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2d 4d

Finger Ratio Predicts Maths Ability?

Some of the studies on the 2D: 4D finger ratios (as measured in the picture above) are interesting when considering what factors possibly affect mathematical ability.  A 2007 study by Mark Brosnan from the University of Bath found that:

“Boys with the longest ring fingers relative to their index fingers tend to excel in math. The boys with the lowest ratios also were the ones whose abilities were most skewed in the direction of math rather than literacy.

With the girls, there was no correlation between finger ratio and numeracy, but those with higher ratios–presumably indicating low testosterone levels–had better scores on verbal abilities. The link, according to the researchers, is that testosterone levels in the womb influence both finger length and brain development.

In men, the ring (fourth) finger is usually longer than the index (second); their so-called 2D:4D ratio is lower than 1. In females, the two fingers are more likely to be the same length. Because of this sex difference, some scientists believe that a low ratio could be a marker for higher prenatal testosterone levels, although it’s not clear how the hormone might influence finger development.”

In the study, Brosnan photocopied the hands of 74 boys and girls aged 6 and 7.  He worked out the 2D:4D finger ratio by dividing the length of the index finger (2D) with the length of the ring finger (4D). They then compared the finger ratios with standardised UK maths and English tests.  The differences found were small, but significant.

2d 4d

Another study of 136 men and 137 women, looked at the link between finger ratio and aggression.  The results are plotted in the graph above – which clearly show this data follows a normal distribution.  The men are represented with the blue line, the women the green line and the overall cohort in red.  You can see that the male distribution is shifted to the left as they have a lower mean ratio.  (Males: mean 0.947, standard deviation 0.029, Females: mean 0.965, standard deviation 0.026).

The 95% confidence interval for average length is 0.889-1.005 for males and 0.913-1.017 for females.  That means that 95% of the male and female populations will fall into these categories.

The correlation between digit ratio and everything from personality, sexuality, sporting ability and management has been studied.  If a low 2D:4D ratio is indeed due to testosterone exposure in the womb (which is not confirmed), then that raises the question as to why testosterone exposure should affect mathematical ability.  And if it is not connected to testosterone, then what is responsible for the correlation between digit ratios and mathematical talent?

I think this would make a really interesting Internal Assessment investigation at either Studies or Standard Level.  Also it works well as a class investigation at KS3 and IGCSE into correlation and scatter diagrams.   Does the relationship still hold for when you look at algebraic skills rather than numeracy?  Or is algebraic talent distinct from numeracy talent?

A detailed academic discussion of the scientific literature on this topic is available here.

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NASA, Aliens and Binary Codes from the Stars

gavel

Amanda Knox and Bad Maths in Courts

This post is inspired by the recent BBC News article, “Amanda Knox and Bad Maths in Courts.”   The article highlights the importance of good mathematical understanding when handling probabilities – and how mistakes by judges and juries can sometimes lead to miscarriages of justice.

A scenario to give to students:

A murder scene is found with two types of blood – that of the victim and that of the murderer.  As luck would have it, the unidentified blood has an incredibly rare blood disorder, only found in 1 in every million men.  The capital and surrounding areas have a population of 20 million – and the police are sure the murderer is from the capital.   The police have already started cataloging all citizens’ blood types for their new super crime-database.  They already have nearly 1 million male samples in there – and bingo – one man, Mr XY, is a match.  He is promptly marched off to trial, there is no other evidence, but the jury are told that the odds are 1 in a million that he is innocent.  He is duly convicted.   The question is, how likely is it that he did not commit this crime? 

Answer:

We can be around 90% confident that he did not commit this crime.  Assuming that there are approximately 10 million men in the capital, then were everyone cataloged on the database we would have on average 10 positive matches.  Given that there is no other evidence, it is therefore likely that he is only a 1 in 10 chance of being guilty.  Even though P(Fail Test/Innocent) = 1/1,000,000,  P(Innocent/Fail test) = 9/10.

