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## Section4.2Riemann Sums

###### Motivating Questions
• How can we use a Riemann sum to estimate the area between a given curve and the horizontal axis over a particular interval?

• What are the differences among left, right, middle, and random Riemann sums?

• How can we write Riemann sums in an abbreviated form?

In Section 4.1, we learned that if we have a moving object with velocity function $v\text{,}$ whenever $v(t)$ is positive, the area between $y = v(t)$ and the $t$-axis over a given time interval tells us the distance traveled by the object over that time period; in addition, if $v(t)$ is sometimes negative and we view the area of any region below the $t$-axis as having an associated negative sign, then the sum of these signed areas over a given interval tells us the moving object's change in position over the time interval.

For instance, for the velocity function given in Figure 4.2.1, if the areas of shaded regions are $A_1\text{,}$ $A_2\text{,}$ and $A_3$ as labeled, then the total distance $D$ traveled by the moving object on $[a,b]$ is

\begin{equation*} D = A_1 + A_2 + A_3\text{,} \end{equation*}

while the total change in the object's position on $[a,b]$ is

\begin{equation*} s(b) - s(a) = A_1 - A_2 + A_3\text{.} \end{equation*}

Because the motion is in the negative direction on the interval where $v(t) \lt 0\text{,}$ we subtract $A_2$ when determining the object's total change in position.

Of course, finding $D$ and $s(b)-s(a)$ for the situation given in Figure 4.2.1 presumes that we can actually find the areas represented by $A_1\text{,}$ $A_2\text{,}$ and $A_3\text{.}$ In most of our work in Section 4.1, such as in Activities 4.1.3 and 4.1.4, we worked with velocity functions that were either constant or linear, so that by finding the areas of rectangles and triangles, we could find the area bounded by the velocity function and the horizontal axis exactly. But when the curve that bounds a region is not one for which we have a known formula for area, we are unable to find this area exactly. Indeed, this is one of our biggest goals in Chapter 4: to learn how to find the exact area bounded between a curve and the horizontal axis for as many different types of functions as possible.

To begin, we expand on the ideas in Activity 4.1.2, where we encountered a nonlinear velocity function and approximated the area under the curve using four and eight rectangles, respectively. In the following preview activity, we focus on three different options for deciding how to find the heights of the rectangles we will use.

###### Preview Activity4.2.1

A person walking along a straight path has her velocity in miles per hour at time $t$ given by the function $v(t) = 0.25t^3-1.5t^2+3t+0.25\text{,}$ for times in the interval $0 \le t \le 2\text{.}$ The graph of this function is also given in each of the three diagrams in Figure 4.2.2.

Note that in each diagram, we use four rectangles to estimate the area under $y = v(t)$ on the interval $[0,2]\text{,}$ but the method by which the four rectangles' respective heights are decided varies among the three individual graphs.

1. How are the heights of rectangles in the left-most diagram being chosen? Explain, and hence determine the value of

\begin{equation*} S = A_1 + A_2 + A_3 + A_4 \end{equation*}

by evaluating the function $y = v(t)$ at appropriately chosen values and observing the width of each rectangle. Note, for example, that

\begin{equation*} A_3 = v(1) \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1\text{.} \end{equation*}
2. Explain how the heights of rectangles are being chosen in the middle diagram and find the value of

\begin{equation*} T = B_1 + B_2 + B_3 + B_4\text{.} \end{equation*}
3. Likewise, determine the pattern of how heights of rectangles are chosen in the right-most diagram and determine

\begin{equation*} U = C_1 + C_2 + C_3 + C_4\text{.} \end{equation*}
4. Of the estimates $S\text{,}$ $T\text{,}$ and $U\text{,}$ which do you think is the best approximation of $D\text{,}$ the total distance the person traveled on $[0,2]\text{?}$ Why?

### Subsection4.2.1Sigma Notation

It is apparent from several different problems we have considered that sums of areas of rectangles is one of the main ways to approximate the area under a curve over a given interval. Intuitively, we expect that using a larger number of thinner rectangles will provide a way to improve the estimates we are computing. As such, we anticipate dealing with sums with a large number of terms. To do so, we introduce the use of so-called sigma notation, named for the Greek letter $\Sigma\text{,}$ which is the capital letter $S$ in the Greek alphabet.

For example, say we are interested in the sum

\begin{equation*} 1 + 2 + 3 + \cdots + 100\text{,} \end{equation*}

which is the sum of the first 100 natural numbers. Sigma notation provides a shorthand notation that recognizes the general pattern in the terms of the sum. It is equivalent to write

\begin{equation*} \sum_{k=1}^{100} k = 1 + 2 + 3 + \cdots + 100\text{.} \end{equation*}

We read the symbol $\sum_{k=1}^{100} k$ as “the sum from $k$ equals 1 to 100 of $k\text{.}$” The variable $k$ is usually called the index of summation, and the letter that is used for this variable is immaterial. Each sum in sigma notation involves a function of the index; for example,

\begin{equation*} \sum_{k=1}^{10} (k^2 + 2k) = (1^2 + 2\cdot 1) + (2^2 + 2\cdot 2) + (3^2 + 2\cdot 3) + \cdots + (10^2 + 2\cdot 10)\text{,} \end{equation*}

and more generally,

\begin{equation*} \sum_{k=1}^n f(k) = f(1) + f(2) + \cdots + f(n)\text{.} \end{equation*}

Sigma notation allows us the flexibility to easily vary the function being used to track the pattern in the sum, as well as to adjust the number of terms in the sum simply by changing the value of $n\text{.}$ We test our understanding of this new notation in the following activity.

###### Activity4.2.2

For each sum written in sigma notation, write the sum long-hand and evaluate the sum to find its value. For each sum written in expanded form, write the sum in sigma notation.

1. $\sum_{k=1}^{5} (k^2 + 2)$

2. $\sum_{i=3}^{6} (2i-1)$

3. $3 + 7 + 11 + 15 + \cdots + 27$

4. $4 + 8 + 16 + 32 + \cdots + 256$

5. $\sum_{i=1}^{6} \frac{1}{2^i}$

### Subsection4.2.2Riemann Sums

When a moving body has a positive velocity function $y = v(t)$ on a given interval $[a,b]\text{,}$ we know that the area under the curve over the interval is the total distance the body travels on $[a,b]\text{.}$ While this is the fundamental motivating force behind our interest in the area bounded by a function, we are also interested more generally in being able to find the exact area bounded by $y = f(x)$ on an interval $[a,b]\text{,}$ regardless of the meaning or context of the function $f\text{.}$ For now, we continue to focus on determining an accurate estimate of this area through the use of a sum of the areas of rectangles, doing so in the setting where $f(x) \ge 0$ on $[a,b]\text{.}$ Throughout, unless otherwise indicated, we also assume that $f$ is continuous on $[a,b]\text{.}$

The first choice we make in any such approximation is the number of rectangles.

If we say that the total number of rectangles is $n\text{,}$ and we desire $n$ rectangles of equal width to subdivide the interval $[a,b]\text{,}$ then each rectangle must have width $\Delta x = \frac{b-a}{n}\text{.}$ We observe further that $x_1 = x_0 + \Delta x\text{,}$ $x_2 = x_0 + 2 \Delta x\text{,}$ and thus in general $x_{i} = a + i\Delta x\text{,}$ as pictured in Figure 4.2.3.

We use each subinterval $[x_i, x_{i+1}]$ as the base of a rectangle, and next must choose how to decide the height of the rectangle that will be used to approximate the area under $y = f(x)$ on the subinterval. There are three standard choices: use the left endpoint of each subinterval, the right endpoint of each subinterval, or the midpoint of each. These are precisely the options encountered in Preview Activity 4.2.1 and seen in Figure 4.2.2. We next explore how these choices can be reflected in sigma notation.

If we now consider an arbitrary positive function $f$ on $[a,b]$ with the interval subdivided as shown in Figure 4.2.3, and choose to use left endpoints, then on each interval of the form $[x_{i}, x_{i+1}]\text{,}$ the area of the rectangle formed is given by

\begin{equation*} A_{i+1} = f(x_i) \cdot \Delta x\text{,} \end{equation*}

as seen in Figure 4.2.4.

If we let $L_n$ denote the sum of the areas of rectangles whose heights are given by the function value at each respective left endpoint, then we see that

\begin{align*} L_n =\mathstrut \amp A_1 + A_2 + \cdots + A_{i+1} + \cdots + A_n\\ =\mathstrut \amp f(x_0) \cdot \Delta x + f(x_1) \cdot \Delta x + \cdots + f(x_i) \cdot \Delta x + \cdots + f(x_{n-1}) \cdot \Delta x\text{.} \end{align*}

In the more compact sigma notation, we have

\begin{equation*} L_n = \sum_{i = 0}^{n-1} f(x_i) \Delta x\text{.} \end{equation*}

Note particularly that since the index of summation begins at $0$ and ends at $n-1\text{,}$ there are indeed $n$ terms in this sum. We call $L_n$ the left Riemann sum for the function $f$ on the interval $[a,b]\text{.}$

There are now two fundamental issues to explore: the number of rectangles we choose to use and the selection of the pattern by which we identify the height of each rectangle. It is best to explore these choices dynamically, and the applet 1 Marc Renault, Geogebra Calculus Applets.found at http://gvsu.edu/s/a9 is a particularly useful one. There we see the image shown in Figure 4.2.5, but with the opportunity to adjust the slider bars for the left endpoint and the number of subintervals.

By moving the sliders, we can see how the heights of the rectangles change as we consider left endpoints, midpoints, and right endpoints, as well as the impact that a larger number of narrower rectangles has on the approximation of the exact area bounded by the function and the horizontal axis.

To see how the Riemann sums for right endpoints and midpoints are constructed, we consider Figure 4.2.6.

