TheAspiringDoc's Math Thread

[lzxnl]

Another way to see why the derivative of sin(x) is cos(x), and why that of cos(x) is -sin(x), is to look at their power series, if you're familiar with what they are (sort of "infinite degree polynomials", roughly speaking).

This is more algebraic than intuitive though.

Few questions.
Firstly, how do you get to these power series? Either you construct these as the sine and cosine functions and derive all of the other known properties later, or you use the Generalised Mean Value Theorem, show convergence for the power series everywhere, show that the remainder term vanishes etc...which all requires knowing what the derivatives are.

For an intuitive grasp, you're better off looking at the graphs. Derivative of sine is cosine can be reasoned as follows.
At x=0, y = sin x is at its steepest -> derivative should be at its greatest at x=0. Function is periodic -> derivative should also be periodic. Function has extrema at pi/2, 3pi/2, 5pi/2 etc -> derivative should be zero here. Etc.

[MathsGuru]

Few questions.
Firstly, how do you get to these power series? Either you construct these as the sine and cosine functions and derive all of the other known properties later, or you use the Generalised Mean Value Theorem, show convergence for the power series everywhere, show that the remainder term vanishes etc...which all requires knowing what the derivatives are.

For an intuitive grasp, you're better off looking at the graphs. Derivative of sine is cosine can be reasoned as follows.
At x=0, y = sin x is at its steepest -> derivative should be at its greatest at x=0. Function is periodic -> derivative should also be periodic. Function has extrema at pi/2, 3pi/2, 5pi/2 etc -> derivative should be zero here. Etc.

Absolutely - the trig graphs or the unit circle are much better for intuitive understanding.
So I'm not trying to put forward a rigorous proof, only a plausibility argument (which, by the way, is about what the graphs provide anyway, since it takes further reasoning to justify why the slope of sine at the origin coincides with the maximum value of cosine - as long as angles are measured in radians!)
For VCE level, I think it's more than enough just to be aware that there is such a thing as a power series, and to do a bit of differentiantion practice with it, to get an alternative algebraic view of the sin-cos relationship to complement the geometric view.
Power series are also a nice way to think about what sort of form a function would have if it equals its own derivative :-)

[TheAspiringDoc]

Which three digit number has the greatest number of different factors?
Thank you!

[MathsGuru]

Which three digit number has the greatest number of different factors?
Thank you!

I think it's 840, with 32 distinct factors

[TheAspiringDoc]

I think it's 840, with 32 distinct factors

Okay, thanks 
But.. Why?

[lzxnl]

Okay, thanks 
But.. Why?

Let's have a look at what constitutes 'distinct factors'.
32 = 2^5. This number has factors 2, 4, 8, 16, 16. So a number a^n, where a is prime, has n-1 factors.
Let's now look at 2000 = 5 * 400 = 5 * 16 * 25 = 2^4 * 5^3
How do we generate the number of factors? Hopefully you can see that the number of factors it has is 18. We get to this by considering the five powers of 2 from 2^0 to 2^4 and the four powers of 5 from 5^0 to 5^3. For each power of 2, there are four possible powers of 5 you can combine with to generate new factors. Then, we subtract the two cases 1 and 2000. That gives 5 x 4 - 2 = 18.

Generalising for a1^(x1) a2^(x2) a3^(x3)...an^(xn), the number of factors would be (x1+1)(x2+2)(x3+1)(x4+1).....(xn+1) - 2.
As an example, 324 = 18^2 = (2*3^2)^2 = 2^2 * 3^4
Here n = 2, number of factors should be 3*5 - 2 = 13
Indeed, the factors of 324 are:
2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, which totals 13 factors.

So to get the largest number of factors, you would ideally want to find a number that has a large number of primes as factors, but not too many as the primes get big. You need a mix of lots of primes and large powers of primes. This suggests that you raise the smaller primes to higher powers first.

