Over its storied and intriguing history, Hollywood has entertained us with many mathematical moments in film. Mathematics inspired the brilliant janitor in Good Will Hunting, the number theorist in Pi, and even Abbott and Costello
just a few examples of mathematics in tv and film...
The Simpsons - Fermat's Last Theorem
Back in 1637, the mathematician Pierre de Fermat scribbled the most famous margin note in history, writing in his geometry text that the equation had no solutions in whole numbers whenever . He boasted that he had a wonderful proof for it which he couldn't fit in the space in the margin. This was later called Fermat's Last Theorem (FLT). Getting to a proof probably wasn't as easy as Fermat thought, since it took the world 358 years to find one! (It was proved by Andrew Wiles in 1994.)
Several of the writers of The Simpsons have backgrounds in mathematics, Pierre de Fermat, ca. 1650 and two episodes of the Simpsons included references to FLT:
1. In the episode Treehouse of Horror VI (3F04), Homer enters a portal into a three-dimensional world rendered with 3D computer graphics. In the background, above a green wireframe mesh, the viewer can see the equation
2. In the episode The Wizard of Evergreen Terrace (5F21), Homer writes the following equation on a chalkboard:
Calculating in Maple
If either of the the equations from the Simpsons were right, they would disprove FLT. Sure enough, Maple can quickly show these equations in fact don't hold:
Notice that the results agree for the first few digits but then diverge. However, if a viewer at the time were to enter these into a standard scientific calculator with only 10 digits of precision, they would not be able to tell them apart!
While calculating with big numbers is no problem for Maple, we don't need that to show these equations do not hold.
Remainders
For the first, since 1782 is even and 1841 is odd, the sum must be odd. However is even, so the equation can't hold.
For the second, we just need to look at the remainder divided by 3: = 0 but = 1.
In fact, if all the powers are 3, it turns out that if you just pick a, b, and c according to a certain rule, you can generate fake FLT "solutions" that are very close and have many digits in common.
With the following app you can generate fake solutions:
Fake FLT "Solution" Generator
Generate F...
Check Answer Mean Girls - Limit Problem
In Mean Girls, Cady Heron (Lindsay Lohan) is a high school student who is punished by her teacher (Tina Fey) for bullying other students at her school. As part of her punishment, Cady joins the Mathletes and has to compete with her team at the Illinois High School Mathletes State Championship. Cady's team, North Shore High, is competing against Marymount Prep. After 87 minutes of play, both teams are tied and have to compete in a tie-breaker: both teams choose a single competitor to compete in a sudden death round in which they are asked one question and the first person to answer wins the competition. Cady is chosen and ends up winning the competition by correctly responding that the limit of
as approaches does not exist.
How did she correctly answer this question?
Calculating with Maple
In Maple, we can calculate the limit easily with a right-click:
That was easy, but why is it undefined? Let's try taking the limit from each side as . In Maple we can do this with the limit command
Another way to see this is to plot the expression:
As you can see the limits from the left goes to positive infinity, while the limit from the right goes to negative infinity. So the two-sided limit doesn't exist! 21 - Monty Hall Problem In 21, Ben Campbell solves his professor's problem about a game show. In fact this is a famous problem called the "Monty Hall problem", named after Monty Hall, the host of the game show Let's Make a Deal.
Here's how is works: You're in a room with three doors. The host, Monty Hall, asks you to choose one of the three doors and recieve the prize behind that door. Behind two doors are goats, and behind the third door is a brand new car.
Before your door is opened, however, Monty opens one of the other two doors, revealing a goat.
He asks you if you want to change your selection to the third (and last remaining door).
The Monty Hall Problem: Do you switch or not?
Answer
The correct answer is that you do want to switch.
1 The simple reason is that if you do not switch, you will only ever have the expected chance of winning the car.
However, once the host opens one of the doors and reveals one of the goats, you obviously do not
improve your chances of winning to better than 1 by sticking with your original choice. Moreover, if you now switch doors, there is a chance you will win the car (counterintuitive though it seems).
Explanation
At the start, the probability of you choosing the right door is .
After one of the options is eliminated, you might reason out that each of the two remaining doors have an equal chance to be hiding the prize. This would seem reasonable, as probability suggests that distribution of chance should be even across your options. However, this is not the case, because you now have more information.
Starting again at the beginning, each door has an equal chance of hiding the prize. You choose a door, 1 and it had chance of being the right one. Let's say you are going to stay with this door. At this point, we don't really care about the other doors anymore, because our choice remains static regardless of any 1 information given. Because we're staying with this door, and if it's the right door, we will win, with a chance of winning. Therefore, staying with the door you first chose has effectively a chance of winning.
Let's say you were in fact wrong in your original door choice. Then what happens is that the show's host eliminates the other wrong choice for you (because he can't reveal the winning door). At this point, you will win the prize if you switch. Now, since there was a chance of being wrong on the first choice, this means you have a chance of winning if you switch!
Still not convinced? Try the Monty Hall game out yourself, below. Or, if that's too slow, you can use the automated game system below the statistics for faster results. Select This Door Select This Door Select This Door
Statistics Automated Gameplay Win Percentage:
1 100
Always Switch Doors
Always Keep Door
Play 20 rounds for me!