For the sum with right endpoints, we see that the area of the rectangle on an arbitrary interval $[x_i, x_{i+1}]$ is given by $B_{i+1} = f(x_{i+1}) \cdot \Delta x\text{,}$ so that the sum of all such areas of rectangles is given by

\begin{align*} R_n =\mathstrut \amp B_1 + B_2 + \cdots + B_{i+1} + \cdots + B_n\\ =\mathstrut \amp f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x + \cdots + f(x_{i+1}) \cdot \Delta x + \cdots + f(x_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(x_i) \Delta x\text{.} \end{align*}

We call $R_n$ the right Riemann sum for the function $f$ on the interval $[a,b]\text{.}$ For the sum that uses midpoints, we introduce the notation

\begin{equation*} \overline{x}_{i+1} = \frac{x_{i} + x_{i+1}}{2} \end{equation*}

so that $\overline{x}_{i+1}$ is the midpoint of the interval $[x_i, x_{i+1}]\text{.}$ For instance, for the rectangle with area $C_1$ in Figure 4.2.6, we now have

\begin{equation*} C_1 = f(\overline{x}_1) \cdot \Delta x\text{.} \end{equation*}

Hence, the sum of all the areas of rectangles that use midpoints is

\begin{align*} M_n =\mathstrut \amp C_1 + C_2 + \cdots + C_{i+1} + \cdots + C_n\\ =\mathstrut \amp f(\overline{x_1}) \cdot \Delta x + f(\overline{x_2}) \cdot \Delta x + \cdots + f(\overline{x}_{i+1}) \cdot \Delta x + \cdots + f(\overline{x}_{n}) \cdot \Delta x\\ =\mathstrut \amp \sum_{i=1}^{n} f(\overline{x}_i) \Delta x\text{,} \end{align*}

and we say that $M_n$ is the middle Riemann sum for $f$ on $[a,b]\text{.}$

When $f(x) \ge 0$ on $[a,b]\text{,}$ each of the Riemann sums $L_n\text{,}$ $R_n\text{,}$ and $M_n$ provides an estimate of the area under the curve $y = f(x)$ over the interval $[a,b]\text{;}$ momentarily, we will discuss the meaning of Riemann sums in the setting when $f$ is sometimes negative. We also recall that in the context of a nonnegative velocity function $y = v(t)\text{,}$ the corresponding Riemann sums are approximating the distance traveled on $[a,b]$ by the moving object with velocity function $v\text{.}$

There is a more general way to think of Riemann sums, and that is to not restrict the choice of where the function is evaluated to determine the respective rectangle heights. That is, rather than saying we'll always choose left endpoints, or always choose midpoints, we simply say that a point $x_{i+1}^*$ will be selected at random in the interval $[x_i, x_{i+1}]$ (so that $x_i \le x_{i+1}^* \le x_{i+1}$), which makes the Riemann sum given by

\begin{equation*} f(x_1^*) \cdot \Delta x + f(x_2^*) \cdot \Delta x + \cdots + f(x_{i+1}^*) \cdot \Delta x + \cdots + f(x_n^*) \cdot \Delta x = \sum_{i=1}^{n} f(x_i^*) \Delta x\text{.} \end{equation*}

At http://gvsu.edu/s/a9, the applet noted earlier and referenced in Figure 4.2.5, by unchecking the “relative” box at the top left, and instead checking “random,” we can easily explore the effect of using random point locations in subintervals on a given Riemann sum. In computational practice, we most often use $L_n\text{,}$ $R_n\text{,}$ or $M_n\text{,}$ while the random Riemann sum is useful in theoretical discussions. In the following activity, we investigate several different Riemann sums for a particular velocity function.

###### Activity4.2.3

Suppose that an object moving along a straight line path has its velocity in feet per second at time $t$ in seconds given by $v(t) = \frac{2}{9}(t-3)^2 + 2\text{.}$

1. Carefully sketch the region whose exact area will tell you the value of the distance the object traveled on the time interval $2 \le t \le 5\text{.}$

2. Estimate the distance traveled on $[2,5]$ by computing $L_4\text{,}$ $R_4\text{,}$ and $M_4\text{.}$

3. Does averaging $L_4$ and $R_4$ result in the same value as $M_4\text{?}$ If not, what do you think the average of $L_4$ and $R_4$ measures?

4. For this question, think about an arbitrary function $f\text{,}$ rather than the particular function $v$ given above. If $f$ is positive and increasing on $[a,b]\text{,}$ will $L_n$ over-estimate or under-estimate the exact area under $f$ on $[a,b]\text{?}$ Will $R_n$ over- or under-estimate the exact area under $f$ on $[a,b]\text{?}$ Explain.

### Subsection4.2.3When the function is sometimes negative

For a Riemann sum such as

\begin{equation*} L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x\text{,} \end{equation*}

we can of course compute the sum even when $f$ takes on negative values. We know that when $f$ is positive on $[a,b]\text{,}$ the corresponding left Riemann sum $L_n$ estimates the area bounded by $f$ and the horizontal axis over the interval.

For a function such as the one pictured in Figure 4.2.7, where in the first figure a left Riemann sum is being taken with 12 subintervals over $[a,d]\text{,}$ we observe that the function is negative on the interval $b \le x \le c\text{,}$ and so for the four left endpoints that fall in $[b,c]\text{,}$ the terms $f(x_i) \Delta x$ have negative function values. This means that those four terms in the Riemann sum produce an estimate of the opposite of the area bounded by $y = f(x)$ and the $x$-axis on $[b,c]\text{.}$

In Figure 4.2.7, we also see evidence that by increasing the number of rectangles used in a Riemann sum, it appears that the approximation of the area (or the opposite of the area) bounded by a curve appears to improve. For instance, in the middle graph, we use 24 left rectangles, and from the shaded areas, it appears that we have decreased the error from the approximation that uses 12. When we proceed to Section 4.3, we will discuss the natural idea of letting the number of rectangles in the sum increase without bound.

For now, it is most important for us to observe that, in general, any Riemann sum of a continuous function $f$ on an interval $[a,b]$ approximates the difference between the area that lies above the horizontal axis on $[a,b]$ and under $f$ and the area that lies below the horizontal axis on $[a,b]$ and above $f\text{.}$ In the notation of Figure 4.2.7, we may say that

\begin{equation*} L_{24} \approx A_1 - A_2 + A_3\text{,} \end{equation*}

where $L_{24}$ is the left Riemann sum using 24 subintervals shown in the middle graph, and $A_1$ and $A_3$ are the areas of the regions where $f$ is positive on the interval of interest, while $A_2$ is the area of the region where $f$ is negative. We will also call the quantity $A_1 - A_2 + A_3$ the net signed area bounded by $f$ over the interval $[a,d]\text{,}$ where by the phrase “signed area” we indicate that we are attaching a minus sign to the areas of regions that fall below the horizontal axis.

Finally, we recall from the introduction to this present section that in the context where the function $f$ represents the velocity of a moving object, the total sum of the areas bounded by the curve tells us the total distance traveled over the relevant time interval, while the total net signed area bounded by the curve computes the object's change in position on the interval.

###### Activity4.2.4

Suppose that an object moving along a straight line path has its velocity $v$ (in feet per second) at time $t$ (in seconds) given by

\begin{equation*} v(t) = \frac{1}{2}t^2 - 3t + \frac{7}{2}\text{.} \end{equation*}
1. Compute $M_5\text{,}$ the middle Riemann sum, for $v$ on the time interval $[1,5]\text{.}$ Be sure to clearly identify the value of $\Delta t$ as well as the locations of $t_0\text{,}$ $t_1\text{,}$ $\cdots\text{,}$ $t_5\text{.}$ In addition, provide a careful sketch of the function and the corresponding rectangles that are being used in the sum.

2. Building on your work in (a), estimate the total change in position of the object on the interval $[1,5]\text{.}$

3. Building on your work in (a) and (b), estimate the total distance traveled by the object on $[1,5]\text{.}$

4. Use appropriate computing technology 2 For instance, consider the applet at http://gvsu.edu/s/a9 and change the function and adjust the locations of the blue points that represent the interval endpoints $a$ and $b\text{.}$to compute $M_{10}$ and $M_{20}\text{.}$ What exact value do you think the middle sum eventually approaches as $n$ increases without bound? What does that number represent in the physical context of the overall problem?

### Subsection4.2.4Summary

• A Riemann sum is simply a sum of products of the form $f(x_i^*) \Delta x$ that estimates the area between a positive function and the horizontal axis over a given interval. If the function is sometimes negative on the interval, the Riemann sum estimates the difference between the areas that lie above the horizontal axis and those that lie below the axis.

• The three most common types of Riemann sums are left, right, and middle sums, plus we can also work with a more general, random Riemann sum. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed in the sum. For a left Riemann sum, we evaluate the function at the left endpoint of each subinterval, while for right and middle sums, we use right endpoints and midpoints, respectively.

• The left, right, and middle Riemann sums are denoted $L_n\text{,}$ $R_n\text{,}$ and $M_n\text{,}$ with formulas

\begin{align*} L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x,\\ R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x,\\ M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x\text{,} \end{align*}

where $x_0 = a\text{,}$ $x_i = a + i\Delta x\text{,}$ and $x_n = b\text{,}$ using $\Delta x = \frac{b-a}{n}\text{.}$ For the midpoint sum, $\overline{x}_{i} = (x_{i-1} + x_i)/2\text{.}$

### SubsectionExercises

###### 4

Consider the function $f(x) = 3x + 4\text{.}$

1. Compute $M_4$ for $y=f(x)$ on the interval $[2,5]\text{.}$ Be sure to clearly identify the value of $\Delta x\text{,}$ as well as the locations of $x_0, x_1, \ldots, x_4\text{.}$ Include a careful sketch of the function and the corresponding rectangles being used in the sum.

2. Use a familiar geometric formula to determine the exact value of the area of the region bounded by $y = f(x)$ and the $x$-axis on $[2,5]\text{.}$

3. Explain why the values you computed in (a) and (b) turn out to be the same. Will this be true if we use a number different than $n = 4$ and compute $M_n\text{?}$ Will $L_4$ or $R_4$ have the same value as the exact area of the region found in (b)?

4. Describe the collection of functions $g$ for which it will always be the case that $M_n\text{,}$ regardless of the value of $n\text{,}$ gives the exact net signed area bounded between the function $g$ and the $x$-axis on the interval $[a,b]\text{.}$

###### 5

Let $S$ be the sum given by

\begin{equation*} S = ((1.4)^2 + 1) \cdot 0.4 + ((1.8)^2 + 1) \cdot 0.4 + ((2.2)^2 + 1) \cdot 0.4 + ((2.6)^2 + 1) \cdot 0.4 +((3.0)^2 + 1) \cdot 0.4\text{.} \end{equation*}
1. Assume that $S$ is a right Riemann sum. For what function $f$ and what interval $[a,b]$ is $S$ this function's Riemann sum? Why?

2. How does your answer to (a) change if $S$ is a left Riemann sum? a middle Riemann sum?

3. Suppose that $S$ really is a right Riemann sum. What is geometric quantity does $S$ approximate?

4. Use sigma notation to write a new sum $R$ that is the right Riemann sum for the same function, but that uses twice as many subintervals as $S\text{.}$

###### 6

A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table.

