowatts per threephase mile. The data presented in these curves has been
corrected for air density factor, 6, by multiplying
the test
voltage by l/6 2/3. Some error might have been introduced
in these curves because in most cases it was necessary to
convert the original data from per-phase measurements.
at the surface of each conductor.
The curves should be
used as an indicator of expected performance
during fair
weather.
For a particular design, reference should be made
to t,he original publications,
and a conversion made for the
design under consideration.
The relation between fair-
Characteristics
of Aerial
Lines
Chapter
Curve ll.4
in. ACSR. Ref. 12. Conductor was washed with gasoline then soap and water. Test configuration: three-phase line,
30-foot flat spacing, 16-foot sag, 30-foot ground clearance, 700 feet
long.
Curve 21.0 in. ACSR. Ref. 11. Conductor weathered by exposure
to air without continuous energization. Test configuration: threephase line, 20-foot flat spacing, 700 feet long.
Curve 31.125 in. hollow copper. Ref. 14. Washed in same manner
as for curve 1. Test configuration: three-phase line, 22-foot flat
spacing.
Curve 41.49 in. hollow copper. Ref. 14. Washed in same manner
as for curve 1. Test configuration: three-phase line, 30-foot flat
spacing, 16-foot sag, 30-foot ground clearance, 700 feet long.
Curve 52.00 in. hollow aluminum. Ref. 14. Washed in same manner as for curve 1. Test configuration: three-phase line, 30-foot
flat spacing, 16 foot sag, 30-foot ground clearance, 700 feet long.
Curve 61.09 in. steel-aluminum.
Ref. 22. Reference curve is average fair-weather corona loss obtained by converting per-phase
measurements to three-phase values, for a line 22.9 foot flat spacing, 32.8 feet high. This conductor used on 220-kv lines in Sweden
which have above dimensions.
Ref. 22 App. A. Plotted curve
Curve 7l.25 in. steel-aluminum.
obtained by estimating average of a number of fair-weather perphase curves given in reference and converting to three-phase loss
for line having 32-foot flat spacing, 50-foot average height.
Curve 81.04 in. steel-aluminum, 24-strand. Ref. 23. Plotted curve
obtained by conversion of per-phase measurements to three-phase
values, using an estimated value for charging kva, to give loss on a
line having 45-foot flat configuration.
Curve 90.91 in. Hollow Copper.
Ref. 11. Conductor washed.
Test configuration: three-phase line, 20-foot flat spacing, 700 feet
long.
Fig. 28--Fair-Weather
Corona-Loss Curves for Stranded
ductors; Air Density Factor, 6= 1.
and foul-weather
corona loss and the variation which can
be expected during fair weather is shown in Fig. 30 for
one conductor.
Corona
loss on a satisfactory
line is primarily caused by
rain. This is shown by the fairly high degree of correlation
between total rainfall and integrated corona loss which has
been noted. (21.28*41)The corona loss at certain points on a
transmission
line can reach high values during bad storm
conditions.
However, such conditions are not likely to occur simultaneously
all along a line. Borgquist
and Vrethem expect only a variation
from 1.6 to 16 kw per mile,
with an average value of 6.5 kw per mile, on their 380-kv
lines now under construction
in Sweden.
The measured
loss on their experimental
line varied from 1.6 to 81 kw per
mile. The calculated fair-weather
corona loss common in
the U.S.A. is generally less than one kw per mile, based on
calculations
using Reference
must be considered, the annual corona loss will not be of
much economic importance20, and the maximum
loss will
not constitute
Corona loss is characterized
on linear coordinates
by a
rather gradual increase in loss with increased voltage up to
the so-called knee and above this voltage, a very rapid
increase in loss. The knee of the fair-weather
loss curve is
generally near the critical disruptive voltage.
A transmis-
Con-
## sion line should be operated at a voltage well below the
voltage at which the loss begins to increase rapidly under
fair-weather
eonditions.
Operation at or above this point
can result in uneconomical
corona loss. A very careful analysis, weighing the annual energy cost and possibly the
maximum
demand against reduced capitalized
line cost,
must be made if operation at a voltage near or above the
knee of the fair-weather
loss curve is contemplated.
Corona loss on a conductor is a function of the voltage
Thus the effect of reduced conductor spacing and lowered height is to increase the corona
loss as a function of the increased gradient.
On transmission lines using a flat conductor configuration,
at the surface of the middle phase conductor is higher than
on the outer conductor.
This results in corona being mo;e
prevalent
on the middle conductor.
Radio influence is probably the factor limiting the choice
of a satisfactory
conductor for a given voltage.
The RI
performance
of transmission
lines has not been as thoroughly investigated
as corona loss. Recent publications
(see references) present most of the information
available.
RI plotted
against
voltage
on linear
graph
paper
is characterized
by a gradual increase in RI up to a vol-
Chapter
Characteristics
of Aerial
59
Lines
Curve l4/0.985/15.7*
(Smooth) Ref. 25. 6 not given, but assumed
1.10, which is average value for Germany.
Reference curve obtained by converting single-phase measurements to three-phase
values on the basis of surface gradient. Dimensions of line used in
making conversion are not given.
Curve 24/0.827/15.7*
(stranded aluminum-steel).
Ref. 25. 6 =
1.092. See discussion of Curve 1.
Curve 33/0.985/11.8*
(Smooth).
Ref. 26. 6 = 1.092. Reference
curve gives single-phase measurements versus line-to-ground voltage, but it is not clear whether actual test voltage or equivalent
voltage at line height is given. Latter was used in making the conIf this is wrong, curve is approximately
version to three-phase.
15 percent low in voltage. Converted to flat configuration of 45
feet.
Curve 42/1.09/17.7*
(Stranded aluminum-steel). 6 = 1.01. Ref. 12,
App. A. Reference curve gives per-phase measurements versus
gradient. Converted to three-phase corona loss on line of 42.5-foot
average height, 39.4-foot flat configuration.
Curve 52/1.25/17.7*
(Stranded aluminum-steel) 6 not given, probably close to unity. Ref. 12. Reference curve, which gives threephase corona loss,- was converted from per-phase measurements.
Dimensions 42.5 feet average height, 39.4 feet flat configuration.
This conductor was selected for use on the Swedish 380-kv system.
Original author probably selected a worse fair-weather condition
than the writer did in plotting curve 4, which could account for
their closeness.
Curve 62/1.04/23.7*
(Stranded aluminum-steel). 6 not given. Ref.
13. Plotted curve is average of two single-phase fair-weather
curves, converted to three-phase loss for 45-foot flat configuration.
See Curve 7.
Curve 72/1.04/15.7*
(Stranded aluminum-steel). 6 not given. Ref.
13. Plotted curve is average of two single-phase fair-weather
curves, converted to three-phase loss for 45-foot flat configuration.
Data for curves 6 and 7 were taken at same time in order to show
effect of sub-conductor separation.
*Bundle-conductor
designation- number of sub-conductors/outside diameter of each sub-conductor
in inches/separation
between
Fig. 29Fair-Weather
Corona-Loss
Three-,
## tage slightly below the minimum
voltage at which measurable corona loss is detected.
Above this voltage, the
increase in the RI is very rapid. The rate of increase in RI
is influenced
by conductor
surface and diameter,
being
higher for smooth conductors
and large-diameter
conductors.
Above a certain voltage, the magnitude
of the
RI field begins to level off. For practical conductors,
the
leveling off value is much too high to be acceptable,
and
where RI is a factor, lines must be designed to operate below the voltage at which the rapid increase starts during fair
weather.
Figures 32 and 33 are characteristic
RI curves.
The relation between fair- and foul-weather
corona performance is shown in Fig. 32.
An evaluation
of RI in the design of a high-voltage
line
must consider not only its magnitude,
but its effect on the
various communication
services which require protection.
Amplitude-modulated
and power-line carrier
are the most common services encountered
but other services such as aviation, marine, ship-to-shore
SOS calls, police and a number of government
services might also have
to be considered.
In determining
the RI performance
of a proposed line,
the magnitude
of the RI factors for the entire frequency
and Four-conductor
Bundles;
## Air Density Factor, 6= 1.00.
range of communication
services likely to be encountered,
should be known. An evaluation
of these factors in terms
of their effect on various communication
services must take
into consideration
many things. These are available signal
intensities
along the line, satisfactory
signal-to-noise
ratios, effect of weather on the RI factors and on the importance
of particular
communication
services, number
and type of receivers in vicinity of the line, proximity of
particular
receivers, transfer of RI to lower-voltage
circuits, the general importance
of particular
communication
services, and means for improvement
of reception at individual receiver locations.21
For extra-high-voltage
and
double-circuit
high-voltage
lines the tolerable limits of RI
might be higher because the number of receivers affected,
the coupling to lower voltage circuits, and the coupling to
Also fewer lines are required
for the same power handling
ability, and wider right-ofways are used which tend to reduce the RI problem.
Although RI increases very rapidly with increased gradient at the surface of a conductor, theoretical considerations
characteristics
of a transmission
line as
spacing is reduced, indicate that the RI from a transmission
line will not be seriously affected by reduced spacing.42
60
Characteristics
of Aerial
Lines
Chapter 3
Standard
meters35,36 can measure the average, quasi-peak, and peak values of the RI field. The average value is the amplitude of the RI field averaged continuously over 1/2 second. For quasi-peak measurements,
a circuit having a short time constant
(0.001-0.01
sec.) for
charging and a long time constant (0.3 to 0.6 sec.) for discharging is used, with the result that the meter indication
is near the peak value of the RI field. Aural tests of radio
reception indicate that quasi-peak readings interpreted
in
field strengths
represent more
accurately the nuisance
value of the RI field. The peak
value is the maximum instantaneous
value during a given
period. The type of measurements
before evaluating
published RI information
conclusions
can be drawn.
