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## Section9.4Graphs of Quadratic Functions Chapter Review

### Subsection9.4.1Introduction to Functions

In Section 9.1 we covered the definitions of a relation, a function, and domain and range. We then discussed function notation and how to evaluate functions at a particular value as well as how to solve an equation with function notation. Last we introduced the vertical line test

###### Example9.4.1Introduction to Functions

One week in Portland, it was rainy. Shocking, I know. Shown is a diagram for the day of the week and how much rain fell on those days. Convert this diagram to a set of ordered pairs where the first coordinate is the day of the week and the second coordinate is the amount of rain that fell in Portland that day, in inches.

Explanation

Since Monday has an arrow to $0.2\text{,}$ that must mean that it rained $0.2$ inches of rain on Monday. That gives us the ordered pair $(\text{Monday},0.2)\text{.}$ Likewise, we also have $(\text{Tuesday},0.4)\text{.}$ To write down all of the ordered pairs, we will use a set:

\begin{equation*} \left\{(\text{Monday},0.2), (\text{Tuesday},0.4), (\text{Wednesday},0.0), (\text{Thursday},0.6), (\text{Friday},0.0)\right\} \end{equation*}
###### Example9.4.3Functions as Predictors

Recall that a a function should predict the output perfectly if you know the input. In Example 9.4.1, we saw that the day of a particular week in Portland was related to the rainfall during that week. Was this relation a function?

Explanation

Yes, that relation was a function. If I tell you the day of the week, you can with certainty “predict” how much rain there was on that day by looking at the data. There was a single answer to the question “How much rain fell on Wednesday during that week in Portland?”

At this point, we should note that all historical weather data can be viewed as a function of some sort: On any particular day in the recent past, we can “predict” (by looking it up somewhere) how much rain fell in any city in the world.

However, the opposite question, “If it rained $0.0$ inches, what day of the year was it?” is not going to represent a function because there will be more than one day of the year that it didn't rain at all. Even in Portland.

###### Example9.4.4Algebraic Functions
1. For the equation $y=2x^2+x\text{,}$ will $y$ be a function of $x\text{?}$

2. For the equation $y^2=x\text{,}$ will $y$ be a function of $x\text{?}$

Explanation
1. This question is equivalent to the question “If you input any $x$ value, will there be only one $y$ value?” The answer is “yes.” If you input any number for $x$ you first square that number and multiply it by two, then add the number. For example, if we substitute $\substitute{-3}\text{,}$ we get:

\begin{align*} y\amp=2(\substitute{x})^2+(\substitute{x})\\ y\amp=2(\substitute{-3})^2+(\substitute{-3})\\ \amp=2(9)-3\\ \amp=18-3\\ \amp=15 \end{align*}

There is no way we could substitute a number for $x$ and get more than one value for $y\text{.}$

2. This question is equivalent to the question “If you input any $x$ value, will there be only one $y$ value?” The answer is “no.” If you input any number for $x$ and solve for $y\text{,}$ you might actually get two solutions. For example, if we substitute $\substitute{1}\text{,}$ we get:

\begin{align*} y^2\amp=\substitute{x}\\ y^2\amp=\substitute{1} \end{align*}
\begin{align*} y\amp=\sqrt{1}\amp\text{ or }\amp\amp y\amp=-\sqrt{1}\\ y\amp=1\amp\text{ or }\amp\amp y\amp=-1 \end{align*}

Since there is more than one $y$ value for the $x$ value $1\text{,}$ the relation $y^2=x$ is not a function.

###### Example9.4.5Function Notation

For the function $V=g(x)\text{,}$ what is the name of the function, what is the input variable, and what is the output variable?

Explanation

For the function $V=g(x)\text{,}$ the name of the function is $g\text{,}$ the input variable is $x\text{,}$ and the output variable is $V\text{.}$

###### Example9.4.6Evaluating Functions

Let $a(x)=x^2-2x\text{,}$ let the function $b$ be defined by the graph in Figure 9.4.7, and let the function $c$ be defined by the table in Table 9.4.8.

 $x$ $c(x)$ $-5$ $7$ $-3$ $-2$ $2$ $2$ $6$ $4$ $9$ $7$
1. Evaluate $a(-3)\text{.}$

2. Evaluate $b(-3)\text{.}$

3. Evaluate $c(-3)\text{.}$

Explanation
1. To evaluate $a(-3)$ we will substitute the value $\substitute{-3}$ in wherever we see $x$ in the equation for $a(x)\text{.}$

\begin{align*} a(\substitute{x})\amp=\substitute{x}^2-2\substitute{x}\\ a(\substitute{-3})\amp=(\substitute{-3})^2-2(\substitute{-3})\\ \amp=9+6\\ \amp=15 \end{align*}
2. To evaluate $b(-3)\text{,}$ we need to look at the graph of $y=b(x)$ and look for the one place on the graph where the $x$-value is $-3\text{.}$ The $y$-value at this point is $1\text{,}$ so we would say that

\begin{equation*} b(-3)=1 \end{equation*}
3. To evaluate $c(-3)$ we examine the table and find the place where the $x$-value is $-3\text{.}$ We conclude that $c(-3)=-2\text{.}$

###### Example9.4.9Solving Equations That Have Function Notation

Still let $a(x)=x^2-2x\text{,}$ let the function $b$ be defined by the graph in Figure 9.4.10, and let the function $c$ be defined by the table in Table 9.4.11.

 $x$ $c(x)$ $-5$ $7$ $-3$ $-2$ $2$ $2$ $6$ $4$ $9$ $7$
1. Solve the equation $a(x)=0\text{.}$

2. Solve the equation $b(x)=2\text{.}$

3. Solve the equation $c(x)=7\text{.}$

Explanation
1. To solve the equation $a(x)=0\text{,}$ we should set the formula for $a(x)$ equal to $0\text{.}$ In this case, factoring will then help.

\begin{align*} x^2-2x\amp=0\\ x(x-2)\amp=0 \end{align*}
\begin{align*} x\amp=0\amp\text{ or }\amp\amp x-2\amp=0\\ x\amp=0\amp\text{ or }\amp\amp x\amp=2 \end{align*}

So, the solution set is $\{0,2\}\text{.}$

2. To solve the equation $b(x)=2\text{,}$ we should examine the graph and find all of the locations where the $y$-value is $2\text{.}$ According to the graph, that happens twice, both when $x=-4$ and when $x=0\text{.}$ So the solution set is $\{-4,0\}.$

3. To solve the equation $c(x)=7\text{,}$ we examine the table and find all the places where the function value is $7\text{.}$ According to the table, that happens twice: once when $x=-5$ and again when $x=9\text{.}$ So the solution set is $\{-5,9\}\text{.}$

###### Example9.4.12Domain and Range

Still let the function $b$ be defined by the graph in Figure 9.4.13, and let the function $c$ be defined by the table in Table 9.4.14.

 $x$ $c(x)$ $-5$ $7$ $-3$ $-2$ $2$ $2$ $6$ $4$ $9$ $7$
1. Write the domain of $b$ in interval notation.

2. Write the domain of $c$ in interval notation.

3. Write the range of $b$ in interval notation.

4. Write the range of $c$ in interval notation.

Explanation

Recall the definitions of domain and range.

1. To find the the domain of $b\text{,}$ look at the graph of $b$ and see what $x$-values are used for the graph. In other words, what $x$-values could you input and expect to get a $y$-value? It looks like the graph extends forever to the right as well as the left. Therefore, the domain is $(-\infty,\infty)\text{.}$

2. To find the the domain of $c\text{,}$ we list the $x$-values in the table in a set. That would be $\{-5,-3,2,6,9\}\text{.}$

3. To find the the range of $b\text{,}$ look at the graph of $b$ and see what $y$-values are on the graph. The lowest $y$-value is $-3$ and all higher $y$-values are possible outputs. Therefore, the range is $[-3,\infty)\text{.}$

4. To find the the range of $c\text{,}$ we list the $y$-values in the table in a set. That would be $\{7,-2,2,4\}\text{.}$

###### Example9.4.15Determining if a Relation is a Function

Is $y$ a function of $x$ in the following graphs?

Explanation

The graph in Figure 9.4.16 fails the vertical line test and so $y$ is not a function of $x\text{.}$ The graph in Figure 9.4.17 passes the vertical line test and so $y$ is a function of $x\text{.}$

In Section 9.2 we covered the definition of a quadratic function, and how to determine the direction, vertex, axis of symmetry, and vertical and horizontal intercepts. We then learned how to algebraically find the vertex, how to graph a parabola using a table, and how to find the domain and range of quadratic functions.

###### Example9.4.18Finding the Vertex and Axis of Symmetry

Algebraically find the vertex of the parabola described by the quadratic function $f(x)=3x^2+8x-7\text{.}$

Explanation

To find the vertex of a parabola algebraically, we use the formula $h=-\frac{b}{2a}$ to find the axis of symmetry first. For our equation, $\substitute{a=3}$ and $\lighthigh{b=8}\text{,}$ so:

\begin{align*} h\amp=-\frac{\lighthigh{8}}{2(\substitute{3})}=-\frac{4}{3} \end{align*}

Next, we evaluate $f\mathopen{}\left(\substitute{-\frac{4}{3}}\right)\mathclose{}$ to find the $y$-coordinate of the vertex.

\begin{align*} f\mathopen{}\left(\substitute{-\frac{4}{3}}\right)\mathclose{}\amp=3\left(\substitute{-\frac{4}{3}}\right)^2+8\left(\substitute{-\frac{4}{3}}\right)-7\\ \amp=3\left(\frac{16}{9}\right)-\frac{32}{3}-7\\ \amp=\frac{16}{3}-\frac{32}{3}-\frac{21}{3}\\ \amp=\frac{37}{3} \end{align*}

So, the parabola has its axis of symmetry at $x=-\frac{4}{3}$ and has its vertex at the point $\left(-\frac{4}{3},-\frac{37}{3}\right)\text{.}$

Identify the key features of the quadratic function $y=-2x^2-4x+6$ shown in Figure 9.4.20.

Explanation

First, we see that this parabola opens downward because the leading coefficient is negative.