1. Plot the given data on a set of axes with time on the horizontal axis and the velocity on the vertical axis.

2. Estimate the total distance traveled during the car the time brakes using a middle Riemann sum with 3 subintervals.

3. Estimate the total distance traveled on $[0,1.8]$ by computing $L_6\text{,}$ $R_6\text{,}$ and $\frac{1}{2}(L_6 + R_6)\text{.}$

4. Assuming that $v(t)$ is always decreasing on $[0,1.8]\text{,}$ what is the maximum possible distance the car traveled before it stopped? Why?

###### 7

The rate at which pollution escapes a scrubbing process at a manufacturing plant increases over time as filters and other technologies become less effective. For this particular example, assume that the rate of pollution (in tons per week) is given by the function $r$ that is pictured in Figure 4.2.9.

1. Use the graph to estimate the value of $M_4$ on the interval $[0,4]\text{.}$

2. What is the meaning of $M_4$ in terms of the pollution discharged by the plant?

3. Suppose that $r(t) = 0.5 e^{0.5t}\text{.}$ Use this formula for $r$ to compute $L_5$ on $[0,4]\text{.}$

4. Determine an upper bound on the total amount of pollution that can escape the plant during the pictured four week time period that is accurate within an error of at most one ton of pollution.


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# Cardinality

{{#invoke:Hatnote|hatnote}} In mathematics, the cardinality of a set is a measure of the "number of elements of the set". For example, the set A = {2, 4, 6} contains 3 elements, and therefore A has a cardinality of 3. There are two approaches to cardinality – one which compares sets directly using bijections and injections, and another which uses cardinal numbers.[1] The cardinality of a set is also called its size, when no confusion with other notions of size[2] is possible.

The cardinality of a set A is usually denoted | A |, with a vertical bar on each side; this is the same notation as absolute value and the meaning depends on context. Alternatively, the cardinality of a set A may be denoted by n(A), A, card(A), or # A.

## Comparing sets

While the cardinality of a finite set is just the number of its elements, extending the notion to infinite sets usually starts with defining the notion of comparison of arbitrary (in particular infinite) sets.

Bijective function from N to E. Although E is a proper subset of N, both sets have the same cardinality.

### Definition 1: | A | = | B |

Two sets A and B have the same cardinality if there exists a bijection, that is, an injective and surjective function, from A to B. Such sets are said to be equipotent, equipollent, or equinumerous.
For example, the set E = {0, 2, 4, 6, ...} of non-negative even numbers has the same cardinality as the set N = {0, 1, 2, 3, ...} of natural numbers, since the function f(n) = 2n is a bijection from N to E.

### Definition 2: | A | ≥ | B |

A has cardinality greater than or equal to the cardinality of B if there exists an injective function from B into A.

### Definition 3: | A | > | B |

A has cardinality strictly greater than the cardinality of B if there is an injective function, but no bijective function, from B to A.
For example, the set R of all real numbers has cardinality strictly greater than the cardinality of the set N of all natural numbers, because the inclusion map i : NR is injective, but it can be shown that there does not exist a bijective function from N to R (see Cantor's diagonal argument or Cantor's first uncountability proof).

If | A | ≥ | B | and | B | ≥ | A | then | A | = | B | (Cantor–Bernstein–Schroeder theorem). The axiom of choice is equivalent to the statement that | A | ≥ | B | or | B | ≥ | A | for every A,B.[3][4]

## Cardinal numbers

{{#invoke:main|main}}

Above, "cardinality" was defined functionally. That is, the "cardinality" of a set was not defined as a specific object itself. However, such an object can be defined as follows.

The relation of having the same cardinality is called equinumerosity, and this is an equivalence relation on the class of all sets. The equivalence class of a set A under this relation then consists of all those sets which have the same cardinality as A. There are two ways to define the "cardinality of a set":

1. The cardinality of a set A is defined as its equivalence class under equinumerosity.
2. A representative set is designated for each equivalence class. The most common choice is the initial ordinal in that class. This is usually taken as the definition of cardinal number in axiomatic set theory.

Assuming AC, the cardinalities of the infinite sets are denoted

${\displaystyle \aleph _{0}<\aleph _{1}<\aleph _{2}<\ldots .}$

The cardinality of the natural numbers is denoted aleph-null (${\displaystyle \aleph _{0}}$), while the cardinality of the real numbers is denoted by "${\displaystyle {\mathfrak {c}}}$" (a lowercase fraktur script "c"), and is also referred to as the cardinality of the continuum. Cantor showed, using the diagonal argument, that ${\displaystyle {\mathfrak {c}}>\aleph _{0}}$. We can show that ${\displaystyle {\mathfrak {c}}=2^{\aleph _{0}}}$, this also being the cardinality of the set of all subsets of the natural numbers. The continuum hypothesis says that ${\displaystyle \aleph _{1}=2^{\aleph _{0}}}$, i.e. ${\displaystyle 2^{\aleph _{0}}}$ is the smallest cardinal number bigger than ${\displaystyle \aleph _{0}}$, i.e. there is no set whose cardinality is strictly between that of the integers and that of the real numbers. The continuum hypothesis is independent of ZFC, a standard axiomatization of set theory; that is, it is impossible to prove the continuum hypothesis or its negation from ZFC (provided ZFC is consistent). See below for more details on the cardinality of the continuum.[5][6][7]

## Finite, countable and uncountable sets

If the axiom of choice holds, the law of trichotomy holds for cardinality. Thus we can make the following definitions:

## Infinite sets

Our intuition gained from finite sets breaks down when dealing with infinite sets. In the late nineteenth century Georg Cantor, Gottlob Frege, Richard Dedekind and others rejected the view of Galileo (which derived from Euclid) that the whole cannot be the same size as the part. One example of this is Hilbert's paradox of the Grand Hotel. Indeed, Dedekind defined an infinite set as one that can be placed into a one-to-one correspondence with a strict subset (that is, having the same size in Cantor's sense); this notion of infinity is called Dedekind infinite. Cantor introduced the cardinal numbers, and showed that (according to his bijection-based definition of size) some infinite sets are greater than others. The smallest infinite cardinality is that of the natural numbers (${\displaystyle \aleph _{0}}$).

### Cardinality of the continuum

{{#invoke:main|main}}

One of Cantor's most important results was that the cardinality of the continuum (${\displaystyle {\mathfrak {c}}}$) is greater than that of the natural numbers (${\displaystyle \aleph _{0}}$); that is, there are more real numbers R than whole numbers N. Namely, Cantor showed that

${\displaystyle {\mathfrak {c}}=2^{\aleph _{0}}>{\aleph _{0}}}$
(see Cantor's diagonal argument or Cantor's first uncountability proof).

The continuum hypothesis states that there is no cardinal number between the cardinality of the reals and the cardinality of the natural numbers, that is,

${\displaystyle {\mathfrak {c}}=\aleph _{1}=\beth _{1}}$
(see Beth one).

However, this hypothesis can neither be proved nor disproved within the widely accepted ZFC axiomatic set theory, if ZFC is consistent.

Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space. These results are highly counterintuitive, because they imply that there exist proper subsets and proper supersets of an infinite set S that have the same size as S, although S contains elements that do not belong to its subsets, and the supersets of S contain elements that are not included in it.

The first of these results is apparent by considering, for instance, the tangent function, which provides a one-to-one correspondence between the interval (−½π, ½π) and R (see also Hilbert's paradox of the Grand Hotel).

The second result was first demonstrated by Cantor in 1878, but it became more apparent in 1890, when Giuseppe Peano introduced the space-filling curves, curved lines that twist and turn enough to fill the whole of any square, or cube, or hypercube, or finite-dimensional space. These curves are not a direct proof that a line has the same number of points as a finite-dimensional space, but they can be used to obtain such a proof.

Cantor also showed that sets with cardinality strictly greater than ${\displaystyle {\mathfrak {c}}}$ exist (see his generalized diagonal argument and theorem). They include, for instance:

• the set of all subsets of R, i.e., the power set of R, written P(R) or 2R
• the set RR of all functions from R to R

Both have cardinality

${\displaystyle 2^{\mathfrak {c}}=\beth _{2}>{\mathfrak {c}}}$
(see Beth two).
${\displaystyle {\mathfrak {c}}^{2}=\left(2^{\aleph _{0}}\right)^{2}=2^{2\times {\aleph _{0}}}=2^{\aleph _{0}}={\mathfrak {c}},}$
${\displaystyle {\mathfrak {c}}^{\aleph _{0}}=\left(2^{\aleph _{0}}\right)^{\aleph _{0}}=2^{{\aleph _{0}}\times {\aleph _{0}}}=2^{\aleph _{0}}={\mathfrak {c}},}$
${\displaystyle {\mathfrak {c}}^{\mathfrak {c}}=\left(2^{\aleph _{0}}\right)^{\mathfrak {c}}=2^{{\mathfrak {c}}\times \aleph _{0}}=2^{\mathfrak {c}}.}$

## Examples and properties

• If X = {a, b, c} and Y = {apples, oranges, peaches}, then | X | = | Y | because { (a, apples), (b, oranges), (c, peaches)} is a bijection between the sets X and Y. The cardinality of each of X and Y is 3.
• If | X | < | Y |, then there exists Z such that | X | = | Z | and ZY.
• If | X | ≤ | Y | and | Y | ≤ | X |, then | X | = | Y |. This holds even for infinite cardinals, and is known as Cantor–Bernstein–Schroeder theorem.
• Sets with cardinality of the continuum

## Union and intersection

If A and B are disjoint sets, then

${\displaystyle \left\vert A\cup B\right\vert =\left\vert A\right\vert +\left\vert B\right\vert \,.}$

From this, one can show that in general the cardinalities of unions and intersections are related by[8]

${\displaystyle \left\vert C\cup D\right\vert +\left\vert C\cap D\right\vert =\left\vert C\right\vert +\left\vert D\right\vert \,.}$

format: plain


text:
Given that $10^{0.48} = x, 10^{0.70} = y,$ and $x^{z} = y^{2}$, then the value of z is close to
[A]3.7
[B]2.9
[C]1.45
[D]1.88

2.9
Given that,
$10^{0.48} = x, 10^{0.70} = y$
$\therefore x^{z} = y^{2}$
$=> (10^{0.48})^{z} = (10^{0.70})^{2}$
$=> 10^{0.48z} = 10^{1.4}$
$=> z = \frac{1.4}{0.48} = 2.9$
Hence option [B] is the right answer.


text:
# Tag Info

## New answers tagged authentication

3

The authentication tag is defined as an output parameter in GCM (see section 7, step 7 of NIST SP 800-38D). In all the API's I've encountered it's appended to the ciphertext. Where it is actually placed is up to the protocol designer. The protocol designer may well consider the place behind the ciphertext as ad hoc default though. The name "tag" of course ...

