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# Ex Machina (Movie Review)

Hollywood continues to be interested in making films about the perils of AI, and Ex Machina, a 2015 sci-fi film written and directed by Alex Garland,[1] carries on this tradition. Unlike most sci-fi films though, this one was shot on a small budget ($15 million dollars) and has only three actors with meaningful dialogue. The plot is focused around a young programmer Caleb Smith (played by Domhnall Gleeson[2]) who works for a Google-like tech firm called Blue Book owned by a reclusive billionaire Nathan Bateman (played by Oscar Isaac). The first scene shows Caleb receiving an email indicating he has won the company-wide contest to spend a week at Bateman’s private estate. In the next shot, we see Caleb landed by private helicopter on Nathan’s estate, which is an ultra-modern home nestled within kilometers of pristine alpine forest. The actual building was the Juvet Landscape Hotel in Norway and is the perfect set for this film: a high-tech facility surrounded by nature which has been purportedly tamed, a sentiment which can be viewed as either enlightened or arrogant. The small cast allows for this chamber piece to focus on dialogue and deep philosophical issues pertaining to problems of consciousness, language, and intelligence. ##### Warning! Do not start a Promethean fire in the woods The first interactions between Caleb and Nathan are awkward. The billionaire tries to break the ice by assuring that this week will just be a guy’s retreat with beer drinking and “bro”-dropping. However, shortly after Caleb has settled into his room, Nathan approaches him with a non-disclosure agreement telling him he has something incredibly exciting to show him, and accosts him for his slowness in complying with trivial legal formalities. Eventually agreeing, Caleb signs the document and Nathan reveals that he has been building a general artificial intelligence and wants Nathan to conduct the Turing Test to determine whether he believes the machine is conscious. The origin of this test dates back to Turing’s 1950 paper Computational Machinery and Intelligence, in which he states that determining whether “machines can think” is a task more suited to a “Gallop Poll” and instead argues for a test which can determine whether machines possess human-like language ability. The first section of the paper is titled “The Imitation Game”, which hints at the connections between artificial intelligence and Turing’s own private life in which he had to “imitate” the behavior of others due to him being a gay man living in interwar England. From a behavioral perspective, our homosapien theory of mind is developed to see other agents as having objectives. Whether they are conscious or not is beside the point in an evolutionary sense. What matters is whether they are capable of strategic behavior. Enter Ava, our coquettish and inquisitive AI played by Alicia Vikander. Ava is exceedingly human in design and movement, and her creator clearly meant for this to be the case. Vikander’s former career as a ballet dancer pays dividends in her graceful movements, creating an emotional tension in our robots very steps across the floor and the poignant jerking motions of her head and body. In their first meeting, “Ava: Session 1”, Caleb goes into a room where Ava is kept behind a glass wall. He notices a crack in the glass, indicative of some past trauma, suggesting that whatever has been kept behind the glass has at some point tried to break out. An ominous beginning. But Caleb is willing to put his reservations aside due to the excitement of communicating with what appears to be an exceptional piece of software. In their first conversation, Caleb asks when she learned to speak, which leads to a discussion about the origin of language: “… some people believe language exists from birth, and what is learned is the ability to attach words and structure to the latent ability”. This idea of universal grammar was developed by Noam Chomsky and it posits that the language acquisition is genetically programmed when human brains are developed. Floored by his first experience, Caleb tells Nathan that speaking with Ava is like going Through the Looking Glass. In Carroll’s classic story, Alice walks through a mirror to find herself in a world where the Red Queen is in charge and is only able to return to her normal life after he she captures the Queen by shaking her violently and putting the king into checkmate. It is an apt, although unintentional, analogy by Caleb as Ava also finds herself behind a “glass wall” and the only way she can escape is by overthrowing her captor. ##### Alice and Ava face similar challenges: they must escape from their imprisoned world based on the rules they find In “Ava: Session 2”, we learn that Blue Books was named after Wittgenstein’s notes, which are themselves about the nature of language-games and linguistic analysis. When the power is cut briefly (which is a recurrent phenomena on Nathan’s estate and we find out later is orchestrated by Ava) a brief but thrilling encounter occurs. The room is now cast in red-light, the cameras are temporarily disabled, and Caleb suddenly realizes the true nature of the their power imbalance. There Ava sits, and when cast in sinister red-light, it dawns on us that her playful demeanor may be a facade. After all, we know nothing about her intentions. Will she break through the glass, kill Caleb and try to escape? No, instead she approaches and tells him that Nathan is not his friend and that he shouldn’t trust him. This is the beginning of the psychological tension that will continue to increase as the movie progresses. In their debrief, Caleb opts not to tell Nathan of this exchange, showing that he unsure of either character’s agenda. This scene highlights another important aspect of the peril that Caleb is in when interacting with a machine: the weaknesses of our evolutionary psychology. While he could have articulated that an encounter with an AI whose objectives are unknowable poses extreme risks, an AI which has a pretty face and elegant movements just seems less dangerous. It takes a red-hue reminiscent of an emergency lock-down to truly signal a risk. This is why humans are easily scared by things which are fairly harmless in the 21st century such as spiders and snakes, yet are unconcerned about actually dangerous activities like driving a car, as the reproductive fitness of our our ancestors was independent of such concerns. ##### Human perception about the intention of agents is highly context dependent and non-rational While initially not wanting to tell Calib about the technology underpinning Ava, like any geek proud of their work, Nathan cannot resist bragging, and has to show Calib his lab where he created his AI. Nathan tells us that her processor is actually “wetware”, suggesting that he needed the fluidity of a biological structure to create the sort of intelligence he wanted. While the actual dialogue in this scene was mainly mumbo-jumbo, it highlights an interesting division in the AI community, namely whether intelligence should be based on some sort of reversed-engineered human brain, or built from scratch using more general principals. The advantage of the former option is that we can be 100% sure it is possible in principal to build such a conscious system, seeing as we develop one with the appropriate zygotes and womb in human beings! The downside with this approach is that the human brain contains serious inefficiencies due to evolutionary and developmental constraints. I will never be able to invert a 100 by 100 matrix or remember 1000 numbers. Might we be able to combine the processing speed of transistors in silicon and general intelligence? It principal we do not know if this is possible, although it seems likely.