Amanda Knox

Eighteen months ago, Amanda Knox and Raffaele Sollecito, who were previously convicted of the murder of British exchange student Meredith Kercher, were acquitted.  The judge at the time ruled out re-testing a tiny DNA sample found at the scene, stating that, “The sum of the two results, both unreliable… cannot give a reliable result.”

This logic however, whilst intuitive is not mathematically correct.   As explained by mathematician Coralie Colmez in the BBC News article, by repeating relatively unreliable tests we can make them more reliable – the larger the pooled sample size, the more confident we can be in the result.

sally clark

Sally Clark

One of the most (in)famous examples of bad maths in the court room is that of Sally Clark – who was convicted of the murder of her two sons in 1999.  It has been described as, “one of the great miscarriages of justice in modern British legal history.”  Both of Sally Clark’s children died from cot-death whilst still babies.  Soon afterwards she was arrested for murder.  The case was based on a seemingly incontrovertible statistic – that the chance of 2 children from the same family dying from cot-death was 1 in 73 million.  Experts testified to this, the jury were suitably convinced and she was convicted.

The crux of the prosecutor’s case was that it was so statistically unlikely that this had happened by chance, that she must have killed her children.  However, this was bad maths – which led to an innocent woman being jailed for four years before her eventual acquittal.

Independent Events

The 1 in 73 million figure was arrived at by simply looking at the probability of a single cot-death (1 in 8500 ) and then squaring it – because it had happened twice.  However, this method only works if both events are independent – and in this case they clearly weren’t.  Any biological or social factors which contribute to the death of a child due to cot-death will also mean that another sibling is also at elevated risk.

Prosecutor’s Fallacy

Additionally this figure was presented in a way which is known as the “prosecutor’s fallacy” – the 1 in 73 million figure (even if correct) didn’t represent the probability of Sally Clark’s innocence, because it should have been compared against the probability of guilt for a double homicide.   In other words, the probability of a false positive is not the same as the probability of innocence.  In mathematical language, P(Fail Test/Innocent) is not equal to P(Innocent/Fail test).

Subsequent analysis of the Sally Clark case by a mathematics professor concluded that rather than having a 1 in 73 million chance of being innocent, actually it was about 4-10 times more likely this was due to natural causes rather than murder.  Quite a big turnaround – and evidence of why understanding statistics is so important in the courts.

This topic has also been highlighted recently by the excellent ToK website, Lancaster School ToK.

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Does it Pay to be Nice?  Game Theory and Evolution

Golden Balls, hosted by Jasper Carrot, is based on a version of the Prisoner’s Dilemma. For added interest, try and predict what the 2 contestants are going to do. Any psychological cues to pick up on?

Game theory is an interesting branch of mathematics with links across a large number of disciplines – from politics to economics to biology and psychology.  The most well known example is that of the Prisoner’s Dilemma.  (Illustrated below).  Two prisoners are taken into custody and held in separate rooms.  During interrogation they are told that if they testify to everything (ie betray their partner) then they will go free and their partner will get 10 years.  However, if they both testify they will both get 5 years, and if they both remain silent then they will both get 6 months in jail.

prisoner dilemma

So, what is the optimum strategy for prisoner A?  In this version he should testify – because whichever strategy his partner chooses this gives prisoner A the best possible outcome.  Looking at it in reverse, if prisoner B testifies, then prisoner A would have been best testifying (gets 5 years rather than 10).   If prisoner B remains silent, then prisoner A would have been best testifying (goes free rather than 6 months).

This brings in an interesting moral dilemma – ie. even if the prisoner and his partner are innocent they are is placed in a situation where it is in his best interest to testify against their partner – thus increasing the likelihood of an innocent man being sent to jail.  This situation represents a form of plea bargaining – which is more common in America than Europe.

Part of the dilemma arises because if both men know that the optimum strategy is to testify, then they both end up with lengthy 5 year jail sentences.  If only they can trust each other to be altruistic rather than selfish – and both remain silent, then they get away with only 6 months each.   So does mathematics provide an amoral framework?  i.e. in this case mathematically optimum strategies are not “nice,” but selfish.

