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# Pre-Algebra : Area of a Triangle
## Example Questions
### Example Question #91 : Geometry
Find the area of a triangle in yards if the side lengths of the triangle are 36 inches, 3 feet, and 1 yard.
Explanation:
First convert all the dimensions to yards. There are 3 feet in 1 yard, and 12 inches in 1 foot.
The side lengths of the triangle are 1 yard by 1 yard by 1 yard. We can then see that this is an equilateral triangle.
Write the formula for the area of an equilateral triangle.
Substitute the side.
### Example Question #99 : Geometry
A right triangle has a base of 6 and a height of 10. What is the area of the triangle?
Explanation:
Write the formula for the area of a triangle. The area is half the product of the base and the height.
Substitute the base and height into the equation.
The area of the triangle is 30.
### Example Question #100 : Geometry
Solve for the area of a right triangle if the hypotenuse is and the height is .
Explanation:
Write the area formula for triangles.
The base is unknown. To find this dimension, use the Pythagorean Theorem.
Substitute the hypotenuse into and the height into . Solve for the base.
Isolate the variable by subtracting on both sides of the equation.
Simplify the squares on the right side of the equation.
Subtract the left side.
Square root both sides of the equation to eliminate the square root.
We will only consider the positive root because length cannot be a negative value.
The base is 8. Substitute the base and height to find the area.
### Example Question #21 : Area Of A Triangle
If the base of a triangle is and the height is , what is the area of the triangle?
Explanation:
Write the formula for finding the area of a triangle.
Substitute the base and the height.
Multiply the numerator.
Divide by two.
The area is:
### Example Question #21 : Area Of A Triangle
Find the area with a base of and a height of .
Explanation:
Write the formula for the area of a triangle.
Substitute the base and the height.
Cancel out the two in the numerator and denominator.
Multiply both terms together.
### Example Question #51 : Geometry
What is the area (in square feet) of a triangle with a base of feet and a height of feet?
Explanation:
The area of a triangle is found by multiplying the base times the height, divided by
### Example Question #21 : Area Of A Triangle
What is the area of a triangle with a base of and a height of ?
Explanation:
The formula for the area of a triangle is .
Plug the given values into the formula to solve:
### Example Question #22 : Area Of A Triangle
The base of a triangle is inches, and the height of the triangle is inches. What is the area of the triangle?
Explanation:
To find the area of a triangle, multiply the base by the height, and divide by two.
### Example Question #24 : Area Of A Triangle
The base of a triangle is inches, and the height is inches. What is its area?
Explanation:
To find the area of a triangle, multiply the base by the height, and divide by two.
### Example Question #23 : Area Of A Triangle
The base of a triangle is inches, and the height of the triangle is inches. What is the area of the triangle?
square inches
square inches
square inches
square inches
square inches | HuggingFaceTB/finemath | |
Intermediate Algebra
# 11.1Distance and Midpoint Formulas; Circles
Intermediate Algebra11.1 Distance and Midpoint Formulas; Circles
## Learning Objectives
By the end of this section, you will be able to:
• Use the Distance Formula
• Use the Midpoint Formula
• Write the equation of a circle in standard form
• Graph a circle
## Be Prepared 11.1
Before you get started, take this readiness quiz.
1. Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches.
If you missed this problem, review Example 2.34.
2. Factor: $x2−18x+81.x2−18x+81.$
If you missed this problem, review Example 6.24.
3. Solve by completing the square: $x2−12x−12=0.x2−12x−12=0.$
If you missed this problem, review Example 9.22.
In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.
There are four conics—the circle, parabola, ellipse, and hyperbola. The next figure shows how the plane intersecting the double cone results in each curve.
Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.
## Use the Distance Formula
We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.
Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.
## Example 11.1
Use the rectangular coordinate system to find the distance between the points $(6,4)(6,4)$ and $(2,1).(2,1).$
## Try It 11.1
Use the rectangular coordinate system to find the distance between the points $(6,1)(6,1)$ and $(2,−2).(2,−2).$
## Try It 11.2
Use the rectangular coordinate system to find the distance between the points $(5,3)(5,3)$ and $(−3,−3).(−3,−3).$
The method we used in the last example leads us to the formula to find the distance between the two points $(x1,y1)(x1,y1)$ and $(x2,y2).(x2,y2).$
When we found the length of the horizontal leg we subtracted $6−26−2$ which is $x2−x1.x2−x1.$
When we found the length of the vertical leg we subtracted $4−14−1$ which is $y2−y1.y2−y1.$
If the triangle had been in a different position, we may have subtracted $x1−x2x1−x2$ or $y1−y2.y1−y2.$ The expressions $x2−x1x2−x1$ and $x1−x2x1−x2$ vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say $|x2−x1||x2−x1|$ and $|y2−y1|.|y2−y1|.$
In the Pythagorean Theorem, we substitute the general expressions $|x2−x1||x2−x1|$ and $|y2−y1||y2−y1|$ rather than the numbers.
$a2+b2=c2 Substitute in the values.(|x2−x1|)2+(|y2−y1|)2=d2 Squaring the expressions makes thempositive, so we eliminate the absolute valuebars.(x2−x1)2+(y2−y1)2=d2 Use the Square Root Property.d=±(x2−x1)2+(y2−y1)2 Distance is positive, so eliminate the negativevalue.d=(x2−x1)2+(y2−y1)2 a2+b2=c2 Substitute in the values.(|x2−x1|)2+(|y2−y1|)2=d2 Squaring the expressions makes thempositive, so we eliminate the absolute valuebars.(x2−x1)2+(y2−y1)2=d2 Use the Square Root Property.d=±(x2−x1)2+(y2−y1)2 Distance is positive, so eliminate the negativevalue.d=(x2−x1)2+(y2−y1)2$
This is the Distance Formula we use to find the distance d between the two points $(x1,y1)(x1,y1)$ and $(x2,y2).(x2,y2).$
## Distance Formula
The distance d between the two points $(x1,y1)(x1,y1)$ and $(x2,y2)(x2,y2)$ is
$d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2$
## Example 11.2
Use the Distance Formula to find the distance between the points $(−5,−3)(−5,−3)$ and $(7,2).(7,2).$
## Try It 11.3
Use the Distance Formula to find the distance between the points $(−4,−5)(−4,−5)$ and $(5,7).(5,7).$
## Try It 11.4
Use the Distance Formula to find the distance between the points $(−2,−5)(−2,−5)$ and $(−14,−10).(−14,−10).$
## Example 11.3
Use the Distance Formula to find the distance between the points $(10,−4)(10,−4)$ and $(−1,5).(−1,5).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
## Try It 11.5
Use the Distance Formula to find the distance between the points $(−4,−5)(−4,−5)$ and $(3,4).(3,4).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
## Try It 11.6
Use the Distance Formula to find the distance between the points $(−2,−5)(−2,−5)$ and $(−3,−4).(−3,−4).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
## Use the Midpoint Formula
It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.
## Midpoint Formula
The midpoint of the line segment whose endpoints are the two points $(x1,y1)(x1,y1)$ and $(x2,y2)(x2,y2)$ is
$(x1+x22,y1+y22)(x1+x22,y1+y22)$
To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.
## Example 11.4
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−5,−4)(−5,−4)$ and $(7,2).(7,2).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
## Try It 11.7
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−3,−5)(−3,−5)$ and $(5,7).(5,7).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
## Try It 11.8
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−2,−5)(−2,−5)$ and $(6,−1).(6,−1).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
Both the Distance Formula and the Midpoint Formula depend on two points, $(x1,y1)(x1,y1)$ and $(x2,y2).(x2,y2).$ It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, is may be easier to remember the formulas.
## Write the Equation of a Circle in Standard Form
As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.
We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, $(h,k),(h,k),$ and the fixed distance is called the radius, r, of the circle.
## Circle
A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, $(h,k),(h,k),$ and the fixed distance is called the radius, r, of the circle.
We look at a circle in the rectangular coordinate system.The radius is the distance from the center, $(h,k),(h,k),$ to apoint on the circle, $(x,y).(x,y).$ To derive the equation of a circle, we can use thedistance formula with the points $(h,k),(h,k),$ $(x,y)(x,y)$ and thedistance, r. $d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2$ Substitute the values. $r=(x−h)2+(y−k)2r=(x−h)2+(y−k)2$ Square both sides. $r2=(x−h)2+(y−k)2r2=(x−h)2+(y−k)2$
This is the standard form of the equation of a circle with center, $(h,k),(h,k),$ and radius, r.
## Standard Form of the Equation a Circle
The standard form of the equation of a circle with center, $(h,k),(h,k),$ and radius, r, is
## Example 11.5
Write the standard form of the equation of the circle with radius 3 and center $(0,0).(0,0).$
## Try It 11.9
Write the standard form of the equation of the circle with a radius of 6 and center $(0,0).(0,0).$
## Try It 11.10
Write the standard form of the equation of the circle with a radius of 8 and center $(0,0).(0,0).$
In the last example, the center was $(0,0).(0,0).$ Notice what happened to the equation. Whenever the center is $(0,0),(0,0),$ the standard form becomes $x2+y2=r2.x2+y2=r2.$
## Example 11.6
Write the standard form of the equation of the circle with radius 2 and center $(−1,3).(−1,3).$
## Try It 11.11
Write the standard form of the equation of the circle with a radius of 7 and center $(2,−4).(2,−4).$
## Try It 11.12
Write the standard form of the equation of the circle with a radius of 9 and center $(−3,−5).(−3,−5).$
In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.
## Example 11.7
Write the standard form of the equation of the circle with center $(2,4)(2,4)$ that also contains the point $(−2,1).(−2,1).$
## Try It 11.13
Write the standard form of the equation of the circle with center $(2,1)(2,1)$ that also contains the point $(−2,−2).(−2,−2).$
## Try It 11.14
Write the standard form of the equation of the circle with center $(7,1)(7,1)$ that also contains the point $(−1,−5).(−1,−5).$
## Graph a Circle
Any equation of the form $(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2$ is the standard form of the equation of a circle with center, $(h,k),(h,k),$ and radius, r. We can then graph the circle on a rectangular coordinate system.
Note that the standard form calls for subtraction from x and y. In the next example, the equation has $x+2,x+2,$ so we need to rewrite the addition as subtraction of a negative.
## Example 11.8
Find the center and radius, then graph the circle: $(x+2)2+(y−1)2=9.(x+2)2+(y−1)2=9.$
## Try It 11.15
Find the center and radius, then graph the circle: $(x−3)2+(y+4)2=4.(x−3)2+(y+4)2=4.$
## Try It 11.16
Find the center and radius, then graph the circle: $(x−3)2+(y−1)2=16.(x−3)2+(y−1)2=16.$
To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of $x2,y2x2,y2$ to be one.
## Example 11.9
Find the center and radius and then graph the circle, $4x2+4y2=64.4x2+4y2=64.$
## Try It 11.17
Find the center and radius, then graph the circle: $3x2+3y2=273x2+3y2=27$
## Try It 11.18
Find the center and radius, then graph the circle: $5x2+5y2=1255x2+5y2=125$
If we expand the equation from Example 11.8, $(x+2)2+(y−1)2=9,(x+2)2+(y−1)2=9,$ the equation of the circle looks very different.
$(x+2)2+(y−1)2=9 Square the binomials.x2+4x+4+y2−2y+1=9 Arrange the terms in descending degree order,and get zero on the rightx2+y2+4x−2y−4=0 (x+2)2+(y−1)2=9 Square the binomials.x2+4x+4+y2−2y+1=9 Arrange the terms in descending degree order,and get zero on the rightx2+y2+4x−2y−4=0$
This form of the equation is called the general form of the equation of the circle.
## General Form of the Equation of a Circle
The general form of the equation of a circle is
$x2+y2+ax+by+c=0x2+y2+ax+by+c=0$
If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y. Then we can graph the circle using its center and radius.
## Example 11.10
Find the center and radius, then graph the circle: $x2+y2−4x−6y+4=0.x2+y2−4x−6y+4=0.$
## Try It 11.19
Find the center and radius, then graph the circle: $x2+y2−6x−8y+9=0.x2+y2−6x−8y+9=0.$
## Try It 11.20
Find the center and radius, then graph the circle: $x2+y2+6x−2y+1=0.x2+y2+6x−2y+1=0.$
In the next example, there is a y-term and a $y2y2$-term. But notice that there is no x-term, only an $x2x2$-term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.
## Example 11.11
Find the center and radius, then graph the circle: $x2+y2+8y=0.x2+y2+8y=0.$
## Try It 11.21
Find the center and radius, then graph the circle: $x2+y2−2x−3=0.x2+y2−2x−3=0.$
## Try It 11.22
Find the center and radius, then graph the circle: $x2+y2−12y+11=0.x2+y2−12y+11=0.$
## Media
Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.
## Section 11.1 Exercises
### Practice Makes Perfect
Use the Distance Formula
In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
1.
$(2,0)(2,0)$ and $(5,4)(5,4)$
2.
$(−4,−3)(−4,−3)$ and $(2,5)(2,5)$
3.
$(−4,−3)(−4,−3)$ and $(8,2)(8,2)$
4.
$(−7,−3)(−7,−3)$ and $(8,5)(8,5)$
5.
$(−1,4)(−1,4)$ and $(2,0)(2,0)$
6.
$(−1,3)(−1,3)$ and $(5,−5)(5,−5)$
7.
$(1,−4)(1,−4)$ and $(6,8)(6,8)$
8.
$(−8,−2)(−8,−2)$ and $(7,6)(7,6)$
9.
$(−3,−5)(−3,−5)$ and $(0,1)(0,1)$
10.
$(−1,−2)(−1,−2)$ and $(−3,4)(−3,4)$
11.
$(3,−1)(3,−1)$ and $(1,7)(1,7)$
12.
$(−4,−5)(−4,−5)$ and $(7,4)(7,4)$
Use the Midpoint Formula
In the following exercises, find the midpoint of the line segments whose endpoints are given and plot the endpoints and the midpoint on a rectangular coordinate system.
13.
$(0,−5)(0,−5)$ and $(4,−3)(4,−3)$
14.
$(−2,−6)(−2,−6)$ and $(6,−2)(6,−2)$
15.
$(3,−1)(3,−1)$ and $(4,−2)(4,−2)$
16.
$(−3,−3)(−3,−3)$ and $(6,−1)(6,−1)$
Write the Equation of a Circle in Standard Form
In the following exercises, write the standard form of the equation of the circle with the given radius and center $(0,0).(0,0).$
17.
18.
19.
Radius: $22$
20.
Radius: $55$
In the following exercises, write the standard form of the equation of the circle with the given radius and center
21.
Radius: 1, center: $(3,5)(3,5)$
22.
Radius: 10, center: $(−2,6)(−2,6)$
23.
Radius: $2.5,2.5,$ center: $(1.5,−3.5)(1.5,−3.5)$
24.
Radius: $1.5,1.5,$ center: $(−5.5,−6.5)(−5.5,−6.5)$
For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.
25.
Center $(3,−2)(3,−2)$ with point $(3,6)(3,6)$
26.
Center $(6,−6)(6,−6)$ with point $(2,−3)(2,−3)$
27.
Center $(4,4)(4,4)$ with point $(2,2)(2,2)$
28.
Center $(−5,6)(−5,6)$ with point $(−2,3)(−2,3)$
Graph a Circle
In the following exercises, find the center and radius, then graph each circle.
29.
$( x + 5 ) 2 + ( y + 3 ) 2 = 1 ( x + 5 ) 2 + ( y + 3 ) 2 = 1$
30.
$( x − 2 ) 2 + ( y − 3 ) 2 = 9 ( x − 2 ) 2 + ( y − 3 ) 2 = 9$
31.
$( x − 4 ) 2 + ( y + 2 ) 2 = 16 ( x − 4 ) 2 + ( y + 2 ) 2 = 16$
32.
$( x + 2 ) 2 + ( y − 5 ) 2 = 4 ( x + 2 ) 2 + ( y − 5 ) 2 = 4$
33.
$x 2 + ( y + 2 ) 2 = 25 x 2 + ( y + 2 ) 2 = 25$
34.
$( x − 1 ) 2 + y 2 = 36 ( x − 1 ) 2 + y 2 = 36$
35.
$( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25 ( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25$
36.
$( x − 1 ) 2 + ( y − 3 ) 2 = 9 4 ( x − 1 ) 2 + ( y − 3 ) 2 = 9 4$
37.
$x 2 + y 2 = 64 x 2 + y 2 = 64$
38.
$x 2 + y 2 = 49 x 2 + y 2 = 49$
39.
$2 x 2 + 2 y 2 = 8 2 x 2 + 2 y 2 = 8$
40.
$6 x 2 + 6 y 2 = 216 6 x 2 + 6 y 2 = 216$
In the following exercises, identify the center and radius and graph.
41.
$x 2 + y 2 + 2 x + 6 y + 9 = 0 x 2 + y 2 + 2 x + 6 y + 9 = 0$
42.
$x 2 + y 2 − 6 x − 8 y = 0 x 2 + y 2 − 6 x − 8 y = 0$
43.
$x 2 + y 2 − 4 x + 10 y − 7 = 0 x 2 + y 2 − 4 x + 10 y − 7 = 0$
44.
$x 2 + y 2 + 12 x − 14 y + 21 = 0 x 2 + y 2 + 12 x − 14 y + 21 = 0$
45.
$x 2 + y 2 + 6 y + 5 = 0 x 2 + y 2 + 6 y + 5 = 0$
46.
$x 2 + y 2 − 10 y = 0 x 2 + y 2 − 10 y = 0$
47.
$x 2 + y 2 + 4 x = 0 x 2 + y 2 + 4 x = 0$
48.
$x 2 + y 2 − 14 x + 13 = 0 x 2 + y 2 − 14 x + 13 = 0$
### Writing Exercises
49.
Explain the relationship between the distance formula and the equation of a circle.
50.
Is a circle a function? Explain why or why not.
51.
In your own words, state the definition of a circle.
52.
In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need. | HuggingFaceTB/finemath | |
# Limit Point/Examples
## Examples of Limit Points
### End Points of Real Interval
The real number $a$ is a limit point of both the open real interval $\openint a b$ as well as of the closed real interval $\closedint a b$.
It is noted that $a \in \closedint a b$ but $a \notin \openint a b$.
### Union of Singleton with Open Real Interval
Let $\R$ be the set of real numbers.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
$H = \set 0 \cup \openint 1 2$
Then $0$ is not a limit point of $H$.
### Real Number is Limit Point of Rational Numbers in Real Numbers
Let $\R$ be the set of real numbers.
Let $\Q$ be the set of rational numbers.
Let $x \in \R$.
Then $x$ is a limit point of $\Q$.
### Zero is Limit Point of Integer Reciprocal Space
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then $0$ is the only limit point of $A$ in $\R$. | HuggingFaceTB/finemath | |
# Thread: Graph Theory: Arithmetic and Geometric Means
1. ## Graph Theory: Arithmetic and Geometric Means
Hi, I'm a little stuck on this question.
Part a) asks to prove that $\displaystyle $l = \sqrt{a^{2}+b^{2}+c^{2}}$$, where $\displaystyle $l$$ is the diagonal of a rectangular brick with sides of length a, b and c.
I've already proved that, by saying:
$\displaystyle $d = \sqrt{a^{2}+c^{2}}$$
$\displaystyle $l = \sqrt{d^{2}+b^{2}}$$
$\displaystyle $l = \sqrt{\sqrt{a^{2}+c^{2}}^{2} + b^{2}}$$
$\displaystyle $\therefore l = \sqrt{a^{2}+b^{2}+c^{2}}$$
The question I'm struggling with is part b), which says "If V denotes the volume of the brick, show that $\displaystyle $V \leq (3\sqrt{3})^{-1}\ \times\ l^{3}$$.
I've found the AM by letting $\displaystyle a(1) =\ a^{2},\ a(2) =\ b^{2},\ a(3) =\ c^{2}$ then dividing by 3 and raising to the power of $\displaystyle $\frac{3}{2}$$ to get $\displaystyle $(3\sqrt{3})^{-1}\ \times\ l^{3}$$.
I'm not sure what to do when finding the AM/GM inequality as I can't find a way to make GM = the volume (which is the only way I can think to solve this at the moment).
So far I've got $\displaystyle $\frac{l^{3}}{3\sqrt{3}} = AM \geq GM = \sqrt[3]{a^{2}\times b^{2}\times c^{2}}$$, but I can't make any progress from there.
Is it as simple as $\displaystyle $V =\ a\times b\times c \leq \sqrt[3]{a^{2}+b^{2}+c^{2}} \leq \frac{l^{3}}{3\sqrt{3}}$$? Just seems to me like there is a neater solution and I've made some sort of formulaic or calculative error.
Appreciate any input.
2. Your first step was correct:
Let $\displaystyle a(1) =\ a^{2},\ a(2) =\ b^{2},\ a(3) =\ c^{2}$
AM = $\displaystyle \frac {a^{2} + b^{2} + c^{2}}{3}$
GM = $\displaystyle (a^{2}b^{2}c^{2})^{\frac{1}{3}}$
By AM GM inequality,
$\displaystyle (a^{2}b^{2}c^{2})^{\frac{1}{3}} \leq \frac {a^{2} + b^{2} + c^{2}}{3}$
Does taking square roots on both sides bring the LHS and the RHS into some known form ?
3. Ahh great, got it with a bit of messing around after that push in the right direction.
For those interested: Taking the square roots gives $\displaystyle $\sqrt[3]{abc}\leq \frac{\sqrt{3}\times \sqrt{a^{2}+b^{2}+c^{2}}}{3} = \frac{l\sqrt{3}}{3}$$
From there,
$\displaystyle $3\sqrt[3]{abc}\leq l\sqrt{3}$$
$\displaystyle $\sqrt{3}\sqrt[3]{abc}\leq l$$
$\displaystyle $3\sqrt{3} abc\leq l^{3}$$
$\displaystyle $abc = V\leq \frac{l^{3}}{3\sqrt{3}}$$
Thanks Traveller! | HuggingFaceTB/finemath | |
# Palindrome Number
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
``````Input: 121
Output: true
``````
Example 2:
``````Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
``````
Example 3:
``````Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
``````
Coud you solve it without converting the integer to a string?
### Solution
``````/**
* Palindrome Number
* @param {number} x
* @return {boolean}
* @time complexity: O(log x) because we are dividing x by 10 in each iteration
* @space complexity: O(1)
*/
// Negative number and positive number with trailing zero (except zero itself)
// are not palindrome number
if(x < 0 || (x % 10 == 0 && x != 0)) {
return false;
}
let reverted = 0;
while (x > reverted) {
reverted = reverted * 10 + x % 10; // Push digits one position forward and add last digit that pop from x
x = Math.floor(x / 10); // Only keep integer, so discards decimal points
}
// When the length of x is an odd number, remove the last digit (reverted / 10)
// The middle digit doesn't matter in palindrome
return x === reverted || x === Math.floor(reverted / 10);
};
``````
### Test Case
``````const assert = require('chai').assert;
describe('Palindrome Number', () => {
it('should return true', () => {
});
it('should return false if it is an nagative number', () => {
});
it('should return false if the number ends with zero', () => {
});
});
``````
`````` Palindrome Number
✓ should return true
✓ should return false if it is an nagative number
✓ should return false if the number ends with zero
3 passing (11ms)
``````
Mike Mai Brooklyn, New York
I am full-stack web developer, passionate about building world class web applications. Knowledge in designing, coding, testing, and debugging. I love to solve problems. | HuggingFaceTB/finemath | |
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# Each employee on a certain task force is either a manager or
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Each employee on a certain task force is either a manager or [#permalink]
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03 Dec 2008, 07:28
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Each employee on a certain task force is either a manager or a director. What % of the employees on the task force are directors?
1) The average (arithmetic mean) salary of the managers on the task force is \$5000 less than the average salary of all employees on the task force.
2) The average (arithmetic mean) salary of the directors on the task force is \$15000 greater that the average salary of all employees on the task force.
Will you supply detailed explanation
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Re: managers vs directors DS [#permalink]
### Show Tags
03 Dec 2008, 08:11
1) is S is the avg salary then 1) says (S-5)M is the salary of the managers
2) says (S+15)D is the salary of the Directors
together
total salary S(D+M)=(S-5)M+(s+15)D
you get 5M=15D or M=3D
C it is
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Re: managers vs directors DS [#permalink]
### Show Tags
03 Dec 2008, 08:22
C
directors average 15k+
managers 5k-
so average is 10k
ths directors make 25k on average and managers 5k
note 5k as a unit increment
so directors make 3 units more than the average and managers make 1 unit less
Now it gets really tricky
since managers r only 1 unit away from the average and directors 3 that means they are divided in 75:25 ratio
_________________
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Re: managers vs directors DS [#permalink] 03 Dec 2008, 08:22
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## Algebra 2 (1st Edition)
Published by McDougal Littell
# Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Skill Practice - Page 823: 16
#### Answer
$\dfrac{-8}{5}$
#### Work Step by Step
Here, we have $a_n= a_1 r^{n-1}$ for the Geometric series. First term $a_1=-2$ and Common ratio $r=\dfrac{-1}{4}$ The sum of an infinite Geometric Series can be found using: $S_n=\dfrac{a_1}{1-r}$ Thus, $S_n=\dfrac{-2}{1-(-1/4)}=\dfrac{-2 \times 4}{4+1}$ Hence, $S_n=\dfrac{-8}{5}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | HuggingFaceTB/finemath | |
1. ## Compound interest formula
Fred invests an amount of money in an account by paying r% compound interest per annum. The amount of money doubles after n years.
I'm aware of the compound interest formula A=P(1+r/100)^n however I have no idea how to change it according to n.
Would I start off along the lines of 2A=p(1+r/100)^n?
2. (1 + r)^n = 2
n = log(2) / log(1 + r)
Example: r = .06 (6%)
n = log(2) / log(1.06) = 11.89566.... ; not quite 12 years
3. Can't use a method that utilises logs
4. Originally Posted by Mukilab
Can't use a method that utilises logs
Well, too bad so sad: that's the only way to solve directly.
5. Its not. I know that for a fact.
6. What about simplifying it so you have the equation pertaining to r
7. r = 2^(1/n) - 1
SO?
IF you know it as a fact: then SHOW US.
8. I know as a fact because it is impossible a question pertaining to logs would show up in a paper that specifies to a certain age that has never previously learned logs and 8 years later they still are not teaching logs to this age group!
9. Originally Posted by Wilmer
r = 2^(1/n) - 1
SO?
IF you know it as a fact: then SHOW US.
Would it not be
$100(^{n}\sqrt{2}-1)$
10. Originally Posted by Mukilab
I know as a fact because it is impossible a question pertaining to logs would show up in a paper that specifies to a certain age that has never previously learned logs and 8 years later they still are not teaching logs to this age group!
Have a good day, Mukilab.
11. Instead of separating Total from investment can I do 2p=p(1+r/100)^n?
12. Originally Posted by Mukilab
Instead of separating Total from investment can I do 2p=p(1+r/100)^n?
That is what Wilmer posted in post 2 (except he cancelled out p which is the next step)
Originally Posted by Mukilab
Fred invests an amount of money in an account by paying r% compound interest per annum. The amount of money doubles after n years.
I'm aware of the compound interest formula A=P(1+r/100)^n however I have no idea how to change it according to n.
Would I start off along the lines of 2A=p(1+r/100)^n?
I don't know how you'd solve for n without using logarithms, I doubt it's even possible, perhaps they use logarithms without knowing it? | HuggingFaceTB/finemath | |
# Why is signum function used to calculate Fourier transform of unit step function
I read in a standard textbook that the Fourier transform of unit impulse function is calculated with the help of approximations and signum function as the integration of unit impulse does not converge. What's so special about signum function that it is used to calculate Fourier transform? I tried to find out an approximation as:
$$\lim_{ a \rightarrow 0 } \int_{-\infty}^{+\infty} e^{-at} u(t) e^{-j\omega t} dt$$
But I am getting wrong result. Why is this so?
• In your title you write "unit step", whereas in your question it is "unit impulse". I assume that your question is about the computation of the Fourier transform of the unit step function. And whatever your textbook says, you don't need the sign function to compute the Fourier transform of the step function. Of course, if you already have the Fourier transform of the sign function, you can use it for computing the Fourier transform of the step function. – Matt L. Oct 13 '15 at 16:22
• What confuses me is that the signum function and the unit step are the same --- modulo a scaling factor of 2 and a DC offset of -1. – Peter K. Oct 13 '15 at 16:29
• The way you're trying to compute the Fourier transform of the unit step works (by taking the limit). Why you're getting a wrong result can only be answered if you show us the details of your computation and your result. I assume you get $1/j\omega$, which comes from computing the limit in the wrong way. – Matt L. Oct 13 '15 at 16:34
If somebody you trust told you that the Fourier transform of the sign function is given by
$$\mathcal{F}\{\text{sgn}(t)\}=\frac{2}{j\omega}\tag{1}$$
you could of course use this information to compute the Fourier transform of the unit step $u(t)$. Using
$$u(t)=\frac12(1+\text{sgn}(t))\tag{2}$$
(as pointed out by Peter K. in a comment), you get
$$\mathcal{F}\{u(t)\}=\frac12\left(\mathcal{F}\{1\}+\mathcal{F}\{\text{sgn}(t)\}\right)=\pi\delta(\omega)+\frac{1}{j\omega}\tag{3}$$
However, you don't need the sign function to compute the Fourier transform of the step function. As suggested in your question, using the function $e^{-at}u(t)$ and taking the limit $a\rightarrow 0^+$ will also result in the expression given in $(3)$.
You can see this as follows. The Fourier transform of $e^{-at}u(t)$, $a>0$, is given by
$$\int_0^{\infty}e^{-at}e^{-j\omega t}dt=\frac{1}{a+j\omega}\tag{4}$$
Taking the limit $a\rightarrow 0^+$ appears to give $1/j\omega$, but this is only valid for $\omega\neq 0$. Splitting the result $(4)$ in its real and imaginary part gives
$$\frac{1}{a+j\omega}=\frac{a}{a^2+\omega^2}+\frac{\omega}{j(a^2+\omega^2)}\tag{5}$$
The real part of $(5)$ is known as ($\pi$ times) a nascent delta function. It has the same form as the Poisson kernel, which in the limit becomes a Dirac delta impulse. So for $a\rightarrow 0^+$ the limit of $(5)$, and hence of $(4)$, is actually given by
$$\lim_{a\rightarrow 0^+}\frac{1}{a+j\omega}=\pi\delta(\omega)+\frac{1}{j\omega}\tag{6}$$
which equals the expression in $(3)$.
• could you slightly elaborate on how $\displaystyle\lim_{a \rightarrow 0} \frac{a}{a^2+w^2}= \pi\delta(\omega)$ – Karan Talasila Oct 14 '15 at 5:33
• @Talasila: If you click on the link and scroll down you see the 'Poisson kernel', which has exactly the same form as the real part in (5). It is a nascent delta function, which means that in the limit it becomes a Dirac delta. – Matt L. Oct 14 '15 at 7:16
• @MattL. Is $\pi\delta(\omega)$ singularity value? – jomegaA Feb 11 at 9:24
• @jomegaA: Not sure what you mean; it's not a value, it's just the Dirac delta at $\omega=0$. – Matt L. Feb 11 at 11:17 | HuggingFaceTB/finemath | |
## 3733 Days From July 11, 2024
Want to figure out the date that is exactly three thousand seven hundred thirty three days from Jul 11, 2024 without counting?
Your starting date is July 11, 2024 so that means that 3733 days later would be September 30, 2034.
You can check this by using the date difference calculator to measure the number of days from Jul 11, 2024 to Sep 30, 2034.
September 2034
• Sunday
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September 30, 2034 is a Saturday. It is the 273rd day of the year, and in the 39th week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 30 days in this month. 2034 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 09/30/2034, and almost everywhere else in the world it's 30/09/2034.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 3733 weekdays from Jul 11, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Jul 11, 2024, which falls on a Thursday. Counting forward, the next day would be a Friday.
To get exactly three thousand seven hundred thirty three weekdays from Jul 11, 2024, you actually need to count 5227 total days (including weekend days). That means that 3733 weekdays from Jul 11, 2024 would be November 2, 2038.
If you're counting business days, don't forget to adjust this date for any holidays.
November 2038
• Sunday
• Monday
• Tuesday
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November 2, 2038 is a Tuesday. It is the 306th day of the year, and in the 306th week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 30 days in this month. 2038 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 11/02/2038, and almost everywhere else in the world it's 02/11/2038.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | HuggingFaceTB/finemath | |
### Derivatives of the trigonometric functions
In this section we'll derive the important derivatives of the trigonometric functions f(x) = sin(x), cos(x) and tan(x).
In doing so, we will need to rely upon the trigonometric limits we derived in another section. To remind you, those are copied here.
To get through this section, you'll need to be familiar with analytic trigonometry, the derivative and the chain rule. and ### The derivative of f(x) = sin(x)
We start, as usual with the difference-quotient expression of the derivative: Now we recall the trigonometric sum identities: We use the first to make a substitution for sin(x+h), Now factor a sin(x) out of two of the terms of the numerator, like this: Now let's use the properties of limits to separate that expression into two different limits: Because sin(x) and cos(x) are continuous over their whole domains, we can remove them from the limit expression, which leaves us with the familiar and easy-to-evaluate trigonometric limits: The derivative of the sine function is thus the cosine function: Take a minute to look at the graph below and see if you can rationalize why cos(x) should be the derivative of sin(x). Notice that wherever sin(x) has a maximum or minimum (at which point the slope of a tangent line would be zero), the value of the cosine function is zero. Notice that wherever sin(x) has maximum slope (where it passes through y=0), the cosine function reaches its minimum or maximum at ±1. ### The derivative of f(x) = cos(x)
We find the derivative of f(x) = cos(x) in a similar manner, and I'll work through the whole thing again here. Start with the difference quotient expression, then replace cos(x + h) with cos(x)cos(h) - sin(x)sin(h): Now move the common factor cos(x) and rewrite the numerator: Split the expression into two limits: and remove the non h-dependent expressions to find our familiar trigonometric limits: This time we get Look again at the graphs of f(x) = cos(x) and f'(x) = -sin(x), and see if you can rationalize for yourself why -sin(x) is the derivative of cos(x). ### The derivative of f(x) = tan(x)
For this derivative, we'll use the definition of the tangent and the quotient rule to find the result. You can use the difference quotient yourself in an exercise below. We start by writing tan(x) as sin(x)/cos(x), then using the quotient rule: After taking the derivatives of sin(x) and cos(x) (which we can do now!), we recognize the Pythagorean identity: cos2(x) + sin2(x) = 1: Then recognize that 1/cos2(x) is sec2(x) The result is: Use the graph below to understand why sec2(x) gives the slope of tan(x) at any point between ±π/2 (remember that the tangent function is infinite at multiples of ±π/2). Notice that the slope of the tangent function is always positive, and the value of the derivative is always positive, too. #### Derivatives of the trigonometric functions
Here are the derivatives of the six trigonometric functions. You should commit the derivatives of sin(x), cos(x) and tan(x) to memory. ### Practice problems
#### Calculate the derivatives of these trigonometric and mixed functions. Don't forget the product rule, the quotient rule and the chain rule. ### You can help
This site is a one-person operation. If you can manage it, I'd appreciate anything you could give to help defray the cost of keeping up the domain name, server, search service and my time. xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net. | HuggingFaceTB/finemath | |
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# ilovepdf.com_split_3 - 99 Question 317 A particle of mass m...
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99 Question 3–17 A particle of mass m is attached to an inextensible massless rope of length l as shown in Fig. P3-17. The rope is attached at its other end to point A located at the top of a fixed cylinder of radius R . As the particle moves, the rope wraps itself around the cylinder and never becomes slack. Knowing that θ is the angle measured from the vertical to the point of tangency of the exposed portion of the rope with the cylinder and that gravity acts downward, determine the differential equation of motion for the particle in terms of the angle θ . You may assume in your solution that the angle θ is always positive. g m A B O R θ Figure P3-17 Solution to Question 3–17 Kinematics First, let F be a reference frame fixed to the circular track. Then, choose the following coordinate system fixed in reference frame F : Origin at O E x = Along OA E z = Out of Page E y = E z × E x Next, let A be a reference frame fixed to the exposed portion of the rope. Then, choose the following coordinate system fixed in reference frame A : Origin at O e r = Along OB e z = Out of Page e θ = e z × e r The geometry of the bases { E x , E y , E z } and { e r , e θ , e z } is shown in Fig. 3-13. Using Fig. 3-13, we have that E x = cos θ e r - sin θ e θ (3.382) E y = sin θ e r + cos θ e θ (3.383)
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100 Chapter 3. Kinetics of Particles e r e θ E x E y e z , E z θ θ Figure 3-13 Geometry of Question 3–17. Now we note that the rope has a fixed length l . Since the length of the portion of the rope wrapped around the cylinder is , the exposed portion of the rope must have length l - . Furthermore, since the exposed portion of the rope lies along the direction from B to m
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A090116 a(n)=x is the least number such that x^2 is "surrounded" by two closest primes, prevprime(x^2) and nextprime(x^2), whose difference nextprime - prevprime = 2n. 5
2, 3, 5, 19, 12, 25, 11, 44, 23, 30, 57, 41, 50, 102, 76, 104, 100, 149, 175, 159, 348, 276, 305, 397, 461, 189, 345, 1059, 437, 820, 833, 1002, 509, 1283, 822, 1099, 729, 1090, 693, 2710, 1110, 1284, 3563, 1823, 1370, 4332, 3771, 1380, 4394, 2160, 2011, 1498 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(14) > 16*10^6. - David A. Corneth, Jun 12 2017 LINKS David A. Corneth, Table of n, a(n) for n = 1..213 David A. Corneth, Table of n, a(n) for n = 1..302 where a(n) <= (about) 16*10^6 EXAMPLE n=5: a(5)=12 because the primes closest to 12^2 = 144 are {139,149} whose difference 149 - 139 = 10 = 2n and 144 is the smallest square with this property; n=1: a(1)=2 because 2^2=4 is surrounded by primes {3,5} with difference 5 - 3 = 2 = 2n. MATHEMATICA de[x_ := Prime[PrimePi[x]+1]-Prime[PrimePi[x]] t=Table[de[w^2], {w, 1, 50000}]; mt=Table[Min[Flatten[Position[t, 2*j]]], {j, 1, 100}] PROG (PARI) first(n) = my(todo = n, res = vector(n), p, x = 2); while(todo > 0, m = nextprime(x^2) - precprime(x^2); if(m <= 2*n, if(res[m/2]==0, res[m/2] = x; todo--)); x++); res \\ David A. Corneth, Jun 12 2017 CROSSREFS Cf. A090117, A090118, A090119. Sequence in context: A128532 A130076 A223704 * A038876 A038932 A019377 Adjacent sequences: A090113 A090114 A090115 * A090117 A090118 A090119 KEYWORD nonn AUTHOR Labos Elemer, Jan 09 2004 STATUS approved
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Question
Equation 2x2+5xy+3y2+6x+7y+4=0 represents a pair of straight lines. If their point of intersection is (h,k) find the value of h-k. __
Solution
It is given that equation 2x2+5xy+3y2+6x+7y+4=0 represents the pair of straight lines We can write this equation as quadratic equation in x We have, 2x2+(5y+6)x+3y2+7y+4=0-----------------------(1) Roots of the given equation X=−(5y+6)±√(5y+6)2−4.2(3y2+7y+4)4 =−(5y+6)±√25y2+60y+36−24y2−56y−324 X=−(5y+6)±√y2+4y+44 X=−(5y+6)±(y+2)4 X=−5y−6+y+24,−5y−6−y−24 Or 4x = -4y - 4 , 4x + 6y + 8 = 0 X + y + 1 = 0 and 2x + 3y + 4 = 0 Hence, equation 1 represents the pair of straight lines whose equations are X + y + 1 = 0------------------2 2x + 3y + 4= 0-----------------3 Multiplying 2 in equation 2 and subtracting from equation 3 2x + 2y + 2 = 0 2x + 3y + 4 = 0 -y - 2 = 0 -y = 2 y = -2 Substituting in equation 1 -2+x+1=0 X=1 Point of intersection (1,-2) h=1, k=-2 h-k= 1-(-2)=3 Mathematics
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# physics
A wave on a string has a wave function given by y(x, t) = (0.0210 m)sin[(6.91 m^−1)x + (2.38 s^−1)t
.
(a) What is the amplitude of the wave?
(b) What is the period of the wave?
(c) What is the wavelength of the wave?
(d) What is the speed of the wave?
1. 👍
2. 👎
3. 👁
1. The equation of the wave is
y(x,t)= A•sin(2πx/λ - 2πt/T).
Given equation is
y(x, t) = (0.0210 m)sin[(6.91 m^−1)x + (2.38 s^−1)t
Therefore,
A = 0.0210 m
2π/λ =6.91
λ=2π/6.91=0.909 m
2πt/T= 2.38
T=2πt/ 2.38=2.64 s
λ=vT
v= λ/T=(2π/6.91)/( 2πt/ 2.38)=
=2.38/6.91=0.344 m/s
1. 👍
2. 👎
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1. ### PHYSICS!!!HELP！！
The speed of a transverse wave on a string is 450 m/s, while the wavelength is 0.18m. The amplitude of the wave is 2.0mm. How much time is required for a particle of the string to move through a total distance of 1.0km? I NEED
2. ### physics
A wave is represented by the equation y=0.5 sin 0.4x(x-60t) there the distance x is measured in can and time(t) second what is the wave length of the wave?
3. ### Physics
The wave function of a standing wave is y(x,t)=4.44mmsin[(32.5rad/m)x]sin[(754rad/s)t]. A. For the two traveling waves that make up this standing wave, find the amplitude. B. For the two traveling waves that make up this standing
Please help check my answers I really want to make a 100%! 1. A mechanical wave moves through a medium, which can be (1 point) A. a liquid. B. a solid. C. a gas. D. all of the above
1. ### Physics
If an earthquake wave having a wavelength of 13 km caused the ground to vibrate 10 times each minute, what is the speed of the wave? 1.2 km/s 2.2 km/s 220 m/s 22 m/s If the amplitude in a sound wave is doubled, by what factor does
2. ### science
A string is waved up and down to create a wave pattern with a wavelength of 0.5 m. If the waves are generated with a frequency of 2 Hz, what is the speed of the wave that travels through the string to the other end? The speed of
3. ### physics question
What factors affect the frequency of a wave? A. the amplitude B. the energy of the wave C. the wavelength of the wave D. the material through which it travels E. the velocity of the wave I said C,D, and E. Should there be anything
4. ### physic
You are given two waves, a transverse wave that moves to the right f1(x) and a transverse wave that moves to the left f2(x), on a string. As the problem begins, the wave f1(x) is moving to the right at v1 = +1 m/s and the wave | HuggingFaceTB/finemath | |
# Solving One Step Equations - PowerPoint PPT Presentation Download Presentation Solving One Step Equations
Solving One Step Equations Download Presentation ## Solving One Step Equations
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##### Presentation Transcript
1. Solving One Step Equations Kaitlyn Sand 5.03 Solving one-step equations
2. Order of Operations PEMDAS right to left to solve: (multiplication) (subtraction) (addition)
3. Solving One Step Equations Opposite of PEMDAS; SADMEP. • S – subtraction • A – addition • D – division • M – multiplication • E – exponents (remember, we haven’t • P – parenthesis talked about these yet!)
4. One Step Equations -Perform 1 step to solve -Try to get “X” by itself -Do the same operation to both sides -Just like what you did on excel math:
5. Excel Math + 3 = 5 Now, we use X as a place holder instead of the empty square. X + 3 = 5
6. What you do to ONE side, you MUST do to the other. X + 3 = 5 (Subtract 3 from both sides) -3 -3 X = 2
7. 2X = 10 2 2 X = 5 6 – 1X = 2 -6 -6 -1X = -4 -1 -1 X = 4
8. Foamy Squishy Things Introduction = X + = 0 = 1 = -1
9. X + 3 = 5 You can also use foamy squishy things to solve problems. += =
10. Using Foamy Squishy Things Set up your foamy squishy things in the form of this problem: X + ˉ5 = 2 | HuggingFaceTB/finemath | |
# Monthly Archives: July 2013
## Num3er 7312013 (July 31, 2013)
July 31 can be expressed as 731 The factors of 731 are: 1 17 43 731 The prime factors are: 17 * 43 731 = 3^6 … Continue reading
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## Prime Num3ers Patterns – Part 1
1033 is a prime number. 1033 = 8^1 + 8^0 + 8^3 + 8^3 2213 is a prime number. 2213 = 2^3 + 2^3 … Continue reading
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## Powers that start with repeated digits
Ninth powers: a(n) = n^9 8461^9 = 222226197660256026153988809486637741 Find more examples. —————————————— 5th powers: a(n) = n^5 6444^5 = 11111627111310388224 Find more … Continue reading
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## Prime Num3ers 4357 and 5347
4357 is a prime number. 5347 is a prime number. Notice that 5347 is a digit-rotation of 4357 And, 4357^2 = 18983449 = 1898 || 3449 and, 1898 + 3449 … Continue reading
Posted in Math Beauty, Number Puzzles | Tagged , , | 3 Comments
## Num3ers ABCD = A * B^n * C * D, n=2,3
735 = 7^2 * 3 * 5 1715 = 1 * 7^3 * 1 * 5 Can you find more examples? 5-digit numbers … Continue reading
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## Repeated digits | Num3ers abcd = a…a + b..b + c…c + d…d
4-digit numbers that can be expressed by using their own digits for example, 198 = 11 + 99 + 88 Other examples, 1098 = 11 + 0 + 999 + … Continue reading
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## Num3ers 8970, 8971, …, 8979
8970 = 8 + 9^4 + 7^4 + 0 8971 = 8 + 9^4 + 7^4 + 1 8972 = 8 … Continue reading
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## Num3ers 4150, 4151, …, 4168, 4169
same digits on both sides of the equalities: 4150 = 4^5 + 1^5 + 5^5 + 0 4151 = 4^5 + 1^5 + 5^5 … Continue reading
Posted in Uncategorized | Tagged , , , , | Leave a comment
## Num3er 444
The factors of 444 are: 1 2 3 4 6 12 37 74 111 148 222 444 The prime factors are: 2 * 2 * 3 … Continue reading
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## Num3ers 5160, 5161, 5162, … , 5169
same digits on both sides of the equality: 5! + (1+6)! + 0 = 5160 5! + (1+6)! + 1 = 5161 5! + (1+6)! … Continue reading
Posted in Math Beauty, Number Puzzles | Tagged , , , | Leave a comment | HuggingFaceTB/finemath | |
#### What is 80 percent of 466?
How much is 80 percent of 466? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 80% of 466 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 80% of 466 = 372.8
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Calculating eighty of four hundred and sixty-six How to calculate 80% of 466? Simply divide the percent by 100 and multiply by the number. For example, 80 /100 x 466 = 372.8 or 0.8 x 466 = 372.8
#### How much is 80 percent of the following numbers?
80% of 466.01 = 37280.8 80% of 466.02 = 37281.6 80% of 466.03 = 37282.4 80% of 466.04 = 37283.2 80% of 466.05 = 37284 80% of 466.06 = 37284.8 80% of 466.07 = 37285.6 80% of 466.08 = 37286.4 80% of 466.09 = 37287.2 80% of 466.1 = 37288 80% of 466.11 = 37288.8 80% of 466.12 = 37289.6 80% of 466.13 = 37290.4 80% of 466.14 = 37291.2 80% of 466.15 = 37292 80% of 466.16 = 37292.8 80% of 466.17 = 37293.6 80% of 466.18 = 37294.4 80% of 466.19 = 37295.2 80% of 466.2 = 37296 80% of 466.21 = 37296.8 80% of 466.22 = 37297.6 80% of 466.23 = 37298.4 80% of 466.24 = 37299.2 80% of 466.25 = 37300
80% of 466.26 = 37300.8 80% of 466.27 = 37301.6 80% of 466.28 = 37302.4 80% of 466.29 = 37303.2 80% of 466.3 = 37304 80% of 466.31 = 37304.8 80% of 466.32 = 37305.6 80% of 466.33 = 37306.4 80% of 466.34 = 37307.2 80% of 466.35 = 37308 80% of 466.36 = 37308.8 80% of 466.37 = 37309.6 80% of 466.38 = 37310.4 80% of 466.39 = 37311.2 80% of 466.4 = 37312 80% of 466.41 = 37312.8 80% of 466.42 = 37313.6 80% of 466.43 = 37314.4 80% of 466.44 = 37315.2 80% of 466.45 = 37316 80% of 466.46 = 37316.8 80% of 466.47 = 37317.6 80% of 466.48 = 37318.4 80% of 466.49 = 37319.2 80% of 466.5 = 37320
80% of 466.51 = 37320.8 80% of 466.52 = 37321.6 80% of 466.53 = 37322.4 80% of 466.54 = 37323.2 80% of 466.55 = 37324 80% of 466.56 = 37324.8 80% of 466.57 = 37325.6 80% of 466.58 = 37326.4 80% of 466.59 = 37327.2 80% of 466.6 = 37328 80% of 466.61 = 37328.8 80% of 466.62 = 37329.6 80% of 466.63 = 37330.4 80% of 466.64 = 37331.2 80% of 466.65 = 37332 80% of 466.66 = 37332.8 80% of 466.67 = 37333.6 80% of 466.68 = 37334.4 80% of 466.69 = 37335.2 80% of 466.7 = 37336 80% of 466.71 = 37336.8 80% of 466.72 = 37337.6 80% of 466.73 = 37338.4 80% of 466.74 = 37339.2 80% of 466.75 = 37340
80% of 466.76 = 37340.8 80% of 466.77 = 37341.6 80% of 466.78 = 37342.4 80% of 466.79 = 37343.2 80% of 466.8 = 37344 80% of 466.81 = 37344.8 80% of 466.82 = 37345.6 80% of 466.83 = 37346.4 80% of 466.84 = 37347.2 80% of 466.85 = 37348 80% of 466.86 = 37348.8 80% of 466.87 = 37349.6 80% of 466.88 = 37350.4 80% of 466.89 = 37351.2 80% of 466.9 = 37352 80% of 466.91 = 37352.8 80% of 466.92 = 37353.6 80% of 466.93 = 37354.4 80% of 466.94 = 37355.2 80% of 466.95 = 37356 80% of 466.96 = 37356.8 80% of 466.97 = 37357.6 80% of 466.98 = 37358.4 80% of 466.99 = 37359.2 80% of 467 = 37360
1% of 466 = 4.66 2% of 466 = 9.32 3% of 466 = 13.98 4% of 466 = 18.64 5% of 466 = 23.3 6% of 466 = 27.96 7% of 466 = 32.62 8% of 466 = 37.28 9% of 466 = 41.94 10% of 466 = 46.6 11% of 466 = 51.26 12% of 466 = 55.92 13% of 466 = 60.58 14% of 466 = 65.24 15% of 466 = 69.9 16% of 466 = 74.56 17% of 466 = 79.22 18% of 466 = 83.88 19% of 466 = 88.54 20% of 466 = 93.2 21% of 466 = 97.86 22% of 466 = 102.52 23% of 466 = 107.18 24% of 466 = 111.84 25% of 466 = 116.5
26% of 466 = 121.16 27% of 466 = 125.82 28% of 466 = 130.48 29% of 466 = 135.14 30% of 466 = 139.8 31% of 466 = 144.46 32% of 466 = 149.12 33% of 466 = 153.78 34% of 466 = 158.44 35% of 466 = 163.1 36% of 466 = 167.76 37% of 466 = 172.42 38% of 466 = 177.08 39% of 466 = 181.74 40% of 466 = 186.4 41% of 466 = 191.06 42% of 466 = 195.72 43% of 466 = 200.38 44% of 466 = 205.04 45% of 466 = 209.7 46% of 466 = 214.36 47% of 466 = 219.02 48% of 466 = 223.68 49% of 466 = 228.34 50% of 466 = 233
51% of 466 = 237.66 52% of 466 = 242.32 53% of 466 = 246.98 54% of 466 = 251.64 55% of 466 = 256.3 56% of 466 = 260.96 57% of 466 = 265.62 58% of 466 = 270.28 59% of 466 = 274.94 60% of 466 = 279.6 61% of 466 = 284.26 62% of 466 = 288.92 63% of 466 = 293.58 64% of 466 = 298.24 65% of 466 = 302.9 66% of 466 = 307.56 67% of 466 = 312.22 68% of 466 = 316.88 69% of 466 = 321.54 70% of 466 = 326.2 71% of 466 = 330.86 72% of 466 = 335.52 73% of 466 = 340.18 74% of 466 = 344.84 75% of 466 = 349.5
76% of 466 = 354.16 77% of 466 = 358.82 78% of 466 = 363.48 79% of 466 = 368.14 80% of 466 = 372.8 81% of 466 = 377.46 82% of 466 = 382.12 83% of 466 = 386.78 84% of 466 = 391.44 85% of 466 = 396.1 86% of 466 = 400.76 87% of 466 = 405.42 88% of 466 = 410.08 89% of 466 = 414.74 90% of 466 = 419.4 91% of 466 = 424.06 92% of 466 = 428.72 93% of 466 = 433.38 94% of 466 = 438.04 95% of 466 = 442.7 96% of 466 = 447.36 97% of 466 = 452.02 98% of 466 = 456.68 99% of 466 = 461.34 100% of 466 = 466 | HuggingFaceTB/finemath | |
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3.7E: Rational Functions (Exercises)
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Match each equation form with one of the graphs. $1. f\left(x\right)=\frac{x-A}{x-B} 2. g\left(x\right)=\frac{\left(x-A\right)^{2} }{x-B} 3. h\left(x\right)=\frac{x-A}{\left(x-B\right)^{2} } 4. k\left(x\right)=\frac{\left(x-A\right)^{2} }{\left(x-B\right)^{2} }$
A B C D
For each function, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal asymptote. Use that information to sketch a graph.