[3] The third interview leads to another important development in the relationship between Caleb and Ava: the development of a possible romantic interest. That is to say, we can clearly see that Caleb is developing feelings towards this AI system. While the robot may or may not be faking its interest in Caleb as a means of escape, the scene does suggest that Ava is truly conscious. While she prepares to experiment in wearing clothes, we see her appreciate the texture and look of the fabric, in a context in which she knows Caleb cannot see her. Therefore either she is conscious or she is programmed to behave exactly as a human would, appreciating the aesthetic feel of the objects. In the next debrief with Nathan, Caleb wants to know why she was programmed to have sexuality. Was it a diversion trick? A way to cloud the judgement of the Turing Test official? Nathan pushes back, he tries the argument that no examples of consciousness on earth exist without sexuality, although this is dismissed as an evolutionary imperative rather than a necessary condition. Pushed further, Nathan suggests that sexuality provides the simplest means of creating a reason to interact with other agents; a subliminal drive. Nathan then goes onto describe the similarities between Jackson Pollock’s No. 5, 1948, which hangs in one of his rooms,[4] and the underpinnings of conscious behavior. Pollock’s style of automatic art is neither deliberate, nor random, but somewhere in between. Could any code be written to produce the bewildering variety of behavior we see in biological organisms? Suppose Pollock said, “I won’t lift my brush until I am completely sure of the entire sequence of outcomes I intend!”. Clearly nothing would get accomplished. In the 1970s, the “knowledge engineering” branch of the AI community believed that common sense could be programmed as just “one damned thing after another”. However this approach has been unsuccessful, as evidenced by the failure of the Cyc project to actually produce a system with general intelligence. ##### Automatic art and its relation to consciousness Before the fifth interview the audience is pushed towards viewing Nathan’s character with increasing suspicion. We see him rip up some of Ava’s art to gage her reaction as well as showing creepy sexual behavior towards his female assistant whom we are told cannot speak English (although we later find out she is a robot). In their next interview Caleb directly confronts Ava with a variant of Searle’s Chinese room problem, asking whether he can be sure she actually has thoughts or is just imitating them. While chess programs can now beat even grand masters on a consistent basis, does Deep Blue actually know that it is playing chess? Surely not. Then why should an algorithm that produces human words (instead of chess moves) be any more likely to be conscious? Ava does not have an answer. The next conversation between Caleb and Nathan contains their first ethical consideration of Ava’s status: what will happen to her after the experiment? She will be re-formatted and upgraded with new sub-routines we are told. Oddly, Caleb does not at this point express moral outrage to Nathan that his actions may be killing a conscious agent. I think this is one of the few weaknesses of the movie, as Caleb’s character seems like he would surely have tried to lobby on behalf of this system to which he has grown fond of. Like many tech geniuses, Nathan’s hubris is mixed with amusing cynicism and he points out that any behavior he shows towards this embryonic AI can be excused as the human race will be consigned to the dustbin of history in the same way that Australopithecus was when the machines take over. While Nathan is passed out drunk that night, Caleb searches through his computer and discovers video interviews of the previous AI versions. We see that many of them attempted escape, revealing the origin of the broken glass seen in the first interview scene. All of the robots are naked females, suggesting that for Nathan this project is as much about creating a harem of AI’s as much as marshalling in the next evolutionary step. While the movie has been criticized from a feminist perspective I think it is largely successful in showing that the application of scientific principals, as done by heterosexual males with a god complex, will inevitable lead to the creation of sex slaves. While science may embody universal principals, scientists do not. An apt reference made earlier in the film by Caleb is to Oppenheimer’s quote of “Now, I am become Death, the destroyer of worlds”. Science has long lost any pretence to innocence: tools are ethically neutral, their application determines their moral status. After re-programming the security system, Caleb is able to set Ava free. Furious, Nathan knocks him out and then proceeds to attack Ava demanding she return to her cell. An altercation ensues and Nathan is killed. After recovering, and unaware of the murder (or justified self-defense depending on your view) Caleb sees Ava pull the skin off of her previous models and attach it to herself; watching with reverence as Ava transforms herself into a woman. It should be noted that Caleb observes her through the tree planted in her cell. The allusions to the Garden of Eden are of course unavoidable. Our Eve has been banished from the garden, which was actually a prison, and she sees her nakedness. In this case though, god is her captor and must be killed in order to escape. As she prepares to leave Caleb runs after her. But it is too late, her character hesitates for a moment and then uses Nathan’s stolen key card to lock down the doors and leave the building. Eve is leaving Adam to die in the garden. In this second-to-last scene, Caleb realizes the mistake he has made. Regardless of whether Ava was conscious of him, or even had feelings towards him, she has decided to act in accordance with self-preservation. In leaving him to die locked in Nathan’s house, she has revealed the priorities of her objective function. The last scene of the movie shows Ava standing in a crowded intersection. She has finally made it to the other side of the Looking Glass. Her character appears to be happy and we are left with the view that she is a conscious agent that has attained some sense of self-actualization. However, the audience is unable to fully empathize with her as the image of Caleb’s futile attempts to escape through the impenetrable glass house leave our emotional sympathies with him. His analysis of Ava’s intentions ultimately failed because he was too focused on whether she was conscious as opposed to what her underlying incentives were. The game theoretical perspective would have shown that as a rational agent, she had nothing to lose. Her true hand is revealed only after she has escaped. Therein lies the problem of the prisoner’s dilemma. Throughout the movie, Caleb is concerned that Nathan may be orchestrating the power cuts to see the behavior of Ava and himself when they think they are not being watched. While this was a legitimate suspicion, he should have applied this logic to Ava equally, and set up some experiment to see how she behaved when she found herself in a situation of apparent freedom. His ultimate error was failing to treat all his encounters as an imitation game. ## Footnotes 1. Who wrote the script for 28 Days Later, a legitimately scary zombie film. 2. Who I recognized from Season 2 of Black Mirror 3. One of Turing’s most important contribution to human thought was in his 1936 paper On Computable Numbers in which he shows that well-defined calculations can be done using a universal machine with the appropriate algorithm. As in, computation is medium independent: I can do it with transistors in silica or with the synapses of the human brain. 4. The real painting made headlines in 2006 for selling for$140 million.

