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# Charles Law (Initial temperature)

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The Charles Law (Initial temperature) equation computes the initial temperature of a fixed mass of gas before it underwent a change in volume.

## INPUTS

Based on Charles Law we can compute the initial temperature give the following:

• T_2: the resultant temperature
• V_1: the initial volume
• V_2: the resultant volume

Remember to input temperatures in degrees Kelvin: K = C + 273.  However, other temperature units are available and will be automatically converted to Kelvin.

# NOTES

Charles law is an experimental gas law which describes how gases tend to expand when heated. Charles law states:  The volume of a given mass of a gas is directly proportional to its temperature (measured in Kelvin the absolute temperature scale). If pressure and the amount of gas remain constant; the law states, the volume of the gas increases or decreases by the same factor as its temperature change.  The law was named after scientist Jacques Charles, who formulated the original law in his unpublished work from the 1780s.

Jacques Charles (1746 - 1823) discovered the relationship between the volume of a gas and the temperature in 1787.  This relationship was discovered again independently by Joseph Louis Gay-Lussac in 1802.

Charles is credited with many discoveries in the area of ballooning; he invented the suspending basket carried on balloons and the valves to regulate the gas in the balloons.  Charles was the first to experiment with small hydrogen filled balloons.


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# Gas Laws: Gay-Lussac's Law, Dalton's Law of Partial Pressures

CHEM-DDKRAU

A container at a pressure of 1.02 atm contains the following amounts of gases at 25$^o$C: 1 mole of hydrogen, 2 moles of nitrogen, and 3 moles of helium.

If the volume of the container remains constant, and the temperature is raised to 75$^o$C, what will be the final partial pressure of the helium gas (in atm)?

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## Key Concepts

• An augmented matrix is one that contains the coefficients and constants of a system of equations.
• A matrix augmented with the constant column can be represented as the original system of equations.
• Row operations include multiplying a row by a constant, adding one row to another row, and interchanging rows.
• We can use Gaussian elimination to solve a system of equations.
• Row operations are performed on matrices to obtain row-echelon form.
• To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions.
• A calculator can be used to solve systems of equations using matrices.
• Many real-world problems can be solved using augmented matrices.

## Glossary

augmented matrix
a coefficient matrix adjoined with the constant column separated by a vertical line within the matrix brackets
coefficient matrix
a matrix that contains only the coefficients from a system of equations
Gaussian elimination
using elementary row operations to obtain a matrix in row-echelon form
main diagonal
entries from the upper left corner diagonally to the lower right corner of a square matrix
row-echelon form
after performing row operations, the matrix form that contains ones down the main diagonal and zeros at every space below the diagonal
row-equivalent
two matrices $A$ and $B$ are row-equivalent if one can be obtained from the other by performing basic row operations
row operations
adding one row to another row, multiplying a row by a constant, interchanging rows, and so on, with the goal of achieving row-echelon form


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# Same Coefficients

Algebra Level 3

Solve the following system for distinct $$x$$, $$y$$, and $$z$$.

$\begin{cases} 5732x+2134y+2134z = 7866 \\ 2134x+5732y+2134z= 670 \\ 2134x+2134y+5732z= 11464 \end{cases}$

What is $$x+y+z=$$ ?

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# 5.2: Unit Circle - Sine and Cosine Functions

Skills to Develop

• Find function values for the sine and cosine of  30° or $$\frac{\pi}{6}$$, 45° or $$\frac{\pi}{4}$$, and  60° or $$\frac{\pi}{3}$$.
• Identify the domain and range of sine and cosine functions.
• Find reference angles.
• Use reference angles to evaluate trigonometric functions.

Looking for a thrill? Then consider a ride on the Singapore Flyer, the world’s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet—a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.

Figure $$\PageIndex{1}$$: The Singapore Flyer is the world’s tallest Ferris wheel. (credit: “Vibin JK”/Flickr)

## Finding Function Values for the Sine and Cosine

To define our trigonometric functions, we begin by drawing a unit circle, a circle centered at the origin with radius 1, as shown in Figure $$\PageIndex{2}$$. The angle (in radians) that $$t$$ intercepts forms an arc of length $$s$$. Using the formula $$s=rt$$, and knowing that $$r=1$$, we see that for a unit circle, $$s=t$$.

For any angle $$t,$$ we can label the intersection of the terminal side and the unit circle by its coordinates, $$(x,y)$$. The coordinates $$x$$ and $$y$$ will be the outputs of the trigonometric functions $$f(t)= \cos t$$ and $$f(t)= \sin t$$, respectively. This means $$x= \cos t$$ and $$y= \sin t$$. The actual function names are "cosine" and "sine", but we use the notation $$\cos(t)$$ and $$\sin(t)$$.

Figure $$\PageIndex{2}$$: Unit circle where the central angle is $$t$$ radians

UNIT CIRCLE

A unit circle has a center at $$(0,0)$$ and radius $$1$$. In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle $$1$$.

Let $$(x,y)$$ be the endpoint on the unit circle of an arc of arc length $$s$$. The $$(x,y)$$ coordinates of this point depend on the angle. Hence, each coordinate can be described as a function of the angle.

### Defining Sine and Cosine Functions

Now that we have our unit circle labeled, we can learn how the $$(x,y)$$ coordinates relate to the arc length and angle. The sine function relates a real number $$t$$ to the y-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle $$t$$ equals the $$y$$-value of the endpoint on the unit circle of an arc of length $$t$$. Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the $$y$$-coordinate of the corresponding point on the unit circle. In Figure $$\PageIndex{3}$$, the sine is equal to $$y$$.

Similarly, the output of the cosine function of an angle $$t$$ equals the $$x$$-value of the endpoint on the unit circle of an arc of length $$t$$. In Figure $$\PageIndex{3}$$, the cosine is equal to $$x$$.

Figure $$\PageIndex{3}$$

Because it is understood that sine and cosine are functions, we do not always write them with parentheses: $$\sin t$$ is the same as $$\sin (t)$$ and $$\cos t$$ is the same as $$\cos (t)$$. Likewise, $$\cos ^2 t$$ is a commonly used shorthand notation for $$( \cos (t))^2$$. Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer.

definition: SINE AND COSINE FUNCTIONS

If $$t$$ is a real number and a point $$(x,y)$$ on the unit circle corresponds to an angle in standard position of  $$t$$ radians, then

\begin{align} \cos t & = x \\ \sin t & = y. \end{align}

Given a point $$P(x,y)$$ on the unit circle corresponding to an angle of $$t$$, find the sine and cosine

1. The sine of $$t$$ is equal to the y-coordinate of point $$P: \sin t=y$$.
2. The cosine of $$t$$ is equal to the x-coordinate of point $$P: \cos t=x$$.

Example $$\PageIndex{1}$$: Finding Function Values for Sine and Cosine

Point $$P$$  is a point on the unit circle corresponding to an angle of $$t$$, as shown in Figure $$\PageIndex{4}$$. Find $$\cos (t)$$  and $$\sin (t)$$.

Figure $$\PageIndex{4}$$

Solution

We know that $$\cos t$$ is the x-coordinate of the corresponding point on the unit circle and $$\sin t$$  is the y-coordinate of the corresponding point on the unit circle. So:

\begin{align} x & = \cos t= \frac{1}{2} \\ y & = \sin t= \frac{\sqrt{3}}{2} \end{align}

Analysis: Because the unit circle has radius 1, we can think of both the $$x$$- and $$y$$-coordinates as ratios of a right triangle, if angle $$t$$ is acute. The length of the hypotenuse is 1, and the legs have length $$x$$ and $$y$$. Thus,

\begin{align \cos t = x &= \frac{x}{1} \\ \sin t = y &= \frac{y}{1}.