MAD

Game theory became quite popular during the Cold War, as the matrix above represented the state of the nuclear stand-off.  The threat of Mutually Assured Destruction (MAD) meant that neither the Americans or the Russians had any incentive to strike, because that would inevitably lead to a retaliatory strike – with catastrophic consequences.  The above matrix uses negative infinity to represent the worst possible outcome, whilst both sides not striking leads to a positive pay off.  Such a game has a very strong Nash Equilibrium – ie. there is no incentive to deviate from the non strike policy.  Could the optimal maths strategy here be said to be responsible for saving the world?

selfish gene

Game theory can be extended to evolutionary biology – and is covered in Richard Dawkin’s The Selfish Gene in some detail.  Basically whilst it is an optimum strategy to be selfish in a single round of the prisoner’s dilemma, any iterated games (ie repeated a number of times) actually tend towards a co-operative strategy.  If someone is nasty to you on round one (ie by testifying) then you can punish them the next time.  So with the threat of punishment, a mutually co-operative strategy is superior.

You can actually play the iterated Prisoner Dilemma game as an applet on the website Game Theory. Alternatively pairs within a class can play against each other.

An interesting extension is this applet, also on Game Theory, which models the evolution of 2 populations – residents and invaders.  You can set different responses – and then see what happens to the respective populations.  This is a good reflection of interactions in real life – where species can choose to live co-cooperatively, or to fight for the same resources.

The first stop for anyone interested in more information about Game Theory should be the Maths Illuminated website – which has an entire teacher unit on the subject – complete with different sections,a video and pdf documents.  There’s also a great article on Plus Maths – Does it Pay to be Nice? all about this topic.  There are a lot of different games which can be modeled using game theory – and many are listed here . These include the Stag Hunt, Hawk/ Dove and the Peace War game.  Some of these have direct applicability to population dynamics, and to the geo-politics of war versus peace.

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grahams number

Graham’s Number – literally big enough to collapse your head into a black hole

Graham’s Number is a number so big that it would literally collapse your head into a black hole were you fully able to comprehend it. And that’s not hyperbole – the informational content of Graham’s Number is so astronomically large that it exceeds the maximum amount of entropy that could be stored in a brain sized piece of space – i.e. a black hole would form prior to fully processing all the data content. This is a great introduction to notation for really big numbers. Numberphile have produced a fantastic video on the topic:

Graham’s Number makes use of Kuth’s up arrow notation (explanation from wikipedia:)

In the series of hyper-operations we have

1) Multiplication:

   \begin{matrix}    a\times b & = & \underbrace{a+a+\dots+a} \\    & & b\mbox{ copies of }a   \end{matrix}

For example,

   \begin{matrix}   4\times 3 & = & \underbrace{4+4+4} & = & 12\\    & & 3\mbox{ copies of }4   \end{matrix}

2) Exponentiation:

   \begin{matrix}    a\uparrow b= a^b = & \underbrace{a\times a\times\dots\times a}\\    & b\mbox{ copies of }a   \end{matrix}

For example,

   \begin{matrix}    4\uparrow 3= 4^3 = & \underbrace{4\times 4\times 4} & = & 64\\    & 3\mbox{ copies of }4   \end{matrix}

3) Tetration:

   \begin{matrix}    a\uparrow\uparrow b & = {\ ^{b}a}  = & \underbrace{a^{a^{{}^{.\,^{.\,^{.\,^a}}}}}} &     = & \underbrace{a\uparrow (a\uparrow(\dots\uparrow a))}  \\       & & b\mbox{ copies of }a     & & b\mbox{ copies of }a   \end{matrix}

For example,

   \begin{matrix}    4\uparrow\uparrow 3 & = {\ ^{3}4}  = & \underbrace{4^{4^4}} &     = & \underbrace{4\uparrow (4\uparrow 4)} & = & 4^{256} & \approx & 1.34078079\times 10^{154}& \\       & & 3\mbox{ copies of }4     & & 3\mbox{ copies of }4   \end{matrix}
3\uparrow\uparrow 2=3^3=27
3\uparrow\uparrow 3=3^{3^3}=3^{27}=7625597484987
3\uparrow\uparrow 4=3^{3^{3^3}}=3^{3^{27}}=3^{7625597484987}
3\uparrow\uparrow 5=3^{3^{3^{3^3}}}=3^{3^{3^{27}}}=3^{3^{7625597484987}}
etc.