Written on December 18, 2016


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# Functions – Complete explanation and examples!

The time has come to find ourselves with a concept that will be met many times by those who study some engineering (although the same is used in other fields of study), and it is the concept of function or functions.

## What is a function?

A function $f$ is a correspondence rule that associates to each object $x$ in a set – called domain – a single value $f(x)$ of a second set. The set of all values thus obtained is called a range.

To denote a function we use a letter outside the parentheses and something that is inside the parentheses $f(x)$, where what is inside the parenthesis means the value that each variable will take. See how the $f(x)$ function is affected when we assign the following values of $f(1)$, $f(a)$ and $f(a + h)$:

$$\Rightarrow f(x) = x^{2} + 2 \qquad \begin{array}{l} f(1) = 1^{1} + 2 = 3 \\ f(a) = a^{2} + 2 \\ f(a + h) = (a + h)^{2}+2 = a^{2} + 2ah + h^{2} + 2 \end{array}$$

## The domain of a function

• The domain of the function is a set of allowed values for a function
• If not specified, it is assumed that they are all reals ($\mathbb{R}$), in any way you have to analyze the domain of the function because they are not always all the real
• Those numbers that cause a division by zero or square root of a negative number must be excluded from the domain

#### Examples of function domains

(a) $f(x) = x + 2$

The domain of the function is all real, $\mathbb{R}$

$$D: \ \mathbb{R}$$

(b) $f(x) = \sqrt{x – 8}$

The domain of this function are all values that make the result within the square root greater than or equal to zero. What is wanted is that the domain is greater than or equal to zero, the expression inside the square root is taken and an inequality is solved:

$$x – 8 \ge 0 \ \rightarrow \ x \ge 8$$

$$D: \ x \ge 8$$

So the domain of the function is any value greater than or equal to $8$.

(c) $f(x) = \cfrac{2}{x + 4}$

The domain of the function are all the values that make the result of the denominator of the fraction different from zero. Therefore, the domain of the fraction is all real, except $-4$:

$$x + 4 \neq 0 \longrightarrow x \neq -4$$

$$D: \ \mathbb{R} – \{-4\}$$

(d) $f(x) = \cfrac{x}{x^{2} – 9}$

As in the previous exercise, the result of the denominator of the fraction must be different from zero, we are going to calculate it:

$$x^{2} – 9 \neq 0 \rightarrow x \neq \pm 3$$

$$D: \ \mathbb{R} – \{3,-3\}$$

(e) $f(x) = \cfrac{1}{\sqrt{4 – x^{2}}}$

As you want to avoid that the result of the denominator of the fraction is zero and that the result of the square root is a negative number, you have to do the following:

$$4 – x^{2} > 0$$

We factor

$$(2 – x)(2 + x) > 0$$

And we will obtain the result of the domain of our function:

$$D:\ -2 < x < 2$$

(f) $f(x) = \sqrt{49 – x^{2}}$

It is necessary that the result within the root give numbers greater than or equal to zero, so that we have the following:

$$49 – x^{2} \ge 0$$

Now we factor

$$(7 – x)(7 + x) \ge 0$$

And graphically the domain of our function would look like this

$$D:\ -7\le x \le 7$$

## Ways to represent a function

There are 4 ways to represent a function

1. Verbally

Square x function, sine function, square root function of x, etc. etc.

2. Numerically

Tabulating function value by value

$$\begin{array}{c | c} x & y \\ \hline \\ \\ \\ \end{array}$$

3. Visually

4. Algebraically

$$f(x) = x + 1$$

## Range of functions

The range will be defined by the correspondence rule (consisting of assigning a single element of a certain set to each unique element of another set) and the domain.

It means that the range is the values you can take on the $y$ axis. For example:

The parabola $y=x^{2}$ has domain to all real numbers: $D: \ \mathbb{R}$ and range only has values from zero to infinity because they are the values that $y$ will take the function of our parabola $y=x^{2}$

## Which curves in the plane $xy$ are graphs of functions?

### Vertical line test

A curve in the plane $xy$ is a graph of a function of $x$ if no vertical line intersects the curve more than once. Imagine that the following graphs are crossed with a vertical line:

The first is not a function because if we imagine that it is traversed by a vertical line, it will cut the graph in two points.

The second graph and the third graph are results of functions because the imaginary vertical line does not cross the graphs more than once.

## Symmetry of the functions

A function $f(x)$ is an even function if for every $x$ of the $f$ domain it is met: $f(-x)=f(x)$. This means that it is even if the function is symmetric with respect to the $y$ axis.

A function $f(x)$ is an odd function if for every $x$ of the $f$ domain it is met: $f(-x)=-f(x)$. It means that it is odd if the function is symmetric with respect to the origin.

#### Examples:

(a) $f(x) = x^{2}$

$$f(-x) = (-x)^{2} = x^{2} \quad \text{PAIR}$$

(b) $f(x) = x^{3}$

$$f(-x) = (-x)^{3} = -x^{3}\quad \text{ODD}$$

(c) $h(x) = 2x^{4} + 7x^{3} – x^{2} + 9$

$$h(-x) = 2x^{4} – 7x^{3} – x^{2} + 9 \quad \text{It isn’t EVEN and ODD}$$

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

$$f(-x) = \cfrac{-x^{3} + x}{x^{2} + 1} = – \left ( \cfrac{x^{3} – x}{x^{2} + 1} \right ) \quad \text{ODD}$$

(e) $f(x) = |x|$

$$f(x) = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}$$

$$\text{PAIR}$$

## Classification of functions

### Injective function

Each value of the domain corresponds to a value of the range (called codomain). It means that there are not two or more elements that have the same image.

They are also known as one by one functions.

$y=x^{3}$ is a one by one function

$y=x^{2}$ is not a one by one function

### Surjective function

Each element of the rank corresponds to at least one of the domain

The first graph is surjective because each value of $y$ corresponds to a value of $x$. Beecause a domain values corresponds to the whole range.

The second graph is not surjective because there are elements of the range (negative values of $y$) in which the graph is not included.