The lateral attenuation
of RI from a transmission
line
depends on the line dimensions and is independent
of voltage. At distances between 40 and 150 feet from the outer
conductor, the attenuation
at 1000 kc varies from 0.1 to 0.3
db per foot, with the lower values applying generally to
high-voltage
lines. Typical lateral attenuation
curves are
shown in Fig. 34. Lateral attenuation
is affected by local
conditions.
Because of the rapid attenuation
of RI laterally from a line, a change of a few hundred feet in the location of a right-of-way
can materially
aid in protecting
a
communication
service.
9. Selection of Conductor
## Fig. 30Corona Loss on 1.09 Inch Stranded Aluminum-Steel
Conductor under Different Weather Conditions.
This conductor is in use on the Swedish 220-kv system. Note variation
in fair-weather corona loss and the relation between fair- and
foul-weather corona loss. Plotted curves obtained by converting per-phase measurements
to three-phase values for a line
having 32-foot flat spacing, 50-foot average height. No correction made for air density factor. Ref. 22, App. A.
## The conductor configuration,
the number of circuits, and
the presence of ground wires affect the radiation from the
line with a given RI voltage on the conductors.
Very little
characteristics
of transmission lines and caution should be exercised in applying data
not taken on a line configuration
closely approximating
the
design under consideration.
The RI field from a transmission
line varies somewhat
as the inverse of the radio frequency measured.
Thus services in the higher-frequency
bands, (television37, frequencymodulated
less apt to be affected.
Directional
antennas
which are
generally
used at these frequencies,
on the average, increase the signal-to-noise
ratio. The lower signal strengths,
and wider band-widths
generally found in the high-frequency bands can alter this picture somewhat.
Frequencymodulated
is inherently
less sensitive
to RI
because of its type of modulation.
## In the selection of a satisfactory
conductor
from the
standpoint
of its corona performance
for voltages up to
230 kv, operating experience and current practice are the
best guide. Experience in this country indicates that the
corona performance
of a transmission
line will be satisfactory when a line is designed so that the fair-weather
corona loss according to Petersons formu1a,16 is less than
one kw per three-phase
mile. Unsatisfactory
corona. performance in areas where RI must be considered has been
reported for lines on which the calculated corona loss is in
excess of this value, or even less in the case of medium highvoltage lines. Figure 31 is based on Petersons formula and
indicates
satisfactory
conductors
which can be used on
high-voltage
lines. For medium high-voltage lines (138 kv)
considerably
more margin below the one kw curve is necessary because of the increased probability
of exposure of receivers to RI from the line, and a design approaching
0.1
kw should be used.
## 10. Bundle Conductors
A bundle conductor
is a conductor made up of two or
more sub-conductors,
and is used as one phase conductor. Bundle conductors are also called duplex, triplex, etc.,
conductors,
referring to the number of sub-conductors
and
are sometimes
referred to as grouped or multiple
conductors.
Considerable
work on bundle conductors
has
been done by the engineers of Siemens-Schuckertwerke27
who concluded that bundle conductors were not economical at 220 kv, but for rated voltages of 400 kv or more, are
the best solution for overhead transmission.
Rusck and
Rathsman46 state that the increase in transmitting
capacity justifies economically
the use of two-conductor
bundles on 220-kv lines.
Fig. 31Quick-Estimating
61
Chapter 3
Corona-Loss
## Curves based on Petersons formula
Curves.
Carrol and Rockwell paper for comparison.
The advantages of bundle conductors are higher disruptive voltage with conductors of reasonable dimensions, reduced surge impedance and consequent higher power capabilities, and less rapid increase of corona loss and RI with
increased voltage. 22,27,28
against increased circuit cost, increased. charging kva if it
cannot be utilized, and such other considerations as the
large amount of power which. would be carried by one circuit. It is possible with a two-conductor bundle composed
of conductors of practical size to obtain electrical characteristics, excepting corona, equivalent to a single conductor
up to eight inches in diameter.
Theoretically there is an optimum sub-conductor separation for bundle conductors that will give minimum crest
gradient on the surface of a sub-conductor and hence highest disruptive voltage. For a two-conductor bundle, the
separation is not very critical, and it is advantageous to
use a larger separation than the optimum which balances
the reduced corona performance and slightly increased circuit cost against the advantage of reduced reactance.
Assuming isolated conductors which are far apart compared to their diameter and have a voltage applied between them, the gradient at the surface of one conductor
is given by:
(79b)
with
a few check
points
from
the
## where the symbols have the same meaning as used in Eq.
(79a). This equation is the same as equation (79a), except
that surface factor, m, and air density factor, S, have been
omitted. These factors should be added to Eqs. 80 and 81
for practical calculations. For a two-conductor bundle, the
equation for maximum gradient at the surface of a subconductor33 is:
(80)
where:
S = separation between sub-conductors
in centimeters.
## Because of the effect of the sub-conductors on each
other, the gradient at the surface of a sub-conductor is not
uniform. It varies in a cosinusoidal manner from a maximum at a point on the outside surface on the line-of-centers, to a minimum at the corresponding point on the inside
surface. This effect modifies the corona performance of a
bundle conductor such that its corona starting point corresponds to the voltage that would be expected from calculations, but the rate of increase of corona with increased
voltage is less than for a single conductor. This effect can
be seen by comparing curve 6 of Fig, 28 with curve 2
of Fig. 29. Cahen and Pelissier2124concluded that the
corona performance of a two-conductor bundle is more
accurately indicated by the mean between the average
## Characteristics of Aerial Lines
62
KILOVOLTS
Fig. 32Radio influence and corona loss measurements
on an experimental test line. Ref. 26.
Chapter
Fig. 33Fair-Weather
Field from a Transmission Line as a Function of Voltage. Measurements made
opposite mid-span
on the 230-kv Covington-Grand
Coulee
Line No. 1 of the Bonneville Power Administration.
RI values
1.108 inch ACSR conductor, 27-foot flat spacare quasi-peak.
ing, 41-foot height, test frequency800 kc.
## and maximum gradient at the surface of a sub-conductor,
which is given by:
(81)
If it is desired to determine the approximate disruptive
voltage of a conductor,
meter rms can be substituted for g and the equations solved
for eO in kv rms. This value neglects air density Factor
and surface factor, which can be as low as 0.80 (consult
references 10 and 16 for more accurate calculations).
380 kv Systems using bundle conductors are being built
or under consideration in Sweden, France, and Germany.
Curve lAverage lateral attenuation for a number of transmission
lines from 138- to 450-kv. O X A l are plotted values which apply
to this curve only. Test frequency 1000 kc. Ref. 21.
Curve 2Lateral Attenuation from the 220-kv Eguzon-Chaingy line
in France. Line has equilateral spacing, but dimensions not given.
Distance measured from middle phase. Test frequency868 kc.
Ref. 24.
Curve 3Lateral Attenuation from 230-kv MidwayColumbia
Line
Conductor height 47.5
feet, test frequency 830 kc. Ref. 42.
## HORIZONTAL DISTANCE FROM OUTSIDE CONDUCTOR-FEET
Fig. 34Lateral Attenuation of Radio Influence in Vicinity of
High-Voltage
Transmission Lines.
Chapter
## Characteristics of Aerial Lines
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3.
4.
,.
5
6.
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A.I.E.E.
Transactions, Vol. 62, 1934, pages 169-72.
Measurements Pertaining to the Coordination of Radio Reception with Power Apparatus and Systems, C. M. Foust, C. W.
Frick, A.I.E.E. Transactions Vol. 62,1943, pages 284-91, disc. 458.
AIEE paper No. 47-140.
Results of Tests Carried out at the 500-kv Experimental Station
of Chevilly (France), Especially on Corona Behavior of Bundle
Conductors, F. Cahen, A.I.E.E.
Transactions, 1948, Vol. 67,
Part II, pages 1118-25.
Radio-Noise Influence of 230-KV Lines, H. L. Rorden, A.I.E.E.
Transaction, Vol. 66, 1947, pages 6778: disc. 682.
Radio Influence from High Voltage Corona, G. R. Slemon, AIEE
paper No. 49-60.
Comparative Investigation of D. C.- and A. C.-Corona on TwoConductor Transmission Lines (In German), R. Strigel, Wissenschaftliche Veroeftentlichungen A us Den Siemens-Werken, Vol.
15, Part 2, 1936, pages 68-91.
The Swedish 380 KV System, A. Rusck, Bo G. Rathsman, Electrical Engineering, Dec. 1949, pages 102.5-9.
Series Capacitor and Double Conductors in the Swedish Transmission System, A. Rusck, Bo G. Rathsman, Electrical Engineering, Jan. 1950, pages 537.
Effect of Earthing on Corona Losses, Conductor Diameter And
Length of Insulator Strings, The Brown Boveri Review, Vol.
XXXV
NOS. 7/8, July/August,
1948, pages 192-201.
The Transmission of Electric Power (a book), by W. A. Lewis,
(1948 Lithoprinted Edition) Illinois Institute of Technology. | 26,115 | 98,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-26 | latest | en | 0.914931 |
https://stdlib.io/develop/docs/api/@stdlib/math/base/dists/erlang/logpdf/ | 1,571,642,535,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00061.warc.gz | 720,319,562 | 54,610 | # Logarithm of Probability Density Function
Evaluate the natural logarithm of the probability density function (PDF) for an Erlang distribution.
The probability density function (PDF) for an Erlang random variable is
where k is the shape parameter and lambda is the rate parameter.