Then we locate the vertex which is the point $(-1,8)\text{.}$ The axis of symmetry is the vertical line $x=-1\text{.}$

The vertical intercept or $y$-intercept is the point $(0,6)\text{.}$

The horizontal intercepts are the points $(-3,0)$ and $(1,0)\text{.}$

###### Example9.4.22Graphing Quadratic Functions by Making a Table

Determine the vertex and axis of symmetry for the quadratic function $g(x)=2x^2+8x+2\text{.}$ Then make a table of values and sketch the graph of the function.

Explanation

To determine the vertex of $g(x)=2x^2+8x+2\text{,}$ we want to find the $x$-value of the vertex first. We will use $h=-\frac{b}{2a}$ for $\substitute{a=2}$ and $\lighthigh{b=8}\text{:}$

\begin{align*} h\amp=-\frac{\lighthigh{b}}{2\substitute{a}}\\ h\amp=-\frac{\lighthigh{8}}{2(\substitute{2})}\\ h\amp=-2 \end{align*}

To find the vertex, we evaluate $g(\substitute{-2})\text{.}$

\begin{align*} g(\substitute{-2})\amp=2(\substitute{-2})^2+8(\substitute{-2})+2\\ \amp=2(4)-16+2\\ \amp=8-16+2\\ \amp=-6 \end{align*}

Now we know that our axis of symmetry is the line $x=-2$ and the vertex is the point $(-2,-6)\text{.}$ We will set up our table with two values on each side of $x=-2\text{.}$ We choose $x= -4, -3, \highlight{-2}, -1,$ and $0\text{.}$ Then, we'll determine the $y$-coordinates by replacing $x$ with each value and we have the complete table as shown in Table 9.4.23. Notice that each pair of $y$-values on either side of the vertex match. This helps us to check that our vertex and $y$-values are correct.

Now that we have our table, we will plot the points and draw in the axis of symmetry as shown in Figure 9.4.24. We complete the graph by drawing a smooth curve through the points and drawing an arrow on each end as shown in Figure 9.4.25

###### Example9.4.26The Domain and Range of Quadratic Functions

Use the graph of $g(x)=2x^2+8x+2$ in from Example 9.4.22 shown in Figure 9.4.25 to identify the domain and range of $g$ in set-builder and interval notation.

Explanation

For the domain, we look horizontally and see the graph is a continuous curve and one arrow points to the left and the other arrow points to the right. The domain is $\{x\mid x\ \textrm{is a real number}\}$ which is equivalent to $(-\infty,\infty)\text{.}$

For the range we look up and down on the graph, which opens upward. Both arrows point upward and the lowest point on the graph is the vertex at the point $(-2,-6)\text{.}$ The range is $\{y \mid y \ge -6\}$ which is equivalent to $[-6,\infty)\text{.}$

###### Example9.4.27Applications of Quadratic Functions Involving the Vertex

The value of FedEx stock, in dollars, between December 28, 2017 and February 9, 2018 can be very closely approximated by a quadratic function, $F(x)=-0.07x^2+2.7x+248\text{.}$ Find the vertex of this parabola and interpret it in the context of the situation.

Explanation

To find the vertex of $F(x)=-0.07x^2+2.7x+248\text{,}$ we first find the axis of symmetry using the formula $h=-\frac{b}{2a}\text{.}$ For our equation, $\substitute{a=-0.07}$ and $\lighthigh{b=2.7}\text{,}$ so:

\begin{align*} h\amp=-\frac{\lighthigh{2.7}}{2(\substitute{-0.07})}\\ h\amp=-\frac{2.7}{-0.14}\\ h\amp\approx19.3 \end{align*}

Next, we substitute the value of the axis of symmetry, $\substitute{19.3}\text{,}$ into the formula for the parabola.

\begin{align*} F(\substitute{x})\amp=-0.07\substitute{x}^2+2.7\substitute{x}+248\\ F(\substitute{19.3})\amp=-0.07\left(\substitute{19.3}\right)^2+2.7\left(\substitute{19.3}\right)+248\\ \amp\approx274 \end{align*}

So, the parabola has its axis of symmetry at $x=19.3$ and has its vertex at about the point $(19.3,274)\text{.}$ In addition, since $a=-0.07\text{,}$ the parabola will be opening downward. Given this information, the vertex of this parabola is its maximum.

In the reality of the situation, the $19.3$ indicates the number of days after December 28, 2017 which would be sometime on January 16. The $274$ indicates the stock price in dollars. In conclusion, FedEx's stock peaked (between December 28, 2017 and February 9, 2018) on January 16 at a value of $\274\text{.}$

In Section 9.3 we covered how to find the vertical and horizontal intercepts of a quadratic function algebraically, how to make a graph of a parabola using key features, applications of quadratic functions and what their real world domains and ranges mean, and how to tell a quadratic function from other types of functions.

###### Example9.4.28Finding the Vertical and Horizontal Intercepts Algebraically
1. Algebraically determine the vertical and horizontal intercepts of the quadratic function $h(x)=6x^2-13x+6\text{.}$

2. Algebraically determine the vertical and horizontal intercepts of the quadratic function $k(x)=2x^2-2x-5\text{.}$

Explanation
1. To find the vertical intercept, we evaluate the function when $x=0\text{.}$

\begin{align*} h(\substitute{0})\amp=6(\substitute{0})^2-13(\substitute{0})+6\\ \amp=6 \end{align*}

So, the vertical intercept of $h$ is $(0,6)\text{.}$

Next, to find the horizontal intercepts, we set the function equal to zero. To solve that equation we will use Algorithm 8.6.2. For this particular example, we will practice factoring using the AC method. Here, $ac=36$ and factor pairs that add up to $-13$ are $-9$ and $-4\text{,}$ as you will see.

\begin{align*} \substitute{h(x)}\amp=6x^2-13x+6\\ \substitute{0}\amp=6x^2\mathbin{\highlight{-}}\mathbin{\highlight{13x}}+6\\ 0\amp=6x^2\mathbin{\highlight{-}}\mathbin{\highlight{9x}}\mathbin{\highlight{-}}\mathbin{\highlight{4x}}+6\\ 0\amp=3x\highlight{(2x-3)}-2\highlight{(2x-3)}\\ 0\amp=(3x-2)\highlight{(2x-3)} \end{align*}
\begin{align*} 0\amp=3x-2\amp\text{ or }\amp\amp0\amp=2x-3\\ x\amp=\frac{2}{3}\amp\text{ or }\amp\amp x\amp=\frac{3}{2} \end{align*}

So, the horizontal intercepts are $\left(\frac{2}{3},0\right)$ and $\left(\frac{3}{2},0\right)\text{.}$

2. To find the vertical intercept, we have to evaluate the function when $x=0\text{.}$

\begin{align*} k(\substitute{0})\amp=2(\substitute{0})^2-2(\substitute{0})-5\\ \amp=-5 \end{align*}

So, the vertical intercept of $k$ is $(0,-5)\text{.}$

Next, to find the horizontal intercepts, we set the function equal to zero. To solve that equation we will use Algorithm 8.6.2. For this particular example, we will use the quadratic formula because the square root method will not work (because there is a linear term) and factoring fails.

\begin{align*} \substitute{k(x)}\amp=2x^2-2x-5\\ \substitute{0}\amp=2x^2-2x-5\\ \end{align*}

We identify that $\substitute{a=2}\text{,}$ $\lighthigh{b=-2}\text{,}$ and $c=-5$

\begin{align*} x\amp=\frac{-\lighthigh{b}\pm\sqrt{\lighthigh{b}^2-4\substitute{a}c}}{2\substitute{a}}\\ x\amp=\frac{-(\lighthigh{-2})\pm\sqrt{(\lighthigh{-2})^2-4(\substitute{2})(-5)}}{2(\substitute{2})}\\ x\amp=\frac{2\pm\sqrt{4+40}}{4}\\ x\amp=\frac{2\pm\sqrt{44}}{4}\\ x\amp=\frac{2\pm\sqrt{\highlight{4}\cdot11}}{4}\\ x\amp=\frac{2\pm\highlight{2}\sqrt{11}}{4}\\ x\amp=\frac{2}{4}\pm\frac{2\sqrt{11}}{4}\\ x\amp=\frac{1}{2}\pm\frac{\sqrt{11}}{2}\\ x\amp=\frac{1\pm\sqrt{11}}{2} \end{align*}

So, the horizontal intercepts are $\left(\frac{1+\sqrt{11}}{2},0\right)$ and $\left(\frac{1-\sqrt{11}}{2},0\right)\text{.}$ If you wanted to graph these points, you would need to approximate them as $\left(2.16,0\right)$ and $\left(-1.16,0\right)\text{.}$

###### Example9.4.29Graphing Quadratic Functions Using Their Key Features

Graph the function $j(x)=-2x^2+6x+8$ by algebraically determining its key features.

Explanation

To start, we'll note that this function will open downward, as the leading coefficient is negative.

To find the $y$-intercept, we'll evaluate $j(0)\text{:}$

\begin{align*} j(\substitute{0})\amp=-2(\substitute{0})^2+6(\substitute{0})+8\\ \amp=8 \end{align*}

The $y$-intercept is $(0,8)\text{.}$

Next, we'll find the horizontal intercepts by setting $j(x)=0$ and solving for $x\text{:}$

\begin{align*} -2x^2+6x+8\amp=0\\ -2(x^2-3x-4)\amp=0\\ 2(x-4)(x+1)\amp=0 \end{align*}
\begin{align*} x-4\amp=0\amp \text{or}\amp\amp x+1\amp=0\\ x\amp=4\amp \text{or}\amp\amp x\amp=-1 \end{align*}

The $x$-intercepts are $(4,0)$ and $(-1,0)\text{.}$

Lastly, we'll determine the vertex. Noting that $\substitute{a=-2}$ and $\lighthigh{b=6}\text{,}$ we have:

\begin{align*} x\amp=-\frac{\lighthigh{b}}{2\substitute{a}}\\ x\amp=-\frac{\lighthigh{6}}{2(\substitute{-2})}\\ \amp=1.5 \end{align*}

Using the $x$-value $\substitute{1.5}$ to find the $y$-coordinate, we have:

\begin{align*} j(\substitute{1.5})\amp=-2(\substitute{1.5})^2+6(\substitute{1.5})+8\\ \amp=12.5 \end{align*}

The vertex is the point $(1.5,12.5)\text{,}$ and the axis of symmetry is the line $x=1.5\text{.}$

We're now ready to graph this function. We'll start by drawing and scaling the axes so all of our key features will be displayed as shown in Figure 9.4.30. Next, we'll plot these key points as shown in Figure 9.4.31. Finally, we'll note that this parabola opens downward and connect these points with a smooth curve, as shown in Figure 9.4.32.

