You are currently browsing the category archive for the ‘Uncategorized’ category.

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?

If you enjoyed this post you might also like:

How Are Prime Numbers Distributed? Twin Primes Conjecture – a discussion about the amazing world of prime numbers.

Wau: The Most Amazing Number in the World? – a post which looks at the amazing properties of Wau


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 2018-03-19 at 5.29.06 PM

IB Maths Revision

I’d strongly recommend starting your revision of topics from Y12 – certainly if you want to target a top grade in Y13.  My favourite revision site is Revision Village – which has a huge amount of great resources – questions graded by level, full video solutions, practice tests, and even exam predictions.  Standard Level students and Higher Level students have their own revision areas.  Have a look!

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?

 

 

Screen Shot 2015-12-09 at 7.03.47 AM

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!

Screen Shot 2015-11-22 at 6.36.36 PM

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.

Screen Shot 2015-11-22 at 6.37.27 PM

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.

Screen Shot 2015-11-22 at 6.51.08 PM

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:

Screen Shot 2015-11-22 at 6.37.03 PM

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.

Screen Shot 2015-11-30 at 9.52.12 PM

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!

Screen Shot 2015-09-30 at 8.13.29 PM

IB HL Calculus Option Videos

For those students studying the IB Maths Higher Level Calculus option, I’ve just finished putting together video playlists to cover the whole option syllabus.  These include both videos teaching the course content and also worked past paper solutions.  Hopefully this should make what is a demanding unit a little bit more manageable.

Calculus Option Part 1:

Limits, limits at infinity, determining if limits exist, differentiating from first principles, L’Hopital’s Rule, Squeeze theorem, Rolle’s theorem, Mean Value theorem.

Calculus Option Part 2:

differential equations, solving differential equations through separating variables, substitution and integrating factor, sketching slope fields.

Calculus Option Part 3:

Improper integrals, Comparison test, Riemann sums, divergence test, comparison test, limit comparison test, alternating series test, absolute convergence, power series, Ratio test.

Calculus Option Part 4:

Power series, Taylor and Maclaurin series,

Other playlists

You can find all the playlists for the:
HL core maths content here.
SL maths here
Studies maths here.

mmd1

International School Code Breaking Challenge

We’re now running a huge school code breaking competition.  There are a total of 7 competitions to enter – each one with a number of codes to crack.  Each time a code is cracked this gives the password to access the next clue.  Students who break all codes will be entered onto the leaderboard of fame.  Levels of codes range from upper KS2 to post 16 – so there should be something for everyone.

Choose your level, and good luck!

codeimage4

Level 1: Easy, Age 11-16.  This is intended to be a gentle introduction to code breaking.  Hints provided throughout.

codeimage3

Level 2: Easy, Age 11-16.  This builds on some of the skills from the Level 1 code and includes a mixture of common codes, from Morse Codes to substitution ciphers.

Screen Shot 2015-02-12 at 3.38.30 PM

Level 3: Murder mystery, Medium, Age 11-16.  A murder has been committed in the Maths Department!  Solve the clues to uncover the murderer, the weapon and the room.
Level 4: Murder mystery, Medium, Age 11-16.  Another murder has been committed!  Who can solve it first?

sherlock

Level 5: Medium, Age 11-16.  This level combines knowledge of a variety of codes and methods of solving them.  Caesar shifts, transposition codes and more.

codeimage6

Level 6: Hard, Age 15-18.  This is a real tough challenge – not for the faint hearted!  It is primarily aimed at post 16 students and you can expect to look at binary codes, RSA codes and modulo arithmetic.

Level Extra: Easy, Age 11-16.  A bonus code breaking challenge in case you solve all the others!

euclidean

Non Euclidean Geometry – Spherical Geometry

This article follow on from Non Euclidean Geometry – An Introduction – read that first!

Most geometers up until the 19th century had focused on trying to prove that Euclid’s 5th (parallel) postulate was true.  The underlying assumption was that Euclidean geometry was true and therefore the 5th postulate must also be true.

The German mathematician Franz Taurinus made huge strides towards developing non-Euclidean geometries when in 1826 he published his work on spherical trigonometry.

euclid14

Spherical trigonometry is a method of working out the sides and angles of triangles which are drawn on the surface of spheres.

One of the fundamental formula for spherical trigonometry, for a sphere of radius k is:

cos(a/k) = cos(b/k).cos(c/k) + sin(b/k).sin(c/k).cosA

So, say for example we have a triangle as sketched above.  We know the radius of the sphere is 1, that the angle A = 60 degrees, the length b = 1, the length c =1, we can use this formula to find out what the length a is:

cos(a) = cos(1).cos(1) + sin(1).sin(1).cos60

a = 0.99996

We can note that for the same triangle sketched on a flat surface we would be able to use the formula:

a2 = b2 + c2 – 2bc.cosA

a2= 1 + 1 – 2cos60

a = 1

Taurinus however wanted to investigate what would happen if the sphere had an imaginary radius (i).  Without worrying too much about what a sphere with an imaginary radius would look like, let’s see what this does to the previous spherical trigonometric equations:

The sphere now has a radius of ik where i = √-1, so:

cos(a/ik) = cos(b/ik).cos(c/ik) + sin(b/ik).sin(c/ik).cosA

But cos(ix) = cosh(x) and sin(ix) = (-1/i)sinh(x)  – where cosh(x) and sinh(x) are the hyperbolic trig functions.   So we can convert the above equation into:

cosh(a/k) = cosh(b/k)cosh(c/k) – sinh(b/k).sinh(c/k).cosA

This equation will give us the relationship between angles and sides on a triangle drawn on a sphere with an imaginary radius.

Now, here’s the incredible part – this new geometry based on an imaginary sphere (which Taurinus called Log-Spherical Geometry) actually agreed with the hypothesis of the acute angle  (the idea that triangles could have an angle sum less than 180 degrees).

Even more incredible, if you take the limit as k approaches infinity of this new equation, you are left with:

a2 = b2 + c2 – 2bc.cosA

What does this mean?  Well, if we have a sphere of infinite imaginary radius it stretches and flattens to be indistinguishable from a flat plane – and this is where our normal Euclidean geometry works.  So, Taurinus had created a geometry for which our own Euclidean geometry is simply a special case.

So what other remarkable things happen in this new geometric world?  Well we have triangles that look like this:

euclid15

This triangle has angle A = 0, angle C = 90 and lines AB and AC are parallel, (they never meet).  This sketch introduces a whole new concept of parallelism far removed from anything Euclid had imagined. The angle  β is called the angle of parallelism – and measures the angle between a perpendicular and parallel line.  Unlike in Euclidean geometry this angle does not have to be 90 degrees.  Indeed the angle  β will now change as we move the perpendicular along AC – as it is dependent on the length of the line a.

So, we are now into some genuinely weird and wonderful realms where normal geometry no longer makes sense.  Be warned – it gets even stranger!  More on that in the next post.

If you enjoyed this post you might also like:

Non Euclidean Geometry IV – New Universes – The fourth part in the non-Euclidean Geometry series.

The Riemann Sphere – The Riemann Sphere is a way of mapping the entire complex plane onto the surface of a 3 dimensional sphere.

Circular Inversion – Reflecting in a Circle The hidden geometry of circular inversion allows us to begin to understand non-Euclidean geometry.

Website Stats

  • 5,850,816 views

Recent Posts

Follow IB Maths Resources from British International School Phuket on WordPress.com