4) Pentation:

   \begin{matrix}    a\uparrow\uparrow\uparrow b= &     \underbrace{a_{}\uparrow\uparrow (a\uparrow\uparrow(\dots\uparrow\uparrow a))}\\     & b\mbox{ copies of }a   \end{matrix}

and so on.

Examples:

3\uparrow\uparrow\uparrow2 = 3\uparrow\uparrow3 = 3^{3^3} = 3^{27}=7,625,597,484,987
   \begin{matrix}     3\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow3\uparrow\uparrow3 = 3\uparrow\uparrow(3\uparrow3\uparrow3) = &     \underbrace{3_{}\uparrow 3\uparrow\dots\uparrow 3} \\    & 3\uparrow3\uparrow3\mbox{ copies of }3   \end{matrix}   \begin{matrix}    = & \underbrace{3_{}\uparrow 3\uparrow\dots\uparrow 3} \\    & \mbox{7,625,597,484,987 copies of 3}   \end{matrix}=\underbrace{3^{3^{3^{3^{\cdot^{\cdot^{\cdot^{\cdot^{3}}}}}}}}}_{7,625,597,484,987}

Which clearly can lead to some absolutely huge numbers very quickly. Graham’s Number – which was arrived at mathematically as an upper bound for a problem relating to vertices on hypercubes is (explanation from Wikipedia)

grahams number

where the number of arrows in each layer, starting at the top layer, is specified by the value of the next layer below it; that is,

G = g_{64},\text{ where }g_1=3\uparrow\uparrow\uparrow\uparrow 3,\  g_n = 3\uparrow^{g_{n-1}}3,

and where a superscript on an up-arrow indicates how many arrows are there. In other words, G is calculated in 64 steps: the first step is to calculate g1 with four up-arrows between 3s; the second step is to calculate g2 with g1 up-arrows between 3s; the third step is to calculate g3 with g2 up-arrows between 3s; and so on, until finally calculating G = g64 with g63 up-arrows between 3s.

So a number so big it can’t be fully processed by the human brain.  This raises some interesting questions about maths and knowledge – Graham’s Number is an example of a number that exists but is beyond full human comprehension – it therefore is an example of a upper bound of human knowledge.  Therefore will there always be things in the Universe which are beyond full human understanding?  Or can mathematics provide a shortcut to knowledge that would otherwise be inaccessible?

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What is the sum of the infinite sequence 1, -1, 1, -1, 1…..?

This is a really interesting puzzle to study – which fits very well when studying geometric series, proof and the history of maths.

The two most intuitive answers are either that it has no sum or that it sums to zero.  If you group the pattern into pairs, then each pair (1, -1) = 0.  However if you group the pattern by first leaving the 1, then grouping pairs of (-1,1) you would end up with a sum of 1.

Firstly it’s worth seeing why we shouldn’t just use our formula for a geometric series:

with r as the multiplicative constant of -1.  This formula requires that the absolute value of r is less than 1 – otherwise the series will not converge.

The series 1,-1,1,-1…. is called Grandi’s series – after a 17th century Italian mathematician (pictured) – and sparked a few hundred years worth of heated mathematical debate as to what the correct summation was.

cesaro summation

Using the Cesaro method (explanation pasted from here )

If an = (−1)n+1 for n ≥ 1. That is, {an} is the sequence

1, -1, 1, -1, \ldots.\,

Then the sequence of partial sums {sn} is

1, 0, 1, 0, \ldots,\,

so whilst the series not converge, if we calculate the terms of the sequence {(s1 + … + sn)/n} we get:

\frac{1}{1}, \,\frac{1}{2}, \,\frac{2}{3}, \,\frac{2}{4}, \,\frac{3}{5}, \,\frac{3}{6}, \,\frac{4}{7}, \,\frac{4}{8}, \,\ldots,

so that

\lim_{n\to\infty} \frac{s_1 + \cdots + s_n}{n} = 1/2.