-1

It seems that you do want to be given the answer and not just hints, so I will do that. But I'll go step by step so that you can stop reading if you want to finish by yourself. It seems to me that both solutions satisfy the first requirement consisting in authenticity. Breaking this property would consist in changing the content of $R$ to fool $A$ (this ...

6

First the theoretical explanations: Integrity and authenticity are different goals to achieve, but both are achieved (for symmetric encryption) with a MAC. You should probably be using encrypt-than-MAC or an authenticated cipher unless you have very good reasons not to. No blanket statements can be made though. HMAC: HMAC is a often used construct. It ...

2

When we have regions of the packet that we only authenticate but not encrypt, that happens because we have data that we want to bind to the encrypted region, but we don't need to include within the encrypted region. Examples of this are: For IPsec, we include the sequence number (as a part of the ESP header). We include that within the authentication ...

0

It's the same situation as for RSA encryption. If you accidentally pick a bad $s$, i.e. one that's not coprime to $N$ then the protocol fails. How likely is this to happen? If you hit a bad $s$, then you can find the factors of $N$, hence you can break the protocol - put another way, finding a bad $s$, whether deliberately or by accident is no easier than ...

2

The way a stream cipher works, traditionally, is that $E_k$ produces a pseudorandom bitstream (the keystream) based solely on the key $k$. The message is then encrypted by XORing the message with the keystream. This has a number of consequences, notably that if you know both the plaintext and ciphertext, it's trivial to compute the keystream (if \$C=M\oplus ...

2

I have found one source that claims to have invented the word "pepper". -- But this doesn't mean that they were the first or the only inventors. I guess it may have been invented and reinvented several times. 1999 Paper Kedem, G., & Ishihara, Y. (1999, August 23-26). Brute Force Attack on UNIX Passwords with SIMD Computer. Paper presented at the 8th ...

1

No, it is not. There are several misconceptions with your scheme. First, your question/answer pairing does simply not contain enough entropy. This is much less entropy than your common "please choose a password" input, because you can't just use all words/numbers/common symbols. You have to stick to answers that make sense. This is what Ricky Demer hinted ...

Top 50 recent answers are included


text:
# Would 2 C₂ be 2 moles of diatomic carbon, or 4 moles of carbon?

Would molecules like $\ce{C2, H2}$, etc., be considered 1 mole of diatomic [element] or 2 moles of [element]?

You can have 1 mole of $\ce{C_2}$ which contains 2 mole of $\ce{C}$ just as you can have for example 1 mole of $\ce{C_2H_5OH}$ which contains 2 mole of $\ce{C}$, 6 mole of $\ce{H}$ and 1 mole of $\ce{O}$.

The point is that you always have to mention whether you are indicating moles of molecule or moles of atoms. That is why a statement like: "I have 1 mole of oxygen" is ambiguous, because it could be referring to oxygen gas (i.e. $\ce{O_2}$) or to atomic oxygen (i.e. $\ce{O}$).

• C2 is available from your regular supplier, but it certainly exists, has been monitored spectroscopically and used in reactions, see my comment to this question – Klaus-Dieter Warzecha Feb 2 '14 at 20:41
• Of course I meant that it is NOT commercially available ;) – Klaus-Dieter Warzecha Feb 2 '14 at 23:34
• @KlausWarzecha good to know! Never heard about it. – Michiel Feb 3 '14 at 6:19

$\ce{C_2}$ is the blue-green Swan band emission of comets' ion tails and the blue in gas flames. The big question is its electronic structure (presumably not $\ce{•C#C•}$). Its paper synthesis in cryogenic gas matrix (UV cleavage) or in vacuum (flash pyrolysis or UV) seems to be easy, from

ROO-C(=O)-C#C-C(=O)-OOR, the peresters of acetylenedicarboxylic acid with R = $\ce{-CF_3}$, $\ce{-C(CH_3)_3}$, or $\ce{-Si(CH_3)_3}$.

Strings of alpha-omega diradical polycarbyne, $\ce{(-C#C-)_n}$ or $\ce{(=C=C=)_n}$ (your choice in LCAO, same in MO), in carbon arcs are suggested to pericyclicly close to fullerenes.

• I'm not quite sure of how this answers my question. Could you explain a little further and use words a high school student could understand? – Gandalf the White Feb 3 '14 at 1:45
• $\ce{C_2}$ is one mole of $\ce{C_2}$ and two moles of carbon atoms. A box of eggs is one dozen eggs and 12 eggs simultaneously. The fun is in the footnotes. How can you make it on a bench top, what can you do with it? The cold synthesis byproduct is squaric acid. It's a hoot of a compound, including derived squarylium dyes, <P> link.springer.com/static-content/images/967/… – Uncle Al Feb 3 '14 at 2:02
• What an eccentric performance. – Leonardo Feb 3 '14 at 2:19
• The diradical recombines with the two liberated alkoxy radicals to an acetylene diether. That antarafacially 2+2 dimerizes to the squaric acid precursor. Aqueous acid triiodide to oxidize and hydrolyze. God save us from the congenitally inconsequential. – Uncle Al Feb 3 '14 at 23:21


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# What’s the proper format for custom commands in local startup?

Hello. I have a R7800 and I’m trying to set the CPU to performance. Not sure where to put the following commands in local startup:

`echo performance > /sys/devices/system/cpu/cpufreq/policy0/scaling_governor`

`echo performance > /sys/devices/system/cpu/cpufreq/policy1/scaling_governor`

Do I erase everything above exit 0 and paste it there with the pound symbol before both commands and do I leave a space between the two commands? Sorry for what may seem like silly questions, still kind of new here.

You leave everything as it is, and just insert your two commands above `exit 0` in `/etc/rc.local`, without touching anything else.

Oh ok. Thanks for the response. I forgot to mention I was trying to do it through luci.

You can do that in Luci. Go to System->Startup and select the 'Local Startup' tab. That allows you to edit rc.local from within Luci.

Oh okay. I get it now. Thanks guys.


text:
# How do you graph, identify the domain, range, and asymptotes for y=sec(1/2)x?

Jul 20, 2018

Domain; $x \in \left(- \infty , \infty\right)$. Range of.
$y = \sec \left(\frac{x}{2}\right) \in \left(- \infty , 1\right] U \left[1 , \infty\right) \Rightarrow y \notin \left(- 1 , 1\right)$.
Asymptotes: x = ( 2k + 1 )pi, k = 0, +-1, +-2, +-3, ...

#### Explanation:

The period of $y = \sec \left(\frac{x}{2}\right)$ is $\frac{2 \pi}{\frac{1}{2}} = 4 \pi$.

Cosine range is $\left[- 1 , 1\right]$ Its reciprocal

secant range is $\left(- \infty , 1\right] U \left[1 , \infty\right)$. So,

$y = \sec \left(\frac{x}{2}\right) \in \left(- \infty , 1\right] U \left[1 , \infty\right) \Rightarrow y \notin \left(- 1 , 1\right)$.

Domain; $x \in \left(- \infty , \infty\right)$

Asymptotes: x = ( 2k + 1 )pi, k = 0, +-1, +-2, +-3, ...,

by solving the denominator $\cos \left(\frac{x}{2}\right) = 0$.

See illustrative graph, with the markings os asymptotes near O and

the out-of- bounds range #( - 1, 1 ):.
graph{(y cos ( x/2 ) - 1)(y^2-1)(x^2-(pi)^2) = 0[-10 10 -5 5]}

Slide the graph $\rightarrow \uparrow \leftarrow \downarrow$, to view the extended graph.


text:
# Asymptotic Weyl Character Formula

Let $G$ be a complex semi-simple group along with a chosen pair of opposite Borel subgroups (so we get all the root-theoretic data we need). Let $\lambda$ be a dominant weight, and let $V(\lambda)$ be the corresponding highest weight representation. I'm interested in a formula for the character of $V(n \lambda)$ as $n \rightarrow \infty$. Of course to have this limit converge, I need to normalize things. So I consider the character of $V(n\lambda)$ as a measure on the Lie algebra of the torus, which I rescale to live on precisely the convex hull of the weights appearing on $V(\lambda)$, and which I scale by a fixed power of $n$. Then I consider the weak limit of this measure. This measure is also the Duistermaat-Heckman measure of the flag variety corresponding to the line bundle $\mathcal{O}(\lambda)$ (perhaps I should restrict to $\lambda$ regular to be correct).

The Fourier transforms of these measure appearing in the limit are precisely the character in the usual sense (given by the Weyl character formula), so the Fourier transform of the limiting measure should be calculable. So I am asking for the Fourier transform of this measure.

In the case of $SL_2$, for $\lambda$ equal to the fundamental weight $\omega$, this is just the Duistermaat-Heckman measure of $\mathbb{P}^1$. This is just the Lebesgue measure supported on the interval $[-\omega,\omega]$, which clearly agrees with what we know about characters of $SL_2$ representations. And it's Fourier transform is (up to some constants) the sinc function $\frac{\sin(x)}{x}$, which can also be seen from the Weyl character formula since $\frac{\sin(x)}{x} = \frac{e^{ix} - e^{-ix}}{2ix}$ However, I do not know how to write down such a formula in other types. I expect that there should be some formula with roughly the form of the Weyl character formula. Does anyone know any such formula?