The Five-hundred-meter Aperture Spherical Telescope 1 en.wikipedia.org/wiki/Five_hundred_meter_Aperture_Spherical_Telescope (FAST) located in SW China is the worlds largest radio telescope. The name is actually incorrect because it is not spherical: it is parabolic! Parabolic telescopic dishes are very common because parabolas have the unique feature that they focus light coming in at a single point where a collecting instrument is placed.

If a scientist somehow managed to climb to the rim of the dish and roll a ball from the top down toward the center, her ball would travel along the parabola $h(x)=0.0018x^2-0.9x\text{,}$ where $x$ is the horizontal distance from the rim of the dish where the ball was released, in meters, and $h(x)$ is the height of the ball above above the rim, in meters (so negative height would mean below the rim). Note that this is an approximation based on real data.

1. Find and interpret the vertical intercept of the graph.

2. Find and interpret the horizontal intercepts of the graph.

3. Find and interpret the vertex of the graph.

Explanation
1. The vertical intercept is found when $\substitute{x=0}\text{.}$

\begin{align*} h(\substitute{x})\amp=0.0018\substitute{x}^2-0.9\substitute{x}\\ h(\substitute{0})\amp=0.0018(\substitute{0})^2-0.9(\substitute{0})\\ \amp=0 \end{align*}

So the vertical intercept is $(0,0)$ which means that the ball was released $0$ meters from the rim (on the rim) at a height of $0$ meters above the rim (again, on the rim).

2. The horizontal intercepts are found when $h(x)=0\text{.}$ This will be a quadratic equation that we can solve using factoring.

\begin{align*} h(x)\amp=0\\ 0.0018x^2-0.9x\amp=0\\ x\left(0.0018x-0.9\right)\amp=0 \end{align*}
\begin{align*} x\amp=0\amp\text{ or }\amp\amp 0.0018x-0.9\amp=0\\ x\amp=0\amp\text{ or }\amp\amp 0.0018x\amp=0.9\\ x\amp=0\amp\text{ or }\amp\amp x\amp=500 \end{align*}

The horizontal intercepts are $(0,0)$ and $(500,0)\text{.}$ The interpretation of $(0,0)$ is the same as before. The interpretation of $(500,0)$ is that the other side of the rim of the dish is $500$ meters away. If the ball were to somehow roll all the way down, and then back up the other side, it would pop up at the opposite rim exactly $500$ meters horizontally away.

3. The vertex is found when the $x$-value is at the axis of symmetry. To find that, we use the formula $x=-\frac{\lighthigh{b}}{2\substitute{a}}$ where $\substitute{a=0.0018}$ and $\lighthigh{b=-0.9}$

\begin{align*} x\amp=-\frac{\lighthigh{b}}{2\substitute{a}}\\ x\amp=-\frac{\lighthigh{-0.9}}{2(\substitute{0.0018})}\\ \amp=250 \end{align*}

Then, to find the $y$-value, we substitute that $x$-value, $\substitute{250}$ into the original function.

\begin{align*} h(\substitute{x})\amp=0.0018\substitute{x}^2-0.9\substitute{x}\\ h(\substitute{250})\amp=0.0018(\substitute{250})^2-0.9(\substitute{250})\\ \amp=-112.5 \end{align*}

The vertex is the point $(250,-112.5)\text{.}$ Since the vale of $a$ is positive, we know that this parabola opens upward. This means that the vertex is the lowest point on the graph. So the lowest point of the telescope dish is $250$ meters from the rim (horizontally) and $112.5$ meters below the height of the rim. That is one big dish!

###### Example9.4.34The Domain and Range of Quadratic Applications

In the FAST telescope example, find the real world domain and range of the function $h\text{.}$

Explanation

Real world domain and range problems are sometimes subjective. The domain in this case would represent all possible $x$-values that make sense in the reality of the situation. Recall that $x$ is the horizontal distance from the rim of the dish where the ball was released, in meters. Since we found out that the dish is $500$ meters across we could say that the domain is $[0,500]\text{.}$

The range will represent all possible $y$ values, that make sense in reality. Recall that $h(x)$ is the height above above the rim, in meters. Since the ball won't roll any higher than the rim of the dish, $0$ will be the largest $y$-value. We found out that the lowest point on the graph is $-112.5$ meters below the rim. So the range is [-112.5,0].

###### Example9.4.36Distinguishing Quadratic Functions from Other Functions and Relations

Decide if the equations represent quadratic functions or something else.

1. $y-1=3(x-2)-5$

2. $y=3(x-2)^2-5$

3. $y=\sqrt{x-2}-5$

4. $y^2=x^2-5$

Explanation

Recall that Definition 9.2.2 says that a quadratic function has the form $f(x)=ax^2+bx+c$ where $a\text{,}$ $b\text{,}$ and $c$ are real numbers, and $a \neq 0\text{.}$

1. The equation $y-1=3(x-2)-5$ is not quadratic. It is in fact a linear equation.

2. The equation $y=3(x-2)^2-5$ is quadratic. We could simplify the right hand side and would have something of the form $y=ax^2+bx+c\text{.}$

3. The equation $y=\sqrt{x-2}-5$ is not quadratic. The $x$ is inside a radical, not squared, so it cannot be converted into the form $y=ax^2+bx+c\text{.}$

4. The equation $y^2=x^2-5$ is not quadratic. It cannot be re-written in the form $y=ax^2+bx+c$ (due to the $y^2$ term).

### Subsection9.4.4Exercises

###### 1

Which of the following graphs show $y$ as a function of $x\text{?}$

###### 2

Which of the following graphs show $y$ as a function of $x\text{?}$

###### 3

Evaluate the function at the given values.

$F(x)={-5x+4}$

1. $F(1)=$

2. $F(-3)=$

3. $F(0)=$

###### 4

Evaluate the function at the given values.

$G(x)={-3x+8}$

1. $G(5)=$

2. $G(-3)=$

3. $G(0)=$

###### 5

Evaluate the function at the given values.

$\displaystyle{H(x)=\frac{{7x}}{{-6x+3}}}$

1. $H(4)=$ .

2. $H(-6)=$ .

###### 6

Evaluate the function at the given values.

$\displaystyle{K(x)=\frac{{2x}}{{3x-3}}}$

1. $K(4)=$ .

2. $K(-1)=$ .

###### 7

Evaluate the function at the given values.

$K(x)={-2x^{2}-5x-6}$

1. $K(2)=$

2. $K(-4)=$

###### 8

Evaluate the function at the given values.

$f(x)={-3x^{2}+3x-1}$

1. $f(4)=$

2. $f(-5)=$

###### 9

Solve for $x\text{,}$ where $g(x)={x^{2}-2x-82}\text{.}$

If $g(x)=-2\text{,}$ then $x=$.

###### 10

Solve for $x\text{,}$ where $h(x)={x^{2}-x-17}\text{.}$

If $h(x)=-5\text{,}$ then $x=$.

###### 11

A function is graphed.

This function has domain and range .

###### 12

A function is graphed.

This function has domain and range .

###### 13

Function $f$ is graphed.

1. $\displaystyle{f(2)={}}$

2. Solve $\displaystyle{f(x)=1}\text{.}$

###### 14

Function $f$ is graphed.

1. $\displaystyle{f(0)={}}$

2. Solve $\displaystyle{f(x)=1}\text{.}$

###### 15

Heather started saving in a piggy bank on her birthday. The function $f(x)={5x+3}$ models the amount of money, in dollars, in Heather’s piggy bank. The independent variable represents the number of days passed since her birthday.

Interpret the meaning of $f(2)=13\text{.}$

• A. Two days after Heather started her piggy bank, there were $\13$ in it.

• B. The piggy bank started with $\2$ in it, and Heather saves $\13$ each day.

• C. Thirteen days after Heather started her piggy bank, there were $\2$ in it.

• D. The piggy bank started with $\13$ in it, and Heather saves $\2$ each day.

###### 16

Haley started saving in a piggy bank on her birthday. The function $f(x)={4x+3}$ models the amount of money, in dollars, in Haley’s piggy bank. The independent variable represents the number of days passed since her birthday.

Interpret the meaning of $f(4)=19\text{.}$

• A. Nineteen days after Haley started her piggy bank, there were $\4$ in it.

• B. The piggy bank started with $\4$ in it, and Haley saves $\19$ each day.

• C. Four days after Haley started her piggy bank, there were $\19$ in it.

• D. The piggy bank started with $\19$ in it, and Haley saves $\4$ each day.

###### 17

The following figure has the graph $y=d(t)\text{,}$ which models a particle’s distance from the starting line in feet, where $t$ stands for time in seconds since timing started.

1. $d(2)=$

2. Interpret the meaning of $d(2)\text{:}$

• A. In the first $8$ seconds, the particle moved a total of $2$ feet.

• B. In the first $2$ seconds, the particle moved a total of $8$ feet.

• C. The particle was $8$ feet away from the starting line $2$ seconds since timing started.

• D. The particle was $2$ feet away from the starting line $8$ seconds since timing started.

3. Solve $d(t)={4}$ for $t\text{.}$ $t=$

4. Interpret the meaning of part c’s solution(s):

• A. The article was $4$ feet from the starting line $1$ seconds since timing started, or $8$ seconds since timing started.

• B. The article was $4$ feet from the starting line $8$ seconds since timing started.

• C. The article was $4$ feet from the starting line $1$ seconds since timing started.

• D. The article was $4$ feet from the starting line $1$ seconds since timing started, and again $8$ seconds since timing started.

###### 18

The following figure has the graph $y=d(t)\text{,}$ which models a particle’s distance from the starting line in feet, where $t$ stands for time in seconds since timing started.

