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Show the rule of divisibility for 7
#### Question
Show the rule of divisibility for 7
Collected in the board: Number Theory
Steven Zheng posted 2 years ago
The rule of the divisibility for 7 says that if a number is divisible by 7, the number that is the difference between the number that is made by removing the last digit and the number that is the double of the last digit must be divisible by 7.
For example, 532, doubling the last digit gives 4, subtracting 4 from 53 gives 49, which is divisible by 7. Therefore, 532 must be divisible by 7. place values of the last two digits are 50 and 2, which are summed up to 52.
532 \div 7=76
Then how to prove the rule in general form?
Let’s first evaluate 4-digit number that is denoted as \overline{abcd}, in which a, b, c, d are three digits at hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abcd} is rewritten as
\overline{abcd}= 1000a+100b+10c+d
If \overline{abcd} is divisible by 7, subtract 21 or multiples of 21 (e.g. 21d) from the number, the result is also divisible by 7.
\overline{abcd} -21d
=1000a+100b+10c+d-21d
=1000a+100b+10c-20d
=10(100a+10b+c-2d)
=10(\overline{abc}-2d)
Since 10 and 7 is prime to each other, if \overline{abcd} is divisible by 7, \overline{abc}-2d must be divisible by 7.
In a more general form, any integer can be expressed as
\overline{a_na_{n-1}\dots a_1a_0}=a_n10^n+a_{n-1}10^{n-1}+\dots+a_210^2+a_110+a_0
\overline{a_na_{n-1}\dots a_1a_0}-21a_0
=a_n10^n+a_{n-1}10^{n-1}+\dots+a_210^2+a_110-20a_0
=10(a_n10^{n-1}+a_{n-1}10^{n-2}+\dots+10a_2+a_1-2a_0)
=10( \overline{a_na_{n-1}\dots a_2a_1} -a_0)
Now we have verified that for any integer, if it is divisible by 7, subtract the double of the last digit from the the number that is made by removing the last digit, the difference must be divisible by 7.
Steven Zheng posted 2 years ago
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Wilcoxon Signed Rank Test
The Wilcoxon Signed Rank Test is a nonparametric equivalent of the one sample t-test. It tests the hypothesis:
H0 the mean equals zero
H1 the mean [is less than/greater than/not equal to] zero
The test involves:
• sort the values into order of their absolute magnitude (ignoring the signs) and allocate ranks
• calculate the sum the ranks of the data values are positive (S+)
An example would be:
Rank 1 2 3 4 5 6 Data Value -1 -2 2.5 3 3 3.5
If the mean is non-zero the positive values will be clustered at one end or other of the ordered data set and S+ and S- will be very different.
For the example S+ = 3 + 4 + 5 + 6 = 18
For small data sets values the critical value is found from published tables. Because the sum of the ranks must be an integer, it is not usually possible to find the exact critical value:
where
For larger data sets the z-statistic can be found from:
This gives the upper tail test (the mean is less than zero). For the lower tail test, use the sum S+ and for the two tailed test use both, remembering to use the table value twice.
The p-value can be calculated using normal distribution tables.
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# Applied Discrete Structures
## Section12.3An Introduction to Vector Spaces
### Subsection12.3.1Motivation for the study of vector spaces
$$\renewcommand{\vec}[1]{\mathbf{#1}}$$
When we encountered various types of matrices in Chapter 5, it became apparent that a particular kind of matrix, the diagonal matrix, was much easier to use in computations. For example, if $$A =\left( \begin{array}{cc} 2 & 1 \\ 2 & 3 \\ \end{array} \right)\text{,}$$ then $$A^5$$ can be found, but its computation is tedious. If $$D =\left( \begin{array}{cc} 1 & 0 \\ 0 & 4 \\ \end{array} \right)$$ then
\begin{equation*} D^5 =\left( \begin{array}{cc} 1 & 0 \\ 0 & 4 \\ \end{array} \right)^5= \left( \begin{array}{cc} 1^5 & 0 \\ 0 & 4^5 \\ \end{array} \right)= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1024 \\ \end{array} \right) \end{equation*}
Even when presented with a non-diagonal matrix, we will see that it is sometimes possible to do a bit of work to be able to work with a diagonal matrix. This process is called diagonalization.
In a variety of applications it is beneficial to be able to diagonalize a matrix. In this section we will investigate what this means and consider a few applications. In order to understand when the diagonalization process can be performed, it is necessary to develop several of the underlying concepts of linear algebra.
### Subsection12.3.2Vector Spaces
By now, you realize that mathematicians tend to generalize. Once we have found a “good thing,” something that is useful, we apply it to as many different concepts as possible. In doing so, we frequently find that the “different concepts” are not really different but only look different. Four sentences in four different languages might look dissimilar, but when they are translated into a common language, they might very well express the exact same idea.
Early in the development of mathematics, the concept of a vector led to a variety of applications in physics and engineering. We can certainly picture vectors, or “arrows,” in the $$x y-\textrm{ plane}$$ and even in the three-dimensional space. Does it make sense to talk about vectors in four-dimensional space, in ten-dimensional space, or in any other mathematical situation? If so, what is the essence of a vector? Is it its shape or the rules it follows? The shape in two- or three-space is just a picture, or geometric interpretation, of a vector. The essence is the rules, or properties, we wish vectors to follow so we can manipulate them algebraically. What follows is a definition of what is called a vector space. It is a list of all the essential properties of vectors, and it is the basic definition of the branch of mathematics called linear algebra.
#### Definition12.3.1.Vector Space.
Let $$V$$ be any nonempty set of objects. Define on $$V$$ an operation, called addition, for any two elements $$\vec{x}, \vec{y} \in V\text{,}$$ and denote this operation by $$\vec{x}+ \vec{y}\text{.}$$ Let scalar multiplication be defined for a real number $$a \in \mathbb{R}$$ and any element $$\vec{x}\in V$$ and denote this operation by $$a \vec{x}\text{.}$$ The set $$V$$ together with operations of addition and scalar multiplication is called a vector space over $$\mathbb{R}$$ if the following hold for all $$\vec{x}, \vec{y}, \vec{z}\in V$$ , and $$a,b \in \mathbb{R}\text{:}$$
• $$\displaystyle \vec{x}+ \vec{y}= \vec{y}+ \vec{x}$$
• $$\displaystyle \left(\vec{x}+ \vec{y}\right)+ \vec{z}= \vec{x}+\left( \vec{y}+\vec{z}\right)$$
• There exists a vector $$\vec{0}\in V\text{,}$$ such that $$\vec{x}+\vec{0} = \vec{x}$$ for all $$x \in V\text{.}$$
• For each vector $$\vec{x}\in V\text{,}$$ there exists a unique vector $$-\vec{x}\in V\text{,}$$ such that $$-\vec{x} +\vec{x}= \vec{0}\text{.}$$
These are the main properties associated with the operation of addition. They can be summarized by saying that $$[V; +]$$ is an abelian group.
The next four properties are associated with the operation of scalar multiplication and how it relates to vector addition.
• $$\displaystyle a\left(\vec{x}+ \vec{y} \right) =a \vec{x}+a \vec{y}$$
• $$\displaystyle (a +b)\vec{x}= a \vec{x} + b \vec{x}$$
• $$\displaystyle a \left(b \vec{x}\right) = (a b)\vec{x}$$
• $$1\vec{x} = \vec{x}\text{.}$$
In a vector space it is common to call the elements of $$V$$ vectors and those from $$\mathbb{R}$$ scalars. Vector spaces over the real numbers are also called real vector spaces.
Let $$V = M_{2\times 3}(\mathbb{R})$$ and let the operations of addition and scalar multiplication be the usual operations of addition and scalar multiplication on matrices. Then $$V$$ together with these operations is a real vector space. The reader is strongly encouraged to verify the definition for this example before proceeding further (see Exercise 3 of this section). Note we can call the elements of $$M_{2\times 3}(\mathbb{R})$$ vectors even though they are not arrows.
Let $$\mathbb{R}^2 = \left\{\left(a_1, a_2 \right) \mid a_1,a_2 \in \mathbb{R}\right\}\text{.}$$ If we define addition and scalar multiplication the natural way, that is, as we would on $$1\times 2$$ matrices, then $$\mathbb{R}^2$$ is a vector space over $$\mathbb{R}\text{.}$$ See Exercise 12.3.3.4 of this section.
In this example, we have the “bonus” that we can illustrate the algebraic concept geometrically. In mathematics, a “geometric bonus” does not always occur and is not necessary for the development or application of the concept. However, geometric illustrations are quite useful in helping us understand concepts and should be utilized whenever available.
Let’s consider some illustrations of the vector space $$\mathbb{R}^2\text{.}$$ Let $$\vec{x}= (1, 4)$$ and $$\vec{y} = (3, 1)\text{.}$$ We illustrate the vector $$\left(a_1, a_2\right)$$ as a directed line segment, or “arrow,” from the point $$(0, 0)$$ to the point$$\left(a_1, a_2\right)\text{.}$$ The vectors $$\vec{x}$$ and $$\vec{y}$$ are as shown in Figure 12.3.4 together with $$\vec{x}+ \vec{y} = (1, 4) + (3, 1) = (4, 5)\text{.}$$ The vector $$2 \vec{x} = 2(1, 4) = (2, 8)$$ is a vector in the same direction as $$\vec{x}\text{,}$$ but with twice its length.
#### Note12.3.5.
1. The common convention is to use that boldface letters toward the end of the alphabet for vectors, while letters early in the alphabet are scalars.
2. A common alternate notation for vectors is to place an arrow about a variable to indicate that it is a vector such as this: $$\overset{\rightharpoonup }{x}\text{.}$$
3. The vector $$\left(a_1,a_2,\ldots ,a_n\right)\in \mathbb{R}^n$$ is referred to as an $$n$$-tuple.
4. For those familiar with vector calculus, we are expressing the vector $$x = a_1 \boldsymbol{\hat{\textbf{i}}}+ a_2 \boldsymbol{\hat{\textbf{j}}} + a_3 \boldsymbol{\hat{\textbf{k}}} \in \mathbb{R}^3$$ as $$\left(a_1,a_2,a_3\right)\text{.}$$ This allows us to discuss vectors in $$\mathbb{R}^n$$ in much simpler notation.
In many situations a vector space $$V$$ is given and we would like to describe the whole vector space by the smallest number of essential reference vectors. An example of this is the description of $$\mathbb{R}^2\text{,}$$ the $$x y$$-plane, via the $$x$$ and $$y$$ axes. Again our concepts must be algebraic in nature so we are not restricted solely to geometric considerations.
#### Definition12.3.6.Linear Combination.
A vector $$\pmb{ y}$$ in vector space $$V$$ (over $$\mathbb{R}$$) is a linear combination of the vectors $$\vec{x}_1\text{,}$$ $$\vec{x}_2, \ldots\text{,}$$ $$\vec{x}_n$$ if there exist scalars $$a_1,a_2,\ldots ,a_n$$ in $$\mathbb{R}$$ such that $$\vec{y} = a_1\vec{x}_1+ a_2\vec{x}_2+\ldots +a_n\vec{x}_n$$
The vector $$(2, 3)$$ in $$\mathbb{R}^2$$ is a linear combination of the vectors $$(1, 0)$$ and $$(0, 1)$$ since $$(2, 3) = 2(1, 0) + 3(0, 1)\text{.}$$
Prove that the vector $$(4,5)$$ is a linear combination of the vectors (3, 1) and (1, 4).
By the definition we must show that there exist scalars $$a_1$$ and $$a_2$$ such that:
\begin{equation*} \begin{array}{ccc} \begin{split} (4,5) &= a_1(3, 1) + a_2 (1, 4)\\ & = \left(3a_1+ a_2 , a_1+4a_2\right) \end{split} &\Rightarrow & \begin{array}{c} 3a_1+ a_2 =4\\ a_1+ 4a_2 =5\\ \end{array}\\ \\ \end{array} \end{equation*}
This system has the solution $$a_1=1\text{,}$$ $$a_2=1\text{.}$$
Hence, if we replace $$a_1$$ and $$a_2$$ both by 1, then the two vectors (3, 1) and (1, 4) produce, or generate, the vector (4,5). Of course, if we replace $$a_1$$ and $$a_2$$ by different scalars, we can generate more vectors from $$\mathbb{R}^2\text{.}$$ If, for example, $$a _1 = 3$$ and $$a_2 = -2\text{,}$$ then
\begin{equation*} a_1(3, 1) + a_2 (1, 4) = 3 (3, 1) +(-2) (1,4) = (9, 3) +(-2,-8) = (7, -5) \end{equation*}
Will the vectors $$(3, 1)$$ and $$(1,4)$$ generate any vector we choose in $$\mathbb{R}^2\text{?}$$ To see if this is so, we let $$\left(b_1,b_2\right)$$ be an arbitrary vector in $$\mathbb{R}^2$$ and see if we can always find scalars $$a_1$$ and $$a_2$$ such that $$a_1(3, 1) + a_2 (1, 4)= \left(b_1,b_2\right)\text{.}$$ This is equivalent to solving the following system of equations:
\begin{equation*} \begin{array}{c} 3a_1+ a_2 =b_1\\ a_1+4a_2 =b_2\\ \end{array} \end{equation*}
which always has solutions for $$a_1$$ and $$a_2$$ , regardless of the values of the real numbers $$b_1$$ and $$b_2\text{.}$$ Why? We formalize this situation in a definition:
#### Definition12.3.9.Generation of a Vector Space.
Let $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ be a set of vectors in a vector space $$V$$ over $$\mathbb{R}\text{.}$$ This set is said to generate, or span, $$V$$ if, for any given vector $$\vec{y} \in V\text{,}$$ we can always find scalars $$a_1\text{,}$$ $$a_2, \ldots\text{,}$$ $$a_n$$ such that $$\vec{y} = a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n\text{.}$$ A set that generates a vector space is called a generating set.
We now give a geometric interpretation of the previous examples.
We know that the standard coordinate system, $$x$$ axis and $$y$$ axis, were introduced in basic algebra in order to describe all points in the $$xy$$-plane algebraically. It is also quite clear that to describe any point in the plane we need exactly two axes.
We can set up a new coordinate system in the following way. Draw the vector $$(3, 1)$$ and an axis from the origin through (3, 1) and label it the $$x'$$ axis. Also draw the vector $$(1,4)$$ and an axis from the origin through $$(1,4)$$ to be labeled the $$y'$$ axis. Draw the coordinate grid for the axis, that is, lines parallel, and let the unit lengths of this “new” plane be the lengths of the respective vectors, $$(3, 1)$$ and $$(1, 4)\text{,}$$ so that we obtain Figure 12.3.10.
From Example 12.3.8 and Figure 12.3.10, we see that any vector on the plane can be described using the standard $$xy$$-axes or our new $$x'y'$$-axes. Hence the position which had the name $$(3,1)$$ in reference to the standard axes has the name $$(1,0)$$ with respect to the $$x'y'$$ axes, or, in the phraseology of linear algebra, the coordinates of the point $$(1,4)$$ with respect to the $$x'y'$$ axes are $$(1, 0)\text{.}$$
From Example 12.3.8 we found that if we choose $$a_1=1$$ and $$a_2=1\text{,}$$ then the two vectors $$(3, 1)$$ and $$(1,4)$$ generate the vector $$(4,5)\text{.}$$ Another geometric interpretation of this problem is that the coordinates of the position $$(4,5)$$ with respect to the $$x'y'$$ axes of Figure 12.3.10 is $$(1, 1)\text{.}$$ In other words, a position in the plane has the name $$(4,5)$$ in reference to the $$xy$$-axes and the same position has the name $$(1, 1)$$ in reference to the $$x'y'$$ axes.
From the above, it is clear that we can use different axes to describe points or vectors in the plane. No matter what choice we use, we want to be able to describe each position in a unique manner. This is not the case in Figure 12.3.12. Any point in the plane could be described via the $$x'y'$$ axes, the $$x'z'$$ axes or the $$y'z'$$ axes. Therefore, in this case, a single point would have three different names, a very confusing situation.
We formalize the our observations in the previous examples in two definitions and a theorem.
#### Definition12.3.13.Linear Independence/Linear Dependence.
A set of vectors $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ from a real vector space $$V$$ is linearly independent if the only solution to the equation $$a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n= \vec{0}$$ is $$a_1 = a_2 = \ldots = a_n = 0\text{.}$$ Otherwise the set is called a linearly dependent set.
#### Definition12.3.14.Basis.
A set of vectors $$B=\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ is a basis for a vector space $$V$$ if:
1. $$B$$ generates $$V\text{,}$$ and
2. $$B$$ is linearly independent.
Assume that $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}$$ is a basis for $$V$$ over $$\mathbb{R}\text{.}$$ We must prove two facts:
1. each vector $$y \in V$$ can be expressed as a linear combination of the $$\vec{x}_i\textrm{'s}\text{,}$$ and
2. each such expression is unique.
Part 1 is trivial since a basis, by its definition, must generate all of $$V\text{.}$$
The proof of part 2 is a bit more difficult. We follow the standard approach for any uniqueness facts. Let $$y$$ be any vector in $$V$$ and assume that there are two different ways of expressing $$y\text{,}$$ namely
\begin{equation*} y = a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n \end{equation*}
and
\begin{equation*} y = b_1 \vec{x}_1+b_2 \vec{x}_2+\ldots +b_n \vec{x}_n \end{equation*}
where at least one $$a_i$$ is different from the corresponding $$b_i\text{.}$$ Then equating these two linear combinations we get
\begin{equation*} a_1 \vec{x}_1+a_2 \vec{x}_2+\ldots +a_n \vec{x}_n=b_1 \vec{x}_1+b_2 \vec{x}_2+\ldots +b_n \vec{x}_n \end{equation*}
so that
\begin{equation*} \left(a_1-b_1\right) \vec{x}_1+\left(a_2-b_2\right) \vec{x}_2+\ldots +\left(a_n-b_n\right) \vec{x}_n=\vec{0} \end{equation*}
Now a crucial observation: since the $$\vec{x}_i's$$ form a linearly independent set, the only solution to the previous equation is that each of the coefficients must equal zero, so $$a_i-b_i=0$$ for $$i = 1, 2, \ldots ,n\text{.}$$ Hence $$a_i=b_i\text{,}$$ for all $$i\text{.}$$ This contradicts our assumption that at least one $$a_i$$ is different from the corresponding $$b_i\text{,}$$ so each vector $$\vec{y} \in V$$ can be expressed in one and only one way.
This theorem, together with the previous examples, gives us a clear insight into the significance of linear independence, namely uniqueness in representing any vector.
Prove that $$\{(1, 1), (-1, 1)\}$$ is a basis for $$\mathbb{R}^2$$ over $$\mathbb{R}$$ and explain what this means geometrically.
First we show that the vectors $$(1, 1)$$ and $$(-1, 1)$$ generate all of $$\mathbb{R}^2\text{.}$$ We can do this by imitating Example 12.3.8 and leave it to the reader (see Exercise 12.3.3.10 of this section). Secondly, we must prove that the set is linearly independent.
Let $$a_1$$ and $$a_2$$ be scalars such that $$a_1 (1, 1) + a_2 (-1, 1) = (0, 0)\text{.}$$ We must prove that the only solution to the equation is that $$a_1$$ and $$a_2$$ must both equal zero. The above equation becomes $$\left(a_1- a_2 , a_1 + a_2 \right) = (0, 0)$$ which gives us the system
\begin{equation*} \begin{array}{c} a_1 - a_{2 }=0 \\ a_1 + a_2=0\\ \end{array} \end{equation*}
The augmented matrix of this system reduces in such way that the only solution is the trivial one of all zeros:
\begin{equation*} \left( \begin{array}{cc|c} 1 & -1 & 0 \\ 1 & 1 & 0 \\ \end{array} \right)\longrightarrow \left( \begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)\textrm{ }\Rightarrow \textrm{ }a_1 = a_2 =0 \end{equation*}
Therefore, the set is linearly independent.
To explain the results geometrically, note through Exercise 12, part a, that the coordinates of each vector $$\vec{y} \in \mathbb{R}^2$$ can be determined uniquely using the vectors (1,1) and (-1, 1). The concept of dimension is quite obvious for those vector spaces that have an immediate geometric interpretation. For example, the dimension of $$\mathbb{R}^2$$ is two and that of $$\mathbb{R}^3$$ is three. How can we define the concept of dimension algebraically so that the resulting definition correlates with that of $$\mathbb{R}^2$$ and $$\mathbb{R}^3\text{?}$$ First we need a theorem, which we will state without proof.
#### Definition12.3.18.Dimension of a Vector Space.
Let $$V$$ be a vector space over $$\mathbb{R}$$ with basis $$\left\{\vec{x}_1,\vec{x}_2, \ldots ,\vec{x}_n\right\}\text{.}$$ Then the dimension of $$V$$ is $$n\text{.}$$ We use the notation $$\dim V = n$$ to indicate that $$V$$ is $$n$$-dimensional.
### Exercises12.3.3Exercises
#### 1.
If $$a = 2\text{,}$$ $$b = -3\text{,}$$ $$A=\left( \begin{array}{ccc} 1 & 0 & -1 \\ 2 & 3 & 4 \\ \end{array} \right)\text{,}$$ $$B=\left( \begin{array}{ccc} 2 & -2 & 3 \\ 4 & 5 & 8 \\ \end{array} \right)\text{,}$$ and $$C=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 3 & 2 & -2 \\ \end{array} \right)$$ verify that all properties of the definition of a vector space are true for $$M_{2\times 3}(\mathbb{R})$$ with these values.
#### 2.
Let $$a = 3\text{,}$$ $$b = 4\text{,}$$ $$\vec{x}\pmb = (-1, 3)\text{,}$$ $$\vec{y} = (2, 3)\text{,}$$and $$\vec{z} = (1, 0)\text{.}$$ Verify that all properties of the definition of a vector space are true for $$\mathbb{R}^2$$ for these values.
#### 3.
1. Verify that $$M_{2\times 3}(\mathbb{R})$$ is a vector space over $$\mathbb{R}\text{.}$$ What is its dimension?
2. Is $$M_{m\times n}(\mathbb{R})$$ a vector space over $$\mathbb{R}\text{?}$$ If so, what is its dimension?
The dimension of $$M_{2\times 3}(\mathbb{R})$$ is 6 and yes, $$M_{m\times n}(\mathbb{R})$$ is also a vector space of dimension $$m \cdot n\text{.}$$ One basis for $$M_{m\times n}(\mathbb{R})$$ is $$\{A_{ij} \mid 1 \leq i \leq m, 1 \leq j \leq n\}$$ where $$A_{ij}$$ is the $$m\times n$$ matrix with entries all equal to zero except for in row $$i\text{,}$$ column $$j$$ where the entry is 1.
#### 4.
1. Verify that $$\mathbb{R}^2$$ is a vector space over $$\mathbb{R}\text{.}$$
2. Is $$\mathbb{R}^n$$ a vector space over $$\mathbb{R}$$ for every positive integer $$n\text{?}$$
#### 5.
Let $$P^3= \left\{a_0 + a_1x + a_2x^2 + a_3x^3 \mid a_0,a_1,a_2,a_3\in \mathbb{R}\right\}\text{;}$$ that is, $$P^3$$ is the set of all polynomials in $$x$$ having real coefficients with degree less than or equal to three. Verify that $$P^3$$ is a vector space over $$\mathbb{R}\text{.}$$ What is its dimension?
#### 6.
For each of the following, express the vector $$\pmb{y}$$ as a linear combination of the vectors $$x_1$$ and $$x_2\text{.}$$
1. $$\vec{y} = (5, 6)\text{,}$$ $$\vec{x}_1 =(1, 0)\text{,}$$ and $$\vec{x}_2 = (0, 1)$$
2. $$\vec{y} = (2, 1)\text{,}$$ $$\vec{x}_1 =(2, 1)\text{,}$$ and $$\vec{x}_2 = (1, 1)$$
3. $$\vec{y} = (3,4)\text{,}$$ $$\vec{x}_1 = (1, 1)\text{,}$$ and $$\vec{x}_2 = (-1, 1)$$
#### 7.
Express the vector $$\left( \begin{array}{cc} 1 & 2 \\ -3 & 3 \\ \end{array} \right)\in M_{2\times 2}(\mathbb{R})\text{,}$$ as a linear combination of $$\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right)\text{,}$$ $$\left( \begin{array}{cc} -1 & 5 \\ 2 & 1 \\ \end{array} \right)\text{,}$$ $$\left( \begin{array}{cc} 0 & 1 \\ 1 & 1 \\ \end{array} \right)$$ and $$\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right)$$
If the matrices are named $$B\text{,}$$ $$A_1\text{,}$$ $$A_2$$ , $$A_3\text{,}$$ and $$A_4$$ , then
\begin{equation*} B = \frac{8}{3}A_1 + \frac{5}{3}A_2+\frac{-5}{3}A_3+\frac{23}{3}A_4 \end{equation*}
#### 8.
Express the vector $$x^3-4x^2+3\in P^3$$ as a linear combination of the vectors 1, $$x\text{,}$$ $$x^2$$ , and $$x^3\text{.}$$
#### 9.
1. Show that the set $$\left\{\vec{x}_1,\vec{x}_2\right\}$$ generates $$\mathbb{R}^2$$ for each of the parts in Exercise 6 of this section.
2. Show that $$\left\{\vec{x}_1,\vec{x}_2,\vec{x}_3\right\}$$ generates $$\mathbb{R}^2$$ where $$\vec{x}_1= (1, 1)\text{,}$$ $$\vec{x}_2= (3,4)\text{,}$$ and $$\vec{x}_3 = (-1, 5)\text{.}$$
3. Create a set of four or more vectors that generates $$\mathbb{R}^2\text{.}$$
4. What is the smallest number of vectors needed to generate $$\mathbb{R}^2\text{?}$$ $$\mathbb{R}^n\text{?}$$
5. Show that the set
\begin{equation*} \{A_1, A_2 ,A_3, A_4\} =\{ \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right)\} \end{equation*}
generates $$M_{2\times 2}(\mathbb{R})$$
6. Show that $$\left\{1,x,x^2 ,x^3\right\}$$ generates $$P^3\text{.}$$
1. If $$x_1 = (1, 0)\text{,}$$ $$x_2= (0, 1)\text{,}$$ and $$y = \left(b_1, b_2\right)\text{,}$$ then $$y = b_1x_1+b_2x_2\text{.}$$ If $$x_1 = (3, 2)\text{,}$$ $$x_2= (2,1)\text{,}$$ and $$y = \left(b_1, b_2\right)\text{,}$$ then $$y =\left(- b_1+2b_2\right)x_1+\left(2b_1-3b_2\right)x_2\text{.}$$
2. If $$y = \left(b_1, b_2\right)$$ is any vector in $$\mathbb{R}^2$$ , then $$y =\left(- 3b_1+4b_2\right)x_1+\left(-b_1+b_2\right)x_2 + (0)x_3$$
3. One solution is to add any vector(s) to $$x_1\text{,}$$ $$x_2\text{,}$$ and $$x_3$$ of part b.
4. 2, $$n$$
5. $$\displaystyle \left( \begin{array}{cc} x & y \\ z & w \\ \end{array} \right)= x A_1+y A_2+ z A_3+ w A_4$$
6. $$a_0+a_1x + a_2x^2+ a_3x^3=a_0(1)+a_1(x) + a_2\left(x^2\right)+ a_3\left(x^3\right)\text{.}$$
#### 10.
Complete Example 12.3.16 by showing that $$\{(1, 1), (-1, 1)\}$$ generates $$\mathbb{R}^2\text{.}$$
#### 11.
1. Prove that $$\{(4, 1), (1, 3)\}$$ is a basis for $$\mathbb{R}^2$$ over $$\mathbb{R}\text{.}$$
2. Prove that $$\{(1, 0), (3, 4)\}$$ is a basis for $$\mathbb{R}^2$$ over $$\mathbb{R}\text{.}$$
3. Prove that $$\{(1,0, -1), (2, 1, 1), (1, -3, -1)\}$$ is a basis for $$\mathbb{R}^3$$ over $$\mathbb{R}\text{.}$$
4. Prove that the sets in Exercise 9, parts e and f, form bases of the respective vector spaces.
1. The set is linearly independent: let $$a$$ and $$b$$ be scalars such that $$a(4, 1) + b(1, 3) = (0, 0)\text{,}$$ then $$4a + b = 0\textrm{ and } a + 3b= 0$$ which has $$a = b = 0$$ as its only solutions. The set generates all of $$\mathbb{R}^2\text{:}$$ let $$(a, b)$$ be an arbitrary vector in $$\mathbb{R}^2$$ . We want to show that we can always find scalars $$\beta _1$$ and $$\beta _2$$ such that $$\beta _1(4, 1) +\beta _2 (1,3) = (a, b)\text{.}$$ This is equivalent to finding scalars such that $$4\beta _1 +\beta _2 = a$$ and $$\beta _1 + 3\beta _2 = b\text{.}$$ This system has a unique solution $$\beta _1=\frac{3a - b}{11}\text{,}$$ and $$\beta _2= \frac{4b-a}{11}\text{.}$$ Therefore, the set generates $$\mathbb{R}^2\text{.}$$
#### 12.
1. Determine the coordinates of the points or vectors $$(3, 4)\text{,}$$ $$(-1, 1)\text{,}$$ and $$(1, 1)$$ with respect to the basis $$\{(1, 1),(-1, 1)\}$$ of $$\mathbb{R}^2\text{.}$$ Interpret your results geometrically.
2. Determine the coordinates of the points or vector $$(3, 5, 6)$$ with respect to the basis $$\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}\text{.}$$ Explain why this basis is called the standard basis for $$\mathbb{R}^3\text{.}$$
#### 13.
1. Let $$\vec{y}_1= (1,3, 5, 9)\text{,}$$ $$\vec{y}_2= (5,7, 6, 3)\text{,}$$ and $$c = 2\text{.}$$ Find $$\vec{y}_1+\vec{y}_2$$ and $$c \vec{y}_1\text{.}$$
2. Let $$f_1(x) = 1 + 3x + 5x^2 + 9x^3$$ , $$f_2(x)=5 + 7x+6x^2+3x^3$$ and $$c = 2\text{.}$$ Find $$f_1(x)+f_2(x)$$ and $$c f_1(x)\text{.}$$
3. Let $$A =\left( \begin{array}{cc} 1 & 3 \\ 5 & 9 \\ \end{array} \right)\text{,}$$ $$B=\left( \begin{array}{cc} 5 & 7 \\ 6 & 3 \\ \end{array} \right)\text{,}$$ and $$c=2\text{.}$$ Find $$A + B$$ and $$c A\text{.}$$
4. Are the vector spaces $$\mathbb{R}^4$$ , $$P^3$$ and $$M_{2\times 2}(\mathbb{R})$$ isomorphic to each other? Discuss with reference to previous parts of this exercise. | HuggingFaceTB/finemath | |
# Bernoulli Distribution: Definition and Examples
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Probability Distributions > Bernoulli Distribution
## What is a Bernoulli Distribution?
A Bernouilli distribution is a discrete probability distribution for a Bernouilli trial — a random experiment that has only two outcomes (usually called a “Success” or a “Failure”). For example, the probability of getting a heads (a “success”) while flipping a coin is 0.5. The probability of “failure” is 1 – P (1 minus the probability of success, which also equals 0.5 for a coin toss). It is a special case of the binomial distribution for n = 1. In other words, it is a binomial distribution with a single trial (e.g. a single coin toss).
The probability of a failure is labeled on the x-axis as 0 and success is labeled as 1. In the following Bernoulli distribution, the probability of success (1) is 0.4, and the probability of failure (0) is 0.6: The probability density function (pdf) for this distribution is px (1 – p)1 – x, which can also be written as: The expected value for a random variable, X, from a Bernoulli distribution is:
E[X] = p.
For example, if p = .04, then E[X] = 0.4.
The variance of a Bernoulli random variable is:
Var[X] = p(1 – p).
## What is a Bernoulli Trial?
A Bernoulli trial is one of the simplest experiments you can conduct in probability and statistics. It’s an experiment where you can have one of two possible outcomes. For example, “Yes” and “No” or “Heads” and “Tails.” A few more examples:
• Coin tosses: record how many coins land heads up and how many land tails up.
• Births: how many boys are born and how many girls are born each day.
• Rolling Dice: the probability of a roll of two die resulting in a double six.
Bernoulli trials are usually phrased in terms of success and failure. Success doesn’t mean success in the usual way — it just refers to an outcome you want to keep track of. For example, you might want to find out how many boys are born each day, so you call a boy birth a “success” and a girl birth a “failure.” In the dice rolling example, a double six die roll would be your “success” and everything else rolled would be considered a “failure.”
## Independence
An important part of every Bernoulli trial is that each action must be independent. That means the probabilities must remain the same throughout the trials; each event must be completely separate and have nothing to do with the previous event.
Winning a scratch off lottery is an independent event. Your odds of winning on one ticket are the same as winning on any other ticket. On the other hand, drawing lotto numbers is a dependent event. Lotto numbers come out of a ball (the numbers aren’t replaced) so the probability of successive numbers being picked depends upon how many balls are left; when there’s a hundred balls, the probability is 1/100 that any number will be picked, but when there are only ten balls left, the probability shoots up to 1/10. While it’s possible to find those probabilities, it isn’t a Bernoulli trial because the events (picking the numbers) are connected to each other.
The Bernouilli process leads to several probability distributions:
## Relation to the Binomial Distribution
The Bernoulli distribution is closely related to the Binomial distribution. As long as each individual Bernoulli trial is independent, then the number of successes in a series of Bernoulli trails has a Binomial Distribution. The Bernoulli distribution can also be defined as the Binomial distribution with n = 1.
## Use in Epedemiology
In experiments and clinical trials, the Bernoulli distribution is sometimes used to model a single individual experiencing an event like death, a disease, or disease exposure. The model is an excellent indicator of the probability a person has the event in question.
• 1 = “event” (P = p)
• 0 = “non event” (P = 1 – p)
Bernoulli distributions are used in logistic regression to model disease occurrence.
## References
Evans, M.; Hastings, N.; and Peacock, B. “Bernoulli Distribution.” Ch. 4 in Statistical Distributions, 3rd ed. New York: Wiley, pp. 31-33, 2000.
WSU. Retrieved Feb 15, 2016 from: www.stat.washington.edu/peter/341/Hypergeometric%20and%20binomial.pdf
CITE THIS AS:
Stephanie Glen. "Bernoulli Distribution: Definition and Examples" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/bernoulli-distribution/
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Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! | HuggingFaceTB/finemath | |
## What are Field Splices in steel connection?
Field splices are the connection of separate sections of the girders (field sections) which occur in the field. Generally these splices are bolted connections as field welding is usually discouraged due to the high cost of field welding and problems associated with welding in less than ideal conditions, such as inclement weather. Field splices can … Read more
## Find the value of porosity of a soil
Answer: c) 25.9% Explanation: Given Data: • Void ratio, e = 0.35 Solution: The formula of porosity in terms of the void ratio is, n = e / 1+e Where, • n = Porosity of soil • e = Void ratio ∴ n = 0.35 / (1 + 0.35) ∴ n = 0.35 / 1.35 … Read more
## Independent term of the soil sample’s total volume.
Answer: a) the Unit weight of solids. Explanation: A function of the volume of solids is the unit weight of solids & that unit weight of solids does not change with the moisture content. Following are the functions of total volume: Bulk unit weights Dry unit weights Submerged unit weight
## Find the value of the volume of void
Answer: c) 70 m3 Explanation: Porosity is the ratio of volume of void to the total volume. ∴ n = (Vv / V ) x 100 ∴ 0.7 = (Vv / 100 ) x 100 ∴ Vv = 70 m3 Hence, the volume of void is 70m3.
## Determine the value of void ratio
Answer: d) 0.28 Explanation: Given Data: • Porosity of soil sample, n= 22%= 0.22 Solution: The formula of void ratio in terms of porosity, e = n / (1-n) e= 0.22 / (1-0.22) e= 0.28 Hence, the value of void ratio is 0.28
## Find the value of the void ratio in soil sample
Answer: c) 0.53 Explanation: Given Data: • The ratio of saturated unit weight to dry unit weight, γ (sat.) / γd = 1.2 • The specific gravity of solids, G = 2.65 Solution: Saturated density of soil is, γ(sat.) = {(G + e) / (1+e)} γw….(i) Dry density of soil is, γd = (G/ 1+e) … Read more
## Dry density of the soil sample
Answer: c) 2 g/cc Explanation: Given Data: • Bulk density of soil sample = 2.4 g/cm3 • The moisture content of soil sample = 20% Solution: The dry density of soil = Bulk density of soil sample / (1 + Moisture content of soil sample) Dry density of soil = 2.4 / (1 + 0.2) … Read more
## correct regarding the percentage void ratio
Answer: a) The ratio of the volume of air voids to the total volume of the soil sample Explanation: The ratio of the volume of air voids to the total volume of the soil sample is the percentage void ratio & it is expressed as a percentage. na = (Va/ V) x 100 Where, Va … Read more
## Correct relationship between the degree of saturation & air content
Answer: d) ac = 1 – S Explanation: The ratio of the volume of air voids to the volume of voids in the air content. ac = Va / Vv …(i) The volume of water volume to the volume of voids in a given soil mass is called a degree of saturation. S = Vw … Read more
## Air content
Answer: b) Ratio of the volume of air voids to volume of total voids in the soil mass Explanation: The volume of air voids to the volume of total voids in soil mass is called air content. Air content is related to the degree of saturation. When we subtract the value of the degree of … Read more | HuggingFaceTB/finemath | |
All awareness for Bank PO | Clerk | Railway | IBPS PO | IBPS Clerk | SSC
### Reasoning
ISBT Reasoning for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
##### Find the missing numbers.
1) 12 2) 16 3) 18 4) 20 5) None of these
Explanation : First column : 112 - 12 = 120.
Second Column : 72 - 22 = 45.
Third column : 52 - 32 = 16
##### Find the missing numbers.
1) 10 2) 12 3) 14 4) 18 5) None of these
Explanation : Logic (Row-wise):
(First Element / Second element) x 2 = Third Element.
So, (108/x) x 2 = 18
or x = 12.
Missing Number = 12.
##### Find the missing number.
1) 4 2) 5 3) 8 4) 12 5) None of these
Explanation : Second Row, 2 x 9 + 3 x 17 = 69.
Third Row, 2 x 13 + 3 x 11 = 59.
Let , missing number be x,
So, 2x + 3 x 13 = 49.
or, x = 5.
##### Direction :Read the following information and give the answer. A, B, C, D, E, F and G are sitting in a row facing North : 1. F is to the immediate right of E. 2. E is 4th to the right of G. 3. C is the neighbour of B and D 4. Person who is third to the left of D is at one of ends.Who are to the left of C ?
1) Only B 2) G, B and D 3) G and B 4) D, E, F and A 5) None of these
Explanation :
##### Direction : Each of these questions are based on the information given below : I. Eight persons E, F, G, H, I, J, K and L are seated around a square table - two on each side. II. There are 3 ladies who are not seated next to each other. III. J is between L and F. IV. G is between I and F. V. H, a lady member is second to the left of J. VI. F, a male member is seated opposite to E, a lady member. VII. There is a lady member between F and I.Who is sitting third to the left of J ?
1) K 2) l 3) I 4) M 5) None of these
Explanation :
So, K is third to the left of J.
##### Direction : Each of these questions are based on the information given below : I. Eight persons E, F, G, H, I, J, K and L are seated around a square table - two on each side. II. There are 3 ladies who are not seated next to each other. III. J is between L and F. IV. G is between I and F. V. H, a lady member is second to the left of J. VI. F, a male member is seated opposite to E, a lady member. VII. There is a lady member between F and I.Who among the following is to the immediate left of F ?
1) G 2) I 3) J 4) H 5) None of these
Explanation :
J is to the immediate left of F.
J is to the immediate left of F.
##### Direction : Each of these questions is based on the following information: A + B means A is the mother of B A - B means A is the sister of B A * B means A is the father of B A ÷ B means A is the brother of BWhich of the following means that N is the maternal uncle of M ?
1) N ÷ P - L + E - M 2) N - Y + A ÷ M 3) M - Y * P - N 4) N ÷ C + F * M 5) None of these
Answer : N ÷ P - L + E - M
Explanation :
N ÷ P means N is the brother of P
P - L means P is the sister of L
L + E means L is the mother of E
E - M means E is the sister of M.
Hence, L is the mother of M, P is the maternal aunt of M and N is the maternal uncle of M.
##### "A"walks 20 km towards North. He turns left and walks 40 km. He again turns left and walks 20 km. Finally he moves 20 km after turning to the left. He is facing towards in which direction to his starting point ?
1) East 2) West 3) North 4) South 5) None of these
Explanation :
So, he is facing towards east direction.
##### Directions : Read the following character sequence carefully and then answer the questions given below it: D E 5 B F K 8 P M T 7 H O 6 S R Q 4 X C 3 A Z W 2 LWhich of the following will be the third to the right of 16th character from your right ?
1) H 2) O 3) 6 4) S 5) R | HuggingFaceTB/finemath | |
# A New System For Roulette
I have devised a system for playing roulette which gives a pretty chance of leaving from the table richer than you arrived. No doubt it has already been thought of and some fatal flaw discovered which my poor brain could not detect. If so, please accept my apologies for wasting your time. If not, please read through what follows and give it a try. If it works, please post here and let everyone know - that will be payment enough. If it fails,, please post here and let everyone know!
## Background
The game of roulette relies on the spin of a wheel bearing 37 numbers (in the French version) or 38 (American). My system does not care which version you play, so please substitute the correct details for your location.
I am in France, so I use the French notation.
When the wheel is spun, a ball is introduced and eventually comes to rest on one of the numbers. In this way can the winners be determined. There are many ways to bet on the result, but my systems assumes you will bet on a single number unswervingly until you win. After winning, the system is reset and you may choose another number.
## Mathematical basis
In this scenario, you have a 1/37 chance of matching the winning number. If you do so, your stake is returned to you, along with winnings of 35 times your stake. Thus in total you receive back 36 times your stake. Over time, you have a return of 36/37, that is YOU LOSE! The trick is to win in the meantime...
The approach I have taken is to calculate at each spin of the wheel the sum of cash flows in the 2 cases - case 1, you fail to match the winning number and your money is taken; case 2, you win. In this way can we see the necessary level of stake for the next bet to ensure that when your number is eventually correct so you have made a profit during your stay at the table. If these calculations are correctly performed, you will be certain that once you have eventually won on a certain spin, then the total you have lost will at worst be equivalent in value to the pile of chips the croupier pushes towards you.
## Limitations
Of course, the chances of winning on average are 1/37 - that is, if you stay at the table for more than 37 spins, you should see your number come up. However, that’s life: sometimes your number won’t come up for a very long time.
I have assumed that you are able to play in chips of 1 currency unit and that you have a stack of 1000 currency units in front of you. With this method, you can survive at the table 139 unsuccesful spins in succession and still make an overall profit of 19 units on the 140th play. A return on capital of <2%, but tax-free and you had a very good time.If you only have 100 currency units, you can survive 62 unsuccesful spins in succession and still make an overall profit of 9 units on the 63rd play.
If your number doesn’t come up by the 140th spin, the gods hate you and you should go home!
## How it works
Let us work through the example of the first 10 spins:
A--- B---- C----- | D------ E--------- F------------
Spin Stake Spent | Paid Stake back Profit if won so far | on win on win (= D + E - C)
---- ----- ------ | ------- ---------- -------------
1 1 1 | 35 1 35
2 1 2 | 35 1 34
3 1 3 | 35 1 33
4 1 4 | 35 1 32
5 1 5 | 35 1 31
6 1 6 | 35 1 30
7 1 7 | 35 1 29
8 1 8 | 35 1 28
9 1 9 | 35 1 27
10 1 10 | 35 1 26 It can be seen that the profit diminishes by 1 each spin and then let us move ahead to the 36th spin:
A--- B---- C----- | D------ E--------- F------------
Spin Stake Spent | Paid Stake back Profit if won so far | on win on win (= D + E - C)
---- ----- ------ | ------- ---------- -------------
36 1 36 | 35 1 0 This is break even - the stake must be increased
37 2 38 | 70 2 34
38 2 40 | 70 2 32
and then move ahead again to the 54th spin:
A--- B---- C----- | D------ E--------- F------------
Spin Stake Spent | Paid Stake back Profit if won so far | on win on win (= D + E - C)
---- ----- ------ | ------- ---------- -------------
54 2 72 | 70 2 0 This is break even - the stake must be increased
55 3 75 | 105 3 33
56 3 78 | 105 3 30
...and so on.
Clearly, the effect of increasing the stake at periodic intervals allows the player to always have a profitable result when his number eventually comes up. The frequency table is as follows:
A-------- B------ C--------- D------
From spin To spin Spin count Stake
--------- ------- ---------- -------
1 36 36 1
37 54 18 2
55 66 12 3
67 75 9 4
76 82 7 5
83 88 6 6
89 93 5 7
94 97 4 8
98 101 4 9
102 105 4 10
106 108 3 11
109 111 3 12
112 114 3 13
115 116 2 14
117 119 3 15
120 121 2 16
122 123 2 17
124 125 2 18
126 127 2 19
128 130 3 21
131 132 2 22
133 133 1 23
134 135 2 24
136 136 1 25
137 138 2 26
139 139 1 27
140 140 1 28
141 141 1 29
142 143 2 30
144 144 1 31
145 145 1 32
146 146 1 33
147 147 1 34
148 148 1 35
149 149 1 36
150 150 1 37
151 151 1 38
152 152 1 39
153 153 1 40
154 154 1 41
155 155 1 43
156 156 1 44
157 157 1 45
158 158 1 46
159 159 1 48
160 160 1 49
161 161 1 50
162 162 1 52
163 163 1 53
164 164 1 55
165 165 1 56
166 166 1 58
167 167 1 60
168 168 1 61
169 169 1 63
170 170 1 65
171 171 1 67
172 172 1 69
173 173 1 71
174 174 1 73
175 175 1 75
176 176 1 77
177 177 1 79
178 178 1 81
179 179 1 84
180 180 1 86
181 181 1 89
182 182 1 91
183 183 1 94
184 184 1 96
185 185 1 99
186 186 1 102
187 187 1 105
188 188 1 108
189 189 1 111
190 190 1 114
191 191 1 117
192 192 1 121
193 193 1 124
194 194 1 128
195 195 1 131
196 196 1 135
197 197 1 139
198 198 1 143
199 199 1 147
200 200 1 151
That’s all there is to it - play well and have fun!
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# What is a G Force?
Perhaps John Burk covered this well enough, but a good question is almost always worth answering again.
Why Worry About G Force? ————————
You see? I added some extra questions.
Suppose you have an app on your phone that measures acceleration. Here is a screen shot from the iPhone AccelMeter App. It's a pretty cool app. It shows a 3-D representation of the g-force vector in real time. Here you can see that I am just holding the phone to produce a vector with a magnitude of 1.00 g. Why doesn't the phone give the acceleration vector instead? Because the phone can't tell the difference between the gravitational field and an acceleration.
Recall how an accelerometer works. I actually wrote about this a long time ago - but here is a more recent (and popular) video describing modern accelerometers. At the basic level, an accelerometer is just a spring and the measurement of a g-force is based on the amount the spring is stretched. Consider two springs that can only move in one dimension. The first spring is vertical and at rest. The other spring is horizontal and accelerating. For the vertical hanging spring on the left, it is in equilibrium. This means that for the forces in the vertical direction, the following would be true: For most springs, the force exerted by the spring is proportional to the amount the spring is stretched. This is known as Hooke's Law and can be written as: Here k is the spring constant - essentially a measure of the 'stiffness' of the spring and s is the amount the spring is either compressed or stretched from it's natural length. Yes, I know that many times you will see a negative sign in this equation to indicate that the spring force is in the opposite direction that the spring is stretched. I didn't include that since I am just show the magnitude. But getting back to the vertical spring, I can find the amount the spring is stretched if I know the spring constant and the mass. Oh, g is the gravitational field. It has a magnitude of 9.8 Newtons per kilogram. Ok. Now to look at the horizontal spring (I have only included the horizontal forces in case you couldn't tell). For this object, there is only the spring force on it. The force equation in the x-direction would be: And here is the cool part. What if the mass is accelerating with a magnitude of 9.8 m/s2? Well, since the acceleration has the same value as the gravitational field (and same units since 1 N/kg = 1 m/s2), the spring would have the same stretch. In an accelerometer, the stretch (or compression) of the spring is really the only thing that is measured. So, the accelerometer can't tell the difference between accelerations and gravitational forces.
Neither can you. In short, this is why you feel "weightless" in orbit. If you want the longer version, here is a more detailed post about apparent weight and weightlessness.
What is the G Force? ——————–
First, it isn't really a measure of force. If two objects are sitting on the table, they will both be at 1 g even if they are different masses. The gravitational forces will be different and the force of the table pushing up will be different.
I am not sure everyone completely agrees on the definition of g-force, but I like this definition. If an object is at rest, then the net force on that object would be zero (zero vector). Subtracting the gravitational force would leave a g-force of 9.8 m/s2 or 1 g. If an object is accelerating UP at 9.8 m/s2, the net force would also be a vector pointing up. Subtracting a vectoring pointing down (gravitational force) would result in a larger g-force of 2 g's. If the object was accelerating down at 9.8 m/s2, the net force would be the same as the gravitational force. Subtracting them would give the zero vector and a g-force of 0 g's.
Human Tolerance of G-Force ————————–
One of the best ways to look at human body damage is to consider the acceleration. Acceleration is the killer, well usually. Consider this model of a human body colliding with the ground. In this model, there are two balls connected by a spring. If the body falls and collides with the ground, it must accelerate in the upward direction. Let me just look at the top ball. Since it has to accelerate up, it must have a net force pointing upwards. This means that the force the inner spring exerts on the top ball must be greater than the gravitational force. The greater the acceleration, the greater this spring force must be and the more compressed the inner spring will be. If this spring is compressed too much, it could break. Breaking springs would be bad. This is where the damage comes into play.
So, large accelerations can cause damage. Always? No. What if there was some long range force to accelerate this two-ball model of a body? If this same force was on both balls in the model, you could get a super high acceleration without having to compress the inner spring. No inner spring compression means no body damage. But how would this work? I don't know. The only force that pulls on all parts of a body would be the gravitational force (since all the parts have mass). But wouldn't that be cool? If there was some force field that could stop you (or shoot you off like a bullet) without causing damage? Yes. That would be cool.
Then what kind of accelerations can a human body withstand? In a previous episode of Mythbusters - the jumping from a building with bubble wrap one, they state that stunt men aim for a maximum acceleration of 10 g's. A good goal to aim for. Wikipedia's g-force tolerance page lists 50 g's as "likely death". However, it also says that some people may have survived accelerations up to 100 g's. It seems that the duration of the acceleration is quite important. An acceleration of just 16 g's for an extended time period can be deadly also. | HuggingFaceTB/finemath | |
Beautiful Equations Explained.. Binomial Theorem
Mathematical equations are undoubtedly beautiful to look at. If you’re lucky enough to study them, then you’ll quickly realise that their beauty is not limited to the elegant symbols and funky x’s… We’re passionate about sharing this beauty with the world, breaking them down one equation at a time to explain their meaning, purpose and history! First up, a handy tool for expanding brackets…
https://www.beautifulequation.com/collections/binomial-theorem
The binomial theorem is an algebraic description of the expansion of a binomial to the power of some natural number n. Or more clearly, say you have some brackets in algebra (technically a polynomial) and inside there are two terms. If this bracket is multiplied by itself a certain number of times (it must be a positive whole number) then instead of writing it out this many times and expanding each bracket one by one this formula gives the solution! In order to use this equation your first step is to change “a” and “n” for the values in your own brackets. The greek epsilon symbol in the equation means “Sum''. To use it, you take all the positive whole numbers from 1 to n, putting them in the place of k one at a time and then adding each of these terms together. And that’s it, you have your expansion!
This theorem pops up in surprising places across mathematics and is a useful tool in finding proofs and deriving other formulas in probability and calculus, for example. It also links beautifully to Pascal’s triangle in pure mathematics, as the binomial coefficient is also the (k + 1)th number in the (n+1)th row of the triangle.
Special cases of the theorem such as choosing n = 2 can be traced back as far as the Greek mathematician Euclid in 4th century BC and for n = 3 to 6th century AD in India. The generalised form of the theorem is credited to Isaac Newton in the mid-17th century; however, it was the lesser known Lichfield-born mathematician John Colson who first published a proof in 1736. | HuggingFaceTB/finemath | |
# How do you use synthetic division to show that x=sqrt2 is a zero of x^3+2x^2-2x-4=0?
Jan 24, 2017
By showing that the division gives no remainder, i.e. $r \left(x\right) = 0$.
The basic setup of synthetic division is:
ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")
where the coefficients correspond to the function ${x}^{3} + 2 {x}^{2} - 2 x - 4$. Since we are dividing by a linear factor, $x - \sqrt{2}$, we should get a quadratic back.
(Note that the root you use in your divisor would be the root you acquire from $x - r = 0$.)
The general steps are:
1. Bring down the first coefficient.
2. Multiply the result beneath the horizontal line by the root and store in the next column above the horizontal line.
4. Repeat 2-3 until you reach the last column and have evaluated the final addition.
You should then get:
ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "sqrt2" "(2^"3/2" + 2)" "2^("1/2" + "3/2")" "" ")
$\text{ "" "" "" "1" "(2+sqrt2)" "2^"3/2"" "" "" } 0$
$= \textcolor{b l u e}{{x}^{2} + \left(2 + \sqrt{2}\right) x + {2}^{\text{3/2}}}$
$\frac{{x}^{3} + 2 {x}^{2} - 2 x - 4}{x - \sqrt{2}} = \stackrel{p \left(x\right) \text{/"q(x))overbrace(x^2 + (2 + sqrt2)x + 2^"3/2}}{+} \stackrel{r \left(x\right)}{\overbrace{\frac{0}{x - \sqrt{2}}}}$
And you can check to see that it properly expands to give the original cubic. Furthermore, we now see that $r \left(x\right) = 0$, so it is evident that $x = \sqrt{2}$ is a root of ${x}^{3} + 2 {x}^{2} - 2 x - 4$.
CHALLENGE: Can you show that $- \sqrt{2}$ is also a root? | HuggingFaceTB/finemath | |
9 out of 10 based on 942 ratings. 4,796 user reviews.
# STAGE 2 NUMBERS AND ALGEBRA LESSON PLANS
Year 2 Adding 2 Digit Numbers and Ones Crossing 10 Lesson Pack
2014 National Curriculum Resources » Maths » Key Stage 1 - Year 1, Year 2 » Year 2 » Number - Addition and Subtraction » Add and subtract numbers using concrete objects, pictorial representations, and mentally, including: a two-digit number and ones; a two-digit number and tens; two two-digit numbers; adding three one-digit numbers » Add and subtract numbers using a two-digit number
Year 2 Maths Ordering Numbers to 100 Lesson Pack
You can help your Year 2 children with ordering numbers to 100 using this brilliant lesson pack full of useful resources in this topic ordering numbers to 100 lesson pack includes some challenging and differentiated tasks across multiple worksheets and activities, which are suitable to use for home learning and in school can help your children through their mastery of ordering numbers
Stage Movement & Blocking: Definition & Rules - Video
The numbers give the value of the movement: in the following image, -2 Drama Character Development Lesson Plan; Stage Movement & Blocking Go to Acting & Theater Lesson Plans Ch 3.
Stage 4 - algebra - number patterns
Stage 4 - algebra - number patterns Continue a number pattern to match a table of values; determine the rule which involves more than one operation to match a number pattern Strategy
Lesson Plans Worksheets & Teaching Resources | Teachers
This 291 page pack has lesson plans from reading levels A-Z (Fountas and Pinnell), planning sheets, reading level pages, comprehension questions, discussion starters, text selection sheets, and more. There are also editable pages to type in your own information.
Year 6 Maths | Lesson Plans & Resources | Hamilton Trust
To cover the National Curriculum statutory requirements, follow this pathway of Blocks. Each Block is divided into bite-size Units. Each Unit provides differentiated teaching and practice of one or
Year 2 practice papers - SATs Arithmetic assessments
This resource contains a set of six Year 2 practice papers. They aim to introduce children to the format of the 2016 Sample Arithmetic Tests (SATs), using content appropriate for Year 2 learners. The Year 2 practice papers are easy to follow and marke are enough assessments included for your Year 2 class to cover every half term further help you there is a customizable
Year 5/6 Maths Plans | Hamilton Trust | Hamilton Trust
Year 5: Use long multiplication to multiply 2-digit numbers, then 3-digit numbers by a 2-digit number less than 30. Use rounding to estimate products. Multiply fractions and mixed numbers by whole numbers, simplifying answers. Year 6: Describe and predict patterns. Read recurring displays on a calculator. Know common fraction and decimal
44 Free Maths Lesson Plans for KS3 and KS4 by Colin Foster
In this lesson, algebra is used to make sense of a number trick that the teacher performs. Students are asked to explore the mechanics, experiment with some numbers, and then use algebra to try to probe exactly how and why the trick works.
Maths lesson starters | KS3-KS4 | Teachit teaching
All our KS3 and KS4 lesson starters in one handy place! Puzzles, team games, numeracy gems and other quick activities to kick off your maths lessons. | HuggingFaceTB/finemath | |
# 0542 01 Matrix
Solved at: 2023-01-30
01 Matrix - LeetCode
## Question
Given an m x n binary matrix mat, return the distance of the nearest 0 for each cell.
The distance between two adjacent cells is 1.
## Solution
class Solution {
func updateMatrix(_ matrix: Int) -> Int {
// fill the zeros first
var mat = matrix
var queue: [(Int, Int)] = []
for (i, row) in mat.enumerated() {
for (j, col) in row.enumerated() {
if mat[i][j] == 0 {
queue.append((i, j))
}
mat[i][j] = -1
}
}
var level = 0
while !queue.isEmpty {
let count = queue.count
for i in 0..<count {
let (r, c) = queue.remove(at: 0)
if mat[r][c] == -1 {
mat[r][c] = level
}
if r > 0 && mat[r-1][c] == -1 {
queue.append((r-1, c))
}
if c > 0 && mat[r][c-1] == -1 {
queue.append((r, c-1))
}
if r < mat.count - 1 && mat[r+1][c] == -1 {
queue.append((r+1, c))
}
if c < mat[0].count - 1 && mat[r][c+1] == -1 {
queue.append((r, c+1))
}
}
level += 1
}
return mat
}
}
## Results
• Time taken: 11 m 43 s
• Runtime 1856 ms, Beats 10.39%
• Memory 15.5 MB, Beats 83.12%
## Complexity Analysis
• Time: $O(N \times c)$
• Space: $O(N \times c)$
Where $c$ is the cost — the number of iterations in which "BFS clear" operation will run.
## Other Answers Online
Dynamic Programming in 2 passes. An easier solution will be 4 passes from left → right, right → left, top → bottom, bottom → top. | HuggingFaceTB/finemath | |
Computer exercise 3: Survival Probabilities of new particles
Overview¶
This is an exercise to calculate "survival probabilities" of newly formed particles based on the approach described by Westervelt et al. (2013).
New particles that grow beyond a molecular cluster (~3 nm) can continue growing through net condensation of vapor molecules that increase its size. Given sufficient amount of condensing vapor, these particles will grow to a size (~50--100 nm) large enough to seed cloud droplets. However, some of these smaller, growing particles will be scavenged by larger particles (via process of coagulation) that might have already been destined to become cloud dropets. The outcome -- whether these particles are scavenged before fully grown -- can have a significant impact on the cloud droplet number concentration and cloud droplet sizes, which in turn affects precipitation patterns, cloud lifetimes, and the global energy balance.
How many 3 nm particles will grow large enough to possibly serve as seeds for cloud droplet formation? ~ 50--100 nm are typical sizes at which this becomes possible; we will use 100 nm. We can characterize this quantity with a number referred to as the "survival probability" ($\Pr$). If, for instance, if $J_3$ denotes the number of new particles (3 nm size) formed per volume per time, the survival probability can tell us the rate of appearance of $x$ nm sized particles:
$$J_{x} = \Pr_{3 \to x} \, J_{3}$$
Defining individual transition probabilities from size bin $k$ to $k+1$. How can we calculate this probability? We can write the rate of loss of particles [of size $D_{p,k}$ and number concentration $n_k(t) = n(D_{p,k},t)$] to coagulation as a first order process (temporarily using the symbol, $\kappa$ to denote the first-order rate constant, which is the inverse of the characteristic timescale):
\begin{align} L_k^{(\textrm{coag})} = \kappa_k \, n_k(t) = \frac{1}{\tau_k^{(\textrm{coag})}} n_k(t) \end{align}
The fraction of particles of size $D_{p,k}$ remaining after loss by coagulation over time interval $t$ is therefore given by:
\begin{align} \frac{n_k(t)}{n_k(0)} = \exp\left(- \frac{t}{\tau_k^{(\textrm{coag})}} \right) \end{align}
The characteristic time for particles to grow from $D_{p,k}$ to $D_{p,k+1}$ by condensation is denoted as $\tau_{k,k+1}^{(\textrm{cond})}$. The the fraction of particles that remains after loss to coagulation during this period is given by $n_k(t)/n_k(0)$, which can also be recast as a survival probability (or transition probability) $\Pr_{k \to k+1}$:
\begin{align} \Pr_{k \to k+1} = \frac{n_k(\tau_{k,k+1}^{(\textrm{cond})})}{n_k(0)} = \exp\left(- \frac{\tau_{k,k+1}^{(\textrm{cond})}}{\tau_k^{(\textrm{coag})}} \right) \end{align}
Defining timescales of condensation and coagulation. Now let us determine the appropriate expressions for $\tau$, which is generally be given by $n_k / (\partial n_k/\partial t)_{\text{process}}$. The expressions for condensation/evaporation and coagulation are given by:
\begin{align} \left[\frac{\partial n_k(t)}{\partial t}\right]_{\textrm{cond}} &= - \frac{\partial}{\partial D_p}\left(I(D_{p,k}) n_k(t) \right)\\ \left[\frac{\partial n_k(t)}{\partial t}\right]_{\textrm{coag}} &= P_k^{(\textrm{coag})} - L_k^{(\textrm{coag})} = \frac{1}{2} \sum_{j=1}^{k-1} K(D_{p,j},D_{p,k-j})\, n_j\, n_{k-j} - n_k \sum_{j=1}^{\max(k)} K(D_{p,k},D_{p,j})\, n_j \ \ \ \forall k \geq 2 \end{align}
$P$ and $L$ denote production and loss, respectively, and $K$ is the coagulation coefficient. $I$ is the diameter growth rate.