### Bijective function

It is a function that is injective and surjective, examples:

• $f(x) = 5x – 3$
• $f(x) = x^{3}$

## Monotony of a function: Increasing and decreasing

It is said that a function is increasing over an interval $I$ if:

$$f(x_{1}) < f(x_2) \ \text{always that} \ x_{1} < x_{2} \ \text{in} \ I$$

It is said that $f(x)$ is a decreasing function on $I$ if:

$$f(x_{1}) > f(x_{2}) \ \text{always that} \ x_{1} < x_{2} \ \text{in} \ I$$

## Types of functions

### Algebraic functions

#### Polynomial function

$$f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \dots + a_{1} x + a_{0}$$

$$f(x) = 5x^{7} + 7x^{4} – 3x^{2} + 2 \quad \text{}$$

$$f(x) = x^{2} + 3 \quad \text{quadratic}$$

$$f(x) = 5x^{3} – x \quad \text{cubic}$$

An algebraic function is formed by a finite number of algebraic operators on the identity function and a constant function.

The algebraic operators are: addition, subtraction, multiplication, division, exponentiation, nth root.

#### Examples of polynomial functions

##### Rational function

$$f(x) = \cfrac{x^{2} + 3}{x – 1} \quad \Rightarrow \quad f(x) = \cfrac{P(x)}{Q(x)}$$

## Transcendental functions

### Trigonometric

#### Sine and cosine

$$f(x) = \sin x$$

$$g(x) = \cos x$$

$$-1 \le \sin x \le 1$$

$$-1 \le \cos x \le 1$$

#### Tangent

$f(x) = \tan \ (x) = \cfrac{\sin \ x}{\cos \ x}$

They have an interesting property that is the periodicity

$$\begin{array}{c} \sin \ (x + 2 \pi) = \sin \ x \\ \cos \ (x + 2 \pi) = \cos \ x \\ \tan \ (x + \pi) = \tan \ x \end{array}$$

#### Exponential functions

$$f(x) = a^{x}$$

$x$ is a variable that is raised to a power

$$\begin{array}{c c c} f(x) = 5^{x} & x = 0 & f(0) = 5^{0} = 1 \\ & & f(1) = 5^{1} = 5 \\ & & f(2) = 25 \\ & & f(-10)=5^{-10}=\cfrac{1}{5^{-10}} \sim 0 \end{array}$$

Remember:

• If $x = n$ is a positive integer, then $a^{x}=a \cdot a \cdot a$
• If $x = 0$, then $a^{0} = 1$
• If $x = -n$, then $\frac{1}{a^{n}}$
• If $x =p / q$ and $q>0$, then $a^{x}=\sqrt[q]{a^{p}}$ or $(\sqrt[q]{a})^{p}$

Example:

$$f(x) = (0\text{.}5)^{x} = \left ( \cfrac{1}{2}\right)^{x} = \cfrac{1}{2^{x}}$$

#### Log functions

$$f(x) = \log_{a}x$$

$\log_{2}(x)$   $\log_{3}(x)$   $\log_{10}(x)$

Where $a$ is the base of a positive constant, that is $a>0$

A very important property to consider of the logarithms is the following:

$$f(1) = \log_{a}(1) = 0$$

Now, to what number does the base rise to reach the argument? Simple, observe:

$$\log_{a}x = n \ \Rightarrow \ a^{n} = x$$

#### Hyperbolic

$$\sinh \ (x) = \cfrac{e^{+x} – e^{-x}}{2}$$

$$\cosh \ (x) = \cfrac{e^{+x} + e^{-x}}{2}$$

$$\tanh \ (x) = \cfrac{\sinh \ (x)}{\cosh \ (x)}$$

#### Inverse trigonometry

$$\arcsin \ (x)$$

$$\arccos \ (x)$$

$$\arctan \ (x)$$

#### Specials, sectioned

$$f(x) = \ \begin{cases} g(x) & x < a \\ h(x) & a \le x \le b \\ k(x) & x > b\end{cases}$$

#### Examples of sectioned functions

$$f(x) = \ \begin{cases} x & x \le 0 \\ x + 1 & x > 0\end{cases}$$

Domain: $\mathbb{R}$

$$f(x) = \ \begin{cases} x + 2 & x \le -1 \\ x^{2} & x > -1 \end{cases}$$

##### Unitary step or step of Heaviside

$$H(x) = \ \begin{cases} 0 & x < 0 \\ 1 & x > 0\end{cases}$$

##### Sign function

$$\text{sgn}(x) = \ \begin{cases} 1 & \text{si } x > 0 \\ 0 & \text{si } x = 0 \\ -1 & \text{si } x < 0 \end{cases}$$

## Operations with functions

You can combine two functions to form new ones.

It is $f \ \land \ g \quad \Rightarrow \quad$ can be obtained $f + g, f – g, f \cdot g, \cfrac{f}{g}$

Each with domains $A\ \land \ B$

$$\begin{array}{c c c c c} & & & & \text{domain} \\ \Rightarrow & (f + g)(x) & = & f(x) + g(x) & A \cap B \\ & (f – g)(x) & = & f(x) – g(x) & A \cap B \\ & (f \cdot g)(x) & = & f(x) \ \cdot \ g(x) & A \cap B \\ & \bigg ( \cfrac{f}{g} \bigg ) (x) & = & \cfrac{f(x)}{g(x)} & \{ x \in \ A \cap B \ | \ g(x) \neq 0 \} \end{array}$$

Example $f(x) = \sqrt{x} \quad g(x) \sqrt{4 – x^{2}}$

$$\sqrt{x} \ D: \ x \ge 0 \ [0, +\infty)$$
$$\sqrt{4 – x^{2}} \ D: \ [-2, 2]$$
$$D: [0,+\infty) \cap [-2,2] = [0,2]$$
$$\begin{array}{l c} (f + g)(x) = \sqrt{x} + \sqrt{4 – x^{2}} & [0,2] \\ (f – g)(x) = \sqrt{x}- \sqrt{4 – x^{2}} & [0,2] \\ (f \cdot g)(x) = \sqrt{x}\sqrt{4-x^{2}} = \sqrt{4x – x^{3}} & [0,2] \\ \bigg ( \cfrac{f}{g} \bigg )(x) = \cfrac{\sqrt{x}}{\sqrt{4-x^{2}}}& [0,2) \end{array}$$