So, using different methods we have shown that this series “should” have a summation of 0 (grouping in pairs), or that it “should” have a sum of 1 (grouping in pairs after the first 1), or that it “should” have no sum as it simply oscillates, or that it “should”  have a Cesaro sum of 1/2 – no wonder it caused so much consternation amongst mathematicians!

This approach can be extended to the complex series, 1 + i + i^2 + i^3 + i^4 + i^5 + \cdots which is looked at in the blog  God Plays Dice

This is a really great example of how different proofs can sometimes lead to different (and unexpected) results. What does this say about the nature of proof?

The Mathematics of Crime and Terrorism

The ever excellent Numberphile have just released a really interesting video looking at what mathematical models are used to predict terrorist attacks and crime.  Whereas a Poisson distribution assumes that events that happen are completely independent, it is actually the case that one (say) burglary in a neighbourhood means that another burglary is much more likely to happen shortly after.  Therefore we need a new distribution to model this.  The one that Hannah Fry talks about in the video is called the Hawkes process – which gets a little difficult.  Nevertheless this is a nice video for showing the need to adapt models to represent real life data.

Screen Shot 2016-05-21 at 7.18.26 AM

This was the last question on the May 2016 Calculus option paper for IB HL.  It’s worth nearly a quarter of the entire marks – and is well off the syllabus in its difficulty.  You could make a case for this being the most difficult IB HL question ever.  As such it was a terrible exam question – but would make a very interesting exploration topic.  So let’s try and understand it!

Part (a)

First I’m going to go through a solution to the question – this was provided by another HL maths teacher, Daniel – who worked through a very nice answer.  For the first part of the question we need to try and understand what is actually happening – we have the sum of an integral – where we are summing a sequence of definite integrals.  So when n = 0 we have the single integral from 0 to pi of sint/t.  When n = 1 we have the single integral from pi to 2pi of sint/t.  The summation of the first n terms will add the answers to the first n integrals together.

Screen Shot 2016-05-21 at 9.16.26 PM

This is the plot of y = sinx/x from 0 to 6pi.  Using the GDC we can find that the roots of this function are n(pi).  This gives us the first mark in the question – as when we are integrating from 0 to pi the graph is above the x axis and so the integral is positive. When we integrate from pi to 2pi the graph is below the x axis and so the integral is negative.  Since our sum consists of alternating positive and negative terms, then we have an alternating series.

Part (b i)

This is where it starts to get difficult!  You might be tempted to try and integrate sint/t – which is what I presume a lot of students will have done.  It looks like integration by parts might work on this.  However this was  a nasty trap laid by the examiners – integrating by parts is a complete waste of time as this function is non-integrable.  This means that there is no elementary function or standard basic integration method that will integrate it.  (We will look later at how it can be integrated – it gives something called the Si(x) function).  Instead this is how Daniel’s method progresses:

Screen Shot 2016-05-21 at 9.50.36 PM

Hopefully the first 2 equalities make sense – we replace n with n+1 and then replace t with T + pi.  dt becomes dT when we differentiate t = T + pi.  In the second integral we have also replaced the limits (n+1)pi and (n+2)pi with n(pi) and (n+1)pi as we are now integrating with respect to T and so need to change the limits as follows:

t = (n+1)(pi)

T+ pi = (n+1)(pi)

T = n(pi).  This is now the lower integral value.

The third integral uses the fact that sin(T + pi) = – sin(T).