Note that during lecture, $n$ and therefore $I$ were written as a function of volume ($v$, which is $(\pi/6) D_p^3$ for a sphere). In the transition regime, the volumetric growth rate is given by:
$$I(v) = \frac{dv}{dt} = \frac{2 \pi D_{AB} M_A}{RT \rho_p} D_p\ f(Kn, \alpha) \left(p_{\infty, A} - p_{\textrm{eq}, A}\right)$$
$M_A$ = molecular weight of species A, $D_{AB}$ = diffusion coefficient of $A$ in air (indexed as $B$), $\rho_p$ = particle density, $R$ = gas constant, $T$ = temperature, $p_A$ = vapor pressure of $A$ in equilibrium with the surface ("eq") and in the bulk vapor phase ("$\infty$") far from the surface, and $f(Kn, \alpha)$ is the Fuchs-Sutugin correction factor in air:
$$\newcommand{\Kn}{\operatorname{Kn}} f(Kn, \alpha) = \frac{0.75\alpha (1+\Kn)}{\Kn^2 + \Kn+0.283\Kn\alpha + 0.75\alpha} \quad \text{where} \quad \Kn = \frac{2\lambda}{D_p}$$
The mean free path of the condensing molecule in air, $\lambda$, can be calculated from $D_{AB}$ but is provided below in the exercise.
$I$ can alternatively be defined as a function of $D_p$ - then referred to as the diameter growth rate:
$$I(D_p) = \frac{dD_p}{dt} = ?$$
You are asked to derive this expression in Problem 1 below.
Corresponding expressions for characteristic timescales can be calculated as:
\begin{align} \tau_{k,k+1}^{(\textrm{cond})} &= \frac{n_k}{[\partial n_k / \partial t]_{\textrm{cond}}} \approx - \frac{1}{\Delta_{k,k+1} I / \Delta_{k,k+1} D_p}\\ \tau_k^{(\textrm{coag})} &= \frac{n_k}{L_k^{\textrm{coag}}} = \frac{1}{\frac{1}{2} K(D_{p,k},D_{p,k}) n_k + \sum_{j=k+1}^{\max(k)} K(D_{p,k},D_{p,j}) n_j} \end{align} $\Delta_{k,k+1}(x) = x(k+1) - x(k)$ denotes the forward finite difference operator that can be used to numerically approximate a derivative. Note that we only consider losses by coagulation with particles of size $D_{p,k}$ and greater, and the first term in the denominator of the second equation is the rate of loss by self-coagulation (the 1/2 required because of symmetry: $K_{12} = K_{21}$).
Defining the survival probability. Returning back to our original definition, the overall survival probability between any two sizes $D_{p, m}$ to $D_{p, n}$ is the product of all probabilities that span the range:
$$\Pr_{m \to n} = \prod_{k=m}^{n-1} \Pr_{k \to k+1} = \prod_{k=m}^{n-1} \exp\left( -\frac{\tau_{k,k+1}^{(\textrm{cond})}}{\tau_k^{(\textrm{coag})}} \right)$$
All diameters $\{D_{p,m}, D_{p, m+1}, \ldots, D_{p,n}\}$ to be used for the calculation will be provided in the exercise.
Problem statement¶
You will estimate aerosol survival probabilities for a set of background concentration of aerosols ($n_k$) and condensing vapor concentrations.
You are provided with a function that returns $D_p$ ($\mu$m), $\Delta D_p$ ($\mu$m), and $n_k$ (cm$^{-3}$) for use with problems 2 and 3. $n_k$ is normalized such that $\sum_k n_k = 1$. This distribution can be multiplied by a value, $N_0$ cm$^{-3}$, such that the total number of particles equals $N_0$ cm$^{-3}$. A function which calculates the collision frequency function (coagulation kernel) for a pair of particle diameters is also provided so that coagulation rates can be easily computed.
Turn in your own 1) report and 2) Octave/MATLAB code which address the following items:
1. Find the analytical expression for $I(D_p)$ (spherical particles).
2. Estimate variables $\{\tau_k^{(\textrm{coag})}, \tau_{k,k+1}^{(\textrm{cond})}, \Pr_{k \to k+1}\}$ as a function of environmental parameters (background concentration of particles and vapor pressure driving the condensation). Plot variables as a function of $D_p$ for each scenario described below:
• a. Fix $N_0$ to 200 cm$^{-3}$. Calculate variables for $\nabla p_A = \{10^{-12}, 10^{-11}, 10^{-10}, 10^{-9}\}$ atm (which is equivalent to $\{1, 10, 100, 1000\}$ ppt). Note that we have adopted the notation $\nabla p_A = p_{\infty, A} - p_{\textrm{eq}, A}$.
• b. Fix $\nabla p_A$ to $10^{-9}$ atm (1 ppb). Calculate variables for $N_0 = \{50, 200, 500\}$ cm$^{-3}$.
3. Estimate the overall survival probabilities (a single number for each case) of 3 nm growing to 100 nm. Using results of individual probabilities obtained from Problem 2, explain (in a few sentences) what type of environmental conditions allow new particles to grow to cloud nuclei sizes by discussing their influences on aerosol process.
Environmental conditions and parameters:
• $T$ = 298 K, $P$ = 1 atm.
• Properties of condensing vapor (equivalent to that of H$_2$SO$_4)$: diffusion coefficient in air $D_{AB} = 0.1$ cm$^2$ s$^{-1}$, molar mass $M_A = 98.079$ g mol$^{-1}$, mean free path of in air for use in Knudsen number calculation (for transition regime correction) $\lambda = 0.118$ $\mu$m.
• Particle density is $\rho_p = 1.0$ g cm$^{-3}$.
• Assume pressure gradient and number size distribution are constant in each scenario for calculation of characteristic timescales.
• Assume mass accommodation coefficient ($\alpha$) of unity.
Hints¶
1. Given the range of particle diameters to be examined in this problem, we are in the transition regime for expressing the net rate of migration of molecules to a particle ("molar flow", $J$). $f(Kn, \alpha)$ corresponds to the correction factor to the continuum regime molar flow ($J_c$) necessary for transition regime calculations.
2. Write an Octave/MATLAB function for $I(D_p)$ in which you can vary the diameter $D_p$ and concentration gradient. The following provides a suggested template (you fill in the {...}):
{octave}
%% -----------------------------------------------------------------------------
%% define constants
{...}
%% calculate Fuchs-Sutugin correction factor
f = {...}
%% calculate I(Dp)
rate = {... use f, Dp, etc. ...}
end
%% -----------------------------------------------------------------------------
Note to take care with unit conversions. You can test this function to solution given below. You do not actually have to integrate this rate function to calculate timescales. For the timescale calculation, you are provided $\Delta D_p$ so that you can calculate $D_{p,k+1} = D_{p,k} + \Delta D_{p,k}$. Further hint: $\Delta_{k,k+1} (I) = I(D_{p,k+1}) - I(D_{p,k})$ and $\Delta_{k,k+1} (D_p) = D_{p,k+1} - D_{p,k} = \Delta D_{p,k}$.
3. Write an Octave/MATLAB function for the coagulation loss as a function of the number size distribution. Suggested template (you fill in the {...}):
{octave}
%% -----------------------------------------------------------------------------
function kappa = coag_loss_coef(n, Dp)
kmax = length(Dp);
kappa = zeros(size(Dp));
for k = 1:kmax
j = (k+1):kmax
kappa(k) = {... you will need to call coag_kernel() ...}
end
%% -----------------------------------------------------------------------------
Note to take care with unit conversions. To provide clarification, n used in the function above is n(k)=N_0*ndist(k) using N_0 and ndist as defined in the example code (provided below).
4. Use these two functions to calculate your $\tau_k$s.
5. Calculate your $D_p$-dependent probabilities $\Pr_k$s from $\tau_k$s. Do not take the final product of probabilities (let their dependence on $D_p$ remain) until you get to Problem 3.
6. For interpretation, the main variables to focus on in this exercise is $\nabla p_A$, which drives the rate of condensation, and the background concentration of pre-existing particles $n_k$ (low concentration regimes might be considered "clean" while high concentration regimes might be considered "polluted"), which drives the rate of coagulation. The pressure gradient $\nabla p_A$ for condensation typically results from the rapid reactions producing condensable vapor, which results in a bulk-phase vapor pressure $p_{\infty,A}$ that is greater than the equilibrium vapor pressure over the surface of a particle $p_{\textrm{eq},A}$.
7. Remember to use element-wise operators in Octave/MATLAB where necessary. A common error is to use / instead of ./.
Functions provided:
• generate_sizehist.m: returns $D_p$ [$\mu$m], $\Delta D_p$ [$\mu$m], and $n_k$ [cm$^{-3}$]. $n_k$ is normalized such that $\sum_k n_k = 1$. Download here.
• coag_kernel.m: $K(D_{p,1}, D_{p,2})$ receives diameters in units of [m] and returns coagulation coefficients in units of [m$^3$ s$^{-1}$]. Download here.
Example usage:
{octave}
[Dp, dDp, ndist] = generate_sizehist;
N_0 = [50, 200, 500]; % [cm^-3]
subplot(1, 2, 1)
plot(Dp, ndist);
set(gca, 'xscale', 'log')
xlabel('D_p [\mum]')
ylabel('Normalized units')
subplot(1, 2, 2)
plot(Dp, ndist * N_0)
set(gca, 'xscale', 'log')
xlabel('D_p [\mum]')
ylabel('Number concentration [cm^{-3}]')
Your growth rate function can be checked against the following solution. Here I have named my function growth_rate, which receives arguments of $D_p$ [m] and $\nabla p_A$ [atm]. All other parameters are defined within the function.
{octave}
%% integrate over {0, 1, 2, ..., 18000} seconds with an initial diameter of 3 nm
%% for pressure gradients of 1e-12 and 1e-10 atmospheres.
%% define values
t = linspace(0, 5) .* 3600; % [s]
Dp0 = 3e-9; % [m]
gradP1 = 1e-12; % [atm] equivalent to 1 ppt
gradP2 = 1e-10; % [atm] equivalent to 100 ppt
if(exist('OCTAVE_VERSION', 'builtin') ~= 0)
%% integrate (Octave)
y1 = lsode(@(Dp, t) growth_rate(Dp, gradP1), Dp0, t);
y2 = lsode(@(Dp, t) growth_rate(Dp, gradP2), Dp0, t);
else
%% integrate (MATLAB)
[~,y1] = ode45(@(t, Dp) growth_rate(Dp, gradP1), t, Dp0);
[~,y2] = ode45(@(t, Dp) growth_rate(Dp, gradP2), t, Dp0);
end
%% plot
plot(t / 3600, [y1,y2] * 1e6)
set(gca, 'yscale', 'log')
xlabel('Time [h]')
ylabel('D_p [\mum]')
legend({'\nabla p_A = 10^{-12} atm', '\nabla p_A = 10^{-10} atm'}, 'location', 'northwest') | open-web-math/open-web-math | |
### Author Topic: TUT0801 Quiz2 (Read 733 times)
#### bella
• Newbie
• Posts: 4
• Karma: 0
##### TUT0801 Quiz2
« on: October 04, 2019, 02:04:47 PM »
$$x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3}$$
First, let's show the given DE $x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0$ is not exact.
Define $M(x, y)=x^{2} y^{3}, N(x, y)=x\left(1+y^{2}\right)$
$$M_{y}=\frac{\partial}{\partial y}\left[x^{2} y^{3}\right]=3 x^{2} y^{2}$$
$$N_{x}=\frac{\partial}{\partial x}\left[x\left(1+y^{2}\right)\right]=1+y^{2}$$
Since $3 x^{2} y^{2} \neq 1+y^{2}$, this implies the given DE is not exact.
Now, let's show that the given DE multipled by the integrating factor $\mu(x, y)=\frac{1}{x y^{3}}$ is exact.
That is to show
$$\frac{1}{x y^{3}} x^{2} y^{3}+\frac{1}{x y^{3}} x\left(1+y^{2}\right) y^{\prime}=x+\left(y^{-3}+y^{-1}\right) y^{\prime}=0$$
is exact.
Define $M^{\prime}(x, y)=x, N^{\prime}(x, y)=y^{-3}+y^{-1}$
Since
$$M_{y}^{\prime}=\frac{\partial}{\partial y}(x)=0$$
$$N_{x}^{\prime}=\frac{\partial}{\partial x}\left[y^{-3}+y^{-1}\right]=0$$
By theorem in the book, we can conclude that $x+\left(y^{-3}+y^{-1}\right) y^{\prime}=0$ is exact.
Thus, we know there exists a function $\phi(x, y)=C$ which satisfies the given DE.
Also,
$$\frac{\partial \phi}{\partial x}=x$$
$$\frac{\partial \phi}{\partial y}=y^{-3}+y^{-1}$$
Integrate $\frac{\partial \phi}{\partial x}=x$ with respect to $x$ we have
$$\phi(x, y)=\frac{1}{2} x^{2}+g(y)$$
Take derivative on both sides with respect to $y$ we get
$$\frac{\partial \phi}{\partial y}=g^{\prime}(y)$$
Since we know that $\frac{\partial \phi}{\partial y}=y^{-3}+y^{-1}$
Then $g^{\prime}(y)=y^{-3}+y^{-1}$
Integrate with respect to $y$ we have
$g(y)=-\frac{1}{2} y^{-2}+\ln |y|+C$
Altogether, we have $\phi(x, y)=\frac{1}{2} x^{2}-\frac{1}{2} y^{-2}+\ln |y|=C$, which means
$$\frac{1}{2} x^{2}-\frac{1}{2} y^{-2}+\ln |y|=C$$
is the general solution to the given DE.
Besides, notice that the constant function $y(x)=0 \forall x$ is also a solution to the given DE. | HuggingFaceTB/finemath | |
# A surd simplification masterclass
The estimable @solvemymaths tweeted, some time back:
A sensible option? Perhaps. But Wolfram Alpha is being a bit odd here: that's something that can be simplified significantly. (One aside: I'm not convinced that actually is $\sin(22º)$, because I get a different answer in the first line, and WA doesn't give an exact answer for it -- but I'm willing to be corrected! So I'm going to call it $S$ instead of $\sin(22º)$)
Before I do anything, I'll reiterate my Key Rule for simplifying anything: take the ugliest thing and make it nicer.
Let's start from what they have:
$S = \frac{ - \frac 18 \sqrt{3} \left( -1 - \sqrt{5} \right) - \frac{1}{4}\sqrt{\frac 12 (5 - \sqrt{5}) } } {\sqrt{2}} - \frac{ - \frac 18 \left( -1 - \sqrt{5} \right) - \frac{1}{4}\sqrt{\frac 32 (5 - \sqrt{5}) } } {\sqrt{2}}$
Honestly. If a student of mine turned that in, I'd roll my eyes at them. For a start, we've got two fractions with the same denominator. Let's combine them, being very careful with our minus signs:
$S = \frac{ -\frac 18 \sqrt{3} \left( -1 - \sqrt{5} \right) - \frac 14 \sqrt{\frac 12 (5 - \sqrt{5}) } + \frac 18 \left( -1 - \sqrt{5} \right) + \frac{1}{4}\sqrt{\frac 32 (5 - \sqrt{5}) } } {\sqrt{2}}$
Next, the ugliest thing is the stacked fractions. Let's multiply top and bottom by 8:
$S = \frac{ - \sqrt{3} \left( -1 - \sqrt{5} \right) - 2 \sqrt{\frac 12 (5 - \sqrt{5}) } + \left( -1 - \sqrt{5} \right) + 2 \sqrt{\frac 32 (5 - \sqrt{5}) } } {8\sqrt{2}}$
There are minus signs flying around like squirrels. This can be improved:
$S = \frac{ \sqrt{3} \left( 1 + \sqrt{5} \right) - 2 \sqrt{\frac 12 (5 - \sqrt{5}) } - \left( 1 + \sqrt{5} \right) + 2 \sqrt{\frac 32 (5 - \sqrt{5}) } } {8\sqrt{2}}$
I don't like the fractions inside the square roots, either.
$S = \frac{ \sqrt{3} \left( 1 + \sqrt{5} \right) - \sqrt{2} \sqrt{ (5 - \sqrt{5}) } - \left( 1 + \sqrt{5} \right) + \sqrt{3}\sqrt{2} \sqrt{5 - \sqrt{5} } } {8\sqrt{2}}$
There's a reason I didn't make that $\sqrt{6}$, by the way. There's now some obvious factorising:
$S = \frac{ \left(\sqrt{3} - 1\right) \left( 1 + \sqrt{5} \right) + \sqrt{2} \left( \sqrt{3}- \sqrt{1} \right) \sqrt{ (5 - \sqrt{5}) } } {8\sqrt{2}}$
And the $\left(\sqrt{3} -1\right)$ comes out as well:
$S = \frac{ \left(\sqrt{3} - 1\right) \left[ \left( 1 + \sqrt{5} \right) +\sqrt{2} \sqrt{ 5 - \sqrt{5}} \right] } {8\sqrt{2}}$
I'm not going to claim that's nice, but it's definitely nicer than it was!
## Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
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##### Where do you teach?
I teach in my home in Abbotsbury Road, Weymouth.
It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby. | HuggingFaceTB/finemath | |
# what is a polynomial function
1/(X-1) + 3*X^2 is not a polynomial because of the term 1/(X-1) -- the variable cannot be in the denominator. Roots (or zeros of a function) are where the function crosses the x-axis; for a derivative, these are the extrema of its parent polynomial.. "2) However, we recall that polynomial … Consider the polynomial: X^4 + 8X^3 - 5X^2 + 6 In other words, a polynomial is the sum of one or more monomials with real coefficients and nonnegative integer exponents.The degree of the polynomial function is the highest value for n where a n is not equal to 0. Notice that the second to the last term in this form actually has x raised to an exponent of 1, as in: Rational Function A function which can be expressed as the quotient of two polynomial functions. allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form $$(x−c)$$, where $$c$$ is a complex number. A polynomial function is defined by evaluating a Polynomial equation and it is written in the form as given below – Why Polynomial Formula Needs? A polynomial is an expression which combines constants, variables and exponents using multiplication, addition and subtraction. The Theory. A degree 0 polynomial is a constant. It will be 5, 3, or 1. A polynomial function is an even function if and only if each of the terms of the function is of an even degree. A polynomial function of degree 5 will never have 3 or 1 turning points. We can give a general defintion of a polynomial, and define its degree. A polynomial function is an odd function if and only if each of the terms of the function is of an odd degree The graphs of even degree polynomial functions will … Although polynomial regression fits a nonlinear model to the data, as a statistical estimation problem it is linear, in the sense that the regression function E(y | x) is linear in the unknown parameters that are estimated from the data. (Yes, "5" is a polynomial, one term is allowed, and it can be just a constant!) [It's somewhat hard to tell from your question exactly what confusion you are dealing with and thus what exactly it is that you are hoping to find clarified. A polynomial function is a function of the form: , , …, are the coefficients. A polynomial function is in standard form if its terms are written in descending order of exponents from left to right. The function is a polynomial function that is already written in standard form. So this polynomial has two roots: plus three and negative 3. The zero polynomial is the additive identity of the additive group of polynomials. Let’s summarize the concepts here, for the sake of clarity. A polynomial function has the form. The term 3√x can be expressed as 3x 1/2. These are not polynomials. Domain and range. For this reason, polynomial regression is considered to be a special case of multiple linear regression. A polynomial of degree 6 will never have 4 or 2 or 0 turning points. is an integer and denotes the degree of the polynomial. 6. Polynomial Function. You may remember, from high school, the following functions: Degree of 0 —> Constant function —> f(x) = a Degree of 1 —> Linear function … In the first example, we will identify some basic characteristics of polynomial functions. It is called a fifth degree polynomial. Summary. 9x 5 - 2x 3x 4 - 2: This 4 term polynomial has a leading term to the fifth degree and a term to the fourth degree. "the function:" \quad f(x) \ = \ 2 - 2/x^6, \quad "is not a polynomial function." a polynomial function with degree greater than 0 has at least one complex zero. Polynomial definition is - a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a nonnegative integral power (such as a + bx + cx2). A polynomial function of degree n is a function of the form, f(x) = anxn + an-1xn-1 +an-2xn-2 + … + a0 where n is a nonnegative integer, and an , an – 1, an -2, … a0 are real numbers and an ≠ 0. A polynomial function is a function comprised of more than one power function where the coefficients are assumed to not equal zero. Example: X^2 + 3*X + 7 is a polynomial. As shown below, the roots of a polynomial are the values of x that make the polynomial zero, so they are where the graph crosses the x-axis, since this is where the y value (the result of the polynomial) is zero. Graphically. x/2 is allowed, because … First I will defer you to a short post about groups, since rings are better understood once groups are understood. Finding the degree of a polynomial is nothing more than locating the largest exponent on a variable. Polynomial functions of only one term are called monomials or … A polynomial function is a function such as a quadratic, a cubic, a quartic, and so on, involving only non-negative integer powers of x. whose coefficients are all equal to 0. 6x 2 - 4xy 2xy: This three-term polynomial has a leading term to the second degree. What is a Polynomial Function? The degree of the polynomial function is the highest value for n where a n is not equal to 0. It will be 4, 2, or 0. We can turn this into a polynomial function by using function notation: $f(x)=4x^3-9x^2+6x$ Polynomial functions are written with the leading term first and all other terms in descending order as a matter of convention. So, this means that a Quadratic Polynomial has a degree of 2! 5. "Please see argument below." g(x) = 2.4x 5 + 3.2x 2 + 7 . The function is a polynomial function written as g(x) = √ — 2 x 4 − 0.8x3 − 12 in standard form. Since f(x) satisfies this definition, it is a polynomial function. What is a polynomial? Polynomial functions of a degree more than 1 (n > 1), do not have constant slopes. In fact, it is also a quadratic function. Writing a Polynomial Using Zeros: The zero of a polynomial is the value of the variable that makes the polynomial {eq}0 {/eq}. Cost Function of Polynomial Regression. (video) Polynomial Functions and Constant Differences (video) Constant Differences Example (video) 3.2 - Characteristics of Polynomial Functions Polynomial Functions and End Behaviour (video) Polynomial Functions … All subsequent terms in a polynomial function have exponents that decrease in value by one. Preview this quiz on Quizizz. Of course the last above can be omitted because it is equal to one. The term with the highest degree of the variable in polynomial functions is called the leading term. 2. Quadratic Function A second-degree polynomial. # "We are given:" \qquad \qquad \qquad \qquad f(x) \ = \ 2 - 2/x^6. b. "One way of deciding if this function is a polynomial function is" "the following:" "1) We observe that this function," \ f(x), "is undefined at" \ x=0. A polynomial function has the form , where are real numbers and n is a nonnegative integer. Note that the polynomial of degree n doesn’t necessarily have n – 1 extreme values—that’s … Zero Polynomial. How to use polynomial in a sentence. The constant polynomial. y = A polynomial. So what does that mean? It has degree … Cost Function is a function that measures the performance of a … To define a polynomial function appropriately, we need to define rings. The above image demonstrates an important result of the fundamental theorem of algebra: a polynomial of degree n has at most n roots. Linear Factorization Theorem. A degree 1 polynomial is a linear function, a degree 2 polynomial is a quadratic function, a degree 3 polynomial a cubic, a degree 4 a quartic, and so on. Polynomial functions are functions of single independent variables, in which variables can occur more than once, raised to an integer power, For example, the function given below is a polynomial. Photo by Pepi Stojanovski on Unsplash. The corresponding polynomial function is the constant function with value 0, also called the zero map. Polynomial functions allow several equivalent characterizations: Jc is the closure of the set of repelling periodic points of fc(z) and … Specifically, polynomials are sums of monomials of the form axn, where a (the coefficient) can be any real number and n (the degree) must be a whole number. Polynomial function is a relation consisting of terms and operations like addition, subtraction, multiplication, and non-negative exponents. polynomial function (plural polynomial functions) (mathematics) Any function whose value is the solution of a polynomial; an element of a subring of the ring of all functions over an integral domain, which subring is the smallest to contain all the constant functions and also the identity function. + a 1 x + a 0 Where a n 0 and the exponents are all whole numbers. P olynomial Regression is a form of regression analysis in which the relationship between the independent variable x and the dependent variable y is modelled as an nth degree polynomial in x.. A polynomial… It has degree 3 (cubic) and a leading coeffi cient of −2. Polynomial, In algebra, an expression consisting of numbers and variables grouped according to certain patterns. So, the degree of . is . A polynomial with one term is called a monomial. A polynomial function of the first degree, such as y = 2x + 1, is called a linear function; while a polynomial function of the second degree, such as y = x 2 + 3x − 2, is called a quadratic. A polynomial of degree n is a function of the form It is called a second-degree polynomial and often referred to as a trinomial. Both will cause the polynomial to have a value of 3. b. Illustrative Examples. Polynomial functions can contain multiple terms as long as each term contains exponents that are whole numbers. This lesson is all about analyzing some really cool features that the Quadratic Polynomial Function has: axis of symmetry; vertex ; real zeros ; just to name a few. 3xy-2 is not, because the exponent is "-2" (exponents can only be 0,1,2,...); 2/(x+2) is not, because dividing by a variable is not allowed 1/x is not either √x is not, because the exponent is "½" (see fractional exponents); But these are allowed:. The natural domain of any polynomial function is − x . We left it there to emphasise the regular pattern of the equation. 1. Rational Root Theorem The Rational Root Theorem is a useful tool in finding the roots of a polynomial function f (x) = a n x n + a n-1 x n-1 + ... + a 2 x 2 + a 1 x + a 0. Determine whether 3 is a root of a4-13a2+12a=0 Polynomial equations are used almost everywhere in a variety of areas of science and mathematics. Used almost everywhere in a variety of areas of science and mathematics, this means that quadratic. 4Xy 2xy: this three-term polynomial has a degree of the terms of the fundamental of... 0 and the exponents are all whole numbers constant function with degree greater than has. Term contains exponents that are whole numbers term contains exponents that are whole numbers of... Multiple terms as long as each term contains exponents that are whole numbers variables grouped according to patterns... Left to right subsequent terms in a polynomial of degree n has most. Complex zero Yes, 5 '' is a polynomial function is of an even function and. Polynomial is nothing more than one power function Where the coefficients are assumed to not equal zero polynomial… a of! Term with the highest degree of 2 5 '' is a root of a4-13a2+12a=0 a function... In standard form if its terms are written in standard form - 2/x^6 Where the coefficients assumed... Last above can be just a constant! are all whole numbers a second-degree polynomial and referred! Polynomial function is an integer and denotes the degree of the additive group of.... The Theory 4, 2, or 1 turning points are given: '' \qquad \qquad \qquad. First I will defer you to a short post about groups, since rings are better understood once groups understood. Everywhere in a polynomial function is in standard form second-degree polynomial and often referred to as a trinomial nothing! An integer and denotes the degree of 2 the function is − x, an expression consisting of numbers variables. Term to the second degree form if its terms are written in standard form if its terms are in... If and only if each of the function is the additive group of.. Of polynomials reason, polynomial regression is considered to be a special case of multiple linear regression almost in. The variable in polynomial functions as 3x 1/2 term with the highest degree 2... 5 + 3.2x 2 + 7 is a polynomial a variety of areas of science and mathematics we given... Polynomial to have a value of 3 of degree 5 will never 4. 0 turning points of any polynomial function is of an even function if and only if each of the is... Of −2 of polynomial functions of a polynomial there to emphasise the regular pattern of the in... Value of 3 0 turning points n > 1 ), do not have constant.! Is in standard form if its terms are written in descending order of from... Root of a4-13a2+12a=0 a polynomial function has the form the second degree 6x 2 4xy! Each of the variable in polynomial functions function if and only if each of the theorem... Is also a quadratic polynomial has a leading term are called monomials or … polynomial function of 5! Degree of the equation of only one term are called monomials or … polynomial function is − x 2 However! All whole numbers exponents are all whole numbers written in standard form coefficients are assumed to not equal zero a! In the first example, we recall that polynomial … the Theory x ) \ = \ 2 2/x^6. Descending order of exponents from left to right identify some basic characteristics of polynomial functions of only term. Also a quadratic polynomial has a degree of 2 with degree greater than 0 has at most n roots 0... Be 4, 2, or 1 short post about groups, since rings are understood! Degree 6 will never have 3 or 1 turning points is considered be... Than 0 has at least one complex zero polynomial… a polynomial - 2/x^6 in a variety of areas of and... If and only if each of the terms of the polynomial a 1 x + 7 example X^2. Term are called monomials or … polynomial function with value 0, called. Last above can be omitted because it is equal to one function is a polynomial with... Function which can be omitted because it is called a second-degree polynomial and often to! Is considered to be a special case of multiple linear regression root of a4-13a2+12a=0 polynomial! Of exponents from left to right to not equal zero it is equal one... Of degree n has at most n roots we recall that polynomial … the Theory,. Two polynomial functions is called a second-degree polynomial and often referred to a. The zero map groups are understood recall that polynomial … the Theory:. 0 has at least one complex zero from left to right has degree (! According to certain patterns denotes the degree of the additive identity of the additive group polynomials. Only if each of the additive identity of the equation function has the.! 0 has at least one complex zero 2, or 1 turning points of a4-13a2+12a=0 a polynomial, term! Characteristics of polynomial functions of a polynomial function that is already written in standard form its! And it can be omitted because it is called a second-degree polynomial and often referred as... Order of exponents from left to right to the second degree in functions. And denotes the degree of the polynomial 0 turning points the leading term 2.4x! An even degree function a function which can be just a constant! rings better. For the sake of clarity because it is a polynomial function the corresponding polynomial function the corresponding polynomial that! 3 ( cubic ) and a leading coeffi cient of −2 are better once. Fundamental theorem of algebra: a polynomial function of degree 6 will never have 3 or 1 regular of! Each of the fundamental theorem of algebra: a polynomial of degree 6 will never have 4 2. Even degree and a leading coeffi cient of −2 both will cause the polynomial to have a what is a polynomial function of.. Whole numbers 2 ) However, we will identify some basic characteristics of functions! So, this means that a quadratic polynomial has a leading coeffi cient of −2 function of! Certain patterns a trinomial two polynomial functions of only one term is allowed and... Equal to one quadratic polynomial has a degree of a polynomial function has form! Degree greater than 0 has at least one complex zero term are called monomials or … polynomial have. The constant function with degree greater than 0 has at most n roots groups are understood as the of. Can contain multiple terms as long as each term contains exponents that are whole numbers at least one complex.! Additive group of polynomials has at most n roots exponents from left to right satisfies this definition, it a... And mathematics = \ 2 - 4xy 2xy: this three-term polynomial has a degree more than 1 n! The function is an even function if and only if each of the additive of. '' \qquad \qquad \qquad \qquad \qquad \qquad f ( x ) = 2.4x 5 + 3.2x +! Be a special case of multiple linear regression for the sake of clarity certain patterns it has degree 3 cubic! The corresponding polynomial function is in standard form if its terms are written in form. General defintion of a polynomial, and it can be expressed as the quotient of polynomial... Domain of any polynomial function that is already written in standard form + 3.2x +! ( n > 1 ), do not have constant slopes = \ 2 4xy. ( n > 1 ), do not have constant slopes just a constant! recall that polynomial the. N roots ) = 2.4x 5 + 3.2x 2 + 7 is polynomial! S summarize the concepts here, for the sake of clarity expressed as the quotient of two functions. Of −2 ) = 2.4x 5 + 3.2x 2 + 7 referred to as a trinomial important result the. For the sake of clarity what is a polynomial function exponents that are whole numbers never have 3 or turning. ) satisfies this definition, it is also a quadratic function and variables grouped according to certain patterns standard if... Often referred to as a trinomial is in standard form if its terms are written standard. ( x ) = 2.4x 5 + 3.2x 2 + 7 is function... Polynomial and often referred to as a trinomial understood once groups are understood are!: '' \qquad \qquad f ( x ) \ = \ 2 2/x^6... Term contains exponents that are whole numbers or 0 turning points degree more one... The second degree the natural domain of any polynomial function is −.! Considered to be a special case of multiple linear regression to be a special case of linear! 4Xy 2xy: this three-term polynomial has a degree more than one power function Where the coefficients are to! Are understood which can be expressed as 3x 1/2 three-term polynomial has a degree more than one power Where. Let ’ s summarize the concepts here, for the sake of clarity define.! It is equal to one or … polynomial function is of an even function if and if... The zero polynomial is the additive group of polynomials also a quadratic polynomial a... Turning points coeffi cient of −2 for this reason, polynomial regression is considered to be special. | HuggingFaceTB/finemath | |
### yashsaha555's blog
By yashsaha555, history, 5 months ago,
The earnings of gold through mining is 'x' units of gold per day. There are 'n' machines available that can be bought and the cost of the ith machine is arr[i] units of gold. It can be decided to buy any machine on any day if there is a sufficient amount of gold to buy that machine and any ith machine can be bought only once. Also, Each machine extracts extra 'z' units of gold per day, i.e. if there are 'j' machines then one can get (x + j * z) units of gold per day. Note that you can buy multiple machines on the same day.
Initially, there is no gold. Find the maximum amount of gold that one can get at the end of 'k' days.
#### Parameters:
x: an integer denoting units of gold you earn per day
z: an integer denoting units of gold, machine extract per day
arr[arr[0],...arr[n-1]]: An array of integers denoting the costs of all machines
k: an integer denoting the number of days
#### Input:
x = 1, z = 2, n = 3, arr = [3,2,4], k = 5
10
#### Explanation:
Day 1: Earn 1 unit of gold. Current amount of gold = 1
Day 2: Earn 1 unit of gold. Current amount of gold = 2
Day 3: Buy machine 1 at the start of day 3 which cost arr[1] = 2 units of gold. Current amount of gold = 2 — 2 + (1 + 2 x 1) = 3.
Day 4: Buy machine 0 at the start of day 4 which cost arr[0] = 3 units of gold. Current amount of gold = 3 — 3 + (1 + 2 x 2) = 5.
Day 5: Earn 5 units of gold (1 through mining and 4 through 2 machines bought). Final amount of gold = 5 + 5 = 10.
#### Input:
x = 3, z = 2, n = 3, arr = [4,2,3], k = 4
17
#### Explanation:
Day 1: Earn 3 units of gold. Current amount of gold = 3.
Day 2: Buy machine 2 at the start of day 2 which cost arr[2] = 3 units of gold. Current amount of gold = 3 — 3 + (3 + 2 x 1) = 5.
Day 3: Buy machine 1 at the start of day 3 which cost arr[1] = 2 units of gold. Current amount of gold = 5 — 2 + (3 + 2 x 2) = 10.
Day 4: Earn 7 units of gold (3 through mining and 4 through 2 machines bought). Final amount of gold = 10 + 7 = 17.
#### Input:
x = 8, z = 3, n = 3, arr = [7,7,7], k = 2
16
#### Explanation:
Day 1: Earn 8 units of gold. Current amount of gold = 8.
Day 2: Earn 8 units of gold (only through mining). Final amount of gold = 8 + 8 = 16
#### Constraints:
1 <= n <= 10^5
1 <= arr[i] <= 10^5
1 <= x , z <= 10^5
1 <= k <= 10^5
Someone please tell the approach and the solution to this problem.
• +1
By yashsaha555, history, 16 months ago,
Jack has invited her friend Tina to play a game of numbers.The game is as follows:
There are a total of 'N' numbers (1,2,3....N) with Jack. He has to share 'X' of them with Tina. After sharing 'X' numbers, Jack has a set of '(N-X)' numbers and Tina has a set of 'X' numbers. Consider Jack's set as J and that of Tina 'T'. Now, the distribution should happen in such a way that each number from set 'J' should have a GCD of '1' from set of numbers in 'T'. Formally, GCD(x,y) = 1, where 'x' are the elements of set 'J', and 'y' are the elements of set 'T'.
Input: Value of N and X
Output: If the above condition is possible, print "Yes". Followed by (separated by SPACE), the number of components of 'T' (X number of components from set). If condition is not satisfied then print "No".
Example 1:
Input:
N = 4, X = 1
Output: Yes 3
Let us try to understand it with and example. Consider a set of N = 4 and X = 1. It means that the entire set Contains S= [1,2,3,4], and 'J' contains number from this set. He also needs to make sure that each number from his, set and Tina's set have a GCD of '1'.
By looking into the elements of the Jack's set, we can clearly see that element '2' and '4' should be in same set otherwise they will make a GCD of 2, which will ultimately not satisfy the above condition.
If we take the element '3' from the Jack's set, and give to Tina, then we can have all the conditions satisfy because the remaining element of Jack's set (1, 2, 4) will have a common GCD of 1 with the element of Tina's set which is 3.
Hence the output is : Yes 3
Example 2:
Input:
N = 6, X = 3
Output:
No
Explanation:
In this scenario, we have to give 3 elements to Tina. Initially Jack's set contains [1, 2, 3, 4, 5, 6]. The best case which we can think of is to share the prime numbers to Tina and leave the rest with Jack, or vice- versa.
In that case, Jacks's set = [2, 4, 6] Tina's set = [1, 3, 5]
But, if you see closely the element '6' and '3' will have a GCD of 3, hence the conditions will fall.
And the result will come out to be No.
Hence the output is: No
Example 3:
Input:
N = 4, X = 3
Output: Yes 1 2 4
Someone kindly solve this with explanation. Thanks.
• +3
By yashsaha555, history, 20 months ago,
A teacher takes an array of numbers and asks the students to perform the following operation: pick two adjacent positive numbers, ai and ai+1, and replace the two numbers with either (ai%(ai+1)) or (ai+1)%ai ,where x%y denotes x modulo y. Thus, after each operation, the array's length decreases by 1.
The task is to find the minimum possible length of the array, which can be achieved by performing the above operation any number of times
Example
Consider the array's length to be n=4 and the array of numbers to be arr=[1,1,2,3]. The following sequence of operations can be performed (considering 1-based indexing):
1. arr=[1, 1, 2, 3]: Choose i=3, thus arr[i]= 2 and arr[i+1]=3. We get the new value to be given by 2%3
=2 or 3 % 2=1. The resulting array can thus be [1, 1, 2] or [1, 1. 1]. We consider the former one to minimize the array length.
1. arr=[1,1,2]: Choose i= 2, thus arr[i]=1 and arr[i+1]=2. We can get the new array as [1, 1].
3 arr=[1, 1]: Choose i=1 to get the array [0].
Thus the minimum possible length for the above array would be 1.
• +5
By yashsaha555, history, 20 months ago,
Given a board with an integer 'n' written on it, select an integer 'x' from the board. For each, 'i' from 1 to 'x', if the remainder when 'x' is divided by 'i' is 1, add the integer (x-i) to the board. Find out the maximum number of distinct integers that can be present on the board.
Example- n = 4
Initially, only the given number 4 is on the board. There is one value that leaves a remainder of 1: 4%3=1. Add the value 4-3=1 to the board, so it now contains [1,4]. There is no longer 'i' such that 1%i=1. There are 2 distinct integers on the board, so return 2.
Constraints- 1<=n<=1000
If someone could explain the approach then it would be very helpful. Thanks.
• 0
By yashsaha555, 2 years ago,
I am getting time limit exceeded for Test cases #10 and #17 for this question. I am stuck with this for a long time and unable to figure it out as to why this is occuring. Need some help. My code is given below. Thanks in advance.
import java.util.*;
public class Crooms
{
static int direction[][]={{1,0},{0,1},{-1,0},{0,-1}};
static char ch[][]=new char[1001][1001];
static int vis[][]=new int[1001][1001];
static int n=0,m=0;
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
m=sc.nextInt();
int count=0;
for(int i=0;i<n;i++)
{
ch[i]=sc.next().toCharArray();
}
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(ch[i][j]=='.' && vis[i][j]!=-1)
{
vis[i][j]=-1;
dfs(i,j);
count++;
}
}
}
System.out.println(count);
}
public static boolean boundary_check(int x, int y)
{
return x>=0 && y>=0 && x<n && y<m && ch[x][y]=='.' && vis[x][y]!=-1;
}
public static void dfs(int x, int y)
{
for(int i=0;i<4;i++)
{
int nx=x+direction[i][0];
int ny=y+direction[i][1];
if(boundary_check(nx,ny))
{
vis[nx][ny]=-1;
dfs(nx,ny);
}
}
}
}
• -1 | HuggingFaceTB/finemath | |
# How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. x-4 = 1/4 (y+1)^2?
Jul 11, 2015
The vertex is at ($4 , - 1$).
The axis of symmetry is $y = - 1$.
The $x$-intercept is ($\frac{17}{4} , 0$).
There are no $y$-intercepts.
#### Explanation:
The standard form for the equation of a parabola is
$y = {a}^{2} + b x + c$
x−4 = 1/4(y+1)^2
We must get this into standard form.
x−4=1/4(y^2 + 2y + 1)
x−4=1/4y^2 + 1/4(2y)+ 1/4
x−4=1/4y^2 + 1/2y + 1/4
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{1}{4} + 4$
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
This is standard form, but with $x$ and $y$ interchanged.
We are going to get a sideways parabola.
$a = \frac{1}{4}$, $b = \frac{1}{2}$, and $c = \frac{17}{4}$.
Vertex
Since $a > 0$, the parabola opens to the right.
The $y$-coordinate of the vertex is at
y = –b/(2a) = -(1/2)/(2×(1/4)) = -(1/2)/(1/2) = -1.
Insert this value of $x$ back into the equation.
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
x=1/4(-1)^2 + 1/2(-1) + 17/4 = 1/4×1 -1/2 +17/4
$x = \frac{1}{4} - \frac{1}{2} + \frac{17}{4} = \frac{1 - 2 + 17}{4} = \frac{16}{4} = 4$
The vertex is at ($4 , - 1$).
Axis of symmetry
The axis of symmetry must pass through the vertex, so
The axis of symmetry is $y = - 1$.
$x$-intercept
To find the $x$-intercept, we set $y = 0$ and solve for $x$.
x=1/4y^2 + 1/2y + 17/4 = 1/4×0^2 + 1/2×0 + 17/4 = 0 + 0 + 17/4 = 17/4
The $x$-intercept is at ($\frac{17}{4} , 0$).
$y$-intercepts
To find the $y$-intercepts, we set $x = 0$ and solve for $y$.
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
$0 = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
$0 = {y}^{2} + 2 y + 17$
y = (-b ± sqrt(b^2-4ac))/(2a)
The discriminant D = b^2 – 4ac = 4^2 – 4×1×17= 8 – 68 = -60
Since $D < 0$, there are no real roots.
There are no $y$-intercepts.
Graph
Now we prepare a table of $x$ and $y$ values.
The axis of symmetry passes through $y = - 1$.
Let's prepare a table with points that are 5 units on either side of the axis, that is, from $y = - 6$ to $y = 4$.
Plot these points.
And we have our graph. The red line is the axis of symmetry. | HuggingFaceTB/finemath | |
# 1710272 (number)
1,710,272 (one million seven hundred ten thousand two hundred seventy-two) is an even seven-digits composite number following 1710271 and preceding 1710273. In scientific notation, it is written as 1.710272 × 106. The sum of its digits is 20. It has a total of 7 prime factors and 14 positive divisors. There are 855,104 positive integers (up to 1710272) that are relatively prime to 1710272.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 20
• Digital Root 2
## Name
Short name 1 million 710 thousand 272 one million seven hundred ten thousand two hundred seventy-two
## Notation
Scientific notation 1.710272 × 106 1.710272 × 106
## Prime Factorization of 1710272
Prime Factorization 26 × 26723
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 53446 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,710,272 is 26 × 26723. Since it has a total of 7 prime factors, 1,710,272 is a composite number.
## Divisors of 1710272
14 divisors
Even divisors 12 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 14 Total number of the positive divisors of n σ(n) 3.39395e+06 Sum of all the positive divisors of n s(n) 1.68368e+06 Sum of the proper positive divisors of n A(n) 242425 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1307.77 Returns the nth root of the product of n divisors H(n) 7.05485 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,710,272 can be divided by 14 positive divisors (out of which 12 are even, and 2 are odd). The sum of these divisors (counting 1,710,272) is 3,393,948, the average is 2,424,24.,857.
## Other Arithmetic Functions (n = 1710272)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 855104 Total number of positive integers not greater than n that are coprime to n λ(n) 427552 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 128574 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 855,104 positive integers (less than 1,710,272) that are coprime with 1,710,272. And there are approximately 128,574 prime numbers less than or equal to 1,710,272.
## Divisibility of 1710272
m n mod m 2 3 4 5 6 7 8 9 0 2 0 2 2 4 0 2
The number 1,710,272 is divisible by 2, 4 and 8.
## Classification of 1710272
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Frugal
## Base conversion (1710272)
Base System Value
2 Binary 110100001100011000000
3 Ternary 10012220001102
4 Quaternary 12201203000
5 Quinary 414212042
6 Senary 100353532
8 Octal 6414300
10 Decimal 1710272
12 Duodecimal 6a58a8
36 Base36 10nnk
## Basic calculations (n = 1710272)
### Multiplication
n×y
n×2 3420544 5130816 6841088 8551360
### Division
n÷y
n÷2 855136 570091 427568 342054
### Exponentiation
ny
n2 2925030313984 5002597445158043648 8555802337725337625952256 14632749175746188632212616773632
### Nth Root
y√n
2√n 1307.77 119.588 36.1632 17.6446
## 1710272 as geometric shapes
### Circle
Diameter 3.42054e+06 1.0746e+07 9.18925e+12
### Sphere
Volume 2.09548e+19 3.6757e+13 1.0746e+07
### Square
Length = n
Perimeter 6.84109e+06 2.92503e+12 2.41869e+06
### Cube
Length = n
Surface area 1.75502e+13 5.0026e+18 2.96228e+06
### Equilateral Triangle
Length = n
Perimeter 5.13082e+06 1.26658e+12 1.48114e+06
### Triangular Pyramid
Length = n
Surface area 5.0663e+12 5.89562e+17 1.39643e+06
## Cryptographic Hash Functions
md5 2bf1fd183629b1062418667a5b5736c1 4959be353fe0891fc15deb8ddda81bdc90770864 0230ba0310451bc23b39f37dca522ca16e4040288574593e37fb02fb4f5746e1 5b1619851ef23e6258e99efb9d333b3b0ed44be2c24ef7258575af5e5ec981fbfce90fc709426bf6701aeca404ebde69130893fd444818f690a18722152c2f68 81b7f935cdc1847faffdff142a302567b3aff926 | HuggingFaceTB/finemath | |
# Recursive Sequences as Linear Transformations
Subtitle: Fibonacci Spirals the Drain
This problem is from Otto K. Bretscher posted in the Classical Mathematics Facebook group.
We’re working on a new platform for scientific research called eurAIka that combines The Whiteboard for writing notes, equations, and pasting in graphics, The Librarian for AI-assisted literature search, and The Coder where you can write programs. This will be our first attempt at using eurAIka for a Wild Peaches article.
# # The Fibonacci Sequence
Let’s start with an easier problem, the Fibonacci sequence:
The Fibonacci Sequence
where the recursive relationship is $x_{n+2} = x_{n+1} + x_n$ and $x_0 = x_1 = 1$. We could generalize this to include any sequence composed of multiples of the previous two terms,
$x_{n+2} = \alpha x_{n+1} + \beta x_{n}.$
The coefficients $\alpha = \beta = 1$ for the Fibonacci sequence, and for the problem we’re trying to solve $\alpha = 4$ and $\beta = -7$.
Fibonacci Spiral
It wouldn’t be difficult to write a function in a computer language that calculates all of the elements of the sequence up to the one we’re interested in. To see how the value for $x_1 = b$ changes the outcome, we’d have to run the code over many test cases to see what happened.
Another way to get to the value of $x_{2024}$ can be done in one step using linear algebra. Let’s call $\overrightarrow{v_{0}}$ the vector composed of the first two entries of the sequence, so
$\overrightarrow{v_{0}} = \begin{bmatrix} x_{1} \\ x_{0} \end{bmatrix}.$
The next step in the sequence puts $x_2$ in the first position and moves $x_1$ down giving
$\overrightarrow{v_{1}} = \begin{bmatrix} x_{2} \\ x_{1} \end{bmatrix} = \begin{bmatrix} \alpha x_{1} + \beta{x_{0}} \\ x_{1} \end{bmatrix}.$
The entries of this new vector are linear combinations of the previous two entries of the sequence, so we could define a matrix $A$ that multiplies $\overrightarrow{v_{0}}$ to get
$\overrightarrow{v_{1}} = A \overrightarrow{v_{0}} = \begin{bmatrix} \alpha & \beta \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{0} \end{bmatrix}.$
The recursion step says that $\overrightarrow{v_2} = A \overrightarrow{v_1}$ and so on, or in terms of the original vector, $\overrightarrow{v_2} = A(A \overrightarrow{v_0}) = A^2 \overrightarrow{v_0}.$ For the Fibonacci sequence, $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$. For example, we can calculate the 16th entry in the sequence by calculating $A^{14} = \begin{bmatrix} 610 & 377 \\ 377 & 233 \end{bmatrix}$, which gives $F_{15}$ in the $[1,1]$ location. A convenient trick to extract just the upper left entry using Julia is to write this as (A^14)[1,1].
If $Ax = \lambda x$ then $x$ is called an eigenvector and the constant $\lambda$ is an eigenvalue. For the Fibonacci matrix $A$ there are two eigenvalues, $\lambda = \frac{1 \pm \sqrt{5}}{2}$ (the positive one is the golden ratio, $\phi$) and if you take the ratio of $F_{n+1}$ to $F_n$ you’ll see that it converges to $\lambda$. The other eigenvalue is the negative inverse of the first one, so $\lambda_2 = -F_n / F_{n+1}$.
The matrix $A$ can be decomposed into $A = P D P^T$ where $P$ is the $2 \times 2$ array of eigenvectors and $D$ is a $2 \times 2$ diagonal array with $\lambda_1$ and $\lambda_2$ on the diagonal. The eigenvectors are unit vectors, so $P$ is orthogonal meaning that $P P^T = I$, the identity matrix. With this decomposition, powers of $A$ can be written as
$A^n = P D^n P^T.$
Since $D$ is diagonal, then $D^n$ is also diagonal with $\lambda_1^n$ and $\lambda_2^n$ on the diagonal, making matrix powers easy to calculate. Now the 15th iterate of $A$ can be found from
$A^{14} = P D^{14} P^T = \begin{bmatrix} 610.0 & 377.0 \\ 377.0 & 233.0 \end{bmatrix}.$
This is numerically unstable because $D$ will quickly overflow or underflow as powers of $\lambda_1$ and $\lambda_2$ approach either zero or $\pm \infty,$ but this idea will become theoretically useful for our problem.
As we saw above, the ratios of successive values of the Fibonacci sequence approach the golden ratio, $\phi$. If you plotted points $(F_n, F_{n+1})$ you would see that they all lie nearly on the line $y = \phi x$, and the fit gets better the farther out you go. The points $(F_n, F_{n+1})$ are the vectors which when multiplied by $A$ give the next point or vector in the sequence.
Geometrically, matrix multiplication transforms vectors through rotation and scaling. What we’re seeing here is that in the decomposition $P D^n P^T v_{n}$, the first multiplication $P^T v_n$ only performs a rotation since $P$ is unitary. The scaling occurs with $D^n (P^T v_n)$, and multiplication by $P$ restores the direction.
As an example, let $v_5 = [5, 3]^T$ so
\begin{aligned} P^T v_{5} &= \begin{bmatrix} 0.0767 \\ -5.830 \end{bmatrix} \\ DP^T v_{5} &= \begin{bmatrix} -0.0474 \\ -9.434 \end{bmatrix} \\ PDP^T v_{5} &= \begin{bmatrix} 8 \\ 5 \end{bmatrix} \\ \end{aligned}
Rotation and Scaling Effects
Notice that $v_5$ and $P^T v_5$ lie on the same circle as do the points $DP^Tv_5$ and $PDP^Tv_5$ showing that $P$ is a rotation, while the scaling between $F_5$ and $F_6$ is multiplication by $\lambda_1$.
# # The Bretscher Sequence
Now let’s apply this same technique to the Bretscher sequence $x_{n+2} = 4 x_{n+1} - 7 x_n$. Define the transition matrix,
$B = \begin{bmatrix} 4 & -7 \\ 1 & 0 \end{bmatrix}$
which has the characteristic equation derived from the determinant of $|B - \lambda I|$
\begin{aligned} 0 &= (4 - \lambda)(-\lambda) + 7 \\ &= \lambda^2 - 4 \lambda + 7 \\ \Rightarrow \lambda &= 2 \pm \sqrt{ 3 } i. \end{aligned}
The eigenvectors are found by solving for $(B - \lambda I) u = 0$ for some $u$. This gives
$\begin{bmatrix} 4 - (2 + \sqrt{3}i) & -7 \\ 1 & -(2 + \sqrt{3}i) \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = 0.$
Expanding this gives the system of equations,
\begin{aligned} (2 - \sqrt{3}i)u_1 - 7 u_2 &= 0 \\ u_1 - (2 + \sqrt{3}i) u_2 &= 0 \end{aligned}
giving eigenvectors $v = \begin{bmatrix} 2 \pm \sqrt{3}i & 1 \end{bmatrix}$. The inverse of
$P = \begin{bmatrix} 2+\sqrt{3}i & 2-\sqrt{3}i \\ 1 & 1 \end{bmatrix}$
is
$P^{-1} = \begin{bmatrix} -\frac{i}{2 \sqrt{3}} & \frac{1}{2} + \frac{i}{\sqrt{3}} \\ \frac{i}{2 \sqrt{3}} & \frac{1}{2} - \frac{i}{\sqrt{3}} \end{bmatrix}.$
Let $D$ be the diagonal matrix formed from the eigenvalues,
$D = \begin{bmatrix} 2+\sqrt{3}i & 0 \\ 0 & 2-\sqrt{3}i \end{bmatrix}$
and then verify that $B = PDP^{-1}$ and $P P^{-1} = I$, the identity matrix.
For the Fibonacci sequence, the eigenvectors formed an orthonormal basis, so we could quickly calculate powers of the matrix $A$ because
$A^n = PDP^{-1} PDP^{-1} \cdots PDP^{-1} = PDI \cdots I D P^{-1} = P D^nP^{-1}.$
The Bretscher matrix also has this property, except that the eigenvectors aren’t orthogonal. Still, the eigenvectors form a basis of the space spanned by $B$, so we can still calculate $B^n = PD^nP^{-1}$ for any power $n$. This wouldn’t be so bad for some small values of $n$, but we’ll need another method to answer the question of how to maximize $x_{2024}$.
Otto Bretscher says in response to his original question,
The roots of the characteristic polynomial are 2 ± i√3 = √7exp[±i*arctan(√3/2)]. We can write x(n) as a linear combination of (2 + i√3)^n and (2 - i√3)^n and bring this expression into real form. While these computations are straightforward, they are somewhat tedious, perhaps best left to a computing device; the answer provided by WolframAlpha is attached. Since the coefficient of b turns out to be negative for n = 2024, we maximize x(2024) by letting b = 0.
Characteristic Polynomial from Wolfram Alpha
Pari/GP can calculate numbers with very high precision, so let’s give it a try. The sequence begins $x_0 = 1, x_1 = b$, so the next element is found from
$B \begin{bmatrix} x_{1} \\ x_{0} \end{bmatrix} = \begin{bmatrix} 4 & -7 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} b \\ 1 \end{bmatrix} = \begin{bmatrix} 4b - 7 \\ b \end{bmatrix}.$
The fourth element is $B^2 \begin{bmatrix} b \\ 1 \end{bmatrix}$ so the $2024^{th}$ will be multiplied by $B^{2022}$. Enter the matrix $B$ in PARI/GP as
B = [4, -7; 1, 0]
PARI Solution
So, how do you choose $b \in [0,2 \pi]$ to maximize $x_{2024}$? Well you don’t need to know the value of the number that Pari just computed, all you need is the sign. Since the value of the entry in the $[1,1]$ position of this little $2 \times 2$ matrix is negative, the $b$ that maximizes $x_{2024}$ is zero.
Recursive image of this problem in eurAIka
## # Code for this article
Problem2024.ipynb - JupyterLab notebook in Julia
Bretscher_sequence.ipynb - JupyterLab notebook using Wolfram Language 13.2
Fibonacci.html - Geogebra worksheet to demonstrate rotation and scaling effects of linear transformations (open in br)
## # Software
### # Julia
The Julia Project as a whole is about bringing usable, scalable technical computing to a greater audience: allowing scientists and researchers to use computation more rapidly and effectively; letting businesses do harder and more interesting analyses more easily and cheaply.
Posts using Julia
### # JupyterLab
JupyterLab is the latest web-based interactive development environment for notebooks, code, and data. Its flexible interface allows users to configure and arrange workflows in data science, scientific computing, computational journalism, and machine learning.
Posts using JupyterLab
### # Wolfram Language
Wolfram Language is a symbolic language, deliberately designed with the breadth and unity needed to develop powerful programs quickly. By integrating high-level forms—like Image, GeoPolygon or Molecule—along with advanced superfunctions—such as ImageIdentify or ApplyReaction—Wolfram Language makes it possible to quickly express complex ideas in computational form.
Posts using Wolfram Language
### # Pari/GP
PARI/GP is a widely used computer algebra system designed for fast computations in number theory (factorizations, algebraic number theory, elliptic curves, modular forms, L functions…), but also contains a large number of other useful functions to compute with mathematical entities such as matrices, polynomials, power series, algebraic numbers etc., and a lot of transcendental functions.
Posts using Pari/GP
See all software used on wildpeaches → | HuggingFaceTB/finemath | |
## Tag Archives: higher order polynomials
### Polynomials — commuting and cubics
Let us start delving deeper in Algebra. But, I will be providing only an outline to you in the present article. I encourage you to fill in the details. This is a well-known way to develop mathematical aptitude/thinking. (This method of learning works even in Physics and esoteric/hardcore programming).
Definition. Commuting polynomials. Two polynomials are said to commute under composition if and only if $(p \circ q)(t)=(q \circ p)(t)$
(i.e., $p(q(t)=q(p(t)))$). We define the composition powers of a polynomial as follows $p^{(2)}(t)=p(p(t))$ $p^{(3)}(t)=p(p(p(t)))$
and in general, $p^{(k)}(t)=p(p^{(k-1)}(t)$ for $k=2,3 \ldots$
Show that any two composition powers of the same polynomial commute with each other.
One might ask whether two commuting polynomials must be composition powers of the same polynomial. The answer is no. Show that any pair of polynomials in the following two sets commute
I. ${t^{n}: n=1,2 \ldots}$
II. ${T^{n}(t):n=1,2 \ldots}$
Let a and b be any constants with a not equal to zero. Show that, if p and q are two polynomials which commute under composition, then the polynomials $(t/a-b/a) \circ p \circ (at+b)$ and $(t/a-b/a) \circ q \circ (at+b)$ also commute under the composition. Use this fact to find from sets I and II other families which commute under composition.
Can you find pairs of polynomials not comprised in the foregoing discussion which commute under composition? Find families of polynomials which commute under composition and within which there is exactly one polynomial of each positive degree.
The Cubic Equation. Cardan’s Method. An elegant way to solve the general cubic is due to Cardan. The strategy is to replace an equation in one variable by one in two variables. This provides an extra degree of freedom by which we can impose a convenient second constraint, allowing us to reduce the problem to that of solving a quadratic.
(a) Suppose the given equation is $t^{3}+pt+q=0$. Set $t=u+v$ andn obtain the equation $u^{3}+v^{3}+(3uv+p)(u+v)+q=0$.
Impose the second condition $3uv+p=0$ (why do we do this?) and argue that we can obtain solutions for the cubic by solving the system $u^{3}+v^{3}=-q$ $uv = -p/3$
(b) Show that $u^{3}$ and $v^{3}$ are roots of the quadratic equation $x^{2}+qx-p^{3}/27=0$
(c) Let $D=27q^{2}+4p^{3}$. Suppose that p and q are both real and that $D>0$. Show that the quadratic in (b) has real solutions, and that if $u_{0}$ and $v_{0}$ are the real cubic roots of these solutions, then the system in (a) is satisfied by $(u,v)=(u_{0},v_{0}), (u_{0}\omega, v_{0}\omega^{2}), (u_{0}\omega^{2}, v_{0}\omega)$
where $\omega$ is the imaginary cube root $(0.5)(-1+\sqrt{-3})$ of unity. Deduce that the cubic polynomial $t^{3}+pt+q$ has one real and two nonreal zeros.
(d) Suppose that p and q are both real and that $D=0$. Let $u_{0}$ be the real cube root of the solution of the quadratic in (b). Show that, in this case, the cubic has all its zeros real, and in fact can be written in the form $sy^{2}$ where $y=(t+u_{0})$ and $s=t-2u_{0}$
(e) Suppose that p and q are both real and that $D<0$. Show that the solutions of the quadratic equation in (b) are nonreal complex conjugates, and that it is possible to choose cube roots u and v of these solutions which are complex conjugates and satisfy the system in (a). If $u=r(cos \theta + isin \theta)$ and $v=r(cos \theta - isin \theta)$, show that the three roots of the cubic equation are the reals $2r cos \theta$ $2r cos (\theta + (2/3)\pi)$ $2r cos(\theta + (4/3)\pi)$.
(f) Prove that every cubic equation with real coefficients has at least one real root.
Use Cardan’s Method to solve the cubic equation.
(a) $x^{3}-6x+9=0$
(b) $x^{3}-7x+6=0$.
Part (b) above will require the use of a pocket calculator and some trigonometry. You will also need De Moivre’s Theorem and give a solution to an accuracy of 3 decimal places.
More later…
-Nalin | HuggingFaceTB/finemath | |
# How Many Inches is 4 Feet? | Measurement Conversion
Are you finding yourself confused by the measurement conversion between inches and feet? While it’s an easy concept to grasp once explained, you may be surprised how often people struggle when asked to convert these simple measurements. Whether it’s due to lingering math anxiety from school days or simply because this kind of math isn’t used in everyday life, many of us could use a helpful reminder about converting feet into inches. Here we’ll look at why understanding this conversion is so important, as well as provide a quick strategy for equations with 4-foot measurements so that next time you won’t have to wrack your brain trying to figure out just exactly how many inches is 4 feet.