1. $d(5)=$

2. Interpret the meaning of $d(5)\text{:}$

• A. The particle was $5$ feet away from the starting line $7.5$ seconds since timing started.

• B. The particle was $7.5$ feet away from the starting line $5$ seconds since timing started.

• C. In the first $7.5$ seconds, the particle moved a total of $5$ feet.

• D. In the first $5$ seconds, the particle moved a total of $7.5$ feet.

3. Solve $d(t)={3}$ for $t\text{.}$ $t=$

4. Interpret the meaning of part c’s solution(s):

• A. The article was $3$ feet from the starting line $8$ seconds since timing started.

• B. The article was $3$ feet from the starting line $1$ seconds since timing started, and again $8$ seconds since timing started.

• C. The article was $3$ feet from the starting line $1$ seconds since timing started, or $8$ seconds since timing started.

• D. The article was $3$ feet from the starting line $1$ seconds since timing started.

###### 19

The function $C$ models the the number of customers in a store $t$ hours since the store opened.

 $t$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $C(t)$ $0$ $41$ $75$ $97$ $97$ $73$ $48$ $0$
1. $C(1)=$

2. Interpret the meaning of $C(1)\text{:}$

• A. There were $41$ customers in the store $1$ hour after the store opened.

• B. In $1$ hour since the store opened, the store had an average of $41$ customers per hour.

• C. In $1$ hour since the store opened, there were a total of $41$ customers.

• D. There was $1$ customer in the store $41$ hours after the store opened.

3. Solve $C(t)=97$ for $t\text{.}$ $t=$

4. Interpret the meaning of Part c’s solution(s):

• A. There were $97$ customers in the store $3$ hours after the store opened.

• B. There were $97$ customers in the store $4$ hours after the store opened.

• C. There were $97$ customers in the store either $4$ hours after the store opened, or $3$ hours after the store opened.

• D. There were $97$ customers in the store $4$ hours after the store opened, and again $3$ hours after the store opened.

###### 20

The function $C$ models the the number of customers in a store $t$ hours since the store opened.

 $t$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $C(t)$ $0$ $43$ $78$ $97$ $95$ $81$ $43$ $0$
1. $C(3)=$

2. Interpret the meaning of $C(3)\text{:}$

• A. There were $97$ customers in the store $3$ hours after the store opened.

• B. There were $3$ customers in the store $97$ hours after the store opened.

• C. In $3$ hours since the store opened, there were a total of $97$ customers.

• D. In $3$ hours since the store opened, the store had an average of $97$ customers per hour.

3. Solve $C(t)=43$ for $t\text{.}$ $t=$

4. Interpret the meaning of Part c’s solution(s):

• A. There were $43$ customers in the store $1$ hours after the store opened, and again $6$ hours after the store opened.

• B. There were $43$ customers in the store either $1$ hours after the store opened, or $6$ hours after the store opened.

• C. There were $43$ customers in the store $1$ hours after the store opened.

• D. There were $43$ customers in the store $6$ hours after the store opened.

###### 21

Find the axis of symmetry and vertex of the quadratic function.

${y}={4x^{2}+32x-3}$

Axis of symmetry:

Vertex:

###### 22

Find the axis of symmetry and vertex of the quadratic function.

${y}={x^{2}-4x+5}$

Axis of symmetry:

Vertex:

###### 23

Find the axis of symmetry and vertex of the quadratic function.

${y}={3x^{2}-15x+2}$

Axis of symmetry:

Vertex:

###### 24

Find the axis of symmetry and vertex of the quadratic function.

${y}={4x^{2}+12x-4}$

Axis of symmetry:

Vertex:

###### 25

For $y=4x^2-8x+5\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 26

For $y=2x^2+4x+7\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 27

For $y=-x^2+4x+2\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 28

For $y=-x^2+2x-5\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 29

You will build a rectangular sheep enclosure next to a river. There is no need to build a fence along the river, so you only need to build on three sides. You have a total of $500$ feet of fence to use. Find the dimensions of the pen such that you can enclose the maximum possible area. One approach is to let $x$ represent the length of fencing that runs perpendicular to the river, and write a formula for a function of $x$ that outputs the area of the enclosure. Then find its vertex and interpret it.

The length of the pen (parallel to the river) should be , the width (perpendicular to the river) should be , and the maximum possible area is .

###### 30

You will build a rectangular sheep enclosure next to a river. There is no need to build a fence along the river, so you only need to build on three sides. You have a total of $400$ feet of fence to use. Find the dimensions of the pen such that you can enclose the maximum possible area. One approach is to let $x$ represent the length of fencing that runs perpendicular to the river, and write a formula for a function of $x$ that outputs the area of the enclosure. Then find its vertex and interpret it.

The length of the pen (parallel to the river) should be , the width (perpendicular to the river) should be , and the maximum possible area is .

###### 31

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={x^{2}+4x+3}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 32

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={-x^{2}-3x-2}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 33

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={x^{2}+x+5}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 34

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={x^{2}+8x+4}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 35

Find the $y$-intercept and any $x$-intercept(s) of the parabola with equation ${y}={16x^{2}-8x+1}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 36

Find the $y$-intercept and any $x$-intercept(s) of the parabola with equation ${y}={25x^{2}-1}\text{.}$

$y$-intercept:

$x$-intercept(s):

Graph each curve by algebraically determining its key features.

###### 37

$y=x^2-7x+12$

###### 38

$y=x^2+5x-14$

###### 39

$y=-x^2-x+20$

###### 40

$y=-x^2+4x+21$

###### 41

$y=x^2+6x$

###### 42

$y=x^2-8x$

###### 43

$y=x^2+4x+7$

###### 44

$y=x^2-2x+6$

###### 45

$y=2x^2-4x-30$

###### 46

$y=3x^2+21x+36$

###### 47

From a clifftop over the ocean ${170\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${264.6\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ On its way down it fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic function $f\text{,}$ where $f(t)={-4.9t^{2}+264.6t+170}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $f(t)$ represents the object’s height (above sea level) in meters.

1. After , this object reached its maximum height of .

2. This object flew for before it landed in the ocean.

3. This object was above sea level ${47\ {\rm s}}$ after its release.

4. This object was ${3663.7\ {\rm m}}$ above sea level twice: once after its release, and again later after its release.

###### 48

From a clifftop over the ocean ${130\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${294\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ On its way down it fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic function $f\text{,}$ where $f(t)={-4.9t^{2}+294t+130}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $f(t)$ represents the object’s height (above sea level) in meters.

1. After , this object reached its maximum height of .

2. This object flew for before it landed in the ocean.

3. This object was above sea level ${18\ {\rm s}}$ after its release.

4. This object was ${967.9\ {\rm m}}$ above sea level twice: once after its release, and again later after its release.


text:
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#### Archived

This topic is now archived and is closed to further replies.

# Engine compilation.. (or something to that effect)

## Recommended Posts

Hello, I have been busy programming a game engine over the past month. It can''t do that much at the moment, but it''s continually evolving. However, so far I have about 25+ files (header and source). I use VC++, and the only way I have been able to get the intellisense thing to work is to include ALL of those files into my project. This is really a pain, and I was wondering if there is an easier way to do it. I have lots of classes (i split up my classes one per file), so I would like to avoid adding all the class descriptions to one big file. I was wondering if I could create, say, a DLL or something, so that I could include all of my files and still get intellisense to work. I have organized my files neatly into folders so it is not that hard to add them to my project, but they still take up a lot of room in VC++ (and yes, I organize them into folders in vc++ too.) Anyone have any ideas? Thanks in advance! Fuzztrek ¬_¬

##### Share on other sites
Intellisense is broken. Get Visual Assist from Whole Tomato.

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hate to sound cheap..but is there another way? or is VS.NET better?

¬_¬

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Yes, vs.net is much better.

The real question is this: Is Intellisense worth $70-$80 (or whatever Visual Assist is), or is it worth the several hundred or more dollars for visual studio .net ?

Send one of your friends to get hired at Microsoft, and tell him to send you a discounted ($20) version of VC++.NET to you! #### Share this post ##### Link to post ##### Share on other sites quote: Original post by Nypyren Send one of your friends to get hired at Microsoft, and tell him to send you a discounted ($20) version of VC++.NET to you!

Nice friend you are!

And to the original post, yes, make DLL''s if its a windows only engine.

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Alright! thanks so much! I actually just got VS.NET yesterday :D

¬_¬

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VS.net''s Intellisense isn''t much better than what was in VC 6. Visual Assist is much more useful IMHO.

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Bases

# Bases

The primary purpose of bases in QuantumOptics.jl is to specify the dimension of the Hilbert space of the system and to make sure that quantum objects associated to distinct bases can't be combined accidentally in an incorrect way. Many of the common types of bases used in quantum mechanics are already implemented in Quantum systems.

## Composite bases

Hilbert spaces of composite systems can be handled with the CompositeBasis which can be created using the tensor function or the equivalent ⊗ operator:

basis_fock = FockBasis(10)
basis_particle = MomentumBasis(0., 10., 50)
basis = tensor(basis_fock, basis_particle)
basis = basis_fock ⊗ basis_particle

Most of the time this will happen implicitly when tensor products of operators are formed.

## Generic bases

If an appropriate basis type is not implemented, the quick and dirty way is to use a GenericBasis, which just needs to know the dimension of the Hilbert space and is ready to go:

b = GenericBasis(5)

However, since operators and states represented in any generic basis can be combined as long as the bases have the same dimension, it might lead to errors that otherwise would have been caught.

## Implementing new bases

The cleaner way is to implement own special purpose bases by deriving from the abstract Basis type. The only mandatory property of all basis types is that they have a field shape which specifies the dimensionality of their Hilbert space. E.g. a spin 1/2 basis could be implemented as:

struct SpinBasis <: Basis
shape::Vector{Int}
SpinBasis() = new(Int[2]) # Constructor
end

The default behavior for new bases is to allow operations for bases of the same type, but reject mixing with other bases. Finer control over the interaction with other bases can be achieved by overloading the == operator as well as the QuantumOpticsBase.multiplicable function.


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I am hoping this really happens
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By displaying the art and culture of the people living tere
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I'd visit the DotGreenway 3 - 5 times a week!
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I would like to see gardens, benches and art.
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### Cannon Balls

How high will a ball taking a million seconds to fall travel?