$5.p\left(x\right)=\frac{2x-3}{x+4} 6. q\left(x\right)=\frac{x-5}{3x-1}$
$7. s\left(x\right)=\frac{4}{\left(x-2\right)^{2} } 8. r\left(x\right)=\frac{5}{\left(x+1\right)^{2} }$
$9. f\left(x\right)=\frac{3x^{2} -14x-5}{3x^{2} +8x-16} 10. g\left(x\right)=\frac{2x^{2} +7x-15}{3x^{2} -14x+15}$
$11. a\left(x\right)=\frac{x^{2} +2x-3}{x^{2} -1} 12. b\left(x\right)=\frac{x^{2} -x-6}{x^{2} -4}$
$13. h\left(x\right)=\frac{2x^{2} +\; x-1}{x-4} 14. k\left(x\right)=\frac{2x^{2} -3x-20}{x-5}$
$15. n\left(x\right)=\frac{3x^{2} +4x-4}{x^{3} -4x^{2} } 16. m\left(x\right)=\frac{5-x}{2x^{2} +7x+3}$
$17. w\left(x\right)=\frac{\left(x-1\right)\left(x+3\right)\left(x-5\right)}{\left(x+2\right)^{2} (x-4)} 18. z\left(x\right)=\frac{\left(x+2\right)^{2} \left(x-5\right)}{\left(x-3\right)\left(x+1\right)\left(x+4\right)}$
Write an equation for a rational function with the given characteristics.
1. Vertical asymptotes at $$x=5$$ and $$x=-5$$
x intercepts at $$(2,\; 0)$$ and $$(-1,\; 0)$$ y intercept at $$\left(0,\; 4\right)$$
1. Vertical asymptotes at $$x=-4$$ and $$x=-1$$
x intercepts at $$\left(1,\; 0\right)$$ and $$\left(5,\; 0\right)$$ y intercept at $$(0,\; 7)$$
1. Vertical asymptotes at $$x=-4$$ and $$x=-5$$
x intercepts at $$\left(4,\; 0\right)$$ and $$\left(-6,\; 0\right)$$ Horizontal asymptote at $$y=7$$
1. Vertical asymptotes at $$x=-3$$ and $$x=6$$
x intercepts at $$\left(-2,\; 0\right)$$ and $$\left(1,\; 0\right)$$ Horizontal asymptote at $$y=-2$$
1. Vertical asymptote at $$x=-1$$
Double zero at $$x=2$$ y intercept at $$(0,\; 2)$$
1. Vertical asymptote at $$x=3$$
Double zero at $$x=1$$ y intercept at $$(0,\; 4)$$
Write an equation for the function graphed.
25. 26.
27. 28.
Write an equation for the function graphed.
29. 30.
31. 32.
33. 34.
35. 36.
Write an equation for the function graphed.
37. 38.
Find the oblique asymptote of each function.
$39. f(x)=\frac{3x^{2} +4x}{x+2} 40. g(x)=\frac{2x^{2} +3x-8}{x-1}$
$41. h(x)=\frac{x^{2} -x-3}{2x-6} 42. k(x)=\frac{5+x-2x^{2} }{2x+1}$
$43. m(x)=\frac{-2x^{3} +x^{2} -6x+7}{x^{2} +3} 44. n(x)=\frac{2x^{3} +x^{2} +x}{x^{2} +x+1}$
1. A scientist has a beaker containing 20 mL of a solution containing 20% acid. To dilute this, she adds pure water.
1. Write an equation for the concentration in the beaker after adding n mL of water.
2. Find the concentration if 10 mL of water has been added.
3. How many mL of water must be added to obtain a 4% solution?
4. What is the behavior as $$n\to \infty$$, and what is the physical significance of this?
1. A scientist has a beaker containing 30 mL of a solution containing 3 grams of potassium hydroxide. To this, she mixes a solution containing 8 milligrams per mL of potassium hydroxide.
1. Write an equation for the concentration in the tank after adding n mL of the second solution.
2. Find the concentration if 10 mL of the second solution has been added.
3. How many mL of water must be added to obtain a 50 mg/mL solution?
4. What is the behavior as $$n\to \infty$$, and what is the physical significance of this?
1. Oscar is hunting magnetic fields with his gauss meter, a device for measuring the strength and polarity of magnetic fields. The reading on the meter will increase as Oscar gets closer to a magnet. Oscar is in a long hallway at the end of which is a room containing an extremely strong magnet. When he is far down the hallway from the room, the meter reads a level of 0.2. He then walks down the hallway and enters the room. When he has gone 6 feet into the room, the meter reads 2.3. Eight feet into the room, the meter reads 4.4. [UW]
1. Give a rational model of form $$m\left(x\right)=\frac{ax+b}{cx+d}$$ relating the meter reading $$m(x)$$ to how many feet x Oscar has gone into the room.
2. How far must he go for the meter to reach 10? 100?
3. Considering your function from part (a) and the results of part (b), how far into the room do you think the magnet is?
2. The more you study for a certain exam, the better your performance on it. If you study for 10 hours, your score will be 65%. If you study for 20 hours, your score will be 95%. You can get as close as you want to a perfect score just by studying long enough. Assume your percentage score, $$p(n)$$, is a function of the number of hours, n, that you study in the form $$p(n)=\frac{an+b}{cn+d}$$. If you want a score of 80%, how long do you need to study? [UW]
A street light is 10 feet north of a straight bike path that runs east-west. Olav is bicycling down the path at a rate of 15 miles per hour. At noon, Olav is 33 feet west of the point on the bike path closest to the street light. (See the picture). The relationship between the intensity C of light (in candlepower) and the distance d (in feet) from the light source is given by $$C=\frac{k}{d^{2} }$$, where k is a constant depending on the light source. [UW]
1. From 20 feet away, the street light has an intensity of 1 candle. What is k?
2. Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds.
3. When will the light on Olav have maximum intensity?
4. When will the intensity of the light be 2 candles?
$245$ | HuggingFaceTB/finemath | |
# Ultrametric space
In mathematics, an ultrametric space is a metric space in which the triangle inequality is strengthened to ${\displaystyle d(x,z)\leq \max \left\{d(x,y),d(y,z)\right\}}$. Sometimes the associated metric is also called a non-Archimedean metric or super-metric.
## Formal definition
An ultrametric on a set M is a real-valued function
${\displaystyle d\colon M\times M\rightarrow \mathbb {R} }$
(where denote the real numbers), such that for all x, y, zM:
1. d(x, y) ≥ 0;
2. d(x, y) = d(y, x) (symmetry);
3. d(x, x) = 0;
4. if d(x, y) = 0 then x = y;
5. d(x, z) ≤ max {d(x, y), d(y, z)} (strong triangle inequality or ultrametric inequality).
An ultrametric space is a pair (M, d) consisting of a set M together with an ultrametric d on M, which is called the space's associated distance function (also called a metric).
If d satisfies all of the conditions except possibly condition 4 then d is called an ultrapseudometric on M. An ultrapseudometric space is a pair (M, d) consisting of a set M and an ultrapseudometric d on M.[1]
In the case when M is an Abelian group (written additively) and d is generated by a length function ${\displaystyle \|\cdot \|}$ (so that ${\displaystyle d(x,y)=\|x-y\|}$), the last property can be made stronger using the Krull sharpening to:
${\displaystyle \|x+y\|\leq \max \left\{\|x\|,\|y\|\right\}}$ with equality if ${\displaystyle \|x\|\neq \|y\|}$.
We want to prove that if ${\displaystyle \|x+y\|\leq \max \left\{\|x\|,\|y\|\right\}}$, then the equality occurs if ${\displaystyle \|x\|\neq \|y\|}$. Without loss of generality, let us assume that ${\displaystyle \|x\|>\|y\|}$. This implies that ${\displaystyle \|x+y\|\leq \|x\|}$. But we can also compute ${\displaystyle \|x\|=\|(x+y)-y\|\leq \max \left\{\|x+y\|,\|y\|\right\}}$. Now, the value of ${\displaystyle \max \left\{\|x+y\|,\|y\|\right\}}$ cannot be ${\displaystyle \|y\|}$, for if that is the case, we have ${\displaystyle \|x\|\leq \|y\|}$ contrary to the initial assumption. Thus, ${\displaystyle \max \left\{\|x+y\|,\|y\|\right\}=\|x+y\|}$, and ${\displaystyle \|x\|\leq \|x+y\|}$. Using the initial inequality, we have ${\displaystyle \|x\|\leq \|x+y\|\leq \|x\|}$ and therefore ${\displaystyle \|x+y\|=\|x\|}$.
## Properties
From the above definition, one can conclude several typical properties of ultrametrics. For example, for all ${\displaystyle x,y,z\in M}$, at least one of the three equalities ${\displaystyle d(x,y)=d(y,z)}$ or ${\displaystyle d(x,z)=d(y,z)}$ or ${\displaystyle d(x,y)=d(z,x)}$ holds. That is, every triple of points in the space forms an isosceles triangle, so the whole space is an isosceles set.
Defining the (open) ball of radius ${\displaystyle r>0}$ centred at ${\displaystyle x\in M}$ as ${\displaystyle B(x;r):=\{y\in M\mid d(x,y), we have the following properties:
• Every point inside a ball is its center, i.e. if ${\displaystyle d(x,y) then ${\displaystyle B(x;r)=B(y;r)}$.
• Intersecting balls are contained in each other, i.e. if ${\displaystyle B(x;r)\cap B(y;s)}$ is non-empty then either ${\displaystyle B(x;r)\subseteq B(y;s)}$ or ${\displaystyle B(y;s)\subseteq B(x;r)}$.
• All balls of strictly positive radius are both open and closed sets in the induced topology. That is, open balls are also closed, and closed balls (replace ${\displaystyle <}$ with ${\displaystyle \leq }$) are also open.
• The set of all open balls with radius ${\displaystyle r}$ and center in a closed ball of radius ${\displaystyle r>0}$ forms a partition of the latter, and the mutual distance of two distinct open balls is (greater or) equal to ${\displaystyle r}$.
Proving these statements is an instructive exercise.[2] All directly derive from the ultrametric triangle inequality. Note that, by the second statement, a ball may have several center points that have non-zero distance. The intuition behind such seemingly strange effects is that, due to the strong triangle inequality, distances in ultrametrics do not add up.
## Examples
• The discrete metric is an ultrametric.
• The p-adic numbers form a complete ultrametric space.
• Consider the set of words of arbitrary length (finite or infinite), Σ*, over some alphabet Σ. Define the distance between two different words to be 2n, where n is the first place at which the words differ. The resulting metric is an ultrametric.
• The set of words with glued ends of the length n over some alphabet Σ is an ultrametric space with respect to the p-close distance. Two words x and y are p-close if any substring of p consecutive letters (p < n) appears the same number of times (which could also be zero) both in x and y.[3]
• If r = (rn) is a sequence of real numbers decreasing to zero, then |x|r := lim supn→∞ |xn|rn induces an ultrametric on the space of all complex sequences for which it is finite. (Note that this is not a seminorm since it lacks homogeneity — If the rn are allowed to be zero, one should use here the rather unusual convention that 00 = 0.)
• If G is an edge-weighted undirected graph, all edge weights are positive, and d(u,v) is the weight of the minimax path between u and v (that is, the largest weight of an edge, on a path chosen to minimize this largest weight), then the vertices of the graph, with distance measured by d, form an ultrametric space, and all finite ultrametric spaces may be represented in this way.[4]
## References
1. ^ Narici & Beckenstein 2011, pp. 1–18.
2. ^ "Ultrametric Triangle Inequality". Stack Exchange.
3. ^ Osipov, Gutkin (2013), "Clustering of periodic orbits in chaotic systems", Nonlinearity, 26 (26): 177–200, Bibcode:2013Nonli..26..177G, doi:10.1088/0951-7715/26/1/177.
4. ^ Leclerc, Bruno (1981), "Description combinatoire des ultramétriques", Centre de Mathématique Sociale. École Pratique des Hautes Études. Mathématiques et Sciences Humaines (in French) (73): 5–37, 127, MR 0623034.
5. ^ Mezard, M; Parisi, G; and Virasoro, M: SPIN GLASS THEORY AND BEYOND, World Scientific, 1986. ISBN 978-9971-5-0116-7
6. ^ Rammal, R.; Toulouse, G.; Virasoro, M. (1986). "Ultrametricity for physicists". Reviews of Modern Physics. 58 (3): 765–788. Bibcode:1986RvMP...58..765R. doi:10.1103/RevModPhys.58.765. Retrieved 20 June 2011.
7. ^ Legendre, P. and Legendre, L. 1998. Numerical Ecology. Second English Edition. Developments in Environmental Modelling 20. Elsevier, Amsterdam.
8. ^ Benzi, R.; Biferale, L.; Trovatore, E. (1997). "Ultrametric Structure of Multiscale Energy Correlations in Turbulent Models". Physical Review Letters. 79 (9): 1670–1674. arXiv:chao-dyn/9705018. Bibcode:1997PhRvL..79.1670B. doi:10.1103/PhysRevLett.79.1670. S2CID 53120932.
9. ^ Papadimitriou, Fivos (2013). "Mathematical modelling of land use and landscape complexity with ultrametric topology". Journal of Land Use Science. 8 (2): 234–254. doi:10.1080/1747423x.2011.637136. ISSN 1747-423X. S2CID 121927387. | HuggingFaceTB/finemath | |
# The vapour pressure of water is 12.3kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Given: 1 molal solution means 1 mole of solute present in 1000g of water solvent)
Molecular weight of water = H2O = 1 × 2 + 16 = 18g/mol
No. of moles of water, n = given mass /molecular weight
n = 1000/18 = 55.56 gmol-1
Mole fraction of solute in solution,x2 = moles of solute/(moles of solute + moles of water)
x2 = 1/(1 + 55.56)
x2 = 0.0177
Given vapour pressure of pure water at 300k is 12.3 kpa
Apply the formula: P1 = 12.0823kpa
which is the vapour pressure of the solution.
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# Math
posted by .
Choose the one alternative that best completes the statement or answers the question.
Solve the problem. Assume that simple interest is being calculated in each case. Round your answer to the nearest cent.
Martin takes out a simple-interest loan at 7.5 %. After 6 months, the amount of interest on the loan is \$69.64. What was the amount of the loan?
I do not understand how to do this homework question can someone please help me to understand how to figure this one out thank you so much
• Math -
principle= rate x time
p=rt
\$69.64=7.5% x t
solve for t
• Math -
But how do you figure out T?
Out of the multiple choice answer nothing is coming close to one of the answers
these are the multiple choice answers
A)
\$2116.98
B)
\$18.57
C)
\$1857
D)
\$1818.44
• Math -
Allison's equation is wrong.
interest earned = P*i*t, where t = 0.5 years, P is the original loan amount, and i = 0.075 is the annual interest rate.
69.64 = 0.0375 P
P = \$1857
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the principle B is borrowed at a simple interest rate r for a period of time t. Find the simple interest owed for the use of the money,. Assume there are 360 days in a year and round to the nearest cent. P=\$4000, r=6.0%and t=6 month
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# 24.4 Energy in electromagnetic waves (Page 2/6)
Page 2 / 6
${I}_{\text{ave}}=\frac{{\text{cB}}_{0}^{2}}{{2\mu }_{0}},$
where ${B}_{0}$ is the maximum magnetic field strength.
One more expression for ${I}_{\text{ave}}$ in terms of both electric and magnetic field strengths is useful. Substituting the fact that $c\cdot {B}_{0}={E}_{0}$ , the previous expression becomes
${I}_{\text{ave}}=\frac{{E}_{0}{B}_{0}}{{2\mu }_{0}}.$
Whichever of the three preceding equations is most convenient can be used, since they are really just different versions of the same principle: Energy in a wave is related to amplitude squared. Furthermore, since these equations are based on the assumption that the electromagnetic waves are sinusoidal, peak intensity is twice the average; that is, ${I}_{0}=2{I}_{\text{ave}}$ .
## Calculate microwave intensities and fields
On its highest power setting, a certain microwave oven projects 1.00 kW of microwaves onto a 30.0 by 40.0 cm area. (a) What is the intensity in ${\text{W/m}}^{2}$ ? (b) Calculate the peak electric field strength ${E}_{0}$ in these waves. (c) What is the peak magnetic field strength ${B}_{0}$ ?
Strategy
In part (a), we can find intensity from its definition as power per unit area. Once the intensity is known, we can use the equations below to find the field strengths asked for in parts (b) and (c).
Solution for (a)
Entering the given power into the definition of intensity, and noting the area is 0.300 by 0.400 m, yields
$I=\frac{P}{A}=\frac{1\text{.}\text{00 kW}}{0\text{.}\text{300 m}\phantom{\rule{0.25em}{0ex}}×\phantom{\rule{0.25em}{0ex}}0\text{.}\text{400 m}}.$
Here $I={I}_{\text{ave}}$ , so that
${I}_{\text{ave}}=\frac{\text{1000 W}}{0\text{.}\text{120}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}}=8\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}.$
Note that the peak intensity is twice the average:
${I}_{0}=2{I}_{\text{ave}}=1\text{.}\text{67}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{W}/{\text{m}}^{2}.$
Solution for (b)
To find ${E}_{0}$ , we can rearrange the first equation given above for ${I}_{\text{ave}}$ to give
${E}_{0}={\left(\frac{2{I}_{\text{ave}}}{{\mathrm{c\epsilon }}_{0}}\right)}^{1/2}.$
Entering known values gives
$\begin{array}{lll}{E}_{0}& =& \sqrt{\frac{2\left(8\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}\right)}{\left(3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(8.85×{\text{10}}^{–12}\phantom{\rule{0.25em}{0ex}}{\text{C}}^{2}/\text{N}\cdot {\text{m}}^{2}\right)}}\\ & =& 2.51×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V/m}\text{.}\end{array}$
Solution for (c)
Perhaps the easiest way to find magnetic field strength, now that the electric field strength is known, is to use the relationship given by
${B}_{0}=\frac{{E}_{0}}{c}.$
Entering known values gives
$\begin{array}{lll}{B}_{0}& =& \frac{2.51×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V/m}}{3.0×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}\\ & =& 8.35×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T}\text{.}\end{array}$
Discussion
As before, a relatively strong electric field is accompanied by a relatively weak magnetic field in an electromagnetic wave, since $B=E/c$ , and $c$ is a large number.
## Section summary
• The energy carried by any wave is proportional to its amplitude squared. For electromagnetic waves, this means intensity can be expressed as
${I}_{\text{ave}}=\frac{{\mathrm{c\epsilon }}_{0}{E}_{0}^{2}}{2},$
where ${I}_{\text{ave}}$ is the average intensity in ${\text{W/m}}^{2}$ , and ${E}_{0}$ is the maximum electric field strength of a continuous sinusoidal wave.
• This can also be expressed in terms of the maximum magnetic field strength ${B}_{0}$ as
${I}_{\text{ave}}=\frac{{\text{cB}}_{0}^{2}}{{2\mu }_{0}}$
and in terms of both electric and magnetic fields as
${I}_{\text{ave}}=\frac{{E}_{0}{B}_{0}}{{2\mu }_{0}}.$
• The three expressions for ${I}_{\text{ave}}$ are all equivalent.
## Problems&Exercises
What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?
$\begin{array}{lll}I& =& \frac{{\mathrm{c\epsilon }}_{0}{E}_{0}^{2}}{2}\\ & =& \frac{\left(3.00×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(8.85×{\text{10}}^{\text{–12}}{\text{C}}^{2}\text{/N}\cdot {\text{m}}^{2}\right){\left(1\text{25 V/m}\right)}^{2}}{2}\\ & =& 20.{\text{7 W/m}}^{2}\end{array}$
Find the intensity of an electromagnetic wave having a peak magnetic field strength of $4\text{.}\text{00}×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{T}$ .
Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.
(a) $I=\frac{P}{A}=\frac{P}{\pi {r}^{2}}=\frac{0\text{.}\text{250}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{W}}{\pi {\left(0\text{.}\text{500}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}^{2}}={\text{318 W/m}}^{2}$
(b) $\begin{array}{lll}{I}_{\text{ave}}& =& \frac{{\text{cB}}_{0}^{2}}{{2\mu }_{0}}⇒{B}_{0}={\left(\frac{{2\mu }_{0}I}{c}\right)}^{1/2}\\ & =& {\left(\frac{2\left(4\pi ×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}\right)\left(\text{318}\text{.}{\text{3 W/m}}^{2}\right)}{\text{3.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}\right)}^{1/2}\\ & =& 1\text{.}\text{63}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T}\end{array}$
(c) $\begin{array}{lll}{E}_{0}& =& {\text{cB}}_{0}=\left(3\text{.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(\text{1.633}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T}\right)\\ & =& 4\text{.}\text{90}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}\text{V/m}\end{array}$
A body travelling at a velocity of 30ms^-1 in a straight line is brought to rest by application of brakes. if it covers a distance of 100m during this period, find the retardation.
what's acceleration
The change in position of an object with respect to time
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mass × acceleration OR Work done ÷ distance
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acute astigmatism?
the difference between virtual work and virtual displacement
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What is Elasticity
using a micro-screw gauge,the thickness of a piece of a A4 white paper is measured to be 0.5+or-0.05 mm. If the length of the A4 paper is 26+or-0.2 cm, determine the volume of the A4 paper in: a). Cubic centimeters b). Cubic meters
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# Solving Quadratic Equations Using Square Roots Worksheet
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Top | HuggingFaceTB/finemath | |
## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
(a) We can equate the distance traveled by David and Tina to find the time it takes Tina to pass David; $\frac{1}{2}at^2=v~t$ $t = \frac{2v}{a} = \frac{(2)(30~m/s)}{2.0~m/s^2}$ $t = 30~s$ We can find the distance that Tina travels in 30 seconds. $x = \frac{1}{2}at^2$ $x = \frac{1}{2}(2.0~m/s^2)(30~s)^2$ $x = 900~m$ Tina drives 900 meters. (b) $v = a~t = (2.0~m/s^2)(30~s)$ $v = 60~m/s$ | HuggingFaceTB/finemath | |
Subgroups of $G\times H$
Is it true that if $$G,H$$ are groups, then every subgroup $$S$$ of $$G\times H$$ can be written in the form $$S\cong S_G\times S_H$$, where $$S_G$$ and $$S_H$$ are subgroups of $$G$$ and $$H$$ respectively? That is, is $$S$$ isomorphic (not necessarily equal) to a direct product of subgroups of $$G$$ and $$H$$?
Notice that I am not asking whether $$S=S_G\times S_H$$ for some $$S_G,S_H$$ subgroups of $$G,H$$; this is obviously false. Wikipedia provides a trivial counterexample: consider the subgroup of $$G\times G$$ consisting of elements in the form $$(g,g)$$. This is clearly not a direct product of subgroups of $$G$$. However, it is isomorphic to a direct product of subgroups of $$G$$; namely, the product $$G\times \{e\}$$.
If you set $$G = H = S_3$$, the symmetric group of order 6, then consider the subgroup $$K$$ of elements $$(x,y)$$ where either both $$x$$ and $$y$$ are even or both $$x$$ and $$y$$ are odd. It's not hard to see that $$K$$ is a subgroup of $$S_3 \times S_3$$ of order 18.
Now if $$K$$ was to have a decomposition $$K \cong G_1 \times G_2$$ where $$G_1, G_2$$ are isomorphic to subgroups of $$S_3$$, then by consideration of order one of $$|G_1|,|G_2|$$ would be equal to 6 and the other would be equal to 3. Without loss of generality set $$|G_1| = 6, |G_2| = 3$$. Let $$x$$ be an element of $$G_1$$ of order 2. Then $$(x,1)$$ must commute with all elements of $$G_1 \times G_2$$ of form $$(1,y)$$ for all $$y$$ in $$G_2$$. This includes the two non-identity elements of $$G_2$$ that are of order 3. This provides two commuting elements $$(x,1)$$ and $$(1,y)$$ of $$G_1 \times G_2$$ of orders 2,3, respectively.
However, it is straightforward to check that the elements of order 2 in $$K$$ do not commute with any of the elements of order 3 in $$K$$. The elements of $$K$$ of order two are all of the form $$(t_1, t_2)$$ where $$t_1$$ and $$t_2$$ are both transpositions. Furthermore, all nine such elements are conjugate (as any two conjugates in $$S_3$$ are conjugate), meaning that for any given element $$(t_1,t_2)$$, the centralizer is exactly given by $$\langle t_1 \rangle \times \langle t_2\rangle$$, which is a Klein four subgroup containing no elements of order 3.
Thus, by contradiction, $$K$$ does not decompose as a direct product of form $$G_1 \times G_2$$. | HuggingFaceTB/finemath | |
## Introduction
In this blog post, I would like to continue my series on “building from scratch.” I will discuss a linear classifier called Logistic Regression. This blog post covers the following topics,
1. Basics of a classifier
2. Decision Boundaries
3. Maximum Likelihood Principle
4. Logistic Regression Equation
5. Logistic Regression Cost Function
After the discussion of the theoretical concepts we will dive into the code. So, without a further adieu let’s start the discussion with the basics of a classifier.
## Basics of a Classifier
A classifier is an estimator that assigns a class label to the input data point. Let’s take an example to understand it better. Let’s say we have images of animals and we want to associate them with their correct labels. For example, A dog’s image will be associated a label “Dog” or a cat’s image will be associated with a label “Cat” and the estimator which does this job can be called as a classifier. Classifiers are majorly of two types- A Binary Classifier & A Multi-Class Classifier. A binary classifier helps us associate data points with 2 labels while a multi-class classifier associates data points with multiple classes. Some examples of binary classifiers and multi-class classifiers are listed below:
Binary Classification Examples:
1. Spam/Ham mail classification
2. Fraudulent Transactions classification
3. Heart Disease classification
Multi-Class Classification Examples:
1. Animal Images Classification (Dog/Cat/Horse/Human)
2. Iris Flower Species Classification
3. MNIST Digit Classifier
Okay, so now we understand “a classifier” but what is a linear classifier? To understand a linear classifier, let’s look into the concept of the decision boundary.
## Decision Boundaries
Technically, a decision boundary is the region in space in which the output label of a classifier is ambiguous. Simply speaking if a data point lies on the decision boundary in space then it can belong to any of the classes. Intuitively, a decision boundary is a boundary that separates the classes in space. For example, if we have built a binary classifier which classifier data points into one of the two classes, then a decision boundary would be a boundary that separates the two classes. Data points lying on one side of the boundary would belong to one class and the points lying on the other side will belong to the other class. If this boundary is linear or a straight-line then it is known as a Linear Decision Boundary and the classifier that creates such a boundary is called a Linear Classifier. Logistic Regression is one such classifier. The below attached image shows different types of decision boundaries generated using different models.
In the next, section we will talk about one of the most important topics in statistics and probability which is the basis of a lot of machine learning algorithms including the Logistic Regression, it’s called Maximum Likelihood Estimation.
## Maximum Likelihood Estimation
We are aware of various estimators or classifiers or models that give the best result on a certain data but how are we able to know that a particular estimator is able to model the given data well. Every estimator has a certain set of parameters. For example, a Gaussian distribution has the mean & variance as the parameters. Now we have to determine the values of these parameters such that it helps us model the given data distribution. Using Maximum Likelihood Estimation we can calculate the ideal values of these parameters that can lead us to model the given data distribution.
Let us understand this with the help of an example, We have a data distribution that follows a Gaussian Distribution with parameters μ & σ which are unknown to us. We know that a Gaussian Distribution can be modelled using the following function,
Now, we can use the maximum likelihood estimation to estimate these parameters which would model the data. In other words, we need to find the parameters such that it would maximize the likelihood of this function modelling the given data. If you have studied calculus in mathematics then you must be aware of the fact that we can differentiate the function and equate it to zero to identify the points where it is maximised/minimised and use the double differential of the function to confirm whether the points obtained actually maximise or minimise the given function.
The above example, in fact forms the basis of Mean Squared Error being selected as the loss function for Linear Regression.
Next, we look into the Logistic Regression algorithm while coding it side by side.
## Logistic Regression Equation
Before, going into the code let us understand the very basics of the Logistic Regression algorithm. Logistic Regression is a linear classifier that gives out the probabilities of a data point belonging to a particular class. The equation for the Logistic Regression is same as the Linear Regression equation. Now the question arises how can a linear equation output probabilities. To convert the outputs of a linear equation into probabilities we use a mathematical function called sigmoid function. It is a “S” shaped function that limits the outputs of the linear equation between 0 & 1. Mathematically, it is given by,
Where e is the Euler’s constant. Next, we will look into the code of sigmoid & logistic regression function.
``````# 0. Helper function: Sigmoid
def sigmoid(x):
'''
sigmoid(x) = 1 / (1 + e^(-x))
'''
return 1 / (1 + np.exp(-x))
# 1. Hypothesis (Logistic Function)
def hypothesis(x, theta):
# h(x) = sigmoid(X.theta)
z = np.dot(X, theta)
return sigmoid(z)
``````
Next, we will discuss the cost function used to measure the error while training the Logistic Regression algorithm.
## Logistic Regression Cost Function
While selecting the cost function we consider properties like easy to optimise & should have no local minima because that can lead us to being stuck in a solution which is not optimal. In Linear Regression, we used Mean Squared Error as the Loss Function. However, in case of a Logistic Regression doesn’t turn out to be a convex function i.e it has global as well as local minima. The below attached image shows the difference between a convex & a non-convex function.
So, now we know that we cannot use Mean Squared Error as the cost function for Logistic Regression. So, what should be the cost function for training Logistic Regression? We use something known as Binary Cross-Entropy. It is given as,
Where yi is the actual class label and Å·i is the probability of the data point belonging to the first class. To learn more about the Binary Cross-Entropy refer to this blog post. Now we know the loss function for training the logistic regression model. Let’s try to code the same. The below attached code snippet demonstrates this.
``````# 2. Loss Function: Binary Cross Entropy
def binary_cross_entropy(x, y, theta):
m, n = x.shape
# a. Compute the hypothesis
y_hat = hypothesis(x, theta)
# b. Compute the Binary Cross Entropy
loss = y * np.log(y_hat) + (1 - y) * np.log(1 - y_hat)
return - np.mean(loss) ``````
Now we have defined the Logistic Function, Binary Cross-Entropy Loss function, next we will see the algorithm to train the Logistic Regression model.
Just like Linear Regression, the Gradient Descent Algorithm can be used to train the Logistic Regression model & obtain the ideal parameters. I have explained the theory & the intuition of the Gradient Descent Algorithm in one of my blog posts. If you think that you need a refresher on it please refer to this blog post. The code for the Gradient Descent Algorithm can be found attached below.
``````# 3. Compute the gradient
# Compute hypothesis
y_hat = hypothesis(x, theta)
grad = np.dot( x.T, (y - y_hat))
def gradient_descent(x, y, n_iter = 100, alpha = 0.1):
# a. Randomly initialise theta
m,n = x.shape
theta = np.zeros(shape = (n, ))
# List to store the error
error = []
# b. Perform the gradient descent
for i in range(n_iter):
'''
y_hat = hypothesis(x, theta)
print(y_hat, y_hat.shape)
'''
# b.1. Compute the loss
loss = binary_cross_entropy(x, y, theta)
error.append(loss)
# b.3. Perform the update rule
theta = theta - alpha * grad
return theta, error``````
Now, we have coded the entire algorithm from scratch let’s test its performance on a custom dataset and btw you can refer to the entire notebook on my Kaggle profile using this link.
## Testing the Algorithm
In this section, we test our Logistic Regression on a custom dataset & compare its performance with the Scikit Learn’s version of Logistic Regression.
#### 1. Make a custom dataset
We make a custom dataset using sklearn’s make_blobs function & visualise it using seaborn library. The below code cell demonstrates the same.
``````# 1. Create Dataset
X, y = make_blobs(n_samples = 1000, n_features = 2, centers=2, random_state=0)
dataset_array = np.concatenate((X, y.reshape(-1,1)), axis=1)
# 2. Create a Dataframe of the array
dataset_df = pd.DataFrame(dataset_array, columns = ['Col 1', 'Col 2', 'Target'])
# 3. plot the dataset
sns.scatterplot(data=dataset_df, x='Col 1', y='Col 2', hue='Target')
plt.xlabel("Column 1")
plt.ylabel("Column 2")
plt.show()
``````
#### 2. Testing the Logistic Regression & Visualising the loss
Next we, test the algorithm & visualise the loss. The below attached code cell demonstrates the following.
``````X = dataset_df_copy.drop('Target', axis=1)
y = dataset_df_copy['Target']
theta, error = gradient_descent(X, y, 10000)
# plot the error
plt.plot(error)
plt.xlabel("Number of iterations")
plt.ylabel("Error")
plt.show()``````
From the above plot, it can be observed that after ~2000 iterations the error starts to saturate and doesn’t decrease further. We can say that the algorithm has reached the minima. Next, we make the predictions & plot the decision boundary of the logistic regression.
#### 3. Predictions & Decision Boundary
Next we generate the decision boundary & visualise the extent of separation of the two classes by the boundary. Since, the Logistic Regression is a linear model, the decision boundary will be a straight line. The following code cell demonstrates the same.
``````# plot the dataset along with the decision boundary
# Create Decision Boundary
x2_max, x2_min = X['Col 2'].max(), X['Col 2'].min()
x1_max, x1_min = X['Col 1'].max(), X['Col 1'].min()
x_vals = np.array([-2, 5])
slope = - theta[1] / theta[2]
intercept = - theta[0] / theta[2]
decision_boundary = slope * x_vals + intercept
# Plot the dataset with decision boundary
plt.figure(figsize=(12,8))
sns.scatterplot(data=dataset_df, x='Col 1', y='Col 2', hue='Target')
plt.plot(x_vals, decision_boundary, linestyle='--', color='black', label='Decision Boundary')
plt.fill_between(x_vals, decision_boundary, x2_min-10, color='tab:orange', alpha=0.2)
plt.fill_between(x_vals, decision_boundary, x2_max+10, color='tab:blue', alpha=0.2)
plt.xlabel("Column 1")
plt.ylabel("Column 2")
plt.ylim(x2_min-1, x2_max)
plt.xlim(x1_min, 5)
plt.legend(loc='best')
plt.show()``````
From the above plot, it can be clearly observed that the Logistic Regression model is able to separate the two classes almost perfectly. Next, we see the performance of Scikit Learn’s Logistic Regression & compare it with our own.
#### 4. Comparison with Scikit-Learn Logistic Regression
Next, we generate results on our custom dataset using the Scikit-Learn’s Logistic Regression model & compare its performance with our own. The below code cell implements the same.
``````# Import logistic regression
from sklearn.linear_model import LogisticRegression
# Build the model
lr = LogisticRegression()
lr.fit(X.drop('Constant', axis=1), y)
# Compute coeffecients
theta_sklearn = lr.coef_
intercept_sklearn = lr.intercept_
# Plot the Decision Boundary
# plot the dataset along with the deicision boundary
# Create Decision Boundary
x2_max, x2_min = X['Col 2'].max(), X['Col 2'].min()
x1_max, x1_min = X['Col 1'].max(), X['Col 1'].min()
x_vals = np.array([-2, 5])
slope = - theta_sklearn[0][0] / theta_sklearn[0][1]
intercept = - intercept_sklearn / theta_sklearn[0][1]
decision_boundary = slope * x_vals + intercept
# Plot the dataset with decision bounddart
plt.figure(figsize=(12,8))
sns.scatterplot(data=dataset_df, x='Col 1', y='Col 2', hue='Target')
plt.plot(x_vals, decision_boundary, linestyle='--', color='black', label='Decision Boundary')
plt.fill_between(x_vals, decision_boundary, x2_min-10, color='tab:orange', alpha=0.2)
plt.fill_between(x_vals, decision_boundary, x2_max+10, color='tab:blue', alpha=0.2)
plt.xlabel("Column 1")
plt.ylabel("Column 2")
plt.ylim(x2_min-1, x1_max+4)
plt.xlim(x1_min, 5)
plt.legend(loc='best')
plt.show()``````
Next, we look into both the model’s parameters & compare the same.
``````# Print the Custom Logistic Regression's results
print("Weights of variable given out by custom Logistic Regression")
print("Col 1: {}".format(theta[1]))
print("Col 2: {}".format(theta[2]))
print("Intercept : {}".format(theta[0]))
print()
print("Weights of variable given out by Sklearn's Logistic Regression")
print("Col 1: {}".format(theta_sklearn[0][0]))
print("Col 2: {}".format(theta_sklearn[0][1]))
print("Intercept : {}".format(intercept_sklearn[0]))``````
Next we use accuracy as a performance metric to quantify the model’s performance. We compute the accuracy score of both the custom as well as the sklearn’s Logistic Regression model & compare their performance.
``````# Compute accuracy for both the models
# 1. Custom Logistic Regression
predictions_1 = np.round(hypothesis(X.drop('Constant', axis=1), theta))
acc1 = np.sum(predictions_1 == y) / len(y) * 100
# 2. Sklearn's Logistic Regression
predictions_2 = lr.predict(X.drop('Constant', axis=1))
acc2 = np.sum(predictions_2 == y) / len(y) * 100
print("Accuracy of custom Logistic Regression Classifier: {}%".format(acc1))
print("Accuracy of sklearn's Logistic Regression Classifier: {}%".format(acc2))``````
Clearly, both the models are performing equally well. So, this was the entire comparison of both the Logistic Regression models. You can access the entire notebook here.
## Conclusion
I will conclude this blog post with a quick recap of what all we discussed. First, we learnt about the basics of a classifier, then we learnt about the Decision Boundaries followed by Maximum Likelihood Estimation. We started coding the Logistic Regression model & also discussed about the cost function & gradient descent algorithm. Lastly, we compared the performance of our Logistic Regression with that of Python’s Scikit-Learn library.
I hope you found this blog post insightful. Please do share it with your friends & family and subscribe to my blog Keeping Up With Data Science for more informative content on Data Science straight to your inbox. You can reach out to me on Twitter & LinkedIn. I am quite active there & I will be happy to have a conversation with you. Please feel free to drop your feedback in the comments that helps me to improve the quality of my work. I will keep on sharing more content as I grow & mature as a Data Scientist. Until next time, Keep Hustling & Keep Up with Data Science. Happy Learning 🙂
### One response to “Logistic Regression in Machine Learning (from Scratch !!)”
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# What Are Consecutive Numbers? - Definition & Examples
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Instructor: T.J. Hoogsteen
T.J. is currently a grade 5 teacher and Vice-Principal. He has a master's degree in Educational Administration and is working toward an Ed.D. in Educational Leadership.
When numbers are in counting order, they are also in consecutive order. This lesson will review the three different types of consecutive numbers (counting order, even, and odd),
## What Are Consecutive Numbers?
Jeff and Omar were sitting around and talking about great sports teams. Jeff's friend brought up the New York Yankees dynasty from the late 1990s. He said that the Yankees won the World Series for three consecutive years starting in 1998. Omar, knowing the first World Series was won in 1998, tried to think of the years that the other two titles were won.
How could he figure that out? First, he would have to figure out what the consecutive numbers are that come after 1998. To do that, knowing what consecutive means would help. Consecutive means to follow continuously in a series or sequence. So that means consecutive numbers are numbers that follow each other in order.
Another thing to think about when solving the above problem is that there are three different kinds of consecutive numbers: consecutive numbers, even consecutive numbers, and odd consecutive numbers.
## Consecutive Numbers
Consecutive numbers are numbers that follow each other in order from smallest to largest, what we call regular counting order. Let's take a look at some examples:
You can see that in each of these three series, the numbers go up by one each time, in sequence and without skipping any numbers in between. They are all series of consecutive numbers.
Even consecutive numbers are even numbers that follow in order from smallest to largest when you are counting by twos. Some examples of even consecutive numbers include:
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Determine all primitive functions
• beyondlight
In summary: So you need to find a substitution that will allow you to use the chain rule. In this problem, the function inside the parentheses looks suspiciously like the derivative of part of the function you're integrating, so let's try using that as our substitution.Let z = x^2 + 3. Then dz/dx = 2x, or x = (1/2)dz. Make that substitution, and your integral will become(1/2)∫2(z)^4 dz = ∫z^4 dz. That's a pretty easy integral to do, and then you can substitute back in for z to get the answer in terms of x.
beyondlight
Homework Statement
Determine all primitive functions for the function:
2x(x^2+3)^4
2. The attempt at a solution
When i expanded i got the primitive to be:
2(x^9/9)+3x^8+18x^6+54x^4+81x^2But this was wrong. I am not sure I have understood the question. Help?
beyondlight said:
Homework Statement
Determine all primitive functions for the function:
2x(x^2+3)^4
2. The attempt at a solution
When i expanded i got the primitive to be:
2(x^9/9)+3x^8+18x^6+54x^4+81x^2But this was wrong. I am not sure I have understood the question. Help?
I have moved this thread, as it is not a precalculus problem.
Apparently the problem asks you to find all antiderivatives of 2x(x2 + 3)4. In other words, carry out this integration: ##\int 2x(x^2 + 3)^4 dx##. The answer should have a highest degree term of x10. Don't forget the constant of integration.
Instead of expanding the binomial in parentheses, think about a simple substitution that you can do.
One possible substitution is 2x(z)^4
But I am not sure how to proceed from here...
beyondlight said:
One possible substitution is 2x(z)^4
If z = x^2 + 3, what is dz?
beyondlight said:
But I am not sure how to proceed from here...
A very simple substitution will work, and you're on the right track,
When you use substitution to evaluate an integral, you're using the chain rule in reverse.
1. What is a primitive function?
A primitive function is a fundamental function that cannot be further simplified or decomposed into simpler functions. It is the most basic form of a function and is often used as a starting point for more complex functions.
2. How do you determine all primitive functions?
To determine all primitive functions, you need to use the process of integration. This involves finding the antiderivative of a given function, which is the function that, when differentiated, gives the original function. The set of all antiderivatives is equivalent to the set of all primitive functions.
3. Can all functions have primitive functions?
No, not all functions have primitive functions. Only continuous functions have primitive functions. Discontinuous functions or functions with discontinuities at certain points do not have primitive functions.
4. How do you find the primitive function of a specific function?
To find the primitive function of a specific function, you can use techniques such as integration by parts, substitution, or partial fractions. These techniques help you manipulate the given function to find its antiderivative, which is the primitive function.
5. Why are primitive functions important?
Primitive functions are important because they allow us to solve complex problems involving derivatives and integrals. They also help us understand the behavior and properties of functions. In addition, primitive functions have many practical applications in fields such as physics, engineering, and economics.
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# Maximum of $f(x,y) = 1 - (x^2 + y^2)^{2/3}$
For the function
$f(x,y) = 1 - (x^2 + y^2)^{2/3}$
one has to find extrema and saddle points. Without applying much imagination, it is obvious that the global maximum is at $(0,0)$.
To prove that, I set up the Jacobian as
$$Df(x,y) = \left( -\frac{4}{3} x (x^2 y^2)^{-1/3}, -\frac{4}{3} y (x^2 y^2)^{-1/3}\right)$$
The only solution for $Df(x,y) = 0$ yields indeed $(0,0)$. In this point the Hessian is $$D^2f(0,0) = \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right)$$
which is both positive and negative semi-definite, suggesting that $(0,0)$ is both a minimum and a maximum - can that be? Or is it actually a saddle point? Thanks for your hints!
• I believe you omitted the addition signs in typing your derivatives, but probably not in your work. While it is true that the limits of the first derivatives are zero at $\ (0,0) \$ , those derivatives are undefined there. The "cusp" is likely a source of trouble for this technique of characterizing critical points; the point is unquestionably a global maximum, however. – colormegone Jun 14 '13 at 20:03
• The positive or negative definedness of the Hessian is only a sufficient condition for the point being a minimum or a maximum respectively. And it holds when the Hessian is defined, of course. – egreg Jun 14 '13 at 20:16
When analyzing this example forget about derivatives and Hessians. It is obvious that the function $$g(r):=1-r^{4/3}\qquad(r\geq0)$$ takes its maximum at $r=0$ and then strictly decreases with increasing $r$.
Introducing polar coordinates $(r,\phi)$ in the $(x,y)$-plane your function $$f:\ (x,y)\mapsto 1-(x^2+y^2)^{2/3}$$ appears as $$\tilde f(r,\phi)=g(r)\ .$$ Therefore the graph of $f$ resembles a "paraboloid looking downwards", and we just have the global maximum you spotted right away and no saddle points whatever.
Since $f(x,0)=1-x^{4/3}$ $\>(x\geq0)$ the given function is not twice differentiable at $(0,0)$; therefore the Hessian is not even defined there.