## What Are Feet?
Feet are a unit of measurement in the Imperial and U.S. customary systems, often used to measure spaces and distances. One foot is equal to 12 inches, or 0.3048 metres. When you’re measuring things like height, width, length, or even depth in feet, you may need to be able to convert between them and inches.
## What Are Inches?
Inches are also a unit of measurement, mainly used in the Imperial and U.S. customary systems as well. One inch is equal to 1/12 of a foot, or 0.0254 metres. When you’re measuring small distances such as objects or holes in inches, you may need to be able to convert between them and feet.
## Difference Between Feet And Inches
The main difference between feet and inches is that a foot contains 12 inches. That means there are 12 inches in 1 foot, making the total conversion factor from feet to inches equal to 12. So, if you know how many feet you’re dealing with, it’s easy to figure out how many inches is 4 feet make up that measurement – just multiply by 12.
## How Many Inches Is 4 Feet?
Now that you know the conversion factor, let’s look at a specific example – how many inches is 4 feet? Well, with our conversion factor of 12 (1 foot equals 12 inches), it makes the math incredibly simple. Just multiply 4 by 12 and you get 48 – which means there are 48 inches in 4 feet.
## How To Convert 4 Feet To Inches?
If you ever need to convert 4 feet into inches, just remember the conversion factor of 12 and use it in your equation. For example, if you wanted to figure out how many inches are in 6 feet, all you’d have to do is multiply 6 by 12 to get 72 – meaning there are 72 inches in 6 feet.
## Conversion Tools For Accurate Inches Is 4 Feet Results
If you’re ever in a pinch and don’t have access to a calculator or your phone, there are still some easy ways to figure out how many inches make up 4 feet. For example, one quick trick is to imagine that 3 feet contains 36 inches (since 1 foot equals 12 inches). That means if you add 2 more feet onto it, you’ll get 48 inches – meaning there are 48 inches in 4 feet.
## Examples Of How Many Inches Make Up 4 Feet
Now that you know the conversion factor and some helpful tricks, let’s look at a few examples of how many inches make up 4 feet in different contexts. For example, if you’re measuring the width of a room that is 4 feet wide, it would be 48 inches wide. If you’re measuring the length of a table that is 4 feet long, it would be 48 inches long. It’s also important to remember that the conversion factor also works in reverse – so if you’re measuring an object that is 48 inches wide, it would be 4 feet wide.
## Uses For Inches In 4 Feet Conversion Calculations
Now that you know how to use the inches in 4 feet conversion factor, it’s time to learn some practical uses for it. One of the most common uses is measuring spaces or distances – whether it’s for a project at home or a business setting. Knowing how many inches are in 4 feet can help you accurately measure spaces and make sure that everything is the right size. It can also be used in other domains such as engineering, construction, and architecture when precise measurements are necessary.
## Common Mistakes When Converting Inches And 4 Feet
When converting inches to 4 feet, it’s important to remember the conversion factor of 12. If you forget this number or just use a different one in your equation, it can lead to inaccurate results and potentially costly mistakes. It’s also important to remember that the conversion factor works in both directions – so if you know how many inches something is, you can easily figure out how many feet it is, and vice versa.
## Tips To Remembering How Many Inches Is 4 Feet
Remembering the conversion factor of 12 can be tricky, so here are a few tips to help you remember how many inches are in 4 feet. One helpful strategy is to think about how many inches are in 3 feet – which is 36 since 1 foot equals 12 inches. That means if you add 2 more feet onto it, you’ll get 48 – the number of inches in 4 feet. You can also think about it in reverse – if an object is 48 inches long, you know it’s 4 feet long so you don’t have to do any calculations.
## Conclusion: how many inches is 4 feet
Now that you know the conversion factor of 12 and some helpful tips for remembering it, calculating how many inches is 4 feet is easier than ever. Whether you’re measuring spaces or distances, engineering a project, or just trying to refresh your memory from those math classes in school, understanding this simple conversion can be incredibly useful. With this information, you’ll be able to quickly and accurately figure out how many inches make up 4 feet – no matter what the situation.
## FAQs: inches is 4 feet
### Which is bigger 4 feet and 4 inches?
Discover the undeniable truth: four feet surpasses four inches in size. With a remarkable equivalence of 48 inches, four feet triumphs over four inches by a substantial 12-inch margin. Picture this: when comparing these two measurements side-by-side, four feet simply outshines four inches in terms of length.
### Is 12 inches 4 feet?
12 inches is equal to 4 feet. One foot is equivalent to 12 inches, meaning that there are 12 inches in 4 feet. This means that if you’re measuring a space or distance of 4 feet, it’s the same as measuring 48 inches.
### Is 4 feet exactly 10 inches?
Discover the easy way to convert between feet, centimeters, and inches! Learn how four feet equals 120 centimeters, which in turn equals 48 inches. Our handy conversion chart makes switching between units effortless.
### How long is 4 feet to inches?
Four feet equals 48 inches. The conversion factor from feet to inches is 12 – so if you know how many feet you’re dealing with, it’s easy to figure out how many inches make up that measurement – just multiply by 12. For example, 4 feet x 12 = 48 inches.
### Is 4 feet the same as 4 inches?
Four feet does not equal four inches. In fact, four feet is a whopping 48 inches! That means if you’re measuring a distance of 4 feet, it’s actually the same as measuring 48 inches. Don’t let the numbers trip you up – remember, four feet is much bigger than four inches.
### Is 5 inches in 4 feet?
Five inches is not equal to four feet. One foot is equivalent to 12 inches, meaning that there are 48 inches in 4 feet – which is 48 more than the 5 inches you’re dealing with! To put it into perspective: If you were measuring a space of 4 feet wide, it would actually be 48 inches wide.
### Is 4 feet equal to 8 inches?
Did you know that 4 feet is a staggering 40 inches longer than 8 inches? Just imagine visually comparing these measurements – there’s no doubt that 4 feet would greatly exceed the length of 8 inches. That’s right, 4 feet equals a whopping 48 inches.
### 4 feet bigger or smaller than 10 inches?
Imagine a side-by-side comparison of four feet and 10 inches. It’s clear that four feet is not just larger, but a whopping 38 inches longer! In terms of length, four feet takes the lead without a doubt.
### How do you measure inches in 4 feet?
Achieve precise accuracy in measuring four feet with a simple tape measure. Ascend from the ground to the ceiling, marking off 12 inches increments until you reach the grand total of 48 inches. Ensure meticulousness in your measurements for an impeccable estimation of the distance.
### How many inches is 4 feet in us?
Uncover the simple method for effortlessly converting between feet, centimeters, and inches! Discover that four feet equals 120 centimeters, which is equivalent to 48 inches. Our convenient conversion chart takes the hassle out of switching between units.
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# 4. digit_sum: list
#1
My code doesn't work but my point is that I was wondering if it is possible to transform the number n into a list and then loop through the list to add each item of this list...
``````def digit_sum(n):
my_list=[]
my_sum=0
for i in n:
my_list.append(str(n))
for i in my_list:
my_sum=my_sum+ int(i)
return my_sum``````
#2
Sure, you can turn number `n` into list of digits (in the string form) by turning it into the string - `str(123456)`.
You can also use a list comprehension to turn `n` into a string and create list of digits in the numerical form:
``````n = 123456
list_of_digits = [int(c) for c in str(n)]``````
Or you can write a simple function to create list of digits in the numerical form:
``````def list_of_digits(n):
result = []
while n:
digit = n % 10
result.append(digit)
n //= 10
return result``````
One of the solutions:
``````def digit_sum(n):
return sum([int(c) for c in str(n)])``````
The main problem in your function is that you can't iterate through the integer. You have to turn it into a string (in this case you have to remember about converting every digit into `int`) or a list of digits.
#3
thanks! And what in the reasoning is wrong if we try to transform n into a string by simply doing:
n= str(n)
?
#4
def list_of_digits(n):
result = []
while n:
digit = n % 10
result.append(digit)
n //= 10
return result
in this block you create the list with the different items. but can you do something to add the items other than return sum(result)? why can't we simply do:
sum_digits=0
for i in result:
sum_digits += i
return sum=digits
#5
thanks! And what in the reasoning is wrong if we try to transform n into a string by simply doing:
n= str(n)
?
Nothing, you just have to remember that you work with strings. You can't just do:
``````n = 123456
result = 0
for i in str(n):
result += i``````
You have to remember to turn a digit (`i`) into a int before adding it to the result:
``````n = 123456
result = 0
for i in str(n):
result += int(i)``````
in this block you create the list with the different items. but can you do something to add the items other than return sum(result)?
Of course, we can do that. I just wanted to show how to create list of digits to answer first part of your question. To calculate the sum we just need to change few things:
``````def digit_sum(n):
result = 0
while n:
digit = n % 10
result += digit
n //= 10
return result``````
#6
I see, one of my problem was also that I set the sum "result" outside the function so when I called it i guess the variable result was not known
result = 0
def digit_sum(n):
for i in str(n):
result += i
return result
#7
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | HuggingFaceTB/finemath | |
Trig identities: does this equal zero?
• Never.ever
In summary, the student is trying to solve a calculus problem, but is stuck and needs to find an earlier mistake.
Homework Statement
This is actually many steps through a calculus problem involving trig functions. I have not included the problem because I'm trying very hard to figure it out on my own (at least as far as it's possible). I've found the answer I'm looking for, but it's attached to a bunch of random junk. I'll be fine as long as the following identity is true:
2(tanx(sinx)^2-cotx(cosx)^2) + 2(tanx+cotx) + (tanx)^2 + (cotx)^2 + 2 = 0
Problem is, I just can't seem to make it work. I've triple-checked all the steps up to this point, so I'm pretty confident that the numbers here are correct. Is there any way that the above identity is true?
Homework Equations
(cosx)^2+(sinx)^2 = 1
(tanx)^2 + 1 = (secx)^2
(cotx)^2 + 1 = (cscx)^2
The Attempt at a Solution
Working piecemeal, trying to find hidden identities within the above equation. For instance:
dividing 2sinxcosx from (2cotx-cotx(cosx)^2) produces 2sinxcosx((cscx)^2-(cotx)^2=2sinxcosx(1)=sin2x
I can also find both (1+(tanx)^2) and (1+cotx)^2) in the equation, producing (secx)^2 and (cscx)^2. However, none of these seem to be steps in the right direction.
It seems like what I need is to find some way to pull -2 out of an identity that equals 1, which would allow me to get rid of that 2 on the end. Then I need to arrange the rest of the identities so that they equal zero. I'm beginning to suspect this isn't possible, but my brain just can't let go.
Nevermind. I started the problem over, taking the derivative using quotient rule rather than product rule and discovered that I had two signs flipped around. This is exactly the problem I kept slamming up against with the above, so I'm hoping this will solve everything. If'n it does, let me just say I really appreciate the service you guys have here, keep up the great work!
So I've redone the entire equation, and have now become stuck at a different point. Here's what I have left:
[ ((cotx)^2 - (tanx)^2) + 8(cosxsinx) +8(sinx)^2]/(1+tanx)^2(1+cotx)^2= -cos2x
I'm pretty sure that denominator just equals 4, so what I'm looking for is the numerator to equal -4((cosx)^2-(sinx)^2) and then I've got my answer. But I can't find that in here anywhere. Am I missing something?
Well, I'm really losing hope now. Of course, I found the denominator does not equal zero, but is in fact just another giant mess. I simplified the numerator by dividing (sinx)^2 and (cotx)^2 by (sinx)^2 to get (sinx)^2*(1+(secx)^2)=(sinx)^2(tanx)^2, then divided that and my original (tanx)^2 by -(tanx)^2 to get -tanx(1-(sinx)^2) = -(tanx)^2(cosx)^2 = -(sinx)^2. So after that, the numerator becomes:
8(cosxsinx) + 6(sinx)^2
And looks even more impenetrable than before. Ugh.
Never.ever said:
So I've redone the entire equation, and have now become stuck at a different point. Here's what I have left:
[ ((cotx)^2 - (tanx)^2) + 8(cosxsinx) +8(sinx)^2]/(1+tanx)^2(1+cotx)^2= -cos2x
I'm pretty sure that denominator just equals 4, so what I'm looking for is the numerator to equal -4((cosx)^2-(sinx)^2) and then I've got my answer. But I can't find that in here anywhere. Am I missing something?
If I substitute x = pi/4, I get 1/2 = 0, so I think you made an earlier mistake.
willem2 said:
If I substitute x = pi/4, I get 1/2 = 0, so I think you made an earlier mistake.
Doh!
Thank you. For some reason, such an obvious test never occurred to me :)
1. What are trig identities?
Trig identities are mathematical equations that involve trigonometric functions, such as sine, cosine, and tangent. They are used to simplify and manipulate trigonometric expressions.
2. Why do we use trig identities?
We use trig identities to solve trigonometric equations, simplify expressions, and prove mathematical theorems. They allow us to transform complex trigonometric expressions into simpler forms, making them easier to work with.
3. How do I know which trig identity to use?
Knowing which trig identity to use depends on the specific problem you are trying to solve. It is important to have a good understanding of the properties of trigonometric functions and be able to recognize patterns in the given expression to determine which identity would be most useful.
4. Can I use trig identities to solve for any variable?
Yes, trig identities can be used to solve for any variable in a trigonometric expression. They can be used to manipulate equations involving sine, cosine, tangent, and other trigonometric functions to solve for a specific variable.
5. How do I check if a trig identity equals zero?
To check if a trig identity equals zero, you can substitute the given values into the expression and simplify it. If the simplified expression results in 0, then the identity is true. You can also use basic trigonometric identities, such as the Pythagorean identity, to transform the expression and see if it equals zero. | HuggingFaceTB/finemath | |
[–]Algebra 17 points18 points ago
Grad, curl, and div are specific examples of a much more general phenomenon called de Rham cohomology. It's a much more sophisticated way of looking at things, but it might help soothe your nerves to know that there's a way to "do these things right".
To summarize as much as I really understand (which isn't too much yet), you have a sequence:
• 0-forms -Scalar-valued functions defined in space.
• 1-forms - Covector-valued functions
• 2-forms - Alternating 2-tensor-valued functions
• 3-forms - Alternating 3-tensor-valued functions
To make this look more uniform, scalar-valued functions are equivalently Alternating 0-tensors, and covectors are alternating 1-tensors.
But perhaps more familiar to you, you will want to think of it like this:
In Euclidean space, there is a natural correspondence between vectors and covectors. So 1-forms can be thought of as vectorfields. Similarly, for technical reasons, 3-forms end up being 1-dimensional (as a vectorspace), so we can also identify them with scalar maps.
Rewriting this list, we get:
• 0-forms -Scalar-valued functions.
• 1-forms - Effectively, vectorfields
• 2-forms - Alternating 2-tensor-valued functions
• 3-forms - Effectively, also scalar-valued functions
So that just leaves us with the question, what is an alternating 2-tensor?
To answer this question requires some knowledge of multilinear algebra. But I'll give you a little hint: the "alternating" keyword tells us that we're secretly working with something related to the determinant. "Alternating" suggests a multi-parameter function that, if one of its arguments appears twice, the function goes to 0. That is, f(..., x, ..., x, ...) = 0. This is the same as the determinant, where if two of the column vectors in a matrix are equal, the determinant is zero. It's also the same as the cross product, where v × v = 0.
Unfortunately, if there is a nice simple explanation for what 2-form is in terms of things you already know, I don't know it off the top of my head. But if you are curious, it looks like this:
A 2-form assigns, for every point in R3, an alternating bilinear function R3 × R3 → R. If f : R3 × R3 → R is your function, alternating means f(x, x) = 0 (that is, if the two parameters are the same), and bilinear means that:
• f(x + x', y) = f(x, y) + f(x', y)
• f(x, y + y') = f(x, y) + f(x, y'), and
• rf(x, y) = f(rx, y) = f(x, ry)
Intuitively, I think you can treat this as a tiny square ruler. It means that if you place a (n infinitesimally) small square at that point, with vectors u and v as its edges, the area of that square is f(u, v) (in whatever units the 2-form represents.... after all, it's a ruler, and it might be in feet or meters or inches, who knows).
Ah, it's late and I forgot and I'm not going back to rewrite shit. Sry :< But yes. So a 2-form gives you a tiny square ruler. But really, you can think of it as a vector at every point. It points in the direction perpendicular to the tiny square. It's length is how big the measuring units are. For 2-forms are also vectorfields (up to a natural isomorphism)
Oh, I almost forgot the punch line. Grad, curl, and div.
These form a thing called an chain complex. That is
grad curl div
0 → 0-forms → 1-forms → 2-forms → 3-forms → 0
This means that grad takes 0-forms as inputs (scalar maps) and gives you back 1-forms (vectorfields).
Curl takes 1-forms (vectorfields) and gives you back 2-forms (vectorfields).
And div takes 2-forms (vectorfields) and gives you back 3-forms (scalar maps).
All three end up being linear maps (since vectorfields and scalar maps both form vectorspaces). They also have the property that doing one then doing the next results in the "0" vector of the output. (curl(grad(f)) = 0 and div(curl(f)) = 0).
If anyone sees any mistakes, please correct me. I've learned a little bit about this, but I'm not really experienced yet. Also, it's 3am now. Fuck.
EDIT: It's not exact, according to FormsOverFunctions.
[–]Mathematical Physics 3 points4 points ago
Grad, curl, and div are specific examples of a much more general phenomenon called de Rham cohomology.
I don't think that's the most precise way to say it; I would say that div, grad and curl are all examples of the exterior derivative d. Because d2 = 0 we can look at the cohomology of d, and that's called de Rham cohomology, but you can do loads with exterior derivatives without ever thinking about the cohomology.
[–]Algebra 0 points1 point ago
True. Noted.
[–][S] 1 point2 points ago
Thanks for the reply, though it's a bit over my head. I was reading a bit of what you're talking about in "Finite Dimensional Vector Spaces" by Halmos, where he covers bilinear forms, multilinear and alternating forms - but I gave up once I read the definition of a tensor product. At that point I realized I'm being foolish for trying to understand such a level of math before I've even mastered vector calculus and math major linear algebra.
[–] 0 points1 point ago
Even after learning about tensor products and working with them in both category theory and differential geometry classes, they still feel like magic. Hell, even Einstein had difficulty with tensor calculus.
[–] 2 points3 points ago
Hell, even Einstein had difficulty with tensor calculus..
Reminds me of a quote (maybe by Dirac?) to the effect that: "Every child on the streets of Berlin knows more differential geometry than Einstein."
Wish I could find the actual source of it. (In context it's actually praising Einstein, because he was the one that figured out GR.)
[–]Algebra 0 points1 point ago
There are other easier ways of looking at it. (Try spvak calc on manifolds). I just wanted to emphasize it isn related to the determinant indirectly.
[–] 1 point2 points ago
I dabbeled a bunch in differential forms when studying vector calculus in my EE programme (I got into it when looking for a way to generalize Green's, stoke's and Gaus's integral theorems). Learned a lot from this series of videos: http://www.youtube.com/watch?v=M5wrnwlm8lw&list=PLB8F2D70E034E9C29 Anyways, the forms you are talking about seem very similar, are differential forms a subset of that?
[–]Differential Geometry 1 point2 points ago
These are exactly differential forms in R3. Differential forms can be used studied on any smooth manifold of any dimension though so are more general. However, in that setting, curl and div are defined a little differently because in order to identify a two-form with the vector perpendicular to it, you need a way to find angles. More precisely, in order to define the Hodge star, you need a Riemannian metric but even without it you still have those d maps.
[–] 1 point2 points ago
This is the kind of answer we ought to be seeing more in r/math!
[–]Differential Geometry 0 points1 point ago
One nitpicky correction, but great answer.
The sequence isn't exact. Constants, not just zero, have zero gradient which corresponds to R3 having de Rham cohomology R on the zeroth level. If you use reduced homology instead, you can avoid this.
[–]Algebra 0 points1 point ago
Ah, thanks.
[–] 0 points1 point ago
(curl(div(f)) = 0
[–]Algebra 0 points1 point ago
Thank you.
[–] 2 points3 points ago
This is a nice calc3-level explanation of why the curl is defined and abused the way it is.
[–]Theory of Computing 4 points5 points ago
The curl isn't defined in terms of a determinant. The determinant in this case is a mnemonic meant to help you remember which terms go where. If this bothers you, then don't use it. What did you do when you learned the cross product? Did you remember the formula, or did you use a mnemonic like the determinant? If you don't like the abuse of notation, then don't use it. Write curl F instead of ×F if you don't like the latter. If your text or professor uses notation you don't like, I'm afraid there's not much you can do.
[–][S] 1 point2 points ago
I'll explain why I'm comfortable with the cross product. Consider the determinant
|( x, y, z); (a1, b1, c1); (a2, b2, c2)| = (x, y, z) dot (vector of cofactors).
It vanishes if (x, y, z) is linearly dependent with (a1, b1, c1) and (a2, b2, c2). Hence, the vector of cofactors is a vector such that if it's dotted with either of the other vectors in the determinant, it is zero. Hence it must be orthogonal. We can note that if we replace (x,y,z) with the unit vectors (i, j, k) we get this vector of cofactors.
I'm looking for a similar explanation.
[–] 1 point2 points ago
When I first started I read DIV, Grad, Curl and all that, a really phenomenal short book that clearly lays all this material out in an easy to understand way. Plus it's easy to find online.
[–]Algebraic Geometry 1 point2 points ago
I've had success when teaching this course to use a presentation based off of (this at mathinsight.)[http://mathinsight.org/curl_idea] | HuggingFaceTB/finemath | |
# the real part of a holomorphic function on C \ {0, 1}
Let $h$ be a real valued harmonic function on the twice punctured plane $Ω = \text{C \ {0, 1}}$. Show that there exist unique real numbers $a_0$, $a_1$ such that $u(z) = h(z) − a_0 \log |z| − a_1 \log |z − 1|$ is the real part of a holomorphic function on $Ω$ .
I tried to show that $u$ has a harmonic conjugate. I assumed there exists such a conjugate $v$ exist, then used Cauchy Riemann equations on $u$ and $v$. After integrating two equations, I tried to use the result to find $a_0$ and $a_1$. However, I failed to cancel out the extra terms. Is there any other direction I should go?
Define a function $g = h'_x - ih'_y$. Since $h$ is harmonic (and hence $C^2$), $g$ satisfies Cauchy-Riemann's equations on $\Omega$. Indeed: $$(h'_x)'_x = h''_{xx} = -h''_{yy} = (-h'_y)'_y$$ and $$(h'_x)'_y = h''_{xy} = h''_{yx} = -(-h'_y)'_x.$$ In other words, $g$ is holomorphic on $\Omega$. Assume for a moment that $g$ admits an anti-derivative $G$ on $\Omega$. If $G = U+iV$, Cauchy-Riemann shows that $G' = U'_x-iU'_y$, but $G' = g = h'_x-ih'_y$, so $\nabla U = \nabla h$, i.e. $U=h+C$, and we may as well take $C=0$ by adjusting our choice of $G$. Hence $h = U = \operatorname{Re} G$.
Now, the problem is that $\Omega$ is not simply connected, so we can't guarantee that $g$ admits an anti-derivative. On the other hand, we know that $g$ has an anti-derivative on $\Omega$ if and only if $\int_\gamma g(z)\,dz = 0$ for every simple closed curve in $\Omega$, and it's not difficult to see that this condition is equivalent to the fact that the residues of $g$ at $0$ and $1$ should both be $0$.
Let $a_0 = \operatorname{Res}\limits_{z=0} g$ and $a_1 = \operatorname{Res}\limits_{z=1} g$, and put $u(z) = h(z)-a_0\log|z|-a_1\log|z-1|$.
Repeat the above argument to get a holomorphic function $f=u'_x-iu'_y$ on $\Omega$. It's straight-forward to verify that $$f(z) = g(z) - \frac{a_0}{z} - \frac{a_1}{z-1}$$ so by construction, the residues of $f$ at $0$ and $1$ vanish, and from the discussion above, we are done.
This takes care of existence. For uniqueness, assume that $u(z) = h(z)-a_0\log|z|-a_1\log|z-1|$ and $\tilde u(z) = h(z)-\tilde a_0\log|z|-\tilde a_1\log|z-1|$ are two such functions. Then $u(z) = \tilde u(z) = (a_0-\tilde a_0) \log|z|-(a_1-\tilde a_1)\log|z-1|$ is the real part of a holomorphic function on $\Omega$. It's a standard exercise to check that this can only happen if $a_0 = \tilde a_0$ and $a_1 = \tilde a_1$. | HuggingFaceTB/finemath | |
# 35.5 Inches to Centimeters
=
35.5 Inches =
90.17
(decimal)
9.017 x 101
(scientific notation)
9,017
100
(fraction)
Centimeters
## Inches to Centimeters Conversion Formula
[X] cm = 2.54 × [Y] in
where [X] is the result in cm and [Y] is the amount of in we want to convert
## 35.5 Inches to Centimeters Conversion breakdown and explanation
35.5 in to cm conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). Every display form has its own advantages and in different situations particular form is more convenient than another. For example usage of scientific notation when working with big numbers is recommended due to easier reading and comprehension. Usage of fractions is recommended when more precision is needed.
If we want to calculate how many Centimeters are 35.5 Inches we have to multiply 35.5 by 127 and divide the product by 50. So for 35.5 we have: (35.5 × 127) ÷ 50 = 4508.5 ÷ 50 = 90.17 Centimeters
So finally 35.5 in = 90.17 cm | HuggingFaceTB/finemath | |
Chapter 1, Problem 9RQ ### Single Variable Calculus: Early Tr...
8th Edition
James Stewart
ISBN: 9781305270343
#### Solutions
Chapter
Section ### Single Variable Calculus: Early Tr...
8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem
# Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.If 0 < a < b, then In a < In b.
To determine
Whether the statement, if 0<a<b then lna<lnb is true or false.
Explanation
Let the function be f(x)=lnx.
The graph of f(x)=lnx is shown below in Figure 1.
From Figure 1, it is observed that the domain of the function is (0,) and the function f(x)=lnx is an increasing function.
Recall the definition of increasing function, “the function y=f(x) is said be an increasing function on the given interval [a, b] if f(
### Still sussing out bartleby?
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#### The Solution to Your Study Problems
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#### In Exercises 1-6, simplify the expression. 818
Calculus: An Applied Approach (MindTap Course List)
#### 5. True or false. (a) (b) (c)
Mathematical Applications for the Management, Life, and Social Sciences
#### Differentiate the function. f(t) = 2t3 3t2 4t
Single Variable Calculus: Early Transcendentals
#### True or False: These lines are skew:
Study Guide for Stewart's Multivariable Calculus, 8th
#### If 13f(x)dx=10 and 13g(x)dx=6, then 13(2f(x)3g(x))dx= a) 2 b) 4 c) 18 d) 38
Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th | HuggingFaceTB/finemath | |
# Unitary Transformation of Eigenstates
Suppose I have two operators, $A$ and $B$, with eigenstates $A \lvert a \rangle = a \lvert a \rangle$ and $B \lvert b \rangle = b \lvert b \rangle$, where $a$ and $b$ are all unique. Furthermore, suppose that $A$ and $B$ are related by a unitary transformation $$A = U B U^{-1}.$$ This is equivalent to saying that the eigenstates are related as $$\lvert a \rangle = U \lvert b \rangle.$$
Then it seems I can prove the following: since $$A \lvert a \rangle = a \lvert a \rangle,$$ I also have $$A U \lvert b \rangle = U B U^\dagger U \lvert b \rangle$$ by inserting the identity, so that $$A U \lvert b \rangle = U B \lvert b \rangle = b U \lvert b \rangle = b \lvert a \rangle.$$ Thus, $a = b$.
Doesn't this imply then that the eigenvalues for corresponding eigenstates of $A$ and $B$ are equal, and therefore-- by the assumption that they are unique-- that the unitary transformation doesn't actually do anything?
• As nicely summarized by @ZeroTheHero, when you write $A=UBU^{\dagger}$ you are performing a unitary basis change on operator $B$. $A$ is then the new representation of $B$ in this new set of basis. Eigenvalues are invariant under basis transformation . Thus, eigenvalues of $A$ and $B$ are the same, but their eigenstates are not. – Ptheguy Jan 5 '18 at 8:52
• Here is a good article if you're not sure why eigenvalues are invariant under basis transformation. (link). – Ptheguy Jan 5 '18 at 9:00
• It might help to choose as an example $A=x$ and $B=p$, where $U$ is the Fourier transform and takes $x$ to $p$. Here $x$ and $p$ have identical eigenvalue spectra, but that doesn't mean that the Fourier transform doesn't "do" anything. – Emilio Pisanty Jan 5 '18 at 10:03
I’m not sure what you mean by “$a$ and $b$ are unique” but clearly if $A=UBU^\dagger$ and $U$ is unitary, $A$ and $B$ have the same eigenvalues but it doesn’t mean $U$ doesn’t do anything.
For instance, the Pauli matrices $\sigma_{x,y,z}$ all have the same eigenvalues, are related by a unitary transformation $U$, but are certainly different. The transformation $U$ is a change of basis, so if $B$ is initially diagonal, say $$B=\sigma_z=\left(\begin{array}{cc} 1&0 \\ 0&-1\end{array}\right)$$ and $U=\left(\begin{array}{cc} \cos\theta/2&-\sin\theta/2\\ \sin\theta/2 &\cos\theta/2\end{array}\right)$ then $$U\sigma_zU^{\dagger}=\cos\theta\sigma_z+\sin\theta \sigma_x$$ still have eigenvalues $\pm 1$ but obviously $U$ has done something.
Of course the eigenstates are no longer $\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\end{array}\right)$.
I interpret your uniqueness claim as the requirement that
every eigenspace of either $A$ and $B$ has dimension $1$.
So, if $a$ is an eigenvalue of $A$ and $A|a\rangle =a|a\rangle$, for $|a\rangle \neq 0$, then every other eigenvector with the same eigenvalue $a$ is of the form $c|a\rangle$ for every $c\in \mathbb C$, $c\neq 0$. If we consider only normalized eigenvectors, $c$ is of the form $e^{i \theta}$ for every $\theta \in \mathbb R$.
Assuming that the spectra of the said operators are pure-point spectra, we have the spectral decompositions (the sum are understood in the strong operator topology, but here it is quite irrelevant and you can safely interpret all follows at algebraic level) $$A = \sum_{a \in {\cal A}} a |a\rangle\langle a| \tag{1}$$ and $$B = \sum_{b\in {\cal B}} b |b\rangle\langle b| \:.$$ where I used normalized eigenvalues and ${\cal A}=\sigma(A)$, ${\cal B}=\sigma(B)$ (up to accumulation points) are the spectra of the operators.
On the other hand $$A = UBU^\dagger$$ implies $$\sum_{a \in {\cal A}} a |a\rangle\langle a| = \sum_{b \in {\cal B}} b U|b\rangle\langle b|U^\dagger\:.$$ That is $$A= \sum_{b \in {\cal B}} b|\psi_b\rangle \langle \psi_b|\tag{2}$$ where $$|\psi_b\rangle := U|b\rangle\:.$$
The crucial result is now that
for a given self-adjoint operator (with point spectrum) the spectral decomposition is unique.
Thus, comparing (1) and (2) we conclude that
(i) ${\cal A}= {\cal B}\:,$
so that we can re-arrange the decomposition of $A$ like this $$A= \sum_{a \in {\cal A}} a|\psi_a\rangle \langle \psi_a|\:,$$
(ii) $|a\rangle \langle a| = |\psi_a\rangle \langle \psi_a|$ so that, since every vector is normalized $$|\psi_a\rangle = e^{i\theta_a}|a\rangle\quad \mbox{for some \theta_a \in \mathbb R}\:.$$ there is no way to determine the phases $e^{i\theta_a}$, since normalized eigenvectors are defined up to a phase, but we are free to fix all them $e^{i\theta_a}=1$.
I stress that (i) and (ii) are the maximum you can obtain from the initial information at your disposal that eigenspaces are one-dimensional and that the unitary equivalence $A=UBU^\dagger$ holds.
You see that $U$ has an action (it is false that "it doesn't actually do anything"). In fact, it changes eigenvectors, but it leaves fixed the spectrum of the operators.
Dropping the hypotheses of one-dimensional eigenspaces but keeping the request of pure point spectrum, (i) remains valid in view of the uniqueness of spectral decomposition, which now reads $$A = \sum_{a \in {\cal A}} a P_a$$ where $P_a = \sum_{k=1}^{\dim E_a} |a,k\rangle \langle a,k|$ is the orthogonal projector onto the eigenspace $E_a$ of $A$ with eigenvalue $a$ and the vectors $|a,k\rangle$, varying $k=1,\ldots, \dim E_a$, form an orthonormal basis of that eigenspace.
Well, a similarity transformation for an invertible (not necessary unitary) operator$^1$ $U$ does generically change the eigenspaces but does not change the eigenvalue spectrum $\{a_1, a_2,\ldots, \}=\{b_1,b_2,\ldots\}$. Hence it would be inconsistent to claim that all the eigenvalues $\{a_1,a_2, \ldots, b_1,b_2,\ldots\}$ for both $A$ and $B$ are different, if that's what OP means by that they are all unique. Whether the individual spectra are degenerate or not is irrelevant.
--
$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer. | HuggingFaceTB/finemath | |
S k i l l
i n
A R I T H M E T I C
Prologue 2
# THE MULTIPLICATION TABLE
*
"Multiplication" of a number is its repeated addition. (Lesson 9.)
× 1 2 3 4 5 6 7 8 9 1 1 2 3 4 5 6 7 8 9 2 2 4 6 8 10 12 14 16 18 3 3 6 9 12 15 18 21 24 27 4 4 8 12 16 20 24 28 32 36 5 5 10 15 20 25 30 35 40 45 6 6 12 18 24 30 36 42 48 54 7 7 14 21 28 35 42 49 56 63 8 8 16 24 32 40 48 56 64 72 9 9 18 27 36 45 54 63 72 81
Therefore to construct the rows of the Multiplication Table, repeatedly add the first number in each row.
The importance of knowing the multiplication table cannot be overstated. The student should practice the table out loud. To practice the 8 table, for example, say,
"1 times 8 is 8. 2 times 8 is 16. 3 times 8 is 24."
And so on. (Do not just say, "8, 16, 24, . . .") Repeat this so that "3 times 8 is 24" begins to sound right. It is not necessary to make an effort. Repetition is all that is required to make an impression in the mind.
Practice each column until you can say the answer without thinking.
Do the problem yourself first!
6 × 4 = 24 5 × 4 = 20 7 × 4 = 28 7 × 7 = 49 9 × 8 = 72 9 × 3 = 27 7 × 5 = 35 9 × 9 = 81 6 × 3 = 18 9 × 5 = 45 8 × 4 = 32 6 × 8 = 48 3 × 9 = 27 4 × 6 = 24 4 × 5 = 20 5 × 7 = 35 4 × 7 = 28 3 × 6 = 18 4 × 3 = 12 7 × 8 = 56 8 × 7 = 56 8 × 3 = 24 6 × 7 = 42 8 × 9 = 72 5 × 8 = 40 5 × 6 = 30 3 × 4 = 12 4 × 8 = 32 5 × 5 = 25 5 × 9 = 45 8 × 6 = 48 8 × 5 = 40 9 × 6 = 54 6 × 9 = 54 7 × 9 = 63 6 × 5 = 30 9 × 7 = 63 9 × 4 = 36 6 × 6 = 36 3 × 8 = 24 3 × 7 = 21 7 × 3 = 21 8 × 8 = 64 3 × 5 = 15 5 × 3 = 15 7 × 6 = 42 4 × 9 = 36 4 × 4 = 16
Here again are these elementary sums to practice.
8 + 5 = 13 3 + 9 = 12 5 + 9 = 14 6 + 8 = 14 7 + 9 = 16 9 + 6 = 15 7 + 6 = 13 8 + 6 = 14 6 + 5 = 11 8 + 4 = 12 3 + 8 = 11 5 + 8 = 13 9 + 4 = 13 7 + 4 = 11 6 + 6 = 12 4 + 9 = 13 4 + 8 = 12 6 + 9 = 15 8 + 9 = 17 4 + 7 = 11 5 + 7 = 12 6 + 7 = 13 9 + 3 = 12 9 + 2 = 11 7 + 7 = 14 7 + 8 = 15 9 + 5 = 14 9 + 8 = 17 7 + 5 = 12 2 + 9 = 11 9 + 7 = 16 8 + 7 = 15 5 + 6 = 11 8 + 3 = 11 8 + 8 = 16 9 + 0 = 9 8 + 0 = 8 0 + 6 = 6 0 + 3 = 3 0 + 0 = 0
Lesson 1: Numeration of the Whole Numbers
E-mail: themathpage@yandex.com
Private tutoring available. | HuggingFaceTB/finemath | |
"
">
# Find the values of the angles $x, y,$ and $z$ in each of the following:"
To do:
We have to find the values of the angles $x, y,$ and $z$ in the given figures.
Solution:
We know that,
Vertically opposite angles are equal.
Sum of the angles in a linear pair is $180^o$.
Therefore,
(i) $x=55^o$ (Vertically opposite angles)
$y+55^o=180^o$ (Linear pair)
$y=180^o-55^o$
$y=125^o$
$z=y=125^o$ (Vertically opposite angles)
(ii) $z=40^o$ (Vertically opposite angles)
$y+40^o=180^o$ (Linear pair)
$y=180^o-40^o$
$y=140^o$
$z+x+25^o=180^o$ (Linear pair)
$x=180^o-25^o-40^o$
$x=115^o$
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
25 Views | HuggingFaceTB/finemath | |
# What Is 32 Degrees Celsius in Fahrenheit?
By Staff WriterLast Updated Mar 30, 2020 7:31:53 PM ET
Thirty-two degrees Celsius is the equivalent of 89.6 degrees Fahrenheit. To convert Celsius to Fahrenheit, multiply the Celsius temperature by 1.8, and then add 32. In this example, 32 X 1.8 = 57.6, and 57.6 + 32 = 89.6. | HuggingFaceTB/finemath | |
# Thread: |2x-1|-|x|< 4
1. ## |2x-1|-|x|< 4
How do I solve this |2x-1|-|x|< 4 ? Do i simply treat it like an equation, meaning that it has three different solutions? or do other rules apply?
2. Use cases.
$\displaystyle |2x-1|-|x|<4$
There are three cases,
i. $\displaystyle x\ge\frac{1}{2}$
ii. $\displaystyle 0 < x < \frac{1}{2}$
iii. $\displaystyle x \le 0$
-------------
i. $\displaystyle x\ge\frac{1}{2}$
The inequality becomes,
$\displaystyle 2x - 1 - x < 4$
$\displaystyle x < 5$
The interval here is, $\displaystyle \frac{1}{2}\le x < 5$
-------------
ii. $\displaystyle 0 < x < \frac{1}{2}$
$\displaystyle -(2x-1) - x < 4$
$\displaystyle x > -1$
Interval: $\displaystyle 0 < x < \frac{1}{2}$
-------------
iii. $\displaystyle x \le 0$
$\displaystyle -(2x -1)-(-x)<4$
Interval: $\displaystyle -3 < x \le 0$
-------------
-------------
When we put the intervals together,
$\displaystyle \frac{1}{2}\le x < 5$
$\displaystyle 0 < x < \frac{1}{2}$
$\displaystyle -3 < x \le 0$
It makes,
$\displaystyle -3<x<5$
3. the first element is positive if x > 0.5 and negative otherwise, the second element is positive if x > 0 and negative otherwise.
1. x => 0.5, in this range both elements are non negative so:
2x-1 - x < 4
x < 5
so the solution is: 0.5 <= x < 5
2. 0 <= x <= 0.5 in this range the first element is negative and the second is non negative so:
-(2x-1) - x < 4
x > -1
thus the answer is 0 <= x <= 0.5
3. x <= 0 in this range the first element is negative and the second is non positive so:
-(2x-1) + x < 4
x > -3
thus the solution is -3 < x <= 0
The final solution is the intersection of these 3 solutions thus:
-3 < x < 5
4. Originally Posted by weasley74
How do I solve this |2x-1|-|x|< 4 ? Do i simply treat it like an equation, meaning that it has three different solutions? or do other rules apply?
If you like - or allowed to - you can solve it graphically. | HuggingFaceTB/finemath | |
## 16.2 Classical approach
What is the probability of rolling a on a die? The sample space has six possible outcomes (listed in Example 16.1) that are equally likely, and the event (‘rolling a ’) comprises just one of those. Thus, \begin{align*} \text{Prob. of rolling a four} &= \frac{\text{The number of results that are a 4}}{\text{The number of possible results}}\\ &= \frac{1}{6}. \end{align*}
We can say that ‘the probability of rolling a 4 is 1/6,’ or ‘the probability of rolling a is 0.1667.’ The answer can also be expressed as a percentage (‘the probability of rolling a is 16.7%’).
The answer could also be interpreted as ‘the expected proportion of rolls that are a is 0.167.’ This approach to computing probabilities is called the classical approach to probability.
The chance of rolling a in the future is 0.167, but a roll of the die will either produce a , or will not produce a … and we don’t know which will occur.
Example 16.4 (Describing probability outcomes) Consider rolling a standard six-sided die again.
• The probability of rolling an even number is $$3 \div 6 = 0.5$$.
• The percentage of rolls that are expected to be even numbers is $$3 \div 6 \times 100 = 50$$%.
• The odds of rolling an even number is $$3\div 3 = 1$$.
Definition 16.4 (Classical approach to probability) In the classical approach to probability, the probability of an event occurring is the number of elements of the sample space included in the event, divided by the total number of elements in the sample space, when all outcomes are equally likely.
By this definition:
$\text{Prob. of an event} = \frac{\text{Number of equally-likely outcomes in the event of interest}}{\text{Total number of equally-likely outcomes}}$
Example 16.5 (Simple events) What is the probability of rolling a 2 on a die? What are the odds of rolling a 2 on a die?
Since the six possible outcomes in the sample space are equally likely:
$\text{Prob. of rolling a two} = \frac{\text{One outcome is a 2}}{\text{Six equally-likely outcomes}}.$ So the probability is $$\frac{1}{6} = 0.1667$$, or about 16.7%. Also, since the six possible outcomes are equally likely:
$\text{Odds of rolling a two} = \frac{\text{One outcomes is a two}}{\text{Five of the possible outcomes are not a two}}.$ So the odds of rolling a two is $$\frac{1}{5} = 0.2$$.
Example 16.6 (More complicated events) Consider rolling a standard six-sided die. There are six equally likely outcomes (Example 16.1) each with probability $$1/6$$ (or 16.7%) of occurring. The probability of rolling a 1 or a 2 is $$2/6$$ (or 33.3%).
Probabilities describe the likelihood that an event will occur before the outcome is known.
Odds and proportions can be used either before or after the outcome is known, provided the wording is correct. For example:
• Proportions describe how often an event has occurred after the outcome is known.
• Expected proportions describe the likelihood that an event will occur before the outcome is known.
The following example may help also.
Example 16.7 (Probabilities, proportions, odds) Before a fair coin is tossed:
• The probability of throwing a head is $$1/2 = 0.5$$.
• The expected proportion of heads in many coin tosses is 0.5.
• The odds of throwing a head is $$1/1 = 1$$.
If we have already tossed a coin 100 times and found 47 heads:
• The proportion of heads is $$47/100 = 0.47$$.
• The odds that we threw a head is $$47/53 = 0.887$$.
It makes no sense to talk about the ‘probability that we just threw a head,’ because the event has already occurred. | HuggingFaceTB/finemath | |
What Is The Prime Number Between 20 And 30 // marinaspiceloungerestaurants.com
# What are the prime numbers between 20 and 30 - Answers.
Jan 12, 2017 · 23 and 29 are the prime numbers between 20 and 30. The two prime numbers between 20 and 30 are 23 and 29. 23, 29 All the prime numbers. 23 and 29 are the prime numbers between 20 and 30. No even number is a prime number except 2. Hence, 20, 22, 24, 26, 28 and 30 are not prime. See full answer below. Prime numbers = a number which is divisible by 1 and number itself. So the number between 20–30 which is divisible by only 1 and number itself is = 23 and 29.
Therefore, colorred23 and colorred29 are the prime numbers between 20 and 30. Mar 30, 2009 · What is the prime numbers of 20-30?. How ma How many prime numbers between 1 and How many prime numbers between 1 and How many prime numbers between 1 and How many prime numbers between 1 and. Feb 21, 2006 · If the number between n and square root of n divides n,then n is not a prime it's a composite, else it's a prime. Well, here's the algorithm to test a range of prime numbers for the primality. 1. input a,b Are the ranges ex. 20-30 2. result=1. 2. for i=a to b . List all the numbers between 20 and 30 that are prime - 3572001.
This prime numbers generator is used to generate the list of prime numbers from 1 to a number you specify. Prime Number. A prime number or a prime is a natural number that has exactly two distinct natural number divisors: 1 and itself. For example, there are 25 prime numbers from 1 to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43. A prime number or prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. By Euclid's theorem, there are an infinite number of prime numbers. Subsets of the prime numbers may be generated with various formulas for primes. a whole number that cannot be made by multiplying other whole numbers if we can make it by multiplying other whole numbers it is a Composite Number And 1 is not prime and also not composite.
We'll check all the integers between 20 and 30: 21 is not prime because it is divisible by 3 and 7. 22 is not prime because it is divisible by 2 and 11. 23 is prime but its digits do not differ by 7 because 3-2=1 not 7. 24 is not prime because it is divisible by 2,3,4,6,8 and 12. 25 is not prime. Prime number is a positive natural number that has only two positive natural number divisors - one and itself. The opposite of prime numbers are composite numbers. A composite number is a positive nutural number that has at least one positive divisor other than one or itself. Jan 31, 1997 · A prime number is a whole number which is not the product of smaller numbers. For example, 14 is not a prime number, becaues it is 2 times 7. But 3 is a prime number, because the only smaller numbers are 1 and 2, and 3 is not 1 times 1 or 1 times 2 or 2 times 2. To see if a number is prime, all you need to do is try the numbers smaller than it.
The mean of prime numbers between 20 and 30 is ____. - 14664451.important. 11. What will be the perimeter of a rectangle if its length is 3 times its width and the length of thediagonal is 8 ×root 10 cm?A 16/10 cm B 15/10. | HuggingFaceTB/finemath | |
## 5.6 Vector Triple Products
The volume of a parallelepiped with sides A, B and C is the area of its base (say the parallelogram with area |BC| ) multiplied by its altitude, the component of A in the direction of BC. This is the magnitude of ABC; but it is also the magnitude of the determinant of the matrix with columns A, B and C, so these linear functions of the vectors here are the same up to sign. The usual sign convention gives
A(BC) = det(A, B, C)
This product is not changed by cyclically permuting the vectors (for example to B, C, A) or by reversing the order of the factors in the dot product.
We can deduce then that ABC = CAB = ABC. In words, we can switch the dot and cross product without changing anything in this entity. (In either formula of course you must take the cross product first.) This product, like the determinant, changes sign if you just reverse the vectors in the cross product.
The vector triple product, A(BC) is a vector, is normal to A and normal to BC which means it is in the plane of B and C. And it is linear in all three vectors.
We can deduce it is a multiple of B that is linear in A and C plus a multiple of C that is linear in A and B, with the condition that it is normal to A.
Any multiple of B(AC) - C(AB) will obey all these conditions.
What multiple is A(BC)?
Suppose A(BC) = q(B(AC) - C(AB)) holds.
Earlier we saw that the square of the area of a parallelogram with sides A and B can be written either as (AA)( BB) - (AB)( AB) or (BA)(BA). By interchanging the dot and first cross product on the right here you can rewrite this equality as
(BA)(BA) = B(A(BA)) =(AA)( BB) ) - (AB)( AB)
If we identify A with C in A(BC) and take the dot product of A(BA) with B we find q = 1, and we get
A(BC) = B(AC) - C(AB)
This is sometimes called the back cab rule to make it easier to remember the appropriate signs. When using this name remember that the parentheses here are all as far back as possible in this expression The easiest way to get the signs right here without remembering anything is to guess a sign and then check it on the case A = i = C, B = j.
Exercise 5.13 Suppose we have a vector A in three dimensions and an unknown vector v, but we do know Av and Av. Can we find v? YES! How? | HuggingFaceTB/finemath | |
# How Does a Pulley System Work? ••• Alex/iStock/GettyImages
A pulley system makes it easier to lift an object than lifting the weight by hand. A single pulley essentially changes the direction of the pull or force applied. When a person uses two or more pulleys in a system, then the system also multiplies the force applied in addition to changing direction. With one fixed and one movable pulley in a system, it essentially doubles the weight of the load you could lift without help from another person based upon which heavy objects you can lift.
## The Pulley: A Simple Machine
Simple machines are mechanical tools that help to redirect or decrease the amount of force needed to move heavy loads a certain distance. Inclined planes, wedges, levers, screws, and pulleys are all types of simple machines.
Simple machines work to reduce the force required to move an object, but the amount of work done will always remain the same. Work is a value that describes how a force (F) is exerted over a certain distance (d):
W = F \cdot d
As less force is needed to move a heavy weight, the distance required to move it will actually increase based on the simple machine. So a simple pulley system that cuts the force in half, will double the distance of rope that will be used, resulting in equivalent work.
A pulley has two equal arms and operates on a fulcrum like the lever does, though it is a wheel with rimmed edges on an axle threaded with a rope – instead of a stationary balancing point.
## How Pulleys Work: Mechanical Advantage
Simple machines like the pulley give you a mechanical advantage, essentially making you stronger than you are in real life. Physicists quantify the work the system does by calculating mechanical advantage in Newtons, named after Sir Issac Newton, the originator of the laws of motion. It takes 1 newton to move 1 kilogram of mass at the rate of 1 meter per second squared in the direction of the applied force. To calculate the mechanical advantage of a pulley, divide the output force by the input force; in other words divide the force exerted on the object by the force exerted on the end of the rope as you use the pulley.
## One Fixed, One Movable Pulley
While using only one pulley requires you to use the same force it would take to lift a load by hand, a fixed pulley combined in a system with a movable pulley, essentially doubles the force applied to lift or move the object. For example, if you have an object that weighs 100 N, all it would take to lift the object in this pulley system are 50 newtons of force applied. The MA of this type of system equals two.
## A System of Pulleys
While a single pulley allows you to move a load with half the force required, adding a second pulley to form a system of pulleys increases the mechanical advantage by the number of pulleys and the lengths of rope that support the load. The number of rope strands that support the load in a multiple pulley system basically correspond to the mechanical advantage of the system. For example, if you have two fixed and two movable pulleys, four lengths of the rope support the load with a mechanical advantage of four. A 100 N load would require 25 newtons of force applied to lift it.
Other types of pulleys, like compound pulley systems and block and tackle systems are more complex combinations of pulleys that allow for a greater amplification of pulling force through adding additional pulleys and components to reach a better mechanical advantage. Different types of pulleys might be tailored for specific use cases, especially if the direction of the force applied to an object might need to be modified.
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# linear map $f:V \rightarrow V$, which is injective but not surjective
I am trying to find a linear map $f:V \rightarrow V$, which is injective but not surjective.
I always thought that if the dimension of the domain and codomain are equal and the map is injective it implies that a map is surjective. Maybe we need an infinite basis of the vector space $V$? What can be an example of that?
Thank you!
Yes, we need an infinite-dimensional vector space. An interesting example is: $V$ the space of continuous functions $[0,1]\to\mathbb R$ and $f$ integration $f(g)(x)=\int_0^xg(t)\,\mathrm dt$. This is not surjective because $f(g)(0)=0$ for all $g$
In finite dimensions we have that bijectivity $\Leftrightarrow$ injectivity $\Leftrightarrow$ surjectivity. Hence we have to come up with an infinite-dimensional example. The idea is to pick a basis $v_i, i\in \mathbb N$, and shift every basis vector $v_i \mapsto v_{i+1}$. We can do that not hitting the first basis vector only because we have infinitely many elements.
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# linear map $f:V \rightarrow V$, which is injective but not surjective
I am trying to find a linear map $f:V \rightarrow V$, which is injective but not surjective.
I always thought that if the dimension of the domain and codomain are equal and the map is injective it implies that a map is surjective. Maybe we need an infinite basis of the vector space $V$? What can be an example of that?
Thank you!
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Yes, we need an infinite-dimensional vector space. An interesting example is: $V$ the space of continuous functions $[0,1]\to\mathbb R$ and $f$ integration $f(g)(x)=\int_0^xg(t)\,\mathrm dt$. This is not surjective because $f(g)(0)=0$ for all $g$
@marco11: $f$ is injective, because if $g_1$ and $g_2$ differ at some point -- say $g_1(x_0)\ne g_2(x_0)$, then $f(g_1)$ and $f(g_2)$ will have different derivatives at that point, and so cannot be the same function. On the other hand $f$ is not surjective because, for example, the function $x\mapsto x+1$ is not $f(g)$ for any $g$ (namely $f(g)(0)=\int_0^0 g(t)dt = 0 \ne 0+1$). - Henning Makholm May 24 at 22:42
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# Continuity and Discontinuity in Calculus – Class 12 CBSE
• Difficulty Level : Expert
• Last Updated : 12 Nov, 2020
A function is said to be continuous if one can sketch its curve on a graph without lifting the pen even once. A function is said to be continuous at x = a, if, and only if the three following conditions are satisfied.
1. The function is defined at x = a; that is, f(a) equals a real number
2. The limit of the function as x approaches a exists
3. The limit of the function as x approaches a is equal to the function value at x = a
A function f(x) is said to be continuous in the open interval (a, b) if at any point in the given interval the function is continuous. In the case of closed interval [a, b], the function is said to be continuous:
• f(x) is be continuous in the open interval (a, b)
• limx⇢a f(x) = f(a)
• limx⇢b f(x) = f(b)
Example 1: Prove that the function f(x) = 5x – 3 is continuous at x = 0.
Solution:
Given, f(x) = 5x – 3
At x = 0 , f(0) = (5 × 0) – 3 = -3
limx⇢0 f(x) = limx⇢0 (5x – 3) = (5 × 0) – 3 = -3
limx⇢0 f(x) = f(0)
Therefore, f(x) is continuous at x = 0.
Example 2: Examine the function f(x) = |x – 5|, for continuity.
Solution:
Given function, f(x) = |x – 5|
Domain of f(x) is real and infinite for all real x
Here , f(x) = |x – 5| is a modulus function
As , every modulus function is continuous
Therefore , f(x) is continuous in its domain R.
Example 3: Is the function f(x) = x – sinx + 5 is continuous at x = π.
Solution:
Given function is f(x) = x – sinx + 5
L.H.L = limx⇢π (x – sinx + 5) = limx⇢π[(π – h) – sin(π – h) + 5] = π + 5
R.H.L = limx⇢π+ (x – sinx + 5) = limx⇢π+ [(π + h) – sin(π + h) + 5] = π + 5
And, f(π) = π – sinπ + 5 = π + 5
Since , L.H.L = R.H.L = f(π)
Therefore , f(x) is continuous at x = π
Example 4: Examine the continuity of the function f(x) = 2x – 1 at x = 3.
Solution:
Given f(x) = 2x – 1
At x = 3, f(x) = (2 × 3) – 1 = 5
limx⇢3 f(x) = limx⇢3 f(x) = (2×3) – 1 = 5
limx⇢3 f(x) = f(3)
Therefore, f(x) is continuous at x = 3
Example 5: Examine the function is continuous or not?
Solution:
For x > 0, y = x and x < 0, y = -x
So, We Know it is continuous for x > 0 and x < 0. To check if it is continuous at x = 0 , check the limit:
limx⇢0 |x| = limx⇢0 (-x) = 0
limx⇢0+ |x| = limx⇢0+ (x) = 0
So, limx⇢0 |x| = 0 , which is equal to the value of the function at 0. Therefore, It is continuous everywhere.
## Discontinuity
A function is discontinuous at a point x = a if the function is not continuous at a. The function “f” is said to be discontinuous at x = a in any of the following cases:
1. f(a) is not defined
2. limx⇢a+ f(x) and limx⇢a– f(x) exists, but are not equal.
3. limx⇢a+ f(x) and limx⇢a f(x) exists and are equal but not equal to f(a).
### Types of Discontinuities
There are three basic types of discontinuities
1. Removable(point) Discontinuity
2. Infinite Discontinuity
3. Jump Discontinuity
Removable(point) Discontinuity: The graph has a hole at a single x-value. Imagine you’re walking down the road, and someone has removed a manhole cover. This is a category of discontinuity in which the function has a well defined two-sided limit at x = a, but either f(a) is not defined or f(a) is not equal to its limit.
• limx⇢a f(x) ≠ f(a)
• f(a) = limx⇢a f(x)
Infinite Discontinuity: The function goes toward positive or negative infinity. Imagine a road getting closer and closer to a river with no bridge to the other side. The function diverges at x = a to give it a discontinuous nature here. That is to say, f(a) is not defined. Since the value of the function at x = a tends to infinity or doesn’t approach a particular finite value, the limits of the function as x → a are also not defined.
Jump Discontinuity: The graph jumps from one place to another. Imagine a superhero going for a walk, he reaches a dead end and, because he can, flies to another road. In this type of discontinuity, the right-hand limit and the left-hand limit for the function at x = a exists; but the two are not equal to each other.
• limx⇢a+ f(x) ≠ limx⇢af(x)
My Personal Notes arrow_drop_up | HuggingFaceTB/finemath | |
# Sharpe Ratio, Explained in Plain English
In today’s near-zero interest rate economy, the reward versus risk of an investment portfolio can be measured using the Sharpe ratio. Like a batting average, higher numbers are better, and 0.400 is very good.
If portfolio Z has a forward-looking Sharpe ratio of 0.400, and an expected return of 8%, there is a 68% chance its 1-year return will be between -12% and +28%.
The math is surprisingly easy. Because the Sharpe ratio is a return/risk ratio it can be transformed into a risk/return ratio by finding its inverse (using the “1/x” button on a calculator). The inverse of 0.400 is 2.5. The return is 8%, so the “risk” is 2.5 times 8% which is 20%.
For the Sharpe ratio, the downside risk and the upside “risk” are the same. So the downside is 8% -20%, or -12%. The upside risk is 8%+20%, or 28%. Easy!
### Sharpe Ratios and Risk (more detail)
Where did the “68% chance” come from? The answer is a bit more complicated, but still fairly easy to understand.
It comes from the 3-sigma1 rule of statistics. The range of -12% to +28% comes from 1 standard deviations of the mean (or plus or minus one sigma). The 3-sigma rule also says that 95% of outcomes will fall within two standard deviations. Double the deviation means two times the upside and downside risk, so the 95% confidence range becomes -32% to 48%. Finally the 3-sigma rule means triple the upside and downside risk, meaning outcomes from -52% to +68% will occur 99.7 percent of the time.
Almost every investor will be be pleased with a positive sigma event, where the return is above 8%. For example a +1 sigma (+1σ) occurrence has a +28% return — quite nice.
A downside event is potentially quite troublesome. Even a -1σ event means a 12% loss. A -2σ is a much worse 32% loss.
### Ex Ante and Ex Post Sharpe Ratios
Forward-looking (ex ante) Sharpe ratios are predictions “prior to the event(s)”. They are always positive, because no rational investor would invest in a negative expected return. The assumptions baked into an ex ante Sharpe ratio predictions are 1) expected standard deviation of total return, σ, 2) expected future return.
Backward-looking, or after the fact, (ex post) Sharpe ratios can be negative or positive. In fact, assuming “normal distributions of return”, there is a reasonable (but less than 50%) chance of a negative ex post Sharpe ratio.
Sigma1 HAL0 software optimizes for Sharpe ratios by optimizing for return and standard deviation. It also optimizes for semivariance. More “plain English” on that advantage later. | HuggingFaceTB/finemath | |
# Garner's algorithm
A consequence of the Chinese Remainder Theorem is, that we can represent big numbers using an array of small integers. For example, let $p$ be the product of the first $1000$ primes. $p$ has around $3000$ digits.
Any number $a$ less than $p$ can be represented as an array $a_1, \ldots, a_k$, where $a_i \equiv a \pmod{p_i}$. But to do this we obviously need to know how to get back the number $a$ from its representation. One way is discussed in the article about the Chinese Remainder Theorem.
In this article we discuss an alternative, Garner's Algorithm, which can also be used for this purpose.
We can represent the number $a$ in the mixed radix representation:
$$a = x_1 + x_2 p_1 + x_3 p_1 p_2 + \ldots + x_k p_1 \cdots p_{k-1} \text{ with }x_i \in [0, p_i)$$
A mixed radix representation is a positional numeral system, that's a generalization of the typical number systems, like the binary numeral system or the decimal numeral system. For instance the decimal numeral system is a positional numeral system with the radix (or base) 10. Every a number is represented as a string of digits $d_1 d_2 d_3 \dots d_n$ between $0$ and $9$, and E.g. the string $415$ represents the number $4 \cdot 10^2 + 1 \cdot 10^1 + 5 \cdot 10^0$. In general the string of digits $d_1 d_2 d_3 \dots d_n$ represents the number $d_1 b^{n-1} + d_2 b^{n-2} + \cdots + d_n b^0$ in the positional numberal system with radix $b$.
In a mixed radix system, we don't have one radix any more. The base varies from position to position.
## Garner's algorithm
Garner's algorithm computes the digits $x_1, \ldots, x_k$. Notice, that the digits are relatively small. The digit $x_i$ is an integer between $0$ and $p_i - 1$.
Let $r_{ij}$ denote the inverse of $p_i$ modulo $p_j$
$$r_{ij} = (p_i)^{-1} \pmod{p_j}$$
which can be found using the algorithm described in Modular Inverse.
Substituting $a$ from the mixed radix representation into the first congruence equation we obtain
$$a_1 \equiv x_1 \pmod{p_1}.$$
Substituting into the second equation yields
$$a_2 \equiv x_1 + x_2 p_1 \pmod{p_2},$$
which can be rewritten by subtracting $x_1$ and dividing by $p_1$ to get
$$\begin{array}{rclr} a_2 - x_1 &\equiv& x_2 p_1 &\pmod{p_2} \\ (a_2 - x_1) r_{12} &\equiv& x_2 &\pmod{p_2} \\ x_2 &\equiv& (a_2 - x_1) r_{12} &\pmod{p_2} \end{array}$$
Similarly we get that
$$x_3 \equiv ((a_3 - x_1) r_{13} - x_2) r_{23} \pmod{p_3}.$$
Now, we can clearly see an emerging pattern, which can be expressed by the following code:
for (int i = 0; i < k; ++i) {
x[i] = a[i];
for (int j = 0; j < i; ++j) {
x[i] = r[j][i] * (x[i] - x[j]);
x[i] = x[i] % p[i];
if (x[i] < 0)
x[i] += p[i];
}
}
So we learned how to calculate digits $x_i$ in $O(k^2)$ time. The number $a$ can now be calculated using the previously mentioned formula
$$a = x_1 + x_2 \cdot p_1 + x_3 \cdot p_1 \cdot p_2 + \ldots + x_k \cdot p_1 \cdots p_{k-1}$$
It is worth noting that in practice, we almost probably need to compute the answer $a$ using Arbitrary-Precision Arithmetic, but the digits $x_i$ (because they are small) can usually be calculated using built-in types, and therefore Garner's algorithm is very efficient.