For a given angle $$t$$, would the $$x, y$$-coordinates change if we used a circle of radius 2, or 3? Yes. Would the ratios change? No.

$$\PageIndex{1}$$:

A certain angle $$t$$  corresponds to a point on the unit circle at $$(−\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$$ as shown in Figure $$\PageIndex{5}$$. Find $$\cos t$$ and $$\sin t$$.

Figure $$\PageIndex{5}$$

$$\cos (t)=−\frac{ \sqrt{2} }{2}, \sin (t)=\frac {\sqrt{2}}{2}$$

### Finding Sines and Cosines of Angles on an Axis

For quadrantral angles, the corresponding point on the unit circle falls on the x- or y-axis. In that case, we can easily calculate cosine and sine from the values of $$x$$ and  $$y$$.

Example $$\PageIndex{2}$$: Calculating Sines and Cosines along an Axis

Find $$\cos (90°)$$ and $$\sin (90°).$$

Solution

Moving $$90°$$ counterclockwise around the unit circle from the positive x-axis brings us to the top of the circle, where the $$(x,y)$$ coordinates are (0, 1), as shown in Figure $$\PageIndex{6}$$.

Figure $$\PageIndex{6}$$

Using our definitions of cosine and sine,

\begin{align} x &= \cos t = \cos (90°) = 0 \\ y &= \sin t = \sin (90°) = 1 \end{align}

The cosine of 90° is 0; the sine of 90° is 1.

$$\PageIndex{2}$$

Find cosine and sine of the angle $$π$$.

$$\cos (π)=−1, \sin (π)=0$$

### The Pythagorean Identity

Now that we have defined sine and cosine, we will learn how they relate to each other and the unit circle. The equation for the unit circle is $$x^2+y^2=1$$.  Because $$x= \cos t$$ and $$y=\sin t$$, we can substitute for $$x$$ and $$y$$ to get $$\cos ^2 t+ \sin ^2 t=1.$$ This equation, $$\cos ^2 t+ \sin ^2 t=1,$$ is known as the Pythagorean Identity. See Figure $$\PageIndex{7}$$.

Figure $$\PageIndex{7}$$

We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, and vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.

PYTHAGOREAN IDENTITY

The Pythagorean Identity states that, for any real number $$t$$,

$\cos^2 t+ \sin^2 t=1$

Given the sine of some angle $$t$$ and its quadrant location, find the cosine of $$t$$.

1. Substitute the known value of $$\sin (t)$$ into the Pythagorean Identity.
2. Solve for $$\cos (t)$$.
3. Choose the solution with the appropriate sign for the $$x$$-values in the quadrant where $$t$$ is located.

Example $$\PageIndex{3}$$: Finding a Cosine from a Sine or a Sine from a Cosine

If  $$\sin (t)=\frac{3}{7}$$ and $$t$$ terminates in the second quadrant, find $$\cos (t)$$.

Solution

If we drop a vertical line from the point on the unit circle corresponding to the terminal side of  $$t$$, we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See Figure $$\PageIndex{8}$$.

Figure $$\PageIndex{8}$$

Substituting the known value for sine into the Pythagorean Identity,

\begin{align} \cos ^2 (t)+ \sin ^2(t) &=1 \\ \cos ^2(t)+\dfrac{9}{49} &=1 \\ \cos ^2(t) & = \dfrac{40}{49} \\ \cos (t)=± \sqrt{\dfrac{40}{49}}=±\dfrac{\sqrt{40}}{7}=±\dfrac{2\sqrt{10}}{7} \end{align}

Because the angle terminates in the second quadrant, we know the x-value is a negative real number, so the cosine is also negative. So

$\cos (t)=−\dfrac{2\sqrt{10}}{7}.$

$$\PageIndex{3}$$

If $$\cos (t)=\frac{24}{25}$$ and tt is in the fourth quadrant, find $$\sin (t)$$.

$$\sin (t)=−\frac{7}{25}$$

## Finding Sines and Cosines of Special Angles

We have already learned some properties of the special angles, such as the conversion from radians to degrees. We can also calculate sines and cosines of the special angles using the Pythagorean Identity and our knowledge of triangles.

### Finding Sines and Cosines of 45° Angles

First, we will look at angles of $$45°$$ or $$\frac{π}{4}$$, as shown in Figure $$\PageIndex{9}$$ . A $$45°–45°–90°$$ triangle is an isosceles triangle, so the x- and y-coordinates of the corresponding point on the circle are the same. Because the x- and y-values are the same, the sine and cosine values will also be equal.

Figure $$\PageIndex{9}$$

At $$t=\frac{π}{4}$$, which is 45 degrees, the radius of the unit circle bisects the first quadrant. This means the radius lies along the line $$y=x$$. A unit circle has a radius equal to 1. So, the right triangle formed below the line $$y=x$$ has sides $$x$$ and $$y\; (y=x),$$ and a radius = 1. See Figure $$\PageIndex{10}$$.

Figure $$\PageIndex{10}$$

From the Pythagorean Theorem we get

$x^2+y^2=1$

Substituting $$y=x$$, we get

$x^2+x^2=1$

Combining like terms we get

$2x^2=1$

And solving for $$x$$, we get

\begin{align} x^2 &=\dfrac{1}{2} \\ x &=±\dfrac{1}{\sqrt{2}} \end{align}

In Quadrant I, $$x=\frac{1}{\sqrt{2}}$$.

At $$t=\frac{π}{4}$$ or $$45^{\circ}$$,

\begin{align} (x,y) & =(x,x)=(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}) \\ x &= \dfrac{1}{\sqrt{2}},\; y=\dfrac{1}{\sqrt{2}} \\ \cos t &= \dfrac{1}{\sqrt{2}},\; \sin t=\dfrac{1}{\sqrt{2}} \end{align}

If we then rationalize the denominators, we get

\begin{align} \cos t &= \dfrac{1}{\sqrt{2}}\dfrac{\sqrt{2}}{\sqrt{2}} \\[2mm] &= \dfrac{\sqrt{2}}{2} \approx 0.707 \\[2mm] \sin t &= \dfrac{1}{\sqrt{2}} \dfrac{\sqrt{2}}{\sqrt{2}} \\[2mm] &=\dfrac{\sqrt{2}}{2} \approx 0.707 \end{align}

Therefore, the $$(x,y)$$ coordinates of a point on a circle of radius $$1$$ at an angle of $$45°$$ are $$\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$$.

### Finding Sines and Cosines of 30° and 60° Angles

Next, we will find the cosine and sine at an angle of  $$30°$$, or $$\frac{π}{6}$$. First, we will draw a triangle inside a circle with one side at an angle of $$30°$$,and another at an angle of $$−30°$$, as shown in Figure $$\PageIndex{11}$$. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be $$60°$$, as shown in Figure $$\PageIndex{12}$$.

Figure $$\PageIndex{11}$$

Figure $$\PageIndex{12}$$

Because all the angles are equal, the sides are also equal. The vertical line has length $$2y$$, and since the sides are all equal, we can also conclude that $$r=2y$$ or $$y=\frac{1}{2}r$$. Since $$\sin t=y$$,

$\sin (\dfrac{π}{6})=\dfrac{1}{2}r$

And since $$r=1$$ in our unit circle,

\begin{align} \sin (\dfrac{π}{6}) & = \dfrac{1}{2}(1) \\ &= \dfrac{1}{2} \end{align}

Using the Pythagorean Identity, we can find the cosine value.