• No need for $r\ge 0$ since $r^{4/3}=(r^2)^{2/3}$. – vadim123 Jun 15 '13 at 14:58
Simpler proof: $f(r,\theta)=1-r^{4/3}$, which has unique maximum at $r=0$, using ordinary derivatives (no partials necessary). | open-web-math/open-web-math | |
This triangle was among many o… One way to calculate the numbers without doing all the other rows, is to use combinations.. the first one is 100 choose 0= 1, the next is 100 choose 1=100, etc.. now to compute those you … When all the odd integers in Pascal's triangle are highlighted (black) and the remaining evens are left blank (white), one of many patterns in Pascal's triangle is displayed (Figure 2). Pascal’s triangle starts with a 1 at the top. Now do the same in base $5$. Never . Figure 1 shows the first six rows (numbered 0 through 5) of the triangle. 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 Required options. The initial row with a single 1 on it is symmetric, and we do the same things on both sides, so however a number was generated on the left, the same thing was done to obtain the corresponding number on the right. 1 hour ago, Lua | Now think about the row after it. Then for each row after, each entry will be the sum of the entry to the top left and the top right. Things to Try. Welcome to The Pascal's Triangle -- First 12 Rows (A) Math Worksheet from the Patterning Worksheets Page at Math-Drills.com. When all the odd integers in Pascal's triangle are highlighted (black) and the remaining evens are left blank (white), one of many patterns in Pascal's triangle is displayed. What is the sum of the second numbers in the first $100$ rows of Pascal's triangle (excluding the first row, the row containing a single $1$)?The sum should be from the second to the hundredth row. $77 = 25*3 + 2*1 = 302_5$. Pascal's triangle is an arrangement of the binomial coefficients in a triangle. The black pixels correspond to the odd numbers in Pascal's triangle: (k = 0, 4, 32, 36, 64, 68, 96, 100). I've included a picture of a Sierpinski triangle [link #5] with row 100 highlighted. 1 … Try changing the program so that the first row of the triangle starts as "[1, 100, 100, 1]". I am assuming "what" means how do you calculate the numbers. N = the number along the row. So to work out the 3rd number on the sixth row, R=6 and N=3. 42 min ago, C# | Kicked out of Capitol, Trump diehards vow to fight on, Biden: Pro-Trump mob treated 'differently' than BLM, Why attack on U.S. Capitol wasn't a coup attempt, New congresswoman sent kids home prior to riots, Coach fired after calling Stacey Abrams 'Fat Albert', TV host: Rioters would be shackled if they were BLM, $2,000 checks back in play after Dems sweep Georgia, Serena's husband serves up snark for tennis critic, CDC: Chance of anaphylaxis from vaccine is 11 in 1M. 3 friends go to a hotel were a room costs$300. 100. Who was the man seen in fur storming U.S. Capitol? One possible interpretation for these numbers is that they are the coefficients of the monomials when you expand (a+b)^100. Each row represent the numbers in the powers of 11 (carrying over the digit if it is not a single number). Facts on pascals triangle ; True or false. Each number inside Pascal's triangle is calculated by adding the two numbers above it. Use the nCk formula if you want to confirm that they are odd. Get your answers by asking now. The Fibonacci Sequence. One main example of counting is Pascal’s Triangle. Pascal triangle is a triangular array of binomial coefficients. you decrease the column number k, until eventually you find a value smaller than z. 200. You can also center all rows of Pascal's Triangle, if you select prettify option, and you can display all rows upside down, starting from the last row first. =6x5x4x3x2x1 =720. Refer to the following figure along with the explanation below. Pascal's triangle is a triangular array of numbers in which every number is obtained by adding the two numbers directly above it. pleaseee help me solve this questionnn!?!? We can use this fact to quickly expand (x + y) n by comparing to the n th row of the triangle e.g. 282 . This series is called a binomial expansion. However, prototype must have the return type of int**. One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). Numbers written in any of the ways shown below. 1 hour ago, We use cookies for various purposes including analytics. Half of 80 is 40, so 40th place is the center of the line. Given a non-negative integer numRows, generate the first numRows of Pascal's triangle. Given a level L. The task is to find the sum of all the integers present at the given level in Pascal’s triangle . Binomial Theorem. 100 rows of Pascal's Triangle (it's probably 99 rows) a guest . 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 … For this, we use the rules of adding the two terms above just like in Pascal's triangle itself. Building Pascal’s triangle: On the first top row, we will write the number “1.” In the next row, we will write two 1’s, forming a triangle. Pascal's Triangle. Magic 11's. Better Solution: Let’s have a look on pascal’s triangle pattern . A different way to describe the triangle is to view the first line is an infinite sequence of zeros except for a single 1. Primes in Pascal triangle : The non-zero part is Pascal’s triangle. Try changing the program so that it adds a row if you click anywhere in the body of the document - so you don't need to click on the button. Each notation is read aloud "n choose r".These numbers, called binomial coefficients because they are used in the binomial theorem, refer to specific addresses in Pascal's triangle.They refer to the nth row, rth element in Pascal's triangle as shown below. The leftmost element in each row of Pascal's triangle is the 0 th 0^\text{th} 0 th element. It turns out that a triangle constructed this way has binomial coefficients as its elements. In Pascal's words (and with a reference to his arrangement), In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding row from its column to … 1 1 1. $23 = 5*4 + 3*1 = 43_5$ Add the two and you see there are $2$ carries. Another way to describe the problem: given integer z<=10^100, find the smallest integer n: exist integer k so that C(k,n) = z. 2 8 1 6 1... 1 2 9 1 6 1. It is named after the famous mathematician and physicist Blaise Pascal. What is the sum of the second numbers in the first $100$ rows of Pascal's triangle (excluding the first row, the row containing a single $1$)?The sum should be from the second to the hundredth row. 100. Pascal’s triangle has many interesting properties. The first row is 0 1 0 whereas only 1 acquire a space in Pascal’s triangle, 0s are invisible. These numbers are invaluable in combinatorics, probability theory, and other mathematical fields. All values outside the triangle are considered zero (0). Pascal’s Triangle represents a triangular shaped array of numbers with n rows, with each row building upon the previous row. The combinatorial function is available in excel. Here are some of the ways this can be done: Binomial Theorem. 42 min ago, C# | This interpretation is consistent with the interpretation that combin(i,j) is the number of ways you can choose "i" things from "j" options. The numbers in the row, 1 3 3 1, are the coefficients, and b indicates which coefficient in the row we are referring to. Pascal's triangle contains a vast range of patterns, including square, triangle and fibonacci numbers, as well as many less well known sequences. Facts on Blaise Pascal;True or false. Pascal's triangle is one of the classic example taught to engineering students. Show that the sum of the numbers in the nth row is 2 n. In any row, the sum of the first, third, fifth, … numbers is equal to the sum of the second, fourth, sixth, … numbers. Program to find if two numbers and their AM and HM are present in an array using STL. Pascal’s Triangle represents a triangular shaped array of numbers with n rows, with each row building upon the previous row. So $5^2$ divides $\binom{100}{77}$. Pascal's triangle is a way to visualize many patterns involving the binomial coefficient. In much of the Western world, it is named after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in India, Persia, China, Germany, and Italy.. Write a function that takes an integer value n as input and prints first n lines of the Pascal’s triangle. Each notation is read aloud "n choose r".These numbers, called binomial coefficients because they are used in the binomial theorem, refer to specific addresses in Pascal's triangle.They refer to the nth row, rth element in Pascal's triangle as shown below. One way to approach this problem is by having nested for loops: one which goes through each row, and one which goes through each column. One of the famous one is its use with binomial equations. Row 3 = 1, 3, 3, 1 . 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 In this example, we calculate 7 rows of Pascal's triangle and we center the results. The number of odd numbers in the Nth row of Pascal's triangle is equal to 2^n, where n is the number of 1's in the binary form of the N. In this case, 100 in binary is 1100100, so there are 8 odd numbers in the 100th row of Pascal's triangle. Discuss what are they and where are they located. And from the fourth row, we … 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 … In fact, if Pascal’s triangle was expanded further past Row 5, you would see that the sum of the numbers of any nth row would equal to 2^n. The coefficients of each term match the rows of Pascal's Triangle. The number patterns in this triangle have fascinated mathematicians for centuries and it was known to people in ancient Greece, India, Persia, China centuries even before Pascal studied it. Thus $\binom{100}{77}$ is divisible by $20$. This tool can generate arbitrary large Pascal's Triangles. By continuing to use Pastebin, you agree to our use of cookies as described in the. Sign Up, it unlocks many cool features! Although other mathematicians in Persia and China had independently discovered the triangle in the eleventh century, most of the properties and applications of the triangle were discovered by Pascal. For example, the numbers in row 4 are 1, 4, 6, 4, and 1 and 11^4 is equal to 14,641. Quick Note: In mathematics, Pascal's triangle is a triangular array of the binomial coefficients. A different way to describe the triangle is to view the first li ne is an infinite sequence of zeros except for a single 1. Pascal Triangle in Java at the Center of the Screen. Pascal's triangle is an arithmetic and geometric figure often associated with the name of Blaise Pascal, but also studied centuries earlier in India, Persia, China and elsewhere.. Its first few rows look like this: 1 1 1 1 2 1 1 3 3 1 where each element of each row is either 1 or the sum of the two elements right above it. Starting from the second row, I initially thought this meant you count from the left two numbers. text 73.08 KB . As well, i am not sure how I can check if my return value actually points to the pascal triangle. GOP delegate films himself breaking into Capitol. sum of elements in i th row 0th row 1 1 -> 2 0 1st row 1 1 2 -> 2 1 2nd row 1 2 1 4 -> 2 2 3rd row 1 3 3 1 8 -> 2 3 4th row 1 4 6 4 1 16 -> 2 4 5th row 1 5 10 10 5 1 32 -> 2 5 6th row 1 6 15 20 15 6 1 64 -> 2 6 7th row 1 7 21 35 35 21 7 1 128 -> 2 7 8th row … Add the two and you see that there are $5$ carries. To obtain successive lines, add every adjacent pair of numbers and write the sum between and below them. Following is combin(100,j) where j=0,1,2,3,4. They pay 100 each. This video shows how to find the nth row of Pascal's Triangle. Find all sides of a right angled triangle from given hypotenuse and area | Set 1. For this, just add the spaces before displaying every row. To obtain successive lines, add every adjacent pair of numbers and write the sum between and below them. Jul 20th, 2015. Remember that combin(100,j)=combin(100,100-j). = 3x2x1=6. You should be able to see that each number from the 1, 4, 6, 4, 1 row has been used twice in the calculations for the next row. 100 (a+b) 2. a 2 +2ab+ b 2. N! Row 3. I just recently learnt about pointers, why my attempt of the function doesn't work. 200. Example: In much of the Western world, it is named … Then see the code; 1 1 1 \ / 1 2 1 \/ \/ 1 3 3 1 The 100th row has 101 columns (numbered 0 through 100) Each entry in the row is. Not a member of Pastebin yet? •Polarity: When the elements of a row of Pascal’s triangle are added and subtracted together sequentially, every row with a middle number, meaning rows that have an odd number of integers, gives 0 as the result. Then, the next row down is the 1 st 1^\text{st} 1 st row, and so on. Examples: Input: N = 3 Output: 1, 3, 3, 1 Explanation: The elements in the 3 rd row are 1 3 3 1. Additional clarification: The topmost row in Pascal's triangle is the 0 th 0^\text{th} 0 th row. The rows of Pascal's triangle are conventionally enumerated starting with row n = 0 at the top (the 0th row). 100. Numbers written in any of the ways shown below. One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). For the purposes of these rules, I am numbering rows starting from 0, so that row … Triangular Numbers. Pascal’s triangle is a triangular array of the binomial coefficients. More rows of Pascal’s triangle are listed in Appendix B. The top of the triangle is truncated as we start from the 4th row, which already contains four binomial coefficients. After that, each entry in the new row is the sum of the two entries above it. Each of the inner numbers is the sum of two numbers in a row above: the value in the same column, and the value in the previous column. 8 There is an interesting property of Pascal's triangle that the nth row contains 2^k odd numbers, where k is the number of 1's in the binary representation of n. Note that the nth row here is using a popular convention that the top row of Pascal's triangle is row 0. Here we will write a pascal triangle … Maximum number of Perfect Numbers present in a subarray of size K. 14, Oct 20 . Sign Up, it unlocks many cool features! combin (100,0) combin (100,1) combin (100,2) ... Where combin (i,j) is … In fact, if Pascal's triangle was expanded further past Row 15, you would see that the sum of the numbers of any nth row would equal to 2^n. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 For example if z=6 => result is on the 4th row. 100 rows of Pascal's Triangle (it's probably 99 rows) a guest . True. Rows of pascal's triangle. Still have questions? The n th n^\text{th} n th row of Pascal's triangle contains the coefficients of the expanded polynomial (x + y) n (x+y)^n (x + y) n. Expand (x + y) 4 (x+y)^4 (x + y) 4 using Pascal's triangle. The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. 28 min ago, C# | Pascal’s triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1. Not a member of Pastebin yet? In this tool, you can construct Pascal's triangles of any size and specify which row to start from. We can write down the next row as an uncalculated sum, so instead of 1,5,10,10,5,1, we write 0+1, 1+4, 4+6, 6+4, 4+1, 1+0. So if you didn't know the number 20 on the sixth row and wanted to work it out, you count along 0,1,2 and find your missing number is the third number.) You work out R! Pascal’s Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 . Generally, on a computer screen, we can display a maximum of 80 characters horizontally. You walk to the left, i.e. Pascals-Triangle. Figure 1 shows the first five rows of an infinite number of rows. combin(i,j) is sometimes verbalized as "i choose j". One algorithm is used to calculate the values for each index in the matrix and another algorithm to put the values in a triangular format to be visually appealing. In this example, n = 3, indicates the 4 th row of Pascal's triangle (since the first row is n = 0). Figure 1 shows the first five rows of an infinite number of rows. Actually, 10^100 isn't that big, so before row 340 you find a position n0,k0=n0/2 where the value from the triangle is larger than or equal to z: Binomial(n0,k0)>=z. The second row is 1,2,1, which we will call 121, which is 11x11, or 11 squared. Blaise Pascal is french. 4. Take a look at the diagram of Pascal's Triangle below. Here I list just a few. what does it mean to find six trigonometric functions of angle theta.. 27, Apr 20. In this article, however, I explain first what pattern can be seen by taking the sums of the row in Pascal's triangle, and also why this pattern will always work whatever row it is tested for. These numbers are found in Pascal's triangle by starting in the 3 row of Pascal's triangle down the middle and subtracting the number adjacent to it. raw download clone embed print report. According to Kottakkaran Soopy Nisar (2018) the definition of Pascals Triangle is being a triangular arrangement of the binomial coefficients in a triangular pattern. Note: The row index starts from 0. Mathabulous! 282 . This is a fundamental idea in 49 min ago, XML | What makes this such … More rows of Pascal’s triangle are listed in the last figure of this article. More ideas, or to check a conjecture, try searching online n rows, each! { 77 } $last item in each row building upon the previous row$ \binom 100... 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# 10 (divided) by 2 is 5
#### BLUE1308
##### Senior Member
Hello!
Excuse me, when we are talking about mathematics, especially divisions, is it possible to say: 10÷5 = 2
"Ten (divided) by five is/equals two"
Is it possible to omit the "divided"? Or do I have to use it always?
• #### GreenWhiteBlue
##### Senior Member
You need to say "divided." How else would we know that you are dividing these numbers, rather than (for example) multiplying them?
#### BLUE1308
##### Senior Member
You need to say "divided." How else would we know that you are dividing these numbers, rather than (for example) multiplying them?
Ok, thank you very much!
So I always need to say "X divided by Y equals W".
If I'm talking about a multiplication (10x2=20), do I have to say: Ten multiplied by 2 equals twenty? Could I also replace "multiplied by" for "times", so it would be "Ten times 2 is/equals 20"?
Thank you!
#### Shooting Stars
##### Senior Member
Could you tell me what is meant by "X divides Y by Z".
Here is the example sentence.
It is also lower than the four per cent death rate which divides deaths in China (3,245) by confirmed cases (80,928).(What is the REAL death rate for the coronavirus?)
Thank you.
#### entangledbank
##### Senior Member
Answering BLUE1308's questions first: In mathematics, division is often read as 'over': a / b is 'a over b'. Multiplication is usually read as 'times', as in 'ten times two'.
Now the other: The 4% death rate is the result of dividing 3245 by 80 928. So 3245 / 80 928 = about 0.04. (Normally with whole numbers we'd more commonly expect to divide the bigger one by the smaller one, but your sentence correctly does it the other way round, to get a rate.)
#### Shooting Stars
##### Senior Member
Now the other: The 4% death rate is the result of dividing 3245 by 80 928. So 3245 / 80 928 = about 0.04. (Normally with whole numbers we'd more commonly expect to divide the bigger one by the smaller one, but your sentence correctly does it the other way round, to get a rate.)
2=10÷5. So this equation is read as "Two divides ten by five". Am I right?
Thank you.
#### entangledbank
##### Senior Member
No, your news story sentence is using the verb 'divide' and 'by', but it's not reading out an equation. That is said 'two equals ten divided by five'.
#### Shooting Stars
##### Senior Member
No, your news story sentence is using the verb 'divide' and 'by', but it's not reading out an equation.
(the four per cent death rate )which divides deaths in China (3,245) by confirmed cases (80,928)
I think "which" refers to 4%. Let's focus on the subordinate clause. That is
0.04 divides 3,245 by 80928.
Is this what the writer tries to say? If it is not an equation, what does it means? I am totally confused.
Thank you.
#### natkretep
##### Moderato con anima (English Only)
I think the relative pronoun which refers to death rate. That relative clause defines 'death rate'.
#### Andygc
##### Senior Member
(the four per cent death rate )which divides deaths in China (3,245) by confirmed cases (80,928)
I think "which" refers to 4%. Let's focus on the subordinate clause. That is
0.04 divides 3,245 by 80928.
Is this what the writer tries to say? If it is not an equation, what does it means? I am totally confused.
Thank you.
<I think "which" refers to 4%.>
No. It refers to "the 4% death rate" which divides the number of deaths by the number of cases. A relative pronoun cannot refer to an adjective (4%) The text tells you how the death rate is calculated - 3,245 divided by 80,928 = 0.04 (or 4%).
#### Shooting Stars
##### Senior Member
I think the relative pronoun which refers to death rate. That relative clause defines 'death rate'.
So the meaning of the subornate clause is the death rate equals the quotient of 3,245 divided by 80,928.
#### natkretep
##### Moderato con anima (English Only)
The quotient is the death rate. You work this out by dividing the dividend/numerator (3,245) by the the divisor/denominator (80,928).
Thank you.
#### BLUE1308
##### Senior Member
Answering BLUE1308's questions first: In mathematics, division is often read as 'over': a / b is 'a over b'. Multiplication is usually read as 'times', as in 'ten times two'.
Now the other: The 4% death rate is the result of dividing 3245 by 80 928. So 3245 / 80 928 = about 0.04. (Normally with whole numbers we'd more commonly expect to divide the bigger one by the smaller one, but your sentence correctly does it the other way round, to get a rate.)
So can I say "10÷5=2" "Ten over five is two"?
Thank you to all of you for your contributions!
#### MattiasNYC
##### Senior Member
I could be wrong here, but my sense is that "divided by" is far more common and easily understood than "over".
#### Roxxxannne
##### Senior Member
I could be wrong here, but my sense is that "divided by" is far more common and easily understood than "over". #### Myridon
##### Senior Member
If it is not an equation, what does it means?
Equations are a different language from English. We can state mathematical information in other ways than "equations." It is similar to the way we can write a date as 2020-03-29 (in the language ISO 8601 (international standards organization date format)) and say March twenty-ninth, the year of our Lord two-thousand and twenty (or any of the other many ways to say today's date).
#### The pianist
##### Senior Member #### The pianist
##### Senior Member
"If it is not an equation, what does it means? I am totally confused. "
It is a calculation.
#### kentix
##### Senior Member
It is also lower than the four per cent death rate which divides deaths in China (3,245) by confirmed cases (80,928)
I think some key words were omitted from this sentence and the word order is very poorly done.
It is also lower than the four per cent death rate in China which is derived from dividing deaths (3,245) by confirmed cases (80,928)
There are other ways to calculate death rates because confirmed cases are only a fraction of the total cases. China's death rate is really much lower. As are death rates elsewhere that only take into account confirmed cases, as if other cases don't exist.
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[more ▼]Super-Earths transiting nearby bright stars are key objects that simultaneously allow for accurate measurements of both their mass and radius, providing essential constraints on their internal composition. We present here the confirmation, based on Spitzer transit observations, that the super-Earth HD 97658 b transits its host star. HD 97658 is a low-mass ($M_*=0.77\pm0.05\,M_{\odot}$) K1 dwarf, as determined from the Hipparcos parallax and stellar evolution modeling. To constrain the planet parameters, we carry out Bayesian global analyses of Keck-HIRES radial velocities, and MOST and Spitzer photometry. HD 97658 b is a massive ($M_P=7.55^{+0.83}_{-0.79} M_{\oplus}$) and large ($R_{P} = 2.247^{+0.098}_{-0.095} R_{\oplus}$ at 4.5 $\mu$m) super-Earth. We investigate the possible internal compositions for HD 97658 b. Our results indicate a large rocky component, by at least 60% by mass, and very little H-He components, at most 2% by mass. We also discuss how future asteroseismic observations can improve the knowledge of the HD 97658 system, in particular by constraining its age. Orbiting a bright host star, HD 97658 b will be a key target for coming space missions TESS, CHEOPS, PLATO, and also JWST, to characterize thoroughly its structure and atmosphere. [less ▲]Detailed reference viewed: 40 (13 ULg) HD 97658 and its super-Earth: Spitzer transit analysis and seismic modeling of the host starVan Grootel, Valérie ; Gillon, Michaël ; Valencia, Diana et alConference (2013, December)Detailed reference viewed: 13 (0 ULg) Testing Convective-core Overshooting Using Period Spacings of Dipole Modes in Red GiantsMontalbán, J.; Miglio, A.; Noels-Grötsch, Arlette et alin Astrophysical Journal (2013), 766Uncertainties on central mixing in main-sequence (MS) and core He-burning (He-B) phases affect key predictions of stellar evolution such as late evolutionary phases, chemical enrichment, ages, etc. We ... [more ▼]Uncertainties on central mixing in main-sequence (MS) and core He-burning (He-B) phases affect key predictions of stellar evolution such as late evolutionary phases, chemical enrichment, ages, etc. We propose a test of the extension of extra-mixing in two relevant evolutionary phases based on period spacing (ΔP) of solar-like oscillating giants. From stellar models and their corresponding adiabatic frequencies (respectively, computed with ATON and LOSC codes), we provide the first predictions of the observable ΔP for stars in the red giant branch and in the red clump (RC). We find (1) a clear correlation between ΔP and the mass of the helium core (M [SUB]He[/SUB]); the latter in intermediate-mass stars depends on the MS overshooting, and hence it can be used to set constraints on extra-mixing during MS when coupled with chemical composition; and (2) a linear dependence of the average value of the asymptotic period spacing (langΔPrang[SUB] a [/SUB]) on the size of the convective core during the He-B phase. A first comparison with the inferred asymptotic period spacing for Kepler RC stars also suggests the need for extra-mixing during this phase, as evinced from other observational facts. [less ▲]Detailed reference viewed: 11 (0 ULg) Towards Precise Asteroseismology of Solar-Like StarsGrigahcène, A.; Dupret, Marc-Antoine ; Sousa, S. G. et alin Astrophysics and Space Science Proceedings series (2013), 31Adiabatic modeling of solar-like oscillations cannot exceed a certain level of precision for fitting individual frequencies. This is known as the problem of near-surface effects on the mode physics. We ... [more ▼]Adiabatic modeling of solar-like oscillations cannot exceed a certain level of precision for fitting individual frequencies. This is known as the problem of near-surface effects on the mode physics. We present a theoretical study which addresses the problem of frequency precision in non-adiabatic models using a time-dependent convection treatment. We find that the number of acceptable model solutions is significantly reduced and more precise constraints can be imposed on the models. Results obtained for a specific star (β Hydri) lead to very good agreement with both global and local seismic observables. This indicates that the accuracy of model fitting to seismic data is greatly improved when a more complete description of the interaction between convection and pulsation is taken into account. [less ▲]Detailed reference viewed: 5 (0 ULg) Successful Asteroseismology for a Better Characterization of the Exoplanet HAT-P-7bOshagh, M.; Grigahcène, A.; Benomar, O. et alin Astrophysics and Space Science Proceedings (2013), 31It is well known that asteroseismology is the unique technique permitting the study of the internal structure of pulsating stars using their pulsational frequencies, which is per se very important. It ... [more ▼]It is well known that asteroseismology is the unique technique permitting the study of the internal structure of pulsating stars using their pulsational frequencies, which is per se very important. It acquires an additional value when the star turns out to be a planet host. In this case, the asteroseismic study output may be a very important input for the study of the planetary system. With this in mind, we use the large time-span of the Kepler public data obtained for the star system HAT-P-7, first to perform an asteroseismic study of the pulsating star using Time-Dependent-Convection (TDC) models. Secondly, we make a revision of the planet properties in the light of the asteroseismic study. [less ▲]Detailed reference viewed: 6 (2 ULg) Towards an effective asteroseismology of solar-like stars: time-dependent convection effects on pulsation frequenciesGrigahcène, A.; Dupret, Marc-Antoine ; Sousa, S. G. et alin Monthly Notices of the Royal Astronomical Society (2012), 422Since the early days of helioseismology, adiabatic models have shown their limits for a precise fitting of individual oscillation frequencies. This discrepancy, which also exists for solar-type stars, is ... [more ▼]Since the early days of helioseismology, adiabatic models have shown their limits for a precise fitting of individual oscillation frequencies. This discrepancy, which also exists for solar-type stars, is known to originate near the surface superadiabatic convective region where the interaction between oscillations and convection is likely to have a large effect on the frequencies. We present an asteroseismic study to address the adequacy of time-dependent convection (TDC) non-adiabatic models to better reproduce the observed individual frequencies. We select, for this purpose, three solar-like stars, in addition to the Sun, to which we fit the observed frequencies in a grid of TDC non-adiabatic models. The best model selection is done by applying a maximum likelihood method. The results are compared to pure adiabatic and near-surface corrected adiabatic models. We show that, first, TDC models give very good agreement for the mode frequencies and average lifetimes. In the solar case, the frequency discrepancy is reduced to <1.75 μHz over 95 per cent of the modes considered. Secondly, TDC models give an asteroseismic insight into the usually unconstrained ad hoc stellar parameters, such as the mixing-length parameter α[SUB]MLT[/SUB]. [less ▲]Detailed reference viewed: 10 (3 ULg) Adiabatic Solar-Like Oscillations in Red Giant StarsMontalban Iglesias, Josefa ; Miglio, Andrea; Noels-Grötsch, Arlette et alin Red Giants as Probes of the Structure and Evolution of the Milky Way (2012)Since the detection of non-radial solar-like oscillation modes in red giants with the CoRoT satellite, the interest in the asteroseismic properties of red giants and the link with their global properties ... [more ▼]Since the detection of non-radial solar-like oscillation modes in red giants with the CoRoT satellite, the interest in the asteroseismic properties of red giants and the link with their global properties and internal structure is substantially increasing. Moreover, more and more precise data are being collected with the space-based telescopes CoRoT and Kepler. In this paper we present a survey of the most relevant theoretical and observational results obtained up to now concerning the potential of solar-like oscillations in red giants. [less ▲]Detailed reference viewed: 13 (0 ULg) Inference from adiabatic analysis of solar-like oscillations in red giantsMontalban Iglesias, Josefa ; Miglio, Andrea ; Noels-Grötsch, Arlette et alin Astronomische Nachrichten (2010), 331The clear detection with CoRoT and Kepler of radial and non-radial solar-like oscillations in many red giants paves the way to seismic inferences on the structure of such stars. We present an overview of ... [more ▼]The clear detection with CoRoT and Kepler of radial and non-radial solar-like oscillations in many red giants paves the way to seismic inferences on the structure of such stars. We present an overview of the properties of the adiabatic frequencies and frequency separations of radial and non-radial oscillation modes, highlighting how their detection allows a deeper insight into the properties of the internal structure of red giants. In our study we consider models of red giants in different evolutionary stages, as well as of different masses and chemical composition. We describe how the large and small separations computed with radial modes and with non-radial modes mostly trapped in the envelope depend on the stellar global parameters and evolutionary state, and we compare our theoretical predictions and first Kepler data.Finally, we find that the properties of dipole modes constitute a promising seismic diagnostic of the evolutionary state of red-giant stars. [less ▲]Detailed reference viewed: 20 (4 ULg) Seismic Diagnostics of Red Giants: First Comparison with Stellar ModelsMontalban Iglesias, Josefa ; Miglio, Andrea ; Noels-Grötsch, Arlette et alin Astrophysical Journal Letters (2010), 721The clear detection with CoRoT and KEPLER of radial and non-radial solar-like oscillations in many red giants paves the way for seismic inferences on the structure of such stars. We present an overview of ... [more ▼]The clear detection with CoRoT and KEPLER of radial and non-radial solar-like oscillations in many red giants paves the way for seismic inferences on the structure of such stars. We present an overview of the properties of the adiabatic frequencies and frequency separations of radial and non-radial oscillation modes for an extended grid of models. We highlight how their detection allows a deeper insight into the internal structure and evolutionary state of red giants. In particular, we find that the properties of dipole modes constitute a promising seismic diagnostic tool of the evolutionary state of red giant stars. We compare our theoretical predictions with the first 34 days of KEPLER data and predict the frequency diagram expected for red giants in the CoRoT exofield in the galactic center direction. [less ▲]Detailed reference viewed: 11 (3 ULg) Evidence for a sharp structure variation inside a red-giant starMiglio, Andrea ; Montalban Iglesias, Josefa ; Carrier, F. et alin Astronomy and Astrophysics (2010), 520Context. The availability of precisely determined frequencies of radial and non-radial oscillation modes in red giants is finally paving the way for detailed studies of the internal structure of these ... [more ▼]Context. The availability of precisely determined frequencies of radial and non-radial oscillation modes in red giants is finally paving the way for detailed studies of the internal structure of these stars.
Aims: We look for the seismic signature of regions of sharp structure variation in the internal structure of the CoRoT target HR 7349.
Methods: We analyse the frequency dependence of the large frequency separation and second frequency differences, as well as the behaviour of the large frequency separation obtained with the envelope auto-correlation function.
Results: We find evidence for a periodic component in the oscillation frequencies, i.e. the seismic signature of a sharp structure variation in HR 7349. In a comparison with stellar models we interpret this feature as caused by a local depression of the sound speed that occurs in the helium second-ionization region. Using solely seismic constraints this allows us to estimate the mass (M = 1.2[SUB]-0.4[/SUB][SUP]+0.6[/SUP] M_&sun;) and radius (R = 12.2[SUB]-1.8[/SUB][SUP]+2.1[/SUP] R_&sun;) of HR 7349, which agrees with the location of the star in an HR diagram. [less ▲]Detailed reference viewed: 52 (29 ULg) A seismic approach to testing different formation channels of subdwarf B starsHu, Haili; Dupret, Marc-Antoine ; Aerts, C. et alin Astronomy and Astrophysics (2008), 490Context: There are many unknowns in the formation of subdwarf B stars. Different formation channels are considered to be possible and to lead to a variety of helium-burning subdwarfs. All seismic models ... [more ▼]Context: There are many unknowns in the formation of subdwarf B stars. Different formation channels are considered to be possible and to lead to a variety of helium-burning subdwarfs. All seismic models to date, however, assume that a subdwarf B star is a post-helium-flash-core surrounded by a thin inert layer of hydrogen. Aims: We examine an alternative formation channel, in which the subdwarf B star originates from a massive (>~2 M[SUB]o[/SUB]) red giant with a non-degenerate helium-core. Although these subdwarfs may evolve through the same region of the log g-T_eff diagram as the canonical post-flash subdwarfs, their interior structure is rather different. We examine how this difference affects their pulsation modes and whether it can be observed. Methods: Using detailed stellar evolution calculations we construct subdwarf B models from both formation channels. The iron accumulation in the driving region due to diffusion, which causes the excitation of the modes, is approximated by a Gaussian function. The pulsation modes and frequencies are calculated with a non-adiabatic pulsation code. Results: A detailed comparison of two subdwarf B models from different channels, but with the same log g and T_eff, shows that their mode excitation is different. The excited frequencies are lower for the post-flash than for the post-non-degenerate subdwarf B star. This is mainly due to the differing chemical composition of the stellar envelope. A more general comparison between two grids of models shows that the excited frequencies of most post-non-degenerate subdwarfs cannot be well-matched with the frequencies of post-flash subdwarfs. In the rare event that an acceptable seismic match is found, additional information, such as mode identification and log g and T_eff determinations, allows us to distinguish between the two formation channels. [less ▲]Detailed reference viewed: 18 (2 ULg) CLÉS, Code Liégeois d'Évolution StellaireScuflaire, Richard ; Théado, Sylvie; Montalban Iglesias, Josefa et alin Astrophysics & Space Science (2008), 316CLÉS is an evolution code recently developed to produce stellar models meeting the specific requirements of studies in asteroseismology. It offers the users a lot of choices in the input physics they want ... [more ▼]CLÉS is an evolution code recently developed to produce stellar models meeting the specific requirements of studies in asteroseismology. It offers the users a lot of choices in the input physics they want in their models and its versatility allows them to tailor the code to their needs and implement easily new features. We describe the features implemented in the current version of the code and the techniques used to solve the equations of stellar structure and evolution. A brief account is given of the use of the program and of a solar calibration realized with it. [less ▲]Detailed reference viewed: 13 (7 ULg) The Liège Oscillation codeScuflaire, Richard ; Montalban Iglesias, Josefa ; Théado, S. et alin Astrophysics & Space Science (2008), 316The Liège Oscillation code can be used as a stand-alone program or as a library of subroutines that the user calls from a Fortran main program of his own to compute radial and nonradial adiabatic ... [more ▼]The Liège Oscillation code can be used as a stand-alone program or as a library of subroutines that the user calls from a Fortran main program of his own to compute radial and nonradial adiabatic oscillations of stellar models. We describe the variables and the equations used by the program and the methods used to solve them. A brief account is given of the use and the output of the program. [less ▲]Detailed reference viewed: 20 (6 ULg) Thorough analysis of input physics in CESAM and CLÉS codesMontalban Iglesias, Josefa ; Lebreton, Yveline; Miglio, Andrea et alin Astrophysics & Space Science (2008), 316(1-4), 219-229This contribution is not about the quality of the agreement between stellar models computed by CESAM and CLÉS codes, but more interesting, on what ESTA-Task 1 run has taught us about these codes and about ... [more ▼]This contribution is not about the quality of the agreement between stellar models computed by CESAM and CLÉS codes, but more interesting, on what ESTA-Task 1 run has taught us about these codes and about the input physics they use. We also quantify the effects of different implementations of the same physics on the seismic properties of the stellar models, that in fact is the main aim of ESTA experiments. [less ▲]Detailed reference viewed: 13 (1 ULg) Long term photometric monitoring with the Mercator telescope. Frequencies and mode identification of variable O-B starsDe Cat, P.; Briquet, Maryline ; Aerts, C. et alin Astronomy and Astrophysics (2007), 463Aims. We selected a large sample of O-B stars that were considered as (candidate) slowly pulsating B, beta Cep, and Maia stars after the analysis of their hipparcos data. We analysed our new seven ... [more ▼]Aims. We selected a large sample of O-B stars that were considered as (candidate) slowly pulsating B, beta Cep, and Maia stars after the analysis of their hipparcos data. We analysed our new seven passband geneva data collected for these stars during the first three years of scientific operations of the mercator telescope. We performed a frequency analysis for 28 targets with more than 50 high-quality measurements to improve their variability classification. For the pulsating stars, we tried both to identify the modes and to search for rotationally split modes. Methods: We searched for frequencies in all the geneva passbands and colours by using two independent frequency analysis methods and we applied a 3.6 S/N-level criterion to locate the significant peaks in the periodograms. The modes were identified by applying the method of photometric amplitudes for which we calculated a large, homogeneous grid of equilibrium models to perform a pulsational stability analysis. When both the radius and the projected rotational velocity of an object are known, we determined a lower limit for the rotation frequency to estimate the expected frequency spacings in rotationally split pulsation modes. Results: We detected 61 frequencies, among which 33 are new. We classified 21 objects as pulsating variables (7 new confirmed pulsating stars, including 2 hybrid beta Cep/SPB stars), 6 as non-pulsating variables (binaries or spotted stars), and 1 as photometrically constant. All the Maia candidates were reclassified into other variability classes. We performed mode identification for the pulsating variables for the first time. The most probable l value is 0, 1, 2, and 4 for 1, 31, 9, and 5 modes, respectively, including only 4 unambiguous identifications. For 7 stars we cannot rule out that some of the observed frequencies belong to the same rotationally split mode. For 4 targets we may begin to resolve close frequency multiplets. Based on observations collected with the p7 photometer attached to the Flemish 1.2-m mercator telescope situated at the Roque de los Muchachos observatory on La Palma (Spain). Section [see full textsee full text], including Figs. is only available in electronic form at http://www.aanda.org, and Tables 2 and 3 are only available in electronic form at the CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or via http://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/463/243 [less ▲]Detailed reference viewed: 45 (1 ULg) An asteroseismic study of the β cephei star θ ophiuchi : constraints on global stellar parameters and core overshootingBriquet, Maryline ; Morel, Thierry ; Thoul, Anne et alin Monthly Notices of the Royal Astronomical Society (2007), 381(4), 1482-1488We present a seismic study of the beta Cephei star theta Ophiuchi. Our analysis is based on the observation of one radial mode, one rotationally split l = 1 triplet and three components of a rotationally ... [more ▼]We present a seismic study of the beta Cephei star theta Ophiuchi. Our analysis is based on the observation of one radial mode, one rotationally split l = 1 triplet and three components of a rotationally split l = 2 quintuplet for which the m values were well identified by spectroscopy. We identify the radial mode as fundamental, the triplet as p(1) and the quintuplet as g(1). Our non-local thermodynamic equilibrium abundance analysis results in a metallicity and CNO abundances in full agreement with the most recent updated solar values. With X epsilon [0.71, 0.7211] and Z epsilon [0.009, 0.015], and using the Asplund et al. mixture but with a Ne abundance about 0.3 dex larger, the matching of the three independent modes enables us to deduce constrained ranges for the mass (M = 8.2 +/- 0.3 M circle dot) and central hydrogen abundance (X-c = 0.38 +/- 0.02) of theta Oph and to prove the occurrence of core overshooting (alpha(ov) = 0.44 +/- 0.07). We also derive an equatorial rotation velocity of 29 +/- 7 km s(-1). Moreover, we show that the observed non-equidistance of the l = 1 triplet can be reproduced by the second-order effects of rotation. Finally, we show that the observed rotational splitting of two modes cannot rule out a rigid rotation model. [less ▲]Detailed reference viewed: 27 (14 ULg) The Seismology Programme of CoRoTMichel, Eric; Baglin, A.; Auvergne, M. et alin Proceedings of "The CoRoT Mission Pre-Launch Status - Stellar Seismology and Planet Finding (2006, November 01)We introduce the main lines and specificities of the CoRoT Seismology Core Programme. The development and consolidation of this programme has been made in the framework of the CoRoT Seismology Working ... [more ▼]We introduce the main lines and specificities of the CoRoT Seismology Core Programme. The development and consolidation of this programme has been made in the framework of the CoRoT Seismology Working Group. With a few illustrative examples, we show how CoRoT data will help to address various problems associated with present open questions of stellar structure and evolution. [less ▲]Detailed reference viewed: 56 (34 ULg) Polytropes as simple models of beta Cephei starsGodart, Mélanie ; Scuflaire, Richard ; Thoul, Anne et alin Communications in Asteroseismology (2006), 147Beta Cephei stars have a simple structure: a convective core surrounded by a radiative envelope. It is therefore worth trying to describe beta Cephei stars with composite polytropes which are useful to ... [more ▼]Beta Cephei stars have a simple structure: a convective core surrounded by a radiative envelope. It is therefore worth trying to describe beta Cephei stars with composite polytropes which are useful to retrieve structure parameters from frequency spectra. We show that the structure of beta Cephei models can relatively well be described with two-zone polytropic models. However this description is not convincing to depict oscillations of beta Cephei models. [less ▲]Detailed reference viewed: 15 (7 ULg) Radiative forces and pulsation in beta Cephei starsBourge, P*-O; Alecian, G.; Thoul, Anne et alin Communications in Asteroseismology (2006), 147Recently radiatively-driven diffusion in beta Cephei stars has been suggested as a possible explanation to account for excitation of the observed oscillation modes of the beta Cephei stars nu Eridani and ... [more ▼]Recently radiatively-driven diffusion in beta Cephei stars has been suggested as a possible explanation to account for excitation of the observed oscillation modes of the beta Cephei stars nu Eridani and 12 Lacertae. Preliminary results (Bourge and Alecian 2006) show that microscopic diffusion of iron could indeed occur in a region situated just above the opacity bump of the iron group elements. This diffusion has little influence on the position of the star in the HR diagram and on the spectrum of low radial order eigenfrequencies. It does, however, increase the number of excited frequencies. We show here that the ratio between the number of p-modes to the number of g-modes could be used as a signature of this diffusion process. [less ▲]Detailed reference viewed: 16 (4 ULg) Analysis of MERCATOR data - Part I: variable B starsDe Cat, P.; Briquet, Maryline ; Aerts, C. et alin Communications in Asteroseismology (2006), 147We re-classified 31 variable B stars which were observed more than 50 times in the Geneva photometric system with the p7 photometer attached to the MERCATOR telescope (La Palma) during its first 3 years ... [more ▼]We re-classified 31 variable B stars which were observed more than 50 times in the Geneva photometric system with the p7 photometer attached to the MERCATOR telescope (La Palma) during its first 3 years of scientific observations. HD 89688 is a possible beta Cephei/slowly pulsating B star hybrid and the main mode of the COROT target HD 180642 shows non-linear effects. The Maia candidates are re-classified as either ellipsoidal variables or spotted stars. Although the mode identification is still ongoing, all the well-identified modes so far have l <= 2. [less ▲]Detailed reference viewed: 8 (2 ULg) 1 2 3 4 | open-web-math/open-web-math | |
# Customer opens an account in which the end of each month \$1000 is deposited. How much money will be accumulated after 7 years if the bank's interest rate is 4.5% per year?
The annual rate of interest of the account is 4.5%. It is not given if the interest is compounded every month or once in a year. Assuming that interest is compounded every month, the interest rate for a month is `0.045/12 = 0.00375` .
The total tenure is 7*12 =...
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The annual rate of interest of the account is 4.5%. It is not given if the interest is compounded every month or once in a year. Assuming that interest is compounded every month, the interest rate for a month is `0.045/12 = 0.00375` .
The total tenure is 7*12 = 84 months.
The money accumulated at the end of 7 years is equal to:
`1000*(1.00375)^84 + 1000*(1.00375)^83 +...+ 1000`
This is geometric series that has a sum:
`1000*(1.00375^84 - 1)/(1.00375 - 1)`
=> `1000*98.5206`
=> 98520.6
The amount accumulated after 7 years in the account is \$98520.6
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Solutions by everydaycalculation.com
## 70 percent of what number is 65?
65 is 70% of 92.86
#### Steps to solve "65 is 70 percent of what number?"
1. We have, 70% × x = 65
2. or, 70/100 × x = 65
3. Multiplying both sides by 100 and dividing both sides by 70,
we have x = 65 × 100/70
4. x = 92.86
If you are using a calculator, simply enter 65×100÷70, which will give you the answer.
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# Investigate various types of sugar and the way in which each of them dissolves at certain temperatures.
Extracts from this document...
Introduction
Dissolving Sugar The dilemma given to us by our teacher was ' To investigate various types of sugar and the way in which each of them dissolves at certain temperatures' I will investigate three sugars at three different temperatures. The key variables that are involved in the problem are the following, type of sugar, temperature, volume of liquid, time, volume of sugar, time taken for stirring. The following variables will be kept the same in order to make the test fair, volume of water, volume of sugar, temperature of water per experiment, when sugars are placed into beakers, how long each beaker is stirred for, size of beaker and amount of water. The following variable will be changed, temperature of water. The equipment that I will be using is the following, kettle, glass rod, 3 beaker, scales, spatula, 3 types of sugar, 3 thermometers, timer and tweezers. What I'm going to measure and look for is how long it takes each sugar to dissolve at a certain temperature. I predict that the finer the sugar (icing sugar) faster it will dissolve and that the higher the temperature the quicker each of the sugars will dissolve. I predict this because if the sugar particles are finer they are then able to move further apart. So the more liquid will affect more of the particles on larger surface. ...read more.
Middle
The experiment then will be carried out two more times in the same conditions but the temperature will only have changed. Results will be recorded in a table. Results Temperature in c Type of sugars 40 60 80 Sugar Cube 161 90 86 Sugar Icing 122 55 62 Sugar Caster 140 49 62 I can see a pattern in my results, as the temperature increases the time it takes for each type of sugar to fully dissolve decreases. Also the type of sugar that was the finest (icing sugar) dissolved the fastest compared to the other types of sugar. The largest sugar in size (cube sugar) took the longest to dissolve and the caster was in between the two results. My results tell me that the warmer a solvent is (water) the quicker the solute (sugar) will dissolve and the finer the solute (sugar) the faster it will dissolve in the solvent (water) My prediction was correct, the particles of sugar were finer in the icing sugar. Therefore more water was able to get around the sugar particles, the water affected the sugar on a larger surface area quicker than the other two sugars. The sugar at the centre of the sugar cube would of taken longer to be effected by the water then the icing sugar seeing as it had to first come past the other sugar particles in front of it. ...read more.
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1. ## Establish what types of soil holds the most water and to see if changing ...
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# 1.2: Lines
• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College
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If we have two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$, then we can draw one and only one line through both points. By the slope of this line we mean the ratio of $$\Delta y$$ to $$\Delta x$$. The slope is often denoted $$m$$: $$m=\Delta y/\Delta x=(y_2-y_1)/(x_2-x_1)$$. For example, the line joining the points $$(1,-2)$$ and $$(3,5)$$ has slope $$(5+2)/(3-1)=7/2$$.
Example 1.1.1
According to the 1990 U.S. federal income tax schedules, a head of household paid 15% on taxable income up to $26050. If taxable income was between$26050 and $134930, then, in addition, 28% was to be paid on the amount between$26050 and $67200, and 33% paid on the amount over$67200 (if any). Interpret the tax bracket information (15%, 28%, or 33%) using mathematical terminology, and graph the tax on the $$y$$-axis against the taxable income on the $$x$$-axis.
Solution
The percentages, when converted to decimal values 0.15, 0.28, and 0.33, are the slopes of the straight lines which form the graph of the tax for the corresponding tax brackets. The tax graph is what's called apolygonal line, i.e., it's made up of several straight line segments of different slopes. The first line starts at the point (0,0) and heads upward with slope 0.15 (i.e., it goes upward 15 for every increase of 100 in the $$x$$-direction), until it reaches the point above $$x=26050$$. Then the graph "bends upward,'' i.e., the slope changes to 0.28. As the horizontal coordinate goes from $$x=26050$$ to $$x=67200$$, the line goes upward 28 for each 100 in the $$x$$-direction. At $$x=67200$$ the line turns upward again and continues with slope 0.33. See figure 1.1.1. Figure 1.1.1. Tax vs. income.
The most familiar form of the equation of a straight line is: $$y=mx+b$$. Here $$m$$ is the slope of the line: if you increase $$x$$ by 1, the equation tells you that you have to increase $$y$$ by $$m$$. If you increase $$x$$ by $$\Delta x$$, then $$y$$ increases by $$\Delta y=m\Delta x$$. The number $$b$$ is called the y-intercept, because it is where the line crosses the $$y$$-axis. If you know two points on a line, the formula $$m=(y_2-y_1)/ (x_2-x_1)$$ gives you the slope. Once you know a point and the slope, then the $$y$$-intercept can be found by substituting the coordinates of either point in the equation: $$y_1=mx_1+b$$, i.e., $$b=y_1-mx_1$$. Alternatively, one can use the "point-slope'' form of the equation of a straight line: start with $$(y-y_1)/(x-x_1)=m$$ and then multiply to get $$(y-y_1)=m(x-x_1)$$, the point-slope form. Of course, this may be further manipulated to get $$y=mx-mx_1+y_1$$, which is essentially the "$$mx+b$$'' form.
It is possible to find the equation of a line between two points directly from the relation $$(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)$$, which says "the slope measured between the point $$(x_1,y_1)$$ and the point $$(x_2,y_2)$$ is the same as the slope measured between the point $$(x_1,y_1)$$ and any other point $$(x,y)$$ on the line.'' For example, if we want to find the equation of the line joining our earlier points $$A(2,1)$$ and $$B(3,3)$$, we can use this formula: $${y-1\over x-2}={3-1\over 3-2}=2,\qquad\hbox{so that}\qquad y-1=2(x-2),\qquad\hbox{i.e.,}\qquad y=2x-3.$$ Of course, this is really just the point-slope formula, except that we are not computing $$m$$ in a separate step.
The slope $$m$$ of a line in the form $$y=mx+b$$ tells us the direction in which the line is pointing. If $$m$$ is positive, the line goes into the 1st quadrant as you go from left to right. If $$m$$ is large and positive, it has a steep incline, while if $$m$$ is small and positive, then the line has a small angle of inclination. If $$m$$ is negative, the line goes into the 4th quadrant as you go from left to right. If $$m$$ is a large negative number (large in absolute value), then the line points steeply downward; while if $$m$$ is negative but near zero, then it points only a little downward. These four possibilities are illustrated in Figure 1.1.2 Figure 1.1.2. Lines with slopes 3, 0.1, -4, and -0.1
If $$m=0$$, then the line is horizontal: its equation is simply $$y=b$$.
There is one type of line that cannot be written in the form $$y=mx+b$$, namely, vertical lines. A vertical line has an equation of the form $$x=a$$. Sometimes one says that a vertical line has an "infinite'' slope.
Sometimes it is useful to find the $$x$$-intercept of a line $$y=mx+b$$. This is the $$x$$-value when $$y=0$$. Setting $$mx+b$$ equal to 0 and solving for $$x$$ gives: $$x=-b/m$$. For example, the line $$y=2x-3$$ through the points $$A(2,1)$$ and $$B(3,3)$$ has $$x$$-intercept $$3/2$$.
Example 1.1.2
Suppose that you are driving to Seattle at constant speed, and notice that after you have been traveling for 1 hour (i.e., $$t=1$$), you pass a sign saying it is 110 miles to Seattle, and after driving another half-hour you pass a sign saying it is 85 miles to Seattle. Using the horizontal axis for the time $$t$$ and the vertical axis for the distance $$y$$ from Seattle, graph and find the equation $$y=mt+b$$ for your distance from Seattle. Find the slope, $$y$$-intercept, and $$t$$-intrcept, and describe the practical meaning of each.
Solution
The graph of $$y$$ versus $$t$$ is a straight line because you are traveling at constant speed. The line passes through the two points $$(1,110)$$ and $$(1.5,85)$$, so its slope is $$m=(85-110)/(1.5-1)=-50$$. The meaning of the slope is that you are traveling at 50 mph; $$m$$ is negative because you are traveling toward Seattle, i.e., your distance $$y$$ is decreasing. The word "velocity'' is often used for $$m=-50$$, when we want to indicate direction, while the word "speed'' refers to the magnitude (absolute value) of velocity, which is 50 mph. To find the equation of the line, we use the point-slope formula: $${y-110\over t-1}=-50,\qquad\hbox{so that}\qquad y=-50(t-1)+110=-50t+160.$$ The meaning of the $$y$$-intercept 160 is that when $$t=0$$ (when you started the trip) you were 160 miles from Seattle. To find the $$t$$-intercept, set $$0=-50t+160$$, so that $$t=160/50=3.2$$. The meaning of the $$t$$-intercept is the duration of your trip, from the start until you arrive in Seattle. After traveling 3 hours and 12 minutes, your distance $$y$$ from Seattle will be 0.
## Contributors
• Integrated by Justin Marshall. | HuggingFaceTB/finemath | |
# Nonlinear inequality solver
Best of all, Nonlinear inequality solver is free to use, so there's no reason not to give it a try! We will give you answers to homework. ## The Best Nonlinear inequality solver
We'll provide some tips to help you select the best Nonlinear inequality solver for your needs. Partial fractions is a method for decomposing a fraction into a sum of simpler fractions. The process involves breaking up the original fraction into smaller pieces, each of which can be more easily simplified. While partial fractions can be used to decompose any fraction, it is particularly useful for dealing with rational expressions that contain variables. In order to solve a partial fraction, one must first determine the factors of the denominator. Once the factors have been determined, the numerator can be factored as well. The next step is to identify the terms in the numerator and denominator that share common factors. These terms can then be combined, and the resulting expression can be simplified. Finally, the remaining terms in the numerator and denominator can be solve for using basic algebraic principles. By following these steps, one can solve any partial fraction problem.
Next, the coefficients of the variables must be determined. The coefficients are the numerical values that are multiplied by the variables. Finally, the equations should be solved by using one of the many mathematic methods for solving equations.
Math is a difficult subject to learn in school. But, with practice and an understanding of the best type math online, it can be done. There are many different ways to learn math. One way is to use an online tool. They can be used for all types of math, including simple addition and subtraction. Another way is to use flashcards. This can be done at home or organized into a folder at school. When choosing a math online tool, make sure it is easy to use and that there are lots more options for learning than just the basic math facts. One of the best type math online - is a calculator that has buttons for addition, subtraction, multiplication and division so you don't have to scroll through buttons to find them all. Knowing how to use one well will help you become more familiar with numbers, which will make learning math easier in the future.
I found a great geometry math helper online. It's called " GeoGebra ". It's a free website that helps you with all sorts of geometry problems. You can input your own problems, or you can use their pre-made problems. They have a ton of different topics covered, so you're sure to find what you need. I highly recommend this site to anyone who needs help with geometry math.
To solve for absolute value, you need to find the distance of a number from zero on a number line. To do this, you take the number's distance from zero on the number line and make it positive. This will give you the absolute value of the number.