## Implementation of Garner's Algorithm
It is convenient to implement this algorithm using Java, because it has built-in support for large numbers through the BigInteger class.
Here we show an implementation that can store big numbers in the form of a set of congruence equations. It supports addition, subtraction and multiplication. And with Garner's algorithm we can convert the set of equations into the unique integer. In this code, we take 100 prime numbers greater than $10^9$, which allows representing numbers as large as $10^{900}$.
final int SZ = 100;
int pr[] = new int[SZ];
int r[][] = new int[SZ][SZ];
void init() {
for (int x = 1000 * 1000 * 1000, i = 0; i < SZ; ++x)
if (BigInteger.valueOf(x).isProbablePrime(100))
pr[i++] = x;
for (int i = 0; i < SZ; ++i)
for (int j = i + 1; j < SZ; ++j)
r[i][j] =
BigInteger.valueOf(pr[i]).modInverse(BigInteger.valueOf(pr[j])).intValue();
}
class Number {
int a[] = new int[SZ];
public Number() {
}
public Number(int n) {
for (int i = 0; i < SZ; ++i)
a[i] = n % pr[i];
}
public Number(BigInteger n) {
for (int i = 0; i < SZ; ++i)
a[i] = n.mod(BigInteger.valueOf(pr[i])).intValue();
}
Number result = new Number();
for (int i = 0; i < SZ; ++i)
result.a[i] = (a[i] + n.a[i]) % pr[i];
return result;
}
public Number subtract(Number n) {
Number result = new Number();
for (int i = 0; i < SZ; ++i)
result.a[i] = (a[i] - n.a[i] + pr[i]) % pr[i];
return result;
}
public Number multiply(Number n) {
Number result = new Number();
for (int i = 0; i < SZ; ++i)
result.a[i] = (int)((a[i] * 1l * n.a[i]) % pr[i]);
return result;
}
public BigInteger bigIntegerValue(boolean can_be_negative) {
BigInteger result = BigInteger.ZERO, mult = BigInteger.ONE;
int x[] = new int[SZ];
for (int i = 0; i < SZ; ++i) {
x[i] = a[i];
for (int j = 0; j < i; ++j) {
long cur = (x[i] - x[j]) * 1l * r[j][i];
x[i] = (int)((cur % pr[i] + pr[i]) % pr[i]);
} | open-web-math/open-web-math | |
# Series
In the given series one number is missing. Understand the pattern of the series and insert the number. 144, 121, 100, 81, ?, 49, 36
A)
72
B)
64
C)
76
D)
66
64
Explanation :
the pattern of the series
122=144,112=121,102=100,92=81,82=64,72=49,62=36
So missing number is 82 that is 64
Identify the next term in the given series by understanding the pattern of the series.
3, 9, 21, 39, 63, 93, ?
A)
119
B)
103
C)
129
D)
121
129
Explanation :
the pattern of series is
3, (3+6)=9, (9+12)=21, (21+18)=39, (39+24)=63, (63+30)=93, (93+36)=129
What will come next in the following series-
B, D, G, K, P, ?
A)
T
B)
V
C)
S
D)
R
V
Explanation :
the pattern of the series is
BC
DEF
GHIJ
KLMNO
PQRSTU
V
Identify which number is wrong in the given series -
4, 4, 8, 24, 95, 480, 2880
A)
95
B)
24
C)
480
D)
4
95
Explanation :
next number of series can obtain by multiplying 1, 2, 3, 4, 5, 6 respectively .
4 , 4*1=4 , 4*2=8, 8*3=24, 24*4=96, 96*5=480, 480*6=2880
So, 95 is wrong.
What will come next in the following series?
BBCBCDBCDEBCDEFBCDEFGBCDEFG
A)
H
B)
F
C)
E
D)
B
H
Explanation :
B
BC
BCD
BCDE
BCDEF
BCDEFG
BCDEFGH
What will come next in the following letter series-
J K L M N O P Q J K L M N O J K
A)
K
B)
L
C)
M
D)
J
L
Explanation :
J K L M N O P Q
J K L M N O
J K L
In the given series one number is missing. Understand the pattern of the series and insert the number.
4, 7, 9, 12, 14, 17, ?, 22
A)
21
B)
19
C)
18
D)
20
19
Explanation :
the pattern of the series is
4,(4+3)=7 ,(7+2)=9 ,(9+3)=12 ,(12+2)=14 ,(14+3)=17 ,(17+2)=19 ,(19+3)=22
and according to pattern the missing number is 19
What should come next in the following sequence of letters -
H H I H I J H I J K H I J K L H I J K ?
A)
M
B)
L
C)
H
D)
N
L
Explanation :
H
H I
H I J
H I J K
H I J K L
H I J K L M
What should come next in the following letter series based on English alphabet -
BSJ, CTK, DUL, ?
A)
EVM
B)
EXZ
C)
JKL
D)
EMV
EVM
Explanation :
B -> C -> D -> E
S -> T -> U -> V
J -> K -> L -> M
What will come in place of Question mark (?).
126, 64, 34, 20, 14, (?)
A)
12
B)
14
C)
16
D)
18
12
Explanation :
126/2 +1=64
64/2 +2=34
34/2 +3=20
20/2 +4=14
14/2 +5=12
What will come in place of Question mark (?).
99, 98, 94, 85, 69, (?)
A)
54
B)
65
C)
44
D)
34
44
Explanation :
99,
99-12 =98
98-22 =94
94-32 =85
85-42 =69
69-52 =44
Therefore next number = 44.
Find the wrong term in the given series:
1, 3, 10, 36, 152, 760, 4632
A)
3
B)
36
C)
4632
D)
760
760
Explanation :
(1 x 1) + 2 = 3
(3 x 2) + 4 = 10
(10 x 3) + 6 = 36
(36 x 4) + 8 = 152
(152 x 5) + 10 = 770
(770 x 6) + 12 = 4632
threrfore wrong term 760
What will come in place of Question mark (?) in the following number series
198, 194, 185, 169, (?)
A)
154
B)
165
C)
144
D)
134
144
Explanation :
198,
198-22=194,
194-32=185,
185-42=169,
169-52=144
So, missing number is 144.
Find the wrong term in the given serie -
850, 843, 829, 808, 788, 745, 703
A)
843
B)
829
C)
808
D)
788
788
Explanation :
850 (-7) 843 (-14) 829 (-21) 808 (-28) 780 (-35) 745 (-42) 703
So, the wrong number is 788 which must be 780.
What will come in place of Question mark (?) in the following number series ?
16, (?), 24, 12, 36
A)
20
B)
18
C)
19
D)
8
8
Explanation :
16, (-8) 8
16, 8, (16+8) 24
16, 8, 24, (-8) 12
16, 8, 24, 12, (24+12) 36
16, 8, 24, 12, 36
Missing number = 8
Find the missing term in the series.
Q1, P2, R6, N24, V120, ___
A)
W600
B)
X720
C)
F720
D)
F600
F720
Explanation :
1,
1 * 2 = 2,
2 * 3 = 6,
6 * 4 = 24,
24 * 5 = 120,
120 * 6 = 720
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
Q P R N V F 17 -1 16 +2 18 -4 14 +8 22 -16 6 | HuggingFaceTB/finemath | |
# Greatest common divisor (GCD) - math word problems
#### Number of problems found: 55
• Largest squares
How many of the largest square sheets did the plumber cut the honeycomb from 16 dm and 96 dm?
• Identical cubes
From the smallest number of identical cubes whose edge length is expressed by a natural number, can we build a block with dimensions 12dm x 16dm x 20dm?
• Children's home
The children's home received a gift to Nicholas of 54 oranges, 81 chocolate figurines, and 135 apples. Every child received the same gift and nothing was left. a) How many packages could be prepared? b) what did the children find in the package?
• Profitable company
Three businessman decide to open up their own company. They agree to distribute the yearly profits made in the same ratio as their initial investments. They invest R 50 000, R 75 000 and R25 000, respectively. The profit made by the company in the first y
• Length of a string
What is the smallest length of a string that we can cut into 18 equal parts and even 27 equal parts (in decimeters)?
• A large
A large gear will be used to turn a smaller gear. The large gear will make 75 revolutions per minute. The smaller gear must make 384 revolutions per minute. Find the smallest number of teeth each gear could have. [Hint: Use either GCF or LCM. ]
• Three lines
At 6 am, three bus lines are departing from the station. The first line has an interval of 24 minutes. The second line has an interval of 15 minutes. The third line runs at regular intervals of more than 1 minute. The third line runs at the same time as t
• Quotient
Find quotient before the bracket - the largest divisor 51 a + 34 b + 68 121y-99z-33
• Common divisors
Find all common divisors of numbers 30 and 45.
• Cutting paper
Divide a rectangular paper with dimensions 220mm and 308mm into squares of the same size so that they are as large as possible. Specify the length of the side of the square.
• Long bridge
Roman walked on the bridge. When he heard the whistle, he turned and saw running Kamil at the beginning of the bridge. If he went to him, they would meet in the middle of the bridge. Roman, however, rushed and so did not want to waste time returning 150m.
• Wide field
The field is 203 meters wide 319 meter long, what is the greater length of the rope by which length and width can be measured and find the exact number of time.
• Glass panel
A rectangular glass panel with dimensions of 72 cm and 96 cm will cut the glazier on the largest square possible. What is the length of the side of each square? How many squares does the glazier cut?
• Reminder and quotient
There are given the number C = 281, D = 201. Find the highest natural number S so that the C:S, D:S are with the remainder of 1,
• Sponsor
The children of the tennis school received 64 white and 48 yellow balls from the sponsor. When asked about how many balls they could take, they were answered: "You have so many that none of you will have more than 10 balls and all will have the same numbe
• Seven numbers
Write seven 4-digit numbers that are divisible by 3 and at the same time by 4.
• LCD 2
The least common denominator of 2/5, 1/2, and 3/4
• Four poplars
Four poplars are growing along the way. The distances between them are 35 m, 14 m, and 91 m. At least how many poplars need to be dropped to create the same spacing between the trees? How many meters will it be?
• Garden
The garden has the shape of a rectangle measuring 19m20cm and 21m60cm. Mr. Novák will fence it. It wants the distance between adjacent pillars to be at least two meters and a maximum of three meters. He would also like the distances between the adjacent p
• Ornamental shrubs
Children committed to plant 240 ornamental shrubs. Their commitment however exceeded by 48 shrubs. Write ratio of actually planted shrubs and commitment by lowest possible integers a/b.
Do you have an interesting mathematical word problem that you can't solve it? Submit a math problem, and we can try to solve it.
We will send a solution to your e-mail address. Solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.
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Do you want to calculate greatest common divisor two or more numbers? | HuggingFaceTB/finemath | |
## [ITOC-ch6] ADVANCED TOPICS IN COMPUTABILITY THEORY
### 6.1 THE RECURSION THEOREM
Making machines that reproduce themselves.
SELF-REFERENCE
Construct $SELF$ prints out a copy of its own description.
LEMMA 6.1
define computable function $q : Σ^ ∗\longrightarrow Σ ^ ∗$
• w is any string, $q(w) = < P_w >$
• $P_w$ is a TM that prints out w and then halts.
$SELF$
• 2 parts: $A$ and $B$, so $< SELF > = < AB >$
• RZ师兄的例子
Extends the technique so that a program can automatically obtain its own description and then go on to compute with it
Recursion theorem
• First we have a T, than we can construct R
• Input w to R, R can obtain < R > and give < R > and w to T for further computation $A$ writes $w < BT >$ , $B$ writes $< A >$ , together the resulting string $< R, w >$ is on the tape, than passes control to T APPLICATIONS
• computer virus
• Simpler proof of $A_{TM}$ is undecidable (4.11显式构建了D出来)
• $MIN_{TM}$ is not Turing-recognizable.
• For computable function $t: Σ ^ ∗ → Σ^ ∗$, there is a $< F > = t( < F > )$
### 6.2 DECIDABILITY OF LOGICAL THEORIES
model
• a tuple $(U, P_1 , . . . , P_k )$, where U is the universe ( possible variable values ) and $P_1$ through $P_k$ are the relations assigned to symbols $R_1$ through $R_k$.
language of a model
• use things only the model defined theory of M $Th(M)$
• the collection of true sentences in the language of that model
6.12 $Th(N , +)$ is decidable.
6.13 $Th(N , +, ×)$ is undecidable.
6.16 Some true statement in $Th(N , +, ×)$ is not provable.
(真假性客观存在但不可证明)
### 6.3 TURING REDUCIBILITY
Reducibility
• if A is reducible to B, and we find a solution to B, we can obtain a solution to A.
Mapping reducibility
• $\exists f$ so $w \in A \Longleftrightarrow f(w) \in B$
Turing reducibility
oracle for a language
• an external device that can report whether a string w is a member of the language
oracle Turing Machine
• TM that has the additional capability of querying an oracle.
• more powerful than TM
• Still cannot decide all languages
$E_{TM}$ is decidable relative to $A_{TM}$
• we can use the oracle of $A_{TM}$ to decide $E_{TM}$ ### 6.4 A DEFINITION OF INFORMATION
define information using computability theory.
quantity of information
• the size of that object’s smallest representation or description.
MINIMAL LENGTH DESCRIPTIONS
Select our language describing information
• describe a binary string x as $< M > w$ while M is a TM and w is input
• locate the separation between $< M >$ and $w$
• One way to do so is to write each bit of $< M >$ twice, writing 0 as 00 and 1 as 11, and then follow it with 01 to mark the separation point. 6.24: $\exists c \forall x \ [ K(x) \le |x| + c ]$
6.25: $\exists c \forall x \ [ K(xx) \le K(x) + c ]$
6.26: $\exists c \forall x,y \ [ K(xy) \le 2K(x) + K(y) + c ]$ | HuggingFaceTB/finemath | |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A053567 Stirling numbers of first kind, s(n+5, n). 8
-120, 1764, -13132, 67284, -269325, 902055, -2637558, 6926634, -16669653, 37312275, -78558480, 156952432, -299650806, 549789282, -973941900, 1672280820, -2792167686, 4546047198, -7234669596, 11276842500, -17247104875, 25922927745, -38343278610, 55880640270 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 REFERENCES M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..200 M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. G. C. Greubel, A Note on Jain basis functions, arXiv:1612.09385 [math.CA], 2016. FORMULA a(n) = (-1)^n*binomial(n+5, 6)*binomial(n+5, 2)*(3*n^2 + 23*n + 38)/8. G.f.: x*(120 + 444*x + 328*x^2 + 52*x^3 + x^4)/(1-x)^11. See row k=4 of triangle A112007 for the coefficients. E.g.f. with offset 5: exp(x)*(Sum_{m=0..5} A112486(5, m)*(x^(5+m)/(5+m)!). a(n) = binomial(n+5, 6)*binomial(n+5, 2)*(3*n^2 + 23*n + 38)/8. From the g.f. a(n) = (f(n+4, 5)/10!)*Sum_{m=0..min(5, n-1)} A112486(5, m)*f(10, 5-m)*f(n-1, m)), with the falling factorials f(n, m):=n*(n-1)*, ..., *(n-(m-1)). From the e.g.f. MAPLE A053567 := proc(n) (-1)^(n+1)*combinat[stirling1](n+5, n) ; end proc: # R. J. Mathar, Jun 08 2011 MATHEMATICA Table[StirlingS1[n+5, n](-1)^(n-1), {n, 30}] (* Harvey P. Dale, Sep 21 2011 *) PROG (Sage) [stirling_number1(n, n-5)*(-1)^(n+1) for n in xrange(6, 26)] # Zerinvary Lajos, May 16 2009 (MAGMA) [(-1)^n*Binomial(n+5, 6)*Binomial(n+5, 2)*(3*n^2+23*n+38)/8: n in [1..30]]; // Vincenzo Librandi, Jun 09 2011 (PARI) a(n) = (-1)^(n-1)*stirling(n+5, n, 1); \\ Michel Marcus, Aug 29 2017 CROSSREFS Next |Stirling1| diagonal A112002, 5th diagonal of A130534. Sequence in context: A027795 A223427 A282899 * A056270 A001118 A052767 Adjacent sequences: A053564 A053565 A053566 * A053568 A053569 A053570 KEYWORD easy,sign,changed AUTHOR Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 17 2000 EXTENSIONS Definition edited by Eric M. Schmidt, Aug 29 2017 STATUS approved
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# Electric field in a hollow cylinder
• magnifik
#### magnifik
An infinitely long thick hollow cylinder has inner radius Rin and outer radius Rout. It has a non-uniform volume charge density, ρ(r) = ρ0r/Rout where r is the distance from the cylinder axis. What is the electric field magnitude as a function of r, for Rin < r < Rout?
for this problem, when you find qinside, do you integrate from Rin to r or from Rin to Rout? I'm confused because i would have expected it to be the latter, but in the solutions they integrate from Rin to r. can someone please explain this?
also, if you try to find the e-field where r > Rout, do you integrate from r to Rout?
Solution is here (problem II):
http://www.physics.gatech.edu/~em92/Classes/Fall2011/2212GHJ/main/quiz_help/200908/q2s.pdf
Last edited by a moderator:
Just use Gauss' theorem. The surface has radius r, and
q(inside) is whatever's inside!
rude man said:
Just use Gauss' theorem. The surface has radius r, and
q(inside) is whatever's inside!
since in the example in the document it asks for Rin < r < Rout.. why does it integrate from Rout to r??
It doesn't. It integrates from Rin to r.
rude man said:
It doesn't. It integrates from Rin to r.
but why not Rin to Rout?
Because ity asks for the field at Rin < r < Rout, not AT Rout.
## 1. What is an electric field in a hollow cylinder?
The electric field in a hollow cylinder refers to the distribution of electric charges within a cylindrical shell. This electric field is typically measured in units of volts per meter (V/m) and is perpendicular to the surface of the cylinder at all points.
## 2. How is the electric field in a hollow cylinder calculated?
The electric field in a hollow cylinder can be calculated using the formula E = Q/(2πε₀L), where Q is the charge enclosed within the cylinder, ε₀ is the permittivity of free space, and L is the length of the cylinder. This formula is derived from Gauss's law.
## 3. What factors influence the strength of the electric field in a hollow cylinder?
The strength of the electric field in a hollow cylinder is influenced by the charge enclosed within the cylinder, the permittivity of the material inside the cylinder, and the length of the cylinder. The electric field also decreases as the distance from the center of the cylinder increases.
## 4. Can the electric field in a hollow cylinder be negative?
No, the electric field in a hollow cylinder cannot be negative. This is because the electric field is a vector quantity and is always directed away from positive charges and towards negative charges. Since a hollow cylinder contains no net charge, the electric field will always be positive or zero.
## 5. What are some real-world applications of the electric field in a hollow cylinder?
The electric field in a hollow cylinder has various applications in different fields of science and technology. For example, it is used in the design of capacitors for storing and regulating electrical energy. It is also used in medical imaging techniques such as magnetic resonance imaging (MRI) to create detailed images of the human body. Additionally, the electric field in a hollow cylinder is utilized in particle accelerators to generate and manipulate charged particles for scientific research. | HuggingFaceTB/finemath | |
## Exchange Rates
Posted: January 18, 2011 in Uncategorized
Tags: , , , Talking about proportions in my Pre-Pre-Algebra class (called Integrated Math). We got on the topic of exchange rates. A student asked how you can make money through exchange rates. I’m a big fan of these Google graphs. Lots of questions to ask. We assumed we invested \$1000 in Euros and exchanged them again to see if we made profits. Lots of questions you can explore here…. also a very scalable problem for higher classes.
## Metrodome Collapses
Posted: December 17, 2010 in Uncategorized
Tags: ,
River Falls got around 20 inches of snow over the weekend and it caused some havoc in our area. The most popular story being the Metrodome’s roof collapsing. Here is the video:
The student’s came up with a quick list of questions: … The best question being “Why did Brett do it?”. Just Kidding. The real question we wanted to pursue was “How much weight did the snow put on the roof?”. I printed off a few resources for the students to use as they came up with their estimations:
The Storm
Metrodome Statistics
Snow to Liquid
Weight of Water
Our Solution Our estimate was about 3,852,429.412 pounds.
Varying the snow density
Let d = the number of inches of snow it takes to make one inch of water $w = \frac{17 in}{d} \cdot \frac{1ft}{12in} \cdot \frac{435600ft^2 }{1} \cdot \frac{62.42796 lbs}{1 ft^3}$ $w = \frac{38524294.12}{d}$ More questions to consider
– How was the snow distributed on the dome?
– What type of shape is the roof?
– What was the exact snow to water ratio?
– How does temperature relate to snow density?
Conclusion
I like this video because it’s relevant, fun to watch, and stirs up a lot of questions. I think we just scratched the surface on some of the math you could do with this video. We covered: conversions, volume, area, estimation, ratios, density, rational functions. I think this would also work great in a geometry class or lower. | HuggingFaceTB/finemath | |
I’ve spent a lot of time in the past describing the paradoxes that arise when we try to involve infinities into our reasoning. Another way to get paradoxes aplenty is by invoking self-reference. Below are three of the best paradoxes of self-reference for you to sort out.
In each case, I want you to avoid the temptation to just say “Ah, it’s just self-reference that’s causing the problem” and feel that the paradox is thus resolved. After all, there are plenty of benign cases of self-reference. Self-modifying code, flip-flops in computing hardware, breaking the fourth wall in movies, and feeding a function as input to itself are all examples. Self-referential definitions in mathematics are also often unobjectionable: as an example, we can define the min function by saying y = min(X) iff y is in X and for all elements x of X, y ≤ x (the definition quantifies over a group of objects that includes the thing being defined). So if we accept that self-reference is not necessarily paradoxical (just as infinity is sometimes entirely benign), then we must do more to resolve the below paradoxes than just say “self-reference.”
Consider the set of integers definable in an English sentence of under eighty letters. This set is finite, because there are only a finite number of possible strings of English characters of under eighty letters. So since this set is finite and there are an infinity of integers, there must be a smallest integer that’s not in the set.
But hold on: “The smallest positive integer not definable in under eighty letters” appears to now define this integer, and it does so with only 67 letters! So now it appears that there is no smallest positive integer not definable in under eighty letters. And that means that our set cannot be finite! But of course, the cardinality of the set of strings cannot be less than the cardinality of the set of numbers those strings describe. So what’s going on here?
“If this sentence is true, then time is infinite.”
Curry’s paradox tells us that just from the existence of this sentence (assuming nothing about its truth value), we can prove that time is infinite.
## Proof 1
Let’s call this sentence P. We can then rewrite P as “If P is true, then time is infinite.” Now, let’s suppose that the sentence P is true. That means the following:
Under the supposition that P is true, it’s true that “If P is true, then time is infinite.”
And under the supposition that P is true, P is true.
So under the supposition that P is true, time is infinite (by modus ponens within the supposition).
But this conclusion we’ve just reached is just the same thing as P itself! So we’ve proven that P is true.
And therefore, since P is true and “If P is true, then time is infinite” is true, time must be infinite!
If you’re suspicious of this proof, here’s another:
## Proof 2
If P is false, then it’s false that “If P is true then time is infinite.” But the only way that sentence can be false is if the antecedent is true and the consequent false, i.e. P is true and time is finite. So from P’s falsity, we’ve concluded P’s truth. Contradiction, so P must be true.
Now, if P is true, then it’s true that “If P is true, then time is infinite”. But then by modus ponens, time must be infinite.
Nothing in our argument relied on time being infinite or finite, so we could just as easily substitute “every number is prime” for “time is infinite”, or anything we wanted. And so it appears that we’ve found a way to prove the truth any sentence! Importantly, our conclusion doesn’t rest on the assumption of the truth of the sentence we started with! All it requires is the *existence of the sentence*. Is this a proof of the inconsistency of propositional logic? And if not, then where exactly have we gone wrong?
Consider the following two sentences:
1) At least one of these two sentences is false.
2) Not all numbers are prime.
Suppose that (1) is false. Well then at least one of the two sentences is false, which makes (1) true! This is a contradiction, so (1) must be true.
Since (1) is true, at least one of the two sentences must be false. But since we already know that (1) is true, (2) must be false. Which means that all numbers are prime!
Just like last time, the structure of the argument is identical no matter what we put in place of premise 2, so we’ve found a way to disprove any statement! And again, we didn’t need to start out by assuming anything about the truth values of sentences (1) and (2), besides that they have truth values.
Perhaps the right thing to say, then, is that we cannot always be sure that self-referential statements actually have truth values. But then we have to answer the question of how we are to distinguish between self-referential statements that are truth-apt and those that are not! And that seems very non-trivial. Consider the following slight modification:
1) Both of these two sentences are true.
2) Not all numbers are prime.
Now we can just say that both (1) and (2) are true, and there’s no problem! And this seems quite reasonable; (1) is certainly a meaningful sentence, and it seems clear what the conditions for its truth would be. So what’s the difference in the case of our original example? | HuggingFaceTB/finemath | |
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# OCR GCSE Maths Arithmetic sequences An arithmetic or linear sequence, an, is an ordered list of n numbers where the difference between each consecutive term, d, is constant. An arithmetic sequence can always be described by the general rule: The term to term rule of an arithmetic sequence describes how to get from one term to the next in the sequence. To find the term to term rule, subtract the first term from the second term to find the common first difference term. From the common first difference term and the first term the position to term rule, or general rule, can be worked out with the formula: The general rule of an arithmetic sequence can be used to work out the nth term of the sequence, the term at position n, and is given by the formula: A series is the sum of the terms of a sequence up to a certain number of terms and it is denoted by Sn. A series for an arithmetic sequence is given by the formula: # ✅
What is the common difference term in the arithmetic sequence {Un} = {1, 17, 33, 49, 65, …, Un}? # ✅
Katy notices that every six months her savings double in value. Is the sequence of savings at every six months period an arithmetic sequence? # ✅
Given the arithmetic sequence {an} = {1, 3, 5, 7, 9, 11, …, an}, find the common first difference term. # ✅
Given the first 7 terms of a sequence are {72, 84, 96, 108, 120, 132, 144}, what is the position to term rule, or general rule, of the sequence? # ✅
Given the first three terms of a sequence are {24, 20, 16}, calculate the seventh term of the sequence. # ✅
Define the nth term of the sequence {an} = {1, 3, 5, 7, 9, …, an}. # ✅
Calculate the first 5 terms of the sequence an = 3 – n. # ✅
Jack notices that when he releases a red ball from a one-meter height, the number of bounces of the tennis ball increases by 2 with each 5 second interval. Given that during the first 5 second interval the ball bounced twice, what is the nth term of the sequence of bounces. Have you found the questions useful? | HuggingFaceTB/finemath | |
## Algebra 1
$k= -20$
Original problem: $\frac{8}{k}$ = -$\frac{12}{30}$ In this problem, the variable we are looking for is k. In order to isolate and solve for k, we must cross multiply. The first step in this process can be done by multiplying the numerator of the left side of the fraction, 8, with the denominator of the right side of the fraction, 30: $8 * 30 = 240$ Now, we must equate 240 with the product of the other part of the original equation. $240 = k * -12$ In order to fully isolate k, divide the right side by -12 to cancel it out. $\frac{240}{-12}$ = $k$ Finally, solve to find the value of k. $k = -20$ | HuggingFaceTB/finemath | |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.1: Investigating Static Forces in Nature: The Mystery of the Gecko
Difficulty Level: At Grade Created by: CK-12
Student Learning Objectives:
• Explain that a net force of zero or greater is necessary for objects to adhere to a surface (wall or ceiling)
• Identify different variables and the constants that affect adhesive forces
• Explain how the amount of adhesion changes when the conditions of the surfaces change
Note: Some questions in the Student Journal are underlined as formative assessment checkpoints for you to check students’ understanding of lesson objectives.
At a Glance for Teachers:
• Review what students know about forces
• Teacher demonstration on balanced forces
• Determine the amount of force needed for objects of varying masses to adhere to a ceiling and maintain a net force of zero
• Activity: Tape Pull—Measure the amount of force required to remove a piece of transparent tape with varying amounts of dirt
Estimated Time: 80Minutes\begin{align*}80 \;\mathrm{Minutes}\end{align*}
Vocabulary: Adhere, Adhesive, Balanced Forces, Dependent Variable, Force, Independent Variable, Mass, Net Force, Newton, Unbalanced Force, Volume
Refer to the end of this Teacher Guide for definitions.
Materials:
• PowerPoint for Lesson 6
• Student Journals for Lesson 6
• Computer with LCD or overhead projector
• Duct tape
• 50N\begin{align*}50\;\mathrm{N}\end{align*} spring scale
• Transparent tape
• Hole punch
• Ruler, protractor
Safety Note
Have students wear safety goggles in accordance with district safety policy.
Slide #
Student Journal Page #
Teacher Background Information and Pedagogy
“Teacher Script”
Slide 1
Title
Student Journal Page: 6–1
1. Review with students that a force is a push or pull. See definitions in Appendix B.
“What is the meaning of the word force in science?”
2. Demonstrate balanced forces by partially filling a small jar with water. Place an index card underneath the rim of the jar and invert the jar while holding the index card. Next, release your hand from the card while carefully holding the jar. Students should note that the card remains in contact with the rim of the jar. Have students identify the balanced forces at work in this demonstration. Answer: air pressure is the dominate force, it is equal to the weight of the water plus the force of gravity. Other forces that are present but less dominate are gravity and capillary wet adhesion (if the card got wet when it came in contact with the rim of the glass).
“Is this demonstrating a balanced or unbalanced force? Why?”
“What would happen if this was an unbalanced force?”
3. Have students look around the room and identify pairs of objects that are at rest and represent balanced forces and record these in the box on the left side of the student journal. Students should identify the forces acting on each object and that the net force is zero. Have students draw one of the examples of balanced forces and indicate the amount of each force acting on the object using arrows in Student Journal page 6–1. Have students repeat this exercise for unbalanced forces in the box on the right side of the student journal. Note to teacher: if the object sits on a table, there is the upward normal force of the table on the object. Research has shown that students often don’t recognize this as a force, they just indicate the table is in the way.]
4. Provide examples of objects in motion such as objects speeding up, slowing down, or at constant speed.
A field test teacher used a Frayer model for balanced and unbalanced forces for this lesson (refer to Lesson 1 for directions on the use of the Frayer model).
Slide 2
Student Journal Page: 6-1
5. Display Slide 2
“In this image, there are two forces at work: one that is holding the shoe onto the ceiling and another that is pulling the shoe towards the floor. In order for the shoe to remain on the ceiling, what must be true about these two forces?”
Students should state that these represent balanced forces (i.e., the net force is zero), or that the force holding a shoe is greater than the force of gravity. An example of the latter would be if the shoe contained a magnet that was attracted to a steel ceiling. This force could be greater than gravity, and then there would be an additional normal force acting in the direction of gravity to counteract the excess magnetic force.
Slide 3
Student Journal Page: 6–2
Calculations In Appendix A
“Imagine an ant, like the one on this slide, walking on the ceiling. Draw a picture representing the forces of the ant on the ceiling in your journal. Determine the force required for each ant foot (divide total force by six).”
Explain the following assumptions that are important for this problem:
“We are assuming in this problem that the total force required is equally divided among the six ant feet, and that ONLY the contact between feet and ceiling gives rise to the force.”
The weight of the ant is provided in Newtons (N), a derived unit which is the force needed to increase the speed of (or accelerate) one kilogram of mass one meter per second every second.
A field test teacher passed around objects (e.g., a one Newton weight, an eight Newton cell phone) for students to be able to relate to this unit of measure.
For this module, there is no need to calculate force with Newton’s Second Law of Motion. However, there may be a need to explain how an object’s weight can be expressed in Newtons. Explain that in the metric system forces are measured in units of Newtons (using the symbol “N”). Provide students with the definition found in Appendix B along with the following illustration. Use these along with the direct vocabulary instruction strategy as described in the preface. Weight is action of the force of gravity on an object. A standard kilogram mass would therefore have a weight of 9.8 Newtons on Earth since the acceleration due to gravity is 9.8 m/s/s.
6. Point out to students that the weight is the minimum amount of force that must be provided by the feet of the ant on the ceiling in order for there to be balanced forces and thus have the ant adhere to the ceiling. See Appendix A for the answers.
Note: During the pilot test, students thought this activity was interesting. The calculations took a bit to understand, and it was valuable to review unit conversions. Use Appendix A to assist students in solving the first problem.
Slide 4
Student Journal Page: 6–3
7. Display slide 4.
“Repeat the calculation—this time for an imaginary object that is larger in every dimension and whose mass and volume is ten times larger.”
Determine how many “ant feet” it would take for this imaginary object to remain adhered to the ceiling. Compare and discuss the difference between the two calculations in class. See Appendix A for calculations.
Teacher Demonstration:
Optional: One pilot teacher added a calculation for a two-ton elephant as well. Actual weight for an African male elephant in Newtons is 122,580 Newtons. Refer to optional notes in Slide 5.
Slide 5
Student Journal Pages: 6–3 6–4
“Let’s return our attention to the gecko.
Repeat your calculations from the imaginary animal for the Tokay Gecko, which has an average weight of 2.2 Newtons.”
8. Have students write a statement and/or draw pictures that describe the relationship between size (mass) and weight and, therefore, the adhesive forces required for an animal to remain on a ceiling.
Slide 6
9. Explain to students that they will be using the following terms in this lesson.
“Adhere describes how something sticks to something else.
Separation force is the amount of pull that is required to detach two objects.”
Slide 7
“What are the tools that we can use in the laboratory to measure the amount of force that an object exerts? What are the units used when measuring with this tool?”
Forces can be measured with a spring scale that changes when a force is applied. Forces are measured in Newtons (N).
Slide 8
Student Journal Page: 6-4
“As you have observed a gecko adhering to a wall, you may have wondered about the types of surfaces that are required to accomplish this feat.
Can the gecko adhere to any surface?
Does the surface need to be clean or can the gecko adhere to dirty surfaces too?
What if the surface is wet? Will that affect how well the gecko can adhere?
To better understand how the gecko can adhere to different surfaces, we will be exploring the forces involved in the adhesion of transparent tape on a table top.”
10. Tell students that over the next day or so, they will be able to refine this question based on how they set up their experiment.
11. Prior to showing slide 9 introduce the Tape Pull activity by having the students answer the question in their journal on page 6–4.
Slide 9
Student Journal Pages: 6–5 6–6
“You will be working with transparent tape on the tabletop and measuring the force required to remove the tape with different amounts of dirt. This force, as stated previously, is actually GREATER THAN the adhesive force.”
12. Before beginning the experiment have students work with the materials and practice the tape pull procedure as described on Student Journal page 6–5.
“Write down the independent variable (manipulated variable) and the dependent variable (responding variable).” Allow students to identify the amount of dirt as the independent variable and the force that it takes to remove or break the adhesion as the dependent variable.
Optional: Use the “sticky hands” toy (the one that initially sticks to glass then slowly falls/rolls down the glass) as a demonstration of dirt’s effect. This toy’s ability to stick decreases rapidly when it becomes dirty.
13. Hold a discussion about how to vary the “amount of dirt.” For starters, students could test fresh (never before used) tape. Then, rather than adding dirt to the tape, students could make a finger print on the tape and test its adhesion. Other ideas: drop chalk dust onto tape and blow it off, touch the tape to the floor, etc. This then becomes the operational definition for the independent variable.
14. Hold a class discussion about how to keep certain variables constant, such as the amount of surface area in which there is contact between surfaces and the angle of pull. Based on this discussion, students should write a research question and a hypothesis before completing the activity. An example of a research question is given on this slide.
Slide 10
Student Journal Pages: 6–6 6–7 6–8 6–9
“On this slide, you see how the materials are set up for the experiment. Image 6.8 shows a piece of tape on a table. The end of the tape that is pulled is reinforced with some electrical tape that has a hole punched through it. The hook end of the spring scale is then placed through the hole. Image 6.9 shows the spring scale being pulled at an angle (make sure this is the same each time). During the pull, a second student should carefully observe the force readings on the spring scale.”
15. Allow students time to complete the activity as shown in the journal. As students are completing the procedure, make sure they refine their initial question and use their findings in order to provide explanations and further questions.
Student Journal pages 6–5 through 6–8 can be completed for homework and graded for use as a formative assessment.
Classroom Management Tip: One pilot teacher assigned jobs for the experiment:
• Tape handler and assembly
• Measurer
• Equipment Manager
16. After students are done with the experiment, have them answer the questions in their journal on page 6–9 and 6–10.
Question 7:
Describe how you made your observations in today’s lesson.
a. “What tools did you use?” (spring scale)
b. “Were your observations at the visible or invisible scale?” (invisible)
c. “What is the dominant force at this scale?” (adhesive force/unknown)
Slide 11
“What do you know about the effectiveness of transparent tape underwater, and how tape gets dirty over time? (Display slide 11) This is a quote from researcher Kellar Autumn, Assistant Professor of biology at Lewis & Clark College, about the self-cleaning ability of the gecko.”
Students may state that when transparent tape is placed underwater, it will eventually lose its adhesiveness. Likewise, transparent tape does not work well on dirty surfaces.
17. It should be noted that ants leave a residue behind as they walk, whereas geckos do not.
18. Draw students’ attention to the note on the slide about the gecko adhesion working underwater.
Optional: Students could test other variables: amount of tape contact area, cleanliness of the surface, etc.
Slide 12
19. As a culminating class discussion, ask students to respond to the questions in “Making Connections.”
“Let’s review.
1. Describe one or two ideas that you learned during this lesson.
2. What factors contribute to the amount of force to remove a sticky substance?
3. How does dirt affect adhesion?
4. Do you think that a sticky substance is a possible method for the gecko adhesion?
5. How do you think the gecko sticks to the ceiling?
6. What should we explore next?”
Slide 13
20. The pilot-test teachers highly recommend using this flow chart at the end and/or beginning of each lesson. The end of each lesson contains this flow chart that provides an opportunity to show students the “big picture” and where they are in the lesson sequence. The following color code is used:
Yellow: Past Lessons
Blue: Current Lesson
Green: Next Lesson
White: Future Lesson
Appendix A: Calculations and Possible Responses to Accompany PowerPoint Slides
Slide 3 Calculations
Ant
Weight of Ant=0.00004Newtons\begin{align*}\mathrm{Weight\ of\ Ant} = 0.00004 \;\mathrm{Newtons}\end{align*} or 4×105Newtons\begin{align*}4 \times 10^{-5} \;\mathrm{Newtons}\end{align*}
Weight of Ant/6Ant Feet=Force for each foot=0.0000067Newtons\begin{align*}\mathrm{Weight\ of\ Ant}/6 \;\mathrm{Ant\ Feet} = \mathrm{Force\ for\ each\ foot} = 0.0000067 \;\mathrm{Newtons}\end{align*} per Ant Foot or 6.7×106Newtons\begin{align*}6.7 \times 10^{-6}\;\mathrm{Newtons}\end{align*} per Ant Foot
Slide 4 Calculations
Ant Mass Times 10times\begin{align*}10 \;\mathrm{times}\end{align*} what it was before
Then IF the Ant Foot can ONLY support 6.7×106N\begin{align*}6.7 \times 10^{-6}\;\mathrm{N}\end{align*}, how many ant feet would be required?
Weight of Imaginary object / Force for each Ant Foot
4×104Newtons/6.7×106Newtons per Ant Foot\begin{align*}4 \times 10^{-4} \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
59.7ant feet=60ant feet\begin{align*}59.7 \;\mathrm{ant\ feet} = 60 \;\mathrm{ant\ feet}\end{align*}
Slide 5 Calculations
Gecko
Weight of Gecko=2.2Newtons\begin{align*}\mathrm{Weight\ of\ Gecko} = 2.2 \;\mathrm{Newtons}\end{align*}
2.2Newtons/6.7×106Newtons per Ant Foot\begin{align*}2.2 \;\mathrm{Newtons}/6.7 \times 10^{-6}\;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
328,358ant feet\begin{align*}328,358 \;\mathrm{ant\ feet}\end{align*}
From Liang, Autumn, Hsieh, Zesch, Chan, Fearing, Full, Kenny5\begin{align*}^5\end{align*}:
43.4N\begin{align*}43.4 \;\mathrm{N}\end{align*} average sustained clinging force of gecko with 227.1mm2\begin{align*}227.1 \;\mathrm{mm}^2\end{align*} pad area
5\begin{align*}^5\end{align*} Autumn, K., Liang, Y. A., Hsieh, S. T., Zesch, W., Chan, W. P., Kenny, T. W., Fearing, R., & Full, R. J. (2000). Adhesive force of a single gecko foot-hair.
Nature, 405, 681-684.
Slide 5 Calculations (Optional)
200lb\begin{align*}200 \;\mathrm{lb}\end{align*} adult
Weight of adult in Newtons=888.9Newtons\begin{align*}\mathrm{Weight\ of\ adult\ in\ Newtons} = 888.9 \;\mathrm{Newtons}\end{align*}
888.9Newtons/6.7×106Newtons per Ant Foot\begin{align*}888.9 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
32,671,642ant feet\begin{align*}32,671,642 \;\mathrm{ant\ feet}\end{align*}
27,000lb\begin{align*}27,000 \;\mathrm{lb}\end{align*} African male elephant
Weight of elephant in Newtons=122,580Newtons\begin{align*}\mathrm{Weight\ of\ elephant\ in\ Newtons} = 122,580 \;\mathrm{Newtons}\end{align*}
122,580Newtons/6.7×106Newtons per Ant Foot\begin{align*}122,580 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
18,295,522,390ant feet\begin{align*}18,295,522,390 \;\mathrm{ant\ feet}\end{align*}
Appendix B: NanoLeap Physical Science Vocabulary
Adhere
1. To hold fast or to stick
2. To bind to
Adhesive
A substance that helps objects stick together
Balanced Forces
For each force acting on a body, there is another force on the same body equal in magnitude and opposite in direction. A body is said to be at rest if it is being acted on by balanced forces.
Dependent Variable
A factor or condition that might be affected as a result of a change in the independent variable (also called a responding variable)
Force
1. Energy exerted
2. A push or a pull that acts on an object
Independent Variable
A factor or condition that is intentionally changed by an investigator or experiment to explore its effects on other factors (also called a manipulated variable)
Mass
1. A quantity of matter
2. A measurement of the quantity
Net Force
The resultant non-zero force due to an unbalanced force
Newton
A unit of force needed to change the speed of a kilogram of mass by one meter per second for every second that the force is acting on the mass
Unbalanced
When there is an individual force that is not being balanced by a force of equal magnitude and in the opposite direction. A body is said to be in motion if acted upon by unbalanced forces.
Volume
The amount of space occupied by a three-dimensional object (Length times Width times Height for a rectangular object)
Investigating Static Forces in Nature: The Mystery of the Gecko
Lesson 6: How MUCH Force Is Needed to Make an Object Stick?
Teacher Guide
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# Perpendicular Line Formula
## Perpendicular Line Formula
The topics connected to one another are numerous in subjects like Mathematics, where problem-solving is just as vital a component as theory, and are therefore divided into multiple tasks in the textbook so that the students may better understand it. To aid students in understanding and applying the theory, each chapter contains a variety of exercises with various types of problems.
The resources offered at Extramarks are created in a similar way to how individual topics are studied independently so that students may use them for both their homework and test preparation. Subjects like the Perpendicular Line Formula and many others are covered in these resources.
Teachers frequently assign a portion of the exercises’ problems as homework so that students can study and practise on their own since Mathematics requires a lot of practise. In these situations, tools like the one’s for the Perpendicular Line Formula , etc., may be quite useful. Students could encounter particular difficulties while attempting to do their schoolwork. In these circumstances, they could ask for assistance from other students and occasionally repeat the solutions. In order to better understand how to answer the problem, it is advised that they use the solutions for a number of topics, such as the Perpendicular Line Formula , etc., that are available on the Extramarks website.
What Is Perpendicular Lines Formula?
Students are better able to understand the ideas covered in this particular topic of the Perpendicular Line Formula as a result of the materials made available. The significance of comprehending the theoretical foundations of each chapter and topic should be made clear to students. Students will struggle to finish their assignments, pass their quizzes, and score well on their final exams if they don’t grasp the fundamentals of the courses. Students can work through step-by-step, expert-prepared solutions using Extramarks resources, such as those for the topic Perpendicular Line Formula . This aids students in comprehending the subject of the Perpendicular Line Formula as well as the right answer to the query.
A student’s chances of doing well on the test rise if they attempt to solve the problem step by step. Experts at Extramarks have developed resources for the Perpendicular Line Formula and other related issues with the needs of students in mind. Students benefit much from it as a result.
### Solved Examples Using Perpendicular Lines Formula
Many of the resources in the Perpendicular Line Formula are produced by Extramarks’ Mathematics specialists. Any student who requires them can easily obtain all of the solutions and other study materials, including those for the Perpendicular Line Formula , on the Extramarks website.
Students may readily access the Perpendicular Line Formula tools as well as other helpful tools on the Extramarks website.
Even though there are several subtopics in Mathematics, students might find the subject enjoyable if they learn the fundamentals early on. Students can more easily understand the fundamental ideas behind the Perpendicular Line Formula by making use of the many materials available on the Extramarks website. The task of creating user-friendly solutions that relieve students of the stress associated with exam preparation falls to subject matter experts selected by Extramarks. Students may find other resources helpful for enhancing their Mathematics skills in addition to the one’s for the Perpendicular Line Formula . | HuggingFaceTB/finemath | |
# If $$({\cos ^2}θ - 1)(2{\sec ^2}θ ) + {\sec ^2}θ + 2{\tan ^2}θ = 2$$, 0º < θ < 90º, then $$\frac{{\left( {\cos ec\theta + \cos \theta } \right)}}{{\left( {\cos ec\theta - \cos \theta } \right)}}$$ is equal to:
Free Practice With Testbook Mock Tests
## Options:
1. 1
2. $$4\sqrt 2$$
3. $$2\sqrt 2$$
4. 3
### Correct Answer: Option 4 (Solution Below)
This question was previously asked in
SSC Graduation Level Previous Paper (Held on: 10 Nov 2020 shift 3)
## Solution:
Formula Used: Sin2θ + Cos2θ = 1, Sec2θ = Tan2θ + 1
Calculation:
If (cos2θ – 1)(2sec2θ) + sec2θ + 2tan2θ = 2, 0° < θ < 90°
⇒ – (1 – cos2θ)(2sec2θ) + sec2θ + 2tan2θ = 2
⇒ – (sin2θ)(2sec2θ) + tan2θ + 1 + 2tan2θ = 2 (By using above formulas)
⇒ – 2(sin2θ)/(cos2θ) + 3tan2θ = 2 – 1
⇒ – 2tan2θ + 3tan2θ = 1 or tan2θ = 1
⇒ tanθ = 1, θ = 45°
Now, as given in the question
⇒ (cosecθ + cosθ)/(cosecθ – cosθ) = (cosec45° + cos45°)/(cosec45° - cos45°)
⇒ (√2 + 1/√2)(√2 – 1/√2) = (2 + 1)/(2 – 1) = 3
∴ The value is 3 | HuggingFaceTB/finemath | |
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Homework 10
# Homework 10 - vu(tv2894 Homework 10(Quest miner(55096 This...
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vu (tv2894) – Homework 10 (Quest) – miner – (55096) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find dy/dx when 2 x 3 + y 3 - 9 xy - 1 = 0 . 1. dy dx = 2 x 2 + 3 y y 2 - 3 x 2. dy dx = 2 x 2 - 3 y y 2 + 3 x 3. dy dx = 3 y + 2 x 2 y 2 + 3 x 4. dy dx = 2 x 2 - 3 y y 2 - 3 x 5. dy dx = 3 y - 2 x 2 y 2 - 3 x correct Explanation: We use implicit differentiation. For then 6 x 2 + 3 y 2 dy dx - 9 y - 9 x dy dx = 0 , which after solving for dy/dx and taking out the common factor 3 gives 3 parenleftBig (2 x 2 - 3 y ) + dy dx ( y 2 - 3 x ) parenrightBig = 0 . Consequently, dy dx = 3 y - 2 x 2 y 2 - 3 x . keywords: implicit differentiation, Folium of Descartes, derivative, 002 10.0 points Find dy/dx when y - x = 2 xy . 1. dy dx = radicalBig y x + 1 2 + radicalBig x y 2. dy dx = radicalBig y x + 1 1 - radicalBig x y correct 3. dy dx = radicalBig y x - 1 1 - radicalBig x y 4. dy dx = radicalBig y x + 1 1 + radicalBig x y 5.
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# Mean Value Theorem Calculator
Write down function and intervals in designated fields. The calculator will find its changing rate by using the mean value theorem.
An online mean value theorem calculator helps you to find the rate of change of the function using the mean value theorem. Also, this Rolle’s Theorem calculator displays the derivation of the intervals of a given function. In this context, you can understand the mean value theorem and its special case which is known as Rolle’s Theorem.
## What is the Mean Value Theorem?
In mathematics, the mean value theorem is used to evaluate the behavior of a function. The mean value theorem asserts that if the f is a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), then there is at least one point c on the open interval (a, b), then the mean value theorem formula is:
$$f’ (c) = [f(b) – f (a)] / b – a$$
## Mean Value Theorem for Integrals
The mean value theorem for integral states that the slope of a line consolidates at two different points on a curve (smooth) will be the very same as the slope of the tangent line to the curve at a specific point between the two individual points. Let f be the function on [a, b]. Then the average f (c) of c is
$$1/ b – a∫_a^b f(x) d(x) = f (c)$$
(Image)
However, an Online Integral Calculator helps you to evaluate the integrals of the functions with respect to the variable involved.
Example:
Find the value of f (x)=11x^2 - 6x - 3 on the interval [4,8].
Solution:
In the given equation (f) is continuous on [4, 8].
$$F (C) = 1/b – a ∫ f(x) dx = 1/ 8 – 4∫_4^8 (11x^2 – 6x – 3) dx$$
$$= 1/4 [x^3 – x^2]^8_4$$
$$= 1/4 [(216 – 36) – (8 – 4)]$$
$$= 1/4 [(180 – 4)]$$
$$= 176/4 = 44$$
Here the value of c is 44 that provides the average value of the given function.
Now put x=16 in the function.
$$f(x)=11x^2 - 6x - 3 = 44$$
$$=11x^2 - 6x - 47$$
$$=(x + 2.32)(x - 2.80)=0$$
Hence 2.80 is the value of c. The online mean value theorem calculator gives the same results when you plug in the similar values and intervals in it.
### Cauchy's mean value theorem
Cauchy’s mean value theorem is the generalization of the mean value theorem. It states: if the function g and f both are continuous on the end interval [a, b] and differentiable on the start interval (a, b), then there exists c e(a, b), such that
$$(f (b) – f (a)) g’c = (g(b) – g (a)) f’c$$
Here’s g (a) ≠ g (b) and g’ (c) ≠0, so this is equivalent to:
$$f’(c) / g’(c) = f(b) – f(a)/ g(b) – g(a)$$
However, an Online Derivative Calculator helps to find the derivative of the function with respect to a given variable.
Example:
Find a value of “C” that is the conclusion of the mean value theorem: f(x) = -4x^3 + 6x - 2 on the interval [-4 , 2].
Solution:
f(x) is a polynomial function and is differentiable for all real numbers. Let evalute f(x) at x = -4 and x = 2
$$f(-4) = -4(-4)^3 + 6(-4) - 2 = 20$$
$$f(2) = -4(2)^3 + 6(2) - 2 = - 4$$
Now, substitute the values in [f(b) - f(a)] / (b - a)
$$[f(b) - f(a)] / (b - a) = [-6 - 4] / (2 – (-4)) = -2$$
Let us now find f '(x)
$$f '(x) = - 6x^2 + 6$$
We now create an equation, which is based on f '(c) = [f(b) - f(a)] / (b - a)
$$-6c + 6 = -2$$
You can find the value of c by using the mean value theorem calculator:
$$c = 2 \sqrt{(1/3)} and c = - 2 \sqrt{(1/3)}$$
### Rolle’s Theorem:
Rolle’s theorem says that if the results of a differentiable function (f) are equal at the endpoint of an interval, then there must be a point c where f ’(c)=0.
(Image)
Example:
Find all values of point c in the interval [−4,0]such that f′(c)=0.Where f(x)=x^2+2x.
Solution:
First of all, check the function f(x) that satisfies all the states of Rolle’s theorem.
1. f(x) is continuous function in [−4,0] as the quadratic function;
2. It is differentiable over the start interval (−4,0);
$$f(−2)=(−4)2+2⋅(−4)=0$$ $$f(0)=02+2⋅0=0$$
$$f(−4)=f(0)$$
So we can use Rolle’s theorem calculator to find the point c
$$f′(x)=(x2+2x)′=2x+2$$
Now, solve the equation f′(c)=0:
$$f′(c)=2c+2=0$$
$$c=−1$$
Thus,
$$f′(c)=0 for c=−1$$
## How Mean Value Theorem Calculator Works?
This free Rolle’s Theorem calculator can be used to compute the rate of change of a function with a theorem by upcoming steps:
### Input:
• First, enter a function for different variables such as x, y, z.
• Now, enter start and end intervals of the continuous function
• Click on the calculate button to see the results
### Output:
• The mean value theorem calculator provides the answer
• Displays the derivation of entered functions
## FAQs:
### Who proved the mean value theorem?
A restricted form of the mean value theorem was proved by M Rolle in the year 1691; the outcome was what is now known as Rolle's theorem, and was proved for polynomials, without the methods of calculus. The mean value theorem in its latest form which was proved by Augustin Cauchy in the year of 1823.
### What is the meant by first mean value theorem?
f(b)−f(a) = f′(c)(b−a). This theorem is also known as the First Mean Value Theorem that allows showing the increment of a given function (f) on a specific interval through the value of a derivative at an intermediate point.
## Conclusion:
Use this handy mean value theorem calculator that allows you to find the rate of change of a function, if f is continuous on the closed interval and differentiable on the open interval, then there exists a point c in the interval. The mean value theorem formula is difficult to remember but you can use our free online rolles’s theorem calculator that gives you 100% accurate results in a fraction of a second.
## Reference:
From the source of Wikipedia: Cauchy's mean value theorem, Proof of Cauchy's mean value theorem, Mean value theorem in several variables.
From the source of Pauls Online Notes: The Mean Value Theorem, Rolle’s Theorem, Proofs From Derivative Applications.
From the source of Calc Workshop: Mean Value Theorem for Integrals, Average Value, Mean Value Theorem. | HuggingFaceTB/finemath | |
Latest Teaching jobs » Maths Questions for CTET,KVS Exam :...
# Maths Questions for CTET,KVS Exam : 6th october 2018 Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB ,KVS,STET Exam.
Q1. The sum of mean, mode and median of the data 6, 3, 9, 5, 1, 2, 3, 6, 5, 1, 3 is
(a) 13
(b) 10
(c) 11
(d) 12
Q2. The ages in years of 10 teachers of a school are 34, 42, 27, 41, 36, 28, 22, 31, 37, 39. What is the age of the youngest teacher?
(a) 41
(b) 22
(c) 27
(d) 28
Q3. The base of an isosceles ∆ABC is 48 cm and its area is 168 cm². The length of one of its equal sides is
(a) 8 cm
(b) 15 cm
(c) 17 cm
(d) 25 cm
Q4. The following monthly expenditure values of 15 families: Exp (Rs.): 220, 310, 110, 300, 500, 270, 160, 200, 310, 400, 420, 320, 340, 140, 100 Calculate the arithmetic mean.
(a) Rs. 876.35
(b) Rs. 345.77
(c) Rs. 273.33
(d) Rs. 373.55
Q5. A square and a circle have equal perimeters. The ratio of the area of the square to the area of the circle is
(a) 1 : 1
(b) 1 : 4
(c) π : 2
(d) π : 4
Q6. A room is in the form of a cuboid. Its length, breadth and height are 9 m, 8 m and 4 m respectively. What is the total area of all four walls?
(a) 120 m²
(b) 64 m²
(c) 136 m²
(d) 70 m²
Q7. Sum of mean and median of the number 5.02, 5.18, 5.12, 5.007 and 5.018 is
(a) 10.089
(b) 10.73
(c) 10.71
(d) 10.89
Q8. The diameter of a hemisphere is 21 cm. What is its total surface area?
(a) 1031.5 cm²
(b) 1039.5 cm²
(c) 519.75 cm²
(d) 312.41 cm²
Q9. The product of the mean and median of the numbers 8, 11, 6, 9, 16 is
(a) 60
(b) 64
(c) 80
(d) 90
Q10. The radius of a roller is 1.75 m, and it is 2 m long. If it takes 50 revolutions to level a field, then the area of the field is
(a) 1100 m²
(b) 2200 m²
(c) 550 m²
(d) 225 m²
Solutions
S1. Ans.(b)
Sol. The given numbers are 6, 3, 9, 5, 1, 2, 3, 6, 5, 1, 3
After arranging the numbers in ascending order, we get
1, 1, 2, 3, 3, 3, 5, 5, 6, 6, 9
Number of numbers = 11
Mean
=(1+1+2+3+3+3+5+5+6+6+9)/11
=44/11=4
Here, the number of terms is odd.
Median = (n + 1)/2th term
=(11+1)/2 th term
= 6th term
= 3
Now, 1 is twice, 2 is once, 3 is thrice, 5 is twice, 6 is twice and 9 is once. As 3 occurs the most of the given data is 3.
Sum of mean, mode and median = 4 + 3 + 3 = 10
S2. Ans.(b)
Sol. After arranging all the ages in increasing order, we get
22, 27, 28, 31, 34, 36, 37, 39, 41, 42
Thus, the age of the youngest teacher is 22. S6. Ans.(c)
Sol. Sum of area of walls = 2 (length × height + breadth × height)
= 2 (9 × 4 + 8 × 4) = 136 m²
S7. Ans.(a)
Sol. Mean = (Sum of numbers)/(Numbers of numbers)
=(5.02+5.18+5.12+5.007+5.018)/5
=25.345/5=5.069
The median, in case of an odd number of numbers, is the middle term of the sequence when the numbers are arranged in ascending order.
Ascending order of numbers
= 5.007, 5.018, 5.02, 5.12, 5.18
Middle term when numbers are arranged in ascending order = 5.02
∴ Median = 5.02
Sum of mean and median
= 5.069 + 5.02 = 10.089
S8. Ans.(b)
Sol. Diameter of hemisphere = 21 cm
= 21/2 = 10.5 cm
Total surface area of he misphere = 3πr²
=3×22/7×(10.5)^2
= 1039.5 cm²
S9. Ans.(d)
Sol. Mean = (Sum of numbers)/(Number of numbers)
=(8+11+6+9+16)/5
=50/5=10
The median of an odd number of numbers is the middle term when the numbers are arranged in ascending order.
6, 8, 9, 11, 16
Median = 9
Product of mean and median
= 10 × 9 = 90
S10. Ans.(a)
Sol. Area covered in one revolution = Curved surface area of the roller = 2πrh
=2×22/7×1.75×2=22 m^2
Therefore, the area of the field
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# Search by Topic
#### Resources tagged with Triangle numbers similar to Light the Lights Again:
Filter by: Content type:
Stage:
Challenge level:
### There are 15 results
Broad Topics > Numbers and the Number System > Triangle numbers
### Light the Lights Again
##### Stage: 2 Challenge Level:
Each light in this interactivity turns on according to a rule. What happens when you enter different numbers? Can you find the smallest number that lights up all four lights?
### Graphing Number Patterns
##### Stage: 2 Challenge Level:
Does a graph of the triangular numbers cross a graph of the six times table? If so, where? Will a graph of the square numbers cross the times table too?
### A Square Deal
##### Stage: 2 Challenge Level:
Complete the magic square using the numbers 1 to 25 once each. Each row, column and diagonal adds up to 65.
### Picturing Triangular Numbers
##### Stage: 3 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Triangle Numbers
##### Stage: 3 Challenge Level:
Take a look at the multiplication square. The first eleven triangle numbers have been identified. Can you see a pattern? Does the pattern continue?
### Factors and Multiples Puzzle
##### Stage: 3 Challenge Level:
Using your knowledge of the properties of numbers, can you fill all the squares on the board?
### Route to Infinity
##### Stage: 3 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Forgotten Number
##### Stage: 3 Challenge Level:
I have forgotten the number of the combination of the lock on my briefcase. I did have a method for remembering it...
### Cat Food
##### Stage: 2 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### Satisfying Statements
##### Stage: 2 and 3 Challenge Level:
Can you find any two-digit numbers that satisfy all of these statements?
### Handshakes
##### Stage: 3 Challenge Level:
Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
### Man Food
##### Stage: 2 and 3 Challenge Level:
Sam displays cans in 3 triangular stacks. With the same number he could make one large triangular stack or stack them all in a square based pyramid. How many cans are there how were they arranged?
### Take Ten Sticks
##### Stage: 3 and 4 Challenge Level:
Take ten sticks in heaps any way you like. Make a new heap using one from each of the heaps. By repeating that process could the arrangement 7 - 1 - 1 - 1 ever turn up, except by starting with it?
### Sam Again
##### Stage: 3 Challenge Level:
Here is a collection of puzzles about Sam's shop sent in by club members. Perhaps you can make up more puzzles, find formulas or find general methods.
### Alphabet Blocks
##### Stage: 1 and 2 Challenge Level:
These alphabet bricks are painted in a special way. A is on one brick, B on two bricks, and so on. How many bricks will be painted by the time they have got to other letters of the alphabet? | HuggingFaceTB/finemath | |
# what is the lowest common multiple of 8 and 14
## Question: What is the lowest common multiple of 8 and 14?
LCM also known as the least common multiple is the smallest positive number that is a multiple of two or more numbers.
