Monday, November 25, 2013

You can teach Calculus without really needing to teach

And you can learn Calculus in about ten hours a week.

Check out these online courses: Precalculus, Calculus One, and Single-Variable Calculus.  Each is about three months (one semester) long, available on Coursera for FREE, and includes weekly video lessons, peer-interactions, testing and evaluation, and a certificate of completion.  I'll tell you when I evaluate other sources of online course materials: udemy, edX, and so on.

If I was assigned to teach a large Calculus class, I would probably just reuse these resources as it would be hard to do better than some of the online courses I've tried.  Even for a small class, I might embed this material, as in modularity, as in a course within a course.
From Calculus Single Variable on Coursera
 When using these online courses or MOOCs as they are now called, the teacher is still useful.  The teacher needs to study several and select which resources to use, and also to monitor and enhance the experience where possible. 

For more elementary mathematics, like primary and secondary school (but also up to Calculus and hobby math), I like the quizzes at Khan Academy for math drill.  Khan also has extremely, extremely good video lessons. 

Wednesday, October 16, 2013

Course within a course

Because it's time I did something new, because I am/was worried that my programming "skills" from the late 90s need updating. Because...
I finally did it. I'm finally using Coursera! I had worried about the video format (my internet plan allows only a few gigabytes of data monthly, so I don't use a lot of video online; then also the speeds could be frustrating, in the tens of kB/sec often, but on good days in the hundreds.)

The thing to do is to start anyway, and I have started, and I am impressed with Coursera so far.

The impetus for this experiment with online learning was my sizable teaching load.  I have some large classes with about 100 students and had to split them into two or three groups to get any chance of doing a good job.  For now, it's not understood in my country that you can teach a course in sections, so if there are 500 1st year students who need to take Physics, you fit them in an auditorium for Physics 101, you should by no means create 5 or 10 groups of Physics 101.  They say it's because of cost, but I think it's because of inertia/habit. 
Doll within a doll: Matryoshka dolls

Anyhow, I needed to get more time away from stressful, low-yield teaching tasks that don't scale well, and still get the students to learn more through fun, interactive methods. So I recommended two Coursera courses as a required part of my "Internet and Web Applications Technology" course this semester. They are: "Internet History, Technology, and Security" which is shaping up to be really exciting, and "Introduction to Programming for Musicians and Digital Artists" which I hope will take the perceived dryness out of programming and be cool enough to get the whole class programming.

Because there is a Certificate of sorts from Coursera for doing the work to a reasonable level, I can give course credit for it: about 30% of course credit for taking these two easy courses.

Using the MOOCs is an amazing teaching solution for many reasons.
First, Charles Severance and team know a bit more internet history than I do, so why not let them do the job?
Second, it's not so easy to cheat/copy in these courses, compared with homework I assign. I mean, it takes almost as much effort to copy as to do the work. And anyhow, they will experience significant pushback against their attempts to plagiarize.  Note, my kids are not bad people, it's just that copying seems to be part of national culture.
Third, I can spend class time doing other things (including taking my own online courses and theirs)
Fourth, they can interact with a global classroom; ask and answer questions, assure themselves that they're learning at a world standard, etc.
It costs a bit to get internet, but hey, if I can afford it, so can they; they can definitely watch in groups to cut the cost.

Besides Coursera, I've also used Khan Academy, with good results. At my present school, most of my engineering students somehow arrive in the third year with poor basic math skills. So I start off each course by assigning a massive precalculus/calculus assignment (functions, trigonometry, complex numbers, derivatives) in Khan.  I would like to assign integral calculus as well, but it seems the problem sets stop at derivatives for now.

One good thing about Khan is that it's very hard to cheat with the randomized questions, so they find they might as well do the work.  I think Coursera might have randomized questions too - not sure - but they should consider that.    Also, the coach/instructor can monitor all their work in the backend of the site and - yippee - assign course credit, since for many students, course credit is the main thing that would cause them to do the work.
Whatever happened to just learning for fun? ? 