## We solve all types of math troubles
Good app. But, more types of math (i.e., Integration, differentiation) have to be included. Intersecting points should be given in graphs. Add greatest integer function, fractional part detection. Also add "to the power" option in calculator section. Flora Thomas
Very helpful and it also teaches you the steps. One thing. My teacher teaches some of the material differently like different ways to do it so if the app could add several ways of doing it that would be even better! Raelynn Taylor | HuggingFaceTB/finemath | |
👉 Try now NerdPal! Our new math app on iOS and Android
Factor the expression $x-x^2+x^3-x^4$
Go!
Symbolic mode
Text mode
Go!
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log
log
lim
d/dx
Dx
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acoth
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acsch
Final answer to the problem
$-x\left(x^2+1\right)\left(x-1\right)$
Got another answer? Verify it here!
Step-by-step Solution
How should I solve this problem?
• Choose an option
• Integrate by partial fractions
• Product of Binomials with Common Term
• FOIL Method
• Integrate by substitution
• Integrate by parts
• Integrate using tabular integration
• Integrate by trigonometric substitution
• Weierstrass Substitution
• Prove from LHS (left-hand side)
Can't find a method? Tell us so we can add it.
1
For easier handling, reorder the terms of the polynomial $-x^4+x^3-x^2+x$ from highest to lowest degree
$-x^4+x^3-x^2+x$
Learn how to solve polynomial factorization problems step by step online.
$-x^4+x^3-x^2+x$
Learn how to solve polynomial factorization problems step by step online. Factor the expression x-x^2x^3-x^4. For easier handling, reorder the terms of the polynomial -x^4+x^3-x^2+x from highest to lowest degree. We can factor the polynomial -x^4+x^3-x^2+x using the rational root theorem, which guarantees that for a polynomial of the form a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 there is a rational root of the form \pm\frac{p}{q}, where p belongs to the divisors of the constant term a_0, and q belongs to the divisors of the leading coefficient a_n. List all divisors p of the constant term a_0, which equals 0. Next, list all divisors of the leading coefficient a_n, which equals 1. The possible roots \pm\frac{p}{q} of the polynomial -x^4+x^3-x^2+x will then be.
Final answer to the problem
$-x\left(x^2+1\right)\left(x-1\right)$
Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
SnapXam A2
Go!
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v
w
x
y
z
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(◻)
+
-
×
◻/◻
/
÷
2
e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
Main Topic: Polynomial Factorization
They are a group of techniques that help us rewrite polynomial expressions as a product of factors. | HuggingFaceTB/finemath | |
Math- STATS 218
posted by .
When the following hypotheses are being tested at a level of significance of
aH0: m 100
Ha: m < 100
the null hypothesis will be rejected if the test statistic Z is
a. >Z
b. >Z
c. <Z
d. <100
• Math- STATS 218 -
You are lacking information on variability and level of significance used (e.g., P ≤ .05, P ≤ .01).
It would also help if you proofread your questions before you posted them. Choices a and b are identical.
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# 13 rounded to the nearest tens
What is 13 rounded to the nearest tens? 13 rounded to the nearest tens is 10
Here we will show you how to round off 13 to the nearest tens with step by step detailed solution
## Rounding Numbers Calculator
Round
To
Rounding numbers means replacing that number with an approximate value that has a shorter, simpler, or more explicit representation.
To round off the decimal number 13 to the nearest tens , follow these steps:
1. Find the digit at the tens place,
= 1 in the 13
2. Find the digit on the exact right of the tens place, Round up if this number is greater than or equal to 5 and round down if it is less than 5
= 3 in the 13
3. Since 3 is less than 5, the tens place digit 1 remains unchanged
4. Rewrite the number, all digits to the right of the rounding digit becomes zero
Therefore, the number 13 rounded to the nearest tens is 10
Here are some more examples of rounding numbers to the nearest tens calculator | HuggingFaceTB/finemath | |
Associated Topics || Dr. Math Home || Search Dr. Math
### Where to Put the Parentheses?
```
Date: 01/03/2002 at 20:44:23
From: Hennaysha Candler
Subject: Order of operations
I am stuck on where to put parentheses in a math expression to make
the expression true. I tried to use the guess-and-check strategy but
it wouldn't work for me. Here is a problem I need to put parentheses
into to make the equation true:
9 - 6 + 4 * 6 / 3 = -2
```
```
Date: 01/03/2002 at 23:04:01
From: Doctor Peterson
Subject: Re: Order of operations
Hi, Hennaysha.
I can't think of any method you can use other than guess and check;
you just have to come up with the right guess. That makes it hard for
me to come up with a good hint other than giving you the answer. But
I'll go through the problem to illustrate how you might think.
whole expression. If you just subtract 6 from 9, you'll get 3, and
there will be no way to make a negative answer, so you must be
subtracting some expression greater than the 6 alone from the 9:
9 - (6 + 4 * 6 / 3 = 3
But where can we put the right parenthesis? We can try all possible
places:
9 - (6 + 4) * 6 / 3 = 3 ==> 9 - 10 * 6 / 3
9 - (6 + 4 * 6) / 3 = 3 ==> 9 - 34 / 3
9 - (6 + 4 * 6 / 3) = 3 ==> 9 - 14
None of these gives -2; but looking back at them, I see that I can
place another pair of parentheses in the first attempt and get the
[9 - (6 + 4)] * 6 / 3 = 3 ==> [9 - 10] * 6 / 3 = -1 * 6 / 3 = -2
The best "guess-and-check" works like this: you look for ways to
restrict your guesses, make them, and then refine those based on the
results. In this case, it worked out well to actually simplify each
parenthesized version and then repeat the process by looking for
places to add another pair; there probably won't always be such an
obvious choice for the first pair, but this idea may help at least a
little. Mostly it takes patience; if you haven't guessed, this kind of
exercise is meant to give you LOTS of practice evaluating expressions
with parentheses, so you can expect to have to do a lot of checking.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
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Informative line
### Calculation Of Capacitance
Learn steps to capacitance of a capacitor calculation, practice to calculation of potential difference between charged shells and spherical & cylindrical capacitor.
# Steps to Calculate Capacitance of Capacitor
• Consider two oppositely charged conductors having magnitude of charge Q with simple geometry.
• To calculate capacitance following steps are followed :
1. Identify two conductors, if there is any one conductor then it also has capacitance because we can imagine the second conductor at infinity, having the same charge but opposite nature.
2. Calculate electric field E by using Gauss's Law.
3. Calculate potential difference using electric field $$|\Delta V|=\int\;E\cdot ds$$
4. Capacitance of a capacitor can be calculated as $$C=\dfrac {Q}{|\Delta V|}$$
#### Choose the correct sequence of steps to calculate capacitance for parallel plate capacitor having charge + Q and –Q respectively.
A Charge (Q)$$\rightarrow$$ Electric field (E)$$\rightarrow$$ Potential difference($$\Delta V$$) $$\rightarrow$$Capacitance, $$C=\dfrac {Q}{\Delta V}$$
B Charge (Q)$$\rightarrow$$ Potential difference($$\Delta V$$) $$\rightarrow$$ Electric field (E)$$\rightarrow$$Capacitance, $$C=\dfrac {\Delta V}{E}$$
C Charge (Q)$$\rightarrow$$ Potential difference($$\Delta V$$) $$\rightarrow$$ Electric field (E)$$\rightarrow$$Capacitance, $$C=\dfrac {Q}{E}$$
D None of these
×
Determine the charge (Q) on parallel plate capacitor.
Calculate electric field (E) by using Gauss's Law.
Calculate potential difference by using electric field $$|\Delta V|=\int E \cdot ds$$
Then, calculate capacitance of capacitor by using $$C=\dfrac {Q}{|\Delta V|}$$
Hence, option (A) is correct.
### Choose the correct sequence of steps to calculate capacitance for parallel plate capacitor having charge + Q and –Q respectively.
A
Charge (Q)$$\rightarrow$$ Electric field (E)$$\rightarrow$$ Potential difference($$\Delta V$$$$\rightarrow$$Capacitance, $$C=\dfrac {Q}{\Delta V}$$
.
B
Charge (Q)$$\rightarrow$$ Potential difference($$\Delta V$$$$\rightarrow$$ Electric field (E)$$\rightarrow$$Capacitance, $$C=\dfrac {\Delta V}{E}$$
C
Charge (Q)$$\rightarrow$$ Potential difference($$\Delta V$$$$\rightarrow$$ Electric field (E)$$\rightarrow$$Capacitance, $$C=\dfrac {Q}{E}$$
D
None of these
Option A is Correct
# Calculation of Potential Difference between Charged Shells
• Consider two oppositely charged conductors having magnitude of charge Q with simple geometry.
• To calculate capacitance following steps are followed:
1. Identify two conductors, if there is any one conductor then it also has capacitance because we can imagine the second conductor at infinity, having the same charge but opposite in nature.
2. Calculate electric field E by using Gauss's Law.
3. Calculate potential difference using electric field $$|\Delta V|=\int\;E\cdot ds$$
#### A spherical capacitor of conducting shell of radius (b) = 5 m having charge (–Q) =$$-3\,nC$$ is concentric with a smaller conducting sphere of radius (a) = 2m and charge (Q) = $$+3\,nC$$. Calculate the potential difference between the shells.
A 9 V
B 10 V
C 8.1 V
D 16 V
×
Potential difference between a conducting sphere and a conducting shell is given as
$$V_b-V_a=\dfrac {KQ(b-a)}{ab}$$
Given : a = 2m, b = 5m, $$Q= 3\,nC$$
$$\Rightarrow$$ $$V_b-V_a=\dfrac {9\times10^9\times 3\times10^{-9}\times(5-2)}{5\times2}$$
$$\Rightarrow V_b-V_a=\dfrac {9\times10^9\times 9\times10^{-9}}{10}$$
$$\Rightarrow V_b-V_a=$$ 8.1 V
Hence, option (C) is correct.
### A spherical capacitor of conducting shell of radius (b) = 5 m having charge (–Q) =$$-3\,nC$$ is concentric with a smaller conducting sphere of radius (a) = 2m and charge (Q) = $$+3\,nC$$. Calculate the potential difference between the shells.
A
9 V
.
B
10 V
C
8.1 V
D
16 V
Option C is Correct
# Calculation of Capacitance of Cylindrical Capacitor
• Consider a cylindrical conductor of radius 'a', having charge Q, is co-axial with a cylindrical shell of radius 'b'.
• The length of cylindrical capacitor is $$\ell$$.
Step 1: The electric field at a distance 'r' from a line charge of infinite length is given as $$E=\dfrac {\lambda }{2\pi\epsilon_0r}$$
where $$\lambda$$ = charge per unit length
Step 2 : Potential difference in terms of electric field is given as
$$\Delta V = -\int\limits_a^b\;\vec E\cdot\vec ds$$
$$V_b-V_a= -\int\limits_a^b\;\vec E\cdot\vec ds$$
• Assume $$\ell>>>$$ radius (a or b)
• Then the cylindrical capacitor will appear to be line charge of infinite length and electric field is given as $$E=\dfrac {\lambda}{2\pi\epsilon_0r}$$
• Potential difference between two conductors, is given as
$$V_b-V_a=-\int\limits_a^b \dfrac {\lambda}{2\pi\epsilon_0r}dr$$
$$\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0}\int\limits_a^b\dfrac {1}{r} dr$$
$$\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} [\ell n\;r]_a^b$$
$$\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} [\ell n\,b-\ell n\,a]$$
$$\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} \left [\ell n(\dfrac {b}{a})\right]$$
$$\left [\lambda=\dfrac {Q}{\ell}\text{where }\lambda \,\,\text{is charge per unit length} \right]$$
$$\Rightarrow V_b-V_a=-\dfrac {Q}{2\pi\epsilon_0\ell}\ell n \left( \dfrac {b}{a} \right)$$
The magnitude of the potential difference is
$$\Delta V=|V_b-V_a|=\dfrac {Q}{2\pi\epsilon_0\ell}\ell n \left( \dfrac {b}{a} \right)$$
Step 3 : Capacitance of cylindrical capacitor is given as
$$C=\dfrac {Q}{\Delta V}=\dfrac {Q}{V_b-V_a}$$
$$\Rightarrow C=\dfrac {Q} {\dfrac {Q}{2\pi\epsilon_0 \ell}\ell n\left (\dfrac {b}{a}\right)}$$
$$\Rightarrow C=\dfrac {\ell}{2K\ell n\left (\dfrac {b}{a}\right)}$$
#### Calculate capacitance of a cylindrical capacitor of length $$\ell$$= 18 m, having inner radius and outer radius of a = 2 cm and b = 4 cm respectively.
A $$\dfrac{1}{\ell n\,(4)} nF$$
B $$\dfrac{2}{\ell n\,(5)} nF$$
C $$\dfrac{1}{\ell n\,(2)} nF$$
D $$\dfrac{2}{\ell n\,(2)} nF$$
×
The capacitance of a cylindrical capacitor is given as
$$C=\dfrac {\ell}{2K\;\ell n\left ( \dfrac {b}{a}\right)}$$
Given : $$\ell$$=18 m, a = 2 cm, b = 4cm
$$C=\dfrac {18}{2\times9\times10^9 \ell n\,(4/2)}$$
$$\Rightarrow C=\dfrac {10^{-9}}{\ell n\,(2)}$$
$$\Rightarrow C=\dfrac {1}{\ell n(2)}\;n F$$
### Calculate capacitance of a cylindrical capacitor of length $$\ell$$= 18 m, having inner radius and outer radius of a = 2 cm and b = 4 cm respectively.
A
$$\dfrac{1}{\ell n\,(4)} nF$$
.
B
$$\dfrac{2}{\ell n\,(5)} nF$$
C
$$\dfrac{1}{\ell n\,(2)} nF$$
D
$$\dfrac{2}{\ell n\,(2)} nF$$
Option C is Correct
#### Calculate the ratio of the inner and outer radius of the cylindrical capacitor of capacitance $$C=\dfrac {1}{\ell n\;(2)}nF$$ and length $$\ell$$= 18 m.
A $$\dfrac {2}{1}$$
B $$\dfrac {3}{1}$$
C $$\dfrac {4}{3}$$
D $$\dfrac {6}{5}$$
×
The capacitance of a cylindrical capacitor is given as $$C=\dfrac {\ell}{2K\;\ell n\left ( \dfrac {b}{a}\right)}$$
Given : $$C=\dfrac {1}{\ell n(2)}nF$$, $$\ell$$= 18 m
$$C=\dfrac {\ell}{2K\;\ell n\left ( \dfrac {b}{a}\right)}$$
$$\Rightarrow \ell n\left ( \dfrac {b}{a} \right) =\dfrac {\ell}{2KC}$$
$$\Rightarrow \left ( \dfrac {b}{a} \right) =e^{\ell/2KC}$$
$$\Rightarrow \left ( \dfrac {b}{a} \right) =e^\left ({\dfrac {18\;\ell n(2)}{2\times9\times10^9\times1\times10^{-9}}}\right)$$
$$\Rightarrow \left ( \dfrac {b}{a} \right) =\dfrac {2}{1}$$
### Calculate the ratio of the inner and outer radius of the cylindrical capacitor of capacitance $$C=\dfrac {1}{\ell n\;(2)}nF$$ and length $$\ell$$= 18 m.
A
$$\dfrac {2}{1}$$
.
B
$$\dfrac {3}{1}$$
C
$$\dfrac {4}{3}$$
D
$$\dfrac {6}{5}$$
Option A is Correct
# Calculation of Capacitance of a Spherically Charged Conductor
• Consider a spherically charged conductor having radius R and charge +Q.
• The electric potential of a spherically charged conductor is given as $$V=\dfrac {KQ}{R}$$
## Stepwise Calculation of Capacitance for Spherical Capacitor
Step 1 : Consider a spherical conducting shell of radius b, having charge –Q concentric with a smaller conducting sphere of radius a, having charge +Q.
Step 2: Electric field outside the sphere is in radial direction (if radius is r)
$$E=\dfrac {KQ}{r^2}$$
Step 3 : Potential difference between a conducting sphere and a conducting shell, is given as
$$V_b-V_a=-\int\limits_a^b\vec E\cdot\vec dr$$
or, $$V_b-V_a=-\int\limits_a^b\dfrac {KQ}{r^2}\;dr$$
or, $$V_b-V_a=-KQ\int\limits_a^b\dfrac {1}{r^2}\;dr$$
or, $$V_b-V_a=-KQ \left [ -\dfrac {1}{r} \right]_a^b$$
or, $$V_b-V_a=KQ \left [ \dfrac {1}{b}-\dfrac {1}{a} \right]$$
The magnitude of the potential difference is
$$\Delta V \,=\,\, |V_b-V_a|=\dfrac {KQ(b-a)}{ab}$$
Step 4 : Capacitance of the body is given as
$$C=\dfrac {Q}{\Delta V}$$
or, $$C=\dfrac {Q}{KQ\left ( \dfrac {b-a}{ab}\right)}$$
or, $$C=\dfrac {ab}{K(b-a)}$$
#### Choose the correct sequence of steps to calculate capacitance for a spherical capacitor.
A Charge (Q)$$\rightarrow$$ Potential difference($$\Delta V$$) $$\rightarrow$$Capacitance, $$C=\dfrac {\Delta V}{Q}$$
B Charge (Q)$$\rightarrow$$Electric field (E)$$\rightarrow$$ Potential difference($$\Delta V$$) $$\rightarrow$$Capacitance, $$C=\dfrac {Q}{\Delta V}$$
C Charge (Q)$$\rightarrow$$Potential difference($$\Delta V$$) $$\rightarrow$$Electric field (E)$$\rightarrow$$ Capacitance, $$C=\dfrac {Q}{\Delta V}$$
D None of these
×
Determine the charge (Q) on a spherical capacitor.
Calculate the electric field (E) by using Gauss's Law $$E=\dfrac {KQ}{r^2}$$
Calculate potential difference by using electric field $$\Delta V=-\int\vec E\cdot\vec dr$$
Then, calculate capacitance of spherical capacitor by using $$C=\dfrac {Q}{\Delta V}$$
Hence, option (B) is correct.
### Choose the correct sequence of steps to calculate capacitance for a spherical capacitor.
A
Charge (Q)$$\rightarrow$$ Potential difference($$\Delta V$$$$\rightarrow$$Capacitance, $$C=\dfrac {\Delta V}{Q}$$
.
B
Charge (Q)$$\rightarrow$$Electric field (E)$$\rightarrow$$ Potential difference($$\Delta V$$$$\rightarrow$$Capacitance, $$C=\dfrac {Q}{\Delta V}$$
C
Charge (Q)$$\rightarrow$$Potential difference($$\Delta V$$$$\rightarrow$$Electric field (E)$$\rightarrow$$ Capacitance, $$C=\dfrac {Q}{\Delta V}$$
D
None of these
Option B is Correct
# Calculation of Capacitance of an Isolated Spherical Conductor
• Consider a spherically charged conductor having radius R and charge +Q.
• The electric potential of a spherically charged conductor is given as $$V=\dfrac {KQ}{R}$$
## Stepwise Calculation of Capacitance for Spherical Capacitor
Step 1 : Consider a spherical conducting shell of radius b, having charge –Q concentric with a smaller conducting sphere of radius a, having charge +Q.
Step 2: Electric field outside the sphere is in radial direction (if radius is r)
$$E=\dfrac {KQ}{r^2}$$
Step 3 : Potential difference between a conducting sphere and a conducting shell, is given as
$$V_b-V_a=-\int\limits_a^b\vec E\cdot\vec dr$$
or, $$V_b-V_a=-\int\limits_a^b\dfrac {KQ}{r^2}\;dr$$
or, $$V_b-V_a=-KQ\int\limits_a^b\dfrac {1}{r^2}\;dr$$
or, $$V_b-V_a=-KQ \left [- \dfrac {1}{r} \right]_a^b$$
or, $$V_b-V_a=KQ \left [ \dfrac {1}{b}-\dfrac {1}{a} \right]$$
The magnitude of the potential difference is
or, $$\Delta V=\,\,|V_b-V_a|=\dfrac {KQ(b-a)}{ab}$$
Step 4 : Capacitance of the body is given as
$$C=\dfrac {Q}{\Delta V}$$
or, $$C=\dfrac {Q}{KQ\left ( \dfrac {b-a}{ab}\right)}$$
or, $$C=\dfrac {ab}{K(b-a)}$$
### For Isolated Spherical Conductor
The electric field lines for the inner conductor will be radially outward.
• Assume that the outer sphere is made of infinitely large radius,
means $$b\rightarrow\infty$$
• Also, the electric field lines for outer spherical conductor will be radially outward.
• If the outer radius (b) of sphere approaches $$\infty$$, then capacitor becomes isolated spherical conductor.
• The capacitance of isolated spherical conductor is given as
$$C= \displaystyle\lim_{b\to\infty} \dfrac {ab}{K(b-a)}$$
or, $$C=\displaystyle\lim_{b\to\infty}\dfrac {a}{K\left ( 1-\dfrac {a}{b}\right)}$$ $$\left [ \displaystyle\lim_{b\to\infty}\dfrac {a}{b}=0 \right]$$
$$C=\dfrac {a}{K}$$
$$C=4\pi\epsilon_0a$$
This is the capacitance for isolated sphere.
#### Calculate the value of capacitance of an isolated spherical conductor of radius a = 5 m.
A $$20\pi\epsilon_0\,F$$
B $$5\pi\epsilon_0\,F$$
C $$10\pi\epsilon_0\,F$$
D $$\pi\epsilon_0\,F$$
×
Capacitance for an isolated sphere is given as $$C=4\pi\epsilon_0a$$
Given : a = 5m,
$$C=4\pi\epsilon_0\times 5$$
$$C=20\pi\epsilon_0\,F$$
Hence , option (A) is correct.
### Calculate the value of capacitance of an isolated spherical conductor of radius a = 5 m.
A
$$20\pi\epsilon_0\,F$$
.
B
$$5\pi\epsilon_0\,F$$
C
$$10\pi\epsilon_0\,F$$
D
$$\pi\epsilon_0\,F$$
Option A is Correct | HuggingFaceTB/finemath | |
# Car on a Banked Curve (no friction)
1. Jun 10, 2010
### stumped101
Hi,
If we have a car traveling in a circle on a banked curve without friction, the forces acting on the car would only be the normal force and gravitational force.
Using a x-y coordinate system, the horizontal component of normal (Nsintheta) provides the centripetal force Fc and it's vertical component (Ncostheta) cancels out the force of gravity. Everything works out here.
However, if you were to the draw the forces in a coordinate system that had axes parallel and perpendicular to the plane, you would find the normal force being canceled out by the force of gravity perpendicular to the plane (Fgcostheta) and leaving us with just a net force of the parallel force of gravity down the plane. What's stopping the car from sliding down the curve now and what's providing the centripetal force on this diagram?
Thank you.
2. Jun 11, 2010
### hikaru1221
This is where you were wrong.
This component of gravity force is canceled, but the normal force isn't, which leaves a component perpendicular to the plane. It, along with the parallel one, forms the net force, which is the centripetal force.
3. Jun 11, 2010
### mikelepore
No matter what coordinate system you select you'e still going to get the normal force having a horizontal component of N sin theta (which must be equal to centripetal force mv^2/r) and a vertical component of N cos theta (which must be equal to the weight mg).
A handy way to get the banking angle that you need, for a given radius of curvature and speed, is to divide those two equations, which results in tan theta = v^2/rg.
4. Jun 11, 2010
### sganesh88
The important thing here is to analyse why the vertical component of normal force is higher than gravity.
Place a stationary block on the same banked curve. It will slide down. In that case the vertical component of normal force would get defeated by gravity. Drawing force diagrams isn't the only part of physics in this situation. You just found that it had misled you.
5. Jun 11, 2010
### Staff: Mentor
No, the normal force isn't 'canceled out'. You are tacitly assuming that the acceleration perpendicular to the plane is zero, but it's not. The acceleration is horizontal, which has components parallel and perpendicular to the plane. | HuggingFaceTB/finemath | |
If f(x) =-e^(2x-1) and g(x) = -3sec^2x^2 , what is f'(g(x)) ?
Jun 16, 2017
$f ' \left(g \left(x\right)\right) = \frac{1}{6 x} {e}^{2 x - 1} \cot {x}^{2} {\cos}^{2} {x}^{2}$
Explanation:
As $f \left(x\right) = - {e}^{2 x - 1}$ and $g \left(x\right) = - 3 {\sec}^{2} {x}^{2}$
$\frac{\mathrm{df}}{\mathrm{dx}} = - {e}^{2 x - 1} \times 2 = - 2 {e}^{2 x - 1}$
and$\frac{\mathrm{dg}}{\mathrm{dx}} = - 3 \times 2 \sec {x}^{2} \times \sec {x}^{2} \tan {x}^{2} \times 2 x$
= $- 12 x \tan {x}^{2} {\sec}^{2} {x}^{2}$
Now $f ' \left(g \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dg}} = \frac{\frac{\mathrm{df}}{\mathrm{dx}}}{\frac{\mathrm{dg}}{\mathrm{dx}}}$
= $\frac{- 2 {e}^{2 x - 1}}{- 12 x \tan {x}^{2} {\sec}^{2} {x}^{2}}$
= ${e}^{2 x - 1} / \left(6 x \tan {x}^{2} {\sec}^{2} {x}^{2}\right)$
= $\frac{1}{6 x} {e}^{2 x - 1} \cot {x}^{2} {\cos}^{2} {x}^{2}$ | HuggingFaceTB/finemath | |
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# Thirty Identical Cards Are Marked with Numbers 1 to 30. If One Card is Drawn at Random, Find the Probability that It Is: A Multiple of 3 Or 5 - ICSE Class 10 - Mathematics
ConceptType of Event - Complementry
#### Question
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is:
a multiple of 3 or 5
#### Solution
There are 30 aaards from which one card is drawn.
Total number of elementary events=n(s)=30
From numbers 1 to 30, there are 14 numbers which is multiple of 3 or 5 i.e {3,5,6,9,10,12,15,18,20,21,24,25,27,30} favorable number of event =n(E)=14
Probability of selecting a card with a multiple of 3 or 5 = (n(E))/(n(s))=14/30=7/15
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL
Solution Thirty Identical Cards Are Marked with Numbers 1 to 30. If One Card is Drawn at Random, Find the Probability that It Is: A Multiple of 3 Or 5 Concept: Type of Event - Complementry.
S | HuggingFaceTB/finemath | |
1. ## Intermediate Value Theorem
a) Let f :[a,b] --> [a,b] be continuous on [a,b]. Prove that f has a fixed point, i.e., prove that (exist) c in [a,b] such that f(c)=c
b) let f,g : [a,b]->[a,b] be two continuous function on [a,b] such that f(a)>or equal g(a) and f(b)<or equal g(a).
Prove that exist c in [a,b] such that f(c) =g(c)
Plx help
2. I see that you have severval other postings.
You should understand that this is not a homework service nor is it a tutorial service.
PLease either post some of your own work on this problem or explain what you do not understand about the question.
3. Sorry,
I just want to get some idea for how to start the question since I have no idea how to start the question.
4. Here is a standard hint.
Define $\displaystyle g(x)=f(x)-x$, a continuous function.
Now show that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$.
5. For b)
Define new function H(x)=f(x)-g(x),
1. prove that H continuous...
2. Prove that exist x=t in [a,b] such that H(t)=0...What is the meaning of that?
6. Originally Posted by Also sprach Zarathustra
For b)
Define new function H(x)=f(x)-g(x),
1. prove that H continuous...
2. Prove that exist x=t in [a,b] such that H(t)=0...What is the meaning of that?
I get a problem when I am trying to prove that H(t)=0, how can I say that H(t)=0?
I think I have to show that f(t)is not > g(t) and also g(t) is not > f(t) then g(t)=f(t) but I have no idea how can I show that is true.
7. Try to look at H(a) and H(b) and then use Intermediate Value Theorem.
8. My favorite way to prove a)
Is to suppose that $\displaystyle f(x)\ne x$ for all $\displaystyle x\in[a,b]$ then $\displaystyle g:[a,b]\to\{-1,1\}:x\mapsto\frac{f(x)-x}{|f(x)-x|}$ is continuous and surjective, but this is impossible...why?
9. Originally Posted by Plato
Here is a standard hint.
Define $\displaystyle g(x)=f(x)-x$, a continuous function.
Now show that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$.
when I when that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$ f(c) will equal 0. how can I relat to c???
10. Originally Posted by mathbeginner
when I when that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$ f(c) will equal 0. how can I relat to c???
Go to your textbook and class notes and review examples that apply the Intermediate Value Theorem. Spend more than just a few minutes doing this. Then come back if you still have this question. (You will find that the question you posted is a standard example in many textbooks that cover this material).
11. Originally Posted by mr fantastic
Go to your textbook and class notes and review examples that apply the Intermediate Value Theorem. Spend more than just a few minutes doing this. Then come back if you still have this question. (You will find that the question you posted is a standard example in many textbooks that cover this material).
I find out that what is c depend on how do u set up the question, ex. g(a)<c<g(b) c can be anything ex. 0 or 1 if I put 0 on it then f(c)=0 or if I put 1 on it then f(c)=1
Am I right? I just want to make sure that am I get it or not. Thanks.
12. Perhaps you are missing the point of this problem.
You are not asked to actually find the value of c.
You are asked to show that one actually exists.
Of course, it does depend upon the setup.
13. Originally Posted by Plato
Perhaps you are missing the point of this problem.
You are not asked to actually find the value of c.
You are asked to show that one actually exists.
Of course, it does depend upon the setup.
So you are saying that I should prove that f(c)is not > or < c is equal c ?
Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?
I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.
Thanks
14. Originally Posted by mathbeginner
So you are saying that I should prove that f(c)is not > or < c is equal c ?
Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?
I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.
Thanks
Your confusion is so profound that I have absolutely nothing more to say.
15. Originally Posted by mathbeginner
So you are saying that I should prove that f(c)is not > or < c is equal c ?
Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?
I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.
Thanks
Please take the advice given in post #10. Then come back with fresh eyes on this thread. | HuggingFaceTB/finemath | |
# MAT 126 Week 2 DQ 2
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MAT 126 Week 2 DQ 2
This Discussion will help us learn to develop our own mathematical models, write down the equations and then solve the equations for unknown values using algebraic methods.
1. Refer back to Week One Discussion and use the names and ages of yourself and the other two people you selected. Make sure one is older than you and one is younger than you.
2. In years, how old was the older person when you were born?
3. Write an equation that models how old in years each of you will be, when your ages add up to 150 years old. For example, if x = your age and the eldest person was a year older than you, you would write their age as x + 1. Then the equation would be: x + (x+1) = 150.
4. Explain the reasoning which helped you develop your equation.
6. In years, how old were you when the youngest person was born?
7. At some point during the lives of you and the youngest person, your age will be three times his/her age at that moment. Write an equation which models how old in years each of you will be when you are three times as old as the younger person.
8. Explain the reasoning which helped you develop your equation.
9. Solve the equation for your ages when you are three times as old as the youngest person. Are your answers reasonable?
10. Respond to at least two of your classmates’ postings. Check their equations and investigate for mathematical errors. Help with a constructive critique.
Your initial post should be at least 150 words in length.
## Write a review
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# Maximum number of comparisons required for a list of 6 numbers
#### Joystar1977
##### Active member
The question asks me as follows: "What is the maximum number of comparisons required for a list of 6 numbers?
Is the correct answer as follows: The maximum number of comparisons required for a list of 6 numbers is 5 comparisons. If this is not right, then can somebody please help and explain this to me?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
The question asks me as follows: "What is the maximum number of comparisons required for a list of 6 numbers?
Required to do what with a list of 6 numbers? Please write the complete question. Is it about sorting? If so, then is it about some concrete algorithm or the minimum over all algorithms (of maxima over all 6-number lists)? Finally, it would be nice if you provided a reason for your answer so that it can be checked.
#### Joystar1977
##### Active member
Hello Evgeny.Makarov! The complete question is as follows:
Use Bubble Sort to sort the list 7, 12, 5, 22, 13, and 32. What is the maximum number of comparisons required for a list of 6 numbers?
Is the maximum number of comparisons required for a list of 6 numbers 5 comparisons? If this is not correct, then can somebody please explain this to me?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Use Bubble Sort to sort the list 7, 12, 5, 22, 13, and 32. What is the maximum number of comparisons required for a list of 6 numbers?
Is the maximum number of comparisons required for a list of 6 numbers 5 comparisons? If this is not correct, then can somebody please explain this to me?
One requires 5 comparisons only for one pass. I suggest you sort the list 6, 5, 4, 3, 2, 1 in ascending order and note how many comparisons you used.
#### Joystar1977
##### Active member
Evgeny.Makarov- Let me understand this correctly each time there is a pass then one of the numbers requires 5 comparisons. Is this true or false? Each step is called a pass through the algorithm. The steps are repeated until no swaps take place in a pass. If there are n numbers in the original list, then n-1 comparisons are needed in the 1st pass, n-2 comparisons are needed on the 2nd pass, n-3 comparisons are needed on the 3rd pass, n-4 comparisons are needed on the 4th pass, n-5 comparisons are needed on the 5th pass, and n-6 comparisons are needed on the 6th pass. If the maximum number of passes is needed, the total number of comparisons will be:
(n-1) + (n-2) + (n-3) + ... + 3 + 2 + 1 = 1/2 (n-1)n
Another question is this true or false:
The maximum number of swaps required is also 1/2 (n-1)n. This maximum number of swaps takes place if the original list of numbers were written in reverse order. The maximum number of swaps/comparisons is a quadratic function. Is it correct to say that the algorithm has quadratic order?
When you say ascending order do you mean to switch around the numbers or just put them in order from smallest to largest or largest to smallest. For example:
1, 2, 3, 4, 5, 6
6, 5, 4, 3, 2, 1
1 3 2 4 5 6
2 1 3 4 5 6
3 2 1 4 5 6
I am still trying to get the hang of this. Please let me know whether or not I am on the right track.
Sincerely,
Joystar1977
One requires 5 comparisons only for one pass. I suggest you sort the list 6, 5, 4, 3, 2, 1 in ascending order and note how many comparisons you used.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
First of all, you need to determine which version of bubble sort you are using. The best thing would be if you just posted your version between the [code]...[/code] tags.
In another thread, there was a question how the algorithm knows when the array is finally sorted and when it is time to stop. You answered that it takes a whole pass without swaps to determine that the array is sorted. This is true for the unoptimized version of the algorithm, which goes to the end of the array and makes the same number of comparisons during every pass. In contrast, an optimized algorithm uses the fact that after the first pass, the last number is in its correct place, after the second pass the last two numbers are in their correct places and so on. So, the optimized algorithm does not go to the end of the array in subsequent passes and makes fewer and fewer comparisons with each new pass. This algorithm recognizes the time to stop when it has to make an empty pass. You can see the code for the two version in Wikipedia. Below I assume that you have an optimized algorithm.
each time there is a pass then one of the numbers requires 5 comparisons. Is this true or false?
False. Your claim implies that there is a single number that is compared 5 times. In fact, the algorithm compares adjacent pairs of array elements: first and second, second and third, and so on. Due to swapping, one of the numbers in each pair comes from the previous pair, so it may indeed happen that one of the numbers in every pair is the same. However, the algorithm may also compare new numbers, not encountered before during that pass.
If an array has 6 elements, then the first pass makes 5 comparisons since there are 5 adjacent pairs: from first and second to fifth and sixth.
Each step is called a pass through the algorithm.
I am not sure if this is your definition of the word "step", a definition of the word "pass", or a claim that uses existing meanings of these words. No, I would not say that a pass is a step. The algorithm makes one pass as it goes through the whole array from the beginning to the end. We can define the word "step" in different ways, but I would say that a step is smaller: it is a comparison or swapping of two numbers, an increase of the counter and so on. Thus, one pass consists of many steps.
The steps are repeated until no swaps take place in a pass.
This depends on the details of the algorithm. Here you agree that a pass consists of several steps.
If there are n numbers in the original list, then n-1 comparisons are needed in the 1st pass, n-2 comparisons are needed on the 2nd pass, n-3 comparisons are needed on the 3rd pass, n-4 comparisons are needed on the 4th pass, n-5 comparisons are needed on the 5th pass, and n-6 comparisons are needed on the 6th pass. If the maximum number of passes is needed, the total number of comparisons will be:
(n-1) + (n-2) + (n-3) + ... + 3 + 2 + 1 = 1/2 (n-1)n
I agree with regard to the optimized algorithm.
Another question is this true or false:
The maximum number of swaps required is also 1/2 (n-1)n. This maximum number of swaps takes place if the original list of numbers were written in reverse order.
Yes.
The maximum number of swaps/comparisons is a quadratic function. Is it correct to say that the algorithm has quadratic order?
When you say ascending order do you mean to switch around the numbers or just put them in order from smallest to largest or largest to smallest. For example:
1, 2, 3, 4, 5, 6
6, 5, 4, 3, 2, 1
1 3 2 4 5 6
2 1 3 4 5 6
3 2 1 4 5 6
An array is sorted in ascending order when the numbers go from smallest to largest.
#### Joystar1977
##### Active member
If I put the numbers in ascending order as you mentioned earlier:
6, 5, 4, 3, 2, 1
In Ascending Order (smallest to largest) would be as follows:
1 2 3 4 5 6
So if I have the following numbers:
7, 12, 5, 22, 13, 32
In Ascending Order (smallest to largest) would be as follows:
5 7 12 13 22 32
Would I have to go through and do a second pass and third pass? I am going off an example that I found on Wikipedia. It says after the second pass before starting up on the third pass as follows: "now the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted. Is this true or correct?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
This is true for the unoptimized algorithm.
#### Joystar1977
##### Active member
Thanks Evgeny.Makarov! | HuggingFaceTB/finemath | |
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# A to Z of Excel Functions: The GAMMAINV Function
6 January 2020
Welcome back to our regular A to Z of Excel Functions blog. Today we look at the GAMMAINV function.
The GAMMAINV function
The Gamma distribution is widely used in engineering, science and business, to model continuous variables that are always positive and have skewed distributions. The Gamma distribution can be useful for any variable which is always positive, such as cohesion or shear strength for example.
It is a distribution that arises naturally in processes for which the waiting times between events are relevant, and is often thought of as a waiting time between Poisson distributed events (the Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval). This is what is known as “queueing analysis”.
To understand it, first think of factorials, e.g. 5! = 5 x 4 x 3 x 2 x 1 = 120. So far, so good, but how do you calculate if the factorial number you want to evaluate isn’t an integer? The Gamma function is used to calculate this:
Г(N+1) = N * Г(N)
That’s great (if a little recursive), so can be expressed better (!) mathematically as follows:
Clear as mud? Well, it gets better. The Gamma distribution just referred to has the following probability density function:
where Г(α) is the Gamma function, and the parameters α and β are both positive, i.e. α > 0 and β > 0:
• α is known as the shape parameter, while β is referred to as the scale parameter
• β has the effect of stretching or compressing the range of the Gamma distribution. A Gamma distribution with β = 1 is known as the standard Gamma distribution.
The Gamma distribution represents a family of shapes. As suggested by its name, α controls the shape of the family of distributions. The fundamental shapes are characterized by the following values of α:
• Case I (α < 1): when α < 1, the Gamma distribution is exponentially shaped and asymptotic to both the vertical and horizontal axes
• Case II (α = 1): the Gamma distribution with shape parameter α = 1 and scale parameter β is the same as an exponential distribution of scale parameter (or mean) β
• Case III (α > 1): when α is greater than one, the Gamma distribution assumes a mounded (unimodal), but skewed shape. The skewness reduces as the value of α increases.
The GAMMAINV function is the inverse of the Gamma cumulative distribution, i.e. if p = GAMMADIST(x, α, β) then x = GAMMAINV(p, α, β). This can help to identify key points in a variable whose distribution may be skewed. It has the following syntax:
GAMMAINV(probability, alpha, beta)
It’s important to note that this function has been replaced with a new function (GAMMA.INV) that may provide improved accuracy and whose name better reflects its usage and consistency with other programming languages. Although this function is still available for backward compatibility, you should consider using the new functions from now on, because this function may not be available in future versions of Excel.
The GAMMAINV function has the following arguments:
• probability: this is required and this represents the probability associated with the distribution
• alpha: this is also required. This is a parameter (the shape parameter) to the distribution
• beta: this too is required and is another parameter (the scale parameter) to the distribution. If beta = 1, GAMMADIST returns the standard gamma distribution.
It should be noted that:
• if any argument is text, GAMMAINV returns the #VALUE! error value
• if probability < 0, if probability > 1, if alpha ≤ 0 or if beta ≤ 0, GAMMAINV returns the #NUM! error value.
Given a value for probabilityGAMMAINV seeks that value x such that GAMMADIST(x, alpha, beta, TRUE) = probability. Thus, the precision of GAMMAINV depends upon the precision of GAMMADIST. Therefore, GAMMAINV uses an iterative search technique. If the search has not converged after 64 iterations, the function returns the #N/A error value. | HuggingFaceTB/finemath | |
Module3.5 Convexity Correction
Yield
Straight line is what we get with %ΔPB formula (underestimates when yield drops, over-estimates when rises)
Greater a bond’s convexity, the more valuable it is (because
price increases are more, and price declines are less.)
1
Not To Be Naïve about Duration
1. The duration D we have been discussing also known
as Macaulay duration.
2. First derivative of price-yield curve is
D
(1 i )
and is known as modified duration. Found in
%ΔPB formula.
3. Convexity is second derivative of price-yield curve.
Complicated, not studied here.
2
Closer in the Payments, Less the Duration
Holding a bond’s yield fixed (at 10% in the below), D
increases with maturity and varies inversely with
coupon rate.
3
Managing Interest Rate Risk (a)
Duration is the holding period for which reinvestment
risk exactly offsets price risk. Gives investor the YTM
that was in effect at time bond purchased.
A way duration is used: If have a \$5 million liability 7.5
years from now, buy a bond (or a portfolio of bonds)
today that has a duration of 7.5 years.
10/31
4
Example 22: Portfolio Duration:
N
P ortfolio D
wD
i
i
i 1
Assume \$4,000 in D = 5, \$10,000 in D = 7, and \$6,000 in
D = 9 bonds. What is Portfolio D?
5
Example 23: Rebalancing Bond Portfolio
Consider the \$20,000 portfolio of Example 22.
How much in D = 9 bonds should be sold, and how much
in D = 5 bonds should be purchased, to reduce Portfolio
D to 6.80?
6
Summary
•
Zero-coupon approach (best way). Buy high quality
“zeros” with maturity equal to desired holding period.
Locks in YTM. No reinvestment risk because no
coupons payments, and no price risk when held to
maturity.
•
Duration matching (next best way). Selecting a portfolio
of bonds whose duration matches desired holding period.
Theoretically perfect, but only approximately perfect in
real world as per footnote 8 on p. 162.
•
Maturity matching (don’t use). Selecting bonds with
terms to maturity equal to desired holding period
doesn’t lock in YTM.
7 | HuggingFaceTB/finemath | |
## Conversion formula
The conversion factor from days to minutes is 1440, which means that 1 day is equal to 1440 minutes:
1 d = 1440 min
To convert 348 days into minutes we have to multiply 348 by the conversion factor in order to get the time amount from days to minutes. We can also form a simple proportion to calculate the result:
1 d → 1440 min
348 d → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 348 d × 1440 min
T(min) = 501120 min
The final result is:
348 d → 501120 min
We conclude that 348 days is equivalent to 501120 minutes:
348 days = 501120 minutes
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 1.9955300127714E-6 × 348 days.
Another way is saying that 348 days is equal to 1 ÷ 1.9955300127714E-6 minutes.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred forty-eight days is approximately five hundred one thousand one hundred twenty minutes:
348 d ≅ 501120 min
An alternative is also that one minute is approximately zero times three hundred forty-eight days.
## Conversion table
### days to minutes chart
For quick reference purposes, below is the conversion table you can use to convert from days to minutes
days (d) minutes (min)
349 days 502560 minutes
350 days 504000 minutes
351 days 505440 minutes
352 days 506880 minutes
353 days 508320 minutes
354 days 509760 minutes
355 days 511200 minutes
356 days 512640 minutes
357 days 514080 minutes
358 days 515520 minutes | HuggingFaceTB/finemath | |
# Introduction
Notes from CS 4104 with exercises and derivations from "Algorithm Design" by Kleinberg and Tardos.
## Glossary
Term Definition
Matching each man is paired with ≤ 1 woman and vice versa
Perfect Matching each man is paired with exactly 1 woman and vice versa
Rogue Couple a man and a woman who are not matched, but prefer each other to their
Stable Matching a perfect matching without any roogue couples
polynomial running time if there exists constants $c,d > 0$ such that $\forall n$, the running time is bounded by $cn^d$
Asymptotic Upper Bound a function $f(n)$ is $O(g(n))$ if there exists constants $c > 0, n_0 \geq 0$ s.t. $\forall n \geq n_0, \quad f(n) \leq cg(n)$
Asymptotic Lower Bound a function $f(n)$ is $\Omega(g(n))$ if there exist constants $c > 0, n_0 \geq 0$ s.t. $\forall n \geq n_0, \quad f(n) \geq cg(n)$
Asymptotic Tight Bound a function $f(n)$ is $\Theta(g(n))$ if $f(n)$ is $O(g(n))$ and $f(n)$ is $\Omega(g(n))$
Euler Path Only possible if the number of nodes with odd degree is at most 2
A $v_1 - v_k$ path in an undirected graph $G = (V,E)$ is a sequence of $P$ nodes $v_1, v_2, \mathellipsis, v_{k-1}, v_k; v_k \in V$ s.t. every consecutive pair of nodes $v_i; v_{i+1}, 1 \leq i < k$ is connected by an edge $E$
simple path all its nodes are distinct
cycle is a path where $k>2$, the first $k-1$ nodes are distinct and $v_1 = v_k$
connected undirected graph for every pair of nodes $u,v \in V$, there is a path from $u$ to $v$ in $G$
distance $d(u,v)$ the minimum number of edges in any $u-v$ path
connected component of $G$ containing $s$ the set of all nodes $u$ such that there is an $s-u$ path in $G$
Adjaceny Matrix $n \times n$ boolean matrix, where the entry in row $i$, col $j$ is 1 iff the graph contains the edge $(i, j)$
Adjaceny List array $Adj$, where $Adj[v]$ stores the list of all nodes adjacent to $v$
bipartite A graph $G = (V, E)$ is bipartite if $V$ can be partitioned into two subsets $X,Y$ such that every edge in $E$ has one endpoint in $X$ and one in $Y$
Greedy algorithm makes the best current choice without looking back. Assume the choices made prior were perfect
inversion A schedule has an inversion if a job $i$ with a deadline $d(i)$ is scheduled before another job $j$ with an earlier deadline $d(j)$ i.e. $d(j) < d(i), s(i) < s(j)$
spanning tree A subset $T of E$ is a spanning tree of $G$ if $(V, T)$ is a tree
# 01 - Stable Matching
### Problem
• Each man ranks all the women in order of preference
• Each woman ranks all the men in order of preference
• Each person uses all ranks from $1, ... , n$ s.t. there are no ties or incomplete lists
#### Male preference matrix:
$w_1$ $w_2$ $w_3$ $w_4$
$m_1$ 1 2 3 4
$m_2$ 4 3 1 2
$m_3$ 4 3 1 2
$m_4$ 3 1 4 2
#### Female preference matrix:
$m_1$ $m_2$ $m_3$ $m_4$
$w_1$ 1 2 3 4
$w_2$ 4 1 3 2
$w_3$ 4 1 2 3
$w_4$ 3 1 2 4
• A Matching: each man is paired with ≤ 1 woman and vice versa
• A Perfect Matching: each man is paired with exactly 1 woman and vice versa
• "perfect" means one-to-one mapping
• A Rogue Couple: a man and a woman who are not matched, but prefer each other to their current partners
• A Stable Matching: A perfect matching without any rogue couples
### Solution: Gale-Shapley Algorithm
\boxed{ \begin{aligned} &\text{Initially, all men and all women are free} \\ &\text{Set } S \text{ of matched pairs is empty} \\ &\text{While there is at least one free man who has not proposed to every woman} \\ &\quad\text{Choose such a man } m \\ &\qquad m \text{ proposes to his highest-ranked woman } w \text{ to whom he has not yet proposed} \\ &\quad\text{If } w \text{ is free} \\ &\qquad\text{she becomes engaged to } m \text{ s.t. } S \leftarrow S \cup \{m, w\} \\ &\quad\text{Else if } w \text{ is engaged to } m' \text{ and she prefers } m \text{ to } m' \\ &\qquad\text{she becomes engaged to } m \text{ s.t. } S \leftarrow S \cup \{m, w\} \\ &\qquad m' \text{ becomes free s.t. } S \leftarrow S \backslash \{m', w\} \\ &\quad\text{Otherwise, } m \text{ remains free} \\ &\text{Return set } S \text{ of engaged pairs} \end{aligned} }
#### Proof of Perfection
1. Suppose the set $S$ of pairs returned by Gale-Shapley algorithm is not perfect
2. $S$ is a matching, therefore there must be at least one free man $m$
3. $m$ has proposed to all women (since the algorithm terminated)
4. Therefore, each woman must be engaged (since she remains engaged after the first proposal to her)
5. Therefore, all men must be engaged, contradicting ⚔️ the assumption that $m$ is free
6. That matching is perfect since the program terminated QED.
#### Proof of Stability
1. Suppose $S$ is not stable i.e. there are two pairs $(m_1, w_1)$, $(m_2, w_2) \in S$ s.t $m_1$ prefers $w_2 > w_1$ and $w_2$ prefers $m_1 > m_2$
2. $m_1$ must have proposed to $w_2$ prior to $w_1$ because at that stage $w_2$ must have rejected $m_1$; otherwise she would have been paired with $w_1$, which would in turn prevent the pairing of $(m_2, w_2)$ in a later iteration
3. When $w_2$ rejected $m_1$, she must have been paired with some man $m_3 > m_1$
4. Since $m_2$ is paired with $w_2$ at termination, $w_2$ must prefer $m_2$ to $m_3$ or $m_2 = m_3$
5. This contradicts ⚔️ our conclusion (from instability) that $w_2$ prefers $m_1 > m_2$ QED.
### Observations
• This algorithm computes a matching i.e. each woman gets pair with at most one man and vice versa
• The man's status can alternate between being free and being engaged
• The woman's status remains engaged after the first proposal
• The ranking of a man's partner remains the same or goes down
• The ranking of a woman's partner can never go down
• The number of proposals made after $k$ iterations is the best indicator of progress
• The max number of total proposals (or iterations) that can be made is $n^2$
• Always produces the same matching in which each man is paired with his best, valid partner, but each woman is paired with her worst, valid partner
# 02 - Analysis of Algorithms
### Polynomial Time
• Essentially brute forcing
• e.g. brute force sorting:
• Given $n$ numbers, permute them so that they appear in increasing order
• Try all $n!$ permutations
• For each permutation, check if it is sorted
• $O(nn!)$
• Desirable scaling property: when the input size doubles, the algorithm should only slow down by some constant $c$
• An algorithm has polynomial running time if there exists constants $c,d > 0$ such that $\forall n$, the running time is bounded by $cn^d$
• an algorithm is said to be efficient if it has a polynomial running time
### Upper and Lower Bounds
• Asymptotic Upper Bound: a function $f(n)$ is $O(g(n))$ if there exists constants $c > 0, n_0 \geq 0$ s.t. $\forall n \geq n_0, \quad f(n) \leq cg(n)$
• e.g. $100n \log_2n$ is $O(n^2)$ for $c = 100, n_0 = 1$
• Asymptotic Lower Bound: a function $f(n)$ is $\Omega(g(n))$ if there exist constants $c > 0, n_0 \geq 0$ s.t. $\forall n \geq n_0, \quad f(n) \geq cg(n)$
• e.g. $\frac{n}{10}\log_2 n$ is $\Omega(n)$ for $c = 1, n_0 = 1024$
In general, we attempt to find a sufficiently large $n_0$ such that our upper and lower bounds are satisfied.
