The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Current Year 12 Thread Mark VI

Scroll to see replies

Original post by Maths_Lover
Revising to The Legend of Zelda music is just epic. :grin:

So motivational. :work:

Koji Kondo - you absolute genius. :love:


speaking of zelda i just watched a speed-run on YouTube of 3 heart challenge on master mode of ocarina of time it was 2 hours and 50 minutes well spent.
Original post by Emissionspectra
speaking of zelda i just watched a speed-run on YouTube of 3 heart challenge on master mode of ocarina of time it was 2 hours and 50 minutes well spent.


:lol: 3 heart challenge is difficult enough in standard mode! :eek3:

Have you watched the abridged series by adamwestslapdog?
Original post by Maths_Lover
:lol: 3 heart challenge is difficult enough in standard mode! :eek3:

Have you watched the abridged series by adamwestslapdog?


No I'll look into it.
http://www.youtube.com/watch?v=InoMBWnTXpg&feature=endscreen&NR=1
Original post by Emissionspectra
I got to play mozart piano sonata in D major for town pianos at RAH it was amazing :biggrin:

:eek::eek::eek::eek:

I wish I was better at piano, but that's amazing!
Original post by Emissionspectra


THANK YOU. :colone:

I was just reading Quantum by Manjit Kumar... more Quantum Physics is always welcome. :biggrin:
Original post by Maths_Lover
Penguins

Yes, well I already have a super expensive Clarinet (Bb), which makes my case seem a little less convincing :frown:

I'm doing Maths, FM, Biology, Chemistry, Physics, English Literature. My school only timetables 4 lessons max for U6th :cry2:
Original post by Emissionspectra
No I'll look into it.


It is hilarious! Even more so if you've played Zelda games. :tongue:

Here's the first episode. :colone:

http://www.youtube.com/watch?v=geWNqJjbtZU
Original post by Llewellyn
Yes, well I already have a super expensive Clarinet (Bb), which makes my case seem a little less convincing :frown:

I'm doing Maths, FM, Biology, Chemistry, Physics, English Literature. My school only timetables 4 lessons max for U6th :cry2:


Ah... extra good luck, then. :frown:

:console:
Original post by Maths_Lover
THANK YOU. :colone:

I was just reading Quantum by Manjit Kumar... more Quantum Physics is always welcome. :biggrin:


Your welcome :smile:. :O your a girl again :P
I wish they could continue the Twilight Princess music/ gameplay. I loved that game, I did hundreds of sketches of some of the action shots :sogood::sogood:

I know I'm being blasphemous by not preferring wind waker/ ocarina of time, but I loved the Twilight Princess so much.
Original post by Llewellyn
I wish they could continue the Twilight Princess music/ gameplay. I loved that game, I did hundreds of sketches of some of the action shots :sogood::sogood:

I know I'm being blasphemous by not preferring wind waker/ ocarina of time, but I loved the Twilight Princess so much.


Majora's mask is where its at with twilight in close second
Original post by wcp100
...


Original post by Llewellyn
...


I thought you guys might be interested in how I solved the inequalities question, as a different take on how you guys went about it. :cute:

Show that, if p>m>0,p>m>0, then

p−mp+m≤x2−2mx+p2x2+2mx+p2≤p+mp−m, ∀x∈R\displaystyle \frac{p-m}{p+m} \leq \frac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \frac{p+m}{p-m}, \ \forall x \in \mathbb{R}

Let y=x2−2mx+p2x2+2mx+p2\displaystyle y = \frac{x^2-2mx+p^2}{x^2+2mx+p^2} .

y(x2+2mx+p2)=x2−2mx+p2\displaystyle y(x^2+2mx+p^2) = x^2-2mx+p^2

(y−1)x2+2m(y+1)x+p2(y−1)=0\displaystyle (y-1)x^2 +2m(y+1)x + p^2(y-1) = 0 - upon multiplying out the brackets and simplifying, so that the L.H.S. is a quadratic in x equal to zero.

