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The limit does not exist!

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What’s more fetch than toaster strüdel, foot cream in a jar and a fertility vase from the Ndebele tribe? Arguably maths. More specifically, the mathematics of Mean Girls. If you haven’t seen Mean Girls, it’s a not-too-bad movie about teenage girls, teenage boys, teenage drama, and (a little bit of) maths.

Stop trying to make fetch happen. It’s not going to happen.

It’s not exactly October 3rd*, but its been 13 years since Mean Girls came out, and

  • $13 \times 2=26$
  • $26+2\times2017=4060$
  • $4060-3010=1050$, and
  • $1050\log(2)+\frac{\pi}{1423211345}$ is… just a number.

What we’ll be looking at in this article is the last 10 minutes of the movie, where Cady and the Mathletes from North Shore compete in a maths state championship. After 87 minutes of play it’s a tie, and so the two teams enter the sudden death round, where the teams choose a member of the opposition (happening to both be female) to battle for the trophy.

T. Pak as in ‘Trang Pak the grotsky little byotch’? No, just her brother.

In one corner, Cady Heron, AKA ‘Africa girl’, AKA the ‘used-to-be maths geek but not so much anymore after she met the Plastics but is kind of a geek again because her attempted sabotage of Regina George backfired and descended into chaos’. She is also the protagonist of the movie.

In the other, is Caroline Krafft, the female member of Marymount Prep, who “seriously needed to pluck her eyebrows, whose skirt looks like it was picked out by a blind Sunday school teacher, and had some 99-cent lipgloss on her snaggletooth”. This was when I realised, making fun of Caroline Krafft wasn’t going make me any better at writing this article (or maths for that matter).

The two were asked to solve the following;

Caroline Krafft hastily states the incorrect answer, $-1$. If Cady manages to state the correct one then the North Shore mathletes ultimately win.

In the midst of all this Cady then has a miraculous flashback to the week that Aaron got his hair cut, and as she takes a moment to see straight pass Aaron’s face and onto the board behind, then has an epiphany (the kind we would like to have in our exams sometimes) and quickly realises the function diverges.

Aaron Samuels is now a spin class instructor, so he won’t be expected to know what a factorial is anymore. Thank God.

So we don’t exactly know how Cady did this in under 10 seconds, but we can try to figure out the question for ourselves.

So the question reads$$ \lim_{x\to 0} \frac{\ln(1-x)-\sin x} {1-\cos^2 x}.$$

We’ll consider L’Hôpital’s Rule, seeing as its the most straightforward way of going about it.

If we substitute $x$ with 0 we will find$$\frac{\ln(1)-\sin(0)} {1-\cos^2(0)} = \frac{0}{0}.$$
This satisfies the criteria for being able to apply L’Hôpital’s Rule, which is as follows,$$ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}.$$

Let’s apply it to our function, starting by considering the limit from the right;

\begin{align}
\lim_{x\to 0+} \frac{\frac{-1}{1-x}-\cos x} {\sin 2x} &= \lim_{x\to 0+} \frac{-x \cos x + \cos x + 1}{ (x-1) \sin 2x} \\
&= \lim_{x\to 0+} [-x \cos x + \cos x + 1] \lim_{x\to 0+}\left[\frac{1}{x-1}\right] \lim_{x\to 0+}\left[\frac{1}{\sin 2x}\right].
\end{align}
Now,\begin{align}
\lim_{x\to 0+} \left[-x\cos x +\cos x +1 \right] &=2, \\
\lim_{x\to 0+} \left[\frac{1}{\sin 2x } \right] &= \infty, \\
\lim_{x\to 0+} \left[\frac{1}{x-1} \right] &= -1,
\end{align}
so by properties of infinity we get
$$\lim_{x\to 0+} [-x \cos x + \cos x + 1] \lim_{x\to 0+}\left[\frac{1}{x-1}\right] \lim_{x\to 0+}\left[\frac{1}{\sin 2x}\right] = 2 \cdot \infty \cdot -1 = -\infty.$$

Repeating this method for the limit from the left, we will obtain the following,
$$ \lim_{x\to 0-} \frac{-x \cos x + \cos x + 1}{ (x-1) \sin 2x} = +\infty.$$

Just checking.

And so since the limit from the right and the left differ, the whole thing diverges. Yay.

Another approach would be to look at the Taylor series approximations for the separate terms at $x$ near 0. We won’t do that though since it’s not as mathsy, but it might have been the quicker way to go, on the assumption that Cady really knows her Taylor series expansions. But

“Cause the next time you see her she’ll be like, Ohh Kevin G!”