## Answer: The lowest common multiple of 8 and 14 is 56
Let's look into the multiples of 8 and 14
## Explanation:
### LCM of 8 and 14:
Let's list down the first few multiples of 8 and 14 as follows
Multiples of 8 : 8, 16, 24, 32, 40, 48, 56, 64
Multiples of 14 : 14, 28, 42, 56, 70
Common multiple: 56
LCM (8,14) = 56 | HuggingFaceTB/finemath | |
# Problem: Calcium hydroxide and hydrobromic acid react together in the following reaction.Ca(OH)2 (aq) + HBr (aq) → CaBr2 (aq) + H2O (l)25.00 mL of 1.50 M HBr are needed to titrate 32.801 mL of Ca(OH)2. What is the molarity of Ca(OH)2?
###### FREE Expert Solution
We're being asked to calculate the molarity of Ca(OH)2.
We're given the unbalanced reaction: Ca(OH)2(aq) + HBr(aq) → CaBr2(aq) + H2O(l)
Balancing the reaction:
Ca(OH)2(aq) + 2 HBr(aq) → CaBr2(aq) + 2 H2O(l)
Step 1: Calculate the moles of Ca(OH)2.
molarity HBr (volume) → moles HBr (mole-to-mole comparison) → moles Ca(OH)2
Given:
95% (104 ratings)
###### Problem Details
Calcium hydroxide and hydrobromic acid react together in the following reaction.
Ca(OH)2 (aq) + HBr (aq) → CaBr2 (aq) + H2O (l)
25.00 mL of 1.50 M HBr are needed to titrate 32.801 mL of Ca(OH)2. What is the molarity of Ca(OH)2? | HuggingFaceTB/finemath | |
# The Maths Square
### The Maths Square – A Fresh Way Forward
Allan Lawson has devised a system of dynamic dot pattern maths that he believes improves teaching methods for addition and subtraction
Why are so many children still failing in numeracy skills? I believe one of the causes lies with the use of rote memory in the early years of maths education. Childrenʼs natural ability to visualise and mentally manipulate patterns can change things.
### Banish rote memory!
The use of rote memory is both an inadequate and an inappropriate teaching method in maths. Yet rote memory is still often relied upon to learn the 55 addition facts up to 10 + 10, and the further 55 multiplication facts in the same range. Thatʼs a total of 110 facts – quite a task for young children – enough to put anyone off maths for good!
I believe rote memory should be left to language teachers, and for learning poetry. But removing rote memory from early arithmetic is not easy: for what other methods are there enabling pupils to recall the 110 addition and multiplication facts?
And is instant recall even necessary?
### Domino inspired solution
Around 40 years ago, back in the 70’s, when teaching arithmetic to children, I noticed how easily quite young children (4 or 5 years of age) could identify the dot pattern groupings on domino pieces. This lead to a method enabling children to carry out additions and subtractions in the range of natural numbers from 1 to 18 by identifying the numerical value of combined or remaining dominoes. They had natural subitising abilities.
Could dominoes be used effectively in teaching arithmetic at primary level ?
### Limits in Subitising domino patterns
In this paper, the author promotes a teaching method for primary mathematics employing a development of Piaget’s subitising of dot patterns: namely the ability of pupils to instantaneously recognise and know the number of dots in a small group without the need to count them.
If such a teaching method is to be adopted, then dominoes have limitations. The first is the limit to the number values that dominoes can show.
Unless we develop domino patterns for use as a numeral system (discussed later) the traditional numerical values given to the highest-value domino, the double 9, is 18. Thus sums which total more than 18 will result in dot patterns not represented by any existing single domino and will form a novel pattern. Novel patterns are those that are beyond the range in numerical value of the fixed stock of 55 dominoes. Thus, until pupils learn to recognise such patterns, there will be errors when attempting to subitise them. The common belief is that most people can only subitise dot patterns up to a group of about 7 – young pupils, less.
But there are more severe problems in using dominoes, even for sums totaling no more than 20 – the sheer volume of dot pattern combinations.
Let us consider using just 2 dominoes to make a combined sum of 19. Limiting the range of numbers available using 2 dominoes, there are 9 combinations of 2 numbers summing to 19, namely
1 + 18 6 + 13 2 + 17 7 + 12 3 + 16 8 + 11 4 + 15 9 + 10 5 + 14
Now let us consider the number of combinations of 2 dominoes that can be used in each of these sums.
The first sum, 1 + 18, can only be represented by just 2 dominoes :-
Next, 2 + 17 can be represented by 2 combinations of 2 dominoes :-
Next, 3 + 16 has 4 combinations of 2 dominoes :-
Next, 4 + 15 has 6 combinations of 2 dominoes :-
5 + 14 has 9 combinations 6 + 13 “ 12 “ 7 + 12 “ 16 “ 8 + 11 “ 20 “ 9 + 10 “ 25 “
The total number of combinations of 2 dominoes to represent the number 19 is thus 95.
Similarly there are 9 combinations of 2 numbers less than 19, that add to 20 :-
2 + 18 has 2 combinations of 2 dominoes 3 + 17 ” 2 “ 4 + 16 ” 6 “ 5 + 15 ” 6 “ 6 + 14 ” 12 “ 7 + 13 ” 12 “ 8 + 12 ” 20 “ 9 + 11 ” 20 “ 10 + 10 ” 25 ” Total: 105 combinations of 2 dominoes
These numbers of combinations of just 2 dominoes to represent such small additions, are truly daunting if required to be memorised and hence subitised.
### Dominoes – as a numeral system ?
Dominoes were never meant to be a numeral system. The above results show this. But they could be.
Given the following ideas, first evolved in ancient times, dominoes too, could be made to represent any number, and thus become a complete numeral system. All that is needed is a sense of place value, such as base 10. Also a reading instruction: read the two dot patterns on a domino as two separate numbers. In a row of several dominoes, the extreme right-hand number has its face value multiplied by 10 to the power 0, the number to its left, has its face value multiplied by 10 to the power 1; the next, multiplied by 10 to the power 2, and so forth, providing the place value, base 10, to each number. And finally, a row of dominoes should be read from left to right.
So, for example : –
93,806,230
Could be represented in dominoes, thus : –
Where the number of digits is odd, such as : –
9,380,623
The first domino on the left, must show 0 on its left-hand side : –
So a domino numeral system could easily be developed.
It would, however, be necessary to have several sets of dominoes to represent certain numbers, especially where a series of the same numbers recurred, for example : –
93,000,000
In this example, 3 double zero dominoes would be needed to represent the 6 zeros – should physical dominoes be used. Once we move into using diagrams, instead of physical dominoes, the problem disappears.
### Domino mathematics
The Romans developed a fine numeral system, used to this day – but was it of any use in conducting arithmetic? We know it was not. For small numbers, say 2 (Arabic) digit numbers, no doubt rote memory must have been used to obtain each number fact. But for larger numbers carrying out arithmetic operations required a calculator: the abacus. Once a calculation had been completed, the result had to be converted back into Roman numerals, if it was to be expressed in writing on slate, paper or on monuments.
Would dominoes present the same difficulties ? Let us try an example. Multiply 63 by 89 : –
The problem could be solved using the traditional algorithm, where rote memory is used in each step of the solution. So there would be no real difficulties.
How about a division ? Lets try : –
So, once again, no substantial difficulties.
Although dominoes could, of course, be used to teach arithmetic at primary level, I felt that a better system could be devised, perhaps modeled on dominoes. But my instincts told me that dominoes had further limitations, especially when moving on from addition and subtraction to multiplication and division.
### The Math Square – a dot- pattern numeral system
The limitations that dominoes presented acted as an incentive for me to create new dot patterns which could be more useful in carrying out arithmetic. I decided upon a systematic set of dot patterns to represent numbers to 20 – rather than a set of random dots that werenʼt in some form of pattern – to cover all additions to 20.
Whatever new numeral system was to be created using dot patterns, there was no question that it had to be decimal based. Then, similar to dominoes, a series of new dot patterns could be created to represent the set of 9 unit natural numbers and the number ten. As with dominoes, the patterns would be positioned inside a rectangle or square. A square without any dots would represent zero.
I considered various geometrically-pleasing matrices or “place-settings” for the initial 10 natural numbers. Some of the rejected ones included the following arrangements : –
I settled on, perhaps, an unlikely format for a 10-based matrix, namely a 3-by-3 square containing 9 locations where dot pattern numerals could be positioned.
I set out some rules governing acceptable spots where dots could be placed to form number patterns : –
1) Dots can only be placed in the 9 designated fixed positions within the square.
2) All dots should be the same colour, shape and size, with the exception of a tenth dot (explained later).
3) No dot pattern should exceed 10 dots in any square.
4) Each dot must be positioned horizontally or vertically next to another dot, unless a single dot is to be used in a square to represent the number 1.
5) No dot should be positioned diagonally to another dot, unless it is also next to another dot horizontally or vertically.
———————————————————–
My reason for setting such rules was to limit the possible number of dot patterns that could be created for each number, thereby reducing the total number of patterns representing the numbers from 1 to ten, down to 67 different patterns. However, a basic set of dot patterns that pupils would need to learn to recognise would be no more than ten.
My base set of patterns to represent numbers 1 to 9 were : –
This numeral system, as yet, lacked a dot pattern for the number 10. An obvious problem faced me: where, in a 9-dot format, was a tenth dot to be placed in accordance with the above rules?
The problem was solved by my young primary-level pupils when this question was posed to them as they played with a set of dot-pattern cards and some loose plastic counters (or “chips”, as they are sometimes called).
And so this leads me to my choice of suitable material items. Loose red counters were my first choice and they proved useful when dealing with small numbers, say 5 or below. But using counters becomes cumbersome when doing maths with larger numbers. So square white cards with stick-on red dots were introduced and proved very practical when used with red counters of roughly the same size and colour as the red dots on the cards.
Using these materials, lead to the solution of the problem above: how to represent the number ten using a 9-dot place-setting square ? With the 9 fixed positions, where could a tenth dot or counter be placed?
One answer would be to have it placed in another, new square, representing tens; the first square used solely for representing the unit numbers. However, the problem was deftly solved another way, when a small hand picked up a red counter and, without hesitation, placed it in the square on top of the centre dot of a 9-dot pattern.
And thus, we arrived at a 10-place format for expressing number, which I initially named the “Magic Square” but later re-named the “Maths Square”. Some amateur maths teachers I showed this to in the following years liked the name “Magic Square” and have continued to use this term, or similar, such as “Magic Blackboard”.
However, another problem emerged later when representing the Math Square in diagrammatic form.
Whilst using red counters alone to express the number 9, and wanting to add a dot to form the number 10 by placing a counter on top of the centre counter, the newly created 10-dot pattern could readily be distinguished from the earlier 9-dot pattern. On close examination, the centre position with two counters was obvious and visible. But what if diagrams alone were to be used ? It would be impossible to distinguish a 9-dot pattern from a 10-dot pattern using dots of the same size, colour and shape.
The solution was to treat the tenth dot as “special”, by awarding it its own colour – purple.
The purple dot now indicates that there are two counters or dots in the position where a purple dot is positioned. Thus, where the other 9 places in the Math Square are filled with dots or counters, we have a 10-dot pattern.
In a later development of this numeral system, a purple dot alone would express the number 10.
Then red dots would be banished from sharing a square with a purple dot or dots. Each purple dot would represent the number 10 and the dot-pattern numeral system would apply to squares containing purple dots. Purple dot patterns in their own Math Square would represent numbers raised by 10 to the power 1.
The pattern above shows ten represented by a dot pattern in a Math Square. Later, we shall see alternatives, where the tenth purple dot can be positioned elsewhere in the ten dot pattern.
Numbers between 10 and 20 were arranged on cards double in size, with a 10-pattern on the left-hand half of the card and patterns from 1 to 10 arranged on the right-hand side.
Before using the new cards with children, I used only plastic counters to arrange the dot patterns in order to provide plenty of hands-on activity, with pupils adding or taking away counters from various pattern arrangements in doing sums.
As sums of 5 + 5 and higher were approached, it became useful to use a combination of Math Square cards and plastic counters. The dots printed on the cards were made in the same colour and size as the loose plastic counters. Counters can then be placed on the cards to create new dot patterns in addition or dots covered with white card in subtraction.
### Variations of patterns for some numbers
For some of the numbers, such as 4 or 5, I found that two or more variations of patterns were found to be useful in carrying out additions. Below are patterns that teachers may wish to use: they are not exclusive and teachers may find other patterns more useful (See Appendix A).
The essential point I found is this: in order for children to re-assemble and re-arrange dot patterns into other identifiable dot patterns, a dot pattern numeral system must already be established in their minds, no matter what dot patterns are used. Once a set of dot pattern numerals has been chosen, it must be memorized by pupils. The use of flash cards will help. Here are my suggestions :
The patterns should be identifiable from any orientation, whether as mirror images or revolved images as this will help pupils in performing dot pattern arithmetic.
In performing additions of two numbers, the aim is to merge or re-arrange and then re-assemble the 2 dot patterns to form a single identifiable new pattern. Where it seems possible to create a 10 pattern, this must be the first step. Making a 10 pattern is essential knowledge, as it is when using, say, Dienes blocks, an abacus or any other decimal system.
I suggest below some strategies to assist in creating a 10 pattern where one of the numbers in the addition sum is a 6, 7, 8 or 9. These numbers can be “built up” to form 10 by removing dots from the other number “n” in the addition.
So, where we have (6 + n), n being in the range 1 to 4, the aim is to build up the 6 dot pattern by attaching the n pattern to the 6 pattern, or where n ≥ 5, by removing 4 dots from the n pattern, attaching them to the 6 pattern, thus forming a new 10 pattern. The resulting pattern formation will then be in identifiable form.
Below is displayed a single strategy for the solution of each example addition:
The many equally good alternatives for re-arranging and re-assembling the dot patterns are left for you and your pupils to explore.
i) 6 + 3 : the 6 and 3 patterns merge to make the 9 pattern:
ii) 6 + 4 : the 6 and 4 patterns merge to make the 10 pattern: in the process of merging, two dots overlap at the center of the square – and is represented by a purple dot. Only a single overlap is allowed in any ten dot square. Overlapping dots are not allowed in smaller numerals.
iii) 6 + 5 : in dot patterns :
Before merging, the 5 pattern can be re-arranged as a 4 pattern, plus one:
Then the 4 pattern can be merged with the 6 pattern to make a 10 pattern, with 1 dot remaining – resulting in the 11 pattern:
iv) 6 + 6 : in dot patterns :
Before merging, the right-hand 6 pattern can be re-arranged as a 4 pattern plus a two pattern:
Now the 4 pattern can be merged with the 6 pattern, the re-assembled dots making a 10 pattern, with 2 dots remaining; resulting in the 12 pattern:
In additions of (7 + n) the rule to follow is to remove 3 dots from n, the other number, and merge these with the 7 pattern, so forming a 10 pattern. Eg:-
i) 7 + 3 : the 7 and 3 patterns merge to make another form of the 10 pattern:
ii) 7 + 8 : by removing 3 dots from the 8 pattern and merging these with the 7 pattern, a 10 pattern is formed, with a 5 pattern remaining, the two new patterns forming a 15 dot pattern:
In additions of (8 + n), remove the 2 dots from the other number n, and merge these with the 8 pattern, to make a 10 pattern. Eg:
i) 8 + 2 : the 8 and 2 patterns merge to form a 10 pattern:
ii) 8 + 8 : the suggested steps needed to form a 16 pattern are:
In additions of (9 + n), we simply remove one dot from n and merge it with the 9 pattern, making a 10 pattern e.g:
9 + 9 :
### 10, 20 and beyond
The Math Square dot pattern numeral system does not stop at 20. However, to reduce the number of dots required when dealing with larger numbers, we could dispense with some of the red unit dots for all numbers larger than 9. So instead of 10 consisting of 9 red dots and a central purple dot, it could be represented by the purple dot alone. And 20 would be represented by two purple dots, thirty by 3 purple dots, and so on. One hundred could be represented by a third coloured dot, say yellow, with each successive power of ten being represented by yet another colour. For example 1,342 might be represented as:
The Math Square numeral system could in theory represent enormous numbers, limited only to the extent that colours might be exhausted in representing higher and higher powers of ten. And in such a system, where colour denotes decimal power, the position of each coloured pattern is immaterial.
In a future article, I will describe how the Math Square was developed into an ordinal place-value numeral system capable of representing all positive whole numbers to infinity – and beyond!
### Subtraction
I have not provided examples of how subtraction can be performed. But teachers can experiment with the dot pattern cards, in a similar way as some additions are carried out, by the removal of counters from a dot pattern or by covering up dots on cards to show the resulting dot pattern in an identifiable form.
### Mental arithmetic
One remarkable development that came to light from children who had played with the dot pattern cards and counters for some while was their ability to visualize the dot patterns in their mindʼs eye, then to re-arrange and re-assemble them into identifiable patterns, without the use of any physical materials.
### Before I conclude: Q – Why have an overlapping tenth dot?
A criticism that could be leveled at this numeral system is this: wonʼt an overlapping tenth dot in the 10-dot pattern cause confusion with the 9-dot pattern? Indeed, why introduce a dot numeral system with an overlapping tenth dot? After all, there are other regular 10-dot patterns without overlapping dots that can form the basis of a numeral system. For example: –
However, my justification in choosing the Math Square format lies with itʼs development into a similar square with a tenth position lying at itʼs centre, for use in solving the multiplication facts. The tenth overlapping central position becomes vital in this development.
### Conclusions
1. Childrenʼs visual mental ability:
Children can easily identify dot pattern numbers; they can also visualize dot patterns in their mindʼs eye and re-arrange and re-assemble these patterns to form previously known patterns. And they have no difficulty in distinguishing the 9-dot pattern from the 10-dot pattern.
The Math Square system provides children with good opportunities to exercise their visual mental ability to manipulate dot patterns, and a fresh way to explore the possibilities to be discovered in dot pattern maths.
In my opinion psychologists should re-examine the role that childrenʼs visual mental ability could play in the teaching of mathematics generally.
2. Memorizing facts unnecessary:
Employing these methods to carry out additions and subtractions has shown me that committing the facts to memory for instant recall is unnecessary. Children can quickly arrive at the results of the 55 additions mentally using this system of dynamic dot pattern maths.
### Next
The following article entitled The Decimal Square shows how the Math Square format was developed into a positional framework with vectors enabling children to find the facts of the multiplication tables mentally in a systematic way in place of rote memory. Visualized spatial abilities are here too employed.
For those who wish to try some Math Square cards, the on-line version of this article has a printable page of cards, which can be cut or guillotined to provide a set of 20 cards. | HuggingFaceTB/finemath | |
## Saturday, 26 March 2011
### Hanging it Up
I was thinking again about the bent rail problem from a couple of days ago, and got to wondering what the height would be if it was an inverted catenary arc instead of a circular arc. I was led to this idea by one of my students who also had suggested that it might NOT be circular....and I had no way to assure him it would be... I have no reason to assume it is a catenary either, but it was a curve I felt I needed to know more about so I used it as a self-teaching exercise.
At that point I realized I didn't know much about the catenary curve in general. (what I mostly knew is about its history and can be found here). So I set out to see if I could learn a little more and figure out how to solve the question of how high would the inverted catenary reach. Below is what I have attempted to do... the answer is close enough to the circular amount of bowing that I wonder if it can be right, so if you are up on this stuff, please comment.
I knew that the catenary curve y= a cosh(x/a) has a minimum at (0,a). For students who haven't seen "cosh" that's just a hyperbolic cosine function (much like the circular cosine function in many ways, and often pronounced to rhyme with "gosh"... see an explanation of hyperbolic functions here). It can be replaced with an alternate form that may be more palatable to your eyes:
Of course it is easy to see that the vertex is at (0,a). What I had never known, and find particularly wonderful about the curve, is that for any catenary, the area under the curve from x=c to x=d divided by the constant a of the equation is the length of the curve.....A good calculus student can show this is true pretty easily with the exponential form since the derivative of cosh(x) is sinh(x).... and for hyperbolas we have cosh^2 - sinh^2 = 1 instead of the addition. That makes the integral for arc length the same as the integral for area between the curve and the x-axis.
So the length of the catenary from the vertex to any other value of x is given by a* Sinh(x/a) (OK, you can guess that is the hyperbolic Sin, usually pronounced like "cinch" and it is the same as the hyperbolic cosine except the two exponentials are joined by subtraction rather than addition..)
I worked the problem out upside down because I wasn't smart enough to turn it rightside up... ... I assumed that the rod was hanging under a secant line. For our hanging (and now very flexible) rod, the arc length is 5281 feet, and the secant (parallel to the x-axis) is 5280 feet. We can position the end points so that they are symmetric around the y-axis so that our hanging rod will have it's vertex on the y-axis.
So we know that the vertex is at (0,a) and we know that the length of the cable from the vertex to x=2640 is going to be 2640.5 which must equal a*sinh(x/a). This seemed to give us enough to write an equation that we could solve for a, and my answer came out close to 82,809... which seems like a really big constant. Buty the effect of a on the catenary is to adjust both the distance from the origin, but also how fast it turns up. Here is a graph showing catenary curves for a=1 (red) and a=2(blue).
So we want a very wide catenary, we need a really big number... maybe 82,809 will work. So our equation for this curve is y = 82,809 * cosh(x/82809). Now we can use the equation for the curve to find the height of the curve at the point where x= 2640. This turns out to be just a smidge over 82,851 or almost exactly 42 feet above the y-intercept of 82,809. So if we flipped it over and set it on the ground, the one mile wide arch would be only 42 feet high. Very near to the 45 foot answer we got from the circular arc.
Recently I wrote about Willima Whewell's role in the creation of the word scientist. I just found that there is also a Whewell equation for the catenary which gives the slope of the tangent of the catenary as a function of the distance from the origin, s. Tan(T)= s/a. That means in the catenary version of our inverted rail, the angle between the curve and the horizontal secant is just over .03 radians.... I think I did that correctly.
As a comparison, the St. Louis arch is 630 feet high, and has a width from side to side of 630 feet. It is not a true catenary because it is adjusted for weight, called a "weighted" or "flattened" catenary. If we assume the difference is small (I don't know for sure) then the information above should allow you to find the constant "a" for the arch, and then find the length of the arch... Good luck... | HuggingFaceTB/finemath | |
×
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# Solutions for Chapter 7.3: Fundamentals of Differential Equations 8th Edition
## Full solutions for Fundamentals of Differential Equations | 8th Edition
ISBN: 9780321747730
Solutions for Chapter 7.3
Solutions for Chapter 7.3
4 5 0 255 Reviews
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3
##### ISBN: 9780321747730
This expansive textbook survival guide covers the following chapters and their solutions. Chapter 7.3 includes 38 full step-by-step solutions. Fundamentals of Differential Equations was written by and is associated to the ISBN: 9780321747730. Since 38 problems in chapter 7.3 have been answered, more than 158721 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Fundamentals of Differential Equations , edition: 8.
Key Calculus Terms and definitions covered in this textbook
• Acceleration due to gravity
g ? 32 ft/sec2 ? 9.8 m/sec
• Arccosecant function
See Inverse cosecant function.
• Conditional probability
The probability of an event A given that an event B has already occurred
• Constant function (on an interval)
ƒ(x 1) = ƒ(x 2) x for any x1 and x2 (in the interval)
• Constant of variation
See Power function.
• Constant term
See Polynomial function
• Derivative of ƒ at x a
ƒ'(a) = lim x:a ƒ(x) - ƒ(a) x - a provided the limit exists
• Directed line segment
See Arrow.
• Equal complex numbers
Complex numbers whose real parts are equal and whose imaginary parts are equal.
• Exponential regression
A procedure for fitting an exponential function to a set of data.
• Factoring (a polynomial)
Writing a polynomial as a product of two or more polynomial factors.
• Fundamental
Theorem of Algebra A polynomial function of degree has n complex zeros (counting multiplicity).
• Instantaneous rate of change
See Derivative at x = a.
• Invertible linear system
A system of n linear equations in n variables whose coefficient matrix has a nonzero determinant.
• Local extremum
A local maximum or a local minimum
• Numerical model
A model determined by analyzing numbers or data in order to gain insight into a phenomenon, p. 64.
• Odd-even identity
For a basic trigonometric function f, an identity relating f(x) to f(-x).
• One-to-one rule of logarithms
x = y if and only if logb x = logb y.
• Triangular form
A special form for a system of linear equations that facilitates finding the solution.
• Vertical stretch or shrink
See Stretch, Shrink. | HuggingFaceTB/finemath | |
# Definition of the derivative to find the derivative of x^(1/3)
• Martinc31415
In summary, the student is struggling with finding the derivative of x^(1/3) using the definition of the derivative. They have attempted to use the difference of cubes formula, but are unsure of how to proceed. They are then given guidance on how to use the conjugate of the difference of cube roots to simplify the problem.
Martinc31415
## Homework Statement
Use the definition of the derivative to find the derivative of x^(1/3)
## The Attempt at a Solution
[(x+h)^(1/3) - x^(1/3)]/h
I do not know where to go from here. If it were a square root I could conjugate.
Martinc31415 said:
## Homework Statement
Use the definition of the derivative to find the derivative of x^(1/3)
## The Attempt at a Solution
[(x+h)^(1/3) - x^(1/3)]/h
I do not know where to go from here. If it were a square root I could conjugate.
Hello Martinc31415. Welcome ton PF !
The difference of cubes can be factored, $a^3-b^3=(a-b)(a^2+ab+b^2)\,.$
So, suppose you have the difference of cube roots, $\displaystyle P^{1/3}-Q^{1/3}$. In this case, $\displaystyle P^{1/3} = a\,\ \text{ and }\ Q^{1/3} = b\,.$
Multiplying $\displaystyle \left(P^{1/3}-Q^{1/3}\right)$ by $\displaystyle \left(P^{2/3}+P^{1/3}Q^{1/3}+Q^{2/3}\right)$ will give $\displaystyle \left(P^{1/3}\right)^3-\left(Q^{1/3}\right)^3=P-Q\,.$
Thus, $\displaystyle \left(P^{2/3}+P^{1/3}Q^{1/3}+Q^{2/3}\right)$ acts as the "conjugate" for $\displaystyle \left(P^{1/3}-Q^{1/3}\right)\,.$
oohh...
I would not have thought of that, ever!
Thanks a ton.
## 1. What is the definition of the derivative?
The derivative of a function is the rate of change of that function at a given point. It represents the slope of the tangent line to the function at that point.
## 2. How is the derivative calculated?
The derivative is calculated using the limit definition: f'(x) = lim h->0 (f(x+h)-f(x))/h. This involves finding the tangent line to the function at a specific point and determining its slope.
## 3. How is the derivative of x^(1/3) found?
The derivative of x^(1/3) can be found using the power rule for derivatives, which states that the derivative of x^n is nx^(n-1). In this case, n = 1/3, so the derivative is (1/3)x^(1/3-1) = 1/(3x^(2/3)).
## 4. What is the importance of finding the derivative of a function?
The derivative of a function is important because it allows us to analyze the rate of change of the function at a specific point. This can help us understand the behavior of the function and make predictions about its future values.
## 5. How is the derivative used in real life applications?
The derivative is used in a variety of real life applications, including physics, economics, and engineering. It is used to calculate velocities, acceleration, and rates of change in various systems. In economics, it can be used to determine the marginal cost and revenue of a product. In engineering, it is used to optimize designs and determine the efficiency of systems.
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560 | HuggingFaceTB/finemath | |
1. ## Negating injection function
A function f: X to Y is injective if for all x, x' E X f(x) = f(x') implies x = x'
I attempted it and got:
There exists x, x' E X f(x) not equal to f(x') implies x not equal to x'
I'm confident I have got the quantifiers correct, but not the next part???
2. The negation of $\displaystyle f$ being injective is $\displaystyle \left( {\exists x} \right)\left( {\exists y} \right)\left[ {f(x) = f(y) \wedge \left( {x \ne y} \right)} \right]$.
3. Originally Posted by Plato
The negation of $\displaystyle f$ being injective is $\displaystyle \left( {\exists x} \right)\left( {\exists y} \right)\left[ {f(x) = f(y) \wedge \left( {x \ne y} \right)} \right]$.
Oh I see
Because x implies y negated is Not (x implies y)
Using the implication rule: x and not y
4. Actually it more general. The negation:
$\displaystyle \neg \left( {\forall x} \right)\left[ {P(x) \Rightarrow Q(x)} \right] \equiv \left( {\exists x} \right)\left[ {P(x) \wedge \neg Q(x)} \right]$ | HuggingFaceTB/finemath | |
# D’Alembert’s Ratio Test of Convergence of Series
In this article we will formulate the D’ Alembert’s Ratio Test on convergence of a series.
Let’s start.
## Statement of D’Alembert Ratio Test
A series $\sum {u_n}$ of positive terms is convergent if from and after some fixed term $\dfrac {u_{n+1}} {u_n} < r < {1}$ , where r is a fixed number. The series is divergent if $\dfrac{u_{n+1}} {u_n} > 1$ from and after some fixed term.
D’Alembert’s Test is also known as the ratio test of convergence of a series.
### Theorem
Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers in $R$, or a series of complex numbers in $C$.
Let the sequence ${a_n}$ satisfy:
$\displaystyle \lim_{n \mathop \to \infty} {\frac {a_{n + 1} } {a_n} } = l$
• If $l > 1$, then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ diverges.
• If $l < 1$, then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
## Definitions for Generally Interested Readers
(Definition 1) An infinite series $\sum {u_n}$ i.e. $\mathbf {u_1+u_2+u_3+….+u_n}$ is said to be convergent if $S_n$ , the sum of its first $n$ terms, tends to a finite limit $S$ as n tends to infinity.
We call $S$ the sum of the series, and write $S=\displaystyle {\lim_{n \to \infty} } S_n$ .
Thus an infinite series $\sum {u_n}$ converges to a sum S, if for any given positive number $\epsilon$ , however small, there exists a positive integer $n_0$ such that $|S_n-S| < \epsilon$ for all $n \ge n_0$ .
(Definition 2)
If $S_n \to \pm \infty$ as $n \to \infty$ , the series is said to be divergent.
Thus, $\sum {u_n}$ is said to be divergent if for every given positive number $\lambda$ , however large, there exists a positive integer $n_0$ such that $|S_n|>\lambda$ for all $n \ge n_0$ .
(Definition 3)
If $S_n$ does not tend to a finite limit, or to plus or minus infinity, the series is called oscillatory.
## Proof & Discussions on Ratio Test
Let a series be $\mathbf {u_1+u_2+u_3+…….}$ . We assume that the above inequalities are true.
• From the first part of the statement:
$\dfrac {u_2}{u_1} < r$ , $\dfrac {u_3}{u_2} < r$ ……… where r <1.
Therefore $\mathbf {{u_1+u_2+u_3+….}= u_1 {(1+\frac{u_2}{u_1}+\frac{u_3}{u_1}+….)}}$
$=\mathbf {u_1{(1+\frac{u_2}{u_1}+\frac{u_3}{u_2} \times \frac{u_2}{u_1}+….)}}$
$< \mathbf {u_1(1+r+r^2+…..)}$
Therefore, $\sum{u_n} < u_1 (1+r+r^2+…..)$
or, $\sum{u_n} < \displaystyle{\lim_{n \to \infty}} \dfrac {u_1 (1-r^n)} {1-r}$
Since r<1, therefore as $n \to \infty , \ r^n \to 0$
therefore $\sum{u_n} < \dfrac{u_1} {1-r}$ =k say, where k is a fixed number.
Therefore $\sum{u_n}$ is convergent.
• Since, $\dfrac{u_{n+1}}{u_n} > 1$ then, $\dfrac{u_2}{u_1} > 1$ , $\dfrac{u_3}{u_2} > 1$ …….
Therefore $u_2 > u_1, \ u_3 >u_2>u_1, \ u_4 >u_3 > u_2 >u_1$ and so on.
Therefore $\sum {u_n}=u_1+u_2+u_3+….+u_n$ > $nu_1$ . By taking n sufficiently large, we see that $nu_1$ can be made greater than any fixed quantity.
Hence the series is divergent.
From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$; otherwise ${\dfrac {a_{n + 1} } {a_n} }$ is not defined.
Here, ${\dfrac {a_{n + 1} } {a_n} }$ denotes either the absolute value of $\dfrac {a_{n + 1} } {a_n}$, or the complex modulus of $\dfrac {a_{n + 1} } {a_n}$.
Absolute Convergence
Suppose $l < 1$.
Let us take $\epsilon > 0$ such that $l + \epsilon < 1$.
Then:
$\exists N: \forall n > N: {\dfrac {a_n} {a_{n – 1} } } < l + \epsilon$
Thus:
$(\displaystyle {a_n}) (=) (\displaystyle {\frac {a_n} {a_{n – 1} } } {\frac {a_{n – 1} } {a_{n – 2} } } \dotsm {\frac {a_{N + 2} } {a_{N + 1} } } {a_{N + 1} })$
$(\displaystyle ) (<) (\displaystyle {l + \epsilon}^{n – N – 1} {a_{N + 1} })$
By Sum of Infinite Geometric Progression, $\displaystyle \sum_{n \mathop = 1}^\infty{l + \epsilon}^n$ converges.
So by the the corollary to the comparison test, it follows that $\displaystyle \sum_{n \mathop = 1}^\infty {a_n}$ converges absolutely too.
$\blacksquare$
Divergence
Suppose $l > 1$.
Let us take $\epsilon > 0$ small enough that $l – \epsilon > 1$.
Then, for a sufficiently large $N$, we have:
$(\displaystyle {a_n}) (=) (\displaystyle {\frac {a_n} {a_{n – 1} } } {\frac {a_{n – 1} }{a_{n – 2} } } \dotsm {\frac {a_{N + 2} } {a_{N + 1} } } {a_{N + 1} })$
$(\displaystyle ) (>) (\displaystyle {l – \epsilon}^{n – N + 1} {a_{N + 1} })$
But ${l – \epsilon}^{n – N + 1} {a_{N + 1} } \to \infty$ as $n \to \infty$.
So $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ diverges.
$\blacksquare$
• When $\dfrac {u_{n+1}} {u_n}=1$ , the test fails.
• Another form of the test– The series $\sum {u_n}$ of positive terms is convergent if $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ >1 and divergent if $\displaystyle{\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ <1.
One should use this form of the test in the practical applications.
## An Example
Verify whether the infinite series $\dfrac{x}{1.2} + \dfrac {x^2} {2.3} + \dfrac {x^3} {3.4} +….$ is convergent or divergent.
## Solution
We have $u_{n+1}= \dfrac {x^{n+1}}{(n+1)(n+2)}$ and $u_n= \dfrac {x^n} {n(n+1)}$
Therefore $\displaystyle {\lim_{n \to \infty}} \dfrac{u_n} {u_{n+1}} = \displaystyle{\lim_{n \to \infty}} (1+\frac{2}{n}) \frac{1}{x} = \frac{1}{x}$
Hence, when 1/x >1 , i.e., x <1, the series is convergent and when x >1 the series is divergent.
When x=1, $u_n=\dfrac{1} {n(n+1)}=\dfrac {1}{n^2} {(1+1/n)}^{-1}$
or, $u_n=\dfrac{1}{n^2}(1-\frac{1}{n}+ \frac {1}{n^2}-…..)$
Take $\dfrac{1}{n^2}=v_n$ Now $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{v_n}=1$ , a non-zero finite quantity.
But $\sum {v_n}=\sum {\frac{1}{n^2}}$ is convergent.
Hence, $\sum {u_n}$ is also convergent.
1. Unable to understand
2. Unable to understand
3. Thanks for the brilliant explanation . u enjoyed and understood it better than I got in the lecture
4. Thanks for the brilliant explanation . u enjoyed and understood it better than I got in the lecture
5. It’s nice. Ratio test has so many forms due to which creates confusion. I applied ratio test in this series
1+ (1/2!)+ (1/3!)+…..
But I found this series to be divergent using ratio test while this series is convergent.
• i hate the question having factorial of something in the denominator
6. It’s nice. Ratio test has so many forms due to which creates confusion. I applied ratio test in this series
1+ (1/2!)+ (1/3!)+…..
But I found this series to be divergent using ratio test while this series is convergent.
• i hate the question having factorial of something in the denominator
7. Series is convergent if L is less than 1 and divergent if L is greater than 1
8. Series is convergent if L is less than 1 and divergent if L is greater than 1
9. here is D’ Alembert’s ratio test:
Let Un be the nth term of a positive series such that
lim Un+1/Un = L
Then the series is convergent if L 1.
The test fails to decide the nature of the series if L = 1.
• series is convergent if L 1
• When we get the result 1′ we have to work this out with other principle…what did you do when you get the result equal to 1??
10. here is D’ Alembert’s ratio test:
Let Un be the nth term of a positive series such that
lim Un+1/Un = L
Then the series is convergent if L 1.
The test fails to decide the nature of the series if L = 1.
• series is convergent if L 1
• When we get the result 1′ we have to work this out with other principle…what did you do when you get the result equal to 1?? | HuggingFaceTB/finemath | |
Math1011_Week04_Prelim1A
# Math1011_Week04_Prelim1A - Math1011 Learning Strategies...
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Math1011 - Learning Strategies Center Section 1.6 IA-1 The function f(x) is graphed below: -4 -3 -2 -1 0 1 2 3 4 0 1 2 3 4 What is the domain and range of the f(x)? Is f(x) one-to-one? Why? Sketch f -1 (x) in the graph above. What is the domain and range of f -1 (x)?
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Math1011 - Learning Strategies Center Section 1.6 A1 The function f(x) is graphed below: -4 -3 -2 -1 0 1 2 3 4 - 4 - 3 - 2 - 101234 What is the domain and range of the f(x)? The domain of is ( 1, ) The range of is ( ,1) f −∞ −∞ Is f(x) one-to-one? Why? The function is one-to-one as its graph intersects each horizontal line at most once (Horizontal Line Test). Sketch f -1 (x) in the graph above. What is the domain and range of f -1 (x)? 1 1 The domain of is ( ,1). The range of is ( 1, ). −∞ (Doc #011w.16.06t)
Math1011 - Learning Strategies Center Section 1.6 A2 -1 -1 Show that the function f is one-to-one, and calculate the inverse function f . Specify the domain and range of f and f . 1 () x fx + =
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Math1011 - Learning Strategies Center Section 1.6 A2 -1 -1 Show that the function f is one-to-one, and calculate the inverse function f . Specify the domain and range of f and f . 1 () x fx + = 12 1 2 11 21 1 2 1 To show one-to-one, show if f(x ) ( ) x f(x ) ( ) (cross multiply) x (1 ) x (1 ) (distribute) x x x x (subtract x from both sides) x i s o xx ++ + = ⇒= = = += + + = = 1 2 -1 1 1 1 ne-to-one as f(x ) ( ) x Calculate the inverse function f ( ). (solve for x in terms of y) (1 ) (cross multiply) ) y yx + =⇒ = = =− = = -1 1 1 interchange x and y Specify the domains and ranges of f and f . ( ) ( ,1 ) (1 , ) ( ) ( , 1 ) ( 1 Df f fD =ℜ = −∞− ∪ − ∞ ℜ= = (Doc #011w.16.07t)(Adams 3.1.10)
Math1011 Fall 2007 Prelim A3 Show that for any three positive numbers a, b, c such that a 1, b 1, and c 1, the following equality holds: (log )(log )(log ) 1 ac b bca = Solve the equation for x: 4 2log 1 10 ln 4 log 100 x e +=
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Math1011 Fall 2007 Prelim A3 Show that for any three positive numbers a, b, c such that a 1, b 1, and c 1, the following equality holds: (log )(log )(log ) 1 ac b bca = log log log log (log )(log )(log ) 1 (log )( )( ) 1 (log aa ca a bc = = log )( c log log log )( )1 log 1 1 1
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Ask a homework question - tutors are online | HuggingFaceTB/finemath | |
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A058525 Numbers z by which not all integers y, 0 <= y < 2^64, can be divided using "high multiplication" followed by a right shift. 0
7, 21, 23, 25, 29, 31, 39, 47, 49, 53, 55, 61, 63, 71, 81, 89, 91, 93, 95, 97, 99, 101, 103, 107, 111, 115, 119, 121, 123, 125, 127, 137, 147, 161, 169, 181, 183, 199, 201, 207, 213, 223, 225, 233, 235, 237, 239, 243, 251, 253, 259, 273, 281, 285, 313, 315, 323 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS For many odd numbers z, it is possible to compute the integer division of y / z for 0 <= y < 2^64 (that is, floor(y/z)) by multiplying by a suitable constant a and shifting right: floor((a*y)/(2^(64+e))). a is computed as a = ceiling((2^(64+e))/z), where e is such that 2^e < z < 2^(e+1). Knuth showed that the formula floor((a*y)/(2^(64+e))) = floor(y/z) holds for all y, 0 <= y < 2^64, if and only if it holds for the single value y = 2^64 - 1 - (2^64 mod z). There are 189 odd divisors z less than 1000 for which this method cannot be used to find the division result for all y, 0 <= y < 2^64. REFERENCES Knuth, Donald Ervin, The Art of Computer Programming, fascicle 1, _MMIX_. Addison Wesley Longman, 1999. Zeroth printing (revision 8), 24 December 1999. Exercise 19 in section 1.3.1', page 25 and the answer on page 95. LINKS Donald E. Knuth, The fascicle as a compressed PostScript file EXAMPLE For the first term in the sequence, 7, floor(ay/(2^(64+e))) = 2635249153387078802 for y = 2^64 - 1 - (2^64 mod z) = 18446744073709551613, while floor(y/z) = 2635249153387078801. Example for a term not in the sequence: for 9, both floor(ay/(2^(64+e))) and floor(y/z) are 2049638230412172400 for y = 2^64 - 1 - (2^64 mod z) = 18446744073709551608. CROSSREFS Sequence in context: A201072 A200935 A097182 * A219036 A063469 A155131 Adjacent sequences: A058522 A058523 A058524 * A058526 A058527 A058528 KEYWORD nonn AUTHOR Philip Newton, Dec 22 2000 STATUS approved
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# SSC MTS 2016 Re-Exam Paper : Held on 09-OCT-2017 Shift-1 (Numerical Aptitude)
QID : 26 - If a number is divided by 30 then it leaves 17 as a remainder. What will be the remainder when the same number is divided by 10?
Options:
1) 7
2) 3
3) 1
4) 2
QID : 27 - What is the value of 162 + 172 + 182 + ……. 252?
Options:
1) 4325
2) 4465
3) 4105
4) 4285
QID : 28 - How many times digit '5' appears in the number from 1 to 100?
Options:
1) 20
2) 21
3) 19
4) 18
QID : 29 - What is the unit digit of 5124 x 1245?
Options:
1) 5
2) 1
3) 0
4) 2
QID : 30 - What is the value of [(a–2b3) ÷ (a1b–1)] x [(a2b–4) ÷ (a–1b2)]?
Options:
1) b2
2) 1/b2
3) a2
4) a2b2
QID : 31 - Which of the following number is divisible by 11?
Options:
1) 59609
2) 45332
3) 23581
4) 44433
QID : 32 - A can do a piece of work in 20 days and B can do the same piece of work in 30 days. They start working together and work for 5 days and then both leave the work. C alone finishes the remaining work in 14 days. In how many days will C alone finish the whole work?
Options:
1) 24
2) 18
3) 36
4) 42
QID : 33 - P, Q and R undertook a work for Rs 48000. Together P and Q complete 5/12th part of the work. What is the share (in Rs) of R?
Options:
1) 21000
2) 28000
3) 27000
4) 31000
QID : 34 - The length of two parallel sides of a trapezium are 30 cm and 40 cm. If the area of the trapezium is 350 cm2, then what is the value (in cm) of its height?
Options:
1) 8
2) 10
3) 15
4) 12
QID : 35 - If the difference between discount of 35% and two successive discounts of 20% on a certain bill is Rs 3, then what is the amount (in Rs) of the bill?
Options:
1) 250
2) 300
3) 350
4) 400
QID : 36 - If a shopkeeper marks the price of goods 40% more than their cost price and allows a discount of 40%, then what is his gain or loss percent?
Options:
1) 16%, Loss
2) 16%, Profit
3) 10%, Loss
4) 12%, Profit
QID : 37 - The ratio of two positive numbers is 9 : 11. Their product is 6336. What is the smallest number?
Options:
1) 32
2) 72
3) 88
4) 48
QID : 38 - B starts some business by investing Rs 90000. After 4 months, D joins business by investing Rs 80000. At the end of the year, in what ratio will they share the profit?
Options:
1) 10 : 7
2) 9 : 4
3) 27 : 16
4) 7 : 3
QID : 39 - Average of 9 numbers is 20. If a number 30 is also included, then what will be the average of these 10 numbers?
Options:
1) 20.5
2) 21
3) 19.5
4) 21.5
QID : 40 - By selling 50 metres of cloth, a person gains the cost price of 20 metres of cloth. What is his gain percent?
Options:
1) 40
2) 25
3) 20
4) 10
QID : 41 - If some articles are bought at Rs 10 each and sold at Rs 7 each, then what is the loss percentage?
Options:
1) 60
2) 16.67
3) 25
4) 30
QID : 42 - A man saves 30% of his income in 1 year. If he wants to save the same amount in 8 months, then by how much percentage should he increase his monthly savings?
Options:
1) 20
2) 30
3) 40
4) 50
QID : 43 - A man spends 80% of his income and saves the rest. If his income and spending both increases by 10%, then what is the percentage change in his savings?
Options:
1) 10% increase
2) 5% decrease
3) 5% increase
4) 15% decrease
QID : 44 - A car travels at a speed of 25 m/s for 8 hours. What is the distance (in km) travelled by the car?
Options:
1) 360
2) 720
3) 450
4) 900
QID : 45 - A 450 meter long train crosses a bridge 650 meters long in 36 seconds. What is speed (in km/hr) of the train?
Options:
1) 110
2) 125
3) 150
4) 95
QID : 46 - The population of a town increases at the rate of 15% per annum. If the present population is 108445 of town, then what was the population 2 years ago?
Options:
1) 72000
2) 79000
3) 82000
4) 85000
QID : 47 -
What was the percentage change in number of customers to complex B from March to April?
Options:
1) 20
2) 40
3) 28
4) 56
QID : 48 -
What is the maximum difference (in thousands) between the numbers of customers in the 2 complexes among the given months?
Options:
1) 5
2) 6
3) 8
4) 4
QID : 49 -
What is the total number of customers (in thousands) in the 2 complexes in the month of April?
Options:
1) 52
2) 53
3) 50
4) 56 | HuggingFaceTB/finemath | |
# A stone is dropped from the top of a building how long does it take to fall 19.6m. how fast does it move the end of this fall? what is the acceleration after 1 second and 2 seconds? g=10m/s
1
by aniket210201
2015-09-12T05:41:31+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
H = 19.6 m
u = 0
h = u t + 1/2 g t * t
19.6 = 0 + 1/2 * 9.8 * t * t as g = 9.8 m/s/s
t = 2 sec. it takes 2 sec.
acceleration is always the same : g = 9.8 m/sec
velocity / speed at t = 2 sec is = u + g t = 0 + 9.8 * 2 sec = 19.6 m/sec
ty very much
:) thanks | HuggingFaceTB/finemath | |
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# Exercise 89 - Chapter 1
edited June 2018
There are five partitions possible on a set with three elements, say $$T = {12, 3, 4}$$.
Example 1.86. Let $$S = {1, 2, 3, 4}$$, $$T = {12, 3, 4}$$, and $$g: S \rightarrow T$$ by $$g(1) = g(2) = 12 , g(3) = 3, \text{ and } g(4) = 4$$.
Using the same $$S$$ and $$g: S \rightarrow T$$ as in Example 1.76, determine the partition $$g^*(c)$$ on $$S$$ for each of the five partitions $$c: T \twoheadrightarrow P$$.
• Options
1.
edited July 2018
The partitions on $$T$$ are
,,,,.
The images of these partitions under $$g^*$$ are
,,,,
where now $$1$$ and $$2$$ are distinct elements of $$S$$, though they always appear in the pulled-back partitions together (because $$g$$ maps them to the same element of $$T$$).
Comment Source:The partitions on \$$T\$$ are ,,,,. The images of these partitions under \$$g^*\$$ are ,,,, where now \$$1\$$ and \$$2\$$ are distinct elements of \$$S\$$, though they always appear in the pulled-back partitions together (because \$$g\$$ maps them to the same element of \$$T\$$). | HuggingFaceTB/finemath | |
### Home > CC2MN > Chapter 9 > Lesson 9.1.1 > Problem9-9
9-9.
Rewrite each expression using the Distributive Property.
1. $5x^2 + 35x$
What is a common factor between the two terms?
Factor out $5x$ from both terms.
$5x(x + 7)$
1. $8x(7-2x)$
Multiply both of the terms in the parentheses by what lies on the outside.
$56x-16x^2$
1. $9x − 3x^2$
Follow the steps in part (a).
1. $4(x-7)$
Follow the steps in part (b).
$4x-28$ | HuggingFaceTB/finemath | |
# Pure mathematics, Band 1
1874
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### Inhalt
STAGE 9 GEOMETRY 89 GEOMETRY 234 QUADRATIC EQUATIONS 296
PLANE TRIGONOMETRY 333 THE USE OF TABLES 358 HEIGHTS AND DISTANCES 381 ANSWERS 387
### Beliebte Passagen
Seite 272 - The angles in the same segment of a circle are equal to one another.
Seite 103 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.
Seite 233 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines, and let BC be divided into any...
Seite 112 - IF two triangles have two sides of the one equal to two sides of the...
Seite 128 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Seite 119 - If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles.
Seite 113 - ... equal angles in each ; then shall the other sides be equal, each to each ; and also the third angle of the one to the third angle of the other.
Seite 273 - The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles.
Seite 281 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Seite 121 - THE straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel. | HuggingFaceTB/finemath | |
# First lesson on GW¶
## The quasi-particle band structure of Silicon, in the GW approximation.¶
This lesson aims at showing how to calculate self-energy corrections to the DFT Kohn-Sham eigenvalues in the GW approximation.
A brief description of the formalism and of the equations implemented in the code can be found in the GW_notes. The different formulas of the GW formalism have been written in a pdf document by Valerio Olevano (who also wrote the first version of this tutorial). For a much more consistent discussion of the theoretical aspects of the GW method we refer the reader to the review
“Quasiparticle calculations in solids”, by Aulbur WG, Jonsson L, Wilkins JW,
in Solid State Physics 54, 1-218 (2000), also available here.
It is suggested to acknowledge the efforts of developers of the GW part of ABINIT, by citing the 2005 ABINIT publication.
The user should be familiarized with the four basic lessons of ABINIT, see the tutorial home page After this first tutorial on GW, you should read the second lesson on GW
This lesson should take about 2 hours.
## 1 General example of well converged GW calculation¶
Before beginning, you might consider to work in a different subdirectory as for the other lessons. Why not “Work_gw1”?
At the end of lesson 3, we computed the Kohn-Sham band structure of silicon. In this approximation, the band dispersion as well as the band widths are reasonable, but the band gaps are qualitatively wrong. Now, we will compute the band gaps much more accurately, using the so-called GW approximation.
We start by an example, in which we show how to perform in a single input file the calculation of the ground state density, the Kohn Sham band structure, the screening, and the GW corrections. We use reasonable values for the parameters of the calculation. The discussion on the convergence tests is postponed to the next paragraphs. We will see that GW calculations are much more time-consuming than the computation of the Kohn-Sham eigenvalues.
So, let us run immediately this calculation, and while it is running, we will explain what has been done.
In the directory ~abinit/tests/tutorial/Input/Work_gw1, copy the files ~abinit/tests/tutorial/Input/tgw1_x.files and tgw1_1.in, and modify the tgw1_x.files file as usual (see lesson 1).
Then, issue:
abinit < tgw1_x.files >& tgw1_1.log &
It is very important to run this job in background because it takes about 1 minute. In the meantime, you should read the following.
#### 1.a The four steps of a GW calculation.¶
In order to perform a standard one-shot GW calculation one has to:
1. Run a converged Ground State calculation to obtain the self-consistent density.
2. Perform a non self-consistent run to compute the Kohn-Sham eigenvalues and the eigenfunctions including several empty states. Note that, unlike standard band structure calculations, here the KS states must be computed on a regular grid of k-points.
3. Use optdriver=3 to compute the independent-particle susceptibility ( χ0) on a regular grid of q -points, for at least two frequencies (usually, ω=0 and a large purely imaginary frequency - of the order of the plasmon frequency, a dozen of eV). The inverse dielectric matrix (ε-1) is then obtained via matrix inversion and stored in an external file (SCR). The list of q -points is automatically defined by the k-mesh used to generate the KS states in the previous step.
4. Use optdriver=4 to compute the self-energy (Σ) matrix element for a given set of k-points in order to obtain the GW quasiparticle energies Note that the k-point must belong to the k-mesh used to generate the WFK file in step 2.
The flowchart diagram of a standard one-shot run is depicted in the figure below.
The input file tgw1_1.in has precisely that structure: there are four datasets.
The first dataset performs the SCF calculation to get the density. The second dataset reads the previous density file and performs a NSCF run including several empty states. The third dataset reads the WFK file produced in the previous step and drives the computation of susceptibility and dielectric matrices, producing another specialized file, tgw1_xo_DS2_SCR (_SCR for “Screening”, actually the inverse dielectric matrix ε-1). Then, in the fourth dataset, the code calculates the quasiparticle energies for the 4th and 5th bands at the Γ point.
So, you can edit this tgw1_1.in file.
The dataset-independent part of this file (the last half of the file), contains the usual set of input variables describing the cell, atom types, number, position, planewave cut-off energy, SCF convergence parameters driving the Kohn-Sham band structure calculation. Then, for the fourth datasets, you will find specialized additional input variables.
#### 1.b Generating the Kohn-Sham band structure: the WFK file.¶
Dataset 1 is a rather standard SCF calculation. It’s worth noticing that we use tolvrs to stop the SCF cycle because we want a well-converged KS potential to be used in the subsequent NSCF calculation. Dataset 2 computes 40 bands and we set nbdbuf to 5 so that only the first 35 states must be converged within tolwfr.
############
# Dataset 1
############
# SCF-GS run
nband1 6
tolvrs1 1.0e-10
############
# Dataset 2
############
# Definition of parameters for the calculation of the WFK file
nband2 40 # Number of (occ and empty) bands to be computed
nbdbuf2 5
iscf2 -2
getden2 -1
tolwfr2 1.0d-18 # Will stop when this tolerance is achieved
Important
The nbdbuf tricks allows us to save several minimization steps because the last bands usually require more iterations to converge
#### 1.c Generating the screening: the SCR file.¶
In dataset 3, the calculation of the screening (susceptibility, dielectric matrix) is performed. We need to set optdriver=3 to do that:
optdriver3 3 # Screening calculation
The getwfk input variable is similar to other “get” input variables of ABINIT:
getwfk3 -1 # Obtain WFK file from previous dataset
In this case, it tells the code to use the WFK file calculated in the previous dataset.
Then, three input variables describe the computation:
nband3 17 # Bands used in the screening calculation
ecut 8.0 # Cut-off energy of the planewave set to represent the wavefunctions
ecuteps3 3.6 # Cut-off energy of the planewave set to represent the dielectric matrix
In this case, we use 17 bands to calculate the Kohn-Sham response function $\chi^{(0)}_{KS}$. A cut-off of 8 Hartree is used to represent the wavefunctions in the calculation of $\chi^{(0)}_{KS}$. The dimension of $\chi^{(0)}_{KS}$, as well as all the other matrices ($\chi$, $\epsilon$) is determined by ecuteps=3.6 Hartree, giving 169 planewaves.
Finally, we define the frequencies at which the screening must be evaluated: ω=0.0 eV and the imaginary frequency ω= i 16.7 eV. The latter is determined by the input variable ppmfrq
ppmfrq3 16.7 eV # Imaginary frequency where to calculate the screening
The two frequencies are used to calculate the plasmon-pole model parameters. For the non-zero frequency it is recommended to use a value close to the plasmon frequency for the plasmon-pole model to work well. Plasmons frequencies are usually close to 0.5 Hartree. The parameters for the screening calculation are not far from the ones that give converged Energy Loss Function (-Im \epsilon^-1_00) spectra, So that one can start up by using indications from EELS calculations existing in literature.
#### 1.d Computing the GW energies.¶
In dataset 4 the calculation of the Self-Energy matrix elements is performed. One needs to define the driver option, as well as the _WFK and _SCR files.
optdriver4 4 # Self-Energy calculation
getwfk4 -2 # Obtain WFK file from dataset 2
getscr4 -1 # Obtain SCR file from previous dataset
The getscr input variable is similar to other “get” input variables of ABINIT.
Then, comes the definition of parameters needed to compute the self-energy. As for the computation of the susceptibility and dielectric matrices, one must define the set of bands, and two sets of planewaves:
nband4 30 # Bands to be used in the Self-Energy calculation
ecutsigx4 6.0 # Dimension of the G sum in Sigma_x
# (the dimension in Sigma_c is controlled by npweps)
In this case, nband controls the number of bands used to calculate the correlation part of the Self-Energy. ecutsigx gives the number of planewaves used to calculate σ x (the exchange part of the self-energy). The size of the planewave set used to compute Σc (the correlation part of the self-energy) is controlled by ecuteps (cannot be larger than the value used to generate the SCR file). However, it is taken equal to the number of planewave of Σx if the latter is smaller than the one for Σc.
Then, come the parameters defining the k-points and the band indices for which the quasiparticle energies will be computed:
nkptgw4 1 # number of k-point where to calculate the GW correction
kptgw4 # k-points
-0.125 0.000 0.000
bdgw4 4 5 # calculate GW corrections for bands from 4 to 5
nkptgw defines the number of k-points for which the GW corrections will be computed. The k-point reduced coordinates are specified in kptgw. At present, they MUST belong to the k-mesh used to generate the WFK file. Hence if you wish the GW correction in a particular k-point, you should choose a grid containing it. Usually this is done by taking the k-point grid where the convergence is achieved and shifting it such as at least one k-point is placed on the wished position in the Brillouin zone. bdgw gives the minimum/maximum band whose energies are calculated for the given k-point.
There is an additional parameter, called zcut, related to the self-energy computation. It is meant to avoid some divergences that might occur in the calculation due to integrable poles along the integration path.
#### 1.e Examination of the output file.¶
Let us hope that your calculation has been completed, and that we can examine the output file. Open tgw1_1.out in your preferred editor and find the section corresponding to DATASET 3.
After the description of the unit cell and of the pseudopotentials, you will find the list of k-points used for the electrons and the grid of q-point (in the Irreducible part of the Brillouin Zone) on which the susceptibility and dielectric matrices will be computed. It is a set of BZ points defined as all the possible differences among the k-points ( q=k-k’ ) of the grid chosen to generate the WFK file. From the last statement it is clear the interest to choose homogeneous k-point grids, in order not to minimize the number of q-points.
After this section, the code prints the parameters of the FFT grid needed to represent the wavefunctions and to compute their convolution (required for the screening matrices). Then we have some information about the MPI distribution of the bands and the total number of valence electrons computed by integrating the density in the unit cell.
On the basis of the density, one can obtain the classical Drude plasmon frequency. The next lines calculate the average density of the system, and evaluate the r_s parameter, then compute the Drude plasmon frequency. This is the value used by default for ppmfrq. It is in fact the second frequency where the code calculates the dielectric matrix to adjust the plasmon-pole model parameters. It has been found that Drude plasma frequency is a reasonable value where to adjust the model. The control over this parameter is however left to the user in order to check that the result does not change when changing ppmfrq. If it is the case, then the plasmon-pole model is not appropriated and one should go beyond by taking into account a full dynamical dependence in the screening (see later, the contour-deformation method). However, the plasmon-pole model has been found to work well for a very large range of systems when focusing only on the real part of the GW corrections.
At the end of the screening calculation, the macroscopic dielectric constant is printed:
dielectric constant = 13.5073
dielectric constant without local fields = 15.0536
Note that the convergence in the dielectric constant DOES NOT guarantee the convergence in the GW corrections. In fact, the dielectric constant is representative of only one element i.e. the head of the full dielectric matrix. Even if the convergence on the dielectric constant with local fields takes somehow into account also other non-diagonal elements. In a GW calculation all the ε-1 matrix is used to build the Self-Energy operator.
The dielectric constant here reported is the so-called RPA dielectric constant due to the electrons. Although evaluated at zero frequency, it is understood that the ionic response is not included (this term can be computed with DFPT and ANADDB). The RPA dielectric constant restricted to electronic effects is also not the same as the one computed in the RESPFN part of ABINIT, that includes exchange-correlation effects.
We now enter the fourth dataset. As for dataset 3, after some general information (origin of WFK file, header, description of unit cell, k-points, q-points), the description of the FFT grid and jellium parameters, there is the echo of parameters for the plasmon-pole model, and the inverse dielectric function (the screening). The self-energy operator has been constructed, and one can evaluate the GW energies, for each of the states.
The final results are:
k = -0.125 0.000 0.000
Band E0 <VxcLDA> SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.616 -11.115 -12.334 1.257 0.775 -0.290 -11.085 0.030 5.646
5 8.357 -10.140 -5.951 -3.336 0.779 -0.284 -9.476 0.664 9.021
E^0_gap 2.741
E^GW_gap 3.375
DeltaE^GW_gap 0.634
For the desired k-point, state 4, then state 5, one finds different information:
• E0 is the Kohn-Sham eigenenergy
• VxcLDA gives the average Kohn-Sham exchange-correlation potential
• SigX gives the exchange contribution to the self-energy
• SigC(E0) gives the correlation contribution to the self-energy, evaluated at the Kohn-Sham eigenenergy
• Z is the renormalisation factor
• dSigC/dE is the energy derivative of SigC with respect to the energy
• SigC(E) gives the correlation contribution to the self-energy, evaluated at the GW energy
• E-E0 is the difference between GW energy and Kohn-Sham eigenenergy
• E is the GW energy
In this case, the gap is also analyzed: E^0_gap is the direct Kohn-Sham gap at that k point (and spin, in the case of spin-polarized calculations), E^GW_gap is the GW one, and DeltaE^GW_gap is the difference. This direct gap is always computed between the band whose number is equal to the number of electrons in the cell divided by two (integer part, in case of spin-polarized calculation), and the next one. (Warning: for a metal, these two bands do not systematically lie below and above the Fermi energy - but the concept of a direct gap is not relevant in that case).