## Usage
var logpdf = require( '@stdlib/math/base/dists/erlang/logpdf' );
#### logpdf( x, k, lambda )
Evaluates the natural logarithm of the probability density function (PDF) for an Erlang distribution with parameters k (shape parameter) and lambda (rate parameter).
var y = logpdf( 0.1, 1, 1.0 );
// returns ~-0.1
y = logpdf( 0.5, 2, 2.5 );
// returns ~-0.111
y = logpdf( -1.0, 4, 2.0 );
// returns -Infinity
If provided NaN as any argument, the function returns NaN.
var y = logpdf( NaN, 1, 1.0 );
// returns NaN
y = logpdf( 0.0, NaN, 1.0 );
// returns NaN
y = logpdf( 0.0, 1, NaN );
// returns NaN
If not provided a nonnegative integer for k, the function returns NaN.
var y = logpdf( 2.0, -2, 0.5 );
// returns NaN
y = logpdf( 2.0, 0.5, 0.5 );
// returns NaN
If provided k = 0, the function evaluates the logarithm of the PDF of a degenerate distribution centered at 0.
var y = logpdf( 2.0, 0.0, 2.0 );
// returns -Infinity
y = logpdf( 0.0, 0.0, 2.0 );
// returns Infinity
If provided lambda <= 0, the function returns NaN.
var y = logpdf( 2.0, 1, 0.0 );
// returns NaN
y = logpdf( 2.0, 1, -1.0 );
// returns NaN
#### logpdf.factory( k, lambda )
Returns a function for evaluating the PDF for an Erlang distribution with parameters k (shape parameter) and lambda (rate parameter).
var mylogpdf = logpdf.factory( 3, 1.5 );
var y = mylogpdf( 1.0 );
// returns ~-0.976
y = mylogpdf( 4.0 );
// returns ~-2.703
## Examples
var randu = require( '@stdlib/random/base/randu' );
var round = require( '@stdlib/math/base/special/round' );
var logpdf = require( '@stdlib/math/base/dists/erlang/logpdf' );
var lambda;
var k;
var x;
var y;
var i;
for ( i = 0; i < 20; i++ ) {
x = randu() * 10.0;
k = round( randu() * 10.0 );
lambda = randu() * 5.0;
y = logpdf( x, k, lambda );
console.log( 'x: %d, k: %d, λ: %d, ln(f(x;k,λ)): %d', x.toFixed( 4 ), k, lambda.toFixed( 4 ), y.toFixed( 4 ) );
} | 734 | 2,201 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-43 | latest | en | 0.256208 |
https://tutorbin.com/questions-and-answers/2-a-pilot-study-is-run-to-investigate-the-feasibility-of-recruiting-pr | 1,660,789,988,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00611.warc.gz | 533,804,472 | 17,399 | Question
Statistics
2. A pilot study is run to investigate the feasibility of recruiting pregnant women into a study of risk factors for preterm delivery. Women are invited to participate at their first clinical visit for prenatal care. The following represent the gestational ages in weeks of women who consent to participate in the study.
10 16 25 24 6 12 15 19
a. Compute the sample mean gestational age.
b. Compute the sample standard deviation of gestational age.
c. Compute the median gestational age
d. What proportion of the sample enroll in first trimester of pregnancy (i.e., between 1-13 weeks, inclusive, of pregnancy)?
Verified
### Question 37038
Statistics
Let {X1, X2,...XN} be a random sample where each random variable X, has a PDF given by
f(x)=\frac{1}{(\lambda+\mu)} e^{\frac{1}{2}[-(\operatorname{sign}(x)+1)(x / \lambda)+(\operatorname{sign}(x)-1)(x / \mu)]}
\operatorname{sign}(x)=\left\{\begin{array}{rr} 1, & x \geq 0 \\ -1, & x<0 \end{array}\right.
where the sign function is defined as
and N denoting an integer, and the parameters , µ satisfying 1 > 0, µ > 0.In the followings provide complete derivations and explanations of results.
a) Find the likelihood function and the log-likelihood function for the parameters 1,µ
b) Find Maximum Likelihood (ML) estimators for the parameters 1, u.
c) Determine if these ML estimators are unbiased for N = 1.
### Question 33214
Statistics
How many ways can 4 pairs of socks be chosen from 16 pairs?
O 1820
O 496
O43680
### Question 33213
Statistics
How many subsets of {1,2,.... 20} have three elements?
O 5814
01140
O 969
O 6840
### Question 33212
Statistics
Suppose there are eight different colors of flowers in a flower vase. How many different ways we select three different colors of flowers from the vase?
0 6720
O 336
06
56O 56
### Question 33209
Statistics
How many eight-digit integers are there that do not contain repeated digits?
O 10-9 8 7-6 5-4-3
09.9.8-7-6 5 4-3
O P(10, 8)
O P(9, 8)
### Question 33207
Statistics
How many 2-permutations of {a, b, c, d)?
O 12
O 24
O 10
O 35
### Question 33206
Statistics
How many permutations of the letters ABCDEFG contain the string "BC" and "FG'as a substrings?
O 120
0720
O 2520
### Question 33205
Statistics
How many permutations of the letters ABCDEFG contain the string "ABC as a substring?
O 5040
0 120
O 24
0720
### Question 33204
Statistics
How many different order can there a set with {a, b. c. 4. 5. Š}?
O 120
O 1040
0720
O 24
### Question 33203
Statistics
How many numbers from the set of integers between 1 and 1000, inclusive, are divisible by 2 or 5.
O 600
O 500
O 300
0400 | 802 | 2,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-33 | longest | en | 0.782967 |
https://www.rempub.com/operations-algebraic-thinking | 1,597,211,764,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738878.11/warc/CC-MAIN-20200812053726-20200812083726-00546.warc.gz | 817,101,863 | 97,140 | # Addition, Subtraction, Multiplication, & Division
Operations & Algebraic Thinking
Model: REM 13A
Grade Level: 2Interest Level: N/ACCSS Math Level: 1*“Solve problems! Color code the answers.” The exercises found here will improve math and motor skills.Example: Twenty ghosts float on one of the worksheets. Inside each goblin is a single addition problem. Once students have completed the math..
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Model: REM GP006
Grade Level: 1-3 Interest Level: N/A Reading Level: N/ABasic addition operations for addition facts through 10. Systematic, first diagnosing skill levels, then practice, periodic review, and testing. Beginning Assessment and Final Assessment Tests provide measurement tool. Beginning addition skills. An ans..
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Model: REM GP015
Grade Level: 3+ Interest Level: N/A Reading Level: N/AHigher level addition skills involving multiple digits and regrouping. Practice, review, and testing included. An answer key is included in this 40-page book.Table of ContentsBEGINNING ASSESSMENT TESTSECTION ONE: No Regrouping Pract..
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Model: REM 0692G
Grade Level: 2-5 Interest Level: N/A Reading Level: N/AThis is the next-step companion to our Straightforward Basic Computational Skills. It builds on concepts learned in Basic Computational Skills, and advances using multiple-digit numbers to perform the operations of addition, subtraction, multiplication..
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Model: REM GP017
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Grade Level: 3+ Interest Level: N/A Reading Level: N/AHigher level subtraction skills involving multiple digits and regrouping. Practice, review, and testing included. An answer key is included in this 40-page book.Table of ContentsBEGINNING ASSESSMENT TESTBASIC FACTS REVIEWBEGINNING ASSE..
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Model: REM GP013
Grade Level: 3-5 Interest Level: N/A Reading Level: N/ABasic division operations for division facts through 10. Systematic, first diagnosing skill levels, then practice, periodic review, and testing. Beginning Assessment and Final Assessment Tests provide measurement tool. Beginning division skills. An ans..
\$7.99
Model: REM 32
Grade Level: 1Interest Level: N/ACCSS Math Level: 1*Solve the problem! Color the answers. The exercises found here will improve math and motor skills.Basic addition computations have been incorporated into bold illustrations. One drawing graces each of the 25 lessons in this learning unit. Afte..
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Model: REM 963
Grade Level: 1-2Interest Level: N/ACCSS Level: 1-2 This 4-book set will add some color to your beginning math program! Your students will absolutely love these delightful activities that involve coloring along with simple math problems. Great for improving motor skills! 24-28 pags.*Cl..
\$17.99
Model: REM 33
Grade Level: 1Interest Level: N/ACCSS Math Level: 1*Solve the problem! Color the answers. The exercises found here will improve math and motor skills.Basic subtraction computations have been incorporated into bold illustrations. One drawing graces each of the 24 lessons in this learning unit. A..
\$4.99
Model: REM 5012E
Grade Level: 1-3Interest Level: N/AReading Level: N/AImprove the speed and accuracy of beginning learners with this simplified version of our best-selling Timed Math Drills! Easy-to-read practice drills with fewer equations per page cover facts 0-9 and feature 36 single-digit equations per drill...
\$29.99
Model: REM 5012A
Grade Level: 1-3Interest Level: N/AReading Level: N/AImprove the speed and accuracy of beginning learners with this simplified version of our best-selling Timed Math Drills! Easy-to-read practice drills with fewer equations per page cover facts 0-9 and feature 36 single-digit equations per drill...
\$7.99
Model: REM 5012D
Grade Level: 1-3Interest Level: N/AReading Level: N/AImprove the speed and accuracy of beginning learners with this simplified version of our best-selling Timed Math Drills! Easy-to-read practice drills with fewer equations per page cover facts 0-9 and feature 36 single-digit equations per drill...
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Model: REM 5012C
Grade Level: 1-3Interest Level: N/AReading Level: N/AImprove the speed and accuracy of beginning learners with this simplified version of our best-selling Timed Math Drills! Easy-to-read practice drills with fewer equations per page cover facts 0-9 and feature 36 single-digit equations per drill...
\$7.99
Model: REM 5012B
Grade Level: 1-3Interest Level: N/AReading Level: N/AImprove the speed and accuracy of beginning learners with this simplified version of our best-selling Timed Math Drills! Easy-to-read practice drills with fewer equations per page cover facts 0-9 and feature 36 single-digit equations per drill...
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Model: REM 104A
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Model: REM GP044
Grade Level: 3-5 Interest Level: N/A Reading Level: N/AThis book is designed to be a measurement tool for the basic mathematical operations of addition, subtraction, multiplication, and division. Mastery Tests will help parents and teachers measure how well these basic math operations are understood by the..
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Model: REM 1130
Grade Level: 3-4Interest Level: N/ACCSS Math Level: 2-3*Loads of Practice + Fun Activities = Better Results!Number searches, dot-to-dots, coloring activities, secret codes, and more offer lots of skill-specific practice with computation. Skills include multi-digit addition and subtraction w..