The fourth integral then uses graphical logic.  y = -sinx/x looks like this:

Screen Shot 2016-05-21 at 10.08.00 PM

This is the same as y = sinx/x but reflected in the x axis.  Therefore the absolute value of the integral of  y = -sinx/x  will be the same as the absolute integral of y = sinx/x.  The fourth integral has also noted that we can simply replace T with t to produce an equivalent integral.  The last integral then notes that the integral of sint/(t+pi) will be less than the integral of sint/t.  This then gives us the inequality we desire.

Don’t worry if that didn’t make complete sense – I doubt if more than a handful of IB students in the whole world got that in exam conditions.  Makes you wonder what the point of that question was, but let’s move on.

Part (b ii)

OK, by now most students will have probably given up in despair – and the next part doesn’t get much easier.  First we should note that we have been led to show that we have an alternating series where the absolute value of u_n+1 is less than the absolute value of u_n.  Let’s check the requirements for proving an alternating series converges:

Screen Shot 2016-05-21 at 10.22.18 PM

We already have shown it’s an absolute decreasing sequence, so we just now need to show the limit of the sequence is 0.

Screen Shot 2016-05-21 at 10.24.05 PM

OK – here we start by trying to get a lower and upper bound for u_n.  We want to show that as n gets large, the limit of u_n = 0.  In the second integral we have used the fact that the absolute value of an integral of a function is always less than or equal to the integral of an absolute value of a function.  That might not make any sense, so let’s look graphically:

Screen Shot 2016-05-21 at 9.29.06 PM

This graph above is y = sinx/x.  If we integrate this function then the parts under the x axis will contribute a negative amount.

Screen Shot 2016-05-22 at 7.36.26 AM

But this graph is y = absolute (sinx/x).  Here we have no parts under the x axis – and so the integral of absolute (sinx/x) will always be greater than or equal to the integral of y = sinx/x.

To get the third integral we note that absolute (sinx) is bounded between 0 and 1 and so the   integral of 1/x will always be greater than or equal to the integral of absolute (sinx)/x.

We next can ignore the absolute value because 1/x is always positive for positive x, and so we integrate 1/x to get ln(x). Substituting the values of the definite integral gives us a function of ln which as n approaches infinity approaches 0.  Therefore as this limit approaches 0, and this function was always greater than or equal to absolute u_n, then the limit of absolute u_n must also be 0.

Therefore we have satisfied the requirements for the Alternating Series test and so the series is convergent.

Part (c)

Part (c) is at least accessible for level 6 and 7 students as long as you are still sticking with the question.  Here we note that we have been led through steps to prove we have an alternating and convergent series.  Now we use the fact that the sum to infinity of a convergent alternating series lies between any 2 successive partial sums.  Then we can use the GDC to find the first few partial sums:

Screen Shot 2016-05-21 at 10.30.29 PMAnd there we are!  14 marks in the bag.  Makes you wonder who the IB write their exams for – this was so far beyond sixth form level as to be ridiculous.  More about the Si(x) function in the next post.

 

Screen Shot 2016-05-21 at 7.18.26 AM

IB HL Calculus P3 May 2016:  The Hardest IB Paper Ever?

IB HL Paper 3 Calculus May 2016 was a very poor paper.  It was unduly difficult and missed off huge chunks of the syllabus.  You can see question 5 posted above. (I work through the solution to this in the next post).  This is so far off the syllabus as to be well into undergraduate maths.  Indeed it wouldn’t look out of place in an end of first year or end of second year undergraduate calculus exam.  So what’s it doing on a sixth form paper for 17-18 year olds?   The examiners completely abandoned their remit to produce a test of the syllabus content – and instead decided that a one hour exam was the time to introduce extensions to that syllabus, whilst virtually ignoring all the core content of the course.

A breakdown of the questions

1) Maclaurin- on the syllabus.  This was reasonable.  As was using it to find the limit of a fraction.  Part (c) requires use of Lagrange error – which students find difficult and forms a very small part of the course.  If this was the upper level of the challenge in the paper then fair enough, but it was far from it.