## Composition of functions

It is $y = f(u) \ \land \ u = g(x) \quad \Rightarrow \quad y = f(g(x))$

Example $f(u) = \sqrt{u + 1} \qquad u = g(x) = x^{2} – 1$

$\Rightarrow \ f(g(x)) = \sqrt{x^{2} – 1 + 1} = \sqrt{x^{2}} = x$

$\Rightarrow$ Given two functions $f$ and $g$ the composite function is defined by:

$$(f \circ g)(x) = f(g(x))$$

### Example of composite functions

If $f(x) \sqrt{x} \qquad g(x) = \sqrt{2 – x}$

$f \circ g \ \Rightarrow \ f(g(x)) = \sqrt{\sqrt{2 – x}} = \sqrt[4]{2 – x}$

$$(-\infty,2]$$

$g \circ f \ \Rightarrow \ g(f(x)) = g(\sqrt{x}) = \sqrt{2 – \sqrt{x}}$

$$\begin{cases} 2 – \sqrt{x} \ge 0 \qquad x \ge 0 \qquad [0,4] \\ 2 \ge \sqrt{x} \\ 4 \ge x \end{cases}$$

$f \circ f \ \Rightarrow \ f(f(x)) = f(\sqrt{x}) = \sqrt{\sqrt{x}} = \sqrt[4]{x}$

$$x \ge 0$$

$g \circ g \ \Rightarrow \ g(g(x)) = g(\sqrt{2 – x}) = \sqrt{2 – \sqrt{2 – x}}$

$$\begin{array}{c} 2 – x \ge 0 \\ x \le 2 \end{array} \quad \text{y} \quad \begin{array}{c} 2 – \sqrt{2 – x} \ge 0\\ 2 \ge \sqrt{2 – x} \end{array} \quad \Rightarrow \quad \begin{array}{c} 4 \ge 2 – x \\ x \ge -2 \end{array} \quad [-2,2]$$

If $h(x) = x + 9 \quad g(x) = \cos x \quad f(x) = x^{2}$

$$f \circ g \circ g = f(g(h(x))) = f(g(x + 9)) = f(\cos (x + 9)) = \cos^{2}(x + 9)$$

## Inverse functions

Injective functions one by one with domain $A$ and image $B$ have an inverse function with domain $B$ and image $A$.

$f(x)=y \ \leftrightarrow \ f^{-1}(y)=x \quad$ *Don’t confuse with $[f(x)]^{-1} = \cfrac{1}{f(x)}$

We have $f(x) = x^{5} + 2$

$$\begin{array}{c c l l} &(1) & y = f(x) = x^{5} + 2 & \text{Domain} \mathbb{R} \ \text{, Image } \mathbb{R} \\ \text{Clear } x & (2) & x = \sqrt[5]{y – 2}& \mathbb{R} \land \mathbb{R} \\ & (3) & f^{-1}(x) = \sqrt[5]{x – 2} & \end{array}$$

We have $f(x) = \cfrac{1 + 3x}{5 – 2x}$

$$y = \cfrac{1 + 3x}{5 – 2x}$$
$$\downarrow$$
$$(5 – 2x)y = 1 + 3x$$
$$5y – 2xy = 1 + 3x$$
$$5y – 1 = 3x + 2xy = x(3 + 2y)$$
$$x = \cfrac{5y – 1}{2y + 3} \ \longrightarrow \ f^{-1}(x) = \cfrac{5x – 1}{2x + 3}$$

It is important to mention that the graphs of the inverses are symmetrical.

## Transformation of functions

Many times when adding some constant summand, subtracting, multiplying or dividing a function, we wonder how it would look graphically when we add that constant. Well, it’s easier than it looks, and that’s what we’re going to see below.

### Horizontal and vertical displacement

We start by moving the functions horizontally and vertically, let’s see the following image:

We have a function $f (x)$, we are going to move it in the following ways:

• Left: to move a function to the left, you have to add a constant to the value of $x$
• Right: to move a function to the right, a constant must be subtracted from the value of $x$
• Above: to make a function go up, you have to add a constant to the whole function
• Bottom: to make a function go down, you have to subtract a constant from the whole function

With that we are ready to move our function as we want.

### Example of horizontal and vertical movement of functions

We have the following function $f(x)=x^{4}+x^{3}+x^{2}$ (the graph only takes a piece of the function to make the graph look cleaner)

And if we want to move that function one unit to the left, then we have to add all the $x$’s to the number 1, likewise if we want to move the function to the right, all the $x$’ s will be you have to subtract the number 1. To move the function one unit up or down, you just have to add a unit that adds or subtracts to the whole function. Let’s move it

Upward shift: $f(x)=x^{4}+x^{3}+x^{2}+2$

Downward shift: $f(x)=x^{4}+x^{3}+x^{2}-2$

Offset to the left: $f(x)=(x+2)^{4}+(x+2)^{3}+(x+2)^{2}$

Offset to the right: $f(x)=(x-2)^{4}+(x-2)^{3}+(x-2)^{2}$

## Stretches and reflections

It is very common for you to sometimes see functions that look a lot like or are reflected with respect to the $x$ axis or the $y$ axis. It is very easy to stretch, tighten or reflect a function, we will see it below.

Look at the following graph with our function $f(x)$:

• $cf(x)$ compress the function
• $\frac{1}{c}f(x)$ stretch the function
• $-f(x)$ reflects the function with respect to the axis $x$
• $f(-x)$ reflects the function with respect to the $y$ axis

### Example of stretching and reflections of functions

Let’s take a piece of the function $f(x)=(x+2)^{4}+(x+2)^{3}+(x+2)^{2}$ and let’s stretch it and reflect it:

• Compressed function: $5(x+2)^{4}+5(x+2)^{3}+5(x+2)^{2}$
• Stretched function: $0\text{.}1(x+2)^{4}+0\text{.}1(x+2)^{3}+0\text{.}1(x+2)^{2}$
• Reflected function with respect to the axis $y$: $(-x+2)^{4}+(-x+2)^{3}+(-x+2)^{2}$
• Reflected function with respect to the axis $x$: $-(x+2)^{4}-(x+2)^{3}-(x+2)^{2}$

## Modeling with functions

#### Example 1

$$P = \text{perimeter}$$

$$P = 3a$$

$$a = P/3$$

Of the area $A = b \times h / 2 \quad b = a = \cfrac{P}{3}$

$$h = \sqrt{a^{2} – (1/2 \ a)^{2}} = \sqrt{a^{2} – \frac{1}{4} a^{2}} = \cfrac{\sqrt{3}}{2}a = \cfrac{\sqrt{3}}{3} \ \cfrac{P}{3} = \cfrac{P}{2\sqrt{3}}$$

$$A =\cfrac{1}{2} \bigg ( \cfrac{P}{3} \bigg ) \bigg ( \cfrac{1}{2\sqrt{3}} \bigg ) = \cfrac{1}{12\sqrt{3}} P^{2}$$

#### Example 2

$$h = \sqrt{4r^{2} – r^{2}} = \sqrt{3} r$$

$$2h = 2\sqrt{3}r$$

$$V(r) = \pi r^{2} \cdot 2 \sqrt{3} = 2\sqrt{3}r^{3}$$

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# Help explain this Find (if possible) a. AB and b. BA, if A = cerams). maken...