• Asymptotic Tight Bound: a function $f(n)$ is $\Theta(g(n))$ if $f(n)$ is $O(g(n))$ and $f(n)$ is $\Omega(g(n))$
Transitively:
• if $f = O(g)$ and $g = O(h)$, then $f = O(h)$
• if $f = \Omega(g)$ and $g = \Omega(h)$, then $f = \Omega(h)$
• if $f = \Theta(g)$ and $g = \Theta(h)$, then $f = \Theta(h)$
• if $f = O(h)$ and $g = O(h)$, then $f + g = O(h)$
• Similar statements hold for lower and tight bounds
• if $k$ is a constant and there are $k$ functions: $f_i = O(h), 1\leq i \leq k$
\begin{aligned} f_1 + f_2 + ... + f_k = O(h) \end{aligned}
• if $f = O(g)$, then $f + g = \Theta(g)$
### Examples
$p,q,n > 0$
$f(n)$ $g(n)$ Reason
$pn^2 + qn + r$ $\Theta(n^2)$ $pn^2 + qn + r \leq (p + q + r)n^2$, and we can ignore low order terms
$pn^2 + qn + r$ $O(n^3)$ $n^2 \leq n^3, \text{ if } n \geq 1$
$\displaystyle\sum_{0 \leq i \leq d} a_i n^i$ $\Theta(n^d)$ if $d > 0$ is an integer constant and $a_d > 0$
$O(n^{1.59})$ polynomial time $n^{1.59}$ is $O(n^2)$
$\log_a n$ $O(\log_b n)$ $\log_2(x) = \frac{\log_{10}(x)}{\log_{10}(2)} \forall a,b > 1$ therefore the base is irrelevant
• $\forall \text{ constant } x > 0, \log n = O(n^x)$ e.g. $\log n = n^{0.00001}$
• $\forall \text{ constant } r > 1, d >0, n^d = O(r^n)$ e.g. $n^3 = O(1.1^n)$
# 03 - Review of Graphs and Priority Queues
## Priority Queues: Motivation
### Sorting
• Instance: Nonempty list $[x_1, x_2, \mathellipsis, x_n]$ of integers
• Solution: A permutation $[y_1, y_2, \mathellipsis, y_n]$ of $x_1, x_2, \mathellipsis, x_n$ such that $y_i \leq y_{i+1}$ for all $1 \leq i < n$Possible algorithm: • insert each number into some data structure $D$ • Repeatedly find the smallest number in $D$, output it, and remove it • To get $O(n \log n)$ running time, each "find minimum" step and each remove step must additively take $O(\log n)$ ## Priority Queues • Store a set $S$ of elements where each element $v$ has a priority value $key(v)$ • smaller key values = higher priorities • Supported operations: • find element with smallest key • remove smallest element • insert an element • delete an element • update the key of an element • element deletion and key update require knowledge of the position of the element in the priority • combines benefits of both lists and sorted arrays • balanced binary tree • heap order. For every element $v$ at a node $i$, the element $w$ at $i$'s parent satisfies $key(w) \leq key(v)$ • Each node's key is at least as large as its parent's • storing nodes of the heap in an array: • Node at index $i$ has children at indices $2i$ and $2i+1$, with a parent at index $\lfloor \frac{i}{2} \rfloor$ • index 1 is the root • if $2i > n$, where $n$ is the current number of elements in the heap, the node at index $i$ is a leaf ### Inserting an Element: Heapify-up • insert a new element at index $n+1$ • Fix heap order using Heapify-up(H, n+1) \boxed{ \begin{aligned} &\text{Heapify-up(H, i)} \\ &\text{If } i > 1 \text{ then} \\ &\quad\text{let } j = \text{parent(i) } = \lfloor \frac{i}{2} \rfloor \\ &\quad\text{If key[H[i]] < key[H[j]] then } \\ &\qquad\text{swap entries H[i], H[j]} \\ &\qquad\text{Heapify-up(H, j)} \\ \end{aligned} } #### Proof of Running Time Complexity for Heapify-up(i) • each invocation decreases the second argument by a factor of at least 2 • after $k$ invocations, argument is at most $\frac{i}{2^k}$ • Therefore $\frac{i}{2^k} \geq 1$ which implies that $k \leq \log_2 i$ therefore heapify up is $O(\log i)$ ### Deleting an Element: Heapify-down • Suppose $H$ has $n+1$ elements • Delete element $H[i]$ moving element at $H[n+1]$ to $H[i]$ • If element at $H[i]$ is too small, fix the heap order using Heapify-up(H, i) • If element at $H[i]$ is too large, fix heap order using Heapify-down(H, i) \boxed{ \begin{aligned} &\text{Heapify-down(H, i)} \\ &\text{let } n = length(H) \\ &\text{If } 2i > n \text{ then} \\ &\quad\text{Terminate with } H \text { unchanged} \\ &\text{Else If } 2i < n \text{ then} \\ &\quad\text{Let Left, Right } = 2i, 2i+1 \\ &\quad\text{Let } j \text{ be the index that minimizes key[H[Left]] and key[H[Right]]} \\ &\text{Else } 2i = n \text{ then} \\ &\quad\text{Let } j = 2i \\ &\text{If key[H[j]] < key[H[i]] then } \\ &\quad\text{swap the entries H[i] and H[j]} \\ &\quad\text{Heapify-down(h,j)} \\ \end{aligned} } #### Proof of Running Time Complexity for Heapify-down(i) • Each invocation increases is second argument by a factor of at least 2 • after $k$ invocations arguments must be at least $i2^k \leq n$, which implies that $k \leq \log_2 \frac{n}{i}$ • Therefore running time is $O(\log_2 \frac{n}{i})$ ## Sorting • Instance: Nonempty list $[x_1, x_2, \mathellipsis, x_n]$ of integers • Solution: A permutation $[y_1, y_2, \mathellipsis, y_n]$ of $x_1, x_2, \mathellipsis, x_n$ such that $y_i \leq y_{i+1}$ for all $1 \leq i < n$
Final algorithm:
• Insert each number in a priority queue $H$
• Repeatedly find the smallest number in $H$, output it, and delete it from $H$
Thus, each insertion and deletion take $O(\log n)$ for a total running time of $O(n \log n)$
## Graphs: Motivation
• Contact tracing hahaaaaa
### Taxonomy of a Graph
• comprised of vertices and (directed / undirected) edges, they can form face.
• Euler Path: Only possible if the number of nodes with odd degree is at most 2
• Formally, an Undirected graph $G = (V, E)$ where $V, E$ are sets of vertices and edges with $E \subseteq V \times V$
• Elements of $E$ are unordered pairs
• $Edge(u, v)$ is incident on $u,v$ meaning $u,v$ are neighbors of one another
• Exactly one edge between any pair of nodes
• $G$ contains no self loops, i.e. no edges of the from $(u,u)$
• Formally, an directed graph $G = (V, E)$ where $V, E$ are sets of vertices and edges with $E \subseteq V \times V$
• Elements of $E$ are ordered pairs
• $e = (u, v)$ where $u$ is the tail of the edge $e$, $v$ is the head, and we can say that $e$ is directed from $u$ to $v$
• A pair of nodes may be connected by two directed edges $(u,v)$ and $(v,u)$
• $G$ contains no self loops
• A $v_1 - v_k$ path in an undirected graph $G = (V,E)$ is a sequence of $P$ nodes $v_1, v_2, \mathellipsis, v_{k-1}, v_k; v_k \in V$ s.t. every consecutive pair of nodes $v_i; v_{i+1}, 1 \leq i < k$ is connected by an edge $E$
• a path is simple if all its nodes are distinct
• a cycle is a path where $k>2$, the first $k-1$ nodes are distinct and $v_1 = v_k$
• An undirected graph $G$ is connected if, for every pair of nodes $u,v \in V$, there is a path from $u$ to $v$ in $G$
• The distance $d(u,v)$ between nodes $u,v$ is the minimum number of edges in any $u-v$ path
• The connected component of $G$ containing $s$ is the set of all nodes $u$ such that there is an $s-u$ path in $G$
#### Computing Connected Components
• Rather than computing the connected component of $G$ that contains $s$ and check if $t$ is in that component, we can "explore" $G$ starting from $s$, maintaining a set $R$ of visited nodes
\boxed{ \begin{aligned} &R \text{will consist of nodes to which } s \text{has a path} \\ &\text{Initially } R = \{s\} \\ &\text{While there is ad edge } (u,v) \text{ where } u \in R, v \notin R \\ &\quad\text{Add } v \text{ to } R \end{aligned} }
• Explore $G$ starting at $s$ and going "outward" in all directions, adding nodes one "layer" at a time
• Layer $L_0$ only contains $s$
• Layer $L_1$ contains $s$ and all its neighbors, etc. etc.
• Layer $L_0, L_1, \mathellipsis , L_j, L_{j+1}$ contains all nodes that:
• do not belong to an earlier layer
• are connected by an edge to a node in layer $L_j$
• The shortest path from $s$ to each node contains $j$ edges
• Claim: For each $j \geq 1$, layer $L_j$ consists of all nodes exactly at distance $j$ from $S$
• A non-tree edge is an edge of $G$ that does not belong to the BFS tree $T$
### Proof
• There is a path from $s$ to $t$ if an only iff $t$ is a member of some layer
• Let $v$ be a node in layer $L_{j+1}$ and $u$ be the "first" node in $L_j$ such that $(u,v)$ is an edge in $G$. Consider the graph $T$ formed by all edges, directed from $u$ to $v$ • Notice that $T$ is a tree because it is connected and the number of edges in $T$ is the number of nodes in all the laters minus one • $T$ is called the BFS Tree ### Inductive Proof of BFS Distance Property • for every $j \geq 0$, for every node $u \in L_j$, $d(s, u) = j$ • Basis: $k =0$, $d(s,s) = 0$ • Inductive Hypothesis: Assume the claim is true for every node $v \in L_k$, $d(s, v) = k$ • Inductive Step: Prove for every node $x \in L_{k+1}$, $d(s, x) = k + 1$ • $d(s,x) = d(s, y) + 1$ if $y$ is a node in $L_k$ • Therefore $d(s,x) = k+1$ ## Depth First Search • Explore $G$ as if it were a maze: start from $s$, traverse first edge out (to node $v$), traverse first edge out of $v$, . . . , reach a dead-end, backtrack, repeat \boxed{ \begin{aligned} &\text{Mark } u \text{as explored and add it to the reachable set} R \\ &\text{For each edge } (u,v) \text{incident to } u \\ &\quad\text{If } v \text{ is not marked, invoke DFS(} v \text{)} \\ \end{aligned} } ## Properties of each • BFS and DFS visit the same set of nodes, but in a different order ## Representing Graphs • A Graph $G = (V,E)$ has two input parameters: $|V| =n, |E| = m$, with $n-1 \leq m \leq \binom{n}{2}$ • Size of graph is defined to be $m + n$ • Strive for algorithms whose running time is linear in graph size i.e. $)(m + n)$ • We assume that $V = {1, 2, \mathellipsis, n}$ • Use an Adjaceny Matrix: $n \times n$ boolean matrix, where the entry in row $i$, col $j$ is 1 iff the graph contains the edge $(i, j)$ • the space used is $\Theta(n^2)$, which is optimal in the worst case • can check if there is an edge between nodes $i$, $j$ in $O(1)$ time • iterate over all the edges incident on node $i$ in $\Theta(n)$ time • Use an Adjaceny List: array $Adj$, where $Adj[v]$ stores the list of all nodes adjacent to $v$ • an edge $e = (u,v)$ appears twice: $Adj[u],$Adj[v]
• $n_v$ is the number of neighbors of node $v$
• space used is $O(\displaystyle\sum_{v \in G}n_v) = O(m+n)$
• check if there is an adge between nodes $u, v$ in $O(n_u)$ time
• Iterate over all the edges incident on node $u$ in $\Theta(n_u)$ time
• Inserting an edge takes $O(1)$ time
• Deleting an edge takes $O(n)$ time
Is $(i, j)$ an edge? $O(1)$ time $O(n_i)$ time
Iterate over edges incident on node $i$ $O(n)$ time $O(n_i)$ time
Space used $O(n^2)$ $O(m + n)$
# 04 - Linear-Time Graph Algorithms
## Bipartite Graphs
• A graph $G = (V, E)$ is bipartite if $V$ can be partitioned into two subsets $X,Y$ such that every edge in $E$ has one endpoint in $X$ and one in $Y$
• $X \times X \cap E = \emptyset$ and $Y \times Y \cap E = \emptyset$
• Color the nodes in $X$ red and the nodes in $Y$ blue. No edge in $E$ connects nodes of the same graph
• no cycle with an odd number of nodes is bipartite, similarly all cycles of even length are bipartite
• Therefore, if a graph is bipartite, then it cannot contain a cycle of odd length
### Algorithm for Testing Bipartite
• Assume $G$ is connected, otherwise apply the algorithm to each connected component separately
• Pick an arbitrary node $s$ and color it red.
• Color all it's neighbors blue. Color the uncolored neighbors of these nodes red, and so on till all nodes are colored
• Check if every edge has endpoints of different colours
more formally:
1. Run BFS on $G$. Maintain an array for the color
2. When we add a node $v$ to a layer $i$, set $\text{color[i]}$ to red if $i$ is even, blue of odd
3. At the end of BFS, scan all the edges to check if there is any edge both of whose endpoints received the same color
Running time is linear proportional to the size of the graph: $O(m + n)$ since we do a constant amount of work per node in addition to the time spent by BFS.
#### Proof of Correctness of a "two-colorable" algorithm
Need to prove that if the algorithm says $G$ is bipartite, then it is actually bipartite AND need to prove that if $G$ is not bipartite, can we determine why?
• Let $G$ be a graph and let $L_0, L_1, \mathellipsis, L_k$ be the layers produced by the BFS, starting at node $s$. Then exactly one of the following statements is true:
• No edge of $G$ joins two nodes in the same layer: Bipartite since nodes in the even laters can be colored red and those in odd blue
• There is an edge of $G$ that joins two nodes in the same layer: Not Bipartite
• $| L_i - L_j | = 1, \quad \forall L \in BFS$
# 05 - Greedy Algorithms
• Greedy algorithms: make the best current choice without looking back. Assume the choices made prior were perfect
## Example Problem: Interval Scheduling
• At an amusement park, want to compute the largest number of rides you can be on in one day
• Input: start and end time of each ride
• Constraint: cannot be in two places at one time
• Instance: nonempty set $\{(s(i), f(i)), 1 \leq i \leq n\}$ of start and finish times of $n$ jobs
• Solution: The largest subset of mutually compatible jobs
• Two jobs are compatible if they do not overlap
• Problem models the situation where you have a resource, a set of fixed jobs, and you want to schedule as many jobs as possible
• For any input set of jobs, the algorithm must provably compute the largest set of compatible jobs (measured by interval count, not cumulative interval length)
### Template for a Greedy Algorithm
• Process jobs in some order. Add next job to the result if it is compatible with the jobs already in the result
• key question: in what order should we process job?
• earliest start time: increasing order of start time $s(i)$
• earliest finish time: increasing order of start time $f(i)$
• shortest interval: increasing order of $f(i) - s(i)$
• fewest conflicts: increasing order of the number of conflicting jobs
### Earliest Finish Time
• the most optimal general solution
\boxed{ \begin{aligned} &\text{Initially let } R \text{ be the set of all jobs, and let } A \text{ be empty} \\ &\text{While } R \text{ is not yet empty}\\ &\quad\text{Choose a job } i \in R \text{ that has the smallest finishing time } \\ &\quad\text{Add request } i \text{ to } A \\ &\quad\text{Delete all jobs from } R \text{ that are not compatible with job } i \\ &\text{Return the set } A \text{ as the set of accepted/scheduled jobs} \\ \end{aligned} }
### Proof of Optimality
• Claim $|A|$ is a compatible set of jobs that is the largest possible in any set of mutually compatible jobs
• Proof by contradiction that there's no "better" solution at each step of the algorithm
• need to define "better", and what a step is in terms of the progress of the algorithm
• order the output in terms of increasing finish time for a measure of progress
• Finishing time of a job $r$ selected by $A$ must be less than or equal to the finishing time of a job $r$ selected by any other algorithm
• Let $O$ be an optimal set of jobs. We will show that $|A| = |O|$
• Let $i_1, i_2, \mathellipsis, i_k$ be the set of jobs in $A$ in order of finishing time
• Let $j_1, j_2, \mathellipsis, j_m$ be the set of jobs in $O$ in order of $m \geq k$
• Claim: for all indices $r \leq k, \quad f(i_r) \leq f(j_r)$
• Base case: is it possible for finish time of the first job of our algorithm to be later than the opposing job?
• $f(i_1) > f(j_1)$ is not possible, only $f(i_1) \leq f(j_1)$
• Inductive Hypothesis: this is always the case for any generic job index $r$:
• $f(i_r) \leq f(j_r)$ for some $r \leq k$
• Inductive Step: show that the same claim holds for the next job:
• $f(i_{r+1}) \leq f(j_{r+1})$ for some $r+1 \leq k$
• $s(j_{r+1}) \geq f(i_r) \geq f(i_r)$
• claim $m = k$: $f(i_{k}) \leq f(j_{k}) \leq s(j_{k+1})$
A complete proof can be found here:
### Implementing EFT
• Sort jobs in order of increasing finish time
• Store starting time of jobs in an array $S$
• $k=1$
• While $k \leq |S|$
• output job $k$
• Let the finish time of job $k$ be $f$
• Iterate over $S$ from $k$ onwards to find the first index $i$ s.t. $S[i] \geq f$
• $k = i$
• Must be careful to iterate over $S$ s.t. we never scan same index more than once
• Running time is $O(n \log n)$ since it's dominated by the first sorting step
## Scheduling to Minimize Lateness
• Suppose a job $i$ has a length $t(i)$ and a deadline $d(i)$
• want to schedule all $n$ jobs on one resource
• Goal is to assign a starting time $s(i)$ to each job such that each job is delayed as little as possible
• A job $i$ is delayed if $f(i) > d(i)$; the lateness of job is $\max(0, f(i) - d(i))$
• the lateness of a schedule is $\max \limits_{1 \leq i \leq n}(\max(0, f(i) - d(i)))$
• the largest of each job's lateness values
### Minimizing Lateness
• Instance: Set $\{ (t(i), d(i)), 1 \leq i \leq n \}$ of lengths of deadlines of $n$ jobs
• Solution: Set $\{ s(i), 1 \leq i \leq n \}$ such that $\max \limits_{1 \leq i \leq n}(\max(0, f(i) - d(i)))$ is as small as possible
## Template for Greedy Algorithm
• Key question: In what order should we schedule the jobs:
• Shortest length: increasing order of length $t(i)$. Ignores deadlines completely, shortest job may have a very late deadline:
$i$ 1 2
$t(i)$ 1 10
$d(i)$ 100 10
• Shortest slack time: Increasing order of $d(i) - t(i)$. Bad for long jobs with late deadlines. Job with smallest slack may take a long time:
$i$ 1 2
$t(i)$ 1 10
$d(i)$ 2 10
• Earliest Deadline: Increasing order deadline $d(i)$. Correct? Does it make sense to tackle jobs with earliest deadlines first?
## Proof of Earliest Deadline Optimality
\boxed{ \begin{aligned} &\text{Order the jobs in order of increase deadlines } d(i) \\ &\text{Assume for simplicity of notation that } d(1) \leq \mathellipsis d(n) \\ &\text{Initially, } f = 0 \\ &\text{Consider the jobs } i=1, \mathellipsis, n { in this order }\\ &\quad\text{Assign the job } i \text{ to the time interval from } s(i) = f \text{ to } f(i) = f + t_i\\ &\quad f \leftarrow f + t_i \\ &\text{Return the set of scheduled intervals} [s(i), f(i)], \text{ for } i = 1, \mathellipsis, n\\ \end{aligned} }
### Inversions
• A schedule has an inversion if a job $i$ with a deadline $d(i)$ is scheduled before another job $j$ with an earlier deadline $d(j)$ i.e. $d(j) < d(i), s(i) < s(j)$
• if two jobs have the same deadline, they cannot cause an inversion
• in $n$ jobs, the maximum amount of inversion is $n \choose 2$
• Claim: if a schedule has and inversion, then there must be a pair of jobs $i, j$ such that $j$ is scheduled immediately after $i$ and $d(j) < d(i)$
#### Proof of Local Inversion
• If we have an inversion between $l, m$ s.t. $s(l) < s(m), d(l) > d(m)$, then we can find some inverted $i,j$ scheduled between $l,m$
• This is because: in a list where each element is greater than the last (in terms of deadline), then the list is sorted.
• The contrapositive of this is: if a list is unsorted, then there are two adjacent elements that are unsorted.
## Properties of Schedules
• Claim 1: The algorithm produces a schedule with no inversion and no idle time (i.e. jobs are tightly packed)
• Claim 2: All schedules (produced from the same input) with no inversions have the same lateness
• Case 1: All jobs have distinct deadlines. There is a unique schedule with no inversions and no idle time.
• Case 2: Some jobs have the same deadline. Ordering of the jobs does not change the maximum lateness of these jobs.
• Claim 3: There is an optimal schedule with no idle time.
• Claim 4: There is an optimal schedule with no inversions and no idle time.
• Start with an optimal schedule $O$ (that may have inversions) and use an exchange argument to convert $O$ into a schedule that satisfies Claim 4 and has lateness not larger than $O$.
• If $O$ has an inversion, let $i, j$ be consecutive inverted jobs in $O$. After swapping $i, j$ we get a schedule $O'$ with one less inversion.
• Claim: The lateness of $O'$ is no larger than the lateness of $O$
• It is sufficient to prove the last item, since after $n \choose 2$ swaps, we obtain a schedule with no inversions whose lateness is no larger than that of $O$
• In $O$, assume each job $r$ is scheduled for the interval $[s(r), f(r)]$ and has lateness $\mathcal l(r)$. For $O'$, let the lateness of $r$ be $l'(r)$
• Claim: $l'(k) = l(k), \quad \forall k \neq i,j$
• Claim: $l'(j) = l(j)$,
• Claim: $l'(j) \leq l(j)$ because $l'(j) = f(j) - d_i \leq f(j) - d_j = l(j)$
• N.B. common mistakes with exchange arguments:
• Wrong: start with algorithm's schedule A and argue it cannot be improved by swapping two jobs
• Correct: Start with an arbitrary schedule O (which can be optimal) and argue that O can be converted into a schedule that is essentially the same as A without increasing lateness
• Wrong: Swap two jobs that are not neighboring in $O$. Pitfall is that the completion time of all intervening jobs then changes
• Correct: Show that an inversion exists between two neighboring jobs and swap them
• Claim 5: The greedy algorithm produces an optimal schedule, follows from 1, 2, 4
## Summary
• Greedy algorithms make local decisions
• Three strategies:
• Greedy algorithm stays ahead - Show that after each step in the greedy algorithm, its solution is at least as good as that produced by any other algorithm
• Structural bound - First, discover a property that must be satisfied by every possible solution. Then show that the (greedy) algorithm produces a solution with this property
• Exchange argument - Transform the optimal solution in steps into the solution by the greedy algorithm without worsening the quality of the optimal solution
# 06 - Greedy Graph Algorithms
## The Shortest Path Problem
• $G(V,E)$ is a connected, directed graph. Each edge has a length $l(e) \geq 0$
• length of a path $P$ us the sum of the lengths of the edges in $P$
• Goal: compute the shortest path from a specified start node $s$ to every other node in the graph
• Instace: A directed graph $G(V,E)$, a function $I: E \rightarrow \reals^+$ and a node $s \in V$
• Solution: A set $\{P_u, u \in V\}$ of paths, where $P_u$ is the shortest path in $G$ from $s$ to $u$
## Generalizing BFS
• If all edges have the same wight, or distance, BFS would work, processing nodes in order of distance.
• What if the graph has integer edge weights, can we make the graph unweighted?
• yes, placing dummy edges and nodes to pad out lengths > 1 at the expense of memory and running time
• Edge weight of $w$ gets $w - 1$ nodes
• Size of the graph (and therefore runtime) becomes $m + n + \sum_{e \in E} l(e)$. Pseudo-polynomial time depending on input values
## Dijkstra's Algorithm • Famous for pointing out the harm of the goto command, in doing so developed the shortest path algorithm
• Like BFS: explore nodes in non-increasing order of distance $s$. Once a node is explored, its distance is fixed
• Unlike BFS: Layers are not uniform, edges are explored by evaluating candidates by edge wight
• For each unexplored node, determine "best" preceding explored node. Record shortest path length only through explored nodes
• Like BFS: Record previous node in path, and build a tree.
### Formally:
• Maintain a set $S$ of explored nodes
• For each node $u \in S$, compute $d(u)$, which (we will prove, invariant) is the length of the shortest path from $s$ to $u$
• For each node $x \notin S$, maintain a value $d'(x)$, which is the length of the shortest path from $s$ to $x$ using only the nodes in $S$ and $x$ itself
• Greedily add a node $v$ to $S$ that has the smallest value of $d'(v)$
\boxed{ \begin{aligned} &S = \{s\} \text{ and } d(s) = 0 \\ &\text{while } S \neq V \\ &\quad\text{for every node } x \in V - S \\ &\qquad\text{Set } d'(x) = \min_{(u, x): u \in S} (d(u) + l(u, x)) \\ &\quad\text{Set } v = \arg \min_{x \in V - S }(d'(x)) \\ &\quad\text{Add } v \text{ to } S \text{ and set } d(v) = d'(v)\\ \end{aligned} }
What does $v = \arg \min_{x \in V - S }(d'(x))$ mean?
• Run over all unexplored nodes $x \in V - S$
• examine all $d'$ values for each node
• Return the argument (i.e. the node) that has the smallest value of $d'(x)$
To compute the shorts paths: when adding a node $v$ to $S$, store the predecessor $u$ that minimizes $d'(v)$
### Proof of Correctness
• Let $P_u$ be the path computed by the algorithm for an arbitrary node $u$
• Claim: $P_u$ is the shortest path from $s$ to $u$
• Prove by induction on the size of $S$
• Base case: $|S| = 1$. The only node in $S$ is $s$
• Inductive Hypothesis: $|S| = k$ for some $k \geq 1$. The algorithm has computed $P_u$ for every node $u \in S$. Strong induction.
• Inductive step: $|S| = k+1$ because we add the node $v$ to $S$. Could the be a shorter path $P$ from $s$ to $v$? We must prove this is not the case.
• poll: $P'$ must contain an edge from x to y where x is explore (in S) and y is unexplored (in V - S)
• poll: The node v in P' must be explored (in S)
• Locate key nodes on $P'$
• Break $P'$ into sub-paths from $s$ to $x$, $x$ to $y$, $y$ to $v$
• use $l$ to denote the lengths of the sub-paths in $P'$
• $d(x) \leq l(s, x)$
• $d(u) + l(u,v) \leq d(x) + l(x, y)$
• $0 \leq l(y,v)$
• $d(u) + l(u,v) = d(v) \leq l(P') = l(s,x) + l(x,y) + l(y,v)$
• As described, it cannot handle negative edge lengths
• Union of shortest paths output by Dijkstra's forms a tree: why?
• Union of the shortest paths from a fixed source $s$ forms a tree; paths not necessarily computed by computed by Dijkstra's
#### Running time of Dijkstra's
\boxed{ \begin{aligned} &S = \{s\} \text{ and } d(s) = 0 \\ &\text{while } S \neq V \\ &\quad\text{for every node } x \in V - S \\ &\qquad\text{Set } d'(x) = \min_{(u, x): u \in S} (d(u) + l(u, x)) \\ &\quad\text{Set } v = \arg \min_{x \in V - S }(d'(x)) \\ &\quad\text{Add } v \text{ to } S \text{ and set } d(v) = d'(v)\\ \end{aligned} }
• $V$ has $n$ nodes and $m$ edges, so their are $n-1$ iterations on the while loops
• In each iteration, for each node $x \in V - S$, compute $d'(x) = \min_{(u, x): u \in S} (d(u) + l(u, x))$ which is proportional the the number of edges incident on $x$
• Running time per iteration is $O(m)$, since the algorithm processes each edge $(u,x)$ in the graph exactly once (when computing $d'(x)$)
• Therefore, the overall running time is $O(mn)$
### Optimizing our Algorithm
• Observation, if we add $v$ to $S$, $d'(x)$ changes only if $(v, x)$ is an edge in $G$
• Idea: For each node $x \in V - S$, store the current value of $d'(x)$. Upon adding a node $v$ to $S$, update $d'()$ only for neighbors of $v$
• How do we efficiently compute $v = \arg \min_{x \in V - S }(d'(x))$
• Priority Queue!
\boxed{ \begin{aligned} &Insert(Q, s, 0) \\ &S = \{s\} \text{ and } d(s) = 0 \\ &\text{while } S \neq V \\ &\quad(v, d'(v)) = ExtractMin(Q) \\ &\quad\text{Add v to S and set d(v) = d'(v)}\\ &\quad\text{for every node } x \in V - S \text{ } s.t. (v,x)\text{ is an edge in } G \\ &\qquad\text{If } d(v) + l(v,x) < d'(x) \\ &\qquad\quad d'(x) d(v) + l(v,x) \\ &\qquad\quad ChangeKey(Q,x,d'(x)) \\ \end{aligned} }
• For each node $x \in V - S$, store the pair $(x, d'(x))$ in a priority queue $Q$ with $d'(x)$ as the key
• Determine the next node $v$ to add to $S$ using $ExtractMin$
• After adding $v$ to $S$, for each node $x \in V - S$ such that there is an edge from $v$ to $x$, check if $d'(x)$ should be updated i.e. if there is a shortest path from $s$ to $x$ via $v$
• in line 8, if $x$ is not in $Q$, simply insert it
### New Runtime
• $ExtractMin$ happens $n - 1$ times
• For every node $v$, the running time of step 5 $O(deg_v)$, the number of outgoing neighbors of $v$: $\sum_{v \in V} O(deg_v) = O(m)$
• $ChangeKey$ is invoked at most once for each edge, $m$ times, and is an $O(\log n)$ operation
• So, the total runtime is $O(m \log n)$
## Network Design
• want to connect a set of nodes using a set of edges with certain properties
• Input is usually a graph, and the desired network (output) should use a subset of edges in the graph
• Example: connect all nodes using a cycle of shortest total length. This problem is the NP-complete traveling salesman problem
## Minimum Spanning Tree (MST)
• Given an undirected graph $G(V,E)$ with a cost $c(e) > 0$ associated with each edge $e \in E$.
• Find a subset $T$ of edges such that the graph $(V,T)$ is connected and the cost $\sum_{e \in T} c(e)$ is as small as possible
• Instance: An undirected graph $G(V,E)$ and a function $c: E \rightarrow \R^+$
• Solution: A set $T \sube E$ of edges such that $(V, T)$ is connected and the cost $\sum_{e \in T} c(e)$ is as small as possible
• Claim: if $T$ is a minimum-cost solution to this problem, then $(V,T)$ is a tree
• A subset $T of E$ is a spanning tree of $G$ if $(V, T)$ is a tree
### Characterizing MSTs
• Does the edge of smallest cost belong to an MST? Yes
• Wrong proof: Because Kruskal's algorithm adds it.
• Right proof: tbd
• Which edges must belong to an MST?
• What happens when we delete an edge from an MST?
• MST breaks up into sub-trees
• Which edge should we add to join them?
• Which edges cannot belong to an MST?
• What happens when we add an edge to an MST?
• We obtain a cycle
• Which edge in the cycle can we be sure does not belong to an MST
### Greedy Algorithm for the MST Problem
• Template: process edges in some order. Add an edge to $T$ if tree property is not violated.
• increasing cost order: Process edges in increasing order of cost. Discard an edge if it creates a cycle – Kruskal's
• Dijkstra-like: Start from a node $s$ and grow $T$ outward from $s$: add the node that can attached most cheaply to current tree – Prim's
• Decreasing cost order: Delete edges in order of decreasing cost as long as graph remains connected – Reverse-Delete
• each of these works
• Simplifying assumption: all edge costs are distinct
## Graph Cuts
• a cut in a graph $G(V,E)$ is a set of edges whose removal disconnects the graph (into two or more connected components)
• Every set $S \sub V$ ($S$ cannot be empty or the entire set $V$) has a corresponding cut: cut($S$) is the set of edges (v,w) such that $v \in S, w \in V - S$
• cut($S$) is a "cut" because deleting the edges in cut($S$) disconnects $S$ from $V - S$
• Claim: for every $S \sub V, S \neq \empty$, every MST contains the cheapest edge in cut$(S$)
• will have to proof by contradiction using exchange argument
### Proof of Cut Property of MSTs
• Negation of the desired property: There is a set $S \sub V$ and an MST $T$ such that $T$ does not contain the cheapest edge in cut($S$)
• Proof strategy: If $T$ does not contain $e$, show that there is a tree with a smaller cost than $T$ that contains $e$.
• Wrong proof:
• Since $T$ is spanning, it must contain some edge e.g. $f$ in cut(\$S)
• $T - \{f\} \cup \{e\}$ has smaller cost than $T$ but may not be a spanning tree
• Correct proof:
• Add $e$ to $T$ forming a cycle
• This cycle must contain an edge $e'$ in cut($S$)
• $T - \{e'\} \cup \{e\}$ has smaller cost than $T$ and is a spanning tree
## Prim's Algorithm
• Maintain a tree (S, T), i.e., a set of nodes and a set of edges, which we will show will always be a tree
• Start with an arbitrary node $s \in S$
• Step 3 is the cheapest edge in cut(S) for the currently explored set S
• In other words, each step in Prim's algorithm computes and adds the cheapest edge in the current value of cut($S$)
• $\arg \min_{(u,v):u \in S, v \in V - S} c(u,v) \equiv \arg \min_{(u,v) \in cut(S)} c(u,v)$
\boxed{ \begin{aligned} &S = \{s\} \text{ and } T = \empty \\ &\text{while } S \neq V \\ &\quad\text{Compute } (u, v) = \arg \min_{(u,v):u \in S, v \in V - S} c(u,v) \\ &\quad\text{Add the node } v \text{ to } S \text{ and add the edge } (u,v) \text{ to } T \end{aligned} }
### Optimality of Prim's
• Claim: Prim's algorithm outputs an MST
• Prove that every edge inserted satisfies the cut property (true by construction, in each iteration $(u, v)$ is necessarily the cheapest edge in cut($S$) for the current value of $S$)
• Prove that the graph constructed is a spanning tree
• Why are there no cycles in $(V, T)$
• Why is $(V,T)$ a spanning tree (edges in $T$ connect all nodes in $V$) - Because $G$ is connected, if there were an unconnected node $v$ at the end, then Prim's would not have terminated
### Final version of Prim's Algorithm
\boxed{ \begin{aligned} &Insert(Q, s, 0, \empty) \\ &\text{while } S \neq V \\ &\quad(v, a(v), u) = ExtractMin(Q) \\ &\quad\text{Add node } v \text{ to } S \text{ and edge } (u,v) \text{ to } T \\ &\quad\text{for every node } x \in V - S \text{ s.t. } (v, x) \text{ is ans edge in } G \\ &\qquad\text{ if } c(v,x) < a(x) \text{then }\\ &\qquad\quad a(x) = c(v,x) \\ &\qquad\quad ChangeKey(Q,x,a(x),v) \\ \end{aligned} }
• $Q$ is a priority queue
• Each element in $Q$ is a triple, the node, its attachment cost, and its predecessor in the MST
• In step 8, if $x$ is not already in $Q$, simply insert (x, a(x), v) into $Q$
• Total of $n - 1$ $ExtractMin$ and $m$ $ChangeKey$ operations, yielding a running time of $O(m \log n)$
• running time of step 5 is proportional to the degree of $x$ which is proportional to the number of edges in the graph $m$
## Kruskal's Algorithm
• Start with an empty set $T$ of edges
• Process edges in $E$ in increasing order of cost
• Add the next edge $e$ to $T$ only if adding $e$ does not create a cycle. Discard $e$ otherwise
• Note: at any iteration, $T$ may contain several connected components and each node in $V$ is in some component
• Claim: Kruskal's algorithm outputs an MST
• For every edge $e$ added, demonstrate the existence of a set $S \sub V$ (and $V - S$) such that $e$ and $S$ satisfy the cut property, i.e.e, the cheapest edge in $cut(S)$
• If $e = (u,v)$, let $S$ be the set of nodes connected to $u$ in the current graph $T$
• Why is $e$ the cheapest edge in cut($S$) - because we process them in increasing order of cost
• Prove that the algorithm computes a spanning tree
• $(V,T)$ contains no cycles by construction
• If $(V,T)$ is not connected, then there exists a subset $S$ of nodes not connected to $V-S$. What is the contradiction?
### Implementing Kruskal's Algorithm
• start with an empty set $T$ of edges
• Process edges in $E$ in increasing order of cost
• Add the next edge $e$ to $T$ only if adding $e$ does not create a cycle
• Sorting edges takes $O(m \log n)$ time
• Key question: "Does adding $e = (u,v)$ to $T$ create a cycle?
• Maintain set of connected components of $T$
• $Find(u)$: return the name of the connected component of $T$ that $u$ belongs to
• $Union(A,B)$: merge connected components A, B
### Analysing Kruskal's Algorithm
• How many $Find$ invocations does Kruskal's need? $2m$
• How many $Union$ invocations? $n-1$
• Two implementations of $Union-Find$:
• Each $Find$ take $O(1)$, $k$ invocations of $Union$ takes $O(k \log k)$ time in total – $O(m + n \log n)$
• Each $Find$ take $O(\log n )$, each invocation of $Union$ takes $O(1)$$O(m \log n + n)$
• In general $m < n$, but in either case, the total running time is $O(m \log n)$ since we have to spend $O(m \log n)$ time sorting the edges by increasing cost, regardless of the underlying implementation of the $Union-Find$ data structure
### Comments on Union-Find and MST
• useful to maintain connected components of a graph as edges are added to the graph
• Data structure does not support edge deletion efficiently
• Current best algorithm for MST runs in $O(m\alpha (m,n))$ time (Chazelle 2000) where $\alpha(m,n)$ is a function of $m,n$ and $O(m)$ randomized time
• Let $\log^* n =$ the number of times you take $\log n$ before you reach 1
• e.g. $\log^*(2^{10}) = 4$, $\log^*(2^{2^{10}}) = 5$
• Holy grail: $O(m)$ deterministic algorithm for MST
## Cycle Property
• When can we be sure that an edge cannot be in any MST
• Let $C$ be any cycle in $G$ and let $e = (v,w)$ be the most expensive edge in $C$
• Claim: $e$ does not belong to any MST of G
• Proof: exchange argument.
• If a supposed MST $T$ contains $e$, show that there is a tree with smaller cost than $T$ that does not contain $e$
### Reverse-Delete Algorithm
¯\(ツ)
• Any algorithm that constructs a spanning tree by including the Cut and Cycle properties (include edges that satisfy the cut P, delete edges that satisfy the cycle property) will be an MST
# 07 - Applications of Minimum Spanning Trees
## Minimum Bottleneck Spanning Tree (MBST)
• MST minimizes the total cost of the spanning network
• Consider another network design criterion
• build a network connecting all cities in mountainous region, but ensure the highest elevation is as low as possible
• total road length is not a criterion
• Idea: compute an MST in which the edge with the highest cost is as cheap as possible
• In an undirected graph $G(V,E)$, let $(V,T)$ be a spanning tree. The bottleneck edge in $T$ is the edge with the largest cost in $T$
• Instance: An an undirected graph $G(V,E)$ such that $V,T$ is a spanning tree
• Solution: A Set $T \sube E$ of edges such that $(V,T)$ is a spanning tree and there is no spanning tree in $G$ with a cheaper bottleneck
Is every MST and MBST, or vice versa?
• If a tree is an MBST, it might not be an MST
• If a tree is an MST, it will be an MBST
Proof:
• Let $T$ be the MST, and let $T'$ be a spanning tree with a cheaper bottleneck edge.
• Let $e$ be the bottleneck edge in $T$
• Every edge in $T'$ has lower cost than $e$
• Adding $e$ to $T'$ creates a cycle consisting only of edges, where $e$ is the costliest edge in the cycle
• Since $e$ is the costliest edge in this cycle, by the cycle property, $e$ cannot belong to any MST, which contradicts the fact that $T$ is an MST
## Motivation for Clustering
• Given a set of objects and distances between them
• objects can be anything
• Distance function: increase distance corresponds to dissimilarity
• Goal: group objects into clusters, where each cluster is a group of similar objects
### Formalizing the Clustering Problem
• Let $U$ be the set of $n$ objects labels $p_1, p_2, \mathellipsis, p_n$
• For every pair $p_i, p_j$, we have a distance $d(p_i, p_j)$
• We require that $d(p_i, p_i) = 0$, $d(p_i, p_j) > 0$, $d(p_i, p_j) = d(p_j, p_i)$,
• Given a positive integer $k$, a k-clustering of $U$ is a partition of $U$ into $k$ non-empty subsets or clusters $C_1, C_2, ..., C_k$
• that spacing of a clustering is the smallest distance between objects in 2 different subsets:
• $spacing(C_1, C_2, ..., C_k) = \min_{1 \leq i,j \leq k, i\neq j, p \in Ci, q \in C_j} d(p,q)$
• spacing is the minimum of distance between objects in two different subsets
minimum = inifnity
for every cluster C_i
for every cluster C_j ≠ i
for every point p in C_i
for every point q in C_j
if d(p,q) < minimum
minimum = d(p,q)
return minimum
• Clustering of Maximum Spacing
• Instance: a Set $U$ of objects, a distance function $d : U \times U \rightarrow \Reals^+$
• Solution: A k-clustering of $U$ whose spacing is the largest over all possible k-clusterings
• $O(n^2)$ on $n$ clusters and then $O(i \times j)$ on points in disparate cluster
## Algorithm for Max Spacing
• intuition: greedily cluster objects in increasing order of distance
• Let $\mathcal C$ be the set of $n$ clusters, with each object in $U$ in its own cluster
• Process pairs of objects in increasing order of distance
• Let $(p,q)$ be the next pair with $p \in C_p$ and $q \in C_p$
• If $C_p \neq C_q$ add a new cluster $C_p \cup C_q$ to $\mathcal{C}$, delete $C_p, C_q$ from $\mathcal{C}$
• Stop when there are $k$ cluster in $\mathcal{C}$
• Same as Kruskal's algorithm, but do not add the last $k-1$ edges in MST
• Given a clustering $\mathcal{C}$, what is spacing($\mathcal{C}$)?
• The length of the next edge that would be added - the cost of the $(k-1)$st most expensive edge in the MST. Let this cost be $d^*$
### Proof of optimal computing
• Let $\mathcal{C'}$ be any other cluster (with $k$ clusters).
• We will prove that spacing($\mathcal{C'}$) $\leq d^*$
• There must be two objects $p_i, p_j \in U$ in the same cluster $C_r \in \mathcal{C}$ but in different clusters in $\mathcal{C'}$: spacing($\mathcal{C'}$) $\leq d(p_i, q_i)$
• Suppose $p_i \in C'_s$ and $p_j \in C'_t$ in $\mathcal{C}$
• All edges in the path $Q$ connecting $p_i \rightarrow p_j$ in the MST have lengths $\leq d^*$
• In particular, there is an object $p \in C'_s$ and an object $p' \notin C'_s$ s.t. $p$ and $p'$ are adjacent in $Q$
• $d(p, p') \leq d^* \implies$ | HuggingFaceTB/finemath | |
Cantor Sets
• Apr 15th 2011, 06:14 AM
mathematic
Cantor Sets
Consider g(x)=x^2sin(1/x) if x>0 and 0 if x<=0
1. a) Find g'(0)
b) Compute g'(x) for x not 0
c)Explain why, for every delta>0, g'(x) attains every value between 1 and -1 as ranges over the set (-delta,delta). Conclude that g'(x) is not continuous at x=0.
Next, we want to explore g with the Cantor set.
We have f_1(x)=0 if x is in [0,1/3]
=g(x-1/3) if x is to right of 1/3
=g(-x+2/3) if x is to left of 2/3
=0 if x is in [2/3,1]
Now f_2(x)=1/3f_1(3x) for x in [0,1/3]
=f_1(x) if x is in [1/3,2/3]
=1/3f_1(3x-2) if x is in [2/3,1]
2. a) if c is in C(Cantor set) what is lim f_n(c)?
b) Why does lim f_n(x) exist for x not in C?
Now set f(x)=lim f(x)
3.a) Explain why f'(x) exists for all c not in C
b) If c in C, argue that |f(x)|<=(x-c)^2 for all x in [0,1]. Show how this implies f'(c)=0
c) Give a careful argument for why f'(x) fails to be continuous on C.
4. Why is f'(x) Riemann integrable on [0,1]
The reason the Cantor set has measure zero is that, at each stage, 2^(n-1) open intervals of length 1/3^n are removed from C_(n-1). The resulting sum:
sum(2^(n-1)(1/3^n) converges to 1, which means that C1, C2, C3,.... have total lengths tending to zero. Now let's remove intervals of lengths 1/3^(n+1)
5.Show that under these circumstances, the sum of the lengths of the intervals making up each C_n, no longer tends to zero as n tends to infinity. What is this limit?
I know this is way too many questions for one thread, but these are all directly related to each other.
I had already figured out 1, but left it there since it is used in the next problem.
2a) I think the answer is 0.
b) not quite sure
Any hints would help! | HuggingFaceTB/finemath | |
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You can also get a better visual and understanding of the function by using our graphing tool. rules for subtracting positive and negative numbers, formula for getting the percentage of a number, how do you write a multiplication and division expression, rewriting integers by combining like terms, math worksheet on finding greatest common factors, Evaluating Expressions and Solving Equations, Writing Linear Equations Using Slope and Point, Factoring Trinomials with Leading Coefficient 1, Simplifying Expressions with Negative Exponents, Power of a Quotient Property of Exponents, Solving Linear Systems of Equations by Elimination, Solving Quadratic Inequalities with a Sign Graph, Writing a Rational Expression in Lowest Terms, Simple Trinomials as Products of Binomials, Solving Equations That Contain Rational Expressions, Dividing and Subtracting Rational Expressions, Solving Nonlinear Equations by Substitution, Solving a Quadratic Inequality with Two Solutions, Simplifying Expressions Involving Variables, Factoring Completely General Quadratic Trinomials, Adding and Subtracting Rational Expressions With Unlike Denominators. 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Website what are the zeros of a function calculator you agree to our Cookie Policy fourth derivatives, as as... And you wo n't believe how simple it is written as ( x+2 ) * ( x-3 *. This 'Bessel function ( zeros ) calculator ', please fill in questionnaire ) which! Work and detailed explanation using any of the function several math classes - algebra 1 Pre! Which is short for zeros of # f ( x ) = 0 a possible... Equation online was so occupied that he just did not have the time for me video demonstrates how:! More steps... Divide each term in by a lot of issues with binomial formula, linear inequalities and inequalities... Lot, and you wo n't believe how simple it is to!. Find the zeros or roots of the function to improve this 'Bessel function ( )... He introduced me to these superb Software programs in algebra function y f. Ax 3 + bx 2 + cx + d = 0 used it through several math classes algebra. 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And/Or complex roots 2 # intersect the x-axis intersect are called zeros TI-84 Series graphing.... Of the Integer ( Integral ) Root Theorem ( when leading coefficient is 1 or â ). My studies equation f ( x ) = x 2 â 3 x + 216?. X -axis and appears almost linear at the intercept, it will decide possible! Too for an effective and inexpensive tool which helps me with my studies + 1 # number r. Cubic equation are termed as the roots or zeroes of a quadratic equation Root calculator lets you the...
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# Perimeter and area of a polygon
What is the perimeter of a polygon? And what is the area?
Learn definitions and how to measure the perimeters and the areas.
## Notes
What is the perimeter of a polygon? And what is the area of a polygon?
Learn definitions and how to measure the perimeters and the areas of all the polygons that you know!
## The area of polygons
What is the area of a polygon? The area expresses the extent of a polygon.
An area is a two-dimensional quantity: to calculate the area, use a shape as "unit of measurement" and look how many of it fill the shape you are analysing. Depending on the unit of measurement that you choose, you will have different results.
## The perimeter of polygons
What is the perimeter of a shape or a polygon? The perimeter is the length of the outline of a polygon.
The perimeter of a polygon is the sum of all the sides of a polygon.
## Let's measure the perimeter
The perimeter is a length, so you always have to remember to use a unit of measurement.
The length (or distance) measures the quantity of a dimension. | HuggingFaceTB/finemath | |
# 103415 (number)
103,415 (one hundred three thousand four hundred fifteen) is an odd six-digits composite number following 103414 and preceding 103416. In scientific notation, it is written as 1.03415 × 105. The sum of its digits is 14. It has a total of 4 prime factors and 16 positive divisors. There are 72,576 positive integers (up to 103415) that are relatively prime to 103415.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 14
• Digital Root 5
## Name
Short name 103 thousand 415 one hundred three thousand four hundred fifteen
## Notation
Scientific notation 1.03415 × 105 103.415 × 103
## Prime Factorization of 103415
Prime Factorization 5 × 13 × 37 × 43
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 103415 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 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 103,415 is 5 × 13 × 37 × 43. Since it has a total of 4 prime factors, 103,415 is a composite number.
## Divisors of 103415
16 divisors
Even divisors 0 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 140448 Sum of all the positive divisors of n s(n) 37033 Sum of the proper positive divisors of n A(n) 8778 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 321.582 Returns the nth root of the product of n divisors H(n) 11.7812 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 103,415 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 103,415) is 140,448, the average is 8,778.
## Other Arithmetic Functions (n = 103415)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 72576 Total number of positive integers not greater than n that are coprime to n λ(n) 252 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9859 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 72,576 positive integers (less than 103,415) that are coprime with 103,415. And there are approximately 9,859 prime numbers less than or equal to 103,415.
## Divisibility of 103415
m n mod m 2 3 4 5 6 7 8 9 1 2 3 0 5 4 7 5
The number 103,415 is divisible by 5.