Reset
The Fast and the Furious - Speed, Distance, and Time
In the sixth Fast and Furious movie, the final scene features the antagonist, Owen Shaw (Luke Evans), is with his gang trying to board an aircraft that is in motion on a runway. At the same time, Dominic Toretto (Vin Diesel), Brian O'Conner (Paul Walker), and Luke Hobbs (Dwayne Johnson), are also in pursuit with their street cars to try and stop them. The scene lasts over 10 minutes and many fans of the movie have asked "How long was that runway?"
The scene itself has multiple phases involving the plane: the plane landing, slowing down so people can board, speeding up to take off, landing again because of too much weight, and finally crashing at the end.
The plane itself fluctuates between and during this scene. It is known that on average it travelled about The scene itself lasts about 15 minutes.
Calculating in Maple We know that speed is defined as
Rearranging this equation to solve for the distance, we end up with
Substituting in the values we know for the speed and time, we end up with:
(4.1) Maple automatically simplifies the units in our calculation and returns an appropriate result. Using the right- click menu we can convert from feet to miles for the final answer.
According to Wikipedia, the longest runway in the world is at Qamdo Bamda Airport in China, and is 18,045 feet (3.42 miles).
Speed - Projectile Motion and the Range Equation
In the movie Speed, a city bus in L.A. has been rigged with a bomb that arms when the speed exceeds and explodes if the speed drops below . While driving on
the freeway, the bus is forced to jump a gap of unfinished road. Being Hollywood, the jump is successful of course, but just how realistic is this?
The bus's jump is simply an example of projectile motion: the motion of an object projected into the air at an angle. Other examples include a soccer ball being kicked or a football being thrown.
The range equation is used to calculate the horizontal distance covered by a projectile (neglecting air resistance):
where is the projectile's initial speed, is the angle the projectile launches at, and is the acceleration due to gravity, . From the movie, we know that the length of the gap is , and the initial speed is . From the equation we can see that if the launch angle is zero, the range will also become zero. It is necessary for the bus to launch at some angle greater than zero to make the jump.
We can isolate the launch angle, , in the range equation:
...assign values to the variables: ...and solve numerically for :
By default Maple works in radians but we can easily convert to degrees:
In theory, then, if the bus left the road at an angle of approximately 5 degrees, it would be possible for the bus to jump the gap!
It is interesting to note that the Wikipedia page about this movie states that a ramp was used to enable the bus to jump the . The bus was to jump over a regular section of highway, which was edited out of the film with some careful CGI.The director mentions in the DVD commentary that as much as possible was removed from the bus to make it as light as possible. On the first attempt, the bus driver missed the landing, crashing the bus. On the second attempt, the bus was only expected to travel , and cameras were placed accordingly. However the bus traveled much further than expected and landed on top of the camera: fortunately a second camera was also filming.
A Beautiful Mind - Nash Equilibrium In A Beautiful Mind, the Nash Equilibrium was described by the character John Nash (played by Russell Crowe) in the context of dating.
The concept can be applied to any situation where two or more players are competing for some goal: a group of players is in a Nash equilibrium if every player has picked a strategy and no one can do any better by switching their current strategy.
Prisoner's Dilemma
A simple example of Nash Equilibrium comes from the so-called Prisoner's Dilemma problem.
Two members of a criminal gang are arrested and imprisoned. Each prisoner is in solitary confinement without a way of communicating with the other. The authorities don't have enough evidence to convict either prisoner on the main charge, so they intend to sentence both to a year in prison on a lesser charge.
However in hopes of getting a conviction, the authorities offer each prisoner a choice: betray his confederate by testifying that the other committed the crime, or to cooperate with the other by remaining silent. This yields 4 potential outcomes: 1. If A and B both betray the other, each of them serves 2 years in prison 2. If A betrays B but B remains silent, A will be set free and B will serve 3 years in prison 3. If B betrays A but A remains silent, B will be set free and A will serve 3 years in prison 4. If A and B both remain silent, both of them will only serve 1 year in prison (on the lesser charge)
Table: Length of sentences depending on what A and B decide
If A chooses to keep silent, then B is better off to betray him and go free. If A chooses to betray B, then B is still better off to betray A in turn, or else B would get the maximum charge. It works the same when B is deciding what to do.
Because betrayal is the better option, it's logical for A and B each to betray the other (outcome #1), even though that result (both doing two years' time) is worse for both than outcome #4 (both spending a year in prison).
So the Nash equilibrium for the Prisoner's Dilemma problem is to always betray.
It gets more complex when A and B have a history together: this is called the Iterated Prisoner's Dilemma.
Iterated Prisoner's Dilemma
The situation gets more interesting if we can play the Prisoner's Dilemma game back to back: the prisoner's have the chance to cooperate with or betray each other repeatedly, and can take the actions of the previous iterations into account when choosing their strategies. This is called iterated prisoners' dilemma.
The iterated prisoners' dilemma game is used in modeling theories of human cooperation and trust. Using the assumption that the game can model transactions between two people requiring trust, cooperative behaviour in populations can be modeled by a multi-player, iterated, version of the game. In this case the Nash equilibrium may be something other than mutual betrayal.
We can study the iterated prisoner's dilemma problem in Maple. A computer player will employ one of several programmed strategies against the human player. Let's try a few possible strategies for the computer and play against it:
Always Cooperate
Always Betray
Random
Tit for Tat