$cos 2 π 6 + sin 2 ( π 6 ) = 1 cos 2 ( π 6 ) + ( 1 2 ) 2 = 1 cos 2 ( π 6 ) = 3 4 Use the square root property . cos ( π 6 ) = ± 3 ± 4 = 3 2 Since $$y$$ is positive, choose the positive root . cos 2 π 6 + sin 2 ( π 6 )=1 cos 2 ( π 6 )+ ( 1 2 ) 2 =1 cos 2 ( π 6 )= 3 4 Use the square root property. cos( π 6 )= ± 3 ± 4 = 3 2 Since y is positive, choose the positive root.$

The $$(x,y)$$ coordinates for the point on a circle of radius $$1$$ at an angle of $$30°$$ are $$(\frac{\sqrt{3}}{2},\frac{1}{2})$$.  At $$t=\frac{π}{3}$$ (or 60°), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, $$BAD,$$ as shown in Figure $$\PageIndex{13}$$. Angle $$A$$  has measure 60°.

Figure $$\PageIndex{13}$$

The measure of angle $$ABD$$ is 30°. Notice that $$AD$$ is the x-coordinate of point $$B$$, which is at the intersection of the terminal side of the 60° angle and the unit circle. This gives us a triangle $$BAD$$ with hypotenuse of 1 and side $$x$$ of length $$\frac{1}{2}$$.

From the Pythagorean Theorem, we get

$x^2+y^2=1$

Substituting $$x=\frac{1}{2}$$, we get

$\left(\dfrac{1}{2}\right)^2+y^2=1$

Solving for $$y$$, we get

\begin{align} \dfrac{1}{4}+y^2 &=1 \\ y^2 &=1−\dfrac{1}{4} \\ y^2 &= \dfrac{3}{4} \\ y &=± \dfrac{\sqrt{3}}{2} \end{align}

Since $$t=\frac{π}{3}$$ has the terminal side in Quadrant I where the y-coordinate is positive, we choose $$y=\frac{\sqrt{3}}{2}$$, the positive value.

At $$t=\frac{π}{3}$$ (60°), the $$(x,y)$$ coordinates for the point on a circle of radius $$1$$ at an angle of $$60°$$ are $$(\frac{1}{2},\frac{\sqrt{3}}{2})$$, so we can find the sine and cosine.

$( x , y ) = ( 1 2 , 3 2 ) x = 1 2 , y = 3 2 cos t = 1 2 , sin t = 3 2 (x,y)=( 1 2 , 3 2 ) x= 1 2 ,y= 3 2 cos t= 1 2 ,sin t= 3 2$

We have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. Table $$\PageIndex{1}$$ summarizes these values.

Table $$\PageIndex{1}$$

 Angle Cosine Sine 0 $$\frac{π}{6}$$, or $$30^{\circ}$$ $$\frac{π}{4}$$, or 45° $$\frac{π}{3}$$, or 60° $$\frac{π}{2}$$, or 90° 1 $$\frac{\sqrt{3}}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{1}{2}$$ 0 0 $$\frac{1}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{\sqrt{3}}{2}$$ 1

Figure shows the common angles in the first quadrant of the unit circle.

Figure $$\PageIndex{14}$$

### Using a Calculator to Find Sine and Cosine

To find the cosine and sine of angles other than the special angles, we turn to a computer or calculator. Be aware: Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When we evaluate $$\cos (30)$$ on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.

Given an angle in radians, use a graphing calculator to find the cosine

1. If the calculator has degree mode and radian mode, set it to radian mode.
2. Press the COS key.
3. Enter the radian value of the angle and press the close-parentheses key ")".
4. Press ENTER.

Example $$\PageIndex{4}$$: Using a Graphing Calculator to Find Sine and Cosine

Evaluate $$\cos (\frac{5π}{3})$$ using a graphing calculator or computer.

Solution

Enter the following keystrokes:

$$\mathrm{COS( 5 × π ÷ 3 ) ENTER}$$

$\cos (\dfrac{5π}{3})=0.5$

Analysis

We can find the cosine or sine of an angle in degrees directly on a calculator with degree mode. For calculators or software that use only radian mode, we can find the sign of $$20°$$, for example, by including the conversion factor to radians as part of the input:

$\mathrm{SIN( 20 × π ÷ 180 ) ENTER}$

Exercise $$\PageIndex{4}$$

Evaluate $$\sin (\frac{π}{3})$$.

Solution

approximately 0.866025403

## Identifying the Domain and Range of Sine and Cosine Functions

Now that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than 0 and angles larger than 2π  2π can still be graphed on the unit circle and have real values of $$x, y$$, and $$r$$, there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may be any real number.

What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the unit circle, as shown in Figure. The bounds of the x-coordinate are $$[−1,1]$$. The bounds of the y-coordinate are also $$[−1,1]$$. Therefore, the range of both the sine and cosine functions is $$[−1,1]$$.

Figure $$\PageIndex{15}$$

## Finding Reference Angles

We have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the y-coordinate on the unit circle, the other angle with the same sine will share the same y-value, but have the opposite x-value. Therefore, its cosine value will be the opposite of the first angle’s cosine value.

Likewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same x-value but will have the opposite y-value. Therefore, its sine value will be the opposite of the original angle’s sine value.

As shown in Figure, angle  $$α$$  has the same sine value as angle $$t$$; the cosine values are opposites. Angle $$β$$ has the same cosine value as angle $$t$$; the sine values are opposites.

$sin ( t ) = sin ( α ) and cos ( t ) = − cos ( α ) sin ( t ) = − sin ( β ) and cos ( t ) = cos ( β ) sin(t)= sin(α) and cos(t)=−cos(α) sin(t)=−sin(β) and cos(t)= cos(β)$

Figure $$\PageIndex{16}$$

Recall that an angle’s reference angle is the acute angle,$$t$$, formed by the terminal side of the angle $$t$$ and the horizontal axis. A reference angle is always an angle between $$0$$ and $$90°$$, or $$0$$ and $$\frac{π}{2}$$ radians. As we can see from Figure, for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.

Figure $$\PageIndex{17}$$

Given an angle between $$0$$ and $$2π$$, find its reference angle

1. An angle in the first quadrant is its own reference angle.
2. For an angle in the second or third quadrant, the reference angle is $$|π−t|$$  or $$|180°−t|$$.
3. For an angle in the fourth quadrant, the reference angle is $$2π−t$$ or $$360°−t.$$
4. If an angle is less than $$0$$ or greater than $$2π,$$ add or subtract $$2π$$ as many times as needed to find an equivalent angle between $$0$$ and $$2π$$.

Example $$\PageIndex{5}$$: Finding a Reference Angle

Find the reference angle of $$225°$$ as shown in Figure.

Figure $$\PageIndex{18}$$

Solution

Because $$225°$$ is in the third quadrant, the reference angle is

$|(180°−225°)|=|−45°|=45°$

Exercise $$\PageIndex{5}$$:

Find the reference angle of $$\frac{5π}{3}$$.

Solution

$$\frac{π}{3}$$

## Using Reference Angles

Now let’s take a moment to reconsider the Ferris wheel introduced at the beginning of this section. Suppose a rider snaps a photograph while stopped twenty feet above ground level. The rider then rotates three-quarters of the way around the circle. What is the rider’s new elevation? To answer questions such as this one, we need to evaluate the sine or cosine functions at angles that are greater than 90 degrees or at a negative angle. Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find $$(x,y)$$ coordinates for those angles. We will use the reference angle of the angle of rotation combined with the quadrant in which the terminal side of the angle lies.

### Using Reference Angles to Evaluate Trigonometric Functions

We can find the cosine and sine of any angle in any quadrant if we know the cosine or sine of its reference angle. The absolute values of the cosine and sine of an angle are the same as those of the reference angle. The sign depends on the quadrant of the original angle. The cosine will be positive or negative depending on the sign of the x-values in that quadrant. The sine will be positive or negative depending on the sign of the y-values in that quadrant.

USING REFERENCE ANGLES TO FIND COSINE AND SINE

Angles have cosines and sines with the same absolute value as cosines and sines of their reference angles. The sign (positive or negative) can be determined from the quadrant of the angle.