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Model: REM 529D
Grade Level: 1-5 Interest Level: N/A Reading Level: N/ATeach math problem solving with these creative math word problems guaranteed to hold your students' attention while building basic math skills! These on-grade-level teaching books are the perfect way to reinforce textbook instruction while improving ..
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Model: REM 15A
Grade Level: 1-2Interest Level: N/ACCSS Math Level: 1-2*Solving basic math problems! This lesson unit focuses on simple, child-friendly addition and subtraction problems.Example: “8 birds in the birdbath. 3 leave. How many left?”Each of the 20 lessons in this book features five word problem..
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Model: REM 529A
Grade Level: 2-3 Interest Level: N/A Reading Level: N/ASolving creative math problems! This lesson unit focuses on basic, child-friendly addition and subtraction problems.Example: “391 children watched cartoons early in the morning. 569 children watched afternoon cartoons. How many children wat..
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Model: REM 529B
Grade Level: 3-4 Interest Level: N/A Reading Level: N/ASolving creative math problems! This lesson unit focuses on basic, child-friendly addition, subtraction, and multiplication problems.Example: “The grapes looked delicious! A bag of grapes was \$3.82. If the apples were \$2.04, how much more were t..
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Model: REM 529C
Grade Level: 4-5 Interest Level: N/A Reading Level: N/ATeach math problem solving with these creative math word problems guaranteed to hold your students' attention while building basic math skills! This on-grade-level teaching book is the perfect way to reinforce textbook instruction while improving math ..
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Model: REM GP007
Grade Level: 3-5 Interest Level: N/A Reading Level: N/AThe Straight Forward Math Series has been designed for parents and teachers to use with their students. It is a ssimple, straightforward way to teach multiplication facts 0-10. This book features a beginning assessment test, drill pages that work on on..
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Model: REM 537B
Grade Level: 1-2Interest Level: N/ACCSS Level: 1-2 Learning basic math! Each of the 24 lessons in the learning unit features a “number line” of 1-10, 1-15, or 1-20. Students are challenged to use these graphic tools to figure out fundamental addition and subtraction problems.“9 + 1..
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Model: REM GP062B
Grade Level: 2-5 Interest Level: 2-12 Reading Level: N/AEach hands-on, self-checking lessons feature a specific math skill, skill practice, and an animal puzzle. Instructions and animal background information are provided in English and Spanish. 32 pages each.Math Skills Include: - Addition (no regrou..
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Model: REM GP049
Grade Level: 2-5 Interest Level: 2-12 Reading Level: N/AEach hands-on, self-checking lesson features a specific math skill, skill practice, and an animal puzzle. Instructions and animal background information are provided in English and Spanish. 32 pages.Addition skills with no regrouping. Performing a..
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New
Model: REM GP051
The Puzzles and Practice Series builds basic math skills and acquaints student with animal science. The Series is also designed to challenge skills associated with following directions, simple logic, visual discrimination (all puzzle assembly skills), and motor skills (cutting and pasting).Table of Content..
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Model: REM GP062
Grade Level: 2-5 Interest Level: 2-12 Reading Level: N/AEach hands-on, self-checking lesson features a specific math skill, skill practice, and an animal puzzle. Instructions and animal background information are provided in English and Spanish. 32 pages.Multiplication skills. Performing multiple digit..
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New
Model: REM GP059
Grade Level: 2-5Interest Level: 2-12Reading Level: N/AEach hands-on, self-checking lesson features a specific math skill, skill practice, and an animal puzzle. Instructions and animal background information are provided in English and Spanish. 32 pages.Division skills. Performing division skills constructs..
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Model: REM GP052
Grade Level: 2-5 Interest Level: 2-12 Reading Level: N/AEach hands-on, self-checking lesson features a specific math skill, skill practice, and an animal puzzle. Instructions and animal background information are provided in English and Spanish. 32 pages.Multiplication skills. Performing multiplication..
\$5.99
Model: REM GP050
Grade Level: 2-5 Interest Level: 2-12 Reading Level: N/AEach hands-on, self-checking lesson features a specific math skill, skill practice, and an animal puzzle. Instructions and animal background information are provided in English and Spanish. 32 pages.Subtraction skills with no regrouping. Performin..
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Model: REM 1129C
Grade Level: 2-4Interest Level: N/ACCSS Math Level: 1-3*Do your students struggle with word problems? Our unique Step-by-Step Solution gives students the tools to understand word problems and actually enjoy working them! With the help of just 6 easy-to-follow steps, students are able to break apart w..
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Grade Level: 2-3Interest Level: N/ACCSS Math Level: 1-2*Do your students struggle with word problems? Our unique Step-by-Step Solution gives students the tools to understand word problems and actually enjoy working them! With the help of just 6 easy-to-follow steps, students are able to break apart w..
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Grade Level: 3-4Interest Level: N/ACCSS Math Level: 2-3*Do your students struggle with word problems? Our unique Step-by-Step Solution gives students the tools to understand word problems and actually enjoy working them! With the help of just 6 easy-to-follow steps, students are able to break apart w..
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New
Model: REM GP031A
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Grade Level: 2-5 Interest Level: N/A Reading Level: N/AThe Beginning Straight Forward Math Series is systematic: first diagnosing skill levels, then practice, periodic review, and testing. Ideal for students working on Grades 2-5 math skills. An answer key is included with each 40- to 95-page book.Re..
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Model: REM 14A
Grade Level: 2Interest Level: N/ACCSS Math Level: 1*Solve the problem! Color the answers. The exercises found here will improve math and motor skills.Example: 18 gum balls are packed into a coin-operated machine on one of the worksheets. Inside each gumball is a single subtraction problem. Once..
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Showing 1 to 40 of 49 (2 Pages) | 3,078 | 12,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-34 | latest | en | 0.832081 |
https://www.hackmath.net/en/math-problem/31341?tag_id=37 | 1,601,177,726,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400250241.72/warc/CC-MAIN-20200927023329-20200927053329-00213.warc.gz | 858,213,260 | 13,087 | # The body
The body has dimensions of 2m 2dm and 10 cm. It weighs 28 kg. What is its density?
Correct result:
h = 700 kg/m3
#### Solution:
We would be very happy if you find an error in the example, spelling mistakes, or inaccuracies, and please send it to us. We thank you!
Tips to related online calculators
Do you know the volume and unit volume, and want to convert volume units?
Tip: Our Density units converter will help you with the conversion of density units.
Do you want to convert mass units?
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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The shape of the observatory dome is close to the hemisphere. Its outer diameter is 11 m. How many kilograms of paint and how many liters of thinner is used for its double coat if you know that 1 kg of paint diluted with 1 deciliter of thinner will paint | 915 | 3,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-40 | latest | en | 0.885566 |
https://quant.stackexchange.com/questions/69355/mean-variance-optimization-max-sharpe-ratio-portfolio | 1,702,115,190,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00352.warc.gz | 509,404,865 | 43,958 | # mean-variance optimization === max sharpe ratio portfolio?
Noobie here. I just wanna ask a simple question:
in the context of portfolio optimization, is Mean-Variance optimization the same as the max sharpe ratio portfolio?
• While they typically yield the same portfolio, they have different objective functions. In some literature you will see an analysis of the sensitivity of the optimal objective with respect to estimation error in the input. This is a different analysis for MVO than for max-Sharpe. Jan 5, 2022 at 17:44
Basically the answer is yes, although we can also give a slightly more complicated answer:
In Mean Variance Optimization we traditionally consider two problems:
First the slightly simpler problem when there are N risky assets. In this case the solution is a curve, the famous "efficient frontier".
Then, in the next chapter of the textbook, we consider that there are N risky assets and one risk-free asset, so a total of N+1 assets. In this case we can go a little further and the solution concept involves a single point on the frontier, the famous "tangency portfolio" which is also the point that achieves the "maximum sharpe ratio". And mixes of risk free and tangency portfolio also have this Sharpe ratio and are valid solutions.
(So in this version of the problem the answer to your question is a definite yes. But you will also find people who will say that Mean Variance Optimization is equivalent to finding the efficient frontier; that is another way to look at it, when you don't assume a risk-free asset).
• (+1) I'd call finding the maximum Sharpe ratio portfolio a special case of mean-variance portfolio optimization. Jan 5, 2022 at 19:37
To be precise, no! Mean-variance optimization and the maximum Sharpe Ratio portfolio are related but different concepts.
• When someone says "mean-variance optimization" I think of someone formulating a portfolio choice problem where the investor chooses a portfolio return $$R_p$$ from a feasible set $$\mathcal{S}$$, and the investor only has preferences over the mean and variance of the portfolio return. As an expected utility maximization problem, it would take the form:
$$$$\begin{array}{*2{>{\displaystyle}r}} \mbox{maximize (over R_p)} & \operatorname{E}[u(\mu, \sigma^2)] \\ \mbox{subject to} & \operatorname{E}[R_p] = \mu \\ & \operatorname{Var}(R_p) = \sigma^2 \\ & R_p \in \mathcal{S} \end{array}$$$$
• If investor preferences are such that they like higher returns and dislike higher variance, then a mean-variance investor will choose a portfolio such that its return is somewhere on the efficient side of the mean-variance frontier. (I define these terms below.)
• The maximum Sharpe Ratio portfolio (aka tangency portfolio) is a particular portfolio on the efficient side of the mean-variance frontier.
• The maximum Sharpe Ratio portfolio comes up a lot, but that an investor only cares about mean and variance does not on its own imply that he/she will buy the maximum Sharpe Ratio portfolio. You need additional assumptions.
• An investor that only cares about mean and variance (likes a higher mean return and dislikes higher variance) will choose a portfolio along the mean-variance frontier: the frontier of minimum volatility for any given expected return. When someone says "mean-variance optimization," they may be referring to optimization problem of finding the mean-variance frontier.
• Note that this kind of mean-variance investor doesn't care about hedging risks from their job, weather, etc...; doesn't care about skewness, maximum loss, etc.... It's a strong assumption.
