they're basically my favourite parts of the coursecircle geo and now probability, how do you know my weaknesses......i vote hax!
do you have another topic youd prefer? a more conventional one lol (Im thinking demoivres/polynomials/complexy things you can get cool questions with those)?
its a cool question to think about. the answers really random anyway, its always 1/2 lol
I think that mightve even been part in that question in our trial, but I can't remember at all whether it had 2 or 3 partsProceed from:
and the 2n+1 roots of
and
So combining the answer is 2.
Now try this question, it's from my trial. It's pretty neat too.
is a polynomial of degree
with real coefficients, where
for each of
.
(In other words,).
By considering the polynomial, find the leading coefficient of
.
its a cool question to think about. the answers really random anyway, its always 1/2 lol
um try this, from q8 in my trial but its considerably easier than it looks I reckon
Factoriseinto quadratic factors with real coefficients, and hence prove that:
sin(\frac{2\pi}{2n+1})sin(\frac{3\pi}{2n+1})...sin(\frac{n\pi}{2n+1}) = \sqrt{2n+1} )
Didn't think to do auxilliary method (though that works nicely)...what I was thinking was basically from the substitution you get:
Did u do ur HSC last year?its a cool question to think about. the answers really random anyway, its always 1/2 lol
um try this, from q8 in my trial but its considerably easier than it looks I reckon
Factorise
Thats how my teacher did it. I dont really like it that much coz its a bit random, like theres no clear motivation to go from line 1 to line 2 other than guesswork. If you do auxiliary angles, you can sorta metagame the question the whole way through.Didn't think to do auxilliary method (though that works nicely)...what I was thinking was basically from the substitution you get:
\,d\theta\\\\=\displaystyle\int_0^{\frac{\pi}{4}}\ln\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)\,d\theta\\\\=\displaystyle\int_0^{\frac{\pi}{4}}\ln\left(1+\dfrac{1-\tan\theta}{1+\tan\theta}\right)\,d\theta\\\\=\displaystyle\int_0^{\frac{\pi}{4}}\ln\left(\dfrac{2}{1+\tan\theta}\right)\,d\theta\\\\\Rightarrow 2\displaystyle\int_0^{\frac{\pi}{4}}\ln(1 + \tan\theta)\,d\theta=\displaystyle\int_0^{\frac{\pi}{4}}\ln 2\,d\theta\\\\\therefore\displaystyle\int_0^1\dfrac{\ln(1+x)}{1+x^2}\,dx=\displaystyle\int_0^{\frac{\pi}{4}}\ln(1 + \tan\theta)\,d\theta=\dfrac{\pi}{8}\ln2)
This is a very nice probability question
Yep thats correct ^^ It's a kinda nice result, there are lots of nice problems out there with similar methods. For a muuuuch harder and more abstract thing a bit like it, if you feel like trying it, try (but I really meant it with much harderr):=(k+1)P(k)-k=0$ for $ k=0,1,...,n\\ \Rightarrow Q(x)=K\prod_{k=0}^n (x-k) \quad $ for some constant $K$, since $ \deg Q=n+1\\ \Rightarrow 1=Q(-1)=K\prod_{k=0}^n (-1-k)=K\prod_{k=0}^n (1+k)=(n+1)!K\\$where the penultimate equality is from the fact that $n$ is odd. Hence $K=1/(n+1)!$. This gives us $Q(n+1)=1$ and hence $P(n+1)=1$.)
Yeah thats right
One of the nicest probability questions ive come across.
Interestingly I guess I got something pretty differentThe solution is fairly similar, except two transformations were required for this one. Don't really have time to write down a complete solution but the end result is:
Some clever manipulations with binomial coefficients could probably put it in a nicer form.
I'm not entirely sure its correct though. I got (after a bit of manipulation):
Oh I realised I got my method wrong, I defined Q incorrectly
The minus one at the end should be multiplied by (x^2+1)...(x^2 + n); i.e. you need