It's a busy month, but I'll be back later with more analyses of games, and more on prime numbers.  

Friday, September 27, 2013

Chuzzle Puzzle Guzzle

Chuzzle is an addictive little game I found on my cellphone.  You move rows and columns of these dollfaces around to match them by colour so they explode and disappear.
It is a rather daft game, with one exception - the Mindbender mode. With Mindbender, you have to arrange the chuzzles into a pre-specified pattern.  It's a nice challenge, so I did all the 100 available patterns.
Having figured out how to replace/substitute colours and all, maybe I'll be able to do a Rubik's cube someday.
Magic Cube - you know how to solve this?
For now, I guess flat / two-dimensional puzzles are enough of a challenge.  Like this picture puzzle, which to me is easy once you've numbered the tiles.
Click to download and play FREE
I'll be back soon with more on primes.

Saturday, August 3, 2013

The difference between two (consecutive) prime numbers

Recall that
The Continuous derivative method of finding a formula for the estimated difference between two prime numbers
Applying the Product Rule for differentiating functions, P'(n) , the derivative of Prime(n),
is ~= α(n.1/n + 1.log n) = α( 1 + log n )
Yes, the derivative is usually for continuous functions, while Prime[n] is discrete, but approximating this way is OK because it's useful. 
What the derivative means is: let Δ be a reasonably small value, then P(n+Δ) ~= P(n) + Δ.P'(n)
This implies that if we want P(n+1), we may put Δ=1 above and the (n+1)th prime number is approximately P(n) + P'(n), i.e. just add α( 1 + log n )  to the nth prime number.
Got it?
The (n+1)th prime number, note α around 1.1
What is this useful for? 
We've already seen that we may estimate the prime numbers with so-so results using the formula for the nth prime.
But we may sometimes get closer approximations when we seek a prime number, the (n+1)th prime for instance, using the difference formula and prior knowledge of the nth prime.   

If we know the 99th prime number to be 523, what is the 100th prime?
Add 1.2(1+log99) to 523 , to get 523+6.714 ~= 530.
Oh well, this is not a prime number and it's still pretty far from the 100th prime (541), but the formula works better sometimes,
e.g. the 101st prime, given the 100th prime is 541 is
estimated to be 541+1.2(1+ln100) = 541 + 6.726 ~= 547.7
and in actual fact is 547

The finite/discrete derivation of the formula for the estimated difference between two consecutive primes
This yields similar/close results to the continuous derivative steps above:
P[n+1] - P[n] ~= α.(n+1).log(n+1) - α.n.logn
 = α [ n.log(n+1) + log(n+1) - n.log(n) ] 
 = α [ log ( (n+1)^n ) - log (n^n) + log (n+1) ] 
=  α [ log ( {(n+1) / n}^n . (n+1) ) ]
= α [ log {(1 + 1/n)^n .(n+1)} ] 
and at this point, you need the fact that (1+ 1/n)^n is approximately e, which has log 1, for reasonably large n.  For example (1+1/5)^5 is about 2.5, while 1.01^100 is about 2.7.  Anyway, moving on...
so P[n+1] - P[n] ~= α.log(e.(n+1)) = α.[1+log(n+1)] 

What should be done about the difference between the continuous and the discrete formulae?
Ignore the fact that the difference has 1+log(n+1) here, but 1+log(n) in the first derivation.
We might have even chosen to use the derivative at n+1/2Δ instead of at n as we did, and got the difference P[n+1] - P[n] ~= α.[1+log(n+ 1/2)]   
Anyway, when n>>1, log(n) is very close to log(n+1) in the sense that it's much smaller than 1 and much smaller than n: how to see this is another short exercise in using the derivative :) i.e. the derivative of log(n) is 1/n which is << 1 << n

Answers to two previous questions
 1. How does one know that there is no largest PRIME number?  Because when we think we've found the biggest, here is an estimate for the next one...add 1+log(n+1) to it. 
2. How do you know if α > 1?  Sorry I thought I knew, but I don't (yet.) 