-

This is easy to get from the WCF directly: the relevant limit is $1 - \exp(-x) \sim x$. The limits of the Weyl character (thought of as a function on $T$) in its two forms $$\sum_{w\in W} w \cdot \frac{t^{\lambda}}{\prod_{\beta \in \Delta_+}(1 - t^{-\beta})} = \frac{\sum_{w\in W} (-1)^w t^{w\cdot(\lambda+\rho)-\rho}}{\prod_{\beta\in\Delta_+}(1-t^{-\beta})}$$ are the functions on $\mathfrak t$ $$\sum_{w\in W} w \cdot \frac{\exp(-\lambda)}{\prod_{\beta\in \Delta_+} \beta} = \frac{\sum_{w\in W} (-1)^w \exp(-w\cdot\lambda)}{\prod_{\beta\in \Delta_+} \beta}$$ The $\rho$-shift disappears entirely!
Instead of thinking of this as a limiting process, you could compare [Atiyah-Bott '65] localization in K-theory to [Atiyah-Bott '84] localization in cohomology, applied to the $\lambda$ Borel-Weil line bundle on the flag manifold or the Chern character thereof.
As for the $\sin(x)$ formula you want, that's special to the case $\lambda$ a multiple of $\rho$. Check out Why do flag manifolds, in the P(V_rho) embedding, look like products of P^1s?


text:
Convex Discrete Optimization

@inproceedings{Onn2007ConvexDO,
title={Convex Discrete Optimization},
author={Shmuel Onn},
booktitle={Encyclopedia of Optimization},
year={2007}
}
• S. Onn
• Published in Encyclopedia of Optimization 1 March 2007
• Computer Science, Mathematics
We develop an algorithmic theory of convex optimization over discrete sets. Using a combination of algebraic and geometric tools we are able to provide polynomial time algorithms for solving broad classes of convex combinatorial optimization problems and convex integer programming problems in variable dimension. We discuss some of the many applications of this theory including to quadratic programming, matroids, bin packing and cutting-stock problems, vector partitioning and clustering…
20 Citations
• K. G. Murty
• Computer Science, Mathematics
Encyclopedia of Optimization
• 2009
Introduction Limitations Outline and Overview of Main Results and Applications Terminology and Complexity Reducing Convex to Linear Discrete Optimization Edge-Directions and Zonotopes Strongly
• E. Nikolova
• Mathematics, Computer Science
APPROX-RANDOM
• 2010
This paper constructs a δ(1 + e)-approximation algorithm for the stochastic problem, which invokes the linear algorithm only a logarithmic number of times in the problem input, for any desired accuracy level e > 0.
• Computer Science, Mathematics
Math. Program.
• 2014
This algorithm combines Graver basis techniques with a proximity result, which allows for the minimization of separable convex objective functions and to use convex continuous optimization as a subroutine.
• Computer Science
Oberwolfach Reports
• 2015
MINLP is one of the most general classes of MP, which is itself a formal language used to describe optimization problems in terms of parameters, decision variables, an objective function to be optimized, and some constraints to be satisfied.
An algorithm that finds an epsilon-approximate solution to a concave mixed-integer quadratic programming problem and is polynomial in the size of the problem and in 1/\epsilon, provided that the number of integer variables and thenumber of negative eigenvalues of the objective function are fixed.
• S. Onn
• Computer Science, Mathematics
ArXiv
• 2009
We overview our recently introduced theory of n-fold integer programming which enables the polynomial time solution of fundamental linear and nonlinear integer programming problems in variable
• Mathematics
Math. Program.
• 2018
The motivation is in geometrically comparing relaxations in the context of mixed-integer linear and nonlinear optimization, with the goal of gaining algorithmic and modeling insights.
This is a survey on the computational complexity of nonlinear mixedinteger optimization. It highlights a selection of important topics, ranging from incomputability results that arise from number
This is a survey on the computational complexity of nonlinear mixedinteger optimization. It highlights a selection of important topics, ranging from incomputability results that arise from number
• Economics
J. Glob. Optim.
• 2013
A selfish bin packing problem, where each item is a selfish player and wants to minimize its cost, is considered, and it is shown that any feasible packing can be converged to a Nash Equilibrium in O(n2) steps without increasing the social cost.

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A preprocessing algorithm is presented to make certain polynomial time algorithms strongly polynomially bounded in the size of the combinatorial structure and which yields the same set of optimal solutions asw.
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Discret. Optim.
• 2008
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SIAM J. Optim.
• 1999
It is shown that when both d and p are fixed, the number of vertices of any shaped partition polytope is O(n^{d{p\choose 2}})\$ and all vertices can be produced in strongly polynomial time.
• A. Schrijver
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Wiley-Interscience series in discrete mathematics and optimization
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Introduction and Preliminaries. Problems, Algorithms, and Complexity. LINEAR ALGEBRA. Linear Algebra and Complexity. LATTICES AND LINEAR DIOPHANTINE EQUATIONS. Theory of Lattices and Linear
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CBMS-NSF regional conference series in applied mathematics
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How to Round Numbers Preliminaries and some Applications in Combinatorics Cuts and Joins Chromatic Number, Cliques and Perfect Graphs Minimizing a Submodular Function.
This paper gives a precise definition for sets of vectors, called test sets, which will include those sets described above arising in the simplex and flow algorithms, and proves that any “improvement process” which searches through a test set at each stage converges to an optimal point in a finite number of steps.
• Computer Science, Mathematics
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It is shown that one can solve every integer programming problem in polynomial time provided one can efficiently solve the directed augmentation problem, and implies that directed augmentmentation is at least as hard as optimization.
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This article shows that when both p,d are fixed, the partition problem is solvable in strongly polynomial time using O(n d(p-1)-1) arithmetic operations, which improves upon the previously known bound of O( ndp 2 ).
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An adaptive algorithm for the vector partition problem that runs in time O(q(L)ċv) and in space O(L), where q is a polynomial function, L is the input size and v is the number of vertices of the associated partition polytope, based on an output-sensitive algorithm for enumerating all vertices.
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This work finds in strongly polynomial time the minimal cost reliability ratio spanning tree of an undirected graph.


text:
# Ratio and Proportion Aptitude Questions and Answers:

#### Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Ratio and Proportion Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

1. If $$x : y = 3 : 2$$ then find the value of $$x^2 - y^2 : x^2 + y^2$$?

1. $$13 : 5$$
2. $$5 : 13$$
3. $$7 : 12$$
4. $$12 : 7$$

Answer: (b) $$5 : 13$$

Solution: Given, $$\frac{x}{y} = \frac{3}{2}$$ then $$\frac{x^2}{y^2} = \frac{3^2}{2^2} = \frac{9}{4}$$ Hence, $$\frac{x^2 - y^2}{x^2 + y^2} = \frac{9 - 4}{9 + 4}$$ $$= \frac{5}{13}$$

1. The ratio between two numbers is $$7 : 8$$. If each number is reduced by $$15$$, then the ratio becomes $$4 : 5$$. Find the numbers?

1. $$28 \ and \ 32$$
2. $$32 \ and \ 28$$
3. $$24 \ and \ 30$$
4. $$30 \ and \ 24$$

Answer: (a) $$28 \ and \ 32$$

Solution: Let ratio of two numbers = $$7k : 8k$$

by reducing 12 from each number, $$\frac{7k - 12}{8k - 12} = \frac{4}{5}$$ $$k = 4$$ Hence the numbers = $$28 : 32$$

1. A mixture of wine and water in the ratio of $$5 : 4$$. If two litres of water is added to the mixture the ratio becomes $$7 : 8$$. Find the quantity of wine and water in the mixture?

1. $$6.2 \ and \ 5.4 \ litres$$
2. $$5.4 \ and \ 6.2 \ litres$$
3. $$5.8 \ and \ 4.6 \ litres$$
4. $$4.6 \ and \ 5.8 \ litres$$

Answer: (c) $$5.8 \ and \ 4.6 \ litres$$

Solution: Let the ratio of wine and water in the mixture = $$5k : 4k$$

on adding two litres of water, $$\frac{5k}{4k + 2} = \frac{7}{8}$$ $$k = 1.16$$ Hence, quantity of wine and water $$= 5k : 4k$$ $$= 5.8 : 4.6$$

1. If $$x : y = 5 : 4$$ then find the value of $$x - y : x + y$$?

1. $$1 : 5$$
2. $$5 : 1$$
3. $$9 : 1$$
4. $$1 : 9$$

Answer: (d) $$1 : 9$$

Solution: Given, $$\frac{x}{y} = \frac{5}{4}$$ then, $$\frac{x - y}{x + y} = \frac{5 - 4}{5 + 4}$$ $$= \frac{1}{9}$$

1. A $$500 \ ml$$ mixtute of tea and water in the ratio of $$5 : 3$$. Find how much more water to be included to get a new mixture of tea and water in the ratio of $$5 : 4$$?

1. $$60.2 \ ml$$
2. $$62.5 \ ml$$
3. $$65.3 \ ml$$
4. $$68.5 \ ml$$

Answer: (b) $$62.5 \ ml$$

Solution: quantity of tea in the mixture $$\frac{500}{8} \times 5$$ $$= 312.5 \ ml$$ quantity of water in the mixture $$\frac{500}{8} \times 3$$ $$= 187.5 \ ml$$ Let $$k \ ml$$ water to be added then $$\frac{312.5}{187.5 + k} = \frac{5}{4}$$ $$k = 62.5 \ ml$$

1. Number of students in math and biology in an institute are in the ratio of $$3 : 5$$ respectively. If $$100$$ more students join math and $$50$$ more students join biology, then the ratio becomes $$2 : 3$$. Find the total students in both subjects?

1. $$100$$
2. $$125$$
3. $$175$$
4. $$200$$

Answer: (d) $$200$$

Solution: Let students in math and biology = $$3k : 5k$$

on adding $$100$$ students in math and $$50$$ students in biology, $$\frac{3k + 100}{5k + 50} = \frac{2}{3}$$ $$k = 200$$

1. The ratio of length and breadth of a rectangle is $$2 : 3$$ and by raising length and breadth the ratio becomes $$3 : 4$$. Find the ratio of previous area and present area of rectangle?

1. $$1 : 2$$
2. $$2 : 1$$
3. $$3 : 2$$
4. $$2 : 3$$

Answer: (a) $$1 : 2$$

Solution: let ratio of previous length and breadth of the rectangle = $$2A : 3A$$

then previous area of rectangle = $$2A \times 3A$$ = $$6A^2$$

now let the ratio of present length and breadth of the rectangle = $$3A : 4A$$

then present area of the rectangle = $$3A \times 4A$$ = $$12A^2$$

Hence, the ratio of previous area and present area of rectangle, $$= 6A^2 : 12A^2$$ $$= 1 : 2$$

1. $$250$$ bananas are divided among five boys, $$10$$ girls and $$15$$ women. If the ratio of one boy, one girl and one women is $$5 : 7 : 9$$, then find the share of one girl?

1. $$6.7$$
2. $$7.6$$
3. $$8.2$$
4. $$9.6$$

Answer: (b) $$7.6$$

Solution: Given, the ratio of one boy, one girl and one women = $$5 : 7 : 9$$

then overall ratio of $$5$$ boys, $$10$$ girls and $$15$$ women $$= 5 \times 5 : 7 \times 10 : 9 \times 15$$ $$= 25 : 70 : 135$$ now the share of $$10$$ girls, $$= \frac{250}{230} \times 70$$ $$= 76 \ bananas$$ Hence, the share of one girl $$= \frac{76}{10}$$ $$= 7.6 \ bananas$$

1. The ratio between two numbers is $$5 : 6$$. If each number is reduced by $$25$$, then the ratio becomes $$7 : 8$$. Find the numbers?