2) Fundamental Theorem of Calculus – barely on the syllabus – and unpredictable in advance as to what is going to be asked on this.  This has never been asked before on any paper, there is no guidance in the syllabus, there was no support in the specimen paper and most textbooks do not cover this in any detail.  This seems like an all or nothing question – students will either get 7 or 0 on this question.  Part (c) for an extra 3 marks seems completely superfluous.

3) Mean Value Theorem – a small part of the syllabus given dispropotionate exam question coverage because the examiners seem to like proof questions.  This seems like an all or nothing question as well – if you get the concept then it’s 7 marks, if not it’ll likely be 0.

4) Differential equations –  This question would have been much better if they had simply been given the integrating factor /separate variables question in part (b), leaving some extra marks to test something else on part (a) – perhaps Euler’s Method?

5) An insane extension to the syllabus which took the question well into undergraduate mathematics – and hid within it a “trap” to make students try to integrate a function that can’t actually be integrated.  This really should have been nowhere near the exam.  At 14 marks this accounted for nearly a quarter of the exam.

Content unassessed

The syllabus is only 48 hours and all schools spend that time ploughing through limits and differentiability of functions, L’Hopital’s rule, Riemann sums, Rolle’s Theorem, standard differential equations, isoclines, slope fields, the squeeze theorem, absolute and conditional convergence, error bounds, indefinite integrals, the ratio test, power series, radius of convergence.  All of these went pretty much unassessed.  I would say that the exam tested around 15% of the syllabus content.  Even the assessment of alternating series convergence was buried inside question 5 – making is effectively inaccessible to all students.

The result of this is that there will be a huge squash in the grade boundaries – perhaps as low as 50-60% for a Level 6 and 25-35% for a level 4.    The last 20 marks on the paper will probably be completely useless – separating no students at all.  This then produces huge unpredictability as dropping 4-5 marks might take from from a level 5 to level 3 or level 6 to level 4.

Teachers no longer have any confidence in the IB HL examiners

One of my fellow HL teachers posted this following the Calculus exam:

At various times throughout the year I joke with my students about how the HL Mathematics examiners must be like a group of comic book villains sitting in a lair, devising new ways to form cruel questions to make students suffer and this exam leads me to believe that this is not too far fetched of a concept.

And I would tend to agree.  Who wants students to be demoralised with low scores and questions they can’t succeed on.  Surely that should not be an aim when creating an exam!

I’ve taught the HL Calculus Option for the last 4 years – I think the course is a good one.  It’s difficult but a rewarding syllabus which introduces some of the tools needed for undergraduate maths.  However I no longer have any confidence in the IB or the IB examiners to produce a fair test to examine this content.  Many other HL teachers feel the same way.  So what choice is left?  Abandon the Calculus option and start again from scratch with another option?  Or continue to put our trust in the IB, when they continue to let teachers (and more importantly the students) down?

 

 

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Alan Turing Cryptography Competition

Manchester University are running their 5th Alan Turing Cryptography Competition this January.  It’s aimed at secondary and post 16 students.  If you are in the UK and in year 11 or below you can register for the official prizes, for everyone else you can still register and see if you make it onto the leaderboard.

Read some of the introduction to the competition below:

Do you like breaking codes and solving ciphers?
Can you, and your friends, unravel the mystery of the Artificial Adventure?
Would you like the chance to use your mathematical skills to win some great prizes?

The competition starts on Monday 25th January, and you can register your team (or join an existing team) here. A team consists of at most 4 members. It is also possible to register as a `non-competing’ team, for instance if you’re a teacher who would like to follow the competition or if some members of your team are too old to take part. Registration opens on Monday 30th November.  

The competition follows the story of two young cipher sleuths, Mike and Ellie, as they get caught up in a crptographic adventure `The Tale of the Artificial Adventure’. Every week or two weeks a new chapter of the story is released, each with a fiendish code to crack.  There are six chapters in total (plus an epilogue to conclude the story). Points can be earned by cracking each code and submitting your answer. The leaderboard keeps track of how well each team is doing. 