### Dangerous Driver?

Was it possible that this dangerous driving penalty was issued in error?

### Speedo

Investigate the relationship between speeds recorded and the distance travelled in this kinematic scenario.

# Board and Spool

##### Stage: 5 Challenge Level:

Suppose that a cylinder is rotating about a point A with an angular speed $\omega$ and use the fact that an angular speed is the same for all points of the rigid body.

There is another way of solving this problem. Look at the motion of spool as two seperate motions: the spool moves horizontally as its all mass is at point B and rotates about the point B.

The angular speed is the rate of rotation $\omega = \frac{\delta \Theta}{\delta t}$.


text:
# 17.4: Relativistic Kinematics

## Velocity Transformations

Consider the two parallel coordinate frames with the primed frame moving at a velocity $$v$$ along the $$x^{\prime}_1$$ axis as shown in Figure $$17.2.1$$. Velocities of an object measured in both frames are defined to be

$u_i = \dfrac{dx_i}{ dt} \label{17.16} \\ u^{\prime}_i = \dfrac{dx^{\prime}_i}{ dt^{\prime}}$

Using the Lorentz transformations $$(17.3.1)$$, $$(17.3.3)$$ between the two frames moving with relative velocity $$v$$ along the $$x_1$$ axis, gives that the velocity along the $$x^{\prime}_1$$ axis is

$u^{\prime}_1 = \dfrac{dx^{\prime}_1}{ dt^{\prime} } = \dfrac{dx_1 − vdt}{ dt − \dfrac{v}{ c^2} dx_1} = \dfrac{u_1 − v}{ 1 − \dfrac{u_1v}{ c^2} } \label{17.17}$

Similarly we get the velocities along the perpendicular $$x^{\prime}_2$$ and $$x^{\prime}_3$$ axes to be

$u^{\prime}_2 = \dfrac{dx^{\prime}_2 }{dt^{\prime}} = \dfrac{u_2}{ 1 − \dfrac{u_1v}{ c^2}} \label{17.18} \\ u^{\prime}_3 = \dfrac{dx^{\prime}_3 }{dt^{\prime}} = \dfrac{u_3}{ 1 − \dfrac{u_1v}{ c^2}}$

When $$\dfrac{u_1v}{ c^2} \rightarrow 0$$ these velocity transformations become the usual Galilean relations for velocity addition. Do not confuse $$\mathbf{u}$$ and $$\mathbf{u}^{\prime}$$ with $$\mathbf{v}$$; that is, $$\mathbf{u}$$ and $$\mathbf{u}^{\prime}$$ are the velocities of some object measured in the unprimed and primed frames of reference respectively, whereas $$\mathbf{v}$$ is the relative velocity of the origin of one frame with respect to the origin of the other frame.

## Momentum

Using the classical definition of momentum, that is $$\mathbf{p} = m\mathbf{u}$$, the linear momentum is not conserved using the above relativistic velocity transformations if the mass $$m$$ is a scalar quantity. This problem originates from the fact that both $$\mathbf{x}$$ and $$t$$ have non-trivial transformations and thus $$\mathbf{u} = \dfrac{d\mathbf{x}}{dt}$$ is frame dependent.

Linear momentum conservation can be retained by redefining momentum in a form that is identical in all frames of reference, that is by referring to the proper time $$\tau$$ as measured in the rest frame of the moving object. Therefore we define relativistic linear momentum as

$\mathbf{p} \equiv m \dfrac{d\mathbf{x}}{ d\tau} = m \dfrac{d\mathbf{x}}{ dt} \dfrac{dt}{ d\tau} \label{17.19}$

But we know the time dilation relation

$dt = \dfrac{d\tau }{\sqrt{ (1 − \dfrac{u^2}{ c^2} )}} = \gamma_u d\tau \label{17.20}$

Note that the $$\gamma_u$$ in this relation refers to the velocity $$u$$ between the moving object and the frame; this is quite different from the $$\gamma = \dfrac{ 1}{\sqrt{ (1− \dfrac{v^2}{ c^2} )}}$$ which refers to the transformation between the two frames of reference. Thus the new relativistic definition of momentum is

\begin{align} \mathbf{p} &\equiv m \dfrac{d\mathbf{x}}{ d\tau} \\[4pt] &= m\gamma_u \dfrac{d\mathbf{x}}{ dt} \\[4pt] &= \gamma_u m\mathbf{u} \label{17.21} \end{align}

The relativistic definition of linear momentum is the same as the classical definition with the rest mass $$m$$ replaced by the relativistic mass $$\gamma m$$.1

## Center of momentum coordinate system

The classical relations for handling the kinematics of colliding objects, carry over to special relativity when the relativistic definition of linear momentum, Equation \ref{17.21}, is assumed. That is, one can continue to apply conservation of linear momentum. However, there is one important conceptual difference for relativistic dynamics in that the center of mass no longer is a meaningful concept due to the interrelation of mass and energy. However, this problem is eliminated by considering the center of momentum coordinate system which, as in the non-relativistic case, is the frame where the total linear momentum of the system is zero. Using the concept of center of momentum incorporates the formalism of classical non-relativistic kinematics.

## Force

Newton’s second law $$\mathbf{F} = \dfrac{d\mathbf{p}}{ dt}$$ is covariant under a Galilean transformation. In special relativity this definition also applies using the relativistic definition of momentum $$\mathbf{p}$$. The fact that the relativistic momentum $$\mathbf{p}$$ is conserved in the force-free situation, leads naturally to using the definition of force to be

$\mathbf{F} = \dfrac{d\mathbf{p}}{ dt} \label{17.22}$

Then the relativistic momentum is conserved if $$\mathbf{F} = 0$$.

## Energy

The classical definition of work done is defined by

$W_{12} = \int^2_1 \mathbf{F}\cdot d\mathbf{r} = T_2 − T_1 \label{17.23}$

Assume $$T_1 = 0$$, let $$d\mathbf{r} = \mathbf{u}dt$$ and insert the relativistic force relation in Equation \ref{17.23}, gives

$W = T = \int^t_0 \dfrac{d}{dt} (\gamma_u m\mathbf{u}) \cdot \mathbf{u}dt = m \int^u_0 ud (\gamma_u u) \label{17.24}$

Integrate by parts, followed by algebraic manipulation, gives

\begin{align} T &= \gamma_{u} m u^{2}-m \int_{0}^{u} \dfrac{u d u}{\sqrt{1-\dfrac{u^{2}}{c^{2}}}} \\[4pt] &= \gamma_{u} m u^{2} + m c^{2} \sqrt{1-\dfrac{u^{2}}{c^{2}}} - m c^{2} \\ &= \dfrac{m u^{2}}{\sqrt{1-\dfrac{u^{2}}{c^{2}}}} + \dfrac{m c^{2}}{\sqrt{1-\dfrac{u^{2}}{c^{2}}}} \left( 1 - \dfrac{u^{2}}{c^{2}} \right) - m c^{2} \\[4pt] &= m c^{2} \left( \gamma_{u} -1 \right) \label{17.25} \end{align}

Define the rest energy $$E_0$$

$E_0 \equiv mc^2 \label{17.26}$

and total relativistic energy $$E$$

$E \equiv \gamma_u mc^2 \label{17.27}$

then Equation \ref{17.25} can be written as

\begin{align} E &= T + E_0 \\[4pt] &= \gamma_u mc^2 \label{17.28} \end{align}

This is the famous Einstein relativistic energy that relates the equivalence of mass and energy. The total relativistic energy $$E$$ is a conserved quantity in nature. It is an extension of the conservation of energy and manifestations of the equivalence of energy and mass occur extensively in the real world.

In nuclear physics we often convert mass to energy and back again to mass. For example, gamma rays with energies greater than $$1.022$$ $$MeV$$, which are pure electromagnetic energy, can be converted to an electron plus positron both of which have rest mass. The positron can then annihilate a different electron in another atom resulting in emission of two $$511$$ $$keV$$ gamma rays in back to back directions to conserve linear momentum. A dramatic example of Einstein’s equation is a nuclear reactor. One gram of material, the mass of a paper clip, provides $$E = 9 \times 10^{13}$$ joules. This is the daily output of a $$1$$ $${GWatt}$$ nuclear power station or the explosive power of the Nagasaki or Hiroshima bombs.

As the velocity of a particle $$v$$ approaches $$c$$, then $$\gamma$$ and the relativistic mass $$\gamma m$$ both approach infinity. This means that the force needed to accelerate the mass also approaches infinity, and thus no particle can exceed the velocity of light. The energy continues to increase not by increasing the velocity but by increase of the relativistic mass. Although the relativistic relation for kinetic energy is quite different from the Newtonian relation, the Newtonian form is obtained for the case of $$u \ll c$$ in that

\begin{align} T &= \dfrac{mc^2}{\sqrt{1 − \dfrac{u^2}{ c^2}}} − mc^2 \\[4pt] &= mc^2(1 + \dfrac{1}{ 2} \dfrac{u^2}{ c^2} + \cdots ) − mc^2 \\[4pt] &= \dfrac{1}{ 2} mu^2 \label{17.29} \end{align}

An especially useful relativistic relation that can be derived from the above is

$E^2 = p^2c^2 + E^2_0 \label{17.30}$

This is useful because it provides a simple relation between total energy of a particle and its relativistic linear momentum plus rest energy.