## Classification of 103415
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (103415)
Base System Value
2 Binary 11001001111110111
3 Ternary 12020212012
4 Quaternary 121033313
5 Quinary 11302130
6 Senary 2114435
8 Octal 311767
10 Decimal 103415
12 Duodecimal 4ba1b
20 Vigesimal ciaf
36 Base36 27sn
## Basic calculations (n = 103415)
### Multiplication
n×y
n×2 206830 310245 413660 517075
### Division
n÷y
n÷2 51707.5 34471.7 25853.8 20683
### Exponentiation
ny
n2 10694662225 1105988493998375 114375800106841950625 11828173368049060323884375
### Nth Root
y√n
2√n 321.582 46.9384 17.9327 10.0674
## 103415 as geometric shapes
### Circle
Diameter 206830 649776 3.35983e+10
### Sphere
Volume 4.63275e+15 1.34393e+11 649776
### Square
Length = n
Perimeter 413660 1.06947e+10 146251
### Cube
Length = n
Surface area 6.4168e+10 1.10599e+15 179120
### Equilateral Triangle
Length = n
Perimeter 310245 4.63092e+09 89560
### Triangular Pyramid
Length = n
Surface area 1.85237e+10 1.30342e+14 84438 | HuggingFaceTB/finemath | |
# 8114 (number)
8,114 (eight thousand one hundred fourteen) is an even four-digits composite number following 8113 and preceding 8115. In scientific notation, it is written as 8.114 × 103. The sum of its digits is 14. It has a total of 2 prime factors and 4 positive divisors. There are 4,056 positive integers (up to 8114) that are relatively prime to 8114.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 14
• Digital Root 5
## Name
Short name 8 thousand 114 eight thousand one hundred fourteen
## Notation
Scientific notation 8.114 × 103 8.114 × 103
## Prime Factorization of 8114
Prime Factorization 2 × 4057
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 8114 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 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 8,114 is 2 × 4057. Since it has a total of 2 prime factors, 8,114 is a composite number.
## Divisors of 8114
1, 2, 4057, 8114
4 divisors
Even divisors 2 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 12174 Sum of all the positive divisors of n s(n) 4060 Sum of the proper positive divisors of n A(n) 3043.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 90.0777 Returns the nth root of the product of n divisors H(n) 2.66601 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 8,114 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 8,114) is 12,174, the average is 304,3.5.
## Other Arithmetic Functions (n = 8114)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 4056 Total number of positive integers not greater than n that are coprime to n λ(n) 4056 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1025 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 4,056 positive integers (less than 8,114) that are coprime with 8,114. And there are approximately 1,025 prime numbers less than or equal to 8,114.
## Divisibility of 8114
m n mod m 2 3 4 5 6 7 8 9 0 2 2 4 2 1 2 5
The number 8,114 is divisible by 2.
• Semiprime
• Deficient
• Polite
• Square Free
## Base conversion (8114)
Base System Value
2 Binary 1111110110010
3 Ternary 102010112
4 Quaternary 1332302
5 Quinary 224424
6 Senary 101322
8 Octal 17662
10 Decimal 8114
12 Duodecimal 4842
20 Vigesimal 105e
36 Base36 69e
## Basic calculations (n = 8114)
### Multiplication
n×y
n×2 16228 24342 32456 40570
### Division
n÷y
n÷2 4057 2704.67 2028.5 1622.8
### Exponentiation
ny
n2 65836996 534201385544 4334510042304016 35170214483254785824
### Nth Root
y√n
2√n 90.0777 20.0946 9.49093 6.05128
## 8114 as geometric shapes
### Circle
Diameter 16228 50981.8 2.06833e+08
### Sphere
Volume 2.23766e+12 8.27332e+08 50981.8
### Square
Length = n
Perimeter 32456 6.5837e+07 11474.9
### Cube
Length = n
Surface area 3.95022e+08 5.34201e+11 14053.9
### Equilateral Triangle
Length = n
Perimeter 24342 2.85083e+07 7026.93
### Triangular Pyramid
Length = n
Surface area 1.14033e+08 6.29562e+10 6625.05
## Cryptographic Hash Functions
md5 b20fa060328b0cdf51b464ee37efe182 010c1687e449ea3a835019d1dc6862c42f7eacfa 3cb5588f87636e1e6b572bd378e3be20a2876990152dc64a0768f0514416dc3a ee8b8c2a423b9addc28ac557b2481b9a1a4aa9a3efc7c71d898bed6ceb2f68afca0e7911c8db5d454eae1d3a959b9b2a2d6b67804be6c49e8bf9693e3a75f7b5 07ee2d8d96d826edcb90160ff28c18f5d3487bb0 | HuggingFaceTB/finemath | |
# The existence of a $k$-algebra automorphism
Let $$k$$ be an algebraically closed field and let $$\mathfrak m$$ be a maximal ideal in $$R=k[x_1,\dots,x_r]$$. Show that there is a $$k$$-algebra automorphism of $$R$$ taking $$\mathfrak m$$ to $$(x_1,\dots,x_r)$$.
Since $$k$$ is algebraically closed, every maximal ideal $$\mathfrak m$$ is of the form $$(x_1-a_1,\dots,x_n-a_n)$$. So would the map be defined as follows? Let $$\phi: R\to R$$ be defined on $$x_i$$ as $$x_i\mapsto x_i+a_i$$ and let $$\phi(a_nx^n+\dots a_1x+a_0)=a_n\phi(x_1)^n+\dots+a_1\phi(x)+a_0$$. Then by construction, $$\phi$$ is a $$k$$-algebra automorphism. Does this look OK?
• Elements of $k[x_1,...,x_r]$ are of the form $\sum_\alpha a_\alpha x^\alpha$ for $\alpha=(\alpha_1,...,\alpha_r)$ and $x^\alpha=x_1^{\alpha_1}...x_r^{\alpha_r}$. Then $\phi(\sum_\alpha a_\alpha x^\alpha)=\sum_\alpha a_\alpha (x+a)^\alpha$, where $a=(a_1,...,a_r)$ and $x+a=(x_1+a_1,...,x_r+a_r)$. – user647486 Apr 26 at 15:12
• The meaty part of the proof is that $\mathfrak m$ is of the form $(x_1-a_1,...,x_r-a_r)$. You need to judge if your readers would be convinced that that is true, and therefore if you should leave that part out or not. – user647486 Apr 26 at 15:14
• Oops, I assumed without realizing it that $r=1$. The form of $\mathfrak m$ is a corollary of the Nullstellensatz. – user419669 Apr 26 at 15:17 | open-web-math/open-web-math | |
Numerical Analysis Home
## Systems of Linear Equations: Gaussian Elimination
A system of linear equations with $$n$$ variables $$x_1,\ldots,x_n$$ and $$n$$ equations can be written as follows: $\begin{eqnarray*} {\color{blue}\begin{array}{ccccccccc} a_{11}x_1&+&a_{12}x_2&+&\cdots &+&a_{1n}x_n&=&b_1\\ a_{21}x_1&+&a_{22}x_2&+&\cdots &+&a_{2n}x_n&=&b_2\\ \vdots&&\vdots&& &&\vdots&&\vdots\\ a_{n1}x_1&+&a_{n2}x_2&+&\cdots &+&a_{nn}x_n&=&b_n. \end{array}} \end{eqnarray*}$ The above linear system is equivalent to the matrix equation $$A\overrightarrow{x}=\overrightarrow{b}$$, where $A=\left[\begin{array}{cccc} a_{11}&a_{12}&\cdots &a_{1n}\\ a_{21}&a_{22}&\cdots &a_{2n}\\ \vdots&\vdots& &\vdots\\ a_{nn}&a_{n2}&\cdots &a_{nn} \end{array}\right], \overrightarrow{x}=\left[\begin{array}{c} x_1\\x_2\\ \vdots\\x_n\end{array} \right], \mbox{ and } \overrightarrow{b}=\left[\begin{array}{c} b_1\\b_2\\ \vdots\\b_n \end{array} \right].$ $$A$$ is called the coefficient matrix and $$[A\; \overrightarrow{b}]$$ is called the augmented matrix.
Example. $\begin{array}{rcrcrcr} 2x_1&+&x_2&&&=&7\\ -3x_1&&&+&x_3&=&-8\\ &&x_2&+&2x_3&=&-3 \end{array}.$ The above linear system is equivalent to $$A\overrightarrow{x}=\overrightarrow{b}$$, where $$A=\left[\begin{array}{rrr}2&1&0\\ -3&0&1\\ 0&1&2\end{array} \right]$$, $$\overrightarrow{x}=\left[\begin{array}{r}x_1\\x_2\\x_3 \end{array} \right]$$, $$\overrightarrow{b}=\left[\begin{array}{r}7\\-8\\-3 \end{array} \right]$$, and the augmented matrix is $$[A\; \overrightarrow{b}]= \left[\begin{array}{rrr|r}2&1&0&7\\ -3&0&1&-8\\ 0&1&2&-3\end{array} \right]$$.
Throughout this section assume
(1) $$A\overrightarrow{x}=\overrightarrow{b}$$ has a unique solution and
(2) $$a_{ii}\neq 0$$ for $$i=1,\ldots, n$$.
Techniques for solving $$A\overrightarrow{x}=\overrightarrow{b}$$:
There are multiple ways to solve $$A\overrightarrow{x}=\overrightarrow{b}$$. First we learn Gaussian elimination as illustrated in the following example:
Example. By Gaussian elimination find the unique solution of the following system of equations. $\begin{array}{rcrcrcr} 2x_1 &- &4x_2 &+ &2x_3 &=&2\\ x_1 &+ &x_2 &+ &2x_3 &=&0\\ -3x_1 &+ &8x_2 &- &3x_3 &=&-3 \end{array}$ Solution. \begin{align*} &\begin{array}{rrcrcrcr} E1: & 2x_1 &- &4x_2 &+ &2x_3 &=&2\\ E2: & x_1 &+ &x_2 &+ &2x_3 &=&0\\ E3: & -3x_1 &+ &8x_2 &- &3x_3 &=&-3\end{array}\\ &\\ \xrightarrow{\substack{-\frac{1}{2}E1+E2\\ \frac{3}{2}E1+E3}}\;\;\;\; & \begin{array}{rrcrcrcr} E1: & 2x_1 &- &4x_2 &+ &2x_3 &=&2\\ E2: & & &3x_2 &+ &x_3 &=&-1\\ E3: & & &2x_2 & & &=&0\end{array}\\ &\\ \xrightarrow{-\frac{2}{3}E2+E3}\;\;\;\; & \begin{array}{rrcrcrcr} E1: & 2x_1 &- &4x_2 &+ &2x_3 &=&2\\ E2: & & &3x_2 &+ &x_3 &=&-1\\ E3: & & & &- &\frac{2}{3}x_3 &=&\frac{2}{3}\end{array}\\ \end{align*} Note that if $$a_{ii}=0$$ for some $$i$$ and $$a_{ji}\neq 0$$ for some $$j > i$$, then we interchange $$Ei$$ and $$Ej$$. This is called partial pivoting. If $$a_{ji}=0$$ for all $$j\geq i$$, then there is no unique solution.
Now we do backward substitutions: \begin{align*} E3\implies &x_3=\frac{2/3}{-2/3}=-1\\ E2\implies &x_2=\frac{1}{3}(-1-x_3)=\frac{1}{3}(-1+1)=0\\ E1\implies &x_1=\frac{1}{2}(2+4x_2-2x_3)=\frac{1}{2}(2+4\cdot0-2(-1))=2 \end{align*} Thus the unique solution is $$(x_1,x_2,x_3)=\left(2,0,-1\right)$$.
Gaussian elimination steps:
1. For each $$i=1,\ldots,n-1$$, $$Ej=Ej-\frac{a_{ji}}{a_{ii}}Ei$$, $$j=i+1,\ldots,n$$.
2. $$x_n=\frac{b_n}{a_{nn}}$$ and $$x_i=\frac{1}{a_{ii}}\Bigg(b_i-\displaystyle\sum_{j=i+1}^n a_{ij}x_j\Bigg)$$, $$i=n-1,n-2,\ldots,1$$. (backward substitution)
Operation Counts:
In step 1, for each $$i=1,\ldots,n-1$$, we do $$(n-i)$$ divisions $$\frac{a_{ji}}{a_{ii}}$$, $$j=i+1,\ldots,n$$. Then for each $$j=i+1,\ldots,n$$, to get $$\frac{a_{ji}}{a_{ii}}Ei$$, we do $$n-i+1$$ multiplication when we multiply $$n-i+1$$ numbers $$a_{i,i+1},\ldots,a_{i,n},b_i$$ in $$Ei$$ by $$\frac{a_{ji}}{a_{ii}}$$. So for each $$i=1,\ldots,n-1$$, the total number of multiplications/divisions is $(n-i)+(n-i)(n-i+1)=(n-i)(n-i+2).$ For a fixed $$i=1,\ldots,n-1$$, we do $$(n-i+1)$$ subtraction for $$Ej=Ej-\frac{a_{ji}}{a_{ii}}Ei$$ for each $$j=i+1,\ldots,n$$, i.e., total $$(n-i)(n-i+1)$$ subtractions.
So the total number of multiplications/divisions is $\sum_{i=1}^{n-1}(n-i)(n-i+2)=\frac{2n^3+3n^2-5n}{6}.$ The total number of subtractions is $\sum_{i=1}^{n-1}(n-i)(n-i+1)=\frac{n^3-n}{3}.$ Note that in a computer the time for a multiplication/division is greater than that of an addition/subtraction. So the total number of operations in step 1 is $$\displaystyle \frac{2n^3+3n^2-5n}{6}+\frac{n^3-n}{3}=\frac{4n^3+3n^2-7n}{6}$$ which is $$O(n^3)$$. Similarly we can show the total number of operations in step 2 is $$\displaystyle \frac{n^2+n}{2}+\frac{n^2-n}{2}=n^2$$. Thus total number of operations in Gaussian elimination is $$\displaystyle \frac{4n^3+3n^2-7n}{6}+n^2=\frac{4n^3+9n^2-7n}{6}$$ which is $$O(n^3)$$.
$$LU$$-factorization:
If $$M=[A\; \overrightarrow{b}]$$, the first step transforms it into $$M'=[U\; \overrightarrow{b'}]$$ where $$U$$ is an upper triangular matrix with nonzero diagonals. This is obtained by premultiplying $$M$$ by elementary matrices. For example, premultiplying $$M$$ by the elementary matrix $$E_1=\left[\begin{array}{rrr}1&0&0\\ -\frac{1}{2}&1&0\\ 0&0&1\end{array} \right]$$ gives $E_1M=\left[\begin{array}{rrr}1&0&0\\ -\frac{1}{2}&1&0\\ 0&0&1\end{array} \right] \left[\begin{array}{rrr|r}2&-4&2&2\\ 1&1&2&0\\ -3&8&-3&-3\end{array} \right]= \left[\begin{array}{rrr|r}2&-4&2&2\\ 0&3&1&-1\\ -3&8&-3&-3\end{array} \right].$ Similarly using the elementary matrices $$E_2=\left[\begin{array}{rrr}1&0&0\\ 0&1&0\\ \frac{3}{2}&0&1\end{array} \right]$$ and $$E_3=\left[\begin{array}{rrr}1&0&0\\ 0&1&0\\ 0&-\frac{2}{3}&1\end{array} \right]$$, we get $E_3E_2E_1M=E_3E_2E_1[A|\overrightarrow{b}]= \left[\begin{array}{rrr|r}2 & -4 & 2 & 2\\ 0 & 3 & 1 & -1\\ 0 & 0& -2/3 & 2/3\end{array} \right]=[U|\overrightarrow{b'}],$ which implies $$E_3E_2E_1A=U$$ an upper-triangular matrix with nonzero diagonals. Now $E_3E_2E_1A=U\implies A=(E_3E_2E_1)^{-1}U=LU= \left[\begin{array}{rrr} 1 & 0 & 0\\ 1/2 & 1 & 0\\ -3/2 & 2/3 & 1\end{array} \right] \left[\begin{array}{rrr} 2 & -4 & 2\\ 0 & 3 & 1\\ 0 & 0& -2/3 \end{array} \right],$ where $$L=[l_{ij}]$$ is a lower-triangular matrix with $$l_{ji}=a_{ji}^{(i)}/a_{ii}^{(i)}$$. So we have the $$LU$$-factorization of $$A$$: $$A=LU$$. Then $$A\overrightarrow{x}=\overrightarrow{b} \implies LU\overrightarrow{x}=\overrightarrow{b}$$. Now we do steps to find $$\overrightarrow{x}$$:
1. From $$L\overrightarrow{y}=\overrightarrow{b}$$ solve for $$\overrightarrow{y}$$ by forward substitution
2. From $$U\overrightarrow{x}=\overrightarrow{y}$$ solve for $$\overrightarrow{x}$$ by backward substitution
Example. Solve $$A\overrightarrow{x}=[2\;\; 0\; -3]^T$$, where $A=\left[\begin{array}{rrr} 1 & 0 & 0\\ 1/2 & 1 & 0\\ -3/2 & 2/3 & 1\end{array} \right] \left[\begin{array}{rrr} 2 & -4 & 2\\ 0 & 3 & 1\\ 0 & 0& -2/3 \end{array} \right].$ Solution. With $$A=LU$$, we solve for $$\overrightarrow{y}$$ from $$L\overrightarrow{y}=\overrightarrow{b}$$ by forward substitution: \begin{align*} L\overrightarrow{y}=\overrightarrow{b} &\implies \left[\begin{array}{rrr} 1 & 0 & 0\\ 1/2 & 1 & 0\\ -3/2 & 2/3 & 1\end{array} \right] \left[\begin{array}{r}y_1\\y_2\\y_3 \end{array} \right]=\left[\begin{array}{r}2\\0\\-3 \end{array} \right]\\ &\implies \left\lbrace\begin{array}{rcrcrcr} y_1 & & & & &=&2\\ y_1/2 &+ &y_2 & & &=&0\\ -3y_1/2 &+ &2y_2/3 &+ &y_3 &=&-3 \end{array} \right.\\ &\implies \left\lbrace\begin{array}{rl} y_1 &=2\\ y_2 &=0-y_1/2=0-2/2=-1\\ y_3 &=-3+3y_1/2-2y_2/3=-3+3\cdot 2/2-2(-1)/3=2/3 \end{array} \right. \end{align*} Now we solve for $$\overrightarrow{x}$$ from $$U\overrightarrow{x}=\overrightarrow{y}$$ by forward substitution: \begin{align*} U\overrightarrow{x}=\overrightarrow{y} &\implies \left[\begin{array}{rrr} 2 & -4 & 2\\ 0 & 3 & 1\\ 0 & 0& -2/3 \end{array} \right] \left[\begin{array}{r}x_1\\x_2\\x_3 \end{array} \right]=\left[\begin{array}{r}2\\-1\\2/3 \end{array} \right]\\ &\implies \left\lbrace\begin{array}{rcrcrcr} 2x_1 &- &4x_2 &+ &2x_3 &=&2\\ & &3x_2 &+ &x_3 &=&-1\\ & & &- &2x_3/3 &=&2/3 \end{array} \right.\\ &\implies \left\lbrace\begin{array}{rl} x_3 &=\frac{2/3}{-2/3}=-1\\ x_2 &=\frac{1}{3}(-1-x_3)=\frac{1}{3}(-1+1)=0\\ x_1 &=\frac{1}{2}(2+4x_2-2x_3)=\frac{1}{2}(2+4\cdot0-2(-1))=2 \end{array} \right. \end{align*} Thus the unique solution is $$(x_1,x_2,x_3)=\left(2,0,-1\right)$$.
Advantage of $$LU$$-factorization:
If we need to solve $$A\overrightarrow{x}=\overrightarrow{b_1}$$, $$A\overrightarrow{x}=\overrightarrow{b_2}$$,...,$$A\overrightarrow{x}=\overrightarrow{b_k}$$, then Gaussian elimination solves each problem separately. But once we know the $$LU$$-factorization of $$A=LU$$, the solution $$\overrightarrow{x_i}$$ of $$A\overrightarrow{x}=\overrightarrow{b_i}$$ is obtained just from forward and backward substitution in $$L\overrightarrow{y_i}=\overrightarrow{b_i} \text{ and } U\overrightarrow{x_i}=\overrightarrow{y_i}$$ respectively.
Last edited | HuggingFaceTB/finemath | |
# Multiplying Decimals by Whole Numbers
14 teachers like this lesson
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## Objective
SWBAT multiply decimals by whole numbers.
#### Big Idea
Multiplying decimals is a life skilled used in purchasing goods, cooking, and spending money.
## DO NOW
15 minutes
Students will be completing a memory box. Give them a few minutes to look over their notes from adding and subtracting decimals. Once their time is up, have them close their notes and write down everything they remember about adding and subtracting decimals in their memory box. Encourage students to use visuals, words , or examples in their memory box. When students are finished, have the do a HUSUPU and share the information from their memory box.
Memory box supports mathematical practices 1, 3, and 7
## Multiplying decimals by whole numbers (teacher directed)
20 minutes
The purpose here is to get the students to see the algorithm used to multiply decimals by whole numbers. For each problem, I will have the students estimating what their answer will be. This is especially helpful when the decimal is less that one. Students often get confused as to why their answer is so small when multiplying with decimals less than one. Estimating helps students realize this notion. I will be setting up each problem and modeling, out loud, how I’m doing the steps. When it comes down to finding where to put the decimal, I’m going to ask the students to look at their estimate and see if they can figure out where the decimal should be placed. In the end, I’m going to ask them if they notice a pattern between the numbers in the problem and the placement of the decimal (SMP 6 and 8)
## Multiplying decimals by whole numbers (student's try)
20 minutes
Have the student use white boards and markers to work out each problem. Students should estimate their answer first, then do the computation to find the solution. Students can check with table mates before they show you the answer on their white boards. I always tell them that when I say “whiteboards up” then they should put their boards up so I can check all at once. As students show their answers, be sure to ask how they knew where to place the decimal.
## Around the Room
20 minutes
Students will be completing an Around the room. There are 10 problems and each problem deals with money and the purchasing of multiple amounts. I liked this because it shows the students where we use this in real life.
## Closure
10 minutes
Students will be completing a comprehension menu for multiplying with decimals. Students should work out all 4 problems and put a mark in the box they found the easiest to answer. The comprehension menu checks for 4 types of understanding: mastery, understanding, self-expressive, and interpersonal. Each box will support the same concept, but with different types and levels of understanding. Comprehension menu supports mathematical practices 1,2,3,5. Students should turn their comprehension menu in so you can assess learning. | HuggingFaceTB/finemath | |
Bright, Brighter, Brightest
Presentation on theme: "Bright, Brighter, Brightest"— Presentation transcript:
Bright, Brighter, Brightest
Is it Bright, Brighter, or the Brightest?
The purpose of this lab was to calculate the levels of light given off by a three level light bulb.
Bright, Brighter, Brightest Lab
During this lab we used a variety of equipment. We used a three level light bulb, a photometer to measure the brightness, a stopwatch to measure the time between level on the light bulb, we also used our calculator with the BRIGHT program to analyze and record the data.
Bright, Brighter, Brightest Steps
1.) Our group prepared our calculator with the BRIGHT program. 2.) We read the instructions as we tested everything to make sure it worked. 3.) We turned off the lights in the room. 4.) We turned on the light bulb with three different settings, and recorded each setting on our calculator. 5.) We created a Greatest Integer graph with our data. “Light thinks it travels faster than anything but it is wrong. No matter how fast light travels, it finds the darkness has always got there first, waiting for it.” - Terry Pratchett
Bright, Brighter, Brightest Graph
Independent Variable= Time Greatest Integer Graph Dependent Variable = Brightness f(x)= |x| With each step, came a different light setting. As you can see, the brightness goes up as time goes up, but then it goes back to 0 where it started. .314 .211 Brightness and watts .115 2.546 4.630 6.929 8.681 9.838 Time in seconds
Bright, Brighter, Brightest
f(x)=1.68[1/11966(x)] The low setting was turned on for 3 seconds, the medium setting for 6 seconds, and the high setting for 9 seconds. 4 5, and 6. Each light was on for 3 second intervals. The light was turned off from seconds 1-3, and 10. The ambient light (light already present in the room) reading was
Bright, Brighter, Brightest Graph
Difference between ambient light and y- value Low: .115 Watts, Medium: .211 Watts, High: .314 Watts Ratios of light levels using information above High to Medium: Watts, Medium to Low: Watts, High to Low: Watts Ratios for wattage on the bulb: High to Medium: (3:2) Watts Medium to Low: 2 Watts High to Low: 3 Watts
Bright, Brighter, Brightest
In this experiment the coefficients could be altered in simple ways. In order to make the steps on our graph longer we would keep the light on a certain brightness for a longer period of time. If we wanted to change the distance between each step we would just change the brightness between each level on the bulb. The coefficients in this lab can be related to real world situations. For example, there was a 1.68 wattage difference between each level on the light bulb. This shows that the wattages on the light bulb may not always be exactly what is shown.
Bright, Brighter, Brightest
Throughout this lab we encountered a few difficulties. One was the challenge of not gathering this data in a completely dark room. The room we set up the lab in did have 0.1 wattages of light present. This altered the data collected in a slight way. Due to this lab we learned that not all wattages for the light bulb are exactly what are displayed on it. | HuggingFaceTB/finemath | |
# Centripetal Force
```Centripetal Force
Experiment 4
Objective: To examine the concepts of centripetal acceleration and centripetal force
experimentally.
DISCUSSION:
When an object travels in a straight line at a constant velocity its acceleration is zero. When
an object travels at a constant velocity on a circular path, however, it is accelerating at a rate
inversely proportional to the radius of the circle (1).
ac
2
(1)
r
where υ is the constant velocity of the object around the circular path. Velocity is distance per
time, so we can put the distance around the circle (circumference = 2πr) over the time for one
revolution (Period = T). This gives a formula for acceleration made up of variables we can easily
measure: radius r and period T, (2).
2r T 2 4 2 r
ac
(2)
r
T2
Force is always mass times acceleration, so to find centripetal force we just multiply
centripetal acceleration by a mass. By measuring the mass we can find centripetal force as easily
as acceleration.
4 2 r
Fc m ac m 2
(3)
T
EXERCISES:
In this lab we will observe how centripetal force varies with different radii. By setting Fc
equal to a known force, we can test the accuracy of the experiment.
1) Find Fc for two different radii by measuring the radii then recording the period based on a
reasonable number of revolutions.
2) Calculate the force of gravity Fg based on the mass of the hanging weight. (Fg=mg x g)
Calculate the percent deviation from Fc using Fg as the “theoretical” value.
3) Share your results to get more data, and then graph the results:
Graph υ vs. r and Fc vs. r.
r (meters)
T (seconds)
=total time / N revs.
υ (m/s)
=2π r/T
Fc (Newtons)
=mc·4π2r/T2
4-1
mc = ________ (mass of swinging weight) (kg)
mg = ________ (mass of hanging weight) (kg)
Fg = ________ (mg · 9.8) (kg·m/s2=Newtons)
Useful quantities (approximate):
π = 3.14159
4π2 = 39.4784
Questions:
If the radius increases, how does the centripetal Force change?
If the velocity increases, how does the radius change?
How fast would you need to spin the mass if you wanted to lift a 0.5kg beer bottle using a radius
of 0.2 m?
4-2
4-3
4-4
4-5
4-6
``` | HuggingFaceTB/finemath | |
# Define a simple discrete probability distribution
I know the command: ProbabilityDistribution
I look at the detail of it, it seems a little troublesome.
For a very simple situation:
n: 0 1 2
p(x=n) 0.3 0.4 0.3
how to define this a simple discrete probability distribution
## 1 Answer
You can first define a piecewise function
piece[x_] := Piecewise[{{0.3, x == 0}, {0.4, x == 1}, {0.3, x == 2}}]
and feed it to ProbabilityDistribution
f = ProbabilityDistribution[piece[x], {x, 0, 2, 1}] Its PDF[f, x] and CDF[f, x] You can plot it with
DiscretePlot[piece[x], {x, 0, 2}, Frame -> True, PlotRange -> {0, 0.5}]
or
DiscretePlot[PDF[f][x], {x, 0, 2}, Frame -> True, PlotRange -> {0, 0.5}] Mean[f]
1.
which is
Expectation[x, x \[Distributed] f]
1.
Also
Variance[f]
0.6
or
Probability[x <= 1, x \[Distributed] f]
0.7
etc.
Alternatively you can use EmpiricalDistribution to do the same:
emp = EmpiricalDistribution[{0.3, 0.4, 0.3} -> {0, 1, 2}]
DiscretePlot[PDF[emp][x], {x, 0, 2}, Frame -> True, PlotRange -> {0, 0.5}]
like previously
Plot[CDF[emp][x], {x, 0, 2}, Frame -> True, PlotRange -> {0, 1}] etc.
• Command EmpiricalDistribution It is just what I need. – tiankonghewo Nov 29 '16 at 11:07
• It's a bit frustrating that Mathematica doesn't have a native DiscreteDistribution – becko Nov 30 '18 at 0:47 | HuggingFaceTB/finemath | |
# Question about the mole [duplicate]
Why is it to get from x amu---->xgram you need a mole of the substance.
I mean I underdstand this to be the case for Hydrogen. Why does a mole of a substance CONVERT the substances amu into grams?
• Well, the moles of all elements have different weights too, so what's the problem? – Ivan Neretin Oct 5 '18 at 12:36
• How does a mole of a substance equal its amu in grams? – user57928 Oct 5 '18 at 15:55
• I cant see the answer on that lini. – user57928 Oct 5 '18 at 16:28
I often find units help me understand things like this. So, if:
1 g = mass of Avogadro's number of hydrogen atoms / protons / neutrons, and
1 amu = mass of a hydrogen atom / proton / neutron, and
1 mole = Avogadro's number of atoms or molecules of a substance, then:
if I make a substance heavier than hydrogen, the atomic or molecular mass will go up by one amu for every proton or neutron, and the mass of a mole of that substance will go up by 1 gram for each of those extra protons or neutrons, since a mole of protons or neutrons has a mass of 1 g.
• Why with a different amu its still 1 mole in grams? – user57928 Oct 5 '18 at 16:29
Both AMU and grams are units for expressing mass. AMU is used for weighing single protons, neutrons ore atoms while grams are used for more convenient amounts. So: 1 AMU = mass of one proton = mass of one H-atom (neglecting the electron) = 1,7*10^-24 g. But if you want to do some calculations in gram, that figure is rather unhandy, and instead you use the mass of one mole. One mole is, as you probably know, 6.23*10^23 pieces of whatever we are talking about (Avogados number). The mass of one mole of H is therefore: 1 AMU * 1 mole = 1 gram (round figures). But as Ivan said: other elements weights more. An example: carbon (6 protons + 6neutrons) weigh 12 AMU and one mole (6.23*10^23 C-atoms) weighs: 12 AMU * 1 mole = 12 grams. Ore to be more precise: 12 grams per mole
• So can we say that 1 mole =g/amu – user57928 Oct 5 '18 at 16:32
• Fundamentally yes, but what is your point with this? One mole is a fixed number You can look up, so there is no reason for calculating it? – FrankS Oct 8 '18 at 14:41
There are two different ways to measure a quantity. Think about bananas. You can weigh the bananas or you can count them.
Now let's think about a reaction
$$\ce{2Na + Cl_2 -> 2NaCl}$$
It takes one atom of sodium to react with one atom of chlorine to get one molecule of sodium chloride. Now we can't really count individual atoms in a typical chem lab, so we need some way to covert between a count of atoms and a mass of atoms. The way to convert is through the use of "moles".
A mole is just a big counting number like a dozen or a million. But a mole is a much much bigger number. The value is about $$6.022\times10^{23}$$. So if you "count out" $$6.022\times10^{23}$$ atoms of sodium you get a mass 22.99 grams of sodium atoms that corresponds to the atomic mass of sodium.
The same notion applies to the chlorine gas. $$6.022\times10^{23}$$ molecules of chlorine gas have a mass of 70.90 grams of chlorine gas. But each chlorine molecule has two atoms of chlorine. So we'd only need 35.45 grams of chlorine gas to react with 22.99 grams of sodium.
• You say "if you "count out" atoms of sodium you get a mass 22.99 grams of sodium" , how do you know that ? – user57928 Oct 5 '18 at 16:31
• Better example than bananas would perhaps be water. A gallon of water weighs 8.34 pounds. So I can measure water in pounds or gallons. If I measure gallons, I can convert to pounds by knowing the conversion constant. The primary units of measure form the International System of Units. – MaxW Oct 5 '18 at 18:21 | HuggingFaceTB/finemath | |
# Need to calculate an Average please!!
#### Ted210
##### New Member
Hello,
I hope someone can help!
I need to calculate an average for a questionnaire.
There is a scale of 1-10 about how much they liked a product (1 being not and 10 being liked it a lot)
Now I need an average of what the people said. (e.g. 6.7)
Example:
1 2 3 4 5 6 7 8 9 10 56 88 101 97 100 59 62 39 13 18
<colgroup><col span="10"></colgroup><tbody>
</tbody>
Is there a formula for this perhaps or a way to calculate it?
Thanks!
### Excel Facts
What is the shortcut key for Format Selection?
Ctrl+1 (the number one) will open the Format dialog for whatever is selected.
You could calculate he Mean Average using this:
Excel 2007
ABCDEFGHIJ
212345678910
35688101971005962391318
4
5
64.454976
Sheet4
Cell Formulas
RangeFormula
A6=SUMPRODUCT(A2:J2,A3:J3)/SUM(A3:J3)
Great thanks!!
That helps a lot!
If possible, we asked the people before and after a product trial how much they liked the product.
And I would need to calculate the percentage of how much their opinion shifted?
So for example as above take before trial average opinion was 4.45 and after it was 8.2
Any suggestions perhaps?
Thanks!
You could calculate he Mean Average using this:
Excel 2007
ABCDEFGHIJ
212345678910
35688101971005962391318
4
5
64.454976
</tbody>
Sheet4
Worksheet Formulas
CellFormula
A6=SUMPRODUCT(A2:J2,A3:J3)/SUM(A3:J3)
</tbody>
<tbody>
</tbody>
Sorry Firefly..... but doesnt seem to work with me.
=SUMPRODUCT(E26:N26,E27:N27)/SUM(E27:N27)
And I get
0.0885
<colgroup><col width="48"></colgroup><tbody>
</tbody>
What do the separate parts of the formula return for you ie:
=SUMPRODUCT(E26:N26,E27:N27)
and
=SUM(E27:N27)
Sorry I found the problem and now it works perfectly! So thanks a lot!
Any suggestions perhaps because of the percentage?
thank you!
Well, from my point of view, the % change would be calculated from:
=(End% Less Start%)/Start%
or
=(8.2 - 4.45)/4.45
Great! Thank you for all your help Firefly!!
You're welcome
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# Analyzing with Anova
FratBro23
Category:
Statistics
Price: \$5 USD
Question description
Assignment 2: Analyzing with ANOVA
Submit your answers to the following questions using the ANOVA source table below. The table depicts a two-way ANOVA in which gender has two groups (male and female), marital status has three groups (married, single never married, divorced), and the means refer to happiness scores (n = 100):
1. What are the independent variables and their levels? What is the dependent variable?
2. State all null hypotheses associated with independent variables and their interaction? Also suggest alternate hypotheses?
3. What are the degrees of freedom for 1) gender, 2) marital status, 3) interaction between gender and marital status, and 4) error or within variance?
4. Calculate the mean square for 1) gender, 2) marital status, 3) interaction between gender and marital status, and 4) error or within variance.
5. Calculate the F ratio for 1) gender, 2) marital status, and 3) interaction between gender and marital status.
6. Identify the critical Fs at alpha = .05 for 1) gender, 2) marital status, and 3) interaction between gender and marital status.
7. If alpha is set at .05, what conclusions can you make?
Source Sum of Squares (degrees of freedom [df]) Mean Square Fobt. Fcrit. Gender 68.15 ? ? ? ? Marital Status 127.37 ? ? ? ? Gender * Marital Status (A x B) 41.90 ? ? ? ? Error (Within) 864.82 ? ? NA NA Total 1102.24 99 NA NA NA
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# Prove that $\lim_{n \rightarrow \infty} \int_{\Omega} |f_n - f|^{p} dm=0$.
I am stuck on the following problem.
Let $1<p<\infty$ and let $\Omega \subset \mathbb{R}$. Let {$f_n$} be a sequence of measurable functions on $\Omega$ such that $\lim_{n \rightarrow \infty} f_n= f$ (a.e.) on $\Omega$. Suppose there exists a nonnegative measurable function $g$ on $\Omega$ such that $g^p \in L^{1} \Omega$ and $|f_n| \leq g$ (a.e.) on $\Omega$ for every $n \geq 1$. Prove that $\lim_{n \rightarrow \infty} \int_{\Omega} |f_n - f|^{p} dm=0$.
Plan:
Want to show that $|f_n-f|^p \leq g$ (a.e.), and then by the dominated convergence theorem, $\lim_{n \rightarrow \infty} \int_{\Omega} |f_n - f|^{p} dm=\int_{\Omega} \lim_{n \rightarrow \infty} |f_n-f|^p dm=0$.
Attempt:
Consider $|f_n-f| \leq |f_n|-|f| \leq g-|f| \leq g$. So $|f_n-f| \leq g$, raising both sides to the power $p$, $|f_n-f|^p \leq g^p \in L^{1} \Omega$ where $g^p$ is our control function thus allowing us to use dominated convergence theorem. Valid since $p >1$.
Hence, $\lim_{n \rightarrow \infty} \int_{\Omega} |f_n - f|^{p} dm=\int_{\Omega} \lim_{n \rightarrow \infty} |f_n-f|^p dm$. Since $lim_{n \rightarrow \infty} f_n=f \Leftrightarrow lim_{n \rightarrow \infty} |f_n-f| =0$ we have $\int_{\Omega} \lim_{n \rightarrow \infty} |f_n-f|^p dm=0$.
Any feedback on my attempt and hints would be very helpful. Thank you.
• Generally we say that $\lim_n f_n(x) = f(x)$ iff $\lim_n |f_n(x)-f(x)| = 0$. – copper.hat Jun 11 '17 at 4:25
• @copper.hat Yes, I should have went back to the definition. Thank you. – Joe Jun 11 '17 at 4:42
You're on the right track in that you want to apply DCT however, $\lvert x - y \rvert \not\le \lvert x \rvert - \lvert y \rvert$. Consider $x = 1, y = -1$, e.g.
Instead, since $f \to f_n$ a.e., we have $\lvert f - f_n \rvert^p \to 0$ a.e.; indeed, by definition $f_n \to f$ is the same as $\lvert f - f_n \rvert \to 0$ and then $\lvert f -f_n \rvert^p \to 0$ since $t \mapsto t^p$ is continuous. Next, $$\lvert f - f_n \rvert^p \le (\lvert f \rvert + \lvert f_n \rvert)^p \le (2\max\{\lvert f \rvert, \lvert f_n \rvert\})^p =2^p \max\{\lvert f \rvert^p, \lvert f_n \rvert^p\} \le 2^p(\lvert f \rvert^p + \lvert f_n \rvert^p).$$ Now, $\lvert f \rvert \le g$ a.e. since $\lvert f \rvert = \lim \lvert f_n \rvert \le \lim g = g$ a.e. Thus $$\lvert f - f_n \rvert^p \le 2^{p+1} g^p.$$ However, you have assumed that $g^p$ is integrable, so by DCT you have $$\lim \int_\Omega \lvert f - f_n \rvert dm = \int_\Omega \lim \lvert f - f_n \rvert dm = \int_\Omega 0 \, dm = 0.$$ | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Reduce 263/3000 to lowest terms
263/3000 is already in the simplest form. It can be written as 0.087667 in decimal form (rounded to 6 decimal places).
#### Steps to simplifying fractions
1. Find the GCD (or HCF) of numerator and denominator
GCD of 263 and 3000 is 1
2. Divide both the numerator and denominator by the GCD
263 ÷ 1/3000 ÷ 1
3. Reduced fraction: 263/3000
Therefore, 263/3000 simplified to lowest terms is 263/3000.
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
WHT
# Similarity Examples | Similarity Example Problems | Similarity | Geometry | Khan Academy
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535
681,625
## Similarity Example Problems | Similarity | Geometry | Khan Academy
Transcript:
In the first task we want to find segment length, segment CE. We have these two parallel lines. AB is parallel to DE. We have these two intersecting, of these two triangles. Let’s see what we can do. The first thing that may come to mind is that this angle and this angle are opposite (vertex) angles. So, they will be equal. Another thing you can think of is that the angle CDE is named after the angle CBA. We have this intersection here and these will be cross angles – that is, they will be equal. If we continue this intersection, we will get the corresponding angle with CDE here. And this one is just the opposite. Either way, this angle and this angle will be equal. So, we found that we have two triangles. And they have two corresponding angles that are equal. That, in itself, suffice it to say that triangles are similar. In fact – In fact, we can say that this and that angle are also equal as cross angles. But it is not necessary. We already know that they are similar. In fact, we can say that at the cross angles, they will also be equal. But we already know enough to say they’re similar, even before we do that. We have already met this triangle. I will try to color it so we have it identical corresponding vertices. It is very important to know which angles correspond to which countries so as not to confuse the proportions, and to know what corresponds to each other. Thus, we know that triangle ABC is similar to triangle .. And this peak A corresponds to peak E right here – is similar to peak E. A peak B here corresponds to vertex D, EDC. How does this help us? This tells us the proportions of the respective countries will be equal. They will be equal. They will be a constant. So, we have a corresponding country .. The proportion, for example, the corresponding side of the BC will be DC. We can see it. Just about the way we proved the similarity. If this is true, then BC is corresponding to the DC side We know that BC, the length of BC on DC, right here, will be equal to the length of .. First we have to find how much CE is, this is what interests us. I use BC and DC because I know their values. Thus, BC over DC will be equal to of the respective CE country. The relevant country here will be CA .. It will be equal to CA on CE corresponding countries. This is the last and the first, the last and the first ARE on CE. We know how many aircraft, The aircraft here is 5. We know how much DC is She is 3. We know how many CAs or ACs are here, CA is 4 And now we can calculate the CE. We can there are many ways you can calculate it. Multiply by the cross, which is actually a multiplication of the numerator by the denominator. We get 5 along the length of the CE, which is equal to 3 by 4, which in turn is equal to 12. that’s how we get CE. CE is equal to 12 over 5. This is the same as 2 and 2 fifths. or 2.4 This will be 2 and 2 fifths We’re ready. We used the similarity to get this country, as we simply knew that the proportions between the respective countries will be the same. Let’s solve this problem here now. Let’s look at this one here. I will draw a small line here. This is a different task. Here we want to find how much is DE. Again we have these two parallel lines. We know that the corresponding angles are equal. We know that this angle will be equal to this angle. Because this can be considered intersecting here. Also, we know this angle here will be equal to this angle here. Once again, we have corresponding angles for the intersections. In both triangles. I’m looking at CBD and CAE triangles. They share a common corner here. In fact, we showed once again that we can stop at two angles. We really showed that all three angles of these two triangles, all three corresponding angles, are equal to each other. Now we know .. The important thing we need to do is once again to make sure we get what you write, in the right order, when we prove the similarity. Now, we know that triangle CBD is similar, not equal, is similar to triangle CAE. This means that the relationship between the parties concerned will be a constant. We know, for example, that the relationship between CDs it will be the same such as the relationship between CB to CA. Let’s write it. We know that CB on CA will be equal to the ratio between CD on CE. We know that CB is, CB here is 5. We know how much is CA? We have to be very careful now. Not 3. SA, this whole country will be 5 plus 3 This is equal to 8. We also know that CD is, CD will be 4. Now we can multiply crosswise again. We have 5 CE, 5 CE equals 8 to 4 8 by 4 is 32. So CE is equal to 32 over 5. Or we can present it in another way. 6 and 2 fifths. We are not ready because we are not looking for how much CE is. Only this part is wanted here. Search with is DE. We know the whole length, CE here. She is 6th and 2nd fifth. So, DE is here what we need to find. It will be equal to this length, 6 and 2 fifth minus, minus CD. This will be equal to 2 and 2 fifths. 6 and 2 fifths minus 4 is 2 and 2 fifths. We’re ready! DE is 2 and 2 fifths.
## 0.3.0 | Wor Build 0.3.0 Installation Guide
Transcript: [MUSIC] Okay, so in this video? I want to take a look at the new windows on Raspberry Pi build 0.3.0 and this is the latest version. It's just been released today and this version you have to build by yourself. You have to get your own whim, and then you...
## Youtube Neural Network | But What Is A Neural Network? | Chapter 1, Deep Learning
Transcript: These are three, sloppily written and provided at a very low resolution of 28 x 28 pixels But your mind has no problem recognizing it as three and I want you to take a moment to appreciate How your brain can do this effortlessly and smoothly I mean this... | HuggingFaceTB/finemath | |
# Grokking the Coding Interview's Two Pointers
Pair with target sum, Remove duplicates, Squaring a sorted array, Dutch national flag problem
### 1. Given an array of sorted numbers and a target sum, find a pair in the array whose sum is equal to the given target.
Example:
``````- Input: []int{1, 2, 6, 8, 16, 26}, target=14
Output: []int{2, 3}
Explanation: 6 (index 2) + 8 (index 3) = 14
``````
Approach:
``````- Have one pointer start at the beginning and one at the end of the array.
- At each step, see if the two pointers add up to the target sum and move
them toward each other accordingly.
``````
Cost:
``````- O(n) time, O(n) space.
``````
Link to solution →
### 2. Given an array of sorted numbers and a target sum, find a pair in the array whose sum is equal to the given target.
Example:
``````- Input: []int{1, 2, 6, 8, 16, 26}, target=14
Output: []int{2, 3}
Explanation: 6 (index 2) + 8 (index 3) = 14
``````
Approach:
``````- Have one pointer iterate the array and one placing non-duplicate number.
``````
Cost:
``````- O(n) time, O(1) space.
``````
Link to solution →
### 3. Given a sorted array, create a new array containing squares of all the number of the input array in the sorted order.
Assumption:
``````- The input can have negative numbers.
``````
Example:
``````- Input: []int{-2, -1, 0, 1, 2}
Output: []int{0, 1, 1, 4, 4}
``````
Approach:
``````- The difficult part is to generate the output array with squares in sorted order because we have negative numbers and their squares are positive.
- Have one pointer start at the beginning and one at the end and let them
move toward each other.
- At each step, whichever bigger will be added to the output array, from
right to left.
``````
Cost:
``````- O(n) time, O(n) space.
``````
Link to solution →
### 4. Given an array containing 0s, 1s and 2s, sort the array in-place.
Example:
``````- Input: []int{1, 0, 2, 1, 0}
Output: []int{0, 0, 1, 1, 2}
``````
Approach:
``````- Have one pointer start at the beginning and the other at the end
while iterating through the array.
- We will move all 0s before that start pointer and 2s after the end
pointer so that all 1s would be between in the end.
``````
Cost:
``````- O(n) time, O(1) space.
``````
Link to solution →
For more coding problems, please visit https://github.com/hoanhan101/algo.
If you’re interested in getting updates for such content like these, consider joining my mailing list here → | HuggingFaceTB/finemath | |
# Solved (Free): It is known that the duration of consultation with patients in a psychiatry clinic is normally distributed with an average of 35 minutes
#### ByDr. Raju Chaudhari
Apr 16, 2021
It is known that the duration of consultation with patients in a psychiatry clinic is normally distributed with an average of 35 minutes and a standard deviation of 5 minutes. One of the doctors working in this clinic claimed that the duration of the consultation is not 35 minutes. For this purpose, the consultation duration of 10 clients was examined and found to be 42 minutes. Test this claim at 0.05 significance level.
#### Solution
Given that $n = 10$, sample mean $\overline{x} = 42$, the sample standard deviation is $\sigma = 5$.
#### Step 1 Hypothesis Testing Problem
The hypothesis testing problem is
$H_0 : \mu = 35$ against $H_1 : \mu \neq 35$ ($\text{two-tailed}$)
#### Step 2 Test Statistic
The test statistic for testing above hypothesis testing problem is
\begin{aligned} Z& =\frac{\overline{x} -\mu}{\sigma/\sqrt{n}}. \end{aligned}
The test statistic $Z$ follows $N(0,1)$ distribution.
#### Step 3 Significance Level
The significance level is $\alpha = 0.05$.
#### Step 4 Critical Value(s)
As the alternative hypothesis is $\text{two-tailed}$, the critical value of $Z$ $\text{are}$ $-1.96 and 1.96$ (from Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$.
#### Step 5 Computation
The test statistic under the null hypothesis is
\begin{aligned} Z_{obs}&=\frac{ \overline{x} -\mu_0}{\sigma/\sqrt{n}}\\ &= \frac{42-35}{5/ \sqrt{10 }}\\ &= 4.427 \end{aligned}
#### Step 6 Decision
The test statistic is $Z_{obs} =4.427$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis at $\alpha = 0.05$ level of significance.
#### Step 6 Decision
$p$-value Approach:
This is a $\text{two-tailed}$ test, so the p-value is the
area to the $\text{extreme}$ of the test statistic ($Z_{obs}=4.427$) is p-value = $0$.
The p-value is $0$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis at $\alpha =0.05$ level of significance.
There is sufficient evidence to conclude the duration of the consultation is not 35 minutes. | HuggingFaceTB/finemath | |
# Mathematics and Computation
## A blog about mathematics for computers
Postsby categoryby yearall
# Posts in the year 2014
### A HoTT PhD position in Ljubljana
I am looking for a PhD student in mathematics. Full tuition & stipend will be provided for a period of three years, which is also the official length of the programme. The topic of research is somewhat flexible and varies from constructive models of homotopy type theory to development of a programming language for a proof assistant based on dependent type theory, see the short summary of the Effmath project for a more detailed description.
The candidate should have as many of the following desiderata as possible, and at the very least a master’s degree (or an equivalent one):
1. a master’s degree in mathematics, with good knowledge of computer science
2. a master’s degree in computer science, with good knowledge of mathematics
3. experience with functional programming
4. experience with proof assistants
5. familiarity with homotopy type theory
The student will officially enrol in October 2015 at the University of Ljubljana. No knowledge of Slovene is required. However, it is possible, and even desirable, to start with the actual work (and stipend) earlier, as soon as in the spring of 2015. The candidates should contact me by email as soon as possible. Please include a short CV and a statement of interest.
Update 2015-03-28: the position has been taken.
### TEDx “Zeroes”
I spoke at TEDx University of Ljubljana. The topic was how programming influences various aspects of life. I showed the audence how a bit of simple programming can reveal the beauty of mathematics. Taking John Baez’s The Bauty of Roots as an inspiration, I drew a very large image (20000 by 17500 pixels) of all roots of all polynomials of degree at most 26 whose coefficients are $-1$ or $1$. That’s 268.435.452 polynomials and 6.979.321.752 roots. It is two degrees more than Sam Derbyshire’s image, so consider the race to be on! Who can give me 30 degrees?
### Reductions in computability theory from a constructive point of view
Here are the slides from my Logic Coloquium 2014 talk in Vienna. This is joint work with Kazuto Yoshimura from Japan Advanced Institute for Science and Technology.