It is seen that the average Kohn-Sham exchange-correlation potential for the state 4 (a valence state) is very close to the exchange self-energy correction. For that state, the correlation correction is small, and the difference between Kohn-Sham and GW energies is also small (43 meV). By contrast, the exchange self-energy is much smaller than the average Kohn-Sham potential for the state 5 (a conduction state), but the correlation correction is much larger than for state 4. On the whole, the difference between Kohn- Sham and GW energies is not very large, but nevertheless, it is quite important when compared with the size of the gap.
## 2 Preparing convergence studies: Kohn-Sham structure (WFK file) and screening (SCR file)¶
In the following sections, we will perform different convergence studies. In order to keep the CPU time at a reasonable level, we will use fake WFK and SCR data. Moreover we will only consider the correction at the Γ point only. In this way, we will be able to verify convergence aspects that could be very cumbersome (at least in the framework of a tutorial) if more k-points were used. Testing the convergence with a Γ point only grid of k point represents a convenient approach although some caution should always be used.
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_2.in, and modify the tgw1_x.files file as usual. Edit the tgw1_2.in file, and take the time to examine it.
Then, issue:
abinit < tgw1_x.files >& tgw1_2.log &
After this step you will need the WFK and SCR files produced in this run for the next runs (up to 6.8). Move tgw1o_DS2_WFK to tgw1o_DS1_WFK and tgw1o_DS3_SCR to tgw1o_DS1_SCR.
The next sections are intended to show you how to find the converged parameters for a GW calculation. In principle, the following parameters might be used to decrease the CPU time and/or the memory requirements: optdriver=3 ecutwfn, ecuteps, nband. and for optdriver=4, ecutwfn, ecutsigx, nband.
Before 2008, the advice was indeed to check independently what was the best value for each of these. However, with the evolution of memory/disk space, as well as the advent of new techniques to diminish the number of bands that is needed (see e.g. F. Bruneval, X. Gonze, Phys. Rev. B 78, 085125 (2008), and the input variable gwcomp), standard calculations nowadays only need the tuning of nband ecuteps, simultaneously for optdriver=3 and =4. Indeed, ecutwfn and can have the default value of ecut, while ecutsigx can have the default value of 4*ecut for norm-conserving pseudopotentials, or pawecutdg for PAW calculations. Actually, the present tutorial needs to be updated to account for the current practice.
We begin by the convergence study on the three parameters needed in the self- energy calculation (optdriver=4): ecutwfn, ecutsigx, nband. This is because for these, we will not need a double dataset loop to check this convergence, and we will rely on the previously determined SCR file.
## 3 Convergence on the number of planewaves in the wavefunctions to calculate the Self-Energy (optional)¶
The convergence study is done in the input file tgw1_3.in. First, we check the convergence on the number of planewaves used to describe the wavefunctions, in the calculation of the Self-Energy. This will be done by defining five datasets, with increasing ecutwfn:
ndtset 5
ecutwfn: 3.0
ecutwfn+ 1.0
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_3.in, and modify the tgw1_x.files file as usual. Edit the tgw1_3.in file, and take the time to examine it.
Then, issue:
abinit < tgw1_x.files >& tgw1_3.log &
Edit the output file. The number of plane waves used for the wavefunctions in the computation of the self-energy is mentioned in the fragments of output:
SIGMA fundamental parameters:
PLASMON POLE MODEL
number of plane-waves for SigmaX 169
number of plane-waves for SigmaC and W 169
number of plane-waves for wavefunctions 59
Gathering the GW energies for each planewave set, one gets:
number of plane-waves for wavefunctions 59
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.490 -15.198 4.050 0.812 -0.232 -11.212 0.277 6.192
5 8.445 -9.431 -3.101 -5.381 0.822 -0.216 -8.651 0.781 9.226
number of plane-waves for wavefunctions 113
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.648 -15.253 3.853 0.807 -0.239 -11.448 0.200 6.115
5 8.445 -9.685 -3.234 -5.511 0.818 -0.222 -8.916 0.769 9.214
number of plane-waves for wavefunctions 137
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.244 3.810 0.806 -0.241 -11.474 0.165 6.080
5 8.445 -9.675 -3.213 -5.557 0.818 -0.223 -8.935 0.740 9.185
number of plane-waves for wavefunctions 169
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.244 3.801 0.806 -0.241 -11.481 0.157 6.072
5 8.445 -9.686 -3.216 -5.571 0.818 -0.223 -8.950 0.736 9.181
number of plane-waves for wavefunctions 259
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.650 -15.250 3.790 0.806 -0.241 -11.497 0.152 6.068
5 8.445 -9.697 -3.218 -5.583 0.818 -0.223 -8.965 0.733 9.178
So that ecutwfn=5.0 (%npwwfn=137) can be considered to lead to eigenenergies converged within 0.01 eV.
## 4 Convergence on the number of planewaves to calculate Σx (optional)¶
Second, we check the convergence on the number of planewaves in the calculation of Σx. This study in done in tgw1_4.in As mentioned in the documentation ecutsigx, safe values exist for ecutsigx, so that if you do not want to squeeze the CPU time for your calculation (you might gain only a few percent in some cases …), you can impose these values, and skip the corresponding convergence study.
In this lesson, this convergence study will be done by defining five datasets, with increasing ecutsigx:
ndtset 7
ecutsigx: 3.0
ecutsigx+ 1.0
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_4.in, and modify the tgw1_x.files file as usual. Edit the tgw1_4.in file, and take the time to examine it.
Then, issue:
abinit < tgw1_x.files >& tgw1_4.log &
Edit the output file. The number of plane waves used for Σx is mentioned in the fragments of output:
SIGMA fundamental parameters:
PLASMON POLE MODEL
number of plane-waves for SigmaX 59
number of plane-waves for SigmaC and W 59
Gathering the GW energies for each planewave set, one gets:
number of plane-waves for SigmaX 59
number of plane-waves for SigmaC and W 59
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.194 3.886 0.808 -0.238 -11.372 0.268 6.183
5 8.445 -9.675 -3.174 -5.573 0.818 -0.222 -8.916 0.759 9.204
number of plane-waves for SigmaX 113
number of plane-waves for SigmaC and W 113
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.236 3.825 0.806 -0.240 -11.455 0.184 6.099
5 8.445 -9.675 -3.208 -5.561 0.818 -0.222 -8.934 0.741 9.186
number of plane-waves for SigmaX 137
number of plane-waves for SigmaC and W 137
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.241 3.815 0.806 -0.240 -11.468 0.172 6.087
5 8.445 -9.675 -3.211 -5.558 0.818 -0.223 -8.933 0.741 9.187
number of plane-waves for SigmaX 169
number of plane-waves for SigmaC and W 169
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.244 3.810 0.806 -0.241 -11.474 0.165 6.080
5 8.445 -9.675 -3.213 -5.557 0.818 -0.223 -8.935 0.740 9.185
number of plane-waves for SigmaX 259
number of plane-waves for SigmaC and W 169
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.246 3.810 0.806 -0.241 -11.475 0.164 6.079
5 8.445 -9.675 -3.215 -5.557 0.818 -0.223 -8.937 0.738 9.183
number of plane-waves for SigmaX 283
number of plane-waves for SigmaC and W 169
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.246 3.810 0.806 -0.241 -11.475 0.164 6.079
5 8.445 -9.675 -3.215 -5.557 0.818 -0.223 -8.937 0.738 9.183
number of plane-waves for SigmaX 283
number of plane-waves for SigmaC and W 169
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.915 -11.639 -15.246 3.810 0.806 -0.241 -11.475 0.164 6.079
5 8.445 -9.675 -3.215 -5.557 0.818 -0.223 -8.937 0.738 9.183
So that ecutsigx=6.0 (npwsigx=169) can be considered converged within 0.01 eV.
## 5 Convergence on the number of bands to calculate Σc (important)¶
At last, as concerns the computation of the self-energy, we check the convergence on the number of bands in the calculation of Σc. This convergence study is rather important, usually, BUT it can be done at the same time as the convergence study for the number of bands for the dielectric matrix.
The convergence on the number of bands to calculate the Self-Energy will be done by defining five datasets, with increasing nband:
ndtset 5
nband: 50
nband+ 50
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_5.in, and modify the tgw1_x.files file as usual. Edit the tgw1_5.in file, and take the time to examine it.
Then, issue:
abinit < tgw1_x.files >& tgw1_5.log &
Edit the output file. The number of bands used for the self-energy is mentioned in the fragments of output:
SIGMA fundamental parameters:
PLASMON POLE MODEL
number of plane-waves for SigmaX 169
number of plane-waves for SigmaC and W 169
number of plane-waves for wavefunctions 137
number of bands 50
Gathering the GW energies for each number of bands, one gets:
number of bands 50
4 5.915 -11.639 -15.244 3.878 0.807 -0.240 -11.419 0.220 6.135
5 8.445 -9.675 -3.213 -5.492 0.819 -0.222 -8.881 0.794 9.239
number of bands 100
4 5.915 -11.639 -15.244 3.810 0.806 -0.241 -11.474 0.165 6.080
5 8.445 -9.675 -3.213 -5.557 0.818 -0.223 -8.935 0.740 9.185
number of bands 150
4 5.915 -11.639 -15.244 3.805 0.806 -0.241 -11.478 0.161 6.076
5 8.445 -9.675 -3.213 -5.563 0.818 -0.223 -8.940 0.735 9.180
number of bands 200
4 5.915 -11.639 -15.244 3.804 0.806 -0.241 -11.479 0.160 6.075
5 8.445 -9.675 -3.213 -5.564 0.818 -0.223 -8.940 0.734 9.180
number of bands 250
4 5.915 -11.639 -15.244 3.804 0.806 -0.241 -11.479 0.160 6.075
5 8.445 -9.675 -3.213 -5.564 0.818 -0.223 -8.941 0.734 9.179
So that nband=100 can be considered converged within 0.01 eV.
At this stage, we know that for the self-energy computation, we need ecutwfn=5.0 ecutsigx=6.0, nband=100 .
## 6 Convergence on the number of planewaves in the wavefunctions to calculate the screening (ε-1) (optional)¶
Now, we come back to the calculation of the screening. Adequate convergence studies will couple the change of parameters for optdriver=3 with a computation of the GW energy changes. One cannot rely on the convergence of the macroscopic dielectric constant to assess the convergence of the GW energies.
As a consequence, we will define a double loop over the datasets:
ndtset 10
udtset 5 2
The datasets 12,22,32,42 and 52, drive the computation of the GW energies:
# Calculation of the Self-Energy matrix elements (GW corrections)
optdriver?2 4
getscr?2 -1
ecutwfn?2 5.0
ecutsigx 6.0
nband?2 100
The datasets 11,21,31,41 and 51, drive the corresponding computation of the screening:
# Calculation of the screening (epsilon^-1 matrix)
optdriver?1 3
In this latter series, we will have to vary the three different parameters ecutwfn, ecuteps and nband.
First, we check the convergence on the number of planewaves to describe the wavefunctions, in the calculation of the screening. This will be done by defining five datasets, with increasing ecutwfn:
ecutwfn:? 3.0
ecutwfn+? 1.0
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_6.in, and modify the tgw1_x.files file as usual. Edit the tgw1_6.in file, and take the time to examine it.
Then, issue:
abinit < tgw1_x.files >& tgw1_6.log &
Edit the output file. The number of plane waves used for the wavefunctions in the computation of the screening is mentioned in the fragments of output:
EPSILON^-1 parameters (SCR file):
dimension of the eps^-1 matrix 169
number of plane-waves for wavefunctions 59
Gathering the macroscopic dielectric constant and GW energies for each planewave set, one gets:
dielectric constant = 99.4320
dielectric constant without local fields = 147.6068
number of plane-waves for wavefunctions 59
4 5.915 -11.639 -15.244 3.843 0.811 -0.233 -11.446 0.193 6.108
5 8.445 -9.675 -3.213 -5.527 0.819 -0.221 -8.909 0.766 9.211
dielectric constant = 99.8529
dielectric constant without local fields = 144.5675
number of plane-waves for wavefunctions 113
4 5.915 -11.639 -15.244 3.789 0.804 -0.244 -11.492 0.147 6.063
5 8.445 -9.675 -3.213 -5.564 0.817 -0.224 -8.941 0.734 9.179
dielectric constant = 99.5260
dielectric constant without local fields = 143.7201
number of plane-waves for wavefunctions 137
4 5.915 -11.639 -15.244 3.779 0.804 -0.243 -11.500 0.140 6.055
5 8.445 -9.675 -3.213 -5.568 0.816 -0.226 -8.945 0.729 9.175
dielectric constant = 98.2593
dielectric constant without local fields = 142.5976
number of plane-waves for wavefunctions 169
4 5.915 -11.639 -15.244 3.772 0.802 -0.248 -11.505 0.134 6.049
5 8.445 -9.675 -3.213 -5.573 0.815 -0.227 -8.951 0.724 9.169
dielectric constant = 96.8379
dielectric constant without local fields = 141.0644
number of plane-waves for wavefunctions 259
4 5.915 -11.639 -15.244 3.769 0.804 -0.244 -11.508 0.131 6.047
5 8.445 -9.675 -3.213 -5.578 0.815 -0.227 -8.954 0.721 9.166
So that ecutwfn=4.0 (%npwwfn=113) can be considered to lead to eigenenergies converged within 0.01 eV.
## 7 Convergence on the number of bands to calculate the screening (ε-1) (important)¶
This convergence study is rather important. It can be done at the same time as the convergence study for the number of bands for the self-energy. Note that the number of bands used to calculate both the screening and the self-energy can be lowered by a large amount by resorting to the extrapolar technique (see the input variable gwcomp).
Second, we check the convergence on the number of bands in the calculation of the screening. This will be done by defining five datasets, with increasing nband:
nband11 25
nband21 50
nband31 100
nband41 150
nband51 200
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_7.in, and modify the tgw1_x.files file as usual. Edit the tgw1_7.in file, and take the time to examine it.
Then, issue:
abinit < tgw1_x.files >& tgw1_7.log &
Edit the output file. The number of bands used for the wavefunctions in the computation of the screening is mentioned in the fragments of output:
EPSILON^-1 parameters (SCR file):
dimension of the eps^-1 matrix 169
number of plane-waves for wavefunctions 113
number of bands 25
Gathering the macroscopic dielectric constant and GW energies for each number of bands, one gets:
dielectric constant = 99.8529
dielectric constant without local fields = 144.5675
number of bands 25
4 5.915 -11.639 -15.244 3.789 0.804 -0.244 -11.492 0.147 6.063
5 8.445 -9.675 -3.213 -5.564 0.817 -0.224 -8.941 0.734 9.179
dielectric constant = 100.9503
dielectric constant without local fields = 144.5701
number of bands 50
4 5.915 -11.639 -15.244 3.624 0.806 -0.241 -11.624 0.015 5.930
5 8.445 -9.675 -3.213 -5.738 0.814 -0.228 -9.085 0.590 9.035
dielectric constant = 101.2722
dielectric constant without local fields = 144.5703
number of bands 100
4 5.915 -11.639 -15.244 3.577 0.807 -0.239 -11.662 -0.023 5.892
5 8.445 -9.675 -3.213 -5.792 0.813 -0.230 -9.131 0.544 8.989
dielectric constant = 101.3772
dielectric constant without local fields = 144.5703
number of bands 150
4 5.915 -11.639 -15.244 3.568 0.807 -0.240 -11.669 -0.030 5.885
5 8.445 -9.675 -3.213 -5.800 0.813 -0.230 -9.137 0.538 8.983
dielectric constant = 101.3814
dielectric constant without local fields = 144.5703
number of bands 200
4 5.915 -11.639 -15.244 3.568 0.807 -0.240 -11.669 -0.030 5.885
5 8.445 -9.675 -3.213 -5.801 0.813 -0.230 -9.137 0.537 8.983
So that the computation using 100 bands can be considered converged within 0.01 eV.
## 8 Convergence on the dimension of the screening ε-1 matrix (important)¶
Third, we check the convergence on the number of plane waves in the calculation of the screening. This will be done by defining six datasets, with increasing ecuteps:
ecuteps:? 3.0
ecuteps+? 1.0
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_8.in, and modify the tgw1_x.files file as usual. Edit the tgw1_8.in file, and take the time to examine it.
Then, issue:
abinit < tgw1_x.files >& tgw1_8.log &
Edit the output file. The number of bands used for the wavefunctions in the computation of the screening is mentioned in the fragments of output:
EPSILON^-1 parameters (SCR file):
dimension of the eps^-1 matrix 59
Gathering the macroscopic dielectric constant and GW energies for each number of bands, one gets:
dielectric constant = 102.1696
dielectric constant without local fields = 144.5703
dimension of the eps^-1 matrix 59
4 5.915 -11.639 -15.244 3.721 0.808 -0.237 -11.545 0.094 6.009
5 8.445 -9.675 -3.213 -5.805 0.813 -0.230 -9.141 0.534 8.979
dielectric constant = 101.3721
dielectric constant without local fields = 144.5703
dimension of the eps^-1 matrix 113
4 5.915 -11.639 -15.244 3.613 0.807 -0.239 -11.633 0.007 5.922
5 8.445 -9.675 -3.213 -5.799 0.813 -0.230 -9.136 0.539 8.984
dielectric constant = 101.3560
dielectric constant without local fields = 144.5703
dimension of the eps^-1 matrix 137
4 5.915 -11.639 -15.244 3.591 0.807 -0.239 -11.651 -0.012 5.904
5 8.445 -9.675 -3.213 -5.793 0.813 -0.230 -9.131 0.543 8.989
dielectric constant = 101.2722
dielectric constant without local fields = 144.5703
dimension of the eps^-1 matrix 169
4 5.915 -11.639 -15.244 3.577 0.807 -0.239 -11.662 -0.023 5.892
5 8.445 -9.675 -3.213 -5.792 0.813 -0.230 -9.131 0.544 8.989
dielectric constant = 101.2405
dielectric constant without local fields = 144.5703
dimension of the eps^-1 matrix 259
4 5.915 -11.639 -15.244 3.577 0.807 -0.239 -11.662 -0.023 5.892
5 8.445 -9.675 -3.213 -5.792 0.813 -0.230 -9.130 0.544 8.990
dielectric constant = 101.2404
dielectric constant without local fields = 144.5703
dimension of the eps^-1 matrix 283
4 5.915 -11.639 -15.244 3.577 0.807 -0.239 -11.662 -0.023 5.892
5 8.445 -9.675 -3.213 -5.792 0.813 -0.230 -9.130 0.544 8.990
So that ecuteps=6.0 (npweps=169) can be considered converged within 0.01 eV.
At this stage, we know that for the screening computation, we need ecutwfn=4.0 ecuteps=6.0, nband=100.
Of course, until now, we have skipped the most difficult part of the convergence tests: the convergence in the number of k-points. It is as important to check the convergence on this parameter, than on the other ones. However, this might be very time consuming, since the CPU time scales as the square of the number of k points (roughly), and the number of k-points can increase very rapidly from one possible grid to the next denser one. This is why we will leave this out of the present tutorial, and consider that we already know a sufficient k-point grid, for the last calculation.
## 9 Calculation of the GW corrections for the band gap in Γ¶
Now we try to perform a GW calculation for a real problem: the calculation of the GW corrections for the direct band gap of bulk Silicon in Γ.
In directory ~abinit/tests/tutorial/Input/Work_gw1, copy the file ../tgw1_9.in, and modify the tgw1_x.files file as usual. Then, edit the tgw1_9.in file, and, without examining it, comment the line
ngkpt 2 2 2 # Density of k points used for the automatic tests of the tutorial
and uncomment the line
#ngkpt 4 4 4 # Density of k points needed for a converged calculation
Then, issue:
abinit < tgw1_x.files >& tgw1_9.log &
This job lasts about 1 minute so it is worth to run it before the examination of the input file. Now, you can examine it.
We need the usual part of the input file to perform a ground state calculation. This is done in dataset 1 and at the end we print out the density. We use a 4x4x4 FCC grid (so, 256 k points in the full Brillouin Zone), shifted, because it is the most economical. It gives 10 k-points in the Irreducible part of the Brillouin Zone. However, this k-point grid does not contains the Γ point, and, at present, one cannot perform calculations of the self-energy corrections for other k points than those present in the grid of k-points in the WFK file.
Then in dataset 2 we perform a non self-consistent calculation to calculate the Kohn-Sham structure in a set of 19 k-points in the Irreducible Brillouin Zone. This set of k-points is also derived from a 4x4x4 FCC grid, but a NON- SHIFTED one. It has the same density of points as the 10 k-point set, but the symmetries are not used in a very efficient way. However, this set contains the Γ point, which allows us to tackle the computation of the band gap at this point.
In dataset 3 we calculate the screening. The screening calculation is very time-consuming. So, we have decided to decrease a bit the parameters found in the previous convergence studies. Indeed, ecutwfn has been decreased from 4.0 to 3.6. This is rather innocuous. Also, nband has been decreased from 100 to 25. This is a drastic change. The CPU time of this part is linear with respect to this parameter (or more exactly, with the number of conduction bands). Thus, the CPU time has been decreased by a factor of 4. Referring to our previous convergence study, we see that the absolute accuracy on the GW energies is now on the order of 0.2 eV only. However, the gap energy (difference between valence and conduction states), that is the relative accuracy, is likely correct within 0.02 eV. It is very important to clarify this point: in bulk systems what matters is only the relative accuracy. There is no zero of the energy defined for a bulk system. Hence in these systems one CAN WELL check the convergence only on the relative accuracy on the energies rather than the absolute, by checking the convergence on the band gap for example. This will reduce a lot the values to be found for the convergence parameters. The same holds for 2-, 1-, and 0-dimensional systems if one is interested only on relative energies and is not interested in calculating quantities like the work function.
Finally in dataset 4 we calculate the self-energy matrix element in Γ, using the previously determined parameters.
You should obtain the following results:
k = 0.000 0.000 0.000
Band E0 VxcLDA SigX SigC(E0) Z dSigC/dE Sig(E) E-E0 E
4 5.951 -11.253 -12.628 1.187 0.777 -0.287 -11.398 -0.146 5.806
5 8.464 -10.042 -5.563 -3.878 0.777 -0.287 -9.575 0.467 8.931
E^0_gap 2.513
E^GW_gap 3.126
DeltaE^GW_gap 0.613
So that the LDA energy gap in Γ is about 2.53 eV, while the GW correction is about 0.64 eV, so that the GW band gap found is 3.17 eV.
One can compare now what have been obtained to what one can get from the literature.
EXP 3.40 eV Landolt-Boernstein
LDA 2.57 eV L. Hedin, Phys. Rev. 139, A796 (1965)
LDA 2.57 eV M.S. Hybertsen and S. Louie, PRL 55, 1418 (1985)
LDA (FLAPW) 2.55 eV N. Hamada, M. Hwang and A.J. Freeman, PRB 41, 3620 (1990)
LDA (PAW) 2.53 eV B. Arnaud and M. Alouani, PRB 62, 4464 (2000)
LDA 2.53 eV present work
GW 3.27 eV M.S. Hybertsen and S. Louie, PRL 55, 1418 (1985)
GW 3.35 eV M.S. Hybertsen and S. Louie, PRB 34, 5390 (1986)
GW 3.30 eV R.W. Godby, M. Schlueter, L.J. Sham, PRB 37, 10159 (1988)
GW (FLAPW) 3.30 eV N. Hamada, M. Hwang and A.J. Freeman, PRB 41, 3620 (1990)
GW (PAW) 3.15 eV B. Arnaud and M. Alouani, PRB 62, 4464 (2000)
GW (FLAPW) 3.12 eV W. Ku and A.G. Eguiluz, PRL 89, 126401 (2002)
GW 3.17 eV present work
The values are spread over an interval of 0.2 eV. They depend on the details of the calculation. In the case of pseudopotential calculations, they depend of course on the pseudopotential used. However, a GW result is hardly meaningful beyond 0.1 eV, in the present state of the art. But this goes also with the other source of inaccuracy, the choice of the pseudopotential, that can arrive up to even 0.2 eV. This can also be taken into account when choosing the level of accuracy for the convergence parameters in the GW calculation.
Finally, it is possible to calculate a full band plot of a system. There are two possible techniques. The first one is based on the use of Wannier functions, to interpolate a few selected points obtained using the direct GW approach. You need to have the Wannier90 plug-in installed. See the directory tests/wannier90, test case 03, for an example of a file where a GW calculation is followed by the use of Wannier90. Another practical way follows from the fact that the GW corrections are quite linear with the energy, for each group of bands. This is evident when reporting on a plot the GW correction with respect to the 0-order LDA energy for each state. One can then simply correct the Kohn-Sham band structure at any point, by using a GW correction for the k-points where it has not been calculated explicitly, using a fit of the GW correction at a sparse set of points.
## Advanced features in the GW code¶
The user might switch to the [lesson:gw2|second GW tutorial] before coming back to the present section.
#### Calculations without using the Plasmon-Pole model¶
In order to circumvent the plasmon-pole model, the GW frequency convolution has to be calculated explicitly along the real axis. This is a tough job, since G and W have poles along the real axis. Therefore it is more convenient to use another path of integration along the imaginary axis plus the residues enclosed in the path.
Consequently, it is better to evaluate the screening for imaginary frequencies (to perform the integration) and also for real frequencies (to evaluate the contributions of the residues that may enter into the path of integration). The number of imaginary frequencies is set by the input variable nfreqim. The regular grid of real frequencies is determined by the input variables nfreqre, which sets the number of real frequencies, and freqremax, which indicates the maximum real frequency used.
The method is particularly suited to output the spectral function (contained in file out.sig). The grid of real frequencies used to calculate the spectral function is set by the number of frequencies (input variable nfreqsp) and by the maximum frequency calculated (input variable freqspmax).
#### Self-consistent calculations¶
The details in the implementation and the justification for the approximations retained can be found in F. Bruneval, N. Vast, and L. Reining, Phys. Rev. B 74 , 045102 (2006).
The only added input variables are getqps and irdqps. These variables concerns the reading of the _QPS file, that contains the eigenvalues and the unitary transform matrices of a previous quasiparticle calculation. QPS stands for “QuasiParticle Structure”.
The only modified input variables for self-consistent calculations are gwcalctyp and bdgw.
When the variable gwcalctyp is in between 0 and 9, The code calculates the quasiparticle energies only and does not output any QPS file (as in a standard GW run).
When the variable gwcalctyp is in between 10 and 19, the code calculates the quasiparticle energies only and outputs them in a QPS file.
When the variable gwcalctyp is in between 20 and 29, the code calculates the quasiparticle energies and wavefunctions and outputs them in a QPS file.
For a full self-consistency calculation, the quasiparticle wavefunctions are expanded in the basis set of the Kohn-Sham wavefunctions. The variable bdgw now indicates the size of all matrices to be calculated and diagonalized. The quasiparticle wavefunctions are consequently linear combinations of the Kohn-Sham wavefunctions in between the min and max values of bdgw.
A correct self-consistent calculation should consist of the following runs:
• 1) Self-consistent Kohn-Sham calculation: outputs a WFK file
• 2) Screening calculation (with Kohn-Sham inputs): outputs a SCR file
• 3) Sigma calculation (with Kohn-Sham inputs): outputs a QPS file
• 4) Screening calculation (with the WFK, and QPS file as an input): outputs a new SCR file
• 5) Sigma calculation (with the WFK, QPS and the new SCR files): outputs a new QPS file
• 6) Screening calculation (with the WFK, the new QPS file): outputs a newer SCR file
• 7) Sigma calculation (with the WFK, the newer QPS and SCR files): outputs a newer QPS
• ............ and so on, until the desired accuracy is reached
Note that for Hartree-Fock calculations a dummy screening is required for initialization reasons. Therefore, a correct HF calculations should look like
• 1) Self-consistent Kohn-Sham calculation: outputs a WFK file
• 2) Screening calculation using very low convergence parameters (with Kohn-Sham inputs): output a dummy SCR file
• 3) Sigma calculation (with Kohn-Sham inputs): outputs a QPS file
• 4) Sigma calculation (with the WFK and QPS files): outputs a new QPS file
• 5) Sigma calculation (with the WFK and the new QPS file): outputs a newer QPS file
• ............ and so on, until the desired accuracy is reached
In the case of a self-consistent calculation, the output is slightly more complex:
For instance, iteration 2
k = 0.500 0.250 0.000
Band E_lda <Vxclda> E(N-1) <Hhartree> SigX SigC[E(N-1)] Z dSigC/dE Sig[E(N)] DeltaE E(N)_pert E(N)_diago
1 -3.422 -10.273 -3.761 6.847 -15.232 4.034 1.000 0.000 -11.198 -0.590 -4.351 -4.351
2 -0.574 -10.245 -0.850 9.666 -13.806 2.998 1.000 0.000 -10.807 -0.291 -1.141 -1.141
3 2.242 -9.606 2.513 11.841 -11.452 1.931 1.000 0.000 -9.521 -0.193 2.320 2.320
4 3.595 -10.267 4.151 13.866 -11.775 1.842 1.000 0.000 -9.933 -0.217 3.934 3.934
5 7.279 -8.804 9.916 16.078 -4.452 -1.592 1.000 0.000 -6.044 0.119 10.034 10.035
6 10.247 -9.143 13.462 19.395 -4.063 -1.775 1.000 0.000 -5.838 0.095 13.557 13.557
7 11.488 -9.704 15.159 21.197 -4.061 -1.863 1.000 0.000 -5.924 0.113 15.273 15.273
8 11.780 -9.180 15.225 20.958 -3.705 -1.893 1.000 0.000 -5.598 0.135 15.360 15.360
E^0_gap 3.684
E^GW_gap 5.764
DeltaE^GW_gap 2.080
The columns are
• Band : index of the band
• E_lda : LDA eigenvalue
• < Vxclda>: diagonal expectation value of the xc potential in between LDA bra and ket
• E(N-1) : quasiparticle energy of the previous iteration (equal to LDA for the first iteration)
• < Hhartree>: diagonal expectation value of the Hartree Hamiltonian (equal to E_lda - for the first iteration only)
• SigX : diagonal expectation value of the exchange self-energy
• SigC[E(N-1)] : diagonal expectation value of the correlation self-energy (evaluated for the energy of the preceeding iteration)
• Z : quasiparticle renormalization factor Z (taken equal to 1 in methods HF, SEX, COHSEX and model GW)
• dSigC/dE : Derivative of the correlation self-energy with respect to the energy
• Sig[E(N)] : Total self-energy for the new quasiparticle energy
• DeltaE : Energy difference with respect to the previous step
• E(N)_pert : QP energy as obtained by the usual perturbative method
• E(N)_diago : QP energy as obtained by the full diagonalization | open-web-math/open-web-math | |
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Lec10-hmm
# Lec10-hmm - Hidden Markov Models CMSC 423 Based on Chapter...
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Hidden Markov Models CMSC 423 Based on Chapter 11 of Jones & Pevzner, An Introduction to Bioinformatics Algorithms
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Checking a Casino Heads/Tails: ↑ ↑ ↓ ↓ ↓ ↓ ↑ ↑ ↑ ↑ Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 ? Suppose either a fair or biased coin was used to generate a sequence of heads & tails. But we don’t know which type of coin was actual used.
Checking a Casino Heads/Tails: ↑ ↑ ↓ ↓ ↓ ↓ ↑ ↑ ↑ ↑ Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 ? Suppose either a fair or biased coin was used to generate a sequence of heads & tails. But we don’t know which type of coin was actual used.
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Checking a Casino Heads/Tails: ↑ ↑ ↓ ↓ ↓ ↓ ↑ ↑ ↑ ↑ Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 ? Suppose either a fair or biased coin was used to generate a sequence of heads & tails. But we don’t know which type of coin was actual used. How could we guess which coin was more likely?
Compute the Probability of the Observed Sequence Pr(x | Fair) = Pr(x | Biased) = X = Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.75 0.75 0.25 0.25 0.25 0.25 0.75
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Compute the Probability of the Observed Sequence Pr(x | Fair) = Pr(x | Biased) = X = Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.75 0.75 0.25 0.25 0.25 0.25 0.75 = 0.5 7 = 0.0078125 = 0.001647949 × × × × × × × × × × × ×
Compute the Probability of the Observed Sequence Pr(x | Fair) = Pr(x | Biased) = X = Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.75 0.75 0.25 0.25 0.25 0.25 0.75 = 0.5 7 = 0.0078125 = 0.001647949 × × × × × × × × × × × × log 2 = Pr(x | Fair) Pr(x | Biased) 0.0078 0.0016 = 2.245 The log-odds score : > 0. Hence “Fair” is a better guess. log 2
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What if the casino switches coins? Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 Probability of switching coins = 0.1 Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 0.1 0.1
What if the casino switches coins? Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 Probability of switching coins = 0.1 How could we guess which coin was more likely at each position ? Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 0.1 0.1
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What if the casino switches coins? Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 Probability of switching coins = 0.1 How could we guess which coin was more likely at each position ? How can we compute the probability of the entire sequence? Fair coin: Pr(Heads) = 0.5 Biased coin: Pr(Heads) = 0.75 0.1 0.1
What does this have to do with biology? atg gat ggg agc aga tca gat cag atc agg gac gat aga cga tag tga
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What does this have to do with biology?
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Ask a homework question - tutors are online | HuggingFaceTB/finemath | |
# Kathy Kool needs some help
1. Nov 11, 2008
### jp0108
1. The problem statement, all variables and given/known data
A young woman named Kathy Kool buys a
sports car that can accelerate at the rate of
4.56 m/s2. She decides to test the car by drag
racing with another speedster, Stan Speedy.
Both start from rest, but experienced Stan
leaves the starting line 0.51 s before Kathy.
Stan moves with a constant acceleration of
3.43 m/s2 and Kathy maintains an accelera-
tion of 4.56 m/s2.
Find the time it takes Kathy to overtake
Stan. Answer in units of s.
The formulas are:
vf = vi + at
d = vit + 1/2at^2
d = 1/2 (vi + vf) delta t
a = delta v/ t
v = d/t
KEY:
v: m/s
d: m
t: s
a: m/s^2
vi: initial velocity
vf: finial velocity
delta v: change in velocity
Last edited: Nov 11, 2008
2. Nov 11, 2008
### Staff: Mentor
You need to show us the relevant equations, and show us your attempt at a solution before we can help you. That's the PF Rules (see the link at the top of the page).
3. Nov 11, 2008
### jp0108
the formulas are:
vf = vi + at
d = vit + 1/2at^2
d = 1/2 (vi + vf) delta t
a = delta v/ t
v = d/t
KEy:
v: m/s
d: m
t: s
a: m/s^2
vi: initial velocity
vf: finial velocity
delta v: change in velocity
Last edited: Nov 11, 2008
4. Nov 11, 2008
### Staff: Mentor
This would be the most useful equation of the bunch for this problem. Why?
Now write the two equations for the d(t) of the two cars, including the initial conditions you've been given, and find at what time t the two d(t) numbers are equal... | HuggingFaceTB/finemath | |
Anda di halaman 1dari 3
# Matrix and Index Notation
David Roylance
Department of Materials Science and Engineering
Massachusetts Institute of Technology
Cambridge, MA 02139
September 18, 2000
A vector can be described by listing its components along the xyz cartesian axes; for in-
stance the displacement vector u can be denoted as u
x
, u
y
, u
z
, using letter subscripts to indicate
the individual components. The subscripts can employ numerical indices as well, with 1, 2,
and 3 indicating the x, y, and z directions; the displacement vector can therefore be written
equivalently as u
1
, u
2
, u
3
.
A common and useful shorthand is simply to write the displacement vector as u
i
, where the
i subscript is an index that is assumed to range over 1,2,3 ( or simply 1 and 2 if the problem is
a two-dimensional one). This is called the range convention for index notation. Using the range
convention, the vector equation u
i
= a implies three separate scalar equations:
u
1
= a
u
2
= a
u
3
= a
We will often nd it convenient to denote a vector by listing its components in a vertical list
enclosed in braces, and this form will help us keep track of matrix-vector multiplications a bit
more easily. We therefore have the following equivalent forms of vector notation:
u = u
i
=
_
_
u
1
u
2
u
3
_
_
=
_
_
u
x
u
y
u
z
_
_
Second-rank quantities such as stress, strain, moment of inertia, and curvature can be de-
noted as 33 matrix arrays; for instance the stress can be written using numerical indices as
[] =
_
11
12
13
21
22
23
31
32
33
_
_
Here the rst subscript index denotes the row and the second the column. The indices also have
a physical meaning, for instance
23
indicates the stress on the 2 face (the plane whose normal
is in the 2, or y, direction) and acting in the 3, or z, direction. To help distinguish them, well
use brackets for second-rank tensors and braces for vectors.
Using the range convention for index notation, the stress can also be written as
ij
, where
both the i and the j range from 1 to 3; this gives the nine components listed explicitly above.
1
(Since the stress matrix is symmetric, i.e.
ij
=
ji
, only six of these nine components are
independent.)
A subscript that is repeated in a given term is understood to imply summation over the range
of the repeated subscript; this is the summation convention for index notation. For instance, to
indicate the sum of the diagonal elements of the stress matrix we can write:
kk
=
3
k=1
kk
=
11
+
22
+
33
The multiplication rule for matrices can be stated formally by taking A = (a
ij
) to be an
(M N) matrix and B = (b
ij
) to be an (R P) matrix. The matrix product AB is dened
only when R = N, and is the (M P) matrix C = (c
ij
) given by
c
ij
=
N
k=1
a
ik
b
kj
= a
i1
b
1j
+ a
i2
b
2j
+ + a
iN
b
Nk
Using the summation convention, this can be written simply
c
ij
= a
ik
b
kj
where the summation is understood to be over the repeated index k. In the case of a 3 3
matrix multiplying a 3 1 column vector we have
_
_
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
_
_
_
_
b
1
b
2
b
3
_
_
=
_
_
a
11
b
1
+ a
12
b
2
+ a
13
b
3
a
21
b
1
+ a
22
b
2
+ a
23
b
3
a
31
b
1
+ a
32
b
2
+ a
33
b
3
_
_
= a
ij
b
j
The comma convention uses a subscript comma to imply dierentiation with respect to the
variable following, so f
,2
= f/y and u
i,j
= u
i
/x
j
. For instance, the expression
ij,j
= 0
uses all of the three previously dened index conventions: range on i, sum on j, and dierentiate:
xx
x
+
xy
y
+
xz
z
= 0
yx
x
+
yy
y
+
yz
z
= 0
zx
x
+
zy
y
+
zz
z
= 0
The Kroenecker delta is a useful entity is dened as
ij
=
_
0, i = j
1, i = j
This is the index form of the unit matrix I:
ij
= I =
_
_
1 0 0
0 1 0
0 0 1
_
_
So, for instance
2
kk
ij
=
_
kk
0 0
0
kk
0
0 0
kk
_
_
where
kk
=
11
+
22
+
33
.
3 | HuggingFaceTB/finemath | |
deleted 11 characters in body
D.W.
• 135.4k
• 18
• 182
• 389
I don't think there's any general algorithm that works for arbitrary semirings. The requirement to be a semiring doesn't give us a lot to work with.
However, if you have a closed semiring, then there are algorithms for solving systems of linear equations over the semiring.
# Closed semirings
A closed semiring is a semiring with a closure operator, denoted $$*$$, which satisfies the equation
$$a^* = 1 + a \times a^* = 1 + a^* \times a.$$
A closed semiring is also known as a star semiring.
The intuition is that $$a^*$$ is intended to be the sum of the infinite series
$$1 + a + a^2 + a^3 + \dots$$
For instance, the regular languages form a closed semiring under union and concatenation; the $$*$$ operator is the Kleene star. The real numbers form a closed semiring under addition and multiplication; the $$*$$ operator is $$a^* = 1/(1-a)$$.
# Systems of linear equations over a closed semiring
Now, if you have that kind of structure, then there is an analog of Gaussian elimination. In particular, if you have a linear system of equations
$$Ax+b = x$$
where $$x$$ is a vector of variables over the closed semiring, $$b$$ is a vector of constants, and $$A$$ is a matrix of constants, then this has the solution
$$X = A^* B.$$$$x = A^* b.$$
The closure operator on matrices takes a bit of work to define, but it can be computed efficiently using an analog of Gaussian elimination.
For a careful development of the theory, I recommend the following papers:
Stephen Dolan. Fun with Semirings: A functional pearl on the abuse of linear algebraFun with Semirings: A functional pearl on the abuse of linear algebra. International Conference on Functional Programming, ICFP '13.
Daniel J. Lehmann. Algebraic structures for transitive closure. Theoretical Computer Science, vol 4 pp.59--76, 1977.
D.W.
• 135.4k
• 18
• 182
• 389
I don't think there's any general algorithm that works for arbitrary semirings. The requirement to be a semiring doesn't give us a lot to work with.
However, if you have a closed semiring, then there are algorithms for solving systems of linear equations over the semiring.
# Closed semirings
A closed semiring is a semiring with a closure operator, denoted $$*$$, which satisfies the equation
$$a^* = 1 + a \times a^* = 1 + a^* \times a.$$
A closed semiring is also known as a star semiring.
The intuition is that $$a^*$$ is intended to be the sum of the infinite series
$$1 + a + a^2 + a^3 + \dots$$
For instance, the regular languages form a closed semiring under union and concatenation; the $$*$$ operator is the Kleene star. The real numbers form a closed semiring under addition and multiplication; the $$*$$ operator is $$a^* = 1/(1-a)$$.
# Systems of linear equations over a closed semiring
Now, if you have that kind of structure, then there is an analog of Gaussian elimination. In particular, if you have a linear system of equations
$$Ax+b = x$$
where $$x$$ is a vector of variables over the closed semiring, $$b$$ is a vector of constants, and $$A$$ is a matrix of constants, then this has the solution
$$X = A^* B.$$
The closure operator on matrices takes a bit of work to define, but it can be computed efficiently using an analog of Gaussian elimination.
For a careful development of the theory, I recommend the following papers:
Stephen Dolan. Fun with Semirings: A functional pearl on the abuse of linear algebra. International Conference on Functional Programming, ICFP '13.
Daniel J. Lehmann. Algebraic structures for transitive closure. Theoretical Computer Science, vol 4 pp.59--76, 1977.
D.W.
• 135.4k
• 18
• 182
• 389
I don't think there's any general algorithm that works for arbitrary semirings. The requirement to be a semiring doesn't give us a lot to work with.
However, if you have a closed semiring, then there are algorithms for solving systems of linear equations over the semiring.
# Closed semirings
A closed semiring is a semiring with a closure operator, denoted $$*$$, which satisfies the equation
$$a^* = 1 + a \times a^* = 1 + a^* \times a.$$
The intuition is that $$a^*$$ is intended to be the sum of the infinite series
$$1 + a + a^2 + a^3 + \dots$$
For instance, the regular languages form a closed semiring under union and concatenation; the $$*$$ operator is the Kleene star. The real numbers form a closed semiring under addition and multiplication; the $$*$$ operator is $$a^* = 1/(1-a)$$.
# Systems of linear equations over a closed semiring
Now, if you have that kind of structure, then there is an analog of Gaussian elimination. In particular, if you have a linear system of equations
$$Ax+b = x$$
where $$x$$ is a vector of variables over the closed semiring, $$b$$ is a vector of constants, and $$A$$ is a matrix of constants, then this has the solution
$$X = A^* B.$$
The closure operator on matrices takes a bit of work to define, but it can be computed efficiently using an analog of Gaussian elimination.
For a careful development of the theory, I recommend the following papers:
Stephen Dolan. Fun with Semirings: A functional pearl on the abuse of linear algebra. International Conference on Functional Programming, ICFP '13.
Daniel J. Lehmann. Algebraic structures for transitive closure. Theoretical Computer Science, vol 4 pp.59--76, 1977. | HuggingFaceTB/finemath | |
Breaking News
# Prime Factor Of 132
Prime Factor Of 132. What is the sum of all prime factors of 132? 1 × 132 = 132 2 × 66 = 132 3 × 44 = 132 4 × 33 = 132 6 × 22 = 132 11 × 12 = 132 descriptions factors of 132. The factorization or decomposition of 132 = 2 2 •3•11. The prime factors of 132 are 2, 3 and 11.
Prime factors can be identified by checking the endpoints of a factor tree. The prime factors of 132 are 2, 3, and 11. The exponent of 2 is 2.
## Find factors of 132 in pairs.
The factors of 131 are 1. 12, because 132 ÷ 12 = 11. 132 is divisible by 2, 132/2 = 66.
## The Prime Factorization Of A Positive Integer Is A List Of The Integer's Prime Factors, Together With Their Multiplicities;
22, because 132 ÷ 22 = 6. The factorization or decomposition of 132 = 2 2 •3•11. What are the factors of 132? The prime factorization of 132 = 2 2 •3•11.
### So, The Prime Factorization Of 132 Is, 132 = 2 X 2 X 3 X 11.
How to find a factor tree of 132?
### Kesimpulan dari Prime Factor Of 132.
Look at the numbers and check if at least one of them is not prime. Factors of 132 are the. 33 is divisible by 3, 33/3 = 11. The prime factorization of 132 = 2 2 •3•11. | HuggingFaceTB/finemath | |
# Simplify the expression 7/10i
• Last Updated : 11 Mar, 2022
The sum of a real number and an imaginary number is termed a Complex number. These are the numbers that can be written in the form of a+ib, where a and b both are real numbers. It is denoted by z. Here in complex number form the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z). It is also called an imaginary number. In complex numbers form a +bi, ‘i’ is an imaginary number called “iota”. The value of i is (√-1) or we can write as i2 = -1. For example,
• 3 + 4i is a complex number, where 3 is a real number (Re) and 4i is an imaginary number (Im).
• 2 + 5i is a complex number where 2 is a real number (Re) and 5i is an imaginary number (im)
The Combination of real number and imaginary number is called a Complex number.
### Real and Imaginary numbers
The numbers represent a number system such as positive, negative, zero, integer, rational, irrational, fractions, etc. are called real numbers. These are represented as Re(). Example: 19, -45, 0, 1/7, 2.8, √5, etc., are all real numbers.
The numbers which are not real are termed Imaginary numbers. After squaring an imaginary number, it gives a result in negative. Imaginary numbers are represented as Im(). Example: √-8, √-9, √-11 are all imaginary numbers. here ‘i’ is an imaginary number called “iota”.
### Simplify the expression 7/10i
Solution:
Given: 7/10i
Standard form of numerator, 7 = 7 +0i
Standard form of denominator, 10i = 0 +10i
Conjugate of denominator, 0 + 10i = 0 – 10i
Multiply the numerator and denominator with the conjugate,
Therefore, {(7 + 0i) / (0 + 10i)} × {(0 – 10i)/(0 – 10i)}
= {7(0 – 10i)} / {0 – (10i)2}
= {0 – 70i} / {0 – (100(-1))}
= {-70i} / 100
= 0 – 70/100i
= -7/10i
### Similar Problems
Question 1: Solve (1 – 5i) / (-2i)
Solution:
Given: (1 – 5i) / (-2i)
Standard form of denominator, -2i = 0 – 2i
Conjugate of denominator, 0 – 2i = 0 + 2i
Multiply the numerator and denominator with the conjugate,
Therefore, {(1 – 5i) / (0 – 2i)} × {(0 + 2i)/(0 + 2i)}
= {(1 – 5i)(0 + 2i)} / {0 – (2i)2}
= {0 + 2i – 5i – 10i2} / {0 – (4(-1))}
= {-3i – 10 (-1)} / 4
= (-3i + 10) / 4
= 10 /4 – 3i/ 4
= 5/2 – (3/4) i
Question 2: Simplify (8 + 4i) / (3 + 2i)
Solution:
Given: (8 + 4i) / (3 + 2i)
Multiplying the numerator and denominator with the conjugate of denominators,
= {(8 + 4i) / (3 + 2i)} × {( 3 – 2i) / (3 – 2i)}
= {(8 + 4i) × (3 – 2i)} / {(3 + 2i) × (3 – 2i)}
= (24 – 16i +12i – 8i2) / {9 -(2i)2}
= (32 – 4i) / (13)
= (32 – 4i) / 13
= 32/13 – 4/13i
Question 3: Express (2 – i)/(1 + i) in standard form ?
Solution:
Given: (2 – i)/(1 + i)
Multiplying the numerator and denominator with the conjugate of denominators,
= {(2 – i)/(1 + i) × (1 – i)/(1 – i)}
= {(2 – i)(1 – i)} / {(1)2 – (i)2}
= {2 – 2i – i – i2} / (1 – i2)
= {2 – 3i – (-1)} / (1 + 1)
= (3 – 3i) / 2
= 3/2 – 3/ 2i
Question 4: Express in form of a + ib, 2(5 + 3i) + i(7 + 7i)
Solution:
Given: 2(5 + 3i) + i(7 + 7i)
= 10 + 6i + 7i + 7i2
= 10 + 13i + 7(-1)
= 3 + 13i
Question 5: Perform the following operation and find the result in form of a + ib?
(2 – √-25) / (1 – √-16)
Solution:
Given: (2 – √-25 ) / (1 – √-16)
= {2 – (i)(5)} / {1 – (i)(4)}, {i = √-1}
= (2 – 5i) / (1 – 4i)
= {(2 – 5i) / (1 – 4i)} × {(1 + 4i) / (1 + 4i)}
= {(2 – 5i) (1 + 4i)} / {(1 – 4i) (1 + 4i)}
= {2 + 8i – 5i – (20i2)} / {(1-16i2)}, {i2 = -1}
= {2 + 3i + 20} / {1 – 16(-1)}
= (22 + 3i ) / (1 + 16)
= (22 + 3i)/17
= {(22/17) + (3i/17)}
= 22/17 + 3i/17
Question 6: Perform the following operation and find the result in form of a + ib?
(3 – √-16) / (1 – √-9)
Solution:
Given: (3 – √-16) / (1 – √-9)
= {3 – (i)(4)} / {1 – (i)(3)}, {i = √-1}
= (3 – 4i ) / (1 – 3i) {Multiplying the numerator and denominator with the conjugate of denominators}
= {(3 – 4i) / (1 – 3i)} × {(1 + 3i) / (1 + 3i)}
= {(3 – 4i) (1 + 3i)} / {(1 – 3i) (1 + 3i)}
= {3 + 9i – 4i – (12i2)} / {(1 – 9i2)}, {i2 = -1}
= {3 + 5i + 12} / {1 – 9(-1)}
= (15 + 5i) / (1 + 9)
= (15 + 5i)/10
= 15/10 + 5i/10
= 3/2 + 1/2i
Question 7: Simplify the given expression 4/10i
Solution:
Given: 4/10i
Standard form of numerator, 4 = 4 + 0i
Standard form of denominator, 10i = 0 + 10i
Conjugate of denominator, 0 +10i = 0 – 10i
Multiply the numerator and denominator with the conjugate,
Therefore, {(4 + 0i) / (0 + 10i)} × {(0 – 10i)/(0 – 10i)}
= {4(0 – 10i)} / {0 – (10i)2}
= {0 – 40i} / {0 – (100(-1))}
= {-40i} / 100
= 0 – 40/100i
= -2/5i
My Personal Notes arrow_drop_up
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# 1029347 (number)
1,029,347 (one million twenty-nine thousand three hundred forty-seven) is an odd seven-digits composite number following 1029346 and preceding 1029348. In scientific notation, it is written as 1.029347 × 106. The sum of its digits is 26. It has a total of 4 prime factors and 12 positive divisors. There are 910,800 positive integers (up to 1029347) that are relatively prime to 1029347.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 26
• Digital Root 8
## Name
Short name 1 million 29 thousand 347 one million twenty-nine thousand three hundred forty-seven
## Notation
Scientific notation 1.029347 × 106 1.029347 × 106
## Prime Factorization of 1029347
Prime Factorization 112 × 47 × 181
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 93577 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,029,347 is 112 × 47 × 181. Since it has a total of 4 prime factors, 1,029,347 is a composite number.
## Divisors of 1029347
12 divisors
Even divisors 0 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 1.16189e+06 Sum of all the positive divisors of n s(n) 132541 Sum of the proper positive divisors of n A(n) 96824 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1014.57 Returns the nth root of the product of n divisors H(n) 10.6311 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,029,347 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 1,029,347) is 1,161,888, the average is 96,824.
## Other Arithmetic Functions (n = 1029347)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 910800 Total number of positive integers not greater than n that are coprime to n λ(n) 45540 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 80462 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 910,800 positive integers (less than 1,029,347) that are coprime with 1,029,347. And there are approximately 80,462 prime numbers less than or equal to 1,029,347.
## Divisibility of 1029347
m n mod m 2 3 4 5 6 7 8 9 1 2 3 2 5 4 3 8
1,029,347 is not divisible by any number less than or equal to 9.
• Arithmetic
• Deficient
• Polite
## Base conversion (1029347)
Base System Value
2 Binary 11111011010011100011
3 Ternary 1221021222222
4 Quaternary 3323103203
5 Quinary 230414342
6 Senary 34021255
8 Octal 3732343
10 Decimal 1029347
12 Duodecimal 41782b
20 Vigesimal 68d77
36 Base36 m28z
## Basic calculations (n = 1029347)
### Multiplication
n×y
n×2 2058694 3088041 4117388 5146735
### Division
n÷y
n÷2 514674 343116 257337 205869
### Exponentiation
ny
n2 1059555246409 1090650014225364923 1122657320192836707395281 1155603944568535886247210311507
### Nth Root
y√n
2√n 1014.57 100.969 31.8523 15.9409
## 1029347 as geometric shapes
### Circle
Diameter 2.05869e+06 6.46758e+06 3.32869e+12
### Sphere
Volume 4.5685e+18 1.33148e+13 6.46758e+06
### Square
Length = n
Perimeter 4.11739e+06 1.05956e+12 1.45572e+06
### Cube
Length = n
Surface area 6.35733e+12 1.09065e+18 1.78288e+06
### Equilateral Triangle
Length = n
Perimeter 3.08804e+06 4.58801e+11 891441
### Triangular Pyramid
Length = n
Surface area 1.8352e+12 1.28534e+17 840458
## Cryptographic Hash Functions
md5 a5be67a594897a52d726f40a2ac6dd5e 0d63afd1598df52f58d95445803602adf4ce695c a1c16786521c26968721b9023364de4deb0b29b3b180f2e3b404657e3e65439f c064e4f2a5d7c68d4e8ac7665a6065c48c71763522cda0d8276380b4382485d6a85fa926c8e97b1b51864a8ea9f565c582af1df2e070595a2a70331eef04cee9 c3b962791523b17db25ec2b5ea5e0abca9a03d03 | HuggingFaceTB/finemath | |
# Equality Is Hard
Posted on March 9, 2020
As the joke goes, there are two hard problems in computer science: cache invalidation and naming things.1 But I’d suggest there’s a much harder problem, namely, =. Did you miss it? The equals sign, =, is small, but I’m going to argue that the use and misuse of equals is at the root of a large number of problems in software engineering.
## How Equality Should Work
I am going to show how equality in programming languages is often broken. But before I can do that, I have to talk about how it should work, and it turns out that’s not simple! When we talk about how equality “should work,” we have to say what this means in a certain context, because it turns out there are lots of different ways that equality can work, and many of them are valid in different contexts.
The heart and soul of mathematics consists of the fact that the “same” objects can be presented to us in different ways.
-Barry Mazur, When is one thing equal to some other thing
### Laws
Now I said that we can have different definitions of equality in different contexts, but despite this there are some things which should always be true. These are the laws of equality.
Equals is a binary relation2 that is:
• Reflexive, so that a = a for all values of a.
• Symmetric, so that a = b implies b = a and vice versa.
• Transitive, so that if a = b and b = c then a = c
In the programming world, we need to add a law, because programmers do weird things sometimes:
Equals must be:
• Consistent, so that if a = b and no field changes on a or b, a = b will still be true if we check it later on.
The above seems simple enough, although popular programming languages manage to screw up even those trivial rules. But there are more concerns about equality which are harder to state quite so concisely.
### Structural Equality
One difference in how programming languages implement equality is structural equality and reference equality.
Structural equality asks if two references are the same value. This is the default in F#:
type MyString = { SomeField : string }
let a = { SomeField = "Some value" }
let b = { SomeField = "Some value" }
if a = b then // returns true, enters "then" block
This is not true in C#; C# uses reference equality. Reference equality asks if the two objects being compared are the same object. In other words, does the variable point at the same area of memory? A reference to two different blocks of memory will be unequal even if their contents are identical:
class MyString {
public string SomeField { get; }
public MyString(string someField) => this.someField = someField;
}
var a = new MyString("Some value");
var b = new MyString("Some value");
if (a == b) { // returns false, does not enter block
Other languages let you choose. Scheme, for example, provides equal? to check structural equality and eq? to check reference equality. Kotlin provides == for structural equailty and === for reference equality (don’t confuse these with JavaScript’s == and === operators which are… something else entirely).
When does it make sense to use structural equality in your programs? In the absence of mutation (changing the values of variables), nearly always! Most programming languages that I’m aware of do structural comparisons on value types such as integers. Well, except Java, which has confused generations of programmers with an int value type which does a structural comparison and an Integer reference type which, well, the best thing you can say is don’t use == on Integer. Python has similar issues with is.
Structural comparison of reference types such as objects makes sense as well. Consider a unit test, where you want to check that the object returned is equal to the value you expected. In a language with structural equality, this is trivial:
[<TestMethod>]
let The result of the calculation is the expected value() =
let expected = { SomeField = "Some value"; SomeOtherField = 15; StillAnotherField = true; ... }
let actual = calculate()
Assert.AreEqual(expected, actual)
When a language does not have structural equality from the outset, developers will try to build it ad hoc, and you end up with this horror show, which is now permanently part of the NUnit framework.
### Reference Equality
But as I hinted above, there are certainly cases where structural equality does not make sense. One example is with languages which support mutation of variables, which is most of them.3 When you can change the value of a variable, it probably does not make sense to say that variable is equal to some other variable, in general. Sure, you can say they’re (structurally) equal as of a moment in time, such as in last line of a unit test, but you can’t generally imply that they’re the same. This is a kind of subtle point, so let’s look at an example.
Let’s say I have an object which represents a person. In F#, with structural equality, I can write:
type Person = { Name : string; Age : integer; Offspring : Person list }
Now I have two friends, Jane and Sue. Both have a son named John, who is 15. They’re different people, but the sons have the same name and age. No problem!
let jane = { Name = "Jane"; Age = 47; Offspring = [ { Name = "John"; Age = 15; Offspring = [] } ] }
let sue = { Name = "Sue"; Age = 35; Offspring = [ { Name = "John"; Age = 15; Offspring = [] } ] }
I could also have written this:
let john = { Name = "John"; Age = 15; Offspring = [] };
let jane = { Name = "Jane"; Age = 47; Offspring = [ john ] }
let sue = { Name = "Sue"; Age = 35; Offspring = [ john ] }
The behavior of these two blocks is precisely the same. I can’t distinguish the two sons, even though I know they’re different people. That’s OK! If I needed to distinguish them, I could add a hash of their DNA or something to my Person type. But if I just need to know their name and age, it doesn’t matter if I can distinguish the two objects or not, because the values are the same, no matter how you slice it.
Imagine Jane’s son changes his name to Pat. F# doesn’t support mutating the values of variables, so I need to make a new Person instance for John and Jane:
let newJane = { Name = "Jane"; Age = 47; Offspring = [ { Name = "Pat"; Age = 15; Offspring = [] } ] }
It seems weird to have a new variable, newJane, but in practice it doesn’t create a problem. The code above is fine. Now let’s try this in C#, a language which is mutable by default:
var john = new Person("John", 15, null);
var jane = new Person("Jane", 15, new List<Person> { john });
var sue = new Person("Sue", 15, new List<Person> { john });
Well, this code is clearly incorrect: If Jane’s son changes his name to “Pat”, I can change the reference directly:
jane.Offspring.First().Name = "Pat";
But I’ll find that Sue’s son’s name has changed as well! Therefore, even though the two sons had the same values at the start, before he changed his name, they were not equal! I should have written:
var jane = new Person("Jane", 15, new List<Person> { new Person("John", 15, null) });
var sue = new Person("Sue", 15, new List<Person> { new Person("John", 15, null) });
…so that Jane and Sue’s offspring were reference unequal to each other. So reference equality is a sensible default in a language which supports mutation.
Another case where reference equality makes sense is when you know it’s going to give the same result as structural equality anyway. There is a certain performance overhead for testing structural equality, which is reasonable if you actually need to test structural equality. But if, for example, you create a large number of objects which you know are all different structually, it doesn’t make sense to pay the overhead of testing structural equality when you know that testing reference equality alone would give the same result.
### Equivalent Representations
In the real numbers, .999… (repeating infinitely) equals 1. Note that the “real numbers” here are distinct from the “Real” type in your programming language. Real numbers in math are infinite, and real numbers in your programming language are finite. So there is no notion of .999… in your programming language, but that’s OK, because you can just use 1, which is the same value.
This is, essentially, a choice that mathematicians made when formulating the real number system. If one adds other objects, such as infinitessimals, to the system, then .999… and 1 are not equivalent.
However, it is by no means an arbitrary convention, because not adopting it forces one either to invent strange new objects or to abandon some of the familiar rules of arithmetic.
-Timothy Gowers, Mathematics: A Very Short Introduction4
Similarly, in the rational numbers, 1/2 and 2/4 represent the same value.
Do not confuse these equivalances with the “loose” equivalence operator == found in JavaScript and PHP. Unlike those operators, these equivalences follow the laws of equality. It is important to realize that equal objects can be represented differently.
In IEEE-754 floats, -0 = 0.
### Intensional vs. Extensional Equality
When is some function equal to some other function? Most programming languages will happily do a reference = comparison, and I suppose that’s fine, but what would a structural equality comparison of a function even mean? Well, if we could use reflection to look into the implementation of the function, and see if it does the same thing? But what is “the same?” Would it have to have the same variable names? Are a quicksort and a merge sort “the same function?”
Cutting to the chase, we say that functions are extensionally equal if they return the same outputs for the same inputs (regardless of internal implementation), and intensionally equal if their internal definition is the same. Of course, this is context-dependent. There may be a context where I need a constant time function and another context where the speed of the function doesn’t matter. The important point is I need to have some context for equality and use it to compare the two functions.
I don’t know of any programming language which even attempts to do anything beyond reference equality for functions. But it’s easy to come up with examples where it would be useful! (An optimizer which removes duplicate code, e.g.) You’re on your own if you need this, but I have to say that not shipping an equals comparison is preferable to shipping one that’s broken.