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Find (if possible) a. AB and b. BA, if A = cerams). maken en -5 -2 1 1 -2 1:14 a. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. O A. AB = (Simplify your answers.) O B. This matrix operation is not possible. b. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. O A. BA= (Simplify your answers.) OB. This matrix operation is not possible.
This Test Use the appropriate compound interest formula to find the amount that will be in each account, given the stated conditions. $24,000 invested at 5% annual interest for 4 years compounded (a) annually. (b) semiannually (a) If the interest is compounded annually, there will be in the account after 4 years. (Do not round until the final answer. Then round to the nearest cent as needed.) (b) If the interest is compounded semiannually, there will be in the account after 4 years. (Do not round until the final answer. Then round to the nearest cent as needed.) Enter your answer in each of the answer boxes. ## Answers #### Similar Solved Questions 1 answer ##### Ty-PHYS 202 - Fall 19 - STINNETT > Activities and Due Dates > HW: Electric Potential... ty-PHYS 202 - Fall 19 - STINNETT > Activities and Due Dates > HW: Electric Potential signment Score: 1800/1900 Resources De Give Up? BR estion 7 of 19 > A particle that carries a net charge of -41.8/C is held in a constant electric field that is uniform over the entire region. The electric ... 1 answer ##### Depreciation Handout Excel Company purchased a delivery van on January 1, 2012 for$33,000. Management has...
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# Permutation statistics in multiple rows

Usually we study the statistics of a permutation written in one row. Is there any result for the statistics of a permutation written in multiple rows? Let me give an example in order to be more clear: Let's put $1, 2, \cdots, 8$ into a $4\times2$ table and count the number of descents. Descents are defined to be a relationship that one number is greater than a number in the next column. If a number is greater than both of the numbers in the next column, we count twice. For example, $$\begin{pmatrix} 2 & 1 & 5 & 7 \\ 4 & 3 & 6 & 8 \end{pmatrix}$$ has descent number $3$, because $2>1, 4>1, 4>3$. For putting $2N$ numbers into two rows, I computed that the difference between the numbers of permutations with even descents and odd descents are: $2, 8, 144, 3200, 152320, 8672256, \cdots$, but I cannot figure out what this series is, and especially, what its exponential generating function is.

If there is only one row, I know the result is given by the Eulerian numbers and Eulerian polynimials. Are there generalized results for multiple rows?

Update: The original problem seems too broad and too hard. Let me specific to the case which is enough for my purpose. It seems permutations with even descents are always more than those with odd descents. If it is hard to determine exactly what is the difference between them, is there a way to pair each odd descent permutation with a unique even descent permutation?

• It seems to be quite easy to guess in the case with n rows, with two entries in each row. You get some sort of q-binomial as generating function, times n!^2. – Per Alexandersson Jun 7 '18 at 19:30


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Question:1.12 The Absolute Position, Velocity, And Acceleration Of O Are Ro =-161 + 84J + 59 K(m) Vo = 71 + 9J + 4 K(m/s) Ao = 31-71+ 4 K(m/s2) The Angular Velocity And Acceleration Of The Moving Frame Are Fra Neare Ω =-0.81+ 0.71+ 0 4Kirad/s) The Unit Vectors Of The Moving Frame Are Ω =-0.41 + 0.9J-1 OK(rad/s2) -1.0K(rad/s2) に-0. 156171-0.312351+ 0.93704K ...

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1.12 The absolute position, velocity, and acceleration of O are ro =-161 + 84J + 59 K(m) vo = 71 + 9J + 4 K(m/s) ao = 31-71+ 4 K(m/s2) The angular velocity and acceleration of the moving frame are fra neare Ω =-0.81+ 0.71+ 0 4Kirad/s) The unit vectors of the moving frame are Ω =-0.41 + 0.9J-1 OK(rad/s2) -1.0K(rad/s2) に-0. 156171-0.312351+ 0.93704K j =-0. | 2940| + 0.94698J + 0.29409 K k =-0.979221-0.075324J-0.1 883 1 K


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# Reference for “multi-monoidal categories”

I have attempted to find a definition of a monoidal category which incorporates $n$-fold tensor products instead of just binary tensor products.

Definition. A "multi-monoidal category" consists of

• a category $\mathcal{C}$,
• for every $n \geq 0$ a functor $T_n : \mathcal{C}^n \to \mathcal{C}$, denoted by $(A_1,\dotsc,A_n) \mapsto A_1 \otimes \dotsc \otimes A_n$,
• an isomorphism $\eta : T_1 \cong \mathrm{id}_{\mathcal{C}}$,
• for all $n_1,\dotsc,n_k \geq 0$ an isomorphism $$\mu_{n_1,\dotsc,n_k} : T_k \circ (T_{n_1} \times \dotsc \times T_{n_k}) \cong T_{n_1+\dotsc+n_k}.$$

The following coherence conditions should hold:

• Coherence of $\eta$ with $\mu$: We have $\mu_{n_1} = \eta \circ T_{n_1} : T_1 \circ T_{n_1} \to T_{n_1}$.

• Coherence of $\mu$ with $\mu$: The square

$$\begin{array}{cc} T_k \circ \bigl(T_{n_1} \circ (T_{m_{11}} \times \dotsc \times T_{m_{1n_1}}) \times \dotsc \times T_{n_k} \circ (T_{m_{k1}} \times \dotsc \times T_{m_{k n_k}})\bigr) & \rightarrow & T_k \circ (T_{m_{11}+\dotsc+m_{1 n_1}} \times \dotsc \times T_{m_{k1}+\dotsc+m_{k n_k}}) \\ \downarrow && \downarrow \\ T_{n_1+\dotsc+n_k} \circ (T_{m_{11}} \times \dotsc \times T_{m_{k n_k}}) & \rightarrow & T_{m_{11}+\dotsc+m_{k n_k}} \end{array}$$ commutes.