The competition starts on January 25th.  Click on the competition website to register – and good luck!

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Crack the Beale Papers and find a $65 Million buried treasure?

The story of a priceless buried treasure of gold, silver and jewels (worth around $65 million in today’s money) began in January 1822. A stranger by the name of Thomas Beale walked into the Washington Hotel Virginia with a locked iron box, which he gave to the hotel owner, Robert Morriss.  Morriss was to look after the box for Beale as he went off on his travels.

In May 1822 Morriss received a letter from Beale which stated that the  box  contained papers of huge value – but that they were encoded for protection.  Beale went on to ask that Morriss continue to look after the box until his return.  He added that if he did not return in the next 10 years then he had instructed a close friend to send the cipher key on June 1832.  After that time Morriss would be able to decipher the code and learn of the box’s secrets.

Well, Beale never returned, nor did Morriss receive the promised cipher key.  Eventually he decided to open the box.  Inside were three sheets of paper written in code, and an explanatory note. The note detailed that Beale had, with a group of friends discovered a seam of gold and other precious metals in Santa Fe. They had mined this over a number of years – burying the treasure in a secret location for safe keeping.  The note then explained that the coded messages would give the precise location of the treasure as well as detailing which men were due a share.

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Morriss devoted many years to trying to decipher the code in vain – before deciding at the age of 84 in 1862 that he should share his secret with a close friend.  That friend would later publish the Beale Papers in 1885.  The pamphlet that was published stirred huge interest in America – inspiring treasure hunters and amateur cryptographers to try and crack the code.  The second of the 3 coded messages was cracked by the author of the pamphlet using what is known as a book code.  The United States Declaration of Independence was used as the book to encode the message above.

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The first number 115 refers to the 115th word in the Declaration of Independence, which is the word “instituted”.  Therefore the first letter of the decoded message is “I”.  The second number is 73, which refers to the 73rd word in the declaration – which is “hold”, so the second letter of the decoded message is “h”.  Following this method, the following message was revealed:

I have deposited in the county of Bedford, about four miles from Buford’s, in an excavation or vault, six feet below the surface of the ground, the following articles, belonging jointly to the parties whose names are given in number three, herewith:

The first deposit consisted of ten hundred and fourteen pounds of gold, and thirty-eight hundred and twelve pounds of silver, deposited Nov. eighteen nineteen. The second was made Dec. eighteen twenty-one, and consisted of nineteen hundred and seven pounds of gold, and twelve hundred and eighty-eight of silver; also jewels, obtained in St. Louis in exchange for silver to save transportation, and valued at thirteen thousand dollars.

The above is securely packed in iron pots, with iron covers. The vault is roughly lined with stone, and the vessels rest on solid stone, and are covered with others. Paper number one describes the exact locality of the vault, so that no difficulty will be had in finding it. Source

After the pamphlet was published there was great interest in cracking the 2 remaining papers, an interest which has persisted into modern times.  One of the uncracked papers is shown below:

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In 1983 2 amateur treasure hunters were jailed for trying to dig up graves in Bedford, sure that they were about to find the missing gold.  In 1989 a professional treasure hunter called Mel Fisher secretly bought a large plot of land after believing that the treasure was buried underneath.  However nothing was found.  Up until now all efforts to crack the code above have  ended in failure.  Perhaps the pamphlet was a giant hoax?  Or perhaps the treasure is still waiting to be found.

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The town of Bedford still receives visitors from around the world, keen to try and crack this centuries old puzzle.  You can hire metal detectors and go looking for it yourself.  The map above from 1891 shows the 4 mile radius from Buford’s tavern which is thought to contain the treasure.  Maybe one day Beale’s papers will finally be cracked.

For more information on this topic read Simon Singh’s excellent The Code Book – which has more details on this case and many other code breaking puzzles throughout history.

If you want to try your own codebreaking skills, head over to our Schoolcodebreaking site – to test your wits against students from schools around the world!

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