Example $$\PageIndex{1}$$: Rocket Propulsion

Consider a rocket, having initial mass $$M$$, is accelerated in a straight line in free space by exhausting propellant at a constant speed $$v_p$$ relative to the rocket. Let $$u$$ be the speed of the rocket relative to it’s initial rest frame $$S$$, when its rest mass has decreased to $$m$$. At this instant the rocket is at rest in the inertial frame $$S^{\prime}$$. At a proper time $$\tau + d\tau$$ the rest mass is $$m − dm$$ and it has acquired a velocity increment $$du$$ relative to $$S^{\prime}$$ and propellant of rest mass $$dm_p$$ has been expelled with velocity $$v_p$$ relative to $$S^{\prime}$$. At proper time $$\tau$$ in $$S^{\prime}$$ the rest mass is $$mc^2$$. At the time $$\tau + d\tau$$, energy conservation requires that

$\gamma_{u^{\prime}} (m − dm) c^2 + \gamma_{v_p} m_p c^2 = mc^2 \nonumber$

At the same instant, conservation of linear momentum requires

$\gamma_{u^{\prime}} (m − dm) du^{\prime} − \gamma_p v_p dm_p = 0 \nonumber$

To first order these two equations simplify to

$dm_p = \sqrt{ 1 − \left(\dfrac{v_p }{c} \right)^2 } dm \nonumber$

$mdu^{\prime} = dm_p\gamma_{v_p} v_p \nonumber$

Therefore

$mdu^{\prime} = v_pdm \label{a}\tag{a}$

The velocity increment $$du^{\prime}$$ in frame $$S^{\prime}$$ can be transformed back to frame $$S$$ using equation $$(17.3.3)$$, that is

$d + du = \dfrac{u + du^{\prime}}{ 1 + \dfrac{ udu^{\prime}}{ c^2}} \approx u + \left( 1 − \left(\dfrac{u}{c} \right)^2 \right) du^{\prime} \label{b}\tag{b}$

Equations \ref{a} and \ref{b} yield a differential equation for $$u(m)$$ of

$\dfrac{du}{ 1 − (\dfrac{u}{c} )^2} = v_p \dfrac{dm}{m} \nonumber$

Integrate the left-hand side between $$0$$ and $$u$$ and the right-hand side between $$M$$ and $$m$$ gives

$\dfrac{1}{ 2} c \ln \left(\dfrac{1 + \dfrac{u}{c} }{1 − \dfrac{u}{c}} \right) = −v_p \ln \left( \dfrac{m}{M} \right) \nonumber$

This reduces to

$\dfrac{u}{c} = \dfrac{1 − ( \dfrac{m}{M} )^{2v_p/c}}{ 1 + ( \dfrac{m}{M} )^{2v_p/c}} \nonumber$

When $$\dfrac{u}{c} \rightarrow 0$$ this equation reduces to the non-relativistic answer given in equation $$(2.12.34)$$.

1Note that, until recently, the rest mass was denoted by $$m_0$$ and the relativistic mass was referred to as $$m$$. Modern texts denote the rest mass by $$m$$ and the relativistic mass by $$\gamma m$$. This book follows the modern nomenclature for rest mass to avoid confusion.

This page titled 17.4: Relativistic Kinematics is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


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# Overview¶

## Installation¶

ELI5 works in Python 2.7 and Python 3.4+. Currently it requires scikit-learn 0.18+. You can install ELI5 using pip:

pip install eli5

or using:

conda install -c conda-forge eli5

## Features¶

ELI5 is a Python package which helps to debug machine learning classifiers and explain their predictions. It provides support for the following machine learning frameworks and packages:

• scikit-learn. Currently ELI5 allows to explain weights and predictions of scikit-learn linear classifiers and regressors, print decision trees as text or as SVG, show feature importances and explain predictions of decision trees and tree-based ensembles.

Pipeline and FeatureUnion are supported.

ELI5 understands text processing utilities from scikit-learn and can highlight text data accordingly. It also allows to debug scikit-learn pipelines which contain HashingVectorizer, by undoing hashing.

• XGBoost - show feature importances and explain predictions of XGBClassifier, XGBRegressor and xgboost.Booster.

• LightGBM - show feature importances and explain predictions of LGBMClassifier and LGBMRegressor.

• lightning - explain weights and predictions of lightning classifiers and regressors.

• sklearn-crfsuite. ELI5 allows to check weights of sklearn_crfsuite.CRF models.

ELI5 also implements several algorithms for inspecting black-box models (see Inspecting Black-Box Estimators):

• TextExplainer allows to explain predictions of any text classifier using LIME algorithm (Ribeiro et al., 2016). There are utilities for using LIME with non-text data and arbitrary black-box classifiers as well, but this feature is currently experimental.
• Permutation Importance method can be used to compute feature importances for black box estimators.

Explanation and formatting are separated; you can get text-based explanation to display in console, HTML version embeddable in an IPython notebook or web dashboards, JSON version which allows to implement custom rendering and formatting on a client, and convert explanations to pandas DataFrame objects.

## Basic Usage¶

There are two main ways to look at a classification or a regression model:

1. inspect model parameters and try to figure out how the model works globally;
2. inspect an individual prediction of a model, try to figure out why the model makes the decision it makes.

For (1) ELI5 provides eli5.show_weights() function; for (2) it provides eli5.show_prediction() function.

If the ML library you’re working with is supported then you usually can enter something like this in the IPython Notebook:

import eli5
eli5.show_weights(clf)

and get an explanation like this:

Note

Depending on an estimator, you may need to pass additional parameters to get readable results - e.g. a vectorizer used to prepare features for a classifier, or a list of feature names.

Supported arguments and the exact way the classifier is visualized depends on a library.

To explain an individual prediction (2) use eli5.show_prediction() function. Exact parameters depend on a classifier and on input data kind (text, tabular, images). For example, you may get text highlighted like this if you’re using one of the scikit-learn vectorizers with char ngrams:

## Why?¶

For some of classifiers inspection and debugging is easy, for others this is hard. It is not a rocket science to take coefficients of a linear classifier, relate them to feature names and show in an HTML table. ELI5 aims to handle not only simple cases, but even for simple cases having a unified API for inspection has a value:

• you can call a ready-made function from ELI5 and get a nicely formatted result immediately;
• formatting code can be reused between machine learning frameworks;
• ‘drill down’ code like feature filtering or text highlighting can be reused;
• there are lots of gotchas and small differences which ELI5 takes care of;
• algorithms like LIME (paper) try to explain a black-box classifier through a locally-fit simple, interpretable classifier. It means that with each additional supported “simple” classifier/regressor algorithms like LIME are getting more options automatically.

## Architecture¶

In ELI5 “explanation” is separated from output format: eli5.explain_weights() and eli5.explain_prediction() return Explanation instances; then functions from eli5.formatters can be used to get HTML, text, dict/JSON or pandas DataFrame representation of the explanation.

It is not convenient to do that all when working interactively in IPython notebooks, so there are eli5.show_weights() and eli5.show_prediction() functions which do explanation and formatting in a single step.

Explain functions are not doing any work by themselves; they call a concrete implementation based on estimator type. So e.g. eli5.explain_weights() calls eli5.sklearn.explain_weights.explain_linear_classifier_weights() if sklearn.linear_model.LogisticRegression classifier is passed as an estimator.


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# City Bridge – Pt IX

 Flange/Cover Plates Flange or Cover Plates form the top of the Main Girder assembly, tying everything together and a major part of the strength of the girder as a cross-section of the girder shows that the cover plates along with the angles and web make up the girde cross-section. This is one of the areas where I quickly turn to what I an call – Imagineering Simplification – or IS. The math is just .. too much for my poor brain and after all I am not going to actually build build a REAL bridge .. but a model that appears to satisfy the requirements. For girders with constant depth back to back of flange angles it is sufficiently accurate to assume the effective depth h constant. The required flange area varies then in direct proportion with the bending movement.1 What they are saying is this (simplified for my little brain). Suppose you lay a yardstick across a gap and then press with your finger on the end (where it is supported) and in the center where it isn’t. Obviously .. it bends in the center. What they are doing then is add additional additional thickness toward the center of the girder – in this case – different thicknesses of flange plate. Note: Some of the engineers liked the idea of adding the thickness to the side of the bridge girders since this allowed the top of the girder to be the same end to end which would enable laying the same tie but *cost* meant that the cheaper ‘vary the Flange Plate’ meant the track crew had to dap the ties differently depending on location. To calculate the lengths of the cover plates a movement curve is drawn which is then divided proportionately to the various parts making up the flange. When the theoretical length of a cover plate is found they add 6″ for light to 12″ for heaver cover plates. Since I am making this up anyway .. I just used the example in the book. The diagram shows half the bridge length. I brought this into Sketchup and then I put in the 6.825″ (1/2 of 13.625″). Once I scaled the drawing to that length (now Sketchup knows the proportions) I easily marked off the three dimensions for the three plates. I added 6″ (.125″ – this is O scale) and the numbers in yellow show 1/2 the length needed for each cover plate.
 Bearings Plate girders should have rived to each end a base plate of a thickness not less than 3/4 in.for railroad (…) bridges. There should be a bedplate between the base plate and masonry, which may be a rolled steel plate of a thickness not less than of the base plate for railroad spans up to 75 ft. (…); it should not project more than 3-1/2 times its thickness beyond the bottom flange angles. Instead of bedplates, cast steel pedestals are now more frequently used (1915).2 – a is the length of a bearing, b the distance from the edge of the bearing to the face of the masonry (Fig 1). – For ordinary bridges b = 3 to 6 in. The length of a bearing a can be determined by a preliminary calculation of the reaction and the required area of bearing as explained on page 190. – For R.R. bridges (E-50 loading and open flooring)”: Single Track plate girder spans $a=0.02l+0.75ft.$3 Therefore for my bridge: $a=.02X54.5ft+.075ft=1.84ft=22in$ – which in O scale is .458 in. Generally, a sole plate 3/4 in. or more in thickness is countersunk-riveted to the girder for a bearing piece, and has a length equal to that of the masonry bridge seat, and a width equal to or a little greater than that of the flange. (…) The width of the sole plate is generally made great enough to receive anchor bolts clear of the girder flange, and the bed plate or pedestal under it is often omitted. When a bed plate is used for short spans it is generally of rolled steel 3/4 to 1 in. thick, and is of about the same width and length as the sole plate. (…) Two or four short vertical anchor bolts generally secure the girder flange and the sole and bed plate to the masonry.4
 Bridge Weight From a table – American Bridge Company’s Single Track Deck Plate Girder Spans – E-40 34,900 lbs. and E-50 38,800 lbs.5
 Bridge Shoe and Bearing Plate The Bearing Plate shown was calculated to be 22 in. long (0.458″). The 0.240 in. width is that of the girder. The calculation was based on E-50 loading. A simple ratio – $\frac{.382}{.458}X50=E4.2$ shows us that a bearing plate 0.382 in. x 0.240 in. would equal a E-4.2 loading. The Bridge Shoe was designed after a Boller & Hodge Fixed End Bearing with a 51,000 lb. / 25 ton capacity. This is way in excess of the calculated E-40 55 ft. Deck Plate Girder weight of 35,000 lbs. The drawing shows a Bearing Plate under the Girder with a blue plate on it. This is simply the aforementioned 0.382 in. x 0.240 in Bearing Plate extended in width to support the lateral brace plate (in blue). Thinking about it that makes no sense. That brace plate needs to ‘stand alone’ separate from the bearing plate. A reinforcing plate above would work. The question I have is how the Bearing Plate is secured to the Bridge Shoe I adjusted the size of the bearing plate so it is the length of the bridge shoe and the width of the girder. I also adjusted the way the lateral brace attaches. The bridge shoes are ones I had 3D printed – that I have ‘on hand’ – and work perfectly for this build. Hooah!