## The mean-variance frontier (minimum variance for any given mean)
Let $$R$$ be a random vector denoting the return of $$n$$ different assets. $$\operatorname{E}[R]$$ is a vector of expected returns. Let $$\Sigma = \operatorname{Var}(R)$$ be the covariance matrix for the assets. Let $$\mathbf{x}$$ be a vector of portfolio weights. Let's say that buying and shorting is allowed so that feasible set $$\mathcal{S} = \left\{ \mathbf{x} \cdot R \; |\; \mathbf{x} \in \mathbb{R}^n \text{ and } \sum x_i = 1 \right\}$$. Note this will achieve portfolio return $$\mathbf{x}' R$$. The expected return of the portfolio is $$\mathbf{x}' \operatorname{E}[R]$$ (i.e. $$\sum_i x_i \operatorname{E}[R_i]$$). The variance for the portfolio return is $$\mathbf{x}'\Sigma \mathbf{x}$$.
The point $$\left( \mu , \mathbf{x}'\Sigma \mathbf{x} \right)$$ lies on the mean-variance frontier if it's a solution to:
$$$$\begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over \mathbf{x})} & \mathbf{x}'\Sigma \mathbf{x} \\ \mbox{subject to} & \sum x_i = 1 \\ & \mathbf{x}'\operatorname{E}[R] = \mu \end{array}$$$$
That is, portfolio return $$\mathbf{x}'R$$ achieves expected return $$\mu$$ at minimum possible variance. It turns out the mean-variance frontier has two sides:
• The so called efficient side where a portfolio has maximum expected return for a given variance.
• The inefficient side where a portfolio has minimum expected return for a given variance.
• Practical comment: Stepping away from theory and back to practical reality, the problem here is of garbage in and garbage out: you don't know $$\operatorname{E}[R]$$ or $$\Sigma$$ with any precision, and garbage inputs lead to garbage portfolio choice outputs. A quant approach to portfolio choice will involve trying to generate somewhat less ridiculously noisy estimates of expected returns and covariance and formulating a more humble portfolio choice problem cognizant of how difficult it is to estimate those objects.
## Mean variance efficient portfolios
A highly related (but different) problem is to achieve maximum expected return for a given variance $$\sigma^2$$. A portfolio is said to be mean-variance efficient if it gives the maximum expected return achievable for a given level of variance:
$$$$\begin{array}{*2{>{\displaystyle}r}} \mbox{maximize (over \mathbf{x})} & \mathbf{x}'\operatorname{E}[R] \\ \mbox{subject to} & \sum x_i = 1 \\ & \mathbf{x}'\Sigma\mathbf{x} = \sigma^2 \end{array}$$$$
Mean-variance efficient portfolios give returns that lie along higher expected return (i.e. efficient) side of the mean-variance frontier.
## Mean-variance frontier constructed from risky assets
If $$R$$ includes all risky assets and $$\Sigma$$ is full rank (i.e. random vector $$R$$ does NOT include the risk-free rate or redundant assets that allow construction of a risk free rate), then the corresponding mean-variance frontier is called the mean-variance frontier of risky assets.
The distinction here is that one is excluding the risk free rate from the allowed investments.
## Maximum Sharpe ratio portfolio (i.e. tangency portfolio)
A particular portfolio on the efficient side of the mean-variance frontier constructed using risky assets is the tangency portfolio. This portfolio lies on the mean-variance frontier of risky assets and achieves the maximum possible Sharpe ratio.
$$$$\begin{array}{*2{>{\displaystyle}r}} \mbox{maximize (over \mathbf{x})} & \frac{\mathbf{x}'\left( \operatorname{E}[R] - r_f \right)}{\sqrt{\mathbf{x}'\Sigma\mathbf{x}}} \\ \mbox{subject to} & \sum x_i = 1 \end{array}$$$$
### Mean-variance efficient frontier of all assets
If there exists a risk free rate, it can be shown that the efficient side of the mean-variance frontier over all assets (as opposed to just risky assets) is produced by investing in varying combinations of the tangency portfolio and the risk free rate.
Max Sharpe portfolio is a special case in Mean Variance optimization. Special in the way that by setting the right risk aversion parameter, you will get the same result from both optimization methods. However it is not right to say they are the same. Mean Variance optimization is much more flexible as the user can tailor the optimization set-up by setting the suitable risk aversion parameter. | 1,871 | 7,899 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 4, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-50 | latest | en | 0.917182 |
https://ocw.mit.edu/courses/physics/8-02-physics-ii-electricity-and-magnetism-spring-2007/problem-solving/index.htm | 1,618,732,521,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00378.warc.gz | 522,655,588 | 9,669 | # Problem Solving
There are two types of problem solving activities for this class.
• Group Problem Solving (Mondays and Wednesdays)
• Problem Solving Sessions (Fridays)
## Group Problem Solving (Mondays and Wednesdays)
These in-class problems are solved in groups and are not graded.
SES # TOPICS
1 Group problem (PDF)
2
Group problem (PDF)
Group: Line of charge (PDF)
Group: Uniformly charged disk (PDF)
4
Group problem: Superposition (PDF)
Group problem: E from V (PDF)
Group problem: Build it (PDF)
7
Group problem: Charge slab (PDF)
Group problem: Charge slab (PDF)
9 Group problem: Spherical shells (PDF)
10 Partially filled capacitor (PDF)
12 Group problem: B field from coil of radius R (PDF)
14
Group problem: Non-uniform cylindrical wire (PDF)
Group problem: Current sheet (PDF)
17 Group problem: Current loop (PDF)
18 Group problem: Circuit (PDF)
20
Group problem: Changing area (PDF)
Group problem: Generator (PDF)
21 Group problem: Solenoid (PDF)
23
Group problem: Coaxial cable (PDF)
Group problem: Circuits (PDF)
28 Group problem (PDF)
30
Superposition principle (PDF)
Group problem: Plane waves (PDF)
31
Group problem: Inductor (PDF)
Group problem: Capacitor (PDF)
33
Group problem: B field generation (PDF)
Group problem: Energy in wave (PDF)
## Problem Solving Sessions (Fridays)
Counts toward 6% of the course grade.
SES # TOPICS
3 Coordinate systems; Gradients; Line and surface integrals
6 Continuous charge distributions (PDF)
8 Gauss's law (PDF)
11 Capacitors (PDF)
16 Ampere's law (PDF)
19 Magnetic fields: Force and torque on a current loop (PDF)
22 Mutual inductance and transformers; Inductors (PDF)
26 RC and RL circuits (PDF)
29 Driven LRC circuits (PDF) | 439 | 1,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-17 | latest | en | 0.633329 |
http://mathhelpforum.com/trigonometry/160517-prove-identity.html | 1,524,627,724,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00024.warc.gz | 208,660,944 | 9,866 | 1. ## Prove this identity
Can anyone help me prove this trig identity:
Cos(4ά)+4cos(2ά)+3=8cos(4ά)
Thanks
2. I think the identity is
$\displaystyle \cos(4\alpha)+4\cos(2\alpha)+3=8\cos^{4}(\alpha),$ right?
3. Yes, thats right. Sorry...
Thanks
4. Ok. So what ideas have you had so far?
5. [IMG]URL=http://img263.imageshack.us/i/64044892.png/]
Thats what I did...I try everything...still nothing
1. $\displaystyle \cos(4\alpha)=\cos^{4}(\alpha)-2\sin^{2}(\alpha)\cos^{2}(\alpha)+\sin^{4}(\alpha)-4\sin^{2}(\alpha)\cos^{2}(\alpha).$
2. In your expression, you still have a $\displaystyle \cos(2\alpha)$ in there. Reduce the angle, and see what you get. | 230 | 660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-17 | latest | en | 0.746792 |
https://jamesmccaffrey.wordpress.com/2013/05/27/cross-validation-and-neural-networks-and-over-fitting/ | 1,500,654,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423787.24/warc/CC-MAIN-20170721162430-20170721182430-00056.warc.gz | 654,115,675 | 33,643 | ## Cross Validation and Neural Networks and Over-Fitting
Most of the information I see on the Internet about the relationship between cross validation and neural networks is either incomplete or just plain incorrect. Let me explain. Most references, including Wikipedia, state that the purpose of performing cross validation is to get an estimate of the accuracy (or error rate) of a neural network model. That is true, but what the majority of references fail to add is that in most cases the purpose of getting an estimate of an error rate is so that you can do model selection — pick the best from several neural networks. And in most cases you have several models and are dealing with over-fitting. The explanation is a bit subtle.
Suppose you want to create a neural network to predict something. You can start by generating several variations of neural networks, with different combinations of learning rate, momentum, number of hidden nodes, and possibly other features. Let’s say you have 12 variations. Next you perform k-fold cross validation on each of the 12 variations and get an error estimate for each of the 12 variations. (I’ll assume that you understand the mechanics of k-fold cross validation). Now you select the one neural network variation that generated the smallest average error on the k training sets. You have selected the best model in some sense. Finally, you use the learning rate, momentum, and number of hidden nodes of the best variation, with the entire data set as training data, to generate the set of weights and biases for the neural network.
This cross validation approach deals with the problem of over-fitting. If instead of the procedure above, you just tried different neural network variations on the entire data set, you would likely find a variation that fit your data very well — perhaps a variation with many hidden nodes — but the neural network would likely over-fit, and perform poorly when trying to predict using new, previously unseen data.
Now, another possibility to the scenario I just described goes as follows. You decide a priori on a set of neural network’s learning rate, number hidden nodes, and so on. You train on all the available data (first rather than at the end as above), and only then do k-fold cross validation in order to get an estimate of how well your neural network will perform on new data. In this scenario, cross validation is being used just to get an estimate of the error/accuracy of your neural network. But you have ignored the over-fitting issue.