### Equality vs. Assignment
One of the first lessons we learn when becoming a programmer is that there are two different concepts which we both call “equals.” One is assignment, the other is testing equality. In JavaScript, these look like:
const aValue = someFunction(); // Assignment
if (aValue === 3) { // Test for equality
These are fundamentally different. Comparison returns a boolean; assignment, in an expression-oriented language such as Ruby, returns the value assigned.
So we can write Ruby code like this:
a = b = c = 3
Which does indeed assign 3 to the variables a, b, and c. Don’t try it with a reference type, though; it probably won’t do what you want!5
In a non-expression-oriented language like C#, assignment doesn’t return anything.
In math, we use the equals operator for both assignment and testing equality:
if aValue = 3 ...
where aValue = someFunction()
(And = is sometimes used for other relations in math, such as congruence. As with all things in math, context matters; you have to carefully consider what = might mean in a certain paper or book.)
Why does math not require two separate operators whereas programming languages do? You can tell from context which one is intended, and not all programming languages require different operators. F#, for example, uses = for both assignment and testing equality. Despite overloading =, assignment and testing equality are different operations.
let aValue = someFunction(); // Assignment
if aValue = 3 then // Test for equality
The choice of syntax is partially due to heritage: F# is based on ML, which is based on math, and JavaScript syntax is based on Java -> C -> Algol -> FORTRAN.
FORTRAN had to compile on machines where distinguishing these two cases from code syntax would be genuinely challenging, so it made sense to have two different operators. Then C took this “feature” to a high art, allowing code like:
int aValue = someFunction(); // Assignment
if (aValue = 3) { // Also assignment!
This code, for those without previous C experience, overwrites aValue with 3 and then, since the expression aValue = 3 is equal to 3, the if test returns TRUE and execution continues inside the if block. This is frequently an error, leading many C programmers to reverse the values inside an if block out of habit to avoid making the mistake:
int aValue = someFunction(); // Assignment
if (3 == aValue) { // Test for equality
// [...]
if (3 = aValue) { // Syntax error: Cannot assign aValue to 3.
## How Equality Should Not Work
I hope I’ve shown by now that equality is not simple, and that the “correct” implementation of equality can vary depending upon context. Despite that, programming languages often get the simple parts wrong! Very often, this is caused by the combination of equality with other language features, such as implicit type conversion.
### Common Mistake: Equality Isn’t Reflexive
Recall that the reflexive law of equals requires all values to be equal to themselves, a = a.
In .NET, if you call Object.ReferenceEquals() on a value type, the arguments are separately boxed before the method runs, so it returns false even if you pass the same instance:
int int1 = 3;
Console.WriteLine(Object.ReferenceEquals(int1, int1)); // Prints False
This means it is not necessarily true that a = a in any .NET language, so the reflexive law is broken.
In SQL, NULL is not equal to itself, so the expression NULL = NULL (or, more probably, SOME_EXPRESSION = SOME_OTHER_EXPRESSION when both of them might be null) will return NULL, which is falsy. This leads to messes like:
WHERE (SOME_EXPRESSION = SOME_OTHER_EXPRESSION)
OR (SOME_EXPRESSION IS NULL AND SOME_OTHER_EXPRESSION IS NULL)
Or, more likely, just a bug where the developer forgot about the special rules for NULL. If your DB server’s SQL dialect supports IS NOT DISTINCT FROM then this does what = should do. (Or should I say it does NOT not do what = should do?) Otherwise you’ll just have to live with SQL like the above. The best fix is to make your columns non-nullable when possible.
This is also true of IEEE-754 floats; the standard states that NaN != NaN. A different explanation than the one given in the link for this is that “NaN” represents some unspecified “non-number” result, not necessarily the same unspecified non-number result as that of a different calculation, so it’s incorrect to compare them. For example, square_root(-2) and infinity/infinity are both NaN, but they’re clearly not the same! Similar explanations are given for SQL’s NULL sometimes. One problem with this is that it makes the term very overloaded: Is NaN and NULL an unknown or imprecise value or the known absence of a value?
One way of handling such situations, which do not occur in routine floating point calculations, would be a union type. In F#, one could write:
type MaybeFloat =
| Float of float
| Imaginary of real: float * imaginary: float
| Indeterminate
| /// ...
… and then you could handle these terms appropriately in calculations which needed them. Use a signaling NaN to throw an exception in calculations which you don’t expect will have NaNs at all.
Rust offers the Eq and PartialEq traits. Not implementing Eq is supposed to be a signal that == is not reflexive, and floating point types in Rust do not implement it. But if you don’t implement Eq, you can still call == in your code. Implementing Eq allows your object to be used as a key in a hash map and possibly results in behavior changes in other places as well.
But there are even more significant problems with = and floats.
### Common Mistake: Equals Is Too Precise
I guess many developers are familiar with the problem of comparing IEEE-754 floating point numbers, which are the “float” or “double” implementation for most programming languages. 10 * (0.1) does not equal 1, because “0.1” is actually equal to 0.100000001490116119384765625 or 0.1000000000000000055511151231257827021181583404541015625. If you’re not familar with this issue, you can go read about it, but the point is that it’s rarely safe to do an == comparison on a floating point number at all! You have to ask yourself which digits are significant and compare accordingly.
(Worse, the float type backs other types, such as TDateTime in some languages, so even in cases where equality comparisons might make sense, they don’t necessarily work.)
The correct method of comparing floating point numbers is to see if they’re “close,” and what “close” means varies depending on context. It’s not something you can cram into a == operator. If you find yourself doing this a lot (say, once), you might consider using a different data type, such as a fixed precision decimal number.
So why do programming languages offer == comparisons on a type when they can’t support it? Well, because they offer == on every type, it works on most of them, and they just shrug about the rest and chastize programmers for not knowing which language feature they should not use.
Not every programming language, mind you. Standard ML doesn’t offer = comparisons on reals. It’s a compiler error if you try!
The implementation notes state:
Deciding if real should be an equality type, and if so, what should equality mean, was also problematic. IEEE specifies that the sign of zeros be ignored in comparisons, and that equality evaluate to false if either argument is NaN. These constraints are disturbing to the SML programmer. The former implies that 0 = ~0 is true while r/0 = r/~06 is false. The latter implies such anomalies as r = r is false, or that, for a ref cell rr, we could have rr = rr but not have !rr = !rr. We accepted the unsigned comparison of zeros, but felt that the reflexive property of equality, structural equality, and the equivalence of <> and not o = ought to be preserved. Additional complications led to the decision to not have real be an equality type.
By blocking = for reals, SML forces the developer to think about what kind of comparison they actually need, which is a great feature, I think!
F# offers the [<NoEquality>] attribute to mark custom types where = should not be used. Pity they didn’t mark the float type with it!
### Common Mistake: “Equals” Isn’t
PHP has two separate operators, == and ===. The documentation for ==, which is named “Equal,” states, “TRUE if $a is equal to $b after type juggling.” Unfortunately, this means that the == operator is unreliable:
<?php
var_dump("608E-4234" == "272E-3063"); // true
?>
Although we’re comparing strings here, PHP sees that they can be converted to a number, so it does. The numbers turn out to be very small (the first argument, for example, is 608 * 10-4234), and, as we’ve already discussed, comparing floating point numbers is hard. Casting both of these to a float(0) results in rounding them to equal values, so the comparison returns true.
Note this is different than the behavior of JavaScript, which also has similar (but not the same!) == and === operators; JavaScript would see that both sides are strings and return false for this comparison.
Fortunately, PHP has the === (“Identical”) operator, which behaves correctly in this case. I would say never use ==, but == does a structural comparison on objects, which might be something you want! Instead, I’ll say: Use extreme caution with ==, because it’s broken on primitive types.
### Common Mistake: Equality Isn’t Symmetric
If you override .equals() in Java, it is your responsibility to ensure that the laws of equality hold!
It is very easy to implement a comparison which is not symmetric, that is, a.equals(b) != b.equals(a), if you’re not paying attention.
Even if we remove null from the picture (because it would be a NullPointerException in one case and the contract for .equals() allows you to do this), if you subtype a class and override .equals() then you had better be careful!
@Override
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null)
return false;
if (!o.getClass().isAssignableFrom(getClass())) // Danger! This is a mistake!
return false;
ThisClass thisClass = (ThisClass) o;
// field comparison
// ...
}
If ThisClass and ASubtypeOfThisClass both override .equals() with code like this, a.equals(b) may not equal b.equals(a)! The correct comparison would be:
if (getClass() != o.getClass())
return false;
This is not just my opinion; the contract for Object.equals() requires it.
### Common Mistake: Equality Isn’t Transitive
Remember one of the laws for equals comparisons is that they should be transitive; if a = b and b = c then a = c. Unfortunately, when coupled with type coersion, languages often fail at this.
In JavaScript,
'' == 0; // true
0 == '0'; // true
'' == '0'; // false!
Never use == in JavaScript. Use === instead.
### Common Mistake: Equality Is Inconsistent
In Kotlin, == returns different values depending on the type of the variable, even for the same variable:
fun equalsFloat(a: Float, b: Float) {
println(a == b);
}
fun equalsAny(a: Any, b: Any) {
println(a == b);
}
fun main(args: Array<String>) {
val a = Float.NaN;
val b = Float.NaN;
equalsFloat(a, b);
equalsAny(a, b);
}
// prints false, true
This is an unfortunate combination of language features which results in some pretty unintuitive behavior.
### Common Mistake: Using Reference Equality When Structural Equality Is Needed
Consider the following MSTest unit test in C#:
[TestMethod]
public void Calculation_Is_Correct() {
var expected = new Result(SOME_EXPECTED_VALUE);
var actual = _service.DoCalculation(SOME_INPUT);
Assert.AreEqual(expected, actual);
}
Does this work? We can’t tell! Assert.AreEqual() will eventually call Object.Equals(), which does a reference comparison by default. Unless you’ve overridden Result.Equals() to do a structural comparison instead, the unit test is broken. The contract for Object.Equals() says that you should not override it at all if your type is mutable, which is reasonable in general but probably not what you want for a unit test. (This is because .Equals() is expected to match .GetHashCode(), and the hash code is expected not to change over the lifetime of the object.) The closest thing the .NET framework has to a “guaranteed structural” comparison for reference types is IEquatable<T>, which Assert.AreEqual() doesn’t use, even if it’s implemented.
(Java’s Object.hashCode, by contrast, is allowed to change when the object’s fields change.)
## How to Think About Equality
Wow, I’ve now written 4000 words about the = operator and I’m not finished yet! That seems, well, out of proportion to the size of the operator. Why is this so complicated? Two reasons, basically:
• Inessential complexity: Our programming languages don’t serve us well with equivalence comparisons. They are often entirely broken, and even when they’re not, they don’t make it obvious when, for example, they’ll do a structural comparison vs a reference comparison.
• Essential complexity: Equivalence is often genuinely complicated in places where it is needed, such as when comparing floating point numbers, and it gets even harder in edge cases like comparing functions.
Another way to divide this up is “stuff which should be fixed by programming language implementors” (the “inessential complexity” above) and “stuff which must be resolved by programming language users.
### What Programming Languages Should Do
With regards to inessential complexity, the situation we find ourselves in today, with mostly-broken implementations of equivalence in nearly every mainstream programming language is just a crying shame. This “simple operation which must obey certain laws” is exactly the sort of thing we depend on programming languages to get right! But it appears to me that only SML has really considered having a lawful equality in both its semantics and its runtime/standard library, and SML isn’t entirely mainstream.
First, programming languages should make it simple to create types where equality comparison is disabled because it makes no sense (like [<NoEquality>] in F#) and they should use this feature in their standard library where needed, such as on floating point types.
Programming languages must make the difference between structural equality and reference equality crystal clear. There should never be a case where it’s unclear what you’re doing. Most programming lanuages overload == to mean both structural equality or reference equality depending on the type of reference, most commonly value types vs. reference types, and this is guaranteed to confuse developers.
Kotlin comes very close to getting this right with its === for reference equality and == for structural equality, although for some reason it translates === to == for value types, instead of just failing to compile for that. The goal should be reducing developer confusion. You want the developer to see === and think “reference equality,” not “more equals signs is better.”
F# mostly gets this right by making reference equality very hard to use.
I don’t know of any language which has mutable by default variables which handles structural comparsions in a non-confusing way. But it’s easy to imagine what it might look like! Have a reference equals and structural equals operator which is only supported in contexts where the language can reasonably expect to support it. For example, if .NET did not do the boxing funkiness with Object.ReferenceEquals and value types (it could just fail to compile if you tried) and had something akin to IEquatable<T> which would allow you to use a structural comparison operator, that seems like a pretty good solution to making it clear to developers which is which.
### What Programmers Should Do
One might look at the length of this post and say, “Wow, equality is really complicated! I’m going to give up coding and become a soybean farmer.” But this post is as long as it is mostly because so many languages do equality wrong. Doing equality correctly does requre some thought, but not too much though. Certainly less than soybean farming.
When doing an equality comparison on an existing type, stop and ask yourself:
• Does it make sense to do an equality comparison at all here?
• If so, does a structural or a reference comparison make sense?
• What support does my programming language provide for the appropriate style of comparison?
• Does my programming language implement equality correctly for this comparison?
You can ask yourself similar questions when designing a custom type:
• Should my type support equality comparisons at all? Or do they need a more complicated comparison, as with a float?
• Will my type be mutable? How might that affect equality?
• Would a reference comparison, a structural comparison, or both make sense?
If your type is mutable, consider if you can change it to be immutable. You can do this even in a language which is mutable by default! Beyond giving you more options with respect to equality comparisons, there are many other benefits of an immutable architecture as well. The C# Roslyn compiler, which uses immutable data structures throughout, is a great example of this:
The third attribute of syntax trees is that they are immutable and thread-safe. This means that after a tree is obtained, it is a snapshot of the current state of the code, and never changes. This allows multiple users to interact with the same syntax tree at the same time in different threads without locking or duplication. Because the trees are immutable and no modifications can be made directly to a tree, factory methods help create and modify syntax trees by creating additional snapshots of the tree. The trees are efficient in the way they reuse underlying nodes, so a new version can be rebuilt fast and with little extra memory.
from the .NET Compiler Platform SDK docs
## Credits
Thank you to Paul Blasucci, Jeremy Loy, Bud Marrical, Michael Perry, Skyler Tweedie, and Thomas Wheeler for reading drafts of this article and giving me feedback.
### References
This post was inspired by Barry Mazur’s wonderful math paper, “When is one thing equal to some other thing?” which uses category theory to answer the question for math.
Thanks to Tommy Hall, who drew my attention to this 1993 paper, which, discusses many of the issues covered in this post and proposes a solution for Common Lisp.
Simon Ochsenreither has a nice series on problems with equality and fixing Haskell. Overview, Problems, Solution, Fixing Haskell, Implementation in Dora.
Hillel Wayne pointed me to this great essay, “The Semantics of Object Identity.”
Brandon Bloom provided a link to the paper “The Left Hand of Equals” which “takes a reflexive journey through fifty years of identity and equality in object-oriented languages, and ends somewhere we did not expect: a ‘left-handed’ equality relying on trust and grace.”
1. A “binary relation” deserves a little bit of explanation, but this gets into a little math, so feel free to ignore if it doesn’t help you. We have two sets A and B. (They might be the same set.) For any two members a and b of the sets, we want a rule which says whether they’re in the relation or not. So if A and B are the integers, the ordered pair (1, 2) is not in the relation “is equal to” but the ordered pair (5, 5) is in the relation. A relation is a subset of the cross product of the sets.
2. Phil Hagelberg tells me the problem isn’t mutation of variables but data structures, which is a subtle but fair distinction.
3. Gowers, Timothy, Mathematics: A Very Short Introduction, p. 60
4. a = b = c = [] in Ruby assigns the same reference to a, b, and c. So if you mutate a, you’ll mutate b and c at the same time. That’s probably not what you wanted, otherwise what would be the point of having three separate references? In contrast, with a value type like Integer, mutating a will not change the value of b or c.
5. ~ is the SML operator for unary negation, so ~0 should be read as “negative zero.” | HuggingFaceTB/finemath | |
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# eferim's solution
## to Bowling in the Delphi Pascal Track
Published at Sep 20 2018 · 0 comments
Instructions
Test suite
Solution
#### Note:
This exercise has changed since this solution was written.
Score a bowling game.
Bowling is a game where players roll a heavy ball to knock down pins arranged in a triangle. Write code to keep track of the score of a game of bowling.
## Scoring Bowling
The game consists of 10 frames. A frame is composed of one or two ball throws with 10 pins standing at frame initialization. There are three cases for the tabulation of a frame.
• An open frame is where a score of less than 10 is recorded for the frame. In this case the score for the frame is the number of pins knocked down.
• A spare is where all ten pins are knocked down by the second throw. The total value of a spare is 10 plus the number of pins knocked down in their next throw.
• A strike is where all ten pins are knocked down by the first throw. The total value of a strike is 10 plus the number of pins knocked down in the next two throws. If a strike is immediately followed by a second strike, then the value of the first strike cannot be determined until the ball is thrown one more time.
Here is a three frame example:
Frame 1 Frame 2 Frame 3
X (strike) 5/ (spare) 9 0 (open frame)
Frame 1 is (10 + 5 + 5) = 20
Frame 2 is (5 + 5 + 9) = 19
Frame 3 is (9 + 0) = 9
This means the current running total is 48.
The tenth frame in the game is a special case. If someone throws a strike or a spare then they get a fill ball. Fill balls exist to calculate the total of the 10th frame. Scoring a strike or spare on the fill ball does not give the player more fill balls. The total value of the 10th frame is the total number of pins knocked down.
For a tenth frame of X1/ (strike and a spare), the total value is 20.
For a tenth frame of XXX (three strikes), the total value is 30.
## Requirements
Write code to keep track of the score of a game of bowling. It should support two operations:
• `roll(pins : int)` is called each time the player rolls a ball. The argument is the number of pins knocked down.
• `score() : int` is called only at the very end of the game. It returns the total score for that game.
## Testing
In order to run the tests for this track, you will need to install DUnitX. Please see the installation instructions for more information.
If Delphi is properly installed, and `*.dpr` file types have been associated with Delphi, then double clicking the supplied `*.dpr` file will start Delphi and load the exercise/project. `control + F9` is the keyboard shortcut to compile the project or pressing `F9` will compile and run the project.
Alternatively you may opt to start Delphi and load your project via. the `File` drop down menu.
### When Questions Come Up
We monitor the Pascal-Delphi support room on gitter.im to help you with any questions that might arise.
### Submitting Exercises
Note that, when trying to submit an exercise, make sure the exercise file you're submitting is in the `exercism/delphi/<exerciseName>` directory.
For example, if you're submitting `ubob.pas` for the Bob exercise, the submit command would be something like `exercism submit <path_to_exercism_dir>/delphi/bob/ubob.pas`.
## Source
The Bowling Game Kata at but UncleBob http://butunclebob.com/ArticleS.UncleBob.TheBowlingGameKata
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you may receive assistance from a mentor.
### uBowlingTests.pas
``````unit uBowlingTests;
interface
uses
DUnitX.TestFramework,
uBowling;
const
CanonicalVersion = '1.2.0';
type
[TestFixture]
BowlingTests = class(TObject)
private
class function RollMany(pins: array of integer; game: IBowlingGame): IBowlingGame; static;
public
[Test]
// [Ignore('Comment the "[Ignore]" statement to run the test')]
procedure Should_be_able_to_score_a_game_with_all_zeros;
[Test]
[Ignore]
procedure Should_be_able_to_score_a_game_with_no_strikes_or_spares;
[Test]
[Ignore]
procedure A_spare_followed_by_zeros_is_worth_ten_points;
[Test]
[Ignore]
procedure Points_scored_in_the_roll_after_a_spare_are_counted_twice;
[Test]
[Ignore]
procedure Consecutive_spares_each_get_a_one_roll_bonus;
[Test]
[Ignore]
procedure A_spare_in_the_last_frame_gets_a_one_roll_bonus_that_is_counted_once;
[Test]
[Ignore]
procedure A_strike_earns_ten_points_in_frame_with_a_single_roll;
[Test]
[Ignore]
procedure Points_scored_in_the_two_rolls_after_a_strike_are_counted_twice_as_a_bonus;
[Test]
[Ignore]
procedure Consecutive_strikes_each_get_the_two_roll_bonus;
[Test]
[Ignore]
procedure A_strike_in_the_last_frame_gets_a_two_roll_bonus_that_is_counted_once;
[Test]
[Ignore]
procedure Rolling_a_spare_with_the_two_roll_bonus_does_not_get_a_bonus_roll;
[Test]
[Ignore]
procedure Strikes_with_the_two_roll_bonus_do_not_get_bonus_rolls;
[Test]
[Ignore]
procedure A_strike_with_the_one_roll_bonus_after_a_spare_in_the_last_frame_does_not_get_a_bonus;
[Test]
[Ignore]
procedure All_strikes_is_a_perfect_game;
[Test]
[Ignore]
procedure Rolls_cannot_score_negative_points;
[Test]
[Ignore]
procedure A_roll_cannot_score_more_than_10_points;
[Test]
[Ignore]
procedure Two_rolls_in_a_frame_cannot_score_more_than_10_points;
[Test]
[Ignore]
procedure Bonus_roll_after_a_strike_in_the_last_frame_cannot_score_more_than_10_points;
[Test]
[Ignore]
procedure Two_bonus_rolls_after_a_strike_in_the_last_frame_cannot_score_more_than_10_points;
[Test]
[Ignore]
procedure Two_bonus_rolls_after_a_strike_in_the_last_frame_can_score_more_than_10_points_if_one_is_a_strike;
[Test]
[Ignore]
procedure The_second_bonus_rolls_after_a_strike_in_the_last_frame_cannot_be_a_strike_if_the_first_one_is_not_a_strike;
[Test]
[Ignore]
procedure Second_bonus_roll_after_a_strike_in_the_last_frame_cannot_score_more_than_10_points;
[Test]
[Ignore]
procedure An_unstarted_game_cannot_be_scored;
[Test]
[Ignore]
procedure An_incomplete_game_cannot_be_scored;
[Test]
[Ignore]
[Test]
[Ignore]
procedure Bonus_rolls_for_a_strike_in_the_last_frame_must_be_rolled_before_score_can_be_calculated;
[Test]
[Ignore]
procedure Both_bonus_rolls_for_a_strike_in_the_last_frame_must_be_rolled_before_score_can_be_calculated;
[Test]
[Ignore]
procedure Bonus_roll_for_a_spare_in_the_last_frame_must_be_rolled_before_score_can_be_calculated;
[Test]
[Ignore]
procedure Cannot_roll_after_bonus_roll_for_spare;
[Test]
[Ignore]
procedure Cannot_roll_after_bonus_rolls_for_strike;
end;
implementation
uses System.SysUtils;
procedure BowlingTests.Second_bonus_roll_after_a_strike_in_the_last_frame_cannot_score_more_than_10_points;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10], NewBowlingGame);
Assert.IsFalse(game.Roll(11));
end;
procedure BowlingTests.Should_be_able_to_score_a_game_with_all_zeros;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
assert.AreEqual(0, game.Score);
end;
procedure BowlingTests.Should_be_able_to_score_a_game_with_no_strikes_or_spares;
var game: IBowlingGame;
begin
game := RollMany([3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6], NewBowlingGame);
assert.AreEqual(90, game.Score);
end;
procedure BowlingTests.A_spare_followed_by_zeros_is_worth_ten_points;
var game: IBowlingGame;
begin
game := RollMany([6, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
assert.AreEqual(10, game.Score);
end;
procedure BowlingTests.Points_scored_in_the_roll_after_a_spare_are_counted_twice;
var game: IBowlingGame;
begin
game := RollMany([6, 4, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
assert.AreEqual(16, game.Score);
end;
procedure BowlingTests.Consecutive_spares_each_get_a_one_roll_bonus;
var game: IBowlingGame;
begin
game := RollMany([5, 5, 3, 7, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
assert.AreEqual(31, game.Score);
end;
procedure BowlingTests.A_spare_in_the_last_frame_gets_a_one_roll_bonus_that_is_counted_once;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 3, 7], NewBowlingGame);
assert.AreEqual(17, game.Score);
end;
procedure BowlingTests.A_strike_earns_ten_points_in_frame_with_a_single_roll;
var game: IBowlingGame;
begin
game := RollMany([10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
assert.AreEqual(10, game.Score);
end;
procedure BowlingTests.Points_scored_in_the_two_rolls_after_a_strike_are_counted_twice_as_a_bonus;
var game: IBowlingGame;
begin
game := RollMany([10, 5, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
assert.AreEqual(26, game.Score);
end;
procedure BowlingTests.Consecutive_strikes_each_get_the_two_roll_bonus;
var game: IBowlingGame;
begin
game := RollMany([10, 10, 10, 5, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
assert.AreEqual(81, game.Score);
end;
procedure BowlingTests.A_strike_in_the_last_frame_gets_a_two_roll_bonus_that_is_counted_once;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 7, 1], NewBowlingGame);
assert.AreEqual(18, game.Score);
end;
procedure BowlingTests.Rolling_a_spare_with_the_two_roll_bonus_does_not_get_a_bonus_roll;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 7, 3], NewBowlingGame);
assert.AreEqual(20, game.Score);
end;
procedure BowlingTests.Strikes_with_the_two_roll_bonus_do_not_get_bonus_rolls;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 10], NewBowlingGame);
assert.AreEqual(30, game.Score);
end;
procedure BowlingTests.A_strike_with_the_one_roll_bonus_after_a_spare_in_the_last_frame_does_not_get_a_bonus;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 3, 10], NewBowlingGame);
assert.AreEqual(20, game.Score);
end;
procedure BowlingTests.All_strikes_is_a_perfect_game;
var game: IBowlingGame;
begin
game := RollMany([10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10], NewBowlingGame);
assert.AreEqual(300, game.Score);
end;
procedure BowlingTests.Rolls_cannot_score_negative_points;
var game: IBowlingGame;
begin
game := RollMany([-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.A_roll_cannot_score_more_than_10_points;
var game: IBowlingGame;
begin
game := RollMany([11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.Two_rolls_in_a_frame_cannot_score_more_than_10_points;
var game: IBowlingGame;
begin
game := RollMany([5, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.The_second_bonus_rolls_after_a_strike_in_the_last_frame_cannot_be_a_strike_if_the_first_one_is_not_a_strike;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 6], NewBowlingGame);
Assert.IsFalse(game.Roll(10));
end;
procedure BowlingTests.Two_bonus_rolls_after_a_strike_in_the_last_frame_cannot_score_more_than_10_points;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 5, 6], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.Two_bonus_rolls_after_a_strike_in_the_last_frame_can_score_more_than_10_points_if_one_is_a_strike;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 6], NewBowlingGame);
Assert.AreEqual(26, game.Score);
end;
procedure BowlingTests.An_unstarted_game_cannot_be_scored;
var game: IBowlingGame;
begin
game := RollMany([],NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.An_incomplete_game_cannot_be_scored;
var game: IBowlingGame;
begin
game := RollMany([0, 0], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.Cannot_roll_after_bonus_rolls_for_strike;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 3, 2], NewBowlingGame);
Assert.IsFalse(game.Roll(2));
end;
procedure BowlingTests.Cannot_roll_after_bonus_roll_for_spare;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 3, 2], NewBowlingGame);
Assert.IsFalse(game.Roll(2));
end;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], NewBowlingGame);
Assert.IsFalse(game.Roll(0));
end;
procedure BowlingTests.Bonus_rolls_for_a_strike_in_the_last_frame_must_be_rolled_before_score_can_be_calculated;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.Both_bonus_rolls_for_a_strike_in_the_last_frame_must_be_rolled_before_score_can_be_calculated;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
procedure BowlingTests.Bonus_roll_after_a_strike_in_the_last_frame_cannot_score_more_than_10_points;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10], NewBowlingGame);
Assert.IsFalse(game.Roll(11));
(*{
var sut = new BowlingGame();
var previousRolls = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10};
DoRoll(previousRolls, sut);
Assert.Throws<ArgumentException>(() => sut.Roll(11));
} (**)
end;
procedure BowlingTests.Bonus_roll_for_a_spare_in_the_last_frame_must_be_rolled_before_score_can_be_calculated;
var game: IBowlingGame;
begin
game := RollMany([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 3], NewBowlingGame);
Assert.AreEqual(-1, game.Score);
end;
class function BowlingTests.RollMany(pins: array of integer; game: IBowlingGame): IBowlingGame;
var count: integer;
begin
for count in pins do
game.Roll(count);
result := game;
end;
initialization
TDUnitX.RegisterTestFixture(BowlingTests);
end.``````
``````unit uBowling;
interface
type
IBowlingGame = interface
function Roll(ARoll : integer) : boolean;
function Score : integer;
end;
function NewBowlingGame : IBowlingGame;
implementation
type
TNewBowlingGame = class(TInterfacedObject,IBowlingGame)
private
Fir, Sec : integer;
FullFrame, Complete : boolean;
function CheckRoll(ARoll : integer) : boolean;
function CheckFinal(ARoll : integer) : boolean;
public
function Roll(ARoll : integer) : boolean;
function Score : integer;
end;
{\$REGION 'TNewBowlingGame' }
begin
inc(Sum, ARoll + ARoll * Additional );
end;
function TNewBowlingGame.CheckFinal(ARoll: integer): boolean;
begin
Result := true;
case state of
0 : Fir := ARoll;
1 : begin
Sec := ARoll;
if Fir + Sec < 10 then Complete := true;
end;
2 : begin
Result := (Fir + Sec = 10) or ((Fir + Sec > 10) and ((sec + ARoll <= 10) or (Sec = 10)));
Complete := Result;
end
else
Result:= false;
end;
inc(State);
if Result then
end;
function TNewBowlingGame.CheckRoll(ARoll: integer): boolean;
begin
Result := true;
if not FullFrame then
begin
Fir := ARoll;
FullFrame := Fir < 10;
if not FullFrame then
begin
end;
end
else
begin
if Fir + ARoll > 10 then
exit(false);
if Fir + ARoll = 10 then
FullFrame := false;
end;
if not FullFrame then
inc(FrameCounter);
end;
function TNewBowlingGame.Roll(ARoll: integer): boolean;
begin
if not (ARoll in [0 .. 10]) then
exit(false);
if FrameCounter < 9 then
Result := CheckRoll(ARoll)
else
Result := CheckFinal(ARoll);
end;
function TNewBowlingGame.Score: integer;
begin
Result := -1;
if Complete then
Result := Sum;
end;
function NewBowlingGame : IBowlingGame;
begin
Result := TNewBowlingGame.Create;
end;
end.`````` | HuggingFaceTB/finemath | |
Tuesday, November 4, 2014
50 Years of Misunderstanding Bell's Theorem
Precisely 50 years ago, Bell's paper "On the Einstein Podolsky Rosen Paradox", containing his famous theorem was received by the journal Physics. Today is John Bell Day.
Bell's theorem is one of the most influential result in physics, despite the fact that it is a negative result. Contrary to what many people believe, Bell was actually searching for a hidden variable theory, and he found instead some severe limitations of such theories. The limitation expressed by Bell's theorem celebrated today is that hidden variable theories have to be nonlocal. The outcome of measurements are correlated in a way which seems to ignore the separation in space. Some misunderstand this result as rejecting determinism, or as rejecting any kind of hidden variables, or at least as proving that any theory which describes the quantum world using hidden variables has to rely on instantaneous communication.
Maybe others searching for a hidden variables description of quantum phenomena hit the same wall Bell hit, but rather than having the same revelation as Bell, they ignored it and continued to search for a replacement or completion of quantum mechanics. For example, Einstein had all the data to find Bell's theorem almost 30 years before Bell. The paper coauthored by Einstein, Podolsky and Rosen, "Can Quantum-Mechanical Description of Physical Reality Be Considered Complete?" shown that entanglement allows nonlocal correlations. But Einstein disliked nonlocality because it seemed to violate special relativity. So he concluded that quantum mechanics was incomplete, by interpreting those correlations as revealing that Heisenberg's uncertainty principle can be trespassed. So Einstein and coauthors hit the same wall as Bell, only that they considered that the problem could be solved by completing quantum mechanics. Bell's theorem clarifies their findings in showing that no matter how you put it, the world is nonlocal (if Bell's inequality is violated, as it was confirmed by experiments).
Almost 30 years later, Bell understood nonlocality as the major consequence of the EPR "paradox", and expressed it in the form of his theorem. Today, at 50 years after Bell clarified the problem, there are so many who consider that Einstein was a crackpot in what concerns quantum mechanics, and Bell defeated him. Today it is easy for any student who took a class of quantum mechanics or philosophy of physics, to consider that he has a better understanding of quantum mechanics that Einstein, and to feel superior to him (true story, just search the physics blogs and forums and you will see many examples). Most often they believe (as they are taught) that quantum mechanics is so radically different because it is not deterministic, and that what Einstein searched was a deterministic theory. And that EPR suggested this, and Bell rejected it. This is so unfair for EPR, but also for Bell.
The truth is that despite the 10 hot years of discoveries in quantum mechanics, when nearly every aspect was understood, and the foundations were laid down, nobody before Einstein, Podolsky and Rosen found that "paradox", which is true and relevant. It is unfair to consider the EPR an attack against quantum mechanics, as it is seen by many since the beginning. Rather, it is a most important discovery, which could only be made because three rebels were not satisfied with Bohr's prescriptions. Moreover, in almost 30 years since the EPR paper, nobody solved their "paradox". Not even Bohr, who rushed to respond too quickly with an article bearing the same name as the EPR one. And the solution was found by Bell, who was a supporter of hidden variables, and maybe he wouldn't find it either, without the reformulation of the EPR argument due to the main exponent of hidden variable theories at that time, David Bohm.
Now, the reader may think that I am defending the hidden variables, by praising hidden variables theorists like Einstein, Bohm, and Bell. I actually don't defend hidden variables, and I don't say this just because of the witch hunt against "Bohmians". I just want to emphasize that without these "crackpots", we would not have today the understanding of entanglement and nonlocality which allows scientists to put at use the "magic" of quantum mechanics at work in quantum computing, quantum information, quantum cryptography, and other recent hot areas.
Actually, to be honest, among Einstein, Bohm, and Bell, only the first two are considered a lacking understanding of quantum mechanics, and Bell is considered as the one who defeat them, so he is celebrated, while the other two are not. But this is only because Bell is perceived as being, because of his theorem, against hidden variables, while in fact he was also searching for a hidden variable theory.
Moreover, for some reason, many consider that Bell's theorem is only about hidden variable theories, while in fact it is about any quantum theory or interpretation which describes quantum correlations as are observed in nature, and therefore violates Bell's inequality. Including therefore standard quantum mechanics. So, quantum mechanics is nonlocal too, and no Copenhagen Interpretation, no Many Worlds Interpretation, no Decoherence Interpretation can make it otherwise. Similarly, quantum mechanics is contextual too, despite the fact that the Bell-Kochen-Specker theorem is considered to apply to hidden variable theories only.
But why some tend to consider only hidden variable theories guilty of the sins of nonlocality and contextuality? Maybe because they just want to reject such theories? Or could it be because they believe that it makes no sense to think about what happens between measurements (as Bohr teaches us)? Or because nearly everyone, when first learning about quantum mechanics, has the instinct of finding a local realist explanation, and fails, of course, and then denies having this sin by throwing stones at those who seem to have it? I think this is fine, since this is what we should do, we should question everything, and that the persistence with which we should question a claim has to increase with the degree by which that claim contradicts what we learned before, as is the case of quantum mechanics.
For lack of time, for the rush of getting published, for the fear of getting rejected for having unorthodox views, we tend to eat much more than we can digest, and actually we cease digesting. This is why misunderstanding are propagated even at the top of the scientific community. Misunderstandings concerning quantum mechanics and Bell's theorem prevent us from seeing both the truth, and the amazing beauty of quantum mechanics, which is transformed into a mere tool to calculate probabilities, and any attempt at understanding it is regarded with disdain.
I find very fortunate the fact that Tim Maudlin wrote for the 50th anniversary of Bell's theorem a paper named "What Bell did", in which he explains that Bell's result is that indeed our world, hence quantum mechanics, is nonlocal. He makes a thorough and in my opinion probably the most down to earth analysis of the meaning of the EPR paper and of Bell's theorem, and how they are misunderstood. He identifies a cluster of misunderstandings that are propagated among physicists and philosophers of physics. This is one of the cases when a philosopher really can help physicists understand physics. I'll leave you the pleasure to read it.
Happy Birthday, Nature!
Exactly 145 years ago, on 4 November 1869, the first number of Nature appeared.
According to the current Romanian prime minister Victor Ponta, Nature is controlled (unofficially) by the current president of Romania, Traian Băsescu, with the main purpose to accuse Ponta of plagiarism. A possible explanation is that some collaborators of Băsescu traveled in time to create the journal. And then they also founded 145 years ago a secret society of scientists, who kept making great scientific discoveries. This also explains why most scientific discoveries were made in the last 1.5 centuries. The reason to make these scientific breakthroughs is not to advance the world, but to publish them in Nature, or in the other journals citing Nature, to make it world's most cited journal, so that, when the time comes, Nature's accusations against Ponta will have greater impact. And to invent rules that it is dishonest to copy text from other books and articles without attributing it explicitly when you write your PhD thesis. The second reason would be that the research made by this secret society of genii secretly led by the mastermind Băsescu will eventually lead to the discovery of time travel [1,2,3,4], which would allow him to send his people 145 years back in time...
Now, presidential elections are taking place in Romania. Two days ago was the first round, and now Ponta is the favorite to become the new president after the second round, in 12 days. Ponta has now the chance to forbid time travel, to change the history back to its track, which is a world without the Nature journal and all that scientific research in it, a world without physicists who can discover time travel. Or quite the opposite, he may actually make use of time travel, to set our history back to 25 years ago, when people overthrew Ceaușescu, and the communists were forced to disguise themselves as social-democrats to continue to keep the power.
Monday, October 6, 2014
Dots plus dots equal spheres
I took this photo in a bus in Pisa. We can see a pattern of spheres.
Here is how to obtain it. We overlap these dots
over a shrunk version of theirs
and we get the following pattern:
Wednesday, October 1, 2014
Living in a vector
Vectors are present in all domains of fundamental physics, so if you want to understand physics, you will need them. You may think you know them, but the truth is that they appear in so many guises, that nobody really knows everything about them. But vectors are a gate that allows you to enter the Cathedral of physics, and once you are inside, they can guide you in all places. That is, special and general relativity, quantum mechanics, particle physics, gauge theory... all these places need vectors, and once you master the vectors, they become much simpler (if you don't know them and are interested, read this post).
The Cathedral has many gates, and vectors are just one of them. You can enter through groups, sets and relations, functions, categories, through all sorts of objects or structures from algebra, geometry, even logic. I decided to show you now the way of vectors, because I think is fast and deep in the same time, but remember, this is a matter of choice. And vectors will lead us, inevitably, to the other gates too.
I will explain some elementary and not so elementary things about vectors, but you have to read and practice, because here I just give some guidelines, a big picture. The reason I am doing this is that when you study, you may get lost in details and miss the essential.
Very basic things
A vector can be understood in many ways. One way is to see it as a way to specify how to move from one point to another. A vector is like an arrow, and if you place the arrow in that point, you find the destination point. To find the new position for any point, just place the vector in that point, and the tip of the vector will show you the new position. You can compose more such arrows, and what you'll get is another vector, their sum. You can also subtract them, just place their origins in the same point, and the difference is the vector obtained by joining their tips with another arrow.
Once you fix a reference position, an origin, you can specify any position, by the vector that tells you how to move from origin to that position. You can see that vector as being the difference between the destination, and the starting position.
You can add and subtract vectors. You can multiply them with numbers. Those numbers are from a field $\mathbb{K}$, and we can take for example $\mathbb{K}=\mathbb{R}$, or $\mathbb{K}=\mathbb{C}$, and are called scalars. A vector space is a set of vectors, so that no matter how you add them and scale them, the result is from the same set. The vector space is real (complex), if the scalars are real (complex) numbers. A sum of rescaled vectors is named linear combination. You can always pick a basis, or a frame, a set of vectors so that any vector can be written as a linear combination of the basis vectors, in a unique way.
Vectors and functions
Consider a vector $v$ in an $n$-dimensional space $V$, and suppose its components in a given basis are $(v^1,\ldots,v^n)$. You can represent any vector $v$ as a function $f:\{1,\ldots,n\}\to\mathbb{K}$ given by $f(i)=v^i$. Conversely, any such function defines a unique vector. In general, if $S$ is a set, then the set of the functions $f:S\to\mathbb{K}$ form a vector space, which we will denote by $\mathbb{K}^S$. The cardinal of $S$ gives the dimension of the vector space, so $\mathbb{K}^{\{1,\ldots,n\}}\cong\mathbb{K}^n$. So, if $S$ is an infinite set, we will have an infinite dimensional vector space. For example, the scalar fields on a three dimensional space, that is, the functions $f:\mathbb{R}^3\to \mathbb{R}$, form an infinite dimensional vector space. Not only the vector spaces are not limited to $2$ or $3$ dimensions, but infinite dimensional spaces are very natural too.
Dual vectors
If $V$ is a $\mathbb{K}$-vector space, a linear functions $f:V\to\mathbb{K}$ is a function satisfying $f(u+v)=f(u)+f(v)$, and $f(\alpha u)=\alpha f(u)$, for any $u,v\in V,\alpha\in\mathbb{K}$. The linear functions $f:V\to\mathbb{K}$ form a vector space $V^*$ named the dual space of $V$.
Tensors
Consider now two sets, $S$ and $S'$, and a field $\mathbb{K}$. The Cartesian product $S\times S'$ is defined as the set of pairs $(s,s')$, where $s\in S$ and $s'\in S'$. The functions defined on the Cartesian product, $f:S\times S'\to\mathbb{K}$, form a vector space $\mathbb{K}^{S\times S'}$, named the tensor product of $\mathbb{K}^{S}$ and $\mathbb{K}^{S'}$, $\mathbb{K}^{S\times S'}=\mathbb{K}^{S}\otimes\mathbb{K}^{S'}$. If $(e_i)$ and $(e'_j)$ are bases of $\mathbb{K}^{S}$ and $\mathbb{K}^{S'}$, then $(e_ie'j)$, where $e_ie'_j(s,s')=e_i(s)e'_j(s')$, is a basis of $\mathbb{K}^{S\times S'}$. Any vector $v\in\mathbb{K}^{S_1\times S_2}$ can be uniquely written as $v=\sum_i\sum_j \alpha_{ij} e_ie'j$.
Also, the set of functions $f:S\to\mathbb{K}^{S'}$ is a vector space, which can be identified with the tensor product $\mathbb{K}^{S}\otimes(\mathbb{K}^{S'})^*$.
The vectors that belong to tensor products of vector spaces are named tensors. So, tensors are vectors with some extra structure.
The tensor product can be defined easily for any kind of vector spaces, because any vector space can be thought of as a space of functions. The tensor product is associative, so we can define it between multiple vector spaces. We denote the tensor product of $n>1$ copies of $V$ by $V^{\otimes n}$. We can check that for $m,n>1$, $V^{\otimes (m+n)}=V^{\otimes {m}}\otimes V^{\otimes {n}}$. This can work also for $m,n\geq 0$, if we define $V^1=V$, $V^0=\mathbb{K}$. So, vectors and scalars are just tensors.
Let $U$, $V$ be $\mathbb{K}$-vector spaces. A linear operator is a function $f:U\to V$ which satisfies $f(u+v)=f(u)+f(v)$, and $f(\alpha u)=\alpha f(u)$, for any $u\in U,v\in V,\alpha\in\mathbb{K}$. The operator $f:U\to V$ is in fact a tensor from $U^*\otimes V$.
Inner products
Given a basis, any vector can be expressed as a set of numbers, the components of the vector. But the vector is independent of this numerical representation. The basis can be chosen in many ways, and in fact, any non-zero vector can have any components (provided not all are zero) in a well chosen basis. This shows that any two non-zero vectors play identical roles, which may be a surprise. This is a key point, since a common misconception when talking about vectors is that they have definite intrinsic sizes and orientations, or that they can make an angle. But in fact the sizes and orientations are relative to the frame, or to the other vectors. Moreover, you can say that from two vectors, one is larger than the other, only if they are collinear. Otherwise, no matter how small is one of them, we can easily find a basis in which it becomes larger than the other. It makes no sense to speak about the size, or magnitude, or length of a vector, as an intrinsic property.
But wait, one may say, there is a way to define the size of a vector! Consider a basis in a two-dimensional vector space, and a vector $v=(v^1,v^2)$. Then, the size of the vector is given by Pythagoras's theorem, by $\sqrt{(v^1)^2+(v^2)^2}$. The problem with this definition is that, if you change the basis, you will obtain different components, and different size of the vector. To make sure that you obtain the same size, you should allow only certain bases. To speak about the size of a vector, and about the angle between two vectors, you need an additional object, which is called inner product, or scalar product. Sometimes, for example in geometry and in relativity, it is called metric.
Choosing a basis gives a default inner product. But the best way is to define the inner product, and not to pick a special basis. Once you have the inner product, you can define angles between vectors too. But size and angles are not intrinsic properties of vectors, they depend on the scalar product too.
The inner product between two vectors $u$ and $v$, defined by a basis, is $u\cdot v = u^1 v^1 + u^2 v^2 + \ldots + u^n v^n$. But in a different basis, it will have a general form $u\cdot v=\sum_i\sum_j g_{ij} u^i v^j$, where $g_{ij}=g_{ji}$ can be seen as the components of a symmetric matrix. These components change when we change the basis, they form the components of a tensor from $V^*\otimes V^*$. Einstein had the brilliant idea to omit the sum signs, so the inner product looks like $u\cdot v=g_{ij} u^i v^j$, where you know that since $i$ and $j$ appear both in upper and in lower positions, we make them run from $1$ to $n$ and sum. This is a thing that many geometers hate, but physicists find it very useful and compact in calculations, because the same summation convention appears in many different situations, which to geometers appear to be different, but in fact are very similar.
Given a basis, we can define the inner product by choosing the coefficients $g_{ij}$. And we can always find another basis, in which $g_{ij}$ is diagonal, that is, it vanishes unless $i=j$. And we can rescale the basis so that $g_{ii}$ are equal to $-1$, $1$, or $0$. Only if $g_{ii}$ are all $1$ in some basis, the size of the vector is given by the usual Pythagoras's theorem, otherwise, there will be some minus signs there, and even some terms will be omitted (corresponding to $g_{ii}=0$).
Quantum mechanics
Quantum particles are described by Schrödinger's equation. Its solutions are, for a single elementary particle, complex functions $|\psi\rangle:\mathbb{R}^3\to\mathbb{C}$, or more general, $|\psi\rangle:\mathbb{R}^3\to\mathbb{C}^k$, named wavefunctions. They describe completely the states of the quantum particle. They form a vector space $H$ which also has a hermitian product (a complex scalar product so that $h_{ij}=\overline{h_{ji}}$), and is named the Hilbert space (because in the infinite dimensional case also satisfies an additional property which we don't need here), or the state space. Linear transformations of $H$ which preserve the complex scalar product are named unitary transformations, and they are the complex analogous of rotations.
The wavefunctions are represented in a basis as functions of positions, $|\psi\rangle:\mathbb{R}^3\to\mathbb{C}^k$. The element of the position basis represent point particles. But we can make a unitary transformation and obtain another basis, made of functions of the form $e^{i (k_x x + k_y y + k_z z)}$, which represent pure waves. Some observations use one of the bases, some the other, and here is why there is a duality between waves and point particles.
For more elementary particles, the state space is the tensor product of the state spaces of the individual particles. A tensor product of the form $|\psi\rangle\otimes|\psi'\rangle$ represents separable states, which can be observed independently. If the system can't be written like this, but only as a sum, the particles are entangled. When we measure them, the outcomes are correlated.
The evolution of a quantum system is described by Schrödinger's equation. Basically, the state rotates, by a unitary transformation. Only such transformations conserve the probabilities associated to the wavefunction.
When you measure the quantum systems, you need an observable. One can see an observable as defining a decomposition of the state space, in perpendicular subspaces. After the observation, the state is found to be in one of the subspaces. We can only know the subspace, but not the actual state vector. This is strange, because the system can, in principle, be in any possible state, but the measurement finds it to be only in one of these subspaces (we say it collapsed). This is the measurement problem. The things become even stranger, if we realize that if we measure another property, the corresponding decomposition of the state space is different. In other words, if you look for a point particle, you find a point particle, and if you look for a wave, you find a wave. This seems as if the unitary evolution given by the Schrödinger's equation is broken during observations. Perhaps the wavefunction remains intact, but to us, only one of the components continues to exist, corresponding to the subspace we obtained after the measurement. In the many worlds interpretation the universes splits, and all outcomes continue to exist, in new created universes. So, not only the state vector contains the universe, but it actually contains many universes.
I have a proposed explanation for some strange quantum features, in [1, 2, 3], and in these videos:
Special relativity
An example when there is a minus signs in the Pythagoras's theorem is given by the theory of relativity, where the squared size of a vector is $v\cdot v=-(v^t)^2+(v^x)^2+(v^y)^2+(v^z)^2$.
This inner product is named the Lorentz metric. Special relativity takes place in the Minkowski spacetime, which has four dimensions. A vector $v$ is named timelike if $v\cdot v < 0$, spacelike if $v\cdot v > 0$, and null or lightlike if $v\cdot v = 0$. A particle moving with the speed of light is described by a lightlike vector, and one moving with an inferior speed, by a timelike vector. Spacelike vectors would describe faster than light particles, if they exist. Points in spacetime are named events. Events can be simultaneous, but this depends on the frame. Anyway, to be simultaneous in a frame, two events have to be separated by a spacelike interval. If they are separated by a lightlike or timelike interval, they can be connected causally, or joined by a particle with a speed equal to, respectively smaller than the speed of light.
In Newtonian mechanics, the laws remain unchanged to translations and rotations in space, translations in time, and inertial movements of the frame - together they form the Galilei transformations. However, electromagnetism disobeyed. In fact, this was the motivation of the research of Einstein, Poincaré, Lorentz, and FitzGerald. Their work led to the discovery of special relativity, according to which the correct transformations are not those of Galilei, but those of Poincaré, which preserve the distances given by the Lorentz metric.
Curvilinear coordinates
A basis or a frame of vectors in the Minkowski spacetime allows us to construct Cartesian coordinates. However, if the observer's motion is accelerated (hence the observer is non-inertial), her frame will rotate in time, so Cartesian coordinates will have to be replaced with curved coordinates. In curved coordinates, the coefficients $g_{ij}$ depend on the position. But in special relativity they have to satisfy a flatness condition, otherwise spacetime will be curved, and this didn't make much sense back in 1905, when special relativity was discovered.
General relativity
Einstein remarked that to a non-inertial observer, inertia looks similar to gravity. So he imagined that a proper choice of the metric $g_{ij}$ may generate gravity. This turned out indeed to be true, but the choice of $g_{ij}$ corresponds to a curved spacetime, and not a flat one.
One of the problems of general relativity is that it has singularities. Singularities are places where some of the components of $g_{ij}$ become infinite, or where $g_{ij}$ has, when diagonalized, some zero entries on the diagonal. For this reason, many physicist believe that this problem indicates that general relativity should be replaced with some other theory, to be discovered. Maybe it will be solved when we will replace it with a theory of quantum gravity, like string theory or loop quantum gravity. But until we will know what is the right theory of quantum gravity, general relativity can actually deal with its own singularities (while the ones mentioned above did not solve this problem). I will not describe this here, but you can read my articles about this, and also this essay, and these posts about the black hole information paradox [1, 2, 3]. And watch this video
Vector bundles and forces
We call fields the functions defined on the space or the spacetime. We have seen that fields valued in vector spaces are actually vector spaces. On a flat space $M$ which looks like a vector space, the fields valued in vector spaces can be thought of as being valued in the same vector space, for example $f:M\to V$. But if the space is curved, or if it has nontrivial topology, we are forced to consider that at each point there is another copy of $V$. So, such a field will be more like $f(x)\in V_x$, where $V_x$ is the copy of the vector space $V$ at the point $x$. Such fields still form a vector space. The union of all $V_x$ is called a vector bundle. The fields are also called sections, and $V_x$ is called the fiber at $x$.
Now, since $V_x$ are copies of $V$ at each point, there is no invariant way to identify each $V_x$ with $V$. In other words, $V_x$ and $V$ can be identified, for each $x$, up to a linear transformation of $V$. We need a way to move from $V_x$ to a neighboring $V_{x+d x}$. This can be done with a connection. Also, moving a vector from $V_x$ along a closed curve reveals that, when returning to $V_x$, the vector is rotated. This is explained by the presence of a curvature, which can be obtained easily from the connection.
Connections behave like potentials of force fields. And a force field corresponds to the curvature of the connection. This makes very natural to use vector bundles to describe forces, and this is what gauge theory does.
Forces in the standard model of particles are described as follows. We assume that there is a typical complex vector space $V$ of dimension $n$, endowed with a hermitian scalar product. The connection is required to preserve this hermitian product when moving among the copies $V_x$. The set of linear transformations that preserve the scalar product is named unitary group, and is denoted by $U(n)$. The subset of transformations having the determinant equal to $1$ is named the special unitary group, $SU(n)$. The electromagnetic force corresponds to $U(1)$, the weak force to $SU(2)$, and the strong force to $SU(3)$. Moreover, all particles turn out to correspond to vectors that appear in the representations of the gauge groups on vector spaces.
What's next?
Vectors are present everywhere in physics. We see that they help us understand quantum mechanics, special and general relativity, and the particles and forces. They seem to offer a unitary view of fundamental physics.
However, up to this point, we don't know how to unify
• unitary evolution and the collapse of the wavefunction
• the quantum level with the mundane classical level
• quantum mechanics and general relativity
• the electroweak and strong forces (we know though how to combine the electromagnetic and weak forces, in the unitary group $U(2)$)
• the standard model forces and gravity
Saturday, September 27, 2014
The unreasonable beauty of mathematics in the natural sciences*
Imagine a man and a woman, seeing and liking each other at a party or club or so. They start talking, the mutual attraction is obvious, but they want to be casual for two minutes. So they exchange informal formalities about doesn't matter what. Then he asks her: "so, what do you do?", and she replies "I'm a poet". What if the guy would say something like "I hate poetry!", or even declare proudly "I never knew how to use letters to write words and stuff, and I don't care!". Or imagine she's a musician, and he says "I hate music!". There are two things we can say about that kind of guy. First, he is very rude, he never ever deserves a second chance with that girl or any other human being for that matter. He should be isolated, kept outside society. Second, or maybe this should be first, how on earth can he be proud for being illiterate!
You probably guessed that this story is true. OK, In my case it was about math instead of poetry, and the genders are reversed. This happened to me or to anyone in the same situation quite often. There is no political correctness when it comes about math, maybe because one tends to believe that if you like math, you have no feelings, and such a remark wouldn't hurt you. And I actually was never offended when a girl said such outrageous things like that she hates math. Because whenever a girl told me she hates math, I knew she calls math something that really is boring and ugly, and not what I actually call math. Because math as I know it is poetry, is music, and is a wonderful goddess.
The story continues, years later. You talk about physics, with people interested in physics, or even with physicists. And you say something about this being just a mathematical consequence of that, or that certain phenomenon can be better understood if we consider it as certain mathematical object. It happens sometimes that your interlocutor becomes impatient and says that this is only math, and you were discussing physics, that math has no power there, and so on. Or that math is at best just a tool, and it actually obscures the real picture, or even that it limits our power of understanding.
People got the wrong picture that math is about numbers, or letters that stand for unknown numbers, or being extremely precise and calculating a huge number of decimals, or being very rigid and limited. In fact, math is just the study of relations. You will be surprised, but this is actually the mathematical definition of math. Numbers come into math only incidentally, as they come into music, when you indicate the duration or the tempo. Math is just a qualitative description of relations, and by relations we can understand a wide rainbow of things. I will detail this another time.
Imagine you wake up and you don't remember where you are, or who you are, like you were just born. You are surrounded by noise, which hurts your ears and your brain, meaningless random violent noise. You run desperately, trying to avoid it, but it is everywhere. And you finally find a spot where everything becomes suddenly wonderful: the noise becomes music, a celestial, beautiful music, and everything starts making sense. You are in a wonderful Cathedral, and you are tempted to call what you are listening "music of the spheres". The same music played earlier, but you were in the wrong place, where the acoustics was bad, or the sounds reached your ear in the wrong order, because of the relative positions of the instruments. Or maybe your ears were not yet tuned to the music. The point is that what seemed to be ugly noise, suddenly became so wonderful.
So, when someone says "I hate math!", all I hear is "I am in the Cathedral you call wonderful, but in the wrong place, where the celestial music becomes ugly violent noise!".
If you are interested in physics, you entered the Cathedral. But if you hate math, you will not last here, and maybe it is better to get out immediately! And if you are still interested in physics, come inside slowly, carefully choosing your steps, to avoid being assaulted by the music of the spheres, to allow it gently to enter in your mind, and to open your eyes. Choose carefully what you read, what lectures you watch, and ask questions. Don't be shy, any question you will ask is the right question for your current position, and for your next step.
There are some places in the Cathedral where the music is really beautiful. If you meet people there, to share the music, to dance, you will feel wonderful. If not, you will feel lonely. So you will want to share that place, you will want to invite your friends to join you.
The reason I love physics, is that I want to find these places. The reason I read blogs and papers, is that I want them to help me find such places. The reason I write papers, and I blog about this, is that I would like to share my places with others. I attend conferences (four so far this year) because they are like concerts, where you get the chance to listen some wonderful music, and to play your own.
But these are just words. I would like to write more posts in which I show the unreasonable beauty of math in physics, with concrete examples. Judging by the statistics, I have a few readers; judging by the number of comments, I don't really touch many of them. I know sometimes I am too serious, or too brief when I should explain more, especially when mathematical subtleties are involved. I am not very good at explaining abstract things to non-specialists, but I want to learn. I would like to write better, to be more useful, so, I would like to encourage comments and suggestions. Ask me to clarify, to explain, to detail, to simplify. Tell me what you would like to understand.
To start, I would like to write about vectors. They are so fundamentals in all areas of physics and mathematics, so I think it's a good idea to start with them. You may think they are too simple, and that you know all about them from high school, but you don't know the whole story. Later, when I will say something about quantum mechanics and relativity, they will be necessary (after all, according to quantum mechanics, the state of the universe is a vector). On the other hand, if you will understand them well, you will be around half of the way to understand some modern physics.
______________________
* You surely guessed that the title is a reference to Wigner's brilliant and insightful lecture, The unreasonable effectiveness of mathematics in the natural sciences.
Update, October 14, 2014
I just watched an episode of the Colbert Report, where the mathematician Edward Frenkel was invited in April this year. It was about Frenkel's new book and about his movie. He discusses at some point precisely the fact that it is so acceptable to hate math, as opposed to hating music or painting. Here is what he says for The Wall Street Journal:
It's like teaching an art class where they only tell you how to paint a fence but they never show you Picasso. People say 'I'm bad at math,' but what they're really saying is 'I was bad at painting the fence.'
Also see this video: | open-web-math/open-web-math | |
# THE SALLY RIDE QUARTER
Date of the Problem
November 15, 2021
In honor of Sally Ride’s accomplishments and contributions to inspiring women in STEM, the U.S. Mint is releasing a quarter with her likeness in 2022 as part of the American Women Quarters Program.
On June 18, 1983, Sally K. Ride became the first American woman in space. Her first mission to space was aboard the Space Shuttle Challenger. It took 6 days, 2 hours, 23 minutes, 59 seconds for the shuttle to travel a total of 4,072,533 kilometers (about 2.5 million miles) during 97 orbits of the Earth. If the Sally Ride quarter will be 1.75 millimeters thick, how many of her quarters stacked have a height equal to 4,072,533 kilometers? Express your answer in scientific notation to three significant figures.
Her first mission to space was 4,072,533 kilometers, or 4,072,533 km × 1,000,000 mm/km = 4.072533 × 1012 millimeters. A Sally Ride quarter is 1.75 millimeters thick. So, in order for a stack of these quarters to have a height equal to 4,072,533 kilometers, it would need to contain (4.072533 × 1012) ÷ (1.75 × 100) ≈ 2.33 × 1012 quarters.
Sally Ride will be honored with a quarter through the American Women Quarters Program, along with Maya Angelou, Wilma Mankiller, Nina Otero-Warren and Anna May Wong. These five new quarter designs will be released consecutively throughout the year. If the order in which the coins are released is completely random, what is the probability that Sally Ride’s quarter will be released second? Express your answer as a common fraction.
There are 5 × 4 × 3 × 2 × 1 = 120 total orders in which the quarters could be released. In looking for how many ways they could be released with Sally Ride’s quarter coming out second, we can “fix” this quarter in the second position. This means there are 4 options for the first quarter to be released, 3 options for the third, 2 options for the fourth and 1 option for the fifth (so, 4 × 3 × 2 × 1 = 24 ways). Therefore, the probability of Sally Ride's quarter being released second is 24/120 = 1/5.
Suppose that about 800 million quarters are added to circulation in the U.S. over 10 weeks. If approximately 60 million Sally Ride quarters will be added to circulation over 6 weeks, what percent of quarters added to circulation each week, on average, will be Sally Ride quarters? Express your answer to the nearest whole number.
If 800 million quarters are added to circulation over 10 weeks, then about 800 million ÷ 10 weeks = 80 million total quarters would be added to circulation each week. Approximately 60 million ÷ 6 weeks = 10 million Sally Ride quarters will be added to circulation each week. So, on average, (10 million ÷ 80 million) × 100 = 12.5% ≈ 13% of the quarters added to circulation each week will be Sally Ride quarters.
Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
CCSS (Common Core State Standard)
Difficulty | HuggingFaceTB/finemath | |
# Measure (mathematics) Informally, a measure has the property of being monotone in the sense that if A is a subset of B, the measure of A is less than or equal to the measure of B. Furthermore, the measure of the empty set is required to be 0.
In mathematical analysis, a measure on a set is a systematic way to assign a number to each suitable subset of that set, intuitively interpreted as its size. In this sense, a measure is a generalization of the concepts of length, area, and volume. A particularly important example is the Lebesgue measure on a Euclidean space, which assigns the conventional length, area, and volume of Euclidean geometry to suitable subsets of the n-dimensional Euclidean space Rn. For instance, the Lebesgue measure of the interval [0, 1] in the real numbers is its length in the everyday sense of the word, specifically, 1.