1. Design of Steel Bridges: Theory and Practice for the Use of Civil …, Volume 1, F. C. Kunz, 1915 – Pg.155 []
2. Design of Steel Bridges: Theory and Practice for the Use of Civil …, Volume 1, F. C. Kunz, 1915 – Pg.166 []
3. Design of Steel Bridges: Theory and Practice for the Use of Civil …, Volume 1, F. C. Kunz, 1915 – Pg.127 []
4. Types and Details of Bridge Construction: Plate girders, Frank Woodward Skinner, 1906 – Pg.124-125 []
5. Design of Steel Bridges: Theory and Practice for the Use of Civil …, Volume 1, F. C. Kunz, 1915 – Pg.221 []


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# An object with a mass of 4 kg is hanging from a spring with a constant of 2 (kg)/s^2. If the spring is stretched by  16 m, what is the net force on the object?

Feb 13, 2016

${F}_{n} = 7 , 24 N$

#### Explanation:

$G = m \cdot g$
$G = 4 \cdot 9 , 81 = 39 , 24 N$
$F = k \cdot \Delta x$
$F = 2 \cdot 16$
$F = 32 N$
${F}_{n} = 39 , 24 - 32$
${F}_{n} = 7 , 24 N$


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# All Questions

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### Fortran: Best way to time sections of your code?

Sometimes while optimizing code it is required to time certain portions of the code, I have been using the following for years but was wondering if there is a simpler/better way to do it? ...
818 views

### What is the correct way of integrating in astronomy simulations?

I'm creating a simple astronomy simulator that should use Newtonian physics to simulate movement of planets in a system (or any objects, for that matter). All the bodies are circles in an Euclidean ...


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# How do you find the critical points and local max and min for y=(x-1)^4?

Dec 6, 2017

Min: $x = 1$

Explained mathematically but easy to see graphically.

#### Explanation:

$y = {\left(x - 1\right)}^{4}$
Power rule + chain rule
$y ' = 4 {\left(x - 1\right)}^{3} \cdot 1$
$y ' = 4 {\left(x - 1\right)}^{3}$

Critical points are when $f ' \left(x\right) = 0$ so...
we set $y ' = 0$

$0 = 4 {\left(x - 1\right)}^{3}$
$\frac{0}{\textcolor{red}{4}} = \frac{4 {\left(x - 1\right)}^{3}}{\textcolor{red}{4}}$

$0 = {\left(x - 1\right)}^{3}$
0^(color(red)(1/3)=((x-1)^3)^color(red)(1/3)

$0 \textcolor{red}{+} \textcolor{red}{1} = x - 1 \textcolor{red}{+} \textcolor{red}{1}$
$1 = x$

We now know that $x = 1$ is a critical point.

Now we look at the original function $y = {\left(x - 1\right)}^{4}$
We can take a look at $x = 1$ and compare that to $x$ slightly smaller than 1 and $x$ slightly larger than 1

$y = {\left(1 - 1\right)}^{4}$
$y = {\left(0\right)}^{4}$
$y = 0$

$y = {\left(0.99 - 1\right)}^{4}$
$y = {\left(- 0.01\right)}^{4}$
$y = 0.00000001$

$y = {\left(1.01 - 1\right)}^{4}$
$y = {\left(.01\right)}^{4}$
$y = 0.00000001$

The points immediately to the left and right of $x = 1$ are both larger than $x = 1$ therefore $x = 1$ must be a minimum


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## Rocky Mountain Journal of Mathematics

### 4-dissections and 8-dissections for some infinite products

#### Abstract

In this paper, we establish 4- and 8-dissections for some infinite products. In particular, we generalize Hirschhorn's formulas for 8-dissections of a continued fraction of Gordon and its reciprocal. Our results also imply a theorem on the periodicity of signs of the coefficients of an infinite product given by Chan and Yesilyurt.

#### Article information

Source
Rocky Mountain J. Math., Volume 44, Number 5 (2014), 1685-1696.

Dates
First available in Project Euclid: 1 January 2015

https://projecteuclid.org/euclid.rmjm/1420071562

Digital Object Identifier
doi:10.1216/RMJ-2014-44-5-1685

Mathematical Reviews number (MathSciNet)
MR3295650

Zentralblatt MATH identifier
1318.11006

#### Citation

Xia, Ernest X.W.; Yao, X.M. 4-dissections and 8-dissections for some infinite products. Rocky Mountain J. Math. 44 (2014), no. 5, 1685--1696. doi:10.1216/RMJ-2014-44-5-1685. https://projecteuclid.org/euclid.rmjm/1420071562

#### References

• K. Alladi and B. Gordon, Vanishing coefficients in the expansion of products of Rogers-Ramanujan type, in The Rademacher legacy to mathematics, Contemp. Math. 166 (1994), 129–139.
• G.E. Andrews, Ramanujan's lost notebook III–The Rogers-Ramanujan continued fraction, Adv. Math. 41 (1981), 186–208.
• G.E. Andrews and D. Bressoud, Vanishing coefficients in infinite product expansions, J. Austr. Math. Soc. 27 (1979), 199–202.
• S.H. Chan and H. Yesilyurt, The periodicity of the signs of the coefficients of certain infinite products, Pacific J. Math. 225 (2006), 13–32.
• G. Gasper and M. Rahman, Basic hypergeometric series, Encycl. Math. Appl. 35, Second edition, Cambridge University Press, Cambridge, 2004.
• B. Gordon, Some continued fractions of the Rogers-Ramanujan type, Duke Math. J. 32 (1965), 741–748.
• M.D. Hirschhorn, On the expansion of Ramanujan's continued fraction, Ramanujan J. 2 (1998), 521–527.
• ––––, On the expansion of a continued fraction of Gordon, Ramanujan J. 5 (2001), 369–375.
• ––––, On the $2$- and $4$-dissections of Ramanujan's continued fraction and its reciprocal, Ramanujan J. 24 (2011), 85–92.
• R.P. Lewis and Z.G. Liu, A conjecture of Hirschhorn on the $4$-dissection of Ramanujan's continued fraction, Ramanujan J. 4 (1991), 347–352.
• S. Ramanujan, The lost notebook and other unpublished papers, Narosa, Delhi, 1988.
• B. Richmond and G. Szehers, The Taylor coefficients of certain infinite products, Acta Sci. Math. (Szeged) 40 (1978), 347–369.
• L.J. Rogers, Second memoir on the expansion of certain infinite products, Proc. Lond. Math. Soc. 25 (1894), 318–343.
• L.J. Slater, Generalized hypergeometric functions, Cambridge University Press, Cambridge, 1966.
• G.N. Waston and E.T. Whittaker, A course of modern analysis, reprint of the fourth edition, Cambridge University Press, Cambridge, 1996.
• E.X.W. Xia and X.M. Yao, The $8$-dissection of the Ramanujan-Göllnitz-Gordon continued fraction by an iterative method, Int. J. Number Theory 7 (2011), 1589–1593.


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We upgraded Indico to version 3.0. The new search is now available as well.

# 7th International Conference on Position Sensitive Detectors

12-16 September 2005
University of Liverpool
Europe/Zurich timezone

## DEPFET Active Pixel Sensors and pnCCDs for room temperature imaging (Xray) Spectroscopy in space missions and terrestrial astronomy

13 Sep 2005, 14:00
15m
University of Liverpool

#### University of Liverpool

Greenbank Conference Park
Contributed Talk Applications in Astronomy and Astrophysics

### Speaker

Dr Johannes Treis (MPI Munich)

### Description

Two dimensional back-illuminated (500 µm sensitive thickness) X-ray pixel detectors with pixel sizes from 36 µm, 51µm, 75 µm to 150 µm have been developed for applications in X-ray astronomy (0.1 keV to 20keV) and for wave front sensing in adaptive optics systems up to 1 µm wavelength in the NIR. In both applications the environmental conditions are such that operation at warm temperatures´´ around -30o to -10o C is highly desired simultaneously to high readout speed and low noise operations. During our tests with device formats of 64×64 to 264×264 frame rates of up to 1.000 per second have been achieved with noise floors of 2.5 electrons (rms). For temperatures around - 20o C single stage Peltier cooler can be used, making the system compact and easy to use. The achieved energy resolutions with a pnCCD in the full imaging mode was 210 eV (FWHM) at the MnKá – line at 5.9 keV at - 100 C and with 75 µm pixels and a format of 256 × 128 being read out at 200 frames per second. The active pixels sensor DEPFET achieved at the same temperature with the same pixel size but with a format of only 64 × 64 and 300 frames per second an energy resolution of 150 eV only. pnCCD for wave front sensing with a format of 264 × 528 were operated at 1.000 frames per second and a noise of less than 3 electrons at - 400 C. The full set of measurements will be shown to taste the comfortable parameter space for applications in heaven and on earth.

### Primary author

Prof. Struder Lothar (Max-Planck-Institut für extraterrestrische Physik)

### Co-author

Dr Johannes Treis (MPI Munich)

### Presentation materials

 PSD7_treis.pdf PSD7_treis.ppt


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# How to recursively factor a symbolic expression?

In integral containing products of bessel functions , the answer after applying the factor method is some expression whose numerator is not "completely factored". For aesthetics or because it is easier to work with, we may want to factor the numerator too.