1. $$62.5 \ and \ 75$$
2. $$75 \ and \ 62.5$$
3. $$68.5 \ and \ 72$$
4. $$72 \ and \ 68.5$$

Answer: (a) $$62.5 \ and \ 75$$

Solution: Let ratio of two numbers = $$5k : 6k$$

each number is increased by $$25$$, then $$\frac{5k + 25}{6k + 25} = \frac{7}{8}$$ $$k = 12.5$$ Hence the numbers = $$62.5 : 75$$

1. A mixture of milk and water in the ratio of $$4 : 3$$. If $$5$$ litres of water is added to the mixture the ratio becomes $$5 : 4$$. Find the quantity of water in the mixture?

1. $$50 \ litres$$
2. $$75 \ litres$$
3. $$80 \ litres$$
4. $$85 \ litres$$

Answer: (b) $$75 \ litres$$

Solution: Let the ratio of milk and water in the mixture = $$4k : 3k$$

on adding $$5$$ litres of water, $$\frac{4k}{3k + 5} = \frac{5}{4}$$ $$k = 25$$ Hence, quantity of water $$= 3k = 75 \ litres$$


text:
## Intermediate Algebra (12th Edition)

The set of irrational numbers and the set of rational numbers are mutually exclusive (there is nothing common to both of them.) Hence, there is no number, $x$, that is both irrational and rational.


text:
# Quotient-pullbackable equals inner

This article gives a proof/explanation of the equivalence of multiple definitions for the term inner automorphism
View a complete list of pages giving proofs of equivalence of definitions
This fact is related to: Extensible automorphisms problem
View other facts related to Extensible automorphisms problemView terms related to Extensible automorphisms problem |

## Statement

The following are equivalent for an automorphism $\sigma$ of a group $G$:

1. The automorphism is a quotient-pullbackable automorphism: For any homomorphism $\rho:H \to G$, there is an automorphism $\varphi$ of $H$, $\rho \circ \varphi = \sigma \circ \rho$.
2. The automorphism is an inner automorphism.

## Definitions used

### Quotient-pullbackable automorphism

An automorphism $\sigma$ of a group $G$ is termed quotient-pullbackable if given any surjective homomorphism $\rho: H \to G$ there is an automorphism $\varphi$ of $H$ such that $\rho \circ \varphi = \sigma \circ \rho$.

### Inner automorphism

Further information: Inner automorphism

An automorphism $\sigma$ of a group $G$ is termed an inner automorphism if there exists $g \in G$ such that $\sigma = c_g = x \mapsto gxg^{-1}$.


text:
Find 5 the 1 eaje 3 SCalcET8 Guriaoz& obtained rotating the curve about the X-axisNeed {Readll1 Joinu]

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Find 5 the 1 eaje 3 SCalcET8 Guriaoz& obtained rotating the curve about the X-axis Need { Readll 1 Joinu]

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# If x is an integer and x^2 < 37, what is the greatest possible value o

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08 Mar 2019, 03:19
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If x is an integer and x^2 < 37, what is the greatest possible value of x minus the least possible value of x?

A. 5
B. 6
C. 10
D. 12
E. 36

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Re: If x is an integer and x^2 < 37, what is the greatest possible value o  [#permalink]

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08 Mar 2019, 04:40
Bunuel wrote:
If x is an integer and x^2 < 37, what is the greatest possible value of x minus the least possible value of x?

A. 5
B. 6
C. 10
D. 12
E. 36

If x is an integer and x^2 < 37

the greatest possible value of x will be 6

the least possible value of x will be -6

the greatest possible value of x minus the least possible value of x 6+6 = 12

Hence D
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Re: If x is an integer and x^2 < 37, what is the greatest possible value o  [#permalink]

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08 Mar 2019, 04:51

-6 & 6 are the least and greatest possible values of x. Hence difference of these two values is 12.
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Re: If x is an integer and x^2 < 37, what is the greatest possible value o  [#permalink]

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08 Mar 2019, 04:54
Bunuel wrote:
If x is an integer and x^2 < 37, what is the greatest possible value of x minus the least possible value of x?

A. 5
B. 6
C. 10
D. 12
E. 36

$$x^2$$ yields positive values all time. we must consider both positive and negative values of x.

x ( highest ) = 6
x ( lowest) = -6.

greatest possible value = 6 - (-6) = 12.

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Re: If x is an integer and x^2 < 37, what is the greatest possible value o  [#permalink]

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08 Mar 2019, 05:03
Bunuel wrote:
If x is an integer and x^2 < 37, what is the greatest possible value of x minus the least possible value of x?

A. 5
B. 6
C. 10
D. 12
E. 36

greatest value = 6
and least value = -6
so 6-(-6) = 12
IMO D
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If x is an integer and x^2 < 37, what is the greatest possible value o  [#permalink]

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08 Mar 2019, 07:34
since x2<37
greatest positive value of x =6
Least value = -6
6-(-6)=12

D
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Re: If x is an integer and x^2 < 37, what is the greatest possible value o  [#permalink]

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10 Mar 2019, 07:55
1
Question:
If x is an integer and x^2 < 37, what is the greatest possible value of x minus the least possible value of x?

A. 5
B. 6
C. 10
D. 12
E. 36

36 is the square of 6
Therefore, the range lies between -6<x<6.
--> 6-(-6)=6+6=12(Ans) D
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Re: If x is an integer and x^2 < 37, what is the greatest possible value o  [#permalink]

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11 Mar 2019, 08:10
Bunuel wrote:
If x is an integer and x^2 < 37, what is the greatest possible value of x minus the least possible value of x?

A. 5
B. 6
C. 10
D. 12
E. 36

The largest integer x could be is 6, and the smallest it could be is -6. So the difference is 6 - (-6) = 12.

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Re: If x is an integer and x^2 < 37, what is the greatest possible value o   [#permalink] 11 Mar 2019, 08:10
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# 9. Depletion¶

When materials in a system are subject to irradiation over a long period of time, nuclides within the material will transmute due to nuclear reactions as well as spontaneous radioactive decay. The time-dependent process by which nuclides transmute under irradiation is known as depletion or burnup. To accurately analyze nuclear systems, it is often necessary to predict how the composition of materials will change since this change results in a corresponding change in the solution of the transport equation. The equation that governs the transmutation and decay of nuclides inside of an irradiated environment can be written as

\begin{aligned} \frac{dN_i(t)}{dt} = &\sum\limits_j \underbrace{\left [ \underbrace{f_{j \rightarrow i} \int_0^\infty dE \; \sigma_j (E, t) \phi(E,t)}_\text{transmutation} + \underbrace{\lambda_{j\rightarrow i}}_\text{decay} \right ] N_j(t)}_{\text{Production of nuclide }i\text{ from nuclide }j} \\ &- \underbrace{\left [\underbrace{\int_0^\infty dE \; \sigma_i (E,t) \phi(E,t)}_\text{transmutation} + \underbrace{\sum\limits_j \lambda_{i\rightarrow j}}_\text{decay} \right ] N_i(t)}_{\text{Loss of nuclide }i} \end{aligned}

where $$N_i$$ is the density of nuclide $$i$$ at time $$t$$, $$\sigma_i$$ is the transmutation cross section for nuclide $$i$$ at energy $$E$$, $$f_{j \rightarrow i}$$ is the fraction of transmutation reactions in nuclide $$j$$ that produce nuclide $$i$$, and $$\lambda_{j \rightarrow i}$$ is the decay constant for decay modes in nuclide $$j$$ that produce nuclide $$i$$. Note that we have not included the spatial dependence of the flux or cross sections. As one can see, the equation simply states that the rate of change of $$N_i$$ is equal to the production rate minus the loss rate. Because the equation for nuclide $$i$$ depends on the nuclide density for possibly many other nuclides, we have a system of first-order differential equations. To form a proper initial value problem, we also need the nuclide densities at time 0:

$N_i(0) = N_{i,0}.$

These equations can be written more compactly in matrix notation as

(1)$\frac{d\mathbf{n}}{dt} = \mathbf{A}(\mathbf{n},t)\mathbf{n}, \quad \mathbf{n}(0) = \mathbf{n}_0$

where $$\mathbf{n} \in \mathbb{R}^n$$ is the nuclide density vector, $$\mathbf{A}(\mathbf{n},t) \in \mathbb{R}^{n\times n}$$ is the burnup matrix containing the decay and transmutation coefficients, and $$\mathbf{n}_0$$ is the initial density vector. Note that the burnup matrix depends on $$\mathbf{n}$$ because the solution to the transport equation depends on the nuclide densities.

## 9.1. Numerical Integration¶

A variety of numerical methods exist for solving Eq. (1). The simplest such method, known as the “predictor” method, is to divide the overall time interval of interest $$[0,t]$$ into smaller timesteps over which it is assumed that the burnup matrix is constant. Let $$t \in [t_i, t_i + h]$$ be one such timestep. Over the timestep, the solution to Eq. (1) can be written analytically using the matrix exponential

$\mathbf{A}_i = \mathbf{A}(\mathbf{n}_i, t_i) \\ \mathbf{n}_{i+1} = e^{\mathbf{A}_i h} \mathbf{n}_i$

where $$\mathbf{n}_i \equiv \mathbf{n}(t_i)$$. The exponential of a matrix $$\mathbf{X}$$ is defined by the power series expansion

$e^{\mathbf{X}} = \sum\limits_{k=0}^\infty \frac{1}{k!} \left ( \mathbf{X} \right )^k$

where $$\mathbf{X}^0 = \mathbf{I}$$. A series of so-called predictor-corrector methods that use multiple stages offer improved accuracy over the predictor method. The simplest of these methods, the CE/CM algorithm, is defined as

$\mathbf{n}_{i+1/2} = e^{\frac{h}{2}\mathbf{A}(\mathbf{n}_i, t_i)} \mathbf{n}_i \\ \mathbf{n}_{i+1} = e^{h \mathbf{A}(\mathbf{n}_{i+1/2},t_{i+1/2})} \mathbf{n}_i$

Here, the value of $$\mathbf{n}$$ at the midpoint is estimated using $$\mathbf{A}$$ evaluated at the beginning of the timestep. Then, $$\mathbf{A}$$ is evaluated using the densities at the midpoint and used to integrate over the entire timestep.