*The date where Aaron Samuels asked Cady what the date was.

CC-BY-SA: Mathletes, Matty McRib. Mean Girls, Flickr. Screenshots copyright of Paramount Pictures.

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Review: Mathematical socks

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This Christmas, I received mathematical socks. A great gift, you might think. But is it good maths or fake maths? Can you wear them and be taken mathematically seriously? Thus I have undertaken this important review.

Unboxing

The socks come beautifully packaged and folded, tied together with a fancy red label, which gives a nifty standing suggestion.

Beautifully packed mathematical socks

Beautifully packed mathematical socks


Unboxing grade: A

Mathematical content

There are five distinct mathematical items on the socks. I have graded them individually. The younger reader may wish to refer to this helpful guide to converting to new grades.

1. Proof of Pythagoras

Sock v Elements

Pythagoras’ Theorem proved on a sock (left) and in the Elements (right)

Continue reading

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Chalkdust Review of the Year 2016

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Another busy year for Chalkdust is gone(well…almost). That is 50 online articles, 2 new issues, 1 advent calendar a few quizzes and loads more. In the remaining few hours of 2016 we look back to some of the amazing articles written by us and our friends. From everyone on the Chalkdust team enjoy this post and look forward to even better blogs next year.

Attributions:

[Pictures: 1 – adapted from Flickr.com – Moscow New Year 2016 by Valeri Fortuna, CC-BY 2.0;  other pictures by Chalkdust]

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24 December

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Today is the final day of the Chalkdust advent calendar.

Click here to open today’s door.

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23 December

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Welcome to the twenty third day of the 2016 Chalkdust Advent Calendar. Today, we have another puzzle for you to enjoy, plus the answer to the puzzles from 12 December.

Here’s today’s puzzle.

Finding Santa

It’s Christmas Eve Eve and Santa is so nervous that he decides to hide in the top floor of a nearby hotel. Santa starts in a random room and every hour moves to one of the rooms next door to his current room. There are 17 rooms on the top floor of the hotel.

The hotel manager is a very busy lady and only has time to allow you to look into one room every hour. It’s 2pm on Christmas Eve Eve, so you have 34 hours to find Santa or Christmas is ruined. Can you find him in time? 

Now for the solution to the puzzle from 12 December.

Odd factors

Pick a number. Call it $n$. Write down all the numbers from $n+1$ to $2n$ (inclusive). Under each of these, write its largest odd factor. What is the sum of these odd factors? 

Incredibly, the result will always be $n^2$.

To see why, imagine writing every number, $n+1\leq k\leq 2n$, in the form $$k=2^ab$$ where $b$ is an odd number and also the $k$’s largest odd factor. The next largest number whose largest odd factor is $b$ will be $2^{a+1}b=2k$. But this will be larger than $2n$, so outside the range. Therefore each number in the range has a different largest odd factor.

Each of the largest odd factors must be one of $1, 3, 5, …, 2n-1$, as they cannot be larger than $2n$. But there are $n$ odd numbers here and $n$ numbers in the range, so each number $1, 3, 5, …, 2n-1$ is the highest odd factor of one of the numbers (as the highest odd factors are all different).

Therefore, the sum of the odd factors is the sum of the first $n$ odd numbers, which is $n^2$.

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The Indisputable Existence of Santa Claus – A Review

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Welcome to the 22nd day of the 2016 Chalkdust Advent Calendar. Today we have for you, a book review!

Christmas, a time of celebration, joy and meeting family you never knew you had. Regardless of how joyous Christmas can be it also is undoubtedly stressful for some. What present could I get person $x$? Why does my Christmas tree look so ugly even though I spent 2 weeks decorating it? The simple answer is because you probably drenched it in tinsel and epileptic seizure inducing lights. The mathematical answer?

The Indisputable Existence of Santa Claus” is a recent release by Dr Hannah Fry and Dr Thomas Oléron Evans, Dr Hannah Fry who has previously written “The Mathematics of Love”. This book arranges itself as a step-by-step guide on how to prepare for the (mathematically) perfect Christmas, covering every detail from how to wrap presents according to their surface area to volume ratio, to using Markov chains as a means of perfecting the Queen’s speech. 

We shall start with the big question of whether Father Christmas himself actually exists. The first chapter gives a ‘seemingly’ valid proof of Santa’s existence, from another ‘seemingly’ valid proof of how 1+1=0. You must be thinking, what? Obviously there must be a flaw in the progression of the proof somewhere, which there is, but you can discover it for yourself by reading the book. 

