Norman Wildberger
Norman is a research mathematician with a strong interest in education. He is the discoverer of Rational Trigonometry, and his YouTube site (user njwildberger) has more than 30K subscribers.
Location UNSW Australia
Activity

Thanks Malcolm.

Thanks Manuel.

Hi Evanthia, Yes that is a good point, thanks for bringing it to our attention. Next time around we will try to let people know when we have made a correction on a test. But good to see you did very well anyway!

Thanks for joining us Michi.

Thanks Jonathan.

Thanks Christine.

Thanks Nick.

Norman Wildberger replied to Dave Hardcastle
Well done Dave for making it through all the hurdles. And remember that the course material is now available to you for review in the future.

Thanks Evanthia, glad you enjoyed the course!

Norman Wildberger replied to John Vie
Thanks John.

Norman Wildberger replied to Graham Gardiner
Well done Graham! And thanks for the lovely comments.

Norman Wildberger replied to Keith Cook
Thanks Keith. Hopefully that log stuff will sink in after a few months of ingestion.

Norman Wildberger replied to Antonio Avellan
Thanks Antonio.

Norman Wildberger replied to Helen Kilgour
Nice to have you with us on the course Helen, and yes you can use the materials as long as you credit us.

Thanks Luis.

Norman Wildberger replied to tahir rasul
Thanks Tahir

Norman Wildberger replied to Jenny Young
Thanks Jenny.

Norman Wildberger replied to Barrie Kane
Nice to have you with us Barrie, glad to hear you enjoyed the course.

Norman Wildberger replied to Mary Harding
Thanks Mary!

It is a really beautiful episode in the history of thoughtNewton's derivation of Kepler's laws through his laws of motion and gravitation plus a geometric understanding of the brand new calculus that he helped invent. It would make a good course. In fact at UNSW we plan on running a PD course next year called Celestial Motion and Space Flightit will be...

Thanks Cesar.

Thanks Jitka for the nice comment and the many contributions to the discussion.

You're welcome John!

Norman Wildberger replied to Dai Middleton
Thanks Dai! You have made a huge contribution to the course, and I know a lot of participants really got a lot out of your many thoughtful contributions. It has been a pleasure having you on the course with us.

Hi Tony  I do have vague plans to put together a Calculus course in the next year or so: but it will be a very Algebraic Calculus, which in my view is the right way of approaching it. I have a wordpress blog: https://njwildberger.com/ where you can keep in touch with my plans on that.

Thanks Ivana for joining us!

Hi Antonio, Sorry to hear about your difficulties, and hope that you find your way through them. Having some interesting other things to work on and think about can be helpful I am sure. May I interest you in my YouTube channel, user njwildberger? You might find lots more there to engage you interest. Thanks for your participation!

It certainly appears to be something of a coincidence. The pair of heavenly objects: Father Sun, Sister Moon etc. roughly comparable at least in size.

Norman Wildberger replied to Simon C
Glad to hear you made it through both courses! Congrats!

Norman Wildberger replied to Gerallt Jones
Thanks Gerallt.

Norman Wildberger replied to Meryl Goulbourne
Well done Meryl.

The better approach is to define ln(x) first as we have done here, in terms of an area under the graph of y=1/x. And then to define exp(x) as the inverse function to ln(x): that means we reflect the graph of y=ln(x) about the line y=x to get the graph of y=exp(x). Algebraically, y=exp(x) precisely when x=ln(y). This is not the way it is usually done it...

Thanks Nanceybut the course is still going!

Exponential functions include ones like y=2^x. That is to be distinguished from polynomial functions, such as y=x^2. The exponential functions grow with x much more rapidly!

The curve with equation (xy)(x^2+y^21)=0 ought to be regarded as a cubic curve, since if we expand it out we get degree three terms in x and y. However geometrically it is the product of a line and a circle. So it does not exactly look like a typical cubic, but nevertheless is one. Hope that clarifies things.

The definition of CxC is a bit subtle. Suppose that C and C' are two points on the conic, very close together. In that case the line C C' is close to being the tangent to the curve at C. Conclusion: when we multipy a point C by itself, we draw the tangent to the curve at C, then draw the line through O parallel to that tangent, and then see where that line...

Think about a tangent to the curve?

Norman Wildberger replied to Tim Curry
Thanks Tim!

Norman Wildberger replied to Ivan Alim
Thanks Ivan, glad you enjoyed the course. You can check out lots more of my YouTube videos from my channel Insights into Mathematics, user njwildberger.

Thanks, now corrected.

Thanks Tim!

The reason is that given a fixed amount of surface area, we want to maximize the volume, to ensure that the cell can be as productive as it can be subject to the limits imposed by the surface area on inputs and outputs. And it is a famous result in mathematics that for a given surface area, the greatest volume is achieved for a sphere. Thus very generally...

Thanks for the data William.

Nice.

Norman Wildberger replied to Simon C
Thanks Simon!

Good point Reino, I have noted that on the question now.

That's quite correct Ian. In complicated situations some visualization and getting the computer to make the calculation is a great tool.

Mathematics is characterised by its careful use of language. It is not so bad to get practice in reading questions carefully. After all the tests are not timed, so you do have an opportunity of thinking about each answer.

Because we have multiple choice questions, we try to arrange the questions so that you can learn from each answer. Not just find the right answer and tick it, but read each answer carefully and understand it.

Norman Wildberger replied to Clare Fletcher
It was not very clear, so I have changed the wording. The idea is that we are going to raise e to both the left and right sides of the equation. For example from x=y we could get e^x=e^y. That is what we are doing here.

In ordinary arithmetic, we multiply two numbers, and get a third number. With this kind of arithmetic, we multiply two points lying on a cubic curve, and get a third point on that cubic curve!

Thanks Ian, it is corrected now.

That's quite right. The continuous (real) field is actually very hard to apply usefully here: but finite fields whose arithmetic are close to that of the rationals is just the ticket to make this elliptic curve cryptography work.

Cool. What about the second biggest city?

Thanks, it is interesting to see the names of these cities. It looks off to top to be quite close to a Zipf's law distribution?

I think data like this is so interesting. Wouldn't it be great if newspapers actually informed us of facts like this on a regular basis?

Wow, that must be a lovely experience. I had no idea there were things like an International Dark Sky Reserve (IDSR) !

No, it has to be of the form y=a/x for x and y to be inversely proportional. That is not to say that relations like y=1/x^2 are not important they are. They fall into the more general camp of "power laws", which we will be learning all about in the last half of the course.

It is not a coincidence, but a consequence of the physics of tidal forces. If the moon is not a perfect sphere, the differences in the gravitational field strength of the earth between near and far sides will cause any mass imbalance to start affecting the moon's rotation till after billions of years it has lined up its rotation around the earth and it axial...

Sorry, that is not possible. But you could register interest in the next run of the course at
https://www.futurelearn.com/courses/mathslinearquadraticrelations

That is the correct answer. Perhaps you have heard of Wolfram Alpha? It is a computational engine available over the internet. You can ask it
integrate 1/x^2 dx from x=1 to infinity
and it will tell you that the answer is indeed 1. A nice tool to know about!

Here is a clue: think three dimensionally.

Hi Jitka, We do hope to rerun the first course at some point if there is sufficient interest. But nothing is scheduled at this point.