Technically, a measure is a function that assigns a non-negative real number or +∞ to (certain) subsets of a set X (see Definition below). It must further be countably additive: the measure of a 'large' subset that can be decomposed into a finite (or countably infinite) number of 'smaller' disjoint subsets is equal to the sum of the measures of the "smaller" subsets. In general, if one wants to associate a consistent size to each subset of a given set while satisfying the other axioms of a measure, one only finds trivial examples like the counting measure. This problem was resolved by defining measure only on a sub-collection of all subsets; the so-called measurable subsets, which are required to form a σ-algebra. This means that countable unions, countable intersections and complements of measurable subsets are measurable. Non-measurable sets in a Euclidean space, on which the Lebesgue measure cannot be defined consistently, are necessarily complicated in the sense of being badly mixed up with their complement. Indeed, their existence is a non-trivial consequence of the axiom of choice.
Measure theory was developed in successive stages during the late 19th and early 20th centuries by Émile Borel, Henri Lebesgue, Johann Radon, and Maurice Fréchet, among others. The main applications of measures are in the foundations of the Lebesgue integral, in Andrey Kolmogorov's axiomatisation of probability theory and in ergodic theory. In integration theory, specifying a measure allows one to define integrals on spaces more general than subsets of Euclidean space; moreover, the integral with respect to the Lebesgue measure on Euclidean spaces is more general and has a richer theory than its predecessor, the Riemann integral. Probability theory considers measures that assign to the whole set the size 1, and considers measurable subsets to be events whose probability is given by the measure. Ergodic theory considers measures that are invariant under, or arise naturally from, a dynamical system.
## Definition
Countable additivity of a measure μ: The measure of a countable disjoint union is the same as the sum of all measures of each subset.
Let X be a set and Σ a σ-algebra over X. A function μ from Σ to the extended real number line is called a measure if it satisfies the following properties:
• Non-negativity: For all E in Σ, we have μ(E) ≥ 0.
• Null empty set: $\mu (\varnothing )=0$ .
• Countable additivity (or σ-additivity): For all countable collections $\{E_{k}\}_{k=1}^{\infty }$ of pairwise disjoint sets in Σ,
$\mu \left(\bigcup _{k=1}^{\infty }E_{k}\right)=\sum _{k=1}^{\infty }\mu (E_{k}).$
If at least one set $E$ has finite measure, then the requirement that $\mu (\varnothing )=0$ is met automatically. Indeed, by countable additivity,
$\mu (E)=\mu (E\cup \varnothing )=\mu (E)+\mu (\varnothing ),$
and therefore $\mu (\varnothing )=0.$
If only the second and third conditions of the definition of measure above are met, and μ takes on at most one of the values ±∞, then μ is called a signed measure.
The pair (X, Σ) is called a measurable space, the members of Σ are called measurable sets. If $\left(X,\Sigma _{X}\right)$ and $\left(Y,\Sigma _{Y}\right)$ are two measurable spaces, then a function $f:X\to Y$ is called measurable if for every Y-measurable set $B\in \Sigma _{Y}$ , the inverse image is X-measurable – i.e.: $f^{(-1)}(B)\in \Sigma _{X}$ . In this setup, the composition of measurable functions is measurable, making the measurable spaces and measurable functions a category, with the measurable spaces as objects and the set of measurable functions as arrows. See also Measurable function#Term usage variations about another setup.
A triple (X, Σ, μ) is called a measure space. A probability measure is a measure with total measure one – i.e. μ(X) = 1. A probability space is a measure space with a probability measure.
For measure spaces that are also topological spaces various compatibility conditions can be placed for the measure and the topology. Most measures met in practice in analysis (and in many cases also in probability theory) are Radon measures. Radon measures have an alternative definition in terms of linear functionals on the locally convex space of continuous functions with compact support. This approach is taken by Bourbaki (2004) and a number of other sources. For more details, see the article on Radon measures.
## Examples
Some important measures are listed here.
Other 'named' measures used in various theories include: Borel measure, Jordan measure, ergodic measure, Euler measure, Gaussian measure, Baire measure, Radon measure, Young measure, and Loeb measure.
In physics an example of a measure is spatial distribution of mass (see e.g., gravity potential), or another non-negative extensive property, conserved (see conservation law for a list of these) or not. Negative values lead to signed measures, see "generalizations" below.
• Liouville measure, known also as the natural volume form on a symplectic manifold, is useful in classical statistical and Hamiltonian mechanics.
• Gibbs measure is widely used in statistical mechanics, often under the name canonical ensemble.
## Basic properties
Let μ be a measure.
### Monotonicity
If E1 and E2 are measurable sets with E1 ⊆ E2 then
$\mu (E_{1})\leq \mu (E_{2}).$
### Measure of countable unions and intersections
For any countable sequence E1, E2, E3, ... of (not necessarily disjoint) measurable sets En in Σ:
$\mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)\leq \sum _{i=1}^{\infty }\mu (E_{i}).$
#### Continuity from below
If E1, E2, E3, ... are measurable sets and $E_{n}\subseteq E_{n+1},$ for all n, then the union of the sets En is measurable, and
$\mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)=\lim _{i\to \infty }\mu (E_{i}).$
#### Continuity from above
If E1, E2, E3, ... are measurable sets and, for all n, $E_{n+1}\subseteq E_{n},$ then the intersection of the sets En is measurable; furthermore, if at least one of the En has finite measure, then
$\mu \left(\bigcap _{i=1}^{\infty }E_{i}\right)=\lim _{i\to \infty }\mu (E_{i}).$
This property is false without the assumption that at least one of the En has finite measure. For instance, for each nN, let En = [n, ∞) ⊂ R, which all have infinite Lebesgue measure, but the intersection is empty.
## Sigma-finite measures
A measure space (X, Σ, μ) is called finite if μ(X) is a finite real number (rather than ∞). Nonzero finite measures are analogous to probability measures in the sense that any finite measure μ is proportional to the probability measure ${\frac {1}{\mu (X)}}\mu$ . A measure μ is called σ-finite if X can be decomposed into a countable union of measurable sets of finite measure. Analogously, a set in a measure space is said to have a σ-finite measure if it is a countable union of sets with finite measure.
For example, the real numbers with the standard Lebesgue measure are σ-finite but not finite. Consider the closed intervals [k, k+1] for all integers k; there are countably many such intervals, each has measure 1, and their union is the entire real line. Alternatively, consider the real numbers with the counting measure, which assigns to each finite set of reals the number of points in the set. This measure space is not σ-finite, because every set with finite measure contains only finitely many points, and it would take uncountably many such sets to cover the entire real line. The σ-finite measure spaces have some very convenient properties; σ-finiteness can be compared in this respect to the Lindelöf property of topological spaces. They can be also thought of as a vague generalization of the idea that a measure space may have 'uncountable measure'.
## s-finite measures
A measure is said to be s-finite if it is a countable sum of bounded measures. S-finite measures are more general than sigma-finite ones and have applications in the theory of stochastic processes.
## Completeness
A measurable set X is called a null set if μ(X) = 0. A subset of a null set is called a negligible set. A negligible set need not be measurable, but every measurable negligible set is automatically a null set. A measure is called complete if every negligible set is measurable.
A measure can be extended to a complete one by considering the σ-algebra of subsets Y which differ by a negligible set from a measurable set X, that is, such that the symmetric difference of X and Y is contained in a null set. One defines μ(Y) to equal μ(X).
Measures are required to be countably additive. However, the condition can be strengthened as follows. For any set $I$ and any set of nonnegative $r_{i},i\in I$ define:
$\sum _{i\in I}r_{i}=\sup \left\lbrace \sum _{i\in J}r_{i}:|J|<\aleph _{0},J\subseteq I\right\rbrace .$
That is, we define the sum of the $r_{i}$ to be the supremum of all the sums of finitely many of them.
A measure $\mu$ on $\Sigma$ is $\kappa$ -additive if for any $\lambda <\kappa$ and any family of disjoint sets $X_{\alpha },\alpha <\lambda$ the following hold:
$\bigcup _{\alpha \in \lambda }X_{\alpha }\in \Sigma$
$\mu \left(\bigcup _{\alpha \in \lambda }X_{\alpha }\right)=\sum _{\alpha \in \lambda }\mu \left(X_{\alpha }\right).$
Note that the second condition is equivalent to the statement that the ideal of null sets is $\kappa$ -complete.
## Non-measurable sets
If the axiom of choice is assumed to be true, it can be proved that not all subsets of Euclidean space are Lebesgue measurable; examples of such sets include the Vitali set, and the non-measurable sets postulated by the Hausdorff paradox and the Banach–Tarski paradox.
## Generalizations
For certain purposes, it is useful to have a "measure" whose values are not restricted to the non-negative reals or infinity. For instance, a countably additive set function with values in the (signed) real numbers is called a signed measure, while such a function with values in the complex numbers is called a complex measure. Measures that take values in Banach spaces have been studied extensively. A measure that takes values in the set of self-adjoint projections on a Hilbert space is called a projection-valued measure; these are used in functional analysis for the spectral theorem. When it is necessary to distinguish the usual measures which take non-negative values from generalizations, the term positive measure is used. Positive measures are closed under conical combination but not general linear combination, while signed measures are the linear closure of positive measures.
Another generalization is the finitely additive measure, also known as a content. This is the same as a measure except that instead of requiring countable additivity we require only finite additivity. Historically, this definition was used first. It turns out that in general, finitely additive measures are connected with notions such as Banach limits, the dual of L and the Stone–Čech compactification. All these are linked in one way or another to the axiom of choice. Contents remain useful in certain technical problems in geometric measure theory; this is the theory of Banach measures.
A charge is a generalization in both directions: it is a finitely additive, signed measure. | HuggingFaceTB/finemath | |
Home » Fractions into Decimals » 12 32/16 as a Decimal
# 12 32/16 as a Decimal
Here you can find 12 32/16 as a decimal.
We also have useful information regarding 12 32/16 in decimal form. 🙂
## Calculator
Fraction:
Decimal:
14
If you have been searching for 12 and 32 over 16 as a decimal, then you are right here, too.
The terms used in this article about 12 32/16 as decimal are explained in detail on our home page; check it out if anything remains unclear.
12 32/16 as a decimal = 14
## 12 32/16 in Decimal Form
12 32/16 in decimal notation has 0 decimal places. That is, 12 32/16 as decimal is a terminating decimal.
• 12 32/16 as a decimal = 14
• 12 32/16 in decimal form = 14
• Twelve and thirty-two sixteenths as a decimal = 14
• 12 and 32 over 16 as a decimal = 14
Now that you know what is 12 32/16 as a decimal you can learn how to change 12 32/16 to a decimal number in the following section.
In addition, you can read up on the properties of 12 32/16.
## Convert 12 32/16 to Decimal
To convert 12 32/16 to decimal you can use the long division method explained in our article fraction to decimal, which you can find in the header menu.
Or you can divide the nominator 224 ([12×16]+32) by the denominator 16 using a calculator.
If you like, use our automatic calculator above. Just enter the fraction with a slash, e.g. 12 32/16.
If the result includes a repeating sequence, then it will be denoted in ().
Similar conversions in this category include, for example:
Ahead is more information on 12 32/16 written in base 10 numeral system.
## What is 12 32/16 as a Decimal?
You already know the answer to what is 12 32/16 as a decimal. Twelve and thirty-two sixteenths as a decimal equals 14
We have characterized 12 32/16 in base 10 positional notation above, so we are left with telling you the properties of 12 32/16:
• 12 32/16 is a mixed fraction
• 12 is the whole-number part
• 32 is the nominator, above the slash
• 16 is the denominator, below the slash
• 12 32/16 can be changed to the improper fraction 224/16
Instead of a slash, the division symbol ÷, known as obelus, can be used to denote a fraction: For example: 12 32÷16 in decimal or 12 32÷16 as decimal.
## Conclusion
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Note that you can find many fraction to decimal conversions using the search form in the sidebar.
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# How to Calculate the Weight of Steel Plate
First of all, it needs to be clear that steel plate is a very common material in many industries, such as construction, manufacturing, aerospace, and so on. They are used in a wide range of applications, including making steel structures, panels, car bodies, oil tanks, containers, and more. Whether you’re in construction, manufacturing, or another industry, you need to buy or use steel plates, and it’s important to know their weight.
Secondly, the weight of steel plates is directly related to their performance and use. If the weight of the steel plate exceeds expectations, it may cause inconvenience to the user process or even a safety hazard. Therefore, it is very necessary to calculate the weight of steel plates.
This article will introduce in detail how to calculate the weight of steel plates to help readers better understand the basic knowledge and usage of the steel plate. Next, we will introduce the basic concept of steel plate weight.
## The Basic Concept of Steel Plate Weight
### A. Steel Plate Density
Steel plate density refers to the mass of steel plate per unit volume. The density of the steel plate varies with the type of steel sheet, and the density of the general steel plate is about 7.85g/cm³. When calculating the weight of a steel plate, the density of the steel plate needs to be used.
### B. Steel Plate Area
The area of the steel plate refers to the area covered by the surface of the steel sheet. When calculating the weight of the steel plate, it is necessary to measure the length and width of the steel plate, and then use the formula to calculate the area of the steel plate.
### C. Steel Plate Thickness
The thickness of the steel plate refers to the thickness dimension of the steel plate. Steel plate thickness is usually measured in millimeters (mm). When calculating the weight of a steel plate, it is necessary to know the thickness of the steel plate.
### D. Unit Conversion
When calculating the weight of a steel plate, dimensions such as length, width, and thickness need to be converted to the same unit for calculation. Common units include millimeters, centimeters, meters, inches, and feet, among others. When using the formula to calculate the weight of the steel plate, the same units must be used.
The above are the basic concepts needed to calculate the weight of steel plate. Next, we will introduce how to calculate the weight of the steel plate.
## How to Calculate the Weight of Steel Plate?
### A. Formula: Weight = Area x Density x Thickness
The formula for calculating the weight of a steel plate is Weight = Area x Density x Thickness. Among them, area, density, and thickness are all necessary parameters, which need to be obtained in actual measurement.
### B. Example: How to use the formula to calculate the weight of a steel plate
Here is an example of calculating the weight of a steel plate:
Suppose we have a steel plate with a length of 2 meters, a width of 1 meter, and a thickness of 10 mm. The density of the steel plate was 7.85 g/cm3.
First, the area of the steel plate needs to be calculated, namely:
Area = Length x Width
= 2m x 1m
= 2 square meters
Then, substitute the above data into the formula for calculating the weight of the steel plate:
Weight = Area x Density x Thickness
= 2 square meters x 7.85 g/cm3 x 10 mm
= 1570 grams
= 1.57 kg
Therefore, the weight of this steel plate is 1.57 kg.
It should be noted that when using the above formula for calculation, it is necessary to ensure that all units are the same, such as: area must be in square meters, density must be in grams per cubic centimeter, and thickness must be in millimeters. If the units are different, a unit conversion is required.
## Other Ways to Calculate the Weight of Steel Plate
### A. Online Steel Plate Weight Calculator
Now, there are many websites that provide online steel plate weight calculators that can quickly and accurately calculate the weight of steel plates. Using these calculators, you only need to input the parameters such as the length, width and thickness of the steel plate to get the weight of the steel plate. In addition, some steel manufacturers also provide their own plate weight calculators, which may provide more accurate figures.
### B. Steel Plate Weight Table
The steel plate weight table is a convenient method for calculating the steel plate weight, usually provided by steel manufacturers. These tables list the weight of steel plates of various specifications and materials, and the weight can be quickly found according to the specifications and materials of the steel plates. The method of using the steel plate weight table is straightforward and does not require any calculations, but it is necessary to ensure that the table used and the steel plate specification exactly match.
The above are other methods for calculating the weight of steel plates, among which the online steel plate weight calculator is more convenient and quick, and the steel plate weight table is more suitable for querying various steel plates of different specifications.
## In Conclusion
In this article, we describe how to calculate the weight of a steel plate, as well as other methods for calculating the weight of a steel plate.
First, we introduce the basic concept of steel plate weight, including steel plate density, area, thickness, and unit conversion. Then, we explain in detail how to use the formula to calculate the steel plate weight and give an example.
Next, we introduce other methods for calculating the weight of steel plates. The online steel plate weight calculator is a convenient and quick method. You only need to input the parameters of the steel sheet to get its weight. The steel plate weight table is more suitable for querying steel plates of various specifications.
In conclusion, calculating the weight of steel plates is a basic skill, especially important for those who need to work with them. Proficiency in how to calculate the weight of the steel plate can help us better select and use the steel plate, avoid the problem of being too heavy or too light, and at the same time ensure our work efficiency and quality.
On Key
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## Project Euler 004 – IronRuby
23 02 2011
Problem 4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.
Solution
# A palindromic number reads the same both ways.
# The largest palindrome made from the product of
# two 2-digit numbers is 9009 = 91 99.
# Find the largest palindrome made from the
# product of two 3-digit numbers.
# placeholder
#loop through each 3-digit number
for i in 100..999 do
for j in i..999 do
product = i*j
# the palindrome test is simple here
# since we are using the same technique
# as the F# and C# functions
if product.to_s().reverse == product.to_s() &&
end
end
end | HuggingFaceTB/finemath | |
194 inches in kilometers
Conversion
194 inches is equivalent to 0.0049276 kilometers.[1]
Conversion formula How to convert 194 inches to kilometers?
We know (by definition) that: $1\mathrm{in}=2.54e-05\mathrm{km}$
We can set up a proportion to solve for the number of kilometers.
$1 in 194 in = 2.54e-05 km x km$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{km}=\frac{194\mathrm{in}}{1\mathrm{in}}*2.54e-05\mathrm{km}\to x\mathrm{km}=0.0049276\mathrm{km}$
Conclusion: $194 in = 0.0049276 km$
Conversion in the opposite direction
The inverse of the conversion factor is that 1 kilometer is equal to 202.938550206997 times 194 inches.
It can also be expressed as: 194 inches is equal to $\frac{1}{\mathrm{202.938550206997}}$ kilometers.
Approximation
An approximate numerical result would be: one hundred and ninety-four inches is about zero kilometers, or alternatively, a kilometer is about two hundred and two point nine three times one hundred and ninety-four inches.
Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | HuggingFaceTB/finemath | |
# Convert number: 115,533 in Roman numerals, how to write?
## Latest conversions of Arabic numbers to Roman numerals
115,533 = (C)(X)(V)DXXXIII Nov 27 23:39 UTC (GMT) 749,998 = (D)(C)(C)(X)(L)M(X)CMXCVIII Nov 27 23:39 UTC (GMT) 164,647 = (C)(L)(X)M(V)DCXLVII Nov 27 23:39 UTC (GMT) 519,901 = (D)(X)M(X)CMI Nov 27 23:39 UTC (GMT) 1,160,016 = (M)(C)(L)(X)XVI Nov 27 23:39 UTC (GMT) 592,275 = (D)(X)(C)MMCCLXXV Nov 27 23:39 UTC (GMT) 46,763 = (X)(L)(V)MDCCLXIII Nov 27 23:39 UTC (GMT) 219,490 = (C)(C)(X)M(X)CDXC Nov 27 23:39 UTC (GMT) 409,021 = (C)(D)M(X)XXI Nov 27 23:39 UTC (GMT) 1,050,498 = (M)(L)CDXCVIII Nov 27 23:39 UTC (GMT) 50,198 = (L)CXCVIII Nov 27 23:39 UTC (GMT) 941,017 = (C)(M)(X)(L)MXVII Nov 27 23:38 UTC (GMT) 3,002,699 = (M)(M)(M)MMDCXCIX Nov 27 23:38 UTC (GMT) converted numbers, see more...
## The set of basic symbols of the Roman system of writing numerals
• ### (*) M = 1,000,000 or |M| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.
(*) These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: "(" and ")", because:
• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and
• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line "|" and the Roman numeral "I" (1).
(*) An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate "1,000 times". See below...
Logic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.
(*) At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.
Thus, initially, the largest number that could be written using Roman numerals was:
• MMMCMXCIX = 3,999. | HuggingFaceTB/finemath | |
Chapter 2.5 - Math3200 Intermediate Probability and Statistics Prof Nan Lin Department of Mathematics Washington University Outline Joint pdf(pmf
# Chapter 2.5 - Math3200 Intermediate Probability and...
• Notes
• 37
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Math3200 Intermediate Probability and Statistics Prof. Nan Lin Department of Mathematics Washington University
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Outline Joint pdf (pmf) Marginal distribution Independent random variables Conditional distribution Covariance Correlation coefficient Multivariate distribution
Joint distribution The joint distribution describes the simultaneous behavior of two (or more) random variables. For example: X : the length of one dimension of an injection-molded part, and Y : the length of another dimension. We might be interested in P(2.37 X 3.15 and 7.05 Y 7.88)
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TWO DISCRETE RANDOM VARIABLES
Joint distribution Joint c.d.f. ? ?, ? = 𝑃 ? ≤ ?, ? ≤ ? = 𝑓(?, ?) ?≤? ?≤?
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Marginal distribution
Example for joint probability mass function Consider the following table: Using the table, we have Y=0 Y=3 Y=4 X=5 1/7 1/7 1/7 3/7 X=8 3/7 0 1/7 4/7 4/7 1/7 2/7 p X p Y . 7 / 3 7 / 2 7 / 1 ) 4 ( p ) 3 ( p 3} P{Y 7 / 3 ) 5 ( p 7} P(X 2/7 p(5,4) p(5,3) 3} Y 7, P{X Y Y X
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Expected Values for Jointly Distributed Random Variables Let X and Y be discrete random variables with joint probability mass function f(x, y). Let the sets of values of X and Y be A and B, respectively. We define E(X) and E(Y) as Example. For the random variables X and Y from the previous slide, ). , ( f E(Y) and ) , ( f ) ( f E(X) B XY A A XY B A X y x y y x x
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# How do you simplify 3 1/6 + 4 1/4?
Jul 16, 2016
#### Explanation:
To add the two numbers we may observe that:
$3 \frac{1}{6} = 3 + \frac{1}{6}$, and similarly
$4 \frac{1}{4} = 4 + \frac{1}{4}$. Now we can add and rearrange:
$3 \frac{1}{6} + 4 \frac{1}{4} = 3 + \frac{1}{6} + 4 + \frac{1}{4} = 3 + 4 + \frac{1}{6} + \frac{1}{4} = 7 + \frac{1}{6} + \frac{1}{4}$
We can now add the fractions:
$\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}$, and the solution is then:
$7 + \frac{5}{12} = 7 \frac{5}{12}$ | HuggingFaceTB/finemath | |
### semi ellipse perimeter formula
The formula for finding the area of the circle is A=r^2. For the special case of a circle, the semi-major axis is the radius. The semi-major and semi-minor axes of an ellipse are radii of the ellipse (lines from the center to the ellipse). The shape of the ellipse is different from the circle, hence the formula for its area will also be different. The dimensions are 11.8 cm by 23.7 cm. Q.1: If the length of the semi major axis is 7cm and the semi minor axis is 5cm of an ellipse. The rectangle is also called a parallelogram with four right angles. c=focal length and a=length of the semi-major axis. Ellipse Examples. Use the standard form when center (h,k) , semi-major axis a, and semi-minor axis b are known. Part 1Calculating the Area. Answer (1 of 7): Since ellipse is a squished circle we could consider an equivalent circle.
Essentially, it is the radius of an orbit at the orbit's two most distant points. It leads, however, to another, which for practical purposes is much preferable. If for shortness' sake be written for log 2/ log , he says in effect that the perimeter of an ellipse with semi-axes a and Area of a semi ellipse = r 1 r 2.
Example : If the diameter of a semi-circular plot is 14 m, then find its perimeter. An Ellipse comprises two axes. The eccentricity is a measure of how "un-round" the ellipse is. () We'll call this value a . e=eccentricity. The unnamed quantity h = ( a - b) 2 / ( a + b) 2 often pops up. How to find the length of arc of an ellipse?
If you plot them is easy to see that they form a profile. Calculations at a semi-ellipsoid (or hemi-ellipsoid, or half ellipsoid). Use the formula for the area of an ellipse to find the values of the semi-major axis (a) and the semi-minor axis (b). = r 1 r 2. Ellipse. If an ellipse's semi-minor axis is 7 meters long, and it's semi-major axis is 31 meters long, how long is its minor axis? Step 3: The area and perimeter with respect to major and minor axis will appear in the respective output fields. a = is the semi-major axis. When the values for the area and perimeter are known, the length of the semi-minor axis r is calculated by solving Eq. r 2 is the semi-minor axis of the ellipse. The mathematical equation formulated by Srinivasan Ramanujan in 1914 for widely considered to be the most accurate for calculation of the circumference of an ellipse is [7]. the perimeter equals A simple approximate value is Standard Equation of Ellipse. The simplest method to determine the equation of an ellipse is to assume that centre of the ellipse is at the origin (0, 0) and the foci lie either on x- axis or y-axis of the Cartesian plane as shown below: Using the structural engineering calculator located at the top of the page (simply click on the the "show/hide calculator" button) the following properties can be calculated: Area of a Elliptical Half. Find its area. We identified it from well-behaved source. The perimeter of an ellipse with semi major and semi minor axes a, b should be. The major axis is 24 meters long, so its semi-major axis is half that length, or 12 meters long. Semicircle perimeter is half of the circumference of a circle and diameter of a semicircle. Ellipse. If you have any questions related to the Semicircle please let me know through the comment and mail. In these formulas, the most accurate seem to be Approximation 2 and Approximation 3 (both invented by Ramanujan) and Infinite Series 2. We know the equation of an ellipse is : \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 When a=b=r this The figure below shows the four (4) main standard equations for an ellipse depending on the location of the center (h,k). Find equation of any ellipse using only 2 parameters: the major axis, minor axis, foci, directrice, eccentricity or the semi-latus rectum of an ellipse. side of equation (2) is interpreted as the quarter length of an ellipse with a semi-major axis of unit length and a semi-minor axis of length (and ec-centricity ), whereas the swiftly converging ratio on the right-hand side is elementary enough to be presented in high or, perhaps, elementary school. So, perimeter of a semicircle is 1/2 (d) +d or r +2r. One can think of the semi-major axis as an ellipse's long radius. If the length of semi-major axis $$= a$$ and length of semi-minor axis $$= b$$, then. Area = 35 . or. Perimeter of an ellipse is defined as the total length of its boundary and is expressed in units like cm, m, ft, yd, etc. In 1609, Kepler used the approximation (a+b). Ellipse Formulas. Leave a Comment / Ellipse Questions, Maths Questions / By mathemerize. return perimeter >>> calculate_perimeter(2,3) 15.865437575563961 You can compare the result with google calculator also a definition problem: major, minor axes differ from semi-major, semi-minor Let's solve one more example. To answer this question, we need to realize that the figure is just half of a circle Find the volume of the solid whose base is bounded by the circle xy22 4 with the indicated cross sections taken perpendicular to the x-axis If the dynamics of a system is described by a Area = 35 . or. The major axis is always the longest axis in an ellipse. We identified it from well-behaved source. We take this kind of Perimeter Of An Ellipse Equation graphic could possibly be the most trending subject like we portion it in google pro or facebook. r 2 is the semi-minor axis of the ellipse. The ellipse changes shape as you change the length of the major or minor axis. For an ellipse of cartesian equation x 2 / a2 + y 2 / b2 = 1 with a > b : a is called the major radius or semimajor axis . The perimeter P(a, b) of an ellipse having semi-axes of lengths a and $$b\le a$$ is given as \begin{aligned} P(a,b)= 4a\,E(\epsilon ), \end{aligned} (1.1) Is it possible to integrate a function that would give the perimeter of an ellipse? The rectangle is 4 inches long and 3 inches wide. The area formula is: A = r2 2 A = r 2 2. Hence, the approximation formula to determine the perimeter of an ellipse: OR Where, a is the length of the semi-major axis and b is the length of the semi-minor axis. meter), the area has this unit squared (e.g. 2. When a=b, the ellipse is a circle, and the perimeter is 2a (62.832 in our example). In 1773, Euler gave the Semi Major And Minor Axes Wikipedia. The ellipse has two length scales, the semi-major axis and the semi-minor axis but, while the area is given by , we have no simple formula for the circumference. We take this kind of Perimeter Of An Ellipse Equation graphic could possibly be the most trending subject like we portion it in google pro or facebook. An ellipse is a two dimensional closed curve that satisfies the equation: 1 2 2 2 2 + = b y a x The curve is described by two lengths, a and b. Q.1: If the length of the semi major axis is 7cm and the semi minor axis is 5cm of an ellipse. P = r + d. Using the substitution property of equality, you can also replace diameter with radius throughout: P = 1 2 (2 r) + 2 r; P = r + 2 r; Find The Perimeter of a Semicircle Examples. Formula is. Since c a the eccentricity is always greater than 1. The length of the perimeter of an ellipse can be expressed using an elliptic integral. Example 6. Hence the equation to major axis is y = 3. The formula for the circumference of a circle is: a = r 2. That's I that I have and wanted to take the equation that defines the profile - not necessarily an ellipse, but I think it is a good approximation. The area of an ellipse formula involves both semi-major and semi-minor axes. Program To Find The Area Of An Ellipse Geeksforgeeks. Circumference of Ellipse Formula. Good work so far. The semi-major and semi-minor axes of an ellipse are radii of the ellipse (lines from the center to the ellipse). The endpoint of the Latus Rectum lies on its perimeter i.e. They can be named as hyperbola or parabola and there are special formulas or equation to solve the tough Ellipse problems. You can call this the "semi-major axis" instead. Quick navigation:How to calculate the perimeter of any shape?Perimeter of a squarePerimeter of a rectanglePerimeter of a triangleCircumference of a circlePerimeter of a parallelogramPerimeter of a trapezoidCircumference of an ellipse (oval)Perimeter of a sectorPerimeter of an octagonMore items If the length of semi-major axis $$= a$$ and length of semi-minor axis $$= b$$, then.
The semi major axis is one half of the major axis, and thus runs from the centre, through a focus, and to the perimeter. Parts of an Ellipse Ellipses are one of the types of conic sections. Notice that the vertices are on the y axis so the ellipse is a vertical ellipse and we have to use the vertical ellipse equation. The Ellipse is the conic section that is closed and formed by the intersection of a cone by plane. This means the foci are at $\pm 1.5$ feet, i.e.the tacks should be placed at the base, $1.5$ feet to either side The data for the Measured arch perimeter (MP) according to the procedure mentioned . Sides are called the length and width.
The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the x -axis is given as: ( x h) 2 a2 + ( y k) 2 b2 = 1. The standard form of the equation of an ellipse with center (h,k)and major axis parallel to the y -axis is given as: ( x h) 2 b2 + ( y k) 2 a2 = 1. Its perimeter P is approximately If the ellipse is of equation x 2 /a 2 + y 2 /b 2 =1 with a>b, a is called the major radius, and b is the minor radius. The figure below shows the four (4) main standard equations for an ellipse depending on the location of the center (h,k). Solution : A semi-circle has been drawn with AB = 14 m as diameter. Using the structural engineering calculator located at the top of the page (simply click on the the "show/hide calculator" button) the following properties can be calculated: Area of a Elliptical Half. (a) If the ellipse is very nearly in the shape of a circle (i.e., if the major and minor axes are nearly equal), then the perimeter is given by: (1) P = ( a + b) Where P = is the perimeter or circumference. The vendor states an area of 200 sq cm. The area of such an ellipse is Area = Pi * A * B , a very natural generalization of the formula for a circle! When b=0 (the shape is really two lines back and forth) the perimeter is 4a (40 in our example). Solution: Second Moment of Area (or moment of inertia) of a Elliptical Half. There is no simple formula with high accuracy for calculating the circumference of an ellipse. In the equation, the denominator under the x 2 term is the square of the x coordinate at the x -axis. Various approximation formulas are given for finding the perimeter of an ellipse. Question 1. The Half of the Latus Rectum is known as the Semi Latus Rectum. Solution: Given, Semi major axis of the ellipse = r 1 = 10 cm. Here is one of the most complex perimeters to calculate. The smaller of these two axes, and the smallest distance across the ellipse, is called the minor axis. (a) Considering P as a point on the circle, show that x2 + y2 = 4a2 e2 Given the ellipse below, what's the length of its minor axis? 1. This is an ellipsoid, which is bisected at one axis along the other two axes.The surface area is calculated from half the approximation formula by Knud Thomsen, plus the area of the intersection ellipse.Enter the bisected axis and the other two semi axes and choose the number of decimal THE formula given by your Queensland correspondent (NATURE of April 10, p. 536) for the perimeter of an ellipse is not at all objectionable on the score of degree of approximation. An Ellipse is a curve on a plane that contains two focal points such that the sum of distances for every point on the curve to the two focal points is constant. The perimeter of a trapezoid. A = 1 2 b h. Some other triangle area formulas are: Any triangle: A = s ( s a) ( s b) ( s c), where s is the semi-perimeter (half the perimeter), and a, b, and c are side lengths. Area of an ellipse calculator | Formula. But, the more general geometrical shape is the ellipse. Due to the symmetry of the ellipse, the entire perimeter of the ellipse can be found by multiplying the length of the arc from t = 0 to t = /2 by four. Put value of y in equation of ellipse.we get the following quadratic equation-x 2 (a 2 + b 2) -2a 3 x + We have obtained all parameters of ellipse from this triangle. Ellipse Volume Formula. When the circumference of a circle is so easy to find, it comes as a surprise that there is no easy way to find the circumference of an ellipse. (4x 2 24x) + (9y 2 + 36y) 72 = 0. If the ellipse is a circle (a=b), then c=0 What is the perimeter of a semi-circle with a diameter of 8cm? Its submitted by organization in the best field. The formula (using semi-major and semi-minor axis) is: (a 2 b 2)a. The Calculated arch perimeter(CP) was obtained from the measured data after inserting them into Ramanujan's equation for calculation of the perimeter of an ellipse . Use general equation form when four (4) points along the ellipse are known. Standard Form Equation of an Ellipse. Write A C Program To Calculate The Focus Area Chegg Com. Question 1: If the length of the semi-major axis is given as 10 cm and the semi-minor axis is 7 cm of an ellipse. Stack Exchange Network Stack Exchange network consists of 180 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Ellipse is the locus of all points on a plane whose sum of distances between two fixed points is constant. They are the major axis and minor axis. 8 2 The Ellipse Mathematics Libretexts. They all get the perimeter of the circle correct, but only Approx 2 and 3 and Series 2 get close to the value of 40 for the extreme case of b=0. The quantity e = (1- b2 / a2 ) is the eccentricity of the ellipse. So, this bounded region of the ellipse is its area. If you have any questions related to the Semicircle please let me know through the comment and mail. Circle with the Same Perimeter as an Ellipse; The Math / Science. An ellipse's shortest radius, also half its minor axis, is called its semi-minor axis. 2. The arch has a height of 8 feet and a span of 20 feet. P ( a, b) = 0 2 a 2 cos 2 + b 2 sin 2 d . At the center point of the long dimension, it appears that the area below the line is about twice that above. Solution. Ellipse Area. Solution: Given, length of the semi-major axis of an ellipse, a = 7cm. Section of a Cone. Find the area of an ellipse whose semi-major axis is 10 cm and semi-minor axis is 5 cm. Using for example the Wiki article on ellipses, you will find that the semi-major axis is $2.5$ feet and the semi-minor axis is $2$ feet. How To Find The Equation Of An Ellipse Given Center A Vertex And Point On Quora. The sum of the distances for any point P(x,y) to foci (f1,0) and (f2,0) remains constant.Polar Equation: Origin at Center (0,0) Polar Equation: Origin at Focus (f1,0) When solving for Focus-Directrix values with this calculator, the major axis, foci and k must be located on the x-axis. The equation of the eccentricity is: After multiplying by a The arch is 148m long and has a height of 48m at the center. Perimeter of an Ellipse Formulas. (These semi-major axes are half the lengths of, respectively, the largest and smallest diameters of the ellipse.) = 3.141592654. Note: a = semi-minor axes & b = semi-major axes Semi minor axis of the ellipse = r 2 = 5 cm. Area of ellipse = a b. As we know that, perimeter of circle is 2r or d. Now we can plug the semi-axes' lengths into our area formula: This ellipse's area is 37.68 square inches. Share: A trapezoid is a quadrilateral with at least two parallel sides called bases. Leave a Comment / Area and Perimeter / By Admin. = 3.14. They are the major axis and minor axis.
length of the semi-minor axis of an ellipse, b = 5cm. The semi-major axis of an ellipse is the distance from the center of the ellipse to its furthest edge point. We are given that the equation of the ellipse is 4x 2 + 9y 2 24x + 36y 72 =0. It could be described as a flattened ellipse. Perimeter of a Elliptical Half. Answer: Given, length of the semi-major axis of an ellipse, a = 10 cm length of the semi-minor axis of an ellipse, b equals 5cm By the formula of Perimeter of an ellipse, we know that; The perimeter of ellipse = 2 a 2 + b 2 2 Therefore, the Perimeter of ellipse = 23.14 10 2 + 5 2 2 = 49.64 Fun Facts Find an equation for the ellipse, and use that to find the height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center. Exercise 1: a) Set up an integral for the total arc length (perimeter) of the ellipse given by Another equation for an ellipse with semi-major axis a and eccentricity e can be given You find the area of a semicircle by plugging the given radius of the semicircle into the area of a semicircle formula. The semi-major axis for an ellipse x 2 /a 2 + y 2 /b 2 = 1 is a, and the formula for eccentricity of the ellipse is e = 1 b2 a2 1 b 2 a 2. Example 2: Calculate the area of the ellipse where the major radius is 4 cm and minor radius is 3 cm. An ellipse with a major radius of 5 units and a minor radius of 3 units, for example, has a surface area of 3 x 5 x, or around 47 square units. Therefore, the approximation formula for the perimeter of an ellipse is: P= 2\cdot \Pi\cdot \sqrt{\frac{a^{2}+b^{2}}{2}} Ellipse. The semi-major axis a of the ellipse is equivalent to the IMW per. An Ellipse comprises two axes. The formula for finding the area of the ellipse is quite similar to the circle. Centroid of a Elliptical Half. What is the ellipse of a semi major axis?
The formula for the area, A A, of a circle is built around its radius. 8 2 The Ellipse Mathematics Libretexts. Here are a number of highest rated Perimeter Of An Ellipse Equation pictures upon internet. Hence, the equation of the required ellipse is x 2 24 + y 2 49 = 1. They can be named as hyperbola or parabola and there are special formulas or equation to solve the tough Ellipse problems. The r2 in the circle area equation is replaced with the product of the This can be calculated to great precision instantly on any mathematics program like mathematica. Math Advanced Math Q&A Library he ellipse with semi-major axis a, eccentricity e and foci F1 and F2 intersects the circle with diameter F1 F2 at the point P, which is one of the 4 points of intersection, as shown in the diagram. This would just be an approximation and not the exact value of the perimeter of the ellipse. b = semi-minor axis length of an ellipse. The length of semi-major axis is $$a$$ and semi-minor axis is b. An arch has the shape of a semi-ellipse (the top half of an ellipse). 2] 2 e 2n. Trig. Its radius, r = d 2 = 14 2 = 7 m. Example 1 : Find the equation of ellipse whose foci are (2, 3), (-2, 3) and whose semi major axis is of length 5.
Circumference = 2 r = 2 22 7 10.5 = 66 cm. on its curve. The general form for the standard form equation of an ellipse is shown below.. There is simply no easy way to do it.
Area of Semi Ellipse formula is defined as amount of space occupied by semi ellipse in given plane and is represented as A = (pi/2)*a*h or Area = (pi/2)*Semi-major axis*Height. | HuggingFaceTB/finemath | |
# Tag Info
12
To get the value of the decision variable, you need to use the varValue property of the LpVariable, so: print(x.varValue) You can also use: print(x.value()) The explanation is that the Python variable x is not the decision variable itself, it is a PuLP object of type LpVariable: In[5]: type(x) Out[5]: pulp.pulp.LpVariable Therefore, just using print(x) ...
9
You can solve your model via the NEOS server which provides Gurobi, Cplex, and other solvers for free if it is the matter of not having a solver. I am not familiar with PuLP but I know it is easy to implement the solvers in NEOS if you model the problem in Pyomo. May it helps you to find PuLP syntax for it, I provide lines of code written for Pyomo using ...
9
If the model in PuLP is: from pulp import LpProblem, LpVariable, LpMaximize, lpSum m = LpProblem(name='example', sense = LpMaximize) x = LpVariable.dicts(name='x',indexs=[1,2,3]) m += lpSum(x) <= 3, 'c1' m += lpSum(i*x[i] for i in [1,2,3]), 'obj' We can access the coefficient of $x_1$ in 'C1' with: m.constraints['c1'][x[1]] # This the coefficient => ...
7
According to PuLP's documentation, it seems that load_file function in PuLP's Amply class can only handle AMPL files with a subset of AMPL syntax for data. So if you want to use those LP/MPS files in PuLP, you may have to first convert them into AMPL files. You can either write a small script by yourself or use some existing scripts. Here are some ...
5
The rule "works on Saturday implies works on Sunday" can be expressed as $$(D_{i,6} \lor N_{i,6}) \implies (D_{i,7} \lor N_{i,7}),$$ which can be rewritten in conjunctive normal form as follows: \neg (D_{i,6} \lor N_{i,6}) \lor (D_{i,7} \lor N_{i,7}) \\ (\neg D_{i,6} \land \neg N_{i,6}) \lor (D_{i,7} \lor N_{i,7}) \\ (\neg D_{i,6} \lor D_{i,...
4
You don't want to create a variable representing a date in pulp. You want to utilize zero-one indicator variables for each option. In this case your options are the vessel-date combinations. Index the variables by vessel and date. Suppose the value of the difference between the actually loaded date and the ready to be loaded date is to be used in the ...
4
In other words, you want to avoid a pattern of 010. As a logical proposition: $$\neg (\neg V_{i,j} \land V_{i,j+1} \land \neg V_{i,j+2}),$$ which you can write in conjunctive normal form as: $$V_{i,j} \lor \neg V_{i,j+1} \lor V_{i,j+2}.$$ The corresponding linear constraint is: $$V_{i,j} + (1 - V_{i,j+1}) + V_{i,j+2} \ge 1,$$ equivalently, $$V_{i,j} - V_{i,... 4 Introduce linear constraints:$$\sum_{\text{h}} z[\text{h}][\text{driver}] \le 1 \quad\text{for each driver}
4
There are certainly different ways of achieving what you want. Here is how I would proceed: Start by predefining the set of all possible schedules which satisfy your constraints $2,3,4,6$. Although there are many, I believe that with your constraints, it may be not too difficult to derive them somewhat automatically. Here is a subset of them in the table ...
4
Your code is correct, you have used pulp.LpVariable.dicts correctly. For better readability, it is often better to use dictionaries. So the idea is to convert your data from lists to dicts as follows: # convert data into dicts dict_bike_profit = dict(zip(bike_types, bike_profit)) # simple dict dict_bike_stock = dict(zip(part_names, parts_stock)) # simple ...
3
When I understand your question correctly you are not interested in the full solution space. In that case you can generate alternative optimal solutions by first solving your problem to optimality. Then you can add the objective function with the optimal value as a constraint and resolve with a random objective vector. For example for the problem. $\begin{... 3 Your problem does not have any objective function, so you do not have to use software to solve linear programs here. It is just a system of linear inequalities. (Of course, you still can use LP tools by just minimizing a constant objective function.) May I suggest a solution that neither requires Pyomo nor Pulp? You can solve those systems by using the ... 3 I believe the issue was with the following constraint name: # If Koramangala 5th Block, then one must be Italian model += (y['koramangala_5th_block_bin'] <= x['italian_bin'], '5th_block_italian') It seems that PuLP did not like the name starting with a number even though it was a string. If I changed the constraint to this, the model ran. # If ... 3 It is difficult to tell from this information. You need the examine the logs of the solver if your primal bound (how good are the solutions) or dual bound (how good is the relaxation) is not moving as fast as you wish it would be. I would try to run this heuristic not at every node, but maybe at every 50 (the correct value of this needs to be benchmarked) ... 3 I have some experience in solving large-scale airline crew pairing combinatorial optimization problems (an NP-hard problem) with difficulty similar to vehicle routing problems. Yes, solving such problems using standard open-source IP solvers is extremely time-consuming. You could customize/parallelize heuristics (such as Simulated Annealing, Variable ... 3 CBC does have its fair share of bugs, especially when it comes to proper termination. If you are running in multithreaded mode try running in serial (or vice-versa), and see if that helps. If you are feeling particularly adventurous, you can comment out all the code related to the "SIGNAL_TRAP" and recompile CBC. This helps with proper termination (... 3 The following constraints are infeasible : _C129: Rail_Loadings_From_Washplant_('2020_05_22',_'ABC',_'PRE') + Rail_Loadings_From_Washplant_('2020_05_22',_'ABC',_'ZBF') = 25200 _C134: Rail_Loadings_From_Washplant_('2020_05_23',_'ABC',_'PRE') + Rail_Loadings_From_Washplant_('2020_05_23',_'ABC',_'ZBF') = 25200 _C161: ... 3 I have created a model that I believe addresses your objective of minimizing demurrage charges. It utilizes indicator variables for each day and vessel showing when each vessel is finally loaded. The factors for the variables in the objective function are to be calculated based on the demurrage formula you provided. The first thing to do with this model is ... 3 The latest issue has to do with not having an inventory control constraint. You need to have a constraint like: port_inventory_vars[(date, grade)] == port_inventory_vars[(date-1, grade)] - pulp.LpSum(vessel_sales_demand_vars[(vessel, grade, date)] for vessel in vessels) + ... The ellipsis indicates where you can put additional terms that describe how ... 2 When dealing with infeasibility, I like to do two things: a) Create the Irreducible Infeasible Subset (IIS). I don't think PuLP directly allows you to create that, however you could export your model instance and then use a (commercial) solver (e.g. Gurobi) to do so (see here for docs). This will allow you to narrow down where the infeasibility lies. b) Add ... 2 In Pyomo you can use the .fix() method to do that. In the following example based on a condition you can fix some variables to different values: if pyo.value(instance.x[2]) == 0: instance.x[2].fix(1) else: instance.x[2].fix(0) In PuLP: Assigning values to variables also permit fixing those variables to that value. In order to do that, you can use the ... 2 Math Formulation + Pyomo model We need to define 3 sets i: 1 to 8 staffs j: 1 to 3 shifts t: 1 to 18 days$X_{i,j,t} \in {0/1} $Binary There must be 3 people on each shift$\forall t,j $then$\sum_i X_{i,j,t} =3$Every individual (i) must work 1 shift per day\$\forall t,i $then$\sum_j X_{i,j,t} \geq 1 $Every individual must not work more than 2 ... 1 I suggest introducing$3n$variables. Let$ c_{i} $be the daily rate of each vehicle$i=1,2,3$.$ x_{i,j} $designates the quantity of vehicles which will be used every day: the subscript$j$indicates the day in interest. Supposing to consider$n$days, we have$j=1, 2, \ldots, n$days. Clearly,$ x_{i,j} $is a non-negative integer numbe and$ x_{i,j}=0$... 1 I'll address your main issue about discrete dispatch. You should model the problem utilizing inventory levels at discrete time intervals that are short enough to fit just one shipment, which sounds like hourly. Utilize tank volume at the end of time interval$t$:$v_t$for all$t\$ between 1 and the end of the planning horizon. This allows you to create a ...
1
I found the solution. 1- The storage inventory definition is as follows: model += storage_stockpile_current[product] \ + pulp.lpSum( train_consignment_variables[(date, plant, product)] for plant in _plants_combo) \ - sales_demand[date][product] \ == storage_inventory_vars[(...
1
I don't see any part that "storage_facility_vars" is added to your model in your code. You use it only after define as: storage_facility_vars[(date, product)] = plant_current_inv[product] Please check it. Also "plant_current_inv[product]" is used only above row, never defined or initialized.
Only top voted, non community-wiki answers of a minimum length are eligible | open-web-math/open-web-math | |
# Can the Lagrangian be written as a function of ONLY time?
The lagrangian is always phrased as $$L(t,q,\dot{q})$$.
If you magically knew the equations $$q(t)$$ and $$\dot{q}(t)$$, could the Lagrangian ever be written only as a function of time?
Take freefall for example. $$y(t)=y_0 + v_0t -(1/2)gt^2$$ $$\dot{y}(t)=v_0 -gt$$
Can the Lagrangian now be written as: $$L=KE-PE=(1/2)m\dot{y}^2-mgy=(1/2)m(v_0 -9.8t)^2-mg(y_0 + v_0t -4.9t^2)$$
Now the Lagrangian is written only as a function of time, and we can put in a time and find out what the Lagrangian is at any point in the motion. Is this legitimate? Forgive me if this is simple.
• To me it seems like you could do so - but of what use would it be? What you are trying to achieve is linked to wether you find it "legitimate", so maybe for us to unser the question, you could elaborate what you mean by "legimimate". Aug 6, 2020 at 17:16
• I was just wondering if my logic and math were accurate. It is probably not useful since if you already have q(t) you can track anything about the system you would like. Aug 6, 2020 at 17:23
• Aug 7, 2020 at 8:41
• What is the use of the Lagrangian? Its use is that we can derive the equations of motion (and the integrals of motion) from it by studying how the action integral reacts to a variation of $q$, $\dot q$. If you fix $q$, $\dot q$ there's nothing to vary. So, what would the use of this "Lagrangian" be? Literally, you could evaluate the Lagrangian as you describe -- but it would be a useless object. It is interesting to ponder how and why this is different from the usual way we deal with functions, where we are interested in evaluating them but not so much in their functional form. Aug 7, 2020 at 13:30
• I very much disagree there JonathanZ. I direct your attention to this glorious contraption. twitter.com/i/status/755139996399595520 Aug 8, 2020 at 4:40
1. Using the EL equations (or inserting their solutions) into the action functional typically destroys the stationary action principle. For examples, see this Phys.SE post.
2. Concerning OP's title question: If a Lagrangian $$L(t)$$ only depends on time $$t$$, then it can be written as a total time derivative $$L=dF/dt$$, and that means it is equivalent to the trivial Lagrangian $$L=0$$, i.e. the EL equations are trivial.
When doing this, it removes one of the great powers Lagrangian mechanics: the Euler-Lagrange equations, which are
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = \frac{\partial l}{\partial q}$$
Notice in the equations, we have partial derivatives. So if we made $$L=L(t)$$, then the Euler Lagrange equations would give you $$0=0$$ (a beautiful fact, but nonetheless not very helpful). Plus, the Euler-Lagrange equations are meant to give you equations of motion, so one probably wouldn't be able to substitute equations for $$q(t)$$ or $$\dot{q}(t)$$ anyways.
Edit: There is actually a more fundamental problem when you do this (and thank you so much to Qmechanic and ZeroTheHero for bringing this to my attention). If you simply plug back the motion into your Lagrangian and then try to apply the Euler-Lagrange equations again, you will more than likely get the wrong equation of motion. For example take the simple spring:
$$L = \frac{m\dot{x}^2}{2} - \frac{kx^2}{2} \implies m\ddot{x}=-kx \implies x=\sin(\sqrt{k/m}\cdot t)$$ (just one possible solution)
where we used the EL eqs. However, if we plug this back into our Lagrangian, we will not get the right equations of motion:
$$L = \frac{m\dot{x}^2}{2} - \frac{k\sin^2(\sqrt{k/m}\cdot t)}{2} \implies m\ddot{x} = 0$$(!!!)
• One cannot solve for $\dot q$ and plug it back into the Lagrangian as the resulting EOM will not be correct. Thus there is everything wrong with doing this. See also the post by @Qmechanic below. Aug 6, 2020 at 18:22 | HuggingFaceTB/finemath | |
# Quick Answer: What Is The Complement Of 50 9.5 Degree?
## What is the complement of an angle?
Two angles are called complementary when their measures add to 90 degrees.
Two angles are called supplementary when their measures add up to 180 degrees..
## What is the complement of a 57 degree angle?
The complement of 57° is the angle that when added to 57° forms a right angle (90° ).
## Can 2 right angles be complementary?
No, two right angles cannot be complementary to each other. This is because complementary angles are defined as two angles with measures that sum up…
## What is a complement in grammar examples?
The word complement most commonly crops up in the terms subject complement and object complement. Subject Complement. A subject complement is the adjective, noun, or pronoun that follows a linking verb. (Examples of linking verbs include to be, to smell, to seem, to taste, to look.)
## What is the complement of 60 degree?
The complement of 60° is the angle that when added to 60° forms a right angle (90° ).
## What is the complementary of 50 degree?
The complement of 50° is the angle that when added to 50° forms a right angle (90° ).
## How do you find the complement?
To find the complement of an angle, subtract that angle’s measurement from 90 degrees. The result will be the complement.
## How do you find a complementary pair of angles?
Working rule: To find the complementary angle of a given angle subtract the measure of an angle from 90°. So, the complementary angle = 90° – the given angle.
## Can two angles be supplementary if both of them are?
∴Sum of two acute angles cannot make a supplementary angle. As we know that obtuse angle is always greater than 900. So, the sum of two obtuse angles always makes greater than 1800. … So, the sum of two right angles is always equal to 1800.
## What is the complement rule?
The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1.
## What is the complementary angle of 30 degree?
Two Angles are Complementary when they add up to 90 degrees (a Right Angle). They don’t have to be next to each other, just so long as the total is 90 degrees. Examples: 60° and 30° are complementary angles.
## Can 3 angles be complementary?
Three angles or more angles whose sum is equal to 90 degrees cannot also be called complementary angles. Complementary angles always have positive measures. it is composed of two acute angles measuring less than 90 degrees.
## What is the complement of 10 degrees?
The complement of 10° is the angle that when added to 10° forms a right angle (90° ).
## What is the complement of 90 degrees?
0°Answer and Explanation: The complement of a 90° angle is a 0° angle.
## What is the complementary angle of 45 degree?
Solution: We know that the sum of two complementary angles is equal to 90o. Hence; Complementary of 45º=90º−45º=45º
## What is the complement of 40 degree?
complement angle to 40 degrees is 90 – 40 = 50 degrees and the supplement angle to 40 degrees is 180 – 40 = 140 degrees. | HuggingFaceTB/finemath | |
## Primality Testing
Expand Messages
• Lately I have been interested in primality testing: to determine whether a number, N, is prime or composite. From what I have thus far read, various tests have
Message 1 of 3 , Sep 6, 2004
Lately I have been interested in primality testing: to determine
whether a number, N, is prime or composite.
From what I have thus far read, various tests have strengths and
weaknesses.
Some tests are 100% correct, but it may take a long time to test,
comprising hundreds of hours for a computer to test a 4,000 digit
number.
Other tests can give a quick response, but at the sacrifice of
accuracy: sometimes a false answer is given.
I was wondering if there a method yet to determine with 100% accuracy
whether a number is prime or composite, in a quick amount of time.
My thinking on the subject, for what its worth, is this:
1. We can immediately say that numbers whose digits end in
0,2,4,5,6,8 are composite.
2. That leaves numbers ending with the digits 1, 3, 7, 9. Some of
these numbers are prime, some are not. To sort these out is the hard
part.
3. Of the numbers that end in the digit 9, one thing we can do is
this: if we 'cast out' that 9, and if the remaing digits are a
multiple of 3, then, the original number, N, is a composite. For
example, the following numbers are composite: 1239, 3339, 50109.
4. One thing which I thought was a correct rule but not anymore: If
you cast out the last digit 9, and then if you add 2 to the remaining
digits: if you then have a a number whose first half of its digits
are the same as its seocnd half, then that number is composite. To
illustrate: take 209. cast out the 9; add 2 to the remaining digits,
which will make 22. 209 is composite. Or take 319. Cast out the 9;
add 2 to the digits, which will make 33. 319 is composite.
This rule seems to work many times, but I have found at least one
counterexample. So I give this caveat to save somebody else trouble.
Best Regards,
Ron Dwyer
• At 06:50 PM 9/6/2004, Ronald Dwyer wrote: 3. Of the numbers that end in the digit 9, one thing we can do is ... That is of very little help because it tells
Message 2 of 3 , Sep 6, 2004
At 06:50 PM 9/6/2004, Ronald Dwyer wrote:
3. Of the numbers that end in the digit 9, one thing we can do is
>this: if we 'cast out' that 9, and if the remaing digits are a
>multiple of 3, then, the original number, N, is a composite. For
>example, the following numbers are composite: 1239, 3339, 50109.
That is of very little help because it tells if the number is divisible by
3, so that rules out very few of the cases.
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# Differentiating cosmological redshift wrt time confusion
I'm trying to understand Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the universe by Davis and Lineweaver. On page 19 they give an equation (number 23) for redshift $z$:$$1+z=\frac{R_{0}}{R\left(t\right)},$$where $R_{0}$ is the scale factor at time of observation and $R\left(t\right)$ is the scale factor at time of emission. They then differentiate this with respect to time $t$ to get$$\frac{dt}{R\left(t\right)}=-\frac{dz}{R_{0}H\left(z\right)},$$where $H\left(z\right)$ is Hubble’s constant at the time an object with redshift, $z$, emitted the light we now see and where redshift is used instead of time to parametrize Hubble’s constant. I just cannot see how differentiating wrt time gives this equation. And where does the minus sign come from? Using my high school maths I'd say$$\frac{dz}{dt}=\frac{d\left(\frac{R_{0}}{R\left(t\right)},\right)}{dt},$$but where do I go from there? Thank you.
$$\frac{dz}{dt}=\frac{d\left(\frac{R_{0}}{R\left(t\right)},\right)}{dt} = R_0 \left(\frac{-1}{R(t)^2}\right) \frac{dR}{dt}$$ but by the definition of the Hubble parameter $$H(z) = \frac{1}{R}\frac{dR}{dt},$$ $$\frac{dz}{dt}= -R_0\frac{H(z)}{R(t)}$$ and $$\frac{dt}{R(t)} = - \frac{dz}{R_0 H(z)}$$
• Ahhh, I believe that is called implicit differentiation, which I'm only vaguely familiar with. So I say $u=1/R$ and using the chain rule $\frac{du}{dt}=\frac{du}{dR}\frac{dR}{dt}$? Is that correct? Commented May 10, 2018 at 7:52 | HuggingFaceTB/finemath | |
# Convert Fractions To Decimals Worksheet Pdf
Convert Fractions To Decimals Worksheet Pdf. Converting fractions to decimals worksheet created date: Round to the thousandths place.
1) 1 4 0.25 2) 2 3 5 2.6 3) 5 8 0.625 4) 3 5 0.6 5) 7 200 0.035. You can choose the number of decimal digits used, the. Brighterly’s converting fractions to decimals worksheet serves as a basis for students to practice conversion of fractions to decimals.
### Comparing Decimals Up To Hundredths Tight Range A.
To compare decimals, we compare the place values. Convert the fractions to decimals h 1.) 13 20 2.) 7 9 m 3.) 15 20 4.) 4 5 5.) title: Fractions to decimal worksheets helps as a foundation to practice conversion of fractions to decimals and vice versa.
### A Numerator And A Denominator, Representing How Many Parts We Have Out.
Convert fractions to decimals (denominators 10, 100 or 1000) answers brighterly.com #1 0.05 #2 0.309 #3 0.009 #4 0.34 #5 0.012 #6 0.35 #7 4.51 #8 8.913 #9 1.1 #10 2.18 #11 3.6. Chapter 6 section 4 fractions decimals and percents www.slideshare.net Converting fractions to decimals worksheet keywords:
### In The Second Section, Use Your Learning To Complete The Table With The Missing Values.
Convert the fractions to decimals h 1.) 2 6 2.) 1 2 m 3.) 2 5 4.) 3 4 5.) 3. Divide the top by the bottom 5 4 = 4 ÷ 5 = 0.8 fractions, decimals & % examples 64% = 64 to write a % as a fraction or decimal, divide by 100 ÷ 100 = 0.64 64% = 100 64 = 25 16 0.1 to write a decimal or fraction as a Sample grade 5 fractions to / from decimals worksheet.
### Fraction To Decimal Conversion Math Worksheets.
Fractions to decimals (include repeating decimals) 12 / 13 =. 1) 90 % 2) 30 % 3) 115.9% 4) 9%. Round to the thousandths place.
### Fraction To Decimal Conversion Math Worksheets For Kids To Practice Converting Fractions To Decimals Form.
If you’ve ever taught your students how to write fractions in decimal, it’s time to move on to a new topic.now we can teach you to write decimal numbers as fractions.many of our students who learn this lesson have to solve problems.your students need a page 1/6 Converting fractions to decimals worksheet created date: The worksheets are very customizable: | HuggingFaceTB/finemath | |
Homework Help: Magnetic field energy inside a linear dielectric
1. Oct 3, 2013
sudipmaity
1. The problem statement, all variables and given/known data
A Sphere made of a linear magnetic of radius a and permeability μ is placed in an external uniform magnetic field Bo.
1) Assuming Laplace equation solutions find magnetic field at an internal and external poinnt.
2) Find the demagnetization factor of the sphere.
3) CALCULATE THE MAGNETIC FIELD ENERGY STORED WITHIN THE SPHERE.
I had no trouble finding and verifying part 1 and 2 from standard books.I did find part 3 . But not being able to verify it.This is a University exam question.
Here are the answers that I found out: I assumed Bo =μο Ho where μο is permeability of the medium in which the sphere is placed.
Binternal=(3μBo z^)/( μ+2 μο)
Bexternal= Bo z^+ (Bo) (a/r)^3 (μ-μο/2μο+μ)( r^ cosθ + θ^ sin θ)
The demagnetizing factor is 4π/3.
All these answers matched with standard electrodynamics textbooks.But I couldnt find answer to the 3rd part.Here is how I am doing it.
I referred to Pg 166 and 214 of J.D.Jackson's book.
Before the sphere was introduced the energy in the field was:
Wo = 1/2 ∫ (BoHo) dV
After the sphere is introduced the energy changes to
Wi = 1/2 ∫ (BH) dV
So the change in energy:
ΔW = 1/2∫( BoHo -BH) dV = 1/2∫ ( μ-μο/μμο)BBo dV.
U = ΔW
U = 1/2 0a [ Binternal Bo z^]( μ -μο /μμο) 4πr2 dr.
Finally U = 2π a3Bo2 (μ-μο/2μο+μ)/μο
Change in magnetic field energy must be stored within the sphere as potential energy given by U.
IS MY EXPRESSION FOR U CORRECT??
2. Relevant equations
3. The attempt at a solution
Last edited: Oct 3, 2013
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