Given an angle in standard position, find the reference angle, and the cosine and sine of the original angle

1. Measure the angle between the terminal side of the given angle and the horizontal axis. That is the reference angle.
2. Determine the values of the cosine and sine of the reference angle.
3. Give the cosine the same sign as the x-values in the quadrant of the original angle.
4. Give the sine the same sign as the y-values in the quadrant of the original angle.

Example $$\PageIndex{6}$$: Using Reference Angles to Find Sine and Cosine

1. Using a reference angle, find the exact value of $$\cos (150°)$$ and $$\sin (150°)$$.
2. Using the reference angle, find $$\cos \frac{5π}{4}$$ and $$\sin \frac{5π}{4}$$.

Solution

This tells us that 150° has the same sine and cosine values as 30°, except for the sign. We know that

1. 150° is located in the second quadrant. The angle it makes with the x-axis is 180° − 150° = 30°, so the reference angle is 30°. $cos ( 30° ) = 3 2 and sin ( 30° ) = 1 2 . cos(30°)= 3 2 and sin(30°)= 1 2 .$

Since 150° is in the second quadrant, the x-coordinate of the point on the circle is negative, so the cosine value is negative. The y-coordinate is positive, so the sine value is positive.

$cos ( 150° ) = − 3 2 and sin ( 150° ) = 1 2 cos(150°)=− 3 2 and sin(150°)= 1 2$
2.  $$\frac{5π}{4}$$ is in the third quadrant. Its reference angle is $$\frac{5π}{4}−π=\frac{π}{4}$$. The cosine and sine of $$\frac{π}{4}$$ are both $$\frac{sqrt{2}}{2}$$. In the third quadrant, both x  x and y  y are negative, so:

$\cos \dfrac{5π}{4}=−\dfrac{\sqrt{2}}{2} \;\; \text{and} \; \; \sin \dfrac{5π}{4}=−\dfrac{\sqrt{2}}{2}$

Exercise $$\PageIndex{6}$$

1. Use the reference angle of $$315°$$ to find $$\cos (315°)$$ and $$\sin (315°)$$.
2. Use the reference angle of $$−\frac{π}{6}$$ to find $$\cos (−\frac{π}{6})$$ and $$\sin (−\frac{π}{6})$$.

Solution

$$\cos (315°)= \frac{\sqrt{2}}{2}, \sin (315°)=\frac{–\sqrt{2}}{2}$$

$$\cos (−\frac{π}{6})=\frac{\sqrt{3}}{2}, \sin (−\frac{π}{6})=−\frac{1}{2}$$

### Using Reference Angles to Find Coordinates

Now that we have learned how to find the cosine and sine values for special angles in the first quadrant, we can use symmetry and reference angles to fill in cosine and sine values for the rest of the special angles on the unit circle. They are shown in Figure. Take time to learn the $$(x,y)$$ coordinates of all of the major angles in the first quadrant.

Figure $$\PageIndex{19}$$: Special angles and coordinates of corresponding points on the unit circle

In addition to learning the values for special angles, we can use reference angles to find $$(x,y)$$ coordinates of any point on the unit circle, using what we know of reference angles along with the identities

\begin{align} x &= \cos t \\ y & = \sin t \end{align}

First we find the reference angle corresponding to the given angle. Then we take the sine and cosine values of the reference angle, and give them the signs corresponding to the y- and x-values of the quadrant.

Given the angle of a point on a circle and the radius of the circle, find the $$(x,y)$$ coordinates of the point.

Find the reference angle by measuring the smallest angle to the x-axis.

1. Find the cosine and sine of the reference angle.
2. Determine the appropriate signs for $$x$$ and $$y$$  in the given quadrant.

Example $$\PageIndex{7}$$: Using the Unit Circle to Find Coordinates

Find the coordinates of the point on the unit circle at an angle of $$\frac{7π}{6}$$.

Solution

We know that the angle $$\frac{7π}{6}$$ is in the third quadrant.

First, let’s find the reference angle by measuring the angle to the x-axis. To find the reference angle of an angle whose terminal side is in quadrant III, we find the difference of the angle and π.  π.

$\dfrac{7π}{6}−π=\dfrac{π}{6}$

Next, we will find the cosine and sine of the reference angle:

$\cos (\dfrac{π}{6})=\dfrac{3}{2} \;\; \sin (\dfrac{π}{6})=\dfrac{1}{2}$

We must determine the appropriate signs for x and y in the given quadrant. Because our original angle is in the third quadrant, where both x  x and y  y are negative, both cosine and sine are negative.

\begin{align} \cos (\dfrac{7π}{6}) &=−\dfrac{\sqrt{3}}{2} \\ \sin (\dfrac{7π}{6}) & =−\dfrac{1}{2} \end{align}

Now we can calculate the $$(x,y)$$ coordinates using the identities $$x= \cos θ$$ and $$y= \sin θ$$.

The coordinates of the point are $$(−\frac{\sqrt{3}}{2},−\frac{1}{2})$$ on the unit circle.

Exercise $$\PageIndex{7}$$:

Find the coordinates of the point on the unit circle at an angle of $$\frac{5π}{3}$$.

Solution

$$(\frac{1}{2},−\frac{\sqrt{3}}{2})$$

media

Access these online resources for additional instruction and practice with sine and cosine functions.

## Key Equations

 Cosine $$\cos t=x$$ Sine $$\sin t=y$$ Pythagorean Identity $$\cos ^2 t+ \sin ^2 t=1$$

## Key Concepts

• Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit.
• Using the unit circle, the sine of an angle t  t equals the y-value of the endpoint on the unit circle of an arc of length t  t whereas the cosine of an angle t   t equals the x-value of the endpoint. See Example.
• The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See Example.
• When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See Example.
• Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See Example.
• The domain of the sine and cosine functions is all real numbers.
• The range of both the sine and cosine functions is $$[−1,1]$$.
• The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.
• The signs of the sine and cosine are determined from the x- and y-values in the quadrant of the original angle.
• An angle’s reference angle is the size angle, $$t$$, formed by the terminal side of the angle $$t$$  and the horizontal axis. See Example.
• Reference angles can be used to find the sine and cosine of the original angle. See Example.
• Reference angles can also be used to find the coordinates of a point on a circle. See Example.

## Glossary

cosine function
the x-value of the point on a unit circle corresponding to a given angle
Pythagorean Identity
a corollary of the Pythagorean Theorem stating that the square of the cosine of a given angle plus the square of the sine of that angle equals 1
sine function
the y-value of the point on a unit circle corresponding to a given angle
unit circle
a circle with a center at $$(0,0)$$ and radius 1.


text:
## Precalculus (6th Edition) Blitzer

The solution is $0$.
Convert $\frac{\pi }{4}$ into degrees. \begin{align} & \frac{\pi }{4}=\frac{\pi }{4}\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{\pi } \right) \\ & =\left( \frac{\pi }{\pi } \right)\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{4} \right) \\ & =45\text{ }\!\!{}^\circ\!\!\text{ } \end{align} The expressions for $\sin \theta$ and $\cos \theta$ are $\sin \theta =\frac{a}{c}$ …… (I) $\cos \theta =\frac{b}{c}$ …… (II) Here, $a$ is the length of the side opposite to $\theta$ , $b$ is the length of the side adjacent to $\theta$ and $c$ is the hypotenuse. In triangle 1, For $\theta =45\text{ }\!\!{}^\circ\!\!\text{ }$ , $a=1$ , $b=1$ and $c=\sqrt{2}$. Substitute $\frac{\pi }{4}$ for $\theta$ , $1$ for $a$ and $\sqrt{2}$ for $c$ in equation (I). $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ …… (III) Substitute $\frac{\pi }{4}$ for $\theta$ , $1$ for $b$ and $\sqrt{2}$ for $c$ in equation (II). $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ …… (IV) Subtract equation (IV) from equation (III). \begin{align} & \sin \frac{\pi }{4}-\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \\ & =0 \end{align} Therefore, the solution is $0$.


text:
# how to normalize DCT (or FFT) coefficients to compare different signals

I have signals that have a broad range of values. I want to compare them in a meaningful way. Just as an example, two of those signals might be: $$f(t)=100\cos(50t)+40\cos(60t) \quad g(t)=2\cos(50t)+0.8\cos(60t)$$

The coefficients of $$f(t)$$ will be much higher than the coefficients of $$g(t)$$, however, for the purpose of my comparison, those two signals are "equivalent", in that the dynamics are the same, and the difference is just in scaling coefficients.