∀x∈R,\forall x \in \mathbb{R}, the discriminant is non-negative.

⇒(2m(y+1))2−4p2(y−1)(y−1)≥0\Rightarrow (2m(y+1))^2 - 4p^2(y-1)(y-1) \geq 0

(my+m+py−p)(my+m−py+p)≥0(my+m +py -p)(my+m-py+p) \geq 0 - after some algebraic manipulation and simplifying (but not fully as it is not completely obvious yet which factorisation to do).

Now, for the inequality to hold true, both bracketed expressions must have the same sign since +ve times +ve = +ve and -ve times -ve = +ve or one or both bracketed expressions is zero.

Considering what happens when both are negative leads to a contradiction (you can try that for yourselves if you wish), so the bracketed expressions must be either zero or positive, which leads to:

my+m+py−p≥0my+m +py -p \geq 0

y(p+m)−(p−m)≥0y(p+m) - (p-m) \geq 0

y≥p−mp+my \geq \dfrac{p-m}{p+m} - we can safely divide by (p+m)(p+m) as we know it is never zero or negative.

Considering the other bracketed expression:

my+m−py+p≥0my+m-py+p \geq 0

−y(p−m)+(p+m)≥0-y(p-m) + (p+m) \geq 0

y≤p+mp−my \leq \dfrac{p+m}{p-m} - again we can safely divide by (p−m)(p-m) as we know it is never zero or negative either.

In conclusion:

p−mp+m≤y≤p+mp−m\dfrac{p-m}{p+m} \leq y \leq \dfrac{p+m}{p-m}

And since y=x2−2mx+p2x2+2mx+p2\displaystyle y = \frac{x^2-2mx+p^2}{x^2+2mx+p^2} , it follows that

p−mp+m≤x2−2mx+p2x2+2mx+p2≤p+mp−m,∀x∈R\dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m}, \forall x \in \mathbb{R} .

I must admit that it took me a while and a whole lot of messy scribbling out to come up with this argument. :lol:
(edited 11 years ago)
Original post by Emissionspectra
Your welcome :smile:. :O your a girl again :P


:biggrin:

I am indeed. :cute:

Original post by Llewellyn
I wish they could continue the Twilight Princess music/ gameplay. I loved that game, I did hundreds of sketches of some of the action shots :sogood::sogood:

I know I'm being blasphemous by not preferring wind waker/ ocarina of time, but I loved the Twilight Princess so much.


Original post by Emissionspectra
Majora's mask is where its at with twilight in close second


No way, guys. For me it's gotta be Ocarina of Time > Windwaker > Twilight Princess. :tongue:

More Twilight Princess would be cool, though. :yep:

Have you guys got/played Skyward Sword? Personally, I love it but haven't played in ages due to work and stuff... I shall complete it this summer, hopefully. :colone:
Original post by Maths_Lover
I thought you guys might be interested in how I solved the inequalities question, as a different take on how you guys went about it. :cute:

Show that, if p>m>0,p>m>0, then

p−mp+m≤x2−2mx+p2x2+2mx+p2≤p+mp−m, ∀x∈R\displaystyle \frac{p-m}{p+m} \leq \frac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \frac{p+m}{p-m}, \ \forall x \in \mathbb{R}

Let y=x2−2mx+p2x2+2mx+p2\displaystyle y = \frac{x^2-2mx+p^2}{x^2+2mx+p^2} .

Unparseable latex formula:

\displystyle y(x^2+2mx+p^2) = x^2-2mx+p^2



(y−1)x2+2m(y+1)x+p2(y−1)=0\displaystyle (y-1)x^2 +2m(y+1)x + p^2(y-1) = 0 - upon multiplying out the brackets and simplifying, so that the L.H.S. is a quadratic in x equal to zero.

∀x∈R,\forall x \in \mathbb{R}, the discriminant is non-negative.