Our aim here is not to exhaustively describe all integration methods but rather to give a few examples that elucidate the main considerations one must take into account when choosing a method. Generally, there is a tradeoff between the accuracy of the method and its computational expense. In the case of transport-coupled depletion, the expense is driven almost entirely by the time to compute a transport solution, i.e., to evaluate $$\mathbf{A}$$ for a given $$\mathbf{n}$$. Thus, the cost of a method scales with the number of $$\mathbf{A}$$ evaluations that are performed per timestep. On the other hand, methods that require more evaluations generally achieve higher accuracy. The predictor method only requires one evaluation and its error converges as $$\mathcal{O}(h)$$. The CE/CM method requires two evaluations and is thus twice as expensive as the predictor method, but achieves an error of $$\mathcal{O}(h^2)$$. An exhaustive description of time integration methods and their merits can be found in the thesis of Colin Josey.

OpenMC does not rely on a single time integration method but rather has several classes that implement different algorithms. For example, the openmc.deplete.PredictorIntegrator class implements the predictor method, and the openmc.deplete.CECMIntegrator class implements the CE/CM method. A full list of the integrator classes available can be found in the documentation for the openmc.deplete module.

## 9.2. Matrix Exponential¶

As we saw in the previous section, numerically integrating Eq. (1) requires evaluating one or more matrix exponentials. OpenMC uses the Chebyshev rational approximation method (CRAM), which was introduced in a series of papers by Pusa (1, 2), to evaluate matrix exponentials. In particular, OpenMC utilizes an incomplete partial fraction (IPF) form of CRAM that provides a good balance of numerical stability and efficiency. In this representation the matrix exponential is approximated as

$e^{\mathbf{A}t} \approx \alpha_0 \prod\limits_{\ell=1}^{k/2} \left ( \mathbf{I} + 2 \text{Re} \left ( \widetilde{\alpha}_\ell \left (\mathbf{A}t - \theta_\ell \mathbf{I} \right )^{-1} \right ) \right )$

where $$k$$ is the order of the approximation and $$\alpha_0$$, $$\widetilde{\alpha}_\ell$$, and $$\theta_\ell$$ are coefficients that have been tabulated for orders up to $$k=48$$. Rather than computing the full approximation and then multiplying it by a vector, the following algorithm is used to incrementally apply the terms within the product (note that the original description of the algorithm presented by Pusa contains a typo):

1. $$\mathbf{n} \gets \mathbf{n_0}$$

2. For $$\ell = 1, 2, \dots, k/2$$

• $$\mathbf{n} \gets \mathbf{n} + 2\text{Re}(\widetilde{\alpha}_\ell (\mathbf{A}t - \theta_\ell)^{-1})\mathbf{n}$$

3. $$\mathbf{n} \gets \alpha_0 \mathbf{n}$$

The $$k$$th order approximation for CRAM requires solving $$k/2$$ sparse linear systems. OpenMC relies on functionality from scipy.sparse.linalg for solving the linear systems.

## 9.3. Data Considerations¶

In principle, solving Eq. (1) using CRAM is fairly simple: just construct the burnup matrix at various times and solve a set of sparse linear systems. However, constructing the burnup matrix itself involves not only solving the transport equation to estimate transmutation reaction rates (in the case of transport-coupled depletion) or to obtain microscopic cross sections (in the case of transport-independent depletion), but also a series of choices about what data to include. In OpenMC, the burnup matrix is constructed based on data inside of a depletion chain file, which includes fundamental data gathered from ENDF incident neutron, decay, and fission product yield sublibraries. For each nuclide, this file includes:

• What transmutation reactions are possible, their Q values, and their products;

• If a nuclide is not stable, what decay modes are possible, their branching ratios, and their products; and

• If a nuclide is fissionable, the fission products yields at any number of incident neutron energies.

### 9.3.1. Transmutation Reactions¶

In transport-coupled depletion, OpenMC will setup tallies in a problem based on what transmutation reactions are available in a depletion chain file, so any arbitrary number of transmutation reactions can be tracked. In transport-independent depletion, OpenMC will calculate reaction rates for every reaction that is present in both the available cross sections and the depletion chain file. The pregenerated chain files that are available on https://openmc.org include the following transmutation reactions: fission, (n,$$\gamma$$), (n,2n), (n,3n), (n,4n), (n,p), and (n,$$\alpha$$).

### 9.3.2. Capture Branching Ratios¶

Some (n,$$\gamma$$) reactions may result in a product being in either the ground or a metastable state. The most well-known example is capture in Am241, which can produce either Am242 or Am242m. Because the metastable state of Am242m has a significantly longer half-life than the ground state, it is important to accurately model the branching of the capture reaction in Am241. This is complicated by the fact that the branching ratio may depend on the incident neutron energy causing capture.

OpenMC’s transport solver does not currently allow energy-dependent capture branching ratios. However, the depletion chain file does allow a transmutation reaction to be listed multiple times with different branching ratios resulting in different products. Spectrum-averaged capture branching ratios have been computed in LWR and SFR spectra and are available at https://openmc.org/depletion-chains.

### 9.3.3. Fission Product Yields¶

Fission product yields (FPY) are also energy-dependent in general. ENDF fission product yield sublibraries typically include yields tabulated at 2 or 3 energies. It is an open question as to what the best way to handle this energy dependence is. OpenMC includes three methods for treating the energy dependence of FPY:

1. Use FPY data corresponding to a specified energy. This is used by default in both transport-coupled and transport-independent depletion.

2. Tally fission rates above and below a specified cutoff energy. Assume that all fissions below the cutoff energy correspond to thermal FPY data and all fission above the cutoff energy correspond to fast FPY data. Only applicable to transport-coupled depletion.

3. Compute the average energy at which fission events occur and use an effective FPY by linearly interpolating between FPY provided at neighboring energies. Only applicable to transport-coupled depletion.

The method for transport-coupled depletion can be selected through the fission_yield_mode argument to the openmc.deplete.CoupledOperator constructor.

### 9.3.4. Power Normalization¶

In transport-coupled depletion, the reaction rates provided OpenMC are given in units of reactions per source particle. For depletion, it is necessary to compute an absolute reaction rate in reactions per second. To do so, the reaction rates are normalized based on a specified power. A complete description of how this normalization can be performed is described in Normalization of Tally Results. Here, we simply note that the main depletion class, openmc.deplete.CoupledOperator, allows the user to choose one of two methods for estimating the heating rate, including:

1. Using fixed Q values from a depletion chain file (useful for comparisons to other codes that use fixed Q values), or

2. Using the heating or heating-local scores to obtain an nuclide- and energy-dependent estimate of the true heating rate.

The method for normalization can be chosen through the normalization_mode argument to the openmc.deplete.CoupledOperator class.


text:
0
Research Papers

# Carotid Bifurcation Hemodynamics in Older Adults: Effect of Measured Versus Assumed Flow Waveform

[+] Author and Article Information
Yiemeng Hoi

Department of Mechanical and Industrial Engineering, Biomedical Simulation Laboratory, University queryof Toronto, Toronto, ON, Canada M5S 3G8

Bruce A. Wasserman

Department of Radiology, Johns Hopkins University, Baltimore, MD 21210

Edward G. Lakatta

Laboratory of Cardiovascular Science, Intramural Research Program, National Institute on Aging, NIH, Baltimore, MD 21224

David A. Steinman1

Department of Mechanical and Industrial Engineering, Biomedical Simulation Laboratory, University of Toronto, Toronto, ON, Canada M5S 3G8steinman@mie.utoronto.ca

1

Corresponding author.

J Biomech Eng 132(7), 071006 (May 26, 2010) (6 pages) doi:10.1115/1.4001265 History: Received October 26, 2009; Revised February 05, 2010; Posted February 15, 2010; Published May 26, 2010; Online May 26, 2010

## Abstract

Recent work has illuminated differences in carotid artery blood flow rate dynamics of older versus young adults. To what degree flow waveform shape, and indeed the use of measured versus assumed flow rates, affects the simulated hemodynamics of older adult carotid bifurcations has not been elucidated. Image-based computational fluid dynamics models of $N=9$ normal, older adult carotid bifurcations were reconstructed from magnetic resonance angiography. Subject-specific hemodynamics were computed by imposing each individual’s inlet and outlet flow rates measured by cine phase-contrast magnetic resonance imaging or by imposing characteristic young and older adult flow waveform shapes adjusted to cycle-averaged flow rates measured or allometrically scaled to the inlet and outlet areas. Despite appreciable differences in the measured versus assumed flow conditions, the locations and extents of low wall shear stress and elevated relative residence time were broadly consistent; however, the extent of elevated oscillatory shear index was substantially underestimated, more by the use of assumed cycle-averaged flow rates than the assumed flow waveform shape. For studies of individual vessels, use of a characteristic flow waveform shape is likely sufficient, with some benefit offered by scaling to measured cycle-averaged flow rates. For larger-scale studies of many vessels, ranking of cases according to presumed hemodynamic or geometric risk is robust to the assumed flow conditions.

<>

## Figures

Figure 1

Measured flow waveforms at the CCA for all nine subjects (dotted lines) compared with characteristic young adult (green/light solid line) and older adult (blue/dark solid line) waveforms. One potential outlier (subject 4), exhibiting a highly damped waveform, is shown as a magenta/dashed line.

Figure 2

Color-coded distributions of OSI, for all nine subjects, for the various waveform boundary conditions indicated at the top row. Black contour lines on each surface identify the area exposed to OSI above its 80th percentile value, as described in Sec. 2. Geometries are clipped at the CCA5, ICA10, and ECA5 locations used to scale the flow rates. CCA3, ICA5, and ECA2 locations, used to bind the surface for disturbed flow calculations, are shown in dark gray.

Figure 3

SA exposed to various disturbed flow indicators (WSS, RRT, and OSI), derived from assumed-waveform models and compared against the corresponding SA from the subject-specific-waveform models. Colored lines are linear regressions through the respectively colored data points. The black dotted line represents the line of unity.

Figure 4

SA exposed to various disturbed flow indicators (WSS, RRT, and OSI), as a function of the waveform boundary condition used. The dashed lines separate SA (vertical axis) into lower, middle, and upper tertiles according to the distributions of exposure levels found by Lee (8). Symbols denote the subject number, and green, yellow, and red are used to highlight Truth cases falling into the lower, middle, and upper tertiles, respectively.

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text:
# Programmatically generated mesh vertex colors not showing up in Unity?

I'm sorry if this is an obvious thing to solve but I just can't figure it out...did I miss something...I've generated a mesh and during vertices generation I've also set its colors (and colors32 just in case) by reading those values from a ply file, and afterwards I've added a number of vertex color shaders but none of them will show the vertex point colors in play mode, it's either solid black or just like I added no material to the mesh (which I've tried doing manually in play mode too).