Before I finish, let me mention hold-out validation. You begin up front by separating the data set into a training set (typically the first 80% of the data) and a test set (the remaining 205). Then the ideas are similar. You can train several neural network variations using the training set only and pick the one variation that performs best on the test set. Or you can just train one variation and then estimate the generalization error using the test set.
Finally, there is another related process called train-validate-test. This is used in conjunction with early-stopping, but that’s another story. | 624 | 3,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-30 | longest | en | 0.940431 |
https://www.dlubal.com/de/downloads-und-infos/beispiele/verifikationsbeispiele?entity=shell | 1,553,536,154,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204086.87/warc/CC-MAIN-20190325174034-20190325200034-00127.warc.gz | 746,879,720 | 21,320 | # Verifikationsbeispiele
#### Buch über FEM und RFEM
In diesem Buch für Ingenieure und Studenten erfahren Sie Grundlegendes zur Finite-Elemente-Methode praxisnah anhand von überschaubaren Beispielen, die mit RFEM berechnet wurden.
## Helicoid
VE 0208 10. Februar 2019
A membrane is stretched by means of isotropic prestress between two radii of two concentric cylinders not lying in a plane parallel to the vertical axis. Find the final minimal shape of the membrane - the so-called helicoid, and determine the surface area of the resulting membrane. The add-on module RF-FORM-FINDING is used for this purpose. Elastic deformations are neglected both in RF-FORM-FINDING and in analytical solution, also self-weight is neglected in this example.
## Catenoid
VE 0207 14. November 2018
A cylindrical membrane is stretched by means of isotropic prestress. Find the final minimal shape of the membrane - catenoid. Determine the maximum radial deflection of the membrane. The add-on module RF-FORM-FINDING is used for this purpose. Elastic deformations are neglected both in RF-FORM-FINDING and in analytical solution, also self-weight is neglected in this example.
## Balloon – Prestressed Membrane
VE 0206 8. Oktober 2018
A spherical balloon membrane is filled with gas with atmospheric pressure and defined volume (these values are used for FE model definition only). Determine the overpressure inside the balloon due to the given isotropic membrane prestress. The add-on module RF-FORM-FINDING is used for this purpose. Elastic deformations are neglected both in RF-FORM-FINDING and in analytical solution, self-weight is also neglected in this example.
## Buckling of a Circular Ring
VE 0094 10. Juli 2018
A thin circular ring of rectangular cross-section is exposed to an external pressure. Determine the critical load and corresponding load factor for in-plane buckling.
## Thin-Walled Conical Vessel with Hydrostatic Pressure
VE 0085 19. Februar 2018
A thin-walled conical vessel is filled with water. Thus, it is loaded by the hydrostatic pressure. While neglecting self-weight, determine the stresses in surface line and circumferential direction. The analytical solution is based on the theory of thin-walled vessels. This theory was introduced in Verification Example 0084.
## Scordelis-Lo Roof
VE 0083 4. Dezember 2017
A shell roof structure under pressure load is modelled, where the straight edges are free, while at the curved edges the y- and z‑translations are constrained. Neglecting self‑weight, compute the maximal (absolute) vertical deflection, and compare the results with COMSOL Multiphysics 4.3.
## Thin-Walled Spherical Vessel
VE 0084 4. Dezember 2017
A thin-walled spherical vessel is loaded by inner pressure. While neglecting self‑weight, determine the von Mises stressand the radial deflection of the vessel.
#### Suchen nach Beispiel
Programm / Modul
Materialmodell
Berechnungsmethode
Model-Typ
Spezial-Features | 703 | 2,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-13 | latest | en | 0.710364 |
http://www.chegg.com/homework-help/questions-and-answers/sinusoidal-waves-combining-medium-described-following-wave-functions-x-centimeters-t-secon-q1247907 | 1,386,255,797,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163046758/warc/CC-MAIN-20131204131726-00051-ip-10-33-133-15.ec2.internal.warc.gz | 284,779,503 | 6,847 | # waves problem
0 pts ended
Two sinusoidal waves combining in a medium are described by the following wave functions, where x is in centimeters and t is in seconds.
y1 = (5.0 cm) sin p(x + 0.10t)
y2 = (5.0 cm) sin p(x - 0.10t)
Determine the maximum transverse position of an element of the medium at the following positions.
(a) x = 0.100 cm
cm
(b) x = 0.460 cm
cm
(c) x = 1.40 cm
cm
(d) Find the three smallest values of x corresponding to antinodes.
cm (smallest)
cm (second smallest)
cm (third smallest) | 159 | 510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2013-48 | latest | en | 0.805375 |
http://www.dailyspeculations.com/wordpress/?p=8848 | 1,606,364,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186414.7/warc/CC-MAIN-20201126030729-20201126060729-00687.warc.gz | 121,656,192 | 15,912 | # Some Preliminary Thoughts on the Moving Median, from Victor Niederhoffer
November 25, 2013 |
Some preliminary thoughts on the running median 2, 3, 4, 1, 7, 8, 9, 3.
A moving median of the first 5 is 3, of the next 5 is 4, of the next 5 is 7, of the next 5 is 8– it's a good indicator of trend. First recommended to me 53 years ago by Fred Mosteller, Chairman of Harvard's first statistics dept.
It is more stable than the moving average as outliers are removed from sample. It is easy to compute fast with computers for small running numbers like 5 or 100 by repeated sorts. For higher numbers, you can form two groups, those below the median and those above. As a new number comes up you place it in one of the two groups if higher or lower and take away the oldest number. Then adjust to make the two groups equal again. It is not used as much as the moving average so it shouldn't be hurt by front running or spikes when cross over occur. It has a defined distribution when the underlying distribution has inordinate extreme values as frequently occurs with Cauchy or similar distributions with infinite variance.
It's probably a good thing to use when using nearest neighbors as predictors, i.e using the median and running median to compute your predictors. It deserves testing in real life markets for real life applications.
## Ralph Vince writes:
It is the indicator of "expectation," as evidenced by human behavior itself, and not the probability-weighted mean.
Moving medians have some distinct advantages.
They represent real values that occur. For example, taking the average of 1, 2 and 5 gives you 4, which never occurred, whereas the median 2 did occur. Continuing with the same series, should subsequent values in the series be less than 5, the value of 5 will not occur as a moving median. Hence, the moving median eliminates outliers.
One of my appliances has three thermometers to measure temperature. The value displayed is the median (and hence a series of moving medians). Should one of the thermometers be broken, or distorted by being in a particularly hot or cold spot, the median will still give me the best estimate. This elimination of outliers is very useful.
Should you have data whose importance relies upon only crediting occurring values and need to eliminate outliers, then you should test moving medians. We ourselves had experimented with them regarding price series and written extensively about them, but do not use them in our current work. Our reason is that we consider the outliers in a price series to be particularly important.
The following is a plot ratio of SP500 (10 week moving average) / (10 week moving median) for the recent 5 years (SP500 weekly close data).
`SELECT * FROM wp_comments WHERE comment_post_ID = '8848' AND comment_approved = '1' ORDER BY comment_date` | 623 | 2,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-50 | latest | en | 0.945701 |
https://cracku.in/blog/linear-arrangement-questions-for-cat-pdf/ | 1,652,693,980,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00701.warc.gz | 246,504,555 | 36,861 | # Linear Arrangement Questions for CAT PDF
0
12876
Linear Arrangement is one of the important topic in Data Interpretation and Logical reasoning section for CAT. The linear arrangement questions will be generally more easy than the circular arrangement questions. In linear arrangement questions, sometimes the direction is also important, some will be facing north and some will be facing south (Arrangement questions with attributes).
Linear Arrangement Questions for CAT PDF:
Instructions:
Directions for the following three questions: Answer the following questions based on the statements given below:
(i) There are three houses on each side of the road.
(ii) These six houses are labeled as P, Q, R, S, T and U.
(iii) The houses are of different colours, namely, Red, Blue, Green, Orange, Yellow and White.
(iv) The houses are of different heights.
(v) T, the tallest house, is exactly opposite to the Red coloured house.
(vi) The shortest house is exactly opposite to the Green coloured house.
(vii) U, the Orange coloured house, is located between P and S.
(viii) R, the Yellow coloured house, is exactly opposite to P.
(ix) Q, the Green coloured house, is exactly opposite to U.
(x) P, the White coloured house, is taller than R, but shorter than S and Q.
Question 1:What is the colour of the house diagonally opposite to the Yellow coloured house?
a) White
b) Blue
c) Green
d) Red
e)none of these
Question 2:Which is the second tallest house?
a) P
b) S
c) Q
d) R
e)cannot be determined
Question 3:What is the colour of the tallest house?
a) Red
b) Blue
c) Green
d) Yellow
e)none of these
Directions for the following three questions: Answer the questions on the basis of the information given below.Instructions:
A, B, C, D, E, and F are a group of friends. There are two housewives, one professor, one engineer, one accountant and one lawyer in the group. There are only two married couples in the group. The lawyer is married to D, who is a housewife. No woman in the group is either an engineer or an accountant. C, the accountant, is married to F, who is a professor. A is married to a housewife. E is not a housewife.
Question 4:Which of the following is one of the married couples?
a) A & B
b) B & E
c) D & E
d) A & D
Question 5:What is E’s profession?
a) Engineer
b) Lawyer
c) Professor
d) Accountant
Question 6:
How many members of the group are males?
a) 2
b) 3
c) 4
d) Cannot be determined
Solutions (1-3):
Before directly trying to answer the question, it is important to gather all the information given by the question.
There are three houses on each side of the road => Draw 6 lines, 3 in each row, to accommodate P, Q, R, S, T and U.
The houses are of different colours and different heights.
T is tallest and is opposite to red house => Let’s number T as 1.
Shortest house is opposite to green house.
U is orange and is between P and S => Two cases arise here. P-U-S is one possibility and the other possibility is S-U-P.
R is yellow and is opposite to P.
Q is green and is opposite to U. We know that green house is opposite to the shortest house. This implies that U is the shortest house => Number of U is 6.