⇒(2m(y+1))2−4p2(y−1)(y−1)≥0\Rightarrow (2m(y+1))^2 - 4p^2(y-1)(y-1) \geq 0

(my+m+py−p)(my+m−py+p)≥0(my+m +py -p)(my+m-py+p) \geq 0 - after some algebraic manipulation and simplifying (but not fully as it is not completely obvious yet which factorisation to do).

Now, for the inequality to hold true, both bracketed expressions must have the same sign since +ve times +ve = +ve and -ve times -ve = +ve or one or both bracketed expressions is zero.

Considering what happens when both are negative leads to a contradiction (you can try that for yourselves if you wish), so the bracketed expressions must be either zero or positive, which leads to:

my+m+py−p≥0my+m +py -p \geq 0

y(p+m)−(p−m)≥0y(p+m) - (p-m) \geq 0

y≥p−mp+my \geq \dfrac{p-m}{p+m} - we can safely divide by (p+m)(p+m) as we know it is never zero or negative.

Considering the other bracketed expression:

my+m−py+p≥0my+m-py+p \geq 0

−y(p−m)+(p+m)≥0-y(p-m) + (p+m) \geq 0

y≤p+mp−my \leq \dfrac{p+m}{p-m} - again we can safely divide by (p−m)(p-m) as we know it is never zero or negative either.

In conclusion:

p−mp+m≤y≤p+mp−m\dfrac{p-m}{p+m} \leq y \leq \dfrac{p+m}{p-m}

And since y=x2−2mx+p2x2+2mx+p2\displaystyle y = \frac{x^2-2mx+p^2}{x^2+2mx+p^2} , it follows that

p−mp+m≤x2−2mx+p2x2+2mx+p2≤p+mp−m,∀x∈R\dfrac{p-m}{p+m} \leq \frac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m}, \forall x \in \mathbb{R} .

I must admit that it took me a while and a whole lot of messy scribbling out to come up with this argument. :lol:


That must have taken ages to Latex
Original post by Maths_Lover
:biggrin:

I am indeed. :cute:





No way, guys. For me it's gotta be Ocarina of Time > Windwaker > Twilight Princess. :tongue:

More Twilight Princess would be cool, though. :yep:

Have you guys got/played Skyward Sword? Personally, I love it but haven't played in ages due to work and stuff... I shall complete it this summer, hopefully. :colone:


No ive heard of it but never played it. Ugh tempted to bunk school tomorrow
Original post by Emissionspectra
That must have taken ages to Latex


It didn't take too long (a little copying and pasting goes a long way :colone: )- I was gone for ages because I did the dishes as well. :tongue:

Original post by Emissionspectra
No ive heard of it but never played it. Ugh tempted to bunk school tomorrow


Say... whaaat?! It's the newest Zelda game out (it came out on 15th November 2011)! :eek3:
Original post by Maths_Lover
meerkats

:holmes:

But on a serious note, nice solution. Your reasoning towards the end is better than mine. I found the discriminant in a completely different way, but your solution is what the French call magnifique. :biggrin:
Original post by Maths_Lover
It didn't take too long (a little copying and pasting goes a long way :colone: )- I was gone for ages because I did the dishes as well. :tongue:



Say... whaaat?! It's the newest Zelda game out (it came out on 15th November 2011)! :eek3:


Ye but it isn't on Gamecube :frown: and i dont like the Wii.
Original post by Maths_Lover
:biggrin:

I am indeed. :cute:





No way, guys. For me it's gotta be Ocarina of Time > Windwaker > Twilight Princess. :tongue:

More Twilight Princess would be cool, though. :yep:

Have you guys got/played Skyward Sword? Personally, I love it but haven't played in ages due to work and stuff... I shall complete it this summer, hopefully. :colone:

Skyward sword is the one zelda DS game I haven't got. Is it any good? I found the spirit tracks one to be a little disappointing.

Having said that, I think the only DS games I have are either pokemon, mario kart or zelda... May have been a waste of money :/

Related discussions

Latest

Trending

Trending

Articles for you