Moving on, let’s have a look at the inconspicuous game of Secret Santa. For those unfamiliar with the concept, the classic approach to this game is simply, you write your name on a small piece of paper, fold it up and throw it into a hat. After a bit of shaking, someone picks out a piece and takes on the enormous responsibility of finding their victim, *coughs* I mean colleague, a present usually around the price range of £5. So why is this a rather inadequate way of organising this game? 

Well first, you risk the chance of picking out your own name. You might think that’s easy enough a problem to solve, just put it back in the hat…But what if you were the last person? Or even the second last person? Everybody knows your name is back in the hat, and they have a greater chance of picking your name, and in some case, 2 people will have each other’s name, Secret Santa is ruined. Goodnight. The book proposes another method of making sure that no one has their own name, and no one else knows who has their name. A clever yet simple solution involving derangements, alas, Christmas is saved, now lets hope your secret Santa isn’t a Scrooge

My favourite chapter in this book has to be the one on the Queen’s Speech. The beginning of the chapter is an analysis of the Queen’s vocabulary score (based solely on the number of unique words in the first 35K words of her speech. Surprisingly, poor old Lizzy scored lower than her counterparts Jay-Z and Shakespeare. Well, it seems as though maths might be able to give her a hand with that, with something special called a Markov chain, that determines the next word to place in a sentence given the word before. I won’t go into too much detail as to give it away; it just so happens that earlier this week I was reading about how Markov chains are used in determining the probability of flipping a coin and getting a certain outcome which is equivalent to another outcome, so this chapter peaked my interest even further. 

What is particularly good about this book is how accessible it is; at the end of every chapter there are endnotes that explain some of the maths mentioned and includes some extra reading material. The only questionable thing in this book might be the chapter about cooking turkey. They used a chicken instead. Enough said.

Overall, a great read, not too technical but with just enough maths to get you thinking. So even if your grumpy aunt Hilda despises anything to do with maths, there’s now a very slight chance she might enjoy it. 

Also the amount of cracker pulling rules I’ve never heard of is remarkable. I am now going to stop pulling both ends of my own cracker, despite my competitive nature.

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21 December

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Welcome to the twenty-first day of the 2016 Chalkdust Advent Calendar. Today, we bring you our final selection of fascinatingT&Cs apply facts, randomly generated by Santa’s elves.  Remember to send us your favourite scientific curiosities via Facebook, Twitter or email and we’ll feature the best in a blog next year.

The number 21

Setter of fiendish conjectures.

Today is the 21st day of the Chalkdust Advent calendar. The numbers 8 and 9 are the only powers of integers ($2^3$ and $3^2$ respectively) that are consecutive. This was conjectured by the Belgian mathematician Eugène Charles Catalan in 1844. It was proved in 2002 by Preda Mihailescu. Unlike Andrew Wiles, who proved Fermat’s last theorem, Mihailescu didn’t shoot to fame. This had absolutely nothing to do with the number 21.

Rudolph the red-nosed she-reindeer

Female. Female. Not male.

Rudolph the red-nosed reindeer is very confused as to why she has a male name. Male reindeers shed their antlers once they’ve finished using them as swords during the mating season in autumn, while females cast them in spring and regrow them in time for Christmas (when they use them as swords to fight other females over holes in the snow). I’m sure she is confused for many other reasons too. Such as why she has a red nose.

Father Christmas and the multiverse

If the multiverse theory is correct, and our universe is just one of an infinite number of universes, there may be one in which Father Christmas was not popularised by Coke adverts of the 1920s.

And that’s that from Santa’s elves. They’re off to find Rudelle.

[Pictures. Eugène Charles Catlan: Public domain; Rudelle: Tristan Ferne, CC BY-2.0]

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19 December

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Find your perfect partner with this wonderful tree diagram!


Attributions:

[Cauchy  – By Public domain – Library of Congress Prints and Photographs Division. From an illustration in: Das neunzehnte Jahrhundert in Bildnissen / Karl Werckmeister, ed. Berlin : Kunstverlag der photographische gesellschaft, 1901, vol. V, no. 581., Public Domain ; Knot – adapted from Flickr.com – knotted by Shelby Steward, CC-BY 2.0; Emmy Noether – By Unknown – Emmy Noether (1882-1935), Public Domain ; Python – adapted from Flickr.com – Python by Jonathan Kriz, CC-BY 2.0; Daniel Bernoulli – By Johann Jakob Haid – Here, Public Domain ; Scrooge – adapted from Flickr.com – Money by Tax Credits, CC-BY 2.0]

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