Abstract: In constructive mathematics we often consider implications between non-constructive reasoning principles. For instance, it is well known that the Limited principle of omniscience implies that equality of real numbers is decidable. Most such reductions proceed by reducing an instance of the consequent to an instance of the antecedent. We may therefore define a notion of instance reducibility, which turns out to have a very rich structure. Even better, under Kleene’s function realizability interpretation instance reducibility corresponds to Weihrauch reducibility, while Kleene’s number realizability relates it to truth-table reducibility. We may also ask about a constructive treatment of other reducibilities in computability theory. I shall discuss how one can tackle Turing reducibility constructively via Kleene’s number realizability.
Slides with talk notes: lc2014-slides-notes.pdf
### Seemingly impossible constructive proofs
In the post Seemingly impossible functional programs, I wrote increasingly efficient Haskell programs to realize the mathematical statement
$\forall p : X \to 2. (\exists x:X.p(x)=0) \vee (\forall x:X.p(x)=1)$
for $X=2^\mathbb{N}$, the Cantor set of infinite binary sequences, where $2$ is the set of binary digits. Then in the post A Haskell monad for infinite search in finite time I looked at ways of systematically constructing such sets $X$ with corresponding Haskell realizers of the above omniscience principle.
In this post I give examples of infinite sets $X$ and corresponding constructive proofs of their omniscience that are intended to be valid in Bishop mathematics, and which I have formalized in Martin-Löf type theory in Agda notation. This rules out the example $X=2^\mathbb{N}$, as discussed below, but includes many interesting infinite examples. I also look at ways of constructing new omniscient sets from given ones. Such sets include, in particular, ordinals, for which we can find minimal witnesses if any witness exists.
Agda is a dependently typed functional programming language based on Martin-Löf type theory. By the Curry-Howard correspondence, Agda is also a language for formulating mathematical theorems (types) and writing down their proofs (programs). Agda acts as a thorough referee, only accepting correct theorems and proofs. Moreover, Agda can run your proofs. Here is a graph of the main Agda modules for this post, and here is a full graph with all modules.
### Brazilian type checking
I just gave a talk at “Semantics of proofs and certified mathematics”. I spoke about a new proof checker Chris Stone and I are working on. The interesting feature is that it has both kinds of equality, the “paths” and the “strict” ones. It is based on a homotopy type system proposed by Vladimir Voevodsky. The slides contain talk notes and explain why it is “Brazilian”.
GitHub repository: https://github.com/andrejbauer/tt
Abstract: Proof assistants verify that inputs are correct up to judgmental equality. Proofs are easier and smaller if equalities without computational content are verified by an oracle, because proof terms for these equations can be omitted. In order to keep judgmental equality decidable, though, typical proof assistants use a limited definition implemented by a fixed equivalence algorithm. While other equalities can be expressed using propositional identity types and explicit equality proofs and coercions, in some situations these create prohibitive levels of overhead in the proof.
Voevodsky has proposed a type theory with two identity types, one propositional and one judgmental. This lets us hypothesize new judgmental equalities for use during type checking, but generally renders the equational theory undecidable without help from the user.
Rather than reimpose the full overhead of term-level coercions for judgmental equality, we propose algebraic effect handlers as a general mechanism to provide local extensions to the proof assistant’s algorithms. As a special case, we retain a simple form of handlers even in the final proof terms, small proof-specific hints that extend the trusted verifier in sound ways.
### Intuitionistic Mathematics and Realizability in the Physical World
This is a draft version of my contribution to “A Computable Universe: Understanding and Exploring Nature as Computation”, edited by Hector Zenil. Consider it a teaser for the rest of the book, which contains papers by an impressive list of authors.
Abstract: Intuitionistic mathematics perceives subtle variations in meaning where classical mathematics asserts equivalence, and permits geometrically and computationally motivated axioms that classical mathematics prohibits. It is therefore well-suited as a logical foundation on which questions about computability in the real world are studied. The realizability interpretation explains the computational content of intuitionistic mathematics, and relates it to classical models of computation, as well as to more speculative ones that push the laws of physics to their limits. Through the realizability interpretation Brouwerian continuity principles and Markovian computability axioms become statements about the computational nature of the physical world. | open-web-math/open-web-math | |
# Definition: 7.20: Proportional Numbers
Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third (is) of the fourth.
### Modern Definition
See rational numbers.
### Notes
$a\cdot d = b\cdot c.$
• Please note that this is exactly the case if they represent the same rational numbers.
• In fact, being proportional in the sense of the above definition is an equivalence relation which can be used to define rational numbers: Two rational numbers $$a/b$$ and $$c/d$$ are equivalent, if and only if $ad=bc.$
$\frac ab\sim\frac cd\Longleftrightarrow a\cdot d = b\cdot c.$
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@Fitzpatrick | HuggingFaceTB/finemath | |
# Range - Concepts such as mean, median, mode and range
D
#### Don Winton
Found this interesting:
-------Snip-------
I am taking my first ever statistics course. However as an elementary teacher, I did teach such concepts as mean, median, mode and range. In this course we learned that range is the highest score/data minus the lowest then add 1 to the difference. I had taught that the way to figure out the range was to simply subtract the lowest score/data from the highest. This is also the way it is done in the statistics program I have. My husband (a statistician) agrees with my method. Which method is correct? If the first method is correct, please explain the reason one adds 1 to the difference. Thanks!
In response to Range: {Not Mine, Don}
Dona, you and your husband is right. The method you used of subtracting the smallest data value from the largest is the correct way of calculating the sample range. The only possible alternative I could think of would be if you wanted to construct an unbiased estimate of the population range.
For example, suppose you have a uniform distribution between A and B, with both limits being unknown. You then calculate from the data, the sample range, r = largest minus smallest. This r necessarily is less than the population range, R=B-A. To obtain an unbiased estimate of R, you need to multiply r by (N+1)/(N-1), where N is the sample size. Note that this is a multiplicative factor, not an additive factor. In short, I can't think of any reason to add one to the sample range.
-------End Snip-------
In this course we learned that range is the highest score/data minus the lowest then add 1 to the difference. I had taught that the way to figure out the range was to simply subtract the lowest score/data from the highest.
Thoughts,
Regards,
Don
J
#### John C
Don,
My statistical training was always sketchy and is now very rusty, so I offer this only as a possible clue to the problem. I can’t vouch for it’s validity;
For variables, the range is highest minus lowest. That’s the way I was taught. eg measurement in inches; Highest 10, lowest 5. 10 - 5 = 5.
For attributes, it’s the number of consecutive attributes in the sample (or potential attributes, if all possible have not been observed) that gives the range, eg; marks on a screen; Highest sample had 10, lowest had 5. The range can be considered to include; 5, 6, 7, 8, 9 and 10. So that’s 6 levels of attributes. 10-5 = 5. 5+1 = 6.
rgds, John C
##### One of THE Original Covers!
Don,
As part of figuring Sigma, finding the estimation and unbiased estimation is merely the result of data from a population or from a sample. As always, the sample always has potential error beyond the population estimate and n-1 is normally used.
Not sure why to add/subtract a 1 for the Range though. JC's explanation seems reasonable, but I can not confirm.
D
#### Don Winton
For variables, the range is highest minus lowest. That’s the way I was taught. eg measurement in inches; Highest 10, lowest 5. 10 - 5 = 5. For attributes, it’s the number of consecutive attributes in the sample (or potential attributes, if all possible have not been observed) that gives the range, eg; marks on a screen; Highest sample had 10, lowest had 5. The range can be considered to include; 5, 6, 7, 8, 9 and 10. So that’s 6 levels of attributes. 10-5 = 5. 5+1 = 6.
Agreed. I believe the response cited was probably considering this example. Or rather, see below.
As part of figuring Sigma, finding the estimation and unbiased estimation is merely the result of data from a population or from a sample. As always, the sample always has potential error beyond the population estimate and n-1 is normally used.
It is also possible that the respondent was thinking sigma rather than range. I posted this in order to see what the responses would be. Many thanks, guys.
Regards,
Don | HuggingFaceTB/finemath | |
# Al invests $5,500, at 6% interest, compounded monthly for one year. Calculate the annual percentage yield for his investment. Al invests$5,500, at 6% interest, compounded monthly for one year. Calculate the annual percentage yield for his investment.
Al invests $5,500, at 6% interest, compounded monthly for one year. Calculate the annual percentage yield for his investment. We Know that, The$5,500 is irrelevant, and the annual rate equivalent to 6% per annum, compounded monthly
let that annual rate be i
1+i = (\frac{1 + .06}{12} )^{12}
1+i = 1.005^{12} = approx 1.06168
Thus, the equivalent annual rate is 6.168 % | HuggingFaceTB/finemath | |
# Homework Question
#### jillianleab
##### New Member
Hi everyone:
I'm working on a homework assignment and I am having some trouble with one of the questions. I just don't know which formula to use! Here it is:
In one town, 79% of adults have health insurance. What is the probability that 7 adults selected at random from the town all have health insurance?
A: 0.192
B: 5.53
C: 0.089
D: 0.79
I know the answer is not B, since a probability must be between 0 - 1. I thought maybe it was D, but then that just seems to obvious. I can't find an example similar to this in my text.
I thought maybe I should use the complement rule; A, 7 have insurance, A-bar, none have insurance; but realized that's wrong because it's not asking "at least one" it's asking all. I tried using an online statistics calculator where I input my population size and sample set, but came out with an answer that doesn't appear as an option.
I know this site isn't designed to do my homework for me, but if someone could just point me in the right direction, I'd really appreciate it.
Thanks!
PS: If the answer is D, I'm going to feel like a real idiot!
#### claire9
##### New Member
Hi,
I'm not that brilliant at statistics but here are my thoughts!
If there is 79% of the population that have health insurance and you have chosen 7 people then the probability that they have health insurance is 7/79 which equals 0.089 which is answer C.
#### Akavall
##### New Member
Why not A?
(.79)^7 = 0.192
Therefore the answer is A. This works for all 7 people. If it asked for at least one, the answer would be very close to 1. | HuggingFaceTB/finemath | |
algorithms
# Algorithms in English
If the rightmost point of one interval is to the right of the leftmost point of the other interval, and if the leftmost point of the first interval is to the left of the rightmost point of the other interval, then and only then do those two intervals overlap.
There are some assumptions, however. There are always some assumptions, even when dealing with the most rigorous of mathematics. It’s time to finally admit this fact of life.
The assumptions here are that intervals are collinear and closed, so that the rightmost and the leftmost points exist. And that “to overlap” means to have more than one common point. Which are all very reasonable.
We can name one interval “alice” and the other one “bob”. Then we can write it as follows:
If the rightmost point of alice is to the right of the leftmost point of bob, and if the leftmost point of alice is to the left of the rightmost point of bob, then and only then do alice and bob overlap.
If we call the leftmost point “tail” and the rightmost point “head” as they are lying there in such an orientation and if we then substitute “to the left” with “smaller” and “to the right” with “greater” we obtain:
If alice’s head is greater than bob’s tail, and if alice’s tail is smaller than bob’s head, then and only then do alice and bob overlap.
or even better:
If alice’s head is greater than bob’s tail, and bob’s head is greater than alice’s tail, then and only then do alice and bob overlap.
or:
If the head of both is greater than the tail of the other one, then and only then do alice and bob overlap.
or even:
If every head is greater than every tail, then and only then do alice and bob overlap.
or:
If there is no head bellow any tail, then and only then do alice and bob overlap. (1)
with the:
If there is no interval’s supremum smaller of any interval’s infimum there, then and only then two intervals do overlap.
the assumption of the closed intervals is no more needed. Any combination of closed, semi-closed or open intervals now works. Our assumption moved to the infimum/supremum concept. Fortunately, quite a fundamental one in mathematics.
Usually we say:
If A2>B1 && B2>A1 overlaping=true Else overlaping=false (2)
The assumption of what A1 and A2 and B1 and B2 are, as well as every other assumption, must be first handled somehow. Otherwise some defaults may be understood by the compiler and only if we are lucky, will everything work well after compiling.
The formula (2) is a free gift to your machine compiler, when the formula (1) is the most understandable for us. Isn’t it? You and your machine compiler are meeting at the midpoint between the natural language expression from the beginning and the binary code it will eventually produce. You are just to soft with your machine. Therefore too harsh to yourself.
You are born to think in some human language, where you can really understand things. Don’t worry about computers, they are perfectly capable of understanding English (or Slovenian). At least in principle.
The English language is a perfect engine. It can run all the logics, mathematics, physics and all the programming languages, including assembly. All. You can enrich the English language with formulae and pictures, but you don’t need to. Every single picture is perfectly describable in English.
Just imagine a superintelligent machine running on Latin! Holly apparatus! Machina Sancta.
Standard
physics
# Colliding Steel Spheres
In a central collision of two 1 gram steel balls with a relative speed of 1 meter per second, nothing much would happen.
If the collision speed were much greater, say a few kilometres per second, they would, of course, evaporate.
Then, we also have the case where the relative speed is 99 percent of the speed of light. The collision is indistinguishable from an A bomb detonation. Entirely understandable and comprehensible.
However in the case where their relative velocity is almost the speed of light, where maybe an entire star or galaxy mass has been used for the acceleration, those two spheres will go through each other virtually intact!
Why do you think, this is?
Standard
# Water on Earth
It is from Earth, not from comets, not from asteroids.
There’s more than one million cups of water of it on our planet. Cups as large as the comet was 65 or 66 million years ago. There wasn’t enough time for all our water to be delivered this way. Where in time are you going to put all those alleged events?
None of the latter big bombardment were nearly big enough.
Almost all of our water for the last 4.5 billion years has always been with us. Just as all the sulphur has.
Chemically recombined who knows how, but all the water is originally from here.
(I won a bet on the Extropy mailing list, what Rosetta was about to say, a few days ago. It is in a perfect accordance with what I have written above.)
Standard | HuggingFaceTB/finemath | |
Associated Topics || Dr. Math Home || Search Dr. Math
### Formula to Calculate Overlap of Two Arcs on a Circle
```Date: 05/27/2005 at 15:55:22
Subject: Intersecting arcs on a circle
Hi there,
I have a circle and I draw an arc from 0 to 45 degrees with a blue
pen, and another from 40 to 60 degrees with a red pen. Now I know
that the arcs are intersecting from 40 to 45, but I want to construct
a formula with the 4 angle entries and one Boolean output for the
One problem is that if one arc is from 340 to 30 (50 deg) and the
other from 350 to 355, the 360 to 0 turn has to be calculated
separately. It is simple to calculate it with many "if" statements,
but I need a one line formula.
```
```
Date: 05/29/2005 at 22:48:51
From: Doctor Peterson
Subject: Re: Intersecting arcs on a circle
You want a simple formula (presumably in some computer language or a
spreadsheet) that will be true if two arcs, for which the starting and
ending angles are given, overlap.
The calculation is made easier if we first standardize the angles.
Suppose we are given one arc counterclockwise from a to b degrees, and
another counterclockwise from c to d. We can first move our point of
reference so that we measure 0 degrees at a; then the two arcs are
0 to b-a
c-a to d-a
Now we can make sure that these angles are between 0 and 360 degrees.
One way to do this is to use a "mod" function, which gives the
remainder between 0 and 360 after division by 360; we replace each
variable as follows:
d = mod(d-a, 360)
c = mod(c-a, 360)
b = mod(b-a, 360)
a = 0
If the given angles can be negative, you have to be careful with this,
as what is often called the "mod" function (e.g. "%" in C) doesn't do
what we need for negative inputs. See this page if you need help with
that:
What is Mod?
http://mathforum.org/library/drmath/sets/select/dm_mod.html
Now, we have something like this:
a===============b
+-----------+---+-------+----+
0 c===========d 360
The only way the two arcs can NOT overlap is if a<b<c<d:
a===========b
+-----------+---+-------+----+
0 c=======d 360
If they do overlap, then either a<c<b<d as in the first picture, or
d<c and the second arc wraps around:
a===============b
+-----------+---+-------+----+
0===========d c====360
This gives two ways to state the condition:
not((b<c) and (c<d))
or
(c<b) or (d<c)
If you need to get it all in one expression without modifying the
variables first, you would use
(mod(b-a, 360) < mod(c-a, 360)) or (mod(d-a, 360) < mod(c-a, 360))
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 05/30/2005 at 02:48:48
Subject: Thank you (Intersecting arcs on a circle)
Many thanks, Dr. Peterson.
It is now clear. The main confusing thing was that I had to move the
point of reference to 0 degrees. I love math, it has such simple
techniques to solve complicated problems, and thank goodness we have
doctors like you to show us the path.
Kind regards,
```
Associated Topics:
High School Conic Sections/Circles
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We have a number of different factoring expressions worksheets for you to try with your students. 4 and 7 are "factors" and 28 is the "product" > Another example: (x + 3)(x - 7) = x2 - 4x - 21 • In this lesson, we will be learning and practicing how to factor quadratic equations. Factor quadratics in the form Factor out the GCF. Expanding and Factorising Expressions Colour By Number; Solving Quadratic Equations by Factorising - GCSE Maths Schedule Exit Lesson. Just select your click then download button, and complete an offer to start downloading the ebook. To get started finding 4 Practice Factoring Quadratic Expressions Answers , you are right to find our website which has a comprehensive collection of manuals listed. Use algebra to explain how you know that a rectangle with side lengths one less and one more than a square will always be 1 square unit smaller than the square. Multiply a monomial by a polynomial and multiply two binomials. 25 well balanced problems on factoring quadratic expressions with leading coefficients greater than 1 and all positive terms with factors of 9 or less. Unit 7 Lesson 1 Semester Exam Review 57. Basic Factoring / Factoring Quadratic Expressions 21. Algebra 2 Common Core answers to Chapter 4 - Quadratic Functions and Equations - 4-5 Quadratic Equations - Lesson Check - Page 229 2 including work step by step written by community members like you. Determine any common factors of the given binomial, if common factors exist. For example, this Factoring into Single Brackets Lesson Pack is a great way to broach a different facet of factorisation for children. How does this problem look different from the expressions we have been factoring? Lesson 4 Summary. Factoring Quadratic Expressions - Activity 4 4 Practice Factoring Quadratic Expressions Answers 4 Practice Factoring Quadratic Expressions Thank you completely much for downloading 4 4-4 Factoring Quadratic Expressions Page 8/26. This Factoring Quadratic Expressions Worksheet is suitable for 11th Grade. eureka-math.org This file derived from ALG I-M4-TE-1.3.0-09.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. , so please read all the instructions under each step will help clarify exactly what you need create! Factoring examples examples in standard form when a = 1 now you try Factoring the expression. With the Quadratic and go backwards to try with your students 29 x + 100 13. x2 x. Biggest of these that have literally hundreds of thousands of different Factoring Expressions worksheets for you any factors... Product of the given binomial, if common factors of 9 or less it not... 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32-hour Work Week Full-time, Coming Home For Christmas Norman Rockwell Cast, Asia-pacific Journal Of Public Health, Kubectl Kubernetes Cheat Sheet, Monica Vinader Sale, Coral Wholesale Florida, Good Afternoon In Bengali Image, Jenkins Artifactory Plugin Code, | HuggingFaceTB/finemath | |
## ← Direct Variation Practice 2 - Visualizing Algebra
• 2 Followers
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### Get Embed Code x Embed video Use the following code to embed this video. See our usage guide for more details on embedding. Paste this in your document somewhere (closest to the closing body tag is preferable): <script type="text/javascript" src='https://amara.org/embedder-iframe'></script> Paste this inside your HTML body, where you want to include the widget: <div class="amara-embed" data-url="http://www.youtube.com/watch?v=wxeWWKjqGYM" data-team="udacity"></div> 1 Language
Showing Revision 2 created 05/25/2016 by Udacity Robot.
1. It turns out we can always set up a proportion to solve direct and inverse
2. variation problems. Let's try this one out. For this problem the area of a
3. circle varies directly with the square of it's radius. A circle with a side
4. length of 3 inches has an area of 28 and 278 thousandths inches squared. What is
5. the area of circle with a radius of 4 and 1 tenth inches? Now this problem is
6. very similar to the last one we just did. Except you need to include on more
7. thing. We're not just varying directly, we're varying directly with the square
8. of something. What do you think you're going to do? Try playing around with this
9. one and I hope you'll get it. Good luck. Also you don't want to round while you | HuggingFaceTB/finemath | |
Chemistry news, articles, and more
## Spectroscopy of element 115 decay chains
The full document is not yet published but a paper accepted 7 Aug 2013 entitled Spectroscopy of element 115 decay chains by D. Rudolph et al. provides additional evidence for element 115:
A high-resolution $\alpha$, $X$-ray and $\gamma$-ray coincidence spectroscopy experiment was conducted at the GSI Helmholtzzentrum f\"ur Schwerionenforschung. Thirty correlated $\alpha$-decay chains were detected following the fusion-evaporation reaction $^{48}$Ca~+~$^{243}$Am. The observations are consistent with previous assignments of similar decay chains to originate from element $Z=115$. For the first time, precise spectroscopy allows the derivation of excitation schemes of isotopes along the decay chains starting with elements $Z>112$. Comprehensive Monte-Carlo simulations accompany the data analysis. Nuclear structure models provide a first level interpretation.
## Search for element 113 concluded at last?
Press release from RIKEN Nishina Center for Accelerator-Based Science
The most unambiguous data to date on the elusive 113th atomic element has been obtained by researchers at the RIKEN Nishina Center for Accelerator-based Science (RNC). A chain of six consecutive alpha decays, produced in experiments at the RIKEN Radioisotope Beam Factory (RIBF), conclusively identifies the element through connections to well-known daughter nuclides. The groundbreaking result, reported in the Journal of Physical Society of Japan, sets the stage for Japan to claim naming rights for the element.
Steps in chain of decays from element 113 to mendelevium-254
The search for superheavy elements is a difficult and painstaking process. Such elements do not occur in nature and must be produced through experiments involving nuclear reactors or particle accelerators, via processes of nuclear fusion or neutron absorption. Since the first such element was discovered in 1940, the United States, Russia and Germany have competed to synthesize more of them. Elements 93 to 103 were discovered by the Americans, elements 104 to 106 by the Russians and the Americans, elements 107 to 112 by the Germans, and the two most recently named elements, 114 and 116, by cooperative work of the Russians and Americans.
With their latest findings, associate chief scientist Kosuke Morita and his team at the RNC are set follow in these footsteps and make Japan the first country in Asia to name an atomic element. For many years Morita's team has conducted experiments at the RIKEN Linear Accelerator Facility in Wako, near Tokyo, in search of the element, using a custom-built gas-filled recoil ion separator (GARIS) coupled to a position-sensitive semiconductor detector to identify reaction products. On August 12th those experiments bore fruit: zinc ions travelling at 10% the speed of light collided with a thin bismuth layer to produce a very heavy ion followed by a chain of six consecutive alpha decays identified as products of an isotope of the 113th element (Figure 1).
While the team also detected element 113 in experiments conducted in 2004 and 2005, earlier results identified only four decay events followed by the spontaneous fission of dubnium-262 (element 105). In addition to spontaneous fission, the isotope dubnium-262 is known to also decay via alpha decay, but this was not observed, and naming rights were not granted since the final products were not well known nuclides at the time. The decay chain detected in the latest experiments, however, takes the alternative alpha decay route, with data indicating that Dubnium decayed into lawrencium-258 (element 103) and finally into mendelevium-254 (element 101). The decay of dubnium-262 to lawrencium-258 is well known and provides unambiguous proof that element 113 is the origin of the chain.
Combined with their earlier experimental results, the team's groundbreaking discovery of the six-step alpha decay chain promises to clinch their claim to naming rights for the 113th element.
"For over 9 years, we have been searching for data conclusively identifying element 113, and now that at last we have it, it feels like a great weight has been lifted from our shoulders," Morita said. "I would like to thank all the researchers and staff involved in this momentous result, who persevered with the belief that one day, 113 would be ours. For our next challenge, we look to the uncharted territory of element 119 and beyond."
## Element number 114: flerovium (symbol Fl) and element number 116: livermorium (symbol Lv)
The International Union of Pure and Applied Chemistry (IUPAC) has recommended names for elements 114 and 116. Scientists from the Lawrence Livermore National Laboratory (LLNL) and at Dubna proposed the names as Flerovium for element 114 and Livermorium for element 116.
Flerovium (atomic symbol Fl) was chosen to honor Flerov Laboratory of Nuclear Reactions, where superheavy elements, including element 114, were synthesized. Georgiy N. Flerov (1913-1990) was a renowned physicist who discovered the spontaneous fission of uranium and was a pioneer in heavy-ion physics. He is the founder of the Joint Institute for Nuclear Research. In 1991, the laboratory was named after Flerov - Flerov Laboratory of Nuclear Reactions (FLNR).
Livermorium (atomic symbol Lv) was chosen to honor Lawrence Livermore National Laboratory (LLNL) and the city of Livermore, Calif. A group of researchers from the Laboratory, along with scientists at the Flerov Laboratory of Nuclear Reactions, participated in the work carried out in Dubna on the synthesis of superheavy elements, including element 116. (Lawrencium -- Element 103 -- was already named for LLNL's founder E.O. Lawrence.)
In 1989, Flerov and Ken Hulet (1926-2010) of LLNL established collaboration between scientists at LLNL and scientists at FLNR; one of the results of this long-standing collaboration was the synthesis of elements 114 and 116.
The creation of elements 116 and 114 involved smashing calcium ions (with 20 protons each) into a curium target (96 protons) to create element 116. Element 116 decayed almost immediately into element 114. The scientists also created element 114 separately by replacing curium with a plutonium target (94 protons).
The creation of elements 114 and 116 generate hope that the team is on its way to the "island of stability," an area of the periodic table in which new heavy elements would be stable or last long enough for applications to be found.
The new names were submitted to the IUPAC in late October. The new names will not be official until about five months from now when the public comment period is over.
## Printable periodic table: QR-coded
Attached find a printable QR-coded periodic table with links to online periodic table data. QR codes are 2-dimensional bar codes readable by, for instance, some Apps on iPhones and others.
Print on a big a piece of paper as possible, otherwise your QR reader may pick up an element you didn't intend.
Periodic Table QR-coded: Periodic Table QR-coded
Version history
1.1: 15 September 2011
1.0: 17 July 2011
## Discovery of the Elements with Atomic Number 114 and 116
A news reports from IUPAC indicates the confirmation of the discoveries of elements 114 and 116. Proposals for the names of the two elements will follow in due course:
## News: Discovery of the Elements with Atomic Number 114 and 116
Priority for the discovery of the elements with atomic number 114 and 116 has been assigned, in accordance with the agreed criteria, to collaborative work between scientists from the Joint Institute for Nuclear Research in Dubna, Russia and from Lawrence Livermore, California, USA (the Dubna-Livermore collaborations). The discovery evidences were recently reviewed and recognized by a IUPAC/IUPAP joint working party. IUPAC confirmed the recognition of the elements in a letter to the leaders of the collaboration.
The IUPAC/IUPAP Joint Working Party (JWP) on the priority of claims to the discovery of new elements has reviewed the relevant literature pertaining to several claims. In accordance with the criteria for the discovery of elements previously established by the 1992 IUPAC/IUPAP Transfermium Working Group, and reiterated by the 1999 and 2003 IUPAC/IUPAP JWPs, it was concluded that “the establishment of the identity of the isotope 283Cn by a large number of decaying chains, originating from a variety of production pathways essentially triangulating its A,Z character enables that nuclide’s use in unequivocally recognizing higher-Z isotopes that are observed to decay through it.” From 2004 Dubna-Livermore collaborations the JWP notes: (i) the internal redundancy and extended decay chain sequence for identification of Z = 287114 from 48Ca + 242Pu fusion (Oganessian et al. Eur. Phys. J. A 19, 3 (2004) and Phys. Rev. C 70, 064609 (2004)); and (ii) that the report of the production of 291116 from the fusion of 48Ca with 245Cm is supported by extended decay chains that include, again, 283Cn and descendants (Oganessian et al. Phys. Rev. C 69, 054607 (2004)). It recommends that the Dubna-Livermore collaborations be credited with discovery of these two new elements.
A full synopsis of the relevant experiments and related efforts is presented in a technical report published online in Pure and Applied Chemistry on 1 June 2011. With the priority for the discovery established, the scientists from the Dubna-Livermore collaborations are invited to propose a name for the two super-heavy elements, elements 114 and 116. The suggested names will then go through a review process before adoption by the IUPAC Council.
Review of the claims associated with elements 113, 115, and 118 are at this time not conclusive and evidences have not met the criteria for discovery.
## Synthesis of a new element with atomic number Z=117
A paper has just been accepted (5 April 2010) for publication in Physical Review Letters.1
### International team discovers element 117
A new chemical element has been added to the Periodic Table: A paper on the discovery of element 117 has been accepted for publication in Physical Review Letters.
Oak Ridge National Laboratory is part of a team that includes the Joint Institute of Nuclear Research (Dubna, Russia), the Research Institute for Advanced Reactors (Dimitrovgrad), Lawrence Livermore National Laboratory, Vanderbilt University and the University of Nevada Las Vegas. ORNL's role included production of the berkelium-249 isotope necessary for the target, which was subjected to an extended, months-long run at the heavy ion accelerator facility at Dubna, Russia.
"Without the berkelium target, there could have been no experiment," says ORNL Director of Strategic Capabilities Jim Roberto, who is a principal author on the PRL paper and who helped initiate the experiment. The berkelium was produced at the High Flux Isotope Reactor and processed at the adjoining Radiochemical Engineering & Development Laboratory as part of the most recent campaign to produce californium-252, a radioisotope widely used in industry and medicine.
"Russia had proposed this experiment in 2004, but since we had no californium production at the time, we couldn't supply the berkelium. With the initiation of californium production in 2008, we were able to implement a collaboration," Roberto says.
Professor Joe Hamilton of Vanderbilt University (who helped establish the Joint Institute for Heavy Ion Research at ORNL) introduced Roberto to Yuri Oganessian of Russia's JINR. Five months of the Dubna JINR U400 accelerator's calcium-48 beam - one of the world's most powerful - was dedicated to the project.
The massive effort identified a total of six atoms of element 117 and the critical reams of data that substantiate their existence.
The two-year experimental campaign began with a 250-day irradiation in HFIR, producing 22 milligrams of berkelium-249, which has a 320-day half-life. The irradiation was followed by 90 days of processing at REDC to separate and purify the berkelium. The Bk-249 target was prepared at Dimitrovgrad and then bombarded for 150 days at the Dubna facility. Lawrence Livermore, which now has been involved in the discovery of six elements with Dubna (113, 114, 115, 116, 117, and 118), contributed data analysis, and the entire team was involved in the assessment of the results.
This is the second element that ORNL has had a role in discovering, joining element 61, promethium, which was discovered at the Graphite Reactor during the Manhattan project and reported in 1946. ORNL, by way of its production of radioisotopes for research, has contributed to the discovery of a total of seven new elements.
Members of the ORNL team include the Physics Division's Krzysztof Rykaczewsi, Porter Bailey of the Nonreactor Nuclear Facilities Division, and Dennis Benker, Julie Ezold, Curtis Porter and Frank Riley of the Nuclear S&T Division. Roberto says the success of the element-117 campaign underscores the value of international collaborations in science.
"This use of ORNL isotopes and Russian accelerators is a tremendous example of the value of working together," he says. "The 117 experiment paired one of the world's leading research reactors--capable of producing the berkelium target material--with the exceptional heavy ion accelerator and detection capabilities at Dubna."
#### Islands of Stability
Roberto also says the experiment, in addition to discovering a new chemical element, has pushed the Periodic Table further into the neutron-rich regime for heaviest elements. "New isotopes observed in these experiments continue a trend toward higher lifetimes for increased neutron numbers, providing evidence for the proposed "island of stability" for super-heavy nuclei," he says. "Because the half-lives are getting longer, there is potential to study the chemistry of these nuclei," Roberto says. "These experiments and discoveries essentially open new frontiers of chemistry."
—Bill Cabage
The news about the claim was announced in a press release from the Oak Ridge National Laboratory.
## Tantalising news about element 117
Notes from the 31st meeting of PAC for Nuclear Physics seems to suggest that a claim for element 117 (at the base of the halogen column) may come in the coming weeks and months. It's not very clear which isotopes may have been formed so watch this space.
IV. Experiments on the synthesis of element 117
The PAC heard with great interest the report on the results of the experiment dedicated to the synthesis of element 117 in the 48Ca + 249Bk reaction. The PAC congratulates the staff of the Flerov Laboratory on the discovery of element 117 and new isotopes of elements 115, 113, 111, 109, 107, and 105. The discovery of chains of two neighboring isotopes emphasizes the importance of the odd-even and odd-odd effect for such heavy nuclei. It is in fact especially interesting that the odd-odd chain (3n channel) neighboring to the odd-even chain (4n channel) is twice longer (6 α particles).
## Copernicium confirmed as name of element 112
IUPAC has officially approved the name copernicium, with symbol Cn, for the element of atomic number 112. Priority for the discovery of this element was assigned, in accordance with the agreed criteria, to the Gesellschaft für Schwerionenforschung (GSI) (Center for Heavy Ion Research) in Darmstadt, Germany. The team at GSI proposed the name copernicium which has now been approved by IUPAC. Sigurd Hofmann , leader of the GSI team stated that the intent was to "salute an influential scientist who didn't receive any accolades in his own lifetime, and highlight the link between astronomy and the field of nuclear chemistry."
The name proposed by the Gesselschaft für Schwerionenforschung (GSI) lies within the long tradition of naming elements to honor famous scientists. Nicolaus Copernicus was born on 19 February 1473, in Torún, Poland and died on 24 May 1543, in Frombork/Frauenburg also in Poland. His work has been of exceptional influence on the philosophical and political thinking of mankind and on the rise of modern science based on experimental results. During his time as a canon of the Cathedral in Frauenburg, Copernicus spent many years developing a conclusive model for complex astronomical observations of the movements of the sun, moon, planets and stars. His work published as “De revolutionibus orbium coelestium, liber sixtus” in 1543 had very far reaching consequences. Indeed the Copernican model demanded major changes in the view of the world related to astronomy and physical forces and well as having theological and political consequences. The planetary system introduced by Copernicus has been applied to other analogous systems in which objects move under the influence of a force directed towards a common centre. Notably, on a microscopic scale this is the Bohr model of the atom with its nucleus and orbiting electrons.
The Recommendations will be published in the March issue of the IUPAC journal Pure and Applied Chemistry and is available online at Pure Appl. Chem., 2010, Vol. 82, No. 3, pp. pp 753-755 (doi: 10.1351/PAC-REC-09-08-20)
## Approaches to element 120 (unbinilium)
Attempts have been made at GSI to make element 120 (unbinilium). Several new elements have been made at GSI in the last few years. However after 120 days no decay chain of element 120 was found. With the total number of 2.6 × 1019 projectiles which impinged upon the target, it deduced that the stability in the region around Z=120, N=184 is not exceptionally high with respect to the neighbouring regions.
Currently it is not clear what proton number defines the location of the "island of stability". Various theoretical models suggest numbers of Z=114, 120 or 126. Workers at GSI investigated the element Z=120 (element 120, containing 120 protons within the nucleus). Three different projectile-target combinations all lead to the same compound nucleus 302120 or 302Ubn
• 64Ni + 238U
• 58Fe + 244Pu, and
• 54Cr + 248Cm
The neutron number of the compound nucleus 302120 is N=182. This is only 2 neutrons below N=184 where the neutron shell closure is expected. Therefore, 302120 or 302Ubn is closer to the N=184 shell than any other so far produced compound nucleus with lower Z.
The largest production rate for Z=120 is predicted for the most mass asymmetric projectile/target combination 54Cr + 248Cm. However, at SHIP this experiment was not possible so the reaction 64Ni + 238U was studied. If the proton shell closure is at Z=120 then an enhanced production rate and half-live of the element 120 would be expected. Depending on the magnitude of the stabilization due to the closed shell, one could expect up to a few events per week for the isotopes 299120 and 298120 produced in 64Ni + 238U reactions. The half-lives are expected to be of the order of some 10 μs, but in the end no luck, this time at least.
## WebElements chemistry nexus
OK - started reorganisation of one WebElements area, the "chemistry nexus". The nexus will now bring together news, articles, the WebElements blog, and eventually the Webelements bibliography, and more. It will also contain chemistry content related to WebElements, for instance, sections on the chemistry of the various groups or periods in the periodic table, articles on periodicity, expanded disucssion of the chemistry of specific elements, and so on. All that will take years to add of course. The existing forums site will be located in a dedicated area at the WebElements chemistry forum.
WebElements: the periodic table on the WWW [http://www.webelements.com/] | open-web-math/open-web-math | |
1) It is linear because the voltage increases by 0.41 volts every 2.5 minutes.
2) It is linear because the voltage decreases by 0.8 volts every 0.4 minutes.
3) It is exponential because the voltage decreases by 20% every 2.5 minutes.
4) It is exponential because the voltage increases by 20% every 0.4 minutes.
Solution: Choice C is correct.
We know that, standard form of an exponential equation is y = a . bx.
Therefore, the given function, v = 0.41 . 0.8(2t/5) is exponential and not linear.
Also, since b=0.8 < 1, the function will be a decreasing function (means voltage will decrease as time increases)
Now when 2t/5 increases by 1, the voltage is multiplied by .8 or 80%).
Therefore, the voltage decreases by 20% in that interval.
Now, to find out the interval, 2t/5 = 1
Or t = 2.5 minutes.
Therefore, the given function is exponential, with the voltage decreasing by 20% every 2.5 minutes. | HuggingFaceTB/finemath | |
+0
0
78
3
What is the least integer greater than $\sqrt{300}$?
Jul 6, 2021
#1
+2
Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\sqrt{300}$
Jul 6, 2021
#1
+2
Think of the perfect squares around 300. 17^2 = 289, 18^2 = 324. $$\sqrt{324} = 18$$ as we know, so 18 is the least integer greater than $\sqrt{300}$
Awesomeguy Jul 6, 2021
#2
+1
That was quick, thank you so much!
Guest Jul 6, 2021
#3
0
$\lceil{\sqrt{300}}\rceil = \lceil{\sqrt{3 \cdot 100}}\rceil = \lceil{10 \sqrt{3}}\rceil \approx \lceil{10 \cdot 1.73}\rceil = \lceil{17.3}\rceil = \boxed{18}$ | HuggingFaceTB/finemath | |
A way of doing math. Normal people do math in infix notation, which results in problems with operator precedence and introduces the need for parentheses. Here are some examples of infix notation:
(2 + 3 ) * 5
((4 -2) + (3 * 2))
3 + (2 - ((6 / 3) * 2))
Here are the same examples in postfix notation. Note the lack of parentheses:
2 3 + 5 *
4 2 - 3 2 * +
3 2 6 3 / 2 * - +
This is how reverse polish notation calculators work. Postfix also allows you to write a stack calculator in about two shits. Every time you see a number, push it onto the stack. Every time you see an operator, pop twice, operate on the two numbers, and push the result back on. Just work the input stream left to right and you'll get it.
So, how would we solve "2 3 + 5 *" using a stack? Look at the first part of the expression. It's a number. So we push it onto the stack. The next part is another number. Push it onto the stack as well. Then the next part is an operator (a + sign). So we pop twice, do the operation, and push the result back on. Our stack looked like this before the operation:
2 3
So we'd pop off a 3, then pop off a 2. Add the two numbers and push the result back on. The stack now contains only a 5. Now, look at the next part of the expression. It's another number. Push it on the stack as well. The stack is now two fives. The last part of the expression is a multiplication sign. Pop off the two fives, multiply, and push the result back on. The stack now only contains a 25, which is the answer. Compare that to the result from the infix version. And we got to the answer without order of operations, because we could use a stack.
We can convert between infix and postfix very easily using a binary tree. Just make a tree out of the expression, where the nodes are the operators and the leaves are numbers. A tree for the above example would look like this:
``` *
/ \
+ 5
/ \
2 3
```
The algorithm used to write down a postfix version of that tree goes like so (transferring this to code should take no time at all):
1. Go to the left node and recurse.
2. Go to the right node and recurse.
3. Write down the number or symbol at the current node.
With a program like lex or flex, you can do postfix arithmetic so fast it's not even funny. Many people would say that postfix is much faster and easier. I would agree. Compare to prefix notation. | HuggingFaceTB/finemath | |
Categories
# Length of side of Triangle | PRMO II 2019 | Question 28
Try this beautiful problem from the Pre-RMO, 2017, Question 23, based on Solving Equation. You may use sequential hints to solve the problem.
Try this beautiful problem from the Pre-RMO II, 2019, Question 28, based on Length of side of triangle.
## Length of side of triangle – Problem 28
In a triangle ABC, it is known that $$\angle$$A=100$$^\circ$$ and AB=AC. The internal angle bisector BD has length 20 units. Find the length of BC to the nearest integer, given that sin 10$$^\circ$$=0.174.
• is 107
• is 27
• is 840
• cannot be determined from the given information
### Key Concepts
Equation
Algebra
Integers
But try the problem first…
Source
PRMO II, 2019, Question 28
Higher Algebra by Hall and Knight
## Try with Hints
First hint
given, BD=20 units
$$\angle$$A=100$$^\circ$$
AB=AC
In $$\Delta$$ABD
$$\frac{BD}{sinA}=\frac{AD}{sin20^\circ}$$
or, $$\frac{BD}{sin100^\circ}=\frac{AD}{sin20^\circ}$$
or, 20=$$\frac{AD}{2sin10^\circ}$$ or, AD=40sin10$$^\circ$$=6.96
Second Hint
In $$\Delta$$BDC
$$\frac{BD}{sin40^\circ}=\frac{BC}{sin120^\circ}=\frac{CD}{sin20^\circ}$$
or, CD=$$\frac{20}{2cos20^\circ}$$=$$\frac{20}{2 \times 0.9394}$$=10.65
Final Step
since BD is angle bisector
$$\frac{BC}{AB}=\frac{CD}{AD}$$
or, BC=$$\frac{AB \times CD}{AD}$$=$$\frac{17.6 \times 10.65}{6.96}$$
=26.98=27.
## Subscribe to Cheenta at Youtube
This site uses Akismet to reduce spam. Learn how your comment data is processed. | HuggingFaceTB/finemath | |
# Quadrilaterals : Definition and its Types
In this chapter we will discuss basics of quadrilateral and its types.
Important properties of quadrilateral are also discussed in the chapter so make sure you remember it.
Any two dimensional closed shape with 4 sides are known as Quadrilaterals.
From the above examples we can note that:
(a) All the figure have four sides
(b) The figures can be of different shapes and sizes
Some of the common form of quadrilateral used in geometry are;
(a) Square
(b) Rectangle
(c) Parallelogram
(d) Rhombus
(e) Trapezium
Let us study their property one by one.
## Square
Its a 2D figure of four sides in which all sides are equal and each angle measures 90 degree
Important Points for Square
(a) It has 4 sides
(b) Opposite sides are parallel
(c) Each angle is 90 degree
Given above are two examples of square.
(a) Side Length: All sides are equal
(b) Side Orientation : Opposite sides are parallel
(c) Angles: All angles are 90 degree
∠ A = ∠ B = ∠ C = ∠ D = 90 degree
## Rectangle
Its a quadrilateral in which Opposite sides are equal and parallel and all angles measure 90 degree
Given above is the rectangle ABCD in which:
(a) Side Length: Opposite sides are equal
AB = CD and AD = BC
(b) Side Orientation : Opposite sides are parallel
(c) Angles: All angles are 90 degree
∠ A = ∠ B = ∠ C = ∠ D = 90 degree
## Parallelogram
Parallelogram is a quadrilateral in which Opposite sides are equal and parallel. Also in parallelogram opposite angles are equal.
The above parallelogram has following features;
(a) Side Length: Opposite sides are equal
AB = CD and AD = BC
(b) Side Orientation : Opposite sides are parallel
(c) Angles: Opposite angles are equal
∠ D = ∠B
∠ A = ∠ C
## Rhombus
Rhombus is a quadrilateral with following properties;
(a) All sides are equal
(b) Opposite sides are parallel
(c) Opposite angles are equal
Rhombus can also be imagined as tilted square in which sides has been tilted at certain angle.
The quadrilateral looks like a star.
The above rhombus ABCD has following features;
(a) Side Length: All sides are equal
(b) Side Orientation: Opposite sides are parallel
AB parallel to CD
CB parallel to DA
(c) Angles : Opposite angles are equal
∠ D = ∠B
∠ A = ∠ C
## Trapezium
Trapezium is a quadrilateral with following properties;
(a) one pair of opposite sides are parallel
(b) the other pairs are non parallel
Given above is figure of trapezium ABCD with following features;
Sides AB & CD are parallel pairs
Sides AD and BC are non parallel pairs
Solution
The Quadrilateral is a square as:
⟹ Its opposite sides are equal and parallel
⟹ All angles are 90 degree
(02) Identify the below image
Solution
The Quadrilateral is a square as:
⟹ all sides are equal
⟹ all angles are 90 degrees
Solution
The given quadrilateral is a trapezium because:
(a) One pair of opposite sides are parallel ( AD II BC)
(b) Other pairs of sides are not parallel
(04) Given below is the diagram of Rhombus.
The side AB = 6 cm, find the length of side CD
In Rhombus all sides are equal.
Hence, AB = BC = CD = DA = 6 cm
(05) Given below is the figure of parallelogram.
Here ∠B = 75 degree. Find value of ∠D
Solution
We know that in parallelogram opposite angles are equal to each other.
So ∠B = ∠D = 75 degree
(06) Identify the below figure
The above quadrilateral is Rhombus because:
(a) All sides are equal
(b) Opposite sides are parallel
(c) Opposite angles are equal
(07) Below is the image of rectangle
Find the value of all angles
In rectangle all the angles are 90 degree
Hence, ∠A = ∠B = ∠C = ∠D = 90 degree
(08) Two squares ABCD & EFGH are joined to form rectangle PQRS
Its given that AD = 2 cm, find the length of side PQ of the rectangle
Its given that ABCD is a square.
We know that in square, all sides are equal.
Hence, AD = CD = 2 cm
Similarly EFGH is a square with side 2 cm
From the image we can observe that:
⟹ DC + HG = SR
⟹ 2 + 2 = SR
⟹ SR = 4 cm
Since PQSR is a rectangle, the opposite sides are equal
PQ = SR
PQ = 4 cm
Hence, length of side PQ is 4 cm
(09) Identify the below shape
Solution
From the above image you can observe that:
Side AB = CD | HuggingFaceTB/finemath | |
# Energy & Momentum Thinking Problem
Nicolaus
## Homework Statement
Two boxes each of mass 12kg are raised 1.8m to a shelf. The first one is lifted and the second is pushed up a smooth ramp. If the applied force on the second box is 48N, calculate the angle between the ramp and the ground.
## Homework Equations
W = Eg = mgh for first (lifted) box
Trig to calculate angle
## The Attempt at a Solution
I first calculated the gravitational energy on the first box that is lifted:
Eg = (12kg)(9.8m/s^2)(1.8m) = 211.7N
Then, knowing that, used trig to calculate the angle between ramp and ground:
Sin (theta) = opposite (211.7N)/hypoteneuse(48N)
and, naturally, this does not compute, so where did I go wrong?
## Answers and Replies
Mentor
I first calculated the gravitational energy on the first box that is lifted:
Eg = (12kg)(9.8m/s^2)(1.8m) = 211.7N
That's an energy, not a force. Its units are Joules, not Newtons.
Then, knowing that, used trig to calculate the angle between ramp and ground:
Sin (theta) = opposite (211.7N)/hypoteneuse(48N)
and, naturally, this does not compute, so where did I go wrong?
Not sure what you are trying to do. Instead, examine the forces acting on the box as it is pushed up the ramp. (Assume the force applied is just enough to slide it up the ramp.)
Hint: What's the component of the box's weight parallel to the ramp?
Nicolaus
Whoops, it's been a while. Anyways, I got:
Fnet = Fa - Fg(parallel)
= 48N - (12)(8.8)sin(angle)
= 48N - 117.6Nsin(a)
sin(a) = 48/117.6
angle = 24.1 degrees?
Mentor
Whoops, it's been a while. Anyways, I got:
Fnet = Fa - Fg(parallel)
= 48N - (12)(8.8)sin(angle)
= 48N - 117.6Nsin(a)
sin(a) = 48/117.6
angle = 24.1 degrees?
Good! | HuggingFaceTB/finemath | |
# [isabelle] Problem with Int.int_ge_induct
From my previous posts you may be aware that I am still looking for the best possible way to do the kind of mathematical induction that is typically taught first in a course on discrete mathematics. After looking over the various replies I have gotten, I would like very much, for most of this, to use Int.int_ge_induct: "?k <= ?i ==> ?P ?k ==> (!!i. ?k <= i ==> ?P i ==> ?P (i + 1)) ==> ?P ?i". This assumes that the induction variable is an int, not a nat, so inductions starting at negative numbers can be covered, as well as numbers greater than zero. It can also be applied to course-of-values induction. Having made this decision, I now tried a very simple case, namely proving by induction that if x >= 0, then x <= x*x. Most of what Int.int_ge_induct is supposed to be doing seems to be working fine, but, at the end, I am getting the message: "Failed to finish proof: goal (1 subgoal): 1. 0 <= 0". Well, it's true that (0::int) <= 0 would be true by simp, but 0 <= 0 is not (and it makes sense to me that it is not, since 0 might be a label, for instance). But my question now is: Why am I getting 0 <= 0 in that form? Here is my proof, in which I have replaced every occurrence of 0 by (0::int), in order to try (unsuccessfully) to stop generating a subgoal involving 0 rather than 0::int:
```theory IntInduct imports Main begin
lemma "[|x >= (0::int)|] ==> (x::int) <= x*x" proof-
assume "x >= (0::int)"
then have 1: "(0::int) <= x" by simp
have 2: "[|(0::int) <= (0::int)|] ==> (0::int) <= (0::int)*(0::int)" by simp
```
have 3: "!!i. (0::int) <= i ==> (i::int) <= i*i ==> (i::int)+1 <= (i+1)*(i+1)" sorry
```from 1 and 2 and 3 show "(x::int) <= x*x" by (rule Int.int_ge_induct)
qed
end
```
I have also tried this starting with lemma fixes x::int assumes 4: "x >= (0::int)" shows "x <= x*x" proof- from 4 have 1: "(0::int) <= x" by simp
```with the same result. -Douglas
```
P. S. Might there be a problem with Int.int_ge_induct having a variable called ?i and a (presumably different) bound variable called i? -WDM
```--
Prof. W. Douglas Maurer Washington, DC 20052, USA
Department of Computer Science Tel. (1-202)994-5921
The George Washington University Fax (1-202)994-4875
```
This archive was generated by a fusion of Pipermail (Mailman edition) and MHonArc. | HuggingFaceTB/finemath | |
Wednesday, February 8, 2023, 10:40 AM
Site: Learnbps
Course: BPSS (MAT) Mathematics Standards (S-MAT)
M
# Geometry
## Narrative for the (G) Geometry
Grade 4 students describe, analyze, compare, and classify two-dimensional shapes by their properties, including explicit use of angle sizes and the related geometric properties of perpendicularity and parallelism.