Then, you're trying to maximise the product (x1+1)(x2+1)(x3+1)(x4+1)....(xn+1) subject to the constraint 2^(x1) 3^(x2) 4^(x3)....(nth prime)^(xn) < 1000
You can get a feel for how this would work by looking at the primes. Intuitively, you don't want primes past about 13 because then your powers aren't going to be too big; 13 is around 2^4 already. So you'd then play around with numbers. That's all I can think of atm.

[MathsGuru]

Okay, thanks 
But.. Why?

lzxnl's post sums it up pretty well, with a couple of minor comments:
- 4^(x3) near the end should of course be 5^(x3) (the bases are the primes in increasing order)
- and normally 1 and 2000 would be counted as factors of 2000, which eliminates the need for the messy -2

So here's how it works for 840:
has 2 factors: , i.e. 1 and 2
has 4 factors: , i.e. 1, 2, 3, 6
has 8 factors: , i.e. 1, 2, 3, ..., 30
So building up a number as a product of distinct primes (i.e. a 'squarefree' number), the number of factors doubles each time another prime is included.
The furthest we can go using as small primes as possible while not exceeding 999 is with factors since it is the product of 4 primes, each of which can be either included or omitted in a factor of 210

We then increase the powers of some of the primes we already have.
We can't put in another 5 or 7 without exceeding 999.
If we increase the power of 3, we can make which will then have 2x3x2x2=24 factors, but we can go no further.
However, we can fit in another two 2s, making with 4x2x2x2=32 factors.
These factors can be enumerated simply as all the integers of the form where a=0,1,2 or 3 and each of b,c,d = 0 or 1

 

[Orson]

It's awesome to see you taking an interest so early! But, try to enjoy school now, and don't get too hung up on this stuff at the moment. Just saying, don't get offended or anything...Anyway, if you are finding particular concepts hard to understand, try to find a lecture on YouTube. Have a look at KhanAcademy, ProfRobBob and PatrickJMT, they both explain concepts really clearly by doing worked examples and going though theory. It's easier to understand something through a lecture that from written words (for me at least). Also, if you are taking notes and stuff...DON'T bother carefully curating them with colours and stuff (I'm speaking from my experiences with Meth and Physics). Just do practice questions...

Anyway...have fun buddy!

[TheAspiringDoc]

Thanks heaps for your explanations MathsGuru and lzxnl. 
Thanks also Orson 
Could anyone offer some advice on how I could go about learning calculus over the upcoming holidays?

[keltingmeith]

Thanks heaps for your explanations MathsGuru and lzxnl. 
Thanks also Orson 
Could anyone offer some advice on how I could go about learning calculus over the upcoming holidays?

Paul's Online Notes are pretty good, do recommend.

[lzxnl]

Paul's Online Notes are pretty good, do recommend.

^

Learnt more than half of the second year UoM subject MAST20009 in a day using that site.

[Orson]

Thanks also Orson 

Anytime.

Could anyone offer some advice on how I could go about learning calculus over the upcoming holidays?

I'd have to say Khan Academy again. Just go through his series on Calculus from the beginning...get an old textbook (I'd recommend this years Essentials, you will be able to pick them up cheap because they are changing the study design) and go for it!

[TheAspiringDoc]

Paul's notes are indeed very good
I've found that other useful resources are Math Is Fun and Maths Methods Podcasts by Justin Vincent if anyone was interested

On a side note, if I have 18 scores on a stem and leaf plot, all of which are different, what should I write if I'm asked to find the mode?

Thanks 

[keltingmeith]

Paul's notes are indeed very good
I've found that other useful resources are Math Is Fun and Maths Methods Podcasts by Justin Vincent if anyone was interested

On a side note, if I have 18 scores on a stem and leaf plot, all of which are different, what should I write if I'm asked to find the mode?

Thanks 

There is no mode.

[TheAspiringDoc]

I'm aware that the derivative of trig functions like sin(x) is only cos(x) if x is in radians (right?), but does that also apply for inverse trig [ I.e. Does the derivative of cos^-1 (x) only end up being -1/sqrt(1-x^2) if x is in radians?]
Thank you