Thursday, August 1, 2013

The nth prime number

What is the relationship between Prime[n] and n?  Prime[n] by the way is a "function" in Mathematica that gives the nth prime number.
For me, the story of finding it did not start with that specific question, it started with play.
I was playing with the tutorial on my Mathematica installation (v7), it had a plot of Prime[n] against n, say for n from 1 to 100.  So of course you extend that say 1 to 10000, wow cool, and extend it again till the program stalls.
The next time you try this, you notice the graph of Prime[n] against n which is many points really close together bears a striking resemblance to...log(n) or something.  So you plot Prime[n] against n.log(n) and sure enough, well it's a little stochastic looking but it sort of stays within a small constant range 1.1 - 1.2.  More precisely, for small n, Prime[n] ~= 1.2 n log(n)...and for the largest n I could compute, Prime[n] ~= 1.1 n log(n), or if you let α vary a little around 1.15, Prime[n] = α . n log (n)

Why won't I just show you the plots?  Well, I've been procrastinating forever writing this post, and just minutes ago found that my Mathematica license expired.  Dang!  I thought the license was forever to be honest :) Fingers crossed it'll be renewed soon.

Cool question: I wonder if there is a limit to what α can be.  It seems to decrease gently from 1.2 for n in the 10-1000 range, to 1.1 for n much larger.  But for n a lot larger than that, is α ever going to go below 1?  Is there a guarantee that the "slope" of the graph Prime[n] against n.log(n) , averagely speaking, stays above 1 even for n very large?  I'll say yes for now, that the asymptote of that slope or ratio is 1 and I have "proof", but I want to know what you think.

In summary,
Clearly useful for estimating the nth prime number
Note: The log used in this law is the natural logarithm (so called ln) not the base 10 logarithm (often denoted log)

Meanwhile, we can use the knowledge we already have, say to compute the 100th prime number as Prime[100] ~= 1.2 * 100 ln 100
~= 1.2 * 100 * 4.60517018599 
= 552.620422319 , so the 100th prime number is around 552 or 553?  
If we instead used α = 1.1 we would get the estimate Prime[100]=506.568720459
Nifty tool: I just typed "what is log 100" in google search box, and a calculator appeared.  Just remember to use ln not log.  
In reality, the 100th prime number is 541, not 553, not 507, but still, not too bad?  

Wednesday, July 31, 2013

How many prime numbers are there?

That's easy - infinitely many. 
Well, you know there are infinitely many counting numbers (also known as natural numbers) right?  I remember watching this kiddies activity video in the eighties and they had a song " can count forever, there'll always be one more..." and the doubter would say when you named the next really big number, "add one to it!"  You could always add one, ergo, there is no largest or maximum counting number.  

But how does one know that there is no largest PRIME number? 

Long term, there are many more non-primes than primes, that is the prime numbers appear less and less frequently.  Think about it: start counting from 1, 2, ... at first half the numbers are non-prime (all the even numbers.)  After 3, all the factors of three are non-prime too.  That is, from then on, not only all the even numbers, but also all the odd numbers divisible by three, will be non-prime.  And so on. To be prime, a number q will have to pass the test: not divisible by 2, not a factor of 3, not a factor of 4 (implied already since 4 is made up of 2's), not a factor of 5, ... AND not a factor of any other number smaller than itself (although we can stop testing when we reach half of the number or q/2 since no larger number - larger than q/2 and smaller than q - will divide q evenly anyway)
Here's a picture of how the factors eliminate the primes, and so the occurrence of primes is not as random as one might first think, at least if one takes a different point of view: that the occurrence of non-primes is extremely regular, beat beat beat beat - can you hear the music? 

Exactly how infrequently the prime numbers occur "over time" - as the numbers get bigger - is a cool question.  We'll come back to it later. 

Maybe the picture is enough to convince you that there is no largest prime number, that no matter how large/high you go there is always a loose end, a number q that hasn't been captured by any of the wanna-be factors in the 2 to q/2 range. 