P is white and is taller than R but shorter than S and Q => Apart from T, S and Q are also taller than P => S and Q can be 2 and 3 in any order => Number of P is 4 and number of R is 5.
We know that P is opposite to R and Q is opposite to U => S is opposite to T
It is given that T is opposite to red house => S is the red house and hence T is the blue house.
So, we know the colours of all houses and heights of P, R, T and U.
In this question, we are asked to find the house that is opposite to yellow house. R is the yellow house, P is opposite to R and S is on the other corner in P’s row. Hence S is the house that is diagonally opposite to yellow house and the colour of S is Red.
We only know that the second tallest house is either Q or S. Hence the answer is cannot be determined.
We know that P is opposite to R and Q is opposite to U => S is opposite to T
It is given that T is opposite to red house => S is the red house and hence T is the blue house.
T is the tallest house and hence the colour of the tallest house is blue.
According to given conditions, we are able to infer following relations
So A and D are married couple. | 1,081 | 4,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-21 | longest | en | 0.943513 |
https://number.academy/26413 | 1,653,381,099,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00410.warc.gz | 483,110,860 | 11,975 | # Number 26413
Number 26,413 spell 🔊, write in words: twenty-six thousand, four hundred and thirteen . Ordinal number 26413th is said 🔊 and write: twenty-six thousand, four hundred and thirteenth. The meaning of number 26413 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 26413. What is 26413 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 26413.
## What is 26,413 in other units
The decimal (Arabic) number 26413 converted to a Roman number is (X)(X)(V)MCDXIII. Roman and decimal number conversions.
#### Weight conversion
26413 kilograms (kg) = 58230.1 pounds (lbs)
26413 pounds (lbs) = 11980.9 kilograms (kg)
#### Length conversion
26413 kilometers (km) equals to 16413 miles (mi).
26413 miles (mi) equals to 42508 kilometers (km).
26413 meters (m) equals to 86656 feet (ft).
26413 feet (ft) equals 8051 meters (m).
26413 centimeters (cm) equals to 10398.8 inches (in).
26413 inches (in) equals to 67089.0 centimeters (cm).
#### Temperature conversion
26413° Fahrenheit (°F) equals to 14656.1° Celsius (°C)
26413° Celsius (°C) equals to 47575.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
26413 seconds equals to 7 hours, 20 minutes, 13 seconds
26413 minutes equals to 2 weeks, 4 days, 8 hours, 13 minutes
### Codes and images of the number 26413
Number 26413 morse code: ..--- -.... ....- .---- ...--
Sign language for number 26413:
Number 26413 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 26413
### Multiplications
#### Multiplication table of 26413
26413 multiplied by two equals 52826 (26413 x 2 = 52826).
26413 multiplied by three equals 79239 (26413 x 3 = 79239).
26413 multiplied by four equals 105652 (26413 x 4 = 105652).
26413 multiplied by five equals 132065 (26413 x 5 = 132065).
26413 multiplied by six equals 158478 (26413 x 6 = 158478).
26413 multiplied by seven equals 184891 (26413 x 7 = 184891).
26413 multiplied by eight equals 211304 (26413 x 8 = 211304).
26413 multiplied by nine equals 237717 (26413 x 9 = 237717).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 26413
Half of 26413 is 13206,5 (26413 / 2 = 13206,5 = 13206 1/2).
One third of 26413 is 8804,3333 (26413 / 3 = 8804,3333 = 8804 1/3).
One quarter of 26413 is 6603,25 (26413 / 4 = 6603,25 = 6603 1/4).
One fifth of 26413 is 5282,6 (26413 / 5 = 5282,6 = 5282 3/5).
One sixth of 26413 is 4402,1667 (26413 / 6 = 4402,1667 = 4402 1/6).
One seventh of 26413 is 3773,2857 (26413 / 7 = 3773,2857 = 3773 2/7).
One eighth of 26413 is 3301,625 (26413 / 8 = 3301,625 = 3301 5/8).
One ninth of 26413 is 2934,7778 (26413 / 9 = 2934,7778 = 2934 7/9).
show fractions by 6, 7, 8, 9 ...
26413
### Advanced math operations
#### Is Prime?
The number 26413 is not a prime number. The closest prime numbers are 26407, 26417.
#### Factorization and factors (dividers)
The prime factors of 26413 are 61 * 433
The factors of 26413 are 1 , 61 , 433 , 26413
Total factors 4.
Sum of factors 26908 (495).
#### Powers
The second power of 264132 is 697.646.569.
The third power of 264133 is 18.426.938.826.997.
#### Roots
The square root √26413 is 162,520768.
The cube root of 326413 is 29,780998.
#### Logarithms
The natural logarithm of No. ln 26413 = loge 26413 = 10,181612.
The logarithm to base 10 of No. log10 26413 = 4,421818.
The Napierian logarithm of No. log1/e 26413 = -10,181612.
### Trigonometric functions
The cosine of 26413 is 0,059729.
The sine of 26413 is -0,998215.
The tangent of 26413 is -16,71229.
### Properties of the number 26413
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 26413 in Computer Science
Code typeCode value
26413 Number of bytes25.8KB
Unix timeUnix time 26413 is equal to Thursday Jan. 1, 1970, 7:20:13 a.m. GMT
IPv4, IPv6Number 26413 internet address in dotted format v4 0.0.103.45, v6 ::672d
26413 Decimal = 110011100101101 Binary
26413 Decimal = 1100020021 Ternary
26413 Decimal = 63455 Octal
26413 Decimal = 672D Hexadecimal (0x672d hex)
26413 BASE64MjY0MTM=
26413 MD5da92ce36d3c841c78a1dc24ea5abcb72
26413 SHA18e144482b35e1184c1dde91a0a51cba7c9c3c8d8
26413 SHA2245b65517cc123070cd34e2624fa1949b3c31a7c75ec73202fd38e5217
26413 SHA256ce93ee00ffa9c1ac582a96b2c241335c5cced5e1d573a3dbb2b551d6f79fc99e
26413 SHA384a4f7a3e3c771e8c4245cc1a98b3cc054c360b7dcb6e4ed00ae26f19b5f648cccb1142c655134254a018c54c19ab0b6d9
More SHA codes related to the number 26413 ...
If you know something interesting about the 26413 number that you did not find on this page, do not hesitate to write us here.
## Numerology 26413
### Character frequency in number 26413
Character (importance) frequency for numerology.
Character: Frequency: 2 1 6 1 4 1 1 1 3 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 26413, the numbers 2+6+4+1+3 = 1+6 = 7 are added and the meaning of the number 7 is sought.
## Interesting facts about the number 26413
### Asteroids
• (26413) 1999 XB62 is asteroid number 26413. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 12/7/1999.
## Number 26,413 in other languages
How to say or write the number twenty-six thousand, four hundred and thirteen in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 26.413) veintiséis mil cuatrocientos trece German: 🔊 (Anzahl 26.413) sechsundzwanzigtausendvierhundertdreizehn French: 🔊 (nombre 26 413) vingt-six mille quatre cent treize Portuguese: 🔊 (número 26 413) vinte e seis mil, quatrocentos e treze Chinese: 🔊 (数 26 413) 二万六千四百一十三 Arabian: 🔊 (عدد 26,413) ستة و عشرون ألفاً و أربعمائة و ثلاثة عشر Czech: 🔊 (číslo 26 413) dvacet šest tisíc čtyřista třináct Korean: 🔊 (번호 26,413) 이만 육천사백십삼 Danish: 🔊 (nummer 26 413) seksogtyvetusinde og firehundrede og tretten Dutch: 🔊 (nummer 26 413) zesentwintigduizendvierhonderddertien Japanese: 🔊 (数 26,413) 二万六千四百十三 Indonesian: 🔊 (jumlah 26.413) dua puluh enam ribu empat ratus tiga belas Italian: 🔊 (numero 26 413) ventiseimilaquattrocentotredici Norwegian: 🔊 (nummer 26 413) tjue-seks tusen, fire hundre og tretten Polish: 🔊 (liczba 26 413) dwadzieścia sześć tysięcy czterysta trzynaście Russian: 🔊 (номер 26 413) двадцать шесть тысяч четыреста тринадцать Turkish: 🔊 (numara 26,413) yirmialtıbindörtyüzonüç Thai: 🔊 (จำนวน 26 413) สองหมื่นหกพันสี่ร้อยสิบสาม Ukrainian: 🔊 (номер 26 413) двадцять шiсть тисяч чотириста тринадцять Vietnamese: 🔊 (con số 26.413) hai mươi sáu nghìn bốn trăm mười ba Other languages ...
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## Comment
If you know something interesting about the number 26413 or any natural number (positive integer) please write us here or on facebook. | 2,482 | 7,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.663709 |
http://www.brainia.com/essays/Eco-204-Ash-Course-Tutorial/320451.html | 1,500,786,555,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424247.30/warc/CC-MAIN-20170723042657-20170723062657-00014.warc.gz | 403,541,073 | 5,634 | # ECO 204 ASH Course Tutorial / TutorialOutlet
## ECO 204 ASH Course Tutorial / TutorialOutlet
• Submitted By: table21345
• Date Submitted: 10/03/2014 3:27 AM
• Words: 333
• Page: 2
• Views: 1
ECO 204 Entire Course (Ash)
For more course tutorials visit
www.tutorialoutlet.com
ECO 204 Week 1 DQ 1 Elasticity of Demand (Ash)
ECO 204 Week 1 DQ 2 Marginal Utility (Ash)
ECO 204 Week 2 DQ 1 Tax Credits and the Labor Market (Ash)
ECO 204 Week 2 DQ 2 Reduction of Costs (Ash)
ECO 204 Week 3 DQ 1 Perfect Competition (Ash)
ECO 204 Week 3 DQ 2 Oligopoly/Monopolistic Competition (Ash)
ECO 204 Week 3 Assignment Manufacturing Industry Evaluation (Ash)
ECO 204 Week 4 DQ 1 Externalities (Ash)
ECO 204 Week 4 DQ 2 Tax Base (Ash)
ECO 204 Week 5 DQ 1 Comparative and Absolute Advantage (Ash)
ECO 204 Week 5 DQ 2 Equity versus Growth (Ash)
ECO 204 Week 5 Final Paper (potato chip industry) (Ash)
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ECO 204 Week 1 DQ 1 Elasticity of Demand (Ash)
For more course tutorials visit
www.tutorialoutlet.com
Elasticity of Demand. Taxicab fares in most cities are regulated. Several years ago taxicab drivers in Boston obtained permission to raise their fares 10 percent, and they anticipated that revenues would increase by about 10 percent as a result. However, when the commissioner granted the 10 percent increase, revenues increased by only about 5 percent. What can you infer about the elasticity of demand for taxicab rides? What were taxicab drivers assuming about the elasticity of demand? Respond to at least two of your fellow students’ postings.