In my problem I have a discrete signal, but I can compute the DCT (of FFT). The question is: how to normalize the coefficients, so that I can compare signals the way it was done above. What I thought of is this:

Let $$a_n$$ be the n coefficients obtained from applying DCT, then one simply normalize by $$\dfrac{a_n}{\sum_n |a_n|}$$. Is this mathematically sound? is there some theory on how to normalize the coefficients to capture the signal dynamics (meaning the relation of certain frequencies with respect to others)?

I'd appreciate if any can share some resources or can provide some insight.

• This related question and answer that discusses various scaling and how to properly account for windowing etc may be helpful to you: dsp.stackexchange.com/questions/75817/… Jun 18 at 23:32

## 1 Answer

Is this mathematically sound?

Yes. It's a common normalization in ML. The idea's to not change relative values, i.e. ratios between any two coefficients (of same transform/signal).

However, one might want to narrow it further, in case there's e.g. noise of different intensities in the two signals - then noise affects norm, which isn't desired. Then noise levels should be estimated and excluded from the normalizing factor (or simply signals denoised first).

For more complicated behaviors, with e.g. amplitude or frequency that varies over time, FFT and DCT are ill suited, and time-frequency transforms should be preferred, like CWT & STFT. Then independent "components" can be extracted and compared. Example with A.M.


text:
# Cosmological evolution of primordial black holes [HEAP]

The cosmological evolution of primordial black holes (PBHs) is considered. A comprehensive view of the accretion and evaporation histories of PBHs across the entire cosmic history is presented, with focus on the critical mass holes. The critical mass of a PBH for current era evaporation is $M_{cr}\sim 5.1\times10^{14}$ g. Across cosmic time such a black hole will not accrete radiation or matter in sufficient quantity to hasten the inevitable evaporation, if the black hole remains within an average volume of the universe. The accretion rate onto PBHs is most sensitive to the mass of the hole, the sound speed in the cosmological fluid, and the energy density of the accreted components. It is easy for a PBH to accrete to $30M_\odot$ by $z\sim0.1$ even outside any overdense region of the universe, so two merging PBHs are a plausible source for the gravitational wave events GW150914 and GW151226. However it is difficult for isolated PBHs to grow to supermassive black holes (SMBHs) at high redshift with masses large enough to fit observational constraints.

J. Rice and B. Zhang
Tue, 28 Feb 17
48/69

Comments: Accepted for publication in the Journal of High Energy Astrophysics; 33 pages, 5 figures, 3 tables


text:
# Cllcl Ple ; PcIk ; YLYclll;ele; Pla ; Ylxmcrhbd)cllcL; EE; R/R;YIY clcl; rlr; R /R; YlxDibybnd)

###### Question:

Cllcl Ple ; PcIk ; YLY clll;ele; Pla ; Ylx mcrhbd) cllcL; EE; R/R;YIY clcl; rlr; R /R; Ylx Dibybnd)

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text:
+1 vote
609 views
                             0
/     \
5        7
/    \      /   \
6       4    1     3
\
9

Tree  given in the form: (node value(left subtree)(right subtree))
For tree given above: (0(5(6()())(4()(9()())))(7(1()())(3()())))
Input format: K Tree
Output format: Sum
For example, for given tree:
Input: 2 (0(5(6()())(4()(9()())))(7(1()())(3()())))
Output: 14

I just want to know the basic logic using push and pop operation of stack ,since I think it is easy to work with stack here

0
why u want to do it by push and pop?

Better to write a simple binary tree and if node exists for the tree call sum function
0
I am not getting any clue of how to check whether the given string is a tree or not ?
+2
So basically it will be prefix evaluation rt?
0
that could be infix or postfix too. then add up all numbers. No need of parse tree for adding . rt?

+1 vote

This problem can be solved easily with recursion on the basis of parentheses count, Have a look at my code and explanation :
sum of all elements at Kth level from root of a tree, represented as string


text:
SMS scnews item created by Bill Unger at Mon 4 Jun 2012 1524
Type: Seminar
Distribution: World
Expiry: 6 Jun 2012
Calendar1: 6 Jun 2012 1505-1600
CalLoc1: Carslaw 535A
Auth: billu@daumier.maths.usyd.edu.au

# Computational Algebra Seminar: Le Borgne -- Effective computations for some Galois representations

Speaker: Jeremy le Borgne
Title: Effective computations for some Galois representations
Time & Place: 3:05-4pm, Wednesday 6 June, Carslaw 535

Galois groups. I will present the results of my thesis on how to
compute with objects involved in this theory. I will briefly recall
notions on Galois representations with coefficients in finite
fields. I will then present our main tool to study them: the theory
of phi-modules. I will describe an algorithm for the reduction of
phi-modules and explain how it can be used to compute invariants of
Galois representations.


text:
## Electronic Journal of Statistics

### Large and moderate deviations for kernel–type estimators of the mean density of Boolean models

#### Abstract

The mean density of a random closed set with integer Hausdorff dimension is a crucial notion in stochastic geometry, in fact it is a fundamental tool in a large variety of applied problems, such as image analysis, medicine, computer vision, etc. Hence the estimation of the mean density is a problem of interest both from a theoretical and computational standpoint. Nowadays different kinds of estimators are available in the literature, in particular here we focus on a kernel–type estimator, which may be considered as a generalization of the traditional kernel density estimator of random variables to the case of random closed sets. The aim of the present paper is to provide asymptotic properties of such an estimator in the context of Boolean models, which are a broad class of random closed sets. More precisely we are able to prove large and moderate deviation principles, which allow us to derive the strong consistency of the estimator of the mean density as well as asymptotic confidence intervals. Finally we underline the connection of our theoretical findings with classical literature concerning density estimation of random variables.

#### Article information

Source
Electron. J. Statist., Volume 12, Number 1 (2018), 427-460.

Dates
First available in Project Euclid: 16 February 2018

https://projecteuclid.org/euclid.ejs/1518750030

Digital Object Identifier
doi:10.1214/18-EJS1397

Mathematical Reviews number (MathSciNet)
MR3765603

Zentralblatt MATH identifier
06841010

#### Citation

Camerlenghi, Federico; Villa, Elena. Large and moderate deviations for kernel–type estimators of the mean density of Boolean models. Electron. J. Statist. 12 (2018), no. 1, 427--460. doi:10.1214/18-EJS1397. https://projecteuclid.org/euclid.ejs/1518750030

#### References

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• [28] Devroye, L., Györfi, L. and Lugosi, G. (1996)., A Probabilistic Theory of Pattern Recognition. Springer Series in Stochastic Modelling and Applied Probability, New York.
• [29] Falconer, K.J. (1986)., The Geometry of Fractal Sets. Cambridge University Press, Cambridge.
• [30] Federer, H. (1969)., Geometric Measure Theory. Springer, Berlin.
• [31] Gao, F. (2003). Moderate deviations and large deviations for kernel density estimators., J. Theoret. Probab., 16, 401–418.
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• [33] Hug, D. and Last, G. (2000). On support measures in Minkowski spaces and contact distributions in stochastic geometry., Ann. Prob., 28, 796–850.
• [34] Kingman, J.F.C. (1993)., Poisson processes. Oxford University Press, Oxford.
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• [43] Villa E. (2010). Mean densities and spherical contact distribution function of inhomogeneous Boolean models., Stoch. An. Appl., 28, 480–504.
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• [45] Zähle, M. (1982). Random processes of Hausdorff rectifiable closed sets., Math. Nachr., 108, 49–72.