Is there a way to factor a symbolic expression recursively?

Take this example:

sage: , var x y z
sage: expr = x*y - x*z + 4*z*y*x
sage: factor(expr2)
(4*y*z + y - z)*x
sage: _.operands()[0].factor()
4*y*z + y - z
(4*y - 1)*z + y
sage: _.factor()
4*y*z + y - z

so it seems that applying recursively the factor command to the expression tree alone doesn't succeed to get ((4*y - 1)*z + y)*x.

edit retag close merge delete

1

This is for my taste not really a factorization, (after factorizing $x$,) rather a way of extracting the coefficients of a polynomial (w.r.t. one of the variables).

The following can be polished to work recursively, if this is the point,

sage: var( 'x,y,z' );
sage: p = x*y - x*z + 4*z*y*x
sage: p.coefficients( x )
[[4*y*z + y - z, 1]]
sage: p.coefficients?
sage: [ [ q.coefficients(y), xdegree ]
....:     for q, xdegree in p.coefficients( x ) ]
[[[[-z, 0], [4*z + 1, 1]], 1]]
sage: [ [ r.coefficients(z), xdegree ]
....:     for r, xdegree in p.coefficients( x ) ]
[[[[y, 0], [4*y - 1, 1]], 1]]

but there is a "human choice" to prefer y or z (in the first factor of a Factorization object). Is this the direction of the question?

( 2017-06-09 14:28:44 -0600 )edit

yes, that seems useful, thanks. i didn't care so much about a particular choice of one or other factorization, just (as a starter) that a command can reduce the number of summands as much as possible, which didn't happen using factor alone.. and for the use case.. i don't know, take for instance some formulas in electrodynamics)...

( 2017-06-11 10:32:59 -0600 )edit


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# Latent Heat

Written by Jerry Ratzlaff. Posted in Thermodynamics

Latent heat is the heat required to change a material without a change in temperature.

## formula

$$L = \frac{Q}{m}$$

Where:

$$L$$  = latent heat of a substance

$$Q$$ = amount of energy transferred for the phase change

$$m$$ = mass of the substance

Solve for:

$$Q = mL$$


text:
Asymptotes of a function

We define an asymptote as a straight line that can be horizontal, vertical or obliquous that goes closer and closer to a curve which is the graphic of a given function.

These asymptotes usually appear if there are points where the function is not defined.

Let's see an example, since it will make it easier to understand.

Let's take the function $$f(x)=\dfrac{1}{x}$$. It is clear that when $$x=0$$ the function is not defined. It is exactly here where the asymptote appears. Let's see the graphic representation of this function:

When we go closer and closer to $$x=0$$ from the right we can observe that the function tends to infinity, increasingly approaching the straight line $$x=0$$. The same happens on the other side, but going to minus infinity when approaching the line $$x=0$$. We say that the line $$x=0$$ is an asymptote.

Let's see a more exact definition of asymptote of a function $$f(x)$$:

Vertical Asymptote

We will say that the straight line $$x=a$$ (where $$a$$ is a number) is a vertical asymptote if there exist some of these two limits:

• $$\displaystyle \lim_{x\rightarrow a^-}f(x)=\pm\infty$$
• $$\displaystyle \lim_{x\rightarrow a^+}f(x)=\pm\infty$$

The function $$f(x)=\dfrac{1}{1+x}$$ has a vertical asymptote in $$x=-1$$, since the limit $$\displaystyle \lim_{x\rightarrow -1^+}f(x)=\pm\infty$$ exists and therefore the asymptote also does.

Horizontal Asymptote

If the following limit exists:

$$\displaystyle \lim_{x\rightarrow \pm\infty}f(x)=a$$$where $$a$$ is a finite value, then we will say that the line $$y=a$$ is an horizontal asymptote. The function $$f(x)=e^x$$ has an horizontal asymptote in $$y=0$$ since: $$\displaystyle \lim_{x\rightarrow -\infty}e^x=0$$$

Obliquous Asymptote

If the following limits exist and are finite:

• $$\displaystyle \lim_{x\rightarrow \infty}\frac{f(x)}{x}=m$$
• $$\displaystyle \lim_{x\rightarrow \infty}(f(x)-mx)=b$$

Then we will say that an obliquous asymptote exists and the line of the asymptote is given by the equation $$y=mx+b$$.

The obliquous asymptotes only exist in rational functions (division of polynomials) where the polynomial of the nominator has a greater degree than the denominator.

The function $$f(x)=\dfrac{x^2+1}{x}$$ has an obliquous asymptote since the following limits exist:

$$\displaystyle m=\lim_{x\rightarrow \infty}\frac{f(x)}{x}=\lim_{x\rightarrow \infty}\frac{x^2+1}{x^2}=1$$$$$\displaystyle b=\lim_{x\rightarrow \infty}(f(x)-mx)=\lim_{x\rightarrow \infty} \Big( \frac{x^2+1}{x}-\frac{x^2}{x}\Big)=\lim_{x\rightarrow \infty}\Big( \frac{1}{x} \Big) =0$$$

so the line $$y=x$$ is the obliquous asymptote.

(we observe the obliquous asymptote in green, the straight line being $$y = x$$).


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<meta http-equiv="refresh" content="1; url=/nojavascript/">

# SSS Similarity

## Triangles are similar if their corresponding sides are proportional.

%
Progress
Practice SSS Similarity
Progress
%
Similarity Test True or False
Teacher Contributed
When you compare two triangles, if the corresponding sides are in ratio of 510$\dfrac{5}{10}$ and 820$\dfrac{8}{20}$, then the triangles are similar by the SSS similarity test.

qid: 100274


text:
These functions signal simple conditions with rlang::signal() (see also base::signalCondition()). They don't have any effect unless a condition handler is installed for conditions of class lifecycle_experimental or lifecycle_superseded.

signal_experimental(when, what)

signal_superseded(when, what, with = NULL)

lifecycle_cnd_data(cnd)

## Arguments

when The version at which the feature was assigned a lifecycle stage. Feature specifications for the feature and optionally its replacement. See deprecate_soft() for details. A condition object thrown by signal_experimental() or signal_superseded().

## Details

In order to be fast, signal_experimental() and signal_superseded() do not validate their arguments. Instead, you need to catch the condition object and pass it to lifecycle_cnd_data() to get structured data.


text:
# convergence radius of two specific power series

I did two calculations that I think are wrong but I am not sure why.

I have to compute the convergence radius of the following power series

a) $$\sum_0^{\infty} \ln(k!)x^k$$,

b) $$\sum_0^{\infty} k8^kx^{3k}$$.

Here's my attempt:

a)

$$\limsup_{k \rightarrow \infty} \sqrt[k]{a_k} = \limsup_{k \rightarrow \infty} \sqrt[k]{\ln(k!)} = \ln (\limsup_{k \rightarrow \infty} \sqrt[k]{k!})= \infty.$$

Therefore the convergence radius should be $$0$$ and the series only converges for the value $$x=0$$. Unfortunately, Wolfram alpha gives me another answer. Where's the mistake ?

b)

$$\limsup_{k \rightarrow \infty} \sqrt[k]{a_k} = \limsup_{k \rightarrow \infty} \sqrt[k]{k8^k} = \limsup_{k \rightarrow \infty} \sqrt[k]{k} \sqrt[k]{8^k} = 8 \limsup_{k \rightarrow \infty} \sqrt[k]{k} = 8.$$

I would now conclude that the convergence radius is $$\frac{1}{8}$$, but it appears to be $$\frac{1}{2}$$, what did I miss ?

• For the first one, use the ratio test and Stirling approximation. – Claude Leibovici Apr 1 at 10:13

For a) you have made the error of assume the log and the exponent commute. That is $$\ln \sqrt[k]{k!} \ne \sqrt[k]{ \ln k!}$$ and hence $$\limsup_{k \to \infty} \sqrt[k]{ \ln k!} \ne \ln \left( \limsup_{n \to \infty} \sqrt[k]{k!} \right)$$ I have to disagree with the other answer on how to evaluate this limit (after the error). Using Stirling's approximation gives $$\sqrt[k]{k!} \sim \sqrt[k]{ \sqrt{2 \pi k} \left( \frac{k}{e} \right)^k } = \left(2 \pi k \right)^{\frac{1}{2k}} \frac{k}{e} \to \infty$$

Of course the actual problem was the one I pointed out before. Here's how I would attack it. $$\lim_{k \to \infty} \sqrt[k]{ \ln k!}=\lim_{k \to \infty} \frac{ \ln (k+1)!}{\ln (k!)}=\lim_{k \to \infty} \frac{\sum_{j=0}^{k+1} \ln j}{\sum_{j=0}^{k} \ln j}=\lim_{k \to \infty} \frac{\ln (k+1)}{\ln k}=1$$ Note I used the Stolz–Cesàro theorem in the third equality.

I agree with the other answer for (b). You missed the $$3k$$ in the exponent

• I see, thank you very much for the detailed answer ! – Alain Apr 1 at 10:44
• No problem. Happy to help – B.Martin Apr 1 at 20:58

Answer for b). What you have calculated is the radius of convergence of $$\sum k8^{x}x^{k}$$. The given series is convergent if $$|x^{3}| < \frac 1 8$$ and divergent for $$|x^{3}| >\frac 1 8$$. Can you see from this that the radius of convergence is $$\frac 1 2$$?.

Part a). It is not true that $$(k!)^{1/k} \to \infty$$ as $$k \to \infty$$. Use Stirling's approximation for this part.

• Thank you for the comment ! Now I understand what I missed. – Alain Apr 1 at 10:44


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## The Basic Practice of Statistics 7th Edition

$b = r\frac{s_{y}}{s_{x}} = (-0.98)\frac{7.0}{14.0} = -0.49$ a = ȳ - bx̄ = 27.2 - (-0.49) (1992) = 1003.28 General least square regression line is: ŷ = a + bx = 1003.28 - 0.49x
$b = r\frac{s_{y}}{s_{x}} = (-0.98)\frac{7.0}{14.0} = -0.49$ a = ȳ - bx̄ = 27.2 - (-0.49) (1992) = 1003.28 General least square regression line is: ŷ = a + bx = 1003.28 - 0.49x