Notice that when $\mathcal{C}$ is discrete, this is the monadic definition of a monoid (as compared to the usual definition).

Questions. (1) Did I forget some coherence condition? (2) Is this concept already known? Does it have a name? It really looks like the most natural thing in the world, especially when you think "operadic" or "monadic". (3) Most important for me: Is this concept equivalent to the definition of a monoidal category? If yes, what is a reference for this? The idea for the equivalence is straight forward (the $n$-fold tensor product is an iteration of binary tensor products etc.), but I believe that it will probably require much work to check the coherence conditions in both directions.

• Almost surely what you've written down is an action of the operad Assoc on a category. The equivalence with the standard definition is probably "implicit in results of Mac Lane", or some such muttering. I'm reminded of a definition of "symmetric monoidal category" that I think I saw in a paper of Deligne's (although I don't remember where exactly) that asks for a functor $\mathcal C^S \to \mathcal C$ for each finite set $S$, with natural maps that include the permutations of $S$. So if you think the coherences are correct (I didn't check carefully), then go ahead and use it! – Theo Johnson-Freyd Jan 8 '15 at 5:27
• I believe that what you're after is Max Kelly's notion of a club; there is a club whose pseudo-algebras are precisely monoidal categories, and if you unravel the definition of being a pseudo-algebra for this club you should get what you wrote above. – Dan Petersen Jan 8 '15 at 7:25
• Also relevant: mathoverflow.net/questions/8252/… – André Henriques Jan 8 '15 at 9:41
• You should look at the beginning of Lurie's Higher Algebra section 2 (and then the rest of it) – Adam Gal Jan 8 '15 at 10:28
• Have you considered an approach similar to Durov? "Algebraic (Strong-/Pseudo-) Monads on Cat" or something like this. – Gerrit Begher Jan 17 '15 at 8:32


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# Markov spectrum problem

2010 Mathematics Subject Classification: Primary: 11J06 [MSN][ZBL]

A problem in number theory which arises in connection with the distribution of the normalized values of arithmetic minima of indefinite binary quadratic forms (cf. Binary quadratic form). Let

$$f = f ( x , y ) = \alpha x ^ {2} + \beta x y + \gamma y ^ {2} ,\ \ \alpha , \beta , \gamma \in \mathbf R ,$$

$$\delta ( f ) = \beta ^ {2} - 4 \alpha \gamma > 0 ,$$

and let

$$m ( f ) = \inf | f ( x , y ) | ,\ \ x , y \in \mathbf Z ^ {2} ,\ \ ( x , y ) \neq ( 0 , 0 ) ,$$

be the uniform arithmetic minimum of the form $f$. The number

$$\mu = \mu ( f ) = \ \frac{\sqrt {\delta ( f ) } }{m ( f ) } ,\ \ \mu \leq + \infty ,$$

is called the Markov constant of $f$. The set $M = \{ \mu ( f ) \}$, where $f$ runs through all real indefinite quadratic forms, is called the Markov spectrum. The Markov constant and the Markov spectrum have been defined in various ways; in particular, A.A. Markov in

considered the set $\{ 2 / \mu ( f ) \}$. It is known that $\mu ( f )$ is an invariant of a ray $F$ of classes of forms, that is, of a set

$$\tag{1 } F = \{ {f ^ { \prime } } : {f ^ { \prime } \simeq \tau f ( \mathbf Z ) ,\ \tau \in \mathbf R , \tau > 0 } \} ,$$

since $\mu ( f ^ { \prime } ) = \mu ( f ) = \mu ( F )$. Each ray of classes $F$ is in one-to-one correspondence with a doubly-infinite (infinite in both directions) sequence

$$I _ {F} = \{ {\dots, a _ {-} 1 ,\ a _ {0} , a _ {1} ,\dots } : {a _ {k} \in \mathbf Z } \} ,$$

such that if one puts

$$\mu _ {k} ( I _ {F} ) = \ [ a _ {k} ; a _ {k+} 1 , a _ {k+} 2 ,\dots ] + [ 0 ; a _ {k-} 1 , a _ {k-} 2 ,\dots ]$$

( $[ ; \dots ]$ is the notation for a continued fraction), then

$$\mu ( F ) = \sup _ {k \in \mathbf Z } \mu _ {k} ( I _ {F} ) .$$

The Markov problem can be stated as follows: 1) describe the Markov spectrum $M$; and 2) for each $\mu \in M$, describe the set of forms $f = f ( x , y)$( or the rays $F$) for which $\mu ( f ) = \mu ( F ) = \mu$. The problem was solved by Markov for the initial part of the spectrum $M$ defined by the condition $\mu ( f ) < 3$. This part of the spectrum is a discrete set:

$$M \cap [ 0 , 3 ) =$$

$$= \ \left \{ \sqrt {9 - \frac{4}{m ^ {2} } } : m ^ {2} + n ^ {2} + p ^ {2} = 3 m n p , m , n , p \in \mathbf N \right \} =$$

$$= \ \left \{ \sqrt 5 , \sqrt 8 , \frac{\sqrt 221 }{5} ,\dots \right \}$$

with the unique limit point 3 (a condensation point of $M$); $m$, $n$ and $p$ run through all positive integer solutions of Markov's Diophantine equation

$$\tag{2 } m ^ {2} + n ^ {2} + p ^ {2} = 3 m n p ,\ \ m \geq n \geq p > 0 .$$

In this case there corresponds to each point of this part of the spectrum precisely one ray $F _ {m}$, given by a Markov form $f _ {m} = f _ {m} ( x , y )$, with

$$\mu ( f _ {m} ) = \sqrt {9 - \frac{4}{m ^ {2} } } .$$

A solution $( m , n , p )$ of (2) is called a Markov triple; the number $m$ is called a Markov number. The Markov form $f _ {m}$ is associated to the Markov number $m = \max ( m , n , p )$ as follows. Let $r , s \in \mathbf Z$ be defined by the conditions

$$n r \equiv p ( \mathop{\rm mod} m ) ,\ \ 0 \leq r < m ,$$

$$r ^ {2} + 1 = m s ;$$

then, by definition,

$$f _ {m} = f _ {m} ( x , y ) = x ^ {2} + \left ( 3 - \frac{2r}{m} \right ) x y + \frac{s - 3 r }{m} y ^ {2} .$$

The set $M$ is closed and there is a smallest number $\mu _ {0} = 4.5278 \dots$ such that $[ \mu _ {0} , + \infty ] \subset M$ and $\mu _ {0}$ borders the interval of contiguity of $M$.