Welcome, it's hard to tell with you not showing any code or explaining further what you've done :)

If not then here's a quick example:

Shader "Unlit Vertex Color"
{
Properties
{
_MainTex ("Texture", 2D) = "white" {}
}
{
Tags { "RenderType"="Opaque" }

Pass
{
CGPROGRAM
#pragma vertex vert
#pragma fragment frag
// make fog work
#pragma multi_compile_fog

#include "UnityCG.cginc"

struct appdata
{
float4 vertex : POSITION;
float2 uv : TEXCOORD0;
float4 color : COLOR;
};

struct v2f
{
float4 color : COLOR;
float2 uv : TEXCOORD0;
UNITY_FOG_COORDS(1)
float4 vertex : SV_POSITION;
};

sampler2D _MainTex;
float4 _MainTex_ST;
fixed _Brightness;
fixed _Gamma;

v2f vert (appdata v)
{
v2f o;
o.vertex = UnityObjectToClipPos(v.vertex);
o.uv = TRANSFORM_TEX(v.uv, _MainTex);
o.color = v.color;
o.color.w = 1.0;
//UNITY_TRANSFER_FOG(o,o.vertex);
return o;
}

fixed4 frag (v2f i) : SV_Target
{
// sample the texture
fixed4 col = tex2D(_MainTex, i.uv);
col *= i.color;
// apply fog
UNITY_APPLY_FOG(i.fogCoord, col);
return col;
}
ENDCG
}
}
}

I just pass COLOR semantic to where necessary and blend the result with the pixel.

You can get started writing your own by templating some of the shaders from the builtin ones:

• For colors I think you should use whichever notation you prefer, e.g. Color is natural since it maps from 0.0 to 1.0, then later, if you notice performance problems by profiling you will want to adjust places in your code to emit Color32 instead. Keep in mind they both have implicit operators to each other so be careful when optimizing. Now for the shaders I learned the hard way: the 'builtin shaders' link I've mentioned. Here's a tip, use Notepad++ folder search on them, it'll be less intimidating and you'll get a better understanding overall on what is used where. – aybe Feb 23 '19 at 22:47
• I do the following by code, get my mesh, add MeshFilter component and set its mesh, add MeshRenderer component and set its material; and obviously if I have a texture I set material.mainTexture. – aybe Feb 23 '19 at 22:52


text:
# weighted.env: Weighted response envelope estimator In Renvlp: Computing Envelope Estimators

 weighted.env R Documentation

## Weighted response envelope estimator

### Description

Compute the weighted response envelope estimator with weights computed from BIC.

### Usage

weighted.env(X, Y, bstrpNum = 0, min.u = 1,
max.u = ncol(as.matrix(Y)), boot.resi = "full")

### Arguments

 X Predictors. An n by p matrix, p is the number of predictors. The predictors can be univariate or multivariate, discrete or continuous. Y Multivariate responses. An n by r matrix, r is the number of responses and n is number of observations. The responses must be continuous variables. bstrpNum Number of bootstrap samples. A positive integer. min.u Lower bound of the range of u to compute bootstrap error. A postive integer between 1 and p. This argument is relevant only when bstrpNum>0. max.u Upper bound of the range of u to compute bootstrap error. A postive integer between 1 and p. This argument is relevant only when bstrpNum>0. boot.resi A string that can be "full" or "weighted" indicating the model from which the residuals are calculated. If the input is "full", then the residuals are obtained using the standard estimators; and if the input is "weighted", then the residuals are obtained using the weighted envelope estimators. This argument is for computing residuals in residual bootstrap, and it is relevant only when bstrpNum>0.

### Details

This function computes the weighted envelope estimator in a standard multivariate linear regression. And the weighted envelope estimator takes the form

\hat{β}_{w}=∑_{j=1}^{r}w_{j}\hat{β}_{j},

where \hat{β}_{j} is the envelope estimator of β with u=j and w_{j}'s are the weights computed from BIC values

w_{j}=\frac{\exp(-b_{j})}{∑_{k=1}^{r}\exp(-b_{k})},

where b_{j} is the BIC criterion evaluated at the envelope estimator \hat{β}_{j}. For details, see Eck and Cook (2017).

The variation of the weighted envelope estimator is estimated by residual bootstrap. The user can specify the range for bootstrap u=(min.u, max.u), if the weights outside of the range are small.

### Value

The output is a list that contains the following components:

 beta The weighted envelope estimator of the regression coefficients. mu The weighted estimated intercept. Sigma The weighted envelope estimator of the error covariance matrix. w Weights computed based on BIC. loglik The log likelihood function computed with weighted envelope estimator. n The number of observations in the data. bootse The standard error for elements in beta computed by residual bootstrap. This output is available only when bstrpNum>0. ratios The boostrap standard error ratio of the standard multivariate linear regression estimator over the weighted envelope estimator for each element in beta. This output is available only when bstrpNum>0. bic_select A table that lists how many times BIC selected each candidate dimension. If BIC never selects a dimension, this dimension does not appear on the table. This output is available only when bstrpNum>0.

### References

Eck, D. J. and Cook, R. D. (2017). Weighted Envelope Estimation to Handle Variability in Model Selection. Biometrika. To appear.

### Examples

data(wheatprotein)
X <- wheatprotein[, 8]
Y <- wheatprotein[, 1:6]
m <- weighted.env(X, Y)
m$w m$beta

## Not run: m2 <- weighted.env(X, Y, bstrpNum = 100, min.u = 1, max.u = 6, boot.resi = "full")
## Not run: m2$bic_select ## Not run: m2$bootse

Renvlp documentation built on Aug. 8, 2022, 1:06 a.m.


text:
# Find all $x \in \mathbb{Z}_{501}$ for which $51x \equiv 36$

I am stuck with one problem from my discrete math class and don't know how to solve it. I will be grateful for any help!

Find all $x \in \mathbb{Z}_{501}$ for which $51x \equiv 36$, where the multiplication is in $\mathbb{Z}_{501}$.

I started solving it like this:
\begin{align} 51x & \equiv 36 \pmod{501}\\ 51x & \equiv 36 + k501\\ 51x + 501y & = 36 \end{align}

After this, I found $\gcd(51,501)$, which is $3$:
\begin{align} 501 & = 9 \cdot 51 + 42\\ 51 & = 1 \cdot 42 + 9\\ 42 & = 4 \cdot 9 + 6\\ 9 & = 1 \cdot 6 + 3\\ 6 & = 2 \cdot 3 + 0 \end{align}

After this, using back-substitution:
\begin{align} 3 & = -6 \cdot 501 + 59 \cdot 51\\ 36 & = 708 \cdot 51 - 72 \cdot 501 \end{align}

Then I divided equation by the $\gcd$ and solved $17x+167y = 0$.

So my final answers are $x = 708 + 167k$ and $y= -72-17k$ (In our case we don't need $y$, though.)

Answers from my book are $x_1 = 40$, $x_2 = 207$ and $x_3 = 374$, and I don't know how can I get them.

What you have done thus far is correct. Observe that $$708 \equiv 40 \pmod{167}$$ since $$708 = 4 \cdot 167 + 40$$ The distinct solutions are all the integers in $\{0, 1, 2, 3, \ldots, 500\}$, the set of residues modulo $501$, that satisfy the congruence $x \equiv 40 \pmod{167}$. They are \begin{align*} x_1 & \equiv 40 + 0 \cdot 167 \equiv 40 \pmod{501}\\ x_2 & \equiv 40 + 1 \cdot 167 \equiv 207 \pmod{501}\\ x_3 & \equiv 40 + 2 \cdot 167 \equiv 374 \pmod{501} \end{align*}

There is an algorithm for this kind of equation. I usually reformulate: $$51 x \equiv 36 \pmod{501} \iff \\ (51 x) \bmod 501 = 36 \iff \\ 51 x = 501 q + 36 \iff \\ 51 x - 501 q = 36 \quad (*)$$ where $x\in \mathbb{Z}_{501}$ and $q \in \mathbb{Z}$.

The last equation $(*)$ can be interpreted as a linear Diophantine equation, where we limit the solutions for $x$ from $\mathbb{Z}$ to $\mathbb{Z}_{501}$.

Number of solutions:

If $g = \gcd(51, 501)$ divides $36$, we have infinite many solutions $(x,q) \in \mathbb{Z}^2$, otherwise none. We have $g=3 \mid 36$, so we got the first case.

Solution of the homogeneous equation:

The reduced equation (divided by $g$) is $$17 x - 167 q = 12$$ The homogeneous equation is $$17 x - 167 q = 0 \quad (**) \iff \\ 17 x = 167 q$$ $\gcd(17,167)=1$ implies $17\mid q$ and $167\mid x$. So the homogeneous equation $(**)$ is solved by $$\{ (167 t, 17 t) \mid t \in \mathbb{Z} \}$$

Finding a particular solution:

The extended Euclidean algorithm for $(51,-501)$ gives numbers $s=59, t=6$ (and $g=3$) such that $$51 s - 501 t = g = 3 \iff \\ 51 (12\cdot 59) - 501 (12\cdot 6) = 12\cdot 3 = 36$$ so a particular solution of $(*)$ is $(x_p, q_p) = (708, 72)$.

The total solutions are $$\{ (708 + 167 t, 72 + 51 t) \mid t \in \mathbb{Z} \}$$

Limiting to $\mathbb{Z}_{501}$:

We have $$708 + 167 t \ge 0 \iff t \ge -708/167 = -4.23\dotsb$$ and $$708 + 167 t \le 500 \iff t \le -208/167 = -1.24\dotsb$$ so $t\in \{ -2,-3,-4\}$ and $x \in \{ 40, 207, 374 \}$.

Note $\,708\bmod{\!167} = 40\$ thus $\,708+167k = 40+167n.\$ Next, to lift this solution from modulus $167$ to $\,\color{#c00}3\cdot 167 = 501$ we divide $n$ by $\color{#c00}3,\,$ so $\, n = i + \color{#c00}3j,\,$ which yields

\begin{align}{\rm mod}\,\ \color{#c00}3\!\cdot\! 167\!:\,\ 40 + 167n &\equiv 40+167(i+\color{#c00}3j), \ {\rm for} \ \,0\le i < 3\\ &\equiv 40+167\,i,\quad {\rm for}\ \ \ i=0,\,1\,,2\\ &\equiv 40,\ 40\!+\!167\!\cdot\! 1,\ 40\!+167\!\cdot\!2\\ &\equiv 40,\ 207,\ 374 \end{align}