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
ECO 204 Week 1 DQ 2 Marginal Utility (Ash)
For more course tutorials visit
www.tutorialoutlet.com
Marginal Utility. Suppose that you observe that total utility rises as more of an item is consumed. What can you say for certain about marginal utility? Can you say for sure... | 517 | 2,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-30 | latest | en | 0.688825 |
https://oeis.org/A144217 | 1,620,557,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988966.82/warc/CC-MAIN-20210509092814-20210509122814-00511.warc.gz | 437,897,426 | 3,783 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A144217 Weight array of A144216: a rectangular array by antidiagonals. 2
0, 1, 1, 2, 0, 2, 3, 0, 0, 3, 4, 0, 0, 0, 4, 5, 0, 0, 0, 0, 5, 6, 0, 0, 0, 0, 0, 6, 7, 0, 0, 0, 0, 0, 0, 7, 8, 0, 0, 0, 0, 0, 0, 0, 8, 9, 0, 0, 0, 0, 0, 0, 0, 0, 9, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 12, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 13, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,4 LINKS FORMULA Start with R(m,n)=(m(m-1)+n(n-1))/2 for m>=1,n>=1. Put R(m,n)=0 if m=0 or n=0. Define w(m,n)=R(m,n)-R(m-1,n-1)-R(m,n-1)-R(m-1,n) for m>=1, n>=1. Then the weight array W={w(m,n)} is A144217. (See A144112.) EXAMPLE Northwest corner: 0 1 2 3 4 5 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 0 0 0 0 CROSSREFS Cf. A144217. Sequence in context: A293578 A098489 A128064 * A187881 A323474 A132814 Adjacent sequences: A144214 A144215 A144216 * A144218 A144219 A144220 KEYWORD nonn,tabl AUTHOR Clark Kimberling, Sep 14 2008 STATUS approved
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Last modified May 9 06:30 EDT 2021. Contains 343692 sequences. (Running on oeis4.) | 721 | 1,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.586077 |
https://polytope.miraheze.org/wiki/Deltoidal_icositetrahedron | 1,709,515,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00727.warc.gz | 466,873,650 | 12,455 | # Deltoidal icositetrahedron
Deltoidal icositetrahedron
Rank3
TypeUniform dual
Notation
Coxeter diagramm4o3m ()
Conway notationoC
Elements
Faces24 kites
Edges24+24
Vertices6+8+12
Vertex figure8 triangles, 6+12 squares
Measures (edge length 1)
Dihedral angle${\displaystyle \arccos \left(-{\frac {7+4{\sqrt {2}}}{17}}\right)\approx 138.11796^{\circ }}$
Central density1
Number of external pieces24
Level of complexity4
Related polytopes
DualSmall rhombicuboctahedron
ConjugateGreat deltoidal icositetrahedron
Abstract & topological properties
Flag count192
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryB3, order 48
ConvexYes
NatureTame
The deltoidal icositetrahedron, also called the strombic icositetrahedron or small lanceal disdodecahedron, is one of the 13 Catalan solids. It has 24 kites as faces, with 6+12 order-4 and 8 order-3 vertices. It is the dual of the uniform small rhombicuboctahedron.
It can also be obtained as the convex hull of a cube, an octahedron, and a cuboctahedron. If the cube has unit edge length, the octahedron's edge length is ${\displaystyle {\frac {4-{\sqrt {2}}}{2}}\approx 1.29289}$ and the cuboctahedron's edge length is ${\displaystyle {\frac {2{\sqrt {2}}-1}{2}}\approx 0.91421.}$
Each face of this polyhedron is a kite with its longer edges ${\displaystyle {\frac {4-{\sqrt {2}}}{2}}\approx 1.29289}$ times the length of its shorter edges. These kites have three angles measuring ${\displaystyle \arccos \left({\frac {2-{\sqrt {2}}}{4}}\right)\approx 81.57894^{\circ }}$ and one angle measuring ${\displaystyle \arccos \left(-{\frac {2+{\sqrt {2}}}{8}}\right)\approx 115.26317^{\circ }}$.
## Vertex coordinates
A deltoidal icositetrahedron with dual edge length 1 has vertex coordinates given by all permutations of:
• ${\displaystyle \left(\pm {\sqrt {2}},\,0,\,0\right),}$
• ${\displaystyle \left(\pm 1,\,\pm 1,\,0\right),}$
• ${\displaystyle \left(\pm {\frac {{\sqrt {2}}+4}{7}},\,\pm {\frac {{\sqrt {2}}+4}{7}},\,\pm {\frac {{\sqrt {2}}+4}{7}}\right).}$ | 680 | 2,033 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.695627 |
https://imendez.me/post/76050/finding-an-electrical-short-short-circuit-on-your-car-an-electrical-wiring-diagram-which-of-these-best-describes-an-electric-circuit.html | 1,558,425,126,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256281.35/warc/CC-MAIN-20190521062300-20190521084300-00157.warc.gz | 508,811,228 | 12,522 | Home » Wiring Diagram » Finding An Electrical Short Short Circuit On Your Car An Electrical Wiring Diagram Which Of These Best Describes An Electric Circuit #307
# Finding An Electrical Short Short Circuit On Your Car An Electrical Wiring Diagram Which Of These Best Describes An Electric Circuit #307
Description: Finding An Electrical Short Short Circuit On Your Car An Electrical Wiring Diagram Which Of These Best Describes An Electric Circuit #307 from the above 2001x1500 resolutions which is part of the "DIY Needed" directory. Download this image for free in HD resolution the choice "Save full resolution" below. If you do not find the exact resolution you are looking for, then go for a native or higher resolution.
Doing electrical wiring by you might be tough. This can be specifically so any time you absence the expertise and the practical experience in electrical stuff. The detail is you can not do a demo and error approach when working with electrical wiring. Faults could price you a fortune or maybe your lifetime. That is certainly why ahead of starting any do-it-yourself electrical repair, it's important to request some electrical wiring issues in order to ensure you are aware of what you are doing. Something about electrical wiring is the fact wires are usually colour coded. Consequently, it really is basically less difficult to know which of them go with which. Underneath will be the most popular electrical wiring issues whose responses commonly entail hues that will help you identify each individual unique wire. - Electrical Schematic Diagram
Pplato Flap Phys 5 4 Ac Circuits And Electrical Oscillations Figure 1 A Simple D C Circuit Understanding Basic Electrical Circuits #4121
Electric Circuits With Home Made Wires And Bulbs Ingridscience Ca Electric Circuits With Home Made Wires And Bulbs Learning Electricity And Circuits Worksheet #8320
How can You Wire a Change?
Amongst essentially the most typical electrical wiring concerns is on how to wire a change. Though utilizing switches in the home is kind of effortless, wiring one particular may not be that straightforward for everyone. An ON-OFF switch is in fact quite simple to wire. You will discover different kinds of switches, but for this instance, let's say that you are installing a single-pole toggle switch, an extremely typical swap (plus the easiest).
You'll find 3 shades of wires inside of a typical single-pole toggle switch: black, white, and eco-friendly. Splice the black wire in two and hook up them within the terminal screws - just one on prime plus the other within the base screw from the change. The white wire serves like a supply of uninterrupted energy and is particularly typically linked into a mild coloured terminal screw (e.g., silver). Connect the environmentally friendly wire for the ground screw of your respective switch.
These techniques would commonly be ample to create an average switch operate without a problem. Nonetheless, if you are usually not confident which you can carry out the endeavor thoroughly and properly you better enable the pros get it done in its place. Just after all, there is a explanation why this undertaking has become the most typical electrical wiring issues questioned by a lot of people.
How do You Wire a Ceiling Fan?
For many reason, how you can wire a ceiling fan can be one among probably the most prevalent electrical wiring concerns. To simplify this job, you can use only one change for a single ceiling fan. To wire the admirer, it can be merely a matter of connecting the black wire on the ceiling fan into the black wire in the change. When there is a lightweight, the blue wire ought to be connected to your black wire from the change also.
How do You Switch a Breaker and exactly how Does one Increase a Sub Panel?
Though many endeavor to carry out these duties them selves, plenty of people are encouraged to hire an electrician alternatively. It's far more sophisticated and therefore dangerous for some people to test to interchange a breaker or increase a panel. To give you an plan about these types of widespread electrical wiring concerns, you'd probably have to work with a sizzling electrical panel. Should you will not even understand what what this means is, that you are simply just not outfitted to perform the work oneself. Even when you really have to expend a lot more by employing a professional electrician, it is really much safer and even more practical to try and do this as an alternative. - Finding An Electrical Short Short Circuit On Your Car An Electrical Wiring Diagram Which Of These Best Describes An Electric Circuit #307
You will discover factors why they're by far the most usually requested electrical wiring inquiries. A single, a lot of think it's easy to try and do, and two, these are the prevalent electrical jobs at your house. But then you definately mustn't put your security at risk as part of your objective to economize. The stakes could even be significantly bigger when you attempt to get monetary savings and do an electrical wiring occupation devoid of enough information or expertise.
Tablets | iPad HD resolutions: 2048 x 2048
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### Digital Thermometer Using Pic Microcontroller
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