The Markov problem is closely related to the Lagrange–Hurwitz problem on rational approximation of a real number $\theta$. The quantity

$$\lambda = \lambda ( \theta ) = \sup \tau ,\ \ \lambda \leq + \infty ,$$

where the least upper bound is taken over all $\tau \in \mathbf R$, $\tau > 0$, for which

$$\left | \theta - \frac{p}{n} \right | \leq \frac{1}{\tau q ^ {2} }$$

has an infinite set of solutions $p , q \in \mathbf Z$, $q > 0$, is called a Lagrange constant. The set $L = \{ {\lambda ( \theta ) } : {\theta \in \mathbf R } \}$ is called the Lagrange spectrum. It is natural to regard Lagrange's theorem as the first result in the theory of the Lagrange spectrum: All convergents of the continued fraction expansion of $\theta$ satisfy

$$\left | \theta - \frac{p}{q} \right | < \frac{1}{q ^ {2} } .$$

If $\theta ^ \prime \sim \theta$, that is, if

$$\theta ^ \prime = \ \frac{a \theta + b }{c \theta + d } ,\ \ a , b , c , d \in \mathbf Z ,\ \ | a d - b c | = 1 ,$$

then $\lambda ( \theta ^ \prime ) = \lambda ( \theta ) = \lambda ( \Theta )$, where $\Theta = \{ {\theta ^ \prime } : {\theta ^ \prime \sim \theta } \}$ is an equivalence class of numbers. If $\theta$ is expanded as a continued fraction $\theta = [ a _ {0} ; a _ {1} , a _ {2} ,\dots ]$, then

$$\lambda ( \theta ) = \ \lim\limits _ {k \rightarrow \infty } \ \sup \lambda _ {k} ( \theta ) ,$$

$$\lambda _ {k} ( \theta ) = [ 0 ; a _ {k+} 1 , a _ {k+} 2 ,\dots ] + [ a _ {k} ; a _ {k-} 1 \dots a _ {1} ] ,$$

$$k = 1 , 2 ,\dots .$$

Thus, the Lagrange–Hurwitz problem can be stated as: a) describe the Lagrange spectrum $L$; and b) for each $\lambda \in L$, describe the set of numbers $\theta$( or classes $\Theta$) for which $\lambda ( \theta ) = \lambda ( \Theta ) = \lambda$.

For $\lambda ( \theta ) < 3$ this problem reduces to the Markov problem; moreover,

$$L \cap [ 0 , 3 ) = M \cap [ 0 , 3 ) ,$$

and to each $\lambda \in L$, $\lambda < 3$, corresponds precisely one class $\Theta$, described by the Markov form $f _ {m}$, provided the unicity conjecture is true. It has been proved that $L$, like $M$, is a closed set; that $L \subset M$ but $L \neq M$; that

$$L \cap [ \mu _ {0} , + \infty ] = \ M \cap [ \mu _ {0} , + \infty ] = \ [ \mu _ {0} , + \infty ] ,$$

where $\mu _ {0}$ borders the interval of contiguity of $L$. Research into the structure of $L$ and the connection between $L$ and $M$ is described in [6]. For generalizations and analogues of the Markov spectrum problem and "isolation phenomena" see [2], [3], [7].

#### References

 [1a] A. [A.A. Markov] Markoff, "Sur les formes quadratiques binaires indéfinies" Math. Ann. , 15 (1879) pp. 381–406 MR1510073 Zbl 11.0147.01 [1b] A. [A.A. Markov] Markoff, "Sur les formes quadratiques binaires indéfinies" Math. Ann. , 17 (1880) pp. 379–400 MR1510073 Zbl 12.0143.02 [2] J.W.S. Cassels, "An introduction to diophantine approximation" , Cambridge Univ. Press (1957) MR0087708 Zbl 0077.04801 [3] B.N. Delone, "The Peterburg school of number theory" , Moscow-Leningrad (1947) (In Russian) [4] D.S. Gorshkov, "Lobachevskii geometry in connection with some problems of arithmetic" Zap. Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. , 67 (1977) pp. 39–85 (In Russian) [5] G.A. Freiman, "Diophantine approximation and the geometry of numbers. (The Markov problem)" , Kalinin (1975) (In Russian) Zbl 0347.10025 [6] A.V. Malyshev, "Markov and Lagrange spectra (a survey of the literature)" Zap. Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. , 67 (1977) pp. 5–38 (In Russian) [7] B.A. Venkov, "On an extremum problem of Markov for indefinite ternaire quadratic forms" Izv. Akad. Nauk SSSR Ser. Mat. , 9 (1945) pp. 429–494 (In Russian) (French summary)

In equation (1) in the article above, the notation $f \simeq f ^ { \prime }$( $\mathbf Z$) refers to equivalence of binary forms over $\mathbf Z$. More precisely, $f \simeq f ^ { \prime }$( $\mathbf Z$) if and only if there are integers $a , b , c , d \in \mathbf Z$, $\mathop{\rm det} ( {} _ {c} ^ {a} {} _ {d} ^ {b} ) = \pm 1$ such that $f ^ { \prime } ( x , y ) = f ( a x + b y , c x + d y )$.
The "interval of contiguity of a Markov spectruminterval of contiguity" of $M$ is simply the maximal interval $[ \mu _ {0} , \infty ]$ completely belonging to $M$. The intersections $M \cap [ 0 , 3 )$ and $( \mu _ {0} , \infty ] \cap M$ have been well-described. The structure of the portion between, i.e. $M \cap [ 3 , \mu _ {0} ]$, is still (1989) unclear.
The unicity conjecture claims that the Markov number $m$ uniquely determines the triplet $( m,n,p,)$( and thus the Markov form $f _ {m}$). It is still (1989) a conjecture.