Mathematical Curiosities. (1 Viewer)

[Carrotsticks]

Hi everybody!

When I was a student, I often had a lot of 'theory questions' that often could not be answered by my teacher, and I often wished that I had some sort of place where I could ask such questions.

Examples of such questions are:

In Auxiliary Angle method, why do we always take R>0? Why don't we take R<0?

I read in one textbook that the Fundamental Theorem of Algebra asserts that every non-constant polynomial with complex coefficients has at least one complex root. But another textbook says that the FTA means that a polynomial of degree N has N roots. Aren't these two completely different statements?

I know for sure that if I were a student, I would have loved for such a thing, so feel free to ask whatever 'curiosities' you have here! We have a plethora of users here who are able to answer them.

 

[asianese]

Why does the Board write progressively easier MX2 papers? This has always been a curiosity for me...

 

[panda15]

Why did the Board of Studies bring in multiple choice for higher level maths?

 

[Carrotsticks]

Why did the Board of Studies bring in multiple choice for higher level maths?

Cheaper and easier to mark.

Also, a nice way to examine those 'small topics' without having to allocate 2-3 marks for them, otherwise it would be difficult to fit everything into 100 marks.

 

[rumbleroar]

Just out of curiosity, which 4U topics are the "hardest" in your opinion? And which ones would I need to work hardest on? I know its different and relative for all people, but I want to get a range of opinions just to see what I'm facing for the next year.

 

[bottleofyarn]

Just out of curiosity, which 4U topics are the "hardest" in your opinion? And which ones would I need to work hardest on? I know its different and relative for all people, but I want to get a range of opinions just to see what I'm facing for the next year.

I'd go with harder 3U though polys can get really tricky (see past HSC q15/16s).

Does the auxilary angle for R<0 have a different angle theta or something which reduces down through odd/even/trig properties to give the same answer?

Question: how do you find the volume of a solid if the curve is being rotated on an oblique line?

 

[seanieg89]

I'd go with harder 3U though polys can get really tricky (see past HSC q15/16s).

Does the auxilary angle for R<0 have a different angle theta or something which reduces down through odd/even/trig properties to give the same answer?

Question: how do you find the volume of a solid if the curve is being rotated on an oblique line?

1. Yeah, shift alpha by pi to get between a negative and positive R. Everything will work out the same of course. Formulae just look nicer with the convention of taking R > 0 I guess.

2. Within MX2 methods, you can use the perpendicular distance formula to find the volume of the infinitesimal annuli/cylindrical shells which sweep out your solid. (Pretty sure a question of this type worked out in a similar way is an example in the Cambridge book.)

Alternatively you could use MX2 knowledge of complex numbers to rotate your curve in the plane so that your "oblique line" is now horizontal. Then do the usual volumes stuff.

 

[Peeik]

Examples of such questions are:
In Auxiliary Angle method, why do we always take R>0? Why don't we take R<0?

I remember asking this

 

[Carrotsticks]

I remember asking this

Haha and a totally valid question! Students are always told to take R>0 but are never told WHY that is the case!

 

[iSplicer]

Haha and a totally valid question! Students are always told to take R>0 but are never told WHY that is the case!

So.. why is it? (srs)

 

[anomalousdecay]

How is this possible:

Skip to 5:34

Is this possible, or is the guy talking just saying random bullcrap?

 

[nightweaver066]

So.. why is it? (srs)

seanieg89 answered above:

1. Yeah, shift alpha by pi to get between a negative and positive R. Everything will work out the same of course. Formulae just look nicer with the convention of taking R > 0 I guess.

 

[asianese]

How is this possible:

Skip to 5:34

Is this possible, or is the guy talking just saying random bullcrap?

Pretty sure its a troll?

Suppose j=a+ib for b nonzero and real, since j is not real.

J^2= a^2-b^2+2abi =1.

Collecting real and imaginary parts, a^2-b^2=1 and ab=0. Since b non zero, a must be 0. Thus -b^2 =1 so b^2=-1 which is not possible since b was assumed to be real. Thus no real b can make the equality true.

 

[bottleofyarn]

How is this possible:

Skip to 5:34

Is this possible, or is the guy talking just saying random bullcrap?

I would also like to know. All I found for it was https://en.wikipedia.org/wiki/Quaternion (see the table on the right). "i^2 = j^2 = k^2 = ijk = −1,"

 

[anomalousdecay]

Pretty sure its a troll?

Suppose j=a+ib for b nonzero and real, since j is not real.

J^2= a^2-b^2+2abi =1.

Collecting real and imaginary parts, a^2-b^2=1 and ab=0. Since b non zero, a must be 0. Thus -b^2 =1 so b^2=-1 which is not possible since b was assumed to be real. Thus no real b can make the equality true.

That proof looks valid, but what does it mean exactly?

 

[obliviousninja]

Why is mechanics beginning to be phased out, as someone said in the forums a while back?

 

[anomalousdecay]

Why is mechanics beginning to be phased out, as someone said in the forums a while back?

What you mean in terms of HSC exams?
Or in terms of Engineering/Science?

You do know that there is more to mechanics than the 4U version?

 

[seanieg89]

That proof looks valid, but what does it mean exactly?

The proof means that 1 has no non-real complex square roots. Similarly we can show that 1 has no non-real quaternion square roots.

If we take the word "extended number" to loosely mean any extension of the complex numbers which still obeys all the nice properties of the real and complex numbers (formally, we require that we are at least an integral domain), then we can show that a number has at most two square roots.
Hence 1 has no non-real square roots in any such "extended" number system.

Of course if we allow extensions of the complex numbers that are less nice, then 1 can have many square roots. (For example, the matrix diag(1,-1) is a non-real square root of 1 in the ring of 2x2 matrices over the real numbers).

 

[anomalousdecay]

Of course if we allow extensions of the complex numbers that are less nice, then 1 can have many square roots. (For example, the matrix diag(1,-1) is a non-real square root of 1 in the ring of 2x2 matrices over the real numbers).

So are you saying that it is not possible for j in the form a+ib ?

 

[seanieg89]

So are you saying that it is not possible for j in the form a+ib ?

No, not when i means what it normally does.