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# Standard Error For Percent Change

## Contents

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## Standard Deviation Of Percentage Change Excel

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## Percent Difference Formula

• Exchange your learning and research experience among peers and get advice and insight. with standard deviation .

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Contents

group theory

# Contents

## Idea

The $p$-compact groups (Dwyer-Wilkerson 94) for some prime number $p$ are a class of ∞-groups that shares crucial properties with the class of compact Lie groups, without actually being compact Lie groups. Hence they are also called homotopy Lie groups (Møller 95).

## Properties

### Classification

The classification of $p$-compact groups states that there is a bijection between isomorphism classes of connected p-compact groups, and isomorphism classes of root data over the p-adic integers (as conjectured by Clarence Wilkerson and others, in various forms, since the early days of the theory).

This is completely analogous to the classification of connected compact Lie groups, under replacing the integers $\mathbb{Z}$ by the p-adic integers $\mathbb{Z}_p$.

Specializing to $p=2$ one gets as a corollary that any classifying space $B X$ of a connected 2-compact group $X$ splits as

$B X \cong B G \times \big(B DI(4)\big)^s$

the Cartesian product of the 2-completion of the classifying space of the compact Lie group $G$, and $s$ copies of the Dwyer-Wilkerson space $B DI(4)$ for some $s$.

$DI(4) =$ G3 corresponds to the finite $\mathbb{Z}_2$-reflection group which is number 24 on the Shepard-Todd list. It is the only irreducible finite complex reflection group which is realizable over $\mathbb{Z}_2$ but not $\mathbb{Z}$.

(Andersen-Grodal 06, see Grodal 10)

## Examples

• Let $G$ be any compact Lie group whose component group $\pi_0(G)$ is a $p$-group. Define $B \hat{G} = (B G)_p$. Then $\hat{G}$ is a $p$-compact group.

• (Sullivan) The $\mathbb{F}_p$-local sphere $(S^{n-1})_p$, where $n \gt 2$ is an integer dividing $p-1$.

• Dwyer-Wilkerson H-space

## References

Due to

Classification:

Review:

• Jesper Møller, Homotopy Lie groups, Bull. Amer. Math. Soc. (N.S.) 32 (1995) 413-428 (arXiv:math/9510218)

• Jesper Grodal, The Classification of $p$–Compact Groups and Homotopical Group Theory, Proceedings of the International Congress of Mathematicians, Hyderabad 2010 (arXiv:1003.4010, pdf, pdf)


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What is the dual of a pre-injective map? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T00:33:10Z http://mathoverflow.net/feeds/question/120120 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120120/what-is-the-dual-of-a-pre-injective-map What is the dual of a pre-injective map? Simone Virili 2013-01-28T15:51:27Z 2013-05-07T00:22:00Z
In [M. Gromov, Endomorphisms of symbolic algebraic varieties, J. Eur. Math. Soc. (JEMS) 1 (1999), 109–197], Gromov introduces the notion of pre-injective map. Recasting this notion in the setting of Abelian groups we get the following:

Definition. Given a finite discrete Abelian group $G$, an (infinite) index set $X$ and taking the product $G^X$ endowed with the product topology, we say that a continuous endomorphism $\phi:G^X\to G^X$ is pre-injective if the restriction of $\phi$ to $G^{(X)}$ (the subgroup of elements with finite support) is injective.

The Pontryagin-Van Kampen dual of $\phi$ is just an endomorphism $\widehat\phi$ of the discrete group $G^{(X)}$. Can we say that $\phi$ is pre-injective just looking at $\widehat\phi$? In other words, what is the algebraic property of $\widehat\phi$ which corresponds to the pre-injectivity of $\phi$?
 http://mathoverflow.net/questions/120120/what-is-the-dual-of-a-pre-injective-map/120146#120146 Answer by Phu Chung for What is the dual of a pre-injective map? Phu Chung 2013-01-28T20:23:38Z 2013-01-28T20:32:32Z
Well, in the case X is also a group, your question somehow relates with cellular automata, i.e your map $\phi :G^X\rightarrow G^X$ is continuous and X-equivariant. When X is amenable,$\phi$ is pre-injective if and only if $\phi$ is surjective. Thus, your dual $\widehat{\phi}$ will be injective in this case. It is not true anymore when $X=F_2$, the free group with 2 generators, see section 5.11 of Ceccherini-Silberstein, Coornaert's book "Cellular Automata and Groups" for an example of pre-injective but not surjective cellular automaton over $F_2$.


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# Funding public goods

Public goods in economics are those things that are desirable, but neither rivalrous nor excludable. That is, one person’s enjoyment of the good doesn’t detract from anyone else’s enjoyment of it, and it’s impossible to prevent people who haven’t paid for it from enjoying it.

For the purposes of this article, the main example of public goods I’ll be referring to is that of free cultural works, such as open source software and public domain audiobooks.

Obviously, a lot of these public goods are being produced, sometimes for altruistic reasons, sometimes for fun, and sometimes because it makes economic sense for the producer of the good, even if no-one else who will enjoy the good contributes.

Now, there’s nothing wrong with altruism, but people can only afford a certain amount of it, so perhaps we could have even more public goods if we could figure out an efficient way of funding them. How well can we do?

One way of funding public goods is through pledges, as on Indiegogo or Kickstarter. There, a would-be producer of a public good can post a campaign, essentially saying “If people pledge a total of at least \$$x$ by such-and-such a date, I’ll produce this specific public good”. People interested in that particular public good can then pledge money to the campaign, knowing that they’ll be refunded if the total isn’t met. (They also allow campaigns where the money is paid to the campaign even if the target isn’t met, but I’m mainly interested in the kind of campaigns I’ve described above.)

The problem with this is that if someone thinks the total is likely to be met without their pledge, then they have no (non-altruistic) incentive to pledge their own money; they’ll be able to enjoy the eventual public good without contributing. And if they think the total is unlikely to be met even with their pledge, then they still have no incentive to go to the trouble of pledging their money, since the good is unlikely to be produced either way.

It’s only in the unlikely scenario that they think their pledge will be crucial in reaching the total that they have an incentive to make their pledge. So this threshold pledge system, or street performer protocol, as it’s sometimes called, still often relies on a degree of altruism.

One interesting alternative is the so-called Wall Street performer protocol, in which bonds are issued for the completion of a specific public good; they pay out to the owners of the bonds when the public good is supplied, so obviously the owners have an incentive to ensure that the good is supplied. Perhaps I’ll discuss this idea further in a future article, but I have another idea to tell you about in this one.

When a public good can be sufficiently well specified, another option is to offer a bounty to anyone who provides that good. This is sometimes used for fixing bugs in open source software; see, for example, Bountysource. Anyone with the desire can pledge money to increase the bounty for fixing a bug that particularly bugs them.

One problem with these bounties is that they encourage unnecessary duplication of effort. Whoever is the first to provide a solution to the bug collects the bounty. If two or three people are simultaneously working on the same bug, the runners-up aren’t compensated for their effort, even if their eventual solution is better in some way than the first solution provided. This risk of non-compensation might reduce the incentive for talented people to try to collect the bounty.

A particular feature of one obscure bounty website caught my eye, though. The only project currently using BidforFix is OsmAnd. Now, the OsmAnd developers can look at which features people are bidding for at BidforFix and indicate their current top priorities.

This got me to thinking: If someone was particularly interested in OsmAnd’s current third priority, and wanted it done sooner, they would have an incentive to bid for that task in an attempt to raise it to the second, or even the first priority, especially if they believed a small additional bid would be sufficient to raise the priority.

