Bored of Studies Trial Discussion Thread. (2 Viewers)

[deswa1]

Anyone got working for showing it's in SHM? Ended up with a cos^2 - sin^2..

I converted it using double angles to cos(nt) etc. and stuff and then differentiated from there.

 

[nightweaver066]

I converted it using double angles to cos(nt) etc. and stuff and then differentiated from there.

*facepalm makes sense. Would simplify the "x" nicely..

 

[bleakarcher]

Same..

For the question about reasoning why f(x) < g(x) or something like that for 1 mark, i can't seem to express myself properly lol.

Could someone post up their reasoning?

Wait, shit that's wrong.

 

[asianese]

I ...differentiated twice HAHAHHA. It worked, though. After doing double angles, and all. 1 page motherfckers!

 

[barbernator]

add Bcos^2(nt/2) and subtract it and then it cancels to the form x=(A-B)cos(nt)+B, you didn't have to differentiate. and from there ii) followed, through subbing in values.

 

[Carrotsticks]

add Bcos^2(nt/2) and subtract it and then it cancels to the form x=acos(nt)+c, you didn't have to differentiate.

Good man.

 

[barbernator]

did anyone get binomial part 3?

 

[Carrotsticks]

Hint: Symmetry property.

 

[asianese]

Nope...

 

[barbernator]

lol just realised. For the Q13 a) exponential decay, I got the answer out with a t0 hanging off the end and i re-did it but it didn't disappear. t0=0 *facepalm*

 

[asianese]

i'm really smart and read the data wrong so I had like x= e^(142.5) or something ridiculous. lol...error was in the initial conditions

 

[barbernator]

Hint: Symmetry property.

ahhh yes! this is very similar to what you were explaining in your MX2 seminar. just do whatever it says (in this case x n/2) and play with it from there. Will post working.

<a href="http://www.codecogs.com/eqnedit.php?latex=from~ii)\\\\ \frac{n}{2}\sum_{r=0}^{n}\binom{n}{r}^{2}=\frac{n}{2}\binom{2n}{n}\\\\ \frac{n}{2}(\binom{n}{0}^{2}@plus;\binom{n}{1}^{2}@plus;...@plus;\binom{n}{n}^{2})=\frac{n}{2}\binom{2n}{n}\\\\ n(\binom{n}{0}^{2}@plus;\binom{n}{1}^{2}@plus;...@plus;\binom{n}{n/2}^{2})=\frac{n}{2}\binom{2n}{n}(by~symmetry)\\\\ 0\binom{n}{0}^{2}@plus;n\binom{n}{n}^{2}@plus;1\binom{n}{1}^{2}@plus;(n-1)\binom{n}{n-1}^{2}@plus;...=\frac{n}{2}\binom{2n}{n}(again~by~symmetry)\\\\\therefore \sum_{r=0}^{n}r\binom{n}{r}^{2}=\frac{n}{2}\binom{2n}{n}~as~required." target="_blank"><img src="http://latex.codecogs.com/gif.latex?from~ii)\\\\ \frac{n}{2}\sum_{r=0}^{n}\binom{n}{r}^{2}=\frac{n}{2}\binom{2n}{n}\\\\ \frac{n}{2}(\binom{n}{0}^{2}+\binom{n}{1}^{2}+...+\binom{n}{n}^{2})=\frac{n}{2}\binom{2n}{n}\\\\ n(\binom{n}{0}^{2}+\binom{n}{1}^{2}+...+\binom{n}{n/2}^{2})=\frac{n}{2}\binom{2n}{n}(by~symmetry)\\\\ 0\binom{n}{0}^{2}+n\binom{n}{n}^{2}+1\binom{n}{1}^{2}+(n-1)\binom{n}{n-1}^{2}+...=\frac{n}{2}\binom{2n}{n}(again~by~symmetry)\\\\\therefore \sum_{r=0}^{n}r\binom{n}{r}^{2}=\frac{n}{2}\binom{2n}{n}~as~required." title="from~ii)\\\\ \frac{n}{2}\sum_{r=0}^{n}\binom{n}{r}^{2}=\frac{n}{2}\binom{2n}{n}\\\\ \frac{n}{2}(\binom{n}{0}^{2}+\binom{n}{1}^{2}+...+\binom{n}{n}^{2})=\frac{n}{2}\binom{2n}{n}\\\\ n(\binom{n}{0}^{2}+\binom{n}{1}^{2}+...+\binom{n}{n/2}^{2})=\frac{n}{2}\binom{2n}{n}(by~symmetry)\\\\ 0\binom{n}{0}^{2}+n\binom{n}{n}^{2}+1\binom{n}{1}^{2}+(n-1)\binom{n}{n-1}^{2}+...=\frac{n}{2}\binom{2n}{n}(again~by~symmetry)\\\\\therefore \sum_{r=0}^{n}r\binom{n}{r}^{2}=\frac{n}{2}\binom{2n}{n}~as~required." /></a>

 

[jackerino]

FUcking woah. You even START OUT guns blazing lol, Realise's prelim paper at least STARTED with some free marks for everyone HAHAHAHA

 

[Trebla]

Same..

For the question about reasoning why f(x) < g(x) or something like that for 1 mark, i can't seem to express myself properly lol.

Could someone post up their reasoning?

I was gonna be like: LOOK AT TEH GRAPH. Not very convincing.



Shet. That doesn't work...maybe b / and or a are 0<a,b<1...awks.

Simply solve f(x) < g(x) and you get a quadratic inequality which solves out to -1 < x < 1 but since x is non-negative then the domain is 0 < x < 1. Likewise, solve f(x) > g(x) given x > 0 to get the other inequality/domain.

 

[medicore]

i'm really smart and read the data wrong so I had like x= e^(142.5) or something ridiculous. lol...error was in the initial conditions

.

 

[asianese]

no, i'm REALLY smart.

 

[alexobern]

no, i'm REALLY smart.

how's probability?

 

[barbernator]

probability part iii) anyone? I couldn't get it because of the double H's

 

[nightweaver066]

Looking through the paper again.. misread so many questions.. Even skipped one by accident

 

[asianese]

I didn't even do that part ... too many cases / I didn't see a quick way...