If that is not convincing enough, in the next post (or two) we'll derive a good estimate for the difference between two consecutive prime numbers.  So, we are assured that if you add "difference" to a prime number, then we'll get the next largest one.  That is to say, "there'll always be one more!"

Monday, May 27, 2013

Prime numbers

So much to write, so little time, no? 
I have an essay on primes that I first worked out last year, and which I will soooon post.  I'm not even an expert in number theory, so I don't know if the essay is saying anything new. 

It'll start with simple stuff like how many prime numbers are there and proceed to something more interesting like what are they then.  Neither of these questions is very simple, neither is entirely complex.  But I think it requires creativity (in the sense of new viewpoints) to give new answers and someday make the complex simple. 

And that is the march of science, n'est-ce pas?
Roland Garros (French Open) tennis is on

Sunday, May 12, 2013

Upside-down t-shirt

 4 out of 3 people struggle with math.  Sure. 

Saturday, March 23, 2013

Teaching ICT in Lagos

My recent teaching website is bellseng on googlesites.  OK, the school is not exactly in Lagos, it's an hour drive from Ikeja.  
I only have one course to teach this semester, and fewer than 100 students.  Objectives include brushing up on math for some, open-ended math analysis for some others, and discussions of information and communication technology for most.  We'll have fun.  

Friday, February 15, 2013

Mahjong - I really didn't expect this high score

First you learn to win the game, then you find tricks to raise your score, but then you start losing because in chasing the high score you get stray off the straight and narrow path that leads to winning.

Hence I started once again to play plain vanilla mahjong titans, and guess what I just did: scored my highest ever in Fortress.
I wasn't even thinking about the scores, Man!
 Last time I looked, a couple of months ago, here were my high scores:
Interestingly, my absolute highest scores in four layouts (all except Crab and Spider) occurred between 16 and 25 July 2012.  For Crab, the second highest score was 18th July, and the 3rd, 4th, and 5th highest were all in the month before.  I recommend playing for high scores with the Crab layout as it responds well to the score-raising tricks.  
 Contrast this with Spider, for which the five highest scores all came before this score-raising season. 
My Mahjong solitaire scores for different layouts, as at late 2012

For an ordering of how well the score-raising tricks seem to have helped:  
Let high scores in June/July 2012 yield 5points for 1st, 4 for 2nd, ..., 1 for 5th, and add them up.  The totals are Turtle 6, Dragon 10, Cat 9, Fortress 12, Crab 10, Spider 0.  Therefore we have: Fortress, Crab/Dragon, Cat, Turtle, Spider

The tricks raise your score alright, but how much have you hurt the bottomline (win rate?)
What do you make of the fact that win percentages in early 2012 are roughly the same as in late 2012 above, except for Crab and Turtle which gained?  I say without the score-raising experiments, the win rates went up.  Then the sre hurt (caused losses) least in these two layouts.  In the other layouts, the sre-related losses possibly canceled out other increases in win rate. 

I'm officially bored done writing about Mahjong solitaire. What is exciting how many have used my tips to aid their play. Love it.

Monday, January 7, 2013

Looks like I'm really bad at playing regular solitaire

You know that I cracked Mahjong Titans, a solitaire game.  tips here.

But on my computer, there's this regular solitaire that I fail 6 out of 7 times.
For as long as I can remember, my success percentage has stayed between 14% and 16% (Ok I don't remember going as high as 16 percent actually) and I've played it over 500 times.
How to improve my play at solitaire?
I don't think it used to be this bad on previous computers, I think I won more frequently then.
I mean, the solitaire I remember was just common sense: focus on the deep piles, don't get carried away with easy wins, plan for the future, try to balance your red and black tiles...
but now it seems more like tough, irreversible, luck.

Do I just have the hard version or something?

Meanwhile, I'll post more on Mahjong/Mahjonng titans later.  I sort of stopped playing it so frequently, discovered some new tips, probably lowered my win rate while experimenting wantonly with the stuff.  

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