Then, I imagined a situation where someone announces: “I’ll provide either public good X or public good Y. X will take twice as much work as Y, so if, before a week from next Tuesday, the public pledges more than twice as much money for X as they do for Y, I’ll take the money pledged for X and provide X; otherwise I’ll take the money pledged for Y and provide Y.” Then, the public knows exactly when a pledge will be crucial in determining which of the public goods is provided.

Suppose there aren’t any pledges yet, and Alice comes along and reasons “Having Y produced instead of X is worth \$6 to me, but a \$1 pledge is sufficient to tip the scales, so I’ll pledge \$1 for Y”.

Then Bob comes along and reasons “I’d pay \$9 for X to be produced instead of Y, but a \$3 pledge is enough to tip the scales, so I’ll pledge \$3 for X”.

Then Alice comes back and discovers that X is leading, so she pledges an additional dollar for Y.

This continues until Bob has pledged \$9 for X, and Alice has pledged \$5 for Y.

But then along comes Cynthia, who’s willing to pay \$4 for X to be produced instead of Y, so she pledges more money towards X, until Alice has pledged all \$6 for Y, and Bob and Cynthia have pledged a total of \$13 for X.

Of course, more people might come along and pledge more for each of X and Y, but I hope you get the idea.

The effect is that the producer of the public good receives some compensation for their effort, according to how much the public actually wants that good, rather than another. And the important point is this: this happens even without any altruism on the part of the people making the pledges.

It’s not perfect, though; people who would enjoy either X or Y, but are indifferent about which one is produced, still have no non-altruistic incentive to make a pledge, though if they are altruistic, they can safely pledge to both campaigns, knowing that they’ll only have to pay for whichever one wins.

And it’s not only indifferent people who lack incentives to reveal the full value to themselves of the public goods. For example, Y might be worth \$150 to Alice, but because X is worth \$144 to her, she’ll bid no more than \$6 to have Y provided instead of X.

But perhaps this idea will, in some situations, improve on existing mechanisms to non-coercively get the public to pay for public goods.


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# Degree choice — Math or physics

I realize that it is of bad form to cross-post. However, on this matter I very much hope to hear arguments from all sides...

I hope to become a physicist focusing mainly on the theoretical side in the future. I am trying to decide whether to go for a physics or math undergrad course.

Assuming that I am capable of doing either, what are the pros and cons of either route?

I know that mathematics is essential to doing physics, and in most math courses, there are applied math modules that are very much related to physics. Also that many research physicists have math degrees. But surely there are reasons for people to choose the physics course over the math course and vice versa? In particular, what are the possible advantages of a physics degree over a math one, given that a math degree could lead to PhD in physics and possibly open more possibilities in say the financial world (though that is not my interest)?

Thank you.

-
if you have cross-posted, also add the link to the original post. –  rrampage Sep 19 '12 at 10:52
@rrampage: Sure, I have added the link. –  Rinaldo Sep 19 '12 at 10:59
A very reasonable choice as an undergrad is to do both. –  user26872 Sep 19 '12 at 11:16
Thank you, @oen . Indeed that would be best, but the institution I have a place at does not really do both... The physics course comes with supplementary math, but definitely much less rigorous than the math course. whereas the math course is very rigorous but offers a very narrow range of physics topics... –  Rinaldo Sep 19 '12 at 11:27
Where are you doing this? In most US universities, even if you major in physics, you can choose to take the mathematics courses intended for math majors in place of the (less rigorous) math for physics majors. Then you will have your physics degree, but also the greater facility in math. (But, perhaps in some other countries physics majors may take only the physics course and not substitute the courses intended for math majors... Then your choice is more crucial.) –  GEdgar Sep 19 '12 at 13:04

My 2 cents on this:

• If you are mainly interested in being a theoretical physicist, going for Math may better, given that theoretical physics requires a very high level understanding of the more abstract mathematics. I would advise you to at least take some basic physics courses so that you may try to find out which topic excites you the most.
• If you are looking to keep as many potential career paths open (as interests may change over time), Physics is definitely not a bad choice. It exposes you to math, basic engineering, material science, statistics, some programming. Not to mention that experiments in physics form a backbone for basic understanding of several physical concepts (After all, seeing is believing). You do not require too much infrastructure for learning maths; but that University Physics lab is vital for supplementing your knowledge in physics. As an after-thought, many of the current financial products and theories have been developed by engineers and physicists too. So Finance will not be a forbidden sector if you take Physics.

Hope this helps.

-
Thank you, rrampage, for the nice analysis! –  Rinaldo Sep 19 '12 at 11:28

I don't know if this is an option for you, but I would suggest you do both. At my university (Utrecht University in the Netherlands) there is a combined mathematics/physics course. Because it is heavier than the separate mathemtics and physics courses, it is a good preparation for a PhD. scholarship in either discipline. Moreover, when you realize that mathematics is the superior discipline, you can always drop physics ;-)

-

And here are mine -

I choose Physics with Astrophysics as my degree course, I had similar interests to you (I think, regardless of the fact that you haven't listed them) and had in the back of my mind the drive to complete a PhD when my undergrad was all done.

I knew I was more interested in the theoretical side rather than the experimental and as such after one year of studying Physics with a LOT of time spent in the lab I converted my degree to Mathematics with Physics. the best choice I made. I studied only mathematics and theoretical physics modules and then in my 3rd year (at a UK institution) I choose as many Mathematical physics modules as I could $wrt$ the $120$ credits on offered per academic year. This allowed me to study Quantum Field theory, General Relativity and String Theory, along with electives from the Physics department on Condensed Matter Theory and Astro-particle Physics.

I thoroughly enjoyed my undergrad and long for lost days getting red-faced in anger at Green's functions for Feynmann diagram propagators. Really, for me, the combined course had it all.

It then prepared me for a PhD in Cosmology (Inflation & Superstrings) and the abstract mathematical modules I studied in my Master's year really became invaluable.

In short, a combination might be useful for you, but do remember that you gain just as much as you lose by Studying one or the other.

-
Very true, I am not too keen on lab work either! –  Rinaldo Sep 19 '12 at 12:42

Take a math undergraduate degree. You must understand the math before you can understand the physics. If you do not understand the math then you will spend most of your time learning the math while taking the physics courses. This will make you less competitive. The top students in theoretical physics always know more math than the other students.

-


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# Definite integral as the limit of a Riemann sum

AP.CALC:
LIM‑5 (EU)
,
LIM‑5.B (LO)
,
LIM‑5.B.1 (EK)
,
LIM‑5.B.2 (EK)
,
LIM‑5.C (LO)
,
LIM‑5.C.1 (EK)
,
LIM‑5.C.2 (EK)
Definite integrals represent the exact area under a given curve, and Riemann sums are used to approximate those areas. However, if we take Riemann sums with infinite rectangles of infinitely small width (using limits), we get the exact area, i.e. the definite integral! Created by Sal Khan.

## Want to join the conversation?

• Has definite integrals something to do with derivatives? Or its completely different thing?
• No no no. Think of definite integration as the REAL integration (use the term "integral" to mean this definite integral). Integration existed LONG LONG LONG LONG before differentiation was even invented. It is a different subject. Archimedes and the greeks would integrate curves, shapes, etc. etc. to find areas.

Differentiation is a different thing. And the opposite of differentiation is antidifferentiation (which also has the UNFORTUNATE name of indefinite integral, but please use "antidifferentiation" to refer to opposite of differentiation).

Now these two are separate fields. Integration is finding areas under curves, between curves, inside shapes, etc. etc. etc.

Differentiation is finding rate of change of functions (i.e., slopes). And antidifferentiation is the reverse (that is, we have a function f, and we try to figure which function's rate of change f represents).

Now fast forward to Newton, Leibniz time. They already knew a lot about integration and differentiation as separate subjects. But these two made a connection.