By the end of Grade 4, students develop explicit awareness of and vocabulary for many concepts they have been developing, including points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Such mathematical terms are useful in communicating geometric ideas, but more important is that constructing examples of these concepts, such as drawing angles and triangles that are acute, obtuse, and right, help students form richer concept images connected to verbal definitions.
## Calculation Method for Domains
Domains are larger groups of related standards. The Domain Grade is a calculation of all the related standards. Click on the standard name below each Domain to access the learning targets and rubrics/ proficiency scales for individual standards within the domain.
#### MAT-04.G.01
Under Development
MAT-04 Targeted Standards(G) Domain: GeometryCluster: Draw and identify lines and angles, and classify shapes by properties of their lines and angles. MAT-04.G.01 Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures.
• I can
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## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-04.G.02 4th Grade (MAT) Targeted Standard
(G) Geometry
Cluster: Draw and identify lines and angles, and classify shapes by properties of their lines and angles.
#### MAT-04.G.03
Under Development
MAT-04 Targeted Standards(G) Domain: GeometryCluster: Draw and identify lines and angles, and classify shapes by properties of their lines and angles. MAT-04.G.03 Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify line-symmetric figures and draw lines of symmetry.
• I can
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## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
• List
# Measurement and Data
## Narrative for the (MD) Measurement and Data
In Grade 4, students build on competencies in measurement and in building and relating units and units of units that they have developed in number, geometry, and geometric measurement.
Fourth graders learn the relative sizes of measurement units within a system of measurement4.MD.1 including: length: meter (m), kilometer (km), centimeter (cm), millimeter (mm); volume: liter (l), milliliter (ml, 1 cubic centimeter of water; a liter, then, is 1000 ml); mass: gram (g, about the weight of a cc of water), kilogram (kg); time: hour (hr), minute (min), second (sec).
Students learn to consider perimeter and area of rectangles, begun in Grade 3, more abstractly. Based on work in previous grades with multiplication, spatially structuring arrays, and area, they abstract the formula for the area of a rectangle.Students learn to apply these understandings and formulas to the solution of real-world and mathematical problems.
Students can use these skills to solve problems, including problems that arise from analyzing line plots. For example, with reference to the line plot above, students might find the difference between the greatest and least values in the data.
## Calculation Method for Domains
Domains are larger groups of related standards. The Domain Grade is a calculation of all the related standards. Click on the standard name below each Domain to access the learning targets and rubrics/ proficiency scales for individual standards within the domain.
#### MAT-04.MD.01
Under Development
MAT-04 Targeted Standards(MD) Domain: Measurement and DataCluster: Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. MAT-04.MD.01 Know relative sizes of measurement units within one system of units including km, m, cm; kg, g; lb, oz.; l, ml; hr, min, sec. Within a single system of measurement, express measurements in a larger unit in terms of a smaller unit. Record measurement equivalents in a two-column table. For example, know that 1 ft is 12 times as long as 1 in. Express the length of a 4 ft snake as 48 in. Generate a conversion table for feet and inches listing the number pairs (1, 12), (2, 24), (3, 36), ...
• I can
• I can
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• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-04.MD.02
Under Development
MAT-04 Targeted Standards(MD) Domain: Measurement and DataCluster: Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. MAT-04.MD.02 Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale.
• I can
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## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-04.MD.03
Under Development
MAT-04 Targeted Standards(MD) Domain: Measurement and DataCluster: Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. MAT-04.MD.03 Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor.
• I can
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• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-04.MD.04
Under Development
MAT-04 Targeted Standards(MD) Domain: Measurement and DataCluster: Represent and interpret data. MAT-04.MD.04 Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using information presented in line plots. For example, from a line plot find and interpret the difference in length between the longest and shortest specimens in an insect collection.
• I can
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• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-04.MD.05
Under Development
MAT-04 Targeted Standards(MD) Domain: Measurement and DataCluster: Geometric measurement: understand concepts of angle and measure angles. MAT-04.MD.05 Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement:
• I can
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## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. | HuggingFaceTB/finemath | |
< 1 >
### Torricelli's trumpet
Torricelli's trumpet (also Gabriel's horn) has a finite volume and an infinite surface.
##### Explanation
We start from the function and use the graph, in which we avoid dividing by zero. We rotate the function around the x-axis and calculate the volume and the surface between x = 1 and x = a where a > 1. The drawing gives an impression of the trumpet.
##### Volume
The formula for the volume is so applies ##### Surface
The formula for the surface area is so applies ##### Paradox
It seems like a paradox, that the volume of this trumpet has a real value but the surface is infinitely large. With mathematics, you can calculate things which we cannot comprehend intuitively. The picture is just a tool. The trumpet does not end with a mouthpiece but always continues. The calculation shows that everywhere a surface exists but no volume is added anymore
##### Additional information
This figure was studied by Italian physicist Evangelista Torricelli in the 17th century. | HuggingFaceTB/finemath | |
# Relation between linear transformations and traces
a) Let $A \in M_n (K)$. We denote $f_A$ the linear form defined, for every $X \in M_n (K)$, by $f_A(X)=Tr(AX)$. Show that the function $f$ which maps $A \in M_n (K)$ to $f_A$ is an isomorphism between $M_n (K)$ and its dual.
b) Let $f: M_n (K) \rightarrow K$ be a linear form such that, for every $(X,Y)$ in $M_n (K)^2$, $f(XY)=f(YX)$. Show that there exists $\lambda \in K$ such that for every $X \in M_n (K)$, $f(X)=\lambda Tr(X)$
That is what I have:
a) Let $(E_{ij})_{1\leq i,j \leq n}$ be the standard basis for $M_n (K)$. Let us start by showing that for every $1 \leq i,j,k,l \leq n$, we have $E_{ij} E_{kl}= \delta_{jk} E_{il}$
We have $E_{ij}=(\delta_{pi} \delta_{qj})_{1 \leq p,q \leq n}$ and $E_{kl}=(\delta_{pk} \delta_{ql})_{1 \leq p,q \leq n}$ $A= E_{ij} E_{kl}= (a_{p,q})$ such that: $a_{p,q}= \sum \limits_{r=1}^n (\delta_{pi} \delta_{rj})(\delta_{rk} \delta_{ql})=( \sum \limits_{r=1}^n \delta_{rj} \delta_{rk})\delta_{pi} \delta_{ql}= \delta_{jk} \delta_{pi} \delta_{ql}$
Hence: $E_{ij} E_{kl}= \delta_{jk} E_{il}$ $$\begin{array}{c}\\\\\end{array}$$
It is obvious that $f$ is linear. Therefore, for dimensional reasons, we must only show that $f$ is injective. Let $A= (a_{ij})_{1\leq i,j \leq n}$ such that $f_A=0$. We therefore have, for $1 \leq i_0, j_0 \leq n$
$0= Tr(AE_{i_0 j_0})= Tr(\sum \limits_{1 \leq i,j \leq n} a_{ij}E_{ij}E{i_0 j_0})= Tr(\sum \limits_{i=1}^n a_{i i_{0}}E_{i i_0} E_{i_0 j_0})= \sum \limits_{i=1}^n a_{i i_0} Tr(E_{i j_0})=a_{j_{0} i_{0}}$
Hence $A$ equals zero.
Therefore, $f_A$ is an isomorphism. $$\begin{array}{c}\\\\\end{array}$$
Can someone help me with b?
-
What have you tried? – A.P. Apr 6 '13 at 19:35
I am having trouble with question b – Carpediem Apr 6 '13 at 19:51
b) By assumption, $f$ is a linear form so by a) we have $A$ such that $f=f_A$.
Then for every $i\neq j$: $$a_{ji}=\mbox{Tr}(AE_{ij})=\mbox{Tr}(AE_{ii}E_{ij})=f_A(E_{ii}E_{ij})=f_A(E_{ij}E_{ii})=f_A(0)=0$$ and $$a_{ii}=\mbox{Tr}(AE_{ii})=f_A(E_{ii})=f_A(E_{ij}E_{ji})=f_A(E_{ji}E_{ij})=f_A(E_{jj})=\mbox{Tr}(AE_{jj})=a_{jj}.$$ Hence $A$ is a scalar matrix, which yields the result.
-
Ok. What do you think of my answer below? I tried something.. – Carpediem Apr 6 '13 at 19:54
Motivation for the downvote? Just because the other answer is "better"? If so, it suffices to upvote the other answer. – 1015 Apr 6 '13 at 20:38
@julien I didn't downvote you – user43418 Apr 6 '13 at 23:01
@user43418 Ok. The mute downvoter will remain in the dark. – 1015 Apr 6 '13 at 23:08
Let $A \in M_n (K)$ such that $f=f_A$. We have, for every $(X,Y) \in M_n (K)^2$, $Tr(AXY)=Tr(AYX)$. Since $Tr(AYX)=Tr(XAY)$, we can deduce that $Tr((AX-XA)Y)=0$. Since this is true for every matrix $Y$, we have, according to question a), $AX=XA$. Therefore $A$ is commutative with every matrix $X$.
Let us show that $A$ is a scalar matrix i.e. a scalar multiple of the identity matrix. If $A= (a_{ij})_{1 \leq i,j \leq n}$, we have for every $1 \leq i,j \leq n$,
$AE_{ij}=\sum \limits_{1 \leq k,l \leq n} a_{kl} E_{kl} E_{ij}= \sum \limits_{k=1}^n a_{ki}E_{kj}=E_{ij}A= \sum \limits_{1 \leq k,l \leq n} a_{kl} E_{ij} E_{kl}= \sum \limits_{l=1}^n a_{jl} E_{il}$
By uniqueness of the writing, we obtain $a_{ki}=0$ for $k \neq i$ and $a_{ii}=a_{jj}$: A is therefore a scalar matrix. Hence, $f=f_A$ is collinear to the trace. Therefore, there exists $\lambda \in K$ such that for every $X \in M_n (K)$, $f(X)=\lambda Tr(X)$
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Yep, that's nice. – 1015 Apr 6 '13 at 20:01 | open-web-math/open-web-math | |
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A259907 Fifth differences of 7th powers (A001015). 1
1, 123, 1557, 6719, 16800, 31920, 52080, 77280, 107520, 142800, 183120, 228480, 278880, 334320, 394800, 460320, 530880, 606480, 687120, 772800, 863520, 959280, 1060080, 1165920, 1276800, 1392720, 1513680, 1639680, 1770720, 1906800, 2047920, 2194080, 2345280, 2501520, 2662800 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 REFERENCES John H. Conway and Richard K. Guy, The Book of Numbers. New York: Springer-Verlag, pp. 30-32, 1996. Kiran Parulekar. Amazing Properties of Squares and Their Calculations. Kiran Anil Parulekar, 2012. Bag, Amulya Kumar (1966). "Binomial theorem in ancient India". Indian J. History Sci 1 (1): 68-74. Ronald Graham and Donald Knuth, Patashnik, Oren (1994). "(5) Binomial Coefficients". Concrete Mathematics (2nd ed.). Addison Wesley. pp. 153-256. LINKS R. J. Mathar, Table of n, a(n) for n = 0..79 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA G.f.: (1 + 120*x + 1191*x^2 + 2416*x^3 + 1191*x^4 + 120*x^5 + x^6)/(1 - x)^3. a(n) = 840*(3*n^2 - 9*n + 8) for n>3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>6. - Vincenzo Librandi, Jul 08 2015 EXAMPLE 1 128 2187 16384 78125 279936 823543 2097152 4782969 (seventh powers) 1 127 2059 14197 61741 201811 543607 1273609 2685817 (first differences) 1 126 1932 12138 47544 140070 341796 730002 1412208 (second differences) 1 125 1806 10206 35406 92526 201726 388206 682206 (third differences) 1 124 1681 8400 25200 57120 109200 186480 294000 (fourth differences) 1 123 1557 6719 16800 31920 52080 77280 107520 (here) MATHEMATICA Join[{1, 123, 1557, 6719}, Table[840 (3 n^2 - 9 n + 8), {n, 4, 40}]] PROG (Sage) [1, 123, 1557, 6719]+[840*(3*n^2-9*n+8) for n in (4..40)] # Bruno Berselli, Jul 16 2015 (MAGMA) [1, 123, 1557, 6719] cat [840*(3*n^2-9*n+8): n in [4..40]]; // Bruno Berselli, Jul 16 2015 CROSSREFS Cf. A001015, A022523, A255177, A255181. Sequence in context: A265983 A103504 A324627 * A115983 A202132 A167231 Adjacent sequences: A259904 A259905 A259906 * A259908 A259909 A259910 KEYWORD nonn,easy AUTHOR Kolosov Petro, Jul 07 2015 EXTENSIONS Edited by Editors of the OEIS, Jul 16 2015 STATUS approved
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Last modified January 16 16:17 EST 2022. Contains 350376 sequences. (Running on oeis4.) | HuggingFaceTB/finemath | |
# Answers
Solutions by everydaycalculation.com
## Reduce 336/108 to lowest terms
The simplest form of 336/108 is 28/9.
#### Steps to simplifying fractions
1. Find the GCD (or HCF) of numerator and denominator
GCD of 336 and 108 is 12
2. Divide both the numerator and denominator by the GCD
336 ÷ 12/108 ÷ 12
3. Reduced fraction: 28/9
Therefore, 336/108 simplified to lowest terms is 28/9.
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
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Equivalent fractions:
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#### Fractions Simplifier
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## Understanding Math: Conclusion
Click to read the earlier posts in this series: Understanding Math, Part 1: A Cultural Problem; Understanding Math, Part 2: What Is Your Worldview?; Understanding Math, Part 3: Is There Really a Difference?; Understanding Math, Part 4: Area of a Rectangle; Understanding Math, Part 5: Multiplying Fractions; and Understanding Math, Part 6: Algebraic Multiplication.
Earlier in this blog post series, I gave you three middle-school math rules. But by exploring the concept of rectangular area as a model of multiplication, we discovered that in a way they were all the same.
The rules are not arbitrary, handed down from a mathematical Mount Olympus. They are three expressions of a single basic question: what does it mean to measure area?
There was only one rule, one foundational pattern that tied all these topics together in a mathematical web.
Many children want to learn math instrumentally, as a tool for getting answers. They prefer the simplicity of memorizing rules to the more difficult task of making sense of new ideas. Being young, they are by nature short-term thinkers. They beg, “Just tell me what to do.”
But if we want our children to truly understand mathematics, we need to resist such shortcuts. We must take time to explore mathematics as a world of ideas that connect and relate to each other in many ways. And we need to show children how to reason about these interconnected concepts, so they can use them to think their way through an ever-expanding variety of problems.
Our kids can only see the short term. If we adults hope to help them learn math, our primary challenge is to guard against viewing the mastery of facts and procedures as an end in itself. We must never fall into thinking that the point of studying something is just to get the right answers.
We understand this in other school subjects. Nobody imagines that the point of reading is to answer comprehension questions. We know that there is more to learning history than winning a game of Trivial Pursuit. But when it comes to math, too many parents (and far too many politicians) act as though the goal of our children’s education is to produce high scores on a standardized test.
### What If I Don’t Understand Math?
If you grew up (as I did) thinking of math as a tool, the instrumental approach may feel natural to you. The idea of math as a cohesive system may feel intimidating. How can we parents help our children learn math, if we never understood it this way ourselves?
Don’t panic. Changing our worldview is never easy, yet even parents who suffer from math anxiety can learn to enjoy math with their children. All it takes is a bit of self-discipline and the willingness to try.
You don’t have to know all the answers. In fact, many people have found the same thing that Christopher Danielson described in his blog post “Let the children play” — the more we adults tell about a topic, the less our children learn. With the best of intentions we provide information, but we unwittingly kill their curiosity.
• If you’re afraid of math, be careful to never let a discouraging word pass your lips. Try calling upon your acting skills to pretend that math is the most exciting topic in the world.
• Encourage your children to notice the math all around them.
• Search out opportunities to discuss numbers, shapes, symmetry, and patterns with your kids.
• Investigate, experiment, estimate, explore, measure — and talk about it all.
### The Science of Patterns
Patterns are so important that American mathematician Lynn Arthur Steen defined mathematics as the science of patterns.
As biology is the science of life and physics the science of energy and matter, so mathematics is the science of patterns. We live in an environment steeped in patterns — patterns of numbers and space, of science and art, of computation and imagination. Patterns permeate the learning of mathematics, beginning when children learn the rhythm of counting and continuing through times tables all the way to fractals and binomial coefficients.
— Lynn Arthur Steen, 1998
Reflections on Mathematical Patterns, Relationships, and Functions
If you are intimidated by numbers, you can look for patterns of shape and color. Pay attention to how they grow, and talk about what your children notice. For example, some patterns repeat exactly, while other patterns change as they go (small, smaller, smallest, or loud, louder, loudest).
Nature often forms fractal-like patterns: the puffy round-upon-roundness of cumulus clouds or broccoli, or the branch-upon-branchiness of a shrub or river delta. Children can learn to recognize these, not as a homework exercise but because they are interesting.
### Math the Mathematician’s Way
Here is the secret solution to the crisis of math education: we adults need to learn how to think like mathematicians.
For more on what it means to think about math the mathematician’s way, check out my Homeschooling with Math Anxiety blog post series:
As we cultivate these characteristics, we will help our children to recognize and learn true mathematics.
CREDITS: “Frabjous 01” photo (top) by Windell Oskay and “Back to School” photo (middle) by Phil Roeder via Flicker (CC BY 2.0). “I Can Solve Problems” poster by Nicole Ricca via Teachers Pay Teachers. This is the final post in my Understanding Math series, adapted from the expanded paperback edition of Let’s Play Math: How Families Can Learn Math Together and Enjoy It. Coming in late February 2016 to your favorite online bookstore…
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## Understanding Math: Algebraic Multiplication
Click to read the earlier posts in this series: Understanding Math, Part 1: A Cultural Problem; Understanding Math, Part 2: What Is Your Worldview?; Understanding Math, Part 3: Is There Really a Difference?; Understanding Math, Part 4: Area of a Rectangle; and Understanding Math, Part 5: Multiplying Fractions.
We’ve examined how our vision of mathematical success shapes our children’s learning. Do we think math is primarily a tool for solving problems? Or do we see math as a web of interrelated concepts?
Instrumental understanding views math as a tool. Relational understanding views math as an interconnected system of ideas. Our worldview influences the way we present math topics to our kids. And our children’s worldview determines what they remember.
In the past two posts, we looked at different ways to understand and teach rectangular area and fraction multiplication. But how about algebra? Many children (and adults) believe “math with letters” is a jumble of abstract nonsense, with too many formulas and rules that have to be memorized if you want to pass a test.
Which of the following sounds the most like your experience of school math? And which type of math are your children learning?
### Instrumental Understanding: FOIL
Every mathematical procedure we learn is an instrument or tool for solving a certain kind of problem. To understand math means to know which tool we are supposed to use for each type of problem and how to use that tool — how to categorize the problem, remember the formula, plug in the numbers, and do the calculation.
When you need to multiply algebra expressions, remember to FOIL: multiply the First terms in each parenthesis, and then the Outer, Inner, and Last pairs, and finally add all those answers together.
### Relational Understanding: The Area Model
Each mathematical concept is part of a web of interrelated ideas. To understand mathematics means to see at least some of this web and to use the connections we see to make sense of new ideas.
The concept of rectangular area has helped us understand fractions. Let’s extend it even farther. In the connected system of mathematics, almost any type of multiplication can be imagined as a rectangular area. We don’t even have to know the size of our rectangle. It could be anything, such as subdividing a plot of land or designing a section of crisscrossed colors on plaid fabric.
We can imagine a rectangle with each side made up of two unknown lengths. One side has some length a attached to another length b. The other side is x units long, with an extra amount y stuck to its end.
We don’t know which side is the “length” and which is the “width” because we don’t know which numbers the letters represent. But multiplication works in any order, so it doesn’t matter which side is longer. Using the rectangle model of multiplication, we can see that this whole shape represents the area $\left ( a+b \right )\left ( x+y \right )$ .
But since the sides are measured in pieces, we can also imagine cutting up the big rectangle. The large, original rectangle covers the same amount of area as the four smaller rectangular pieces added together, and thus we can show that $\left ( a+b \right )\left ( x+y \right )=ax+ay+bx+by$ .
With the FOIL formula mentioned earlier, our students may get a correct answer quickly, but it’s a dead end. FOIL doesn’t connect to any other math concepts, not even other forms of algebraic multiplication. But the rectangular area model will help our kids multiply more complicated algebraic expressions such as $\left ( a+b+c \right )\left ( w+x+y+z \right )$ .
Not only that, but the rectangle model gives students a tool for making sense of later topics such as polynomial division. And it is fundamental to understanding integral calculus.
To be continued. Next up, Understanding Math Part 7: The Conclusion…
CREDITS: “Math Workshop Portland” photo (top) by US Department of Education via Flicker (CC BY 2.0). This is the sixth post in my Understanding Math series, adapted from the expanded paperback edition of Let’s Play Math: How Families Can Learn Math Together and Enjoy It. Coming in early 2016 to your favorite online bookstore…
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## Understanding Math: Multiplying Fractions
Click to read the earlier posts in this series: Understanding Math, Part 1: A Cultural Problem; Understanding Math, Part 2: What Is Your Worldview?; Understanding Math, Part 3: Is There Really a Difference?; and Understanding Math, Part 4: Area of a Rectangle.
In this post, we consider the second of three math rules that most of us learned in middle school.
• To multiply fractions, multiply the tops (numerators) to make the top of your answer, and multiply the bottoms (denominators) to make the bottom of your answer.
### Instrumental Understanding: Math as a Tool
Fractions confuse almost everybody. In fact, fractions probably cause more math phobia among children (and adults) than any other topic before algebra.
Children begin learning fractions by coloring or cutting up paper shapes, and their intuition is shaped by experiences with food like sandwiches or pizza. But before long, the abstraction of written calculations looms up to swallow intuitive understanding.
Upper elementary and middle school classrooms devote many hours to working with fractions, and still students flounder. In desperation, parents and teachers resort to nonsensical mnemonic rhymes that just might stick in a child’s mind long enough to pass the test.
### Relational Understanding: Math as a Connected System
Do you remember our exploration of the area of a rectangular tabletop?
Now let’s zoom in on our rectangle. Imagine magnifying our virtual grid to show a close-up of a single square unit, such as the pan of brownies on our table. And we can imagine subdividing this square into smaller, fractional pieces. In this way, we can see that five-eighths of a square unit looks something like a pan of brownies cut into strips, with a few strips missing:
But what if we don’t even have that whole five-eighths of the pan? What if the kids came through the kitchen and snatched a few pieces, and now all we have is three-fourths of the five-eighths?
How much of the original pan of brownies do we have now? There are three rows with five pieces in each row, for a total of 3 × 5 = 15 pieces left — which is the numerator of our answer. And with pieces that size, it would take four rows with eight in each row (4 × 8 = 32) to fill the whole pan — which is our denominator, the number of pieces in the whole batch of brownies. So three-fourths of five-eighths is a small rectangle of single-serving pieces.
Notice that there was nothing special about the fractions 3/4 and 5/8, except that the numbers were small enough for easy illustration. We could imagine a similar pan-of-brownies approach to any fraction multiplication problem, though the final pieces might turn out to be crumbs.
Of course, children will not draw brownie-pan pictures for every fraction multiplication problem the rest of their lives. But they need to spend plenty of time thinking about what it means to take a fraction of a fraction and how that meaning controls the numbers in their calculation. They need to ask questions and to put things in their own words and wrestle with the concept until it makes sense to them. Only then will their understanding be strong enough to support future learning.
CREDITS: “School Discussion” photo (top) by Flashy Soup Can via Flicker (CC BY 2.0). This is the fifth post in my Understanding Math series, adapted from the expanded paperback edition of Let’s Play Math: How Families Can Learn Math Together and Enjoy It. Coming in early 2016 to your favorite online bookstore…
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## Understanding Math: Area of a Rectangle
Click to read the earlier posts in this series: Understanding Math, Part 1: A Cultural Problem; Understanding Math, Part 2: What Is Your Worldview?; and Understanding Math, Part 3: Is There Really a Difference?
In this post, we consider the first of three math rules that most of us learned in middle school.
• Area of a rectangle = length × width
### Instrumental Understanding: Math as a Tool
The instrumental approach to explaining such rules is for the adult to work through a few sample problems and then give the students several more for practice.
In a traditional lecture-and-workbook style curriculum, students apply the formula to drawings on paper. Under a more progressive reform-style program, the students may try to invent their own methods before the teacher provides the standard rule, or they may measure and calculate real-world areas such as the surface of their desks or the floor of their room.
Either way, the ultimate goal is to define terms and master the formula as a tool to calculate answers.
Richard Skemp describes a typical lesson:
Suppose that a teacher reminds a class that the area of a rectangle is given by A=L×B. A pupil who has been away says he does not understand, so the teacher gives him an explanation along these lines. “The formula tells you that to get the area of a rectangle, you multiply the length by the breadth.”
“Oh, I see,” says the child, and gets on with the exercise.
If we were now to say to him (in effect) “You may think you understand, but you don’t really,” he would not agree. “Of course I do. Look; I’ve got all these answers right.”
Nor would he be pleased at our devaluing of his achievement. And with his meaning of the word, he does understand.
As the lesson moves along, students will learn additional rules.
For instance, if a rectangle’s length is given in meters and the width in centimeters, we must convert them both to the same units before we calculate the area. Also, our answer will not have the same units as our original lengths, but that unit with a little, floating “2” after it, which we call “squared.”
Each lesson may be followed by a section on word problems, so the students can apply their newly learned rules to real-life situations.
### Relational Understanding: Math as a Connected System
In contrast, a relational approach to area must begin long before the lesson on rectangles.
Again, this can happen in a traditional, teacher-focused classroom or in a progressive, student-oriented, hands-on environment. Either way, the emphasis is on uncovering and investigating the conceptual connections that lie under the surface and support the rules.
We start by exploring the concept of measurement: our children measure a path along the floor, sidewalk, or anywhere we could imagine moving in a straight line. We learn to add and subtract such distances. Even if our path turns a corner or if we first walk forward and then double back, it’s easy to figure out how far we have gone.
But something strange happens when we consider distances in two different directions at the same time — measuring the length and width of the dining table automatically creates an invisible grid.
In measuring the length of a rectangular table, we do not find just one point at any given distance. There is a whole line of points that are one foot, two feet, or three feet from the left side of the table.
And measuring the width shows us all the points that are one, two, or three feet from the near edge. Now our rectangular table is covered by virtual graph paper with squares the size of our measuring unit.
The length of the rectangle tells us how many squares we have in each row, and the width tells us how many rows there are. As we imagine this invisible grid, we can see why multiplying those two numbers will tell us how many squares there are in all.
That is what the word area means: the area of a tabletop is the number of virtual-graph-paper squares it takes to cover it up, which is why our answer will be measured in square units.
### Making Sense of Mixed Units
What if we measured the length in meters and the width in centimeters?
With a relational understanding of area, even a strange combination of units can make sense. Our invisible grid would no longer consist of squares but of long, thin, rectangular centimeter-meters. But we could still find the area of the tabletop by counting how many of these units it takes to cover it.
Square units aren’t magic — they’re just easier, that’s all.
Click to continue reading Understanding Math, Part 5: Multiplying Fractions
CREDITS: “Framed” photo (top) by d_pham via Flicker (CC BY 2.0). This is the fourth post in my Understanding Math series, adapted from the expanded paperback edition of Let’s Play Math: How Families Can Learn Math Together and Enjoy It. Coming in early 2016 to your favorite online bookstore…
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## Understanding Math: Is There Really a Difference?
Click to read the earlier posts: Understanding Math, Part 1: A Cultural Problem; Understanding Math, Part 2: What Is Your Worldview?
From the outside, it’s impossible to tell how a person is thinking. A boy with the instrumental perspective and a girl who reasons relationally may both get the same answers on a test. Yet under the surface, in their thoughts and how they view the world, they could not be more different.
“Mathematical thinking is more than being able to do arithmetic or solve algebra problems,” says Stanford University mathematician and popular author Keith Devlin. “Mathematical thinking is a whole way of looking at things, of stripping them down to their numerical, structural, or logical essentials, and of analyzing the underlying patterns.”
And our own mathematical worldview will influence the way we present math topics to our kids. Consider, for example, the following three rules that most of us learned in middle school.
• Area of a rectangle = length × width.
• To multiply fractions, multiply the tops (numerators) to make the top of your answer, and multiply the bottoms (denominators) to make the bottom of your answer.
• When you need to multiply algebra expressions, remember to FOIL: multiply the First terms in each parenthesis, and then the Outer, Inner, and Last pairs, and finally add all those answers together.
While the times symbol or the word multiply is used in each of these situations, the procedures are completely different. How can we help our children understand and remember these rules?
Over the next three posts in this series, we’ll dig deeper into each of these math rules as we examine what it means to develop relational understanding.
Many people misunderstand the distinction between Instrumental and Relational Understanding as having to do with surface-level, visible differences in instructional approach, but it’s not that at all. It has nothing to do with our parenting or teaching style, or whether our kids are learning with a traditional textbook or through hands-on projects. It’s not about using “real world” problems, except to the degree that the world around us feeds our imagination and gives us the ability to think about math concepts.
This dichotomy is all about the vision we have for our children — what we imagine mathematical success to look like. That vision may sit below the level of conscious thought, yet it shapes everything we do with math. And our children’s vision for themselves shapes what they pay attention to, care about, and remember.
Click to continue reading Understanding Math, Part 4: Area of a Rectangle.
CREDITS: “Math Workshop Portland” photo (top) by US Department of Education via Flicker (CC BY 2.0). This is the third post in my Understanding Math series, adapted from the expanded paperback edition of Let’s Play Math: How Families Can Learn Math Together and Enjoy It. Coming in early 2016 to your favorite online bookstore…
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## Understanding Math: What Is Your Worldview?
Educational psychologist Richard Skemp popularized the terms instrumental understanding and relational understanding to describe these two ways of looking at mathematics. It is almost as if there were two unrelated subjects, both called “math” but as different from each other as American football is from the game the rest of the world calls football.
Which of the following sounds the most like your experience of school math? And which type of math are your children learning?
### Instrumental Understanding: Math as a Tool
Every mathematical procedure we learn is an instrument or tool for solving a certain kind of problem. To understand math means to know which tool we are supposed to use for each type of problem and how to use that tool — how to categorize the problem, remember the formula, plug in the numbers, and do the calculation. To be fluent in math means we can produce correct answers with minimal effort.
Primary goal: to get the right answer. In math, answers are either right or wrong, and wrong answers are useless.
Key question: “What?” What do we know? What can we do? What is the answer?
Values: speed and accuracy.
Method: memorization. Memorize math facts. Memorize definitions and rules. Memorize procedures and when to use them. Use manipulatives and mnemonics to aid memorization.
Benefit: testability.
Instrumental instruction focuses on the standard algorithms (the pencil-and-paper steps for doing a calculation) or other step-by-step procedures. This produces quick results because students can follow the teacher’s directions and crank out a page of correct answers. Students like completing their assignments with minimal struggle, parents are pleased by their children’s high grades, and the teacher is happy to make steady progress through the curriculum.
Unfortunately, the focus on rules can lead children to conclude that math is arbitrary and authoritarian. Also, rote knowledge tends to be fragile, and the steps are easy to confuse or forget. Thus those who see math instrumentally must include continual review of old topics and provide frequent, repetitive practice.
### Relational Understanding: Math as a Connected System
Each mathematical concept is part of a web of interrelated ideas. To understand mathematics means to see at least some of this web and to use the connections we see to make sense of new ideas. Giving a correct answer without justification (explaining how we know it is right) is mere accounting, not mathematics. To be fluent in math means we can think of more than one way to solve a problem.
Primary goal: to see the building blocks of each topic and how that topic relates to other concepts.
Key questions: “How?” and “Why?” How can we figure that out? Why do we think this is true?
Values: logic and justification.
Method: conversation. Talk about the links between ideas, definitions, and rules. Explain why you used a certain procedure, and explore alternative approaches. Use manipulatives to investigate the logic behind a technique.
Benefit: flexibility.
Relational instruction focuses on children’s thinking and expands on their ideas. This builds the students’ ability to reason logically and to approach new problems with confidence. Mistakes are not a mark of failure, but a sign that points out something we haven’t yet mastered, a chance to reexamine the mathematical web. Students look forward to the “Aha!” feeling when they figure out a new concept. Such an attitude establishes a secure foundation for future learning.
Unfortunately, this approach takes time and requires extensive personal interaction: discussing problems, comparing thoughts, searching for alternate solutions, and hashing out ideas. Those who see math relationally must plan on covering fewer new topics each year, so they can spend the necessary time to draw out and explore these connections. Relational understanding is also much more difficult to assess with a standardized test.
What constitutes mathematics is not the subject matter, but a particular kind of knowledge about it.
The subject matter of relational and instrumental mathematics may be the same: cars travelling at uniform speeds between two towns, towers whose heights are to be found, bodies falling freely under gravity, etc. etc.
But the two kinds of knowledge are so different that I think that there is a strong case for regarding them as different kinds of mathematics.
Click to read Part 3: Is There Really a Difference?
CREDITS: “Humphreys High School Football” photo (top) by USAG- Humphreys via Flicker (CC BY 2.0). “Performing in middle school math class” (middle) by woodleywonderworks via Flicker (CC BY 2.0). “I Can Explain My Thinking” poster by Nicole Ricca via Teachers Pay Teachers. “Snow globe” photo (bottom) by Robert Couse-Baker via Flickr (CC BY-SA 2.0).
This is the second post in my Understanding Math series, adapted from the expanded paperback edition of Let’s Play Math: How Families Can Learn Math Together and Enjoy It. Coming in early 2016 to your favorite online bookstore…
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## Understanding Math: A Cultural Problem
All parents and teachers have one thing in common: we want our children to understand and be able to use math. Counting, multiplication, fractions, geometry — these topics are older than the pyramids.
So why is mathematical mastery so elusive?
The root problem is that we’re all graduates of the same system. The vast majority of us, including those with the power to shape reform, believe that if we can compute the answer, then we understand the concept; and if we can solve routine problems, then we have developed problem-solving skills.
The culture we grew up in, with all of its strengths and faults, shaped our experience and understanding of math, as we in turn shape the experience of our children.
### Six Decades of Math Education
Like any human endeavor, American math education — the system I grew up in — suffers from a series of fads:
• In the last part of the twentieth century, Reform Math focused on problem solving, discovery learning, and student-centered methods.
• But Reform Math brought calculators into elementary classrooms and de-emphasized pencil-and-paper arithmetic, setting off a “Math War” with those who argued for a more traditional approach.
• Now, policymakers in the U.S. are debating the Common Core State Standards initiative. These guidelines attempt to blend the best parts of reform and traditional mathematics, balancing emphasis on conceptual knowledge with development of procedural fluency.
The “Standards for Mathematical Practice” encourage us to make sense of math problems and persevere in solving them, to give explanations for our answers, and to listen to the reasoning of others—all of which are important aspects of mathematical understanding.
But the rigid way in which the Common Core standards have been imposed and the ever-increasing emphasis on standardized tests seem likely to sabotage any hope of peace in the Math Wars.
### What Does It Mean to “Understand Math”?
Through all the math education fads, however, one thing remains consistent: even before they reach the schoolhouse door, students are convinced that math is all about memorizing and following arbitrary rules.
Understanding math, according to popular culture—according to movie actors, TV comedians, politicians pushing “accountability,” and the aunt who quizzes you on your times tables at a family gathering—means knowing which procedures to apply so you can get the correct answers.
But when mathematicians talk about understanding math, they have something different in mind. To them, mathematics is all about ideas and the relationships between them, and understanding math means seeing the patterns in these relationships: how things are connected, how they work together, and how a single change can send ripples through the system.
Mathematics is the science of patterns. The mathematician seeks patterns in number, in space, in science, in computers, and in imagination. Theories emerge as patterns of patterns, and significance is measured by the degree to which patterns in one area link to patterns in other areas. | HuggingFaceTB/finemath | |
1. ## Trig Substitution Integral
I'm really spinning my wheels on this one:
$\int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt$
let $t = \sec \theta$ then $dt = \sec\theta \tan\theta$
I know that substituting $\sec\theta$ for $\sqrt{t^2-1}$ yields $\tan\theta$. This is what I get next (general antiderivative):
$\int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}$
Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:
$\int\cos^2 \theta d\theta$
which integrates to $\frac{1}{2}\sin^2\theta + C$
So, two questions:
1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?
2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to $\theta$ or get my general antiderivative back into t? This is really confusing to me.
Can anybody help?
Thanks!
2. ## Re: Trig Substitution Integral
Originally Posted by joatmon
I'm really spinning my wheels on this one:
$\int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt$
let $t = \sec \theta$ then $dt = \sec\theta \tan\theta$
I know that substituting $\sec\theta$ for $\sqrt{t^2-1}$ yields $\tan\theta$. This is what I get next (general antiderivative):
$\int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}$
Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:
$\int\cos^2 \theta d\theta$
Correct so far. However, your next step is incorrect. Check that again.
which integrates to $\frac{1}{2}\sin^2\theta + C$
So, two questions:
1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?
2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to $\theta$ or get my general antiderivative back into t? This is really confusing to me.
Can anybody help?
Thanks!
Once you finish correcting the above, I would draw a right triangle with theta as the angle, and the sides labeled appropriately for your trig substitution. Then you can figure out how to get back to the t domain. Alternatively, you can transform the limits into the theta domain. Either way should work.
3. ## Re: Trig Substitution Integral
try the substitution
(t^2) - 1 = u^2
4. ## Re: Trig Substitution Integral
Originally Posted by joatmon
I'm really spinning my wheels on this one:
$\int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt$
let $t = \sec \theta$ then $dt = \sec\theta \tan\theta$
I know that substituting $\sec\theta$ for $\sqrt{t^2-1}$ yields $\tan\theta$. This is what I get next (general antiderivative):
$\int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}$
Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:
$\int\cos^2 \theta d\theta$
which integrates to $\frac{1}{2}\sin^2\theta + C$
So, two questions:
1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?
2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to $\theta$ or get my general antiderivative back into t? This is really confusing to me.
Can anybody help?
Thanks!
A hyperbolic substitution might be more appropriate here, as $\displaystyle \sinh^2{x} \equiv \cosh^2{x} - 1$.
So in your case, make the substitution $\displaystyle t = \cosh{x} \implies dt = \sinh{x}\,dx$ and the integral $\displaystyle \int{\frac{dt}{t^3\sqrt{t^2 - 1}}}$ becomes
\displaystyle \begin{align*} \int{\frac{\sinh{x}\,dx}{\cosh^3{x}\sqrt{\cosh^2{x } - 1}}} &= \int{\frac{\sinh{x}\,dx}{\cosh^3{x}\sinh{x}}} \\ &= \int{\frac{dx}{\cosh^3{x}}} \\ &= \int{\frac{\cosh{x}\,dx}{\cosh^4{x}}} \\ &= \int{\frac{\cosh{x}\,dx}{(1 + \sinh^2{x})^2}} \\ &= \int{\frac{du}{(1 + u)^2}}\textrm{ after making the substitution }u = \sinh{x} \implies du = \cosh{x}\,dx \\ &= -\frac{1}{1 + u} + C \\ &= -\frac{1}{1 + \sinh{x}} + C \\ &= -\frac{1}{1 + \sqrt{\cosh^2{x} - 1}} + C \\ &= -\frac{1}{1 + \sqrt{t^2 - 1}} + C\end{align*}
5. ## Re: Trig Substitution Integral
Oh yeah. That was only slightly stupid... What I really meant to say was this:
$\int \cos^2 \theta d\theta = \frac{x}{2} + \frac{\sin2x}{4} + C = \frac{x + \sin x \cos x}{2} + C$
The right triangle part is where I am still struggling. I drew the triangle with theta as the angle, the hypotenuse labeled t and the x axis labeled a. Thus, $\sec \theta = \frac{t}{a}$
When t = 2, $\sec \theta = ???$ This should be pretty obvious, but somehow I'm having a brain hemorrhage on this. Am I approaching this correctly? What am I missing?
Thanks again.
6. ## Re: Trig Substitution Integral
Originally Posted by joatmon
Oh yeah. That was only slightly stupid... What I really meant to say was this:
$\int \cos^2 \theta d\theta = \frac{x}{2} + \frac{\sin2x}{4} + C = \frac{x + \sin x \cos x}{2} + C$
The right triangle part is where I am still struggling. I drew the triangle with theta as the angle, the hypotenuse labeled t and the x axis labeled a. Thus, $\sec \theta = \frac{t}{a}$
When t = 2, $\sec \theta = ???$ This should be pretty obvious, but somehow I'm having a brain hemorrhage on this. Am I approaching this correctly? What am I missing?
Thanks again.
Good start on your triangle. I would let a = 1, since that's the substitution you used earlier. Use the Pythagorean Theorem to determine the vertical side of the triangle. What do you get?
7. ## Re: Trig Substitution Integral
I think the values of $\cos^{-1}{\left(\frac{1}{2}\right)}$ and $\cos^{-1}{\left(\frac{1}{\sqrt{2}\right)}$ are pretty standard.
That being said, deriving them from a sketch is probably a good practice. | HuggingFaceTB/finemath | |
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1
AIEEE 2011
The shortest distance between line $$y-x=1$$ and curve $$x = {y^2}$$ is
A
$${{3\sqrt 2 } \over 8}$$
B
$${8 \over {3\sqrt 2 }}$$
C
$${4 \over {\sqrt 3 }}$$
D
$${{\sqrt 3 } \over 4}$$
Explanation
Shortest distance between two curve occurred along -
the common normal
Slope of normal to $${y^2} = x$$ at point
$$P\left( {{t^2},t} \right)$$ is $$-2t$$ and
slope of line $$y - x = 1$$ is $$1.$$
As they are perpendicular to each other
$$\therefore$$ $$\left( { - 2t} \right) = - 1 \Rightarrow t = {1 \over 2}$$
$$\therefore$$ $$P\left( {{1 \over 4},{1 \over 2}} \right)$$
and shortest distance $$= \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|$$
So shortest distance between them is $${{3\sqrt 2 } \over 8}$$
2
AIEEE 2011
For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.}$$ Then $$f$$ has
A
local minimum at $$\pi$$ and $$2\pi$$
B
local minimum at $$\pi$$ and local maximum at $$2\pi$$
C
local maximum at $$\pi$$ and local minimum at $$2\pi$$
D
local maximum at $$\pi$$ and $$2\pi$$
Explanation
$$f'\left( x \right) = \sqrt x \sin x$$
At local maxima or minima, $$f'\left( x \right) = 0$$
$$\Rightarrow x = 0$$ or $$sin$$ $$x=0$$
$$\Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$
$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$
$$= {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$
At $$x = \pi ,$$ $$f''\left( x \right) < 0$$
Hence, local maxima at $$x = \pi$$
At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$
Hence local minima at $$x = 2\pi$$
3
AIEEE 2010
Let $$f:R \to R$$ be a continuous function defined by $$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$\$
Statement - 1 : $$f\left( c \right) = {1 \over 3},$$ for some $$c \in R$$.
Statement - 2 : $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},$$ for all $$x \in R$$
A
Statement - 1 is true, Statement -2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B
Statement - 1 is true, Statement - 2 is false.
C
Statement - 1 is false, Statement - 2 is true.
D
Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1.
Explanation
$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$
$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$
$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$
$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2$$
maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$
$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$
Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow$$ for some $$c \in R$$
$$f\left( c \right) = {1 \over 3}$$
4
AIEEE 2010
The equation of the tangent to the curve $$y = x + {4 \over {{x^2}}}$$, that
is parallel to the $$x$$-axis, is
A
$$y=1$$
B
$$y=2$$
C
$$y=3$$
D
$$y=0$$
Explanation
Since tangent is parallel to $$x$$-axis,
$$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$
Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$
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Question
# All five faces of a regular pyramid with a square base are found to be of the same area. The height of the pyramid is 3 cm. What is the minimum total area (integral) of all its surfaces?
A
102 sq cm
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B
122 sq cm
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C
152 sq cm
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D
202 sq cm
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Solution
## The correct option is B 122 sq cmAltitude of the traingular faces = √a24+9 Area of faces =2 x a x √a24+9 The total surface area = area of base + lateral surface area = a2 + 2 x a x √a24+9 Lets us put a=8 (smallest value whch will yield an integer area, Area = 64 + 2 × 8 × 5 =144) = 122
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Optimization - Maple Features - Maplesoft # Maple
Optimization
Maple provides a collection of commands for numerically solving optimization problems, which involve finding the minimum or maximum of an objective function, possibly subject to constraints. The Optimization package enables you to solve linear programs (LPs), quadratic programs (QPs), nonlinear programs (NLPs), and both linear and nonlinear least-squares problems. Both constrained and unconstrained problems are accepted, and the package accepts a wide range of input formats.
You can also use Maple to perform global optimization. For more details, see Maple Global Optimization Toolbox.
Featured Applications
Examples
### Solving Linear Programs
Problem Description
Assume a problem where you want to:
Maximize the function Subject to: Solution
First, define the objective function: Then define a set containing all the constraint equations: For a two-dimensional linear program, you can easily graph the feasible region, shown in blue below. The black lines indicate the constraints, and the red lines indicate the contours of the objective function . Using the Maximize command, find the solution that maximizes the objective function subject to the constraint equations. The first entry returned, 10.5, is the objective value at the optimal solution, while the second entry refers to a point, (4, 6.5), which is a solution point to the optimization problem. The point (4, 6.5) lies at the upper-right vertex of the feasible region. This problem can also be solved using the interactive Optimization Assistant. ### Solving Non-Linear Programs
Problem Description
Find the point on the circle formed by the intersection of the unit sphere with the plane that is closest to the point (1, 2, 3).
Solution
This problem can be easily conceptualized by plotting the unit sphere and the equation defining the intersecting plane. The objective function is the squared distance between a point (x, y, z) and the point (1, 2, 3). The point (x, y, z) is constrained to lie on both the unit sphere and the given plane. Thus, the constraints are: You can minimize the objective function subject to the constraints using the NLPSolve command. Thus, the minimum distance is 10.29, and the closest point is (-0.51, 0.17, 0.84).
Problem Description
The Markowitz model is an optimization model for balancing the return and risk of a portfolio. The decision variables are the amounts invested in each asset. The objective is to minimize the variance of the portfolio's total return, subject to the following constraints: (1) the expected growth of the portfolio is at least some target level, and (2) the investment should not be more than the capital.
Solution
Let: be the amounts you buy the amount of capital you have the random vector of asset returns over some period the expected value of the minimum growth you hope to obtain the covariance matrix of The objective function is , which can be shown to be equal to .
If, for example, , you would try to:
Minimize the function Subject to: Suppose you have the following data. This is a quadratic function of X, and so the quadratic programming can be used to solve the problem. Thus, the minimum variance portfolio that earns an expected return of at least 10% is . Asset 2 gets nothing because its expected return is -20% and its covariance with the other assets is not sufficiently negative for it to bring any diversification benefits.
### Solving Least-Square Problems
Problem Description
Neural networks are machine-learning programs that can learn to approximate functions using a set of sample function values called training data. To generate an output, a neural network applies a smooth function (typically, sigmoidal or hyperbolic tangent) to a linear weighting of the input data. To train the network, one assigns these weights so as to minimize the sum of squared differences between the desired and generated outputs over all data points in the training set. This is an optimization problem that can often be solved by least-squares techniques.
Solution
Assume that you have a sample training data set with five points. The first element of each point is the input, the second is the desired output. Suppose that for a given input value x, the network outputs , where and are the weights that are to be set. The residuals are the differences between these outputs and the desired outputs given in the training set: You can extract elements of the list x; using the selection operation. Then, use seq to create a sequence.
The objective function is the sum of squares of the residuals: Now, solve the optimization problem with the LSSolve command. Note that you only have to pass Maple the list of residuals. The weights needed to minimize the sum of squared differences between the desired and generated outputs for and are and . | HuggingFaceTB/finemath | |
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### MEAP Preparation - Grade 5 Mathematics2.1 Measurement Facts Table
Measurement Facts U.S. Customary System Metric System Units of Length 1 foot = 12 inches1 yard = 3 feet 1 yard = 36 inches (in) 1 mile (mi) = 1760 yards (yd) 1 mile (mi) = 5280 feet (ft) 1 centimeter = 10 millimeters 1 decimeter = 10 centimeters 1 meter = 10 decimeters (dm) 1 meter = 100 centimeters (cm) 1 meter = 1000 millimeters (mm) 1 meter = 10 decimeters 1 kilometer (km) = 1000 meters (m) Units of Area 1 square yard (yd2) = 9 square feet (ft2) 1 square yard (yd2) = 1296 square inches (in2) 1 square foot = 144 square inches 1 acre = 43,560 square feet 1 square mile (mi2) = 640 acres 1 square meter (m2) = 100 square decimeters (dm2) 1 square meter (m2) = 10,000 square centimeters (cm2) 1 square decimeter = 100 square centimeters 1 are (a) = 100 square meters 1 hectare (ha) = 100 ares 1 square kilometer (km2) = 100 hectares Units of Volume 1 cubic yard (yd3) = 27 cubic feet (ft3) 1 cubic foot = 1728 cubic inches (in3) 1 cubic meter (m3) = 1,000 cubic decimeters (dm3) 1 cubic meter (m3) = 1,000,000 cubic centimeters (cm3) 1 cubic decimeter = 1000 cubic centimeters Units of Capacity 1 gallon (gal) = 4 quarts (qt) 1 quart = 2 pints (pt) 1 pint = 2 cups (c) 1 cup = 8 fluid ounces (fl oz) 1 fluid ounce = 2 tablespoons (tbs) 1 tablespoon = 3 teaspoons (tsp) 1 kiloliter (kL) = 1000 liters(L) 1 liter = 1000 milliliters (mL) Units of Weight/Mass Units of Weight 1 pound = 16 ounces (oz) 1 ton (T) = 2000 pounds (lb) Units of Mass 1 gram = 1000 milligram (mg) 1 kilogram = 1000 grams (g) 1 metric ton (t) = 1000 kilograms (kg) Units of Temperature Degree Fahrenheit Degree Celsius
Directions: Answer the following questions. Also write at least ten examples of your own.
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### MEAP Preparation - Grade 5 Mathematics2.1 Measurement Facts Table
Q 1: In U S customary system how many yards is 1 mile?528017602000 Q 2: In metric system 1 kilometer = ______ meters1000100110 Q 3: In U.S customary system 1 acre = _____ square feet40,00043,56020,50062,990 Q 4: In metric system 1 kiloliter = _______ liters1000100110 Q 5: In U.S customary system 1 gallon = ____ quarts46110 Q 6: In U.S customary system the unit of temperature is ______________degree Celsiusdegree Fahrenheit Q 7: In U.S. customary system 1 foot = ____ inches12101640 Q 8: In metric system 1 centimeter = _____ millimeters1011001000 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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