Integration (definite) OF CONTINUOUS function is related to Antidifferentiation of that function

So the connection is a big theorem, but it is for continuous functions. Integration(area finding) can be done on discontinuous functions (i.e. jump discontinuities) but the theorem doesn't apply to these. It only connects integration of continuous functions to antidifferentiation. Learn the two subjects like they are totally different. Then later connect integration of cont. functions with differential calculus.
• Why wouldn't people use circles to approximate areas under curves?
• Technically you could, assuming that you'd be using circles small enough to meet your requirements for accuracy. Think about this in three dimensions- you can fill a volume with spheres or balls and then summing up the volume of the spheres. Circles don't fit together very neatly, and this is actually pretty mathematically hard to figure out if you wanted to be rigorous about it, but give it a try both ways. In any case, if you can use circles to approximate the area under a curve you'll pretty much know this subject well enough to do anything.
• Let's say you have a function y=x^2 and you want to find the area under the curve along the interval -3 to 3 and you're delta x was 6 (or any finite number). Could you get not just an approximation, but the exact area under the curve if you took the left side approximation and added it to the right side approximation and then divided that sum by 2? In other words, would the amount that you overestimated the area under the curve be exactly the same amount that you underestimated the area under the curve? Would the value of the overestimation (in either the left-side approximation, or right-side approximation) be equal to the value of the underestimation? My guess is that if the function has a vertical line of symmetry AND the interval along the x-axis extends an equal distance away from the line of symmetry in both directions (in our example -3 to 3 with the line of symmetry being the y-axis) that the amount of overestimation and underestimation would be the same and when you took the average of the two amounts you would be left with the exact value of the area under the curve. Thoughts?
• The quick answer is no, that wouldn't work. :-)

You are correct that it would work for a very limited number of cases on a very limited number of functions, but it wouldn't actually work for the example you give, because the function is curved. That means the average of the right approximation and the left approximation is NOT the same as the actual value of the area -- we can see this with the function x^2 by averaging the value at 0 (which is 0) and the value at 1 (which is 1). The average of 1/2 is not the same as the area under the function, which is 1/3.

Because the number of cases this would work for is so limited, it's easier just to use integration for everything -- and this approach has the advantage that it always works.
• Apart from the rectangular bars, there is some area left under the curve. What about that area?
• If you use finitely many Riemann rectangles, you have an approximation to the area under the curve. It is only through taking a limit as the number of rectangles increases ad infinitum (and their width shrinks to zero) that you will obtain the exact area under the curve.
• Can we replace n appraoch infinity by delta x approch zero? If not then why so?
• The real Reimann integral does it this way. It allows any for number of intervals , and each interval can be any size. Then we take the limit of the min_interval_size to 0.
• Wouldn't an infinitely small `Δx` be notated with `δx` or just `δ`?
• The Leibniz notation is dx.
Why? - years on convention.
The x must remain since we need to know with respect to what variable the function is being integrated. This is more obvious once you get into multi variable calculus.
• What is the correct way of saying that delta x becomes extremely small (approaches 0) in Calculus context - delta x becomes infinitely small or delta x becomes infinitesimally small? I know it's a trivial thing, but I wish to use the correct terminology.
• It's neither. We allow Δx to become arbitrarily small. That is, we can take Δx as close to 0 as you wish. The term 'infinitesimal' can help guide intuition, but it's a defining property of the real numbers that nothing is infinitesimal.
• How did Bernard Riemann get all the properties of the integral from the Riemann sum? For example how did he know the definite integral from a to b of f(x) is F(b)-F(a). from the sum because it is infinite. Or if you take the derivative of the indefinite integral you get the integrand (how do you take the derivative of a limit and sigma notation)?
• Those are some very good questions. Any good book on calculus or one on elementary real analysis treating the Riemann integral should answer your questions (the details are too lengthy for me to type up here).

Let me point out two subtle facts. Firstly, a function may possess an anti-derivatve, yet fail to be (Riemann) integrable. This fact is often overlooked, especially at the elementary level. What is more, even if `ƒ` is an integrable function on `[a, b]`, and we define the function `F` on `[a,`` b]` by

`F(x) = ∫ [a, x] ƒ(t) dt,`

the integral going from `a` to `x`, `F` need not be differentiable. In other words, it need not be the case that `F'(x) = ƒ(x)` for all `x` - one can show that `F` is differentiable at `x` if and only if `ƒ` is continuous at `x`. Moreover, one can show that the set of points in `[a, b]` at which `ƒ` is discontinuous has measure zero, so `F` is differentiable almost everywhere (in a technical sense).
• is it possible to use Simpson's rule even when n is not an even number?
• "any" excludes unbounded functions, or functions with "too many" discontinuities and other such degeneracies (but for now, assume you can definite integrate any function that doesn't blow up in size or anything like that)
(1 vote)
• I still just don't get how Riemann went from having the sigma notation to translating it to a definite integral? How did he know that the process of definitely integrating a function (integrating using the formulas and then replacing the bounds in the integral and subtracting them) would lead to the same answer? This question is really getting on my nerves because nowhere online have I found an explanation.
• Hi Christina
You have two questions here, one of notation, how did we go from Σ to ∫, and the other of how did definite integral come about.

Notation
If you have made it to integral calculus, you must have come through algebra and differential calculus, and if so, you have already seen a change in notation, so I’ll start there.

When you were in algebra you calculated the slope of a line as Δy/Δx. When you got to differential calculus the problem was not the slope of a line, but the slope of a curve at some point. The concept was developed by using a secant line, or average slope. If we wanted to estimate the slope at a point c, we drew a line between two points on either side of c that were – Δy/Δx and Δy/Δx from c. The we asked, what happens as Δy/Δx gets smaller and smaller, that is, the endpoints of the secant line get closer and closer to c. We decided that once the Δ of the Δy/Δx was infinitesimally small, we would change the notation from Δy/Δx to dy/dx, to remind us that we are dealing with infinitesimals (also called differentials). So for the slope of a straight line or secant line we used Δy/Δx but for the slope of a tangent line we use dy/dx.

It is pretty much the same deal on how we went from Σ to ∫. The Riemann sum is a sum of sections whose width is Δx, so we have, in general, Σf(x)Δx. As we make Δx smaller and smaller, until it is infinitesimal, we again change the notation from Δx to dx AND we change the notation of Σ to ∫, that is Σf(x)Δx to ∫f(x)dx. It really is just sort of a visual reminder that we are dealing with infinitesimal changes in x, that is, dx. In case you didn’t know, the integral symbol ∫ is just an elongated S, which stand for sum, so yes, the Riemann sum is the same as the Riemann integral, the only difference is the Δx is infinitesimally small.

The Definite Integral
As far as how the definite integral came about, that happened way before Riemann. The 2 theorems are called the Fundamental Theorems of Calculus. They are talked about here on Khan. This Wikipedia page has proofs of them that do not require math skills above what you should have by now - it will clearly show how the definite integral was "discovered". Here are a few links to get you going:
https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
http://mathforum.org/library/drmath/view/53374.html

I sure hope that helped. If not send me a comment on what you need clarified.
Stefen


text:
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Example

# Problem: A fireman of 80 kg slides down a pole. When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s. By how much has thermal energy increased during his slide?

###### FREE Expert Solution

Change in thermal energy:

$\overline{)\begin{array}{rcl}\mathbf{∆}\mathbf{T}\mathbf{E}& {\mathbf{=}}& \mathbf{P}\mathbf{E}\mathbf{-}\mathbf{K}\mathbf{E}\\ & {\mathbf{=}}& \mathbf{m}\mathbf{g}\mathbf{h}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{\mathbf{v}}^{\mathbf{2}}\end{array}}$

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###### Problem Details

A fireman of 80 kg slides down a pole. When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s.

By how much has thermal energy increased during his slide?